keirp/tiny_math · Datasets at Hugging Face

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http://physics.stackexchange.com/questions/17134/inertial-mass-of-a-scalar-field?answertab=active	# Inertial Mass of a scalar field Does it make sense to talk of the inertial mass of a scalar field? By the equivalence principle, it must be equal to its gravitational mass. We know that the scalar field contributes towards the stress-energy tensor, so, shouldn't it have an inertial mass too? - ## 2 Answers Yes, of course, if you produce a localized concentration of energy carried by a scalar field, it exhibits all the properties that this total energy $E=mc^2$ should exhibit. It will enter the right hand side of Einstein's equations so it will curve the surrounding spacetime and create a gravitational field. It will be able to convert to other forms of energy so that the total energy conservation law, including the energy of the scalar field, holds. And finally, it will also act as inertia. For example, a packet of energy carried by a field behaves as a particle. For example, a magnetic monopole may be imagined as a nontrivial localized solution to some field equations including scalar and gauge fields. A magnetic force $F=q_m B$ may act upon the magnetic monopole and the acceleration will indeed be given by $F=ma$ where $m=E/c^2$ is the inertial mass of the magnetic monopole, stored in the energy density of the fields. - Yes, indeed This is exactly why it is assumed in the Standard Model that mass is the result of an interaction with a scalar "world potential" (The Higgs field) while for instance the changes in energy/momentum of a charge are due to the electromagnetic vector potential $A^\mu$ Regards, Hans -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 5, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9120468497276306, "perplexity_flag": "head"}
http://www.reference.com/browse/plus	Definitions Nearby Words Related Questions # Plus-minus sign ± The plus-minus sign (±) is a mathematical symbol commonly used to indicate the precision of an approximation, or as a convenient shorthand for a quantity which has two possible values opposite in sign. In mathematics, the sign is pronounced "plus or minus" and indicates that there are exactly two possible answers, one of which is positive and one of which is negative. In most experimental sciences however, the sign is pronounced "give or take" and it indicates an inclusive range of values that a reading might have. The plus-minus sign does not mean "approximately". ## Usage for indicating precision The use of ± for an approximation is most commonly encountered for presenting the numerical value of a quantity together with its tolerance or its statistical margin of error. For example, "5.7 ± 0.2" denotes a quantity that is specified or estimated to be within 0.2 units of 5.7; it may be anywhere in the range from 5.7 − 0.2 to 5.7 + 0.2. More precisely, in scientific usage it usually comes with a probability of being within the interval, usually that of 2 standard deviations, or 95.4%. A percentage may also be used to indicate the error margin. For example, 230 V ± 10% refers to a voltage within 10% of either side of 230 V (207 V to 253 V). Separate values for the upper and lower bounds may also be used. For example, to indicate that a value is most likely 5.7 but may be as high as 5.9 or as low as 5.6, one could write $5.7^\left\{+0.2\right\}_\left\{-0.1\right\}$. ## Usage as shorthand for two values of opposite signs In mathematical equations, the use of ± may be found as shorthand, to present two equations in one formula: equation+ OR equation- represented with equation± The best-known example is offered by the formula for the solutions of quadratic equations: If $displaystyle ax^2 + bx + c = 0,$ then $displaystyle x = frac\left\{-b pm sqrt\left\{b^2-4ac\right\}\right\}\left\{2a\right\}.$ Written out in full, this states that there are two solutions to the equation, namely $displaystyle x = frac\left\{-b + sqrt \left\{b^2-4ac\right\}\right\}\left\{2a\right\}$ and $displaystyle x = frac\left\{-b - sqrt \left\{b^2-4ac\right\}\right\}\left\{2a\right\}.$ Another example is found in the trigonometric identity $sin\left(x pm y\right) = sin\left(x\right) cos\left(y\right) pm cos\left(x\right) sin\left(y\right).,$ This stands for two identities: one with + on both sides of the equation, and one with − on both sides. A somewhat different use is found in this presentation of the formula for the Taylor series of the sine function: $sinleft\left(x right\right) = x - frac\left\{x^3\right\}\left\{3!\right\} + frac\left\{x^5\right\}\left\{5!\right\} - frac\left\{x^7\right\}\left\{7!\right\} + cdots pm frac\left\{1\right\}\left\{\left(2n+1\right)!\right\} x^\left\{2n+1\right\} + cdots.$ This mild abuse of notation is meant to indicate that the sign of the terms alternate, where (starting the count at 0) the terms with an even index $displaystyle n$ are added while those with an odd index are subtracted. A less ambiguous presentation in this case would use the quantity $\left(-1\right)^n$, which gives $+1$ when $displaystyle n$ is even and −1 when $displaystyle n$ is odd. ## Minus-plus sign There is another character, the minus-or-plus sign (∓), which is seen less often. It only takes on significant meaning when used in conjunction with the "±" sign. It can be used alongside "±" in such expressions as "x ± y ∓ z", which can be interpreted as "x + y − z" or/and "x − y + z", but neither "x + y + z" nor "x − y − z". The upper "−" in "∓" is considered attached to the "+" of "±" (and the lower symbols work in the same way) even though there is no visual indication of the dependency. The original expression can be rewritten as "x ± (y − z)" to avoid confusion, but cases such as the trigonometric identity $cos\left(x pm y\right) = cos\left(x\right) cos\left(y\right) mp sin\left(x\right) sin\left(y\right)$ are most neatly written using the "∓" sign. ## Encodings • In ISO 8859-1, -7, -8, -9, -13, -15, and -16, the plus-minus symbol is given by the code 0xB1hex Since the first 256 code points of Unicode are identical to the contents of ISO-8859-1 this symbol is also at Unicode code point 00B1. • The symbol also has a HTML entity representation of `±`. The rarer minus-plus sign (∓) is not generally found in legacy encodings and does not have a named HTML entity but is available in Unicode with codepoint U+2213 and so can be used in HTML using `∓` • In TeX 'plus-or-minus' and 'minus-or-plus' symbols are encoded as `pm` and `mp` entities, respectively. • These characters are also seen written as the (highly un-semantic) underlined or overlined + symbol. ( + or ).	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 13, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9269683361053467, "perplexity_flag": "middle"}
http://mathhelpforum.com/differential-geometry/120453-normed-vector-space.html	Thread: 1. normed vector space show that in the vector space $<br /> \ell _1 = \left\{ {r_k \,\,:\,\sum {\left\| {r_k } \right\|} \,converges} \right\}<br />$ (i) $<br /> \left\\| {r_k } \right\\| = \sum {r_k } <br />$ is a norm (ii) $<br /> P = \left\{ {\left\{ {x_k } \right\}\,:\,x_k \geqslant 0\,\forall k \in \mathbb{N}} \right\}<br />$ has en empty interior (iii) $<br /> B\left( {\vec 0,1} \right)<br />$ is not compact please help im trying to learn this by myself and i would like help on how to do this problems thanks! 2. Originally Posted by mms show that in the vector space $<br /> \ell _1 = \left\{ {r_k \,\,:\,\sum {\left\| {r_k } \right\|} \,converges} \right\}<br />$ (i) $<br /> \left\\| {r_k } \right\\| = \sum {r_k } <br />$ is a norm (ii) $<br /> P = \left\{ {\left\{ {x_k } \right\}\,:\,x_k \geqslant 0\,\forall k \in \mathbb{N}} \right\}<br />$ has en empty interior (iii) $<br /> B\left( {\vec 0,1} \right)<br />$ is not compact please help im trying to learn this by myself and i would like help on how to do this problems thanks! For (i) just remmeber that for all $n\in \mathbb{N}$ we have $\vert x_n + y_n \vert \leq \vert x_n\vert + \vert y_n \vert$ so adding we get $\sum_{i=1}^{n} \vert x_i+y_i \vert \leq \sum_{i=1}^{n} \vert x_i \vert + \vert y_i \vert \leq \sum_{i=1}^{\infty } \vert x_i \vert + \vert y_i \vert < \infty$ For (ii) take a sequence $y^k=(y_{n}^k) \in \ell _1$ such that $y_{n}^k = x_n$ if $n\neq k$ and $y_{n}^k= -\frac{1}{k}$ if $n=k$ then $\Vert x-y^k \Vert = \vert x_k + \frac{1}{k} \vert \rightarrow 0$ as $k\rightarrow \infty$ (since $x_n \rightarrow 0$ ) For (iii) what can you say about the sequence $x^k=(x_{n}^k)$ where $x_{n}^k=0$ if $n\neq k$ and $x_{n}^k=\frac{1}{2}$ if $n=k$ (Notice that $\Vert x^k - x^l \Vert = 1$ for all $k\neq l$)	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 26, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9239837527275085, "perplexity_flag": "head"}
http://mathhelpforum.com/advanced-algebra/112760-eigenvalues.html	# Thread: 1. ## Eigenvalues I need to find the characteristic polynomial, eigenvalues, and eigenvectors or the following matrix. [1 1] [1 1] So far, this is what I have done: det(Lamda-A) And got [lamda - 1 -1] [-1 lamda - 1] The det of that gives (lamda - 1)(lamda-1) + 1 which is lamda squared - 2 lamda +2 I used the quad formula on that and got 1 plus/minus i I dont know where to go from here. I think Now i do this: A-(1+i)I = [i 1] [1 i] i have no clue where to go from there 2. Hi! There really is no need to calculate the characteristic polynomial that way. First, you can easily observe that 0 is an eigenvalue of geometric multiplicty 1 (since $rank(A) = 1$. Now, note that $trace(A) = 1+1 = 2$ and A only has two eigenvalues, so the second one must be 2. Another way to know that 2 is an eigenvalue is to note that the sum of all rows is 2, therefore, according to a theorem which you should have learned, 2 is an eigenvalue with corresponding eigenvector $(1 \ 1)^T$. So we know that 2 is an eigenvalue of geometric multiplicity $\geq 1$ and 0 is an eigenvector with geometric multiplicity 1. Since algebraic multiplicity $\geq$ geometric multiplicity for each eigenvalue, we get that the algebraic and geometric multiplicities for each eigenvector are 1 and 1. Therefore: $\Delta_A(x) = x(x-2)$ With 0,2 as eigenvalues and $(1\ -1)^T, (1 \ 1)^T$ the corresponding eigenvectors. However, if you still want to know what you did wrong -- $det(\lambda I-A) = (\lambda-1)(\lambda-1) - (-1)(-1) = (\lambda-1)^2 -1 \neq (\lambda-1)^2+1$ $=\Rightarrow det(\lambda I-A) = \lambda^2 -2\lambda +1 -1 = \lambda^2 -2\lambda = \lambda(\lambda-2)$ 3. ## Thanks Thanks a bunch. It would be nice if i could add :P 4. So I was trying to do it the way i was. I got the (1,1), but instead of (1,-1) I got (-1,1). 0lamda - A = -1 -1 -1 -1 then rref that and got 1 1 0 0 x1= -r x2 = r -1, 1 5. That is fine. The eigenvector can be any non-zero scalar multiple of $(1 \ -1)^T$. In this case, $(-1)(1 \ -1)^T = (-1 \ 1)^T$. From this, $(2\ -2)^T, \ (-8\ 8)^T,...$ could all be eigenvectors as well.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 12, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9159992337226868, "perplexity_flag": "head"}
http://mathhelpforum.com/differential-geometry/92431-examples.html	# Thread: 1. ## Examples Give an example or say why it is impossible: 1. A function which is discontinuous at every point of the set $\{\frac{1}{n}:n\in\mathbb{N}\}$ but continuous elsewhere on $\mathbb{R}$ I think it's possible since that set could be written as a countable union of closed sets, but I can't think of a particular example. 2. An infinite subset of [0,1] with no limit points. I think this is impossible but I'm not sure about my reasoning - is this correct?: Any subset of [0,1] is bounded so by the Bolzano Weierstrass theorem any sequence in this subset must have a convergent subsequence, so it has a limit point. 3. A closed set whose supremum is not a limit point of this set. not sure... 4.A power series that is absolutely convergent at only one point. Probably something with the point being 0... not sure though Thanks for any help 2. 1. Hint: Spoiler: $f(x)=\begin{cases}x&x\ne\frac1n\\[1mm]0&\mbox{otherwise}\end{cases}$ 3. Hint: Spoiler: Any finite set that is bounded above. 4. Hint: Spoiler: $\sum_{n\,=\,1}^\infty\frac{(-1)^n}nx^n$ 3. Thanks Abstractionist. Regarding your answer to question 4, is it always true that if $\sum a_n$ diverges, then $\sum a_nx^n$ also diverges, for any x besides 0? And if not, what makes your example special? Thanks again 4. Originally Posted by Aileys. is it always true that if $\sum a_n$ diverges, then $\sum a_nx^n$ also diverges, for any x besides 0? No. For example, $\sum_{n\,=\,1}^\infty\frac1n$ diverges but $\sum_{n\,=\,1}^\infty\frac{x^n}n$ converges for $-1\le x\le0.$ However, the convergence is only conditional, not absolute. 5. Ahhh OK so then is it always true that if $\sum a_n$ diverges then the only x for which $\sum a_nx^n$ converges absolutely is zero? Also while you're here ( ) was my answer to #2 correct?	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 13, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9410819411277771, "perplexity_flag": "head"}
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I asked this question for many people/professors without getting a sufficient answer, why in QM Lebesgue spaces of second degree are assumed to be the one that corresponds to the Hilbert vector space ... 3answers 375 views ### Existence of creation and annihilation operators In a multiple particle Hilbert space (any space of any multi-particle system), is it sufficient to define creation and annihilation operators by their action (e.g. mapping an n-particle state to an ... 4answers 198 views ### How to apply an algebraic operator expression to a ket found in Dirac's QM book? I've been trying to learn quantum mechanics from a formal point of view, so I picked up Dirac's book. In the fourth edition, 33rd page, starting from this:$$\xi\|\xi'\rangle=\xi'\|\xi'\rangle$$ (Where ... 2answers 300 views ### Uniqueness of eigenvector representation in a complete set of compatible observables Sakurai states that if we have a complete, maximal set of compatible observables, say $A,B,C...$ Then, an eigenvector represented by $\|a,b,c....>$, where $a,b,c...$ are respective eigenvalues, is ... 2answers 191 views ### QM formalism is one big confusion - lack of geometrical explaination with images I have been trying to learn QM and it went well (all untill harmonic oscilator) until i had to face the formalism: Hilbert space- As a novice to QM i am very sad that in none of the books i have ... 2answers 399 views ### Observing the exponential growth of Hilbert space? One of the weirdest things about quantum mechanics (QM) is the exponential growth of the dimensions of Hilbert space with increasing number of particles. This was already discussed by Born and ... 1answer 128 views ### How does a state in quantum mechanics evolve? I have a question about the time evolution of a state in quantum mechanics. The time-dependent Schrodinger equation is given as $$i\hbar\frac{d}{dt}\|\psi(t)\rangle = H\|\psi(t)\rangle$$ I am ... 1answer 83 views ### State space of QFT, CCR and quantization, and the spectrum of a field operator? In the canonical quantization of fields, CCR is postulated as (for scalar boson field ): $$[\phi(x),\pi(y)]=i\delta(x-y)\qquad\qquad(1)$$ in analogy with the ordinary QM commutation relation: ... 3answers 160 views ### Banach Space representations of physical systems I think most physicists mostly model physical systems as some kind of Hilbert space. Hilbert spaces are a strict subset of Banach spaces. Questions: Can physical systems really have non-compact ... 3answers 352 views ### Can we have discontinuous wavefunctions in the Infinite Square well? The energy eigenstates of the infinite square well problem look like the Fourier basis of L2 on the interval of the well. So then we should be able to for example make square waves that are an ... 1answer 311 views ### Quantum mechanic newbie: why complex amplitudes, why Hilbert space? I'm just starting learning quantum mechanics by myself (2 "lectures" so far) and I was wondering why we need to define quantum states in a complex vector space rater than a real one? Also I was ... 1answer 343 views ### Where does the wave function of the universe live? Please describe its home Where does the wave function of the universe live? Please describe its home. I think this is the Hilbert space of the universe. (Greater or lesser, depending on which church you belong to.) Or maybe ... 3answers 248 views ### If I go to the church of the greater Hilbert space, can I have Unitary Collapse? Actually, unitary pseudo-collapse? Von Neuman said quantum mechanics proceeds by two processes: unitary evolution and nonunitary reduction, also now called projection, collapse and splitting. ... 2answers 114 views ### Vector representation of wavefunction in quantum mechanics? I am new to quantum mechanics, and I just studied some parts of "wave mechanics" version of quantum mechanics. But I heard that wavefunction can be represented as vector in Hilbert space. In my eye, ...	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 39, "mathjax_display_tex": 7, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9121148586273193, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/143928/proof-that-ab-ac-is-bounded-for-bounded-b-and-c?answertab=active	# Proof that $ab/(a+c)$ is bounded for bounded $b$ and $c$ Is there a concise way to prove that $\frac{ab}{a+c} \in [0, 1]$ for all $a > 0$, $b \in [0, 1]$, and $c \in [0, 1]$? - 5 Note that $0\leq ab\leq a\leq a+c$. Then you can get the conclusion. – molan May 11 '12 at 16:08 @molan Perfect, thanks! Make it an answer and it gets my vote. – ezod May 11 '12 at 16:15 No thanks. But I think it don't deserve that. – molan May 11 '12 at 16:20 ## 1 Answer $$0\leq \frac{ab}{a+c}=\frac{b}{1+\frac{c}{a}}\leq b\leq 1$$ - You could add $0 \le$ on the left hand side and $\le 1$ on the right – Henry May 11 '12 at 16:31 You also should assume that $a\not=0$. – molan May 12 '12 at 2:40 It is given that $a>0$ – Julius May 12 '12 at 5:08	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 12, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9267320036888123, "perplexity_flag": "middle"}
http://mathhelpforum.com/advanced-algebra/119587-determinants-matrix-inverses.html	# Thread: 1. ## Determinants & Matrix Inverses If A and B are n by n, AB = -BA, and n is odd, show that either A or B are not invertible. Show that no 3 x 3 matrix A exists such that A^2 + I = 0. Any help is appreciated. 2. Originally Posted by BrownianMan If A and B are n by n, AB = -BA, and n is odd, show that either A or B are not invertible. Show that no 3 x 3 matrix A exists such that A^2 + I = 0. Any help is appreciated. Since $\det(AB)=\det(A)\det(B)$ , we get $AB=-BA\Longrightarrow \det(AB)=\det(-BA)=\det(-B)\det(A)=(-1)^n\det(B)\det(A)$. Now suppose both matrices are invertible (i.e., their determinant is non-zero), and get a huge contradiction. Tonio 3. You may have another constraint on your matrices you haven't mentioned yet - the matrix (where i=square root of -1): i,0,0 0,i,0 0,0,i satisfies the 2nd equation. If you demand that the matrix eigenvalues are real, the argument Tonio mentions should work.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9127157926559448, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/118892/on-the-fundamental-lemma	## on the fundamental lemma ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I consider the fundamental lemma for the spherical Hecke algebra. Let $G$ a connected reductive quasisplit group on $F$, a local field of equal characteristic $p$. and $H$ an endoscopic group. Can we reduce the fundamental lemma for (G,H) to the fundamental lemma for $(H,G)$ with $G_{der}=G_{sc}$ or even better to (H, G) where both G and H have a simply connected derived group? -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 6, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.7991892695426941, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/tagged/lie-algebras?sort=unanswered&pagesize=15	# Tagged Questions For questions about Lie algebras, an algebraic structure whose main use is in studying geometric objects such as Lie groups and differentiable manifolds. 0answers 204 views ### Abelian Cartan subalgebras If a Lie algebra is semisimple or reductive, its Cartan subalgebras are Abelian, and their elements semisimple. Are there non-reductive algebras with Abelian Cartan subalgebras all of whose ... 0answers 78 views ### Trivial summand of a representation's symmetric power The following comes from Exercise 13.17 of Fulton and Harris's book, Representation Theory: A First Course. 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I used to think that things called "parabolic" must have something to do with parabolas or their defining quadratic equations. In fact, terms like parabolic coordinate, parabolic partial differential ... 0answers 73 views ### Quantum Adjoint Action of the Coordinate Algebra on the Enveloping Algebra As is well known, any Lie group $G$ has a canonical action on its Lie algebra $\frak{g}$, namely the adjoint action $Ad$. Firstly, let me ask, does this extend to an action of $G$ on its enveloping ... 0answers 123 views ### Generating function for characters of representations One example of such a generating function that I know how to derive is for $SU(2)$, $\frac{1}{(1-tx)(1-\frac{t}{x})}$. The coefficient of $t^n$ in the above function is the character in the $n+1$ ... 0answers 89 views ### Pullback of a 3-form to SU2 I have a left invariant 3-form, $\sigma$ on an simply connected Lie group, $G$ whose value at the identity is $\sigma=\langle[x,y],z\rangle$, where $\langle\cdot,\cdot\rangle$ denotes an invariant ... 0answers 109 views ### An exercise in Serre's Lie algebra book Let $k$ be a commutative ring. Prove that a Lie $k$-algebra $\mathfrak{g} = 0$ iff $U\mathfrak{g} = k$. Use the adjoint representaion. Here is my attempt at it: The only non-trivial statement is ... 0answers 158 views ### Root Systems of Lie Groups. Let $G$ be a compact Lie group assumed to be a subgroup of $U(n)$. Also, let $T$ be a maximal torus of G. Then there exists a basis $\{v_1, \ldots ,v_d\}$ of the Lie algebra of $G$, $\mathfrak{g}$, ... 0answers 25 views ### Relationship between representations of $\mathfrak{sl}_{2n}\mathbb{C}$ and $\mathfrak{sp}_{2n}\mathbb{C}$ If $V=\mathbb{C}^{2n}$ denotes the standard representation of $\mathfrak{sl}_{2n}\mathbb{C}$, what can we say about $\wedge^kV$ in terms of the standard representation $W$ of ... 0answers 33 views ### Exercise in Erdmann's Intro to Lie algebras I'm working on question 4.8 on page 36 of Erdmann's book called Introduction to Lie Algebras. The question is as follows: Let $L$ be a Lie algebra over a field $F$, such that $[a,b],b]=0$ for all ... 0answers 29 views ### Is it possible to further simplify the product of three exponentials $e^A e^B e^C$ when $[A,C]=kB$ (k is a scalar) The background is calculation of the little group elements of Poincare group for massless particles. 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Following structure is given $\left(\mathbb{C}^d\right)^{\otimes n}$. Consider irreducible representations of $U(d)$. And consider the fully symmetric subspace $T_{\alpha}$ in ...	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 143, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8927125334739685, "perplexity_flag": "head"}
http://mathoverflow.net/questions/22359/why-havent-certain-well-researched-classes-of-mathematical-object-been-framed-by/22401	## why haven’t certain well-researched classes of mathematical object been framed by category theory? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Category theory is doing/has done a stellar job on Set, FinSet, Grp, Cob, Vect, cartesian closed categories provide a setting for $\lambda$-calculus, and Baez wrote a paper (Physics, Topology, Logic and Computation: A Rosetta Stone) with Mike Stay about many of the interconnections between them. But there are mathematical objects that aren't thought of in a category-theoretic fashion, at least the extant literature doesn't tend to treat them as such. For instance nobody talks about Series, Products, IndefInt as being categories in their own right. (infinite series, infinite products, and indefinite integrals, respectively). (google searches for the phrase "the category of infinite series" in both the web and book databases have no hits whatsoever). I suppose my question is: why not? - 39 One of the first things to learn about category theory is that not everything in mathematics is a category. (-: – Mike Shulman Apr 23 2010 at 15:02 39 "To a man with a hammer, everything looks like a nail." Mark Twain – Andrey Rekalo Apr 23 2010 at 15:19 39 Some version of this question is asked here on a monthly basis. The category-theorists always respond in a reasonable way, explaining that their subject, like any other, has limitations. To be blunt: no amount of adjoint-this or colimit-that will tell you whether a sequence converges or whether a PDE has a solution. Huge swathes of current mathematics depends on proving convergence and solving PDE. – Tim Perutz Apr 23 2010 at 15:48 5 @Yemon: Maybe someone whose only tool is a claw hammer would try to use it to unscrew things, despite the absurdity of even trying. The mathematical analogue is evident. – BCnrd Apr 23 2010 at 15:50 12 Tim Perutz's comment reminds me of a point made in the preface of Evans's PDE book. "PDE is not a branch of functional analysis. ... [T]he insistence on an overly abstract viewpoint, and consequent ignoring of deep calculus and measure theoretic estimates, is ultimately limiting." Even functional analytifying is often too much in analysis and differential equations, let alone categorifying. – Jonas Meyer Apr 23 2010 at 16:28 show 5 more comments ## 6 Answers Fundamentally I agree with Mike Shulman's comment and I do not really want to claim the following fancy language is at all necessary to answer this question, but you may (or may not) find it illuminating. From the standpoint of higher category theory, categories (i.e., 1-categories) are just one level among many in a family of mathematical structures. Typically a mathematical object will "naturally" exist as an n-category for some particular n. For example, Set is naturally a 1-category, while Cat is naturally a 2-category. Your examples Series and so on seem to just be 0-categories, i.e., sets, since as Pete explained in his answer, there is no obvious natural notion of morphism between infinite series. Asking why Series is not a 1-category is like asking why Set is not a 2-category; these are just not the natural categorical levels that these objects live at. - ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. To turn a class into a category, you need a notion of morphisms between objects in the class. That's the long and short of it. Consider for instance the class of infinite real series, say viewed as the set $S = \mathbb{R}^{\aleph_0}$ of sequences of real numbers. (There is often some notational and "ontological" confusion between the terms of an infinite series, its associated sequence of partial sums, and its sum, if it has one. Which one of these "is" the series? But such considerations are not relevant here and indeed are usually viewed as antithetical to the categorical point of view.) To get a category, you need to identify a set of morphisms between any two elements of this set. This can certainly be done in any number of ways -- for instance one could use the ordering induced from the standard ordering on $\mathbb{R}$ and the lexicographic ordering of the sequence, and then $S$ is a totally ordered set. We could then define a category by having $\operatorname{Hom}(s,t)$ to be a one point set if $s \leq t$ and the empty set otherwise (and then take the unique composition law of morphisms, defined when $s \leq t \leq u$). But the question is: what does this category have to do with any aspect of the theory of infinite series? Apparently nothing. You could create any number of other categories with underlying set $S$ but you run into the same problem: the very old and extremely well-developed area of mathematics which studies the convergence and divergence of real infinite series simply does not have anything evident to do with any notion of "morphisms" between infinite series. Similarly for the other examples you mention. Categorical structure is a very fundamental kind of mathematical structure; it's a great way of thinking and unifies and conceptualizes the study of many kinds of mathematical objects in highly disparate fields. But it doesn't explain everything, and it is frankly a bit weird to think it should. - 1 There are morphisms between different series of a given type -- there are variety of <a href="en.wikipedia.org/wiki/… transformations</a>, such as Eulerian series acceleration, there's scalar multiplication, which turns one given series into another, and there's the equivalence class of all series that sum to the same thing (pi and e formulas for instance.) – deoxygerbe Apr 23 2010 at 20:41 the link doesn't seem to be working right. The full url is here: en.wikipedia.org/wiki/Series_acceleration – deoxygerbe Apr 23 2010 at 20:45 8 I notice you made a point of mentioning convergence. On the other hand, if you consider formal series without regard to convergence, then we have some nice category theory going on. For a long time, combinatorics resisted a categorical treatment, but then along came combinatorial species, and lots of interesting combinatorial structures turned out to be functors. And at that point, many properties of formal power series, considered as generating functions, suddenly acquired category theoretical meaning. Resistance is futile. :-) en.wikipedia.org/wiki/Combinatorial_species – Dan Piponi Apr 23 2010 at 21:14 +1 for the Borg reference. I often have the fealing, category theory is assimilating the rest of the mathematical galaxy. ;-) – Johannes Hahn Apr 24 2010 at 0:06 It might be worth noting that the problems of computing Feynmann integrals in quantum field theory is one that is traditionally phrased as one of analysis, but is now studied by pure mathematicians using categorical techniques (among others). - Paul Taylor's "Abstract Stone Duality" http://www.paultaylor.eu/ASD/ is an attempt to recast elementary real analysis (including sequences) involving categorical ideas. - 1 As is Peter Freyd's 'Algebraic Real Analysis' tac.mta.ca/tac/volumes/20/10/20-10abs.html – David Corfield Apr 26 2010 at 7:45 A Google scholar search for "category theory" "power series" brings up the paper: Elements of Stream Calculus::(An Extensive Exercise in Coinduction). So series can be usefully thought of in a category-theoretic fashion, and although this involves formal series the methods can be used to find solutions to differential equations. - For just another point, we could contemplate the hilarious line "why can't a woman... be more like a man", from you-know-where. That is, the style of (much) "analysis" reflects as much the personalities of the participants as it does the "mathematical reality"... the latter being self-referentially "defined" by the personalities. That is, perhaps the issue addressed by the question resides in the implicit assumptions of the questioner? :) In my own experience, "(naive?) category theory", in the sense of paying attention more to interactions among objects than to their internal structures, has been extremely helpful, at the very least in showing that various seeming "choices" or "constructions" were irrelevant... But I do recognize that other people thing other-ly... - I would phrase Paul's sentence a bit differently, in saying the notion of for example colimit allows for the description of an internal structure in terms of the way the object is put together from external objects, and indeed that the colimit structure is defined by all its interrelationships with other objects. In a technical sense, proving something is a colimit allows for the elegant proof by "verification of the universal property", which often is the "best" proof, since that is how the notion of colimit is defined. – Ronnie Brown Jan 4 at 14:28	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 8, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9355809092521667, "perplexity_flag": "middle"}
http://en.wikipedia.org/wiki/Ordinal_number	# Ordinal number Representation of the ordinal numbers up to ωω. Each turn of the spiral represents one power of ω In set theory, an ordinal number, or just ordinal, is the order type of a well-ordered set. They are usually identified with hereditarily transitive sets. Ordinals are an extension of the natural numbers different from integers and from cardinals. Like other kinds of numbers, ordinals can be added, multiplied, and exponentiated. Ordinals were introduced by Georg Cantor in 1883 to accommodate infinite sequences and to classify sets with certain kinds of order structures on them.[1] He derived them by accident while working on a problem concerning trigonometric series—see Georg Cantor. Two sets S and S' have the same cardinality if there is a bijection between them (ie there exists a function f that is both injective and surjective, that is it maps each element x of S to a unique element y = f(x) of S' and each element y of S' comes from exactly one such element x of S.) If a partial order < is defined on set S, and a partial order <' is defined on set S' , then the posets (S,<) and (S' ,<') are order isomorphic if there is a bijection f that preserves the ordering. That is, f(a) <' f(b) if and only if a < b. Every well ordered set (S,<) is order isomorphic to the set of ordinals less than one specific ordinal number (the order type of (S,<)) under their natural ordering. The finite ordinals (and the finite cardinals) are the natural numbers: 0, 1, 2, …, since any two total orderings of a finite set are order isomorphic. The least infinite ordinal is ω, which is identified with the cardinal number $\aleph_0$. However in the transfinite case, beyond ω, ordinals draw a finer distinction than cardinals on account of their order information. Whereas there is only one countably infinite cardinal, namely $\aleph_0$ itself, there are uncountably many countably infinite ordinals, namely ω, ω + 1, ω + 2, …, ω·2, ω·2 + 1, …, ω2, …, ω3, …, ωω, …, ωωω, …, ε0, …. Here addition and multiplication are not commutative: in particular 1 + ω is ω rather than ω + 1 and likewise, 2·ω is ω rather than ω·2. The set of all countable ordinals constitutes the first uncountable ordinal ω1, which is identified with the cardinal $\aleph_1$ (next cardinal after $\aleph_0$). Well-ordered cardinals are identified with their initial ordinals, i.e. the smallest ordinal of that cardinality. The cardinality of an ordinal defines a many to one association from ordinals to cardinals. In general, each ordinal α, is the order type of the set of ordinals strictly less than the ordinal, α itself. This property permits every ordinal to be represented as the set of all ordinals less than it. Ordinals may be categorized as: zero, successor ordinals, and limit ordinals (of various cofinalities). Given a class of ordinals, one can identify the α-th member of that class, i.e. one can index (count) them. Such a class is closed and unbounded if its indexing function is continuous and never stops. The Cantor normal form uniquely represents each ordinal as a finite sum of ordinal powers of ω. However, this cannot form the basis of a universal ordinal notation due to such self-referential representations as ε0 = ωε0. Larger and larger ordinals can be defined, but they become more and more difficult to describe. Any ordinal number can be made into a topological space by endowing it with the order topology; this topology is discrete if and only if the ordinal is a countable cardinal, i.e. at most ω. A subset of ω + 1 is open in the order topology if and only if either it is cofinite or it does not contain ω as an element. ## Ordinals extend the natural numbers A natural number (which, in this context, includes the number 0) can be used for two purposes: to describe the size of a set, or to describe the position of an element in a sequence. When restricted to finite sets these two concepts coincide; there is only one way to put a finite set into a linear sequence, up to isomorphism. When dealing with infinite sets one has to distinguish between the notion of size, which leads to cardinal numbers, and the notion of position, which is generalized by the ordinal numbers described here. This is because, while any set has only one size (its cardinality), there are many nonisomorphic well-orderings of any infinite set, as explained below. Whereas the notion of cardinal number is associated with a set with no particular structure on it, the ordinals are intimately linked with the special kind of sets that are called well-ordered (so intimately linked, in fact, that some mathematicians make no distinction between the two concepts). A well-ordered set is a totally ordered set (given any two elements one defines a smaller and a larger one in a coherent way) in which there is no infinite decreasing sequence (however, there may be infinite increasing sequences); equivalently, every non-empty subset of the set has a least element. Ordinals may be used to label the elements of any given well-ordered set (the smallest element being labelled 0, the one after that 1, the next one 2, "and so on") and to measure the "length" of the whole set by the least ordinal that is not a label for an element of the set. This "length" is called the order type of the set. Any ordinal is defined by the set of ordinals that precede it: in fact, the most common definition of ordinals identifies each ordinal as the set of ordinals that precede it. For example, the ordinal 42 is the order type of the ordinals less than it, i.e., the ordinals from 0 (the smallest of all ordinals) to 41 (the immediate predecessor of 42), and it is generally identified as the set {0,1,2,…,41}. Conversely, any set (S) of ordinals that is downward-closed—meaning that for any ordinal α in S and any ordinal β < α, β is also in the set—is (or can be identified with) an ordinal. So far we have mentioned only finite ordinals, which are the natural numbers. But there are infinite ones as well: the smallest infinite ordinal is ω, which is the order type of the natural numbers (finite ordinals) and that can even be identified with the set of natural numbers (indeed, the set of natural numbers is well-ordered—as is any set of ordinals—and since it is downward closed it can be identified with the ordinal associated with it, which is exactly how we define ω). A graphical “matchstick” representation of the ordinal ω². Each stick corresponds to an ordinal of the form ω·m+n where m and n are natural numbers. Perhaps a clearer intuition of ordinals can be formed by examining a first few of them: as mentioned above, they start with the natural numbers, 0, 1, 2, 3, 4, 5, … After all natural numbers comes the first infinite ordinal, ω, and after that come ω+1, ω+2, ω+3, and so on. (Exactly what addition means will be defined later on: just consider them as names.) After all of these come ω·2 (which is ω+ω), ω·2+1, ω·2+2, and so on, then ω·3, and then later on ω·4. Now the set of ordinals we form in this way (the ω·m+n, where m and n are natural numbers) must itself have an ordinal associated with it: and that is ω2. Further on, there will be ω3, then ω4, and so on, and ωω, then ωω², and much later on ε0 (epsilon nought) (to give a few examples of relatively small—countable—ordinals). We can go on in this way indefinitely far ("indefinitely far" is exactly what ordinals are good at: basically every time one says "and so on" when enumerating ordinals, it defines a larger ordinal). The smallest uncountable ordinal is the set of all countable ordinals, expressed as ω1. ## Definitions ### Well-ordered sets Further information: Ordered set In a well-ordered set, every non-empty subset has a smallest element. Given the axiom of dependent choice, this is equivalent to just saying that the set is totally ordered and there is no infinite decreasing sequence, something perhaps easier to visualize. In practice, the importance of well-ordering is justified by the possibility of applying transfinite induction, which says, essentially, that any property that passes on from the predecessors of an element to that element itself must be true of all elements (of the given well-ordered set). If the states of a computation (computer program or game) can be well-ordered in such a way that each step is followed by a "lower" step, then you can be sure that the computation will terminate. Now we don't want to distinguish between two well-ordered sets if they only differ in the "labeling of their elements", or more formally: if we can pair off the elements of the first set with the elements of the second set such that if one element is smaller than another in the first set, then the partner of the first element is smaller than the partner of the second element in the second set, and vice versa. Such a one-to-one correspondence is called an order isomorphism and the two well-ordered sets are said to be order-isomorphic, or similar (obviously this is an equivalence relation). Provided there exists an order isomorphism between two well-ordered sets, the order isomorphism is unique: this makes it quite justifiable to consider the two sets as essentially identical, and to seek a "canonical" representative of the isomorphism type (class). This is exactly what the ordinals provide, and it also provides a canonical labeling of the elements of any well-ordered set. So we essentially wish to define an ordinal as an isomorphism class of well-ordered sets: that is, as an equivalence class for the equivalence relation of "being order-isomorphic". There is a technical difficulty involved, however, in the fact that the equivalence class is too large to be a set in the usual Zermelo–Fraenkel (ZF) formalization of set theory. But this is not a serious difficulty. We will say that the ordinal is the order type of any set in the class. ### Definition of an ordinal as an equivalence class The original definition of ordinal number, found for example in Principia Mathematica, defines the order type of a well-ordering as the set of all well-orderings similar (order-isomorphic) to that well-ordering: in other words, an ordinal number is genuinely an equivalence class of well-ordered sets. This definition must be abandoned in ZF and related systems of axiomatic set theory because these equivalence classes are too large to form a set. However, this definition still can be used in type theory and in Quine's axiomatic set theory New Foundations and related systems (where it affords a rather surprising alternative solution to the Burali-Forti paradox of the largest ordinal). ### Von Neumann definition of ordinals Rather than defining an ordinal as an equivalence class of well-ordered sets, we will define it as a particular well-ordered set that (canonically) represents the class. Thus, an ordinal number will be a well-ordered set; and every well-ordered set will be order-isomorphic to exactly one ordinal number. The standard definition, suggested by John von Neumann, is: each ordinal is the well-ordered set of all smaller ordinals. In symbols, λ = [0,λ).[2] Formally: A set S is an ordinal if and only if S is strictly well-ordered with respect to set membership and every element of S is also a subset of S. Note that the natural numbers are ordinals by this definition. For instance, 2 is an element of 4 = {0, 1, 2, 3}, and 2 is equal to {0, 1} and so it is a subset of {0, 1, 2, 3}. It can be shown by transfinite induction that every well-ordered set is order-isomorphic to exactly one of these ordinals, that is, there is an order preserving bijective function between them. Furthermore, the elements of every ordinal are ordinals themselves. Whenever you have two ordinals S and T, S is an element of T if and only if S is a proper subset of T. Moreover, either S is an element of T, or T is an element of S, or they are equal. So every set of ordinals is totally ordered. Further, every set of ordinals is well-ordered. This generalizes the fact that every set of natural numbers is well-ordered. Consequently, every ordinal S is a set having as elements precisely the ordinals smaller than S. For example, every set of ordinals has a supremum, the ordinal obtained by taking the union of all the ordinals in the set. This union exists regardless of the set's size, by the axiom of union. The class of all ordinals is not a set. If it were a set, one could show that it was an ordinal and thus a member of itself, which would contradict its strict ordering by membership. This is the Burali-Forti paradox. The class of all ordinals is variously called "Ord", "ON", or "∞". An ordinal is finite if and only if the opposite order is also well-ordered, which is the case if and only if each of its subsets has a maximum. ### Other definitions There are other modern formulations of the definition of ordinal. For example, assuming the axiom of regularity, the following are equivalent for a set x: • x is an ordinal, • x is a transitive set, and set membership is trichotomous on x, • x is a transitive set totally ordered by set inclusion, • x is a transitive set of transitive sets. These definitions cannot be used in non-well-founded set theories. In set theories with urelements, one has to further make sure that the definition excludes urelements from appearing in ordinals. ## Transfinite sequence If α is a limit ordinal and X is a set, an α-indexed sequence of elements of X is a function from α to X. This concept, a transfinite sequence or ordinal-indexed sequence, is a generalization of the concept of a sequence. An ordinary sequence corresponds to the case α = ω. ## Transfinite induction Main article: Transfinite induction ### What is transfinite induction? Transfinite induction holds in any well-ordered set, but it is so important in relation to ordinals that it is worth restating here. Any property that passes from the set of ordinals smaller than a given ordinal α to α itself, is true of all ordinals. That is, if P(α) is true whenever P(β) is true for all β<α, then P(α) is true for all α. Or, more practically: in order to prove a property P for all ordinals α, one can assume that it is already known for all smaller β<α. ### Transfinite recursion Transfinite induction can be used not only to prove things, but also to define them. Such a definition is normally said to be by transfinite recursion – the proof that the result is well-defined uses transfinite induction. Let F denote a (class) function F to be defined on the ordinals. The idea now is that, in defining F(α) for an unspecified ordinal α, one may assume that F(β) is already defined for all β < α and thus give a formula for F(α) in terms of these F(β). It then follows by transfinite induction that there is one and only one function satisfying the recursion formula up to and including α. Here is an example of definition by transfinite recursion on the ordinals (more will be given later): define function F by letting F(α) be the smallest ordinal not in the class {F(β) \| β < α}, that is, the class consisting of all F(β) for β < α. This definition assumes the F(β) known in the very process of defining F; this apparent vicious circle is exactly what definition by transfinite recursion permits. In fact, F(0) makes sense since there is no ordinal β < 0, and the class {F(β) \| β < 0} is empty. So F(0) is equal to 0 (the smallest ordinal of all). Now that F(0) is known, the definition applied to F(1) makes sense (it is the smallest ordinal not in the singleton class {F(0)} = {0}), and so on (the and so on is exactly transfinite induction). It turns out that this example is not very exciting, since provably F(α) = α for all ordinals α, which can be shown, precisely, by transfinite induction. ### Successor and limit ordinals Any nonzero ordinal has the minimum element, zero. It may or may not have a maximum element. For example, 42 has maximum 41 and ω+6 has maximum ω+5. On the other hand, ω does not have a maximum since there is no largest natural number. If an ordinal has a maximum α, then it is the next ordinal after α, and it is called a successor ordinal, namely the successor of α, written α+1. In the von Neumann definition of ordinals, the successor of α is $\alpha\cup\{\alpha\}$ since its elements are those of α and α itself. A nonzero ordinal that is not a successor is called a limit ordinal. One justification for this term is that a limit ordinal is indeed the limit in a topological sense of all smaller ordinals (under the order topology). When $\langle \alpha_{\iota} \| \iota < \gamma \rangle$ is an ordinal-indexed sequence, indexed by a limit γ and the sequence is increasing, i.e. $\alpha_{\iota} < \alpha_{\rho}\!$ whenever $\iota < \rho,\!$ we define its limit to be the least upper bound of the set $\{ \alpha_{\iota} \| \iota < \gamma \},\!$ that is, the smallest ordinal (it always exists) greater than any term of the sequence. In this sense, a limit ordinal is the limit of all smaller ordinals (indexed by itself). Put more directly, it is the supremum of the set of smaller ordinals. Another way of defining a limit ordinal is to say that α is a limit ordinal if and only if: There is an ordinal less than α and whenever ζ is an ordinal less than α, then there exists an ordinal ξ such that ζ < ξ < α. So in the following sequence: 0, 1, 2, ... , ω, ω+1 ω is a limit ordinal because for any smaller ordinal (in this example, a natural number) we can find another ordinal (natural number) larger than it, but still less than ω. Thus, every ordinal is either zero, or a successor (of a well-defined predecessor), or a limit. This distinction is important, because many definitions by transfinite induction rely upon it. Very often, when defining a function F by transfinite induction on all ordinals, one defines F(0), and F(α+1) assuming F(α) is defined, and then, for limit ordinals δ one defines F(δ) as the limit of the F(β) for all β<δ (either in the sense of ordinal limits, as we have just explained, or for some other notion of limit if F does not take ordinal values). Thus, the interesting step in the definition is the successor step, not the limit ordinals. Such functions (especially for F nondecreasing and taking ordinal values) are called continuous. We will see that ordinal addition, multiplication and exponentiation are continuous as functions of their second argument. ### Indexing classes of ordinals We have mentioned that any well-ordered set is similar (order-isomorphic) to a unique ordinal number $\alpha$, or, in other words, that its elements can be indexed in increasing fashion by the ordinals less than $\alpha$. This applies, in particular, to any set of ordinals: any set of ordinals is naturally indexed by the ordinals less than some $\alpha$. The same holds, with a slight modification, for classes of ordinals (a collection of ordinals, possibly too large to form a set, defined by some property): any class of ordinals can be indexed by ordinals (and, when the class is unbounded in the class of all ordinals, this puts it in class-bijection with the class of all ordinals). So we can freely speak of the $\gamma$-th element in the class (with the convention that the “0-th” is the smallest, the “1-th” is the next smallest, and so on). Formally, the definition is by transfinite induction: the $\gamma$-th element of the class is defined (provided it has already been defined for all $\beta<\gamma$), as the smallest element greater than the $\beta$-th element for all $\beta<\gamma$. We can apply this, for example, to the class of limit ordinals: the $\gamma$-th ordinal, which is either a limit or zero is $\omega\cdot\gamma$ (see ordinal arithmetic for the definition of multiplication of ordinals). Similarly, we can consider additively indecomposable ordinals (meaning a nonzero ordinal that is not the sum of two strictly smaller ordinals): the $\gamma$-th additively indecomposable ordinal is indexed as $\omega^\gamma$. The technique of indexing classes of ordinals is often useful in the context of fixed points: for example, the $\gamma$-th ordinal $\alpha$ such that $\omega^\alpha=\alpha$ is written $\varepsilon_\gamma$. These are called the "epsilon numbers". ### Closed unbounded sets and classes A class $C$ of ordinals is said to be unbounded, or cofinal, when given any ordinal $\alpha$, there is a $\beta$ in $C$ such that $\alpha < \beta$ (then the class must be a proper class, i.e., it cannot be a set). It is said to be closed when the limit of a sequence of ordinals in the class is again in the class: or, equivalently, when the indexing (class-)function $F$ is continuous in the sense that, for $\delta$ a limit ordinal, $F(\delta)$ (the $\delta$-th ordinal in the class) is the limit of all $F(\gamma)$ for $\gamma<\delta$; this is also the same as being closed, in the topological sense, for the order topology (to avoid talking of topology on proper classes, one can demand that the intersection of the class with any given ordinal is closed for the order topology on that ordinal, this is again equivalent). Of particular importance are those classes of ordinals that are closed and unbounded, sometimes called clubs. For example, the class of all limit ordinals is closed and unbounded: this translates the fact that there is always a limit ordinal greater than a given ordinal, and that a limit of limit ordinals is a limit ordinal (a fortunate fact if the terminology is to make any sense at all!). The class of additively indecomposable ordinals, or the class of $\varepsilon_\cdot$ ordinals, or the class of cardinals, are all closed unbounded; the set of regular cardinals, however, is unbounded but not closed, and any finite set of ordinals is closed but not unbounded. A class is stationary if it has a nonempty intersection with every closed unbounded class. All superclasses of closed unbounded classes are stationary, and stationary classes are unbounded, but there are stationary classes that are not closed and stationary classes that have no closed unbounded subclass (such as the class of all limit ordinals with countable cofinality). Since the intersection of two closed unbounded classes is closed and unbounded, the intersection of a stationary class and a closed unbounded class is stationary. But the intersection of two stationary classes may be empty, e.g. the class of ordinals with cofinality ω with the class of ordinals with uncountable cofinality. Rather than formulating these definitions for (proper) classes of ordinals, we can formulate them for sets of ordinals below a given ordinal $\alpha$: A subset of a limit ordinal $\alpha$ is said to be unbounded (or cofinal) under $\alpha$ provided any ordinal less than $\alpha$ is less than some ordinal in the set. More generally, we can call a subset of any ordinal $\alpha$ cofinal in $\alpha$ provided every ordinal less than $\alpha$ is less than or equal to some ordinal in the set. The subset is said to be closed under $\alpha$ provided it is closed for the order topology in $\alpha$, i.e. a limit of ordinals in the set is either in the set or equal to $\alpha$ itself. ## Arithmetic of ordinals Main article: Ordinal arithmetic There are three usual operations on ordinals: addition, multiplication, and (ordinal) exponentiation. Each can be defined in essentially two different ways: either by constructing an explicit well-ordered set that represents the operation or by using transfinite recursion. Cantor normal form provides a standardized way of writing ordinals. The so-called "natural" arithmetical operations retain commutativity at the expense of continuity. ## Ordinals and cardinals ### Initial ordinal of a cardinal Each ordinal has an associated cardinal, its cardinality, obtained by simply forgetting the order. Any well-ordered set having that ordinal as its order-type has the same cardinality. The smallest ordinal having a given cardinal as its cardinality is called the initial ordinal of that cardinal. Every finite ordinal (natural number) is initial, but most infinite ordinals are not initial. The axiom of choice is equivalent to the statement that every set can be well-ordered, i.e. that every cardinal has an initial ordinal. In this case, it is traditional to identify the cardinal number with its initial ordinal, and we say that the initial ordinal is a cardinal. Cantor used the cardinality to partition ordinals into classes. He referred to the natural numbers as the first number class, the ordinals with cardinality $\aleph_0$ (the countably infinite ordinals) as the second number class and generally, the ordinals with cardinality $\aleph_{n-2}$ as the n-th number class.[3] The α-th infinite initial ordinal is written $\omega_\alpha$. Its cardinality is written $\aleph_\alpha$. For example, the cardinality of ω0 = ω is $\aleph_0$, which is also the cardinality of ω2 or ε0 (all are countable ordinals). So (assuming the axiom of choice) we identify ω with $\aleph_0$, except that the notation $\aleph_0$ is used when writing cardinals, and ω when writing ordinals (this is important since, for example, $\aleph_0^2$ = $\aleph_0$ whereas $\omega^2>\omega$). Also, $\omega_1$ is the smallest uncountable ordinal (to see that it exists, consider the set of equivalence classes of well-orderings of the natural numbers: each such well-ordering defines a countable ordinal, and $\omega_1$ is the order type of that set), $\omega_2$ is the smallest ordinal whose cardinality is greater than $\aleph_1$, and so on, and $\omega_\omega$ is the limit of the $\omega_n$ for natural numbers n (any limit of cardinals is a cardinal, so this limit is indeed the first cardinal after all the $\omega_n$). See also Von Neumann cardinal assignment. ### Cofinality The cofinality of an ordinal $\alpha$ is the smallest ordinal $\delta$ that is the order type of a cofinal subset of $\alpha$. Notice that a number of authors define cofinality or use it only for limit ordinals. The cofinality of a set of ordinals or any other well ordered set is the cofinality of the order type of that set. Thus for a limit ordinal, there exists a $\delta$-indexed strictly increasing sequence with limit $\alpha$. For example, the cofinality of ω² is ω, because the sequence ω·m (where m ranges over the natural numbers) tends to ω²; but, more generally, any countable limit ordinal has cofinality ω. An uncountable limit ordinal may have either cofinality ω as does $\omega_\omega$ or an uncountable cofinality. The cofinality of 0 is 0. And the cofinality of any successor ordinal is 1. The cofinality of any limit ordinal is at least $\omega$. An ordinal that is equal to its cofinality is called regular and it is always an initial ordinal. Any limit of regular ordinals is a limit of initial ordinals and thus is also initial even if it is not regular, which it usually is not. If the Axiom of Choice, then $\omega_{\alpha+1}$ is regular for each α. In this case, the ordinals 0, 1, $\omega$, $\omega_1$, and $\omega_2$ are regular, whereas 2, 3, $\omega_\omega$, and ωω·2 are initial ordinals that are not regular. The cofinality of any ordinal α is a regular ordinal, i.e. the cofinality of the cofinality of α is the same as the cofinality of α. So the cofinality operation is idempotent. ## Some “large” countable ordinals For more details on this topic, see Large countable ordinal. We have already mentioned (see Cantor normal form) the ordinal ε0, which is the smallest satisfying the equation $\omega^\alpha = \alpha$, so it is the limit of the sequence 0, 1, $\omega$, $\omega^\omega$, $\omega^{\omega^\omega}$, etc. Many ordinals can be defined in such a manner as fixed points of certain ordinal functions (the $\iota$-th ordinal such that $\omega^\alpha = \alpha$ is called $\varepsilon_\iota$, then we could go on trying to find the $\iota$-th ordinal such that $\varepsilon_\alpha = \alpha$, “and so on”, but all the subtlety lies in the “and so on”). We can try to do this systematically, but no matter what system is used to define and construct ordinals, there is always an ordinal that lies just above all the ordinals constructed by the system. Perhaps the most important ordinal that limits a system of construction in this manner is the Church–Kleene ordinal, $\omega_1^{\mathrm{CK}}$ (despite the $\omega_1$ in the name, this ordinal is countable), which is the smallest ordinal that cannot in any way be represented by a computable function (this can be made rigorous, of course). Considerably large ordinals can be defined below $\omega_1^{\mathrm{CK}}$, however, which measure the “proof-theoretic strength” of certain formal systems (for example, $\varepsilon_0$ measures the strength of Peano arithmetic). Large ordinals can also be defined above the Church-Kleene ordinal, which are of interest in various parts of logic. ## Topology and ordinals For more details on this topic, see Order topology. Any ordinal can be made into a topological space in a natural way by endowing it with the order topology. See the Topology and ordinals section of the "Order topology" article. ## Downward closed sets of ordinals A set is downward closed if anything less than an element of the set is also in the set. If a set of ordinals is downward closed, then that set is an ordinal—the least ordinal not in the set. Examples: • The set of ordinals less than 3 is 3 = { 0, 1, 2 }, the smallest ordinal not less than 3. • The set of finite ordinals is infinite, the smallest infinite ordinal: ω. • The set of countable ordinals is uncountable, the smallest uncountable ordinal: ω1. ## Notes 1. Thorough introductions are given by Levy (1979) and Jech (2003). 2. Levy (1979, p. 52) attributes the idea to unpublished work of Zermelo in 1916 and several papers by von Neumann the 1920s. 3. Dauben (1990:97) ## References • Cantor, G., (1897), Beitrage zur Begrundung der transfiniten Mengenlehre. II (tr.: Contributions to the Founding of the Theory of Transfinite Numbers II), Mathematische Annalen 49, 207-246 English translation. • Conway, J. H. and Guy, R. K. "Cantor's Ordinal Numbers." In The Book of Numbers. New York: Springer-Verlag, pp. 266–267 and 274, 1996. • Dauben, Joseph Warren, (1990), Georg Cantor: his mathematics and philosophy of the infinite. Chapter 5: The Mathematics of Cantor's Grundlagen. ISBN 0-691-02447-2 • Hamilton, A. G. (1982), Numbers, Sets, and Axioms : the Apparatus of Mathematics, New York: Cambridge University Press, ISBN 0-521-24509-5 See Ch. 6, "Ordinal and cardinal numbers" • Kanamori, A., Set Theory from Cantor to Cohen, to appear in: Andrew Irvine and John H. Woods (editors), The Handbook of the Philosophy of Science, volume 4, Mathematics, Cambridge University Press. • Levy, A. (1979), Basic Set Theory, Berlin, New York: Springer-Verlag Reprinted 2002, Dover. ISBN 0-486-42079-5 • • Sierpiński, W. (1965). Cardinal and Ordinal Numbers (2nd ed.). Warszawa: Państwowe Wydawnictwo Naukowe. Also defines ordinal operations in terms of the Cantor Normal Form. • Suppes, P. (1960), Axiomatic Set Theory, D.Van Nostrand Company Inc., ISBN 0-486-61630-4	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 98, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9277245998382568, "perplexity_flag": "head"}
http://wikipedia.sfstate.us/Battle_of_the_sexes_(game_theory)	edit # Battle of the sexes (game theory) For other uses, see Battle of the sexes (disambiguation). \| \| \| \| \|-----------------------\|-------\|----------\| \| \| Opera \| Football \| \| Opera \| 3,2 \| 0,0 \| \| Football \| 0,0 \| 2,3 \| \| Battle of the Sexes 1 \| \| \| \| \| \| \| \|-----------------------\|-------\|----------\| \| \| Opera \| Football \| \| Opera \| 3,2 \| 1,1 \| \| Football \| 0,0 \| 2,3 \| \| Battle of the Sexes 2 \| \| \| In game theory, battle of the sexes (BoS), also called Bach or Stravinsky,1 is a two-player coordination game. Imagine a couple that agreed to meet this evening, but cannot recall if they will be attending the opera or a football match (and the fact that they forgot is common knowledge). The husband would most of all like to go to the football game. The wife would like to go to the opera. Both would prefer to go to the same place rather than different ones. If they cannot communicate, where should they go? The payoff matrix labeled "Battle of the Sexes (1)" is an example of Battle of the Sexes, where the wife chooses a row and the husband chooses a column. In each cell, the first number represents the payoff to the wife and the second number represents the payoff to the husband. This representation does not account for the additional harm that might come from not only going to different locations, but going to the wrong one as well (e.g. he goes to the opera while she goes to the football game, satisfying neither). To account for this, the game is sometimes represented as in "Battle of the Sexes (2)". ## Equilibrium analysis This game has two pure strategy Nash equilibria, one where both go to the opera and another where both go to the football game. For the first game, there is also a Nash equilibrium in mixed strategies, where the players go to their preferred event more often than the other. For the payoffs listed above, each player attends their preferred event with probability 3/5. This presents an interesting case for game theory since each of the Nash equilibria is deficient in some way. The two pure strategy Nash equilibria are unfair; one player consistently does better than the other. The mixed strategy Nash equilibrium (when it exists) is inefficient. The players will miscoordinate with probability 13/25, leaving each player with an expected return of 6/25 (less than the return one would receive from constantly going to one's less favored event). One possible resolution of the difficulty involves the use of a correlated equilibrium. In its simplest form, if the players of the game have access to a commonly observed randomizing device, then they might decide to correlate their strategies in the game based on the outcome of the device. For example, if the couple could flip a coin before choosing their strategies, they might agree to correlate their strategies based on the coin flip by, say, choosing football in the event of heads and opera in the event of tails. Notice that once the results of the coin flip are revealed neither the husband nor wife have any incentives to alter their proposed actions – that would result in miscoordination and a lower payoff than simply adhering to the agreed upon strategies. The result is that perfect coordination is always achieved and, prior to the coin flip, the expected payoffs for the players are exactly equal. ### Working out the above Let us calculate the four probabilities for the actions of the individuals (Man and Woman), which depend on their expectations of the behaviour of the other, and the relative payoff from each action. The Man either goes to the Football or the Opera (and not both or neither), and likewise the Woman. The Probability that the man goes to the football game, $Pmf$, equals the payoff if he does (whether or not the woman does), divided by the same payoff plus the payoff if he goes to the opera instead: • $Pmf = \cfrac{1 Pwo + 3 Pwf}{1 Pwo + 3 Pwf + 2 Pwo + 0 Pwf} = \cfrac{1 Pwo + 3 Pwf}{3 Pwo + 3 Pwf}$ We know that she either goes to one or the other, so $Pwo + Pwf = 1$, so: • $Pmf = \cfrac{1 Pwo + 3 Pwf}{3}$ Similarly: • $Pmo = \cfrac{2 Pwo}{3}$ • $Pwf = \cfrac{2 Pmf}{3}$ • $Pwo = \cfrac{3 Pmo + 1 Pmf}{3}$ This forms a set of simultaneous equations. We can solve these, starting with $Pmf$ for example, by substituting in the equations above: • $Pmf = \cfrac{1 \cfrac{3 Pmo + 1 Pmf}{3} + 3 \cfrac{2 Pmf}{3}}{3}$ • $Pmf = \cfrac{1 \cfrac{3 Pmo + 1 Pmf}{3} + 2 Pmf}{3}$ • $Pmf = \cfrac{Pmo + \tfrac{1}{3}Pmf + 2 Pmf}{3}$ Remembering that $Pmo + Pmf = 1$, we can make this an equation where the only unknown is $Pmf$: • $Pmf = \cfrac{1 - Pmf + \tfrac{1}{3}Pmf + 2 Pmf}{3}$ And then rearrange so that $Pmf$ is only on one side: • $3 Pmf = 1 - Pmf + \tfrac{1}{3}Pmf + 2 Pmf$ • $3 Pmf = 1 + \tfrac{4}{3}Pmf$ • $3 Pmf - \tfrac{4}{3}Pmf = 1$ • $\tfrac{9}{3} Pmf - \tfrac{4}{3}Pmf = 1$ • $\cfrac{9 Pmf - 4 Pmf}{3} = 1$ • $\cfrac{5 Pmf}{3} = 1$ • $5 Pmf = 3$ • $Pmf = \tfrac{3}{5}$ Knowing that $Pmf + Pmo = 1$, we deduce: • $Pmo = 1 - \dfrac{3}{5} = \tfrac{2}{5}$ • $Pwf = \cfrac{2 Pmf}{3} = \tfrac{6}{15} = \tfrac{2}{5}$ • $Pwo = 1 - \dfrac{2}{5} = \tfrac{3}{5}$ Then we can calculate the probability of coordination $Pc$ (that M and W do the same thing, independently), as: • $Pc = Pmf Pwf + Pmo Pwo$ • $Pc = \tfrac{3}{5} \tfrac{2}{5} + \tfrac{2}{5} \tfrac{3}{5}$ • $Pc = 2 \tfrac{6}{25} = \tfrac{12}{25}$ And the probability of miscoordination $Pm$ (that M and W do different things, independently): • $Pm = Pmf Pwo + Pmo Pwf$ • $Pm = \tfrac{3}{5} \tfrac{3}{5} + \tfrac{2}{5} \tfrac{2}{5}$ • $Pm = \tfrac{9}{25} + \tfrac{4}{25} = \tfrac{13}{25}$ And just to check our probability working: • $Pc + Pm = \tfrac{12}{25} + \tfrac{13}{25} = \tfrac{25}{25} = 1$ So the probability of miscoordination is $\tfrac{13}{25}$ as stated above. The expected payoff E for each individual ($Em$ and $Ew$) is the probability of each event multiplied by the payoff if it happens. For example, the Probability that the man goes to football and the Woman goes to football multiplied by the Expected payoff to the man if that happens ($Emmfwf$): • $Em = Pmf Pwf Emmfwf + Pmf Pwo Emmfwo + Pmo Pwo Emmowo + Pmo Pwf Emmowf$ • $Em = \tfrac{3}{5} \tfrac{2}{5} 3 + \tfrac{3}{5} \tfrac{3}{5} 1 + \tfrac{2}{5} \tfrac{3}{5} 2 + \tfrac{2}{5} \tfrac{2}{5} 0$ • $Em = \tfrac{18}{25} + \tfrac{9}{25} + \tfrac{12}{25} + 0$ • $Em = \tfrac{39}{25}$ Which is not the same as the $\tfrac{6}{5}$ stated above! For comparison, let us assume that the man always goes to football and the woman, knowing this, chooses what to do based on revised probabilities and expected values to her: • $Pmf = 1$ • $Pmo = 0$ • $Pwf = \cfrac{2 Pmf}{3} = \tfrac{2}{3}$ • $Pwo = 1 - \dfrac{2}{3} = \tfrac{1}{3}$ • $Em = Pmf Pwf Emmfwf + Pmf Pwo Emmfwo + Pmo Pwo Emmowo + Pmo Pwf Emmowf$ • $Em = 1 \tfrac{2}{3} 3 + 1 \tfrac{1}{3} 1 + 0 \tfrac{1}{3} 2 + 0 \tfrac{2}{3} 0$ • $Em = \tfrac{6}{3} + \tfrac{1}{3} = \tfrac{7}{3}$ This is symmetric for $Ew$ if the woman always goes to the opera and the man chooses randomly with probabilities based on the expected outcome, due to the symmetry in the value table. But if both players always do the same thing (both have simple strategies), the payoff is just 1 for both, from the table above. ## Burning money \| \| \| \| \|----------\|-------\|----------\| \| \| Opera \| Football \| \| Opera \| 4,1 \| 0,0 \| \| Football \| 0,0 \| 1,4 \| \| Unburned \| \| \| \| \| \| \| \|----------\|-------\|----------\| \| \| Opera \| Football \| \| Opera \| 2,1 \| -2,0 \| \| Football \| -2,0 \| -1,4 \| \| Burned \| \| \| Interesting strategic changes can take place in this game if one allows one player the option of "burning money" – that is, allowing that player to destroy some of her utility. Consider the version of Battle of the Sexes pictured here (called Unburned). Before making the decision the row player can, in view of the column player, choose to set fire to 2 points making the game Burned pictured to the right. This results in a game with four strategies for each player. The row player can choose to burn or not burn the money and also choose to play Opera or Football. The column player observes whether or not the row player burns and then chooses either to play Opera or Football. If one iteratively deletes weakly dominated strategies then one arrives at a unique solution where the row player does not burn the money and plays Opera and where the column player plays Opera. The odd thing about this result is that by simply having the opportunity to burn money (but not actually using it), the row player is able to secure her favored equilibrium. The reasoning that results in this conclusion is known as forward induction and is somewhat controversial. For a detailed explanation, see [1] p8 Section 4.5. In brief, by choosing not to burn money, the player is indicating she expects an outcome that is better than any of the outcomes available in the "burned" version, and this conveys information to the other party about which branch she will take. ## References • Luce, R.D. and Raiffa, H. (1957) Games and Decisions: An Introduction and Critical Survey, Wiley & Sons. (see Chapter 5, section 3). • Fudenberg, D. and Tirole, J. (1991) Game theory, MIT Press. (see Chapter 1, section 2.4) 1. Osborne, Rubinstein (1994). A course in game theory. The MIT Press.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 53, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9344313740730286, "perplexity_flag": "middle"}
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How its notions are used in the field of ... 1answer 131 views ### Expressibility and numbering A predicate $P$ is expressible (in PA) if there exists a formula $\phi(x_1,\ldots, x_n)$ of $L_A$ such that for all $m_1,\ldots, m_n$ elements of $\mathbb{N}$, we have that $P(m_1,\ldots, m_n)$ holds ... 1answer 151 views ### Second incompleteness and Model theorey If we let $T$ be a consistent theory in the language of arithmetic $\mathcal{L}_A$ theory extending Peano Arithmetic — with specified numbering of formulas $\left[\cdot\right]$ and suppose that ...	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 104, "mathjax_display_tex": 10, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9138002395629883, "perplexity_flag": "middle"}
http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.jam/1331817623	### Orthogonal Multiwavelet Frames in ${L}^{2}({R}^{d})$ Liu Zhanwei, Hu Guoen, and Wu Guochang Source: J. Appl. Math. Volume 2012 (2012), Article ID 846852, 18 pages. #### Abstract We characterize the orthogonal frames and orthogonal multiwavelet frames in ${L}^{2}({R}^{d})$ with matrix dilations of the form $(Df)(x)=\sqrt{\|\text{d}\text{e}\text{t}A\|}f(Ax)$, where $A$ is an arbitrary expanding $d×d$ matrix with integer coefficients. Firstly, through two arbitrarily multiwavelet frames, we give a simple construction of a pair of orthogonal multiwavelet frames. Then, by using the unitary extension principle, we present an algorithm for the construction of arbitrarily many orthogonal multiwavelet tight frames. Finally, we give a general construction algorithm for orthogonal multiwavelet tight frames from a scaling function. First Page: We're sorry, but we are unable to provide you with the full text of this article because we are not able to identify you as a subscriber. If you have a personal subscription to this journal, then please login. If you are already logged in, then you may need to update your profile to register your subscription. Read more about accessing full-text	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 5, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8538500666618347, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/103394?sort=votes	## Fundamental Solutions with compact support (distributions) ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Assume that we have a differential operator such as $-\frac{\partial}{\partial x^2} + id$ on $\mathbb{R}^1$ We also then argue that if a fundamental solution has compact support, then it is supported on the origin. My follow up question is how can one then show assuming that the fundamental solution is compactly supported - that the differential operator must be of order 0? - I myself am confused by the context in which an argument proceeds by saying that if a fundamental solution for a differential operator were compactly supported then it would have support $\{0\}$, leading to deducing that the operator is order $0$, so apparently is a multiplication operator? Or is this meant to be about pseudo-differential operators? Would you clarify the context? – paul garrett Jul 28 at 19:01 I'm not sure if this includes pseudo-differential operators. The question wasn't well stated to me, I got the impression it only involved differential operators (with constant coefficients). Apparently this can be shown by an appeal to the Fourier transform of distributions... – Peadar Coyle Nov 7 at 0:52 ## 2 Answers For a constant coefficient partial differential operator P(D), the fundamental solution of P can never belong to $\epsilon'(\mathbb{R}^{n})$,i.e.have compact support. In fact,assume we have $P(D)u=f$,where u is a distribution,then u have compact support $\Leftrightarrow$ $\frac{f}{P(\xi)}$ is analytic(The result can be found in Hormander's ALPDO, volume 1,ch7.) Now,if we have $$P(D)u=\delta$$,obviously $\frac{1}{P(\xi)}$ is never an analytic function for a polynomia P.So the fundamental solution of P can not be compact supported. - ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. If the fundamental solution is supported at the origin, it must be a finite combination of derivatives of the Dirac distribution. This means that your original operator was of nonpositive order. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 9, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9402008056640625, "perplexity_flag": "middle"}
http://mathhelpforum.com/pre-calculus/54256-parallel-perpendicular-lines-help.html	# Thread: 1. ## parallel & perpendicular lines help hey everyone, this is my first post...i hope i can find some help here, as i have searched my math book up and down, only to find no answer. i am in college algebra, and the section is parallel & perpendicular lines. the problem i am totally stuck on is this... Code: ``` x - y x + y ----- = ----- - 1 3 2 AND 7 = - ( x - y ) + 4y``` They want me to rewrite both in slope-intercept form, then find out if they are parallel or perpendicular. I would really and truly appreciate any and all help with this...i just don't know what to do thanks so much, ryan 2. Originally Posted by shipwreck hey everyone, this is my first post...i hope i can find some help here, as i have searched my math book up and down, only to find no answer. i am in college algebra, and the section is parallel & perpendicular lines. the problem i am totally stuck on is this... Code: ``` x - y x + y ----- = ----- - 1 3 2 AND 7 = - ( x - y ) + 4y``` They want me to rewrite both in slope-intercept form, then find out if they are parallel or perpendicular. I would really and truly appreciate any and all help with this...i just don't know what to do thanks so much, ryan Hello Ryan, $\frac{x-y}{3}=\frac{x+y}{2}-1$ First, let's multiply everything by the LCD of 6 to get rid of those pesky denominators. $6\left(\frac{x-y}{3}\right)=6\left(\frac{x+y}{2}\right)-6(1)$ $2x-2y=3x+3y-6$ Simplify to $-5y=x-6$ $\boxed{y=\frac{-1}{5}x+\frac{6}{5}}$ Now for the other one... $7=-(x-y)+4y$ $7=-x+y+4y$ Simplifying.... $5y=x+7$ $\boxed{y=\frac{1}{5}x+\frac{7}{5}}$ Compare the slopes of the two boxed equations. What do you think? 3. well i know that if both slopes are identical, than they are considered parallel....but these aren't identical, although i don't know if there is a particular rule or not about them being opposites, but NOT being reciprocal negative opposites... 1/5 x & 1/5x would be parallel, but i am just not 100% certain about 1/5x & -1/5x. i would say they are not parallel...right? 4. anyone able to confirm if i am right or not about these being parallel or not? 5. Originally Posted by shipwreck anyone able to confirm if i am right or not about these being parallel or not? You answered your question yourself. If the slopes are not the same then they aren't the parallel	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 9, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9525296688079834, "perplexity_flag": "middle"}
http://en.wikisource.org/wiki/On_some_Dynamical_Conditions/IV	On some Dynamical Conditions/IV From Wikisource Preston, Samuel Tolver (1878), “The Bearing of the Kinetic Theory of Gravitation on the Phenomena of "Cohesion" and "Chemical Action," together with the important connected Inferences regarding the existence of Stores of Motion in Space, No. IV”, Philosophical Magazine 5: 297-311 Samuel Tolver Preston No. IV[1] 1. IT would be natural to expect that any theory competent to explain the effects of gravity ought to be able to throw some light upon the subsidiary effects of molecules exhibited in "cohesion," "chemical action," &c. Before proceeding to consider this question, and in order to have a clear conception of the point we have to deal with, we will recapitulate in a few words the physical conditions involved in the case of gravity as already dealt with. It has been our object to point out that the molecules of a gas within the range of free path are moving in precisely the right way to produce [298] gravity in two masses immersed in the gas within the range of free path. For since it has been proved from the kinetic theory that the particles of a gas adjust their motions so as to move uniformly or equally in all directions, and since the particles within the range of free path are moving in unbroken streams, it follows that two masses immersed in the gas at a distance apart within this range will (owing to the one sheltering the other) be struck with more particles on their remote (unsheltered) sides than on their adjacent (sheltered) sides, so that the two masses will be urged together. This, therefore, fulfils Le Sage's fundamental idea without the necessity for accepting any of his postulates. We need not accept the scarcely realizable postulates of streams of particles coming from indefinite distances in space (at uniform angles), each stream moving continuously in one direction; but we can substitute for this the natural conception of the normal motion of the particles of a gas within the range of free path, where, although each particle is continually changing the direction of its motion, yet the general character of the motion of the system as a whole remains unchanged; or the system of particles automatically correct their motions so as to continue to move uniformly or equally in all directions, as demonstrated in connexion with the kinetic theory of gases. This movement of the particles equally in all directions is the condition required to produce equal gravific effect in all directions. Thus all we require to admit in order to produce all the effects of gravity as necessary results, is the existence of a gas in space. This gas differs from an ordinary gas only as to scale, i. e. in the proximity, velocity, and extreme minuteness of its particles, whereby a length of free path commensurate with the greatest observed range of gravity is insured, the extreme minuteness of the particles being at the same time adapted to that high velocity which the effects of gravity require, and which also necessarily renders the medium itself impalpable or concealed from the senses. The range of free path, though great in one sense, may be considered small and suitable for a gas that pervades the vast range of the visible universe. 2. In applying these principles to cohesion, or the approach of molecules in chemical reactions, it is so far easy to see that when two molecules of matter come very close together, or if we suppose them actually to come into contact, then they will cut off the entire stream of particles of the gravific medium from between the parts in contact; and therefore, as the gravific particles now only strike against the remote sides of the two molecules, the latter will be urged together with very great [299] force, thus explaining "cohesion".[2] But then a difficulty at once presents itself here. When two masses (or molecules) are gradually approached towards each other, instead of the tendency to approach gradually increasing up to a maximum ( as we should expect from the theory), they begin to repel at a certain distance, and very considerable force is in general required to overcome this first repulsion, when the masses then unite into one. Thus two freshly cut pieces of lead may be made to unite with some pressure, also glass, or various metals, with more or less pressure. There is therefore a neutral point which has to be passed, when the tendency to recede changes into a tendency to approach. The same thing is exhibited ( conversely) when a substance is broken into two parts by tension. If pulled (nearly) up to the neutral point, the two parts recoil or return into their old positions. If pulled beyond the neutral point, the parts repel and will not return into their old positions, i. e. they separate permanently. The thing, therefore, to be explained is the existence of this neutral point, or, in other words, the repulsion that exists at a certain distance from the surfaces. 3. The explanation we have to offer here depends upon quite recent investigations. It must be observed first that facts prove the existence of a second medium in space besides the gravific medium, viz. the heat- or light-conveying medium (the aether). If we admit the existence of one medium in space constituted according to the kinetic theory (the gravific medium), it would be natural to conclude that the second medium (or aether) was constituted in an analogous manner. We shall give independent reasons afterwards that lead to infer this constitution, and endeavour to answer possible objections; but in the mean time it is only necessary to suppose it to be so constituted (in the absence of proof to the contrary); and if this supposition serves to explain in general principle a number of facts, this will be one argument for its truth. On account of the extreme shortness of the waves of light and heat, it would be reasonable to suppose that the length of free [300] path of the aether particles was contained within compact limits, or was, at any rate, shorter than the length of the wave itself. It has been proved recently, in investigations by Mr. Johnstone Stoney in connexion with the radiometer,[3] that a medium constituted according to the kinetic theory has a special power of propagating a pressure unequal in various directions, or that, when a layer of the medium (such as a layer of air) is intercepted between two surfaces whose distance apart is a small multiple of the length of free path of the particles of air, the layer can then transmit a pressure in the line perpendicular to the surfaces which is in excess of the transverse pressure; and thus a repulsion is produced, accounting for the spheroidal state, the motion of the radiometer, &c. In fact it is evident (as pointed out) that, since in a medium constituted according to the kinetic theory the particles move in straight lines, the particles (when the distance of the opposed surfaces approximates to the range of free path) get reflected backwards and forwards repeatedly between the opposed surfaces, the increments of energy received by the particles accumulating by successive reflections, so that the particles produce a bombardment tending to separate the two opposed surfaces The increments of velocity imparted by the heated[4] [301] surfaces are also mainly received in the line joining the surfaces (not so much transversely); so that this conduces to the pressure on the surfaces, or repulsion. 4. This is precisely what we have to put forward, in its application to the aether, as an explanation of the repulsion in the cases referred to, such as for example the repulsion of two lenses or glass surfaces placed together in such proximity as to exhibit "Newton's rings," the repulsion of two molecules &c.; for if the tether be constituted according to the kinetic theory, we shall inevitably have the same phenomena here, though on an infinitely more energetic scale; for the particles of tether come into direct contact with the vibrating molecules of matter, whose energy of vibration is known to be enormous at normal temperature; and the layer of aether is very thin, and the motion of the aether particles very rapid,[5] so that the successive increments of velocity imparted by the vibrating molecules accumulate by successive reflections (backwards and forwards) between the opposed surfaces, producing a forcible repulsion. These results have been theoretically demonstrated to follow on the basis of the kinetic theory, and have been established by experimental facts. It is a point of great importance to observe that it is specially the kinetic theory that explains this otherwise most curious fact of an excess of pressure in a medium in one direction (producing a repulsion), with normal pressure existing in transverse directions, which otherwise it would be so difficult to explain, and which must be explained in order to account in a realizable manner for the phenomena observed. It is difficult to conceive how any other means of explaining this curious fact could be afforded than that supplied by the kinetic theory. Moreover it is generally admitted that heat has the property of producing repulsion. The "heat" of the molecules in the cases mentioned is known to consist in their vibrations, by which they generate waves of heat in the aether. We have therefore to explain under what particular constitution of a medium vibrations can (within certain limits) produce repulsion. The kinetic theory of the constitution of the medium solves completely this peculiarly difficult problem. [302] 5. When the two surfaces (or two molecules) are pushed up closer to each other, then the energy of the gravific medium directed against the remote sides of the molecules prevails more and more, since the mutual sheltering-power of the molecules increases in an enormously rapid ratio as contact is neared, and so the unbalanced energy of the gravific medium directed with full force against the remote sides of the opposed molecules at length outweighs the action of the intercepted aether particles, and the two molecules are propelled together ( or unite). 6. We may allude to a few examples serving to illustrate the application of the above principles. Supposing we take the common case of the ignition of a gas jet. Then when the gas is turned on, the molecules of gas and air mingle with each other and are known to be exchanging motion and rebounding from each other, and yet they do not unite. According to the above principles the molecules, as they approach each other in their encounters, are kept apart by the forcible vibrations (which the molecules are known to possess[6]) which, through the increments of velocity imparted to the particles of the intervening aether, produce a repulsion in the manner described, as soon as the molecules in their encounters have approached nearly within range of the mean path of the aether particles. When a flame is applied to the jet, the rapidly moving gaseous molecules of which the flame consists naturally produce a disturbance, jostling some of the molecules of the mixture of gas and air against each other, so that, the neutral point is passed, whereby the molecules are brought into such proximity that their mutual sheltering action causes the gravific medium to impinge with full energy upon their remote sides, thus urging the molecules together (producing combination). The molecules are thrown into forcible vibration by the shock of approach, and become luminous through the energy of the waves thus generated by them in the surrounding aether. These vibrations of the compound molecules after combination naturally cause the forcible rebound of any other molecules that happen to be in their proximity, the disturbance thus set up sufficing to effect the successive ( practically instantaneous) combination of the entire jet of gas. The same considerations of course apply to the practically instantaneous combination (explosion) of a mixture (in definite proportions) of gas and air, by an initial disturbance [303] produced by a flame. In the case of solid bodies, where the molecules are fixed or under control, a forcible pressure or concussion may serve to bring the molecules over the neutral point (and thus effect combination), as illustrated by the effect of the blow struck in "percussion" powders. It would not appear that matter in the gaseous state could ever be exploded by pressure (so long as the gaseous state was retained); for the molecules of gases cannot be pressed against each other by any amount of pressure, since, the molecules being in free translatory motion among themselves, the only effect of pressure would evidently be to put a greater number of molecules into unit of volume, without thereby causing the molecules in their encounters to approach nearer to each other than before. The degree of approach of the molecules (in their encounters) depends evidently on their momentum or velocity; and this remains the same whatever the pressure. 7. Heat could not apparently be said to augment the energy of chemical combination, since, in general, heat is known to possess the exactly opposite effect, or to disintegrate matter. The part played by heat in effecting chemical combination would seem to consist simply in producing a molecular disturbance, whereby unavoidably some molecules are urged towards each other so as to pass the outer neutral point, which is the necessary preliminary to combination. No doubt, when heated elements combine, the original heat adds itself to the work thus to be derived, as the heat cannot be destroyed, though it cannot increase the work of combination. Heat may (as is known) entirely prevent chemical combination, and oven dissociate combined elements. The action of heat in preventing chemical combination and producing dissociation would on the above principles consist in the fact that, when the vibratory motion of the molecules becomes excessive, this vibratory motion generates such a pressure in the intervening layer of aether on the approach of the molecules as to prevent them from passing the neutral point: or, indeed, no neutral point may exist, provided the pressure or repulsion thus generated be such as to outweigh the action of the gravific medium, as appears actually to take place in the dissociation of matter by excessive heat. Thus it would appear probable from this, that when combination ensues in the case of a mixture of gases previously considerably heated (but not so much so as to produce dissociation), the molecules on combination do not at once settle clown into that full proximity (which belongs to a lower temperature), but they do so gradually as the temperature falls. Thus the work of combination is prolonged over the falling temperature, and the cooling thereby [304] somewhat retarded. Precisely the same thing is illustrated in the aggregation of groups of molecules (to form masses), as in the aggregation of single molecules to form compound molecules. Thus when a bar of iron is welded by heat, the molecules (though aggregated or combined) do not settle down into their final positions of proximity until the bar cools, the bar being observed to contract on cooling. In this instance also the cooling of the bar is somewhat retarded by the approach of the molecules in the act of cooling. 8. In the case of the ignition of a solid body, the same considerations no doubt apply as in the case of a gas. Thus, for example, the molecules of oxygen are impinging against the surface of a piece of coal, but do not produce ignition. To effect this a certain number of the molecules must be impelled with sufficient energy against the coal so as to carry them over the neutral point (i. e. beyond the initial repulsion). The application of a flame, which consists of matter in a state of violent agitation, suffices to effect this, and, no doubt by loosening some of the molecules of carbon (of the coal) and giving them translatory motion and mixing them with the air, facilitates the process. 9. As a further illustration of the exact similarity of behaviour of single molecules and groups of molecules (masses) as regards the existence of the above-mentioned neutral point, we may take the case of the substance iodine. This substance gives off a visible vapour at normal temperatures. The single molecules of iodine composing the vapour rebound from each other without uniting; and this can only be due to the existence of the above-mentioned neutral point, outside which there is a repulsion. If the colliding molecules were to approach within the neutral point, they would unite and form solid iodine. No doubt some of the molecules of the vapour (as their velocities are known to be very diverse) do pass beyond the neutral point; and thus molecules of vapour striking against the fragments of solid iodine in the bottle, will sometimes unite with the solid iodine and form part of it, while, on the other hand, other molecules of the solid which happen to possess excessive vibrating energy are thrown off, this being the known way in which the balance in evaporation is maintained. The masses of iodine have the same neutral point as the single molecules, since two masses of the substance when pressed together will not readily unite; i. e. the neutral point, where the outer repulsion terminates, must be passed first.[7] [305] 10. Just as increase of vibrating energy (temperature) tends, by the increase of pressure thus produced in the intervening film of the medium, to dissociate molecules, so reduction of vibrating energy (attendant on reduction of temperature) tends to facilitate the approach of molecules, on account of the reduction of the pressure or repulsive action of the intervening film. Thus the molecules of a vapour when their vibrating energy is reduced (by a fall of temperature) may by the simple momentum of their own encounters, carry themselves over the neutral point, and thus effect the condensation of the vapour. Numerous other cases might be cited illustrative of the application of the above principles, as, indeed, the molecular effects are very similar in their fundamental aspects. The molecular phenomena, however diverse, may be all correlated in one fundamental respect, viz. as consisting in phenomena of approach and recession. The fundamental conditions to be explained, therefore, are the conditions capable of producing the approach and recession of molecules. Whatever may be said of the above deductions, it is at least so far certain that the conditions investigated, and based upon experimental facts, are competent to produce these fundamental movements of approach and recession in the case of molecules, and to do so in the simplest manner, the constitution of media according to the kinetic theory being admittedly the simplest conceivable. To look therefore to other conditions than the simplest would be to imply that the same results are brought about by a superfluity of mechanism. This superfluity is known not to be the characteristic of nature; and all the teaching of mechanism points to the fact that superfluity or unnecessary complication entirely prevents the attainment of precision and certainty in the mechanical effects. The great precision and unfailing certainty of the molecular effects would therefore render it necessary to infer that the regulating mechanism was simple, or that there was no unnecessary superfluity. 11. The fundamental conclusion above drawn regarding the mechanism concerned in the approach of molecules is grounded upon the only explanation of the mechanism of gravity that has withstood criticism and received support by competent judges, viz. the kinetic theory of gravity, of which Le Sage's ingenious idea forms the fundamental basis, and is at once the simplest explanation of gravity conceivable. The [306] application of this theory to molecules in close contact ("cohesion" &c.), is necessary and inevitable, and it serves to correlate the molecular effects generally under one cause. The explanation of the fundamental condition capable of producing the recession of molecules, as above given, rests upon experimental facts recently established, and upon a basis for the constitution of the aether which is the simplest conceivable. 12. We now propose to show some independent reasons in support of this constitution for the aether, in addition to the argument afforded by the numerous molecular effects which this constitution, in principle, serves to explain. First, if the subject be reflected on, it will be apparent that, in principle, u. movement of the component particles of the medium in straight lines is the only possible constitution for the ultimate medium in space. For a particle of matter cannot move in a curved line unless it have a medium about it to control its motion. Thus a planet can move in a curve because it has a medium about it (the gravific medium) to cause it to move in a curve. It is a known principle that a particle of matter cannot of itself change the direction of its motion. The particles of the ultimate medium in space must therefore move in straight lines. This deduction is surely of great importance in the inquiry as to the constitution of the physical media in space. Also in addition to this, the observed facts of gravity prove that the particles of the gravific medium move in straight lines, since no other motion than this can harmonize with the observed effects of gravity. It would therefore surely be a strange thing if the particles of the aether, as a second medium immersed in the gravific medium, did not move in straight lines. To suppose this would be very like supposing that when the particles of a second gas are immersed among those of another, the particles of the first gas acted upon those of the second to make them move otherwise than in straight lines, which is known to be impossible. Moreover the fact of the kinetic theory representing the simplest conceivable constitution for a medium would by itself be a strong argument for this constitution in the case of the aether. The very fact of the great precision and delicacy of the operations performed by the author as the mechanism for the transmission of the varied phenomena of colour &c. would point to a simple constitution; just as the complex effects of sound with all its intricate and varied gradations of tone are known to be transmitted by a medium (the air) of the simplest conceivable constitution, viz. that represented by the beautiful kinetic theory of gases.[8] The more intricate the functions of a mechanism, [307] the more is simplicity indispensable, and superfluity incompatible with precision and certainty in the results. To assume a constitution for the aether that could not be realized or clearly explained would surely be futile, since the explanation or clear conception forms the logical support of any theory, without which the theory resembles a mere dogmatic statement incapable of being sustained by reason. 13. There is one other point which we would notice in connexion with this subject. The idea would appear to be to a certain extent prevalent that the aether must have a constitution essentially different from the air, because the vibrations producing light are transverse, while those producing sound are longitudinal. It seems to be sometimes inferred from this that the vibrations of the Bather are only transverse, and those of the air only longitudinal. There would be no warrant for this conclusion; and we think that it has done harm and greatly hindered any rational idea from being formed of the nature of the aether. According to the kinetic theory, which is known to represent the constitution of the air, the vibrations of the particles of air disturbed by a vibrating body and propagated in the form of waves, are not only longitudinal; for since according to the kinetic theory the particles of air in their normal state are moving equally in all directions, it follows that these particles are accelerated and retarded both in transverse and in longitudinal directions at the passage of waves. It is true that the transverse component of the motion probably may not affect the ear, on account of its special structure. It would be wrong, however, to infer from this that the transverse component of the motion did not exist. So in the case of the aether, it would be unwarranted to infer that the longitudinal component of the motion did not exist, because this component was incapable of affecting the eye. The eye and the ear may be very differently constituted; and a motion that affects the one might not affect the other. Sir John Herschel says regarding this point in his essay "On Light" (' Popular Lectures on Scientific Subjects,' page 358): — "According to any conception we can form of an elastic medium, its particles must be conceived free to move ( within certain limits greater or less according to the coercive forces which restrain them) in every direction." He then goes on to explain how the efficacy of the transverse component of the movement in the case of light, and the longitudinal component of the movement in the case of sound, may be accounted for by the diverse structure of the eye and ear. Any inference which is not valid, invariably does some harm; and this idea of a forward movement being propagated in a medium by only transverse vibrations, being almost inconceivable, has [308] naturally led to some incongruous ideas regarding the structure of the aether, in the effort to explain it. Thus some have supposed the aether to resemble a solid, which is in direct opposition to the teaching of the senses; for we move about so freely in this "solid" as to be unconscious of its existence. Another supposition has been that "lines of tension," behaving somewhat in analogy to stretched chords, exist in the aether. Such a mechanism would be, to say the least, somewhat deranged by the passage of a planet through the aether. Indeed it is sufficiently evident that these are the hopeless attempts made to surmount an impossible condition, or a difficulty for whose existence there is really no warrant. If the aether be not a solid, or a liquid (for liquids oppose enormous resistances to the passage of bodies through them at high speeds), then what other resource have we than to conclude that it is a gas? 14. A gaseous constitution of the aether according to the kinetic theory would perfectly satisfy the two fundamental conditions of a medium highly elastic in all directions, and opposing no appreciable resistance to the free movement of bodies (the planets &c.) through its substance. For it is a known fact that the resistance opposed by a medium constituted according to the kinetic theory to the passage of bodies through it diminishes as the normal velocity of the particles of the medium increases. The high normal velocity of the particles of the aether, proved by the velocity of light, therefore necessarily renders the resistance inappreciable, and the medium itself impalpable and undetected by the senses. 15. A difficulty has been raised in the way of the aether being constituted as a gas on the following grounds, which, being only anxious for truth, we are bound to consider.[9] It has been argued that if the aether be constituted as a gas, the specific heat of unit of volume of the aether would be the same as that of any ordinary gas at the same pressure, and that therefore it would appear that the presence of the aether could not fail to be detected in the experiments on the specific heat of ordinary gases. We have to offer the following as a means of meeting this difficulty. It will be admitted that the detection of the aether in the experiments on specific heat will depend, not on the specific capacity for heat possessed bv the aether, but on the rate at which the heat passes from the gas experimented on to the aether. The molecules of the gas are moving through the aether with their normal translatory motion, this motion of the molecules representing the "heat" of the gas. It will be evident that the rate at which the [309] motion ("heat") of the molecules of the gas passes to the aether will depend on the resistance the aether offers to the passage of these molecules through it. But we have shown that this resistance may (on account of the high normal velocity of the aether particles) be inappreciable. Hence the rate of passage of the heat from the gas to the tether will be inappreciable. This, we submit, removes the difficulty in question. It is clear that, if the aether opposes no appreciable resistance to the passage of a planet through it (moving at several miles per second), it cannot be affected by the passage of a molecule of a gas through it, which in its relatively slow rate of translatory motion may be considered at rest compared with the aether particles. The high normal velocity of the aether particles is only appropriate to their minute mass. 16. It must be apparent to any reflecting observer, that in physical science we have a vast array of facts accumulated through years of experiment, but a great paucity of causes; or the number of facts known is quite out of all proportion to the number of causes known, these latter being replaced by more or less vague and unsubstantial theories. As, therefore, we have no paucity of facts as a basis to reason upon, it surely cannot be too soon to make an effort to correct this anomalous state of things, and to replace the above unsubstantial theories by rational conceptions of the processes of nature, Clearness of conception is the test of truth, and constitutes its real dignity; and theories, however elaborated, if vague, have no real dignity.[10] Since there is nothing occult about the physical media in space, in so far as they differ in no way from ordinary matter excepting in the mere scale or dimensions of their parts, and since it is obviously just as easy to reason of matter of one dimension as of another, any hesitation in entering upon this course of study would be wholly uncalled for; indeed, surely there is reason for a rational interest in realizing the admirable adaptation of these media in a mechanical point of view for their special functions; and the question as to the utilization of the stores of motion enclosed by them to the best advantage may present a problem of the highest practical interest and importance.[11] It should be observed that these stores of [310] motion simply consist in small particles of matter in a state of rapid motion; or there is nothing occult about the subject at all, as indeed obviously principles of reasoning are independent of size. The minute size (and consequent invisibility) of the particles is necessary to the efficiency of the media as powerful motive agents, since minuteness of size is necessary to render a high velocity possible for the particles, without producing disturbing effects among the matter immersed in these media. There is one very noteworthy point that cannot be too distinctly kept in view in connexion with this subject. It is the fact that the high intensity of the stores of motion possessed by these media, and which renders them so important, serves to conceal their existence from the senses. Thus the higher the intensity of the store of motion enclosed by these media, and consequently the greater their capacity for practical utility, the more likely (if the mere evidence of the senses were relied on) is their existence to be forgotten. For it may be proved beforehand, by the kinetic theory of gases, that the greater the velocity of the component particles of a medium, and consequently the greater the value of the store of energy enclosed (which may even reach an explosive intensity), the more does the presence of the medium elude detection, because the resistance opposed by the medium to the passage of bodies through it diminishes as the velocity of the particles increases. The less indication the mere senses (unaided by reason) afford of the existence of such media, the higher, therefore, should we be warranted in inferring their importance to be. Even independently of all question of the existence of these media, it may be proved beforehand that, if media did exist and enclose stores of motion to an enormous intensity, they would be concealed. This is, no doubt, a remarkable fact, and contrary to preconceived ideas, as it would doubtless appear on the first thought that the higher the intensity of a store of energy existing in space, the more likely would it be to make itself apparent to the senses, whereas precisely the contrary is found to be the fact. This forms a notable instance of one of those cases where analysis completely reverses preconceived ideas. It is possibly the absence of appreciation of this fact that may in some way account for the failure of the most striking proof's of nature to carry their practical teaching, as for example, the sudden setting free of concealed motion in the explosion of a mass of gunpowder. Here to the mere bodily senses, we have apparently an actual creation of motion. Something [311] more, however, than the evidence of the mere bodily senses may be required, to appreciate the truths of nature, as it is a notorious fact that the most important truths generally lie below the surface. It should be noted that these media would not be efficient as working agents unless they were concealed; for concealment (as observed) is the necessary condition to the enclosure of a store of motion to a high intensity. Possibly the absence of realization of this fact, and perhaps that prejudice which besets every new path, may in some degree account for what must otherwise appear an extraordinary indifference and absence of inquiry in a subject of great mechanical interest and involving possibly issues of the highest importance and practical utility. When this, like every other illogical prejudice to change, comes to be broken down by the light of reason and reflection, there may be just ground for surprise at the previous delay, and at the shallow and unsubstantial character of the theories which so long supplanted rational conceptions of the processes of nature. London, March 13, 1878. 1. The three previous papers treating of the subject of gravitation are in the Philosophical Magazine for September and November 1877, and February 1878. 2. The spectroscope proves molecules to be complex bodies, on account of the number of different periods of vibration they can take up; and it was pointed out in the last paper that there are grounds for inferring them to possess interstices, or a more or less open structure. It is evident, therefore, that the shapes of molecules, as to whether their parts fitted over each other or not (and thus afforded more or less shelter from the impinging particles of the gravific medium), would have some influence on the behaviour of molecules as to the energy of their approach (reactions). This might account in some degree for the varied behaviour termed "chemical affinity," though possibly there are, besides this, other modifying physical conditions. 3. Philosophical Magazine, December 1877. 4. There is another point in connexion with the motion of the particles, which no doubt, however, has been already noticed. Under normal conditions, a body vibrating opposite to another tends (as is known) to produce rarefaction in the intervening medium; but in the case of a film whose thickness is near the range of mean path of the particles, there would appear to be a special cause tending greatly to reduce this effect, and even perhaps to produce the contrary effect, viz. a condensation ( which would greatly increase the repulsion). Thus, under the increments of velocity received, there is a tendency for the molecules of the gaseous film to be turned round so as to move more normal to the film. Suppose, for instance, an elastic sphere to be rebounding obliquely between two planes. Suppose increments of velocity to be given to the sphere by vibrating one of the planes. Then these increments of velocity given to the sphere will evidently make it rebound more normal to the surfaces. So in the case of the molecules of a gaseous film, rebounding backwards and forwards between two surfaces (such as the air film which supports a drop in the so-called "spheroidal" state, the air film which supports a grain of powder in some experiments of Professor Barrett, referred to by Mr. Johnstone Stoney), the molecules of the film will tend, by the increments of velocity given them, to turn round so as to move in a direction more normal to the surfaces. This evidently makes the lateral pressure exerted by the film less, and consequently its lateral expansion (or rarefaction) less. If we imagine the extreme case where the molecules of the film are all turned round so as to move exactly normal to the surface of the film, then whatever the velocities of the molecules of the film (i. e. whatever the longitudinal pressure, or repulsion, exerted by them), the film would exert no lateral pressure at all. There would consequently be a lateral inrush of air, increasing the density of the film, and therefore increasing the repulsion (since those molecules which enter the film become themselves available for producing repulsion). Possibly, from this cause, these films may be actually denser than normal density. At all events the above cause makes their density greater than it otherwise would be, and the repulsion exerted by them greater. 5. It may be noted that, if the aether be constituted according to the kinetic theory, the normal velocity of its particles is $\frac{3}{\sqrt{5}}$ X velocity of a wave of light. See appendix to paper "On the Mode of Propagation of Sound" (Phil. Mag. June 1877), added by Prof. Maxwell. 6. The molecules of matter in the gaseous state are known to possess, in addition to the translatory motion peculiar to that state, a vibratory motion, in virtue of which the molecules generate waves of regular periods iu the tether (these periods having in many cases been measured by the spectroscope). 7. The above effects were described in a little book ' Physics of the Ether' (E. & F. N. Spon), published by me in 1875; but the cause of the reduction of the pressure of the medium, which determines the approach of molecules, was there wrongly stated, the error having arisen from a seeming analogy between the approach of bodies to masses (tuning-forks &c.) vibrating in air — in the absence of the knowledge recently acquired of the repulsion of gaseous layers. Much of the main principles of the book, however, remain as they were — to be supplemented by the investigations contained in the present papers. 8. There would surely be nothing to admire in complication in itself. The whole aim of mechanical design is directed towards the attainment of simplicity, which being unique, entails intellectual labour to find it. 9. See paper "On the Dynamical Evidence of the Molecular Constitution of Bodies," by Prof. Maxwell (' Nature,' March 11, 1875). 10. Vagueness, paradox, and mystery surely belong rather to those intellects which are incapable of rising to clear and definite conceptions. 11. As an instance of the change of views on the most practical subjects that the acceptance of these principles entails, we may cite the case of the employment of coal, which by the recognition of the existence of the stores of motion in space, becomes a mechanism or machine for deriving motion. The expenditure of coal, therefore, represents the expenditure of mechanism or machinery. Hence in deriving motion through coal we expend a quantity of machinery proportional to the power derived. Without asking the question whether it is necessary in every case, in deriving motion from a source, to expend machinery proportional to the power derived (i. e. that the work done should be the equivalent of the machinery expended), it is at least so far certain that no remedy for this could be discovered unless the physical conditions of the case were recognized. This work is in the public domain in the United States because it was published before January 1, 1923. The author died in 1917, so this work is also in the public domain in countries and areas where the copyright term is the author's life plus 80 years or less. This work may also be in the public domain in countries and areas with longer native copyright terms that apply the rule of the shorter term to foreign works.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 1, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9557612538337708, "perplexity_flag": "middle"}
http://www.physicsforums.com/showthread.php?t=335881	Physics Forums Stacked blocks 1. The problem statement, all variables and given/known data Imagine that there is a slab (call it block 1) on top of which sits a block (call it block 2). The slab connects to a cable that passed through a pulley and attaches to another block (call it block 3), hanging freely. Assuming that the coefficient of kinetic friction is the same for both blocks (and the same with static friction): a) Find the acceleration of the system. b) Verify the familiar limits. c) Under what conditions does block 2 stay on top of block 1? 2. Relevant equations/The attempt at a solution If blocks 1 and 2 move together, then all three blocks have the same acceleration: $$a=\frac{m_3g-(m_1+m_2)g\mu_k}{(m_1+m_2+m_3}$$ If block 2 begins to slip, then we still take all the accelerations as the same (block 2 is just beginning to slip, so it is still pretty much moving with block 1 (i.e. same acceleration) and it's frictional coefficient will be static) TO MAKE A LONG STORY SHORT: DID I SET UP NEWTON'S 2ND LAW CORRECTLY? then Newtons 2nd Law for block 1, block 2 and block 3(respectfully) are: [tex] \sum(1,x): T+F_{f,s}-F_{f,k}=(m_1+m_2)a [/tex] [tex] \sum(1,y): F_{n,1}-(m_1+m_2)g=0 [/tex] [tex] \sum(2,x): F_{f,s}=m_2a[/tex] $$\sum(2,y): F_{n,2}-m_2g=0$$ $$\sum(3,x): 0=0$$ $$\sum(3,y): m_3g-T=m_3a$$ Where the frictional forces take on the familiar value of the coefficient times the normal. Solving, I get: [tex]a= \frac{m_3g+\mu_sm_2g-\mu_kg(m_1+m_2)}{m_1+m_1+m_3}[/tex] For part b) while testing the familiar limits, i found that as $$m_3 \rightarrow \infty$$ then $$a=g$$, which makes sense. I found that if $$m_2=0$$, then we get what we expect. And when I calculate when $$m_2 \rightarrow \infty$$, $$a=0$$. BUT, when I try to calculate as $$m_1\rightarrow \infty$$, $$a=\frac{-\infty}{\infty}$$, which doesn't make sense. Did I screw up Newtons 2nd law? For part c) all I did was use newton's 2nd law for the second body and solved for $$a$$, the maximum acceleration: $$a=\mu_sg$$, which doesn't make sense either since it should depend on mass. Thoughts?!?! PhysOrg.com science news on PhysOrg.com >> Front-row seats to climate change>> Attacking MRSA with metals from antibacterial clays>> New formula invented for microscope viewing, substitutes for federally controlled drug Blog Entries: 7 Recognitions: Gold Member Homework Help Part (a) is OK. It is the same result as if blocks 1 and 2 were glued together to form a single block of mass m1+m2. Part (b) is not OK. The force that accelerates m2 (net force) is the force of static friction, fs. Newton's 2nd Law says that it must be equal to m2 times the acceleration from part (a). As you increase m3, you increase the acceleration which means that fs must increase accordingly. But fs cannot increase forever. At the point where fs can no longer increase, and ony then, can you say fs = μsN. You have all the ingredients to you can put it together from this point on. Thanks for the reply kuruman. So my equation for the acceleration is correct then? You say that I cannot include fs = μsN until the point in which the static friction can not keep up with the acceleration of blocks 1 and 3, so then it shouldn't be in my equation for the acceleration? What about part c)? Under what condition does block 2 stay on top of block 1? I understand that this only happens if fs = μsN, eg if μs=fs/N (or less), but in this equation what is fs? Blog Entries: 7 Recognitions: Gold Member Homework Help Stacked blocks The force of static friction fs is whatever is necessary to provide the observed acceleration. Because it is the net force on mass m2, you can write $$f_{s}=m_{2}a=m_{2}\frac{m_{3}g-\mu_{k}g(m_{1}+m_{2})}{m_{1}+m_{2}+m_{3}}$$ That is always true unless the top mass starts sliding. At the threshold of when sliding is just about going to happen, the force of static friction has reached its maximum value and can no longer "provide the observed acceleration" past that value. What does the above equation become then? What is the maximum value of the force of static friction? When: $$\mu_s\geq \frac{F_{f,s}}{F_n}=\frac{am_2}{g}\left(\frac{m_{3}g-\mu_{k}g(m_{1}+m_{2})}{m_{1}+m_{2}+m_{3}}\right)$$ ? Also what about the problem with negative acceleration when m_2 -> infinity? Blog Entries: 7 Recognitions: Gold Member Homework Help $$f_{s}=m_{2}a \leq \mu_{s}F_{n}$$ Replace a and Fn with the appropriate quantities. Also what about the problem with negative acceleration when m_2 -> infinity? What about it? Do you really think that if you make mass m2 large enough, say the size of a battleship, the masses will accelerate in the opposite direction? I didn't think so. What is more likely to happen instead? Remember, the acceleration you have has been derived assuming that the masses accelerate so that the hanging mass drops. Thread Tools \| \| \| \| \|-------------------------------------\|-------------------------------\|---------\| \| Similar Threads for: Stacked blocks \| \| \| \| Thread \| Forum \| Replies \| \| \| Introductory Physics Homework \| 17 \| \| \| Introductory Physics Homework \| 27 \| \| \| Introductory Physics Homework \| 0 \| \| \| Introductory Physics Homework \| 4 \| \| \| Introductory Physics Homework \| 4 \|	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 16, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9120947122573853, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/33079/matrix-structure-maybe-you-can-see-something-i-cant?answertab=votes	# Matrix structure; maybe you can see something I can't… I am attempting to find a 'smarter' way to solve a matrix, in the form $Ax=B$, where $B_{i}=F_i*N$ $A_{i,j}=-F_i/K_{j,i}$ where $N$ is constant, $K$ is a constant matrix, and $F$ is a vector of; $F_i=Y_i/K_{i,i}$ where $Y$ is a vector constant. I'm not hugely mathematical, and my linear algebra skills could be put on the back of a napkin, but seeing this kind of repetition indicates to me that there must be a shortcut for this. Does anyone have any insights? - 1 You can cancel the $F_i$'s. – Yuval Filmus Apr 15 '11 at 1:20 I was thinking more about the structure of the resultant matrix but yeah, I wanted a second opinion on that too (I have no real understanding of the weirdnesses of linalg...) – Andrew Bolster Apr 15 '11 at 1:23 Think of it as a system of equations. For each $i$ you have an equation of the form $\sum_j A_{ij} x_i = B_i$. In your case, this is $\sum_j -F_i/K_{ji} = NF_i$, so you can cancel the $F_i$'s. Otherwise it's almost completely arbitrary (other than the fact that $-1/K_{ji}$ cannot be zero). – Yuval Filmus Apr 15 '11 at 1:44 You can also cancel the $N$ if you want. Solve if for RHS which is constant $1$, then multiply the result by $N$. – Yuval Filmus Apr 15 '11 at 1:45 There's a typo above; it should read $\sum_j A_{ij} x_j = B_i$. – joriki Apr 15 '11 at 7:50	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 18, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9341250061988831, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/175263/gradient-and-hessian-of-general-2-norm	# Gradient And Hessian Of General 2-Norm Given $f(\mathbf{x}) = \\|\mathbf{Ax}\\|_2 = (\mathbf{x}^\mathrm{T} \mathbf{A}^\mathrm{T} \mathbf{Ax} )^{1/2}$, $\nabla f(\mathbf{x}) = \frac {\mathbf{A}^\mathrm{T} \mathbf{Ax}} {\\|\mathbf{Ax}\\|_2} = \frac {\mathbf{A}^\mathrm{T} \mathbf{Ax}} {(\mathbf{x}^\mathrm{T} \mathbf{A}^\mathrm{T} \mathbf{Ax} )^{1/2}}$ $\nabla^2 f(\mathbf{x}) = \frac { (\mathbf{x}^\mathrm{T} \mathbf{A}^\mathrm{T} \mathbf{Ax} )^{1/2} \cdot \mathbf{A}^\mathrm{T} \mathbf{A} - (\mathbf{A}^\mathrm{T} \mathbf{Ax})^\mathrm{T} (\mathbf{x}^\mathrm{T} \mathbf{A}^\mathrm{T} \mathbf{Ax} )^{-1/2} \mathbf{A}^\mathrm{T} \mathbf{Ax} } {(\mathbf{x}^\mathrm{T} \mathbf{A}^\mathrm{T} \mathbf{Ax} ) } = \frac { \mathbf{A}^\mathrm{T} \mathbf{A} } { (\mathbf{x}^\mathrm{T} \mathbf{A}^\mathrm{T} \mathbf{Ax} )^{1/2}} - \frac {\mathbf{x}^\mathrm{T} \mathbf{A}^\mathrm{T} \mathbf{A} \mathbf{A}^\mathrm{T} \mathbf{Ax} } { (\mathbf{x}^\mathrm{T} \mathbf{A}^\mathrm{T} \mathbf{Ax} )^{3/2} }$ I guess I am looking for confirmation that I have done the above correctly. The dimensions match up except for the second term of the Hessian is a scalar, which makes me think that something is missing. Edit: Also, the last equality reduces to $\nabla^2 f(\mathbf{x}) = \frac {\mathbf{A}^\mathrm{T} \mathbf{A} - \nabla f(\mathbf{x})^\mathrm{T} \nabla f(\mathbf{x})} {\\|\mathbf{Ax}\\|_2}$ - 1 My advice is to do it first without the square root. Given your $f,$ define $g = f^2$ and find the gradient and Hessian. Just to confirm your suspicions, the gradient really is a vector, the Hessian really is a square (symmetric) matrix. – Will Jagy Jul 25 '12 at 23:26 1 Try squaring and expanding $f^2(x+\delta)$. The gradient and Hessian (of $f^2$) should be fairly clear. – copper.hat Jul 25 '12 at 23:27 Also, do this in agonizing detail for the case of just two variables, you could call them $x_1,x_2$ if you like, but you could also stick with $x,y.$ The gradient becomes a vector with just two functions as entries, the Hessian a 2 by 2 matrix. You are getting a bit lost in the formalism. – Will Jagy Jul 25 '12 at 23:38 1 It looks to me like everything is fine except that the numerator should be $A^TA - (\nabla f)(\nabla f)^T$. (Also, $\nabla^2$ usually denotes the Laplacian, which is the trace of the Hessian.) – Rahul Narain Jul 26 '12 at 4:46 ## 1 Answer It is easier to work with $\phi(x) = \frac{1}{2} f^2(x)$. Just expand $\phi$ around $x$. $\phi(x+\delta) = \frac{1}{2} (x + \delta)^T A^T A (x + \delta) = \phi(x) + x^TA^TA \delta + \frac{1}{2} \delta^T A^T A \delta$. It follows from this that the gradient $\nabla \phi(x) = A^T A x$, and the Hessian is $H = A^TA$. To finish, let $g(x) = \sqrt{2x}$, and note that $f = g \circ \phi$. To get the first derivative, use the composition rule to get $D f(x) = Dg(\phi(x)) D \phi(x)$, which gives $Df(x) = \frac{1}{\sqrt{2 \phi(x)}} x^T A^T A = \frac{1}{\\|Ax\\|} x^T A^T A$. Let $\eta(x) = \frac{1}{\\|Ax\\|}$, and $\gamma(x) = x^T A^T A$, and note that $D f(x) = \eta(x) \cdot \gamma(x)$, so we can use the product rule. Let $h(x) = Df(x)$ then the product rule gives $D h(x) (\delta) = (D \eta(x) (\delta)) \gamma(x) + \eta(x) D \gamma(x) (\delta)$. Expanding this yields: $Dh(x)(\delta) = (- \frac{1}{\\|Ax\\|^2} \frac{1}{\\|Ax\\|} x^T A^T A \delta) x^T A^T A + \frac{1}{\\|Ax\\|} \delta^T A^T A$. Noting that $x^T A^T A \delta = \delta^T A^T A x$, we can write this as: $$Dh(x)(\delta) = \delta^T(\frac{1}{\\|Ax\\|} A^T A - \frac{1}{\\|Ax\\|^3 } A^T A x x^T A^T A),$$ or alternatively: $$D^2 f(x) = \frac{1}{\\|Ax\\|} A^T A - \frac{1}{\\|Ax\\|^3 } A^T A x x^T A^T A .$$ The only difference with the formula given in the question is that the latter dyad was written incorrectly (instead of the dyad $g g^T$, you have the scalar $g^T g$). - This is helpful but not complete. To get to the answer, one now needs the chain rule to find the gradient and Hessian of $f(\mathbf{x})=2\sqrt{\phi(\mathbf{x})}$. – AnonSubmitter85 Jul 26 '12 at 2:10 @AnonSubmitter85: I elaborated my answer a little more. – copper.hat Jul 26 '12 at 5:36 Thank you. This has helped me tremendously. – AnonSubmitter85 Jul 26 '12 at 18:18 You're very welcome. – copper.hat Jul 26 '12 at 18:30	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 32, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9304863810539246, "perplexity_flag": "head"}
http://mathoverflow.net/questions/51494?sort=newest	## Why the name ‘separable’ space? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) It is well known that a separable space is a topological space that has a countable dense subset. I am wondering how is this related to the name 'separable'? Any intuition where the name come from? - 27 My understanding is it comes from the special case of R, where it means that any two real numbers can be separated by, say, a rational number. – Qiaochu Yuan Jan 8 2011 at 20:39 Ha, that's a nice observation! – minimax Jan 8 2011 at 20:47 Maybe, it refers to the separation of points by means of countable collection of small open sets (where separation means that for any two points we may find their disjoint neighborhoods in our collection). It is probably not formally equivalent to the existence of a countable dense subset, but something similar. – Fedor Petrov Jan 8 2011 at 22:23 ## 3 Answers As far as I know the word separable was introduced by M. Fréchet in Sur quelques points du calcul fonctionnel, Rend. Circ. Mat. Palermo 22 (1906), 1-74. The paper can be obtained via this link (Springer). It's the famous paper in which he introduced metric spaces. He considers first slightly more general objects which he calls classes (V): where (V) stands for voisinage — neighborhood. Remark: Metrics are introduced under the name écart in no 49 on page 30. It is peculiar that the symmetry condition is not explicitly mentioned but it seems to be understood as Fréchet immediately mentions that metric spaces generalize classes (V) cf. no 27 on page 17f. However, I couldn't find an instance where he actually uses it, he is always careful to respect the order — I may have missed something since I haven't read the paper in detail. I quote the relevant passage [from no 37 on page 23f]: Nous appellerons ensuite classe séparable une classe qui puisse être considérée d'au moins une façon comme l'ensemble dérivé d'un ensemble dénombrable de ses propres éléments. [...] Ceci étant, nous nous bornerons maintenant à l'étude des classes (V) NORMALES, c'est-à-dire parfaites, séparables et admettant une généralisation du théorème de CAUCHY. Cette limitation n'a du reste rien d'artificiel, elle provient directement de la comparaison des classes (V) avec les ensembles linéaires [...] [...] Passons maintenant aux classes séparables. On peut qualifier ainsi les ensembles linéaires en considérant la droite indéfinie comme l'ensemble dérivé de l'ensemble des points d'abscisses rationnelles. Mais il n'en est pas de même pour toute classe parfaite (V). I am unable to translate this in a reasonable way (but see Amit's comment below for a translation). Very roughly: Fréchet defines separable spaces as we do it today and says that in the following he will restrict attention to complete, perfect and separable metric spaces. The last quoted paragraph indeed confirms Qiaochu's comment. - 5 We will henceforth call a class separable if it can be considered in at least one way as the derived set of a denumerable set of its elements. [...] We restrict ourselves now to the study of (V) NORMAL classes, that is to say perfect, separable and admitting a generalization of Cauchy's theorem. This limitation has nothing artificial, it comes directly from the comparison of the classes (V) with the linear sets. [...] We pass now to separable classes. [Not sure about this part] as the derived set of the set of rational points. But it isn't so for all perfect (V) classes. – Amit Kumar Gupta Jan 9 2011 at 17:57 Thank you very much! I'd translate the second to last sentence by "One may describe the 'subsets of the real line' (ensembles linéaires) in this way by considering the 'real line' (droite indéfinie) as the derived set of the set of the points with rational abscissae." – Theo Buehler Jan 9 2011 at 19:02 ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. Well "séparer" means just "disjoin", "split up" etc. There are few meanings for "séparation" in french. For exemple "séparer par des fonctions". If you have a space $X$ and a set of functions ${\cal F}$ from $X$ to ${\bf R}$ (or anything else), you say that "${\cal F}$ sépare les points de $X$" iff for two different point $x$ and $x'$ there exists a function $f \in {\cal F}$ such that $f(x) \neq f(x')$, this is a common use of the word "séparer" in french, nothing mysterious. And this vocabulary can be applied to any analogous situations, in topology or whatever else context. I heard the first time this wording (when I was a student) in the case I mention above, long before I have heard using this wording in a topology course. - This does not explain the terminological connection between "separable" and "existence of countable dense subset" – Martin Brandenburg Jan 9 2011 at 10:02 No, I just mean that separable has been used for many different situations, the common origin is separability of points by a mean or another. I guess there is no more to look for. – Patrick I-Z Jan 9 2011 at 12:37 I have a question: do you use in english the verb "to separate" in the same broad meaning than in french? – Patrick I-Z Jan 9 2011 at 12:43 2 Concerning your specific example: I have seen the phrase "$\mathcal{F}$ separates the points of $X$", yes. I believe it's standard terminology. – Zhen Lin Jan 9 2011 at 13:05 According to this part of Hausdorff's collected works, the name "separable" was coined by Fréchet. Hausdorff writes this denotation wouldn't be very suggestive but established. The earliest use in the German Zentralblatt, where the word occurs in a review by Hahn, dates back to 1918. I am sure one can find the first use of the word in the given context in Fréchet's works but I doubt that he will explain his motivation for the use of this particular word there. So we will (presumably) never know... -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 12, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8073470592498779, "perplexity_flag": "middle"}
http://quant.stackexchange.com/questions/7082/are-there-any-derivatives-which-pay-amount-ap-b2-c-where-p-is-the-price	# Are there any derivatives which pay amount $a(p-b)^{2}-c$ where $p$ is the price of underling asset? Are there any derivatives which pay amount $a(p-b)^{2}-c$ where $p$ is the price of underling asset ? (or in the case of options $max(0,a(p-b)^{2}-c)$) I'm not very strict here but I only want to know if there are any derivatives which profile of profits is quadratic function of price of underling assets (quadratic on some subset of set of all $p$ )? - 2 They're sometimes called "power contracts" and though I've known them to be coded into the pricing libraries of at least 2 big investment banks, I've never seen one on the books. – Brian B Jan 23 at 19:16 @Brian B very interesting, thanks – Qbik Jan 23 at 21:57 ## 3 Answers You can ask for a quote from a bank as I am sure they will create it for you. If you want to create this kind of payoff yourself, you can use the following paper from Peter Carr where he introduces the spanning formula for replicating any twice differentiable payoff. http://www.math.nyu.edu/research/carrp/papers/pdf/twrdsfig.pdf - Variance swaps are arguably a little like that. As mentioned, power contracts too, but sometimes only by virtue of correlation/cross-gammas between volatilities and underlyings. I haven't personally seen a contract with an exponent in it. - • There are probably currently none out there right now with the exact same pay off structure (not sure how you would set b and c and what rational you would apply thus I said variants may exist but probably not the exact same ones) • Having said that you can request a bank to price you ANY derivative you like, they will quote you for sure. You better be very confident that you are able to price them yourself correctly or you will get ripped off royally. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 6, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9616979956626892, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/295516/complex-numbers-negative-absolute-value-radius	# Complex numbers: negative absolute value (radius)? I need to find a complex number that represented by the following poler representation: ($\mathbb r$, $\theta$) = ($-5$, $\pi \over 2$) My question is: how is a negative radius (absolute value) possible? What does it mean \ sign? - It is not possible, at least not under the usual, standard definitions. Check this carefully. – DonAntonio Feb 5 at 15:37 I've seen things like this before; when I've encountered it, it has just meant to take the radius in the opposite direction. So, normally we would expect this to correspond to the point $5i$ in the complex plane (or $(0,5)$ if we identify it with $\Bbb R^2$. However, with the negative, it will correspond to $-5i$. I'm not going to say this is how it is for you, just when I've encountered it previously. – Clayton Feb 5 at 15:39 @Clayton The $\theta$ signs the angle with the positive X Axis, doesn't it? – user1798362 Feb 5 at 15:44 @user1798362: I'm not sure what you mean by signs the angle with the positive $x$-axis. The angle begins at $0^\circ$ there and rotates counterclockwise through an angle of $\pi/2$ in your case. – Clayton Feb 5 at 15:47 @Clayton Ok, thank you!! I have addition question and I'll better and it here instead of another ask: if $\theta$ = 45, than `a = b` in case $\mathbb z$ = `a + bi`? – user1798362 Feb 5 at 15:49 show 7 more comments ## 1 Answer $$z=r\cos\theta+ir\sin\theta$$ -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 14, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9364437460899353, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/252908/maximal-unramified-extension-of-mathbbf-pt	# Maximal Unramified Extension of $\mathbb{F}_p((t))$ The maximal unramified extension of $\mathbb{Q}_p$ can be described quite explicitly: add all roots of unity of order prime to $p$. This is done by the correspondence between finite unramified extensions of $\mathbb Q_p$ and finite extensions of $\mathbb F_p$. Similarly, can one find an explicit description for the maximal unramified extension of $\mathbb F_p((t))$? - ## 1 Answer For $\mathbb{F}_p((t))$, this is in a sense even simpler: the maximal unramified extension is the direct limit over $\mathbb{F}_{p^n}((t))$, $n\in \mathbb{N}$. This is very easy to see, since unramified extensions of a local field are in bijection with extensions of the residue field. In other words, the maximal unramified extension is again generated by roots of unity of order prime to $p$. It's good fun to try and classify, using class field theory, the maximal tamely ramified extension of $\mathbb{F}_p((t))$. I once did that in my Part III essay if you are interested. - 1 Technically, $\overline{\mathbb{F}_p}((t))$ is transcendant over $\mathbb{F}_p((t))$, you want $\cup F_{p^n}((t))$ instead. – mercio Dec 7 '12 at 9:08 @mercio, you are right, of course, I have corrected it. Thanks. – Alex B. Dec 7 '12 at 9:15 Your answer is correct. However you omit the fact that $\mathbb{F}_p((t))$ is not a perfect field. So there could be unramified, non-separable extensions; your approach is not treating them. Luckily such extensions do not exist, but this is not easy to prove. In fact the valuation theory of $\mathbb{F}_p((t))$ is more complicated than that of $\mathbb{Q}_p$. – Hagen Dec 7 '12 at 11:55 @Hagen I don't think I omit anything. It is in fact quite easy to prove that there is a bijection between unramified extensions of a local field (with perfect residue field) and extensions of the residue field, and this does take care of any separability issues. In one direction, given an unramified extension of local fields, reduce modulo the (unchanged) uniformiser to get an extension of residue fields. Conversely, given an extension of residue fields, take a primitive element of this extension, lift it using Hensel, adjoin the lift to the local field to get an unramified extension. – Alex B. Dec 7 '12 at 15:24 Ok, but suppose there exists a proper, finite extension $K$ of $\mathbb{F}_p((t))$ such that $e=1$ (ramification index) and $f=1$ (inertia degree). Then the correspondence between unramified extensions and extensions of the residue field is not injective. Such an extension must necessarily be inseparable. – Hagen Dec 7 '12 at 16:14 show 2 more comments	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 21, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9154986143112183, "perplexity_flag": "head"}
http://crypto.stackexchange.com/questions/710/how-to-forge-schnorr-signatures-if-you-can-guess-the-challenge/713	# How to forge Schnorr signatures if you can guess the challenge Underlying the Schnorr signature is an identification protocol: let $G$ be a cyclic group where discrete log is "hard" and choose $g$ as a generator of $G$. Now have Alice pick a random (secret) exponent $s$ and publish $v=g^s$. To identify herself to Bob, it goes like this: 1. Alice chooses a random exponent $r$ and sends $g^r$ (commits $r$) 2. Bob sends random exponent $e$ (challenge $e$) 3. Alice sends $r + se$ 4. Bob makes sure $g^r = g^{r + se}v^{-e}$ That's all well and good, but what if a cheating prover can guess $e$? Then how does he fool Bob? - ## 4 Answers In the other answers, you'll find how to simulate a proof if you know $e$. This answer is meant to provide some "color commentary" on the other answers. It is a companion piece. Notation • In step 1, Alice sends $g^r$. Call this value $a=g^r$. • In step 3, Alice sends $r+se$. Call this value $b=r+se$. • In step 1-3, one value is sent in each step: {$a,e,b$}. We'll call these three values a transcript for public key $v$. • In step 4, with the new notation, Bob checks: $a=g^bv^{-e}$. Basic question With this notation, it may be easier to see how knowing $e$ can help. Generate a random $b$ and then compute $a$ so that the equation holds and {$a,e,b$} will accept. Another thing to note Other answers noted that since you choose $a$ directly, you do not really know an $r$ such that $a=g^r$ (by the discrete log problem). This is one thing that distinguishes a true proof from a simulated one. A second thing to note, which is very important, is that to simulate a proof, you start by computing (choosing) $b$ and then you compute $a$ using $b$ and $e$. In a real execution, you compute everything in order. The only way to forge? The other answers have shown a way to simulate the proof. Is it the only way? For example, is it possible to choose an $a$ value knowing $e$ and then compute the right $b$ to make {$a,e,b$} accept (short of knowing $s$)? The answer is no. Pretend you have a black box that could do this: it takes {$a,e$} as input and returns the proper $b$ without knowing $s$. If such as box existed, you could query {$a,e_1$) and get $b_1$, and then query {$a,e_2$} with the same $a$ and different $e$, and get $b_2$. However one can verify that this is sufficient to compute $s$: in fact, $s=\frac{b_1-b_2}{e_1-e_2}$. So there is a contradiction: the box doesn't know $s$ by definition and yet it does "know" $s$ (in the sense that it can compute it). Therefore, by contradiction, such a box cannot exist, and neither can a forgery {$a,e,b$} where $a$ and $e$ are chosen. Enforcing only true transcripts If we put all of the above together, it basically says that if {$a,e,b$} accepts, it must have either been computed by someone who truly knows $s$, or it was computed backward by choosing/knowing $b$ and $e$ before computing $a$. Can we eliminate the second case? One way is to be Bob. Another might be to be there in person to see that $a$ is sent before $e$: however do you really know that Alice doesn't know $e$? Can you really be sure? The answer is yes. If you can ensure $a$ is computed before $e$, then we are done. A simple technique (called Fiat-Shamir) is to set $e=\mathcal{H}(a)$, where $\mathcal{H}$ is a hash function (technically a random oracle). If you choose/know $e$ first, you cannot find an $a$ that will hash to it (by preimage-resistance). - If Alice guesses $e$ then she chooses a random value $x$ and computes $h = g^x v^{-e}$, a value which she sends to Bob at step 1. At step 3, Alice sends $x$. When Bob does step 4, he recomputes $g^x v^{-e}$ and finds $h$, and he is happy. However, Alice does not know $s$. The "commitment" at step 1 is a way for Alice to say: "I know a $r$ corresponding to this $g^r$, which allows me to do the computation for any challenge $e$". But if Alice can guess the challenge in advance, she needs not really know $r$. Here, Alice sends a value $h$ which is part of the group and thus is equal to $g^r$ for some value $r$, but Alice does not know that $r$. - Well, if fake-Alice guesses the challenge exponent $e$ in step 1, then she can guess the value of $v^{-e}$. That means she can pick an arbitrary value to stand in for $r+se$, compute $g^{r+se}v^{-e}$, and send that as her commitment in step 1. Then, assuming Bob sends the guessed exponent in step 1, fake-Alice sends the value $r+se$ that she picked above. That means that, in step 4, Bob will validate that $g^r = g^{r+se}v^{-e}$, because he is actually validating ${step1} = g^{step3}v^{-e}$, and fake-Alice picked the $step1$ and $step3$ values to make this equation work. - Suppose that we have Eve, that knows what $e$ is going to be, and does not need to know the prover's private key $a$, just the public one $v$. She then sends $g^k \cdot v^{-e}$ as her first "move", where she can choose her own $k$ (you can modify the $k$ in different plays to make it all look nice and random...). The verifier sends $e$ as expected, of course, and finally Eve sends $k$ as her final move. The verifier only checks that her last move times $v^{-e}$ equals her first move (see step 4.) and this is true by construction. Note that the verifier has no way of seeing that the first move wasn't of the form $g^r$ for some $r$ known to the sender, or that $r + se$ is sent, etc. He only sees 2 numbers by the prover, that have to satisfy a (known in advance) relation.... -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 103, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9460291266441345, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/27241/what-shannon-channel-capacity-bound-is-associated-to-two-coupled-spins/27243	# What Shannon channel capacity bound is associated to two coupled spins? The question asked is: What is the Shannon channel capacity $C$ that is naturally associated to the two-spin quantum Hamiltonian $H = \boldsymbol{L\cdot S}$? This question arises with a view toward providing a well-posed and concrete instantiation of Chris Ferrie's recent question titled Decoherence and measurement in NMR. It is influenced too by the guiding intuition of Anil Shaji and Carlton Caves' Qubit metrology and decoherence (arXiv:0705.1002) that "To make the analysis [of quantum limits] meaningful we introduce resources." And finally, it is reasonable to hope that so simple and natural a question might have a rigorous answer that is simple and natural too---but to the best of my (imperfect) knowledge, no such answer is given in the literature. ## Definitions Let Alice measure-and-control by arbitrary local operations a spin-$j_\text{S}$ particle on a local Hilbert space $\mathcal{S}$ having $\dim \mathcal{S} = 2j_\text{S}+1$, upon which spin operators $\{S_1,S_2,S_3\}$ are defined satisfying $[S_1,S_2] = i S_3$ as usual. Similarly let Bob measure-and-control by arbitrary local operations a spin-$j_\text{L}$ particle on local Hilbert space $\mathcal{L}$ having $\dim \mathcal{L} = 2j_\text{L}+1$ upon which spin operators $\{L_1,L_2,L_3\}$ are defined satisfying $[L_1,L_2] = i L_3$ as usual. Let the sole dynamical interaction between the spins — and thus the primary resource constraint acting upon the communication channel — be the Hamiltonian $H = \boldsymbol{L\cdot S}$ defined on the product space $\mathcal{S}\otimes \mathcal{L}$. Further allow Bob to communicate information to Alice by a classical communication channel of unbounded capacity, but let Alice have no channel of communication to Bob, other than the channel that is naturally induced by $H$. Then the question asked amounts to this: what is the maximal Shannon information rate $C(j_\text{S},j_\text{L})$ (in bits-per-second) at which Alice can communicate (classical) information to Bob over the quantum channel induced by $H$? ## Narrative In practical effect, this question asks for rigorous and preferably tight bounds on the channel capacity associated to single-spin microscopy. The sample-spin $S$ can be regarded as a sample spin that can be modulated in any desired fashion, and the receiver-spin $L$ can be regarded variously as a tuned circuit, a micromechanical resonator, or ferromagnetic resonator, as shown below: The analysis of the PNAS survey Spin Microscopy's Heritage, Achievements, and Prospects (2009) can be readily extended to yield the following conjectured asymptotic form: $$\lim_{j_\text{S}\ll j_\text{L}} C(j_\text{S},j_\text{L})=\frac{j_\text{S}\,(j_\text{L})^{1/2}}{(2\pi)^{1/2}\log 2}$$ Note in particular that the dimensionality of Bob's receiver-spin Hilbert space $\mathcal{L}$ is $\mathcal{O}(\,j_\text{L})$; thus a Hilbert-space having exponentially large dimension is not associated to Bob's receiver. However it is perfectly admissible for Alice and Bob to (for example) collaborate in squeezing their respective spin states; in particular the question is phrased such that Alice may receive real-time instruction of unbounded complexity from Bob in doing so. ## Preferred Form of the Answer A closed-form answer giving a tight bound $C(j_\text{S},j_\text{L})$ is preferred, however a demonstration that (e.g.) $\mathcal{O}(C)$ is given by some closed asymptotic expression (as above) is acceptable. It would also be very interesting, both from a fundamental physics point-of-view and from a medical research point-of-view, to have a better appreciation of whether the above conjectured capacity bound on spin imaging and spectroscopy can be substantially improved by any means whatsoever. - 3 I downvoted this question as it seems like a nice question but it is not written in the clearest way. You talk a lot around the question and I am not sure why you included that image. I think if it is stated in a simple motivation-definition-question format it'd be a nice question. I am still not completely clear on why one should care about such an Hamiltonian for greater than spin-1/2 systems. I'll be happy to remove my downvote if you think I am being unreasonable. – Matty Hoban Nov 7 '11 at 0:28 Hoban, downvotes trouble me far less than indifference! :) Yes, physics hovers uneasily between beautiful mathematics and beautiful experiments. Spins larger than 1/2 are motivated mainly experimentally, as mentioned in my comment appended to Aram Harrow's answer (below): it is that nature provides us with large-spin particles (in the form of ferromagnets) having a spatial spin density billions of times greater than can be achieved by (e.g.) trapped-ion condensates. The price to be paid for this spin density is that the effective Hilbert space dimension is far lower than said condensates. – John Sidles Nov 7 '11 at 0:57 ## 2 Answers This is an open question. The capacity of some related Hamiltonians was computed in quant-ph/0207052, and an upper bound was derived in 0704.0964, but the Hamiltonian you describe is an example of one for which we don't know the exact answer. However, quant-ph/0207052 also contains a conjecture (eq 35) about the capacity you're interested in. Their conjecture is supported by numerical experiments, but they can't rule out that some block coding strategy could do better. Edited in light of our conversation this afternoon: From our discussion, it sounds like you've proven a lower bound of $\Omega(j_S j_L)$ using some cat-like states, based on principles similar to those in quant-ph/0605013. I think I can prove an asymptotically matching upper bound of $O(j_S j_L)$. So the exact constant is still open (and I think a hard problem), but at least we know the scaling. For this bound, first argue that the interaction can be modelled as $2j_S$ qubits for Alice and $2j_L$ qubits for Bob, each in a symmetric state. To prove an upper bound, we can relax the restriction that the qubits be in a symmetric state. Then the Hamiltonian you describe can be expressed as a sum of interactions between each qubit of Alice's and each qubit of Bob's, each of which has constant strength. These two-qubit Hamiltonians have capacities upper-bounded by a constant. One reference for this last claim is quant-ph/0205057, but it was probably known earlier. - Aram, these are wonderful references for which I am very grateful! With specific regard to quant-ph/0207052 (eq. 35) this result (AFAICT) applies solely to coupled qubits (that is, spin-1/2 particles), whereas when we look into the coming decades of quantum spin microscopy, we foresee spins coupled to ferromagnetic spheres having spin-j ~ 10^5. These spheres have (effectively) a state-space of dimensionality much larger than one qubit, yet much small than 10^5 qubits; in brief their state-space is large enough to squeeze, but too small for (e.g.) quantum computation. More please! :) – John Sidles Nov 7 '11 at 0:05 Aram, as a followup, I am beginning to appreciate more of the reasons why this simple-to-state question is difficult to answer. For example, the case $j_S = 1/2$ and $j_L \gg 1$ large can be regarded as effectively describing a cavity QED experiment, with $\mathcal{S}$ the state-space of the transmitting two-level atom and $\mathcal{L}$ the state-space of the receiving high-quantum-efficiency detector; these systems are known to be tough to analyze theoretically and tough to build experimentally. Conversely, this means that a closed-form tight capacity bound would be quite valuable to know. – John Sidles Nov 7 '11 at 10:48 @AramHarrow: This update sounds an awful lot like metrology. It seems like the Heisenberg limit may apply. – Joe Fitzsimons Nov 10 '11 at 9:50 – Aram Harrow Nov 21 '11 at 8:42 @AramHarrow: Thank you for these wonderful comments! I've been busy running an experiment for the past couple of weeks (hence the slowness in responding), but I concur with the general thrust of your remarks, and have thereby gained a better appreciation of (what is likely to be) the order of the upper and lower bounds to the channel capacity that are (respectively) achieved by cat-state sensors and coherent-state sensors, and thus have rated it as an "answer". – John Sidles Dec 14 '11 at 19:14 Aram's answer seems perfect, but since you are also asking about the case for higher dimensional systems, let me add that there is a simply way to get somewhat non-trivial upper and lower bounds on $C(j_S,j_L)$. As a lower bound, you can simply synthesize an arbitrary gate which implements communication between the quantum systems (for an explicit algorithm on constructing arbitrary gates see Nielsen et al, Phys. Rev. A 66, quant-ph/0109064). A non-trivial upper bound is given by the Margolus-Levitin theorem. Their paper (quant-ph/9710043) gives the maximum number of orthogonal states a quantum system can pass through in a given period of time, or conversely, a minimum time (as a function of energy) required to from the initial state to an orthogonal state (which is necessarily a lower bound on the time required to perfectly transmit one bit). - Thank you Joe ... it will take a couple of days to process these references. A challenge with the Nielsen construction is "gate to rate", that is, the construction does not limit the energy associated to Hamiltonian resources; thus is not evident how to extract a channel capacity bound of any kind. A challenge with the Margolus-Levitan theorem is that the operations associated to the bound are (apparently) not local; thus the upper bound (that I get from applying it) is unphysically optimistic. Perhaps there are ways around these obstructions; this is what will take awhile to think about. – John Sidles Nov 7 '11 at 22:03 Progress report: preliminary efforts to construct concrete two-spin communication channels that approach the Margolus-Levitin bound have (so far) foundered on the local-operation restriction. This contributes to an appreciation (on my part) that Aram Harrow's answer is correct in asserting that "This [two-qudit channel capacity] is an open question", largely because the locality-of-operation restriction has deep-and-subtle consequences. One emerging benefit is an increased appreciation that fundamental QIT constraints are associated to single-photon sources and efficient local detectors. – John Sidles Nov 8 '11 at 13:18 2 The Bravyi paper I mentioned (0704.0964) can be thought of as applying the Margolus-Levitin principle to upper bound the entangling rate. The bound is constant * norm of Hamiltonian * log dimension. (Note that this is an upper bound for the communication rate. Also dimensional factors are inevitable for both communication rate and entanglement rate, because transmitting an n-bit message can mean a single pi/2 rotation.) Unfortunately his argument is incomplete because we don't know how to control how much ancillas can increase the capacity. – Aram Harrow Nov 9 '11 at 7:55 1 @AramHarrow: Good point. However, it seems like you can bound the effective dimensionality via the number of splittings in the eigenvalues of local Hamiltonians. – Joe Fitzsimons Nov 9 '11 at 8:17 1 @AramHarrow: To explain what I mean: Imagine you have two spin-1/2 systems. Then your coupling induces at most 4 distinct eigenspaces which pick up relative phase, independent of the number of ancillae, or how they are prepared prior to coupling. Given that the information transfer in periods of free evolution is determined entirely by the relative phases between these 4 spaces, this is identical to what can be achieved with two spins. The only subtlety then comes from how you can combine multiple uses of such a channel. – Joe Fitzsimons Nov 9 '11 at 8:30 show 4 more comments	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 32, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 5, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9363108277320862, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/77650/a-volterra-type-equation	## A Volterra-type equation ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Consider the following integral equation $\phi(x) = f(x) + \frac{1}{x}\int_0^x N(x,y)\phi(y)\;dy$, where $f$ and $N$ are continuous and bounded functions. Are solutions $\phi$ of the above equation unique? If so, can one get an estimate of the form $\sup_{(0,x)} \|\phi\| \leq C \sup_{(0,x)}\|f\|$ ? Additional info: Assume that $\phi(0)=\phi(1)=f(0)=0$ and that $\\|N\\|_\infty\geq 1$. I am only interested in a solution (or lack thereof) on the interval $[0,1]$. As a note, without the $\frac{1}{x}$ multiplier, this is a Volterra equation of the second kind and existence/uniqueness of a solution $\phi$ is well-known and the above estimate is indeed satisfied. - ## 1 Answer Not in general, since if $N(x,y)=a>1$ then the equation with $f=0$ has the solution $\phi(x)=x^{a-1}$. If, however, $N(0,0)<1$ (assuming as in the question that $N$ is continuous) then one can use the Banach fixed-point theorem (on short intervals) to get a unique solution. - I assume you mean $\|N(0,0)\| < 1$. You can of course get a stronger existence result if you have $\\| N \\|_{\infty} < 1$. – Christopher A. Wong Oct 10 2011 at 5:45 Thanks for the replies. I had noticed the contraction argument for $\\|N\\|_\infty < 1$, but the case I am interested in, I have $\\|N\\|_\infty > 1$, but I also have $\phi(0)=\phi(1)=0$ (and of course then $f(0)=0$). I should have added these to the problem statement (which I will do now). I would be happy with either uniqueness or non-existence. – Jeff Oct 10 2011 at 13:08	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 21, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9317253232002258, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/40849/how-to-check-if-a-symmetric-4-times4-matrix-is-positive-semi-definite	# How to check if a symmetric $4\times4$ matrix is positive semi-definite? 1. How does one check whether symmetric $4\times4$ matrix is positive semi-definite? 2. What if this matrix has also rank deficiency: is it rank 3? - 2 You can use the determinant criterion: the upper-left $1\times 1$, $2\times 2$, $3\times 3$ and $4 \times 4$ squares should all have non-negative determinant. – Yuval Filmus May 23 '11 at 16:00 1 But that determinant criterion isn't enough in general, as $\begin{bmatrix}0&0&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&-1\end{bmatrix}$ shows. – Jonas Meyer May 23 '11 at 16:07 2 @Yuval - I believe the determinant criterion holds for positive definite matrices but not necessarily for positive semidefinite ones. In other words, if some of the principal minors are zero, it does not necessarily imply the matrix is positive semidefinite. – svenkatr May 23 '11 at 16:16 3 In that case, you can add $\epsilon > 0$ (to the diagonal) and then rerun all your computations (just when the matrix doesn't have full rank). A suitable $\epsilon$ can be found by looking at the magnitude of the entries (we want to guarantee that we don't miss any small negative eigenvalue). – Yuval Filmus May 23 '11 at 16:48 ## 5 Answers Since the matrix is symmetric, the eigenvalues will be real. Calculate the eigenvalues and see if they are all $\geq 0$. If this is true,the matrix is positive semidefinite. - 1 You don't even need the eigenvalues. Once you have the polynomial invariants (coefficients of the characteristic polynomial), you can use Descarte's rule of signs. – Willie Wong♦ Aug 24 '11 at 16:22 1 Even computing the characteristic polynomial here is overkill. Robert's suggestion takes less effort... – J. M. Aug 25 '11 at 4:21 Another method is to check there are no negative pivots in row reduction (after taking into account the possibility of 0's on the diagonal). The procedure can be written recursively as follows: 1) If $A$ is $1 \times 1$, then it is positive semidefinite iff $A_{11} \ge 0$. Otherwise: 2) If $A_{11} < 0$, then $A$ is not positive semidefinite. 3) If $A_{11} = 0$, then $A$ is positive semidefinite iff the first row of $A$ is all 0 and the submatrix obtained by deleting the first row and column is positive semidefinite. 4) If $A_{11} > 0$, for each $j > 1$ subtract $A_{j1}/A_{11}$ times row 1 from row $j$, and then delete the first row and column. Then $A$ is positive semidefinite iff the resulting matrix is positive semidefinite. - As stated above, Sylvester's criterion doesn't work in this case, so you can't simply check the four leading principal minors. However, it does suffice to check that all 15 of the principal minors are nonnegative. See here for a reference. Another basic approach involves symmetric row reduction. This involves operations of the following type: 1. Perform a row operation, and then 2. Immediately perform the corresponding column operation. For example, you can multiply any row by a constant, as long as you immediately multiply the corresponding column by a constant. Note that this multiplies the diagonal entry by the square of the constant. Using these operations, you can use a variant of Gaussian elimination to reduce any symmetric matrix to a diagonal matrix with 1's, 0's, and -1's along the diagonal. A matrix is positive semidefinite if and only if the resulting diagonal entries are all 0's and 1's. - Let's say your matrix is $A$. You can check the eigenvalues. If all eigenvalues $\geq 0$, the matrix is positive semi-definite (if all eigenvalues $>0$ it is positive definite). It might be possible to use the Gershgorin circle theorem instead of calculating the eigenvalues explicitly. If all the diagonal elements are positive and are larger than or equal to the sum of the absolute values of the other elements in same row (or column) (for every diagonal element), then the matrix is positive semi-definite. You can try to find a simpler semi-definite matrix $B$ such that $B^2 = A$ ($B$ is unique). This is in general done using the diagonalization of the matrix, so it will probably be easier just calculating the eigenvalues. You can not use a modification of Sylvester's criterion ("all leading principal minors are non-negative") to determine positive semidefiniteness. If $A$ is $4 \times 4$ and rank 3, it has 0 as an eigenvalue (since there exists a vector $v$ such that $Av = 0v$). This does not affect the positive semi-definiteness of the matrix, but it will not be positive definite. - One quick first check, if any element $A_{ii}$ on the diagonal is negative, the matrix has a negative eigenvalue (for real symmetric matrices only of course) -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 34, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8858813047409058, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/27279/a-nice-overview-and-maybe-derivation-of-the-poincare-transformations-of-the-ve?answertab=votes	# A nice overview (and maybe derivation) of the Poincaré transformations of the Vector Spherical Harmonics With $Y_{lm}(\vartheta,\varphi)$ being the Spherical Harmonics and $z_l^{(j)}(r)$ being the Spherical Bessel functions ($j=1$), Neumann functions ($j=2$) or Hankel functions ($j=3,4$) defining $$\psi_{lm}^{(j)}(r,\vartheta,\varphi)=z_l^{(j)}(r)Y_{lm}(\vartheta,\varphi),$$ what are representations of the Poincaré transformations applied to the Vector Spherical Harmonics $$\vec L_{lm}^{(j)} = \vec\nabla \psi_{lm}^{(j)},\\ \vec M_{lm}^{(j)} = \vec\nabla\times\vec r \psi_{lm}^{(j)},\\ \vec N_{lm}^{(j)} = \vec\nabla\times\vec M_{lm}^{(j)}$$ ? Does any publication cover all Poincaré-transformations, i.e. not only translations and rotations but also Lorentz boosts? I'd prefer one publication covering all transformations at once due to the different normalizations sometimes used. - – Tobias Kienzler Mar 1 '12 at 12:19 ## 1 Answer The problem with Poincaré group is in the fact that it is not compact. That's why this question is non-trivial. Though, properly formulated search gives few papers on this topic. Try to find the answer in this paper http://arxiv.org/abs/math-ph/0507056 . The paper itself may be not that interesting, but there is a nice introduction with a number of useful references. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 5, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8554949760437012, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/61275?sort=newest	## Analysis and finitely generated groups ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Dear all, this is perhaps a bit a vague question, but some references would already be very helpfull. So let $G$ be a finitely generated group and choose some finite set of generators. This allows to define the length of a group element as usual, by the minimal number of generators needed to write it as a product of these generators. Using this length you can define all sort of analytic functions by replacing the "n" in various summation formulas by the $\mathrm{Length}(g)$ and the summation is now over the group elements $g$. To have a concrete example in mind, the "exponential series" is now e.g. \begin{equation} \mathrm{lexp}(z) = \sum_{g \in G} \frac{z^{\mathrm{Length}(g)}}{\mathrm{Length}(g)!}. \end{equation} Since for a given length there are at most exponentially many group elements, the length-exponential function is entire. For $G = \mathbb{Z}$ and $1$ as generator this reproduces the usual exponential series up to a factor $2$ as negative and positive $n \in \mathbb{Z}$ contribute with the same power of $z$. So my question is: what is known about the analytic features of such functions (depending on the choice of generators, depending on the group itself, etc). I guess there should be some literature on the market, but I'm really not in this buisness... - 1 While I've never seen the particular example you give, there is a vast literature on related generating functions (usually called growth functions or something to that effect). Have look at Chapters VI and VII of Pierre de la Harpe's book Topics in geometric group theory (Chicago lectures in mathematics) for a discussion of many examples and a survey of the main results. The extensive reference list in the book should provide many pointers to the literature. – Theo Buehler Apr 11 2011 at 10:52 @Theo: thanks a lot. I will check out this book in the library, googlebook gives only a couple of pages. But they look nice indeed. – Stefan Waldmann Apr 12 2011 at 8:31 This sounds highly related to Geometric Group Theory. I would check in books on that area, in particular De La Harpe's Geometric Group Theory, where he actually computes some of these if I recall correctly. – Benjamin Hayes May 7 2011 at 15:03 Dear Benjamin: thanks! I had a look at the book (Theo mentioned it already) and it indeed looks very inspiring. It seems that I have to work a bit more in that direction. Very interesting... – Stefan Waldmann May 9 2011 at 9:28 Just a vague comment: I haven't seen exactly what you describe, but I know that some people study things like the "Ihara zeta function of a graph". If the graph you consider is the Cayley graph of your group, then you get formulas looking a bit like yours... – Sylvain Bonnot Jun 10 2011 at 22:11 show 1 more comment ## 1 Answer One special case of groups, where one certainly gets rather quickly explicit and nontrivial expressions should be finite or affine Coxeter groups, that are finite/infinite and defined by involution generators and relations from their Dynkin diagrams - more my topic than graphs ;-) If the Coxeter-diagram belongs to a finite/affine Lie algebra root system resp. Weyl group (which is the generic case!), these rather strong structures should give you enough informations to control the elements of the group ordered by their length. I want to be specific in two cases (finite/infinite) I found rather quickly in the relevant literature, but are without factorial in the numerator: For a finite Weyl-group $W$ acting as reflection group on an $n$-dimensional vectorspace the well-known Chevalley-Solomon theorem (often used "the-other-way-around") asserts, that: $$W(z):=\sum_{g\in G} z^{Length(g)}=\prod_{\alpha\in \Delta^+}\frac{1-z^{Height(\alpha)+1}}{1-z^{Height(\alpha)}} = \prod_{k=1}^n\frac{1-z^{d_k}}{1-z}$$ where $d_k$ are the fundamental degrees of the reflection action, i.e. the degrees of a homogenious basis of the invariant part of the polynomial ring over $V$ (having $n$ variables). The middle term is crucial for the proof (and pretty), but the product over the entire root system $\Delta$ is not helpful for our question ;-) Example For $S_n$ (root system $A_{n-1}$) we have $d_k=k$ (each elementary symmetric polynomial), thus: $$\sum_{g\in S_n} z^{Length(g)}=\prod_{k=1}^n\frac{1-z^{k}}{1-z}$$ For an affine Weyl group $\tilde{W}$ i.e. with Dynkin-diagram as some finite $W$ (suppose irreducible) with one node added, Bott's theorem states that (omitting $d_k=1$-Terms): $$\tilde{W}(z)=W(z)\prod_{k=1}^n\frac{1}{1-z^{d_k-1}}$$ Example $\tilde{A}_n$ is derived from closing the $n$-chain $A_n$ (i.e. $S_{n+1}$) with an additional $x$. Hence is generated very similar to the symmetric group but infinite (all non-mentioned pairs elements commute!): $$G=\langle t_1,t_2,...t_n,x\rangle\qquad (t_it_{i+1})^3=1\quad (t_1x)^3=(xt_n)^3=1$$ Hence the length-generating function now has a pole: $$\tilde{W}(z)=W(z)\prod_{k=2}^n\frac{1}{1-z^{k-1}}=\frac{1}{(1-z)^n}\prod_{k=2}^n\frac{1-z^k}{1-z^{k-1}}$$ Finally I must of course mention that the beatifully exotic root systems of Nichols algebras, that would even count the lengths of Coxeter gruppoids ;-) This was all written down without much further thought, but if there's still interest in the topic (?) I'd be happy about a further discussion! Maybe concerning factorial or something else.... SOURCES: • Definitely Humphreys "Reflection Groups and Coxeter groups" • the formulas also directly online e.g. the "survey" www.math.umn.edu/~reiner/Papers/SteinbergNotes.ps). • Some infinite worked-out examples e.g. in dml.ms.u-tokyo.ac.jp/PSRT/PSRT_26/PSRT_26_093-102.pdf and many more) -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 25, "mathjax_display_tex": 5, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9088295698165894, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/45839/why-are-navier-stokes-equations-needed/45993	# Why are Navier-Stokes equations needed? Can't we picture air or water molecules individually? Then, why are Navier-Stokes equations needed, after all? Can't we just aggregate individual ones? Or is it computationally difficult, or inefficient to generate the picture of whole flow? - 4 – Emilio Pisanty Dec 4 '12 at 2:33 ## 3 Answers Consider a standard volume of $1\textrm{ m}^3$ of air. This contains on the order of $10^{25}$ molecules of O2 and N2. If you needed to simulate or explain the physics occurring in that volume of air, would you want to model $10^{25}$ molecules and all the interactions between them or, say, 100x100x100 cells based on the Navier-Stokes equations? Theoretically, it is possible to simulate every fluid flow ever by tracking every single molecule. But direct simulation of turbulence using the Eulerian Navier-Stokes equations requires $Re^{9/4}$ grid points and is thus totally impractical for Reynolds numbers larger than a few thousand. So simulating something with $10^{25}$ things to track is completely impossible. - 3 Just recording the position coordinates as doubles would take 1000000000000000 GB of RAM. – user1504 Dec 4 '12 at 3:31 2 A quibble, but a cubic metre contains more than a mole (6.023e23 molecules). At STP a mole of an ideal gas is 22.4 litres. – John Rennie Dec 4 '12 at 6:59 You're right, it's been a long time since I had to think about those relationships. I'll fix it – tpg2114 Dec 4 '12 at 19:55 1 I don't think numerical simulation was the reason Navier and Stokes derived their equations. – Bernhard Dec 5 '12 at 11:59 Just to add to this correct answer: even if you could simulate every molecule, you would gain little by it unless for some reason you knew the initial position of every molecule to start with. Even then, it would only help you to predict the future behaviour of that particular volume of fluid at that particular time - it wouldn't tell you anything about what would happen if you repeated the same experiment again, but with different molecules that weren't in exactly the same starting places. – Nathaniel Dec 5 '12 at 13:00 show 3 more comments Apart from the experimental and numerical difficulties to track each particle in a macroscopic piece of matter, there are physical reasons. Theoretically, we can show that in the hydrodynamic regime the exact equations of motion for the collection of atoms/molecules reduce to the Navier-Stokes equations. This is shown in statistical mechanics. Due to complex molecular effects, including molecular chaos, most degrees of freedom self-cancel and only some few dynamical modes survive. Those modes are collective --i.e. they describe the fluid as a whole-- and robust --i.e., they survive in macroscopic scales of space and time--. That is, even if you were to trace the individual motion of each particle and next average to describe the collective motion of the whole fluid, you would see that your description is practically indistinguishable [] from that given by solving the Navier-Stokes equations. [] The differences are of the order of the inverse of the size of the system and vanish for a macroscopic system. - "Theoretically, we can show that in the hydrodynamic ..." I believe this paragraph is wrong at the current time. I thought you can derive NS from the Boltzmann equation, so techinically you can derive NS to a certain extent only for gases, but not for liquids. Or do you know statistical NS derivation for liquids? – Yrogirg Dec 5 '12 at 14:47 @Yrogirg: There is a kinetic theory for dense fluids as well, but is not of the Boltzmann type, of course. Moreover, you can obtain the hydrodynamic equations without using any kinetic equation, for instance using the Mori formalism for the microscopic dynamical variables. – juanrga Dec 6 '12 at 11:32 As apparent from the other answers, the continuum approach of the Navier-Stokes equations makes life easier, by reducing the degrees of freedom. For example, Poisseuile flow (laminar flow between flat plates or in a pipe), is easily solved using the Navier-Stokes equations, but I have no clue how to solve it using just molecular behavior (on the back of an envelope that is). If you calculate the Knudsen number of a specific problem, which is just the molecular mean free path divided by some macroscopic length scale, you can see if the continuum approach is a valid for the considered system. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 5, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9328474998474121, "perplexity_flag": "middle"}
http://www.purplemath.com/learning/viewtopic.php?t=2138	# The Purplemath Forums Helping students gain understanding and self-confidence in algebra. ## finding the area of parallelogram / combination/ permutation Sequences, counting (including probability), logic and truth tables, algorithms, number theory, set theory, etc. 4 posts • Page 1 of 1 ### finding the area of parallelogram / combination/ permutation by jigoku_snow on Wed Aug 03, 2011 1:56 pm 1) ABCD is a parallelogram and the coordinates of A, B, C are (-8, -11), (1, 2) and (4, 3) respectively. Find the area of the parallelogram. [i]i manage to find that D =( -5.-10) . area of parallelogram = AD x height i'm stuck here. how to find the height? i tried it by finding the angle between AB and AD first , (37.32 degree) then use it to find the vertical line from B. but i dont get the right answer. 2). Points A and C have coordinates (-1, 2) and (9, 7) respectively. A rectangle ABCD has AC as a diagonal. Calculate the possible coordinates of B and D if the length of AB is 10 units. from the question, i know that AB=CD=10. the equation of line AB = x^2 +y^2+ 2x -4y-97=0. the equation of line CD= a^2 +b^2-18 a- 14b +30=0. then what should i do next? 3)If the five letters a, b, c, d, e are put into a hat, in how many ways could you draw two letters? must we use permutation or combination? if permutation, why? why can't i use combination? 5C2? 4))3 balls are to be placed in 3 different boxes, not necessarily with 1 ball in each box. any box can hold up to 3 balls. find the number of ways the balls can placed if i) they are all of the same colour ii) they are all of different colours i) what is wrong with my workings? 3C3 + (3C2 x 1C1) + 3C1? ii) i have no clue at all for this. isnt it suppose to be the same as the first one? 5)GOLD MEDAL- if 2 D letters come first and the 2 L letters come last, find how many different arrangement there are D D _ _ _ _ _ L L 5P5 x 2P2 x 2P2. why i dont need to permute the letter D and the L? 6) 3 letters are selected at random from the word "BIOLOGY" . Find the number that the selection contains one letter O 2C1 x 5C2 = 20 , the answer given is 5C2 = 10 , why i dont have to write 2C1? 7)The sum of the first 5 terms of a G.P. is twice the sum of the terms from the 6th to the 15th inclusive. Show that r^5 = 1/2 (SQRT3 -1). S5 = 2 (S15 - S5) a(1-r^5)/ i-r = 2a/ 1-r [(1-r^15)-(1-r^5)] 1 - r^5 = 2(r^5 -r^15) 2r^15-3r^5+1=0 let u=r^5 2u^3 - 3u +1=0 i'm stuck here. what to do next? jigoku_snow Posts: 2 Joined: Wed Aug 03, 2011 1:46 pm Sponsor Sponsor ### Re: finding the area of parallelogram / combination/ permuta by nona.m.nona on Thu Aug 04, 2011 1:55 am jigoku_snow wrote:1) ABCD is a parallelogram and the coordinates of A, B, C are (-8, -11), (1, 2) and (4, 3) respectively. Find the area of the parallelogram. Use the Distance Formula to find the lengths of each base. Find the slope of the line containing either base; use this to find the perpendicular slope. Find the line equation for the perpendicular through one of the corners; find the point where this line intersects the other base (or the other base, extended). Find the distance between the two bases, and thus the height. Plug the base lengths and the height into the area formula. jigoku_snow wrote:2). Points A and C have coordinates (-1, 2) and (9, 7) respectively. A rectangle ABCD has AC as a diagonal. Calculate the possible coordinates of B and D if the length of AB is 10 units. from the question, i know that AB=CD=10. the equation of line AB = x^2 +y^2+ 2x -4y-97=0. the equation of line CD= a^2 +b^2-18 a- 14b +30=0. then what should i do next? The line AB will be linear; what you have posted is a quadratic. How did you derive this? (One of the links in the previous response explains how to find linear equations from two points.) As for finding the coordinates, try drawing the two points you have now, along with the line containing them. The rectangle is either to the one side of that line, or else to the other. So where must the other two points be? jigoku_snow wrote:3)If the five letters a, b, c, d, e are put into a hat, in how many ways could you draw two letters? must we use permutation or combination? if permutation, why? why can't i use combination? 5C2? Since nothing in the exercise says anything about ordering, combinations should be sufficient. However, you can see your book and your class notes; if your book, by default, defines "drawing" as "one at a time, with order taken into account", then you should use permutations. jigoku_snow wrote:4))3 balls are to be placed in 3 different boxes, not necessarily with 1 ball in each box. any box can hold up to 3 balls. find the number of ways the balls can placed if i) they are all of the same colour ii) they are all of different colours i) what is wrong with my workings? 3C3 + (3C2 x 1C1) + 3C1? You seem to be counting the ways to choose the balls, but not the ways in which you could then distribute them (in the chosen groupings) amongst the boxes. jigoku_snow wrote:ii) i have no clue at all for this. isnt it suppose to be the same as the first one? If you have boxes A, B, and C and if you are placing three red balls in them, then you have the first situation. However, if you have boxes A, B, and C and if you are placing a red, a green, and a yellow ball in them, then you have the second situation. In the first case, "Box A has a red ball" is the same as "Box A has a red ball", since the balls are indistinguishable. In the second case, "Box A has a red ball" is different from "Box A has a green ball". jigoku_snow wrote:5)GOLD MEDAL- if 2 D letters come first and the 2 L letters come last, find how many different arrangement there are D D _ _ _ _ _ L L 5P5 x 2P2 x 2P2. why i dont need to permute the letter D and the L? How does "D" differ from "D"? How are they distinguishable? jigoku_snow wrote:6) 3 letters are selected at random from the word "BIOLOGY" . Find the number that the selection contains one letter O 2C1 x 5C2 = 20 , the answer given is 5C2 = 10 , why i dont have to write 2C1? What was your logic for including "2C1"? jigoku_snow wrote:7)The sum of the first 5 terms of a G.P. is twice the sum of the terms from the 6th to the 15th inclusive. Show that r^5 = 1/2 (SQRT3 -1). [i]S5 = 2 (S15 - S5) a(1-r^5)/ i-r = 2a/ 1-r [(1-r^15)-(1-r^5)] How did you arrive at this equation? Using the formulation explained here: $a\left(\frac{1\, -\, r^5}{1\, -\, r}\right)\, =\, 2a\left(\frac{1\, -\, r^{15}}{1\, -\, r}\, -\, \frac{1\, -\, r^5}{1\, -\, r}\right)$ Multiply by (1 - r)/a to get: $1\, -\, r^5\, =\, 2\left(1\, -\, r^{15}\right)\, -\, \left(1\, -\, r^5\right)$ $1\, -\, r^5\, =\, 2\, -\, 2r^{15}\, -\, 2\, +\, 2r^5$ $2r^{15}\, -\, 3r^5\, +\, 1\, =\, 0$ Then solve the polynomial for its rational root (which clearly does not apply, as a value for the common ratio, to this series). Solve the resulting quadratic for the other root, using the definition of "geometric progression" to pick the correct value. nona.m.nona Posts: 198 Joined: Sun Dec 14, 2008 11:07 pm ### Re: finding the area of parallelogram / combination/ permuta by jigoku_snow on Sat Aug 06, 2011 8:50 am The line AB will be linear; what you have posted is a quadratic. How did you derive this? since i cant find the gradient so i use the distance formula. i let B=(x,y) and D=(a.b) What was your logic for including "2C1"? well, is because there are 2 'O' s. but i think i know why i dont have to write that, is because they are indistinguishable. How did you arrive at this equation? from a(1-r^5)/ 1-r = 2a [ (1-r^15/ 1-r) - ( 1-r^5 / 1-r )] i simplify it . Then solve the polynomial for its rational root i think this question is way too far for me as i not yet learn polynomial. jigoku_snow Posts: 2 Joined: Wed Aug 03, 2011 1:46 pm ### Re: finding the area of parallelogram / combination/ permuta by nona.m.nona on Sat Aug 06, 2011 2:40 pm The line AB will be linear; what you have posted is a quadratic. How did you derive this? jigoku_snow wrote:since i cant find the gradient so i use the distance formula. i let B=(x,y) and D=(a.b) You have two points on a straight line. You can definitely find the slope ("gradient"). Please study the lesson in the link, provided earlier, to learn how. Then solve the polynomial for its rational root jigoku_snow wrote:i think this question is way too far for me as i not yet learn polynomial. You are working with polynomials, such as the quadratic you created for the first exercise and "1 - r15" here, so I am not sure what you mean when you say that you haven't yet learned about polynomials. Please study the lesson provided in the link to learn how to solve polynomials. nona.m.nona Posts: 198 Joined: Sun Dec 14, 2008 11:07 pm 4 posts • Page 1 of 1 Return to Discrete Math • Board index • The team • Delete all board cookies • All times are UTC	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 4, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9184966683387756, "perplexity_flag": "middle"}
http://unapologetic.wordpress.com/2008/01/22/the-differential-mean-value-theorem/?like=1&source=post_flair&_wpnonce=3fdfac1636	# The Unapologetic Mathematician ## The Differential Mean Value Theorem Let’s say we’ve got a function $f$ that’s continuous on the closed interval $\left[a,b\right]$ and differentiable on $(a,b)$. We don’t even assume the function is defined outside the interval, so we can’t really set up the limit for differentiability at the endpoints, but they don’t matter much in the end. Anyhow, if we look at the graph of $f$ we could just draw a straight line from the point $(a,f(a))$ to the point $(b,f(b))$. The graph itself wanders away from this line and back, but the line tells us that on average we’re moving from $f(a)$ to $f(b)$ at a certain rate — the slope of the line. Since this is an average behavior, sometimes we must be going faster and sometimes slower. The differential mean value theorem says that there’s at least one point where we’re going exactly that fast. Geometrically, this means that the tangent line will be parallel to the secant we drew between the endpoints. In formulas we say there is a point $c\in(a,b)$ with $f'(c)=\frac{f(b)-f(a)}{b-a}$. First let’s nail down a special case, called “Rolle’s theorem”. If $f(a)=0=f(b)$, we’re asserting that there is some point $c\in(a,b)$ with $f'(c)=0$. Since $\left[a,b\right]$ is compact and $f$ is continuous, the extreme value theorem tells us that $f$ must take a maximum and a minimum. If these are both zero, then we’re looking at the constant function $f(x)=0$, and any point in the middle satisfies $f'(c)=0$. On the other hand, if either the maximum or minimum is nonzero, then we have a local extremum at a point $c\in(a,b)$ where $f$ is differentiable (since it’s differentiable all through the open interval). Now Fermat’s theorem tells us that $f'(c)=0$ since $c$ is a local extremum! Thus Rolle’s theorem is proved. Now for the general case. Start with the function $f$ and build from it the function $g(x)=f(x)-\frac{f(b)-f(a)}{b-a}(x-a)-f(a)$. On the graph, this corresponds to applying an “affine transformation” (which sends straight lines in the plane to other straight lines in the plane) to pull both $f(a)$ and $f(b)$ down to zero. In fact, it’s a straightforward calculation to see that $g(a)=0=g(b)$. Thus Rolle’s theorem applies and we find a point $c$ with $g'(c)=0$. But applying our laws of differentiation, we see that $g'(c)=f'(c)-\frac{f(b)-f(a)}{b-a}$. And so $f'(c)=\frac{f(b)-f(a)}{b-a}$, as desired. ### Like this: Posted by John Armstrong \| Analysis, Calculus ## 21 Comments » 1. The “trick” behind proving the Mean Value theorem is basically to write $g(x) = f(x) - h(x)$, where $y = h(x)$ is the equation of the straight line connecting the points $(a, f(a))$ and $(b, f(b))$. And this is what you did above. I know it might not sound very interesting, but I have always wondered if we can obtain another useful result (just like the mean value theorem, for instance) if we let $y = h(x)$ be the equation of a parabola passing through the points $(a, f(a))$ and $(b, f(b))$. And, do we obtain more “useful” results if $y = h(x)$ is some function other than a straight line or a parabola? (That was just a random thought.) Comment by \| January 22, 2008 \| Reply 2. Indeed you do, Vishal, but you ask that the function be twice-differentiable and you match another parameter: the derivative of $f$ at $a$. Then those three parameters specify a unique parabola, which has constant second derivative. And there’s some point in between $a$ and $b$ at which the function actually has that second derivative. There’s a whole sequence of these “generalized mean value theorems”, which can be used as Taylor series remainder approximations. Your exercise is to work out the analogues of Rolle’s theorem you guessed were around, and to see if you can prove them in general. I’ll come back to this when I’m doing Taylor series. Comment by \| January 22, 2008 \| Reply 3. I had a hunch earlier that we could indeed derive more of those “generalized mean value theorems” and somehow relate those to the Taylor series approximations. Your comment, it seems, has confirmed that hunch. I will be looking forward to your future post(s) on Taylor series. Comment by \| January 22, 2008 \| Reply 4. This theorem is also known as the Lagrange mean value theorem. Comment by \| January 23, 2008 \| Reply 5. It also has an interesting corollary that says that the derivative of any function that is differentiable on an interval has the mean value property, i.e. it hits all its intermediate values between any 2 of the values that it assumes, even if that derivative is not continuous. In partcular, a simple step function H(x) that is 0 for x1 is not a derivative of any function F differentiable on (-1,1), so H(x) has no primitive on (-1,1). Comment by \| January 23, 2008 \| Reply 6. In comment #5 H(x)=0 for x0, also known as the Heaviside function. Comment by \| January 23, 2008 \| Reply 7. There must be a bug in the system, that messed up my posts, again (now using LaTex): $H(x)=0$ for $x \le 0$ and $0$ for $x>0$. Comment by \| January 23, 2008 \| Reply 8. There must be a bug in the system, that messed up my posts, again (now using LaTex): $H(x)=0$ for $x \le 0$ and $=0$ for $x>0$. Comment by \| January 23, 2008 \| Reply 9. Michael, you’re having trouble with the < symbol. That’s not a bug in the system. The bug in the system is that I can’t seem to log in to WordPress at all. Whether this has to do with a comment flood or not, I have no idea. Just in case, could you try in the future to think carefully about what you want to say and then say it all in one comment? Comment by \| January 23, 2008 \| Reply 10. Sorry, I got troubles with > that seems to have some special meaning on this site and also with LaTex preprocessor here that messed up my formula. I wish we could edit our entries. I doubt that your trouble logging in is related to the number of comments on your page. Comment by \| January 23, 2008 \| Reply 11. It’s not this site. > has a special meaning in HTML, which is sort of the basis of the web. WordPress’ processor can’t tell the difference between using it to mean “greater than” and “close a tag”. But still, you made three separate comments before the litany of trying to fix the problem with the > sign. Slow down, take a deep breath, and decide what you want to say before saying it. This isn’t Twitter. Comment by \| January 23, 2008 \| Reply 12. Actually I wanted to say that H(x) is the Heaviside function that jumps from 0 to 1 at x=0. Comment by \| January 23, 2008 \| Reply 13. [...] of the Mean Value Theorem So now that we have the mean value theorem what can we do with it? First off, we can tell something that seems intuitively obvious. We know [...] Pingback by \| January 24, 2008 \| Reply 14. [...] to get back on track. Today, we’ll see a theorem about integrals that’s similar to the Differential Mean Value Theorem. Specifically, it states that if we have a continuous function then there is some so [...] Pingback by \| February 12, 2008 \| Reply 15. [...] the Differential Mean Value Theorem tells us that in each subinterval there is some so that . We stick this into the [...] Pingback by \| March 6, 2008 \| Reply 16. [...] an extension of the Fundamental Theorem of Calculus we’ll see it’s an extension of the Differential Mean Value Theorem. And this despite the most strenuous objections of my resident gadfly that the DMVT and FToC have [...] Pingback by \| October 1, 2008 \| Reply 17. [...] a nice technical result we may have call for from time to time: a higher-dimensional version of the differential mean value theorem. Remember that this says that if we’ve got a function continuous on the closed interval and [...] Pingback by \| October 13, 2009 \| Reply 18. [...] we can apply the mean value theorem to [...] Pingback by \| October 15, 2009 \| Reply 19. [...] Taylor’s theorem. Specifically, the version of Taylor’s theorem that resembles the mean value theorem. And we’ll even use an approach like we did for the extension of that result to higher [...] Pingback by \| October 20, 2009 \| Reply 20. [...] is some number between and that exists by the differential mean value theorem. Now to find the derivative, we take the limit of the difference [...] Pingback by \| January 13, 2010 \| Reply 21. [...] let denote the sum of the terms and , the remainder after terms, of the series (2), so that . By Lagrange’s mean value theorem, we have , . We shall now consider . So far we have taken as an arbitrary but we shall now choose [...] Pingback by \| July 7, 2011 \| Reply « Previous \| Next » ## About this weblog This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”). I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 51, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9431554079055786, "perplexity_flag": "head"}
http://mathoverflow.net/questions/19638/infinite-collection-of-elements-of-a-number-field-with-very-similar-annihilating/19774	## Infinite collection of elements of a number field with very similar annihilating polynomials ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Hello all, let $n$ be an integer $\geq 2$ and let $\alpha$ be an algebraic number of degree $n$. Let $R$ be the ring of algebraic integers in ${\mathbb Q}(\alpha)$, and let $B$ be the subset of $R$ containing the elements whose degree is exactly $n$. Any $\beta \in B$ has a minimal polynomial $X^n+b_{n-1}X^{n-1}+ \ldots + b_1X+b_0$. Identifying this latter polynomial with the uple $(b_0,b_1, \ldots ,b_{n-1})$ allows us to view $B$ as a subset of ${\mathbb Z}^n$. I define a combinatorial subvariety $V$ of dimension at most $r$ of ${\mathbb Z}^n$ to be a subset of $Z^n$ such that there is a set of indices $I \subseteq \lbrace 1,2, \ldots , n \rbrace$ with $\|I\|=n-r$ and the projection $p:V \to {\mathbb Z}^{n-r}, (v_1,v_2, \ldots ,v_n) \mapsto (v_i)_{i\in I}$ is constant. My question is : what is the smallest $r$ such that there is an infinite subset $B' \subset B$ corresponding to a subvariety of dimension at most $r$ ? In other words, we are asking for infinitely many elements in $B$, whose minimal polynomials are as similar as possible". An easy case is when $\alpha=a^{\frac{1}{n}}$ for some $a \in {\mathbb Q}$, because the rational multiples of $\mathbb \alpha$ correspond to a subvariety of dimension 1, so that $r=1$ in this case. - Should "annulating" (in the title) be "annihilating"? – Michael Lugo Mar 28 2010 at 17:21 I suspect there is a uniform bound on such r: given an r, there is the hypersurface $D(b_0,...,b_{n-1})/D_0 = y^2$, where $D$ is the discriminant of the monic polynomial, and $D_0$ will be the discriminant of a fixed monic polynomial. Not all solutions to this equation will be in the same field, but probably infinitely many. I believe there are conjectures on the number of integer solutions to hypersurfaces in $r$ variables, can anyone who knows shed more light? – Dror Speiser Mar 28 2010 at 17:28 Thanks Michael, I corrected the title. – Ewan Delanoy Mar 28 2010 at 19:52 "...allows us to view B as a subset of Z^n". Not quite, because distinct elements of B might have the same min poly. – Kevin Buzzard Mar 28 2010 at 19:57 @ Kevin : this does not matter because at most $n$ many elements share the same min poly. So an infinite subset of $B$ will always yield an infinite subset of $Z^n$. – Ewan Delanoy Mar 29 2010 at 3:52 show 1 more comment ## 2 Answers For $n>4$, almost all fields of degree $n$ will have $r>1$: Fix a field $K$ with discriminant $D_0$. Fix the $n-1$ coefficients $b_{n-1},...,b_{i+1}, b_{i-1},..., b_0$. The discriminant of the polynomial $x^n+b_{n-1}x^{n-1}+...$ is a polynomial $D(b_i)$ in the single variable $b_i$, and is of degree at least $4$. If this polynomial is squarefree, as it will be for almost all $n-1$ fixed coefficients, then the hypersurface $D_0y^2 = D(b_i)$ has genus at least $1$, and hence finitely many integer points. But, every polynomial defining the same field must have the same discriminant up to a square factor, and hence $r > 1$. Going back on my comment above: since the degree of the discriminant (multivariate) polynomial is large (linear in the number of variables) the equation $D(b_0,...,b_{n-1}) = D_0y^2$ will probably have only a finite number of solutions for most $D_0$, if $r$ is much smaller than $n$. Therefore, my new pessimistic conjecture is that for almost all fields you will have $r \gg n$. Note: $r \le n-1$ - in any number field there are always an infinite number of algebraic integers with trace 0. - @Dror: the heart of this answer is surely right but I'm not convinced that D(b_i) is always a polynomial of degree greater than 4 [try quintics with i=0 for example], and even if it is then the hypersurface might be singular and have lots of rational points, like y^2=x^101 or something. But in some sense this is a moraly correct approach, and it will surely suffice to find one explicit number field with r>1. – Kevin Buzzard Mar 29 2010 at 21:56 Sorry sorry, I meant genus at least 1, Siegel's theorem covering 1. The "almost all" takes out the singular hypersurfaces. I believe some very easy sieve theory can prove this since, mod p, most assignments of the other coefficients should give a squarefree polynomial. – Dror Speiser Mar 29 2010 at 22:10 I follow the argument and believe you. – Kevin Buzzard Mar 29 2010 at 22:16 @ Dror : very nice. I think that we can always take r=1 when n=3, because Kevin's computation on $x^3-x+1$ generalizes. – Ewan Delanoy Mar 30 2010 at 5:17 ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. The answer "is" that the smallest $r$ is what it is, and what it is could well depend on $\alpha$. Let me also raise the possibility that there might be no simple "formula" relating $r$ to $\alpha$. This in some sense is the "problem" with questions like this ("given some data, compute some number $r$: what 'is' $r$?")---they are not really questions (in my mind, at least). Who knows though, perhaps someone can find some extra structure. For example can one always take $r=1$? That's a proper question ;-) I'd be surprised though! But on a more positive note let me say that in my (rather long) answer to http://mathoverflow.net/questions/12486/integers-not-represented-by-2-x2-x-y-3-y2-z3-z I show in passing that if $\alpha$ is a root of $z^3-z+1$ then there are infinitely many integers $C$ such that $z^3-z+C$ is irreducible and has a root in $\mathbf{Q}(\alpha)$, giving a perhaps slightly less trivial example. The integers $C$ are the odd solutions to $27C^2-4=23D^2$ and there are infinitely many of these (the smallest two being 1 and 599). - 1 I completely disagree with your comment that the question is not meaningful. For any given $\alpha$, the (smallest) $r$ is a defnite value. It may be that $r$ is very hard to compute in terms of $\alpha$ in general, but you cannot say that the question is meaningless. Thanks for your example with $z^3-z-1$. I guess there is a parametrization of the solutions of $27C^2-4=23D^2$ by some linear-recurrence sequence. – Ewan Delanoy Mar 29 2010 at 3:50 Well OK :-) Yes, I agree that the smallest $r$ is a definite value. My point is that if the question is a question, then an answer to it could be "the value of $r$ is whatever it comes out to be". The parametrisation of the solutions to 27C^2-4=23D^2 can be obtained by the theory of Pell's equation: the solutions grow exponentially but there are infinitely many of them. – Kevin Buzzard Mar 29 2010 at 6:46 @Ewan: aah yes, I understand what you're saying: yes, the solutions to the equation are generated by a degree 2 linear recurrence relation, and 2/3 of them are odd. – Kevin Buzzard Mar 29 2010 at 6:49	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 80, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9317443370819092, "perplexity_flag": "head"}
http://gilkalai.wordpress.com/2009/03/15/colorful-caratheodory-revisited/?like=1&source=post_flair&_wpnonce=e5382928c8	Gil Kalai’s blog ## Colorful Caratheodory Revisited Posted on March 15, 2009 by Janos Pach wrote me: ”I saw that you several times returned to the colored Caratheodory and Helly theorems and related stuff, so I thought that you may be interested in the enclosed paper by Holmsen, Tverberg and me, in which – to our greatest surprise – we found that the right condition in the colored Caratheodory theorem is not that every color class contains the origin in its convex hull, but that the union of every pair of color classes contains the origin in its convex hull. This already guarantees that one can pick a point of each color so that the simplex induced by them contains the origin. A similar version of the colored Helly theorem holds. Did you know this?” I did not know it. This is very surprising! The paper of Holmsen, Pach and Tverberg mentions that this extension was discovered independently by J. L. Arocha, I. B´ar´any, J. Bracho, R. Fabila and L. Montejano. Let me just mention the colorful Caratheodory agai. (we discussed it among various Helly-type theorems in the post on Tverberg’s theorem.) The Colorful Caratheodory Theorem: Let $S_1, S_2, \dots S_{d+1}$ be $d+1$ sets in $R^d$. Suppose that $x \in conv(S_1) \cap conv (S_2) \cap conv (S_3) \cap \dots \cap conv (S_{d+1})$. Then there are $x_1 \in S_1$, $x_2 \in S_2$, $\dots x_{d+1} \in S_{d+1}$ such that $x \in conv (x_1,x_2,\dots,x_{d+1})$. And the strong theorem is: The Strong Colorful Caratheodory Theorem: Let $S_1, S_2, \dots S_{d+1}$ be $d+1$ sets in $R^d$. Suppose that $x \in conv(S_i \cup S_j)$ for every $1 \le i <j \le d+1$. Then there are $x_1 \in S_1$, $x_2 \in S_2$, $\dots x_{d+1} \in S_{d+1}$ such that $x \in conv (x_1,x_2,\dots,x_{d+1})$. Janos, whom I first met thirty years ago, and who gave the second-most surprising introduction to a talk I gave, started his email with the following questions: “Time to time I visit your lively blog on the web, although I am still not quite sure what a blog is… What is wordpress? Do you need to open an account with them in order to post things? Is there a special software they provide online which makes it easy to include pictures etc? How much time does it take to maintain such a site?” These are excellent questions that may interest others and I promised Janos that I will reply on the blog. So I plan comments on these questions in some later post. Meanwhile any comments from the floor are welcome. ### Like this: This entry was posted in Convexity and tagged Helly type theorems. Bookmark the permalink. ### 2 Responses to Colorful Caratheodory Revisited 1. Edna says: What is the first most surprising introduction? I hope it’s not the answer I am afraid you will give… 2. Gil says: Dear Edna, there is nothing to be afraid of. Indeed Avi’s introduction was the most surprising ever, and quite hilarious. As Avi promised to put it in writing for the back cover of my book, it will reach the large audience it deserves. Even now people who listened to that introduction ask me about the imortal translation to English to “where is Pluto” and some subtelties of the “cut its head/cut its tail” game. • ### Blogroll %d bloggers like this:	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 17, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9397937059402466, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/96372/rate-of-change-of-mass-of-a-parameterized-region	## Rate of change of mass of a parameterized region ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Let $R_t$ be a family of compact, simply connected regions in the plane defined by `$R_t = \{x\in\mathbb{R}^2 : h(x) \leq t\}$` for all $t$, where $h(x)$ is some nicely behaved smooth function. Suppose $f(x)$ is a probability density on $\mathbb{R}^2$ and define $M(t) = \iint_{R_t} f(x) dA$ for all $t$. Is it true that $\frac{dM}{dt}\|_{t=t_0} = \int_{\partial R_{t_0}} f(x) /\\|\nabla h\\| ds$ where $ds$ denotes integration with respect to arc length? If not, what is the right expression for $\frac{dM}{dt}$? I assume this is some well-known first-year calculus-type problem but I can't find it stated in any context (it may very well be a common homework problem though I've not seen it). - 2 This is not a good question for MO. See en.wikipedia.org/wiki/Coarea_formula – Anton Petrunin May 9 2012 at 0:51 This is a special case of the coarea formula. – Liviu Nicolaescu May 9 2012 at 17:58 Thanks to both of you, I wasn't familiar with that result! That clears this question up for me nicely. – Jennifer Gao May 11 2012 at 0:26 ## 1 Answer With $H$ the Heaviside function (characteristic function of $\mathbb R_+$), you have $$M(t)=\int_{\mathbb R^n} f(x) H(t-h(x)) dx$$ and thus, at least formally, $$\dot M(t)=\int_{\mathbb R^n} f(x) \delta_0(t-h(x)) dx= \int_{\mathbb R^n} f(x)\Vert\nabla h(x)\Vert ^{-1} \underbrace{\delta_0(t-h(x))\Vert\nabla h(x)\Vert dx}_{d\sigma} ,$$ where $d\sigma$ is the Euclidean surface measure on {$x, h(x) =t$}. Here, it is important to assume that $h$ is say $C^1$ such that $dh\not=0$ at $h=t$ for $t$ in neighborhood of some distinguished value $t_0$. As a result, $$\dot M(t)=\int_{\partial R(t)}f(x)\Vert\nabla h(x)\Vert ^{-1} d\sigma.$$ The previous computation can be justified by Green's formula: for $X$ a $C^1_c$ vector field on $\Omega$ ( an open subset of $\mathbb R^n$ with a $C^1$ boundary) $$\int_\Omega \text{div} X\ dx=\int_{\partial \Omega} X\cdot \nu\ d\sigma,$$ where $\nu$ is the exterior unit normal to $\partial \Omega$, and $d\sigma$ is the Euclidean surface measure on $\partial \Omega$. That formula can be proven by showing $$d\sigma=\lim_{\epsilon\rightarrow 0} \phi(\frac{\rho(x)}{\epsilon})\epsilon^{-1}\Vert\nabla \rho(x)\Vert\quad\text{(distribution sense),}$$ for $\phi\in C^\infty_c(\mathbb R),\int \phi(t) dt=1$, $\Omega=${$x, \rho(x)<0$}, $d\rho\not=0$ at $\rho=0$. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 34, "mathjax_display_tex": 5, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9266827702522278, "perplexity_flag": "head"}
http://mathhelpforum.com/math-topics/17246-need-help-please-physics.html	# Thread: 1. ## need help please (physics) In college homecoming competition, 16 students lift sports car while holding the car off the ground, each student exerts an upward force of 400 N. (a) what is the mass of the car in kilograms? (b) what is its weight in pounds? 2. Originally Posted by twistedmexican In college homecoming competition, 16 students lift sports car while holding the car off the ground, each student exerts an upward force of 400 N. (a) what is the mass of the car in kilograms? (b) what is its weight in pounds? a) Each of the 16 students is exerting an upward force on the car and (I presume) the car is stationary and off the ground. So the students are supplying the only upward force, which is counteracting the weight of the car. Thus $w = 16F = 16 \cdot (400~N) = 6400~N$ (where w is the weight of the car.) So the mass of the car is $m = \frac{w}{g} = \frac{6400~N}{9.8~m/s^2} = 653.061~kg$ b) 1 lb = 4.45 N, so $\frac{6400~N}{1} \cdot \frac{1~lb}{4.45~N} = 1439.75~lb$ -Dan	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 3, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9389955997467041, "perplexity_flag": "middle"}
http://unapologetic.wordpress.com/2008/07/06/the-carter-gelsinger-eversion/?like=1&source=post_flair&_wpnonce=eb8d9d939c	The Unapologetic Mathematician The Carter-Gelsinger Eversion I’ve mentioned Outside In before. That video shows a way of turning a sphere inside out. It’s simpler than the first explicit eversions to be discovered, but the simplicity is connected to a high degree of symmetry. This leads to very congested parts of the movie, where it’s very difficult to see what’s going on. Further, many quadruple points — where four sections of the surface pass through the same point — occur simultaneously, and even higher degree points occur. We need a simpler version. What would constitute “simple” for us, then? We want as few multiple points as possible, and as few at a time as possible. In fact, it would be really nice if we could write it down algebraically, in some sense? But what sense? Go back to the diagrammatics of braided monoidal categories with duals. There we could draw knots and links to represent morphisms from the monoidal identity object to itself. And topologically deformed versions of the same knot encoded the same morphism. This is the basic idea of the category $\mathcal{T}ang$ of tangles. But if we shift our perspective a bit, we consider the 2-category of tangles. Instead of saying that deformations are “the same” tangle, we consider explicit 2-isomorphisms between tangles. We’ve got basic 2-isomorphisms for each of the Reidemeister moves, and a couple to create or cancel caps and cups in pairs (duality) and to pull crossings past caps or cups (naturality). Just like we can write out any link diagram in terms of a small finite collection of basic tangles, we can write out any link diagram isotopy in terms of a small finite collection of basic moves. What does a link diagram isotopy describe? Links (in our picture) are described by collections of points moving around in the plane. As we stack up pictures of these planes the points trace out a link. So now we’ve got links moving around in space. As we stack up pictures of these spaces, the links trace out linked surfaces in four-dimensional space. And we can describe any such surface in terms of a small collection of basic 2-morphisms in the braided monoidal 2-category of 2-tangles. These are analogous to the basic cups, caps, and crossings for tangles. Of course the natural next step is to consider how to deform 2-tangles into each other. And we again have a small collection of basic 3-morphisms that can be used to describe any morphisms of 2-tangles. These are analogous to the Reidemeister moves. Any deformation of a surface (which is written in terms of the basic 2-morphisms) can be written out in terms of these basic 3-morphisms. We can simplify our picture a bit. Instead of knotting surfaces in four-dimensional space, let’s just let them intersect each other in three-dimensional space. To do this, we need to use a symmetric monoidal 3-category with duals, since there’s no distinction between two types of crossings. And now we come back to eversions. We write the sphere as a 2-dimensional cup followed by a 2-dimensional cap. Since we have duals, we can consider one side to be “painted red” and one side “painted blue”. One way of writing the sphere has the outside painted red and the other side is painted blue. An eversion in our language will be an explicit list of 3-morphisms that run from one of these spheres to the other. Scott Carter and Sarah Gelsinger have now created just such an explicit list of directions to evert a sphere. And, what’s more, they’ve rendered pictures of it! Here, for the first time in public, is a 50MB PDF file showing the Carter-Gelsinger eversion. First they illustrate the basic pieces of a diagram of knotted surfaces (pp. 1-4). Then they illustrate the basic 2-morphisms that build up surfaces (pp. 5-6), and write out a torus as an example (p. 7). Then come a few more basic 2-morphisms that involve self-intersections (pp. 8-9) and a more complicated immersed sphere (pp. 10-11). Each of these is written out also as a “movie” of self-intersecting loops in the plane. Next come the “movie moves” — the 3-morphisms connecting the 2-morphism “movies” (pp. 12-17). These are the basic pieces that let us move from one immersed surface to another. Finally, the eversion itself, consisting of the next 79 pages. Each one consists of an immersed sphere, rendered in a number of different ways. On the left is a movie of immersed plane curves. On the top are three views of the sphere as a whole — a “solid” view on the right, a sketch of the double-point curves in the middle, and a “see-through” view on the left. The largest picture on each page is a more schematic view I don’t want to say too much about. The important thing to see here is that between each two frames of this movie is exactly one movie move. Everything here is rendered into pictures, but we could write out the movie on each page as a sequence of 2-morphisms form the top of the page to the bottom. Then moving from one page to the next we trace out a sequence of 3-morphisms, writing out the eversion explicitly in terms of the basic 3-morphisms. As an added bonus, there’s only ever one quadruple point — where we pass from Red 26 to Blue 53 — and no higher degree points. I’d like to thank Scott for not only finishing off this rendering he’s been promising for ages, but for allowing me to host its premiere weblog appearance. I, for one, am looking forward to the book, although I’m not sure this one will be better than the movie. [UPDATE] Some people have been having trouble with the whole 50MB PDF (and more people might as the Carnival comes to see this page. Scott Carter broke the file up into five pieces, and I’ve put them up here in a new post. There’s a glitch in part 4, but I’ll have that one up as soon as I can. Like this: Posted by John Armstrong \| Category theory, Knot theory, Topology 7 Comments » 1. Just like the enemy’s gate, the link is down. Comment by \| July 6, 2008 \| Reply 2. bah! Comment by \| July 6, 2008 \| Reply 3. better now Comment by \| July 6, 2008 \| Reply 4. John, I’m really excited about the overview you wrote describing the logic leading to the tools used for the Carter-Gelsinger eversion. I’m not a mathematician, so I truly appreciate your top-down “systems engineering” (hope you don’t find that offensive!)overview showing how how to extend known concepts into the problem-solution domain. To me, this is an amazing example of how an understanding of n-categories helps to derive the solution of a specific problem. It is even more understandable because the basic tools being extended don’t require an advanced mathematics degree to understand (at least the elementary versions). You have sketched out all the steps required to have a preliminary understanding of what is really going on here, a road map of how to proceed. Outstanding! Now I can more fully appreciate the accomplishment of Carter and Gelsinger (two smart people whose work I never thought I’d be able to understand). There’s no way I could have appreciated the basis for their achievement without your contribution. Thank you for this excellent overview! Charlie C. Comment by Charlie C \| July 6, 2008 \| Reply 5. [...] on the C-G Eversion Some people had trouble grabbing the whole 50MB file that I posted, so Scott Carter broke it into pieces. He also included these comments: The red, blue, and purple [...] Pingback by \| July 10, 2008 \| Reply 6. [...] this, me droogs Scott Carter recently took a break from turning spheres inside out and made a bunch of videos for YouTube. Since I’m stuck all day tomorrow learning which [...] Pingback by \| August 14, 2008 \| Reply 7. this is pretty cool Comment by prashant sharma \| November 1, 2008 \| Reply « Previous \| Next » About this weblog This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”). I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 1, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9309076070785522, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/110505/is-there-a-simple-test-to-determine-whether-a-polytope-is-integral	## Is there a simple test to determine whether a polytope is integral? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) It is known that any rational convex polytope expressed as $\{ x\in\mathbb{R}^d : Ax \ge b \}$, where $A\in\mathbb{Z}^{k\times d}$ and $b\in\mathbb{Z}^k$, can be written as the convex hull of finitely many points. My question is, given the above representation in terms of hyperplanes, one can determine (easily) whether the polytope is integral ---that is, whether the polytope can be written as the convex hull of points in the integer lattice. - Are you looking for a theoretical solution for a class of examples, or a computational solution for specific instances? – gordon-royle Oct 24 at 5:28 I am currently interested in figuring out whether, under some reasonable conditions, certain polytopes I am studying are integral. The class of polytopes I am looking at must satisfy some specific symmetry conditions. If for certain parameters I could argue that the polytope must be totally unimodular (as someone pointed out below), then that would be nice ... – Steven Collazos Oct 27 at 18:46 ## 2 Answers An incomplete solution: There is a polynomial-time test for total unimodularity, and if $A$ is totally unimodular, then the polytope is integral. - What if $A$ is not unimodular? E.g., $A=\begin{pmatrix}1 & 0\\ 0 & 1\\ 2 & 1 \end{pmatrix}$ and $b=\begin{pmatrix}0 \\ 0 \\ 2\end{pmatrix}$? – J.C. Ottem Oct 24 at 4:52 Obviously he doesn't (yet) have an answer for that case. $\:$ – Ricky Demer Oct 24 at 4:53 (the example should have been $A=\begin{pmatrix}1 & 0\\ 0 &1\\ 2 & -1\end{pmatrix}$ and $b=\begin{pmatrix}0 \\ 0 \\ -2\end{pmatrix}$.) – J.C. Ottem Oct 24 at 4:56 Right, if $A$ is not totally unimodular then this test is inconclusive. I'm not aware of an efficient test that works for all $A,b$, though I'd certainly be interested to see one! – Michael Biro Oct 24 at 5:29 ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. This paper shows that testing integrality is coNP-complete. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 12, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9280717372894287, "perplexity_flag": "head"}
http://mathoverflow.net/questions/16888/when-do-divisors-pull-back/59694	## When do divisors pull back? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) If we have a nonconstant map of nonsingular curves $\varphi:X\rightarrow Y$, then Hartshorne defines a map $\varphi^* Div(Y)\rightarrow Div(X)$ using the fact that codimension one irreducibles are just points, and looking at `$\mathcal{O}_{Y,f(p)} \rightarrow \mathcal{O}_{X,p}$`. My question is if we don't have a nice map of curves, what conditions can we put on the morphism so that we may pull divisors back? Clearly it's not true in general, since we can take a constant map and then topologically the inverse image doesn't even have the right codim. Thinking about this in terms of Cartier divisors (and assuming the schemes are integral), it seems like we just need a way to transport functions in $K(Y)$ to functions in $K(X)$. If $\varphi$ is dominant, then we'll get such a map. Is this sufficient? Also is there something we can say when $\varphi$ is not dominant? Something like we have a way to map divisors with support on $\overline{\varphi(X)}$ to divisors on $X$? - ## 5 Answers If you want to pull back a Cartier divisor $D$, you can do that provided the image of $f$ is not contained in the support of $D$: just pull back the local equations for $D$. If this does not happen, on an integral scheme, you can just pass to the associated line bundle $\mathcal{O}_X(D)$ and pull back that, obtaining `$f^{} \mathcal{O}_X(D)$`; of course you lose some information because a line bundle determines a Cartier divisor only up to linear equivalence. Fulton invented a nice way to avoid this distinction. Define a pseudodivisor on $X$ to be a triple $(Z, L, s)$ where $Z$ is a closed subset of $X$, $L$ a line bundle and $s$ a nowhere vanishing section on $X \setminus Z$, hence a trivialization on that open set. Then you can simply define the pullback of this triple as `$(f^{-1}(Z), f^{} L, f^{} s)$`, so you can always pull back pseudo divisors, whatever $f$ is. The relation with Cartier divisors is the following: to a Cartier divisor $D$ you can associate a pseudodivisor $(\|D\|, \mathcal{O}_X(D), s)$, where $s$ is the section of $\mathcal{O}_X(D)$ which gives a local equation for $D$. This correspondence is not bijective. First, a pseudodivisor $(Z, L, s)$ determines a Cartier divisor if $Z \subsetneq X$; note that in this case enlarging $Z$ will not change the associated Cartier divisor, so to obtain a bijective correspondence with Cartier divisors you have to factor out pseudodivisors by an equivalence relation, which I leave to you to formulate. But if $Z = X$, you only obtain a line bundle on $X$, and you have no way to get back a Cartier divisor. - ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. If you have a morphism $f:X\rightarrow Y$ of schemes and $D$ is a Cartier divisor on $Y$, then you can pull back the invertible sheaf $\mathcal{O}(D)$ to $X$ and it will be an invertible subsheaf of the sheaf of total quotients $\mathcal{K}_X$. This sheaf defines a linear equivalence class of Cartier divisors but there's no clear choice of a specific divisor. If however you furthermore assume that $X$ is reduced, and $D$ is given by sections $f_i$ on an open covering $U_i$ of $Y$ whose support doesn't contain the image of any of the irreducible components of $X$, then you can define a Cartier divisor $f^D$ on $X$ as $(f^{-1}(U_i), f_i\circ f)$ and this lies in the class of $f^\mathcal{O}(D)$. This kind of "support-avoidance" is quite common with these arguments. - Did not ask X to be integral？ Why should the pull back be a subsheaf of the sheaf of total quotients？ – MZWang May 7 2012 at 1:46 It seems to me that you are approaching this in the wrong way. First of all you have to decide whether you want to pull-back divisors or divisor classes Pulling back divisors means pulling back cycles and you run into all sorts of problems. If your map is flat, then this is OK. See Fulton's Intersection theory book for more on this. (There is a section called "flat pull-back of cycles" somewhere early). This is actually why Hartshorne did not have to worry about pulling back divisors or divisor classes. Any non-constant morphism between non-singular curves is flat. Of course, I am not saying this is the most general condition under which you can pull-back divisors, just that this is the "natural" one. Then again, this is a little too much, because it works for any dimensional cycles, not just divisors. Pulling back divisor classes is somewhat easier (and harder at the same time). It is easier, because then you don't have to worry about the support. As already pointed out by Frank, you just "convert" your divisor class to a line bundle, pull that back and "convert" it back to a Cartier divisor. Actually for this you need $X$ to be integral, but you seem to be OK with assuming that. Anyway, another, perhaps more interesting question is how to pull-back Weil divisors. The main problem there is that they are not defined locally by a single equation, so their associated sheaf is not an line bundle, only a reflexive sheaf of rank $1$. Unfortunately pulling back those does not necessarily give you a reflexive sheaf. One solution is to still do that and then take the reflexive hull, but this will likely not give a group homomorphism. Another partial way to handle Weil divisors is to restrict to $\mathbb Q$-Cartier divisors, i.e., a divisor that itself may not be Cartier, but a multiple of which is Cartier. Then you just take that multiple, pull that back and then divide by the same number you multiplied with. Interestingly this can result in fractional coefficients, so it will only preserve linear equivalence mod $\mathbb Q$. The simplest example of a $\mathbb Q$-Cartier, but not Cartier divisor is a ruling of a quadratic cone (look at intersection numbers to prove this). A little more interesting example is included in this MO answer. For an example of a non-$\mathbb Q$-Cartier divisor consider the cone over a quadric surface in $\mathbb P^3$ and let the divisor be the cone over one line on the surface. - Assume that both $X$ and $Y$ are locally noetherian and take a Cartier divisor $D$ on Y. I claim that $D$ pullbacks to a Cartier divisor on $X$ if and only if $f(Ass\ X) \cap \|D\|=\emptyset$. Clearly this doesn't make sense unless I define what is a pullback of $D$: a Cartier divisor $E$ on $X$ is a pullback of $D$, written $E=f^D$, if 1) $\|E\|\subseteq f^{-1}(\|D\|)$. 2) $U=X-f^{-1}(\|D\|)$ is schematically dense, i.e. $Ass\ X \subseteq U$, and there exists an isomorphism $\phi : f^\mathcal{O}_Y(D)\longrightarrow \mathcal{O}_X(E)$ such that $\phi(f^1)=1$ on $U$. Condition $1)$ relates the supports $\|D\|$ and $\|f^* D\|$ while condition $2)$ states that $f^\mathcal{O}_Y(D)\simeq \mathcal{O}_X(f^ D)$ and assures that $f^* D$ is uniquely determined. The idea behind the above definition is that if $\mathcal{L}$ is an invertible sheaf on a scheme $X$ then $\mathcal{L}\simeq \mathcal{O}_X(E)$ for a Cartier divisor $E$ if and only if there exists a schematically dense open subset $U\subseteq X$ and a nowhere zero section $s\in \mathcal{L}(U)$. Moreover given $(U,s)$ there exists a unique $E$ s.t. $\|E\|\cap U = \emptyset$ and there exists an isomorphism $\mathcal{L}\simeq \mathcal{O}_X(E)$ sending $s$ to $1$ on $U$. Note that in general if $C$ is a divisor $X-\|C\|$ is schematically dense and $1$ is a nowhere zero section of $\mathcal{O}_X(C)$ on it. In our case $f(Ass\ X) \cap \|D\|=\emptyset$ means that $U=X-f^{-1}(\|D\|)$ is schematically dense. Since $f^* 1$ is a nowhere zero section of $f^\mathcal{O}_Y(D)$ over $U$ we obtain a divisor $f^D$ satisfying conditions $1),2)$. For example if $f(Ass\ X) \subseteq Ass Y$ you can pullback any Cartier divisor on $Y$. This happens if $f$ is flat or a dominant map between integral schemes. Finally if $D$ is effective $f^D$ exists, i.e. $f(Ass\ X) \cap \|D\|=\emptyset$, if and only if $f^{-1}(D)$ is an effective divisor and in this case they coincide. Moreover the exact sequence defining $D$ on $Y$ pullbacks to the exact sequence defining $f^{-1}(D)$ on $X$, which is very useful. (For the effective case see [Stacks project, 22.3]). - You shouldn't need to worry about whether or not the support intersects the closure $\overline{\varphi(X)}$. If it isn't, then your divisor simply pulls back to the zero divisor on X. - The problem is that the support may contain $\varphi(X)$ (or the image of a component). In that case, the thing you want the pullback to be ends up having the wrong codimension. – Anton Geraschenko♦ Nov 12 2010 at 3:20 Ah, that's fair. For example, given the inclusion $i :D \hookrightarrow X$, then if we do this naively, then the "pullback" $iD = D$ which is certainly not a divisor. – Simon Rose Nov 12 2010 at 15:54	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 116, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9214378595352173, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/54413/what-conservation-law-corresponds-to-this-local-u1-symmetry-of-the-ccr?answertab=oldest	What conservation law corresponds to this local $U(1)$ symmetry of the CCR? It is known that canonical commutation relations do not fix the form of momentum operator. That means that if canonical commutation relations (CCR) are given by $$[\hat{x}^i,\hat{p}_j]~=~i\hbar~\delta^i_j~ {\bf 1}$$ they can be satisfied by the following choice of momentum operators: $p_x = -ih\frac{∂}{∂x}+\frac{∂f}{∂x}$ $p_y = -ih\frac{∂}{∂y}+\frac{∂f}{∂y}$ $p_z = -ih\frac{∂}{∂z}+\frac{∂f}{∂z}$ where $f(x,y,z)$ - arbitrary function. On the other hand, for any choice of $f(x,y,z)$ momentum operators can be transformed to their most frequently used form $(-ih\frac{∂}{∂x})$ (etc for $y$ and $z$) by the following transformation of the wave function $\psi$ and operators $p$: $\psi'=e^{-\frac{i}{h}f(x,y,z)}\psi$ $p^{'}_x =e^{-\frac{i}{h}f(x,y,z)}p_x e^{+\frac{i}{h}f(x,y,z)}=-ih\frac{∂}{∂x}$ Hence, we obtain $U(1)$ gauge transformation using only canonical commutation relations for momentum and position operators. Does this mean that $U(1)$ gauge invariance corresponds to conservation of momentum rather than to conservation of electric charge? - 1 Answer 1) If we interpret OP's transformation$^1$ as a passive transformation, i.e. a mere change of coordinates/description that doesn't alter the system, then there is no conservation law. 2) So in the following, let us interpret OP's transformation as an active transformation. For a system to have a symmetry, its action $S$ (or more precisely, in this quantum mechanical context, its Hamiltonian operator $\hat{H}$) must respect this symmetry. The corresponding conservation law would depend on the specific form of Hamiltonian $\hat{H}$. This is all we have to say about QM. 3) Finally, in a field theoretic context, we can interpret OP's transformation $$\tag{1} \frac{\hbar}{i} {\bf \nabla} ~\longrightarrow~ e^{-i\Lambda({\rm r})}\frac{\hbar}{i} {\bf\nabla} e^{i\Lambda({\rm r})}$$ as a pure gauge transformation in an electromagnetic theory with zero electromagnetic field strength. When interpreted as an EM theory, on one hand, global gauge symmetry leads to electric charge conservation, cf. Noether's first Theorem. See also this Phys.SE question. On the other hand, there is no conserved quantity associated with local gauge symmetry per se, cf. Noether's second Theorem. (Its off-shell Noether identity is a triviality. See also this Phys.SE question.) -- $^1$ OP's transformation is related to the choice of phase factors in overlaps between position and momentum eigenstates in QM, cf. this Phys.SE post. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 20, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8899617791175842, "perplexity_flag": "head"}
http://stats.stackexchange.com/tags/moments/info	# Tag info ## About moments For questions about the moments of a random variable, that is, the expectation of $X^k$ where $X$ is a random variable and $k$ an integer (or a real number with $X$ non-negative).	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 4, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8091872334480286, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/193872/prove-1-2-4-8-dots-1	# Prove $1 + 2 + 4 + 8 + \dots = -1$ [duplicate] Possible Duplicate: Infinity = -1 paradox I was told by a friend that $1 + 2 + 4 + 8 + \dots$ equaled negative one. When I asked for an explanation, he said: Do I have to? Okay so, Let $x = 1+2+4+8+\dots$ $2x-x=x$ $2(1+2+4+8+\dots) - (1+2+4+8+\dots) = (1+2+4+8+\dots)$ Therefore, $(2+4+8+16+\dots) + (-1-2-4-8+\dots) = (1+2+4+8+\dots)$. Now $-2$ and $2$, $-4$ and $4$, $-8$ and $8$ and so on, cancel out, and the only thing left is $-1$. Therefore, $1+2+4+8+\dots = -1$. I feel that this conclusion is not right, but I cannot express it. Can anyone tell if this proof is wrong, and if it is, how it is wrong? - 9 The algebraic operations are not valid if the sum doesn't exist (which it doesn't). Also, some infinite series (conditionally convergent) are very sensitive to rearrangement, so the rearrangements must be valid on the level of partial sums in order to be valid for the infinite sums. However, these very same algebraic manipulations can be used to give analytic continuations which give the regularized value of $-1$ for the divergent sum. (Alternatively you can switch out the Euclidean topology for the $2$-adic topology, in which case this is completely valid.) – anon Sep 11 '12 at 1:00 1 If your friend had told you that this sum was $-\frac{1}{12}$.. now that would've been much more interesting! – student Sep 11 '12 at 1:04 1 (Yes, there is another divergent series $1+2+3+\cdots$ whose zeta-regularized value is $-1/12$.) – anon Sep 11 '12 at 1:05 5 – MJD Sep 11 '12 at 1:48 3 +1 MJD. This series converges (to -1) in the 2-adic metric. – GEdgar Sep 11 '12 at 1:51 show 2 more comments ## marked as duplicate by Argon, Austin Mohr, William, Jyrki Lahtonen, mt_Sep 11 '12 at 10:15 This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question. ## 5 Answers The first mistake is right at the beginning, in writing "Let $x=1+2+4+\cdots$." This builds in the assumption that there is such an object as $1+2+4+\cdots$. The second mistake lies in treating this supposed object as if it were a finite but maybe very long sum, to which the sensible rules for manipulating finite sums apply. Remark: Think about the "sum" $1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots$. If we call this $x$, and use a manipulation analogous to the one that you made, we end up with the conclusion that the "sum" is $2$. When infinite series are formally defined, it turns out that the answer is indeed $2$. So some "natural" manipulations yield nonsense, and some yield correct results. Of course that is not a tolerable state of affairs: we cannot use manipulational techniques that sometimes yield a correct result, and sometimes don't. This sort of issue, at a more sophisticated level, led mathematicians in the second half of the $19$th century to look for very careful definitions of the fundamental objects of mathematics, and rigorous proofs of their basic properties. - And if the sum is finite, say length $n$, the proof is flawed anyway, as the multiplication of the LHS sum by two yields a term of the sum that doesn't exist in the RHS, namely $2^n$ - so the two sums do not fully cancel out and the net result is $2^n - 1$ which is the expected result. – Thomas Sep 11 '12 at 2:41 While true, I consider this answer incomplete, since there are generalized summations that allow these kinds of manipulations even for divergent series. And even when those do not work, one has further methods like zeta-regularization and p-adic interpretation. Or in other words, there are many definitions of summation (and arguably much more useful than the standard one in many situations). – Marek Sep 11 '12 at 9:51 @Marek: Yes, it is incomplete. My aim was to present a short fairly vigorously expressed answer that would be locally useful. – André Nicolas Sep 11 '12 at 14:39 The infinite series $\displaystyle 1 + 2 + 4 + 8 + \ldots$ diverges. However, the sum $f(z) = 1 + z + z^2 + z^3 + \ldots$, which converges to $1/(1-z)$ for $\|z\| < 1$, has an analytic continuation to the complex plane with the point $1$ removed, and indeed $f(2) = -1$. So in that sense you could regard $-1$ as the value of the divergent series. For more on this, see http://en.wikipedia.org/wiki/1_%2B_2_%2B_4_%2B_8_%2B_%C2%B7_%C2%B7_%C2%B7 and references there. - Look at Calculus textbook, undergraduate level. When we treat infinite sum, we cannot change the order to compute. Example. $1-1+1-\cdots$ $$(1-1)+(1-1)+\cdots=0+0+\cdots=0$$ $$1+(-1+1)+(-1+1)+\cdots=1.$$ - That is conditional convergence, not relevant in this case, for example $1-\frac{1}{2^2}+\frac{1}{3^2}-\frac{1}{4^2}+..$ you can change the order. – Arjang Sep 11 '12 at 5:47 @Arjang: True, but the naive mistake is the same - that you can do whatever to an infinite sum and everything is fine. – user1729 Sep 11 '12 at 9:15 @user1729 : everything is fine, what is not fine is the naive use of convergence where the more complete forms of convergence need to be considered, cesaro summibility etc.. And this example converges using extended version of convergence. – Arjang Sep 11 '12 at 11:24 1 @Arjang: But surely that is the point - that naive convergence doesn't work. – user1729 Sep 11 '12 at 12:36 Thanks user1729. – Tom Sep 19 '12 at 13:28 show 1 more comment It is similar to "proving" that 1 = 2 by saying that 1+infinity = 2+infinity. - What is wrong is that $1+2+4+8+\cdots$'' is not a number, and so you cannot treat it like one. - try saying that to Euler. – Arjang Sep 11 '12 at 5:48 2 @Arjang: I think Euler is dead for quite some time now. – Asaf Karagila Sep 11 '12 at 9:20 1 @AsafKaragila : His works make him more alive than many alive people. – Arjang Sep 11 '12 at 11:26 2 @Arjang: I think you and modern science have very different definition of "alive". – Asaf Karagila Sep 11 '12 at 12:04 1 @Arjang: the fact that he was genius doesn't preclude that he did make mistakes. – Martin Argerami Sep 11 '12 at 20:24	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 39, "mathjax_display_tex": 2, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9452962875366211, "perplexity_flag": "middle"}
http://nrich.maths.org/2370/solution	### Instant Insanity Given the nets of 4 cubes with the faces coloured in 4 colours, build a tower so that on each vertical wall no colour is repeated, that is all 4 colours appear. ### Magic Caterpillars Label the joints and legs of these graph theory caterpillars so that the vertex sums are all equal. ### Plum Tree Label this plum tree graph to make it totally magic! # Limiting Probabilities ##### Stage: 5 Challenge Level: Congratulations to Angus from Bexhill College; Richard from Bournemouth School and Andrei from Tudor Vianu National College, Bucharest, Romania for their clearly explained and excellent solutions to this problem. The numbers on the edges of this graph give the probabilities of a particle travelling along those edges in the direction given by the arrow. The same information is given by the entry $a_{ij}$ in the following matrix which gives the probability of travelling from vertex $i$ to vertex $j$. $$A=\left (\begin{array}{cccc} 0 &1 &0 &0\\ 0 &0 &0.5 &0.5\\ 1 &0 &0 &0\\ 0 &0 &0 &1 \end{array} \right)$$ This is Angus's solution. Firstly, we see that when we multiply matrices : $$\left ( \begin{array}{ccc} A &B &C \\ D &E &F \\ G &H &I \end{array} \right) \left ( \begin{array}{ccc} a &b &c \\ d &e &f \\ g &h &i \end{array} \right) = \left ( \begin{array}{ccc} Aa+Bd+Cg & Ab+Be+Ch &Ac+Bf+Ci\\ Da+Ed+Fg &Db+Ee+Fh &Dc+Ef+Fi \\ Ga+Hd+Ig &Gb+He+Ih &Gc+Hf+Ii \end{array} \right)$$ and similarly for larger matrices. Hence when the matrix gives probabilities: $$A= \left( \begin{array}{cccc} P(1\to 1) &P(1\to 2) &P(1\to 3) &P(1\to 4)\\ P(2\to 1) &P(2\to 2) &P(2\to 3) &P(2\to 4)\\ P(3\to 1) &P(3\to 2) &P(3\to 3) &P(3\to 4)\\ P(4\to 1) &P(4\to 2) &P(4\to 3) &P(4\to 4) \end{array} \right)$$ where $P(1\to 2)$ is the probability of travelling to $2$ when at $1$. Then the first column of $A^2$ is: $$\left ( \begin{array}{c} P(1\to 1)P (1\to 1) + P (1\to 2) P(2\to 1) + P(1\to 3)P(3\to 1) + P(1\to 4) P(4\to 1)\\ P(2\to 1)P (1\to 1) + P (2\to 2) P(2\to 1) + P(2\to 3)P(3\to 1) + P(2\to 4) P(4\to 1)\\ P(3\to 1)P (1\to 1) + P (3\to 2) P(2\to 1) + P(3\to 3)P(3\to 1) + P(3\to 4) P(4\to 1) \\ P(4\to 1)P (1\to 1) + P(4\to 2) P(2\to 1) + P(4\to 3)P(3\to 1) + P(4\to 4) P(4\to 1) \end{array} \right )$$ and so on for the four columns. Now, the probability of getting from $1$ to $2$ in $x$ steps equals the sum of the probabilities of all the ways of getting from 1 to 2 in x steps : $P(1\to 2 \text{ (2 steps)}) = P(1\to 1)P(1\to 2)+ P(1\to 2)P (2\to 2)+ P(1\to 3)P(3\to 2)+\ldots$. Thus we have in $A^2$ the probability of getting from one point to another in two steps, as each entry in the matrix is the sum of probabilities of getting between the two points in two steps. Similarly, we find that if we cube the matrix, we are adding the probabilities of three-stage routes between two points, and so on for higher powers. Now, we look at what happens for higher powers of $A$. $$A^{20}= \left( \begin{array}{cccc} 0 &0 &0.008 &0.992 \\ 0.008 &0 &0 &0.992 \\ 0 &0016 &0 &0.984 \\ 0 &0 &0 &1 \end{array} \right)$$ $$A^{21}= \left( \begin{array}{cccc} 0.008 &0 &0 &0.992 \\ 0 &0.008 &0 &0.992 \\ 0 &0 &0.008 &0.992 \\ 0 &0 &0 &1 \end{array} \right)$$ $$A^{22}= \left( \begin{array}{cccc} 0 &0.008 &0 &0.992 \\ 0 &0 &0.004 &0.996 \\ 0.008 &0 &0 &0.992 \\ 0 &0 &0 &1 \end{array} \right)$$ First (and easiest) we explain the fourth row. In the network, an object at vertex (4) has a probability of 1 of returning to (4) at the next journey. It therefore always returns to (4) and never reaches any other vertex. The fourth row will remain at $0$ $0$ $0$ $1$ as an object at (4) will remain at (4). Next we come to the cycling of the three values in the loop at the left. At vertices (3) and (1), an object may only travel to one other vertex ((1) and (2) respectively), and at vertex (2) it travels to (3) (with a probability of $0.5$ which we ignore to explain the cycling). Therefore, an object at vertex (1) can travel to (2), then (3) then (1) again and so on. However, it moves on one vertex at a time so, if it cycles through the vertices, after n journeys the object will be at vertex $n (\text{mod } 3)$. Similarly, if it starts at (2), it will be at vertex $(n + 1)(\text{mod } 3)$ and, from (3), at $(n + 2)(\text{mod } 3)$. Lastly, we explain the numerical values of the probabilities. Every time an object is at vertex (2), it travels to (4) with a probability p where p = 0.5 . Therefore, it will only avoid going to (4) with a probability of $(1 - p)^m$ , where $m$ is the number of times the object is at vertex (2) and $m$ is roughly $n/3$. Therefore, the probability of an object starting at vertices (1), (2) or (3) arriving at (4) is $1 - (1 - p)^m$ . As once the object is at (4) it cannot travel to any other vertex, the probability of being at (4) after n journeys can only increase. With this information, we can predict the behaviour of $A^n$ as we let $n$ increase. All values in the fourth column will tend to one as $n$ tends to infinity, as they are given (roughly) by $1 - 0.5^{n/3}$ and $0.5^{n/3}$ tends to 0. The other nine values will cycle as shown above, in the manner : $$\left( \begin{array}{ccc} 0 &a &0 \\ 0 &0 &b \\ c &0 &0 \end{array} \right) \to \left( \begin{array}{ccc} 0 &0 &b \\ c/2 &0 &0 \\ 0 &a &0 \end{array} \right) \to \left( \begin{array}{ccc} c/2 &0 &0 \\ 0 &a/2 &0 \\ 0 &0 &b \end{array} \right) \to \left( \begin{array}{ccc} 0 &a/2 &0 \\ 0 &0 &b/2 \\ c/2 &0 &0 \end{array} \right)$$ and so on for $n=3x+1,\ 3x+2,\ 3x+3,\ 3(x+1)+1$ respectively where $x$ is an integer. The values of $a,\ b$ and $c$ will tend to zero as they are given by $0.5^m$ for $n$ stages where $m$ is of the order of $n/3$. Hence $$A^{100}= \left ( \begin{array}{cccc} 1 &1\times 10^{-10} &0 &0.9999999999 \\ 0 &0 &5\times 10^{-11} &0.99999999995\\ 1\times 10^{-10} &0 &0 &0.9999999999 \\ 0 &0 &0 &1 \end{array} \right)$$ The NRICH Project aims to enrich the mathematical experiences of all learners. To support this aim, members of the NRICH team work in a wide range of capacities, including providing professional development for teachers wishing to embed rich mathematical tasks into everyday classroom practice. More information on many of our other activities can be found here.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 39, "mathjax_display_tex": 9, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9360402822494507, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/30330/fast-algorithm-for-solving-system-of-linear-equations/30802	# fast algorithm for solving system of linear equations I have a system of linear equations, $Ax=b$, with $N$ equations and $N$ unknowns ($N$ large) If I am interested in the solution for only one of the unknowns, what are the best approaches? for example, Assume N=50,000 and we want the solution for $x_1$ through $x_{100}$ only. is there any trick that does not require $O(n^{3})$ (or O(matrix inversion) ) for that ? - 1 Is $A$ full or sparse? Are you doing this once or many times with the same $A$? – joriki Apr 1 '11 at 16:56 A is not sparse and has many nonzero elements, however, the coefficients themselves are derived from a smaller set of variables. – mghandi Apr 2 '11 at 3:37 About your second question, yes I'm doing this many times. I am familiar with the LU decomposition trick to speed up when the coeffs matrix is unchanged. Is there any other tricks to do that even more efficiently? – mghandi Apr 2 '11 at 3:51 – Dirk Dec 15 '12 at 17:43 ## 2 Answers Unless your matrix is sparse or structured (e.g. Vandermonde, Hankel, or those other named matrix families that admit a fast solution method), there is not much hope of doing things better than $O(n^3)$ effort. Even if one were to restrict himself to solving for just one of the 50,000 variables, Cramer will demand computing two determinants for your answer, and the effort for computing a determinant is at least as much as decomposing/inverting a matrix to begin with. - [this should be a comment to @Juan Joder's answer or to the OP's question, but not enough repo] strang gilbert - linear algebra and its applications 3rd edition page 16 at the footer mentions that complexity had fallen already fallen below $O(Cn^{2.376})$ at the date of writing 1988, altough $C$ is so large that makes the algorithm impractical for most matrix sizes found in practice today. It does not mention the name of the algorithm though. personal guess: not sure about solving for single variables, but I don't think you can reduce complexity like that since the entire system is coupled. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 12, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9590341448783875, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/63187?sort=votes	## prime powers between n and 2n ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Is there a result in the spirit of Bertrand-Chebyshev which talks about the existence of prime powers between n and 2n (or 3n or something like that) for n large? - ## 3 Answers For fixed $k$, the existence of a prime power $p^k$ between $n$ and $2n$ is (asymptotically) equivalent to the existence of a prime $p$ between $m$ and $\sqrt[k]{2} m$ where $m = \sqrt[k]{n}$. Bachraoui gives an elementary proof here that there exists a prime between $m$ and $\frac{3}{2} m$ for sufficiently large $m$. I remember reading on MO that it is known that similar elementary proofs exist for showing the existence of primes between $m$ and $(1 + \epsilon) m$ for any $\epsilon > 0$ and, furthermore, that the existence of these proofs itself depends on the PNT. However, I can't track down where I read this. - 3 Possibly it is this question you have in mind: mathoverflow.net/questions/53498/… – quid Apr 27 2011 at 19:00 Yes, that's the one. – Qiaochu Yuan Apr 27 2011 at 20:11 ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. Actually there is a power of 2. It goes to show the power of binary arithmetic ... : write 2n in binary and write zeroes after the initial one. - 1 In other words, if $2^{k - 1}$ is the largest power of 2 less than n, then $n \leq 2^k < 2n$? – Ryan Reich Apr 27 2011 at 18:14 Yes, you said it. – Charles Matthews Apr 27 2011 at 18:43 It follows from the prime number theorem that for fixed $k$, provided $n$ is sufficiently large, there is a prime $p$ such that $p^k$ is between $n$ and $2n$. Does this answer your question? - Thank you very much for your quick answer. Yes, it answers the question, and it goes to show the power of the PNT. – Chebolu Apr 27 2011 at 16:01 1 I wonder if there is an elementary proof of this without using PNT, like Erdős' proof of Bertrand's Postulate? – Tony Huynh Apr 27 2011 at 17:24	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 22, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9496862292289734, "perplexity_flag": "head"}
http://mathhelpforum.com/pre-calculus/150765-equation-line-parallel.html	# Thread: 1. ## Equation of a line which is parallel I know it sounds funny but please help i got headache from this equation even thou i know its easy. Write down the equation of a line which is a) parallel to y=4x+2, and passes through 4 on the y- axis b) parallel to y=10x -5, passes through -1 on the y-axis Thanks For The Help In Advance 2. Write down the equation of a line which is a) parallel to y=4x+2, and passes through 4 on the y- axis b) parallel to y=10x -5, passes through -1 on the y-axis[/QUOTE] the equation of a line which is a) parallel to y=mx+b, and passes through c on the y-axis is y=mx+c. 3. Originally Posted by MrGenuouse a) parallel to y=4x+2, and passes through 4 on the y- axis Headaches are bad so i'll help you out with the first question and you can apply the same method to the second. We seek a line in the form of $y=mx+c$ . We need to find m (the gradient) and c (the y-intercept) . The gradient in the given equation is $m= 4$ and as the line we are after is parallel we can simply substitute this value into the new line giving $y=4x+c$ Now passing through 4 on the y-axis means the y-intercept is 4 so we can substitute that in for c and we are finished $y=4x+4$ How easy was that? 4. Thank You Very much for your help! My headache is over so I am really thankful about that too! i hope i will be able to gain more knowledge from this community as i am 16 years old boy who is really interested in maths.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 4, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9468458294868469, "perplexity_flag": "middle"}
http://ams.org/bookstore?fn=20&arg1=chelsealist&ikey=CHEL-131-H	New Titles \| FAQ \| Keep Informed \| Review Cart \| Contact Us Quick Search (Advanced Search ) Browse by Subject General Interest Logic & Foundations Number Theory Algebra & Algebraic Geometry Discrete Math & Combinatorics Analysis Differential Equations Geometry & Topology Probability & Statistics Applications Mathematical Physics Math Education Return to List The Theory of Matrices, Volume 1 SEARCH THIS BOOK: AMS Chelsea Publishing 1959; 374 pp; hardcover Volume: 131 Reprint/Revision History: second AMS printing 2000 ISBN-10: 0-8218-1376-5 ISBN-13: 978-0-8218-1376-8 List Price: US\$55 Member Price: US\$49.50 Order Code: CHEL/131.H This item is also sold as part of the following set: CHELGANTSET This treatise, by one of Russia's leading mathematicians, gives in easily accessible form a coherent account of matrix theory with a view to applications in mathematics, theoretical physics, statistics, electrical engineering, etc. The individual chapters have been kept as far as possible independent of each other, so that the reader acquainted with the contents of Chapter 1 can proceed immediately to the chapters of special interest. Much of the material has been available until now only in the periodical literature. Reviews "This is an excellent textbook." -- Zentralblatt MATH From a review of the original Russian edition ... "The first part (10 chapters; "General theory") gives in satisfactory detail, and with more than customary completeness, the topics which belong to the main body of the ... subjects ... The point of view is broad and includes much abstract treatment ... "The number of subjects which the book treats well is great ... would appeal to a wide audience." -- Mathematical Reviews From a review of the English translation ... "The work is an outstanding contribution to matrix theory and contains much material not to be found in any other text." -- Mathematical Reviews Volume 1 • I. Matrices and operations on matrices: 1. Matrices. Basic notation; 2. Addition and multiplication of rectangular matrices; 3. Square matrices; 4. Compound matrices. Minors of the inverse matrix • II. The algorithm of Gauss and some of its applications: 1. Gauss's elimination method; 2. Mechanical interpretation of Gauss's algorithm; 3. Sylvester's determinant identity; 4. The decomposition of a square matrix into triangular factors; 5. The partition of a matrix into blocks. The technique of operating with partitioned matrices. The generalized algorithm of Gauss • III. Linear operators in an $$n$$-dimensional vector space: 1. Vector spaces; 2. A linear operator mapping an $$n$$-dimensional space into an $$m$$-dimensional space; 3. Addition and multiplication of linear operators; 4. Transformation of coordinates; 5. Equivalent matrices. The rank of an operator. Sylvester's inequality; 6. Linear operators mapping an $$n$$-dimensional space into itself; 7. Characteristic values and characteristic vectors of a linear operator; 8. Linear operators of simple structure • IV. The characteristic polynomial and the minimal polynomial of a matrix: 1. Addition and multiplication of matrix polynomials; 2. Right and left division of matrix polynomials; 3. The generalized Bézout theorem; 4. The characteristic polynomial of a matrix. The adjoint matrix; 5. The method of Faddeev for the simultaneous computation of the coefficients of the characteristic polynomial and of the adjoint matrix; 6. The minimal polynomial of a matrix • V. Functions of matrices: 1. Definition of a function of a matrix; 2. The Lagrange-Sylvester interpolation polynomial; 3. Other forms of the definition of $$f(A)$$. The components of the matrix $$A$$; 4. Representation of functions of matrices by means of series; 5. Application of a function of a matrix to the integration of a system of linear differential equations with constant coefficients; 6. Stability of motion in the case of a linear system • VI. Equivalent transformations of polynomial matrices. Analytic theory of elementary divisors: 1. Elementary transformations of a polynomial matrix; 2. Canonical form of a $$\lambda$$-matrix; 3. Invariant polynomials and elementary divisors of a polynomial matrix; 4. Equivalence of linear binomials; 5. A criterion for similarity of matrices; 6. The normal forms of a matrix; 7. The elementary divisors of the matrix $$f(A)$$; 8. A general method of constructing the transforming matrix; 9. Another method of constructing a transforming matrix • VII. The structure of a linear operator in an $$n$$-dimensional space: 1. The minimal polynomial of a vector and a space (with respect to a given linear operator); 2. Decomposition into invariant subspaces with co-prime minimal polynomials; 3. Congruence. Factor space; 4. Decomposition of a space into cyclic invariant subspaces; 5. The normal form of a matrix; 6. Invariant polynomials. Elementary divisors; 7. The Jordan normal form of a matrix; 8. Krylov's method of transforming the secular equation • VIII. Matrix equations: 1. The equation $$AX=XB$$; 2. The special case $$A=B$$. Commuting matrices; 3. The equation $$AX-XB=C$$; 4. The scalar equation $$f(X)=O$$; 5. Matrix polynomial equations; 6. The extraction of $$m$$-th roots of a non-singular matrix; 7. The extraction of $$m$$-th roots of a singular matrix; 8. The logarithm of a matrix • IX. Linear operators in a unitary space: 1. General considerations; 2. Metrization of a space; 3. Gram's criterion for linear dependence of vectors; 4. Orthogonal projection; 5. The geometrical meaning of the Gramian and some inequalities; 6. Orthogonalization of a sequence of vectors; 7. Orthonormal bases; 8. The adjoint operator; 9. Normal operators in a unitary space; 10. The spectra of normal, hermitian, and unitary operators; 11. Positive-semidefinite and positive-definite hermitian operators; 12. Polar decomposition of a linear operator in a unitary space. Cayley's formulas; 13. Linear operators in a euclidean space; 14. Polar decomposition of an operator and the Cayley formulas in a euclidean space; 15. Commuting normal operators • X. Quadratic and hermitian forms: 1. Transformation of the variables in a quadratic form; 2. Reduction of a quadratic form to a sum of squares. The law of inertia; 3. The methods of Lagrange and Jacobi of reducing a quadratic form to a sum of squares; 4. Positive quadratic forms; 5. Reduction of a quadratic form to principal axes; 6. Pencils of quadratic forms; 7. Extremal properties of the characteristic values of a regular pencil of forms; 8. Small oscillations of a system with $$n$$ degrees of freedom; 9. Hermitian forms; 10. Hankel forms • Bibliography • Index AMS Home \| Comments: webmaster@ams.org © Copyright 2012, American Mathematical Society Privacy Statement	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 16, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.808344841003418, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/tagged/classical-mechanics+action	# Tagged Questions 1answer 152 views ### What's the motivation behind the action principle? [closed] What's the motivation behind the action principle? Why does the action principle lead to Newtonian law? If Newton's law of motion is more fundamental so why doesn't one derive Lagrangians and ... 4answers 454 views ### Is the principle of least action a boundary value or initial condition problem? Here is a question that's been bothering me since I was a sophomore in university, and should have probably asked before graduating: In analytic (Lagrangian) mechanics, the derivation of the ... 2answers 206 views ### What is the significance of action? What is the physical interpretation of $$\int_{t_1}^{t_2} (T -V) dt$$ where, $T$ is Kinetic Energy and $V$ is potential energy. How does it give trajectory? 2answers 615 views ### Hamilton-Jacobi Equation In the Hamilton-Jacobi equation, we take the partial time derivative of the action. But the action comes from integrating the Lagrangian over time, so time seems to just be a dummy variable here and ... 2answers 281 views ### Is it circular reasoning to derive Newton's laws from action minimization? Usually, a typical example of the use of the action principle that I've read a lot is the derivation of Newton's equation (generalized to coordinate $q(t)$). However, in the classical mechanics ... 2answers 1k views ### Deriving the Lagrangian for a free particle I'm a newbie in physics. Sorry, if the following questions are dumb. I began reading "Mechanics" by Landau and Lifshitz recently and hit a few roadblocks right away. Proving that a free particle ... 4answers 725 views ### Why can't any term which is added to the Lagrangian be written as a total derivative (or divergence)? All right, I know there must be an elementary proof of this, but I am not sure why I never came across it before. Adding a total time derivative to the Lagrangian (or a 4D divergence of some 4 ... 2answers 197 views ### How do I show that there exists variational/action principle for a given classical system? We see variational principles coming into play in different places such as Classical Mechanics (Hamilton's principle which gives rise to the Euler-Lagrange equations), Optics (in the form of Fermat's ... 6answers 2k views ### Why the Principle of Least Action? I'll be generous and say it might be reasonable to assume that nature would tend to minimize, or maybe even maximize, the integral over time of $T-V$. Okay, fine. You write down the action ... 1answer 203 views ### What variables does the action $S$ depend on? Action is defined as, $$S ~=~ \int L(q, q', t) dt,$$ but my question is what variables does $S$ depend on? Is $S = S(q, t)$ or $S = S(q, q', t)$ where $q' := \frac{dq}{dt}$? In ... 1answer 484 views ### Does Action in Classical Mechanics have a Interpretation? [duplicate] Possible Duplicate: Hamilton's Principle The Lagrangian formulation of Classical Mechanics seem to suggest strongly that "action" is more than a mathematical trick. I suspect strongly ...	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 9, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.932561457157135, "perplexity_flag": "middle"}
http://mathoverflow.net/revisions/68599/list	## Return to Answer 2 edited body; deleted 1 characters in body This is now an old thread, but I just came across it. The question you are asking was asked by Behrends a couple of years ago, and he knew about the counter example on the plane that Gjergji points out. Behrends was specifically asking for an incomplete metric space that is a subset of the line that and has the fixed point property. I came up with a partition of $\mathbb R$ into two dense sets $X_0$ and $X_1$ such that for all $i\in{0,1}$, every map $f:X_i\to X_i$ that satisfies $$\forall x,y\in X_i(x\not=y\rightarrow\|f(x)-f(y)\|<\|x-y\|)$$ is actually constant (and in particular has a fixed point). The $X_i$ can be constructed by an easy transfinite recursion and there is no control of how complicated the sets are. This is the huge advantage of Elekes' examples which are Borel. The notes with my proof are here: http://www.hausdorff-center.uni-bonn.de/people/geschke/papers/Behrends.dvihttp://www.hausdorff-center.uni-bonn.de/people/geschke/papers/Behrends.pdf 1 This is now an old thread, but I just came across it. The question you are asking was asked by Behrends a couple of years ago, and he knew about the counter example on the plane that Gjergji points out. Behrends was specifically asking for an incomplete metric space that is a subset of the line that has the fixed point property. I came up with a partition of $\mathbb R$ into two dense sets $X_0$ and $X_1$ such that for all $i\in{0,1}$, every map $f:X_i\to X_i$ that satisfies $$\forall x,y\in X_i(x\not=y\rightarrow\|f(x)-f(y)\|<\|x-y\|)$$ is actually constant (and in particular has a fixed point). The $X_i$ can be constructed by an easy transfinite recursion and there is no control of how complicated the sets are. This is the huge advantage of Elekes' examples which are Borel. The notes with my proof are here: http://www.hausdorff-center.uni-bonn.de/people/geschke/papers/Behrends.dvi	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 12, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9594699740409851, "perplexity_flag": "head"}
http://mathhelpforum.com/calculus/36478-need-help-few-important-homework-questions.html	# Thread: 1. ## Need help with a few important homework questions Graph y^2=2xy The thing with this is that I just don't know how to solve for Y. Apparently I'm not supposed to divide Y from both sides? Also I have no idea how to do this next one. A machine is rolling a metal cylinder under pressure. The radius of the cylinder is decreasing at a constant rate of 0.05 inches per second and the volume is a constant 128(pi) cubic inches. At what rate is the length h changing when the radius r is 1.8 inches? And I'll just throw in this third one cuz I'm not sure if I did it right. What Values of c are guaranteed by the Mean Value Theorem for f(x)=(x-2)^3 on the interval [-1,2]? I got 2-(the square root of)3. Thanks 2. Originally Posted by sean1058 Graph y^2=2xy The thing with this is that I just don't know how to solve for Y. Apparently I'm not supposed to divide Y from both sides? Also I have no idea how to do this next one. A machine is rolling a metal cylinder under pressure. The radius of the cylinder is decreasing at a constant rate of 0.05 inches per second and the volume is a constant 128(pi) cubic inches. At what rate is the length h changing when the radius r is 1.8 inches? And I'll just throw in this third one cuz I'm not sure if I did it right. What Values of c are guaranteed by the Mean Value Theorem for f(x)=(x-2)^3 on the interval [-1,2]? I got 2-(the square root of)3. Thanks For the first one just plot points..if you have to for the mean value one did you solve for c in the following $3(c-2)^2=\frac{0+9}{2+1}$? 3. No I have this $3(c-2)^2=\frac{0+27}{2+1}$ cuz the 3 is to the third power and I can't just plot points for that one, I have to solve it first. 4. Originally Posted by sean1058 Graph y^2=2xy The thing with this is that I just don't know how to solve for Y. Apparently I'm not supposed to divide Y from both sides? Also I have no idea how to do this next one. A machine is rolling a metal cylinder under pressure. The radius of the cylinder is decreasing at a constant rate of 0.05 inches per second and the volume is a constant 128(pi) cubic inches. At what rate is the length h changing when the radius r is 1.8 inches? And I'll just throw in this third one cuz I'm not sure if I did it right. What Values of c are guaranteed by the Mean Value Theorem for f(x)=(x-2)^3 on the interval [-1,2]? I got 2-(the square root of)3. Thanks $y^2-2xy=0\Rightarrow{(y-x)^2-x^2=0}$ so $(y-x)^2=x^2\Rightarrow{\|y-x\|=\|x\|}\Rightarrow{y=x\pm\|x\|}$ 5. Originally Posted by sean1058 Graph y^2=2xy The thing with this is that I just don't know how to solve for Y. Apparently I'm not supposed to divide Y from both sides? Thanks Well, you can't divide each side by y because it's $y^2$, so to undo that you would have to take the square root to get plain old y. Do you know polar coordinates? That could help possibly. Otherwise, the comment above this one is the best you will get. 6. thanks if anybody can help me with that middle one it would be most appreciated 7. nevermind i got it	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 5, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9570817947387695, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/52821/for-an-isolated-system-can-the-entropy-decrease-or-increase/52823	# For an isolated system, can the entropy decrease or increase? In any sizable system, the number of equilibrium states are much, much greater then the number of non-equilibrium states. Since each accessible micro state is equally probably, it is overwhelmingly likely that we will find the system in an equilibrium state. However, for a closed system, one that does not interact with any external system, the number of micro states is fixed. Therefore the entropy is fixed. At any given time, the system will be in one of its micro states. Even if the system is in a micro state that does not look like equilibrium from a macroscopic point of view, its entropy will remain the same, since entropy is a property of all the micro states, not just a given micro state. So, are the number of accessible micro states (and hence entropy) of an isolated system constant? - ## 5 Answers Isolated system: Since the matter, energy, and momentum is fixed, the total number of microstates available that satisfy these constraints is fixed/constant. So is the entropy constant? Yes, if the system is in equilibrium. No, if the system is not in equilibrium. What does that mean in terms of microstates? If the system is in equilibrium, all these microstates are equally probable and the system visits each of these microstates over the course of time (also known as ergodic hypothesis). Therefore the each micostate has equal probability and $S=k ln\Omega$. In such a state there is no more increase in entropy possible. If the system is in non-equilibrium the system doesn't have equal probability of being in every microstate. In fact if the system is stuck in non-equilibrium ( e.g., a hot part and cold part in the box separated by a thermally insulating wall) it cannot access some microstates at all. Hence the system is restricted to fewer microstates. Technically, there is no unique global thermodyanmic state for the whole system and you cannot define entropy. But you can calculate entropy by summing up entropy of different local equilibrium parts (e.g., entropy of the hold part and cold part separately). Since the total number of microstates you can access is small, entropy is lesser than what it could be if you remove that thermally insulating wall letting the system equilibrate. Thus, when you equilibrate more microstates become accessible. Think of the system spreading in the phase space. Thus entropy increases. Once the system has reached complete thermodynamic equilibrium no more entropy increase in possible. All allowable microstates have been made accessible and equally probable! - Your first sentence, answers, I think, the question. The rest is a subject for another question. – yalis Feb 2 at 2:20 +1 I just had a long conversation with a grad student friend of mine, and we came to essentially the same conclusion as this answer. This is a much better answer than mine. – joshphysics Feb 2 at 2:23 Glad to be of help. – Sankaran Feb 2 at 16:15 Since entropy is $$S=k_b \ln \Omega,$$ it is synonymous to a certain phase space volume $\Omega$ which we have to fix somehow. Phase space volume is something like a counter of possible states of the system (in the most simple case, the positions and momenta of all particles). This is (normally) done by setting some macroscopic parameters, like temperature, pressure, number of particles etc. So, if you have a closed system (i.e. a box where nothing can get in or out), you can e.g. fix volume, energy and the number of particles. You end up with $S(E,V,N)=k_b \ln \Omega(E,V,N)$. This can be understood as the number of all possible configurations inside the box which have the total energy $E$, the total possible volume $V$ and consist of $N$ particles - including all the stuff in the box on the left side, on the right side, evenly distributed, half of it moving and the other half standing still etc. $S$ is a constant number because $E$, $V$ and $N$ are fixed in a closed system. The problem is that you can make no statement about the system whatsoever. But - instead of $S(E,V,N)$, you can look at other dependencies of $S$. Let's say you divide up the box into two equally sized parts, a left part and a right part. $N_1$ is the number of particles in the left part, and $N_2$ is the number of particles in the right part (also $N_1+N_2=N$). Now, you can calculate $S(E,V,N_1,N_2)$ and choose $N_1$ and $N_2$. That way, $S$ can change because you can choose $N_1$ and $N_2$ (particles may go from the left to the right part and vice versa). The important point is that $S$ is much greater for $N_1\approx N_2$ than for any other values. This means that most of the possible configurations have approximately the same number of particles on the left and on the right side. Because every configuration is equally probable, most of the time, the system will look like this. But not all the time, though - even when $S(E,V,N_1,N_2)=0$, there is exactly one microscopic configuration which fulfils the macroscopic parameters, and it will occur at some time (if the system is ergodic, but that is a different story). In the given example, the entropy will increase and decrease all the time when particles go from the left half to the right half. But by definition, the percentage of time when the system is in a specific configuration $N_1$,$N_2$ is proportional to $\Omega(E,V,N_1,N_2)=\exp\left( \frac {S(E,V,N_1,N_2)} {k_b}\right)$ (I just thought of this last formula, please correct me if I'm wrong). In conclusion: The term entropy is meaningless without the parameters one wants to fix/vary. Funny example: Let's calculate $S(N_{\textrm{people alive on earth}})$ for the whole universe which describes the number of possible configurations of the universe for a given number of living people on earth. $S(0)$ will probably be much bigger than for any other $N_{paoe}$, which means that, most of the time, there will be no people on earth and the entropy will be huge (this is my interpretation of the the heat death of the universe). Then, they might come back and the entropy will be very small. - ok, but in your example, you are dividing the isolated system into two non-isolated, interacting subsystems (left and right). The entropy of either system can increase or decrease but on the whole we are just sampling different micro states. Entropy should not be dependent on the particular micro state that the system is currently sampling. It only depends on the total number of micro states. In a way that seems to contradict intuition. The universe is closed (?) but it seems that its entropy will increase until heat death is reached. Yet, the number of micro states of the universe is fixed? – yalis Feb 2 at 1:17 The division is necessary for the entropy to be able to change. If the entropy only depends on constant parameters (e.g. E, V, N), it is constant. This also applies to a closed universe. For other parameters (e.g. $N_1$, rotation speed of mars, number of posts on physics.SE), the entropy changes (see my last example). – Rafael Reiter Feb 2 at 10:01 Yes the entropy of an isolated system can increase. Here's an example. Take a system consisting of an ideal gas in a thermally insulated box. Let the initial condition of the system be such that the gas occupies the left half of the box with the right side being vacuum. After the initial time, the gas will expand freely to fill the box. Thermal insulation and free expansion guarantee that $Q = 0$ and $W = 0$, so by the first law $\Delta U = 0$. The entropy of the system will be larger in this final state because of this equation (which gives ideal gas entropy) and the fact that $\Delta U = 0$ and $\Delta N$ = 0 while $\Delta V > 0$. Hope that helps! Cheers! - 2 I actually wrote up this example in the initial draft of the question, but deleted it. The thing here is this: Are we not considering the left half of the system as a sub-system of the closed the system. This subsystem interacts with the whole system by expanding. You are "adding volume" to the subsystem and thus increasing the number of micro states. Thus, I would think that a truly isolated system cannot expand. – yalis Feb 2 at 0:49 1 Another angle. Say gas particles uniformly distributed at some initial time. It's quite possible, though extremely unlikely, that at at some point in the future, all the particles were on one side of the box. Has the entropy of the system decreased? Not sure how it can, since entropy is associated with the number of micro states, which has not changed. – yalis Feb 2 at 0:50 Yea you make a very good point that I was actually thinking about as well when I was writing up this example. To put it another way, viewed as a quantum system, the Von-Neumann entropy cannot change since isolated systems undergo unitary time evolution and the Von-Neumann entropy remains invariant under such an evolution of the density matrix of the system. I've heard that a resolution of such issues has to do with quantum "decoherence," but I'm by no means an expert on such things. – joshphysics Feb 2 at 1:26 I'm almost tempted to start another thread asking about this specific example of the gas in the box and how it can be reconciled with the quantum/microscopic considerations we are both concerned about. I'm not entirely sure such questions have even been resolved to be honest. – joshphysics Feb 2 at 1:36 Yes, if we say the entropy is just the log of the number of accessible microstates, then the entropy cannot change for an isolated system as it evolves. But this contradicts common sense. For an isolated system, we must introduce a notion of coarse graining for entropy to be a useful concept. Coarse graining can be done by dividing the system into subsystems, and then adding the entropies if the subsystems. This coarse grained entropy can and will change over time, as energy is exchanged between the different subsystems. The exact value of the coarse grained entropy will in general depend on the details of how you form the subsystems. The coarse grained entropy should eventually approach the original fine grained entropy as the system reaches equilibrium. - Yes, the idea of breaking an isolated system into subsystems. A given subsystem can have its entropy increase or decrease, until such time as all subsystems have reached maximum entropy. This is in line with the the following statement attributed to Rudolf Clausius on wikipedia: "The entropy of an isolated system not in equilibrium will tend to increase over time, approaching a maximum value at equilibrium" Suggesting that an isolated system can be away from equilibrium. Yet the information content in an isolated system cannot change, so there's still a bit of fuzziness here. – yalis Feb 2 at 2:32 the entropy can decrease as well. On average the entropy will increases (answer above), but it is possible that the entropy decreases. - You are right, but if it is a macroscopic system, don't hold your breath waiting for it to happen. – Paul J. Gans Feb 2 at 1:16	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 37, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9348180890083313, "perplexity_flag": "head"}
http://www.nag.com/numeric/CL/nagdoc_cl23/html/E05/e05sac.html	NAG Library Function Documentnag_glopt_bnd_pso (e05sac) Note: this function uses optional arguments to define choices in the problem specification and in the details of the algorithm. If you wish to use default settings for all of the optional arguments, you need only read Sections 1 to 9 of this document. If, however, you wish to reset some or all of the settings please refer to Section 10 for a detailed description of the algorithm and to Section 11 for a detailed description of the specification of the optional arguments. 1 Purpose nag_glopt_bnd_pso (e05sac) is designed to search for the global minimum or maximum of an arbitrary function, using Particle Swarm Optimization (PSO). Derivatives are not required, although these may be used by an accompanying local minimization function if desired. nag_glopt_bnd_pso (e05sac) is essentially identical to nag_glopt_nlp_pso (e05sbc), but with a simpler interface and with various optional arguments removed; otherwise most arguments are identical. In particular, nag_glopt_bnd_pso (e05sac) does not handle general constraints. 2 Specification #include <nag.h> #include <nage05.h> void nag_glopt_bnd_pso (Integer ndim, Integer npar, double xb[], double fb, const double bl[], const double bu[], void (objfun)(Integer mode, Integer ndim, const double x[], double objf, double vecout[], Integer nstate, Nag_Comm comm), void (monmod)(Integer ndim, Integer npar, double x[], const double xb[], double fb, const double xbest[], const double fbest[], const Integer itt[], Nag_Comm comm, Integer inform), Integer iopts[], double opts[], Nag_Comm comm, Integer itt[], Integer inform, NagError fail) Before calling nag_glopt_bnd_pso (e05sac), nag_glopt_opt_set (e05zkc) must be called with optstr set to ‘Initialize = e05sac’. Optional arguments may also be specified by calling nag_glopt_opt_set (e05zkc) before the call to nag_glopt_bnd_pso (e05sac). 3 Description nag_glopt_bnd_pso (e05sac) uses a stochastic method based on Particle Swarm Optimization (PSO) to search for the global optimum of a nonlinear function $F$, subject to a set of bound constraints on the variables. In the PSO algorithm (see Section 10), a set of particles is generated in the search space, and advances each iteration to (hopefully) better positions using a heuristic velocity based upon inertia, cognitive memory and global memory. The inertia is provided by a decreasingly weighted contribution from a particles current velocity, the cognitive memory refers to the best candidate found by an individual particle and the global memory refers to the best candidate found by all the particles. This allows for a global search of the domain in question. Further, this may be coupled with a selection of local minimization functions, which may be called during the iterations of the heuristic algorithm, the interior phase, to hasten the discovery of locally optimal points, and after the heuristic phase has completed to attempt to refine the final solution, the exterior phase. Different options may be set for the local optimizer in each phase. Without loss of generality, the problem is assumed to be stated in the following form: $minimize x∈Rndim Fx subject to ℓ ≤ x ≤ u ,$ where the objective $F\left(\mathbf{x}\right)$ is a scalar function, $\mathbf{x}$ is a vector in ${R}^{\mathit{ndim}}$ and the vectors $\mathbf{\ell }\le \mathbf{u}$ are lower and upper bounds respectively for the $\mathit{ndim}$ variables. The objective function may be nonlinear. Continuity of $F$ is not essential. For functions which are smooth and primarily unimodal, faster solutions will almost certainly be achieved by using Chapter e04 functions directly. For functions which are smooth and multi-modal, gradient dependent local minimization functions may be coupled with nag_glopt_bnd_pso (e05sac). For multi-modal functions for which derivatives cannot be provided, particularly functions with a significant level of noise in their evaluation, nag_glopt_bnd_pso (e05sac) should be used either alone, or coupled with nag_opt_simplex_easy (e04cbc). The $\mathit{ndim}$ lower and upper box bounds on the variable $\mathbf{x}$ are included to initialize the particle swarm into a finite hypervolume, although their subsequent influence on the algorithm is user determinable (see the option ${\mathbf{Boundary}}$ in Section 11). It is strongly recommended that sensible bounds are provided for all variables. nag_glopt_bnd_pso (e05sac) may also be used to maximize the objective function (see the option ${\mathbf{Optimize}}$). Due to the nature of global optimization, unless a predefined target is provided, there is no definitive way of knowing when to end a computation. As such several stopping heuristics have been implemented into the algorithm. If any of these is achieved, nag_glopt_bnd_pso (e05sac) will exit with NW_SOLUTION_NOT_GUARANTEED, and the parameter inform will indicate which criteria was reached. See inform for more information. In addition, you may provide your own stopping criteria through monmod and objfun. nag_glopt_nlp_pso (e05sbc) provides a comprehensive interface, allowing for the inclusion of general nonlinear constraints. 4 References Gill P E, Murray W and Wright M H (1981) Practical Optimization Academic Press Kennedy J and Eberhart R C (1995) Particle Swarm Optimization Proceedings of the 1995 IEEE International Conference on Neural Networks 1942–1948 Koh B, George A D, Haftka R T and Fregly B J (2006) Parallel Asynchronous Particle Swarm Optimization International Journal for Numerical Methods in Engineering 67(4) 578–595 Vaz A I and Vicente L N (2007) A Particle Swarm Pattern Search Method for Bound Constrained Global Optimization Journal of Global Optimization 39(2) 197–219 Kluwer Academic Publishers 5 Arguments Note: for descriptions of the symbolic variables, see Section 10. 1: ndim – IntegerInput On entry: $\mathit{ndim}$, the number of dimensions. Constraint: ${\mathbf{ndim}}\ge 1$. 2: npar – IntegerInput On entry: $\mathit{npar}$, the number of particles to be used in the swarm. Assuming all particles remain within bounds, each complete iteration will perform at least npar function evaluations. Otherwise, significantly fewer objective function evaluations may be performed. Suggested value: ${\mathbf{npar}}=10×{\mathbf{ndim}}$. Constraint: ${\mathbf{npar}}\ge 5$. 3: xb[ndim] – doubleOutput On exit: the location of the best solution found, $\stackrel{~}{\mathbf{x}}$, in ${R}^{\mathit{ndim}}$. 4: fb – double Output On exit: the objective value of the best solution, $\stackrel{~}{f}=F\left(\stackrel{~}{\mathbf{x}}\right)$. 5: bl[ndim] – const doubleInput 6: bu[ndim] – const doubleInput On entry: ${\mathbf{bl}}$ is $\mathbf{\ell }$, the array of lower bounds, bu is $\mathbf{u}$, the array of upper bounds. The ndim entries in bl and bu must contain the lower and upper simple (box) bounds of the variables respectively. These must be provided to initialize the sample population into a finite hypervolume, although their subsequent influence on the algorithm is user determinable (see the option ${\mathbf{Boundary}}$ in Section 11). If ${\mathbf{bl}}\left[i-1\right]={\mathbf{bu}}\left[i-1\right]$ for any $i\in \left\{1,\dots ,{\mathbf{ndim}}\right\}$, variable $i$ will remain locked to ${\mathbf{bl}}\left[i-1\right]$ regardless of the ${\mathbf{Boundary}}$ option selected. It is strongly advised that you place sensible lower and upper bounds on all variables, even if your model allows for variables to be unbounded (using the option ${\mathbf{Boundary}}=\mathrm{ignore}$) since these define the initial search space. Constraints: • ${\mathbf{bl}}\left[\mathit{i}-1\right]\le {\mathbf{bu}}\left[\mathit{i}-1\right]$, for $\mathit{i}=1,2,\dots ,{\mathbf{ndim}}$; • ${\mathbf{bl}}\left[i-1\right]\ne {\mathbf{bu}}\left[i-1\right]$ for at least one $i\in \left\{1,\dots ,{\mathbf{ndim}}\right\}$. 7: objfun – function, supplied by the userExternal Function objfun must, depending on the value of mode, calculate the objective function and/or calculate the gradient of the objective function for a $\mathit{ndim}$-variable vector $\mathbf{x}$. Gradients are only required if a local minimizer has been chosen which requires gradients. See the option ${\mathbf{Local Minimizer}}$ for more information. The specification of objfun is: void objfun (Integer mode, Integer ndim, const double x[], double objf, double vecout[], Integer nstate, Nag_Comm comm) 1: mode – Integer Input/Output On entry: indicates which functionality is required. ${\mathbf{mode}}=0$ $F\left(\mathbf{x}\right)$ should be returned in objf. The value of objf on entry may be used as an upper bound for the calculation. Any expected value of $F\left(\mathbf{x}\right)$ that is greater than objf may be approximated by this upper bound; that is objf can remain unaltered. ${\mathbf{mode}}=1$ ${\mathbf{Local Minimizer}}='\mathrm{e04ucc}'$ only First derivatives can be evaluated and returned in vecout. Any unaltered elements of vecout will be approximated using finite differences. ${\mathbf{mode}}=2$ ${\mathbf{Local Minimizer}}='\mathrm{e04ucc}'$ only $F\left(\mathbf{x}\right)$ must be calculated and returned in objf, and available first derivatives can be evaluated and returned in vecout. Any unaltered elements of vecout will be approximated using finite differences. ${\mathbf{mode}}=5$ $F\left(\mathbf{x}\right)$ must be calculated and returned in objf. The value of objf on entry may not be used as an upper bound. ${\mathbf{mode}}=6$ ${\mathbf{Local Minimizer}}='\mathrm{e04dgc}'$ only All first derivatives must be evaluated and returned in vecout. ${\mathbf{mode}}=7$ ${\mathbf{Local Minimizer}}='\mathrm{e04dgc}'$ only $F\left(\mathbf{x}\right)$ must be calculated and returned in objf, and all first derivatives must be evaluated and returned in vecout. On exit: if the value of mode is set to be negative, then nag_glopt_bnd_pso (e05sac) will exit as soon as possible with NE_USER_STOP and ${\mathbf{inform}}={\mathbf{mode}}$. 2: ndim – IntegerInput On entry: the number of dimensions. 3: x[ndim] – const doubleInput On entry: $\mathbf{x}$, the point at which the objective function and/or its gradient are to be evaluated. 4: objf – double Input/Output On entry: the value of objf passed to objfun varies with the argument mode. ${\mathbf{mode}}=0$ objf is an upper bound for the value of $F\left(\mathbf{x}\right)$, often equal to the best value of $F\left(\mathbf{x}\right)$ found so far by a given particle. Only objective function values less than the value of objf on entry will be used further. As such this upper bound may be used to stop further evaluation when this will only increase the objective function value above the upper bound. ${\mathbf{mode}}=1$, $2$, $5$, $6$ or $7$ objf is meaningless on entry. On exit: the value of objf returned varies with the argument mode. ${\mathbf{mode}}=0$ objf must be the value of $F\left(\mathbf{x}\right)$. Only values of $F\left(\mathbf{x}\right)$ strictly less than objf on entry need be accurate. ${\mathbf{mode}}=1$ or $6$ Need not be set. ${\mathbf{mode}}=2$, $5$ or $7$ $F\left(\mathbf{x}\right)$ must be calculated and returned in objf. The entry value of objf may not be used as an upper bound. 5: vecout[ndim] – doubleInput/Output On entry: if ${\mathbf{Local Minimizer}}=\mathrm{e04ucc}$, the values of vecout are used internally to indicate whether a finite difference approximation is required. See nag_opt_nlp (e04ucc). On exit: the required values of vecout returned to the calling function depend on the value of mode. ${\mathbf{mode}}=0$ or $5$ The value of vecout need not be set. ${\mathbf{mode}}=1$ or $2$ vecout can contain components of the gradient of the objective function $\frac{\partial F}{\partial {x}_{i}}$ for some $i=1,2,\dots {\mathbf{ndim}}$, or acceptable approximations. Any unaltered elements of vecout will be approximated using finite differences. ${\mathbf{mode}}=6$ or $7$ vecout must contain the gradient of the objective function $\frac{\partial F}{\partial {x}_{i}}$ for all $i=1,2,\dots {\mathbf{ndim}}$. Approximation of the gradient is strongly discouraged, and no finite difference approximations will be performed internally (see nag_opt_conj_grad (e04dgc)). 6: nstate – IntegerInput On entry: nstate indicates various stages of initialization throughout the function. This allows for permanent global arguments to be initialized the least number of times. For example, you may initialize a random number generator seed. ${\mathbf{nstate}}=2$ objfun is called for the very first time. You may save computational time if certain data must be read or calculated only once. ${\mathbf{nstate}}=1$ objfun is called for the first time by a NAG local minimization function. You may save computational time if certain data required for the local minimizer need only be calculated at the initial point of the local minimization. ${\mathbf{nstate}}=0$ Used in all other cases. 7: comm – Nag_Comm Pointer to structure of type Nag_Comm; the following members are relevant to objfun. user – double * iuser – Integer * p – Pointer The type Pointer will be void . Before calling nag_glopt_bnd_pso (e05sac) you may allocate memory and initialize these pointers with various quantities for use by objfun when called from nag_glopt_bnd_pso (e05sac) (see Section 3.2.1 in the Essential Introduction). 8: monmod – function, supplied by the userExternal Function A user-specified monitoring and modification function. monmod is called once every complete iteration after a finalization check. It may be used to modify the particle locations that will be evaluated at the next iteration. This permits the incorporation of algorithmic modifications such as including additional advection heuristics and genetic mutations. monmod is only called during the main loop of the algorithm, and as such will be unaware of any further improvement from the final local minimization. If no monitoring and/or modification is required, monmod may be NULLFN. The specification of monmod is: void monmod (Integer ndim, Integer npar, double x[], const double xb[], double fb, const double xbest[], const double fbest[], const Integer itt[], Nag_Comm comm, Integer inform) 1: ndim – IntegerInput On entry: the number of dimensions. 2: npar – IntegerInput On entry: the number of particles. 3: x[${\mathbf{ndim}}×{\mathbf{npar}}$] – doubleInput/Output Note: the $i$th component of the $j$th particle, ${x}_{j}\left(i\right)$, is stored in ${\mathbf{x}}\left[\left(j-1\right)×{\mathbf{ndim}}+i-1\right]$. On entry: the npar particle locations, ${\mathbf{x}}_{j}$, which will currently be used during the next iteration unless altered in monmod. On exit: the particle locations to be used during the next iteration. 4: xb[ndim] – const doubleInput On entry: the location, $\stackrel{~}{\mathbf{x}}$, of the best solution yet found. 5: fb – doubleInput On entry: the objective value, $\stackrel{~}{f}=F\left(\stackrel{~}{\mathbf{x}}\right)$, of the best solution yet found. 6: xbest[${\mathbf{ndim}}×{\mathbf{npar}}$] – const doubleInput Note: the $i$th component of the position of the $j$th particle's cognitive memory, ${\stackrel{^}{x}}_{j}\left(i\right)$, is stored in ${\mathbf{xbest}}\left[\left(j-1\right)×{\mathbf{ndim}}+i-1\right]$. On entry: the locations currently in the cognitive memory, ${\stackrel{^}{\mathbf{x}}}_{\mathit{j}}$, for $\mathit{j}=1,2,\dots ,{\mathbf{npar}}$ (see Section 10). 7: fbest[npar] – const doubleInput On entry: the objective values currently in the cognitive memory, $F\left({\stackrel{^}{\mathbf{x}}}_{\mathit{j}}\right)$, for $\mathit{j}=1,2,\dots ,{\mathbf{npar}}$. 8: itt[$6$] – const IntegerInput On entry: iteration and function evaluation counters (see description of itt below). 9: comm – Nag_Comm Pointer to structure of type Nag_Comm; the following members are relevant to monmod. user – double * iuser – Integer * p – Pointer The type Pointer will be void . Before calling nag_glopt_bnd_pso (e05sac) you may allocate memory and initialize these pointers with various quantities for use by monmod when called from nag_glopt_bnd_pso (e05sac) (see Section 3.2.1 in the Essential Introduction). 10: inform – Integer Input/Output On entry: ${\mathbf{inform}}=0$ On exit: setting ${\mathbf{inform}}<0$ will cause near immediate exit from nag_glopt_bnd_pso (e05sac). This value will be returned as inform with NE_USER_STOP. You need not set inform unless you wish to force an exit. 9: iopts[$\mathit{dim}$] – IntegerCommunication Array Note: the dimension of the array iopts, corresponding to the array length liopts (see e05zkc), must be at least $\mathrm{100}$. On entry: optional parameter array as generated and possibly modified by calls to nag_glopt_opt_set (e05zkc). The contents of iopts MUST NOT be modified directly between calls to nag_glopt_bnd_pso (e05sac), nag_glopt_opt_set (e05zkc) or nag_glopt_opt_get (e05zlc). 10: opts[$\mathit{dim}$] – doubleCommunication Array Note: the dimension of the array opts, corresponding to the array length lopts (see e05zkc), must be at least $\mathrm{100}$. On entry: optional parameter array as generated and possibly modified by calls to nag_glopt_opt_set (e05zkc). The contents of opts MUST NOT be modified directly between calls to nag_glopt_bnd_pso (e05sac), nag_glopt_opt_set (e05zkc) or nag_glopt_opt_get (e05zlc). 11: comm – Nag_Comm Communication Structure The NAG communication argument (see Section 3.2.1.1 in the Essential Introduction). 12: itt[$6$] – IntegerOutput On exit: integer iteration counters for nag_glopt_bnd_pso (e05sac). ${\mathbf{itt}}\left[0\right]$ Number of complete iterations. ${\mathbf{itt}}\left[1\right]$ Number of complete iterations without improvement to the current optimum. ${\mathbf{itt}}\left[2\right]$ Number of particles converged to the current optimum. ${\mathbf{itt}}\left[3\right]$ Number of improvements to the optimum. ${\mathbf{itt}}\left[4\right]$ Number of function evaluations performed. ${\mathbf{itt}}\left[5\right]$ Number of particles reset. 13: inform – Integer Output On exit: indicates which finalization criterion was reached. The possible values of inform are: inform Meaning $<0$ Exit from a user-supplied subroutine. 0 nag_glopt_bnd_pso (e05sac) has detected an error and terminated. 1 The provided objective target has been achieved. (${\mathbf{Target Objective Value}}$). 2 The standard deviation of the location of all the particles is below the set threshold (${\mathbf{Swarm Standard Deviation}}$). If the solution returned is not satisfactory, you may try setting a smaller value of ${\mathbf{Swarm Standard Deviation}}$, or try adjusting the options governing the repulsive phase (${\mathbf{Repulsion Initialize}}$, ${\mathbf{Repulsion Finalize}}$). 3 The total number of particles converged (${\mathbf{Maximum Particles Converged}}$) to the current global optimum has reached the set limit. This is the number of particles which have moved to a distance less than ${\mathbf{Distance Tolerance}}$ from the optimum with regard to the ${L}^{2}$ norm. If the solution is not satisfactory, you may consider lowering the ${\mathbf{Distance Tolerance}}$. However, this may hinder the global search capability of the algorithm. 4 The maximum number of iterations without improvement (${\mathbf{Maximum Iterations Static}}$) has been reached, and the required number of particles (${\mathbf{Maximum Iterations Static Particles}}$) have converged to the current optimum. Increasing either of these options will allow the algorithm to continue searching for longer. Alternatively if the solution is not satisfactory, re-starting the application several times with ${\mathbf{Repeatability}}=\mathrm{OFF}$ may lead to an improved solution. 5 The maximum number of iterations (${\mathbf{Maximum Iterations Completed}}$) has been reached. If the number of iterations since improvement is small, then a better solution may be found by increasing this limit, or by using the option ${\mathbf{Local Minimizer}}$ with corresponding exterior options. Otherwise if the solution is not satisfactory, you may try re-running the application several times with ${\mathbf{Repeatability}}=\mathrm{OFF}$ and a lower iteration limit, or adjusting the options governing the repulsive phase (${\mathbf{Repulsion Initialize}}$, ${\mathbf{Repulsion Finalize}}$). 6 The maximum allowed number of function evaluations (${\mathbf{Maximum Function Evaluations}}$) has been reached. As with ${\mathbf{inform}}=5$, increasing this limit if the number of iterations without improvement is small, or decreasing this limit and running the algorithm multiple times with ${\mathbf{Repeatability}}=\mathrm{OFF}$, may provide a superior result. 14: fail – NagError *Input/Output The NAG error argument (see Section 3.6 in the Essential Introduction). nag_glopt_bnd_pso (e05sac) will return NW_SOLUTION_NOT_GUARANTEED if no error has been detected, and a finalization criterion has been achieved which cannot guarantee success. This does not indicate that the function has failed, merely that the returned solution cannot be guaranteed to be the true global optimum. The value of inform should be examined to determine which finalization criterion was reached. 6 Error Indicators and Warnings NE_ALLOC_FAIL Dynamic memory allocation failed. NE_BAD_PARAM On entry, argument $⟨\mathit{\text{value}}⟩$ had an illegal value. NE_BOUND On entry, ${\mathbf{bl}}\left[i\right]={\mathbf{bu}}\left[i\right]$ for all $i$. Constraint: ${\mathbf{bu}}\left[i\right]>{\mathbf{bl}}\left[i\right]$ for at least one $i$. On entry, ${\mathbf{bl}}\left[⟨\mathit{\text{value}}⟩\right]=⟨\mathit{\text{value}}⟩$ and ${\mathbf{bu}}\left[⟨\mathit{\text{value}}⟩\right]=⟨\mathit{\text{value}}⟩$. Constraint: ${\mathbf{bu}}\left[i\right]\ge {\mathbf{bl}}\left[i\right]$ for all $i$. NE_DERIV_ERRORS Derivative checks indicate possible errors in the supplied derivatives. NE_INT On entry, ${\mathbf{ndim}}=⟨\mathit{\text{value}}⟩$. Constraint: ${\mathbf{ndim}}\ge 1$. On entry, ${\mathbf{npar}}=⟨\mathit{\text{value}}⟩$. Constraint: ${\mathbf{npar}}\ge 5$. NE_INTERNAL_ERROR An internal error has occurred in this function. Check the function call and any array sizes. If the call is correct then please contact NAG for assistance. NE_INVALID_OPTION Either the option arrays have not been initialized for nag_glopt_bnd_pso (e05sac), or they have become corrupted. The option arrays are not consistent. The option arrays have not been initialized for nag_glopt_bnd_pso (e05sac). NE_USER_STOP User requested exit $⟨\mathit{\text{value}}⟩$ during call to monmod. User requested exit $⟨\mathit{\text{value}}⟩$ during call to objfun. NW_FAST_SOLUTION Target $⟨\mathit{\text{value}}⟩$ achieved after the first iteration. ${\mathbf{fb}}=⟨\mathit{\text{value}}⟩$. NW_SOLUTION_NOT_GUARANTEED A finalization criterion was reached that cannot guarantee success. On exit, ${\mathbf{inform}}=⟨\mathit{\text{value}}⟩$. 7 Accuracy If NE_NOERROR (or NW_FAST_SOLUTION) or NW_SOLUTION_NOT_GUARANTEED on exit, either a ${\mathbf{Target Objective Value}}$ or finalization criterion has been reached, depending on user selected options. As with all global optimization software, the solution achieved may not be the true global optimum. Various options allow for either greater search diversity or faster convergence to a (local) optimum (See Sections 10 and 11). Provided the objective function and constraints are sufficiently well behaved, if a local minimizer is used in conjunction with nag_glopt_bnd_pso (e05sac), then it is more likely that the final result will at least be in the near vicinity of a local optimum, and due to the global search characteristics of the particle swarm, this solution should be superior to many other local optima. Caution should be used in accelerating the rate of convergence, as with faster convergence, less of the domain will remain searchable by the swarm, making it increasingly difficult for the algorithm to detect the basin of attraction of superior local optima. Using the options ${\mathbf{Repulsion Initialize}}$ and ${\mathbf{Repulsion Finalize}}$ described in Section 11 will help to overcome this, by causing the swarm to diverge away from the current optimum once no more local improvement is likely. On successful exit with guaranteed success, NE_NOERROR. This may only happen if a ${\mathbf{Target Objective Value}}$ is assigned and is reached by the algorithm. On successful exit without guaranteed success, NW_SOLUTION_NOT_GUARANTEED is returned. This will happen if another finalization criterion is achieved without the detection of an error. In both cases, the value of inform provides further information as to the cause of the exit. 8 Further Comments The memory used by nag_glopt_bnd_pso (e05sac) is relatively static throughout. As such, nag_glopt_bnd_pso (e05sac) may be used in problems with high dimension number (${\mathbf{ndim}}>100$) without the concern of computational resource exhaustion, although the probability of successfully locating the global optimum will decrease dramatically with the increase in dimensionality. Due to the stochastic nature of the algorithm, the result will vary over multiple runs. This is particularly true if arguments and options are chosen to accelerate convergence at the expense of the global search. However, the option ${\mathbf{Repeatability}}=\mathrm{ON}$ may be set to initialize the internal random number generator using a preset seed, which will result in identical solutions being obtained. 9 Example Note: a modified example is supplied with the NAG Library for SMP & Multicore and links have been supplied in the following subsections. This example uses a particle swarm to find the global minimum of the Schwefel function: $minimize x∈Rndim f = ∑ i=1 ndim xi sinxi$ $xi ∈ -500,500 , for i=1,2,…,ndim .$ In two dimensions the optimum is ${f}_{\mathrm{min}}=-837.966$, located at $\mathbf{x}=\left(-420.97,-420.97\right)$. The example demonstrates how to initialize and set the options arrays using nag_glopt_opt_set (e05zkc), how to query options using nag_glopt_opt_get (e05zlc), and finally how to search for the global optimum using nag_glopt_bnd_pso (e05sac). The function is minimized several times to demonstrate using nag_glopt_bnd_pso (e05sac) alone, and coupled with local minimizers. This program uses the nondefault option ${\mathbf{Repeatability}}=\mathrm{ON}$ to produce repeatable solutions. 9.1 Program Text Program Text (e05sace.c) None. 9.3 Program Results Program Results (e05sace.r) 10 Algorithmic Details The following pseudo-code describes the algorithm used with the repulsion mechanism. $INITIALIZE for j=1, npar xj = R ∈ Uℓbox,ubox x^ j = R∈ Uℓbox,ubox vj = R ∈ U -V max ,Vmax f^j = F x^ j initialize wj wj = Wmax Weight Initialize=MAXIMUM Wini Weight Initialize=INITIAL R ∈ U W min , W max Weight Initialize=RANDOMIZED end for x~ = 12 ℓbox + ubox f~ = F x~ Ic = Is = 0 SWARM while (not finalized), Ic = Ic + 1 for j = 1 , npar xj = BOUNDARYxj,ℓbox,ubox fj = F xj if fj < f^j f^j = fj , x^j = xj if fj < f~ f~ = fj , x~ = xj end for if new f~ LOCMINx~,f~,Oi , Is=0 [see note on repulsion below for code insertion] else Is = Is + 1 for j = 1 , npar vj = wj vj + Cs D1 x^j - xj + Cg D2 x~ - xj xj = xj + vj if xj - x~ < dtol reset xj, vj, wj; x^j = xj else update wj end for if (target achieved or termination criterion satisfied) finalized=true monmod xj end LOCMINx~,f~,Oe$ The definition of terms used in the above pseudo-code are as follows. $\mathit{npar}$ the number of particles, npar ${\mathbf{\ell }}_{\mathrm{box}}$ array of ndim lower box bounds ${\mathbf{u}}_{\mathrm{box}}$ array of ndim upper box bounds ${\mathbf{x}}_{j}$ position of particle $j$ ${\stackrel{^}{\mathbf{x}}}_{j}$ best position found by particle $j$ $\stackrel{~}{\mathbf{x}}$ best position found by any particle ${f}_{j}$ $F\left({\mathbf{x}}_{j}\right)$ ${\stackrel{^}{f}}_{j}$ $F\left({\stackrel{^}{\mathbf{x}}}_{j}\right)$, best value found by particle $j$ $\stackrel{~}{f}$ $F\left(\stackrel{~}{\mathbf{x}}\right)$, best value found by any particle ${\mathbf{v}}_{j}$ velocity of particle $j$ ${w}_{j}$ weight on ${\mathbf{v}}_{j}$ for velocity update, decreasing according to ${\mathbf{Weight Decrease}}$ ${V}_{\mathrm{max}}$ maximum absolute velocity, dependent upon ${\mathbf{Maximum Variable Velocity}}$ ${I}_{c}$ swarm iteration counter ${I}_{s}$ iterations since $\stackrel{~}{\mathbf{x}}$ was updated ${\mathbf{D}}_{1}$,${\mathbf{D}}_{2}$ diagonal matrices with random elements in range $\left(0,1\right)$ ${C}_{s}$ the cognitive advance coefficient which weights velocity towards ${\stackrel{^}{\mathbf{x}}}_{j}$, adjusted using ${\mathbf{Advance Cognitive}}$ ${C}_{g}$ the global advance coefficient which weights velocity towards $\stackrel{~}{\mathbf{x}}$, adjusted using ${\mathbf{Advance Global}}$ $\mathit{dtol}$ the ${\mathbf{Distance Tolerance}}$ for resetting a converged particle $\mathbf{R}\in U\left({\mathbf{\ell }}_{\mathrm{box}},{\mathbf{u}}_{\mathrm{box}}\right)$ an array of random numbers whose $i$-th element is drawn from a uniform distribution in the range $\left({{\mathbf{\ell }}_{\mathrm{box}}}_{\mathit{i}},{{\mathbf{u}}_{\mathrm{box}}}_{\mathit{i}}\right)$, for $\mathit{i}=1,2,\dots ,{\mathbf{ndim}}$ ${O}_{i}$ local optimizer interior options ${O}_{e}$ local optimizer exterior options $\mathrm{LOCMIN}\left(\mathbf{x},f,O\right)$ apply local optimizer using the set of options $O$ using the solution $\left(\mathbf{x},f\right)$ as the starting point, if used (not default) monmod monitor progress and possibly modify ${\mathbf{x}}_{j}$ BOUNDARY apply required behaviour for ${\mathbf{x}}_{j}$ outside bounding box, (see ${\mathbf{Boundary}}$) new ($\stackrel{~}{f}$) true if $\stackrel{~}{\mathbf{x}}$, $\stackrel{~}{\mathbf{c}}$, $\stackrel{~}{f}$ were updated at this iteration Additionally a repulsion phase can be introduced by changing from the default values of options ${\mathbf{Repulsion Finalize}}$ (${r}_{f}$), ${\mathbf{Repulsion Initialize}}$ (${r}_{i}$) and ${\mathbf{Repulsion Particles}}$ (${r}_{p}$). If the number of static particles is denoted ${n}_{s}$ then the following can be inserted after the new($\stackrel{~}{f}$) check in the pseudo-code above. $else if ( ns ≥ rp and ri ≤ Is ≤ ri + rf ) LOCMINx~,f~,Oi use -Cg instead of Cg in velocity updates if Is = ri + rf Is = 0$ 11 Optional Arguments This section can be skipped if you wish to use the default values for all optional arguments, otherwise, the following is a list of the optional arguments available and a full description of each optional argument is provided in Section 11.1. 11.1 Description of the Optional Arguments For each option, we give a summary line, a description of the optional argument and details of constraints. The summary line contains: • the keywords; • a parameter value, where the letters $a$, $i\text{ and }r$ denote options that take character, integer and real values respectively; • the default value, where the symbol $\epsilon $ is a generic notation for machine precision (see nag_machine_precision (X02AJC)), and $\mathit{Imax}$ represents the largest representable integer value (see nag_max_integer (X02BBC)). All options accept the value ‘DEFAULT’ in order to return single options to their default states. Keywords and character values are case insensitive, however they must be separated by at least one space. For nag_glopt_bnd_pso (e05sac) the maximum length of the argument cvalue used by nag_glopt_opt_get (e05zlc) is $15$. Advance Cognitive $r$ Default$\text{}=2.0$ The cognitive advance coefficient, ${C}_{s}$. When larger than the global advance coefficient, this will cause particles to be attracted toward their previous best positions. Setting $r=0.0$ will cause nag_glopt_bnd_pso (e05sac) to act predominantly as a local optimizer. Setting $r>2.0$ may cause the swarm to diverge, and is generally inadvisable. At least one of the global and cognitive coefficients must be nonzero. Advance Global $r$ Default$\text{}=2.0$ The global advance coefficient, ${C}_{g}$. When larger than the cognitive coefficient this will encourage convergence toward the best solution yet found. Values $r\in \left(0,1\right)$ will inhibit particles overshooting the optimum. Values $r\in \left[1,2\right)$ cause particles to fly over the optimum some of the time. Larger values can prohibit convergence. Setting $r=0.0$ will remove any attraction to the current optimum, effectively generating a Monte–Carlo multi-start optimization algorithm. At least one of the global and cognitive coefficients must be nonzero. Boundary $a$ Default$\text{}=\text{FLOATING}$ Determines the behaviour if particles leave the domain described by the box bounds. This only affects the general PSO algorithm, and will not pass down to any NAG local minimizers chosen. This option is only effective in those dimensions for which ${\mathbf{bl}}\left[i-1\right]\ne {\mathbf{bu}}\left[i-1\right]$, $i=1,2,\dots ,{\mathbf{ndim}}$. IGNORE The box bounds are ignored. The objective function is still evaluated at the new particle position. RESET The particle is re-initialized inside the domain. ${\stackrel{^}{\mathbf{x}}}_{j}$ and ${\stackrel{^}{f}}_{j}$ are not affected. FLOATING The particle position remains the same, however the objective function will not be evaluated at the next iteration. The particle will probably be advected back into the domain at the next advance due to attraction by the cognitive and global memory. HYPERSPHERICAL The box bounds are wrapped around an $\mathit{ndim}$-dimensional hypersphere. As such a particle leaving through a lower bound will immediately re-enter through the corresponding upper bound and vice versa. The standard distance between particles is also modified accordingly. FIXED The particle rests on the boundary, with the corresponding dimensional velocity set to $0.0$. Distance Scaling $a$ Default$\text{}=\mathrm{ON}$ Determines whether distances should be scaled by box widths. ON When a distance is calculated between $\mathbf{x}$ and $\mathbf{y}$, a scaled ${L}^{2}$ norm is used. $L2 x,y = ∑ i \| ui ≠ ℓi , i≤ndim xi - yi ui - ℓi 2 1 2 .$ OFF Distances are calculated as the standard ${L}^{2}$ norm without any rescaling. $L2 x,y = ∑ i=1 ndim xi - yi 2 1 2 .$ Distance Tolerance $r$ Default$\text{}={10}^{-4}$ This is the distance, $\mathit{dtol}$ between particles and the global optimum which must be reached for the particle to be considered converged, i.e., that any subsequent movement of such a particle cannot significantly alter the global optimum. Once achieved the particle is reset into the box bounds to continue searching. Constraint: $r>0.0$. Function Precision $r$ Default$\text{}={\epsilon }^{0.9}$ The argument defines ${\epsilon }_{r}$, which is intended to be a measure of the accuracy with which the problem function $F\left(\mathbf{x}\right)$ can be computed. If $r<\epsilon $ or $r\ge 1$, the default value is used. The value of ${\epsilon }_{r}$ should reflect the relative precision of $1+\left\|F\left(\mathbf{x}\right)\right\|$; i.e., ${\epsilon }_{r}$ acts as a relative precision when $\left\|F\right\|$ is large, and as an absolute precision when $\left\|F\right\|$ is small. For example, if $F\left(\mathbf{x}\right)$ is typically of order $1000$ and the first six significant digits are known to be correct, an appropriate value for ${\epsilon }_{r}$ would be ${10}^{-6}$. In contrast, if $F\left(\mathbf{x}\right)$ is typically of order ${10}^{-4}$ and the first six significant digits are known to be correct, an appropriate value for ${\epsilon }_{r}$ would be ${10}^{-10}$. The choice of ${\epsilon }_{r}$ can be quite complicated for badly scaled problems; see Chapter 8 of Gill et al. (1981) for a discussion of scaling techniques. The default value is appropriate for most simple functions that are computed with full accuracy. However when the accuracy of the computed function values is known to be significantly worse than full precision, the value of ${\epsilon }_{r}$ should be large enough so that no attempt will be made to distinguish between function values that differ by less than the error inherent in the calculation. Local Boundary Restriction $r$ Default$\text{}=0.5$ Contracts the box boundaries used by a box constrained local minimizer to, $\left[{\beta }_{l},{\beta }_{u}\right]$, containing the start point $x$, where $∂i = r × ui - ℓi βli = maxℓi, xi - ∂i2 βui = minui, xi + ∂i2 , i=1,…,ndim .$ Smaller values of $r$ thereby restrict the size of the domain exposed to the local minimizer, possibly reducing the amount of work done by the local minimizer. Constraint: $0.0\le r\le 1.0$. Local Interior Iterations ${i}_{1}$ Local Interior Major Iterations ${i}_{1}$ Local Exterior Iterations ${i}_{2}$ Local Exterior Major Iterations ${i}_{2}$ The maximum number of iterations or function evaluations the chosen local minimizer will perform inside (outside) the main loop if applicable. For the NAG minimizers these correspond to: \| \| \| \| \| \|-------------------------------\|------------------------------------------------------------------------------\|------------------------------------------------------------------------------\|------------------------------------------------------------------------------\| \| Minimizer \| Argument/option \| Default Interior \| Default Exterior \| \| nag_opt_simplex_easy (e04cbc) \| maxcal \| ${\mathbf{ndim}}+10$ \| $2×{\mathbf{ndim}}+15$ \| \| nag_opt_conj_grad (e04dgc) \| ${\mathbf{max_iter}}$ \| $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(30,3×{\mathbf{ndim}}\right)$ \| $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(50,5×{\mathbf{ndim}}\right)$ \| \| nag_opt_nlp (e04ucc) \| ${\mathbf{max_iter}}$ \| $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(10,2×{\mathbf{ndim}}\right)$ \| $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(30,3×{\mathbf{ndim}}\right)$ \| Unless set, these are functions of the arguments passed to nag_glopt_bnd_pso (e05sac). Setting $i=0$ will disable the local minimizer in the corresponding algorithmic region. For example, setting ${\mathbf{Local Interior Iterations}}=0$ and ${\mathbf{Local Exterior Iterations}}=30$ will cause the algorithm to perform no local minimizations inside the main loop of the algorithm, and a local minimization with upto $30$ iterations after the main loop has been exited. Constraint: ${i}_{1}\ge 0$, ${i}_{2}\ge 0$. Local Interior Tolerance ${r}_{1}$ Default$\text{}={10}^{-4}$ Local Exterior Tolerance ${r}_{2}$ Default$\text{}={10}^{-4}$ This is the tolerance provided to a local minimizer in the interior (exterior) of the main loop of the algorithm. Constraint: ${r}_{1}>0.0$, ${r}_{2}>0.0$. Local Interior Minor Iterations ${i}_{1}$ Local Exterior Minor Iterations ${i}_{2}$ Where applicable, the secondary number of iterations the chosen local minimizer will use inside (outside) the main loop. Currently the relevant default values are: \| \| \| \| \| \|----------------------\|------------------------------------------------------------------------------\|------------------------------------------------------------------------------\|------------------------------------------------------------------------------\| \| Minimizer \| Argument/option \| Default Interior \| Default Exterior \| \| nag_opt_nlp (e04ucc) \| ${\mathbf{minor_max_iter}}$ \| $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(10,2×{\mathbf{ndim}}\right)$ \| $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(30,3×{\mathbf{ndim}}\right)$ \| Constraint: ${i}_{1}\ge 0$, ${i}_{2}\ge 0$. Local Minimizer $a$ Default$\text{}=\mathrm{OFF}$ Allows for a choice of Chapter e04 functions to be used as a coupled, dedicated local minimizer. OFF No local minimization will be performed in either the INTERIOR or EXTERIOR sections of the algorithm. e04cbc Use nag_opt_simplex_easy (e04cbc) as the local minimizer. This does not require the calculation of derivatives. On a call to objfun during a local minimization, ${\mathbf{mode}}=5$. e04dgc Use nag_opt_conj_grad (e04dgc) as the local minimizer. Accurate derivatives must be provided, and will not be approximated internally. Additionally, each call to objfun during a local minimization will require either the objective to be evaluated alone, or both the objective and its gradient to be evaluated. Hence on a call to objfun, ${\mathbf{mode}}=5$ or $7$. e04ucc Use nag_opt_nlp (e04ucc) as the local minimizer. This operates such that any derivatives of the objective function that you cannot supply, will be approximated internally using finite differences. Either, the objective, objective gradient, or both may be requested during a local minimization, and as such on a call to objfun, ${\mathbf{mode}}=1$, $2$ or $5$. The box bounds forwarded to this function from nag_glopt_bnd_pso (e05sac) will have been acted upon by ${\mathbf{Local Boundary Restriction}}$. As such, the domain exposed may be greatly smaller than that provided to nag_glopt_bnd_pso (e05sac). Maximum Function Evaluations $i$ Default $=\mathit{Imax}$ The maximum number of evaluations of the objective function. When reached this will return NW_SOLUTION_NOT_GUARANTEED and ${\mathbf{inform}}=6$. Constraint: $i>0$. Maximum Iterations Completed $i$ Default$\text{}=1000×{\mathbf{ndim}}$ The maximum number of complete iterations that may be performed. Once exceeded nag_glopt_bnd_pso (e05sac) will exit with NW_SOLUTION_NOT_GUARANTEED and ${\mathbf{inform}}=5$. Unless set, this adapts to the parameters passed to nag_glopt_bnd_pso (e05sac). Constraint: $i\ge 1$. Maximum Iterations Static $i$ Default$\text{}=100$ The maximum number of iterations without any improvement to the current global optimum. If exceeded nag_glopt_bnd_pso (e05sac) will exit with NW_SOLUTION_NOT_GUARANTEED and ${\mathbf{inform}}=4$. This exit will be hindered by setting ${\mathbf{Maximum Iterations Static Particles}}$ to larger values. Constraint: $i\ge 1$. Maximum Iterations Static Particles $i$ Default$\text{}=0$ The minimum number of particles that must have converged to the current optimum before the function may exit due to ${\mathbf{Maximum Iterations Static}}$ with NW_SOLUTION_NOT_GUARANTEED and ${\mathbf{inform}}=4$. Constraint: $i\ge 0$. Maximum Particles Converged $i$ Default $=\mathit{Imax}$ The maximum number of particles that may converge to the current optimum. When achieved, nag_glopt_bnd_pso (e05sac) will exit with NW_SOLUTION_NOT_GUARANTEED and ${\mathbf{inform}}=3$. This exit will be hindered by setting ‘Repulsion’ options, as these cause the swarm to re-expand. Constraint: $i>0$. Maximum Particles Reset $i$ Default $=\mathit{Imax}$ The maximum number of particles that may be reset after converging to the current optimum. Once achieved no further particles will be reset, and any particles within ${\mathbf{Distance Tolerance}}$ of the global optimum will continue to evolve as normal. Constraint: $i>0$. Maximum Variable Velocity $r$ Default$\text{}=0.25$ Along any dimension $j$, the absolute velocity is bounded above by $\left\|{\mathbf{v}}_{j}\right\|\le r×\left({\mathbf{u}}_{j}-{\mathbf{\ell }}_{j}\right)={\mathbf{V}}_{\mathrm{max}}$. Very low values will greatly increase convergence time. There is no upper limit, although larger values will allow more particles to be advected out of the box bounds, and values greater than $4.0$ may cause significant and potentially unrecoverable swarm divergence. Constraint: $r>0.0$. Optimize $a$ Default$\text{}=\mathrm{MINIMIZE}$ Determines whether to maximize or minimize the objective function. MINIMIZE The objective function will be minimized. MAXIMIZE The objective function will be maximized. This is accomplished by minimizing the negative of the objective. Repeatability $a$ Default$\text{}=\mathrm{OFF}$ Allows for the same random number generator seed to be used for every call to nag_glopt_bnd_pso (e05sac). ${\mathbf{Repeatability}}=\mathrm{OFF}$ is recommended in general. OFF The internal generation of random numbers will be nonrepeatable. ON The same seed will be used. Repulsion Finalize $i$ Default $=\mathit{Imax}$ The number of iterations performed in a repulsive phase before re-contraction. This allows a re-diversified swarm to contract back toward the current optimum, allowing for a finer search of the near optimum space. Constraint: $i\ge 2$. Repulsion Initialize $i$ Default $=\mathit{Imax}$ The number of iterations without any improvement to the global optimum before the algorithm begins a repulsive phase. This phase allows the particle swarm to re-expand away from the current optimum, allowing more of the domain to be investigated. The repulsive phase is automatically ended if a superior optimum is found. Constraint: $i\ge 2$. Repulsion Particles $i$ Default$\text{}=0$ The number of particles required to have converged to the current optimum before any repulsive phase may be initialized. This will prevent repulsion before a satisfactory search of the near optimum area has been performed, which may happen for large dimensional problems. Constraint: $i\ge 0$. Swarm Standard Deviation $r$ Default$\text{}=0.1$ The target standard deviation of the particle distances from the current optimum. Once the standard deviation is below this level, nag_glopt_bnd_pso (e05sac) will exit with NW_SOLUTION_NOT_GUARANTEED and ${\mathbf{inform}}=2$. This criterion will be penalized by the use of ‘Repulsion’ options, as these cause the swarm to re-expand, increasing the standard deviation of the particle distances from the best point. Constraint: $r\ge 0.0$. Target Objective $a$ Default$\text{}=\mathrm{OFF}$ Target Objective Value $r$ Default$\text{}=0.0$ Activate or deactivate the use of a target value as a finalization criterion. If active, then once the supplied target value for the objective function is found (beyond the first iteration if ${\mathbf{Target Warning}}$ is active) nag_glopt_bnd_pso (e05sac) will exit with NE_NOERROR and ${\mathbf{inform}}=1$. Other than checking for feasibility only (${\mathbf{Optimize}}=\mathrm{CONSTRAINTS}$), this is the only finalization criterion that guarantees that the algorithm has been successful. If the target value was achieved at the initialization phase or first iteration and ${\mathbf{Target Warning}}$ is active, nag_glopt_bnd_pso (e05sac) will exit with NW_FAST_SOLUTION. This option may take any real value $r$, or the character ON/OFF as well as DEFAULT. If this option is queried using nag_glopt_opt_get (e05zlc), the current value of $r$ will be returned in rvalue, and cvalue will indicate whether this option is ON or OFF. The behaviour of the option is as follows: $r$ Once a point is found with an objective value within the ${\mathbf{Target Objective Tolerance}}$ of $r$, nag_glopt_bnd_pso (e05sac) will exit successfully with NE_NOERROR and ${\mathbf{inform}}=1$. OFF The current value of $r$ will remain stored, however it will not be used as a finalization criterion. ON The current value of $r$ stored will be used as a finalization criterion. DEFAULT The stored value of $r$ will be reset to its default value ($0.0$), and this finalization criterion will be deactivated. Target Objective Safeguard $r$ Default$\text{}=100.0\epsilon $ If the magnitude of ${\mathbf{Target Objective Value}}$ is below $r$, then the absolute error between the optimum and the target will be used. Otherwise the relative error will be used. If a value entered is below $100.0\epsilon $, the default value will be used. Target Objective Tolerance $r$ Default$\text{}=0.0$ The optional tolerance to a user-specified target value. Constraint: $r\ge 0.0$. Target Warning $a$ Default$\text{}=\mathrm{OFF}$ Activates or deactivates the error exit associated with the target value being achieved before entry into the main loop of the algorithm, NW_FAST_SOLUTION. OFF No error will be returned, and the function will exit normally. ON An error will be returned if the target objective is reached prematurely, and the function will exit with NW_FAST_SOLUTION. Verify Gradients $a$ Default$\text{}=\mathrm{ON}$ Adjusts the level of gradient checking performed when gradients are required. Gradient checks are only performed on the first call to the chosen local minimizer if it requires gradients. There is no guarantee that the gradient check will be correct, as the finite differences used in the gradient check are themselves subject to inaccuracies. OFF No gradient checking will be performed. ON A cheap gradient check will be performed on both the gradients corresponding to the objective through objfun. OBJECTIVE FULL A more expensive gradient check will be performed on the gradients corresponding to the objective objfun. Weight Decrease $a$ Default$\text{}=\mathrm{INTEREST}$ Determines how particle weights decrease. OFF Weights do not decrease. INTEREST Weights decrease through compound interest as ${w}_{\mathit{IT}+1}={w}_{\mathit{IT}}\left(1-{W}_{\mathit{val}}\right)$, where ${W}_{\mathit{val}}$ is the ${\mathbf{Weight Value}}$ and $\mathit{IT}$ is the current number of iterations. LINEAR Weights decrease linearly following ${w}_{\mathit{IT}+1}={w}_{\mathit{IT}}-\mathit{IT}×\left({W}_{\mathit{max}}-{W}_{\mathit{min}}\right)/{\mathit{IT}}_{\mathit{max}}$, where $\mathit{IT}$ is the iteration number and ${\mathit{IT}}_{\mathit{max}}$ is the maximum number of iterations as set by ${\mathbf{Maximum Iterations Completed}}$. Weight Initial $r$ Default$\text{}={W}_{\mathit{max}}$ The initial value of any particle's inertial weight, ${W}_{\mathit{ini}}$, or the minimum possible initial value if initial weights are randomized. When set, this will override ${\mathbf{Weight Initialize}}=\mathrm{RANDOMIZED}$ or $\mathrm{MAXIMUM}$, and as such these must be set afterwards if so desired. Constraint: ${W}_{\mathit{min}}\le r\le {W}_{\mathit{max}}$. Weight Initialize $a$ Default$\text{}=\mathrm{MAXIMUM}$ Determines how the initial weights are distributed. INITIAL All weights are initialized at the initial weight, ${W}_{\mathit{ini}}$, if set. If ${\mathbf{Weight Initial}}$ has not been set, this will be the maximum weight, ${W}_{\mathit{max}}$. MAXIMUM All weights are initialized at the maximum weight, ${W}_{\mathit{max}}$. RANDOMIZED Weights are uniformly distributed in $\left({W}_{\mathit{min}},{W}_{\mathit{max}}\right)$ or $\left({W}_{\mathit{ini}},{W}_{\mathit{max}}\right)$ if ${\mathbf{Weight Initial}}$ has been set. Weight Maximum $r$ Default$\text{}=1.0$ The maximum particle weight, ${W}_{\mathit{max}}$. Constraint: $1.0\ge r\ge {W}_{\mathit{min}}$ (If ${W}_{\mathit{ini}}$ has been set then $1.0\ge r\ge {W}_{\mathit{ini}}$.) Weight Minimum $r$ Default$\text{}=0.1$ The minimum achievable weight of any particle, ${W}_{\mathit{min}}$. Once achieved, no further weight reduction is possible. Constraint: $0.0\le r\le {W}_{\mathit{max}}$ (If ${W}_{\mathit{ini}}$ has been set then $0.0\le r\le {W}_{\mathit{ini}}$.) Weight Reset $a$ Default$\text{}=\mathrm{MAXIMUM}$ Determines how particle weights are re-initialized. INITIAL Weights are re-initialized at the initial weight if set. If ${\mathbf{Weight Initial}}$ has not been set, this will be the maximum weight. MAXIMUM Weights are re-initialized at the maximum weight. RANDOMIZED Weights are uniformly distributed in $\left({W}_{\mathit{min}},{W}_{\mathit{max}}\right)$ or $\left({W}_{\mathit{ini}},{W}_{\mathit{max}}\right)$ if ${\mathbf{Weight Initial}}$ has been set. Weight Value $r$ Default$\text{}=0.01$ The constant ${W}_{\mathit{val}}$ used with ${\mathbf{Weight Decrease}}=\mathrm{INTEREST}$. Constraint: $0.0\le r\le \frac{1}{3}$.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 460, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.7673513889312744, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/252629/generalization-of-bernoullis-inequality	# Generalization of Bernoulli's Inequality. When you're first taking an Introduction to Proofs class, you eventually learn about induction. One of the first induction proofs I did was to prove for $n \geq 1$ that $(1+x)^n \geq 1+nx$. You first prove the base case, $n=1$, which yields $(1+x)^1 \geq 1+1x$, which is true. Then you assume that the inequality holds for for the arbitrary $n$ case. Then you try to prove that the inequality holds for $n+1$, i.e. $(1+x)^(n+1) \geq 1+nx$ as well. Using our assumption that $(1+x)^n \geq 1+nx$, we can multiply both sides by $1+x$, which yields $(1+x)(1+x)^n \geq (1+nx)(1+x).$ Since $1+nx^2+nx +x \geq 1+(n+1)x$, the inequality is proven. Now what I’m trying to do is prove a generalization of Bernoulli’s inequality, i.e. the following. “Find the least possible $\alpha < 0$ so that $\forall x \geq \alpha$ and $n \in \mathbb{N}$, that $(1+x)^n \geq 1+nx$. Prove the inequality as well as prove that your $\alpha$ is best possible.” What I’m trying to figure out is what is more special about this question rather than the typical inductive Bernoulli’s inequality proof? What is the purpose of introducing this alpha restricted to less than zero? There must be a reason for it. I read a few different textbooks and they all say that it (it referring to Bernoulli's inequality) works for $x > -1$, so I’m assuming that the alpha would also be -1 because then for any $x \in \mathbb{R}$ such that $x \geq \alpha$, the normal Bernoulli’s inequality holds. Am I completely going off the deep end? Any input would be very much appreciated. - Does the inequality hold when $x=-1$ ? – Ju'x Dec 6 '12 at 22:55 ## 2 Answers We find the answer, using basic calculus. The magic $\alpha$ is not $-1$. Consider the function $f(x)=(1+x)^n -(1+nx)$ for $n \gt 1$. Then $$f'(x)=n((1+x)^{n-1}-1).$$ For $n\gt 1$ and even, the derivative vanishes only at $x=0$. For $n\gt 1$ and odd, the derivative vanishes at $x=-2$ and $x=0$. The only interesting case is $n$ odd. By the usual increasing/dcreasing analysis, we find that $f(x)\ge 0$ when $x\ge -2$. For any $w\lt -2$, by taking a suitably large $n$, we can make $f(w)\lt 0$. The part about $f(x)\ge 0$ for $x\ge -2$ can be pushed through by induction. - So the critical points of f(x) occur at x = 0 for all n and x = -2 for n odd. So x = -2 is not a critical point for n even; therefore, how can x = -2 be a minimum for all n? -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 37, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9576002359390259, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/87104/shouldnt-the-coaddition-on-mathbb-zx-be-a-morphism-of-rings	# Shouldn't the coaddition on $\mathbb Z[x]$ be a morphism of rings? $\operatorname{Spec} \mathbb Z[x]$ is a group scheme. Then there is a morphism of schemes: $\operatorname{Spec} \mathbb Z[x]\times\operatorname{Spec} \mathbb Z[x]\rightarrow\operatorname{Spec} \mathbb Z[x]$ induced by the coaddition $a:\mathbb Z[x]\rightarrow \mathbb Z[x] \otimes\mathbb Z[x]$ mapping $x$ to $1\otimes x+x\otimes 1$. Now, the coaddition is not a morphism of rings, since it does not respect multiplication: $a(x^2)=(1\otimes x^2+x^2\otimes 1)\neq(1\otimes x + x\otimes 1)^2=a(x)\cdot a(x)$. But $a$ should be a morphism of rings, since the addition on $\operatorname{Spec} \mathbb Z[x]$ is a morphism of affine schemes. - ## 1 Answer I think you are being confused by the notation. There is a canonical isomorphism $\mathbb{Z}[t] \otimes_{\mathbb{Z}} \mathbb{Z}[t] \cong \mathbb{Z}[x, y]$ such that $t \otimes 1 \mapsto x$ and $1 \otimes t \mapsto y$. In that light, coaddition is the unique ring homomorphism $\alpha : \mathbb{Z}[t] \to \mathbb{Z}[x, y]$ such that $t \mapsto x + y$, since $\mathbb{Z}[t]$ is freely generated by $t$. In particular, $\alpha(t^2) = x^2 + 2 x y + y^2 = \alpha(t)^2$. - Thank you very much! – Felix Wellen Dec 1 '11 at 15:04	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 17, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9572391510009766, "perplexity_flag": "head"}
http://mathhelpforum.com/calculus/62706-integrate.html	# Thread: 1. ## Integrate I have to integrate K(1-t)^4 dt between 0 and y I get: K(1/5)(1-t)^5 the book has k(1-t)^5/(-5) i dont know how they got -5 then when i evaluate at 0 and y i get: (k/5)[(1-y)^5 -1] and at this spot the book has (k/5)[1-(1-y)^5] thanks 2. You're integrating $k(1-t)^4$, right? Be sure to account for the coefficient of t, which is negative one. That's the only difference between your answer and the correct one. And whenever in doubt, just try differentiating the answer you determined along with the answer given by the textbook. You'll probably notice the difference. 3. Yes, that is what I am integrating. I guess I was just thinking that I could raise the power and divide by the new power. I just ignored what was in the parenthesis. Thanks.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 1, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9671754240989685, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/31143/counting-degrees-of-freedom-of-gauge-bosons	# Counting degrees of freedom of gauge bosons Gauge bosons are represented by $A_{\mu}$, where $\mu = 0,1,2,3$. So in general there are 4 degrees of freedom. But in reality, a photon (gauge boson) has two degrees of freedom (two polarization states). So, when someone asks about on-shell and off-shell degrees of freedom, I thought they are 2 and 4. But I read that the off-shell d.of. are 3. And my question is how to see this? - ## 2 Answers On-shell dof: Consider the constraints of "Equation of Motion" plus "Gauge Condition"(Lorentz gauge, Coulomb gauge,...), #dof=4-2. Off-Shell dof: Consider the constraints of "Gauge Condition" (Lorentz gauge, Coulomb gauge,...) only, #dof=4-1. - One way to see this is the fact that if you write out the Lagragian in terms of the $A^0$ and $A^i$ explicitly, you will find there is no time derivatives for the $A^0$. The only candidate term $F^{00} = \partial^0 A^0 - \partial^0 A^0$ vanishes. This means we shouldn't count $A^0$ as a degree of freedom, and in fact, we should integrate it out. Now the resulting Lagrangian explicitly only has 3 DOF. If you now take this Lagrangian and vary it you will get the EOM which puts one constraint on the 3 DOF which knocks you down to the 2 DOF on-shell. - No $\partial_0 A_0$ term may be interpreted that $A_0$ is not a dynamical dof rather than dof. Three dimensional gravity for example, the gravity has a dof which is not dynamical. – Craig Thone Jul 2 '12 at 16:12 Craig - sorry, I don't understand your comment. Did I make an error in my statement? – DJBunk Jul 2 '12 at 17:25 No. I just want to emphasize $A_0$ is not a dynamical degree of freedom. In Lorentz gauge, I am not sure whether $A_0$ is a degree of freedom. – Craig Thone Jul 3 '12 at 1:42	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 11, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9115456938743591, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/15836/oneupsmanship-and-publishing-etiquette/15837	## Oneupsmanship and Publishing Etiquette ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Apologies for the anonymous (and, in hindsight, mildly longish) question. I have an etiquette question on publishing an improvement of another author's work, but I figured a thread on the matter to address generalities might be nice to have as well. Another paper in my field reports to have solved an open problem by finding a necessary and sufficient condition for deciding if an object $X$ has property $P$. As it turns out, there are cases in which this theorem reduces to a tautology, and not in the good "every theorem is a tautology" way -- it reduces to the statement $X$ has property $P$ if and only if it has property $P$. This happens frequently enough that it seems at the very least a little disingenuous to have called it a necessary and sufficient condition. In writing up a stronger version of the result, I have to decide how to address the previous work, leading to some fairly general questions about publishing etiquette to which the MO audience at large might have some insightful responses: 1) Usually when improving on someone's work, there are a variety of diplomatic words to use -- "We generalize soandso's work" being a standard one, providing both a nod to their efforts while also emphasizing your improvement. Or, if a result is plain wrong, no amount of diplomacy will salvage their result (though could possibly soften the blow), so you can just provide the counter-example and move on. In my specific example, it feels a little harder to address -- their result isn't incorrect, just not as complete as they seem to take credit for. I know I've had both colleagues who positively light up with joy at the thought of besting someone else's paper, and those who I think would likely contact the original author to give them a chance to contribute to the discussion in the paper. Ignoring it altogether doesn't seem like an option given that they imply that the problem is now closed. Let's take the point of view that I am a recent graduate and thus the prestige of improving a published result presumably conveys a non-trivial benefit to me -- any heuristics on how to make decisions of this form? 2) On a logistical level, their techniques are completely different from ours. What's the extent to which we should re-introduce all of their (particularly burdensome, in my opinion) notation and terminology in order to prove that our result is an objective improvement? It's a fairly simple matter to give an example which our result provides an answer to that theirs does not, but it's rather technical (maybe longer than the proof of the result itself) to show that our result handles every case that theirs does (non-tautologically). In addition to the extra clutter in the paper, the prolonged emphasis on provably improving their work seems to work against the diplomacy hoped for in the previous question. - 2 One question is, how important this diplomacy is to you? For example, you are worried about spending time in the paper to show that you provably improve their work. I don't think that mathematical community in general will find this "undiplomatic". They may or may not find your improvements interesting, but just because you prove that you improved somebody else's work will not bother anyone (again, assuming the improvements are actually interesting and you are not dwelling on them just to show how much smarter you are). The only person who could be bothered by this is the original author. – sheldon-cooper Feb 19 2010 at 20:21 So the question is, do you really care about the feelings of that one person enough to modify your paper significantly and maybe even withhold some interesting results? – sheldon-cooper Feb 19 2010 at 20:23 3 @sheldon-cooper: You make a valid point, but I can certainly imagine contexts where this would antagonize more than just one person. – François G. Dorais♦ Feb 19 2010 at 20:29 2 I'm not going to submit an "answer," per se, given the response to Andrew's below. But I personally have grown more and more disillusioned over the years with the lack "humanness" (for lack of a better term) in how we do things. I go out of my way to at the very least acknowledge people somewhere. A recent paper of mine was the result of a pub discussion with 7 or 8 people - I acknowledged them all (and the paper won an award, so it obviously wasn't considered bad form). – Ian Durham Feb 20 2010 at 3:30 1 @ID: I don't quite take your point. It is always good form to acknowledge help given by others, however indirect. (And it doesn't hurt you to do so.) So having given acknowledgments in your situation seems unquestionably appropriate. But it doesn't make sense to acknowledge people with whom you have not had any relevant interactions. OP doesn't need to acknowledge the author whose result he is improving: he needs to cite that person's work. And I don't see anything inhumane about improving on another person's work. That is most of what we do, after all. – Pete L. Clark Feb 20 2010 at 5:23 show 1 more comment ## 6 Answers I am not sure I see what exactly is the ethical issue. You need to be very honest and very clear - you are writing a paper not for yourself but for other people to read and understand, so that should be your first priority. Write something like this. "We prove property P. This result is strongly related to the result in [..] which proves property P' and extends it in the following sense ... Let us note that in all cases our property P is at least as strong as P' (see Thm~?), while in some special cases our property P favorably compares to property P'. Below we give some examples which emphasize the connection." If the matter is easy - work out carefully some examples right in the introduction. If this is more delicate as it seems to be the case, make a new section where you can work out several examples so the reader can see that P' is saying in some cases P' implies P', while your P is saying something stronger. If you stick to math, no one I can think of will get upset over this, and there is no need to make short broad characterizations in the introduction which might upset somebody. At the same time you "besting somebody" will come across in exactly the way you desire. The only downside of this approach I can see is that the paper might get a little longer due to this examples section. But so what - the reader will appreciate the clarity, and an extra couple of pages cannot possibly affect the publication chances. - ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. Igor Pak gave good advice on what to write in your paper. Additionally, I would advise you to contact the author of the original paper and tell them about your work before it is published. This can be a good opportunity to get to know someone who shares common interests with you and it opens possibility for future joint work. I'll bet if you talk to the other author the two of you together are going to have a new idea which is even better than his or yours alone. And another thing. If you write to the author and start talking to him, you will have a chance to find out about the background of his work and you won't have to imagine useless things (such as "this other guy is so stupid"). Perhaps he knows that his theorem sometimes reduces to a tautology and will readily agree with you. Perhaps he thinks it only happens occasionally and will be grateful to you for explaining that this is not so. In any case, just talk to the other guy. - It's hard to judge without details, but this is perhaps a question of metatheory. (This is what item 2 sounds like to me, but this is just an impression.) It is not a requirement in mathematical papers to carefully explain aspects of the metatheory, but it is not prohibited to do so. In your case, it seems that their result is trivial in your metatheory while yours isn't. By clarifying your metatheory, you can carefully explain that in your paper. In many cases, you really don't need to say much about their result since the issue is likely to be with their assumptions or methodology, not with their conclusions. Such a critique should sound much tamer than an attack on their result. Clarifying your metatheory will also separate your result from theirs in other ways. It is possible that their metatheory (presumably not carefully explained in their paper) does not see their result as trivial. Maybe it even sees your result as trivial! A reader that shares their point of view and not yours will immediately see the difference when reading your paper and will be able to appropriately interpret your work regardless of their personal stance. This could avoid further confusion of the same kind. PS: Being a logician, I prefer the term "metatheory" for what many other mathematicians would call "context," "framework," "perspective," "methodology," or whatever. Don't read too much technicality into the term, the metatheory is always informal. (Only logicians occasionally need to make the metatheory formal in order to formally prove metatheorems.) If this is indeed the kind of scenario you're looking at, it should be relatively clear to you exactly what and how much you need to clarify about your own metatheory. - I largely agree with Igor Pak's response, and I think he has the right perspective on the matter. Two comments: [X is the author of the first paper, Y is the OP!] 1) I'm not sure it's critical to show that your result completely subsumes the result of X if that verification is going to, say, double the length of your paper and require you to introduce a lot of terminology and notation that is far away from your approach to the problem. It is not clear what value you are adding to the community by doing this: is your proof going to be any more pleasant to read or insightful than X's? On the other hand, I find it slightly odd that it should be as hard or harder to demonstrate that your results imply X's than it is to prove your (in fact more general) result. I'm having trouble thinking of an example of this phenomenon from my own experience. Is it possible that there's more insight to gained here, and perhaps a(n obvious!) common generalization of your two approaches? 2) A good rule about describing your own work -- borrowed from the creative writing community -- is: show, don't tell. Specifically, you should seek to minimize the number of times in which you tell the reader how she should feel about your work. Rather, you want to present the mathematical information which brings the reader to this conclusion. In this case, I would recommend putting a lot of effort into the writing of the example(s) of the novelty of your approach. The more clear and detailed your writing is, the smaller your sales pitch needs to be. Some direct comparison may be necessary, but if you stick to comparing one mathematical statement to another, then "stronger than" has an objective meaning. It seems that there is little in such a practice that could reasonably cause offense. Unfortunately 1) and 2) are somewhat at cross-purposes to each other. Regarding 1), in lieu of writing out all the gory details it is tempting to write something like "It can be shown that in every case where X's theorem applies, so does mine". If this takes pages and pages to show, then the author might well not see / believe it at first, and it could be annoying to have to work for hours to verify someone's claim that they have trumped you. Maybe the best solution -- although not the easiest -- is for you to carefully write up the implication that your result implies X's result in a separate document, which you submit along with the paper itself and post a copy on your webpage. This also places some of the judgment of whether this gory extra part should be included in the paper in the hands of the referee and the editor, which is perhaps as it should be. - This suggestion may only work in certain situations, but is it possible to split your paper into two? One could be for your new results in the most elegant form, and the other could be primarily for comparing your results to the previous work. You can sent the former to a good journal and throw the latter in a lesser journal or just post on the arXiv. - See if the other author wants to be a co-author. This is playing nice, and also, it eliminates the chance that they will be a referee. - 21 I'd advise against that. First, you will be sharing credit with other people for something you already did all by yourself. More importantly, I suspect your contribution may go completely unnoticed in such case. If a person X is already known for working on this problem, them most of the credit will go to them. It will be just another paper in a series "X with this student", "X with another student", "X with third student", and most of the credit goes to X. – sheldon-cooper Feb 19 2010 at 19:53 4 I see by my draining reputation that this one is not too popular. But consider the original questioner has pointed out that his result is not completely new - he is simply patching up the lack of generality of previous work. Drawing lines in the sand about credit for that seems a bit of a stretch. – Andrew Mullhaupt Feb 19 2010 at 22:04 6 Downvotes on community wiki answers won't actually drain your reputation. – Noah Snyder Feb 19 2010 at 23:27 23 "On appropriate occasions, it may be desirable to offer or accept joint authorship when independent researchers find that they have produced identical results. All the authors listed for a paper, however, must have made a significant contribution to its content, and all who have made such a contribution must be offered the opportunity to be listed as an author." That is from ams.org/secretary/ethics.html and should be taken seriously. If the other person's contribution is in an already published paper, having his or her name on the new paper looks to me to be a violation. – Bill Johnson Feb 20 2010 at 0:08 10 In that case, "See if the other author wants to collaborate" seems to be a better way to phrase your suggestion. – stankewicz Feb 20 2010 at 17:05 show 5 more comments	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 5, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9707450866699219, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/279838/is-the-point-at-0-0-of-the-graph-z-x3-y3-considered-a-saddle-point	# Is the point at $(0,0)$ of the graph $z=x^3 + y^3$ considered a saddle point? Is the point at $(0,0)$ of the graph $z=x^3 + y^3$ considered a saddle point? I was given that function, and I used the second-order derivative test only to find that the $Hf(0,0) = 0$, with the critical point being at 0,0. Is anything that is not a local minimum or a local maximum considered a saddle point? What the graph looks like - ## 1 Answer 1. In your case, the Hessian is a null matrix. This does not guarantee that the point is a saddle point. Although I cannot come up with an example but I think you can have a null matrix as hessian at the minima. But don't quote me for that. 2. Such a point, AFAIK, is given the name "Degenerate Critical Point". Whether it is considered saddle or not, I am not sure. 3. If I am standing at (0,0,0) and you give me an $\epsilon$, I can step $(\epsilon,0)$ and make the function positive or $(-\epsilon,0)$ and make the function negative. So, no matter how small epsilon you give me, I can always move to points that increase the function value and reduce it. This makes the (0,0,0) a saddle point. 4. Does this question help : Classifying singular points as local min, max or saddle points? - Standard example of a local min with vanishing Hessian is $x^4 + y^4$. The common definition of saddle point is that it is neither a local min nor local max, and this is defined without any reference to derivatives or Hessian. – Erick Wong Jan 16 at 5:31 @ErickWong, Fair enough. But given that the gradient and hessian is both 0 (in their own dimensions) at that point, how do you know that it is not maxima or minima? (One can say so by observation or by computing the function at other points but that's not robust). Is there any other way I am missing? – Inquest Jan 16 at 5:34 @Inquest You're not missing anything. Things get harder when the second derivative vanishes, except in simple cases such as these. This homework problem is meant to point out a limitation of the method. – Michael E2 Jan 17 at 12:43	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 9, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9321974515914917, "perplexity_flag": "head"}
http://mathoverflow.net/questions/21031/ultrafilters-vs-well-orderings/21034	## Ultrafilters vs Well-orderings ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) This question was actually asked by John Stillwell in a comment to an answer to this question. I thought I would advertise it as a separate question since no one has yet answered and I am also curious about it. Question: Is the existence of a non-principal ultra-filter on $\omega$ a weaker assumption than the existence of a well-ordering of $\mathbb{R}$? - ## 2 Answers It is consistent that there exists a non-principal ultrafilter over $\omega$ while $\mathbb{R}$ is not well-ordered. To see this, suppose that the partition relation $\omega \to (\omega)^{\omega}$ holds in $L(\mathbb{R})$. Then forcing with $\mathbb{P}= [\omega]^{\omega}$ adjoins a selective ultrafilter $\mathcal{U}$ over $\omega$ and $\mathcal{P}(\omega)$ cannot be well-ordered in $L(\mathbb{R})[\mathcal{U}]$. (See Eisworth's paper: Selective ultrafilters and $\omega \to (\omega)^{\omega}$.) Thus $L(\mathbb{R})[\mathcal{U}]$ is a model of $ZF$ which contains the nonprincipal ultrafilter $\mathcal{U}$ and yet $\mathbb{R}$ cannot be well-ordered in $L(\mathbb{R})[\mathcal{U}]$. An update: it is perhaps also interesting to note that in $L(\mathbb{R})[\mathcal{U}]$, the ultraproduct $\prod_{\mathcal{U}} \bar{\mathbb{F}}_{p}$ of the algebraic closures of the fields of prime order $p$ is not isomorphic to $\mathbb{C}$. - How about relative consistency rather than direct implication? – Harry Gindi Apr 11 2010 at 21:55 Thanks a lot for the answer, Simon. – John Stillwell Apr 12 2010 at 10:52 1 It was a very interesting question. While it is well-known that the existence of non-principal ultrafilters is in general strictly weaker than the axiom of choice, this special case is probably the one which would interest most mathematicians. Plus I now understand a conversation that I recently had with a much more knowledgeable set theorist ... – Simon Thomas Apr 12 2010 at 12:45 ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. Historically it was proved that the ultrafilter lemma is independent from the axiom of choice by showing that there is a model in which there is an infinite Dedekind-finite set of real numbers, but every filter can be extended to an ultrafilter. Where an infinite Dedekind-finite set is an infinite set which does not have a countably infinite subset. The existence of infinite Dedekind-finite sets negates not only the axiom of choice, but also the [much] weaker axiom of countable choice. These sets cannot be well-ordered, and since the real numbers have such subset they cannot be well-ordered themselves in such model. The proof was given by Halpern and Levy in 1964. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 21, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9586628079414368, "perplexity_flag": "head"}
http://www.scholarpedia.org/article/Numerical_analysis	# Numerical analysis From Scholarpedia Kendall E. Atkinson (2007), Scholarpedia, 2(8):3163. Curator and Contributors 1.00 - Kendall E. Atkinson Numerical analysis is the area of mathematics and computer science that creates, analyzes, and implements algorithms for solving numerically the problems of continuous mathematics. Such problems originate generally from real-world applications of algebra, geometry, and calculus, and they involve variables which vary continuously. These problems occur throughout the natural sciences, social sciences, medicine, engineering, and business. Beginning in the 1940's, the growth in power and availability of digital computers has led to an increasing use of realistic mathematical models in science, medicine, engineering, and business; and numerical analysis of increasing sophistication has been needed to solve these more accurate and complex mathematical models of the world. The formal academic area of numerical analysis varies from highly theoretical mathematical studies to computer science issues involving the effects of computer hardware and software on the implementation of specific algorithms. ## Areas of numerical analysis A rough categorization of the principal areas of numerical analysis is given below, keeping in mind that there is often a great deal of overlap between the listed areas. In addition, the numerical solution of many mathematical problems involves some combination of some of these areas, possibly all of them. There are also a few problems which do not fit neatly into any of the following categories. ### Systems of linear and nonlinear equations • Numerical solution of systems of linear equations. This refers to solving for $$x$$ in the equation $$Ax=b$$ with given matrix $$A$$ and column vector $$b\ .$$ The most important case has $$A$$ a square matrix. There are both direct methods of solution (requiring only a finite number of arithmetic operations) and iterative methods (giving increased accuracy with each new iteration). This topic also includes the matrix eigenvalue problem for a square matrix $$A\ ,$$ solving for $$x$$ and $$\lambda$$ in the equation $$Ax=\lambda x\ .$$ • Numerical solution of systems of nonlinear equations. This refers to rootfinding problems which are usually written as $$f(x)=0$$ with $$x$$ a vector with $$n$$ components and $$f(x)$$ a vector with $$m$$ components. The most important case has $$n=m\ .$$ • Optimization. This refers to minimizing or maximizing a real-valued function $$f(x)\ .$$ The permitted values for $$x=(x_{1},\dots,x_{n})$$ can be either constrained or unconstrained. The 'linear programming problem' is a well-known and important case; $$f(x)$$ is linear, and there are linear equality and/or inequality constraints on $$x\ .$$ ### Approximation theory Use computable functions $$p(x)$$ to approximate the values of functions $$f(x)$$ that are not easily computable or use approximations to simplify dealing with such functions. The most popular types of computable functions $$p(x)$$ are polynomials, rational functions, and piecewise versions of them, for example spline functions. Trigonometric polynomials are also a very useful choice. • Best approximations. Here a given function $$f(x)$$ is approximated within a given finite-dimensional family of computable functions. The quality of the approximation is expressed by a functional, usually the maximum absolute value of the approximation error or an integral involving the error. Least squares approximations and minimax approximations are the most popular choices. • Interpolation. A computable function $$p(x)$$ is to be chosen to agree with a given $$f(x)$$ at a given finite set of points $$x\ .$$ The study of determining and analyzing such interpolation functions is still an active area of research, particularly when $$p(x)$$ is a multivariate polynomial. • Fourier series. A function $$f(x)$$ is decomposed into orthogonal components based on a given orthogonal basis $$\left\{ \varphi _{1},\varphi_{2},\dots\right\} \ ,$$ and then $$f(x)$$ is approximated by using only the largest of such components. The convergence of Fourier series is a classical area of mathematics, and it is very important in many fields of application. The development of the Fast Fourier Transform in 1965 spawned a rapid progress in digital technology. In the 1990s wavelets became an important tool in this area. • Numerical integration and differentiation. Most integrals cannot be evaluated directly in terms of elementary functions, and instead they must be approximated numerically. Most functions can be differentiated analytically, but there is still a need for numerical differentiation, both to approximate the derivative of numerical data and to obtain approximations for discretizing differential equations. ### Numerical solution of differential and integral equations These equations occur widely as mathematical models for the physical world, and their numerical solution is important throughout the sciences and engineering. • Ordinary differential equations. This refers to systems of differential equations in which the unknown solutions are functions of only a single variable. The most important cases are initial value problems and boundary value problems, and these are the subjects of a number of textbooks. Of more recent interest are 'differential-algebraic equations', which are mixed systems of algebraic equations and ordinary differential equations. Also of recent interest are 'delay differential equations', in which the rate of change of the solution depends on the state of the system at past times. • Partial differential equations. This refers to differential equations in which the unknown solution is a function of more than one variable. These equations occur in almost all areas of engineering, and many basic models of the physical sciences are given as partial differential equations. Thus such equations are a very important topic for numerical analysis. For example, the Navier-Stokes equations are the main theoretical model for the motion of fluids, and the very large area of 'computational fluid mechanics' is concerned with solving numerically these and other equations of fluid dynamics • Integral equations. These equations involve the integration of an unknown function, and linear equations probably occur most frequently. Some mathematical models lead directly to integral equations; for example, the radiosity equation is a model for radiative heat transfer. Another important source of such problems is the reformulation of partial differential equations, and such reformulations are often called 'boundary integral equations'. ## Some common viewpoints and concerns in numerical analysis Most numerical analysts specialize in small sub-areas of the areas listed above, but they share some common concerns and perspectives. These include the following. • The mathematical aspects of numerical analysis make use of the language and results of linear algebra, real analysis, and functional analysis. • If you cannot solve a problem directly, then replace it with a 'nearby problem' which can be solved more easily. This is an important perspective which cuts across all types of mathematical problems. For example, to evaluate a definite integral numerically, begin by approximating its integrand using polynomial interpolation or a Taylor series, and then integrate exactly the polynomial approximation. • All numerical calculations are carried out using finite precision arithmetic, usually in a framework of floating-point representation of numbers. What are the effects of using such finite precision computer arithmetic? How are arithmetic calculations to be carried out? Using finite precision arithmetic will affect how we compute solutions to all types of problems, and it forces us to think about the limits on the accuracy with which a problem can be solved numerically. Even when solving finite systems of linear equations by direct numerical methods, infinite precision arithmetic is needed in order to find a particular exact solution. • There is concern with 'stability', a concept referring to the sensitivity of the solution of a given problem to small changes in the data or the given parameters of the problem. There are two aspects to this. First, how sensitive is the original problem to small changes in the data of the problem? Problems that are very sensitive are often referred to as 'ill-conditioned' or 'ill-posed', depending on the degree of sensitivity. Second, the numerical method should not introduce additional sensitivity that is not present in the original mathematical problem being solved. In developing a numerical method to solve a problem, the method should be no more sensitive to changes in the data than is true of the original mathematical problem. • There is a fundamental concern with error, its size, and its analytic form. When approximating a problem, a numerical analyst would want to understand the behaviour of the error in the computed solution. Understanding the form of the error may allow one to minimize or estimate it. A 'forward error analysis' looks at the effect of errors made in the solution process. This is the standard way of understanding the consequences of the approximation errors that occur in setting up a numerical method of solution, e.g. in numerical integration and in the numerical solution of differential and integral equations. A 'backward error analysis' works backward in a numerical algorithm, showing that the approximating numerical solution is the exact solution to a perturbed version of the original mathematical problem. In this way the stability of the original problem can be used to explain possible difficulties in a numerical method. Backward error analysis has been especially important in understanding the behaviour of numerical methods for solving linear algebra problems. • In order to develop efficient means of calculating a numerical solution, it is important to understand the characteristics of the computer being used. For example, the structure of the computer memory is often very important in devising efficient algorithms for large linear algebra problems. Also, parallel computer architectures lead to efficient algorithms only if the algorithm is designed to take advantage of the parallelism. • Numerical analysts are generally interested in measuring the efficiency of algorithms. What is the cost of a particular algorithm? For example, the use of Gaussian elimination to solve a linear system $$Ax=b$$ containing $$n$$ equations will require approximately $$2n^{3}/3$$ arithmetic operations. How does this compare with other numerical methods for solving this problem? This topic is a part of the larger area of 'computational complexity'. • Use information gained in solving a problem to improve the solution procedure for that problem. Often we do not fully understand the characteristics of a problem, especially very complicated and large ones. Such a solution process is sometimes referred to as being an 'adaptive procedure', and it can also be viewed as a feedback process. ## Development of numerical methods Numerical analysts and applied mathematicians have a variety of tools which they use in developing numerical methods for solving mathematical problems. An important perspective, one mentioned earlier, which cuts across all types of mathematical problems is that of replacing the given problem with a 'nearby problem' which can be solved more easily. There are other perspectives which vary with the type of mathematical problem being solved. ### Numerical solution of systems of linear equations Linear systems arise in many of the problems of numerical analysis, a reflection of the approximation of mathematical problems using linearization. This leads to diversity in the characteristics of linear systems, and for this reason there are numerous approaches to solving linear systems. As an example, numerical methods for solving partial differential equations often lead to very large 'sparse' linear systems in which most coefficients are zero. Solving such sparse systems requires methods that are quite different from those used to solve more moderate sized 'dense' linear systems in which most coefficients are non-zero. There are 'direct methods' and 'iterative methods' for solving all types of linear systems, and the method of choice depends on the characteristics of both the linear system and on the computer hardware being used. For example, some sparse systems can be solved by direct methods, whereas others are better solved using iteration. With iteration methods, the linear system is sometimes transformed to an equivalent form that is more amenable to being solved by iteration; this is often called 'pre-conditioning' of the linear system. With the matrix eigenvalue problem $$Ax=\lambda x\ ,$$ it is standard to transform the matrix $$A$$ to a simpler form, one for which the eigenvalue problem can be solved more easily and/or cheaply. A favorite choice are 'orthogonal transformations' because they are a simple and stable way to convert the given matrix $$A\ .$$ Orthogonal transformations are also very useful in transforming other problems in numerical linear algebra. Of particular importance in this regard is the least squares solution of over-determined linear systems. The linear programming problem was solved principally by the 'simplex method' until new approaches were developed in the 1980s, and it remains an important method of solution. The simplex method is a direct method that uses tools from the numerical solution of linear systems. ### Numerical solution of systems of nonlinear equations With a single equation $$f(x)=0\ ,$$ and having an initial estimate $$x_{0}$$ of the root $$\alpha\ ,$$ approximate $$f(x)$$ by its tangent line at the point $$\left(x_{0},f(x_{0})\right) \ .$$ Find the root of this tangent line as an approximation to the root $$\alpha$$ of the original equation $$f(x)=0\ .$$ This leads to 'Newton's iteration method', $x_{n+1}=x_{n}-\frac{f(x_{n})}{f^{\prime}(x_{n})},\quad\quad n=0,1,\dots$ Other linear and higher degree approximations can be used, and these lead to alternative iteration methods. An important derivative-free approximation of Newton’s method is the 'secant method'. For a system of $$m$$ nonlinear equations for a solution vector $$x$$ in $$R^m\ ,$$ we approximate $$f(x)$$ by its linear Taylor approximation about the initial estimate $$x_{0}\ .$$ This leads to Newton's method for nonlinear systems, $x_{n+1}=x_{n}-[f^{\prime}(x_{n})]^{-1}f(x_{n}),\quad\quad n=0,1,\dots$ in which $$f^{\prime}(x)$$ denotes the Jacobian matrix, of order $$m\times m$$ for $$f(x)\ .$$ In practice, the Jacobian matrix for $$f(x)$$ is often too complicated to compute directly; instead the partial derivatives in the Jacobian matrix are approximated using 'finite differences'. This leads to a 'finite difference Newton method'. As an alternative strategy and in analogy with the development of the secant method for the single variable problem, there is a similar rootfinding iteration method for solving nonlinear systems. It is called 'Broyden’s method' and it uses finite difference approximations of the derivatives in the Jacobian matrix, avoiding the evaluation of the partial derivatives of $$f(x)\ .$$ ### Numerical methods for solving differential and integral equations With such equations, there are usually at least two general steps involved in obtaining a nearby problem from which a numerical approximation can be computed; this is often referred to as 'discretization' of the original problem. The given equation will have a domain on which the unknown function is defined, perhaps an interval in one dimension and maybe a rectangle, ellipse, or other simply connected bounded region in two dimensions. Many numerical methods begin by introducing a mesh or grid on this domain, and the solution is to be approximated using this grid. Following this, there are several common approaches. One approach approximates the equation with a simpler equation defined on the mesh. For example, consider approximating the boundary value problem $u^{\prime\prime}(s)=f\left( s,u(s)\right) ,\quad0\leq s\leq1$ $u(0)=u(1)=0.$ Introduce a set of mesh points $$s_{j}=jh\ ,$$ $$j=0,1,\dots,n\ ,$$ with $$h=1/n$$ for some given $$n\geq2\ .$$ Approximate the boundary value problem by $\frac{1}{h^{2}}\left[ \tilde{u}_{n}(s_{j+1})-2\tilde{u}_{n}(s_{j})+\tilde {u}_{n}(s_{j-1})\right] =f\left( s_{j},\tilde{u}_{n}\left( s_{j}\right) \right) ,\quad j=1,\dots,n-1$ $\tilde{u}_{n}(0)=\tilde{u}_{n}(1)=0$ The second derivative in the original problem has been replaced by a numerical approximation to the second derivative. The new problem is a finite system of nonlinear equations, presumably amenable to solution by known techniques. The solution to this new problem is $$\tilde{u}_{n}\ ,$$ and it is defined on only the mesh points $$\left\{ s_{j}\right\} \ .$$ A second approach to discretizing differential and integral equations is as follows. Choose a finite-dimensional family of functions, denoted here by $$\mathcal{F}\ ,$$ with which to approximate the unknown solution function $$u\ .$$ Write the given differential or integral equation as $$L\left( u\right) =0\ ,$$ with $$L(v)$$ a function for any function $$v\ ,$$ perhaps over a restricted class of functions $$v\ .$$ The numerical method consists of selecting a function $$\tilde{u}\in\mathcal{F}$$ such that $$L(\tilde{u})$$ is a small function in some sense. The various ways of doing this lead to 'Galerkin methods', 'collocation methods', and 'least square methods'. Yet another approach is to reformulate the equation $$L\left( u\right) =0$$ as an optimization problem. Such reformulations are a part of the classical area of mathematics known as the 'calculus of variations', a subject that reflects the importance in physics of minimization principles. The well-known 'finite element method' for solving elliptic partial differential equations is obtained in this way, although it often coincides with a Galerkin method. The approximating functions in $$\mathcal{F}$$ are often chosen as piecewise polynomial functions which are polynomial over the elements of the mesh chosen earlier. Such methods are sometimes called 'local methods'. When the approximating functions $$p\in\mathcal{F}$$ are defined without reference to a grid, then the methods are sometimes called 'global methods' or 'spectral methods'. Examples of such $$\mathcal{F}$$ are sets of polynomials or trigonometric functions of some finite degree or less. With all three approaches to solving a differential or integral equations, the intent is that the resulting solution $$\tilde{u}$$ be close to the desired solution $$u\ .$$ The business of theoretical numerical analysis is to analyze such an algorithm and investigate the size of $$u-\tilde{u}\ .$$ ## References For an historical account of early numerical analysis, see • Herman Goldstine. A History of Numerical Analysis From the 16th Through the19th Century, Springer-Verlag, New York, 1977. For a current view of numerical analysis as taught at the undergraduate level, see • Cleve Moler. Numerical Computing with MATLAB, SIAM Pub., Philadelphia, 2004. For a current view of numerical analysis as taught at the advanced undergraduate or beginning graduate level, see • Alfio Quarteroni, Riccardo Sacco, and Fausto Saleri. Numerical Mathematics, Springer-Verlag, New York, 2000. • Christoph W. Ueberhuber. Numerical Computation: Vol. 1: Methods, Software, and Analysis, Vol. 2: Methods, Software, and Analysis, Springer-Verlag, New York, 1997. For one perspective on a theoretical framework using functional analysis for studying many problems in numerical analysis, see • Kendall Atkinson and Weimin Han. Theoretical Numerical Analysis: A Functional Analysis Framework, 2nd ed., Springer-Verlag, New York, 2005. As references for numerical linear algebra, see • Gene Golub and Charles Van Loan. Matrix Computations, 3rd ed., Johns Hopkins University Press, 1996. • Nicholas Higham. Accuracy and Stability of Numerical Algorithms, SIAM Pub., Philadelphia, 1996. For an introduction to practical numerical analysis for solving ordinary differential equations, see • Lawrence Shampine, Ian Gladwell, Skip Thompson. Solving ODEs with Matlab, Cambridge University Press, Cambridge, 2003. For information on computing aspects of numerical analysis, see • Michael Overton. Numerical computing with IEEE floating point arithmetic, SIAM Pub., Philadelphia, 2001. • Suely Oliveira and David Stewart. Writing Scientific Software: A Guide to Good Style, Cambridge University Press, Cambridge, 2006. Internal references • Olaf Sporns (2007) Complexity. Scholarpedia, 2(10):1623. • Skip Thompson (2007) Delay-differential equations. Scholarpedia, 2(3):2367. • James Meiss (2007) Dynamical systems. Scholarpedia, 2(2):1629. • Eugene M. Izhikevich (2007) Equilibrium. Scholarpedia, 2(10):2014. • Lawrence F. Shampine and Skip Thompson (2007) Initial value problems. Scholarpedia, 2(3):2861. • Mark Aronoff (2007) Language. Scholarpedia, 2(5):3175. • Philip Holmes and Eric T. Shea-Brown (2006) Stability. Scholarpedia, 1(10):1838.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 6, "mathjax_display_tex": 74, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.917739748954773, "perplexity_flag": "head"}
http://stats.stackexchange.com/questions/45174/voting-system-that-uses-confidence-for-each-voter	# Voting system that uses confidence for each voter Let's say, we have simple "yes/no" question that we want to know answer to. And there are N people "voting" for correct answer. Every voter has a history - list of 1's and 0's, showing whether they were right or wrong about this kind of questions in the past. If we assume history as a binomial distribution, we can find voters' mean performance on such questions, their variation, CI and any other kind of confidence metrics. Basically, my question is: how to incorporate confidence information into voting system? For example, if we consider only mean performance of each voter, then we can construct simple weighted voting system: $$result = sign(\sum_{v \in voters}\mu_v \times (-1)^{1-vote})$$ That is, we can just sum voters' weights multiplied either by $+1$ (for "yes") or by $-1$ (for "no"). It makes sense: if voter 1 has average of correct answers equal to $.9$, and voter 2 has only $.8$, than, probably, 1st person's vote should be considered as more important. On other hand, if 1st person have answered only 10 questions of this kind, and 2nd person have answered 1000 such questions, we are much more confident about 2nd person's skill level than about those of the 1st - it's just possible that 1st person was lucky, and after 10 relatively successful answers he will continue with much worse results. So, more precise question may sound like this: is there statistical metric that incorporates both - strength and confidence about some parameter? - ## 1 Answer You should consider the expertise of a voter as a latent variable of your system. You may then be able to solve your problem with bayesian inference. A representation as graphical model could be like this : Let's denote the variables $A$ for the true answer, $V_i$ for the vote of the voter $i$ and $H_i$ for its history. Say that you also have an "expertise" parameter $\mu_i$ such that $\Pr(A=V_i) = \mu_i$. If you put some prior on these $\mu_i$ -for example a Beta prior- you should be able to use the Bayes theorem to infer $\Pr(\mu_i \mid H_i)$, and then integrate over $\mu_i$ to compute $$\Pr(A \mid V_i, H_i) = \int_{\mu_i} \Pr(A, \mu_i \mid A_i, H_i)~ \mathrm{d}\mu_i$$ These systems are difficult to solve. You can use the EM algorithm as an approximation, or use complete likelihood maximisation scheme to perform exact Bayesian inference. Take a look on this paper Variational Inference for Crowdsourcing, Liu, Peng and Ihler 2012 (presented yesterday at NIPS !) for detailed algorithms for solving this task. - 1 Thanks for your answer, but could you please be a bit more specific about it? In particular, what you mean by expertise? If it's just probability that the person will answer correctly, then we already have its estimate as an average of previous answers, so it's not latent. If you mean than expertise incorporates both average and confidence about our estimate, then how can we propagate probabilities to get expertise and result? – ffriend Dec 6 '12 at 8:21 Yes, you can represent both average and confidence with this "expertise" variable and Bayesian inference. I have added a few explanations and a reference to my answer. Hope that helps ! – Emile Dec 7 '12 at 1:04 Great! This is exactly what I need. Thank you! – ffriend Dec 10 '12 at 8:46	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 13, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9487717151641846, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/255048/suppose-that-we-have-a-p-1-cdot-p-2-cdot-p-3-q-1-cdot-q-2-cdots-q-s-where-p	# Suppose that we have $a=p_1 \cdot p_2\cdot p_3=q_1\cdot q_2\cdots q_s$ where $p_1,p_2,p_3$ and $q_1,q_2,…,q_s$ are primes. Explain why $s=3$. Suppose that we have $a=p_1 \cdot p_2\cdot p_3=q_1\cdot q_2\cdots q_s$ where $p_1,p_2,p_3$ and $q_1,q_2,...,q_s$ are primes. Explain why $s=3$. This seems like a pointless question, but is it $3$ because $p_n$ doesn't go past $p_3$? I'm still having a hard time with this question. I'm mainly not understanding what it is asking. - Please check to see that I didn't alter the intended meaning with my edit. – Cameron Buie Dec 10 '12 at 2:10 Yes, there was an edit that changed the meaning. – Dmitri.Mendeleev Dec 10 '12 at 2:13 Aha! Now it makes sense. – MJD Dec 10 '12 at 2:14 ## 3 Answers I don't think this question is posed correctly, because it's definitely not true: $$30=2\cdot3\cdot5=13+17\Rightarrow s=2$$ and $$42=2\cdot3\cdot7=5+7+13+17\Rightarrow s=4$$ are just two counterexamples. For the corrected post: We want to show that if $a=p_1p_2p_3=q_1q_2\dots q_s$, then $s=3$. Clearly, $a$ divides $p_1$, so by the theorem on prime division, one of the $q_i$ divides $p_1$. Since $q_i$ is prime, then, $q_i=p_1$ and we can cancel it to get $p_2p_3=r_1r_2\dots r_{s-1}$ with $r_i$ prime. Repeating this process (assuming $s\geq 3$), we get $1=r_1r_2\dots r_{s-3}$, and all of the $r_i$ are integers, so either $s-3=0$ or $\|r_i\|=1$. But $r_i\geq 2$, so $s=3$. If $s<3$, then we will instead end up as $p_2p_3=1$ or similar. In this case, $p_i>1$ implies a contradiction. Thus $s=3$ is the only possible conclusion. - 2 Please see corrected post. – amWhy Dec 10 '12 at 2:15 I might be wrong, but isn't this just an application of the Fundamental Theorem of Arithmetic (FTA)? This theorem states that for any given positive integer, there is only one way to express it as a product of primes. The solution of the problem is then obvious. Of course, for a more complete solution you would have to first prove the FTA, but I won't go into that here.. - If you're worried that the question seems pointless, notice that the corresponding claim for addition is not true, since: $$p_1 + p_2 + p_3 = q_1 + \cdots + q_s$$ does not require that $s=3$—for example, $$10 + 11 + 12 = 1 + 2 + 3 + 4 + 5 + 18.$$ Even for multiplication the claim is not true unless you require the $p$s and $q$s to be prime: $$20 \cdot 25 \cdot 30 = 2 \cdot 10\cdot 15 \cdot 50$$ So this claim depends in a crucial way on special properties of multiplication and of prime numbers, and that is why it is not a silly question. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 35, "mathjax_display_tex": 5, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9574447870254517, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/tagged/magnetic-fields?page=2&sort=newest&pagesize=15	# Tagged Questions The magnetic-fields tag has no wiki summary. 1answer 52 views ### What geological properties of the earth could we deduce by measuring magnetic field strength and direction? I wonder if it is of any use for a geophysicist, if we measure the magnetic field strength and direction of the earth. Could we make valid statements about the composition of the the earth, earth ... 1answer 186 views ### If the Moon had gravity as good as the Earth and a magnetic field could it have supported life? If the Moon had gravity as good as Earth and a magnetic field could it have supported life? Because if the Moon had gravity, it could have retained water more than is present today on the surface. ... 0answers 72 views ### Is there an equation for the magnetic field of a conductor attached to a magnet? 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It is said that the line of magnets aligned in certain way results in cancellation of magnetic field on one side of the array, and ... 1answer 288 views ### Physical meaning of magnetic length What is the physical meaning of magnetic length $\ell_B=\frac{\hbar c}{e B}$ in 2D electron system under magnetic field? When $\ell_B \longrightarrow a$, where $a$ is the lattice constant, does that ... 1answer 131 views ### Point where resulting magnetic field is equal to 0 I am trying to solve the following task Given 3 infinitely long current carrying wires, as shown on image ($I_1=I_2$ , $I_3=2I_1$) and $r=5 cm$ calculate the point between them (in the same plane; ... 1answer 88 views ### How does a magnet work? I'm having trouble understanding how a magnet (not the field that is generated as a result but the material itself) work. The particles are aligned in a specific direction to give rise to force but I ... 0answers 68 views ### Changing the charging properties of an inductive charging coil I am attempting to modify an existing inductive charging coil (Palm Touchstone) so that it outputs the correct magnetic flux for a different wireless charging standard (Qi). I am doing this as an ... 2answers 118 views ### Force on a bar magnet A bar magnet is freely falling vertically inside a conducting ring placed in horizontal plane. Will it fall with an acceleration equal to g, less than g or greater than g? 1answer 117 views ### (Earth's) magnetic field Always when we want to represent the magnetic field of the earth we see a similar image: My question is, what exactly does this show? What are the blue and orange lines, what do they represent? Why ... 0answers 84 views ### Angle of Inclination How do I calculate the angle of Inclination for a specific location on the Earth? The only Information I've got is the longitude and latitude in degree, so I do not understand. I now ... 1answer 237 views ### Mutual Inductance and the Dot Convention Can anyone please explain me, the dot convention in coil systems (Mutual and self inductance) with some related images to understand..? 2answers 2k views ### Why does electricity flowing through a copper coil generate a magnetic field? Can some one please explain to me why electricity flowing though a copper coil generates a magnetic field or where I could possibly find that information? Are there other materials that produce a ... 4answers 2k views ### Earth's Magnetic Field, Human Brain and Sleep I remember reading somewhere: when you sleep in a way subjecting your body to cut the geomagnetic field at right angles, you become highly emotional whereas when it is Parallel, it cools your ... 1answer 71 views ### Does a celestial system exhibit a collective magnetic field? 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I was wondering if it would be possible to make a small tokamak; not one ...	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 7, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9256295561790466, "perplexity_flag": "middle"}
http://mathforum.org/mathimages/index.php?title=Markus-Lyapunov_Fractals&diff=25118&oldid=18868	# Markus-Lyapunov Fractals ### From Math Images (Difference between revisions) \| \| \| \| \| \|----------------------------------------\|-----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------\|------------------------------------------------------\|--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------\| \| \| \| Current revision (16:17, 14 July 2011) (edit) (undo) \| \| \| (20 intermediate revisions not shown.) \| \| \| \| \| Line 1: \| \| Line 1: \| \| \| - \| {{Image Description \| + \| {{Image Description Ready \| \| \| \|ImageName=Markus-Lyapunov Fractal \| \| \|ImageName=Markus-Lyapunov Fractal \| \| \| \|Image=Markus-Lyapunov1.gif \| \| \|Image=Markus-Lyapunov1.gif \| \| - \| \|ImageIntro=A representation of the regions of chaos and stability over the space of two population growth rates. \| + \| \|ImageIntro=Markus-Lyapunov fractals are representations of the regions of chaos and stability over the space of two population growth rates. \| \| - \| \|ImageDescElem=The Markus-Lyapunov fractal is much more than a pretty picture; it is a map. The curving bodies and sweeping arms of the image are a color-coded plot that shows us how a population changes as its rate of growth moves between two values. All the rich variations of color in the fractal come from the different levels of stability and chaos possible in such change. \| + \| \|ImageDescElem=The Markus-Lyapunov fractal is much more than a pretty picture – it is a chart. The curving bodies and sweeping arms of the image are a color-coded plot that shows us how a population changes as its rate of growth moves between two values. All the rich variations of color in the fractal come from the different levels of stability and chaos possible in such change. \| \| \| \| \| \| \| - \| The [[Logistic Bifurcation\|logistic map]] is one of the simplest mathematical representations of population growth. Depending on the rate of fecundity used in the map, it will generate either a neutral system, a stable oscillating system, or a [[chaos\|chaotic]] system. To help determine which of these outcomes would occur, the mathematician Aleksandr Lyapunov developed a method for comparing changes in growth and time in order to calculate what has been dubbed the {{EasyBalloon\|Link=Lyapunov exponent\|Balloon=The Lyapunov exponent represents the overall rate of change of a system over many iterations, expressed logarithmically.}}. This is a useful indicator, and here's why: \| + \| The [[Logistic Bifurcation#Jump2\|logistic map]] is one of the most concise mathematical representations of population growth. Depending on the rate of fecundity used in the map, it will generate either a neutral system, a stable oscillating system, or a [[chaos\|chaotic]] system. To help determine which of these outcomes will occur, the mathematician Aleksandr Lyapunov developed a method for comparing changes in growth rate in order to calculate a value called the {{EasyBalloon\|Link=Lyapunov exponent\|Balloon=The Lyapunov exponent represents the overall rate of change of a system over many iterations, expressed logarithmically.}}. This is a useful indicator because, for the logistic map, \| \| - \| If it is zero, the population change is neutral; at some point in time, it reaches a fixed point and remains there. \| + \| \| \| - \| If it is less than zero, the population will become {{EasyBalloon\|Link=stable\|Balloon=Stability is different from a fixed point; A system that oscillates between two values is stable, and a system that oscillates between sixteen values is still stable.}}. The lower the number, the faster and more thoroughly the population will stabilize. \| + \| If the Lyapunov exponent is zero, the population change is neutral; the population size begins and remains at a constant, fixed point. \| \| - \| If it is positive, the population will become chaotic. \| + \| If the Lyapunov exponent is less than zero, the population will become {{EasyBalloon\|Link=stable\|Balloon=Stability is different from a fixed point. A system that oscillates between two values is stable, and a system that oscillates between sixteen values is still stable.}}. The more negative the number, the faster and more thoroughly the population will stabilize. \| \| \| \| + \| If the Lyapaunov exponent is positive, the population will become chaotic. \| \| \| \| \| \| \| \| [[Image:Markus-Lyapunov2.jpg\|left\|frame\|Another example of a Markus-Lyapunov fractal, this one with chaos in black and stability in gold.]] \| \| [[Image:Markus-Lyapunov2.jpg\|left\|frame\|Another example of a Markus-Lyapunov fractal, this one with chaos in black and stability in gold.]] \| \| - \| What does all this have to do with the fantastical shapes of the Markus-Lyapunov fractal? The scientist Mario Markus wanted a way to visualize the potential represented by the Lyapunov exponent as a population moved between ''two'' different rates of growth. So he created a graphical space with one rate of growth measured along the ''x''-axis and the other along the ''y''. Thus for any point, (''x'',''y''), there is one specific Lyapunov exponent that predicts how a population with those rates of change will behave. Markus then created a color scheme to represent different Lyapunov exponents – one color represents positive numbers, and another represents negative numbers and zero. This second color he placed on a gradient from light to dark, so that lower negative numbers are lighter and those closer to zero are darker. The bands of black that appear in many fractals therefore show where the Lyapunov exponent is exactly zero, and bands of white indicate {{EasyBalloon\|Link=superstable\|Balloon=<math>\lambda=-\infty</math>, as the lowest possible Lyapunov exponent, indicates the fastest possible approach to stability.}} points. By this code, Markus could color every point on his graph space based on its Lyapunov exponent. \| + \| What does all this have to do with the fantastical shapes of the Markus-Lyapunov fractal? The scientist Mario Markus wanted a way to visualize the potential represented by the Lyapunov exponent as a population moved between ''two'' different rates of growth. So he created a graphical space with one rate of growth measured along the ''x''-axis and the other along the ''y''. Thus for any point (''x'',''y'') there is one specific Lyapunov exponent that predicts how a population with those rates of change will behave. \| \| \| \| + \| \| \| \| \| + \| Markus then created a color scheme to represent different Lyapunov exponents – one color represents positive numbers, and another represents negative numbers and zero. This second color he placed on a gradient from light to dark, so that lower negative numbers are lighter and those closer to zero are darker. The bands of black that appear in many fractals therefore show where the Lyapunov exponent is exactly zero, and bands of white indicate {{EasyBalloon\|Link=superstable\|Balloon=<math>\lambda=-\infty</math>, as the lowest possible Lyapunov exponent, indicates the fastest possible approach to stability.}} points. By this code, Markus could color every point on his graph space based on its Lyapunov exponent. \| \| \| \| \| \| \| \| Consider the main image on this page. The blue "background" shows all the points where the combination of the rates of change on the ''x'' and ''y'' axes will result in chaotic population growth. The "floating" yellow shapes show where the population will move toward stability. The lighter the yellow, the more stable the population. \| \| Consider the main image on this page. The blue "background" shows all the points where the combination of the rates of change on the ''x'' and ''y'' axes will result in chaotic population growth. The "floating" yellow shapes show where the population will move toward stability. The lighter the yellow, the more stable the population. \| \| \| \| + \| \| \| \| \| + \| Based on this color assignment, if a logistic system with rates of change ''x'' = 2 and ''y'' = 1.6 has a Lyapunov exponent of -0.3, there will be a dark yellow pixel at the graph location (2, 1.6), showing that the system moves slowly toward stability. If, instead, that logistic system had a Lyapunov exponent of 1, that same pixel would be blue, showing chaos. \| \| \| <br style="clear: both" /> \| \| <br style="clear: both" /> \| \| \| \|ImageDesc= \| \| \|ImageDesc= \| \| \| ===The Lyapunov Exponent=== \| \| ===The Lyapunov Exponent=== \| \| - \| The discrete form of the Lyapunov exponent is \| + \| [[Image:Dynamical1.gif\|right\|frame\|As a logistic system progresses through ''n'' iterations, the distance, d''x''<sub>0</sub>, between two arbitrarily close points entered into the system will become a new distance, d''x''<sub>''n''</sub>.]] \| \| - \| {{EquationRef2\|4}}<math>\lambda=\lim_{N \to \infty}\frac{1}{N}\sum_{n=1}^N \log_2 \frac{dx_{(n+1)}}{dx_n}</math> \| + \| [[Image:LambdaGraphs.gif\|right\|frame\|In both graphs, we see the evolution of <math>{\operatorname{d}x_n\over\operatorname{d}x_0}</math> from ''n'' = 0 to ''n'' = 10. The upper graph shows this evolution for λ = -1, while the lower shows this for λ = 1. Notice how quickly we observe convergence in the upper graph and divergence in the lower.]] \| \| - \| In other words, the Lyapunov exponent <math>\lambda</math> represents the [[Limit\|limit]] of the {{EasyBalloon\|Link=mean\|Balloon=The presence of <math>\frac{1}{N}\sum_{n=1}^N ...</math> shows that this is a mean.}} of the {{EasyBalloon\|Link=exponential\|Balloon=The presence of <math>\log_2</math> makes the relationship between <math>\lambda</math> and the original equation exponential.}} {{EasyBalloon\|Link=rates of change\|Balloon=<math>\frac{dx_{(n+1)}}{dx_n}</math> defines the rate of change.}} that occur in each transition, ''x''<sub>''n''</sub> <math>\rightarrow</math> ''x''<sub>(''n''+1)</sub>, as the number of transitions approaches infinity. \| + \| The Lyapunov exponent is a measure of the rate of divergence of two infinitesimally close points in a dynamic system. For a single-variable system such as the logistic map, we can consider two points at an arbitrarily small distance, d''x''<sub>0</sub> from each other. After ''n'' iterations of the system, they will be at distance d''x''<sub>''n''</sub> from each other. The Lyapunov exponent λ represents this change with the approximation: \| \| \| \| + \| {{EquationRef2\|Eq. 1}}<math>{\operatorname{d}x_n\over\operatorname{d}x_0}\approx 2^{\lambda n}</math> \| \| \| \| + \| Or, isolating the Lyapunov exponent λ: \| \| \| \| + \| :::<math>\frac{1}{n}\log_2{\operatorname{d}x_n\over\operatorname{d}x_0}\approx \lambda</math> \| \| \| \| + \| Generalizing this for all ''n'', we use [[Summation Notation\|summation notation]] to consider every step of iteration: \| \| \| \| + \| {{EquationRef2\|Eq. 2}}<math>\frac{1}{n}\sum_{1}^n \log_2 {\operatorname{d}x_{(n+1)}\over\operatorname{d}x_n} \approx \lambda</math> \| \| \| \| + \| In order for this to be accurate, however, it must measure the system's divergence over an infinite period of time, so we define λ as the limit of {{EquationNote\|Eq. 2}} as ''n'' approaches infinity: \| \| \| \| + \| {{EquationRef2\|Eq. 3}}<math>\lambda=\lim_{N \to \infty}\frac{1}{N}\sum_{n=1}^N \log_2 {\operatorname{d}x_{(n+1)}\over\operatorname{d}x_n}</math> \| \| \| \| + \| This is the discrete form of the Lyapunov exponent. \| \| \| \| \| \| \| - \| What does this have to do with stability? The key is the log<sub>2</sub> component, which renders numbers under 1 negative and those over 1 positive. This is what yields the properties of Lyapunov exponents laid out in the "Basic Explanation" – those mean overall rates of change that make the system {{EasyBalloon\|Link=finite\|Balloon=A geometric series or sequence is finite if is multiplied by a factor, '''r''' < 1, that makes it converge to a discrete value or set of values.}} must be less than 1, giving us a negative Lyapunov exponent, while those rates of change that expand the system to the point of chaos must be greater than one, giving us a positive exponent. When the mean overall rate of change is zero, the logarithmic component no longer exists, showing exactly what happens in a superstable system; the rate of change ceases to exist. \| + \| To consider the implications of λ, let us return to {{EquationNote\|Eq. 1}}: \| \| \| \| + \| :::<math>{\operatorname{d}x_n\over\operatorname{d}x_0}\approx 2^{\lambda n}</math> \| \| \| \| + \| We can see that, for λ < 0, the difference between d''x''<sub>0</sub> to d''x''<sub>''n''</sub> will disappear as ''n'' grows. (Indeed, the lower the value of λ, the more quickly this change will disappear.) Similarly, for λ = 0, there will be no difference at all. For λ > 0, however, the difference between the initial distance between the points and the final distance between the points will expand exponentially as ''n'' grows. In other words, a positive Lyapunov exponent indicates a system in which an infinitesimal change in initial conditions can result in massively different final conditions. A negative Lyapunov exponent, on the other hand, indicates a system in which the effects of such an initial change will fade over time. \| \| \| \| \| \| \| - \| In other words, the Lyapunov exponent is a method for examining the rate of change of a system considered over infinite iterations, then taking that rate of change and making it easily identifiable as a value that induces either chaos or stability. \| + \| Recall that the phenomenon captured by a positive Lyapunov exponent – wide variation resulting from infinitely small initial changes – is one of the conditions for a system to be chaotic. Thus a positive Lyapunov, in the presence of other conditions of chaos, implies a chaotic system. \| \| \| \| \| \| \| - \| ====In the Logistic Formula==== \| + \| ====In the Logistic Map==== \| \| - \| Basic differentiation shows us that, for the logistic formula ({{EquationNote\|3}}): \| + \| Recall that the [[Logistic Bifurcation#Jump3\|logistic map]] is: \| \| - \| {{EquationRef2\|5}}<math>\frac{dx_{(n+1)}}{dx_n}=\mathbf{r}-2\mathbf{r}x_0^2</math> \| + \| {{EquationRef2\|Eq. 4}}<math>x_{(n+1)}=\mathbf{r}x_n(1-x_n)</math> \| \| - \| Using this and a sufficiently large ''N'' number of iterations, we can approximate the Lyapunov exponent for the logistic formula to be: \| + \| From which we find \| \| - \| {{EquationRef2\|6}}<math>\lambda \approx \frac{1}{N}\sum_{n=1}^N \log_2 \mathbf{r}-2\mathbf{r}x_0^2</math> \| + \| :::<math>{\operatorname{d}x_{(n+1)}\over\operatorname{d}x_n}=\mathbf{r}-2\mathbf{r}x_n</math> \| \| - \| Here we can see much more clearly a property that we have been assuming -- the variable that has the greatest impact on the stability of the logistic equation is '''r''', not ''x''<sub>''n''</sub>. This is clear here because, as we differentiate the logistic formula in order to examine its overall rate of change, ''x''<sub>''n''</sub> is reduced to the constant value ''x''<sub>0</sub> (the starting volume of the population). It is therefore no longer a variable when we look at the formula on the level of its Lyapunov exponent, and so changing the value of ''x''<sub>''n''</sub> does not change whether the logistic function yields chaos or stability. \| + \| Inserting this into the Lyapunov exponent, {{EquationNote\|Eq. 3}}, we have the Lyapunov exponent for the logistic map: \| \| \| \| + \| {{EquationRef2\|Eq. 5}}<math>\lambda=\lim_{N \to \infty}\frac{1}{N}\sum_{n=1}^N \log_2 \mathbf{r}-2\mathbf{r}x_n</math> \| \| \| \| + \| As noted above, negative values of λ here indicate stability in the logistic map. Also, in the case of the logistic map, any system with a Lyapunov exponent greater than zero is a chaotic system. We can see this when we compare the [[Logistic Bifurcation\|logistic bifurcation]] diagram with a graph of the Lyapunov exponents for the logistic maps of changing '''r''' values: \| \| \| \| + \| [[Image:Lyap Bifurc1.gif\|center\|frame\|Here a graph of the Lyapunov exponents for logistic maps with 0 < '''r''' < 4 is overlaid in red on the [[Logistic Bifurcation\|logistic bifurcation]] diagram. The points of the red graph were calculated directly in Matlab using {{EquationNote\|Eq. 5}}.]] \| \| \| \| \| \| \| \| ===Forcing the Rates of Change=== \| \| ===Forcing the Rates of Change=== \| \| - \| Mathematically, the important part of Markus's contribution to understanding this type of system was not his method for generating fractals, but his use of periodic rate-of-change forcing. We have been discussing the great impact of the '''r''' value in determining the output of the logistic formula, but this value can have still greater impact if we do not choose to keep it constant. Anyone who has studied biology, as Markus has, knows that the rates of change of a populations size do not simply fluctuate with changing supplies of food and space, but also often alternate between two or more specific potential rates of change depending on such things as weather and mating seasons.[[Image:AB.gif\|frame\|A Markus-Lyapunov fractal with rate-of-change pattern ''ab'']] \| + \| Mathematically, the important part of Markus's contribution to understanding this type of system was not his method for generating fractals, but his use of periodic rate-of-change forcing. The value of '''r''' has a great impact on the output of the logistic map, but this value can have still greater impact if we do not choose to keep it constant. \| \| \| \| + \| [[Image:AB.gif\|frame\|A Markus-Lyapunov fractal with rate-of-change pattern ''ab'']] \| \| \| \| \| \| \| - \| In terms of the logistic formula, this means we choose a set of rates of change, '''r'''<sub>1</sub>, '''r'''<sub>2</sub>, '''r'''<sub>3</sub>,..., '''r'''<sub>''p''</sub>, where ''p'' is the period over which the rates of change loop. When we force the rates of change to follow such a loop, we have a new, [[Modular arithmetic\|modular]] logistic equation ({{EquationNote\|3}}): \| + \| In terms of the logistic map, this means we choose a set of rates of change, '''r'''<sub>1</sub>, '''r'''<sub>2</sub>, '''r'''<sub>3</sub>,..., '''r'''<sub>''p''</sub>, where ''p'' is the period over which the rates of change loop. When we force the rates of change to follow such a loop, we have a new, [[Modular arithmetic\|modular]] logistic map ({{EquationNote\|Eq. 4}}): \| \| \| \| \| \| \| - \| {{EquationRef2\|7}}<math>x_{(n+1)}=\mathbf{r}_{n \text{mod} p}x_n(1-x_n)</math> \| + \| {{EquationRef2\|Eq. 6}}<math>x_{(n+1)}=\mathbf{r}_{n \text{mod} p}x_n(1-x_n)</math> \| \| \| \| \| \| \| - \| It is in these forced alterations in rates of change that the fascinating shapes of the Markus-Lyapunov fractal come out. Each of the fractals is formed from some pattern of two rates of change, ''a'' and ''b''. So a pattern ''aba'' would mean each point on the fractal is colored based on the Lyapunov exponent of the logistic formula {{EquationNote\|7}}, where '''r'''<sub>1</sub> = ''a'', '''r'''<sub>2</sub> = ''b'', and '''r'''<sub>3</sub> = ''a''. That is, the '''r''' values would cycle ''a,b,a,a,b,a,a,b,a...''. \| + \| It is in these forced alterations in rates of change that the fascinating shapes of the Markus-Lyapunov fractal come out. Each of the fractals is formed from some pattern of two rates of change, ''a'' and ''b''. So a pattern ''aba'' would mean each point on the fractal is colored based on the Lyapunov exponent of the logistic map {{EquationNote\|Eq. 5}}, where '''r'''<sub>1</sub> = ''a'', '''r'''<sub>2</sub> = ''b'', and '''r'''<sub>3</sub> = ''a''. That is, the '''r''' values would cycle ''a,b,a,a,b,a,a,b,a...''. \| \| \| \| \| \| \| - \| Because the axes used to map these fractals are measurements of changes in ''a'' and ''b'', the pattern ''a'' would simply yield a set of vertical bars, just as the pattern ''b'' would yield horizontal bars. However, once the patterns start to become mixed, more interesting results come out. The image to the right shows an ''ab'' pattern. Note that it is much simpler than other images shown on this page; the main image, for instance, is a ''bbbbbbaaaaaa'' pattern. \| + \| Because the axes used to map these fractals are measurements of changes in ''a'' and ''b'', the pattern ''a'' would simply yield a set of vertical bars, just as the pattern ''b'' would yield horizontal bars. However, once the patterns start to become mixed, more interesting results come out. The image to the right shows an ''ab'' pattern. Note that it is much simpler, in the quantity of spires and crossing arms, than other images shown on this page; the main image, for instance, is a ''bbbbbbaaaaaa'' pattern. \| \| \| <br style="clear: both" /> \| \| <br style="clear: both" /> \| \| - \| \|AuthorName=BernardH, using Mathematica 5 \| + \| \|other=understanding [[Logistic Bifurcation]] \| \| \| \| + \| \|AuthorName=BernardH \| \| \| \|SiteURL=http://en.wikipedia.org/wiki/File:Lyapunov-fractal.png \| \| \|SiteURL=http://en.wikipedia.org/wiki/File:Lyapunov-fractal.png \| \| \| \|Field=Dynamic Systems \| \| \|Field=Dynamic Systems \| \| \| \|Field2=Fractals \| \| \|Field2=Fractals \| \| - \| \|WhyInteresting=[[Image:MarkusLyapunovZoom.gif\|200px\|frame\|An enlargement of a section of "Zircon Zity," showing self-similarity.]] \| + \| \|WhyInteresting= \| \| \| \| + \| <br style="clear: both" /> \| \| \| \| + \| [[Image:MLBlowup.gif\|frame\|An enlargement of a section of "Zircon Zity," showing self-similarity.]] \| \| \| ===Fractal Properties=== \| \| ===Fractal Properties=== \| \| - \| The movements from light to dark and the dramatic curves of the boundaries between stability and chaos here create an astonishing 3D effect. But the image is striking not only for its beauty but also for its self-similarity. Self-similarity is that trait that makes [[fractals]] what they are – zooming in on the image reveals smaller and smaller parts that resemble the whole. Consider the image to the right, enlarged from a section of the main image above. Here we see several shapes that repeat in smaller and smaller iterations. Perhaps ironically, this type of pattern is a common property of chaos. \| + \| The movements from light to dark and the dramatic curves of the boundaries between stability and chaos here create an astonishing 3D effect. But the image is striking not only for its beauty but also for its '''self-similarity'''. Self-similarity is that trait that makes [[Field:Fractals\|fractals]] what they are – zooming in on the image reveals smaller and smaller parts that resemble the whole. Consider the image to the right, showing an enlarged section of the main image above. Here we see several shapes that repeat in smaller and smaller iterations. Perhaps ironically, this type of pattern is a common property of chaos. \| \| \| \| \| \| \| - \| For more images of the fractal properties of chaotic systems, see the [[Henon Attractor]], the [[Harter-Heighway Dragon]] Curve, and [[Julia Sets]]. \| + \| For more images of the fractal properties of chaotic systems, see the [[Blue Fern]], the [[Henon Attractor]], the [[Harter-Heighway Dragon]] Curve, and [[Julia Sets]]. \| \| \| \| \| \| \| \| <br style="clear: both" /> \| \| <br style="clear: both" /> \| \| Line 59: \| \| Line 81: \| \| \| \| \|References=<references /> \| \| \|References=<references /> \| \| \| \| \| \| \| - \| ;Other Sources Consulted: \| + \| \|InProgress=No \| \| - \| : Elert, G. (2007). ''The Chaos Hypertextbook''. http://hypertextbook.com/chaos/ \| + \| \| \| - \| \|InProgress=Yes \| + \| \| \| \| }} \| \| }} \| ## Current revision Markus-Lyapunov Fractal Fields: Dynamic Systems and Fractals Image Created By: BernardH Website: [1] Markus-Lyapunov Fractal Markus-Lyapunov fractals are representations of the regions of chaos and stability over the space of two population growth rates. # Basic Description The Markus-Lyapunov fractal is much more than a pretty picture – it is a chart. The curving bodies and sweeping arms of the image are a color-coded plot that shows us how a population changes as its rate of growth moves between two values. All the rich variations of color in the fractal come from the different levels of stability and chaos possible in such change. The logistic map is one of the most concise mathematical representations of population growth. Depending on the rate of fecundity used in the map, it will generate either a neutral system, a stable oscillating system, or a chaotic system. To help determine which of these outcomes will occur, the mathematician Aleksandr Lyapunov developed a method for comparing changes in growth rate in order to calculate a value called the Lyapunov exponentThe Lyapunov exponent represents the overall rate of change of a system over many iterations, expressed logarithmically.. This is a useful indicator because, for the logistic map, • If the Lyapunov exponent is zero, the population change is neutral; the population size begins and remains at a constant, fixed point. • If the Lyapunov exponent is less than zero, the population will become stableStability is different from a fixed point. A system that oscillates between two values is stable, and a system that oscillates between sixteen values is still stable.. The more negative the number, the faster and more thoroughly the population will stabilize. • If the Lyapaunov exponent is positive, the population will become chaotic. Another example of a Markus-Lyapunov fractal, this one with chaos in black and stability in gold. What does all this have to do with the fantastical shapes of the Markus-Lyapunov fractal? The scientist Mario Markus wanted a way to visualize the potential represented by the Lyapunov exponent as a population moved between two different rates of growth. So he created a graphical space with one rate of growth measured along the x-axis and the other along the y. Thus for any point (x,y) there is one specific Lyapunov exponent that predicts how a population with those rates of change will behave. Markus then created a color scheme to represent different Lyapunov exponents – one color represents positive numbers, and another represents negative numbers and zero. This second color he placed on a gradient from light to dark, so that lower negative numbers are lighter and those closer to zero are darker. The bands of black that appear in many fractals therefore show where the Lyapunov exponent is exactly zero, and bands of white indicate superstable$\lambda=-\infty$, as the lowest possible Lyapunov exponent, indicates the fastest possible approach to stability. points. By this code, Markus could color every point on his graph space based on its Lyapunov exponent. Consider the main image on this page. The blue "background" shows all the points where the combination of the rates of change on the x and y axes will result in chaotic population growth. The "floating" yellow shapes show where the population will move toward stability. The lighter the yellow, the more stable the population. Based on this color assignment, if a logistic system with rates of change x = 2 and y = 1.6 has a Lyapunov exponent of -0.3, there will be a dark yellow pixel at the graph location (2, 1.6), showing that the system moves slowly toward stability. If, instead, that logistic system had a Lyapunov exponent of 1, that same pixel would be blue, showing chaos. # A More Mathematical Explanation Note: understanding of this explanation requires: * [Click to view A More Mathematical Explanation] ### The Lyapunov Exponent [[Image:Dynamical1.gif\|right\|frame\|As a logistic system progresses throug [...] [Click to hide A More Mathematical Explanation] ### The Lyapunov Exponent As a logistic system progresses through n iterations, the distance, dx0, between two arbitrarily close points entered into the system will become a new distance, dxn. In both graphs, we see the evolution of ${\operatorname{d}x_n\over\operatorname{d}x_0}$ from n = 0 to n = 10. The upper graph shows this evolution for λ = -1, while the lower shows this for λ = 1. Notice how quickly we observe convergence in the upper graph and divergence in the lower. The Lyapunov exponent is a measure of the rate of divergence of two infinitesimally close points in a dynamic system. For a single-variable system such as the logistic map, we can consider two points at an arbitrarily small distance, dx0 from each other. After n iterations of the system, they will be at distance dxn from each other. The Lyapunov exponent λ represents this change with the approximation: ${\operatorname{d}x_n\over\operatorname{d}x_0}\approx 2^{\lambda n}$ Or, isolating the Lyapunov exponent λ: $\frac{1}{n}\log_2{\operatorname{d}x_n\over\operatorname{d}x_0}\approx \lambda$ Generalizing this for all n, we use summation notation to consider every step of iteration: $\frac{1}{n}\sum_{1}^n \log_2 {\operatorname{d}x_{(n+1)}\over\operatorname{d}x_n} \approx \lambda$ In order for this to be accurate, however, it must measure the system's divergence over an infinite period of time, so we define λ as the limit of Eq. 2 as n approaches infinity: $\lambda=\lim_{N \to \infty}\frac{1}{N}\sum_{n=1}^N \log_2 {\operatorname{d}x_{(n+1)}\over\operatorname{d}x_n}$ This is the discrete form of the Lyapunov exponent. To consider the implications of λ, let us return to Eq. 1: ${\operatorname{d}x_n\over\operatorname{d}x_0}\approx 2^{\lambda n}$ We can see that, for λ < 0, the difference between dx0 to dxn will disappear as n grows. (Indeed, the lower the value of λ, the more quickly this change will disappear.) Similarly, for λ = 0, there will be no difference at all. For λ > 0, however, the difference between the initial distance between the points and the final distance between the points will expand exponentially as n grows. In other words, a positive Lyapunov exponent indicates a system in which an infinitesimal change in initial conditions can result in massively different final conditions. A negative Lyapunov exponent, on the other hand, indicates a system in which the effects of such an initial change will fade over time. Recall that the phenomenon captured by a positive Lyapunov exponent – wide variation resulting from infinitely small initial changes – is one of the conditions for a system to be chaotic. Thus a positive Lyapunov, in the presence of other conditions of chaos, implies a chaotic system. #### In the Logistic Map Recall that the logistic map is: $x_{(n+1)}=\mathbf{r}x_n(1-x_n)$ From which we find ${\operatorname{d}x_{(n+1)}\over\operatorname{d}x_n}=\mathbf{r}-2\mathbf{r}x_n$ Inserting this into the Lyapunov exponent, Eq. 3, we have the Lyapunov exponent for the logistic map: $\lambda=\lim_{N \to \infty}\frac{1}{N}\sum_{n=1}^N \log_2 \mathbf{r}-2\mathbf{r}x_n$ As noted above, negative values of λ here indicate stability in the logistic map. Also, in the case of the logistic map, any system with a Lyapunov exponent greater than zero is a chaotic system. We can see this when we compare the logistic bifurcation diagram with a graph of the Lyapunov exponents for the logistic maps of changing r values: Here a graph of the Lyapunov exponents for logistic maps with 0 < r < 4 is overlaid in red on the logistic bifurcation diagram. The points of the red graph were calculated directly in Matlab using Eq. 5. ### Forcing the Rates of Change Mathematically, the important part of Markus's contribution to understanding this type of system was not his method for generating fractals, but his use of periodic rate-of-change forcing. The value of r has a great impact on the output of the logistic map, but this value can have still greater impact if we do not choose to keep it constant. A Markus-Lyapunov fractal with rate-of-change pattern ab In terms of the logistic map, this means we choose a set of rates of change, r1, r2, r3,..., rp, where p is the period over which the rates of change loop. When we force the rates of change to follow such a loop, we have a new, modular logistic map (Eq. 4): $x_{(n+1)}=\mathbf{r}_{n \text{mod} p}x_n(1-x_n)$ It is in these forced alterations in rates of change that the fascinating shapes of the Markus-Lyapunov fractal come out. Each of the fractals is formed from some pattern of two rates of change, a and b. So a pattern aba would mean each point on the fractal is colored based on the Lyapunov exponent of the logistic map Eq. 5, where r1 = a, r2 = b, and r3 = a. That is, the r values would cycle a,b,a,a,b,a,a,b,a.... Because the axes used to map these fractals are measurements of changes in a and b, the pattern a would simply yield a set of vertical bars, just as the pattern b would yield horizontal bars. However, once the patterns start to become mixed, more interesting results come out. The image to the right shows an ab pattern. Note that it is much simpler, in the quantity of spires and crossing arms, than other images shown on this page; the main image, for instance, is a bbbbbbaaaaaa pattern. # Why It's Interesting An enlargement of a section of "Zircon Zity," showing self-similarity. ### Fractal Properties The movements from light to dark and the dramatic curves of the boundaries between stability and chaos here create an astonishing 3D effect. But the image is striking not only for its beauty but also for its self-similarity. Self-similarity is that trait that makes fractals what they are – zooming in on the image reveals smaller and smaller parts that resemble the whole. Consider the image to the right, showing an enlarged section of the main image above. Here we see several shapes that repeat in smaller and smaller iterations. Perhaps ironically, this type of pattern is a common property of chaos. For more images of the fractal properties of chaotic systems, see the Blue Fern, the Henon Attractor, the Harter-Heighway Dragon Curve, and Julia Sets. One artist superimposed and edited several real Markus-Lyapunov fractals to create this piece of art. ### Artistic Extensions After Markus saw the incredible beauty and intriguing three-dimensionality of the images generated by his plotting system, he immediately sent the images to a gallery in the hopes that it would display his images in an exhibition.[1] It's easy to see why he did so, and in fact, pictures based on these fractals have become a large part of what is called "fractalist" art. As with all domains of fractalist art, there is a great deal of debate in the art community over whether these images are truly "art" given their intrinsic reliance on a purely scientific, algorithmically-generated chart. One could say that such a process is devoid of creativity, but it is equally valid to say that the identification and presentation of the beauty in the science is an art in itself – a concept that is critical in modern art. Either way, there has been an undeniable artistic fascination with Markus-Lyapunov fractals; if the image seems familiar, you have likely seen it on posters, t-shirts, or any other canvas for graphic design. # Teaching Materials There are currently no teaching materials for this page. Add teaching materials. # References 1. ↑ Dewdney, A. K. (1991). Leaping into Lyapunov Space. Scientific American, (130-132). 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http://en.wikipedia.org/wiki/Optimal_control	# Optimal control Optimal control theory, an extension of the calculus of variations, is a mathematical optimization method for deriving control policies. The method is largely due to the work of Lev Pontryagin and his collaborators in the Soviet Union[1] and Richard Bellman in the United States. ## General method Optimal control deals with the problem of finding a control law for a given system such that a certain optimality criterion is achieved. A control problem includes a cost functional that is a function of state and control variables. An optimal control is a set of differential equations describing the paths of the control variables that minimize the cost functional. The optimal control can be derived using Pontryagin's maximum principle (a necessary condition also known as Pontryagin's minimum principle or simply Pontryagin's Principle[2]), or by solving the Hamilton–Jacobi–Bellman equation (a sufficient condition). We begin with a simple example. Consider a car traveling on a straight line through a hilly road. The question is, how should the driver press the accelerator pedal in order to minimize the total traveling time? Clearly in this example, the term control law refers specifically to the way in which the driver presses the accelerator and shifts the gears. The "system" consists of both the car and the road, and the optimality criterion is the minimization of the total traveling time. Control problems usually include ancillary constraints. For example the amount of available fuel might be limited, the accelerator pedal cannot be pushed through the floor of the car, speed limits, etc. A proper cost functional is a mathematical expression giving the traveling time as a function of the speed, geometrical considerations, and initial conditions of the system. It is often the case that the constraints are interchangeable with the cost functional. Another optimal control problem is to find the way to drive the car so as to minimize its fuel consumption, given that it must complete a given course in a time not exceeding some amount. Yet another control problem is to minimize the total monetary cost of completing the trip, given assumed monetary prices for time and fuel. A more abstract framework goes as follows. Minimize the continuous-time cost functional $J=\Phi\,[\,\textbf{x}(t_0),t_0,\textbf{x}(t_f),t_f\,] + \int_{t_0}^{t_f} \mathcal{L}\,[\,\textbf{x}(t),\textbf{u}(t),t\,] \,\operatorname{d}t$ subject to the first-order dynamic constraints $\dot{\textbf{x}}(t) = \textbf{a}\,[\,\textbf{x}(t),\textbf{u}(t),t\,],$ the algebraic path constraints $\textbf{b}\,[\,\textbf{x}(t),\textbf{u}(t),t\,] \leq \textbf{0},$ and the boundary conditions $\boldsymbol{\phi}\,[\,\textbf{x}(t_0),t_0,\textbf{x}(t_f),t_f\,] = 0$ where $\textbf{x}(t)$ is the state, $\textbf{u}(t)$ is the control, $t$ is the independent variable (generally speaking, time), $t_0$ is the initial time, and $t_f$ is the terminal time. The terms $\Phi$ and $\mathcal{L}$ are called the endpoint cost and Lagrangian, respectively. Furthermore, it is noted that the path constraints are in general inequality constraints and thus may not be active (i.e., equal to zero) at the optimal solution. It is also noted that the optimal control problem as stated above may have multiple solutions (i.e., the solution may not be unique). Thus, it is most often the case that any solution $[\textbf{x}^(t^),\textbf{u}^(t^),t^*]$ to the optimal control problem is locally minimizing. ## Linear quadratic control A special case of the general nonlinear optimal control problem given in the previous section is the linear quadratic (LQ) optimal control problem. The LQ problem is stated as follows. Minimize the quadratic continuous-time cost functional $J=\tfrac{1}{2} \textbf{x}^{\text{T}}(t_f)\textbf{S}_f\textbf{x}(t_f) + \tfrac{1}{2} \int_{t_0}^{t_f} [\,\textbf{x}^{\text{T}}(t)\textbf{Q}(t)\textbf{x}(t) + \textbf{u}^{\text{T}}(t)\textbf{R}(t)\textbf{u}(t)\,]\, \operatorname{d}t$ Subject to the linear first-order dynamic constraints $\dot{\textbf{x}}(t)=\textbf{A}(t) \textbf{x}(t) + \textbf{B}(t) \textbf{u}(t),$ and the initial condition $\textbf{x}(t_0) = \textbf{x}_0$ A particular form of the LQ problem that arises in many control system problems is that of the linear quadratic regulator (LQR) where all of the matrices (i.e., $\textbf{A}$, $\textbf{B}$, $, \textbf{Q}$, and $\textbf{R}$) are constant, the initial time is arbitrarily set to zero, and the terminal time is taken in the limit $t_f\rightarrow\infty$ (this last assumption is what is known as infinite horizon). The LQR problem is stated as follows. Minimize the infinite horizon quadratic continuous-time cost functional $J=\tfrac{1}{2} \int_{0}^{\infty}[\,\textbf{x}^{\text{T}}(t)\textbf{Q}\textbf{x}(t) + \textbf{u}^{\text{T}}(t)\textbf{R}\textbf{u}(t)\,]\, \operatorname{d}t$ Subject to the linear time-invariant first-order dynamic constraints $\dot{\textbf{x}}(t)=\textbf{A} \textbf{x}(t) + \textbf{B} \textbf{u}(t),$ and the initial condition $\textbf{x}(t_0) = \textbf{x}_0$ In the finite-horizon case the matrices are restricted in that $\textbf{Q}$ and $\textbf{R}$ are positive semi-definite and positive definite, respectively. In the infinite-horizon case, however, the matrices $\textbf{Q}$ and $\textbf{R}$ are not only positive-semidefinite and positive-definite, respectively, but are also constant. These additional restrictions on $\textbf{Q}$ and $\textbf{R}$ in the infinite-horizon case are enforced to ensure that the cost functional remains positive. Furthermore, in order to ensure that the cost function is bounded, the additional restriction is imposed that the pair $(\textbf{A},\textbf{B})$ is controllable. Note that the LQ or LQR cost functional can be thought of physically as attempting to minimize the control energy (measured as a quadratic form). The infinite horizon problem (i.e., LQR) may seem overly restrictive and essentially useless because it assumes that the operator is driving the system to zero-state and hence driving the output of the system to zero. This is indeed correct. However the problem of driving the output to a desired nonzero level can be solved after the zero output one is. In fact, it can be proved that this secondary LQR problem can be solved in a very straightforward manner. It has been shown in classical optimal control theory that the LQ (or LQR) optimal control has the feedback form $\textbf{u}(t)=-\textbf{K}(t)\textbf{x}(t)$ where $\textbf{K}(t)$ is a properly dimensioned matrix, given as $\textbf{K}(t)=\textbf{R}^{-1}\textbf{B}^{\text{T}}\textbf{S}(t),$ and $\textbf{S}(t)$ is the solution of the differential Riccati equation. The differential Riccati equation is given as $\dot{\textbf{S}}(t) = -\textbf{S}(t)\textbf{A}-\textbf{A}^{\text{T}}\textbf{S}(t)+\textbf{S}(t)\textbf{B}\textbf{R}^{-1}\textbf{B}^{\text{T}}\textbf{S}(t)-\textbf{Q}$ For the finite horizon LQ problem, the Riccati equation is integrated backward in time using the terminal boundary condition $\textbf{S}(t_f) = \textbf{S}_f$ For the infinite horizon LQR problem, the differential Riccati equation is replaced with the algebraic Riccati equation (ARE) given as $\textbf{0} = -\textbf{S}\textbf{A}-\textbf{A}^{\text{T}}\textbf{S}+\textbf{S}\textbf{B}\textbf{R}^{-1}\textbf{B}^{\text{T}}\textbf{S}-\textbf{Q}$ Understanding that the ARE arises from infinite horizon problem, the matrices $\textbf{A}$, $\textbf{B}$, $\textbf{Q}$, and $\textbf{R}$ are all constant. It is noted that there are in general multiple solutions to the algebraic Riccati equation and the positive definite (or positive semi-definite) solution is the one that is used to compute the feedback gain. The LQ (LQR) problem was elegantly solved by Rudolf Kalman.[3] ## Numerical methods for optimal control Optimal control problems are generally nonlinear and therefore, generally do not have analytic solutions (e.g., like the linear-quadratic optimal control problem). As a result, it is necessary to employ numerical methods to solve optimal control problems. In the early years of optimal control (circa 1950s to 1980s) the favored approach for solving optimal control problems was that of indirect methods. In an indirect method, the calculus of variations is employed to obtain the first-order optimality conditions. These conditions result in a two-point (or, in the case of a complex problem, a multi-point) boundary-value problem. This boundary-value problem actually has a special structure because it arises from taking the derivative of a Hamiltonian. Thus, the resulting dynamical system is a Hamiltonian system of the form $\begin{array}{lcl} \dot{\textbf{x}} & = & \partial H/\partial\boldsymbol{\lambda} \\ \dot{\boldsymbol{\lambda}} & = & -\partial H/\partial\textbf{x} \end{array}$ where $H=\mathcal{L}+\boldsymbol{\lambda}^{\text{T}}\textbf{a}-\boldsymbol{\mu}^{\text{T}}\textbf{b}$ is the augmented Hamiltonian and in an indirect method, the boundary-value problem is solved (using the appropriate boundary or transversality conditions). The beauty of using an indirect method is that the state and adjoint (i.e., $\boldsymbol{\lambda}$) are solved for and the resulting solution is readily verified to be an extremal trajectory. The disadvantage of indirect methods is that the boundary-value problem is often extremely difficult to solve (particularly for problems that span large time intervals or problems with interior point constraints). A well-known software program that implements indirect methods is BNDSCO.[4] The approach that has risen to prominence in numerical optimal control over the past two decades (i.e., from the 1980s to the present) is that of so-called direct methods. In a direct method, the state and/or control are approximated using an appropriate function approximation (e.g., polynomial approximation or piecewise constant parameterization). Simultaneously, the cost functional is approximated as a cost function. Then, the coefficients of the function approximations are treated as optimization variables and the problem is "transcribed" to a nonlinear optimization problem of the form: Minimize $F(\textbf{z})\,$ subject to the algebraic constraints $\begin{array}{lcl} \textbf{g}(\textbf{z}) & = & \textbf{0} \\ \textbf{h}(\textbf{z}) & \leq & \textbf{0} \end{array}$ Depending upon the type of direct method employed, the size of the nonlinear optimization problem can be quite small (e.g., as in a direct shooting or quasilinearization method) or may be quite large (e.g., a direct collocation method[5]). In the latter case (i.e., a collocation method), the nonlinear optimization problem may be literally thousands to tens of thousands of variables and constraints. Given the size of many NLPs arising from a direct method, it may appear somewhat counter-intuitive that solving the nonlinear optimization problem is easier than solving the boundary-value problem. It is, however, the fact that the NLP is easier to solve than the boundary-value problem. The reason for the relative ease of computation, particularly of a direct collocation method, is that the NLP is sparse and many well-known software programs exist (e.g., SNOPT[6]) to solve large sparse NLPs. As a result, the range of problems that can be solved via direct methods (particularly direct collocation methods which are very popular these days) is significantly larger than the range of problems that can be solved via indirect methods. In fact, direct methods have become so popular these days that many people have written elaborate software programs that employ these methods. In particular, many such programs written in FORTRAN include DIRCOL,[7] SOCS,[8] OTIS,[9] GESOP/ASTOS[10] and DITAN.[11] In recent years, due to the advent of the MATLAB programming language, optimal control software in MATLAB has become more common. Examples of academically developed MATLAB software tools implementing direct methods include RIOTS,[12]DIDO,[13] DIRECT,[14] and GPOPS,[15] while an example of an industry developed MATLAB tool is PROPT.[16] These software tools have increased significantly the opportunity for people to explore complex optimal control problems both for academic research and industrial-strength problems. Finally, it is noted that general-purpose MATLAB optimization environments such as TOMLAB have made coding complex optimal control problems significantly easier than was previously possible in languages such as C and FORTRAN. ## Discrete-time optimal control The examples thus far have shown continuous time systems and control solutions. In fact, as optimal control solutions are now often implemented digitally, contemporary control theory is now primarily concerned with discrete time systems and solutions. The Theory of Consistent Approximations[17] provides conditions under which solutions to a series of increasingly accurate discretized optimal control problem converge to the solution of the original, continuous-time problem. Not all discretization methods have this property, even seemingly obvious ones. For instance, using a variable step-size routine to integrate the problem's dynamic equations may generate a gradient which does not converge to zero (or point in the right direction) as the solution is approached. The direct method RIOTS is based on the Theory of Consistent Approximation. ## Examples A common solution strategy in many optimal control problems is to solve for the costate (sometimes called the shadow price) $\lambda(t)$. The costate summarizes in one number the marginal value of expanding or contracting the state variable next turn. The marginal value is not only the gains accruing to it next turn but associated with the duration of the program. It is nice when $\lambda(t)$ can be solved analytically, but usually the most one can do is describe it sufficiently well that the intuition can grasp the character of the solution and an equation solver can solve numerically for the values. Having obtained $\lambda(t)$, the turn-t optimal value for the control can usually be solved as a differential equation conditional on knowledge of $\lambda(t)$. Again it is infrequent, especially in continuous-time problems, that one obtains the value of the control or the state explicitly. Usually the strategy is to solve for thresholds and regions that characterize the optimal control and use a numerical solver to isolate the actual choice values in time. ### Finite time Consider the problem of a mine owner who must decide at what rate to extract ore from his mine. He owns rights to the ore from date $0$ to date $T$. At date $0$ there is $x_0$ ore in the ground, and the instantaneous stock of ore $x(t)$ declines at the rate the mine owner extracts it u(t). The mine owner extracts ore at cost $u(t)^2/x(t)$ and sells ore at a constant price $p$. He does not value the ore remaining in the ground at time $T$ (there is no "scrap value"). He chooses the rate of extraction in time u(t) to maximize profits over the period of ownership with no time discounting. 1. Discrete-time version The manager maximizes profit $\Pi$: $\Pi = \sum_{t=0}^{T-1} \left[ pu_t - \frac{u_t^2}{x_t} \right]$ subject to the law of evolution for the state variable $x_t$ $x_{t+1} - x_t = - u_t\!$ Form the Hamiltonian and differentiate: $H = pu_t - \frac{u_t^2}{x_t} - \lambda_{t+1} u_t$ $\frac{\partial H}{\partial u_t} = p - \lambda_{t+1} - 2\frac{u_t}{x_t} = 0$ $\lambda_{t+1} - \lambda_t = -\frac{\partial H}{\partial x_t} = -\left( \frac{u_t}{x_t} \right)^2$ As the mine owner does not value the ore remaining at time $T$, $\lambda_T = 0\!$ Using the above equations, it is easy to solve for the $x_t$ and $\lambda_t$ series $\lambda_t = \lambda_{t+1} + \frac{(p-\lambda_{t+1})^2}{4}$ $x_{t+1} = x_t \frac{2 - p + \lambda_{t+1}}{2}$ and using the initial and turn-T conditions, the $x_t$ series can be solved explicitly, giving $u_t$. 2. Continuous-time version The manager maximizes profit $\Pi$: $\Pi = \int_0^T \left[ pu(t) - \frac{u(t)^2}{x(t)} \right] dt$ subject to the law of evolution for the state variable $x(t)$ $\dot x(t) = - u(t)$ Form the Hamiltonian and differentiate: $H = pu(t) - \frac{u(t)^2}{x(t)} - \lambda(t) u(t)$ $\frac{\partial H}{\partial u} = p - \lambda(t) - 2\frac{u(t)}{x(t)} = 0$ $\dot\lambda(t) = -\frac{\partial H}{\partial x} = -\left( \frac{u(t)}{x(t)} \right)^2$ As the mine owner does not value the ore remaining at time $T$, $\lambda(T) = 0$ Using the above equations, it is easy to solve for the differential equations governing $u(t)$ and $\lambda(t)$ $\dot\lambda(t) = -\frac{(p-\lambda(t))^2}{4}$ $u(t) = x(t) \frac{p- \lambda(t)}{2}$ and using the initial and turn-T conditions, the functions can be solved numerically. ## References 1. L. S. Pontryagin, 1962. The Mathematical Theory of Optimal Processes. 2. I. M. Ross, 2009. A Primer on Pontryagin's Principle in Optimal Control, Collegiate Publishers. ISBN 978-0-9843571-0-9. 3. Kalman, Rudolf. A new approach to linear filtering and prediction problems. Transactions of the ASME, Journal of Basic Engineering, 82:34–45, 1960 4. Oberle, H. J. and Grimm, W., "BNDSCO-A Program for the Numerical Solution of Optimal Control Problems," Institute for Flight Systems Dynamics, DLR, Oberpfaffenhofen, 1989 5. Betts, J. T., Practical Methods for Optimal Control Using Nonlinear Programming, SIAM Press, Philadelphia, Pennsylvania, 2001 6. Gill, P. E., Murray, W. M., and Saunders, M. A., User's Manual for SNOPT Version 7: Software for Large-Scale Nonlinear Programming, University of California, San Diego Report, 24 April 2007 7. von Stryk, O., User's Guide for DIRCOL (version 2.1): A Direct Collocation Method for the Numerical Solution of Optimal Control Problems, Fachgebiet Simulation und Systemoptimierung (SIM), Technische Universität Darmstadt (2000, Version of November 1999). 8. Betts, J.T. and Huffman, W. P., Sparse Optimal Control Software, SOCS, Boeing Information and Support Services, Seattle, Washington, July 1997 9. Hargraves, C. R. and Paris, S. W., "Direct Trajectory Optimization Using Nonlinear Programming and Collocation", Journal of Guidance, Control, and Dynamics, Vol. 10, No. 4., 1987, pp. 338–342 10. Gath, P.F., Well, K.H., "Trajectory Optimization Using a Combination of Direct Multiple Shooting and Collocation", AIAA 2001–4047, AIAA Guidance, Navigation, and Control Conference, Montréal, Québec, Canada, 6–9 August 2001 11. Vasile M., Bernelli-Zazzera F., Fornasari N., Masarati P., "Design of Interplanetary and Lunar Missions Combining Low-Thrust and Gravity Assists", Final Report of the ESA/ESOC Study Contract No. 14126/00/D/CS, September 2002 12. Schwartz, Adam, Theory and Implementation of Methods based on Runge–Kutta Integration for Solving Optimal Control Problems, University of California at Berkeley, PhD Dissertation, 1996. 13. Ross, I. M. and Fahroo, F., User's Manual for DIDO: A MATLAB Package for Dynamic Optimization, Dept. of Aeronautics and Astronautics, Naval Postgraduate School Technical Report, 2002 14. Williams, P., User's Guide to DIRECT, Version 2.00, Melbourne, Australia, 2008 15. Rao, A. V., Benson, D. A., Huntington, G. T., Francolin, C., Darby, C. L., and Patterson, M. A., , University of Florida Report, August 2008. 16. Rutquist, P. and Edvall, M. M, PROPT – MATLAB Optimal Control Software," 1260 S.E. Bishop Blvd Ste E, Pullman, WA 99163, USA: Tomlab Optimization, Inc. 17. E. Polak, On the use of consistent approximations in the solution of semi-infinite optimization and optimal control problems Math. Prog. 62 pp. 385–415 (1993). ## Further reading Books • Athans, M. A. and Falb, P. L., Optimal Control, McGraw–Hill, New York, 1966. • Becerra, V.M., 2008, Optimal control. Scholarpedia, 3(1):5354 • Bryson, A. E., 1969. Applied Optimal Control: Optimization, Estimation, & Control. • Bryson, A.E and Ho, Y., "Applied Optimal Control: Optimization, Estimation and Control (Revised Printing)", John Wiley and Sons, New York, 1975. • Evans, L.C., An Introduction to Optimal Control Theory (available free online) • Ross, I. M. A Primer on Pontryagin's Principle in Optimal Control, Collegiate Publishers, 2009. ISBN 978-0-9843571-0-9. (http://www.ElissarGlobal.com free chapter available online) • H. O. Fattorini and S. S. Sritharan, "Existence of Optimal Controls for Viscous Flow Problems," Proceedings of the Royal Society of London Series A, Vol. 439, 1992, pp. 81–102. • H. O. Fattorini and S. S. Sritharan, "Necessary and Sufficient Conditions for Optimal Controls in Viscous Flow," Proceedings of the Royal Society of Edinburgh, Series A, Vol. 124A, 1994, pp. 211–251. • H. O. Fattorini and S. S. Sritharan, "Optimal chattering controls for viscous flow," Nonlinear analysis, Theory, Methods and Applications, Vol. 25, No. 8, pp. 763–797, 1995. • H. O. Fattorini and S. S. Sritharan,"Optimal control problems with state constraints in fluid mechanics and combustion," Applied Mathematics and Optimization, Vol. 38(2), 1998, pp. 159–192. • Kirk, D. E., 2004. Optimal Control Theory: An Introduction. • Lebedev, L. P., and Cloud, M. J., 2003. The Calculus of Variations and Functional Analysis with Optimal Control and Applications in Mechanics. World Scientific. Especially chpt. 2. • Lewis, F. L., and Syrmos, V. L., 19nn. Optimal Control, 2nd ed. John Wiley & Sons. • S. S. Sritharan, "Optimal Control of Viscous Flow", SIAM, 1998. ( http://www.nps.edu/academics/schools/gseas/sri/Sritharan-Optimal_Control_of_Viscous_Flow.pdf ) • S. S. Sritharan, "An Optimal Control Problem in Exterior Hydrodynamics," Proceedings of the Royal Society of Edinburgh, Series 121A, 1992, pp. 5–32. • S. S. Sritharan,"Deterministic and stochastic control of viscous flow with linear, monotone and hyper viscosities," Applied Mathematics and Optimization, Vol. 41(2), pp. 255–308, 2000. • Stengel, R. F., 1994. Optimal Control and Estimation. Dover. • Sethi, S. P., and Thompson, G. L., 2000. Optimal Control Theory: Applications to Management Science and Economics, 2nd edition, Springer (ISBN 0387280928 and ISBN 0-7923-8608-6). Slides are available at http://www.utdallas.edu/~sethi/OPRE7320presentation.html • Sontag, Eduardo D. Mathematical Control Theory: Deterministic Finite Dimensional Systems. Second Edition. Springer. (ISBN 0-387-984895) (available free online) • Brogan, William L. 1990. Modern Control Theory. ISBN 0-13-589763-7 • Bryson, A.E.; Ho, Y.C. (1975). Applied optimal control. Washington, DC: Hemisphere. Journals • Optimal Control Applications and Methods. John Wiley & Sons, Inc. • SIAM Journal of Control and Optimization.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 86, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8806440234184265, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/241735/the-roots-of-the-cubic-x3qxr-0-are-a-b-c-how-can-i-find-the-equation-who	# The roots of the cubic $x^3+qx+r=0$ are $a,b,c$. How can I find the equation whose roots are $la+mbc,lb+mca,lc+mab$ The roots of the cubic $x^3+qx+r=0$ are $a,b,c$. How can I find the equation whose roots are $la+mbc,lb+mca,lc+mab$? Can anyone help me to solve this problem? - ## 2 Answers The long, brute-force way: Expand out $\left(x-(la+mbc)\right)\left(x-(lb+mca)\right)\left(x-(lc+mab)\right)$ to get its full form as a cubic; you should find that all of the coefficients are symmetric functions of $(a,b,c)$. Then use the Fundamental Theorem Of Symmetric Polynomials to express those coefficients in terms of the basic symmetric polynomials $S_1(a,b,c) = a+b+c$, $S_2(a,b,c) = ab+bc+ca$, and $S_3(a,b,c)=abc$. Finally, use the traditional theorems on expressing the coefficients of a polynomial in terms of its roots (e.g., $r=-abc$) to express the coefficients in terms of $q$ and $r$. - $la+mbc,lb+mca,lc+mab$ as our roots means that From the given equation we have $a+b+c = 0$ since coefficient of $x^2$ is 0. and $ab+ba+bc =q$ $x^3 + px^2+ qx + r$ (independent of the equation in your question) $p = l(a+b+c) + m(bc+ca+ab)$ $p = l(0) + m(q) = mq$ where q is the coefficient of x in your expression Similarly (I haven't really expressed it in terms of the coefficients of your given equation), $q = (la+mbc)(lb+mca)+(la+mbc)(lc+mba)+lb(mca)(lc+mba)$ $r = (la+mbc)(lb+mca)(lc+mba)$ The cubic polynomial with these roots is of the form $x^3 - (l(a+b+c) + m(bc+ca+ab)))x^2 + ((la+mbc)(lb+mca)+(la+mbc)(lc+mba)+lb(mca)(lc+mba)) x - (la+mbc)(lb+mca)(lc+mba)$ -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 24, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9470086097717285, "perplexity_flag": "head"}
http://mathhelpforum.com/advanced-math-topics/33063-solar-funnel-trig-problem.html	# Thread: 1. ## solar funnel trig problem I posted this in a high school forum yesterday, but didn't get any response, so sorry if you see this twice. I'm trying to work out a formula that solves theta given x and y. I'll attach a diagram. What I'm trying to work out is the smallest theta (widest funnel) where light hitting the top of the funnel just enters the oven, and doesn't hit the funnel on the other side. any help? Attached Thumbnails 2. Originally Posted by nervousnelly I posted this in a high school forum yesterday, but didn't get any response, so sorry if you see this twice. I'm trying to work out a formula that solves theta given x and y. I'll attach a diagram. What I'm trying to work out is the smallest theta (widest funnel) where light hitting the top of the funnel just enters the oven, and doesn't hit the funnel on the other side. any help? You need to solve: $\tan(180-2\theta)=\frac{x+y \cos(\theta)}{y \sin(\theta)}$ use trig identities to reduce to something containing $\sin$s and $\cos$s of $\theta$ only. then see what you can do RonL 3. Thanks CaptainBlack, I appreciated the time you took to getting me closer to the solution. The trouble is high school trig was 16 years ago, and even if I learnt trig identities, i forgot them 2 hours after the last exam. This is for a personal project I'm working on, a solar funnel. I'm not in any classes, it's a one off problem I googled trig identities, and it may as well have been written in greek. (sight pun there) 4. Originally Posted by nervousnelly Thanks CaptainBlack, I appreciated the time you took to getting me closer to the solution. The trouble is high school trig was 16 years ago, and even if I learnt trig identities, i forgot them 2 hours after the last exam. This is for a personal project I'm working on, a solar funnel. I'm not in any classes, it's a one off problem I googled trig identities, and it may as well have been written in greek. (sight pun there) Don't bother, now you will learn the dreadfull secret of maths, when we have a practical problem we don't usually bother with all that. Just post the values of x and y and we will solve this numerically for you (without tidying up the trig) If you need to find theta for many x's and y's what we can do is observe that theta depends on z=x/y, so we compute a table of theta against z. RonL 5. thanks, what I'm hoping for is a little web page with javascript or php function to calculate theta. so x might be 40 and y might be 50, but could be anything. i think if x = y, theta is 90 - 45 / 2 6. Originally Posted by nervousnelly thanks, what I'm hoping for is a little web page with javascript or php function to calculate theta. so x might be 40 and y might be 50, but could be anything. i think if x = y, theta is 90 - 45 / 2 I think the formula you are looking for is $\boxed{\cos\theta = \frac{\sqrt{y^2+8x^2}-y}{4x}}$. When x = y, this gives θ = 60°. This is certainly correct, because if x=y then the triangle formed by the sides labelled x and y, together with the long dashed red line of incident light, is isosceles (another long-forgotten word from high school?), and it's then easy to verify geometrically that its angles must be 30°, 30° and 120°. 7. that's awesome opalg, thank you very much. i made 50 = x = y and got cos t = 1/2 not sure what to do from there? (you're right, no idea what an isosceles is ) edit: i worked it out (cos^-1), thank you very much captainblack and opalg. I appreciate it. 8. I think something is a miss. I seem to be getting numbers between 45 and 90, rather than 0 and 90. If I make y=2 and x=200, I should get a figure close to 0, but I get 45.something. any ideas? 9. Originally Posted by nervousnelly I think something is amiss. I seem to be getting numbers between 45 and 90, rather than 0 and 90. If I make y=2 and x=200, I should get a figure close to 0, but I get 45.something. any ideas? The angle θ can never be less than 45°. If you look at your diagram, you'll see that if θ=45° then the vertical ray of incident light gets reflected horizontally across to the opposite rim of the funnel. If θ is less than 45° then the ray will be reflected back upwards and certainly cannot enter the funnel. Attached Thumbnails 10. you're absolutely right, i see it now. thank you very much, i appreciate it. 11. I might as well show this done numerically: Code: ```>function bi(z1) $ global z $ ll=length(z1);rv=[]; $ ss="tan(pi-2*x)-(z+cos(x))/sin(x)"; $ for idx=1 to ll $ z=z1(idx); $ rv=rv_[z,bisect(ss,pi/4,pi/2)]; $ end $ return rv $endfunction > >bi([0.1:0.1:3]) 0.1 1.47256 0.2 1.38356 0.3 1.30822 0.4 1.24641 0.5 1.19606 0.6 1.15483 0.7 1.12072 0.8 1.09215 0.9 1.06794 1 1.0472 1.1 1.02925 1.2 1.01358 1.3 0.999785 1.4 0.987552 1.5 0.976632 1.6 0.966827 1.7 0.957974 1.8 0.949944 1.9 0.942625 2 0.935929 2.1 0.92978 2.2 0.924112 2.3 0.918873 2.4 0.914014 2.5 0.909497 2.6 0.905286 2.7 0.901352 2.8 0.897667 2.9 0.894209 3 0.890958 >``` RonL	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 5, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9528131484985352, "perplexity_flag": "middle"}
http://nrich.maths.org/792/note	### Prompt Cards These two group activities use mathematical reasoning - one is numerical, one geometric. ### Consecutive Numbers An investigation involving adding and subtracting sets of consecutive numbers. Lots to find out, lots to explore. ### Exploring Wild & Wonderful Number Patterns EWWNP means Exploring Wild and Wonderful Number Patterns Created by Yourself! Investigate what happens if we create number patterns using some simple rules. # Dodecamagic ### Why do this problem? Visualising is a very important mathematical skill. Representing 3d shapes in a 2d form is a sophisticated form of visualising. This problem offers opportunities to visualise and to use deductive reasoning with small numbers. ### Possible approach If your pupils are used to visualising, you may wish to go straight into the problem and see how far they can get. Display the pictures and ensure that everyone understands the problem, then after a little time draw the pupils together and ask for any useful statements or observations they can make. At this stage you may want to suggest that they could choose to use a 3d model if that would help, but you might also want to challenge them to do it 'in their heads'. If your pupils are not used to visualising, you may want to begin by organising them into pairs to make a 3-d model from a net, or from other plastic shapes you have in the classroom, such as Polydron. The main challenge is to match the information from one diagram onto the other. Once the children realise how the two diagrams are connected, the arithmetic is relatively trivial. Labelling the vertices and then opening up the net of the solid can help make the connections too. ### Key questions There are $9$s on both diagrams. Does that help? How? What about the $25$? What number does the F represent? How does the name of the shape help you to know? ### Possible extension You may wish to offer some other model-making activities which serve to consolidate 2d representation of 3d objects. There is a paper folding example of a dodecahedron here, which dextrous children may be able to do, perhaps with a little support. You might want to try making one yourself first! ### Possible support Support children who find visualising difficult with the latest in digital technology - use a digital camera to take photographs of the front and back of different 3d shapes including polyhedra and ask them to match them up, identifying common edges or vertices. The NRICH Project aims to enrich the mathematical experiences of all learners. To support this aim, members of the NRICH team work in a wide range of capacities, including providing professional development for teachers wishing to embed rich mathematical tasks into everyday classroom practice. More information on many of our other activities can be found here.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9359856247901917, "perplexity_flag": "middle"}
http://mathhelpforum.com/calculus/42264-limits-continuity.html	# Thread: 1. ## Limits and continuity... Hello... Q1) Find the limit of the following using L'Hospital rule.... a) lim x-> 0+ [sin(x) - x] / [e^x - 1] b) lim x-> pi/2 [pi/2 - x] . tan(x) Q2) Prove using the Mean value theorem: (x-1)/x < ln(x) < x-1, when x>1 Thanks! 2. Originally Posted by Vedicmaths Hello... Q1) Find the limit of the following using L'Hospital rule.... a) lim x-> 0+ [sin(x) - x] / [e^x - 1] b) lim x-> pi/2 [pi/2 - x] . tan(x) Q2) Prove using the Mean value theorem: (x-1)/x < ln(x) < x-1, when x>1 Thanks! Must you use L'hopital's? Ok for the first one seeing that if we let $sin(x)-x=f(x)$ and $g(x)=e^x-1$ and $g(x)=f(x)=0$ Then $\lim_{x\to{0}}\frac{f(x)}{g(x)}=\lim_{x\to{0}}\fra c{f'(x)}{g'(x)}$ 3. Edit: duplicated mathstuds fine work, sorry L'Hopital's rule states that $\lim_{x\to x_0}\frac{f(x)}{g(x)} = \lim_{x\to x_0}\frac{f'(x)}{g'(x)}$ if $f(x_0)=0$ and $g(x_0) = 0$ or $f(x_0) = \pm \infty$ and $g(x_0) = \pm \infty$ Go for it. Question 1 is easy. For question 2 start by taking the derivative of all parts of the inequality. 4. Originally Posted by Vedicmaths Hello... Q1) Find the limit of the following using L'Hospital rule.... a) lim x-> 0+ [sin(x) - x] / [e^x - 1] b) lim x-> pi/2 [pi/2 - x] . tan(x) Q2) Prove using the Mean value theorem: (x-1)/x < ln(x) < x-1, when x>1 Thanks! The first one is of the form $\frac{0}{0}$ so now using L'hospitals rule we take a derivative of the numerator and the denominator to get $\lim_{x \to 0^+}\frac{\cos(x)-1}{e^x}=\frac{\cos(0)-1}{e^{0}}=\frac{1-1}{1}=0$ For the 2nd rewrite it as $\frac{[x-\pi/2]\sin(x)}{\cos(x)}$ now as $x\to \frac{\pi}{2}$ this is of the form $\frac{0}{0}$ Using L.H again we get $\lim_{x \to \frac{\pi}{2}}\frac{[x-\pi/2]\cos(x)+\sin(x)}{-\sin(x)}=\frac{[0]0+1}{-1}=-1$ For the 2nd question consider the function $f(x)=\ln(x)$ f is continous on $[1,\infty)$ and differentiable on $(1,\infty)$ so it satisfies the hypothesis of the MVT by the MVT $f(b)-f(a)=f'(c)(b-a)$ where $c \in (a,b)$ Let a=1 and b=x and we get $\ln(x)-\ln(1)=\frac{1}{c}\left( x-1 \right)$ simplifying we get $\ln(x) =\frac{x-1}{c} \iff c\ln(x)=x-1$ Since $c \in (1,x) \mbox{ or } 1< c < x$ $\ln(x) < x-1$ Also by the same reasoning $c < x \iff \frac{1}{x} < \frac{1}{c}$ so $\frac{x-1}{x}< \frac{x-1}{c}=\ln(x)$ so finally we get $\frac{x-1}{x} < \ln(x) < x-1$ Yeah!!! 5. Is it necessary L'Hôpital? Q2 can also be killed as follows: Since $\ln x=\int_{1}^{x}{\frac{dy}{y}},$ we have $\frac{1}{x}\le \frac{1}{y}\le 1\implies \frac{x-1}{x}\le \ln x\le x-1.\quad\blacksquare$ (This holds for $x>0.$) 6. Thanks a lot! That really helps alot! 7. Dear Empty Set! In the second problem you suggested to rewrite as ....but how can we say that...If we use (pi/2) form we will end up getting 1. Ans you are getting -1. Will that make any difference? Thanks! 8. Originally Posted by Vedicmaths Dear Empty Set! In the second problem you suggested to rewrite as ....but how can we say that...If we use (pi/2) form we will end up getting 1. Ans you are getting -1. Will that make any difference? Thanks! $\lim_{x \to \frac{\pi}{2}}\frac{[x-\frac{\pi}{2}]\sin(x)}{\cos(x)}=\frac{[\frac{\pi}{2}-\frac{\pi}{2}]\sin(\frac{\pi}{2})}{\cos(\frac{\pi}{2})}=\frac{[0]\cdot 1}{0}=\frac{0}{0}$ So now we can use L'hospitals rule. Then we can take the derivatives to get what I did above. I hope this helps. Good luck. 9. ## confusion! Yes Sir! I got what you were trying to do after using the L'hospital Rule but I was asking that why are you manipulating the term? Since, in the problem it was given as ( pi/2 - x) sinx / cosx...but you have mentioned that we can rewrite the function as So why are you changing the sign here? And I guess that will change the answer. I got 1 from the function given in the original question and you are getting -1 from this one. I am sorry to bother you. But I am just a little bit confused and want to learn if something I do not know. Thanks so much once again! Regards, Originally Posted by Vedicmaths b) lim x-> pi/2 [pi/2 - x] . tan(x) If we take the limit as $x \to \frac{\pi}{2}$ we get $0\cdot \pm \infty$ This doesnt quite fit the form to use L'hospitals rule. We need to manipulate the equation into the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$ Hence I changed $[\frac{\pi}{2}-x]\tan(x)=\frac{[\frac{\pi}{2}-x]\sin(x)}{\cos(x)}$ Now this is of the form $\frac{0}{0}$ So from here we take the derivative of the numerator and the denomiator as per L.H's rule. I guess I don't understand what you mean changing the sign? 11. Originally Posted by TheEmptySet If we take the limit as $x \to \frac{\pi}{2}$ we get $0\cdot \pm \infty$ This doesnt quite fit the form to use L'hospitals rule. We need to manipulate the equation into the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$ Hence I changed $[\frac{\pi}{2}-x]\tan(x)=\frac{[\frac{\pi}{2}-x]\sin(x)}{\cos(x)}$ Now this is of the form $\frac{0}{0}$ So from here we take the derivative of the numerator and the denomiator as per L.H's rule. I guess I don't understand what you mean changing the sign? You did Originally Posted by TheEmptySet For the 2nd rewrite it as $\frac{[x-\pi/2]\sin(x)}{\cos(x)}$ Which is the negative of the original problem. -Dan 12. Sorry, I didn't even notice that I switched them. Thanks	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 44, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9096285700798035, "perplexity_flag": "middle"}
http://www.physicsforums.com/showthread.php?t=100606	Physics Forums ## Complex integration I have the function $$f(z)=\frac{e^z}{1+e^{4z}}$$. and the loop $$\gamma_{r_1,r_2}=I_{r_1,r_2}+II_{r_2}+III_{r_1,r_2}+IV_{r_1},\quad r_1,r_2>0,$$ which bounds the domain $$A_{r_1,r_2}=\{z\in\mathbb{C}\mid -r_1<\Re(z)< r_2\wedge0<\Im(z)<\pi\}.$$ Now I have to show that $$\int_{\gamma_{r_1,r_2}}f(z)\,dz=\tfrac{\sqrt2}{2}\pi.$$ I know that the integral along $\gamma_{r_1,r_2}$ is the sum of the integrals along the four countours $I_{r1,r_2},\dots,IV_{r_1}$. I also know that if $f$ is continuous in a domain D and has an antiderivative $F$ throughout D, then the integral along a contour lying in D is $F(z_T)-F(z_I)$, where $z_T$ is the terminal point, and $z_I$ the initial point, of the countour. However, using this, I am not able to show the aforementioned result, i.e, I get $\int_{\gamma_{r_1,r_2}}f(z)\,dz=0$. Could anyone try to calculate the integral, and then report back what they've found. PhysOrg.com science news on PhysOrg.com >> Front-row seats to climate change>> Attacking MRSA with metals from antibacterial clays>> New formula invented for microscope viewing, substitutes for federally controlled drug Recognitions: Homework Help Science Advisor Quote by sigmund I have the function $$f(z)=\frac{e^z}{1+e^{4z}}$$. and the loop $$\gamma_{r_1,r_2}=I_{r_1,r_2}+II_{r_2}+III_{r_1,r_2}+IV_{r_1},\quad r_1,r_2>0,$$ which bounds the domain $$A_{r_1,r_2}=\{z\in\mathbb{C}\mid -r_1<\Re(z)< r_2\wedge0<\Im(z)<\pi\}.$$ Now I have to show that $$\int_{\gamma_{r_1,r_2}}f(z)\,dz=\tfrac{\sqrt2}{2}\pi.$$ I know that the integral along $\gamma_{r_1,r_2}$ is the sum of the integrals along the four countours $I_{r1,r_2},\dots,IV_{r_1}$. I also know that if $f$ is continuous in a domain D and has an antiderivative $F$ throughout D, then the integral along a contour lying in D is $F(z_T)-F(z_I)$, where $z_T$ is the terminal point, and $z_I$ the initial point, of the countour. However, using this, I am not able to show the aforementioned result, i.e, I get $\int_{\gamma_{r_1,r_2}}f(z)\,dz=0$. Could anyone try to calculate the integral, and then report back what they've found. Ok, your contour as stated is tough for me to follow but looks like you're going around the origin in such a was as to encompass poles of the integrand. Right? When is the denominator zero in the complex plane? Thus can't rely on Cauchy's Theorem but rather the Residue Theorem. Edit: Alright, suppose Cauchy's Theorem is a special case of the Residue Theorem but you know what I mean. Quote by saltydog Ok, your contour as stated is tough for me to follow, but looks like you're going around the origin in such a was as to encompass poles of the integrand. Right? When is the denominator zero in the complex plane? Thus can't rely on Cauchy's Theorem but rather the Residue Theorem. Edit: Alright, suppose Cauchy's Theorem is a special case of the Residue Theorem but you know what I mean. The function f has singularities in $z=\frac{i}{4}\left(\pi+k2\pi\right),~k=0,1,2,\dots$, and the contour is a rectangle with vertices at $z=-r_1$, $z=r_2$, $z=r_2+i\pi$, and $z=-r_1+i\pi$, where $r_1,r_2>0$. It is also oriented counter-clockwise. Moreover, in the first question of the exercise, I am asked to calculate the residues at the singularities. Comparing Cauchy's Integral Theorem and Cauchy's Residue Theorem, I clearly see that the Residue Theorem can be used here, because the Integral Theorem requires the domain, which contains the contour, to be simply connected, while the Residue Theorem allows singularities inside the contour. Thus the Residue Theorem would be the right one to use here. I have not tried to apply this yet, but thanks for pointing me towards this. EDIT: I have just used Maple to quickly calculate the integral, using the Residue Theorem, and I got the right result. Recognitions: Homework Help Science Advisor ## Complex integration Quote by sigmund EDIT: I have just used Maple to quickly calculate the integral, using the Residue Theorem, and I got the right result. Hey Sigmund, you do know how to calculate the residues by hand right? That is, how to show: $$\mathop\lim\limits_{z\to \pi i/4} (z-\pi i/4) \frac{e^z}{1+e^{4z}}=-\frac{1/ \sqrt{2}+i/ \sqrt{2}}{4}$$ Quote by saltydog Hey Sigmund, you do know how to calculate the residues by hand right? That is, how to show: $$\mathop\lim\limits_{z\to \pi i/4} (z-\pi i/4) \frac{e^z}{1+e^{4z}}=-\frac{1/ \sqrt{2}+i/ \sqrt{2}}{4}$$ Just factor the denominator by determining the poles : $$z = \frac{i( \pi + 2k \pi)}{4}$$ and k = 0,1,2,3 So you get 4 factors, one for each k. Then you will see that the $$(z-\pi i/4)$$ term will also appear in the denominator. Be sure that this only goes for first order poles marlon Quote by marlon Just factor the denominator by determining the poles : $$z = \frac{i( \pi + 2k \pi)}{4}$$ and k = 0,1,2,3 So you get 4 factors, one for each k. Then you will see that the $$(z-\pi i/4)$$ term will also appear in the denominator. Be sure that this only goes for first order poles marlon I am not sure how to do this. The denominator of f has a zero of order 1 at $z=i\pi/4$, whence it can be written as $(z-i\pi/4)g(z)$, where g(z) is analytic at $i\pi/4$ and $g(i\pi/4)\neq0$. I do not know how to do this. Couldn't you give me a hint to how to solve this? Recognitions: Homework Help Science Advisor Quote by sigmund I am not sure how to do this. The denominator of f has a zero of order 1 at $z=i\pi/4$, whence it can be written as $(z-i\pi/4)g(z)$, where g(z) is analytic at $i\pi/4$ and $g(i\pi/4)\neq0$. I do not know how to do this. Couldn't you give me a hint to how to solve this? Yea, I'd like to know too. I was referring to the relation: $$\text{Res}_{z=z_0} \left(\frac{p(z)}{q(z)}\right)=\frac{p(z_0)}{q^{'}(z_0)}$$ Note: Only for a simple pole. Quote by saltydog Yea, I'd like to know too. I was referring to the relation: $$\text{Res}_{z=z_0} \left(\frac{p(z)}{q(z)}\right)=\frac{p(z_0)}{q^{'}(z_0)}$$ Note: Only for a simple pole. The function f has a simple pole at $z=z_0$ (z_0 has been stated several times in this thread, so I omit it here). Hence the residue at that point is $\text{Res}(f;z_0)=\lim_{z\to z_0}(z-z_0)f(z)$. If we let $f(z)=p(z)/q(z)$, where p and q are both analytic at z_0, and q has a simple zero at z_0, while $p(z_0)\neq0$, we can do the following simplification: $$\text{Res}(f;z_0)=\lim_{z\to z_0}(z-z_0)f(z)=\lim_{z\to z_0}(z-z_0)\frac{p(z)}{q(z)}=\lim_{z\to z_0}\frac{p(z)}{\frac{q(z)-q(z_0)}{z-z_0}}=\frac{\lim_{z\to z_0}p(z)}{\lim_{z\to z_0}\frac{q(z)-q(z_0)}{z-z_0}}=\frac{p(z_0)}{q'(z_0)}$$ This can be used in the actual problem, because the requirements about analyticity at z_0, and that p(z) is nonzero at z_0, are met. Using this, it is easy to calculate the residue at z_0, while factoring the denominator undoubtedly could be done, although difficult, i presume. As Albert Einstein said: "Everything should be made as simple as possible, but no simpler." Thread Tools \| \| \| \| \|------------------------------------------\|-------------------------------\|---------\| \| Similar Threads for: Complex integration \| \| \| \| Thread \| Forum \| Replies \| \| \| Introductory Physics Homework \| 6 \| \| \| Calculus \| 1 \| \| \| Calculus & Beyond Homework \| 1 \| \| \| Calculus & Beyond Homework \| 19 \| \| \| Calculus \| 6 \|	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 34, "mathjax_display_tex": 17, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9519866108894348, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/100518/closed-form-for-sum-over-orbit-of-roots-of-unity	# Closed form for sum over orbit of roots of unity Let $\sigma$ act on the $m$th roots of unity by $\sigma \zeta = \zeta^r$ for $1\le r <m, (r,m)=1.$ I'm looking to evaluate $\displaystyle\sum_{i=0}^{t(\zeta )-1} \zeta^{r^i},$ where $t(\zeta )$ is the size of the orbit of $\zeta$ under $\sigma .$ I keep running into this pesky little sum and was wondering if anyone on this forum might be able to point me in the direction of converting it to something simple (or at least more amenable for further computation.) It wouldn't surprise me if I'm overlooking something elementary, but it's already frustrated me to the point of posting here, so any guidance whatsoever would be greatly appreciated. - Where does $\sigma$ come from? Isn't $r$ fixed, but arbitrary? – user21436 Jan 19 '12 at 18:17 $\sigma$ is just a name I gave to the map $\zeta \mapsto \zeta^r.$ $r$ is fixed but arbitrary. – Tim Duff Jan 19 '12 at 18:22 ## 1 Answer You're looking at a sum consisting of values of an additive character on ${\mathbb Z}_m$, where the inputs run over a multiplicative subgroup of ${\mathbb Z}_m^\times$. You can use a linear combination of Dirichlet characters to detect that multiplicative subgroup, and this converts your problem to a linear combination of Gauss sums. The relevant Dirichlet characters will be those $\chi$ such that $\chi(r)=1$. - 2 PS: Welcome back Wikipedia :) – Greg Martin Jan 19 '12 at 18:25 Looks like I'll be busy, thank you! – Tim Duff Jan 19 '12 at 18:36	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 17, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9222597479820251, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/9551/applications-of-the-spectral-theorem-to-quantum-mechanics?answertab=active	# Applications of the Spectral Theorem to Quantum Mechanics I'm currently learning some basic functional analysis. Yesterday I arrived at the spectral theorem of self-adjoint operators. I've heard that this theorem has lots of applications in Quantum Mechanics. But let me first state the formulation of the theorem that I'm using: Let $H$ be a Hilbert space. There's a 1-to-1-correspondence between self-adjoint operators $A$ on $H$ and spectral measures $P^{A}$ given by $$A~=~\int_{\mathbb{R}} \lambda ~dP^{A}.$$ ($\lambda$ denotes a constant, $\mathbb{R}$ denotes the real numbers.) A corollary is: Let $g:\mathbb{R}\to\mathbb{R}$ be a function. (Again: $\mathbb{R}$ denotes the set of real numbers.) Then: $$g(A)~:=~\int_{\mathbb{R}} g(\lambda)~ dP^{g(A)}$$ $$P^{g(A)}(\Delta) ~=~ P^{A}(g^{-1}(\Delta))$$ where $\Delta$ denotes a set in the $\sigma$-algebra of $\mathbb{R}$. Okay. Now this is the theorem. First I don't really the application of the corollary in Quantum mechanics. I've heard that suppose you're given an operator $A$ this means that it's easy for you to define operators like $\exp(A)$, especially on infinite dimensional Hilbert spaces. This indeed could be useful in quantum mechanics. Especially when thinking about the "time-evolution operator" of a system. However then I say: Why do you make things so complicated? Suppose you want to calculate $\exp(A)$. Why don't you define $$\exp(A)~:=~1+A+1/2 A^2 + \ldots$$ and require convergence with respect to the operator norm. An example: Consider the vectorspace spanned by the monomials $1,x,x^2,\ldots$ and let $A=d/dx$. Then you can perfectly define $$\exp(d/dx)~:=~1+ d/dx + 1/2 d^2/dx^2 + \ldots$$ and require convergence with respect to the operator norm. In addition to that I've heard that the spectral theorem gives a full description of all self-adjoint operators. Now why is that the case? I mean okay..there's a one to one correspondence between self-adjoint operators and spectral measures...but why does this give me any information about "the inner structure of the operator"? (And why is there this $\lambda$ in the integral? Looks somehow like an eigenvalue of $A$? But I'm just guessing) I'd me more than happy, if you could provide me with some intuition and ideas of how the theorem can be used. - 1 Dear @Matt_Quantum. Two comments: 1) Would you be willing to restrict to separable Hilbert spaces only? 2) The two examples $A=\frac{d}{dx}$ and $\exp(\frac{d}{dx})$ are not self-adjoint operators, if $x$ is supposed to be a real variable. – Qmechanic♦ May 7 '11 at 8:52 1) Okay 2) Okay. Sorry for that. But assume A is some arbitrary self-adjoint operator, one could still define exp(A)=1+A+1/2 A^2 +... and require convergence with respect to the operator norm. – Matt_Quantum May 7 '11 at 12:36 Dear @Matt_Quantum. But the operator norm $\|\|A\|\|$ of an arbitrary self-adjoint operator $A$ could be $\infty$. In the question formulation (v1) you never restricted yourself to only bounded operators, which is actually good, because in quantum mechanics there are lots of unbounded operators. For instance, $\|\| \frac{d}{dx} \|\|=\infty$. – Qmechanic♦ May 7 '11 at 14:56 I may try to work this up into an answer, but for now, I'll just comment: in a way, the spectral measure determines the operator up to a change of variables. Every operator whose spectrum is the real line can be transformed to $id\over dx$ acting on some subspace of the functions on the real line, after a change of variable, for example. It's almost true that the spectrum determines the operator (up to a change of basis) but you do actually need a little more info: multiplicites of eigenvavlues, whether they are part of a continuous set of eigenvalues, etc., spectral measure type thingies. – joseph f. johnson Jan 7 '12 at 22:09 ## 2 Answers Meta: It is true that a lot of quantum mechanics can be taught and understood without much knowledge of the mathematical foundations, and usually it is. Since QM is a mandatory class at many faculties that future experimental physicists have to attend, too, this also makes sense. But for future theoretical and mathematical physicists, it may pay off to learn a little bit about the math, too. A little anecdote: John von Neumann once said to Werner Heisenberg that mathematicians should be grateful for QM, because it led to the invention of a lot of beautiful mathematics, but that mathematicians repaid this by clarifying, e.g., the difference between a selfadjoint and a symmetric operator. Heisenberg asked: "What is the difference?" Suppose you want to calculte exp(A). Why don't you define exp(A):=1+A+1/2 A^2 + ... and require convergence with respect to the operator norm. That's correct. The benefit of the spectral theorem is that you can define f(A) for any selfadjoint (or more generally, normal) operator for any bounded Borel function. This comes in handy in many proofs in operator theory. In addition to that I've heard that the spectral theorem gives a full description of all self-adjoint operators. Now why is that the case? I mean okay..there's a one to one correspondence between self-adjoint operators and spectral measures.. That's correct, too. Spectral measures are much much simpler objects than selfadoint operators, that's why. Futhermore, you can use the spectral theorem to prove that every selfadjoint operator is unitarily equivalent to a multiplication operator (multiply f(x) by x). From an abstract viewpoint, this is a very satisfactory characterization. It does not help much for concrete calcuations in QM, though. BTW: On a more advanced level, you'll need to understand the spectral theorem to understand what a mass gap is in Yang-Mills Theory (millenium problem). Hint: In QFT in Minkowski-Spacetime, one usually assumes that there is a continuous representation of the Poincaré group, especially of the commutative subgroup of translations, on the Hilbert space that contains all physical states. The operators that form the representation have a common spectral measure, this is an application of the SNAG-theorem. The support of this spectral measure is bounded away from zero, that's the definition of the mass gap. - Dear Constantin, the $A=\int_R \lambda\,dP^A$ is just a continuous version of spectral decomposition. Here $dP^A$ is a differential version of the projection operators that define the Hermitian operator. For a discrete spectrum, the corresponding equation would be $$A = \sum_{i} \lambda_i P_{\lambda_{i}}$$ where the sum goes over the eigenvalues $\lambda_i$ and $P_{\lambda_i}$ are the projection operators on the subspace of the Hilbert space that contains the eigenvectors with the eigenvalue $\lambda_i$. Indeed, $\lambda$ is always meant to be a possible eigenvalue of the operator. And indeed, a Hermitian operator is fully determined by its spectrum and the corresponding eigenvectors (and multiplicities) for each eigenvalue, which is why the formula above is an equivalent way to rewrite a Hermitian operator. When the spectrum of $A$ is continuous, the summation over $i$ has to be replaced by an integral, and the corresponding differential $d$ is added in front of $dP_{\lambda}$: it's really the differential of the projection operator on the space of eigenstates with eigenvalues in the interval $[-\infty, \lambda]$; $dP_\lambda = dP_\lambda / d\lambda \cdot d\lambda$, if you wish. But it's really morally the same thing as in the case of the discrete spectrum (which produces delta functions in $dP_\lambda / d\lambda$ if we adopt this terminology). Also, some of your additional proofs involving $\Delta$ are just trivial substitutions under the integral sign. One would need to know lots of details of your mathematical axioms - lots of the particular "math culture" you're coming from - to figure out what could exactly be difficult for a mathematician about the substitutions under the integral sign. There are no difficulties from a physicist's perspective - it's high school maths. http://en.wikipedia.org/wiki/Spectral_theorem#Hermitian_matrices Mathematicians may worry about boundedness and well-definedness of all these things for most of their careers but these things are totally vacuous from the viewpoint of physics. If a physicist finds out that the answer to a physics question requires him to calculate $f(A)$, a function of an operator - an observable - he just has to calculate it whether or not it looks hard or well-defined. In particular, the Taylor expansion for functions such as the exponential is always assumed to hold. You discuss the function $g(A)$ of the operator as an example. The procedures you outline physically mean that one diagonalizes $A$ - which brings the projection operators to a simple form (only one number $1$ on the diagonal) - and then he simply applies the function $g$ to the eigenvalues. In other words, $$A = U D U^{-1} \quad \Rightarrow \quad g(A) = U g(D) U^{-1}$$ where $g(D)$ is simply a diagonal matrix with entries $g(D_{ii})$ on the diagonal. The formula above works because $U^{-1} U$ cancel everywhere in the middle if we write $g(A)$ e.g. as a Taylor expansion - and by generalization, we just declare the formula above to be right even if the Taylor expansion is not appropriate for a mathematician. The Taylor expansion for the exponential is always OK from a physicist's viewpoint. All these objects - Hilbert spaces, operators, their functions (especially exponentials), spectra, eigenvalues - and all these operations - exponentiation, search for projection operators etc. - are important in physics, indeed. And that's what mathematicians often declare when they want their students to listen. But it's equally true that all the points that mathematicians actually focus on most of their lives are totally uninteresting from a scientific viewpoint. This is why the comments that the "material is important in physics" is morally wrong. Mathematicians shouldn't try to support the attractiveness of their teachings by physical applications - especially because their real goal (and the real goal of pure maths) is to make them as independent of natural science as possible. One can't have it both ways. Doing physics or science means that one is allowed to "prove" all these things - such as $g(A)=g(A)$ which is really the content of the "difficult" proof you sketched) - much more elegantly and arguably naively than in maths. One is only worried about the lack of rigor if he can actually find a contradiction with the experiment or other calculations. If it doesn't exist, the science is just perfectly OK. On the other hand, mathematicians are usually looking for problems even if no problems from a scientist's viewpoint exist. This implies a totally different set of priorities and it is unlikely that a student is going to be excited about both. One either prefers to measure the truth according to nitpicking based on predetermined sets of axioms - which is the mathematician's viewpoint - or one is ready to adjust his methods, axioms, and precise definitions of objects (and invent or learn totally new branches of physics) as needed to agree with the empirical data and other accurate calculations - which is the physicist's viewpoint. They're different attitudes and this is why I think that your question should have been posted on a math forum because it's not really following physicist's way of thinking and it is not really motivated by the desire to understand Nature. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 50, "mathjax_display_tex": 7, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.942338764667511, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/41646/golden-ratio-powers-tend-to-integer-values	# Golden ratio powers tend to integer values If $G$ is the golden ratio, then $\lim_{n \to \infty}G^n$ tends ever nearer to integer values that approach $\infty$. Can it therefore be proved that $\infty$ is itself an integer? If not, why not? - 3 – Qiaochu Yuan May 27 '11 at 12:48 1 That is more to do with whether $\infty$ is a number. I am certain it is because, given any really big number n, we know that n+1 is one step closer to our goal of $\infty$ and all the steps we take to get there (albeit an infinite number of steps), are all considered numbers. Even if we never reach our goal of $\infty$ we can certainly say that we have only ever encountered numbers on the way. Surely our destination will ultimately be a number even if we never reach it! – Graphic Equaliser May 27 '11 at 13:02 4 Why are you assuming that "integerness" respects limits? (What does this discussion even mean? Mathematics is nothing without definitions and you have not provided a definition of anything.) – Qiaochu Yuan May 27 '11 at 13:30 2 Let's take a look at your argument that $\infty$ is a number. In your argument, I believe your use of "number" is equivalent to "real number" and the real numbers can be identified 1-1 with points on a Euclidean line. The process you describe is moving in some direction along that line. For every point on that line, there is at least one point on each side of it, so for every real number, there is at least one real number greater than it (and one less than it). If the "goal" is $\infty$, then there cannot be any real number past $\infty$, so $\infty$ cannot be a number. – Isaac May 27 '11 at 13:56 1 @lhf We're attempting to get rid of the [number] tag. I am forced to revive some dead questions as a part of this. :-/ – Srivatsan Dec 1 '11 at 0:04 show 2 more comments ## 3 Answers At least your observation that the powers of the golden ratio $G$ seem to approach large integers is true and interesting. The fundamental equation defining $G$ is $G^2=G+1$. From it you can deduce that $G^n = G \cdot F_n + F_{n-1}$, where $F_n$ is the $n$-th Fibonacci number. Since for large $n$ we have $G \approx F_{n+1}/F_n$, we get $G^n \approx F_{n+1}+F_{n-1}$, an integer. A more precise statement of this approximation is $G^n+H^n = (G+H)F_n +2F_{n-1}=F_{n+1}+F_{n-1}$, where $H=-1/G=1-G$ is the other root of the equation defining $G$. Since $\|H\|<1$, its powers go to zero. Wikipedia mentions this in almost integer. The golden ratio is a Pisot–Vijayaraghavan number, whose characteristic property is that their powers approach integers at an exponential rate. - What does the golden ratio have to do with it? The integer values 1, 2, 3, ..., approach infinity, too, but that doesn't prove infinity is an integer. - Because the numbers 1.5, 2.5, 3.5, ... also approach $\infty$ but they are not integer. However, $G^\infty$ is definitely $\infty$ and also approaches an integer value the nearer it gets to $\infty$ so you can see my argument that $\infty$ is an integer. – Graphic Equaliser May 27 '11 at 12:57 1 @Graphic: $\infty$ is not a number. You cannot say that it is integer, rational, real, imaginary. – Beni Bogosel May 27 '11 at 13:04 6 @Graphic: I don't understand your argument. What do you make of $\lim_{n\to \infty} (G^n+.5)$? It tends to $\infty$ too, but is not integral. – Ross Millikan May 27 '11 at 13:04 $\infty$ can't be a natural number. The natural numbers satisfy: 1. There is exactly one element called '$0$' with no predecessor($-1$ isn't a natural number). 2. Every element has a unique successor, different from itself. You could argue that we excluded by definition in (1) by using 'exactly one' and change it, but you don't get the natural numbers - you get your own custom-made structure. The integers can be defined very conservatively using the natural numbers, so you would have to change this if you wanted "\infty" to be a number. But if you change (1), you lose out on the Principle of Induction which is needed to prove anything useful about the natural numbers. And if you change (2) to remove "different from itself", then {0,1,2,3,4,5,6} with 6+1 = 6 could be considered "the natural numbers" because it satisfies our defining properties. It's not easy to change the definitions to allow $\infty$, and it doesn't seem to give much benefit. It's perfectly possible to define your own structure where $\infty$ exists, but you're asking whether it meets the criteria for a specific already-existing structures and it doesn't. 3,3.1,3.14,3.141,3.1415, etc. is a sequence of rational numbers tending to $\pi$, but $\pi$ is not rational. You can't always argue that because $P(n)$ holds for every $n$, that it must also hold for the limit.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 41, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9401800036430359, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/45737/directionality-of-angular-momentum?answertab=votes	# Directionality of angular momentum I was told that the sum of linear and angular momentum is conserved. Given that angular momentum's direction as a vector is completely arbitrary (I believe there is no physical reason for choosing the cross product going the way it does (that is, perpendicular up or down, it just matters that it's perpendicular), correct me if I'm wrong), and linear momentum is often interchanged into angular momentum, surely the vector sum cannot be conserved? Or am I wrong, and the meaning is that the magnitude is conserved? As a sidenote, I suppose this would be a good place to ask why such an odd system was devised for calculating the direction (not magnitude) of angular momentum. Why not have the arrow pointing in the direction of motion, with its base at the point that we are considering angular momentum about? Is it useful simply to compute the conservation of it about a point, as it just involves addition? - 1 Did you considered that your system should be closed to have them conserved? in sense no outer forces are applied? – TMS Dec 2 '12 at 22:19 1 It was an implicit assumption, as with any momentum conservation questions. – Alyosha Dec 2 '12 at 22:21 The conserved quantities is the vectors, and thus there magnitude too, and there sum is conserved, and if interchanging happens it will always happen in away that will keep the sum intact-ed, and of course there is a reasons to take the momentum as a cross product, one of them is to make the sum conservative! – TMS Dec 2 '12 at 22:28 The problem I was having is that he seemed to imply that the value [sum of all linear momentum] could grow or shrink to [sum of all angular momentum]'s loss or gain (losses/gains are equal in magnitude). On reflection, I think he may have just been wrong, but I wanted to check here that I wasn't being an ignoramus. – Alyosha Dec 2 '12 at 22:31 You should be more specific, who is he? if that a book you should mention it in your question to make it clear, anyway hope you got it. – TMS Dec 2 '12 at 22:33 show 2 more comments ## 2 Answers Linear momentum and angular momentum are each conserved separately. It is not meaningful to talk about their sum. To talk about angular momentum using a concrete example, let's consider a wheel spinning about its axis (whose center we will call the origin). What really matters in this case is the plane in which the rotation is taking place; let's call this the x-y plane. In three dimensions, it is possible to uniquely specify a plane by giving a vector normal (i.e. perpendicular) to the plane (in the z direction, in this case). (In four dimensions, though, a plane cannot be uniquely specified using a vector; one would need to use a wedge product instead.) The magnitude of the vector is the magnitude of the angular momentum, but there are still two directions in which to choose the vector normal to the plane of rotation. By convention, we choose this using the right-hand rule. You could use a left-hand rule instead if you really wished, but nobody does it this way, and it is crucial to be consistent in the definition of the cross product (or you will find contradictions). As an aside, any "vector" which results from the cross product of two physical vectors is actually a pseudovector. This means that if you look at the system in a mirror, the "vector" will actually point the other way. For instance, if a wheel is spinning counterclockwise (angular momentum pointing toward the observer), then a mirror placed beside the wheel would show it spinning clockwise (i.e. with the angular momentum pointing away from the observer). Physical vectors do not act this way under a mirror ("parity") transformation. - If $x$ and $y$ are two conserved quantities, then $Ax + By$ is also a conserved quantity, for any nonzero real constants $A$ and $B$. This includes both $x+y$ and $x-y$, so it is true that their sum is conserved, no matter which way around you define the direction of angular momentum. However, it is not a very interesting or fundamental result that the sum of linear and angular momentum is conserved, because they have different units, so their sum isn't a physically meaningful quantity. It certainly isn't true that one can be converted into the other. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 7, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9593061804771423, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/23305/derivation-of-fourier-series	# Derivation of Fourier Series? Can someone point me to the full derivation of the Fourier Series? I'm having problems understanding how the a's and b's coeffients are worked out. - 1 This isn't really a physics question. I'll flag for the mods to migrate it to math.stackexchange. – Mark Eichenlaub Feb 17 '11 at 20:54 I don't quite understand the question: the Fourier Series is a definition: for any integrable function on a closed interval, its coefficients are given by a particular formula: see e.g. en.wikipedia.org/wiki/Fourier_series#Definition. So what exactly is it that you want to work out? – Pete L. Clark Feb 23 '11 at 2:16 I think the OP wants to see a proof that if $f(x) = \sum a_n\cos nx + b_n\sin nx$, where the convergence is (at least) pointwise, then the coefficients $a_n$ and $b_n$ are given by those standard formulas. – Jesse Madnick Feb 23 '11 at 2:43 ## 2 Answers If you already believe that any function can be written as an infinite series of sines and cosines with appropriate coefficients, and all you are wondering about is how we go about actually computing the coefficients, then you're most of the way there. The tricky part is proving that the Fourier series works at all; the derivation of the coefficients is pretty straightforward. It's still not exactly intuitive and involves some steps that you would never think of, but it's easy enough to follow. A good derivation is here: http://cnx.org/content/m10733/latest/ - Imagine that $f(x) = \sum_n a_n \cos n x + \sum_n b_n \sin n x$ (for $x \in [0,2\pi]$, say). The idea for computing the $a_n$s and $b_n$s is that when you write down integrals of the form $\int_0^{2\pi} \cos m x \cos n x,$ or $\int_0^{2\pi} \sin m x \sin n x$, or $\int_0^{2 \pi} \cos m x \sin n x$, then the integrals vanish (just compute them!) unless $m = n$ and the functions coincide; and in the cases when they don't vanish, their values are easily computed. So taking $f$, and then computing $\int_0^{2\pi} f(x) \cos n x$ or $\int_0^{2 \pi} f(x) \sin n x$, one exactly reads off $a_n$ or $b_n$ (for the value of $n$ you chose). This is where the formulas come from. The way people normally think about this is as a kind of orthogonal projection: the functions $\cos n x$ and $\sin n x$ are like orthogonal basis vectors in a vector space, and the integral is like an inner product. So to find the coefficient of a given basis vector (i.e. an $a_n$ or a $b_n$) one takes the inner product against that particular basis vector. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 21, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9432012438774109, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/135056/radius-of-convergence-of-power-series-or-geometric-series	# Radius of convergence of power series or geometric series To find radius of convergence of geometric series $$\sum_{n=1}^\infty a_n$$ I need to use ratio/root test to find $\|L\|<1$ To find radius of convergence of power series $$\sum_{n=1}^\infty c_n (x-a)^n$$ I am supposed to find the limit $L$ of just the constant term $c_n$? $$\sum_{n=1}^\infty \frac{(2x-5)^n}{n^2}, \qquad c_n = \frac{2^n}{n^2}, \qquad R = 2^{-1}=1/2$$ How do I know what is a constant? Free of $n$? Or I should focus on getting it into the form $\sum_{n=1}^\infty c_n (x-a)^n$. Thats what I did below $$\sum_{n=1}^\infty \frac{(x-1)^{n-3} + (x-1)^{n-1}}{4^n + 2^{2n-1}}$$ I first factorized what I can $$= \sum_{n=1}^\infty (x-1)^n \cdot \frac{(x-1)^{-3} + (x-1)^{-1}}{4^n(1 + 2^{-1})}$$ So I guess $$c_n=\frac{(x-1)^{-3} + (x-1)^{-1}}{4^n(1 + 2^{-1})}$$ Then $$L=\|\frac{c_{n+1}}{c_n}\|$$ But what do I do with the $x$? Correct method is supposed to be knowing its a geometric series of common ratio $\frac{x-1}{4}$. Then radius is \|x-1\|<4. But here, don't I need to get the radius of $x$ alone, without the $-1$? Anyways, I think the main thing I am confused about is when to use which method. For geometric series, I am doing the ratio/root test for the whole $a_n$ while in power series, I am "separating" the $a_n$ into $c_n (x-a)^n$ form? And doing test on the constant terms ($c_n$)? - ## 2 Answers The ratio test is a more general approach than the "recognize it as a geometric sum test" - obviously because the latter won't work if the sum is not, in fact, a geometric series - but they are both valid. One thing to note is that if you compute $\|c_{n+1}/c_n\|$ using your definitions of $c_n$ in the example you give, the numerator $(x-1)^{-3}+(x-1)^{-1}$ does not depend on $n$ and so it will disappear when we compute the ratio of the two coefficients. Same with $(1+2^{-1})$ in the denominator. Define $f_n:= b_n(x-a)^n$ and consider the power series $$f(x)=\sum_{n=0}^\infty f_n=\sum_{n=0}^\infty b_n(x-a)^n.$$ If $\lim\limits_{n\to\infty} \|b_{n+1}/b_n\|=L$ then our "second" test says the radius of convergence is $R=1/L$. Why? Well, $$\lim_{n\to\infty}\left\|\frac{f_{n+1}}{f_n}\right\| =\lim_{n\to\infty}\left\|\frac{b_{n+1}(x-a)^{n+1}}{b_n(x-a)^n}\right\|=L\|x-a\|,$$ so the series $\sum\limits_{n=0}^\infty f_n$ converges if $\big\|L\|x-a\|\big\|<1$ by our "original" ratio test, equivalently $\|x-a\|<R$, which defines a circle of radius $R$ (note: $a$ is the center of the circle and tells us nothing about the radius $R$). This exhibits how the "secondary" test, of just comparing the coefficients of $(x-a)^n$, is actually a shortcut to applying the original ratio test, where we test the sequence $f_n$, so to speak. - Without using ratio test u can also find radius of convergence in this example, $\sum_{n=1}^\infty c_n$ $\frac{(2x-5)^n} {n^2}$ Comparing the given series with power series, $\sum_{n=1}^{\infty}c_n(x-x_0)^n$ We have, $c_n=\frac{1}{2^n n^2}$ and $x_0=(\frac{5}{2})^n$ $R=\lim_{n\to \infty}\|c_n\|^\frac{1}{n}$ $\ \ \ =\lim_{n\to \infty}\|\frac{1}{2[(n)^\frac{1}{n}]^2}\|$ $\ \ \ R=\frac{1}{2}$ $\ \ \ R=\frac{1}{r}$ $\ \ \ r=2$ $\therefore$ radius of convergence is 2 Moreover,interval of convergence $(x_0-r,x_0+r)$ In this example root test is very useful because here constant having power n -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 41, "mathjax_display_tex": 9, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9299858808517456, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/43545/disappearance-of-moduli-for-condensate-of-open-strings?answertab=active	# Disappearance of moduli for condensate of open strings Consider a Dp-brane. Compactify $d$ spatial dimensions over a torus $T^d$. Suppose $d\geqslant p$, and that the Dp-brane is completely wrapped around the compactified dimensions. Look at the open string modes ending on this wrapped D-brane. There is a zero energy open string mode associated with each spatial dimension. That corresponds to the orientation of the worldsheet field excitation. If the direction is along an uncompactified spatial dimension, that corresponds to quanta of brane displacements along that direction. These cases needn't concern us. If the direction is normal to the brane but lies along a compactified dimension, that corresponds to quanta of brane displacements along that direction. If it's tangent to the brane, it corresponds to quanta of the Wilson line of the brane gauge field along that wrapped direction. Here's the question. Suppose the wrapped D-brane has a total mass of $M$. Suppose there is a compactified spatial dimension of radius $R$ along which the brane isn't wrapped, i.e. $p<d$. The brane has Kaluza-Klein momenta along that direction of value $n/R$ where n is integral. The energy spectrum is given by $$\sqrt {M^2 +n^2/R^2} ~\approx~ M + \frac{n^2}{2MR^2} +\mathcal{O}(M^{-3}).$$ A condensate of zero energy open strings with orientation along that direction ought to give a continuous moduli? Why is there no moduli then, and why is the energy spectrum discretized? Or consider a direction in which the brane is wrapped. We ought to have a continuous modulus of Wilson line of the brane gauge field along that dimension? Once again, we have discretization. Why? Anyway, how can a completely wrapped D-brane have KK momentum? In the string worldsheet picture, we have open strings ending at the D-brane background at a fixed position. Along those compactified dimensions along which the brane isn't wrapped, we have Dirichlet boundary conditions. Such open strings can only have winding numbers, but no KK momentum either. Even then, we're dealing with a condensate of open strings with zero winding number. This discretization doesn't occur if the brane remains unwrapped under at least two uncompactified spatial dimensions because the brane now has infinite mass. If it remains unwrapped only along one uncompactified spatial dimension, there's the Mermin-Wagner theorem, meaning there’s no fixed brane position or Wilson line. PS: Maybe this question can be rephrased in terms of BPS. We have a wrapped BPS brane. But nonperturbatively, somehow, the BPS state has to be delocalized along the compactified dimension or its Wilson line in a superposition over all possible values? Open strings with no energy are also BPS. So, we can have any condensate of them and still remain BPS? This clearly isn't the case with a lifted modulus, and a discretization of the energy spectrum. PPS: How do you even express the KK modes of the D-brane, or the dual to its Wilson lines, in terms of a condensate of open strings? Suppose you have the lowest energy state with zero KK momentum. Then, disregarding transverse displacements along the uncompactified spatial dimensions, there is an energy gap to the next energy state. However, a condensate of open strings with internal mode excitations along the compactified dimensions would naively give no energy gap. PPPS: Is the number of open string modes corresponding to these "disappearing moduli" even a well-defined operator? This is precisely because at the perturbative level, such open string modes have no energy. If it's not a well defined operator, just how do you even express this in terms of open string worldsheets? - ## 1 Answer By T-dualities along the finite directions on which the D-brane is wrapped, your first problem is equivalent to the problem of a D0-brane moving on a circle. The massless open-string scalar corresponding to the direction of the circle produces the field $X(\tau)$ living on the world line of the D0-brane. And indeed, by assumption, the global topology of the configuration space for this field $X(\tau)$ includes the periodic identification $X(\tau)\sim X(\tau)+2\pi R$. This is equivalent to the quantization for the complementary momentum, $P=N/R$. But none of these two equivalent signs of the periodicity/quantization of the transverse location/momentum is visible at the level of the spectrum of the open strings. In the perturbative treatment of the D-brane dynamics, we effectively expand around a classical shape of the "heavy" D-brane, i.e. around a classical solution. Imagine $X(\tau)=0$. The open-string excitation corresponding to the transverse field is a quantum of a "wave" that only deviates "infinitesimally" from $X(\tau)=0$. The quantitative distance scale at which $X$ deviates is parameterically shorter than then string scale. So the open-strings know about all the potential of the D-brane to change its shape and vibrate but a finite number of open-string excitations still leaves you near $X(\tau)=0$ so it isn't capable of probing the global properties of the $X$ configuration space. To reach the point $X=2\pi R$, you would need an infinite number of open strings – the number would scale like $1/g$ or something like that – and the perturbative expansion would break down or would become insufficient at that point because it's an expansion around the point $g=0$ at which point the required number of open strings is strictly infinite. So at the leading order, the open-string calculations will be totally insensitive to the global properties of the D-brane configuration space. This is also reflected by your $n^2/2MR^2$ kinetic term from the D-brane. Because $M\sim 1/g$, this kinetic energy scales like $g$ and vanishes in the $n=0$ limit. But yes, once you go to the subleading order in $g_s$ when you're calculating the energy or evolution, you will be able to see that the field on the D0-brane is periodic and the coefficient of this $1/2M$-like kinetic term is a quantized $n^2/R^2$. Analogously, you may solve the other universal problem with the Wilson line which is just the T-dual of the D0-brane above performed along the $X$ axis. You get a D1-brane with a Wilson line that plays exactly the role of the periodic $X$ for the D0-brane. The dual quantity to this periodic Wilson line is the electric flux, $\int d \tilde X\,F_{01}$, along the D-string, which is quantized much like the momentum of the D0-brane – they're dual descriptions of each other. Again, the open-string excitations only produce "infinitesimal variations" (or Planckian, substringy variations) of the Wilson line that are too small to see the periodicity of the Wilson line or the quantization of the complementary electric flux. It's helpful, at least for me, to look at this problem via Matrix theory which picks some low-energy limit of the D-brane dynamics. Indeed, for D0-branes on a circle, one may also describe the situation by D1-branes on the dual circle. The matrix model, matrix string theory, will automatically give you the right gauge theory with the right periodicity conditions of the Wilson line and the quantization for the electric flux. You may be asking how one derives that the gauge theory has the right identifications of the configuration space. One may be more or less rigorous but it's clear that it has to have this identification. This identification boils down to the fact that e.g. the D0-brane at places $X$ and $X+2\pi R$ is really the same state – it's the same state from the perturbative string theory viewpoint as well because the D0-brane locations define two totally identical (i.e. one) boundary conditions for the world sheet, the same boundary CFT. The configuration space for D-branes is the space of different boundary conditions and because $X$ and $X+2\pi R w$ give the same boundary conditions, they're identified. - I know this is a nonperturbative effect. You are trying to reexpress it in terms of matrix theory, but how do you even deal with this question at the level of a condensate of open strings? Or is the description in terms of open strings incomplete, with matrix theory being the complete theory? – Hecles Nov 6 '12 at 14:36 If path integrals over string worldsheets is fundamental, surely there ought to be a way to answer my question purely at the level of open string worldsheets, shouldn't there? A failure to do so is an indictment of the completeness of the string worldsheet description. – Hecles Nov 6 '12 at 14:38 @Hecles: Yes, the perturbative string picture is incomplete. – user1504 Nov 6 '12 at 14:57 Dear @Hecles, yes, I confirm user1504, perturbative string theory is an incomplete description of string theory. It only describes things that exist at $g=0$, and what they're doing at a general $g$ is expressed as a Taylor expansion - which isn't the most general function even among functions that are smooth near $g=0$. Things that are singular or nonperturbative near $g=0$ are invisible in perturbative string theory. Moduli spaces whose volumes go like $1/g$ look infinite in perturbative expansion, and so on... D-branes give you a way to "extend" perturbative expansions if done right. – Luboš Motl Nov 16 '12 at 10:38	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 38, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9171316623687744, "perplexity_flag": "head"}
http://mathoverflow.net/questions/43538?sort=newest	## Wonderful applications of the Vandermonde determinant ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) This semester I am assisting my mentor teaching a first-year undergraduate course on linear algebra in Peking University, China. And now we have come to the famous Vandermonde determinant, which has many useful applications. I wonder if there are some applications of the Vandermonde determinant that are suitable for students without much math background. For example, using the Vandermonde determinant, we can prove that a vector space $V$ over a field $F$ of characteristic 0 cannot be expressed as a finite union of its nontrivial subspaces, i.e., there do not exist subspaces $V_1,\ldots,V_m$ that satisfy $$V_1\cup \cdots\cup V_m=V,$$ where $V_i\ne {0}$ and $V_i\ne V$ for all $i=1,2,\ldots,m$. This can be proved as follows: choose $v_1,\ldots,v_n$ as a basis of $V$, and consider the infinite series $$\alpha_i = v_1 + iv_2+\cdots+i^{n-1}v_n.$$ Using our knowledge of the Vandermonde determinant, one can show that every subset of the $\alpha$'s having $n$ vectors in it consists of a basis of $V$, hence each of the $V_i$'s can contain at most $n-1$ of the $\alpha$'s in it, so there must be infinitely many $\alpha$'s not contained in any of the $V_i$'s. - 2 You should probably add the 'big-list' and 'example' tags, and set this as Community Wiki. – Mariano Suárez-Alvarez Oct 25 2010 at 16:20 The basic proof seems however more simple : take $x\in V_1 \setminus \bigcup_{i>1}V_i$ and $y\in V_2\setminus V_1$; then for some $\lambda \neq \mu$ in $F$, we have $x+\lambda y, x+\mu y \in V_j$ for some $j$, absurd. – Henri Oct 25 2010 at 16:27 yes, but here the set has an extra property: each subset with n elements is a base of $V$, which is a bit more interesting. – zhaoliang Oct 25 2010 at 16:35 2 I think the Vandermonde can also be used to compute the discriminant of a cyclotomic number field. Can an algebraic NTist step forward and elaborate on this? – darij grinberg Oct 26 2010 at 0:12 ## 17 Answers I found some remarkable property of the Vandermonde determinant. Based on this property we are able to introduce a notion "difference between n>2 quantities". Below I give the abstract of this result. The notion of difference between two quantities plays a basic role in mathematics, consequently in all branches of human activity where the mathematics is applied. However the long stand question is: what is {\it the difference between three (or more) quantities}? The binary operation $[a,b]=(a-b)$ possesses the following principal feature: with respect to the third quantity $c$ this operation is decomposed into a sum of the same operations between $a$ and $c$, and $c$ and $b$, i.e., $$[a,b]=[a,c]+[c,b].$$ Denote by $[a,b,c]$ difference between three quantities $a,b,c$. With respect to additional quantity $d$ this definition of the difference has to possess with the following property $$[a,b,c]=[d,b,c]+[a,d,c]+[a,b,d].$$ We prove that this property of difference between three (or $n\geq 2$) quantities is satisfied by one of the features of Vandermonde determinant. - ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. Especially for students with just very basic background, it might be a fun fact that companion matrices are diagonalized by the vandermonde matrix corresponding to the zeros of the characteristic polynomial they encode - provided, of course, that the roots are all distinct. - The Vandermonde determinant is used to prove that cyclic polytopes maximize the number of $i$-dimensional faces for each $i$ among all triangulations of a $(d-1)$-dimensional sphere having exactly n vertices. The cyclic polytope $C(n,d)$ is the convex hull of any $n$ distinct points on the moment curve {$(t,t^2,\dots ,t^d) \| t\in {\mathbf R}$} $\subseteq {\mathbf R}^d$. I like the discussion of cyclic polytopes in G"unter Ziegler's book "Lectures on Polytopes". The wikipedia article en.wikipedia.org/wiki/Cyclic_polytope also seems to give a nice quick summary. - Probably not an application for just any audience, but I thought I'd share... The Vandermonde determinant shows up in matrix models of quantum field theories. Roughly speaking, in these we consider integrals of the form $\int dM f(M)$, over the space of Hermitian matrices, where $f$ is invariant under conjugation by unitary matrices, and $dM$ is the (also conjugation invariant) measure $dM = (\prod_i dM_{ii})(\prod_{i\lt j} dM_{ij})$. We want to calculate the integral by gauge fixing. In other words, to integrate over a judiciously chosen set of representatives of each $U(N)$ orbit. Usually the best set of representatives to take are the diagonal matrices. The procedure for evaluating the integral in that case is exactly that of the Weyl integration formula. In doing so we get what physicists call the Faddeev-Popov determinant, which turns out to be the Vandermonde determinant! In other words, we get an equivalent integral $\int d\lambda_1 ... d\lambda_N \Delta(\lambda_1, ...,\lambda_N) f(diag(\lambda_1,...,\lambda_N))$, where $\Delta$ is the Vandermonde determinant. - The vectors corresponding to any $n$ distinct points on the rational normal curve $x\to (x,x^2,\dots,x^n) \subset \mathbb{A}^n$ span $\mathbb{A}^n$. - The Vandermonde determinant plays a role in the proof by Kempf and Kleiman-Laksov of the existence portion of the famous Brill-Noether theorem (formulated, but not proved, by Brill and Noether): a general, genus $g$ projective curve has an algebraic line bundle of degree $d$ and $(r+1)$-dimensional space of global sections if and only if the "naive parameter count" for the dimension of such, $\rho(g,r,d) = g-(r+1)(g-r+d)$, is nonnegative. The point is that they set up the enumerative formula to count the number of such line bundles (satisfying some appropriate additional conditions). Miraculously the formula comes out to a Vandermonde determinant which can be explicitly evaluated as being nonzero (as opposed to many similar enumerative problems in algebraic geometry which have no such closed formula). - Maybe this should be a comment under Darij Grinberg's or Terry Tao's answer, but anyway: the discriminant of a monic polynomial is the square of the Vandermonde determinant evaluated at roots of the polynomial. Undergraduate students who advanced to a linear algebra course must have encountered at least the discriminant of a quadratic--although in this case the relation between the Vandermondian and the discriminant is not so wonderful... This relation between Vandermondian and discriminant also determines the relation between Euler class and Pontryagin class: http://en.wikipedia.org/wiki/Splitting_principle - The Discrete Fourier Transform, which sends a vector $x=\left(x_j\right)_{j=0}^{N-1}$ to $y=\mathrm{DFT}(x)$ such that $$y_k=\frac{1}{\sqrt{N}}\sum_{j=0}^{N-1}e^{2\pi i \times jk/N}x_j$$ has a matrix representation $$\mathrm{DFT}_{jk}=e^{2\pi i \times jk/N}=\left(e^{2\pi i /N}\right)^{j\times k},$$ which is in fact a doubly Vandermonde matrix: both it and its transpose are Vandermonde matrices. With this you can use the Vandermonde determinant to prove that $\mathrm{DFT}$ is nonsingular, and if you prove using other means that it is unitary (rather easy) then you will get, I think, a nontrivial expression for 1 as a product of differences of roots of unity. - There is an elegant (and short!) application of the (generalized) Vandermonde determinant to the famous problem of D. H. Lehmer in the article [D.C. Cantor and E.G. Straus, Acta Arith. 42 (1982/83), no. 1, 97-–100]. I put here a scan of the article together with corrections given by the authors later. - Maybe it is not really suitable to undergrads (unless they are really problem solving-oriented), but there is a nice proof that $$\prod_{1\leq i \lt j\leq n} \frac{x_j-x_i}{j-i}$$ is integer for all integer sequences $(x_k)_{k=1}^n$ using Vandermonde determinants. The idea is reducing the thesis to the fact that the matrix $$\begin{bmatrix}1 & 1 & \dots \newline \binom{x_1}{1} & \binom{x_2}{1} & \dots \newline \binom{x_1}{2} & \binom{x_2}{2} & \dots \newline \vdots & \vdots & \ddots \end{bmatrix}$$ has integer entries, and thus integer determinant. After clearing the denominators (which give the factor $\prod \frac{1}{j-i}$), one can transform the resulting determinant to a Vandermonde with basic row operations. - The Vandermonde determinant formula implies that Vandermonde matrices are maximum distance separable and can therefore be used to construct error-correcting codes with good properties (BCH codes in particular). - As for applications in symmetric function theory... well, open a book on symmetric function theory at a random page and stare at the formulas. Usually something looking like a Vandermonde determinant will stare back at you. For example: Crawley-Boevey, Lectures on Representation Theory uses it on page 18. Oh, and of course he uses the Cauchy determinant too. While he derives it from a geometric argument, it can also be proven purely algebraically, and the proof uses Vandermonde. (Again, I'm not claiming the proof is new. In fact I believe I have seen it somewhere, but I couldn't find it again...) - I wanted to put this online long ago but somehow never came to actually doing it. Now is a good occasion: http://mit.edu/~darij/www/index.html#hyperfactorial or, directly, the PDF file: http://mit.edu/~darij/www/hyperfactorialBRIEF.pdf This is about a theorem by MacMahon stating that for any three nonnegative integers $a$, $b$, $c$, the number $\frac{H\left(a\right)H\left(b\right)H\left(c\right)H\left(a+b+c\right)}{H\left(b+c\right)H\left(c+a\right)H\left(a+b\right)}$ is an integer, where $H\left(m\right)$ means $0!\cdot 1!\cdot ...\cdot \left(m-1\right)!$. There are various proofs of this now, some of them combinatorial (see the references in the note), but the simplest one is probably the one I give using the Vandermonde determinant (I don't think it's new...). The note is a bit long (10 pages), but the proof ends at page 6. Also note that I prove Vandermonde itself, which takes up some space as well. Another application of Vandermonde appears on page 9: If $a_1$, $a_2$, ..., $a_m$ are $m$ integers, then $\prod\limits_{1\leq i < j\leq m}\left(a_i-a_j\right)$ is divisible by $H\left(m\right)$. This is very well-known (and so is the proof). Finally, a little question - slightly offtopic, I know. Back to the $\frac{H\left(a\right)H\left(b\right)H\left(c\right)H\left(a+b+c\right)}{H\left(b+c\right)H\left(c+a\right)H\left(a+b\right)}$ problem, one might try proving that this is an integer by showing that every prime $p$ divides $H\left(a\right)H\left(b\right)H\left(c\right)H\left(a+b+c\right)$ at least as often as it divides $H\left(b+c\right)H\left(c+a\right)H\left(a+b\right)$. This can be easily shown equivalent to the following: Any nonnegative integers $a$, $b$, $c$ satisfy $\sum\limits_{k=0}^{a-1} \lfloor \frac{k}{p} \rfloor + \sum\limits_{k=0}^{b-1} \lfloor \frac{k}{p} \rfloor + \sum\limits_{k=0}^{c-1} \lfloor \frac{k}{p} \rfloor + \sum\limits_{k=0}^{a+b+c-1} \lfloor \frac{k}{p} \rfloor$ $\geq \sum\limits_{k=0}^{b+c-1} \lfloor \frac{k}{p} \rfloor + \sum\limits_{k=0}^{c+a-1} \lfloor \frac{k}{p} \rfloor + \sum\limits_{k=0}^{a+b-1} \lfloor \frac{k}{p} \rfloor$. (Yes, it can be shown that the $p^2$, $p^3$, ... terms can be ignored.) Is there an easy way to see this? Or any way at all, without going back to the Vandermonde determinant? - (1) Not really an application, but the paper by I. Gessel, Tournaments and Vandermonde's determinant, J. Graph Theory 3 (1979), 305-308, gives a nice connection with tournaments. See also Exercise 2.16 of my book Enumerative Combinatorics, vol. 1 (equivalent to Exercise 2.35 at http://math.mit.edu/~rstan/ec/ec1.pdf). (2) Probably not suitable for an undergraduate course, but if $s_{(n-1,n-2,\dots,1)}$ denotes the Schur function of the staircase shape $(n-1,n-2,\dots,1)$, then the evaluation $$s_{(n-1,n-2,\dots,1)}(x_1,\dots,x_n)=\prod_{1\leq i \lt j\leq n} (x_i+x_j)$$ follows immediately from the bialternant formula for Schur functions, since it reduces to the quotient of two Vandermonde's: $\prod (x_i^2-x_j^2)/\prod(x_i-x_j)$. See Exercise 7.30 of Enumerative Combinatorics, vol. 2. - Vandermonde determinants + Cramer's rule = Lagrange interpolation. (EDIT: Also, there is a qualitative version of the above identity: just from knowing that the Vandermonde determinant is non-vanishing when the $x_i$ are distinct, one can already deduce that polynomial interpolation is theoretically possible, though to get the precise formula one still needs to go through the above identity.) - When I was typing this item, I had Lagrange interpolation in my mind :P I think there should be some applications in combinatorics or symmetric function theory – zhaoliang Oct 25 2010 at 17:54 Of course, what is "wonderful" is quite subjective. One simple application that I like is showing that the functions $e^{cx}$ are linearly independent over ${\mathbb R}$. Another is that if $tr (A^n)=0$ for all $n$, then $A$ is nilpotent. - this reminds me the Linderman-Weierstrass theorem on the transcendence of e. – zhaoliang Oct 25 2010 at 17:50 1 @zhaoliang: In fact, one uses the Vandermonde determinant in the standard proof of Baker's theorem. – Daniel Litt Oct 25 2010 at 23:02 The independence of $e^{c x}$ is a rather trivial fact. To prove it using Vandermond determinant, just differentiate any linear dependence multiple times and pack it into a matrix equality. – Keivan Karai Oct 26 2010 at 13:43 1 @Keivan. A beautiful application of this characterization of nilpotent matrices is Amitsur-Levitski's polynomial identity in $M_n(k)$. – Denis Serre Oct 26 2010 at 16:52 Another application is to prove that for any $0<\alpha_1<\cdots<\alpha_n$ the family $$\sin \alpha_1x,\cdots,\sin\alpha_nx$$ is linearly independent in $C^{4(n-1)}(\mathbb R)$. – Hany Oct 26 2010 at 19:20 I loved encountering the following as a student: The Vandermonde determinant plays a role in the proof of Hilbert's Theorem 90 in Section 9.6 of Schilling and Piper's Basic Abstract Algebra. I certainly should take the time to type this argument in a bit. I just wanted to get the reference out there. - gigapedia.com/… – zhaoliang Oct 25 2010 at 17:18	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 95, "mathjax_display_tex": 10, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9163573980331421, "perplexity_flag": "head"}
http://unapologetic.wordpress.com/2009/12/29/the-mean-value-theorem-for-multiple-integrals/?like=1&source=post_flair&_wpnonce=a832ba55c5	# The Unapologetic Mathematician ## The Mean Value Theorem for Multiple Integrals As in the single variable case, multiple integrals satisfy a mean value property. First of all, we should note that, like one-dimensional Riemann-Stieltjes integrals with increasing integrators, integration preserves order. That is, if $f$ and $g$ are both integrable over a Jordan-measurable set $S$, and if $f(x)\leq g(x)$ at each point $x\in S$, then we have $\displaystyle\int\limits_Sf(x)\,dx\leq\int\limits_Sg(x)\,dx$ This is a simple consequence of the definition of a multiple integral as the limit of Riemann sums, since every Riemann sum for $f$ will be smaller than the corresponding sum for $g$. Now if $f$ and $g$ are integrable on $S$ and $g(x)\geq0$ for every $x\in S$, then we set $m=\inf f(S)$ and $M=\sup f(S)$ — the infimum and supremum of the values attained by $f$ on $S$. I assert that there is some $\lambda$ in the interval $[m,M]$ so that $\displaystyle\int\limits_Sf(x)g(x)\,dx=\lambda\int\limits_Sg(x)\,dx$ In particular, we can set $g(x)=1$ and find $\displaystyle mc(S)\leq\int\limits_Sf(x)\,dx\leq Mc(S)$ giving bounds on the integral in terms of the Jordan content of $S$. Incidentally, $g(x)\,dx$ here is serving a similar role to the integrator $d\alpha$ in the integral mean value theorem for Riemann-Stieltjes integrals. Okay, so since $g(x)\geq0$ we have $mg(x)\leq f(x)g(x)\leq Mg(x)$ for every $x\in S$. Since integration preserves order, this yields $\displaystyle m\int\limits_Sg(x)\,dx\leq\int\limits_Sf(x)g(x)\,dx\leq M\int\limits_Sg(x)\,dx$ If the integral of $g$ is zero, then our result automatically holds for any value of $\lambda$. Otherwise we can divide through by this integral and set $\displaystyle\lambda=\frac{\displaystyle\int\limits_Sf(x)g(x)\,dx}{\displaystyle\int\limits_Sg(x)\,dx}$ which will be between $m$ and $M$. One particularly useful case is when $S$ has Jordan content zero. In this case, we find that any integral over $S$ is itself automatically zero. ### Like this: Posted by John Armstrong \| Analysis, Calculus ## 3 Comments » 1. [...] in for integrals are Jordan measurable, and their boundaries have zero Jordan content, so we know changing things along these boundaries in an integral will make no [...] Pingback by \| December 30, 2009 \| Reply 2. [...] infinitesimal pieces of -dimensional volume. Now, with the change of variables formula and the mean value theorem in hand, we can pull out a macroscopic [...] Pingback by \| January 8, 2010 \| Reply 3. [...] analogue of the integral mean value theorem that holds not just for single integrals, not just for multiple integrals, but for integrals over any measure [...] Pingback by \| June 14, 2010 \| Reply « Previous \| Next » ## About this weblog This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”). I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 36, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8875944018363953, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/15907/pressure-inside-an-ideal-gas	# Pressure Inside an Ideal Gas A non-relativistic ideal gas exerts a pressure at the surface of its container $p = \frac13 \rho \langle v^2 \rangle$ where $\rho$ is the mass density of the gas and $\langle v^2 \rangle$ is the average square of the Maxwell velocity distribution. This is the relation for the pressure at the boundary of the vessel containing the ideal gas. However, if one were to place an infinitesimal "test" area inside the boundary of the vessel the momentum flux across that area would be 0 since the distribution of velocities is symmetric. That is, as many particles would cross one way across the area as cross the opposite way. This suggests that the pressure inside an ideal gas is 0. In general relativity an ideal gas is usually presented as an example of a perfect fluid, that is one with stress energy tensor equal to $T_{\mu \nu} = \left( \begin{array}{cccc} \rho & 0 & 0 & 0 \\ 0 & p & 0 & 0 \\ 0 & 0 & p &0 \\ 0& 0 & 0 & p \end{array} \right)$ Since the stress energy tensor should be a local function it seems strange to assign to $p$ (in the frame in which the velocity distribution is isotropic) in the above equation the value of the pressure at the boundary of the vessel. I can see why this is done: one wants the stress energy tensor to be a smooth function, and also it should be that a gas at finite temperature should gravitate more than a gas at zero temperature (dust)... however this is point is not articulated, in for instance MTW. Perhaps I am missing something elementary? - 1 Why does a net momentum flux of zero for some area suggest a pressure of zero? You completely lost me at that point. – AlanSE Oct 19 '11 at 5:30 Pressure is defined to be the force per unit area. If when the particles collide with the test area they do so elastically, the force (integrated over a small time window) exerted will be the momentum flux (integrated over that time window). Basically as many particles bounce off on one side as they do on the other... the net force is zero... the pressure is zero – Ian Oct 19 '11 at 5:42 1 Your wording is proposing a closed area integral - a Gaussian surface. The integral over any such surface for an idealized gas model should give zero, of course. If I put any object into a gas, I do not expect a net force (aside from buoyancy, drag, etc). In order to define "pressure" you need to take a once-sided integral, and only count those particles going in one direction, like the case of a boundary surface. If I have a jar in a vacuum with a gas in it, the basic idea of pressure is that I count the collisions on one side of the wall only. – AlanSE Oct 19 '11 at 6:19 I think Z is missing the more general definition of pressure: perpendicular momentum transported across an area. In the case where the area sits on the wall of the container, that definition can be restated as the usual "particles bouncing off of the wall" argument. But his question still stands in the case that the area is within the gas. An interstellar gas has no containing wall. Is pressure a sensible concept in this case? Of course it is! Is the pressure zero? Certainly not! I'll give an answer below... – garyp Oct 20 '11 at 1:43 The momentum flux is not zero. If it were, there would be no stress. @Zassounotsukushi: this is the definition of stress. – Ron Maimon Oct 20 '11 at 6:01 ## 2 Answers The stress is the current of the conserved momentum. If momentum is going from one place to another, you have a stress, because the momentum has to go through all the places inbetween. If your intuition comes from action-at-a-distance physics, this can be confusing, because in Newtonian gravity, momentum can jump across empty space between two gravitating objects, leading to a nonlocal force. But this is no longer true in relativity. When you have a gas compressed by a container, the x-momentum flows from one x-end of the container, where a positive x-force is applied (x-momentum pumped into the gas) to the other x-end, going in the x-direction. This means that there is a flow of x-momentum in the x-direction, a pressure. If you place an x-wall somewhere, the force exerted by the left half of the gas on the right half is equal and opposite to the force exerted by the right half on the left half. But this does not mean that there is zero stress. The rightward force is a flow of x momentum in the positive x direction, and the leftward force is also a flow of x momentum in the positive x direction. This is a bit confusing because of the vector nature of momentum, and the tensor nature of stress, so consider a scalar conserved charge. Suppose you have a wire with a nonzero flow of electric charge, a nonzero current, and you cut the wire somewhere with a plane. The current through the plane is the flow of charge from left to right, and if 8 coulombs per second are flowing from left to right, then it is obvious that -8 coulombs per second are flowing from right to left, in order for charge to be conserved. But does this mean that the current is always zero? Obviously not--- the current is defined by one or the other of these quantities, not the sum. Similarly, in a gas, the momentum currents are defined by the force one part of the gas exerts on another, both through particles leaving the region carrying a certain amount of momentum with them, and collisions right by the boundary transferring momentum from particles inside to particles outside. The net momentum flow is defined by the force of one half on the other, or by the negative of the force the other way, not by the sum of the two. - Somebody gets a kick from downvoting all my answers. This answer is the correct resolution of the confusion, regardless. – Ron Maimon Oct 20 '11 at 9:33 1 Well, I up-voted it. – garyp Oct 20 '11 at 12:17 [see my correction in the comments. Sorry, Ian.] Ian is missing the fact that the area segment that he is considering is a directed area. In our imagination we think of the area segment as a tiny transparent, infinitely thin rectangle, and if we flip it 180 degrees it looks the same. We can't be imposing that limitation on our area. We can only reasonably insist that it returns to its original state after a rotation of 360 degrees. Imagine that the little area has an arrow pointing perpendicularly from the surface. Now the momentum that crosses the surface has to be reckoned according to whether or not your atom is crossing in the positive direction or the negative direction. When two atoms cross, one from the left the other from the right, the momentum that crosses the directed area is 2mv. Suddenly the pressure is no longer zero. A directed area is something we don't ordinarily think about, but we do use them without naming them when we use the integral form of Gauss' Law. On the other hand, a directed line segment is quite familiar; we call it a vector instead of "directed line segment". The directed area is more naturally described mathematically by an even more unfamiliar concept: the bivector, which naturally accounts for the directionality of an area segment without having to glue perpendicular arrows on to it. - I don't believe I was missing that fact.... the test area is of course directed. Consider the test area to have unit normal $\hat{n} = (1,0,0)$. A sensible definition of the momentum flux would be $\vec{p} \cdot \hat{n}$. A particle traveling from the "left" would have momentum $\vec{p} = (mv ,0 ,0)$ and would contribute a flux $mv$. A particle traveling from the "right" would have momentum $\vec{p} = ( -mv ,0,0)$ and would contribute a flux of $-mv$. These cancel. – Ian Oct 20 '11 at 2:40 The source of my confusion was in how the Stress tensor (or Stress energy tensor) is defined. It turns out that it is defined such that given a test area one only considers the force that would be exerted on the area from the side opposite the direction of the unit normal. This means that the pressure inside the gas is identical to the pressure at the boundary. – Ian Oct 20 '11 at 3:50 Yes. (lesson learned: I posted an answer because I struggled with the same question some years ago, and came up with a somewhat satisfying answer. I forgot to make sure I correctly remembered the answer.) What I neglected was that in considering the momentum added by the two particles ... one from the right, the other from the left ... one has to ask how much momentum the two particles carry to the right (for instance). They both carry $mv$ to the right, giving $2mv$. I'll go out on another limb and suggest that the factor of two is important. But Ron's answer is better anyway. – garyp Oct 20 '11 at 12:45	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 13, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9479723572731018, "perplexity_flag": "head"}
http://mathforum.org/mathimages/index.php/Hyperbolic_Tilings	# Hyperbolic Tilings ### From Math Images Hyperbolic Tilings Field: [[Field:\| ]] Image Created By: [[Author:\| ]] Hyperbolic Tilings This image is a hyperbolic tiling made from alternating two shapes: heptagons and triangles. # Basic Description In addition to tessellations found in the traditional Euclidean geometry, there are also tilings that can be created in non-tradition or non-Euclidean geometry. Hyperbolic tiling is a type of tessellation that is visualized in a non-Euclidean geometry called hyperbolic geometry. This particular tessellation is viewed in the Poincaré disk model, where straight lines are arcs that form right angles at the boundary of the disk. A hyperbolic tiling The polygons in a hyperbolic tiling grow smaller as they reach the edge of the disk. This is due to the representation of the tiling in the Poincaré disk model that we use to visualize hyperbolic tilings. The Poincaré disk model does not show traditional distances, so that as we get closer and closer to the edge of the disk, the distances increase and the boundary of the disk is in fact infinite. # A More Mathematical Explanation [Click to view A More Mathematical Explanation] ## Schläfli symbol The Schläfli symbol is a notation that can be used to describe the properties o [...] [Click to hide A More Mathematical Explanation] ## Schläfli symbol The Schläfli symbol is a notation that can be used to describe the properties of polygons in a tessellation. The symbol is in the form {n , k}, where n refers to the number of edges of a regular polygon and k refers to the number of regular polygons that meet at each vertex. Thus, if the summation of all the angles at a vertex is 360 degrees: the angle at each vertex is $\frac{360}{k}$ and the polygon has a summation of angles $\frac{n360}{k}$. Regarding the image shown to the right, the tessellation is made up solely of triangles, so that n = 3. At each vertex in the tessellation, there is always a meeting of six triangles, giving us k = 8. Thus, the Schläfli symbol for this tessellation is {3,8}. Also, the angle at each vertex is 35 degrees, and the summation of angles of the polygon is 135 degrees (which is less than the traditional 180 degrees, a characteristic of hyperbolic tilings). ## Types of Tilings Most tessellations are made from regular polygons, which are polygons with edges of equal side length and vertices of equal angle. This allows the tessellation to have the same number of polygons meeting at each vertex. ### Quasi-regular Tilings You will notice that some hyperbolic tilings utilize two polygons instead of one polygon. These types of tilings are called quasi-regular tilings. The two polygons will alternate meeting at each vertex and are usually colored differently for emphasis. To describe a quasi-regular tiling, the Schläfli symbol is still used with an prefix quasi and with n being the number of edges of one polygon and k being the number of edges of one polygon: quasi-{n,k}. It's interesting to note that every regular tessellation can be made into a quasi-regular tessellation. You simply connect the midpoints of all the edges of the polygons in the tessellation. In the image gallery below, the first tessellation shown is a regular tessellation {5,4}. The second image shows how to connect the midpoints of the edges to create a quasi-{5,4} tessellation like the one shown in the third image of the gallery. The n in the quasi-regular tessellation now refers to the the four-sided squares, while the k now represents the five-sided pentagons alternating within the tessellation. ### Dual Tessellations There is another type of tessellation that occurs when the Schläfli symbols {n,k} are switched to {k,n}. This is called a dual tessellation, and an example can be found in the final image in the gallery: {4,5}. (Note, although the Schläfli symbols for the quasi- and duel tessellations are the same for this particular set of examples, they are very different tessellations). Hyperbolic Tiling {5,4} Making Quasi-{5,4} Quasi-{5,4} Dual {4,5} ## Tilings in Different Geometries There are various geometries that can be used to draw tilings, including in the Euclidean, hyperbolic, and elliptic planes. It is interesting to note that Euclidean geometry only allows for three regular tilings (using the regular polygons squares, triangles, and hexagons), while non-Euclidean geometry allows for an infinite amount of regular tilings. A simple way to determine which geometry a tiling can be draw in is by using its Schläfli symbol. Every regular tessellation can be completely broken up into triangles, because every regular polygon is made up of n-2 triangles, where n is the number of sides of the polygon. The Three Regular Euclidean Tessellations In Euclidean tessellations, the summation of angles in all triangles is 180 degrees, so the summation of angles of a n-gon is $(n-2)180\,$. and $\frac{n360}{k}$ is the summation of the angles of the polygon in a tilling then the following must be true: $\frac{1}{n} + \frac{1}{k} = \frac{1}{2}$ [Show Proof][Hide Proof] $\frac{n360}{k} = (n-2)180$ $\frac{n360}{k} = n180 - 360$ $\left( \frac{n360}{k} = n180 - 360 \right)\frac{1}{n360}$ $\frac{n360}{k (n360)} = \frac{n180}{(n360)} - \frac{360}{(n360)}$ $\frac{1}{k} = \frac{1}{2} - \frac{1}{n}$ $\frac{1}{n} + \frac{1}{k} = \frac{1}{2}$ Elliptic Tilings In hyperbolic tessellations, triangles always have a summation of angles less than 180 degrees, so that: $\frac{1}{n} + \frac{1}{k} < \frac{1}{2}$ In elliptic tessellations (also called spherical tessellations), triangles always have a summation of angles greater than 180 degrees, so that: $\frac{1}{n} + \frac{1}{k} > \frac{1}{2}$ # Teaching Materials There are currently no teaching materials for this page. Add teaching materials. Leave a message on the discussion page by clicking the 'discussion' tab at the top of this image page. [[Category:]] [[Category:]] \|other=Hyperbolic Geometry \|AuthorName=Jos Leys \|AuthorDesc=Jos Leys creates images from mathematics using programs such as Ultrafractal and Povray. He has created numerous tessellations, fractals, and other images. \|SiteName=Mathematical Imagery \|SiteURL=http://www.josleys.com/show_gallery.php?galid=262 \|Field=Geometry \|Field2=Topology \|FieldLinks=:Jos Ley's hyperbolic tilings gallery at http://www.josleys.com/show_gallery.php?galid=262 Also, Jos Ley's page on M.C. Echer's artistic works http://www.josleys.com/show_gallery.php?galid=325. Make Your Own Hyperbolic Tilings Applet at http://aleph0.clarku.edu/~djoyce/poincare/PoincareApplet.html \|References= David E. Joyce, Hyperbolic Tessellations Wilhelm Magnus, Non-Euclidean Tessellations and their Groups (www.cims.nyu.edu/vigrenew/ug_research/JohnAdamski05.pdf) Wolfram MathWorld, Wolfram MathWorld Wikipedia, Uniform Tilings in Hyperbolic Plane \|ImageRelates=This image is a quasi-regular hyperbolic tiling with Schläfli symbol of quasi-{7,3}, created using heptagons and triangles. This tessellation was created in the hyperbolic plane, so it must be visualized using the Poincaré disk model. \|ToDo=*The Tilings in Different Geometries section should be expanded so that separate section for each geometry should be created and perhaps merged with the Tessellations page. • An applet for the allowing users to make their own tilings by specifying the Schläfli symbol (similar to this applet) \|InProgress=No \|HideMME=No }}	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 13, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8718791604042053, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/102959/predicate-logic-notation-where-to-put-the-parentheses-etc	# Predicate logic notation: where to put the parentheses, etc. My math professor tends to write $\exists x\in\mathbf{X} \ni P(x)$. Is this a correct use of the such that symbol $\ni$? If not, what is the use of that symbol? Isn't it better to write $\exists x\in\mathbf{X} (P(x))$ assuming $P(x)$ is a complex proposition? Also, while we're at it, is $\forall x\in\mathbf{X} : P(x)$ an acceptable notation or should it really be $\forall x\in\mathbf{X} (P(x))$ for precision? - 2 I'm guessing he means it as "such that." Often we just use a colon: $\exists x\in\mathbf{X}: P(x)$. You could put parentheses around it, as you do. Incidentally, even that is shorthand, since what it really means is: $\exists x(x\in\mathbf{X} \text{ and } P(x))$. – Thomas Andrews Jan 27 '12 at 13:54 This is semi-formal notation, at some distance from the standard ways of expressing the idea in a formal language. – André Nicolas Jan 27 '12 at 15:15 I would describe $\exists x\in\mathbf{X}\ni P(x)$, $\exists x\in\mathbf{X}: P(x)$, and $\exists x\in\mathbf{X}$ s.t. $P(x)$ as shorthand for an English sentence, There is an x in ... . $\exists x\in\mathbf{X}\big(P(x)\big)$ and $(\exists x\in\mathbf{X}) P(x)$, on the other hand, are expressions in slightly different formal languages, abbreviating the even more formal $\exists x\big(x\in\mathbf{X}\land P(x)\big)$ or $(\exists x)\big(x\in\mathbf{X}\land P(x)\big)$. – Brian M. Scott Jan 27 '12 at 15:29 ## 1 Answer The backwards epsilon means "such that", but in this context it's slightly bizarre since the usual set membership symbol appears symmetrically before the $\mathbf{X}$ and of course means something quite different. After all, if $P$ were a function symbol rather than a relation symbol then we might parse the statement quite differently, as $\exists{x} (x \in \mathbf{X} \wedge P(x) \in \mathbf{X})$. We can't see inside your professor's mind, but with usual usage in mind, your expansion is correct: it abbreviates $\exists{x \in \mathbf{X}} (P(x))$, which is in turn an abbreviation for $\exists{x} (x \in \mathbf{X} \wedge P(x))$. Universal quantifiers expand similarly: $\forall{y \in Y} \; \varphi(y)$ is short for $\forall{y} (y \in Y \rightarrow \varphi(y))$. Both colons and full stops (periods) are acceptable separators: $\exists{y \in Y} : \varphi(y)$ and $\forall{z \in Z} \; . \; \psi(z)$ are reasonably common. The key to using abbreviations effectively is to employ them to increase clarity. They increase brevity, but if using one might be ambiguous, go for the expanded version instead. I recommend reading things like Knuth on mathematical writing for further explanations of good practice in mathematical writing. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 25, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9382893443107605, "perplexity_flag": "middle"}
http://crypto.stackexchange.com/questions/tagged/group-theory+pairings	# Tagged Questions 2answers 156 views ### when do we need composite order groups for bilinear maps and when prime order? Why we need bilinear groups of composite order? What's the special security property of the composite order group in comparison with one of prime order?To put it in another way when do we need ... 1answer 126 views ### Must the order of the groups in a bilinear map be the same? I've been reading up on bilinear maps and their application to cryptography and one thing I keep seeing hasn't yet clicked. If $e:G_1\times G_2\to G_n$ is a bilinear map, $G_1,G_2,G_n$ are always ...	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.905463695526123, "perplexity_flag": "middle"}
http://stats.stackexchange.com/questions/4768/amoeba-interview-question/4773	# Amoeba Interview Question I was asked this question during an interview for a trading position with a proprietary trading firm. I would very much like to know the answer to this question and the intuition behind it. Thank you! Amoeba Question: A population of amoebas starts with 1. After 1 period that amoeba can divide into 1, 2, 3, or 0 (it can die) with equal probability. What is the probability that the entire population dies out eventually? - are we to suppose it does each of these with probability $1/4$? – shabbychef Nov 21 '10 at 5:40 yes, each outcome has a 1/4 probability. – AME Nov 21 '10 at 6:10 8 from a biological point of view, that chance is 1. The environment is bound to change to a point that no population can survive, given that in x billion years the sun is to explode. But I guess that's not really the answer he was looking for. ;-) The question doesn't make sense either. An amoebe can only divide into 2 or 0. Moral: traders shouldn't ask questions about biology. – Joris Meys Nov 21 '10 at 12:34 5 – mbq♦ Nov 21 '10 at 14:02 This is a cute question as Mike mentions. The intuition here is that the eventual survival/extinction probability is the same between two generations. A more creative version could be thought of when the survival probability itself varies as a function of the number amoeba present. I've added it to my site blog. – broccoli Aug 23 '12 at 2:33 ## 3 Answers Cute problem. This is the kind of stuff that probabilists do in their heads for fun. The technique is to assume that there is such a probability of extinction, call it $P$. Then, looking at a one-deep decision tree for the possible outcomes we see--using the Law of Total Probability--that $P=\frac{1}{4} + \frac{1}{4}P + \frac{1}{4}P^2 + \frac{1}{4}P^3$ assuming that, in the cases of 2 or 3 "offspring" their extinction probabilities are IID. This equation has two feasible roots, $1$ and $\sqrt{2}-1$. Someone smarter than me might be able to explain why the $1$ isn't plausible. Jobs must be getting tight -- what kind of interviewer expects you to solve cubic equations in your head? - 1 The reason 1 is not a root is easily seen by considering the expected number of Amoeba after $k$ steps, call it $E_k$. One can easily show that $E_k = E_1^k$. Because the probability of each outcome is $1/4,$ we have $E_1 = 3/2$, and so $E_k$ grows without bound in $k$. This clearly does not gibe with $P = 1$. – shabbychef Nov 21 '10 at 18:03 3 @shabbychef It's not so obvious to me. You can have the expectation grow exponentially (or even faster) while the probability of dying out still approaches unity. (For example, consider a stochastic process in which the population either quadruples in each generation or dies out entirely, each with equal chances. The expectation at generation n is 2^n but the probability of extinction is 1.) Thus there is no inherent contradiction; your argument needs something additional. – whuber♦ Nov 21 '10 at 21:02 @shabbychef -- thanks for the edit. I didn't realize we could use embedded TeX for math! @whuber -- shabbychef's statement $E_k = E_1^k$ is just a variation on my statement about the extinction probability, just add expectations instead of multiplying probabilities. Nice work, shab. – Mike Anderson Nov 21 '10 at 23:04 That's clear, Mike, but what's your point? Aren't we talking about how to rule out 1 as a solution? By the way, it's obvious (by inspection and/or by understanding the problem) that 1 will be a solution. This reduces it to a quadratic equation which one can easily solve on the spot. That's not usually the point of an interview question, though. The asker is probably probing to see what the applicant actively knows about stochastic processes, Brownian motion, the Ito calculus, etc., and how they go about solving problems, not whether they can solve this particular question. – whuber♦ Nov 22 '10 at 15:18 2 @shabbychef: One way to rule out P=1 is to study the evolution of the probability generating function. The pgf is obtained by starting with t (representing an initial population of 1) and iteratively replacing t by (1+t+t^2+t^3)/4. For any starting value of t less than 1, a graphic easily shows the iterates converge to Sqrt(2)-1. In particular, the pgf is staying away from 1, showing it cannot converge to 1 everywhere, which would represent complete extinction. This is why "the 1 isn't plausible." – whuber♦ Nov 22 '10 at 21:53 Some back of the envelope calculation (litterally - I had an envelope lying around on my desk) gives me a probability of 42/111 (38%) of never reaching a population of 3. I ran a quick Python simulation, seeing how many populations had died off by 20 generations (at which point they usually either died out or are in the thousands), and got 4164 dead out of 10000 runs. So the answer is 42%. - 3 $\sqrt{2}-1$ is 0.4142, so it is in agreement with Mike's analytical result. And +1, because I like simulations ;-) – mbq♦ Nov 21 '10 at 14:04 Also +1 because I like simulations. Which would have been my answer ;). – EpiGrad Aug 23 '12 at 4:22 This sounds related to the Galton Watson process, originally formulated to study the survival of surnames. The probability depends on the expected number of sub-amoebas after a single division. In this case that expected number is $3/2,$ which is greater than the critical value of $1$, and thus the probability of extinction is less than $1$. By considering the expected number of amoeba after $k$ divisions, one can easily show that if the expected number after one division is less than $1$, the probability of extinction is $1$. The other half of the problem, I am not so sure about. - ## protected by whuber♦Aug 23 '12 at 13:45 This question is protected to prevent "thanks!", "me too!", or spam answers by new users. To answer it, you must have earned at least 10 reputation on this site.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 22, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9489349126815796, "perplexity_flag": "middle"}
http://www.physicsforums.com/showthread.php?t=106556	Physics Forums Thread Closed Page 1 of 2 1 2 > ## Universe Expansion If our universe is expanding and light travels at a finite speed, then, will there be a time in future, where we won't see any light from stars and our whole sky will be black, since all the lights have not reach us? PhysOrg.com science news on PhysOrg.com >> Front-row seats to climate change>> Attacking MRSA with metals from antibacterial clays>> New formula invented for microscope viewing, substitutes for federally controlled drug We should always see the current and possibly new stars in the future. I'm not sure but I beleive that near the beginning (when ever that was) there might have been no stars in the night sky. I think this because the universe has expanded faster then the speed of light. (Which I do not fully understand) The way I understand it; Almost all galaxies are recceding from us (very nearby galaxies excluded). The galaxies themselves are not moving as such, the space between the galaxies is increasing, and appears to be so at an ever faster rate. So in theory, one day we wont be able to see any distant galaxies. The stars in our own galaxy are gravitationally bound to the galaxy, and hence these stars are not receeding from us. So we will still be able to see the stars in our galaxy, at a time when we cannot see any other distant gallaxies. ## Universe Expansion If this were the case, wouldn't that mean that only "space" is expanding (as opposed to galactic regions)? If gravity can eventually overcome spacetime expansion, then doesn't that mean that the expansion is not a spacetime one but an addition of "outer space"? Conversely, if the entirety of spacetime is expanding, wouldn't it be correct to say that, at some point in the future (provided a long enough period has elapsed) my big toe will have moved further away from my pinky toe? Recognitions: Gold Member Science Advisor Note: It is the space-like hyper-surface slice through space-time that expands as the cosmological parameter t increases. Space-time does not 'expand' as it is static, time having been already accounted for within the continuum. The question is; "If space is expanding then what expands with it?" It is generally accepted that gravitational attraction of the local gravitational fields, of the Earth, Sun, Milky Way galaxy, and possibly the Local Group, overwhelm the cosmological expansion and these bodies do not expand with the universe. Einstein himself wrote a paper in the 1940's to prove that the solar system was not co-expanding with the universe. He did so by cutting out a spherical volume from the cosmological model and replacing it with a void with a spherical mass inserted in the middle; thus embedding a Schwarzschild solution inside a cosmological one. The question is how do you take the limit of the Schwarzschild metric as $r \rightarrow \infty$ It could be argued that the Pioneer Anomaly indicates that something might be wrong with this understanding. Garth Blog Entries: 9 Recognitions: Science Advisor Quote by touqra If our universe is expanding and light travels at a finite speed, then, will there be a time in future, where we won't see any light from stars and our whole sky will be black, since all the lights have not reach us? Let’s consider we are located in a spatially infinite universe which contains a homogeneous distribution of light sources in space and let’s assume that no new light sources are created nor destroyed as time passes. Consider a spherical volume with a given radius. Due to expansion of space, it is true that this volume will contain a decreasing number of light sources as time increases and that it will be void sometime. However, the light we are receiving is not determined by the number of light sources contained in this volume as $t \rightarrow \infty$, but by the shape of our past lightcone as $t \rightarrow \infty$. In my opinion the technical answer to your question is based on the fact that our past light cone will always intersect all past spatial hypersurfaces. This means that we are going to receive always light from every epoch in the universe (in our model). Consider a light source A from which we are receiving light emitted N years ago. As $t \rightarrow \infty$, we may not receive the light this source is currently emitting, if this source is located outside our current cosmological event horizon. However, we will receive light from some source located within our cosmological event horizon today (in the same spatial hypersurface than A). The problem is that, as $t \rightarrow \infty$, this source will be located very far in our past lightcone and therefore its light will be very redshifted. Summary: some radiation will always reach us. However, this radiation will not be visible light, due to the increasing redshift of all sources. So if I (being able to live forever without aging) were standing a meter away from a lamp (which also could go unchanging forever) - PRESUMING LOCAL GRAVITATION DID NOT INTERFERE WITH UNIVERSE EXPANSION - would the light from the lamp appear dimmer to me, say, a million or five hundred million years later? If it does, is this because the lamp is further away from me or because the strength of the light "dims" (which I guess would mean redshifts) over the course of millions of years? And would it thus be correct to say that light "dims" (in a non-relativistic or subjective sense) over time (again, observed time - not the actual light itself)? More over, the universe can only expand if it also redshifts, because to expand without redshifting (or contract without blueshifting, I guess) would be a violation of the total conservation of a finite universe's energy? Recognitions: Gold Member Science Advisor I totally agree with hellfire. In a casually connected universe, all light cones, once connected, will forever remain connected [albeit they may redshift beyond detectability over time]. And all light cones too remote to make initial causal contact with our universe prior to inflation will forever remain disconnected. Quote by jhe1984 So if I (being able to live forever without aging) were standing a meter away from a lamp (which also could go unchanging forever) - PRESUMING LOCAL GRAVITATION DID NOT INTERFERE WITH UNIVERSE EXPANSION - would the light from the lamp appear dimmer to me, say, a million or five hundred million years later? If it does, is this because the lamp is further away from me or because the strength of the light "dims" (which I guess would mean redshifts) over the course of millions of years? I don't know whether this have been implicitly answered anywhere, but I sure want to know the opinion to this. Recognitions: Gold Member Science Advisor Staff Emeritus Quote by jhe1984 So if I (being able to live forever without aging) were standing a meter away from a lamp (which also could go unchanging forever) - PRESUMING LOCAL GRAVITATION DID NOT INTERFERE WITH UNIVERSE EXPANSION - would the light from the lamp appear dimmer to me, say, a million or five hundred million years later? That question can't really be answered because it depends primarily on the local distribution of matter instead of the Hubble flow. If you were instead to ask about a "lamp" one hundred million light years away, then yes, it would appear dimmer. If we only consider cosmology, its apparent brightness will depend on its redshift. The more redshifted its light is: 1) The less total energy it carries. This is just because red photons are less energetic than blue ones. 2) The slower the photons arrive at our telescope. This is a GR effect that leads to the lamp's clock ticking slower than ours. 3) The further it will have travelled after one hundred million years. All of these things add up to a dimmer source. Remember, the faster the universe is expanding, the more the object is redshifted. Thus, the fact that the universe is accelerating means that the lamp will appear dimmer in a hundred million years. More over, the universe can only expand if it also redshifts, because to expand without redshifting (or contract without blueshifting, I guess) would be a violation of the total conservation of a finite universe's energy? Actually, energy is not always conserved in GR. The fact that light redshifts to begin with could be considered a violation of the energy conservation law, depending on your definition of "energy". See here for more details: Is Energy Conserved in General Relativity? Recognitions: Gold Member Science Advisor I feel compelled to vote in that quasi-poll, ST! I unwaveringly vote yes, energy is conserved when all is said and done . . . albeit with difficulty. Quote by SpaceTiger That question can't really be answered because it depends primarily on the local distribution of matter instead of the Hubble flow. If you were instead to ask about a "lamp" one hundred million light years away, then yes, it would appear dimmer. If we only consider cosmology, its apparent brightness will depend on its redshift. Actually, energy is not always conserved in GR. The fact that light redshifts to begin with could be considered a violation of the energy conservation law, depending on your definition of "energy". See here for more details: Is Energy Conserved in General Relativity? According to SR, the [Energy,momentum] pair could be thought as another way to describe the physical world by the other pair [displacement,velocity] "Energy" should be regarded as an operator, which should be conserved, if we are describing the physical model correctly. A physical model is "correct" because it has its own invariant, and the term "energy" could be defined as one of those invariants Recognitions: Science Advisor Staff Emeritus The FAQ and Space Tiger are correct on the issue of energy conservation in GR being problematic. Aside from the FAQ, there is a discussion of the history of energy conservation and Emily Noether's contribution in http://arxiv.org/PS_cache/physics/pdf/9807/9807044.pdf In general relativity, on the other hand, it has no meaning to speak of a definite localization of energy. One may define a quantity which is divergence free analogous to the energy-momentum density tensor of special relativity , but it is gauge dependent: i.e., it is not covariant under general coordinate transformations. Consequently the fact that it is divergence free does not yield a meaningful law of local energy conservation. Thus one has, as Hilbert saw it, in such theories ‘improper energy theorems’. A key feature for physics of Noether’s I.V. paper is the clarity her theorems brought to our understanding of the principle of energy conservation. As Feza Gursey wrote [18]: “Before Noether’s Theorem the principle of conservation of energy was shrouded in mystery, leading to the obscure physical systems of Mach and Ostwald. Noether’s simple and profound mathematical formulation did much to demystify physics.” Noether showed in her theorem I that the principle of energy conservation follows from symmetry under time translations. This applies to theories having a finite continuous symmetry group; theories that are Galilean or Poincare invariant, for example. In general relativity, on the other hand, energy conservation takes a different form as will be shown below. Noether’s theorem II applies in the case of general relativity and one sees that she has proved Hilbert’s assertion that in this case one has ‘improper energy theorems’, and that this is a characteristic feature of the theory. It is owing to the fact that the theory is a gauge theory; i.e., that it has an infinite continuous group of symmetries of which time translations are a subgroup. Indeed generally she defines as “improper” divergence relationships, which vanish when the field equations are satisfied, which correspond to a finite continuous subgroup of an infinite continuous group. Generally they do not have the required invariance or covariance properties under the larger group. For example, in general relativity a divergence free energy-momentum (pseudo) tensor can be constructed but it is gauge dependent (see below). Because it is not covariant under general coordinate transformations, it is more properly called a pseudotensor. Such pseudotensors are covariant with respect to the linear transformations of the Poincare group and may be used in asymptotic spacetime regions far from gravitating sources to derive a principle of energy conservation. The bottom line is that in GR we have theories of energy conservation that apply to static space-times, and theories of energy conservation that apply to asymptotically flat space-times, but there is no truly _general_ theorem of energy conservation in GR for arbitrary space-times. The FRW metric is neither asymptotically flat nor is it static, as a consequence there is no conserved "energy of the universe". This point is also made in most GR textbooks, including MTW's "Gravitation". Recognitions: Gold Member Science Advisor Quote by pervect In general relativity, on the other hand, it has no meaning to speak of a definite localization of energy. One may define a quantity which is divergence free analogous to the energy-momentum density tensor of special relativity , but it is gauge dependent: i.e., it is not covariant under general coordinate transformations. Therein lies the problem of a Quantum Gravity, may energy be represented by an operator or not? Does this not require a preferred foliation of space-time in contradiction to the principle of relativity? In order to be able to define such a locally conserved energy one has to select out a preferred foliation, slice, of space-time. SR is formulated for an empty, i.e. curvature free, universe, and there is nothing 'to hang' a selected frame on. However, once one has introduced curvature, i.e. proceeded from the SR limit into the more general GR situation, then matter or energy has been introduced into the system. Such matter/energy may be used to define a specific frame of reference - the Centre of Mass/momentum of the system or the frame in which radiation is globally isotropic. Thus Mach's Principle ought to be important in the local conservation of energy. This is the basis of Self Creation Cosmology. Garth Hi! Do scientists know why universe expands and even accelerates? Something is pushing or pulling? Recognitions: Gold Member Science Advisor hejs In the standard General Relativity (GR) cosmological theory the universe either has to expand or contract, it cannot 'stay still'. The Hubble red shift indicates that distant galaxies are all isotropically moving away from us. Either we live in a very special position in space or the universe as a whole is expanding. It is generally accepted that it is the universe as a whole that is expanding and that agrees with the prediction of GR. But what has it expanded from? If you follow GR all the way back you come to the singularity of the Big Bang in which the volume of the whole universe reduces to zero. Although before you reach that point quantum effects must come into play and they may well prevent the initial singularity from happening, instead the universe expanded from a very small and ultra dense volume, or 'bounced' from the collapse of a previous universe. What made it expand in the first place? The energy of the Big Bang was such that it had to expand, even in the face of a fierce cosmological gravitational field. Interestingly, in GR the kind of energy you need is provided by a negative pressure. This negative pressure shows up at two other places in the mainstream theory. 1. At an early stage of the Big Bang, 10-35 sec after BB, it is thought the universe underwent a burst or very powerful expansion called Inflation. 2. At a later stage of its expansion (at z ~ 1) it is thought the universe underwent an extended period of accelerated expansion called cosmic acceleration. Both would have been caused by two different kinds of negative pressure. Garth What do you mean by negative pressure? Could you give an analogy perhaps? Thread Closed Page 1 of 2 1 2 > Thread Tools \| \| \| \| \|-----------------------------------------\|------------------------------\|---------\| \| Similar Threads for: Universe Expansion \| \| \| \| Thread \| Forum \| Replies \| \| \| Astrophysics \| 8 \| \| \| Special & General Relativity \| 12 \| \| \| General Astronomy \| 13 \| \| \| General Astronomy \| 6 \| \| \| General Astronomy \| 6 \|	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 5, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9394441246986389, "perplexity_flag": "middle"}
http://mathematica.stackexchange.com/questions/tagged/polynomials	# Tagged Questions Questions on the functionality operating on polynomials learn more… \| top users \| synonyms (1) 3answers 102 views ### List of Tribonacci Polynomials with Mathematica? [duplicate] I want to list top ten of Tribonacci polynomials. I have following algorithm, but it doesnt work. ... 3answers 99 views ### The plot of roots of polynomials I have polynomial equation like Tribonacci Polynomials for example: $T_3(x)=x^4+x$. After finding the roots of this polynomial, I want to show these roots in the complex plane. I have tried lots of ... 1answer 48 views ### Table of Polynomials [closed] I made a function makePolynomial that creates a random Polynomial of a certain degree, e.g. 2+T+3T^2+8T^3-5T^4... now I ... 2answers 88 views ### What are Root objects with multiple polynomials? In Mathematica 9 a new flavor of Root object with multiple polynomials was introduced. For example, ... 2answers 76 views ### How to define a polynomial/function from an array of coefficients? I have the coefficients of my desired polynomial in an array CoefArr (I'm new to mathematica, so I think of everything as arrays, it is actually a list I believe) starting with the constant at index ... 3answers 142 views ### Symbolic manipulation of functional form I have a functional polynomial expression of the form: ... 2answers 82 views ### Evaluating Polynomials at Grid Points I am continuing my quest on B-splines. The code below builds a 5x5 matrix out of B-splines, using the BSplineBasis[] routine. I now want to evaluate the polynomials that are stored in each matrix ... 0answers 57 views ### Apart may use Padé method: what's that? How does Apart work? The page tutorial/SomeNotesOnInternalImplementation#7441 says, "Apart ... 5answers 284 views ### Series expansion in terms of Hermite polynomials I am trying to expand a polynomial in terms of orthogonal polynomials (in my case, Hermite). Maple has a nice built-in function for this, ChangeBasis. Is there a ... 2answers 165 views ### Calculating Taylor polynomial of an implicit function given by an equation I'd like to write a procedure that will take an equation: F(x,y,z) = 0 chosen variable: x a point: ... 4answers 133 views ### How to collect terms with positive powers in polynomial I am trying to collect all terms with non-negative powers of $x$ in polynomials like $\frac{1}{x^2}\left(a x^2+x^{\pi }+x+z\right)^2$ First expand the polynomial ... 7answers 215 views ### Defining a function that completes the square given a quadratic polynomial expression How can I write a function that would complete the square in a quadratic polynomial expression such that, for example, CompleteTheSquare[5 x^2 + 27 x - 5, x] ... 1answer 92 views ### Is there any way to force Mathematica to collect a symbol in a polynomial? Let's say that I have a polynomial like this: a + b + c Is there any way that I can get Mathematica to transform it to: ... 0answers 102 views ### Negative power instead of fraction Solve returns a solution in the form {{x->y/a^2 + y^2/a^7}}. Since I want to process the input (with another program) in terms of Laurent polynomials, I would ... 1answer 46 views ### Factorize and find the null space of a polynomial in several variables [duplicate] I've been asked to factor the following polynomial: poly = 6 x^3 + x^2 y - 11 xy^2 - 6 y^3 - 5 x^2 z + 11 xyz + 11 y^2 z - 2 xz^2 - 6 yz^2 + z^3 And to solve for z so that poly = 0 Can anyone help ... 3answers 118 views ### Convert coefficients of polynomials into a matrix I have several sets of 5 polynomials of the form: ... 0answers 90 views ### Computing Ehrhart's polynomial for a convex polytope Is there a Mathematica implementation for computing the Ehrhart polynomial of a convex polytope which is specified either by its vertices or by a set of inequalities? I am interested in knowing this ... 1answer 72 views ### Integrating polynomial functions over polytopes with an add-on package There is a Mathematica package to evaluate integrals over polytopes: http://library.wolfram.com/infocenter/Books/3652/ In the documentation (Functions.nb file) I ... 3answers 158 views ### How to evaluate all the essentially distinct polynomials in 4 variables over $F_2$ on points of $F_2 ^ 4$ I am a beginner with Mathematica. For my research purpose I would like to get a list of all the polynomials in $F_2[x,y,z,w]$ and for each polynomial I would like to know the result that it gives then ... 1answer 55 views ### FindFit and Integration errors First off, appologies for what may sound like a newbie question, as I am very new to using Mathematica. I am trying to find a way to get Mathematica to give me an expression that would describe the ... 1answer 175 views ### How do I get a two-term polynomial with a leading negative sign to display in the correct (i.e. textbook) order? The first three expressions evaluate as expected and the polynomial is displayed in what I would call "textbook" form. The last expression, however, switches the order of terms. Mathematica employs ... 1answer 93 views ### How to transform an expression using algebraical instead of pattern rules [duplicate] I would like to transform rules algebraically. A very simple example would be: - k^2 - 2 k x + x^2 /. {2k -> 1} This transforms to: - $$k^2-2 k x+x^2$$ ... 1answer 124 views ### Expanding a polynomial with fractional powers Given an expression like a + by + cy^2 + dSqrt[f + gy + hy^2] How can I programatically, expand this to a quartic without any fractional powers? Right ... 1answer 89 views ### How can I prevent a polynomial from being simplified? I'm having a problem with polynomials. Let's say I have a polynomial "2x^2 - 5x + 6 - 3x^2" .. How can I check that this expression is not simplified ? Additionally, I would like to locate the ... 1answer 90 views ### How can I get an exponent vector from monomials? I am trying to get an exponent vector from a list of monomials. I am using the CoefficientRules command; however, it is returning a list that includes the ... 3answers 436 views ### Solving cubic equation for real roots I'm looking to solve the following cubic equation for x: $\beta\, x^3 - \gamma \,x = c$. I have plugged in some sample values ($\beta = 2$, $\gamma = 5$ and $c = 2$). When I try to solve this ... 1answer 117 views ### Implementation of Decompose I'm curious as to how Decompose works so I decided to use Trace with the option ... 2answers 190 views ### How can I make the output from Solve look nice? I have a problem with presenting solutions. Roots of 4th order polynomials are big expressions. Is there a way to present the roots, s2 and s3, in normal form with some substitutions? Maybe a way to ... 0answers 25 views ### Handling “Solve was unable to solve the system with inexact coefficient” errors [duplicate] Possible Duplicate: How to get rid of warnings when using Solve on an equation with inexact coefficients? I've been trying to calculate the following in Mathematica: ... 4answers 511 views ### How to get exact roots of this polynomial? The equation $$64x^7 -112x^5 -8x^4 +56x^3 +8x^2 -7x - 1 = 0$$ has seven solutions $x = 1$, $x = -\dfrac{1}{2}$ and $x = \cos \dfrac{2n\pi}{11}$, where $n$ runs from $1$ to $5$. With ... 3answers 375 views ### First positive root Simple question but problem with NSolve. I need help how to extract first positive root? For example eq=-70.5 + 450.33 x^2 - 25 x^4; NSolve[eq== 0, x] If I ... 0answers 78 views ### Know the degree of the equation with the radicals expanded I have an equation with radicals. I would like to know what would be the degree of the polynomials if I'd move the terms and square the equation a number of times sufficient to remove all the ... 1answer 155 views ### How to do the polynomial stuff over finite fields extensions fast? This question is raised from the problem of package FiniteFields being very slow (please, see the corresponding question): I have had an evidence that Mathematica ... 2answers 196 views Equations ... 3answers 257 views ### How to keep Collect[] result in order? For example, Collect[(1 + x + Cos[s] x^2)^3, x] gives the result ... 1answer 187 views ### Generating lots of Examples in Polynomials Rings I'm studying polynomial rings and i would like to know some tricks for generating lots of examples. For instance, suppose i'm interested in polynomials over the integers mod (2,x^3 + 1). To get a ... 4answers 315 views ### How do I find the degree of a multivariable polynomial automatically? I have a very simple question which appears not to have already been answered on this forum. Is there built-in functionality that returns the degree of a multivariable polynomial? For example if the ... 1answer 92 views ### Factorize Parametric Polynomials Is there a possibility to factorize a parametric polynomial expression - meaning that the coefficients are defined as parameters, and not as specific numbers? My example - a polynomial in ... 1answer 157 views ### Small Issue with Chebyshev Derivative Appoximation I am trying to get approximate the derivative of a function from its Chebyshev expansion. I start out with the following random function ... 3answers 977 views ### Get polynomial interpolation formula I'm attempting to get a polynomial interpolation formula out of Mathematica but I am absolutely lost. I stared out using ... 2answers 265 views ### How to deduce a generator formula for a polynomial sequence? Consider a polynomial sequence $\{p_n\}$ generated by some (simple) rule: \begin{array}{l} p_1(x)=x \\ p_2(x)=2 x-x^2 \\ p_3(x)= x^3-3 x^2+3 x \\ p_4(x)=-x^4+4 x^3-6 x^2+4 x \\ p_5(x)= x^5-5 ... 1answer 324 views ### Polynomial Approximation from Chebyshev coefficients I would like to expand a function $f(r)$ in the domain $[0,R]$, around the points $r =0$, and $r = R$ in the following manner $f(r = 0) = \Sigma_{i=0,i = even}^{imax} f_i (r/R)^i$ and \$f(r = R) = ... 2answers 242 views ### 3D Plot: Number of Roots in x of a polynomial in x, a, b and c I have a polynomial in four variables x,a,b and c. The number of roots of the polynomial in x depends of the choice of a, b and c. I would like to have a 3D-Plot with a, b and c on the axes, while the ... 2answers 188 views ### How can I compute the representation matrices of a point group under given basis functions? Take the $C_{3v}$ point group for example: ... 3answers 211 views ### Is it possible to use Composition for polynomial composition? I want to do this: $P = (x^3+x)$ $Q = (x^2+1)$ $P \circ Q = P \circ (x^2+1) = (x^2+1)^3+(x^2+1) = x^6+3x^4+4x^2+2$ I used Composition for testing if that could ... 5answers 367 views ### How do I reassign canonical ordering of symbols? I have a big polynomial that evaluates to: $$A^2 e^2 \phi ^- \phi ^++A e \phi ^- \phi ^+ c_{2 w} g_Z+\frac{1}{2} A e g h W^- \phi ^+ +\ll13\gg,$$ which is supposed to represent some terms in the ... 2answers 139 views ### Strange integration result In Mathematica 8, the Integrate command sometimes strangely integrates polynomials yielding unsimplified (and unexpected) fractional results. As an example, the line: ... 1answer 131 views ### How to express the original ideal elements in the Groebner basis? Suppose I call GroebnerBasis[{f1, f2, ...}, {x1,x2, ...}] The output is a list {g1,g2,...} For each $g_j$, there should be ... 4answers 299 views ### “Evaluating” polynomials of functions (Symbols) I want to implement the following type evaluation symbolically $$(f^2g + fg + g)(x) \to f(x)^2 g(x) + f(x) g(x) + g(x)$$ In general, on left hand side there is a polynomial in an arbitrary number of ... 3answers 460 views ### What function can I use to evaluate $(x+y)^2$ to $x^2 + 2xy + y^2$? What function can I use to evaluate $(x+y)^2$ to $x^2 + 2xy + y^2$? I want to evaluate It and I've tried to use the most obvious way: simply typing and evaluating $(x+y)^2$, But it gives me only ...	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 33, "mathjax_display_tex": 4, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.902289092540741, "perplexity_flag": "middle"}
http://unapologetic.wordpress.com/2011/08/05/chains/?like=1&source=post_flair&_wpnonce=420e8d64d7	# The Unapologetic Mathematician ## Chains We can integrate on the standard cube, and on singular cubes in arbitrary manifolds. What if it’s not very easy to describe a region as the image of a singular cube? This is where chains come in. So a chain is actually pretty simple; it’s just a formal linear combination of singular $k$-cubes. That is, for each $k$ we build the free abelian group $C_k(M)$ generated by the singular $k$-cubes in $M$. If we have a formal sum $c=a_1c_1+\dots+a_lc_l$ — the $c_i$ are all singular $k$-cubes and the $a_i$ are all integers — then we define integrals over the chain by linearity: $\displaystyle\int\limits_c\omega=a_1\int\limits_{c_1}\omega+\dots+a_l\int\limits_{c_l}\omega$ And that’s all there is to it; just cover the $k$-dimensional region you’re interested in with singular $k$-cubes. If there are some overlaps, those areas will get counted twice, so you’ll have to cover them with their own singular $k$-cubes with negative multipliers to cancel them out. Take all the integrals — by translating each one back to the standard $k$-cube — and add (or subtract) them up to get the result! ### Like this: Posted by John Armstrong \| Differential Topology, Topology ## 4 Comments » 1. [...] that we’re armed with chains — formal sums — of singular cubes we can use them to come up with a homology theory. [...] Pingback by \| August 9, 2011 \| Reply 2. [...] anyway, on to the theorem! We know how to integrate a differential -form over a -chain . We also have a differential operator on differential forms [...] Pingback by \| August 17, 2011 \| Reply 3. [...] defined how to integrate forms over chains made up of singular cubes, but we still haven’t really defined integration on manifolds. [...] Pingback by \| September 5, 2011 \| Reply 4. [...] sensible to identify an orientation-preserving singular cube with its image. When we write out a chain, a positive multiplier has the sense of counting a point in the domain more than once, while a [...] Pingback by \| September 8, 2011 \| Reply « Previous \| Next » ## About this weblog This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”). I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 14, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9125106334686279, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/63129/how-do-you-calculate-the-solid-angle-of-a-rectangular-axis-aligned-section-of-a	## How do you calculate the solid angle of a rectangular, axis aligned section of a surface defined by a two dimensional function? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I have $f(x,y) = \frac{1}{2} (1 - x^2 - y^2)$, which is a paraboloid centered around the origin (plot). Now I want to calculate the solid angle (with the origin as the viewpoint) of the surface area defined by f(x,y) with a rectangular, axis aligned section of the xy plane as its input, e.g. $-1 \leq x \leq 1$ and $-1 \leq y \leq 1$. So each point of the surface area has the coordinates `$\vec f(x,y) = ( x, y, f(x, y)) = ( x, y, \frac{1 - x^2 - y^2}{2} )$`. The problem is that I don't know how I convert the Integral over the Surface $\iint_S dS$ in `$\Omega = \iint_S \frac { \vec{r} \cdot \hat{n} \,dS }{r^3}$` into an Integral over $x$ and $y$: $\int_X \int_Y dx dy$. ### If I... ...simply replace it with $\int_{-1}^{1} \int_{-1}^{1} dx dy$, replacing the other values accordingly with • each point on the surface: `$\vec r = \vec f(x, y) = (x, y, \frac{1 - x^2 - y^2}{2})$`, • normal at each point on the surface: `$\hat{n} = \frac{\vec f_x \times \vec f_y} {\|\vec f_x \times \vec f_y\|} = (x, y, 1) $` with `$\vec f_x(x, y) = (1, 0, -x)$` and `$\vec f_y(x,y) = (0, 1, -y)$`, I receive `$\Leftrightarrow \int_x \int_y \frac{ \vec f(x, y) \cdot (x,y,1) } { \|\vec f(x, y)\|^3 \cdot \|(x, y, 1)\|} $` `$\Leftrightarrow \int_x \int_y \frac{ (x, y, \frac{1 - x^2 - y^2}{2}) \cdot (x, y 1)} {\|(x, y, \frac{1 - x^2 - y^2}{2})\|^3 \cdot \|(x, y, 1)\|}$` `$\Leftrightarrow \int_x \int_y \frac{ x^2 + y^2 + \frac{1 - x^2 - y^2}{2} } { ( x^2 + y^2 + (\frac{1 - x^2 - y^2}{2})^2)^{\frac{3}{2}} \cdot (x^2 + y^2 + 1)^{\frac{1}{2}} }$` If I let wolframalpha calculate[1] that, the result is $5.87$. This is clearly wrong though, because the paraboloid covers more than the hemisphere, and thus needs to have a solid angle of more than $2\pi$. So what do I need to change? ### Background I'm using this paraboloid as a mapping to project geometry into a texture, and for the next step I need to find out what portion of the hemisphere each pixel covers. So ideally I need a way to calculate this as fast as possible - I might need to search for a similar, faster function for actual usage. [1] http://www.wolframalpha.com/input/?i=Integrate+(x%5e2+%2b+y%5e2+%2b+(1-x%5e2-y%5e2)%2f2)+%2f+((x%5e2+%2b+y%5e2+%2b+((1-x%5e2-y%5e2)%2f2)%5e2)%5e(3%2f2)++(x%5e2+%2b+y%5e2+%2b+1)%5e(1%2f2))+from+x%3d-1+to+1+y%3d-1+to+1&incParTime=true - I edited the tags - this doesn't fall under topology or math. phys. as far as I can see. – David Roberts Apr 27 2011 at 11:34 ## 3 Answers You need to include the differential surface area in your parametrized version of the integral. In effect, you replace the $\hat{n} dS$ term with $$\frac{\vec{f}_x\times\vec{f}_y}{\\|\vec{f}_x\times\vec{f}_y\\|} {\\|\vec{f}_x\times\vec{f}_y\\|} dxdy.$$ Although, really the two norms just cancel, so you needn't calculate them. Your integral then becomes $$\int_x \int_y \frac{( x,y,\frac{1-x^2-y^2}{2})}{(x^2+y^2+(\frac{1-x^2-y^2}{2})^2)^{(3/2)}}\cdot( x,y,1) dx dy.$$ This simplifies to $$\int_x \int_y \frac{4}{(1+x^2+y^2)^2} dx dy.$$ WolframAlpha gives the value of this as about 6.96336. - ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. This is not an answer, just another way of viewing the calculation. You need only compute the area of the roughly one-eighth of the sphere that falls below the equator (green below) and beneath the line formed by the origin and the curve $(x,1,-x^2 /2)$ for $x\in[-1,1]$, tracing out one boundary curve. The portion outside your solid angle is composed of four of these regions. - I'm afraid I need to calculate the solid angle for arbitrary sections/intervals. – hrehf Apr 27 2011 at 16:39 The situation differs little with arbitrary intervals. Still there are curves on the sphere bounding the area you want to measure. If you need an exact answer, you will need to explicitly compute those curves. If you are content with a numerical approximation, you could estimate that area in many ways. – Joseph O'Rourke Apr 27 2011 at 18:03 The normal to the paraboloid plays no role in this. A "surface element" ${\rm d}(x,y)$ at the point $(x,y)$ in the $(x,y)$-parameter plane produces via $\vec f$ (or rather $\vec f_$) a surface element $dS$ at the point $\vec f(x,y)$ on your paraboloid $S$, and then this surface element $dS$ casts a shadow $d\omega$ on the unit sphere $S^2$ through central projection from $O$, i.e., via normalization of $\vec f$. Since $$\|\vec f(x,y)\|^2=x^2+y^2+{1\over4}(1-x^2-y^2)^2={1\over4}(1+x^2+y^2)^2$$ it follows that the shadow on $S^2$ is produced by the map $$\vec g: \quad (x,y) \mapsto {2\over 1+x^2+y^2} \bigl(x,y,{1\over2}(1-x^2-y^2)\bigr)\ .$$ This $\vec g$ is nothing else but an (unusual) parametric representation of $S^2$. In order to compute the area of the shadowed part of $S^2$ one has to compute $d\omega=\|g_x\times g_y\|{\rm d}(x,y)$ and to integrate this over the intended rectangle in the $(x,y)$-plane. The computation gives, as already remarked by Ben, $$d\omega={4\over(1+x^2+y^2)^2}{\rm d}(x,y)\ .$$ Transforming to polar coordinates one finds for the $[-1,1]^2$-rectangle the exact value $8\sqrt 2\ \arctan(1/\sqrt 2)\doteq 6.96366$. - The shadowing makes sense to me and fits my view of the problem. I see that an alternate (?) view of the problem is that we need to integrate over the area of the projection of $S$ onto the unit sphere. However this confuses me even more, because I'm not sure what we need to do in order to represent the projection/transformation/etc. in the first place. Also, I do not see how you arrived at $\frac{2}{1 + x^2 + y^2} \cdot \vec f$, can you explain some more? Is this the complete integral that gives us the solid angle? – hrehf Apr 27 2011 at 16:37 @href: You were lucky: x^2+y^2+{1/over4}(1-x^2-y^2)^2={1/over4}(1+x^2+y^2)^2\$. – Christian Blatter Apr 27 2011 at 22:33	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 37, "mathjax_display_tex": 6, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9129917025566101, "perplexity_flag": "head"}
http://www.oilfieldwiki.com/wiki/Archimedes'_principle	# Archimedes' principle From Oilfield Wiki - the Oilfield Encyclopedia Archimedes (287 BC - 212 BC), the discoverer of this principle Archimedes' principle is a law of physics stating that the upward force (buoyancy) exerted on a body immersed in a fluid is equal to the weight of the amount of fluid the body displaces. In other words, an immersed object is buoyed up by a force equal to the weight of the fluid it actually displaces. Archimedes' principle is an important and underlying concept in the field of fluid mechanics. This principle is named after its discoverer, Archimedes of Syracuse.[1] ## Principle Archimedes' treatise "On Floating Bodies" states that: Any object, wholly or partially immersed in a fluid, is buoyed up by a force equal to the weight of the fluid displaced by the object. — Archimedes of Syracuse with the clarifications that for a sunken object the volume of displaced fluid is the volume of the object. Thus, in short, buoyancy = weight of displaced fluid. Archimedes' principle is true of liquids and gases, both of which are fluids. If an immersed object displaces 1 kilogram of fluid, the buoyant force acting on it is equal to the weight of 1 kilogram (technically, as a kilogram is unit of mass and not of force, the buoyant force is the weight of 1 kg, which is 9.8 Newtons.) It is important to note that the term immersed refers to an object that is either completely or partially submerged. If a sealed 1-liter container is immersed halfway into the water, it will displace a half-liter of water and be buoyed up by a force equal to the weight of a half-liter of water, no matter what is in the container. If such an object is completely immersed (submerged), it will be buoyed up by a force equivalent to the weight of a full liter of water (1 kilogram of mass). If the container is completely submerged and does not compress, the buoyant force will equal to the weight of 1 kilogram of water at any depth. This is due to the fact that at any depth, the container can displace no greater volume of water than its own volume. The weight of this displaced water (not the weight of the submerged object) equals the buoyant force. Thus, for objects floating or sunken, Archimedes' principle may be stated in terms of forces: ### Floatation Archimedes' principle relates buoyant force and displacement of fluid. However, the concept of Archimedes' principle can be applied when considering why objects float. Proposition 5 of Archimedes' treatise On Floating Bodies states that: Any floating object displaces its own weight of fluid. — Archimedes of Syracuse[2] In other words, for a floating object on a liquid, the weight of the displaced liquid is the weight of the object. Thus, only in the special case of floating does the buoyant force acting on an object equal the objects weight. Consider a 1-ton block of solid iron. As iron is nearly eight times denser than water, it displaces only 1/8 ton of water when submerged, which is not enough to keep it afloat. Suppose the same iron block is reshaped into a bowl. It still weighs 1 ton, but when it is put in water, it displaces a greater volume of water than when it was a block. The deeper the iron bowl is immersed, the more water it displaces, and the greater the buoyant force acting on it. When the buoyant force equals 1 ton, it will sink no farther. When any boat displaces a weight of water equal to its own weight, it floats. This is often called the "principle of floatation": A floating object displaces a weight of fluid equal to its own weight. Every ship, submarine, and dirigible must be designed to displace a weight of fluid equal to its own weight. A 10,000-ton ship must be built wide enough to displace 10,000 tons of water before it sinks too deep in the water. The same is true for vessels in air (as air is a fluid): a dirigible that weighs 100 tons displaces at least 100 tons of air. If it displaces more, it rises; if it displaces less, it falls. If the dirigible displaces exactly its weight, it hovers at a constant altitude. ## Refinements Archimedes' principle does not consider the surface tension (capillarity) acting on the body.[3] ## Formula The weight of the displaced fluid is directly proportional to the volume of the displaced fluid (if the surrounding fluid is of uniform density). In simple terms, the principle states that the buoyant force on an object is going to be equal to the weight of the fluid displaced by the object, or the density of the fluid multiplied by the submerged volume times the gravitational constant, g. Thus, among completely submerged objects with equal masses, objects with greater volume have greater buoyancy. Suppose a rock's weight is measured as 10 newtons when suspended by a string in a vacuum with gravity acting upon it. Suppose that when the rock is lowered into water, it displaces water of weight 3 newtons. The force it then exerts on the string from which it hangs would be 10 newtons minus the 3 newtons of buoyant force: 10 − 3 = 7 newtons. Buoyancy reduces the apparent weight of objects that have sunk completely to the sea floor. It is generally easier to lift an object up through the water than it is to pull it out of the water. Assuming Archimedes' principle to be reformulated as follows, $\text{apparent immersed weight} = \text{weight} - \text{weight of displaced fluid}\,$ then inserted into the quotient of weights, which has been expanded by the mutual volume $\frac { \text{density}} { \text{density of fluid} } = \frac { \text{weight}} { \text{weight of displaced fluid} }$ yields the formula below. The density of the immersed object relative to the density of the fluid can easily be calculated without measuring any volumes: $\frac { \text {density of object}} { \text{density of fluid} } = \frac { \text{weight}} { \text{weight} - \text{apparent immersed weight}}\,$ (This formula is used for example in describing the measuring principle of a dasymeter and of hydrostatic weighing.) Example: If you drop wood into water, buoyancy will keep it afloat. Example: A helium balloon in a moving car. In increasing speed or driving a curve, the air moves in the opposite direction of the car's acceleration. The balloon however, is pushed due to buoyancy "out of the way" by the air, and will actually drift in the same direction as the car's acceleration. When an object is immersed in a liquid the liquid exerts an upward force which is known as buoyant force and it is proportional to the weight of displaced liquid. The sum force acting on the object, then, is proportional to the difference between the weight of the object ('down' force) and the weight of displaced liquid ('up' force), hence equilibrium buoyancy is achieved when these two weights (and thus forces) are equal. ## See also • Buoyancy • "Eureka", reportedly exclaimed by Archimedes upon discovery that the volume of displaced fluid is equal to the volume of the submerged object ## References 1. {{#invoke:Citation/CS1\|citation \|CitationClass=journal }} 2. "The works of Archimedes". p. 257. Retrieved 11 March 2010. "Any solid lighter than a fluid will, if placed in the fluid, be so far immersed that the weight of the solid will be equal to the weight of the fluid displaced." ar:مبدأ أرخميدس ca:Principi d'Arquimedes cs:Archimédův zákon de:Archimedisches Prinzip et:Archimedese seadus el:Αρχή του Αρχιμήδη es:Principio de Arquímedes eu:Arkimedesen printzipioa fr:Poussée d'Archimède hy:Արքիմեդի օրենք hr:Arhimedov zakon it:Principio di Archimede he:חוק ארכימדס ka:არქიმედეს კანონი kk:Архимед заңы sw:Kanuni ya Archimedes lv:Arhimēda spēks lt:Archimedo dėsnis hu:Arkhimédész törvénye nl:Wet van Archimedes ja:アルキメデスの原理 no:Arkimedes' prinsipp nn:Arkimedes-lova pms:Prinsipi d'Archimede pl:Prawo Archimedesa ro:Forță arhimedică ru:Закон Архимеда sq:Ligji i Arkimedit si:ආකිමිඩීස් නියමය simple:Archimedes' principle sk:Archimedov zákon sl:Arhimedov zakon sr:Архимедов закон sv:Arkimedes princip ta:ஆர்க்கிமிடீசு தத்துவம் th:หลักการของอาร์คิมิดีส tr:Arşimet prensibi uk:Закон Архімеда vi:Lực đẩy Archimedes wo:Dalu Archiede zh:阿基米德浮體原理	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 3, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8568761944770813, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/47523?sort=oldest	Example of: K-regular graph with girth K, for a given K Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I've been reading some papers about (n,g)cages, but it doesn't seem trivial to me that there is, indeed, for example, a K-regular graph with girth K, for large Ks. So: Can you give me an example of a K-regular graph (it may be a graph with minimum degree K, if it's easier) and girth K (it may be >= K, if it's easier) for a given integer K. Thanks in advance. - 3 Answers In this paper http://onlinelibrary.wiley.com/doi/10.1002/jgt.3190070210/abstract even more strict theorem is proved. If I remember well, the example you are asking about belongs to Tutte. - I did not look closely, but it seems that that article starts with a given graph have a certain degree and minimal girth and uses it to make something meeting more conditions (on even and odd girth) – Aaron Meyerowitz Dec 19 2010 at 0:26 You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. It is one of the problems for which Random Graph theory seems amazingly powerful. Bollobas gives a very short (non-deterministic) construction of graphs with "minimum degree at least K" and "girth at least g". He's actually creating a large random graph with few short cycles, then removing an edge from each of them to break them all. As the average degree of his graph is more than 2K, he then removes vertices of degree less than K until the minimum degree is large enough. You can read all about that in "Extremal Graph Theory (Bollobas)", chapter 3.1 : "Graphs with large minimum degree and large girth" (he also gives a construction of K-regular graphs of girth "at least something") Nathann - I am not sure this construction originates with Bollobas; a closely related construction (high chromatic number, high girth) goes back to Erdos. – Qiaochu Yuan Dec 18 2010 at 3:39 The easiest way to do this is as said the probabilistic method. However, for those who prefer non-random constructions, here is a greedy method. Choose some large $n$ (you can calculate what is needed easily enough). Start with the empty $n$-vertex graph. Add edges greedily between vertices subject to two conditions. First, the vertices you join should always have distance at least $K$ in the current graph (if not connected, then assume distance is infinite). Second, both vertices should have degree at most $K-1$. When this procedure is forced to terminate for lack of such pairs, you have a graph with maximum degree $K$ and girth at least $K$. Now take any vertex $v$ of degree less than $K$. Look at all the vertices at distance less than $K$ from $v$ (including $v$). This set must include all the vertices of degree less than $K$, or you would not have terminated. But the set has at most $1+K+K(K-1)+K(K-1)^2+..+K(K-1)^{K-1}=C$ vertices, since the maximum degree is $K$. Similarly, the set of vertices at distance at most $2K$ from $v$ is bounded, and we can presume there are at least $KC^2$ vertices at distance more than $2K$ from $v$. Now joining greedily vertices of degree less than $K$ to these far-away vertices greedily without creating short cycles must succeed (each edge added blocks at most $C$ vertices, and there are certainly not more than $CK$ edges required). To make this have girth exactly $K$, start with a $K$-cycle. To make it regular is a little harder: one option is to run the first procedure (starting with a $K$-cycle which we insist on preserving forever, to fix the girth) with a much higher distance requirement to join two edges (say $3K$), then after termination, identify a low-degree vertex $u$ and adding an edge to some far-away $v$ (as before) then removing some edge $vw$. Now $w$ cannot have been (before edge removal) a low degree vertex (it is too far away from the first low degree vertex) and furthermore it cannot be within distance $K$ of any remaining low degree vertex, so you can join it to a remaining low degree vertex. Rinse, repeat, assuming $n$ satisfies the parity condition for a $K$-regular graph to exist (and is large enough) you will succeed. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 33, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9498964548110962, "perplexity_flag": "head"}
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From what I understand the change from a liquid to a gas requires energy and therefore ...	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 31, "mathjax_display_tex": 4, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9362846612930298, "perplexity_flag": "middle"}
http://stats.stackexchange.com/questions/20870/is-there-a-fast-algorithm-to-check-for-arp-stationarity/26822	# Is there a fast algorithm to check for AR(p) stationarity? It is well-known that an AR(p) process $$x_t=\sum_{i=1}^p \varrho_i x_{t-i} + \epsilon_t \,,$$ is causal and stationary if and only if the roots of the polynomial $$\mathcal{P}(u) = 1 - \sum_{i=1}^p \varrho_i u^i$$ are all outside the unit circle in the complex plane. (Here is a Cross Validated discussion on the topic.) My question is whether or not there exists a faster algorithm than the naïve one that consists in (a) finding the roots of $\mathcal{P}$ and (b) checking none one them is inside the unit circle. - This is a mathematical question. – vinux Jan 10 '12 at 12:53 1 @vinux: It is more of a computational question in that I am looking for the smallest power $d$ such that the answer can be produced in a time $\text{O}(p^d)$... – Xi'an Jan 10 '12 at 14:23 1 The process, as it's written is always causal. The condition about the roots tells if the system is stable - which implies stationarity. But it's more common to speak about stability conditions, rather than stationarity conditions, IMO. – leonbloy Apr 20 '12 at 18:39 ## 2 Answers The Schur-Cohn algorithm has $d=2$; this is what I learned in a computational statistics class at Berkeley some years ago. - 1 Thank you, this is exactly what I was looking for. – Xi'an Jan 10 '12 at 15:43 For comparison purposes, do you also know of the order of an efficient root finder? – Xi'an Jan 10 '12 at 19:35 1 I think they are also $d=2$, using the following thought process. Evaluation of a p-degree polynomial is $O(p)$. Finding a single root should be independent of the order of the polynomial. After the root is found, the polynomial can be reduced to a polynomial of degree $p-1$, but this doesn't affect the order. So each of the $p$ roots takes $O(p)$ to find, leading to an $O(p^2)$ algorithm. – jbowman Jan 10 '12 at 20:49 1 – jbowman Jan 10 '12 at 20:55 2 – jbowman Jan 11 '12 at 15:22 show 1 more comment It is unnecessary to find the $p$ complex roots as far as these are not to be used for themselves. Moreover, most (if not all) root finding processes can fail for large $p$. Another solution is as follows. The $\mathrm{AR}(p)$ model can be reparametrised thanks to its $p$ partial autocorrelations (PACs) $\zeta_k$ for $1 \le k \le p$. The PACs are often denoted as $\phi_{k,k}$ because their computation involves an array $\phi_{k,\ell}$. There is a one-to-one correspondence between the vector $\boldsymbol{\rho}$ of the $p$ coefficients in the stationarity region and the vector $\boldsymbol{\zeta}$ in the region defined by the $p$ conditions $\|\zeta_k \| < 1$ for $1 \le k \le p$. The transformation "AR to PAC" $\boldsymbol{\rho} \mapsto \boldsymbol{\zeta}$ is quite simple and is given as a pretty recursion formula due to Barndorff-Nielsen and Schou. Testing stationarity from the vector of coefficients boils down to computing the $\zeta_k$ and stop as soon as one condition $\| \zeta_k \| \ge 1$ is found. The less well-known inverse transform "PAC to AR" $\boldsymbol{\zeta} \mapsto \boldsymbol{\rho}$ is available explicitly (due to Monahan), and is as simple as is the direct one. It might also be useful in some cases. The two transformations are implemented (in R) in a CRAN R package named FitAR which provides as well an efficient `invertibleQ` function to test stationarity. The package is described in the following article where list of references is provided. McLeod, A.I. and Zhang Y., "Improved Subset Autoregression: With R Package" Journal of Statistical Software, vol. 28, Issue 2, Oct 2008. http://www.jstatsoft.org/v28/i02 - 2 – leonbloy Apr 20 '12 at 18:34	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 28, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9421729445457458, "perplexity_flag": "head"}
http://mathoverflow.net/revisions/122725/list	## Return to Question 2 added 102 characters in body I believe that the following questions are very basic, but I don't know how to get a reference. Consider a curve in the plane $C\in \mathbb C^2$ with a singularity at $0$ and suppose it is unibranch at zero (i.e. analytically irreducible). Then I guess one should be able to define "arithmetic genus defect" of the curve at $0$. Namely if one smooths analytically $C$, its geometric genus will grow by a positive number (in case of the cusp $x^2=y^3$ it will grow by one), and let us call this number the defect. Question 1. Is this defect well defined (independent of a smoothing)? How is it called and how one should calculate it (say it terms of the local ring of $C$ at $0$)? Question 2. Suppose we have an explicit local parametrisation of $C$ at $0$, say by two holomorphic functions $f(t), g(t)$ (polynomials if you wish). Is it possible to find this "defect" as a certain invariant of this pair of functions at $t=0$? Question 1 is settled in the answer of unknown and Question 2 in comments to it by Roy and Vivek 1 # "Arithmetic genus" of a plane curve singularity. I believe that the following questions are very basic, but I don't know how to get a reference. Consider a curve in the plane $C\in \mathbb C^2$ with a singularity at $0$ and suppose it is unibranch at zero (i.e. analytically irreducible). Then I guess one should be able to define "arithmetic genus defect" of the curve at $0$. Namely if one smooths analytically $C$, its geometric genus will grow by a positive number (in case of the cusp $x^2=y^3$ it will grow by one), and let us call this number the defect. Question 1. Is this defect well defined (independent of a smoothing)? How is it called and how one should calculate it (say it terms of the local ring of $C$ at $0$)? Question 2. Suppose we have an explicit local parametrisation of $C$ at $0$, say by two holomorphic functions $f(t), g(t)$ (polynomials if you wish). Is it possible to find this "defect" as a certain invariant of this pair of functions at $t=0$?	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 22, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9495062828063965, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/248770/simplify-frac5-sqrt206-sqrt20-to-frac5-sqrt58/248771	# Simplify $\frac{5+ \sqrt{20}}{6+ \sqrt{20}}$ to $\frac{5+ \sqrt{5}}{8}$ I would like to know how the following expression was simplified? From this $$\frac{5+ \sqrt{20}}{6+ \sqrt{20}}$$ to this $$\frac{5+ \sqrt{5}}{8}$$ - 1 Multiply the numerator and denominator by $6-\sqrt{20}$ and simplify – Simon Hayward Dec 1 '12 at 20:26 Thank you. I totally forgot this trick! – Mohammed Saleh Alorayed Dec 4 '12 at 3:27 ## 2 Answers You could simplify it a step prematurely for kicks and giggles. $\sqrt{20}=\sqrt{4\cdot 5}=2\sqrt{5}$. Now, $\frac{5+\sqrt{20}}{6+\sqrt{20}}=\frac{5+2\sqrt{5}}{6+2\sqrt{5}}$. Then, of course, multiply by the fraction by $\frac{6-2\sqrt{5}}{6-2\sqrt{5}}$. Multiplying by the conjugate to remove the radical is a common practice. It comes from the fact that $(a-b)(a+b)=a^2+ab-ba-b^2=a^2-b^2.$ So, in the case of $\frac{\text{something}}{u+\sqrt{v}}$, we multiply by $\frac{u-\sqrt{v}}{u-\sqrt{v}}$ since we know that $(u+\sqrt{v})(u-\sqrt{v})=u^2-v^2$. Similarly, if it is $u-\sqrt{v}$ in the denominator, we simply multiply by $\frac{u+\sqrt{v}}{u+\sqrt{v}}$. - You can multiply by the "conjugate": $$\frac{(5 +\sqrt{20})}{(6+\sqrt{20})}\frac{(6 - \sqrt{20})}{(6 - \sqrt{20})} = \dots$$ -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 12, "mathjax_display_tex": 3, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9485697746276855, "perplexity_flag": "middle"}
http://mathhelpforum.com/advanced-algebra/207579-graph-matrices.html	# Thread: 1. ## As graph matrices? I would like to please guide me on this question: Let $S_+$ denote the set of semi positive definite matrices in $\mathbb{R}^{2\times 2}$ is known that $S_+\subseteq Sym \simeq\mathbb{R}^{3}$,wherein $Sym$ are the matrices symmetric. But Is it possible to give a geometric interpretation of $S_+$ in $\mathbb{R}^{3}$? Can be graphed? Thank you very much for your attention, any suggestions are welcome.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 6, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.929343581199646, "perplexity_flag": "head"}
http://en.wikipedia.org/wiki/Universal_algebra	# Universal algebra Universal algebra (sometimes called general algebra) is the field of mathematics that studies algebraic structures themselves, not examples ("models") of algebraic structures. For instance, rather than take particular groups as the object of study, in universal algebra one takes "the theory of groups" as an object of study. ## Basic idea From the point of view of universal algebra, an algebra (or algebraic structure) is a set A together with a collection of operations on A. An n-ary operation on A is a function that takes n elements of A and returns a single element of A. Thus, a 0-ary operation (or nullary operation) can be represented simply as an element of A, or a constant, often denoted by a letter like a. A 1-ary operation (or unary operation) is simply a function from A to A, often denoted by a symbol placed in front of its argument, like ~x. A 2-ary operation (or binary operation) is often denoted by a symbol placed between its arguments, like x * y. Operations of higher or unspecified arity are usually denoted by function symbols, with the arguments placed in parentheses and separated by commas, like f(x,y,z) or f(x1,...,xn). Some researchers allow infinitary operations, such as $\textstyle\bigwedge_{\alpha\in J} x_\alpha$ where J is an infinite index set, thus leading into the algebraic theory of complete lattices. One way of talking about an algebra, then, is by referring to it as an algebra of a certain type $\Omega$, where $\Omega$ is an ordered sequence of natural numbers representing the arity of the operations of the algebra. ### Equations After the operations have been specified, the nature of the algebra can be further limited by axioms, which in universal algebra often take the form of identities, or equational laws. An example is the associative axiom for a binary operation, which is given by the equation x * (y * z) = (x * y) * z. The axiom is intended to hold for all elements x, y, and z of the set A. ## Varieties Main article: Variety (universal algebra) An algebraic structure that can be defined by identities is called a variety, and these are sufficiently important that some authors consider varieties the only object of study in universal algebra, while others consider them an object.[citation needed] Restricting one's study to varieties rules out: • Predicate logic, notably quantification, including existential quantification ( $\exists$ ) and universal quantification ($\forall$) • Relations, including inequalities, both $a \neq b$ and order relations In this narrower definition, universal algebra can be seen as a special branch of model theory, in which we are typically dealing with structures having operations only (i.e. the type can have symbols for functions but not for relations other than equality), and in which the language used to talk about these structures uses equations only. Not all algebraic structures in a wider sense fall into this scope. For example ordered groups are not studied in mainstream universal algebra because they involve an ordering relation. A more fundamental restriction is that universal algebra cannot study the class of fields, because there is no type in which all field laws can be written as equations (inverses of elements are defined for all non-zero elements in a field, so inversion cannot simply be added to the type). One advantage of this restriction is that the structures studied in universal algebra can be defined in any category which has finite products. For example, a topological group is just a group in the category of topological spaces. ### Examples Most of the usual algebraic systems of mathematics are examples of varieties, but not always in an obvious way – the usual definitions often involve quantification or inequalities. #### Groups To see how this works, let's consider the definition of a group. Normally a group is defined in terms of a single binary operation , subject to these axioms: • Associativity (as in the previous section): x (y * z) = (x * y) * z. • Identity element: There exists an element e such that for each element x, e * x = x = x * e. • Inverse element: It can easily be seen that the identity element is unique. If we denote this unique identity element by e then for each x, there exists an element i such that x * i = e = i * x. (Sometimes you will also see an axiom called "closure", stating that x * y belongs to the set A whenever x and y do. But from a universal algebraist's point of view, that is already implied when you call * a binary operation.) Now, this definition of a group is problematic from the point of view of universal algebra. The reason is that the axioms of the identity element and inversion are not stated purely in terms of equational laws but also have clauses involving the phrase "there exists ... such that ...". This is inconvenient; the list of group properties can be simplified to universally quantified equations if we add a nullary operation e and a unary operation ~ in addition to the binary operation , then list the axioms for these three operations as follows: • Associativity: x (y * z) = (x * y) * z. • Identity element: e * x = x = x * e. • Inverse element: x * (~x) = e = (~x) * x. (Of course, we usually write "x −1" instead of "~x", which shows that the notation for operations of low arity is not always as given in the second paragraph.) What has changed is that in the usual definition there are: • a single binary operation (signature (2)) • 1 equational law (associativity) • 2 quantified laws (identity and inverse) ...while in the universal algebra definition there are • 3 operations: one binary, one unary, and one nullary (signature (2,1,0)) • 3 equational laws (associativity, identity, and inverse) • no quantified laws It is important to check that this really does capture the definition of a group. The reason that it might not is that specifying one of these universal groups might give more information than specifying one of the usual kind of group. After all, nothing in the usual definition said that the identity element e was unique; if there is another identity element e', then it is ambiguous which one should be the value of the nullary operator e. Proving that it is unique is a common beginning exercise in classical group theory textbooks. The same thing is true of inverse elements. So, the universal algebraist's definition of a group is equivalent to the usual definition. At first glance this is simply a technical difference, replacing quantified laws with equational laws. However, it has immediate practical consequences – when defining a group object in category theory, where the object in question may not be a set, one must use equational laws (which make sense in general categories), and cannot use quantified laws (which do not, as objects in general categories do not have elements). Further, the perspective of universal algebra insists not only that the inverse and identity exist, but that they be maps in the category. The basic example is of a topological group – not only must the inverse exist element-wise, but the inverse map must be continuous (some authors also require the identity map to be a closed inclusion, hence cofibration, again referring to properties of the map). ## Basic constructions We assume that the type, $\Omega$, has been fixed. Then there are three basic constructions in universal algebra: homomorphic image, subalgebra, and product. A homomorphism between two algebras A and B is a function h: A → B from the set A to the set B such that, for every operation fA of A and corresponding fB of B (of arity, say, n), h(fA(x1,...,xn)) = fB(h(x1),...,h(xn)). (Sometimes the subscripts on f are taken off when it is clear from context which algebra your function is from) For example, if e is a constant (nullary operation), then h(eA) = eB. If ~ is a unary operation, then h(~x) = ~h(x). If * is a binary operation, then h(x * y) = h(x) * h(y). And so on. A few of the things that can be done with homomorphisms, as well as definitions of certain special kinds of homomorphisms, are listed under the entry Homomorphism. In particular, we can take the homomorphic image of an algebra, h(A). A subalgebra of A is a subset of A that is closed under all the operations of A. A product of some set of algebraic structures is the cartesian product of the sets with the operations defined coordinatewise. ## Some basic theorems • The Isomorphism theorems, which encompass the isomorphism theorems of groups, rings, modules, etc. • Birkhoff's HSP Theorem, which states that a class of algebras is a variety if and only if it is closed under homomorphic images, subalgebras, and arbitrary direct products. ## Motivations and applications This section does not cite any references or sources. Please help improve this section by adding citations to reliable sources. Unsourced material may be challenged and removed. (April 2010) In addition to its unifying approach, universal algebra also gives deep theorems and important examples and counterexamples. It provides a useful framework for those who intend to start the study of new classes of algebras. It can enable the use of methods invented for some particular classes of algebras to other classes of algebras, by recasting the methods in terms of universal algebra (if possible), and then interpreting these as applied to other classes. It has also provided conceptual clarification; as J.D.H. Smith puts it, "What looks messy and complicated in a particular framework may turn out to be simple and obvious in the proper general one." In particular, universal algebra can be applied to the study of monoids, rings, and lattices. Before universal algebra came along, many theorems (most notably the isomorphism theorems) were proved separately in all of these fields, but with universal algebra, they can be proven once and for all for every kind of algebraic system. The 1956 paper by Higgins referenced below has been well followed up for its framework for a range of particular algebraic systems, while his 1963 paper is notable for its discussion of algebras with operations which are only partially defined, typical examples for this being categories and groupoids. This leads on to the subject of higher dimensional algebra which can be defined as the study of algebraic theories with partial operations whose domains are defined under geometric conditions. Notable examples of these are various forms of higher dimensional categories and groupoids. ### Category theory and operads Further information: Category theory and Operad theory A more generalised programme along these lines is carried out by category theory. Given a list of operations and axioms in universal algebra, the corresponding algebras and homomorphisms are the objects and morphisms of a category. Category theory applies to many situations where universal algebra does not, extending the reach of the theorems. Conversely, many theorems that hold in universal algebra do not generalise all the way to category theory. Thus both fields of study are useful. A more recent development in category theory that generalizes operations is operad theory – an operad is a set of operations, similar to a universal algebra. ## History In Alfred North Whitehead's book A Treatise on Universal Algebra, published in 1898, the term universal algebra had essentially the same meaning that it has today. Whitehead credits William Rowan Hamilton and Augustus De Morgan as originators of the subject matter, and James Joseph Sylvester with coining the term itself.[1] At the time structures such as Lie algebras and hyperbolic quaternions drew attention to the need to expand algebraic structures beyond the associatively multiplicative class. In a review Alexander Macfarlane wrote: "The main idea of the work is not unification of the several methods, nor generalization of ordinary algebra so as to include them, but rather the comparative study of their several structures." At the time George Boole's algebra of logic made a strong counterpoint to ordinary number algebra, so the term "universal" served to calm strained sensibilities. Whitehead's early work sought to unify quaternions (due to Hamilton), Grassmann's Ausdehnungslehre, and Boole's algebra of logic. Whitehead wrote in his book: "Such algebras have an intrinsic value for separate detailed study; also they are worthy of comparative study, for the sake of the light thereby thrown on the general theory of symbolic reasoning, and on algebraic symbolism in particular. The comparative study necessarily presupposes some previous separate study, comparison being impossible without knowledge."[2] Whitehead, however, had no results of a general nature. Work on the subject was minimal until the early 1930s, when Garrett Birkhoff and Øystein Ore began publishing on universal algebras. Developments in metamathematics and category theory in the 1940s and 1950s furthered the field, particularly the work of Abraham Robinson, Alfred Tarski, Andrzej Mostowski, and their students (Brainerd 1967). In the period between 1935 and 1950, most papers were written along the lines suggested by Birkhoff's papers, dealing with free algebras, congruence and subalgebra lattices, and homomorphism theorems. Although the development of mathematical logic had made applications to algebra possible, they came about slowly; results published by Anatoly Maltsev in the 1940s went unnoticed because of the war. Tarski's lecture at the 1950 International Congress of Mathematicians in Cambridge ushered in a new period in which model-theoretic aspects were developed, mainly by Tarski himself, as well as C.C. Chang, Leon Henkin, Bjarni Jónsson, Roger Lyndon, and others. In the late 1950s, Edward Marczewski[3] emphasized the importance of free algebras, leading to the publication of more than 50 papers on the algebraic theory of free algebras by Marczewski himself, together with Jan Mycielski, Władysław Narkiewicz, Witold Nitka, J. Płonka, S. Świerczkowski, K. Urbanik, and others. ## Footnotes 1. Grätzer, George. Universal Algebra, Van Nostrand Co., Inc., 1968, p. v. 2. Quoted in Grätzer, George. Universal Algebra, Van Nostrand Co., Inc., 1968. 3. Marczewski, E. "A general scheme of the notions of independence in mathematics." Bull. Acad. Polon. Sci. Ser. Sci. Math. Astronom. Phys. 6 (1958), 731–736. ## References • Bergman, George M., 1998. An Invitation to General Algebra and Universal Constructions (pub. Henry Helson, 15 the Crescent, Berkeley CA, 94708) 398 pp. ISBN 0-9655211-4-1. • Birkhoff, Garrett, 1946. Universal algebra. Comptes Rendus du Premier Congrès Canadien de Mathématiques, University of Toronto Press, Toronto, pp. 310–326. • Brainerd, Barron, Aug–Sep 1967. Review of Universal Algebra by P. M. Cohn. American Mathematical Monthly, 74(7): 878–880. • Burris, Stanley N., and H.P. Sankappanavar, 1981. A Course in Universal Algebra Springer-Verlag. ISBN 3-540-90578-2 Free online edition. • Cohn, Paul Moritz, 1981. Universal Algebra. Dordrecht, Netherlands: D.Reidel Publishing. ISBN 90-277-1213-1 (First published in 1965 by Harper & Row) • Freese, Ralph, and Ralph McKenzie, 1987. Commutator Theory for Congruence Modular Varieties, 1st ed. London Mathematical Society Lecture Note Series, 125. Cambridge Univ. Press. ISBN 0-521-34832-3. Free online second edition. • Grätzer, George, 1968. Universal Algebra D. Van Nostrand Company, Inc. • Higgins, P. J. Groups with multiple operators. Proc. London Math. Soc. (3) 6 (1956), 366–416. • Higgins, P.J., Algebras with a scheme of operators. Mathematische Nachrichten (27) (1963) 115–132. • Hobby, David, and Ralph McKenzie, 1988. The Structure of Finite Algebras American Mathematical Society. ISBN 0-8218-3400-2. Free online edition. • Jipsen, Peter, and Henry Rose, 1992. Varieties of Lattices, Lecture Notes in Mathematics 1533. Springer Verlag. ISBN 0-387-56314-8. Free online edition. • Pigozzi, Don. General Theory of Algebras. • Smith, J.D.H., 1976. Mal'cev Varieties, Springer-Verlag. • Whitehead, Alfred North, 1898. A Treatise on Universal Algebra, Cambridge. (Mainly of historical interest.)	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 7, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.90040123462677, "perplexity_flag": "middle"}
http://cameroncounts.wordpress.com/2011/07/20/the-field-with-one-element/	always busy counting, doubting every figured guess . . . ## The field with one element Posted on 20/07/2011 by Two weeks ago I was at the British Combinatorial Conference in Exeter. It was smaller than many recent BCCs, but a really good conference: the invited talks were uniformly excellent. I want to focus here on one talk, that of Koen Thas. He covered a lot of ground, mentioning for example the conjecture that finite projective planes necessarily have prime power order, and the conjecture (due to Jacques Tits, on which I inadvertently made a contribution) that there is no generalised quadrangle with a finite number (greater than 2) of points on a line and an infinite number of lines through a point. (I settled the first case, where there are three points on a line; the case of four points on a line was done by Kantor and independently Brouwer, and five points on a line by Cherlin, and that is the limit of our knowledge.) I will focus even more narrowly, on the last part of his talk, and the last of seventeen sections in his paper, concerning “the field with one element”. Some explanation for non-mathematicians is in order. A field is an algebraic system in which addition and multiplication are defined, and “the usual rules apply”: addition and multiplication are commutative and associative, there are identity elements and inverses for both operations, and the distributive law (a.k.a. expanding brackets) holds. In a field, we can’t divide by zero. An absolutely standard part of the definition is that the additive identity element 0 and the multiplicative identity 1 are unequal: 0≠1. So, by definition, there is no “field with one element”: every field has at least two elements. Thus, the field with one element is essentially a poetic concept (I am not sure which figure of speech best describes it). But, undaunted, I will try to explain. ### A tale of three qs The Gaussian, or q-binomial, coefficient, which I will write here as Gauss(n,k)q, is given by the formula (qn−1)(qn-1−1)…(qn-k+1−1)/(qk−1)(qk-1−1)…(q−1). Despite appearances, it is a polynomial in q, of degree k(n−k). This can be seen by showing, from the definition, that the Gaussian coefficients can also be specified by Gauss(n,0)q = Gauss(n,n)q = 1 and the “Pascal-like” recurrence Gauss(n,k)q = Gauss(n−1,k−1)q+qkGauss(n−1,k)q. For example, Gauss(4,2)q = q4+q3+2q2+q+1. A simple property is that, if we substitute q=1 in the Gaussian coefficient Gauss(n,k)q, we obtain the corresponding binomial coefficient Bin(n,k). The easiest way to see this is from the recurrence; alternatively, use the fact that the limit of (qn−1)/(qk−1) as q→1 is n/k (either by l’Hôpital’s rule, or by simply dividing top and bottom by q−1). #### Fixed sets for powers of a cycle So are the Gaussian coefficients nothing more than polynomials which happen to take interesting values when q=1? To see that this is not so, turn to another of the invited speakers at the BCC, Bruce Sagan, who talked about a recent discovery which has produced a lot of activity, the cyclic sieving phenomenon. Incidentally, Bruce used remarkably well the opportunity given by the publication of the main speakers’ talks in advance: his talk was slow and beautifully clear, while his paper is densely packed with interesting mathematics. The binomial coefficient Bin(n,k) counts the k-element subsets of an n-element set. Now suppose we have a cyclic permutation σ of the set, and we wish to count its orbits on k-subsets. By the Orbit-counting Lemma, we have to count the subsets fixed by any power of σ. Now it turns out that the number of sets fixed by a power of σ of order d is the result of substituting a primitive dth root of unity for q in Gauss(n,k)q. This fact is only the tip of a very large iceberg, for which see Bruce’s paper. But I want to speak of two other, completely different, occurrences of the Gaussian coefficients, discovered much earlier. In both cases, the letter q is traditionally used: is this just coincidence? (A question for a historian, maybe.) #### Finite vector spaces A finite field necessarily has a prime power number of elements; Galois showed that for any prime power q there is a unique field with q elements, up to isomorphism. This is now called a Galois field, and denoted by GF(q), or Fq. Now let V be an n-dimensional vector space over GF(q). Then V can be identified with the set of all n-tuples of coordinates (relative to a basis), so \|V\|=qn. The number of k-dimensional subspaces of V is the Gaussian coefficient Gauss(n,k)q. The standard proof of this fact involves noticing that the number of linearly independent k-tuples of vectors in V is (qn−1)(qn−q)…(qn−qk-1), while each k-dimensional subspace contains (qk−1)(qk−q)…(qk−qk-1) such tuples; dividing these numbers gives the Gaussian coefficient. There is an alternative counting proof. Take V=Fn, where F=GF(q), the space of all n-tuples. Now every k-dimensional subspace has a unique basis consisting of k vectors in reduced echelon form with no zero vectors: this means that, if the vectors are the rows of a k×n matrix, then • the first non-zero entry in each row is a 1; • these “leading 1s” occur further to the right as we go down the matrix; • all other entries in the column of a “leading 1″ are zero. Now it is easy to verify the Pascal-like recurrence for the number of matrices in reduced echelon form: the two terms on the right correspond to matrices where the leading 1 in the last row is in the last position (so deleting the last row and column gives a reduced echelon matrix) or in an earlier position (so the last column is arbitrary, and deleting it gives a reduced echelon matrix). Indeed the definition of “reduced echelon form” makes sense over any alphabet containing elements called 0 and 1; and the argument shows that the number of k×n matrices in reduced echelon form is the Gaussian coefficient, where q is the alphabet size (not necessarily a prime power). #### Area under lattice paths Consider lattice points (with integer coordinates) in the plane. You are standing at the origin, and have to move to the point (n−k,k); you are allowed to step from any point to its neighbour to the right or above. Clearly you must take n steps, of which k should be vertical. The vertical steps can be chosen anywhere in the sequence of moves. So the number of possible walks is Bin(n,k). Now suppose that we are interested in the area under such a path. This can take any value between 0 (for the path that takes all the horizontal steps first) to k(n−k) (for the path that takes the vertical steps first). A solution to this problem could be a generating function, a polynomial in an indeterminate q where the coefficient of qm is the number of paths with area m. It turns out that this generating function is Gauss(n,k)q. This can be seen most easily from the recurrence relation. The first term accounts for paths with the last step vertical (this step adds nothing to the area); the second counts paths with the last step horizontal (the last step adds k to the area, so multiplies the generating function by qk. Remarkably, we have seen three occurrences of the Gaussian coefficient: in the first, q is typically a root of unity; in the second, a prime power; and in the third, an indeterminate. ### Geometry over the one-element field, 1 In 1957, Jacques Tits suggested an analogy between the combinatorics of sets and subsets and the geometry of vector spaces and subspaces: roughly speaking, the combinatorics was the case q=1 of the geometry, where q is the order of the field. The (n−1)-dimensional projective space over a field F is the geometry whose points, lines, planes, … are the subspaces of dimension 1, 2, 3, … of an n-dimensional vector space over F. (Caution: dimension shift!) Axiomatically, the geometries are very similar. According to the Veblen–Young axioms, the projective space is characterised by the properties • Two points lie on a unique line. • If a line meets two sides of a triangle, not at their intersection, then it also meets the third. • A subspace is a set of points containing the line through any two of its points. We have to make some low-dimensional exceptions: a 1-dimensional projective space (according to these axioms) is just an arbitrary set of points forming a line, with no algebraic structure; and there are non-Desarguesian projective planes which are not obtained from vector spaces. But the correspondence is exact for higher dimensions, assuming that any line has at least three points. (If the lines are finite, then the number of points on a line is q+1, where q is the size of the coordinatising field.) By contrast, if every line has just two points, then an arbitrary set trivially satisfies the axioms, with the lines being all the 2-element subsets and the subspaces arbitrary subsets. Thus, the subsets of an arbitrary set “are” the projective spaces over the “field with one element”. What is an affine space? The number qn of points in an n-dimensional affine space reduces to 1 when q=1; so the affine space is a single point; but it has n ghostly directions through the point (lines with only one point). Recall that we obtain a projective space from an affine space by adding a point at infinity on each parallel class of lines; thus to get from the n-dimensional affine space to the projective space, we add n points, obtaining n+1 altogether, in agreement with our previous description. If this analogy is to be fruitful, it should suggest new insights; and indeed this one does. The “general linear group” in n dimensions over the field of one element is the symmetric group Sn; this reflects the BN-pair structure of the general linear group. Indeed, this extends to the other groups of Lie type: the group over the field of one element is the Weyl group of the BN-pair. There is interesting combinatorics too. In fact, one of my earliest papers exploited this analogy. A theorem of Fisher (for t=1), Petrenjuk (for t=2), and Ray-Chaudhuri and Wilson (in general) says that a 2t-design on v points with block size at most v−t has at least Bin(v,t) blocks. (Such a design is a collection of k-subsets, with 2t≤k≤v−t, such that any 2t-set is contained in a constant number of these sets.) I extended the result by replacing “set” and “subset” by “vector space over finite field” and “subspace”; the binomial lower bound is replaced by Gauss(v,t)q, where q is the field order. This result was almost immediately superseded by the much more general results of Philippe Delsarte. Several similar results exist. Probably the most famous is the vector space Ramsey theorem of Graham, Leeb and Rothschild, an exact translation of the classical Ramsey theorem. ### Geometry over the field of one element, 2 In recent years, this enterprise has become much more serious. In 2009, Javier Lopez talked about this in our Pure Mathematics seminar: I wish I had got more out of it than I did. (The abstract is here.) Enough to say that, if a geometer wants to make sense of the field with one element, it is necessary to be able to define varieties, schemes, etc. over it. For some particular varieties, we know how to proceed: we have already dealt with projective and affine spaces over the 1-element field, and for algebraic groups we just get the Weyl group. The situation for other varieties is more complicated. Soulé proposed a definition, which led to some results disagreeing with the naive interpretation. A revised version due to Connes and Consani seems to have fared better. Manin took things even further, proposing a particularly simple form for the zeta function of a variety over the 1-element field F1. (This presupposes that we have a notion of an extension of degree n of F1; even though this is another 1-element field, it is not the same as F1 itself!) Soulé proposed a zeta-function associated with any suitable counting function: for the Gaussian coefficient, the result is in agreement with Manin’s suggestion. ### So what is the one-element field? It seems that the 1-element field should have multiplication but not addition. But addition has been restored with the concept of a hyperfield. An abelian hypergroup is a set H with a “hyperoperation” +: this means that u+v is a non-empty subset, rather than an element, of H, and an appropriate collection of axioms is satisfied. A typical hypergroup is the set of orbits of an automorphism group of an abelian group, the hyperoperation giving the set of all orbits hit by sums of elements in the chosen orbits. Now a hyperfield is a set with operations of addition and multiplication, so that addition is a hypergroup, multiplication is a group on the non-zero elements, and the distributive laws hold. The Krasner hyperfield has elements 0 and 1: hyperaddition is given by 1+1 = {0,1} (all other instances follow from the axioms), and multiplication is obvious. This is the algebraic object which might play the role of the 1-element field. There is more to the story: the adèle ring of a global field modulo multiplication by non-zero field elements forms a hyperfield which is an extension of the Krasner hyperfield. Moreover, one can do geometry over the Krasner hyperfield, in a way which seems more satisfying than the hints and guesses we have had hitherto. Connes and Consani show that finite commutative hyperfield extensions of the Krasner hyperfield corresond to “incidence groups” of finite projective spaces, together with some low-dimensional extensions (including projective planes with sharply point-transitive groups, a notorious problem). Clearly a very productive analogy, when we have made sense of it! See Thas’ paper for more insights. ## About Peter Cameron I count all the things that need to be counted. This entry was posted in events, exposition and tagged affine geometry, algebraic geometry, binomial coefficient, Bruce Sagan, cyclic sieving, Gaussian coefficient, hyperfield, hypergroup, Koen Thas, projective geometry, q-analogue, scheme. Bookmark the permalink. ### 8 Responses to The field with one element 1. Peter Cameron says: Read more about the field with one element here: http://matrix.cmi.ua.ac.be/XTRA/ncg.pdf 2. Pingback: Do We Need Mysticism In Theory? « Gödel’s Lost Letter and P=NP 3. Jon Awbrey says: Hendiadys? — or maybe the converse of that? 4. Jon Awbrey says: This reminds me of some things I used to think about — I always loved working playing with generating functions and I can remember a time in the 80s when I was very pleased with myself for working out the q-analogue of integration by parts — but it will probably take me a while to warm up those old gray cells today. Let me first see if I can get LaTex to work in these comment boxes … The Gaussian coefficient, also know as the q-binomial coefficient, is notated as Gauss(n, k)q and given by the following formula: (qn−1)(qn−1−1) … (qn−k+1−1) / (qk−1)(qk−1−1) … (q−1). $\text{Gauss}(n, k)_q = \frac{(q^n - 1)(q^{n-1} - 1) \ldots (q^{n-k+1} - 1)}{(q^k - 1)(q^{k-1} - 1) \ldots (q - 1)}$ • Jon Awbrey says: Not sure what happened with the HTML version … • Jon Awbrey says: I tried a test copy on my blog and it came out like this. • Peter Cameron says: I haven’t figured out what the WordPress rules actually are. Even in posts, it is not proper HTML: for example, it obeys line ends in the input file. But rules for titles are different, and comments are different again. • Jon Awbrey says: Yes, it even seems to vary with the same theme used on different sides of the Atlantic. Maybe my use of strikeouts on “working” interfered with the later sub-&-superscript formatting? Feel free to delete or edit any of this discussion that is too junky or might mislead readers.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 1, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9300742149353027, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/29209/can-every-particle-be-regarded-as-being-a-combination-of-black-holes-and-white-h	# Can every particle be regarded as being a combination of Black holes and White holes? Can the statement be regarded as true? That every particle, or element in the universe can be regarded as a combination of black hole and white hole in variable proportion. - 4 Except for colorful objects that also include red holes, green holes, blue holes. Huh!? There aren't any white holes in the Universe because they're the time-reversed histories to black holes that would violate the second law of thermodynamics. Each microstate one could call a "white hole microstate" is actually a black hole microstate. And black holes only deserve to be called in this way if they're large enough - heavier than the Planck mass and compact objects. Normal objects in the Universe are composed of elementary particles which are much lighter and they're not (usefully) black holes. – Luboš Motl May 30 '12 at 7:06 The Compton wavelength of elementary particles is too large to allow them to be black holes (or white holes for that matter). Also, white holes don't exist as Lubos says correctly. Which you could have found out easily by Googling. – WIMP May 30 '12 at 7:32 ## 3 Answers You first have to understand what a "white hole" is. It's the time reverse of a black hole. It was rightly pointed out in previous answers that white holes violate the second law of thermodynamics. Now, like anything in thermodynamics, this makes them unlikely but not impossible (unlikely here usually means unlikely even in an astronomical number of universes ...). But if you keep that aside, there is a more beautiful picture of what a white hole is: it is a quantum superposition of all possible black holes of about the same size. There are so many ways to make a black hole, and they come in so many microstates, that the quantum superposition corresponding to a white hole is astronomically unlikely, ... but not impossible. Such considerations could be important if you ask what the relation is between elementary particles and black holes. This has been answered already: if you try to make such a comparison you would have to realise first that all known particles are so light that those black holes would be tiny, in fact billions of times smaller than the smallest distance conceivable in physics: the Planck length. Because of this, comparing known particles with black holes, or imagining them as being built from black holes, black, white or otherwise, is not considered to be a fruitful exercise in theoretical physics. - so, from the perspective of observers that cross a Kerr black hole and exit after a while through the corresponding white hole; how do they explain away to the observers they might find outside, the astronomical improbability of having found a white hole? will the outside observers agree that this is just a fortituous fluctuation of an otherwise thermal membrane? – user56771 Aug 21 '12 at 2:11 @user56771: Your comment is why the extended Kerr is not taken seriously--- it can't possibly describe travel to another universe. Penrose said that you just get cooked at the Cauchy horizon, but I think you come out in this universe. – Ron Maimon Aug 21 '12 at 6:12 1 @RonMaimon, good, i want to believe that too, but suppose we say that in this universe we have an excess of kerr black holes relative to kerr white holes given by some astronomical statistical number of $10^{10}$ against 1. Where all those extra kerr black holes are connecting to? if the white hole is an analytic continuation of the black hole then there can be no excess: the mapping needs to be one-to-one – user56771 Aug 21 '12 at 12:53 @user56771: It is impossible to have an "excess" of Kerr black holes--- the two concepts of Kerr black hole and Kerr white hole are the same. You can't come out of another black hole, you just come out of the same one you fell into. This is strongly suggested by AdS/CFT, but the gluing is tricky, as it is easy to get traversable CTC's in a wrong gluing, and you get no hints from the classical solution as to how it's supposed to be done. – Ron Maimon Aug 21 '12 at 15:48 ok @RonMaimon, but if all these observers/travelers can really exit in the same universe, they might end up afterwards and compare notes, and they'll have to agree that white holes are more abundant than what quantum gravity theory predicts. How do they reconcile this? – user56771 Aug 22 '12 at 0:29 show 3 more comments Over the years there have been suggestions that elementary particles may be black holes. However no-one has ever been able to make this quantitative and I doubt anyone believes it these days. There was some discussion of this in what is the difference between a blackhole and a point particle, and Googling will find you lots of hits on this subject. I've never heard of the idea that particles may be a combination of a black hole and a white hole. It's hard to see how such an arrangement could be possible. The term "white hole" is widely used in science fiction, but SF has the wrong idea about them. The equations of general relativity are time symmetric, and if you reverse the direction of time the black hole solution turns into a white hole. However there is no evidence that the time reversed solution has any physical significance, and as the comments above suggest, no-one (in the mainstream physics community) believes that white holes exist. The nearest we come go to a white hole existing are suggestions that the Big Bang was a white hole. Have a look at John Baez's article on this for more info. - A "White hole" is the time reverse of a black hole, as discovered by Hawking and explained by 'tHooft. The reason is that black holes can be in thermal equilibrium with radiation, and the time reverse of a thermal equilibrium state is still thermal equilibrium. This intuition is confirmed in AdS/CFT, where a thermal black hole in AdS can be described by a time-reversal invariant thermal boundary state, and so one is safe to conclude Hawking's argument is valid in quantum generality. The question of whether particles are black holes is to my mind entirely answered by string theory. The first point is that black holes are classical, and you are talking about quantum particles, so the question is only whether the particles can be interpreted as black holes in an appropriate limit. The F-strings from which the particles in the standard model are made are dual to branes, either by going to strong coupling if you are in a heterotic string theory, where the strings are revealed to be 2-branes wrapping a dimension, or else by going to strong coupling in IIB string theory, where you find that the F-string swaps with a D-brane. In either case, M2 branes or D1 branes are both classically intepretable as black holes, because when you stack them up, their supergravity solution is exactly a black hole. So within string theory, it is not wrong to say that every particle is a particularly highly quantum black hole, in the smallest quantum in which it comes. This doesn't mean the classical GR picture works for these objects, it doesn't. The other answers explain that it fails completely. But the quantum picture that all matter is extremal black holes (excluding orbifolds and other exotica) is the basic picture of modern string theory. It unifies particle physics and black hole physics in a fundamental way. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9466248750686646, "perplexity_flag": "head"}

http://physics.stackexchange.com/questions/17134/inertial-mass-of-a-scalar-field?answertab=active

# Inertial Mass of a scalar field Does it make sense to talk of the inertial mass of a scalar field? By the equivalence principle, it must be equal to its gravitational mass. We know that the scalar field contributes towards the stress-energy tensor, so, shouldn't it have an inertial mass too? - ## 2 Answers Yes, of course, if you produce a localized concentration of energy carried by a scalar field, it exhibits all the properties that this total energy $E=mc^2$ should exhibit. It will enter the right hand side of Einstein's equations so it will curve the surrounding spacetime and create a gravitational field. It will be able to convert to other forms of energy so that the total energy conservation law, including the energy of the scalar field, holds. And finally, it will also act as inertia. For example, a packet of energy carried by a field behaves as a particle. For example, a magnetic monopole may be imagined as a nontrivial localized solution to some field equations including scalar and gauge fields. A magnetic force $F=q_m B$ may act upon the magnetic monopole and the acceleration will indeed be given by $F=ma$ where $m=E/c^2$ is the inertial mass of the magnetic monopole, stored in the energy density of the fields. - Yes, indeed This is exactly why it is assumed in the Standard Model that mass is the result of an interaction with a scalar "world potential" (The Higgs field) while for instance the changes in energy/momentum of a charge are due to the electromagnetic vector potential $A^\mu$ Regards, Hans -

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http://www.reference.com/browse/plus

Definitions Nearby Words Related Questions # Plus-minus sign ± The plus-minus sign (±) is a mathematical symbol commonly used to indicate the precision of an approximation, or as a convenient shorthand for a quantity which has two possible values opposite in sign. In mathematics, the sign is pronounced "plus or minus" and indicates that there are exactly two possible answers, one of which is positive and one of which is negative. In most experimental sciences however, the sign is pronounced "give or take" and it indicates an inclusive range of values that a reading might have. The plus-minus sign does not mean "approximately". ## Usage for indicating precision The use of ± for an approximation is most commonly encountered for presenting the numerical value of a quantity together with its tolerance or its statistical margin of error. For example, "5.7 ± 0.2" denotes a quantity that is specified or estimated to be within 0.2 units of 5.7; it may be anywhere in the range from 5.7 − 0.2 to 5.7 + 0.2. More precisely, in scientific usage it usually comes with a probability of being within the interval, usually that of 2 standard deviations, or 95.4%. A percentage may also be used to indicate the error margin. For example, 230 V ± 10% refers to a voltage within 10% of either side of 230 V (207 V to 253 V). Separate values for the upper and lower bounds may also be used. For example, to indicate that a value is most likely 5.7 but may be as high as 5.9 or as low as 5.6, one could write $5.7^\left\{+0.2\right\}_\left\{-0.1\right\}$. ## Usage as shorthand for two values of opposite signs In mathematical equations, the use of ± may be found as shorthand, to present two equations in one formula: equation+ OR equation- represented with equation± The best-known example is offered by the formula for the solutions of quadratic equations: If $displaystyle ax^2 + bx + c = 0,$ then $displaystyle x = frac\left\{-b pm sqrt\left\{b^2-4ac\right\}\right\}\left\{2a\right\}.$ Written out in full, this states that there are two solutions to the equation, namely $displaystyle x = frac\left\{-b + sqrt \left\{b^2-4ac\right\}\right\}\left\{2a\right\}$ and $displaystyle x = frac\left\{-b - sqrt \left\{b^2-4ac\right\}\right\}\left\{2a\right\}.$ Another example is found in the trigonometric identity $sin\left(x pm y\right) = sin\left(x\right) cos\left(y\right) pm cos\left(x\right) sin\left(y\right).,$ This stands for two identities: one with + on both sides of the equation, and one with − on both sides. A somewhat different use is found in this presentation of the formula for the Taylor series of the sine function: $sinleft\left(x right\right) = x - frac\left\{x^3\right\}\left\{3!\right\} + frac\left\{x^5\right\}\left\{5!\right\} - frac\left\{x^7\right\}\left\{7!\right\} + cdots pm frac\left\{1\right\}\left\{\left(2n+1\right)!\right\} x^\left\{2n+1\right\} + cdots.$ This mild abuse of notation is meant to indicate that the sign of the terms alternate, where (starting the count at 0) the terms with an even index $displaystyle n$ are added while those with an odd index are subtracted. A less ambiguous presentation in this case would use the quantity $\left(-1\right)^n$, which gives $+1$ when $displaystyle n$ is even and −1 when $displaystyle n$ is odd. ## Minus-plus sign There is another character, the minus-or-plus sign (∓), which is seen less often. It only takes on significant meaning when used in conjunction with the "±" sign. It can be used alongside "±" in such expressions as "x ± y ∓ z", which can be interpreted as "x + y − z" or/and "x − y + z", but neither "x + y + z" nor "x − y − z". The upper "−" in "∓" is considered attached to the "+" of "±" (and the lower symbols work in the same way) even though there is no visual indication of the dependency. The original expression can be rewritten as "x ± (y − z)" to avoid confusion, but cases such as the trigonometric identity $cos\left(x pm y\right) = cos\left(x\right) cos\left(y\right) mp sin\left(x\right) sin\left(y\right)$ are most neatly written using the "∓" sign. ## Encodings • In ISO 8859-1, -7, -8, -9, -13, -15, and -16, the plus-minus symbol is given by the code 0xB1hex Since the first 256 code points of Unicode are identical to the contents of ISO-8859-1 this symbol is also at Unicode code point 00B1. • The symbol also has a HTML entity representation of `±`. The rarer minus-plus sign (∓) is not generally found in legacy encodings and does not have a named HTML entity but is available in Unicode with codepoint U+2213 and so can be used in HTML using `∓` • In TeX 'plus-or-minus' and 'minus-or-plus' symbols are encoded as `pm` and `mp` entities, respectively. • These characters are also seen written as the (highly un-semantic) underlined or overlined + symbol. ( + or ).

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http://mathhelpforum.com/differential-geometry/120453-normed-vector-space.html

Thread: 1. normed vector space show that in the vector space $ \ell _1 = \left\{ {r_k \,\,:\,\sum {\left| {r_k } \right|} \,converges} \right\} $ (i) $ \left\| {r_k } \right\| = \sum {r_k } $ is a norm (ii) $ P = \left\{ {\left\{ {x_k } \right\}\,:\,x_k \geqslant 0\,\forall k \in \mathbb{N}} \right\} $ has en empty interior (iii) $ B\left( {\vec 0,1} \right) $ is not compact please help im trying to learn this by myself and i would like help on how to do this problems thanks! 2. Originally Posted by mms show that in the vector space $ \ell _1 = \left\{ {r_k \,\,:\,\sum {\left| {r_k } \right|} \,converges} \right\} $ (i) $ \left\| {r_k } \right\| = \sum {r_k } $ is a norm (ii) $ P = \left\{ {\left\{ {x_k } \right\}\,:\,x_k \geqslant 0\,\forall k \in \mathbb{N}} \right\} $ has en empty interior (iii) $ B\left( {\vec 0,1} \right) $ is not compact please help im trying to learn this by myself and i would like help on how to do this problems thanks! For (i) just remmeber that for all $n\in \mathbb{N}$ we have $\vert x_n + y_n \vert \leq \vert x_n\vert + \vert y_n \vert$ so adding we get $\sum_{i=1}^{n} \vert x_i+y_i \vert \leq \sum_{i=1}^{n} \vert x_i \vert + \vert y_i \vert \leq \sum_{i=1}^{\infty } \vert x_i \vert + \vert y_i \vert < \infty$ For (ii) take a sequence $y^k=(y_{n}^k) \in \ell _1$ such that $y_{n}^k = x_n$ if $n\neq k$ and $y_{n}^k= -\frac{1}{k}$ if $n=k$ then $\Vert x-y^k \Vert = \vert x_k + \frac{1}{k} \vert \rightarrow 0$ as $k\rightarrow \infty$ (since $x_n \rightarrow 0$ ) For (iii) what can you say about the sequence $x^k=(x_{n}^k)$ where $x_{n}^k=0$ if $n\neq k$ and $x_{n}^k=\frac{1}{2}$ if $n=k$ (Notice that $\Vert x^k - x^l \Vert = 1$ for all $k\neq l$)

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http://mathhelpforum.com/advanced-algebra/112760-eigenvalues.html

# Thread: 1. ## Eigenvalues I need to find the characteristic polynomial, eigenvalues, and eigenvectors or the following matrix. [1 1] [1 1] So far, this is what I have done: det(Lamda-A) And got [lamda - 1 -1] [-1 lamda - 1] The det of that gives (lamda - 1)(lamda-1) + 1 which is lamda squared - 2 lamda +2 I used the quad formula on that and got 1 plus/minus i I dont know where to go from here. I think Now i do this: A-(1+i)I = [i 1] [1 i] i have no clue where to go from there 2. Hi! There really is no need to calculate the characteristic polynomial that way. First, you can easily observe that 0 is an eigenvalue of geometric multiplicty 1 (since $rank(A) = 1$. Now, note that $trace(A) = 1+1 = 2$ and A only has two eigenvalues, so the second one must be 2. Another way to know that 2 is an eigenvalue is to note that the sum of all rows is 2, therefore, according to a theorem which you should have learned, 2 is an eigenvalue with corresponding eigenvector $(1 \ 1)^T$. So we know that 2 is an eigenvalue of geometric multiplicity $\geq 1$ and 0 is an eigenvector with geometric multiplicity 1. Since algebraic multiplicity $\geq$ geometric multiplicity for each eigenvalue, we get that the algebraic and geometric multiplicities for each eigenvector are 1 and 1. Therefore: $\Delta_A(x) = x(x-2)$ With 0,2 as eigenvalues and $(1\ -1)^T, (1 \ 1)^T$ the corresponding eigenvectors. However, if you still want to know what you did wrong -- $det(\lambda I-A) = (\lambda-1)(\lambda-1) - (-1)(-1) = (\lambda-1)^2 -1 \neq (\lambda-1)^2+1$ $=\Rightarrow det(\lambda I-A) = \lambda^2 -2\lambda +1 -1 = \lambda^2 -2\lambda = \lambda(\lambda-2)$ 3. ## Thanks Thanks a bunch. It would be nice if i could add :P 4. So I was trying to do it the way i was. I got the (1,1), but instead of (1,-1) I got (-1,1). 0lamda - A = -1 -1 -1 -1 then rref that and got 1 1 0 0 x1= -r x2 = r -1, 1 5. That is fine. The eigenvector can be any non-zero scalar multiple of $(1 \ -1)^T$. In this case, $(-1)(1 \ -1)^T = (-1 \ 1)^T$. From this, $(2\ -2)^T, \ (-8\ 8)^T,...$ could all be eigenvectors as well.

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http://mathhelpforum.com/differential-geometry/92431-examples.html

# Thread: 1. ## Examples Give an example or say why it is impossible: 1. A function which is discontinuous at every point of the set $\{\frac{1}{n}:n\in\mathbb{N}\}$ but continuous elsewhere on $\mathbb{R}$ I think it's possible since that set could be written as a countable union of closed sets, but I can't think of a particular example. 2. An infinite subset of [0,1] with no limit points. I think this is impossible but I'm not sure about my reasoning - is this correct?: Any subset of [0,1] is bounded so by the Bolzano Weierstrass theorem any sequence in this subset must have a convergent subsequence, so it has a limit point. 3. A closed set whose supremum is not a limit point of this set. not sure... 4.A power series that is absolutely convergent at only one point. Probably something with the point being 0... not sure though Thanks for any help 2. 1. Hint: Spoiler: $f(x)=\begin{cases}x&x\ne\frac1n\\[1mm]0&\mbox{otherwise}\end{cases}$ 3. Hint: Spoiler: Any finite set that is bounded above. 4. Hint: Spoiler: $\sum_{n\,=\,1}^\infty\frac{(-1)^n}nx^n$ 3. Thanks Abstractionist. Regarding your answer to question 4, is it always true that if $\sum a_n$ diverges, then $\sum a_nx^n$ also diverges, for any x besides 0? And if not, what makes your example special? Thanks again 4. Originally Posted by Aileys. is it always true that if $\sum a_n$ diverges, then $\sum a_nx^n$ also diverges, for any x besides 0? No. For example, $\sum_{n\,=\,1}^\infty\frac1n$ diverges but $\sum_{n\,=\,1}^\infty\frac{x^n}n$ converges for $-1\le x\le0.$ However, the convergence is only conditional, not absolute. 5. Ahhh OK so then is it always true that if $\sum a_n$ diverges then the only x for which $\sum a_nx^n$ converges absolutely is zero? Also while you're here ( ) was my answer to #2 correct?

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http://physics.stackexchange.com/questions/tagged/hilbert-space?sort=votes&pagesize=50

# Tagged Questions The hilbert-space tag has no wiki summary. learn more… | top users | synonyms 2answers 210 views ### Physical interpretation of different selfadjoint extensions Given a symmetric (densely defined) operator in a Hilbert space, there might be quite a lot of selfadjoint extensions to it. This might be the case for a Schrödinger operator with a "bad" potential. ... 2answers 86 views ### What Shannon channel capacity bound is associated to two coupled spins? The question asked is: What is the Shannon channel capacity $C$ that is naturally associated to the two-spin quantum Hamiltonian $H = \boldsymbol{L\cdot S}$? This question arises with a view ... 1answer 370 views ### Intuitive meaning of Hilbert Space formalism I am totally confused about the Hilbert Space formalism of Quantum Mechanics. 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Dual spaces are home to bras in quantum mechanics; cotangent spaces are home to linear maps in the tensor formalism of general relativity. After taking courses in these two subjects, I've still never ... 2answers 80 views ### Quantum Mechanical Operators in the argument of an exponential In Quantum Optics and Quantum Mechanics, the time evolution operator $$U(t,t_i) = \exp\left[\frac{-i}{\hbar}H(t-t_i)\right]$$ is used quite a lot. Suppose $t_i =0$ for simplicity, and say the ... 3answers 250 views ### Takhatajan's mathematical formulation of quantum mechanics So I began skimming L. Takhatajan's Quantum Mechanics For Mathematicians, and saw the mathematical formulation of QM that he uses (page 51). (The PDF file is available here.) I've only taken a basic ... 1answer 251 views ### Rigged Hilbert space and QM Are there any comprehensive texts that discuss QM using the notion of rigged Hilbert spaces? It would be nice if there were a text that went through the standard QM examples using this structure. 3answers 483 views ### What is a basis for the Hilbert space of a 1-D scattering state? Suppose I have a massive particle in non-relativistic quantum mechanics. Its wavefunction can be written in the position basis as $$\vert \Psi \rangle = \Psi_x(x,t)$$ or in the momentum basis as ... 1answer 272 views ### Why is the Haar measure times the volume of the eigenvalue simplex considered a good measure of Hilbert space volume? In particular, why do we need both of these to find the volume? And should I be thinking of it as an actual volume or not? This Hilbert space volume is talked about in this paper. It says There ... 3answers 686 views ### Don't understand the integral over the square of the Dirac delta function In Griffiths' Introduction to Quantum Mechanics he gives the eigenfunctions of the Hermitian operator $\hat{x}=x$ as being $$g_{\lambda}\left(x\right)~=~B_{\lambda}\delta\left(x-\lambda\right).$$ ... 2answers 112 views ### Space of states in quantum mechanics A state in quantum mechanics I think is just a vector in a complex Hilbert space. As the physical properties are defined up to a phase $e^{i\theta}$ then this Hilbert space is invariant under the ... 2answers 129 views ### Uniqueness of eigenvector representation in a complete set of compatible observables [duplicate] Possible Duplicate: Uniqueness of eigenvector representation in a complete set of compatible observables Sakurai states that if we have a complete, maximal set of compatible observables, ... 1answer 136 views ### Scattering states of Hydrogen atom in non-relativistic perturbation theory In doing second order time-independent perturbation theory in non-relativistic quantum mechanics one has to calculate the overlap between states E^{(2)}_n ~=~ \sum_{m \neq n}\frac{|\langle m | H' ... 1answer 94 views ### Spontaneous symmetry breaking: How can the vacuum be infinitly degenerate? In classical field theories, it is with no difficulty to imagine a system to have a continuum of ground states, but how can this be in the quantum case? Suppose a continuous symmetry with charge $Q$ ... 1answer 265 views ### Born-Oppenheimer Approximation equivalent to Tensor-product ? If you have a wave function $\Psi$ of a system consisting of an electron and the vibrational modes of the crystal, THEN we represent the wavefunction $\Psi%$ to be in the Hilbert Space formed by the ... 1answer 190 views ### Existence of adjoint of an antilinear operator, time reversal The time reversal operator $T$ is an antiunitary operator, and I saw $T^\dagger$ in many places (for example when some guy is doing a "time reversal" $THT^\dagger$), but I wonder if there is a ... 2answers 145 views ### Kugo and Ojima's Canonical Formulation of Yang-Mills using BRST I am trying to study the canonical formulation of Yang-Mills theories so that I have direct access to the $n$-particle of the theory (i.e. the Hilbert Space). To that end, I am following Kugo and ... 0answers 77 views +50 ### Coherent U(N) intertwiners in LQG and a measure on the Grassmanian This is a detailed question about $U(N)$ intertwiners in LQG, and it comes from the the paper by Freidel and Livine (2011 - archive). It is very specific but related to finding a measure on a quotient ... 0answers 67 views ### Shape of the state space under different tensor products I am currently studying generalized probabilistic theories. Let me roughly recall how such a theory looks like (you can skip this and go to "My question" if you are familiar with this). Recall: In a ... 1answer 104 views ### The issue on existence of inverse operations of $a$ and $a^{\dagger}$ I have asked a question at math.stackexchange that have a physical meaning. My assumption: Suppose $a$ and $a^\dagger$ is Hermitian adjoint operators and $[a,a^\dagger]=1$. 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In the fourth edition, 33rd page, starting from this:$$\xi|\xi'\rangle=\xi'|\xi'\rangle$$ (Where ... 2answers 300 views ### Uniqueness of eigenvector representation in a complete set of compatible observables Sakurai states that if we have a complete, maximal set of compatible observables, say $A,B,C...$ Then, an eigenvector represented by $|a,b,c....>$, where $a,b,c...$ are respective eigenvalues, is ... 2answers 191 views ### QM formalism is one big confusion - lack of geometrical explaination with images I have been trying to learn QM and it went well (all untill harmonic oscilator) until i had to face the formalism: Hilbert space- As a novice to QM i am very sad that in none of the books i have ... 2answers 399 views ### Observing the exponential growth of Hilbert space? One of the weirdest things about quantum mechanics (QM) is the exponential growth of the dimensions of Hilbert space with increasing number of particles. This was already discussed by Born and ... 1answer 128 views ### How does a state in quantum mechanics evolve? I have a question about the time evolution of a state in quantum mechanics. The time-dependent Schrodinger equation is given as $$i\hbar\frac{d}{dt}|\psi(t)\rangle = H|\psi(t)\rangle$$ I am ... 1answer 83 views ### State space of QFT, CCR and quantization, and the spectrum of a field operator? In the canonical quantization of fields, CCR is postulated as (for scalar boson field ): $$[\phi(x),\pi(y)]=i\delta(x-y)\qquad\qquad(1)$$ in analogy with the ordinary QM commutation relation: ... 3answers 160 views ### Banach Space representations of physical systems I think most physicists mostly model physical systems as some kind of Hilbert space. Hilbert spaces are a strict subset of Banach spaces. Questions: Can physical systems really have non-compact ... 3answers 352 views ### Can we have discontinuous wavefunctions in the Infinite Square well? The energy eigenstates of the infinite square well problem look like the Fourier basis of L2 on the interval of the well. So then we should be able to for example make square waves that are an ... 1answer 311 views ### Quantum mechanic newbie: why complex amplitudes, why Hilbert space? I'm just starting learning quantum mechanics by myself (2 "lectures" so far) and I was wondering why we need to define quantum states in a complex vector space rater than a real one? Also I was ... 1answer 343 views ### Where does the wave function of the universe live? Please describe its home Where does the wave function of the universe live? Please describe its home. I think this is the Hilbert space of the universe. (Greater or lesser, depending on which church you belong to.) Or maybe ... 3answers 248 views ### If I go to the church of the greater Hilbert space, can I have Unitary Collapse? Actually, unitary pseudo-collapse? 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http://math.stackexchange.com/questions/143928/proof-that-ab-ac-is-bounded-for-bounded-b-and-c?answertab=active

# Proof that $ab/(a+c)$ is bounded for bounded $b$ and $c$ Is there a concise way to prove that $\frac{ab}{a+c} \in [0, 1]$ for all $a > 0$, $b \in [0, 1]$, and $c \in [0, 1]$? - 5 Note that $0\leq ab\leq a\leq a+c$. Then you can get the conclusion. – molan May 11 '12 at 16:08 @molan Perfect, thanks! Make it an answer and it gets my vote. – ezod May 11 '12 at 16:15 No thanks. But I think it don't deserve that. – molan May 11 '12 at 16:20 ## 1 Answer $$0\leq \frac{ab}{a+c}=\frac{b}{1+\frac{c}{a}}\leq b\leq 1$$ - You could add $0 \le$ on the left hand side and $\le 1$ on the right – Henry May 11 '12 at 16:31 You also should assume that $a\not=0$. – molan May 12 '12 at 2:40 It is given that $a>0$ – Julius May 12 '12 at 5:08

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http://mathhelpforum.com/advanced-algebra/119587-determinants-matrix-inverses.html

# Thread: 1. ## Determinants & Matrix Inverses If A and B are n by n, AB = -BA, and n is odd, show that either A or B are not invertible. Show that no 3 x 3 matrix A exists such that A^2 + I = 0. Any help is appreciated. 2. Originally Posted by BrownianMan If A and B are n by n, AB = -BA, and n is odd, show that either A or B are not invertible. Show that no 3 x 3 matrix A exists such that A^2 + I = 0. Any help is appreciated. Since $\det(AB)=\det(A)\det(B)$ , we get $AB=-BA\Longrightarrow \det(AB)=\det(-BA)=\det(-B)\det(A)=(-1)^n\det(B)\det(A)$. Now suppose both matrices are invertible (i.e., their determinant is non-zero), and get a huge contradiction. Tonio 3. You may have another constraint on your matrices you haven't mentioned yet - the matrix (where i=square root of -1): i,0,0 0,i,0 0,0,i satisfies the 2nd equation. If you demand that the matrix eigenvalues are real, the argument Tonio mentions should work.

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http://mathoverflow.net/questions/118892/on-the-fundamental-lemma

## on the fundamental lemma ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I consider the fundamental lemma for the spherical Hecke algebra. Let $G$ a connected reductive quasisplit group on $F$, a local field of equal characteristic $p$. and $H$ an endoscopic group. Can we reduce the fundamental lemma for (G,H) to the fundamental lemma for $(H,G)$ with $G_{der}=G_{sc}$ or even better to (H, G) where both G and H have a simply connected derived group? -

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http://math.stackexchange.com/questions/tagged/lie-algebras?sort=unanswered&pagesize=15

# Tagged Questions For questions about Lie algebras, an algebraic structure whose main use is in studying geometric objects such as Lie groups and differentiable manifolds. 0answers 204 views ### Abelian Cartan subalgebras If a Lie algebra is semisimple or reductive, its Cartan subalgebras are Abelian, and their elements semisimple. Are there non-reductive algebras with Abelian Cartan subalgebras all of whose ... 0answers 78 views ### Trivial summand of a representation's symmetric power The following comes from Exercise 13.17 of Fulton and Harris's book, Representation Theory: A First Course. Let $V$ denote the standard representation of $\mathfrak{sl}_3\mathbb{C}$, with weights ... 0answers 75 views ### Reference for l-adic Lie algebras I don't know much at all about Lie algebras or representation theory, and I'm trying to read Ribet's `Review of Abelian l-adic Representations and Elliptic Curves'. 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Let $\Delta \subset \Phi$ be a base. I already ... 0answers 44 views ### finding highest weight of dual of a representation of a semisimple lie algebra If $V$ is an irreducible representation of a semi simple lie algebra having highest weight $\lambda$ then what will be the highest weight of the corresponding irreducible representation $V^*$ (Dual of ... 0answers 288 views ### Weyl Character formula applied to Sp$(4,\mathbb{C})\cap$ U$(4)$. I posted a question a short while ago on this but got no response. I have worked on this more and so now have a more specific question. To start with we work with the $\mathbb{Q}$ version of ... 0answers 113 views ### Inscrutable proof in Humphrey's book on Lie algebras and representations This is a question pertaining to Humphrey's Introduction to Lie Algebras and Representation Theory Is there an explanation of the lemma in §4.3-Cartan's Criterion? I understand the proof given there ... 0answers 82 views ### The Nambu bracket Does anybody know how to show the Jacobi identity for the Nambu bracket in $\mathbb{R}^3$? The Nambu bracket with respect to $c \in \mathcal{F}(\mathbb{R}^3)$ is defined as \{F,G\}_c = \langle\nabla ... 0answers 111 views ### Integral forms of loop algebras. The question following is about integral forms for semisimple Lie algebras and loop algebras constructed from them. Let $\frak g$ a finite-dimensional Lie algebra over $\mathbb C$ and \$L(\frak ... 0answers 84 views ### The Weyl Group of $F_4$ The Weyl Group of $F_4$ is of order $1152=2^{7} \cdot 3^{2}$. By Burnside's theorem the group is solvable. Is there a way to see solvability from the root system? Is it possible to see the order of ... 0answers 191 views ### Invariant inner product $\langle\,,\rangle$ on a Lie algebra Let $\mathfrak{g}$ be a complex semisimple Lie algebra and $\mathfrak{h}$ be a Cartan subalgebra of $\mathfrak{g}$. We can use the Killing form to identify $\mathfrak{h}$ and $\mathfrak{h}^*$ ... 0answers 101 views ### On the root systems Let $\Phi$ be a root system of $E$. $\alpha,\beta\in \Phi$. Let $\lbrace \beta+i\alpha | i\in \mathbb{Z}\rbrace\cap \Phi$, $\alpha$-string through $\beta$, be $\beta-r\alpha,\ldots,\beta+q\alpha$, ... 0answers 16 views ### Centralizers of connected linear group and its Lie algebra If we have that $G$ is a connected linear group and $H<G$ with $\mathfrak{h}$ the lie algebra of $H$ and we define the centralizers of the elements in the following way: \$Z(H):=\{a\in G| ... 0answers 23 views ### The normalizer of a proper sub-algebra properly contains the sub-algebra in a nilpotent lie algebra. So I am just making my way in to the theory of Lie Algebras. The question at hand comes from page 14 of Humphreys' "Introduction to Lie Algebra and Representation Theory" Given a finite dimensional ... 0answers 28 views ### To what extent are formulas obtained in one Lie group valid in another Lie group with an isomorphic Lie algebra? In quantum optics, I am trying to explore the group generated by squeezing and rotation operators. These are closely related to area-preserving linear transforms, which they induce on the phase space, ... 0answers 29 views ### Character of half-spin representation Let $S^\pm$ be the half-spin representations of $\mathfrak{so}_{2n}\mathbb{C}$. Fulton-Harris's Representation Theory says on page 378 that the character $D^\pm$ of $S^\pm$ is the sum \sum x_1^{\pm ... 0answers 29 views ### Radical of $\mathfrak{gl}_n$ I find it intuitive enough that the radical of $\mathfrak{gl}_n\mathbb F$ is the scalar matrices, but I have trouble finding an easy, but complete proof: Proof. Let $\mathfrak s$ denote the scalar ... 0answers 68 views ### Lie bracket of vector fields definition equivalence Lie bracket of vector fields is defined in two ways: Let $\Phi^X_t$ be the flow associated with the vector field $X$, and let $d$ denote the tangent map derivative operator. Then the Lie ... 0answers 35 views ### Regarding the definition of vector field flow To make the connection to the Lie derivative, let $t \mapsto \Phi^X_t$ be the 1-parameter group of diffeomorphisms (or flow) generated by the vector field $X$. The differential \$ d\Phi^X_t ... 0answers 22 views ### Finding the Lie map Suppose I have a group homomorphism $\rho:SL(2,\mathbb{C})\to SO_0(3,1)$ given by $\rho(a)X=aXa^*$ and I want to see how the corresponding Lie map $L\rho$ looks like. By definition ... 0answers 44 views ### The enveloping algebra of a finite dimensional Lie algebra has no zero divisor Let $L$ be a complex, finite dimensional Lie algebra. It is well-known that the graded associative algebra of the enveloping algebra $U(L)$ is isomorphic to the symmetric algebra $S(L)$. Therefore ... 0answers 16 views ### Integer domain enveloping algebra I must prove that if $L$ is a Lie algebra and denoting $U(L)$ the enveloping algebra, then $U(L)$ hasn't zero divisions (e.g. if $ab=0 \,\,\, a,b \in U(L)$ then $a=0$ or $b=0$). Some ideas? 0answers 26 views ### When is the Killing form null? When is the Killing form $\kappa$ of a Lie algebra $\mathfrak g$ null, i.e. $\kappa(\cdot,\cdot)=0$? Surely this is true for any Lie algebra with trivial bracket, but is this the only case? I can't ... 0answers 75 views ### When is the adjoint representation self-dual? Let $G$ be an algebraic group (say, connected). Given a rep. $\rho:G\to GL(V)$ there is a dual rep. $\rho^{\vee}:G\to GL(V^{\vee})$ defined by $\rho^{\vee}(g)\phi =\phi\circ \rho(g^{-1})$. My question ... 0answers 48 views ### Weyl group of $\mathfrak{sl}(2,\mathbb{C})$ $\mathfrak{g}$ is a complex semisimple lie algebra which is a subalgebra of some $\mathfrak{gl}(n,\mathbb{C})$, we have chosen a compact real form $\mathfrak{l}$ of $\mathfrak{g}$ and let $K$ be the ... 0answers 71 views ### Commutator formula in infinite dimensions The commutator formula states that for $A,B$ elements of a Lie algebra, \lim_{n\to \infty}\left\{ ... 0answers 38 views ### Are there any infinite dimensional subalgebras of the Witt algebra? The Lie bracket of elements of the Witt algebra is given by: $[L_m,L_n]=(m-n)L_{m+n}$ Are there any infinite dimensional subalgebras of the Witt algebra that are not isomorphic to the Witt algebra ... 0answers 54 views ### Standard parabolic Lie subalgebras and conjugacy Let $\mathfrak g$ be a given semisimple Lie algebra with corresponding adjoint Lie group $G$. A parabolic subalgebra is any subalgebra containing a Borel subalgebra. We can pick a Borel ... 0answers 45 views ### ADE type root lattice Let $\Phi$ be a root system of ADE type, $L$ is the corresponding root lattice, show that $\Phi=\{\alpha\in L:(\alpha,\alpha)=2\}$, where $(,)$ is the normalized non-degenerate symmetric bilinear form ... 0answers 98 views ### Decomposing products of spinor representations into anti-symmetric tensors There is always a natural $2^{[\frac{d}{2}]}$ dimensional spinorial representation of $SO(d-1,1)$ (..induced from a representation of the related Clifford algebra..) and if $[m]$ denote the space of ... 0answers 55 views ### Basis of the Engel algebra If I have a connected, simply connected nilpotent lie group given by the commutators between the elements of a basis of its Lie algebra how can I recover the left invariant vector fields? For ... 0answers 40 views ### Representations of $U(d)$. Calculation of Gelfand-Zeitlin patterns for particular vectors. Following structure is given $\left(\mathbb{C}^d\right)^{\otimes n}$. Consider irreducible representations of $U(d)$. And consider the fully symmetric subspace $T_{\alpha}$ in ...

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http://mathoverflow.net/questions/22359/why-havent-certain-well-researched-classes-of-mathematical-object-been-framed-by/22401

## why haven’t certain well-researched classes of mathematical object been framed by category theory? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Category theory is doing/has done a stellar job on Set, FinSet, Grp, Cob, Vect, cartesian closed categories provide a setting for $\lambda$-calculus, and Baez wrote a paper (Physics, Topology, Logic and Computation: A Rosetta Stone) with Mike Stay about many of the interconnections between them. But there are mathematical objects that aren't thought of in a category-theoretic fashion, at least the extant literature doesn't tend to treat them as such. For instance nobody talks about Series, Products, IndefInt as being categories in their own right. (infinite series, infinite products, and indefinite integrals, respectively). (google searches for the phrase "the category of infinite series" in both the web and book databases have no hits whatsoever). I suppose my question is: why not? - 39 One of the first things to learn about category theory is that not everything in mathematics is a category. (-: – Mike Shulman Apr 23 2010 at 15:02 39 "To a man with a hammer, everything looks like a nail." Mark Twain – Andrey Rekalo Apr 23 2010 at 15:19 39 Some version of this question is asked here on a monthly basis. The category-theorists always respond in a reasonable way, explaining that their subject, like any other, has limitations. To be blunt: no amount of adjoint-this or colimit-that will tell you whether a sequence converges or whether a PDE has a solution. Huge swathes of current mathematics depends on proving convergence and solving PDE. – Tim Perutz Apr 23 2010 at 15:48 5 @Yemon: Maybe someone whose only tool is a claw hammer would try to use it to unscrew things, despite the absurdity of even trying. The mathematical analogue is evident. – BCnrd Apr 23 2010 at 15:50 12 Tim Perutz's comment reminds me of a point made in the preface of Evans's PDE book. "PDE is not a branch of functional analysis. ... [T]he insistence on an overly abstract viewpoint, and consequent ignoring of deep calculus and measure theoretic estimates, is ultimately limiting." Even functional analytifying is often too much in analysis and differential equations, let alone categorifying. – Jonas Meyer Apr 23 2010 at 16:28 show 5 more comments ## 6 Answers Fundamentally I agree with Mike Shulman's comment and I do not really want to claim the following fancy language is at all necessary to answer this question, but you may (or may not) find it illuminating. From the standpoint of higher category theory, categories (i.e., 1-categories) are just one level among many in a family of mathematical structures. Typically a mathematical object will "naturally" exist as an n-category for some particular n. For example, Set is naturally a 1-category, while Cat is naturally a 2-category. Your examples Series and so on seem to just be 0-categories, i.e., sets, since as Pete explained in his answer, there is no obvious natural notion of morphism between infinite series. Asking why Series is not a 1-category is like asking why Set is not a 2-category; these are just not the natural categorical levels that these objects live at. - ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. To turn a class into a category, you need a notion of morphisms between objects in the class. That's the long and short of it. Consider for instance the class of infinite real series, say viewed as the set $S = \mathbb{R}^{\aleph_0}$ of sequences of real numbers. (There is often some notational and "ontological" confusion between the terms of an infinite series, its associated sequence of partial sums, and its sum, if it has one. Which one of these "is" the series? But such considerations are not relevant here and indeed are usually viewed as antithetical to the categorical point of view.) To get a category, you need to identify a set of morphisms between any two elements of this set. This can certainly be done in any number of ways -- for instance one could use the ordering induced from the standard ordering on $\mathbb{R}$ and the lexicographic ordering of the sequence, and then $S$ is a totally ordered set. We could then define a category by having $\operatorname{Hom}(s,t)$ to be a one point set if $s \leq t$ and the empty set otherwise (and then take the unique composition law of morphisms, defined when $s \leq t \leq u$). But the question is: what does this category have to do with any aspect of the theory of infinite series? Apparently nothing. You could create any number of other categories with underlying set $S$ but you run into the same problem: the very old and extremely well-developed area of mathematics which studies the convergence and divergence of real infinite series simply does not have anything evident to do with any notion of "morphisms" between infinite series. Similarly for the other examples you mention. Categorical structure is a very fundamental kind of mathematical structure; it's a great way of thinking and unifies and conceptualizes the study of many kinds of mathematical objects in highly disparate fields. But it doesn't explain everything, and it is frankly a bit weird to think it should. - 1 There are morphisms between different series of a given type -- there are variety of <a href="en.wikipedia.org/wiki/… transformations</a>, such as Eulerian series acceleration, there's scalar multiplication, which turns one given series into another, and there's the equivalence class of all series that sum to the same thing (pi and e formulas for instance.) – deoxygerbe Apr 23 2010 at 20:41 the link doesn't seem to be working right. The full url is here: en.wikipedia.org/wiki/Series_acceleration – deoxygerbe Apr 23 2010 at 20:45 8 I notice you made a point of mentioning convergence. On the other hand, if you consider formal series without regard to convergence, then we have some nice category theory going on. For a long time, combinatorics resisted a categorical treatment, but then along came combinatorial species, and lots of interesting combinatorial structures turned out to be functors. And at that point, many properties of formal power series, considered as generating functions, suddenly acquired category theoretical meaning. Resistance is futile. :-) en.wikipedia.org/wiki/Combinatorial_species – Dan Piponi Apr 23 2010 at 21:14 +1 for the Borg reference. I often have the fealing, category theory is assimilating the rest of the mathematical galaxy. ;-) – Johannes Hahn Apr 24 2010 at 0:06 It might be worth noting that the problems of computing Feynmann integrals in quantum field theory is one that is traditionally phrased as one of analysis, but is now studied by pure mathematicians using categorical techniques (among others). - Paul Taylor's "Abstract Stone Duality" http://www.paultaylor.eu/ASD/ is an attempt to recast elementary real analysis (including sequences) involving categorical ideas. - 1 As is Peter Freyd's 'Algebraic Real Analysis' tac.mta.ca/tac/volumes/20/10/20-10abs.html – David Corfield Apr 26 2010 at 7:45 A Google scholar search for "category theory" "power series" brings up the paper: Elements of Stream Calculus::(An Extensive Exercise in Coinduction). So series can be usefully thought of in a category-theoretic fashion, and although this involves formal series the methods can be used to find solutions to differential equations. - For just another point, we could contemplate the hilarious line "why can't a woman... be more like a man", from you-know-where. That is, the style of (much) "analysis" reflects as much the personalities of the participants as it does the "mathematical reality"... the latter being self-referentially "defined" by the personalities. That is, perhaps the issue addressed by the question resides in the implicit assumptions of the questioner? :) In my own experience, "(naive?) category theory", in the sense of paying attention more to interactions among objects than to their internal structures, has been extremely helpful, at the very least in showing that various seeming "choices" or "constructions" were irrelevant... But I do recognize that other people thing other-ly... - I would phrase Paul's sentence a bit differently, in saying the notion of for example colimit allows for the description of an internal structure in terms of the way the object is put together from external objects, and indeed that the colimit structure is defined by all its interrelationships with other objects. In a technical sense, proving something is a colimit allows for the elegant proof by "verification of the universal property", which often is the "best" proof, since that is how the notion of colimit is defined. – Ronnie Brown Jan 4 at 14:28

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http://en.wikipedia.org/wiki/Ordinal_number

# Ordinal number Representation of the ordinal numbers up to ωω. Each turn of the spiral represents one power of ω In set theory, an ordinal number, or just ordinal, is the order type of a well-ordered set. They are usually identified with hereditarily transitive sets. Ordinals are an extension of the natural numbers different from integers and from cardinals. Like other kinds of numbers, ordinals can be added, multiplied, and exponentiated. Ordinals were introduced by Georg Cantor in 1883 to accommodate infinite sequences and to classify sets with certain kinds of order structures on them.[1] He derived them by accident while working on a problem concerning trigonometric series—see Georg Cantor. Two sets S and S' have the same cardinality if there is a bijection between them (ie there exists a function f that is both injective and surjective, that is it maps each element x of S to a unique element y = f(x) of S' and each element y of S' comes from exactly one such element x of S.) If a partial order < is defined on set S, and a partial order <' is defined on set S' , then the posets (S,<) and (S' ,<') are order isomorphic if there is a bijection f that preserves the ordering. That is, f(a) <' f(b) if and only if a < b. Every well ordered set (S,<) is order isomorphic to the set of ordinals less than one specific ordinal number (the order type of (S,<)) under their natural ordering. The finite ordinals (and the finite cardinals) are the natural numbers: 0, 1, 2, …, since any two total orderings of a finite set are order isomorphic. The least infinite ordinal is ω, which is identified with the cardinal number $\aleph_0$. However in the transfinite case, beyond ω, ordinals draw a finer distinction than cardinals on account of their order information. Whereas there is only one countably infinite cardinal, namely $\aleph_0$ itself, there are uncountably many countably infinite ordinals, namely ω, ω + 1, ω + 2, …, ω·2, ω·2 + 1, …, ω2, …, ω3, …, ωω, …, ωωω, …, ε0, …. Here addition and multiplication are not commutative: in particular 1 + ω is ω rather than ω + 1 and likewise, 2·ω is ω rather than ω·2. The set of all countable ordinals constitutes the first uncountable ordinal ω1, which is identified with the cardinal $\aleph_1$ (next cardinal after $\aleph_0$). Well-ordered cardinals are identified with their initial ordinals, i.e. the smallest ordinal of that cardinality. The cardinality of an ordinal defines a many to one association from ordinals to cardinals. In general, each ordinal α, is the order type of the set of ordinals strictly less than the ordinal, α itself. This property permits every ordinal to be represented as the set of all ordinals less than it. Ordinals may be categorized as: zero, successor ordinals, and limit ordinals (of various cofinalities). Given a class of ordinals, one can identify the α-th member of that class, i.e. one can index (count) them. Such a class is closed and unbounded if its indexing function is continuous and never stops. The Cantor normal form uniquely represents each ordinal as a finite sum of ordinal powers of ω. However, this cannot form the basis of a universal ordinal notation due to such self-referential representations as ε0 = ωε0. Larger and larger ordinals can be defined, but they become more and more difficult to describe. Any ordinal number can be made into a topological space by endowing it with the order topology; this topology is discrete if and only if the ordinal is a countable cardinal, i.e. at most ω. A subset of ω + 1 is open in the order topology if and only if either it is cofinite or it does not contain ω as an element. ## Ordinals extend the natural numbers A natural number (which, in this context, includes the number 0) can be used for two purposes: to describe the size of a set, or to describe the position of an element in a sequence. When restricted to finite sets these two concepts coincide; there is only one way to put a finite set into a linear sequence, up to isomorphism. When dealing with infinite sets one has to distinguish between the notion of size, which leads to cardinal numbers, and the notion of position, which is generalized by the ordinal numbers described here. This is because, while any set has only one size (its cardinality), there are many nonisomorphic well-orderings of any infinite set, as explained below. Whereas the notion of cardinal number is associated with a set with no particular structure on it, the ordinals are intimately linked with the special kind of sets that are called well-ordered (so intimately linked, in fact, that some mathematicians make no distinction between the two concepts). A well-ordered set is a totally ordered set (given any two elements one defines a smaller and a larger one in a coherent way) in which there is no infinite decreasing sequence (however, there may be infinite increasing sequences); equivalently, every non-empty subset of the set has a least element. Ordinals may be used to label the elements of any given well-ordered set (the smallest element being labelled 0, the one after that 1, the next one 2, "and so on") and to measure the "length" of the whole set by the least ordinal that is not a label for an element of the set. This "length" is called the order type of the set. Any ordinal is defined by the set of ordinals that precede it: in fact, the most common definition of ordinals identifies each ordinal as the set of ordinals that precede it. For example, the ordinal 42 is the order type of the ordinals less than it, i.e., the ordinals from 0 (the smallest of all ordinals) to 41 (the immediate predecessor of 42), and it is generally identified as the set {0,1,2,…,41}. Conversely, any set (S) of ordinals that is downward-closed—meaning that for any ordinal α in S and any ordinal β < α, β is also in the set—is (or can be identified with) an ordinal. So far we have mentioned only finite ordinals, which are the natural numbers. But there are infinite ones as well: the smallest infinite ordinal is ω, which is the order type of the natural numbers (finite ordinals) and that can even be identified with the set of natural numbers (indeed, the set of natural numbers is well-ordered—as is any set of ordinals—and since it is downward closed it can be identified with the ordinal associated with it, which is exactly how we define ω). A graphical “matchstick” representation of the ordinal ω². Each stick corresponds to an ordinal of the form ω·m+n where m and n are natural numbers. Perhaps a clearer intuition of ordinals can be formed by examining a first few of them: as mentioned above, they start with the natural numbers, 0, 1, 2, 3, 4, 5, … After all natural numbers comes the first infinite ordinal, ω, and after that come ω+1, ω+2, ω+3, and so on. (Exactly what addition means will be defined later on: just consider them as names.) After all of these come ω·2 (which is ω+ω), ω·2+1, ω·2+2, and so on, then ω·3, and then later on ω·4. Now the set of ordinals we form in this way (the ω·m+n, where m and n are natural numbers) must itself have an ordinal associated with it: and that is ω2. Further on, there will be ω3, then ω4, and so on, and ωω, then ωω², and much later on ε0 (epsilon nought) (to give a few examples of relatively small—countable—ordinals). We can go on in this way indefinitely far ("indefinitely far" is exactly what ordinals are good at: basically every time one says "and so on" when enumerating ordinals, it defines a larger ordinal). The smallest uncountable ordinal is the set of all countable ordinals, expressed as ω1. ## Definitions ### Well-ordered sets Further information: Ordered set In a well-ordered set, every non-empty subset has a smallest element. Given the axiom of dependent choice, this is equivalent to just saying that the set is totally ordered and there is no infinite decreasing sequence, something perhaps easier to visualize. In practice, the importance of well-ordering is justified by the possibility of applying transfinite induction, which says, essentially, that any property that passes on from the predecessors of an element to that element itself must be true of all elements (of the given well-ordered set). If the states of a computation (computer program or game) can be well-ordered in such a way that each step is followed by a "lower" step, then you can be sure that the computation will terminate. Now we don't want to distinguish between two well-ordered sets if they only differ in the "labeling of their elements", or more formally: if we can pair off the elements of the first set with the elements of the second set such that if one element is smaller than another in the first set, then the partner of the first element is smaller than the partner of the second element in the second set, and vice versa. Such a one-to-one correspondence is called an order isomorphism and the two well-ordered sets are said to be order-isomorphic, or similar (obviously this is an equivalence relation). Provided there exists an order isomorphism between two well-ordered sets, the order isomorphism is unique: this makes it quite justifiable to consider the two sets as essentially identical, and to seek a "canonical" representative of the isomorphism type (class). This is exactly what the ordinals provide, and it also provides a canonical labeling of the elements of any well-ordered set. So we essentially wish to define an ordinal as an isomorphism class of well-ordered sets: that is, as an equivalence class for the equivalence relation of "being order-isomorphic". There is a technical difficulty involved, however, in the fact that the equivalence class is too large to be a set in the usual Zermelo–Fraenkel (ZF) formalization of set theory. But this is not a serious difficulty. We will say that the ordinal is the order type of any set in the class. ### Definition of an ordinal as an equivalence class The original definition of ordinal number, found for example in Principia Mathematica, defines the order type of a well-ordering as the set of all well-orderings similar (order-isomorphic) to that well-ordering: in other words, an ordinal number is genuinely an equivalence class of well-ordered sets. This definition must be abandoned in ZF and related systems of axiomatic set theory because these equivalence classes are too large to form a set. However, this definition still can be used in type theory and in Quine's axiomatic set theory New Foundations and related systems (where it affords a rather surprising alternative solution to the Burali-Forti paradox of the largest ordinal). ### Von Neumann definition of ordinals Rather than defining an ordinal as an equivalence class of well-ordered sets, we will define it as a particular well-ordered set that (canonically) represents the class. Thus, an ordinal number will be a well-ordered set; and every well-ordered set will be order-isomorphic to exactly one ordinal number. The standard definition, suggested by John von Neumann, is: each ordinal is the well-ordered set of all smaller ordinals. In symbols, λ = [0,λ).[2] Formally: A set S is an ordinal if and only if S is strictly well-ordered with respect to set membership and every element of S is also a subset of S. Note that the natural numbers are ordinals by this definition. For instance, 2 is an element of 4 = {0, 1, 2, 3}, and 2 is equal to {0, 1} and so it is a subset of {0, 1, 2, 3}. It can be shown by transfinite induction that every well-ordered set is order-isomorphic to exactly one of these ordinals, that is, there is an order preserving bijective function between them. Furthermore, the elements of every ordinal are ordinals themselves. Whenever you have two ordinals S and T, S is an element of T if and only if S is a proper subset of T. Moreover, either S is an element of T, or T is an element of S, or they are equal. So every set of ordinals is totally ordered. Further, every set of ordinals is well-ordered. This generalizes the fact that every set of natural numbers is well-ordered. Consequently, every ordinal S is a set having as elements precisely the ordinals smaller than S. For example, every set of ordinals has a supremum, the ordinal obtained by taking the union of all the ordinals in the set. This union exists regardless of the set's size, by the axiom of union. The class of all ordinals is not a set. If it were a set, one could show that it was an ordinal and thus a member of itself, which would contradict its strict ordering by membership. This is the Burali-Forti paradox. The class of all ordinals is variously called "Ord", "ON", or "∞". An ordinal is finite if and only if the opposite order is also well-ordered, which is the case if and only if each of its subsets has a maximum. ### Other definitions There are other modern formulations of the definition of ordinal. For example, assuming the axiom of regularity, the following are equivalent for a set x: • x is an ordinal, • x is a transitive set, and set membership is trichotomous on x, • x is a transitive set totally ordered by set inclusion, • x is a transitive set of transitive sets. These definitions cannot be used in non-well-founded set theories. In set theories with urelements, one has to further make sure that the definition excludes urelements from appearing in ordinals. ## Transfinite sequence If α is a limit ordinal and X is a set, an α-indexed sequence of elements of X is a function from α to X. This concept, a transfinite sequence or ordinal-indexed sequence, is a generalization of the concept of a sequence. An ordinary sequence corresponds to the case α = ω. ## Transfinite induction Main article: Transfinite induction ### What is transfinite induction? Transfinite induction holds in any well-ordered set, but it is so important in relation to ordinals that it is worth restating here. Any property that passes from the set of ordinals smaller than a given ordinal α to α itself, is true of all ordinals. That is, if P(α) is true whenever P(β) is true for all β<α, then P(α) is true for all α. Or, more practically: in order to prove a property P for all ordinals α, one can assume that it is already known for all smaller β<α. ### Transfinite recursion Transfinite induction can be used not only to prove things, but also to define them. Such a definition is normally said to be by transfinite recursion – the proof that the result is well-defined uses transfinite induction. Let F denote a (class) function F to be defined on the ordinals. The idea now is that, in defining F(α) for an unspecified ordinal α, one may assume that F(β) is already defined for all β < α and thus give a formula for F(α) in terms of these F(β). It then follows by transfinite induction that there is one and only one function satisfying the recursion formula up to and including α. Here is an example of definition by transfinite recursion on the ordinals (more will be given later): define function F by letting F(α) be the smallest ordinal not in the class {F(β) | β < α}, that is, the class consisting of all F(β) for β < α. This definition assumes the F(β) known in the very process of defining F; this apparent vicious circle is exactly what definition by transfinite recursion permits. In fact, F(0) makes sense since there is no ordinal β < 0, and the class {F(β) | β < 0} is empty. So F(0) is equal to 0 (the smallest ordinal of all). Now that F(0) is known, the definition applied to F(1) makes sense (it is the smallest ordinal not in the singleton class {F(0)} = {0}), and so on (the and so on is exactly transfinite induction). It turns out that this example is not very exciting, since provably F(α) = α for all ordinals α, which can be shown, precisely, by transfinite induction. ### Successor and limit ordinals Any nonzero ordinal has the minimum element, zero. It may or may not have a maximum element. For example, 42 has maximum 41 and ω+6 has maximum ω+5. On the other hand, ω does not have a maximum since there is no largest natural number. If an ordinal has a maximum α, then it is the next ordinal after α, and it is called a successor ordinal, namely the successor of α, written α+1. In the von Neumann definition of ordinals, the successor of α is $\alpha\cup\{\alpha\}$ since its elements are those of α and α itself. A nonzero ordinal that is not a successor is called a limit ordinal. One justification for this term is that a limit ordinal is indeed the limit in a topological sense of all smaller ordinals (under the order topology). When $\langle \alpha_{\iota} | \iota < \gamma \rangle$ is an ordinal-indexed sequence, indexed by a limit γ and the sequence is increasing, i.e. $\alpha_{\iota} < \alpha_{\rho}\!$ whenever $\iota < \rho,\!$ we define its limit to be the least upper bound of the set $\{ \alpha_{\iota} | \iota < \gamma \},\!$ that is, the smallest ordinal (it always exists) greater than any term of the sequence. In this sense, a limit ordinal is the limit of all smaller ordinals (indexed by itself). Put more directly, it is the supremum of the set of smaller ordinals. Another way of defining a limit ordinal is to say that α is a limit ordinal if and only if: There is an ordinal less than α and whenever ζ is an ordinal less than α, then there exists an ordinal ξ such that ζ < ξ < α. So in the following sequence: 0, 1, 2, ... , ω, ω+1 ω is a limit ordinal because for any smaller ordinal (in this example, a natural number) we can find another ordinal (natural number) larger than it, but still less than ω. Thus, every ordinal is either zero, or a successor (of a well-defined predecessor), or a limit. This distinction is important, because many definitions by transfinite induction rely upon it. Very often, when defining a function F by transfinite induction on all ordinals, one defines F(0), and F(α+1) assuming F(α) is defined, and then, for limit ordinals δ one defines F(δ) as the limit of the F(β) for all β<δ (either in the sense of ordinal limits, as we have just explained, or for some other notion of limit if F does not take ordinal values). Thus, the interesting step in the definition is the successor step, not the limit ordinals. Such functions (especially for F nondecreasing and taking ordinal values) are called continuous. We will see that ordinal addition, multiplication and exponentiation are continuous as functions of their second argument. ### Indexing classes of ordinals We have mentioned that any well-ordered set is similar (order-isomorphic) to a unique ordinal number $\alpha$, or, in other words, that its elements can be indexed in increasing fashion by the ordinals less than $\alpha$. This applies, in particular, to any set of ordinals: any set of ordinals is naturally indexed by the ordinals less than some $\alpha$. The same holds, with a slight modification, for classes of ordinals (a collection of ordinals, possibly too large to form a set, defined by some property): any class of ordinals can be indexed by ordinals (and, when the class is unbounded in the class of all ordinals, this puts it in class-bijection with the class of all ordinals). So we can freely speak of the $\gamma$-th element in the class (with the convention that the “0-th” is the smallest, the “1-th” is the next smallest, and so on). Formally, the definition is by transfinite induction: the $\gamma$-th element of the class is defined (provided it has already been defined for all $\beta<\gamma$), as the smallest element greater than the $\beta$-th element for all $\beta<\gamma$. We can apply this, for example, to the class of limit ordinals: the $\gamma$-th ordinal, which is either a limit or zero is $\omega\cdot\gamma$ (see ordinal arithmetic for the definition of multiplication of ordinals). Similarly, we can consider additively indecomposable ordinals (meaning a nonzero ordinal that is not the sum of two strictly smaller ordinals): the $\gamma$-th additively indecomposable ordinal is indexed as $\omega^\gamma$. The technique of indexing classes of ordinals is often useful in the context of fixed points: for example, the $\gamma$-th ordinal $\alpha$ such that $\omega^\alpha=\alpha$ is written $\varepsilon_\gamma$. These are called the "epsilon numbers". ### Closed unbounded sets and classes A class $C$ of ordinals is said to be unbounded, or cofinal, when given any ordinal $\alpha$, there is a $\beta$ in $C$ such that $\alpha < \beta$ (then the class must be a proper class, i.e., it cannot be a set). It is said to be closed when the limit of a sequence of ordinals in the class is again in the class: or, equivalently, when the indexing (class-)function $F$ is continuous in the sense that, for $\delta$ a limit ordinal, $F(\delta)$ (the $\delta$-th ordinal in the class) is the limit of all $F(\gamma)$ for $\gamma<\delta$; this is also the same as being closed, in the topological sense, for the order topology (to avoid talking of topology on proper classes, one can demand that the intersection of the class with any given ordinal is closed for the order topology on that ordinal, this is again equivalent). Of particular importance are those classes of ordinals that are closed and unbounded, sometimes called clubs. For example, the class of all limit ordinals is closed and unbounded: this translates the fact that there is always a limit ordinal greater than a given ordinal, and that a limit of limit ordinals is a limit ordinal (a fortunate fact if the terminology is to make any sense at all!). The class of additively indecomposable ordinals, or the class of $\varepsilon_\cdot$ ordinals, or the class of cardinals, are all closed unbounded; the set of regular cardinals, however, is unbounded but not closed, and any finite set of ordinals is closed but not unbounded. A class is stationary if it has a nonempty intersection with every closed unbounded class. All superclasses of closed unbounded classes are stationary, and stationary classes are unbounded, but there are stationary classes that are not closed and stationary classes that have no closed unbounded subclass (such as the class of all limit ordinals with countable cofinality). Since the intersection of two closed unbounded classes is closed and unbounded, the intersection of a stationary class and a closed unbounded class is stationary. But the intersection of two stationary classes may be empty, e.g. the class of ordinals with cofinality ω with the class of ordinals with uncountable cofinality. Rather than formulating these definitions for (proper) classes of ordinals, we can formulate them for sets of ordinals below a given ordinal $\alpha$: A subset of a limit ordinal $\alpha$ is said to be unbounded (or cofinal) under $\alpha$ provided any ordinal less than $\alpha$ is less than some ordinal in the set. More generally, we can call a subset of any ordinal $\alpha$ cofinal in $\alpha$ provided every ordinal less than $\alpha$ is less than or equal to some ordinal in the set. The subset is said to be closed under $\alpha$ provided it is closed for the order topology in $\alpha$, i.e. a limit of ordinals in the set is either in the set or equal to $\alpha$ itself. ## Arithmetic of ordinals Main article: Ordinal arithmetic There are three usual operations on ordinals: addition, multiplication, and (ordinal) exponentiation. Each can be defined in essentially two different ways: either by constructing an explicit well-ordered set that represents the operation or by using transfinite recursion. Cantor normal form provides a standardized way of writing ordinals. The so-called "natural" arithmetical operations retain commutativity at the expense of continuity. ## Ordinals and cardinals ### Initial ordinal of a cardinal Each ordinal has an associated cardinal, its cardinality, obtained by simply forgetting the order. Any well-ordered set having that ordinal as its order-type has the same cardinality. The smallest ordinal having a given cardinal as its cardinality is called the initial ordinal of that cardinal. Every finite ordinal (natural number) is initial, but most infinite ordinals are not initial. The axiom of choice is equivalent to the statement that every set can be well-ordered, i.e. that every cardinal has an initial ordinal. In this case, it is traditional to identify the cardinal number with its initial ordinal, and we say that the initial ordinal is a cardinal. Cantor used the cardinality to partition ordinals into classes. He referred to the natural numbers as the first number class, the ordinals with cardinality $\aleph_0$ (the countably infinite ordinals) as the second number class and generally, the ordinals with cardinality $\aleph_{n-2}$ as the n-th number class.[3] The α-th infinite initial ordinal is written $\omega_\alpha$. Its cardinality is written $\aleph_\alpha$. For example, the cardinality of ω0 = ω is $\aleph_0$, which is also the cardinality of ω2 or ε0 (all are countable ordinals). So (assuming the axiom of choice) we identify ω with $\aleph_0$, except that the notation $\aleph_0$ is used when writing cardinals, and ω when writing ordinals (this is important since, for example, $\aleph_0^2$ = $\aleph_0$ whereas $\omega^2>\omega$). Also, $\omega_1$ is the smallest uncountable ordinal (to see that it exists, consider the set of equivalence classes of well-orderings of the natural numbers: each such well-ordering defines a countable ordinal, and $\omega_1$ is the order type of that set), $\omega_2$ is the smallest ordinal whose cardinality is greater than $\aleph_1$, and so on, and $\omega_\omega$ is the limit of the $\omega_n$ for natural numbers n (any limit of cardinals is a cardinal, so this limit is indeed the first cardinal after all the $\omega_n$). See also Von Neumann cardinal assignment. ### Cofinality The cofinality of an ordinal $\alpha$ is the smallest ordinal $\delta$ that is the order type of a cofinal subset of $\alpha$. Notice that a number of authors define cofinality or use it only for limit ordinals. The cofinality of a set of ordinals or any other well ordered set is the cofinality of the order type of that set. Thus for a limit ordinal, there exists a $\delta$-indexed strictly increasing sequence with limit $\alpha$. For example, the cofinality of ω² is ω, because the sequence ω·m (where m ranges over the natural numbers) tends to ω²; but, more generally, any countable limit ordinal has cofinality ω. An uncountable limit ordinal may have either cofinality ω as does $\omega_\omega$ or an uncountable cofinality. The cofinality of 0 is 0. And the cofinality of any successor ordinal is 1. The cofinality of any limit ordinal is at least $\omega$. An ordinal that is equal to its cofinality is called regular and it is always an initial ordinal. Any limit of regular ordinals is a limit of initial ordinals and thus is also initial even if it is not regular, which it usually is not. If the Axiom of Choice, then $\omega_{\alpha+1}$ is regular for each α. In this case, the ordinals 0, 1, $\omega$, $\omega_1$, and $\omega_2$ are regular, whereas 2, 3, $\omega_\omega$, and ωω·2 are initial ordinals that are not regular. The cofinality of any ordinal α is a regular ordinal, i.e. the cofinality of the cofinality of α is the same as the cofinality of α. So the cofinality operation is idempotent. ## Some “large” countable ordinals For more details on this topic, see Large countable ordinal. We have already mentioned (see Cantor normal form) the ordinal ε0, which is the smallest satisfying the equation $\omega^\alpha = \alpha$, so it is the limit of the sequence 0, 1, $\omega$, $\omega^\omega$, $\omega^{\omega^\omega}$, etc. Many ordinals can be defined in such a manner as fixed points of certain ordinal functions (the $\iota$-th ordinal such that $\omega^\alpha = \alpha$ is called $\varepsilon_\iota$, then we could go on trying to find the $\iota$-th ordinal such that $\varepsilon_\alpha = \alpha$, “and so on”, but all the subtlety lies in the “and so on”). We can try to do this systematically, but no matter what system is used to define and construct ordinals, there is always an ordinal that lies just above all the ordinals constructed by the system. Perhaps the most important ordinal that limits a system of construction in this manner is the Church–Kleene ordinal, $\omega_1^{\mathrm{CK}}$ (despite the $\omega_1$ in the name, this ordinal is countable), which is the smallest ordinal that cannot in any way be represented by a computable function (this can be made rigorous, of course). Considerably large ordinals can be defined below $\omega_1^{\mathrm{CK}}$, however, which measure the “proof-theoretic strength” of certain formal systems (for example, $\varepsilon_0$ measures the strength of Peano arithmetic). Large ordinals can also be defined above the Church-Kleene ordinal, which are of interest in various parts of logic. ## Topology and ordinals For more details on this topic, see Order topology. Any ordinal can be made into a topological space in a natural way by endowing it with the order topology. See the Topology and ordinals section of the "Order topology" article. ## Downward closed sets of ordinals A set is downward closed if anything less than an element of the set is also in the set. If a set of ordinals is downward closed, then that set is an ordinal—the least ordinal not in the set. Examples: • The set of ordinals less than 3 is 3 = { 0, 1, 2 }, the smallest ordinal not less than 3. • The set of finite ordinals is infinite, the smallest infinite ordinal: ω. • The set of countable ordinals is uncountable, the smallest uncountable ordinal: ω1. ## Notes 1. Thorough introductions are given by Levy (1979) and Jech (2003). 2. Levy (1979, p. 52) attributes the idea to unpublished work of Zermelo in 1916 and several papers by von Neumann the 1920s. 3. Dauben (1990:97) ## References • Cantor, G., (1897), Beitrage zur Begrundung der transfiniten Mengenlehre. II (tr.: Contributions to the Founding of the Theory of Transfinite Numbers II), Mathematische Annalen 49, 207-246 English translation. • Conway, J. H. and Guy, R. K. "Cantor's Ordinal Numbers." In The Book of Numbers. New York: Springer-Verlag, pp. 266–267 and 274, 1996. • Dauben, Joseph Warren, (1990), Georg Cantor: his mathematics and philosophy of the infinite. Chapter 5: The Mathematics of Cantor's Grundlagen. ISBN 0-691-02447-2 • Hamilton, A. G. (1982), Numbers, Sets, and Axioms : the Apparatus of Mathematics, New York: Cambridge University Press, ISBN 0-521-24509-5 See Ch. 6, "Ordinal and cardinal numbers" • Kanamori, A., Set Theory from Cantor to Cohen, to appear in: Andrew Irvine and John H. Woods (editors), The Handbook of the Philosophy of Science, volume 4, Mathematics, Cambridge University Press. • Levy, A. (1979), Basic Set Theory, Berlin, New York: Springer-Verlag Reprinted 2002, Dover. ISBN 0-486-42079-5 • • Sierpiński, W. (1965). Cardinal and Ordinal Numbers (2nd ed.). Warszawa: Państwowe Wydawnictwo Naukowe. Also defines ordinal operations in terms of the Cantor Normal Form. • Suppes, P. (1960), Axiomatic Set Theory, D.Van Nostrand Company Inc., ISBN 0-486-61630-4

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http://wikipedia.sfstate.us/Battle_of_the_sexes_(game_theory)

edit # Battle of the sexes (game theory) For other uses, see Battle of the sexes (disambiguation). | | | | |-----------------------|-------|----------| | | Opera | Football | | Opera | 3,2 | 0,0 | | Football | 0,0 | 2,3 | | Battle of the Sexes 1 | | | | | | | |-----------------------|-------|----------| | | Opera | Football | | Opera | 3,2 | 1,1 | | Football | 0,0 | 2,3 | | Battle of the Sexes 2 | | | In game theory, battle of the sexes (BoS), also called Bach or Stravinsky,1 is a two-player coordination game. Imagine a couple that agreed to meet this evening, but cannot recall if they will be attending the opera or a football match (and the fact that they forgot is common knowledge). The husband would most of all like to go to the football game. The wife would like to go to the opera. Both would prefer to go to the same place rather than different ones. If they cannot communicate, where should they go? The payoff matrix labeled "Battle of the Sexes (1)" is an example of Battle of the Sexes, where the wife chooses a row and the husband chooses a column. In each cell, the first number represents the payoff to the wife and the second number represents the payoff to the husband. This representation does not account for the additional harm that might come from not only going to different locations, but going to the wrong one as well (e.g. he goes to the opera while she goes to the football game, satisfying neither). To account for this, the game is sometimes represented as in "Battle of the Sexes (2)". ## Equilibrium analysis This game has two pure strategy Nash equilibria, one where both go to the opera and another where both go to the football game. For the first game, there is also a Nash equilibrium in mixed strategies, where the players go to their preferred event more often than the other. For the payoffs listed above, each player attends their preferred event with probability 3/5. This presents an interesting case for game theory since each of the Nash equilibria is deficient in some way. The two pure strategy Nash equilibria are unfair; one player consistently does better than the other. The mixed strategy Nash equilibrium (when it exists) is inefficient. The players will miscoordinate with probability 13/25, leaving each player with an expected return of 6/25 (less than the return one would receive from constantly going to one's less favored event). One possible resolution of the difficulty involves the use of a correlated equilibrium. In its simplest form, if the players of the game have access to a commonly observed randomizing device, then they might decide to correlate their strategies in the game based on the outcome of the device. For example, if the couple could flip a coin before choosing their strategies, they might agree to correlate their strategies based on the coin flip by, say, choosing football in the event of heads and opera in the event of tails. Notice that once the results of the coin flip are revealed neither the husband nor wife have any incentives to alter their proposed actions – that would result in miscoordination and a lower payoff than simply adhering to the agreed upon strategies. The result is that perfect coordination is always achieved and, prior to the coin flip, the expected payoffs for the players are exactly equal. ### Working out the above Let us calculate the four probabilities for the actions of the individuals (Man and Woman), which depend on their expectations of the behaviour of the other, and the relative payoff from each action. The Man either goes to the Football or the Opera (and not both or neither), and likewise the Woman. The Probability that the man goes to the football game, $Pmf$, equals the payoff if he does (whether or not the woman does), divided by the same payoff plus the payoff if he goes to the opera instead: • $Pmf = \cfrac{1 Pwo + 3 Pwf}{1 Pwo + 3 Pwf + 2 Pwo + 0 Pwf} = \cfrac{1 Pwo + 3 Pwf}{3 Pwo + 3 Pwf}$ We know that she either goes to one or the other, so $Pwo + Pwf = 1$, so: • $Pmf = \cfrac{1 Pwo + 3 Pwf}{3}$ Similarly: • $Pmo = \cfrac{2 Pwo}{3}$ • $Pwf = \cfrac{2 Pmf}{3}$ • $Pwo = \cfrac{3 Pmo + 1 Pmf}{3}$ This forms a set of simultaneous equations. We can solve these, starting with $Pmf$ for example, by substituting in the equations above: • $Pmf = \cfrac{1 \cfrac{3 Pmo + 1 Pmf}{3} + 3 \cfrac{2 Pmf}{3}}{3}$ • $Pmf = \cfrac{1 \cfrac{3 Pmo + 1 Pmf}{3} + 2 Pmf}{3}$ • $Pmf = \cfrac{Pmo + \tfrac{1}{3}Pmf + 2 Pmf}{3}$ Remembering that $Pmo + Pmf = 1$, we can make this an equation where the only unknown is $Pmf$: • $Pmf = \cfrac{1 - Pmf + \tfrac{1}{3}Pmf + 2 Pmf}{3}$ And then rearrange so that $Pmf$ is only on one side: • $3 Pmf = 1 - Pmf + \tfrac{1}{3}Pmf + 2 Pmf$ • $3 Pmf = 1 + \tfrac{4}{3}Pmf$ • $3 Pmf - \tfrac{4}{3}Pmf = 1$ • $\tfrac{9}{3} Pmf - \tfrac{4}{3}Pmf = 1$ • $\cfrac{9 Pmf - 4 Pmf}{3} = 1$ • $\cfrac{5 Pmf}{3} = 1$ • $5 Pmf = 3$ • $Pmf = \tfrac{3}{5}$ Knowing that $Pmf + Pmo = 1$, we deduce: • $Pmo = 1 - \dfrac{3}{5} = \tfrac{2}{5}$ • $Pwf = \cfrac{2 Pmf}{3} = \tfrac{6}{15} = \tfrac{2}{5}$ • $Pwo = 1 - \dfrac{2}{5} = \tfrac{3}{5}$ Then we can calculate the probability of coordination $Pc$ (that M and W do the same thing, independently), as: • $Pc = Pmf Pwf + Pmo Pwo$ • $Pc = \tfrac{3}{5} \tfrac{2}{5} + \tfrac{2}{5} \tfrac{3}{5}$ • $Pc = 2 \tfrac{6}{25} = \tfrac{12}{25}$ And the probability of miscoordination $Pm$ (that M and W do different things, independently): • $Pm = Pmf Pwo + Pmo Pwf$ • $Pm = \tfrac{3}{5} \tfrac{3}{5} + \tfrac{2}{5} \tfrac{2}{5}$ • $Pm = \tfrac{9}{25} + \tfrac{4}{25} = \tfrac{13}{25}$ And just to check our probability working: • $Pc + Pm = \tfrac{12}{25} + \tfrac{13}{25} = \tfrac{25}{25} = 1$ So the probability of miscoordination is $\tfrac{13}{25}$ as stated above. The expected payoff E for each individual ($Em$ and $Ew$) is the probability of each event multiplied by the payoff if it happens. For example, the Probability that the man goes to football and the Woman goes to football multiplied by the Expected payoff to the man if that happens ($Emmfwf$): • $Em = Pmf Pwf Emmfwf + Pmf Pwo Emmfwo + Pmo Pwo Emmowo + Pmo Pwf Emmowf$ • $Em = \tfrac{3}{5} \tfrac{2}{5} 3 + \tfrac{3}{5} \tfrac{3}{5} 1 + \tfrac{2}{5} \tfrac{3}{5} 2 + \tfrac{2}{5} \tfrac{2}{5} 0$ • $Em = \tfrac{18}{25} + \tfrac{9}{25} + \tfrac{12}{25} + 0$ • $Em = \tfrac{39}{25}$ Which is not the same as the $\tfrac{6}{5}$ stated above! For comparison, let us assume that the man always goes to football and the woman, knowing this, chooses what to do based on revised probabilities and expected values to her: • $Pmf = 1$ • $Pmo = 0$ • $Pwf = \cfrac{2 Pmf}{3} = \tfrac{2}{3}$ • $Pwo = 1 - \dfrac{2}{3} = \tfrac{1}{3}$ • $Em = Pmf Pwf Emmfwf + Pmf Pwo Emmfwo + Pmo Pwo Emmowo + Pmo Pwf Emmowf$ • $Em = 1 \tfrac{2}{3} 3 + 1 \tfrac{1}{3} 1 + 0 \tfrac{1}{3} 2 + 0 \tfrac{2}{3} 0$ • $Em = \tfrac{6}{3} + \tfrac{1}{3} = \tfrac{7}{3}$ This is symmetric for $Ew$ if the woman always goes to the opera and the man chooses randomly with probabilities based on the expected outcome, due to the symmetry in the value table. But if both players always do the same thing (both have simple strategies), the payoff is just 1 for both, from the table above. ## Burning money | | | | |----------|-------|----------| | | Opera | Football | | Opera | 4,1 | 0,0 | | Football | 0,0 | 1,4 | | Unburned | | | | | | | |----------|-------|----------| | | Opera | Football | | Opera | 2,1 | -2,0 | | Football | -2,0 | -1,4 | | Burned | | | Interesting strategic changes can take place in this game if one allows one player the option of "burning money" – that is, allowing that player to destroy some of her utility. Consider the version of Battle of the Sexes pictured here (called Unburned). Before making the decision the row player can, in view of the column player, choose to set fire to 2 points making the game Burned pictured to the right. This results in a game with four strategies for each player. The row player can choose to burn or not burn the money and also choose to play Opera or Football. The column player observes whether or not the row player burns and then chooses either to play Opera or Football. If one iteratively deletes weakly dominated strategies then one arrives at a unique solution where the row player does not burn the money and plays Opera and where the column player plays Opera. The odd thing about this result is that by simply having the opportunity to burn money (but not actually using it), the row player is able to secure her favored equilibrium. The reasoning that results in this conclusion is known as forward induction and is somewhat controversial. For a detailed explanation, see [1] p8 Section 4.5. In brief, by choosing not to burn money, the player is indicating she expects an outcome that is better than any of the outcomes available in the "burned" version, and this conveys information to the other party about which branch she will take. ## References • Luce, R.D. and Raiffa, H. (1957) Games and Decisions: An Introduction and Critical Survey, Wiley & Sons. (see Chapter 5, section 3). • Fudenberg, D. and Tirole, J. (1991) Game theory, MIT Press. (see Chapter 1, section 2.4) 1. Osborne, Rubinstein (1994). A course in game theory. The MIT Press.

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http://math.stackexchange.com/questions/tagged/computer-science

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### Orthogonal Multiwavelet Frames in ${L}^{2}({R}^{d})$ Liu Zhanwei, Hu Guoen, and Wu Guochang Source: J. Appl. Math. Volume 2012 (2012), Article ID 846852, 18 pages. #### Abstract We characterize the orthogonal frames and orthogonal multiwavelet frames in ${L}^{2}({R}^{d})$ with matrix dilations of the form $(Df)(x)=\sqrt{|\text{d}\text{e}\text{t}A|}f(Ax)$, where $A$ is an arbitrary expanding $d×d$ matrix with integer coefficients. Firstly, through two arbitrarily multiwavelet frames, we give a simple construction of a pair of orthogonal multiwavelet frames. Then, by using the unitary extension principle, we present an algorithm for the construction of arbitrarily many orthogonal multiwavelet tight frames. Finally, we give a general construction algorithm for orthogonal multiwavelet tight frames from a scaling function. First Page: We're sorry, but we are unable to provide you with the full text of this article because we are not able to identify you as a subscriber. If you have a personal subscription to this journal, then please login. If you are already logged in, then you may need to update your profile to register your subscription. Read more about accessing full-text

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http://mathoverflow.net/questions/103394?sort=votes

## Fundamental Solutions with compact support (distributions) ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Assume that we have a differential operator such as $-\frac{\partial}{\partial x^2} + id$ on $\mathbb{R}^1$ We also then argue that if a fundamental solution has compact support, then it is supported on the origin. My follow up question is how can one then show assuming that the fundamental solution is compactly supported - that the differential operator must be of order 0? - I myself am confused by the context in which an argument proceeds by saying that if a fundamental solution for a differential operator were compactly supported then it would have support $\{0\}$, leading to deducing that the operator is order $0$, so apparently is a multiplication operator? Or is this meant to be about pseudo-differential operators? Would you clarify the context? – paul garrett Jul 28 at 19:01 I'm not sure if this includes pseudo-differential operators. The question wasn't well stated to me, I got the impression it only involved differential operators (with constant coefficients). Apparently this can be shown by an appeal to the Fourier transform of distributions... – Peadar Coyle Nov 7 at 0:52 ## 2 Answers For a constant coefficient partial differential operator P(D), the fundamental solution of P can never belong to $\epsilon'(\mathbb{R}^{n})$,i.e.have compact support. In fact,assume we have $P(D)u=f$,where u is a distribution,then u have compact support $\Leftrightarrow$ $\frac{f}{P(\xi)}$ is analytic(The result can be found in Hormander's ALPDO, volume 1,ch7.) Now,if we have $$P(D)u=\delta$$,obviously $\frac{1}{P(\xi)}$ is never an analytic function for a polynomia P.So the fundamental solution of P can not be compact supported. - ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. If the fundamental solution is supported at the origin, it must be a finite combination of derivatives of the Dirac distribution. This means that your original operator was of nonpositive order. -

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http://mathhelpforum.com/pre-calculus/54256-parallel-perpendicular-lines-help.html

# Thread: 1. ## parallel & perpendicular lines help hey everyone, this is my first post...i hope i can find some help here, as i have searched my math book up and down, only to find no answer. i am in college algebra, and the section is parallel & perpendicular lines. the problem i am totally stuck on is this... Code: ``` x - y x + y ----- = ----- - 1 3 2 AND 7 = - ( x - y ) + 4y``` They want me to rewrite both in slope-intercept form, then find out if they are parallel or perpendicular. I would really and truly appreciate any and all help with this...i just don't know what to do thanks so much, ryan 2. Originally Posted by shipwreck hey everyone, this is my first post...i hope i can find some help here, as i have searched my math book up and down, only to find no answer. i am in college algebra, and the section is parallel & perpendicular lines. the problem i am totally stuck on is this... Code: ``` x - y x + y ----- = ----- - 1 3 2 AND 7 = - ( x - y ) + 4y``` They want me to rewrite both in slope-intercept form, then find out if they are parallel or perpendicular. I would really and truly appreciate any and all help with this...i just don't know what to do thanks so much, ryan Hello Ryan, $\frac{x-y}{3}=\frac{x+y}{2}-1$ First, let's multiply everything by the LCD of 6 to get rid of those pesky denominators. $6\left(\frac{x-y}{3}\right)=6\left(\frac{x+y}{2}\right)-6(1)$ $2x-2y=3x+3y-6$ Simplify to $-5y=x-6$ $\boxed{y=\frac{-1}{5}x+\frac{6}{5}}$ Now for the other one... $7=-(x-y)+4y$ $7=-x+y+4y$ Simplifying.... $5y=x+7$ $\boxed{y=\frac{1}{5}x+\frac{7}{5}}$ Compare the slopes of the two boxed equations. What do you think? 3. well i know that if both slopes are identical, than they are considered parallel....but these aren't identical, although i don't know if there is a particular rule or not about them being opposites, but NOT being reciprocal negative opposites... 1/5 x & 1/5x would be parallel, but i am just not 100% certain about 1/5x & -1/5x. i would say they are not parallel...right? 4. anyone able to confirm if i am right or not about these being parallel or not? 5. Originally Posted by shipwreck anyone able to confirm if i am right or not about these being parallel or not? You answered your question yourself. If the slopes are not the same then they aren't the parallel

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http://en.wikisource.org/wiki/On_some_Dynamical_Conditions/IV

On some Dynamical Conditions/IV From Wikisource Preston, Samuel Tolver (1878), “The Bearing of the Kinetic Theory of Gravitation on the Phenomena of "Cohesion" and "Chemical Action," together with the important connected Inferences regarding the existence of Stores of Motion in Space, No. IV”, Philosophical Magazine 5: 297-311 Samuel Tolver Preston No. IV[1] 1. IT would be natural to expect that any theory competent to explain the effects of gravity ought to be able to throw some light upon the subsidiary effects of molecules exhibited in "cohesion," "chemical action," &c. Before proceeding to consider this question, and in order to have a clear conception of the point we have to deal with, we will recapitulate in a few words the physical conditions involved in the case of gravity as already dealt with. It has been our object to point out that the molecules of a gas within the range of free path are moving in precisely the right way to produce [298] gravity in two masses immersed in the gas within the range of free path. For since it has been proved from the kinetic theory that the particles of a gas adjust their motions so as to move uniformly or equally in all directions, and since the particles within the range of free path are moving in unbroken streams, it follows that two masses immersed in the gas at a distance apart within this range will (owing to the one sheltering the other) be struck with more particles on their remote (unsheltered) sides than on their adjacent (sheltered) sides, so that the two masses will be urged together. This, therefore, fulfils Le Sage's fundamental idea without the necessity for accepting any of his postulates. We need not accept the scarcely realizable postulates of streams of particles coming from indefinite distances in space (at uniform angles), each stream moving continuously in one direction; but we can substitute for this the natural conception of the normal motion of the particles of a gas within the range of free path, where, although each particle is continually changing the direction of its motion, yet the general character of the motion of the system as a whole remains unchanged; or the system of particles automatically correct their motions so as to continue to move uniformly or equally in all directions, as demonstrated in connexion with the kinetic theory of gases. This movement of the particles equally in all directions is the condition required to produce equal gravific effect in all directions. Thus all we require to admit in order to produce all the effects of gravity as necessary results, is the existence of a gas in space. This gas differs from an ordinary gas only as to scale, i. e. in the proximity, velocity, and extreme minuteness of its particles, whereby a length of free path commensurate with the greatest observed range of gravity is insured, the extreme minuteness of the particles being at the same time adapted to that high velocity which the effects of gravity require, and which also necessarily renders the medium itself impalpable or concealed from the senses. The range of free path, though great in one sense, may be considered small and suitable for a gas that pervades the vast range of the visible universe. 2. In applying these principles to cohesion, or the approach of molecules in chemical reactions, it is so far easy to see that when two molecules of matter come very close together, or if we suppose them actually to come into contact, then they will cut off the entire stream of particles of the gravific medium from between the parts in contact; and therefore, as the gravific particles now only strike against the remote sides of the two molecules, the latter will be urged together with very great [299] force, thus explaining "cohesion".[2] But then a difficulty at once presents itself here. When two masses (or molecules) are gradually approached towards each other, instead of the tendency to approach gradually increasing up to a maximum ( as we should expect from the theory), they begin to repel at a certain distance, and very considerable force is in general required to overcome this first repulsion, when the masses then unite into one. Thus two freshly cut pieces of lead may be made to unite with some pressure, also glass, or various metals, with more or less pressure. There is therefore a neutral point which has to be passed, when the tendency to recede changes into a tendency to approach. The same thing is exhibited ( conversely) when a substance is broken into two parts by tension. If pulled (nearly) up to the neutral point, the two parts recoil or return into their old positions. If pulled beyond the neutral point, the parts repel and will not return into their old positions, i. e. they separate permanently. The thing, therefore, to be explained is the existence of this neutral point, or, in other words, the repulsion that exists at a certain distance from the surfaces. 3. The explanation we have to offer here depends upon quite recent investigations. It must be observed first that facts prove the existence of a second medium in space besides the gravific medium, viz. the heat- or light-conveying medium (the aether). If we admit the existence of one medium in space constituted according to the kinetic theory (the gravific medium), it would be natural to conclude that the second medium (or aether) was constituted in an analogous manner. We shall give independent reasons afterwards that lead to infer this constitution, and endeavour to answer possible objections; but in the mean time it is only necessary to suppose it to be so constituted (in the absence of proof to the contrary); and if this supposition serves to explain in general principle a number of facts, this will be one argument for its truth. On account of the extreme shortness of the waves of light and heat, it would be reasonable to suppose that the length of free [300] path of the aether particles was contained within compact limits, or was, at any rate, shorter than the length of the wave itself. It has been proved recently, in investigations by Mr. Johnstone Stoney in connexion with the radiometer,[3] that a medium constituted according to the kinetic theory has a special power of propagating a pressure unequal in various directions, or that, when a layer of the medium (such as a layer of air) is intercepted between two surfaces whose distance apart is a small multiple of the length of free path of the particles of air, the layer can then transmit a pressure in the line perpendicular to the surfaces which is in excess of the transverse pressure; and thus a repulsion is produced, accounting for the spheroidal state, the motion of the radiometer, &c. In fact it is evident (as pointed out) that, since in a medium constituted according to the kinetic theory the particles move in straight lines, the particles (when the distance of the opposed surfaces approximates to the range of free path) get reflected backwards and forwards repeatedly between the opposed surfaces, the increments of energy received by the particles accumulating by successive reflections, so that the particles produce a bombardment tending to separate the two opposed surfaces The increments of velocity imparted by the heated[4] [301] surfaces are also mainly received in the line joining the surfaces (not so much transversely); so that this conduces to the pressure on the surfaces, or repulsion. 4. This is precisely what we have to put forward, in its application to the aether, as an explanation of the repulsion in the cases referred to, such as for example the repulsion of two lenses or glass surfaces placed together in such proximity as to exhibit "Newton's rings," the repulsion of two molecules &c.; for if the tether be constituted according to the kinetic theory, we shall inevitably have the same phenomena here, though on an infinitely more energetic scale; for the particles of tether come into direct contact with the vibrating molecules of matter, whose energy of vibration is known to be enormous at normal temperature; and the layer of aether is very thin, and the motion of the aether particles very rapid,[5] so that the successive increments of velocity imparted by the vibrating molecules accumulate by successive reflections (backwards and forwards) between the opposed surfaces, producing a forcible repulsion. These results have been theoretically demonstrated to follow on the basis of the kinetic theory, and have been established by experimental facts. It is a point of great importance to observe that it is specially the kinetic theory that explains this otherwise most curious fact of an excess of pressure in a medium in one direction (producing a repulsion), with normal pressure existing in transverse directions, which otherwise it would be so difficult to explain, and which must be explained in order to account in a realizable manner for the phenomena observed. It is difficult to conceive how any other means of explaining this curious fact could be afforded than that supplied by the kinetic theory. Moreover it is generally admitted that heat has the property of producing repulsion. The "heat" of the molecules in the cases mentioned is known to consist in their vibrations, by which they generate waves of heat in the aether. We have therefore to explain under what particular constitution of a medium vibrations can (within certain limits) produce repulsion. The kinetic theory of the constitution of the medium solves completely this peculiarly difficult problem. [302] 5. When the two surfaces (or two molecules) are pushed up closer to each other, then the energy of the gravific medium directed against the remote sides of the molecules prevails more and more, since the mutual sheltering-power of the molecules increases in an enormously rapid ratio as contact is neared, and so the unbalanced energy of the gravific medium directed with full force against the remote sides of the opposed molecules at length outweighs the action of the intercepted aether particles, and the two molecules are propelled together ( or unite). 6. We may allude to a few examples serving to illustrate the application of the above principles. Supposing we take the common case of the ignition of a gas jet. Then when the gas is turned on, the molecules of gas and air mingle with each other and are known to be exchanging motion and rebounding from each other, and yet they do not unite. According to the above principles the molecules, as they approach each other in their encounters, are kept apart by the forcible vibrations (which the molecules are known to possess[6]) which, through the increments of velocity imparted to the particles of the intervening aether, produce a repulsion in the manner described, as soon as the molecules in their encounters have approached nearly within range of the mean path of the aether particles. When a flame is applied to the jet, the rapidly moving gaseous molecules of which the flame consists naturally produce a disturbance, jostling some of the molecules of the mixture of gas and air against each other, so that, the neutral point is passed, whereby the molecules are brought into such proximity that their mutual sheltering action causes the gravific medium to impinge with full energy upon their remote sides, thus urging the molecules together (producing combination). The molecules are thrown into forcible vibration by the shock of approach, and become luminous through the energy of the waves thus generated by them in the surrounding aether. These vibrations of the compound molecules after combination naturally cause the forcible rebound of any other molecules that happen to be in their proximity, the disturbance thus set up sufficing to effect the successive ( practically instantaneous) combination of the entire jet of gas. The same considerations of course apply to the practically instantaneous combination (explosion) of a mixture (in definite proportions) of gas and air, by an initial disturbance [303] produced by a flame. In the case of solid bodies, where the molecules are fixed or under control, a forcible pressure or concussion may serve to bring the molecules over the neutral point (and thus effect combination), as illustrated by the effect of the blow struck in "percussion" powders. It would not appear that matter in the gaseous state could ever be exploded by pressure (so long as the gaseous state was retained); for the molecules of gases cannot be pressed against each other by any amount of pressure, since, the molecules being in free translatory motion among themselves, the only effect of pressure would evidently be to put a greater number of molecules into unit of volume, without thereby causing the molecules in their encounters to approach nearer to each other than before. The degree of approach of the molecules (in their encounters) depends evidently on their momentum or velocity; and this remains the same whatever the pressure. 7. Heat could not apparently be said to augment the energy of chemical combination, since, in general, heat is known to possess the exactly opposite effect, or to disintegrate matter. The part played by heat in effecting chemical combination would seem to consist simply in producing a molecular disturbance, whereby unavoidably some molecules are urged towards each other so as to pass the outer neutral point, which is the necessary preliminary to combination. No doubt, when heated elements combine, the original heat adds itself to the work thus to be derived, as the heat cannot be destroyed, though it cannot increase the work of combination. Heat may (as is known) entirely prevent chemical combination, and oven dissociate combined elements. The action of heat in preventing chemical combination and producing dissociation would on the above principles consist in the fact that, when the vibratory motion of the molecules becomes excessive, this vibratory motion generates such a pressure in the intervening layer of aether on the approach of the molecules as to prevent them from passing the neutral point: or, indeed, no neutral point may exist, provided the pressure or repulsion thus generated be such as to outweigh the action of the gravific medium, as appears actually to take place in the dissociation of matter by excessive heat. Thus it would appear probable from this, that when combination ensues in the case of a mixture of gases previously considerably heated (but not so much so as to produce dissociation), the molecules on combination do not at once settle clown into that full proximity (which belongs to a lower temperature), but they do so gradually as the temperature falls. Thus the work of combination is prolonged over the falling temperature, and the cooling thereby [304] somewhat retarded. Precisely the same thing is illustrated in the aggregation of groups of molecules (to form masses), as in the aggregation of single molecules to form compound molecules. Thus when a bar of iron is welded by heat, the molecules (though aggregated or combined) do not settle down into their final positions of proximity until the bar cools, the bar being observed to contract on cooling. In this instance also the cooling of the bar is somewhat retarded by the approach of the molecules in the act of cooling. 8. In the case of the ignition of a solid body, the same considerations no doubt apply as in the case of a gas. Thus, for example, the molecules of oxygen are impinging against the surface of a piece of coal, but do not produce ignition. To effect this a certain number of the molecules must be impelled with sufficient energy against the coal so as to carry them over the neutral point (i. e. beyond the initial repulsion). The application of a flame, which consists of matter in a state of violent agitation, suffices to effect this, and, no doubt by loosening some of the molecules of carbon (of the coal) and giving them translatory motion and mixing them with the air, facilitates the process. 9. As a further illustration of the exact similarity of behaviour of single molecules and groups of molecules (masses) as regards the existence of the above-mentioned neutral point, we may take the case of the substance iodine. This substance gives off a visible vapour at normal temperatures. The single molecules of iodine composing the vapour rebound from each other without uniting; and this can only be due to the existence of the above-mentioned neutral point, outside which there is a repulsion. If the colliding molecules were to approach within the neutral point, they would unite and form solid iodine. No doubt some of the molecules of the vapour (as their velocities are known to be very diverse) do pass beyond the neutral point; and thus molecules of vapour striking against the fragments of solid iodine in the bottle, will sometimes unite with the solid iodine and form part of it, while, on the other hand, other molecules of the solid which happen to possess excessive vibrating energy are thrown off, this being the known way in which the balance in evaporation is maintained. The masses of iodine have the same neutral point as the single molecules, since two masses of the substance when pressed together will not readily unite; i. e. the neutral point, where the outer repulsion terminates, must be passed first.[7] [305] 10. Just as increase of vibrating energy (temperature) tends, by the increase of pressure thus produced in the intervening film of the medium, to dissociate molecules, so reduction of vibrating energy (attendant on reduction of temperature) tends to facilitate the approach of molecules, on account of the reduction of the pressure or repulsive action of the intervening film. Thus the molecules of a vapour when their vibrating energy is reduced (by a fall of temperature) may by the simple momentum of their own encounters, carry themselves over the neutral point, and thus effect the condensation of the vapour. Numerous other cases might be cited illustrative of the application of the above principles, as, indeed, the molecular effects are very similar in their fundamental aspects. The molecular phenomena, however diverse, may be all correlated in one fundamental respect, viz. as consisting in phenomena of approach and recession. The fundamental conditions to be explained, therefore, are the conditions capable of producing the approach and recession of molecules. Whatever may be said of the above deductions, it is at least so far certain that the conditions investigated, and based upon experimental facts, are competent to produce these fundamental movements of approach and recession in the case of molecules, and to do so in the simplest manner, the constitution of media according to the kinetic theory being admittedly the simplest conceivable. To look therefore to other conditions than the simplest would be to imply that the same results are brought about by a superfluity of mechanism. This superfluity is known not to be the characteristic of nature; and all the teaching of mechanism points to the fact that superfluity or unnecessary complication entirely prevents the attainment of precision and certainty in the mechanical effects. The great precision and unfailing certainty of the molecular effects would therefore render it necessary to infer that the regulating mechanism was simple, or that there was no unnecessary superfluity. 11. The fundamental conclusion above drawn regarding the mechanism concerned in the approach of molecules is grounded upon the only explanation of the mechanism of gravity that has withstood criticism and received support by competent judges, viz. the kinetic theory of gravity, of which Le Sage's ingenious idea forms the fundamental basis, and is at once the simplest explanation of gravity conceivable. The [306] application of this theory to molecules in close contact ("cohesion" &c.), is necessary and inevitable, and it serves to correlate the molecular effects generally under one cause. The explanation of the fundamental condition capable of producing the recession of molecules, as above given, rests upon experimental facts recently established, and upon a basis for the constitution of the aether which is the simplest conceivable. 12. We now propose to show some independent reasons in support of this constitution for the aether, in addition to the argument afforded by the numerous molecular effects which this constitution, in principle, serves to explain. First, if the subject be reflected on, it will be apparent that, in principle, u. movement of the component particles of the medium in straight lines is the only possible constitution for the ultimate medium in space. For a particle of matter cannot move in a curved line unless it have a medium about it to control its motion. Thus a planet can move in a curve because it has a medium about it (the gravific medium) to cause it to move in a curve. It is a known principle that a particle of matter cannot of itself change the direction of its motion. The particles of the ultimate medium in space must therefore move in straight lines. This deduction is surely of great importance in the inquiry as to the constitution of the physical media in space. Also in addition to this, the observed facts of gravity prove that the particles of the gravific medium move in straight lines, since no other motion than this can harmonize with the observed effects of gravity. It would therefore surely be a strange thing if the particles of the aether, as a second medium immersed in the gravific medium, did not move in straight lines. To suppose this would be very like supposing that when the particles of a second gas are immersed among those of another, the particles of the first gas acted upon those of the second to make them move otherwise than in straight lines, which is known to be impossible. Moreover the fact of the kinetic theory representing the simplest conceivable constitution for a medium would by itself be a strong argument for this constitution in the case of the aether. The very fact of the great precision and delicacy of the operations performed by the author as the mechanism for the transmission of the varied phenomena of colour &c. would point to a simple constitution; just as the complex effects of sound with all its intricate and varied gradations of tone are known to be transmitted by a medium (the air) of the simplest conceivable constitution, viz. that represented by the beautiful kinetic theory of gases.[8] The more intricate the functions of a mechanism, [307] the more is simplicity indispensable, and superfluity incompatible with precision and certainty in the results. To assume a constitution for the aether that could not be realized or clearly explained would surely be futile, since the explanation or clear conception forms the logical support of any theory, without which the theory resembles a mere dogmatic statement incapable of being sustained by reason. 13. There is one other point which we would notice in connexion with this subject. The idea would appear to be to a certain extent prevalent that the aether must have a constitution essentially different from the air, because the vibrations producing light are transverse, while those producing sound are longitudinal. It seems to be sometimes inferred from this that the vibrations of the Bather are only transverse, and those of the air only longitudinal. There would be no warrant for this conclusion; and we think that it has done harm and greatly hindered any rational idea from being formed of the nature of the aether. According to the kinetic theory, which is known to represent the constitution of the air, the vibrations of the particles of air disturbed by a vibrating body and propagated in the form of waves, are not only longitudinal; for since according to the kinetic theory the particles of air in their normal state are moving equally in all directions, it follows that these particles are accelerated and retarded both in transverse and in longitudinal directions at the passage of waves. It is true that the transverse component of the motion probably may not affect the ear, on account of its special structure. It would be wrong, however, to infer from this that the transverse component of the motion did not exist. So in the case of the aether, it would be unwarranted to infer that the longitudinal component of the motion did not exist, because this component was incapable of affecting the eye. The eye and the ear may be very differently constituted; and a motion that affects the one might not affect the other. Sir John Herschel says regarding this point in his essay "On Light" (' Popular Lectures on Scientific Subjects,' page 358): — "According to any conception we can form of an elastic medium, its particles must be conceived free to move ( within certain limits greater or less according to the coercive forces which restrain them) in every direction." He then goes on to explain how the efficacy of the transverse component of the movement in the case of light, and the longitudinal component of the movement in the case of sound, may be accounted for by the diverse structure of the eye and ear. Any inference which is not valid, invariably does some harm; and this idea of a forward movement being propagated in a medium by only transverse vibrations, being almost inconceivable, has [308] naturally led to some incongruous ideas regarding the structure of the aether, in the effort to explain it. Thus some have supposed the aether to resemble a solid, which is in direct opposition to the teaching of the senses; for we move about so freely in this "solid" as to be unconscious of its existence. Another supposition has been that "lines of tension," behaving somewhat in analogy to stretched chords, exist in the aether. Such a mechanism would be, to say the least, somewhat deranged by the passage of a planet through the aether. Indeed it is sufficiently evident that these are the hopeless attempts made to surmount an impossible condition, or a difficulty for whose existence there is really no warrant. If the aether be not a solid, or a liquid (for liquids oppose enormous resistances to the passage of bodies through them at high speeds), then what other resource have we than to conclude that it is a gas? 14. A gaseous constitution of the aether according to the kinetic theory would perfectly satisfy the two fundamental conditions of a medium highly elastic in all directions, and opposing no appreciable resistance to the free movement of bodies (the planets &c.) through its substance. For it is a known fact that the resistance opposed by a medium constituted according to the kinetic theory to the passage of bodies through it diminishes as the normal velocity of the particles of the medium increases. The high normal velocity of the particles of the aether, proved by the velocity of light, therefore necessarily renders the resistance inappreciable, and the medium itself impalpable and undetected by the senses. 15. A difficulty has been raised in the way of the aether being constituted as a gas on the following grounds, which, being only anxious for truth, we are bound to consider.[9] It has been argued that if the aether be constituted as a gas, the specific heat of unit of volume of the aether would be the same as that of any ordinary gas at the same pressure, and that therefore it would appear that the presence of the aether could not fail to be detected in the experiments on the specific heat of ordinary gases. We have to offer the following as a means of meeting this difficulty. It will be admitted that the detection of the aether in the experiments on specific heat will depend, not on the specific capacity for heat possessed bv the aether, but on the rate at which the heat passes from the gas experimented on to the aether. The molecules of the gas are moving through the aether with their normal translatory motion, this motion of the molecules representing the "heat" of the gas. It will be evident that the rate at which the [309] motion ("heat") of the molecules of the gas passes to the aether will depend on the resistance the aether offers to the passage of these molecules through it. But we have shown that this resistance may (on account of the high normal velocity of the aether particles) be inappreciable. Hence the rate of passage of the heat from the gas to the tether will be inappreciable. This, we submit, removes the difficulty in question. It is clear that, if the aether opposes no appreciable resistance to the passage of a planet through it (moving at several miles per second), it cannot be affected by the passage of a molecule of a gas through it, which in its relatively slow rate of translatory motion may be considered at rest compared with the aether particles. The high normal velocity of the aether particles is only appropriate to their minute mass. 16. It must be apparent to any reflecting observer, that in physical science we have a vast array of facts accumulated through years of experiment, but a great paucity of causes; or the number of facts known is quite out of all proportion to the number of causes known, these latter being replaced by more or less vague and unsubstantial theories. As, therefore, we have no paucity of facts as a basis to reason upon, it surely cannot be too soon to make an effort to correct this anomalous state of things, and to replace the above unsubstantial theories by rational conceptions of the processes of nature, Clearness of conception is the test of truth, and constitutes its real dignity; and theories, however elaborated, if vague, have no real dignity.[10] Since there is nothing occult about the physical media in space, in so far as they differ in no way from ordinary matter excepting in the mere scale or dimensions of their parts, and since it is obviously just as easy to reason of matter of one dimension as of another, any hesitation in entering upon this course of study would be wholly uncalled for; indeed, surely there is reason for a rational interest in realizing the admirable adaptation of these media in a mechanical point of view for their special functions; and the question as to the utilization of the stores of motion enclosed by them to the best advantage may present a problem of the highest practical interest and importance.[11] It should be observed that these stores of [310] motion simply consist in small particles of matter in a state of rapid motion; or there is nothing occult about the subject at all, as indeed obviously principles of reasoning are independent of size. The minute size (and consequent invisibility) of the particles is necessary to the efficiency of the media as powerful motive agents, since minuteness of size is necessary to render a high velocity possible for the particles, without producing disturbing effects among the matter immersed in these media. There is one very noteworthy point that cannot be too distinctly kept in view in connexion with this subject. It is the fact that the high intensity of the stores of motion possessed by these media, and which renders them so important, serves to conceal their existence from the senses. Thus the higher the intensity of the store of motion enclosed by these media, and consequently the greater their capacity for practical utility, the more likely (if the mere evidence of the senses were relied on) is their existence to be forgotten. For it may be proved beforehand, by the kinetic theory of gases, that the greater the velocity of the component particles of a medium, and consequently the greater the value of the store of energy enclosed (which may even reach an explosive intensity), the more does the presence of the medium elude detection, because the resistance opposed by the medium to the passage of bodies through it diminishes as the velocity of the particles increases. The less indication the mere senses (unaided by reason) afford of the existence of such media, the higher, therefore, should we be warranted in inferring their importance to be. Even independently of all question of the existence of these media, it may be proved beforehand that, if media did exist and enclose stores of motion to an enormous intensity, they would be concealed. This is, no doubt, a remarkable fact, and contrary to preconceived ideas, as it would doubtless appear on the first thought that the higher the intensity of a store of energy existing in space, the more likely would it be to make itself apparent to the senses, whereas precisely the contrary is found to be the fact. This forms a notable instance of one of those cases where analysis completely reverses preconceived ideas. It is possibly the absence of appreciation of this fact that may in some way account for the failure of the most striking proof's of nature to carry their practical teaching, as for example, the sudden setting free of concealed motion in the explosion of a mass of gunpowder. Here to the mere bodily senses, we have apparently an actual creation of motion. Something [311] more, however, than the evidence of the mere bodily senses may be required, to appreciate the truths of nature, as it is a notorious fact that the most important truths generally lie below the surface. It should be noted that these media would not be efficient as working agents unless they were concealed; for concealment (as observed) is the necessary condition to the enclosure of a store of motion to a high intensity. Possibly the absence of realization of this fact, and perhaps that prejudice which besets every new path, may in some degree account for what must otherwise appear an extraordinary indifference and absence of inquiry in a subject of great mechanical interest and involving possibly issues of the highest importance and practical utility. When this, like every other illogical prejudice to change, comes to be broken down by the light of reason and reflection, there may be just ground for surprise at the previous delay, and at the shallow and unsubstantial character of the theories which so long supplanted rational conceptions of the processes of nature. London, March 13, 1878. 1. The three previous papers treating of the subject of gravitation are in the Philosophical Magazine for September and November 1877, and February 1878. 2. The spectroscope proves molecules to be complex bodies, on account of the number of different periods of vibration they can take up; and it was pointed out in the last paper that there are grounds for inferring them to possess interstices, or a more or less open structure. It is evident, therefore, that the shapes of molecules, as to whether their parts fitted over each other or not (and thus afforded more or less shelter from the impinging particles of the gravific medium), would have some influence on the behaviour of molecules as to the energy of their approach (reactions). This might account in some degree for the varied behaviour termed "chemical affinity," though possibly there are, besides this, other modifying physical conditions. 3. Philosophical Magazine, December 1877. 4. There is another point in connexion with the motion of the particles, which no doubt, however, has been already noticed. Under normal conditions, a body vibrating opposite to another tends (as is known) to produce rarefaction in the intervening medium; but in the case of a film whose thickness is near the range of mean path of the particles, there would appear to be a special cause tending greatly to reduce this effect, and even perhaps to produce the contrary effect, viz. a condensation ( which would greatly increase the repulsion). Thus, under the increments of velocity received, there is a tendency for the molecules of the gaseous film to be turned round so as to move more normal to the film. Suppose, for instance, an elastic sphere to be rebounding obliquely between two planes. Suppose increments of velocity to be given to the sphere by vibrating one of the planes. Then these increments of velocity given to the sphere will evidently make it rebound more normal to the surfaces. So in the case of the molecules of a gaseous film, rebounding backwards and forwards between two surfaces (such as the air film which supports a drop in the so-called "spheroidal" state, the air film which supports a grain of powder in some experiments of Professor Barrett, referred to by Mr. Johnstone Stoney), the molecules of the film will tend, by the increments of velocity given them, to turn round so as to move in a direction more normal to the surfaces. This evidently makes the lateral pressure exerted by the film less, and consequently its lateral expansion (or rarefaction) less. If we imagine the extreme case where the molecules of the film are all turned round so as to move exactly normal to the surface of the film, then whatever the velocities of the molecules of the film (i. e. whatever the longitudinal pressure, or repulsion, exerted by them), the film would exert no lateral pressure at all. There would consequently be a lateral inrush of air, increasing the density of the film, and therefore increasing the repulsion (since those molecules which enter the film become themselves available for producing repulsion). Possibly, from this cause, these films may be actually denser than normal density. At all events the above cause makes their density greater than it otherwise would be, and the repulsion exerted by them greater. 5. It may be noted that, if the aether be constituted according to the kinetic theory, the normal velocity of its particles is $\frac{3}{\sqrt{5}}$ X velocity of a wave of light. See appendix to paper "On the Mode of Propagation of Sound" (Phil. Mag. June 1877), added by Prof. Maxwell. 6. The molecules of matter in the gaseous state are known to possess, in addition to the translatory motion peculiar to that state, a vibratory motion, in virtue of which the molecules generate waves of regular periods iu the tether (these periods having in many cases been measured by the spectroscope). 7. The above effects were described in a little book ' Physics of the Ether' (E. & F. N. Spon), published by me in 1875; but the cause of the reduction of the pressure of the medium, which determines the approach of molecules, was there wrongly stated, the error having arisen from a seeming analogy between the approach of bodies to masses (tuning-forks &c.) vibrating in air — in the absence of the knowledge recently acquired of the repulsion of gaseous layers. Much of the main principles of the book, however, remain as they were — to be supplemented by the investigations contained in the present papers. 8. There would surely be nothing to admire in complication in itself. The whole aim of mechanical design is directed towards the attainment of simplicity, which being unique, entails intellectual labour to find it. 9. See paper "On the Dynamical Evidence of the Molecular Constitution of Bodies," by Prof. Maxwell (' Nature,' March 11, 1875). 10. Vagueness, paradox, and mystery surely belong rather to those intellects which are incapable of rising to clear and definite conceptions. 11. As an instance of the change of views on the most practical subjects that the acceptance of these principles entails, we may cite the case of the employment of coal, which by the recognition of the existence of the stores of motion in space, becomes a mechanism or machine for deriving motion. The expenditure of coal, therefore, represents the expenditure of mechanism or machinery. Hence in deriving motion through coal we expend a quantity of machinery proportional to the power derived. Without asking the question whether it is necessary in every case, in deriving motion from a source, to expend machinery proportional to the power derived (i. e. that the work done should be the equivalent of the machinery expended), it is at least so far certain that no remedy for this could be discovered unless the physical conditions of the case were recognized. This work is in the public domain in the United States because it was published before January 1, 1923. The author died in 1917, so this work is also in the public domain in countries and areas where the copyright term is the author's life plus 80 years or less. This work may also be in the public domain in countries and areas with longer native copyright terms that apply the rule of the shorter term to foreign works.

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http://www.physicsforums.com/showthread.php?t=335881

Physics Forums Stacked blocks 1. The problem statement, all variables and given/known data Imagine that there is a slab (call it block 1) on top of which sits a block (call it block 2). The slab connects to a cable that passed through a pulley and attaches to another block (call it block 3), hanging freely. Assuming that the coefficient of kinetic friction is the same for both blocks (and the same with static friction): a) Find the acceleration of the system. b) Verify the familiar limits. c) Under what conditions does block 2 stay on top of block 1? 2. Relevant equations/The attempt at a solution If blocks 1 and 2 move together, then all three blocks have the same acceleration: $$a=\frac{m_3g-(m_1+m_2)g\mu_k}{(m_1+m_2+m_3}$$ If block 2 begins to slip, then we still take all the accelerations as the same (block 2 is just beginning to slip, so it is still pretty much moving with block 1 (i.e. same acceleration) and it's frictional coefficient will be static) TO MAKE A LONG STORY SHORT: DID I SET UP NEWTON'S 2ND LAW CORRECTLY? then Newtons 2nd Law for block 1, block 2 and block 3(respectfully) are: [tex] \sum(1,x): T+F_{f,s}-F_{f,k}=(m_1+m_2)a [/tex] [tex] \sum(1,y): F_{n,1}-(m_1+m_2)g=0 [/tex] [tex] \sum(2,x): F_{f,s}=m_2a[/tex] $$\sum(2,y): F_{n,2}-m_2g=0$$ $$\sum(3,x): 0=0$$ $$\sum(3,y): m_3g-T=m_3a$$ Where the frictional forces take on the familiar value of the coefficient times the normal. Solving, I get: [tex]a= \frac{m_3g+\mu_sm_2g-\mu_kg(m_1+m_2)}{m_1+m_1+m_3}[/tex] For part b) while testing the familiar limits, i found that as $$m_3 \rightarrow \infty$$ then $$a=g$$, which makes sense. I found that if $$m_2=0$$, then we get what we expect. And when I calculate when $$m_2 \rightarrow \infty$$, $$a=0$$. BUT, when I try to calculate as $$m_1\rightarrow \infty$$, $$a=\frac{-\infty}{\infty}$$, which doesn't make sense. Did I screw up Newtons 2nd law? For part c) all I did was use newton's 2nd law for the second body and solved for $$a$$, the maximum acceleration: $$a=\mu_sg$$, which doesn't make sense either since it should depend on mass. Thoughts?!?! PhysOrg.com science news on PhysOrg.com >> Front-row seats to climate change>> Attacking MRSA with metals from antibacterial clays>> New formula invented for microscope viewing, substitutes for federally controlled drug Blog Entries: 7 Recognitions: Gold Member Homework Help Part (a) is OK. It is the same result as if blocks 1 and 2 were glued together to form a single block of mass m1+m2. Part (b) is not OK. The force that accelerates m2 (net force) is the force of static friction, fs. Newton's 2nd Law says that it must be equal to m2 times the acceleration from part (a). As you increase m3, you increase the acceleration which means that fs must increase accordingly. But fs cannot increase forever. At the point where fs can no longer increase, and ony then, can you say fs = μsN. You have all the ingredients to you can put it together from this point on. Thanks for the reply kuruman. So my equation for the acceleration is correct then? You say that I cannot include fs = μsN until the point in which the static friction can not keep up with the acceleration of blocks 1 and 3, so then it shouldn't be in my equation for the acceleration? What about part c)? Under what condition does block 2 stay on top of block 1? I understand that this only happens if fs = μsN, eg if μs=fs/N (or less), but in this equation what is fs? Blog Entries: 7 Recognitions: Gold Member Homework Help Stacked blocks The force of static friction fs is whatever is necessary to provide the observed acceleration. Because it is the net force on mass m2, you can write $$f_{s}=m_{2}a=m_{2}\frac{m_{3}g-\mu_{k}g(m_{1}+m_{2})}{m_{1}+m_{2}+m_{3}}$$ That is always true unless the top mass starts sliding. At the threshold of when sliding is just about going to happen, the force of static friction has reached its maximum value and can no longer "provide the observed acceleration" past that value. What does the above equation become then? What is the maximum value of the force of static friction? When: $$\mu_s\geq \frac{F_{f,s}}{F_n}=\frac{am_2}{g}\left(\frac{m_{3}g-\mu_{k}g(m_{1}+m_{2})}{m_{1}+m_{2}+m_{3}}\right)$$ ? Also what about the problem with negative acceleration when m_2 -> infinity? Blog Entries: 7 Recognitions: Gold Member Homework Help $$f_{s}=m_{2}a \leq \mu_{s}F_{n}$$ Replace a and Fn with the appropriate quantities. Also what about the problem with negative acceleration when m_2 -> infinity? What about it? Do you really think that if you make mass m2 large enough, say the size of a battleship, the masses will accelerate in the opposite direction? I didn't think so. What is more likely to happen instead? Remember, the acceleration you have has been derived assuming that the masses accelerate so that the hanging mass drops. Thread Tools | | | | |-------------------------------------|-------------------------------|---------| | Similar Threads for: Stacked blocks | | | | Thread | Forum | Replies | | | Introductory Physics Homework | 17 | | | Introductory Physics Homework | 27 | | | Introductory Physics Homework | 0 | | | Introductory Physics Homework | 4 | | | Introductory Physics Homework | 4 |

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http://math.stackexchange.com/questions/33079/matrix-structure-maybe-you-can-see-something-i-cant?answertab=votes

# Matrix structure; maybe you can see something I can't… I am attempting to find a 'smarter' way to solve a matrix, in the form $Ax=B$, where $B_{i}=F_i*N$ $A_{i,j}=-F_i/K_{j,i}$ where $N$ is constant, $K$ is a constant matrix, and $F$ is a vector of; $F_i=Y_i/K_{i,i}$ where $Y$ is a vector constant. I'm not hugely mathematical, and my linear algebra skills could be put on the back of a napkin, but seeing this kind of repetition indicates to me that there must be a shortcut for this. Does anyone have any insights? - 1 You can cancel the $F_i$'s. – Yuval Filmus Apr 15 '11 at 1:20 I was thinking more about the structure of the resultant matrix but yeah, I wanted a second opinion on that too (I have no real understanding of the weirdnesses of linalg...) – Andrew Bolster Apr 15 '11 at 1:23 Think of it as a system of equations. For each $i$ you have an equation of the form $\sum_j A_{ij} x_i = B_i$. In your case, this is $\sum_j -F_i/K_{ji} = NF_i$, so you can cancel the $F_i$'s. Otherwise it's almost completely arbitrary (other than the fact that $-1/K_{ji}$ cannot be zero). – Yuval Filmus Apr 15 '11 at 1:44 You can also cancel the $N$ if you want. Solve if for RHS which is constant $1$, then multiply the result by $N$. – Yuval Filmus Apr 15 '11 at 1:45 There's a typo above; it should read $\sum_j A_{ij} x_j = B_i$. – joriki Apr 15 '11 at 7:50

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http://math.stackexchange.com/questions/175263/gradient-and-hessian-of-general-2-norm

# Gradient And Hessian Of General 2-Norm Given $f(\mathbf{x}) = \|\mathbf{Ax}\|_2 = (\mathbf{x}^\mathrm{T} \mathbf{A}^\mathrm{T} \mathbf{Ax} )^{1/2}$, $\nabla f(\mathbf{x}) = \frac {\mathbf{A}^\mathrm{T} \mathbf{Ax}} {\|\mathbf{Ax}\|_2} = \frac {\mathbf{A}^\mathrm{T} \mathbf{Ax}} {(\mathbf{x}^\mathrm{T} \mathbf{A}^\mathrm{T} \mathbf{Ax} )^{1/2}}$ $\nabla^2 f(\mathbf{x}) = \frac { (\mathbf{x}^\mathrm{T} \mathbf{A}^\mathrm{T} \mathbf{Ax} )^{1/2} \cdot \mathbf{A}^\mathrm{T} \mathbf{A} - (\mathbf{A}^\mathrm{T} \mathbf{Ax})^\mathrm{T} (\mathbf{x}^\mathrm{T} \mathbf{A}^\mathrm{T} \mathbf{Ax} )^{-1/2} \mathbf{A}^\mathrm{T} \mathbf{Ax} } {(\mathbf{x}^\mathrm{T} \mathbf{A}^\mathrm{T} \mathbf{Ax} ) } = \frac { \mathbf{A}^\mathrm{T} \mathbf{A} } { (\mathbf{x}^\mathrm{T} \mathbf{A}^\mathrm{T} \mathbf{Ax} )^{1/2}} - \frac {\mathbf{x}^\mathrm{T} \mathbf{A}^\mathrm{T} \mathbf{A} \mathbf{A}^\mathrm{T} \mathbf{Ax} } { (\mathbf{x}^\mathrm{T} \mathbf{A}^\mathrm{T} \mathbf{Ax} )^{3/2} }$ I guess I am looking for confirmation that I have done the above correctly. The dimensions match up except for the second term of the Hessian is a scalar, which makes me think that something is missing. Edit: Also, the last equality reduces to $\nabla^2 f(\mathbf{x}) = \frac {\mathbf{A}^\mathrm{T} \mathbf{A} - \nabla f(\mathbf{x})^\mathrm{T} \nabla f(\mathbf{x})} {\|\mathbf{Ax}\|_2}$ - 1 My advice is to do it first without the square root. Given your $f,$ define $g = f^2$ and find the gradient and Hessian. Just to confirm your suspicions, the gradient really is a vector, the Hessian really is a square (symmetric) matrix. – Will Jagy Jul 25 '12 at 23:26 1 Try squaring and expanding $f^2(x+\delta)$. The gradient and Hessian (of $f^2$) should be fairly clear. – copper.hat Jul 25 '12 at 23:27 Also, do this in agonizing detail for the case of just two variables, you could call them $x_1,x_2$ if you like, but you could also stick with $x,y.$ The gradient becomes a vector with just two functions as entries, the Hessian a 2 by 2 matrix. You are getting a bit lost in the formalism. – Will Jagy Jul 25 '12 at 23:38 1 It looks to me like everything is fine except that the numerator should be $A^TA - (\nabla f)(\nabla f)^T$. (Also, $\nabla^2$ usually denotes the Laplacian, which is the trace of the Hessian.) – Rahul Narain Jul 26 '12 at 4:46 ## 1 Answer It is easier to work with $\phi(x) = \frac{1}{2} f^2(x)$. Just expand $\phi$ around $x$. $\phi(x+\delta) = \frac{1}{2} (x + \delta)^T A^T A (x + \delta) = \phi(x) + x^TA^TA \delta + \frac{1}{2} \delta^T A^T A \delta$. It follows from this that the gradient $\nabla \phi(x) = A^T A x$, and the Hessian is $H = A^TA$. To finish, let $g(x) = \sqrt{2x}$, and note that $f = g \circ \phi$. To get the first derivative, use the composition rule to get $D f(x) = Dg(\phi(x)) D \phi(x)$, which gives $Df(x) = \frac{1}{\sqrt{2 \phi(x)}} x^T A^T A = \frac{1}{\|Ax\|} x^T A^T A$. Let $\eta(x) = \frac{1}{\|Ax\|}$, and $\gamma(x) = x^T A^T A$, and note that $D f(x) = \eta(x) \cdot \gamma(x)$, so we can use the product rule. Let $h(x) = Df(x)$ then the product rule gives $D h(x) (\delta) = (D \eta(x) (\delta)) \gamma(x) + \eta(x) D \gamma(x) (\delta)$. Expanding this yields: $Dh(x)(\delta) = (- \frac{1}{\|Ax\|^2} \frac{1}{\|Ax\|} x^T A^T A \delta) x^T A^T A + \frac{1}{\|Ax\|} \delta^T A^T A$. Noting that $x^T A^T A \delta = \delta^T A^T A x$, we can write this as: $$Dh(x)(\delta) = \delta^T(\frac{1}{\|Ax\|} A^T A - \frac{1}{\|Ax\|^3 } A^T A x x^T A^T A),$$ or alternatively: $$D^2 f(x) = \frac{1}{\|Ax\|} A^T A - \frac{1}{\|Ax\|^3 } A^T A x x^T A^T A .$$ The only difference with the formula given in the question is that the latter dyad was written incorrectly (instead of the dyad $g g^T$, you have the scalar $g^T g$). - This is helpful but not complete. To get to the answer, one now needs the chain rule to find the gradient and Hessian of $f(\mathbf{x})=2\sqrt{\phi(\mathbf{x})}$. – AnonSubmitter85 Jul 26 '12 at 2:10 @AnonSubmitter85: I elaborated my answer a little more. – copper.hat Jul 26 '12 at 5:36 Thank you. This has helped me tremendously. – AnonSubmitter85 Jul 26 '12 at 18:18 You're very welcome. – copper.hat Jul 26 '12 at 18:30

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http://mathoverflow.net/questions/51494?sort=newest

## Why the name ‘separable’ space? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) It is well known that a separable space is a topological space that has a countable dense subset. I am wondering how is this related to the name 'separable'? Any intuition where the name come from? - 27 My understanding is it comes from the special case of R, where it means that any two real numbers can be separated by, say, a rational number. – Qiaochu Yuan Jan 8 2011 at 20:39 Ha, that's a nice observation! – minimax Jan 8 2011 at 20:47 Maybe, it refers to the separation of points by means of countable collection of small open sets (where separation means that for any two points we may find their disjoint neighborhoods in our collection). It is probably not formally equivalent to the existence of a countable dense subset, but something similar. – Fedor Petrov Jan 8 2011 at 22:23 ## 3 Answers As far as I know the word separable was introduced by M. Fréchet in Sur quelques points du calcul fonctionnel, Rend. Circ. Mat. Palermo 22 (1906), 1-74. The paper can be obtained via this link (Springer). It's the famous paper in which he introduced metric spaces. He considers first slightly more general objects which he calls classes (V): where (V) stands for voisinage — neighborhood. Remark: Metrics are introduced under the name écart in no 49 on page 30. It is peculiar that the symmetry condition is not explicitly mentioned but it seems to be understood as Fréchet immediately mentions that metric spaces generalize classes (V) cf. no 27 on page 17f. However, I couldn't find an instance where he actually uses it, he is always careful to respect the order — I may have missed something since I haven't read the paper in detail. I quote the relevant passage [from no 37 on page 23f]: Nous appellerons ensuite classe séparable une classe qui puisse être considérée d'au moins une façon comme l'ensemble dérivé d'un ensemble dénombrable de ses propres éléments. [...] Ceci étant, nous nous bornerons maintenant à l'étude des classes (V) NORMALES, c'est-à-dire parfaites, séparables et admettant une généralisation du théorème de CAUCHY. Cette limitation n'a du reste rien d'artificiel, elle provient directement de la comparaison des classes (V) avec les ensembles linéaires [...] [...] Passons maintenant aux classes séparables. On peut qualifier ainsi les ensembles linéaires en considérant la droite indéfinie comme l'ensemble dérivé de l'ensemble des points d'abscisses rationnelles. Mais il n'en est pas de même pour toute classe parfaite (V). I am unable to translate this in a reasonable way (but see Amit's comment below for a translation). Very roughly: Fréchet defines separable spaces as we do it today and says that in the following he will restrict attention to complete, perfect and separable metric spaces. The last quoted paragraph indeed confirms Qiaochu's comment. - 5 We will henceforth call a class separable if it can be considered in at least one way as the derived set of a denumerable set of its elements. [...] We restrict ourselves now to the study of (V) NORMAL classes, that is to say perfect, separable and admitting a generalization of Cauchy's theorem. This limitation has nothing artificial, it comes directly from the comparison of the classes (V) with the linear sets. [...] We pass now to separable classes. [Not sure about this part] as the derived set of the set of rational points. But it isn't so for all perfect (V) classes. – Amit Kumar Gupta Jan 9 2011 at 17:57 Thank you very much! I'd translate the second to last sentence by "One may describe the 'subsets of the real line' (ensembles linéaires) in this way by considering the 'real line' (droite indéfinie) as the derived set of the set of the points with rational abscissae." – Theo Buehler Jan 9 2011 at 19:02 ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. Well "séparer" means just "disjoin", "split up" etc. There are few meanings for "séparation" in french. For exemple "séparer par des fonctions". If you have a space $X$ and a set of functions ${\cal F}$ from $X$ to ${\bf R}$ (or anything else), you say that "${\cal F}$ sépare les points de $X$" iff for two different point $x$ and $x'$ there exists a function $f \in {\cal F}$ such that $f(x) \neq f(x')$, this is a common use of the word "séparer" in french, nothing mysterious. And this vocabulary can be applied to any analogous situations, in topology or whatever else context. I heard the first time this wording (when I was a student) in the case I mention above, long before I have heard using this wording in a topology course. - This does not explain the terminological connection between "separable" and "existence of countable dense subset" – Martin Brandenburg Jan 9 2011 at 10:02 No, I just mean that separable has been used for many different situations, the common origin is separability of points by a mean or another. I guess there is no more to look for. – Patrick I-Z Jan 9 2011 at 12:37 I have a question: do you use in english the verb "to separate" in the same broad meaning than in french? – Patrick I-Z Jan 9 2011 at 12:43 2 Concerning your specific example: I have seen the phrase "$\mathcal{F}$ separates the points of $X$", yes. I believe it's standard terminology. – Zhen Lin Jan 9 2011 at 13:05 According to this part of Hausdorff's collected works, the name "separable" was coined by Fréchet. Hausdorff writes this denotation wouldn't be very suggestive but established. The earliest use in the German Zentralblatt, where the word occurs in a review by Hahn, dates back to 1918. I am sure one can find the first use of the word in the given context in Fréchet's works but I doubt that he will explain his motivation for the use of this particular word there. So we will (presumably) never know... -

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http://quant.stackexchange.com/questions/7082/are-there-any-derivatives-which-pay-amount-ap-b2-c-where-p-is-the-price

# Are there any derivatives which pay amount $a(p-b)^{2}-c$ where $p$ is the price of underling asset? Are there any derivatives which pay amount $a(p-b)^{2}-c$ where $p$ is the price of underling asset ? (or in the case of options $max(0,a(p-b)^{2}-c)$) I'm not very strict here but I only want to know if there are any derivatives which profile of profits is quadratic function of price of underling assets (quadratic on some subset of set of all $p$ )? - 2 They're sometimes called "power contracts" and though I've known them to be coded into the pricing libraries of at least 2 big investment banks, I've never seen one on the books. – Brian B Jan 23 at 19:16 @Brian B very interesting, thanks – Qbik Jan 23 at 21:57 ## 3 Answers You can ask for a quote from a bank as I am sure they will create it for you. If you want to create this kind of payoff yourself, you can use the following paper from Peter Carr where he introduces the spanning formula for replicating any twice differentiable payoff. http://www.math.nyu.edu/research/carrp/papers/pdf/twrdsfig.pdf - Variance swaps are arguably a little like that. As mentioned, power contracts too, but sometimes only by virtue of correlation/cross-gammas between volatilities and underlyings. I haven't personally seen a contract with an exponent in it. - • There are probably currently none out there right now with the exact same pay off structure (not sure how you would set b and c and what rational you would apply thus I said variants may exist but probably not the exact same ones) • Having said that you can request a bank to price you ANY derivative you like, they will quote you for sure. You better be very confident that you are able to price them yourself correctly or you will get ripped off royally. -

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http://math.stackexchange.com/questions/295516/complex-numbers-negative-absolute-value-radius

# Complex numbers: negative absolute value (radius)? I need to find a complex number that represented by the following poler representation: ($\mathbb r$, $\theta$) = ($-5$, $\pi \over 2$) My question is: how is a negative radius (absolute value) possible? What does it mean \ sign? - It is not possible, at least not under the usual, standard definitions. Check this carefully. – DonAntonio Feb 5 at 15:37 I've seen things like this before; when I've encountered it, it has just meant to take the radius in the opposite direction. So, normally we would expect this to correspond to the point $5i$ in the complex plane (or $(0,5)$ if we identify it with $\Bbb R^2$. However, with the negative, it will correspond to $-5i$. I'm not going to say this is how it is for you, just when I've encountered it previously. – Clayton Feb 5 at 15:39 @Clayton The $\theta$ signs the angle with the positive X Axis, doesn't it? – user1798362 Feb 5 at 15:44 @user1798362: I'm not sure what you mean by signs the angle with the positive $x$-axis. The angle begins at $0^\circ$ there and rotates counterclockwise through an angle of $\pi/2$ in your case. – Clayton Feb 5 at 15:47 @Clayton Ok, thank you!! I have addition question and I'll better and it here instead of another ask: if $\theta$ = 45, than `a = b` in case $\mathbb z$ = `a + bi`? – user1798362 Feb 5 at 15:49 show 7 more comments ## 1 Answer $$z=r\cos\theta+ir\sin\theta$$ -

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http://math.stackexchange.com/questions/252908/maximal-unramified-extension-of-mathbbf-pt

# Maximal Unramified Extension of $\mathbb{F}_p((t))$ The maximal unramified extension of $\mathbb{Q}_p$ can be described quite explicitly: add all roots of unity of order prime to $p$. This is done by the correspondence between finite unramified extensions of $\mathbb Q_p$ and finite extensions of $\mathbb F_p$. Similarly, can one find an explicit description for the maximal unramified extension of $\mathbb F_p((t))$? - ## 1 Answer For $\mathbb{F}_p((t))$, this is in a sense even simpler: the maximal unramified extension is the direct limit over $\mathbb{F}_{p^n}((t))$, $n\in \mathbb{N}$. This is very easy to see, since unramified extensions of a local field are in bijection with extensions of the residue field. In other words, the maximal unramified extension is again generated by roots of unity of order prime to $p$. It's good fun to try and classify, using class field theory, the maximal tamely ramified extension of $\mathbb{F}_p((t))$. I once did that in my Part III essay if you are interested. - 1 Technically, $\overline{\mathbb{F}_p}((t))$ is transcendant over $\mathbb{F}_p((t))$, you want $\cup F_{p^n}((t))$ instead. – mercio Dec 7 '12 at 9:08 @mercio, you are right, of course, I have corrected it. Thanks. – Alex B. Dec 7 '12 at 9:15 Your answer is correct. However you omit the fact that $\mathbb{F}_p((t))$ is not a perfect field. So there could be unramified, non-separable extensions; your approach is not treating them. Luckily such extensions do not exist, but this is not easy to prove. In fact the valuation theory of $\mathbb{F}_p((t))$ is more complicated than that of $\mathbb{Q}_p$. – Hagen Dec 7 '12 at 11:55 @Hagen I don't think I omit anything. It is in fact quite easy to prove that there is a bijection between unramified extensions of a local field (with perfect residue field) and extensions of the residue field, and this does take care of any separability issues. In one direction, given an unramified extension of local fields, reduce modulo the (unchanged) uniformiser to get an extension of residue fields. Conversely, given an extension of residue fields, take a primitive element of this extension, lift it using Hensel, adjoin the lift to the local field to get an unramified extension. – Alex B. Dec 7 '12 at 15:24 Ok, but suppose there exists a proper, finite extension $K$ of $\mathbb{F}_p((t))$ such that $e=1$ (ramification index) and $f=1$ (inertia degree). Then the correspondence between unramified extensions and extensions of the residue field is not injective. Such an extension must necessarily be inseparable. – Hagen Dec 7 '12 at 16:14 show 2 more comments

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http://crypto.stackexchange.com/questions/710/how-to-forge-schnorr-signatures-if-you-can-guess-the-challenge/713

# How to forge Schnorr signatures if you can guess the challenge Underlying the Schnorr signature is an identification protocol: let $G$ be a cyclic group where discrete log is "hard" and choose $g$ as a generator of $G$. Now have Alice pick a random (secret) exponent $s$ and publish $v=g^s$. To identify herself to Bob, it goes like this: 1. Alice chooses a random exponent $r$ and sends $g^r$ (commits $r$) 2. Bob sends random exponent $e$ (challenge $e$) 3. Alice sends $r + se$ 4. Bob makes sure $g^r = g^{r + se}v^{-e}$ That's all well and good, but what if a cheating prover can guess $e$? Then how does he fool Bob? - ## 4 Answers In the other answers, you'll find how to simulate a proof if you know $e$. This answer is meant to provide some "color commentary" on the other answers. It is a companion piece. Notation • In step 1, Alice sends $g^r$. Call this value $a=g^r$. • In step 3, Alice sends $r+se$. Call this value $b=r+se$. • In step 1-3, one value is sent in each step: {$a,e,b$}. We'll call these three values a transcript for public key $v$. • In step 4, with the new notation, Bob checks: $a=g^bv^{-e}$. Basic question With this notation, it may be easier to see how knowing $e$ can help. Generate a random $b$ and then compute $a$ so that the equation holds and {$a,e,b$} will accept. Another thing to note Other answers noted that since you choose $a$ directly, you do not really know an $r$ such that $a=g^r$ (by the discrete log problem). This is one thing that distinguishes a true proof from a simulated one. A second thing to note, which is very important, is that to simulate a proof, you start by computing (choosing) $b$ and then you compute $a$ using $b$ and $e$. In a real execution, you compute everything in order. The only way to forge? The other answers have shown a way to simulate the proof. Is it the only way? For example, is it possible to choose an $a$ value knowing $e$ and then compute the right $b$ to make {$a,e,b$} accept (short of knowing $s$)? The answer is no. Pretend you have a black box that could do this: it takes {$a,e$} as input and returns the proper $b$ without knowing $s$. If such as box existed, you could query {$a,e_1$) and get $b_1$, and then query {$a,e_2$} with the same $a$ and different $e$, and get $b_2$. However one can verify that this is sufficient to compute $s$: in fact, $s=\frac{b_1-b_2}{e_1-e_2}$. So there is a contradiction: the box doesn't know $s$ by definition and yet it does "know" $s$ (in the sense that it can compute it). Therefore, by contradiction, such a box cannot exist, and neither can a forgery {$a,e,b$} where $a$ and $e$ are chosen. Enforcing only true transcripts If we put all of the above together, it basically says that if {$a,e,b$} accepts, it must have either been computed by someone who truly knows $s$, or it was computed backward by choosing/knowing $b$ and $e$ before computing $a$. Can we eliminate the second case? One way is to be Bob. Another might be to be there in person to see that $a$ is sent before $e$: however do you really know that Alice doesn't know $e$? Can you really be sure? The answer is yes. If you can ensure $a$ is computed before $e$, then we are done. A simple technique (called Fiat-Shamir) is to set $e=\mathcal{H}(a)$, where $\mathcal{H}$ is a hash function (technically a random oracle). If you choose/know $e$ first, you cannot find an $a$ that will hash to it (by preimage-resistance). - If Alice guesses $e$ then she chooses a random value $x$ and computes $h = g^x v^{-e}$, a value which she sends to Bob at step 1. At step 3, Alice sends $x$. When Bob does step 4, he recomputes $g^x v^{-e}$ and finds $h$, and he is happy. However, Alice does not know $s$. The "commitment" at step 1 is a way for Alice to say: "I know a $r$ corresponding to this $g^r$, which allows me to do the computation for any challenge $e$". But if Alice can guess the challenge in advance, she needs not really know $r$. Here, Alice sends a value $h$ which is part of the group and thus is equal to $g^r$ for some value $r$, but Alice does not know that $r$. - Well, if fake-Alice guesses the challenge exponent $e$ in step 1, then she can guess the value of $v^{-e}$. That means she can pick an arbitrary value to stand in for $r+se$, compute $g^{r+se}v^{-e}$, and send that as her commitment in step 1. Then, assuming Bob sends the guessed exponent in step 1, fake-Alice sends the value $r+se$ that she picked above. That means that, in step 4, Bob will validate that $g^r = g^{r+se}v^{-e}$, because he is actually validating ${step1} = g^{step3}v^{-e}$, and fake-Alice picked the $step1$ and $step3$ values to make this equation work. - Suppose that we have Eve, that knows what $e$ is going to be, and does not need to know the prover's private key $a$, just the public one $v$. She then sends $g^k \cdot v^{-e}$ as her first "move", where she can choose her own $k$ (you can modify the $k$ in different plays to make it all look nice and random...). The verifier sends $e$ as expected, of course, and finally Eve sends $k$ as her final move. The verifier only checks that her last move times $v^{-e}$ equals her first move (see step 4.) and this is true by construction. Note that the verifier has no way of seeing that the first move wasn't of the form $g^r$ for some $r$ known to the sender, or that $r + se$ is sent, etc. He only sees 2 numbers by the prover, that have to satisfy a (known in advance) relation.... -

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http://physics.stackexchange.com/questions/27241/what-shannon-channel-capacity-bound-is-associated-to-two-coupled-spins/27243

# What Shannon channel capacity bound is associated to two coupled spins? The question asked is: What is the Shannon channel capacity $C$ that is naturally associated to the two-spin quantum Hamiltonian $H = \boldsymbol{L\cdot S}$? This question arises with a view toward providing a well-posed and concrete instantiation of Chris Ferrie's recent question titled Decoherence and measurement in NMR. It is influenced too by the guiding intuition of Anil Shaji and Carlton Caves' Qubit metrology and decoherence (arXiv:0705.1002) that "To make the analysis [of quantum limits] meaningful we introduce resources." And finally, it is reasonable to hope that so simple and natural a question might have a rigorous answer that is simple and natural too---but to the best of my (imperfect) knowledge, no such answer is given in the literature. ## Definitions Let Alice measure-and-control by arbitrary local operations a spin-$j_\text{S}$ particle on a local Hilbert space $\mathcal{S}$ having $\dim \mathcal{S} = 2j_\text{S}+1$, upon which spin operators $\{S_1,S_2,S_3\}$ are defined satisfying $[S_1,S_2] = i S_3$ as usual. Similarly let Bob measure-and-control by arbitrary local operations a spin-$j_\text{L}$ particle on local Hilbert space $\mathcal{L}$ having $\dim \mathcal{L} = 2j_\text{L}+1$ upon which spin operators $\{L_1,L_2,L_3\}$ are defined satisfying $[L_1,L_2] = i L_3$ as usual. Let the sole dynamical interaction between the spins — and thus the primary resource constraint acting upon the communication channel — be the Hamiltonian $H = \boldsymbol{L\cdot S}$ defined on the product space $\mathcal{S}\otimes \mathcal{L}$. Further allow Bob to communicate information to Alice by a classical communication channel of unbounded capacity, but let Alice have no channel of communication to Bob, other than the channel that is naturally induced by $H$. Then the question asked amounts to this: what is the maximal Shannon information rate $C(j_\text{S},j_\text{L})$ (in bits-per-second) at which Alice can communicate (classical) information to Bob over the quantum channel induced by $H$? ## Narrative In practical effect, this question asks for rigorous and preferably tight bounds on the channel capacity associated to single-spin microscopy. The sample-spin $S$ can be regarded as a sample spin that can be modulated in any desired fashion, and the receiver-spin $L$ can be regarded variously as a tuned circuit, a micromechanical resonator, or ferromagnetic resonator, as shown below: The analysis of the PNAS survey Spin Microscopy's Heritage, Achievements, and Prospects (2009) can be readily extended to yield the following conjectured asymptotic form: $$\lim_{j_\text{S}\ll j_\text{L}} C(j_\text{S},j_\text{L})=\frac{j_\text{S}\,(j_\text{L})^{1/2}}{(2\pi)^{1/2}\log 2}$$ Note in particular that the dimensionality of Bob's receiver-spin Hilbert space $\mathcal{L}$ is $\mathcal{O}(\,j_\text{L})$; thus a Hilbert-space having exponentially large dimension is not associated to Bob's receiver. However it is perfectly admissible for Alice and Bob to (for example) collaborate in squeezing their respective spin states; in particular the question is phrased such that Alice may receive real-time instruction of unbounded complexity from Bob in doing so. ## Preferred Form of the Answer A closed-form answer giving a tight bound $C(j_\text{S},j_\text{L})$ is preferred, however a demonstration that (e.g.) $\mathcal{O}(C)$ is given by some closed asymptotic expression (as above) is acceptable. It would also be very interesting, both from a fundamental physics point-of-view and from a medical research point-of-view, to have a better appreciation of whether the above conjectured capacity bound on spin imaging and spectroscopy can be substantially improved by any means whatsoever. - 3 I downvoted this question as it seems like a nice question but it is not written in the clearest way. You talk a lot around the question and I am not sure why you included that image. I think if it is stated in a simple motivation-definition-question format it'd be a nice question. I am still not completely clear on why one should care about such an Hamiltonian for greater than spin-1/2 systems. I'll be happy to remove my downvote if you think I am being unreasonable. – Matty Hoban Nov 7 '11 at 0:28 Hoban, downvotes trouble me far less than indifference! :) Yes, physics hovers uneasily between beautiful mathematics and beautiful experiments. Spins larger than 1/2 are motivated mainly experimentally, as mentioned in my comment appended to Aram Harrow's answer (below): it is that nature provides us with large-spin particles (in the form of ferromagnets) having a spatial spin density billions of times greater than can be achieved by (e.g.) trapped-ion condensates. The price to be paid for this spin density is that the effective Hilbert space dimension is far lower than said condensates. – John Sidles Nov 7 '11 at 0:57 ## 2 Answers This is an open question. The capacity of some related Hamiltonians was computed in quant-ph/0207052, and an upper bound was derived in 0704.0964, but the Hamiltonian you describe is an example of one for which we don't know the exact answer. However, quant-ph/0207052 also contains a conjecture (eq 35) about the capacity you're interested in. Their conjecture is supported by numerical experiments, but they can't rule out that some block coding strategy could do better. Edited in light of our conversation this afternoon: From our discussion, it sounds like you've proven a lower bound of $\Omega(j_S j_L)$ using some cat-like states, based on principles similar to those in quant-ph/0605013. I think I can prove an asymptotically matching upper bound of $O(j_S j_L)$. So the exact constant is still open (and I think a hard problem), but at least we know the scaling. For this bound, first argue that the interaction can be modelled as $2j_S$ qubits for Alice and $2j_L$ qubits for Bob, each in a symmetric state. To prove an upper bound, we can relax the restriction that the qubits be in a symmetric state. Then the Hamiltonian you describe can be expressed as a sum of interactions between each qubit of Alice's and each qubit of Bob's, each of which has constant strength. These two-qubit Hamiltonians have capacities upper-bounded by a constant. One reference for this last claim is quant-ph/0205057, but it was probably known earlier. - Aram, these are wonderful references for which I am very grateful! With specific regard to quant-ph/0207052 (eq. 35) this result (AFAICT) applies solely to coupled qubits (that is, spin-1/2 particles), whereas when we look into the coming decades of quantum spin microscopy, we foresee spins coupled to ferromagnetic spheres having spin-j ~ 10^5. These spheres have (effectively) a state-space of dimensionality much larger than one qubit, yet much small than 10^5 qubits; in brief their state-space is large enough to squeeze, but too small for (e.g.) quantum computation. More please! :) – John Sidles Nov 7 '11 at 0:05 Aram, as a followup, I am beginning to appreciate more of the reasons why this simple-to-state question is difficult to answer. For example, the case $j_S = 1/2$ and $j_L \gg 1$ large can be regarded as effectively describing a cavity QED experiment, with $\mathcal{S}$ the state-space of the transmitting two-level atom and $\mathcal{L}$ the state-space of the receiving high-quantum-efficiency detector; these systems are known to be tough to analyze theoretically and tough to build experimentally. Conversely, this means that a closed-form tight capacity bound would be quite valuable to know. – John Sidles Nov 7 '11 at 10:48 @AramHarrow: This update sounds an awful lot like metrology. It seems like the Heisenberg limit may apply. – Joe Fitzsimons Nov 10 '11 at 9:50 – Aram Harrow Nov 21 '11 at 8:42 @AramHarrow: Thank you for these wonderful comments! I've been busy running an experiment for the past couple of weeks (hence the slowness in responding), but I concur with the general thrust of your remarks, and have thereby gained a better appreciation of (what is likely to be) the order of the upper and lower bounds to the channel capacity that are (respectively) achieved by cat-state sensors and coherent-state sensors, and thus have rated it as an "answer". – John Sidles Dec 14 '11 at 19:14 Aram's answer seems perfect, but since you are also asking about the case for higher dimensional systems, let me add that there is a simply way to get somewhat non-trivial upper and lower bounds on $C(j_S,j_L)$. As a lower bound, you can simply synthesize an arbitrary gate which implements communication between the quantum systems (for an explicit algorithm on constructing arbitrary gates see Nielsen et al, Phys. Rev. A 66, quant-ph/0109064). A non-trivial upper bound is given by the Margolus-Levitin theorem. Their paper (quant-ph/9710043) gives the maximum number of orthogonal states a quantum system can pass through in a given period of time, or conversely, a minimum time (as a function of energy) required to from the initial state to an orthogonal state (which is necessarily a lower bound on the time required to perfectly transmit one bit). - Thank you Joe ... it will take a couple of days to process these references. A challenge with the Nielsen construction is "gate to rate", that is, the construction does not limit the energy associated to Hamiltonian resources; thus is not evident how to extract a channel capacity bound of any kind. A challenge with the Margolus-Levitan theorem is that the operations associated to the bound are (apparently) not local; thus the upper bound (that I get from applying it) is unphysically optimistic. Perhaps there are ways around these obstructions; this is what will take awhile to think about. – John Sidles Nov 7 '11 at 22:03 Progress report: preliminary efforts to construct concrete two-spin communication channels that approach the Margolus-Levitin bound have (so far) foundered on the local-operation restriction. This contributes to an appreciation (on my part) that Aram Harrow's answer is correct in asserting that "This [two-qudit channel capacity] is an open question", largely because the locality-of-operation restriction has deep-and-subtle consequences. One emerging benefit is an increased appreciation that fundamental QIT constraints are associated to single-photon sources and efficient local detectors. – John Sidles Nov 8 '11 at 13:18 2 The Bravyi paper I mentioned (0704.0964) can be thought of as applying the Margolus-Levitin principle to upper bound the entangling rate. The bound is constant * norm of Hamiltonian * log dimension. (Note that this is an upper bound for the communication rate. Also dimensional factors are inevitable for both communication rate and entanglement rate, because transmitting an n-bit message can mean a single pi/2 rotation.) Unfortunately his argument is incomplete because we don't know how to control how much ancillas can increase the capacity. – Aram Harrow Nov 9 '11 at 7:55 1 @AramHarrow: Good point. However, it seems like you can bound the effective dimensionality via the number of splittings in the eigenvalues of local Hamiltonians. – Joe Fitzsimons Nov 9 '11 at 8:17 1 @AramHarrow: To explain what I mean: Imagine you have two spin-1/2 systems. Then your coupling induces at most 4 distinct eigenspaces which pick up relative phase, independent of the number of ancillae, or how they are prepared prior to coupling. Given that the information transfer in periods of free evolution is determined entirely by the relative phases between these 4 spaces, this is identical to what can be achieved with two spins. The only subtlety then comes from how you can combine multiple uses of such a channel. – Joe Fitzsimons Nov 9 '11 at 8:30 show 4 more comments

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http://mathoverflow.net/questions/77650/a-volterra-type-equation

## A Volterra-type equation ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Consider the following integral equation $\phi(x) = f(x) + \frac{1}{x}\int_0^x N(x,y)\phi(y)\;dy$, where $f$ and $N$ are continuous and bounded functions. Are solutions $\phi$ of the above equation unique? If so, can one get an estimate of the form $\sup_{(0,x)} |\phi| \leq C \sup_{(0,x)}|f|$ ? Additional info: Assume that $\phi(0)=\phi(1)=f(0)=0$ and that $\|N\|_\infty\geq 1$. I am only interested in a solution (or lack thereof) on the interval $[0,1]$. As a note, without the $\frac{1}{x}$ multiplier, this is a Volterra equation of the second kind and existence/uniqueness of a solution $\phi$ is well-known and the above estimate is indeed satisfied. - ## 1 Answer Not in general, since if $N(x,y)=a>1$ then the equation with $f=0$ has the solution $\phi(x)=x^{a-1}$. If, however, $N(0,0)<1$ (assuming as in the question that $N$ is continuous) then one can use the Banach fixed-point theorem (on short intervals) to get a unique solution. - I assume you mean $|N(0,0)| < 1$. You can of course get a stronger existence result if you have $\| N \|_{\infty} < 1$. – Christopher A. Wong Oct 10 2011 at 5:45 Thanks for the replies. I had noticed the contraction argument for $\|N\|_\infty < 1$, but the case I am interested in, I have $\|N\|_\infty > 1$, but I also have $\phi(0)=\phi(1)=0$ (and of course then $f(0)=0$). I should have added these to the problem statement (which I will do now). I would be happy with either uniqueness or non-existence. – Jeff Oct 10 2011 at 13:08

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http://math.stackexchange.com/questions/40849/how-to-check-if-a-symmetric-4-times4-matrix-is-positive-semi-definite

# How to check if a symmetric $4\times4$ matrix is positive semi-definite? 1. How does one check whether symmetric $4\times4$ matrix is positive semi-definite? 2. What if this matrix has also rank deficiency: is it rank 3? - 2 You can use the determinant criterion: the upper-left $1\times 1$, $2\times 2$, $3\times 3$ and $4 \times 4$ squares should all have non-negative determinant. – Yuval Filmus May 23 '11 at 16:00 1 But that determinant criterion isn't enough in general, as $\begin{bmatrix}0&0&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&-1\end{bmatrix}$ shows. – Jonas Meyer May 23 '11 at 16:07 2 @Yuval - I believe the determinant criterion holds for positive definite matrices but not necessarily for positive semidefinite ones. In other words, if some of the principal minors are zero, it does not necessarily imply the matrix is positive semidefinite. – svenkatr May 23 '11 at 16:16 3 In that case, you can add $\epsilon > 0$ (to the diagonal) and then rerun all your computations (just when the matrix doesn't have full rank). A suitable $\epsilon$ can be found by looking at the magnitude of the entries (we want to guarantee that we don't miss any small negative eigenvalue). – Yuval Filmus May 23 '11 at 16:48 ## 5 Answers Since the matrix is symmetric, the eigenvalues will be real. Calculate the eigenvalues and see if they are all $\geq 0$. If this is true,the matrix is positive semidefinite. - 1 You don't even need the eigenvalues. Once you have the polynomial invariants (coefficients of the characteristic polynomial), you can use Descarte's rule of signs. – Willie Wong♦ Aug 24 '11 at 16:22 1 Even computing the characteristic polynomial here is overkill. Robert's suggestion takes less effort... – J. M. Aug 25 '11 at 4:21 Another method is to check there are no negative pivots in row reduction (after taking into account the possibility of 0's on the diagonal). The procedure can be written recursively as follows: 1) If $A$ is $1 \times 1$, then it is positive semidefinite iff $A_{11} \ge 0$. Otherwise: 2) If $A_{11} < 0$, then $A$ is not positive semidefinite. 3) If $A_{11} = 0$, then $A$ is positive semidefinite iff the first row of $A$ is all 0 and the submatrix obtained by deleting the first row and column is positive semidefinite. 4) If $A_{11} > 0$, for each $j > 1$ subtract $A_{j1}/A_{11}$ times row 1 from row $j$, and then delete the first row and column. Then $A$ is positive semidefinite iff the resulting matrix is positive semidefinite. - As stated above, Sylvester's criterion doesn't work in this case, so you can't simply check the four leading principal minors. However, it does suffice to check that all 15 of the principal minors are nonnegative. See here for a reference. Another basic approach involves symmetric row reduction. This involves operations of the following type: 1. Perform a row operation, and then 2. Immediately perform the corresponding column operation. For example, you can multiply any row by a constant, as long as you immediately multiply the corresponding column by a constant. Note that this multiplies the diagonal entry by the square of the constant. Using these operations, you can use a variant of Gaussian elimination to reduce any symmetric matrix to a diagonal matrix with 1's, 0's, and -1's along the diagonal. A matrix is positive semidefinite if and only if the resulting diagonal entries are all 0's and 1's. - Let's say your matrix is $A$. You can check the eigenvalues. If all eigenvalues $\geq 0$, the matrix is positive semi-definite (if all eigenvalues $>0$ it is positive definite). It might be possible to use the Gershgorin circle theorem instead of calculating the eigenvalues explicitly. If all the diagonal elements are positive and are larger than or equal to the sum of the absolute values of the other elements in same row (or column) (for every diagonal element), then the matrix is positive semi-definite. You can try to find a simpler semi-definite matrix $B$ such that $B^2 = A$ ($B$ is unique). This is in general done using the diagonalization of the matrix, so it will probably be easier just calculating the eigenvalues. You can not use a modification of Sylvester's criterion ("all leading principal minors are non-negative") to determine positive semidefiniteness. If $A$ is $4 \times 4$ and rank 3, it has 0 as an eigenvalue (since there exists a vector $v$ such that $Av = 0v$). This does not affect the positive semi-definiteness of the matrix, but it will not be positive definite. - One quick first check, if any element $A_{ii}$ on the diagonal is negative, the matrix has a negative eigenvalue (for real symmetric matrices only of course) -

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http://physics.stackexchange.com/questions/27279/a-nice-overview-and-maybe-derivation-of-the-poincare-transformations-of-the-ve?answertab=votes

# A nice overview (and maybe derivation) of the Poincaré transformations of the Vector Spherical Harmonics With $Y_{lm}(\vartheta,\varphi)$ being the Spherical Harmonics and $z_l^{(j)}(r)$ being the Spherical Bessel functions ($j=1$), Neumann functions ($j=2$) or Hankel functions ($j=3,4$) defining $$\psi_{lm}^{(j)}(r,\vartheta,\varphi)=z_l^{(j)}(r)Y_{lm}(\vartheta,\varphi),$$ what are representations of the Poincaré transformations applied to the Vector Spherical Harmonics $$\vec L_{lm}^{(j)} = \vec\nabla \psi_{lm}^{(j)},\\ \vec M_{lm}^{(j)} = \vec\nabla\times\vec r \psi_{lm}^{(j)},\\ \vec N_{lm}^{(j)} = \vec\nabla\times\vec M_{lm}^{(j)}$$ ? Does any publication cover all Poincaré-transformations, i.e. not only translations and rotations but also Lorentz boosts? I'd prefer one publication covering all transformations at once due to the different normalizations sometimes used. - – Tobias Kienzler Mar 1 '12 at 12:19 ## 1 Answer The problem with Poincaré group is in the fact that it is not compact. That's why this question is non-trivial. Though, properly formulated search gives few papers on this topic. Try to find the answer in this paper http://arxiv.org/abs/math-ph/0507056 . The paper itself may be not that interesting, but there is a nice introduction with a number of useful references. -

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http://mathoverflow.net/questions/61275?sort=newest

## Analysis and finitely generated groups ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Dear all, this is perhaps a bit a vague question, but some references would already be very helpfull. So let $G$ be a finitely generated group and choose some finite set of generators. This allows to define the length of a group element as usual, by the minimal number of generators needed to write it as a product of these generators. Using this length you can define all sort of analytic functions by replacing the "n" in various summation formulas by the $\mathrm{Length}(g)$ and the summation is now over the group elements $g$. To have a concrete example in mind, the "exponential series" is now e.g. \begin{equation} \mathrm{lexp}(z) = \sum_{g \in G} \frac{z^{\mathrm{Length}(g)}}{\mathrm{Length}(g)!}. \end{equation} Since for a given length there are at most exponentially many group elements, the length-exponential function is entire. For $G = \mathbb{Z}$ and $1$ as generator this reproduces the usual exponential series up to a factor $2$ as negative and positive $n \in \mathbb{Z}$ contribute with the same power of $z$. So my question is: what is known about the analytic features of such functions (depending on the choice of generators, depending on the group itself, etc). I guess there should be some literature on the market, but I'm really not in this buisness... - 1 While I've never seen the particular example you give, there is a vast literature on related generating functions (usually called growth functions or something to that effect). Have look at Chapters VI and VII of Pierre de la Harpe's book Topics in geometric group theory (Chicago lectures in mathematics) for a discussion of many examples and a survey of the main results. The extensive reference list in the book should provide many pointers to the literature. – Theo Buehler Apr 11 2011 at 10:52 @Theo: thanks a lot. I will check out this book in the library, googlebook gives only a couple of pages. But they look nice indeed. – Stefan Waldmann Apr 12 2011 at 8:31 This sounds highly related to Geometric Group Theory. I would check in books on that area, in particular De La Harpe's Geometric Group Theory, where he actually computes some of these if I recall correctly. – Benjamin Hayes May 7 2011 at 15:03 Dear Benjamin: thanks! I had a look at the book (Theo mentioned it already) and it indeed looks very inspiring. It seems that I have to work a bit more in that direction. Very interesting... – Stefan Waldmann May 9 2011 at 9:28 Just a vague comment: I haven't seen exactly what you describe, but I know that some people study things like the "Ihara zeta function of a graph". If the graph you consider is the Cayley graph of your group, then you get formulas looking a bit like yours... – Sylvain Bonnot Jun 10 2011 at 22:11 show 1 more comment ## 1 Answer One special case of groups, where one certainly gets rather quickly explicit and nontrivial expressions should be finite or affine Coxeter groups, that are finite/infinite and defined by involution generators and relations from their Dynkin diagrams - more my topic than graphs ;-) If the Coxeter-diagram belongs to a finite/affine Lie algebra root system resp. Weyl group (which is the generic case!), these rather strong structures should give you enough informations to control the elements of the group ordered by their length. I want to be specific in two cases (finite/infinite) I found rather quickly in the relevant literature, but are without factorial in the numerator: For a finite Weyl-group $W$ acting as reflection group on an $n$-dimensional vectorspace the well-known Chevalley-Solomon theorem (often used "the-other-way-around") asserts, that: $$W(z):=\sum_{g\in G} z^{Length(g)}=\prod_{\alpha\in \Delta^+}\frac{1-z^{Height(\alpha)+1}}{1-z^{Height(\alpha)}} = \prod_{k=1}^n\frac{1-z^{d_k}}{1-z}$$ where $d_k$ are the fundamental degrees of the reflection action, i.e. the degrees of a homogenious basis of the invariant part of the polynomial ring over $V$ (having $n$ variables). The middle term is crucial for the proof (and pretty), but the product over the entire root system $\Delta$ is not helpful for our question ;-) Example For $S_n$ (root system $A_{n-1}$) we have $d_k=k$ (each elementary symmetric polynomial), thus: $$\sum_{g\in S_n} z^{Length(g)}=\prod_{k=1}^n\frac{1-z^{k}}{1-z}$$ For an affine Weyl group $\tilde{W}$ i.e. with Dynkin-diagram as some finite $W$ (suppose irreducible) with one node added, Bott's theorem states that (omitting $d_k=1$-Terms): $$\tilde{W}(z)=W(z)\prod_{k=1}^n\frac{1}{1-z^{d_k-1}}$$ Example $\tilde{A}_n$ is derived from closing the $n$-chain $A_n$ (i.e. $S_{n+1}$) with an additional $x$. Hence is generated very similar to the symmetric group but infinite (all non-mentioned pairs elements commute!): $$G=\langle t_1,t_2,...t_n,x\rangle\qquad (t_it_{i+1})^3=1\quad (t_1x)^3=(xt_n)^3=1$$ Hence the length-generating function now has a pole: $$\tilde{W}(z)=W(z)\prod_{k=2}^n\frac{1}{1-z^{k-1}}=\frac{1}{(1-z)^n}\prod_{k=2}^n\frac{1-z^k}{1-z^{k-1}}$$ Finally I must of course mention that the beatifully exotic root systems of Nichols algebras, that would even count the lengths of Coxeter gruppoids ;-) This was all written down without much further thought, but if there's still interest in the topic (?) I'd be happy about a further discussion! Maybe concerning factorial or something else.... SOURCES: • Definitely Humphreys "Reflection Groups and Coxeter groups" • the formulas also directly online e.g. the "survey" www.math.umn.edu/~reiner/Papers/SteinbergNotes.ps). • Some infinite worked-out examples e.g. in dml.ms.u-tokyo.ac.jp/PSRT/PSRT_26/PSRT_26_093-102.pdf and many more) -

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http://physics.stackexchange.com/questions/45839/why-are-navier-stokes-equations-needed/45993

# Why are Navier-Stokes equations needed? Can't we picture air or water molecules individually? Then, why are Navier-Stokes equations needed, after all? Can't we just aggregate individual ones? Or is it computationally difficult, or inefficient to generate the picture of whole flow? - 4 – Emilio Pisanty Dec 4 '12 at 2:33 ## 3 Answers Consider a standard volume of $1\textrm{ m}^3$ of air. This contains on the order of $10^{25}$ molecules of O2 and N2. If you needed to simulate or explain the physics occurring in that volume of air, would you want to model $10^{25}$ molecules and all the interactions between them or, say, 100x100x100 cells based on the Navier-Stokes equations? Theoretically, it is possible to simulate every fluid flow ever by tracking every single molecule. But direct simulation of turbulence using the Eulerian Navier-Stokes equations requires $Re^{9/4}$ grid points and is thus totally impractical for Reynolds numbers larger than a few thousand. So simulating something with $10^{25}$ things to track is completely impossible. - 3 Just recording the position coordinates as doubles would take 1000000000000000 GB of RAM. – user1504 Dec 4 '12 at 3:31 2 A quibble, but a cubic metre contains more than a mole (6.023e23 molecules). At STP a mole of an ideal gas is 22.4 litres. – John Rennie Dec 4 '12 at 6:59 You're right, it's been a long time since I had to think about those relationships. I'll fix it – tpg2114 Dec 4 '12 at 19:55 1 I don't think numerical simulation was the reason Navier and Stokes derived their equations. – Bernhard Dec 5 '12 at 11:59 Just to add to this correct answer: even if you could simulate every molecule, you would gain little by it unless for some reason you knew the initial position of every molecule to start with. Even then, it would only help you to predict the future behaviour of that particular volume of fluid at that particular time - it wouldn't tell you anything about what would happen if you repeated the same experiment again, but with different molecules that weren't in exactly the same starting places. – Nathaniel Dec 5 '12 at 13:00 show 3 more comments Apart from the experimental and numerical difficulties to track each particle in a macroscopic piece of matter, there are physical reasons. Theoretically, we can show that in the hydrodynamic regime the exact equations of motion for the collection of atoms/molecules reduce to the Navier-Stokes equations. This is shown in statistical mechanics. Due to complex molecular effects, including molecular chaos, most degrees of freedom self-cancel and only some few dynamical modes survive. Those modes are collective --i.e. they describe the fluid as a whole-- and robust --i.e., they survive in macroscopic scales of space and time--. That is, even if you were to trace the individual motion of each particle and next average to describe the collective motion of the whole fluid, you would see that your description is practically indistinguishable [*] from that given by solving the Navier-Stokes equations. [*] The differences are of the order of the inverse of the size of the system and vanish for a macroscopic system. - "Theoretically, we can show that in the hydrodynamic ..." I believe this paragraph is wrong at the current time. I thought you can derive NS from the Boltzmann equation, so techinically you can derive NS to a certain extent only for gases, but not for liquids. Or do you know statistical NS derivation for liquids? – Yrogirg Dec 5 '12 at 14:47 @Yrogirg: There is a kinetic theory for dense fluids as well, but is not of the Boltzmann type, of course. Moreover, you can obtain the hydrodynamic equations without using any kinetic equation, for instance using the Mori formalism for the microscopic dynamical variables. – juanrga Dec 6 '12 at 11:32 As apparent from the other answers, the continuum approach of the Navier-Stokes equations makes life easier, by reducing the degrees of freedom. For example, Poisseuile flow (laminar flow between flat plates or in a pipe), is easily solved using the Navier-Stokes equations, but I have no clue how to solve it using just molecular behavior (on the back of an envelope that is). If you calculate the Knudsen number of a specific problem, which is just the molecular mean free path divided by some macroscopic length scale, you can see if the continuum approach is a valid for the considered system. -

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http://www.purplemath.com/learning/viewtopic.php?t=2138

# The Purplemath Forums Helping students gain understanding and self-confidence in algebra. ## finding the area of parallelogram / combination/ permutation Sequences, counting (including probability), logic and truth tables, algorithms, number theory, set theory, etc. 4 posts • Page 1 of 1 ### finding the area of parallelogram / combination/ permutation by jigoku_snow on Wed Aug 03, 2011 1:56 pm 1) ABCD is a parallelogram and the coordinates of A, B, C are (-8, -11), (1, 2) and (4, 3) respectively. Find the area of the parallelogram. [i]i manage to find that D =( -5.-10) . area of parallelogram = AD x height i'm stuck here. how to find the height? i tried it by finding the angle between AB and AD first , (37.32 degree) then use it to find the vertical line from B. but i dont get the right answer. 2). Points A and C have coordinates (-1, 2) and (9, 7) respectively. A rectangle ABCD has AC as a diagonal. Calculate the possible coordinates of B and D if the length of AB is 10 units. from the question, i know that AB=CD=10. the equation of line AB = x^2 +y^2+ 2x -4y-97=0. the equation of line CD= a^2 +b^2-18 a- 14b +30=0. then what should i do next? 3)If the five letters a, b, c, d, e are put into a hat, in how many ways could you draw two letters? must we use permutation or combination? if permutation, why? why can't i use combination? 5C2? 4))3 balls are to be placed in 3 different boxes, not necessarily with 1 ball in each box. any box can hold up to 3 balls. find the number of ways the balls can placed if i) they are all of the same colour ii) they are all of different colours i) what is wrong with my workings? 3C3 + (3C2 x 1C1) + 3C1? ii) i have no clue at all for this. isnt it suppose to be the same as the first one? 5)GOLD MEDAL- if 2 D letters come first and the 2 L letters come last, find how many different arrangement there are D D _ _ _ _ _ L L 5P5 x 2P2 x 2P2. why i dont need to permute the letter D and the L? 6) 3 letters are selected at random from the word "BIOLOGY" . Find the number that the selection contains one letter O 2C1 x 5C2 = 20 , the answer given is 5C2 = 10 , why i dont have to write 2C1? 7)The sum of the first 5 terms of a G.P. is twice the sum of the terms from the 6th to the 15th inclusive. Show that r^5 = 1/2 (SQRT3 -1). S5 = 2 (S15 - S5) a(1-r^5)/ i-r = 2a/ 1-r [(1-r^15)-(1-r^5)] 1 - r^5 = 2(r^5 -r^15) 2r^15-3r^5+1=0 let u=r^5 2u^3 - 3u +1=0 i'm stuck here. what to do next? jigoku_snow Posts: 2 Joined: Wed Aug 03, 2011 1:46 pm Sponsor Sponsor ### Re: finding the area of parallelogram / combination/ permuta by nona.m.nona on Thu Aug 04, 2011 1:55 am jigoku_snow wrote:1) ABCD is a parallelogram and the coordinates of A, B, C are (-8, -11), (1, 2) and (4, 3) respectively. Find the area of the parallelogram. Use the Distance Formula to find the lengths of each base. Find the slope of the line containing either base; use this to find the perpendicular slope. Find the line equation for the perpendicular through one of the corners; find the point where this line intersects the other base (or the other base, extended). Find the distance between the two bases, and thus the height. Plug the base lengths and the height into the area formula. jigoku_snow wrote:2). Points A and C have coordinates (-1, 2) and (9, 7) respectively. A rectangle ABCD has AC as a diagonal. Calculate the possible coordinates of B and D if the length of AB is 10 units. from the question, i know that AB=CD=10. the equation of line AB = x^2 +y^2+ 2x -4y-97=0. the equation of line CD= a^2 +b^2-18 a- 14b +30=0. then what should i do next? The line AB will be linear; what you have posted is a quadratic. How did you derive this? (One of the links in the previous response explains how to find linear equations from two points.) As for finding the coordinates, try drawing the two points you have now, along with the line containing them. The rectangle is either to the one side of that line, or else to the other. So where must the other two points be? jigoku_snow wrote:3)If the five letters a, b, c, d, e are put into a hat, in how many ways could you draw two letters? must we use permutation or combination? if permutation, why? why can't i use combination? 5C2? Since nothing in the exercise says anything about ordering, combinations should be sufficient. However, you can see your book and your class notes; if your book, by default, defines "drawing" as "one at a time, with order taken into account", then you should use permutations. jigoku_snow wrote:4))3 balls are to be placed in 3 different boxes, not necessarily with 1 ball in each box. any box can hold up to 3 balls. find the number of ways the balls can placed if i) they are all of the same colour ii) they are all of different colours i) what is wrong with my workings? 3C3 + (3C2 x 1C1) + 3C1? You seem to be counting the ways to choose the balls, but not the ways in which you could then distribute them (in the chosen groupings) amongst the boxes. jigoku_snow wrote:ii) i have no clue at all for this. isnt it suppose to be the same as the first one? If you have boxes A, B, and C and if you are placing three red balls in them, then you have the first situation. However, if you have boxes A, B, and C and if you are placing a red, a green, and a yellow ball in them, then you have the second situation. In the first case, "Box A has a red ball" is the same as "Box A has a red ball", since the balls are indistinguishable. In the second case, "Box A has a red ball" is different from "Box A has a green ball". jigoku_snow wrote:5)GOLD MEDAL- if 2 D letters come first and the 2 L letters come last, find how many different arrangement there are D D _ _ _ _ _ L L 5P5 x 2P2 x 2P2. why i dont need to permute the letter D and the L? How does "D" differ from "D"? How are they distinguishable? jigoku_snow wrote:6) 3 letters are selected at random from the word "BIOLOGY" . Find the number that the selection contains one letter O 2C1 x 5C2 = 20 , the answer given is 5C2 = 10 , why i dont have to write 2C1? What was your logic for including "2C1"? jigoku_snow wrote:7)The sum of the first 5 terms of a G.P. is twice the sum of the terms from the 6th to the 15th inclusive. Show that r^5 = 1/2 (SQRT3 -1). [i]S5 = 2 (S15 - S5) a(1-r^5)/ i-r = 2a/ 1-r [(1-r^15)-(1-r^5)] How did you arrive at this equation? Using the formulation explained here: $a\left(\frac{1\, -\, r^5}{1\, -\, r}\right)\, =\, 2a\left(\frac{1\, -\, r^{15}}{1\, -\, r}\, -\, \frac{1\, -\, r^5}{1\, -\, r}\right)$ Multiply by (1 - r)/a to get: $1\, -\, r^5\, =\, 2\left(1\, -\, r^{15}\right)\, -\, \left(1\, -\, r^5\right)$ $1\, -\, r^5\, =\, 2\, -\, 2r^{15}\, -\, 2\, +\, 2r^5$ $2r^{15}\, -\, 3r^5\, +\, 1\, =\, 0$ Then solve the polynomial for its rational root (which clearly does not apply, as a value for the common ratio, to this series). Solve the resulting quadratic for the other root, using the definition of "geometric progression" to pick the correct value. nona.m.nona Posts: 198 Joined: Sun Dec 14, 2008 11:07 pm ### Re: finding the area of parallelogram / combination/ permuta by jigoku_snow on Sat Aug 06, 2011 8:50 am The line AB will be linear; what you have posted is a quadratic. How did you derive this? since i cant find the gradient so i use the distance formula. i let B=(x,y) and D=(a.b) What was your logic for including "2C1"? well, is because there are 2 'O' s. but i think i know why i dont have to write that, is because they are indistinguishable. How did you arrive at this equation? from a(1-r^5)/ 1-r = 2a [ (1-r^15/ 1-r) - ( 1-r^5 / 1-r )] i simplify it . Then solve the polynomial for its rational root i think this question is way too far for me as i not yet learn polynomial. jigoku_snow Posts: 2 Joined: Wed Aug 03, 2011 1:46 pm ### Re: finding the area of parallelogram / combination/ permuta by nona.m.nona on Sat Aug 06, 2011 2:40 pm The line AB will be linear; what you have posted is a quadratic. How did you derive this? jigoku_snow wrote:since i cant find the gradient so i use the distance formula. i let B=(x,y) and D=(a.b) You have two points on a straight line. You can definitely find the slope ("gradient"). Please study the lesson in the link, provided earlier, to learn how. Then solve the polynomial for its rational root jigoku_snow wrote:i think this question is way too far for me as i not yet learn polynomial. You are working with polynomials, such as the quadratic you created for the first exercise and "1 - r15" here, so I am not sure what you mean when you say that you haven't yet learned about polynomials. Please study the lesson provided in the link to learn how to solve polynomials. nona.m.nona Posts: 198 Joined: Sun Dec 14, 2008 11:07 pm 4 posts • Page 1 of 1 Return to Discrete Math • Board index • The team • Delete all board cookies • All times are UTC

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http://unapologetic.wordpress.com/2008/01/22/the-differential-mean-value-theorem/?like=1&source=post_flair&_wpnonce=3fdfac1636

# The Unapologetic Mathematician ## The Differential Mean Value Theorem Let’s say we’ve got a function $f$ that’s continuous on the closed interval $\left[a,b\right]$ and differentiable on $(a,b)$. We don’t even assume the function is defined outside the interval, so we can’t really set up the limit for differentiability at the endpoints, but they don’t matter much in the end. Anyhow, if we look at the graph of $f$ we could just draw a straight line from the point $(a,f(a))$ to the point $(b,f(b))$. The graph itself wanders away from this line and back, but the line tells us that on average we’re moving from $f(a)$ to $f(b)$ at a certain rate — the slope of the line. Since this is an average behavior, sometimes we must be going faster and sometimes slower. The differential mean value theorem says that there’s at least one point where we’re going exactly that fast. Geometrically, this means that the tangent line will be parallel to the secant we drew between the endpoints. In formulas we say there is a point $c\in(a,b)$ with $f'(c)=\frac{f(b)-f(a)}{b-a}$. First let’s nail down a special case, called “Rolle’s theorem”. If $f(a)=0=f(b)$, we’re asserting that there is some point $c\in(a,b)$ with $f'(c)=0$. Since $\left[a,b\right]$ is compact and $f$ is continuous, the extreme value theorem tells us that $f$ must take a maximum and a minimum. If these are both zero, then we’re looking at the constant function $f(x)=0$, and any point in the middle satisfies $f'(c)=0$. On the other hand, if either the maximum or minimum is nonzero, then we have a local extremum at a point $c\in(a,b)$ where $f$ is differentiable (since it’s differentiable all through the open interval). Now Fermat’s theorem tells us that $f'(c)=0$ since $c$ is a local extremum! Thus Rolle’s theorem is proved. Now for the general case. Start with the function $f$ and build from it the function $g(x)=f(x)-\frac{f(b)-f(a)}{b-a}(x-a)-f(a)$. On the graph, this corresponds to applying an “affine transformation” (which sends straight lines in the plane to other straight lines in the plane) to pull both $f(a)$ and $f(b)$ down to zero. In fact, it’s a straightforward calculation to see that $g(a)=0=g(b)$. Thus Rolle’s theorem applies and we find a point $c$ with $g'(c)=0$. But applying our laws of differentiation, we see that $g'(c)=f'(c)-\frac{f(b)-f(a)}{b-a}$. And so $f'(c)=\frac{f(b)-f(a)}{b-a}$, as desired. ### Like this: Posted by John Armstrong | Analysis, Calculus ## 21 Comments » 1. The “trick” behind proving the Mean Value theorem is basically to write $g(x) = f(x) - h(x)$, where $y = h(x)$ is the equation of the straight line connecting the points $(a, f(a))$ and $(b, f(b))$. And this is what you did above. I know it might not sound very interesting, but I have always wondered if we can obtain another useful result (just like the mean value theorem, for instance) if we let $y = h(x)$ be the equation of a parabola passing through the points $(a, f(a))$ and $(b, f(b))$. And, do we obtain more “useful” results if $y = h(x)$ is some function other than a straight line or a parabola? (That was just a random thought.) Comment by | January 22, 2008 | Reply 2. Indeed you do, Vishal, but you ask that the function be twice-differentiable and you match another parameter: the derivative of $f$ at $a$. Then those three parameters specify a unique parabola, which has constant second derivative. And there’s some point in between $a$ and $b$ at which the function actually has that second derivative. There’s a whole sequence of these “generalized mean value theorems”, which can be used as Taylor series remainder approximations. Your exercise is to work out the analogues of Rolle’s theorem you guessed were around, and to see if you can prove them in general. I’ll come back to this when I’m doing Taylor series. Comment by | January 22, 2008 | Reply 3. I had a hunch earlier that we could indeed derive more of those “generalized mean value theorems” and somehow relate those to the Taylor series approximations. Your comment, it seems, has confirmed that hunch. I will be looking forward to your future post(s) on Taylor series. Comment by | January 22, 2008 | Reply 4. This theorem is also known as the Lagrange mean value theorem. Comment by | January 23, 2008 | Reply 5. It also has an interesting corollary that says that the derivative of any function that is differentiable on an interval has the mean value property, i.e. it hits all its intermediate values between any 2 of the values that it assumes, even if that derivative is not continuous. In partcular, a simple step function H(x) that is 0 for x1 is not a derivative of any function F differentiable on (-1,1), so H(x) has no primitive on (-1,1). Comment by | January 23, 2008 | Reply 6. In comment #5 H(x)=0 for x0, also known as the Heaviside function. Comment by | January 23, 2008 | Reply 7. There must be a bug in the system, that messed up my posts, again (now using LaTex): $H(x)=0$ for $x \le 0$ and $0$ for $x>0$. Comment by | January 23, 2008 | Reply 8. There must be a bug in the system, that messed up my posts, again (now using LaTex): $H(x)=0$ for $x \le 0$ and $=0$ for $x>0$. Comment by | January 23, 2008 | Reply 9. Michael, you’re having trouble with the < symbol. That’s not a bug in the system. The bug in the system is that I can’t seem to log in to WordPress at all. Whether this has to do with a comment flood or not, I have no idea. Just in case, could you try in the future to think carefully about what you want to say and then say it all in one comment? Comment by | January 23, 2008 | Reply 10. Sorry, I got troubles with > that seems to have some special meaning on this site and also with LaTex preprocessor here that messed up my formula. I wish we could edit our entries. I doubt that your trouble logging in is related to the number of comments on your page. Comment by | January 23, 2008 | Reply 11. It’s not this site. > has a special meaning in HTML, which is sort of the basis of the web. WordPress’ processor can’t tell the difference between using it to mean “greater than” and “close a tag”. But still, you made three separate comments before the litany of trying to fix the problem with the > sign. Slow down, take a deep breath, and decide what you want to say before saying it. This isn’t Twitter. Comment by | January 23, 2008 | Reply 12. Actually I wanted to say that H(x) is the Heaviside function that jumps from 0 to 1 at x=0. Comment by | January 23, 2008 | Reply 13. [...] of the Mean Value Theorem So now that we have the mean value theorem what can we do with it? First off, we can tell something that seems intuitively obvious. We know [...] Pingback by | January 24, 2008 | Reply 14. [...] to get back on track. Today, we’ll see a theorem about integrals that’s similar to the Differential Mean Value Theorem. Specifically, it states that if we have a continuous function then there is some so [...] Pingback by | February 12, 2008 | Reply 15. [...] the Differential Mean Value Theorem tells us that in each subinterval there is some so that . We stick this into the [...] Pingback by | March 6, 2008 | Reply 16. [...] an extension of the Fundamental Theorem of Calculus we’ll see it’s an extension of the Differential Mean Value Theorem. And this despite the most strenuous objections of my resident gadfly that the DMVT and FToC have [...] Pingback by | October 1, 2008 | Reply 17. [...] a nice technical result we may have call for from time to time: a higher-dimensional version of the differential mean value theorem. Remember that this says that if we’ve got a function continuous on the closed interval and [...] Pingback by | October 13, 2009 | Reply 18. [...] we can apply the mean value theorem to [...] Pingback by | October 15, 2009 | Reply 19. [...] Taylor’s theorem. Specifically, the version of Taylor’s theorem that resembles the mean value theorem. And we’ll even use an approach like we did for the extension of that result to higher [...] Pingback by | October 20, 2009 | Reply 20. [...] is some number between and that exists by the differential mean value theorem. Now to find the derivative, we take the limit of the difference [...] Pingback by | January 13, 2010 | Reply 21. [...] let denote the sum of the terms and , the remainder after terms, of the series (2), so that . By Lagrange’s mean value theorem, we have , . We shall now consider . So far we have taken as an arbitrary but we shall now choose [...] Pingback by | July 7, 2011 | Reply « Previous | Next » ## About this weblog This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”). I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.

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http://mathoverflow.net/questions/19638/infinite-collection-of-elements-of-a-number-field-with-very-similar-annihilating/19774

## Infinite collection of elements of a number field with very similar annihilating polynomials ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Hello all, let $n$ be an integer $\geq 2$ and let $\alpha$ be an algebraic number of degree $n$. Let $R$ be the ring of algebraic integers in ${\mathbb Q}(\alpha)$, and let $B$ be the subset of $R$ containing the elements whose degree is exactly $n$. Any $\beta \in B$ has a minimal polynomial $X^n+b_{n-1}X^{n-1}+ \ldots + b_1X+b_0$. Identifying this latter polynomial with the uple $(b_0,b_1, \ldots ,b_{n-1})$ allows us to view $B$ as a subset of ${\mathbb Z}^n$. I define a combinatorial subvariety $V$ of dimension at most $r$ of ${\mathbb Z}^n$ to be a subset of $Z^n$ such that there is a set of indices $I \subseteq \lbrace 1,2, \ldots , n \rbrace$ with $|I|=n-r$ and the projection $p:V \to {\mathbb Z}^{n-r}, (v_1,v_2, \ldots ,v_n) \mapsto (v_i)_{i\in I}$ is constant. My question is : what is the smallest $r$ such that there is an infinite subset $B' \subset B$ corresponding to a subvariety of dimension at most $r$ ? In other words, we are asking for infinitely many elements in $B$, whose minimal polynomials are as similar as possible". An easy case is when $\alpha=a^{\frac{1}{n}}$ for some $a \in {\mathbb Q}$, because the rational multiples of $\mathbb \alpha$ correspond to a subvariety of dimension 1, so that $r=1$ in this case. - Should "annulating" (in the title) be "annihilating"? – Michael Lugo Mar 28 2010 at 17:21 I suspect there is a uniform bound on such r: given an r, there is the hypersurface $D(b_0,...,b_{n-1})/D_0 = y^2$, where $D$ is the discriminant of the monic polynomial, and $D_0$ will be the discriminant of a fixed monic polynomial. Not all solutions to this equation will be in the same field, but probably infinitely many. I believe there are conjectures on the number of integer solutions to hypersurfaces in $r$ variables, can anyone who knows shed more light? – Dror Speiser Mar 28 2010 at 17:28 Thanks Michael, I corrected the title. – Ewan Delanoy Mar 28 2010 at 19:52 "...allows us to view B as a subset of Z^n". Not quite, because distinct elements of B might have the same min poly. – Kevin Buzzard Mar 28 2010 at 19:57 @ Kevin : this does not matter because at most $n$ many elements share the same min poly. So an infinite subset of $B$ will always yield an infinite subset of $Z^n$. – Ewan Delanoy Mar 29 2010 at 3:52 show 1 more comment ## 2 Answers For $n>4$, almost all fields of degree $n$ will have $r>1$: Fix a field $K$ with discriminant $D_0$. Fix the $n-1$ coefficients $b_{n-1},...,b_{i+1}, b_{i-1},..., b_0$. The discriminant of the polynomial $x^n+b_{n-1}x^{n-1}+...$ is a polynomial $D(b_i)$ in the single variable $b_i$, and is of degree at least $4$. If this polynomial is squarefree, as it will be for almost all $n-1$ fixed coefficients, then the hypersurface $D_0y^2 = D(b_i)$ has genus at least $1$, and hence finitely many integer points. But, every polynomial defining the same field must have the same discriminant up to a square factor, and hence $r > 1$. Going back on my comment above: since the degree of the discriminant (multivariate) polynomial is large (linear in the number of variables) the equation $D(b_0,...,b_{n-1}) = D_0y^2$ will probably have only a finite number of solutions for most $D_0$, if $r$ is much smaller than $n$. Therefore, my new pessimistic conjecture is that for almost all fields you will have $r \gg n$. Note: $r \le n-1$ - in any number field there are always an infinite number of algebraic integers with trace 0. - @Dror: the heart of this answer is surely right but I'm not convinced that D(b_i) is always a polynomial of degree greater than 4 [try quintics with i=0 for example], and even if it is then the hypersurface might be singular and have lots of rational points, like y^2=x^101 or something. But in some sense this is a moraly correct approach, and it will surely suffice to find one explicit number field with r>1. – Kevin Buzzard Mar 29 2010 at 21:56 Sorry sorry, I meant genus at least 1, Siegel's theorem covering 1. The "almost all" takes out the singular hypersurfaces. I believe some very easy sieve theory can prove this since, mod p, most assignments of the other coefficients should give a squarefree polynomial. – Dror Speiser Mar 29 2010 at 22:10 I follow the argument and believe you. – Kevin Buzzard Mar 29 2010 at 22:16 @ Dror : very nice. I think that we can always take r=1 when n=3, because Kevin's computation on $x^3-x+1$ generalizes. – Ewan Delanoy Mar 30 2010 at 5:17 ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. The answer "is" that the smallest $r$ is what it is, and what it is could well depend on $\alpha$. Let me also raise the possibility that there might be no simple "formula" relating $r$ to $\alpha$. This in some sense is the "problem" with questions like this ("given some data, compute some number $r$: what 'is' $r$?")---they are not really questions (in my mind, at least). Who knows though, perhaps someone can find some extra structure. For example can one always take $r=1$? That's a proper question ;-) I'd be surprised though! But on a more positive note let me say that in my (rather long) answer to http://mathoverflow.net/questions/12486/integers-not-represented-by-2-x2-x-y-3-y2-z3-z I show in passing that if $\alpha$ is a root of $z^3-z+1$ then there are infinitely many integers $C$ such that $z^3-z+C$ is irreducible and has a root in $\mathbf{Q}(\alpha)$, giving a perhaps slightly less trivial example. The integers $C$ are the odd solutions to $27C^2-4=23D^2$ and there are infinitely many of these (the smallest two being 1 and 599). - 1 I completely disagree with your comment that the question is not meaningful. For any given $\alpha$, the (smallest) $r$ is a defnite value. It may be that $r$ is very hard to compute in terms of $\alpha$ in general, but you cannot say that the question is meaningless. Thanks for your example with $z^3-z-1$. I guess there is a parametrization of the solutions of $27C^2-4=23D^2$ by some linear-recurrence sequence. – Ewan Delanoy Mar 29 2010 at 3:50 Well OK :-) Yes, I agree that the smallest $r$ is a definite value. My point is that if the question is a question, then an answer to it could be "the value of $r$ is whatever it comes out to be". The parametrisation of the solutions to 27C^2-4=23D^2 can be obtained by the theory of Pell's equation: the solutions grow exponentially but there are infinitely many of them. – Kevin Buzzard Mar 29 2010 at 6:46 @Ewan: aah yes, I understand what you're saying: yes, the solutions to the equation are generated by a degree 2 linear recurrence relation, and 2/3 of them are odd. – Kevin Buzzard Mar 29 2010 at 6:49

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http://gilkalai.wordpress.com/2009/03/15/colorful-caratheodory-revisited/?like=1&source=post_flair&_wpnonce=e5382928c8

Gil Kalai’s blog ## Colorful Caratheodory Revisited Posted on March 15, 2009 by Janos Pach wrote me: ”I saw that you several times returned to the colored Caratheodory and Helly theorems and related stuff, so I thought that you may be interested in the enclosed paper by Holmsen, Tverberg and me, in which – to our greatest surprise – we found that the right condition in the colored Caratheodory theorem is not that every color class contains the origin in its convex hull, but that the union of every pair of color classes contains the origin in its convex hull. This already guarantees that one can pick a point of each color so that the simplex induced by them contains the origin. A similar version of the colored Helly theorem holds. Did you know this?” I did not know it. This is very surprising! The paper of Holmsen, Pach and Tverberg mentions that this extension was discovered independently by J. L. Arocha, I. B´ar´any, J. Bracho, R. Fabila and L. Montejano. Let me just mention the colorful Caratheodory agai. (we discussed it among various Helly-type theorems in the post on Tverberg’s theorem.) The Colorful Caratheodory Theorem: Let $S_1, S_2, \dots S_{d+1}$ be $d+1$ sets in $R^d$. Suppose that $x \in conv(S_1) \cap conv (S_2) \cap conv (S_3) \cap \dots \cap conv (S_{d+1})$. Then there are $x_1 \in S_1$, $x_2 \in S_2$, $\dots x_{d+1} \in S_{d+1}$ such that $x \in conv (x_1,x_2,\dots,x_{d+1})$. And the strong theorem is: The Strong Colorful Caratheodory Theorem: Let $S_1, S_2, \dots S_{d+1}$ be $d+1$ sets in $R^d$. Suppose that $x \in conv(S_i \cup S_j)$ for every $1 \le i <j \le d+1$. Then there are $x_1 \in S_1$, $x_2 \in S_2$, $\dots x_{d+1} \in S_{d+1}$ such that $x \in conv (x_1,x_2,\dots,x_{d+1})$. Janos, whom I first met thirty years ago, and who gave the second-most surprising introduction to a talk I gave, started his email with the following questions: “Time to time I visit your lively blog on the web, although I am still not quite sure what a blog is… What is wordpress? Do you need to open an account with them in order to post things? Is there a special software they provide online which makes it easy to include pictures etc? How much time does it take to maintain such a site?” These are excellent questions that may interest others and I promised Janos that I will reply on the blog. So I plan comments on these questions in some later post. Meanwhile any comments from the floor are welcome. ### Like this: This entry was posted in Convexity and tagged Helly type theorems. Bookmark the permalink. ### 2 Responses to Colorful Caratheodory Revisited 1. Edna says: What is the first most surprising introduction? I hope it’s not the answer I am afraid you will give… 2. Gil says: Dear Edna, there is nothing to be afraid of. Indeed Avi’s introduction was the most surprising ever, and quite hilarious. As Avi promised to put it in writing for the back cover of my book, it will reach the large audience it deserves. Even now people who listened to that introduction ask me about the imortal translation to English to “where is Pluto” and some subtelties of the “cut its head/cut its tail” game. • ### Blogroll %d bloggers like this:

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http://mathoverflow.net/questions/96372/rate-of-change-of-mass-of-a-parameterized-region

## Rate of change of mass of a parameterized region ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Let $R_t$ be a family of compact, simply connected regions in the plane defined by `$R_t = \{x\in\mathbb{R}^2 : h(x) \leq t\}$` for all $t$, where $h(x)$ is some nicely behaved smooth function. Suppose $f(x)$ is a probability density on $\mathbb{R}^2$ and define $M(t) = \iint_{R_t} f(x) dA$ for all $t$. Is it true that $\frac{dM}{dt}|_{t=t_0} = \int_{\partial R_{t_0}} f(x) /\|\nabla h\| ds$ where $ds$ denotes integration with respect to arc length? If not, what is the right expression for $\frac{dM}{dt}$? I assume this is some well-known first-year calculus-type problem but I can't find it stated in any context (it may very well be a common homework problem though I've not seen it). - 2 This is not a good question for MO. See en.wikipedia.org/wiki/Coarea_formula – Anton Petrunin May 9 2012 at 0:51 This is a special case of the coarea formula. – Liviu Nicolaescu May 9 2012 at 17:58 Thanks to both of you, I wasn't familiar with that result! That clears this question up for me nicely. – Jennifer Gao May 11 2012 at 0:26 ## 1 Answer With $H$ the Heaviside function (characteristic function of $\mathbb R_+$), you have $$M(t)=\int_{\mathbb R^n} f(x) H(t-h(x)) dx$$ and thus, at least formally, $$\dot M(t)=\int_{\mathbb R^n} f(x) \delta_0(t-h(x)) dx= \int_{\mathbb R^n} f(x)\Vert\nabla h(x)\Vert ^{-1} \underbrace{\delta_0(t-h(x))\Vert\nabla h(x)\Vert dx}_{d\sigma} ,$$ where $d\sigma$ is the Euclidean surface measure on {$x, h(x) =t$}. Here, it is important to assume that $h$ is say $C^1$ such that $dh\not=0$ at $h=t$ for $t$ in neighborhood of some distinguished value $t_0$. As a result, $$\dot M(t)=\int_{\partial R(t)}f(x)\Vert\nabla h(x)\Vert ^{-1} d\sigma.$$ The previous computation can be justified by Green's formula: for $X$ a $C^1_c$ vector field on $\Omega$ ( an open subset of $\mathbb R^n$ with a $C^1$ boundary) $$\int_\Omega \text{div} X\ dx=\int_{\partial \Omega} X\cdot \nu\ d\sigma,$$ where $\nu$ is the exterior unit normal to $\partial \Omega$, and $d\sigma$ is the Euclidean surface measure on $\partial \Omega$. That formula can be proven by showing $$d\sigma=\lim_{\epsilon\rightarrow 0} \phi(\frac{\rho(x)}{\epsilon})\epsilon^{-1}\Vert\nabla \rho(x)\Vert\quad\text{(distribution sense),}$$ for $\phi\in C^\infty_c(\mathbb R),\int \phi(t) dt=1$, $\Omega=${$x, \rho(x)<0$}, $d\rho\not=0$ at $\rho=0$. -

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http://mathhelpforum.com/math-topics/17246-need-help-please-physics.html

# Thread: 1. ## need help please (physics) In college homecoming competition, 16 students lift sports car while holding the car off the ground, each student exerts an upward force of 400 N. (a) what is the mass of the car in kilograms? (b) what is its weight in pounds? 2. Originally Posted by twistedmexican In college homecoming competition, 16 students lift sports car while holding the car off the ground, each student exerts an upward force of 400 N. (a) what is the mass of the car in kilograms? (b) what is its weight in pounds? a) Each of the 16 students is exerting an upward force on the car and (I presume) the car is stationary and off the ground. So the students are supplying the only upward force, which is counteracting the weight of the car. Thus $w = 16F = 16 \cdot (400~N) = 6400~N$ (where w is the weight of the car.) So the mass of the car is $m = \frac{w}{g} = \frac{6400~N}{9.8~m/s^2} = 653.061~kg$ b) 1 lb = 4.45 N, so $\frac{6400~N}{1} \cdot \frac{1~lb}{4.45~N} = 1439.75~lb$ -Dan

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http://unapologetic.wordpress.com/2008/07/06/the-carter-gelsinger-eversion/?like=1&source=post_flair&_wpnonce=eb8d9d939c

The Unapologetic Mathematician The Carter-Gelsinger Eversion I’ve mentioned Outside In before. That video shows a way of turning a sphere inside out. It’s simpler than the first explicit eversions to be discovered, but the simplicity is connected to a high degree of symmetry. This leads to very congested parts of the movie, where it’s very difficult to see what’s going on. Further, many quadruple points — where four sections of the surface pass through the same point — occur simultaneously, and even higher degree points occur. We need a simpler version. What would constitute “simple” for us, then? We want as few multiple points as possible, and as few at a time as possible. In fact, it would be really nice if we could write it down algebraically, in some sense? But what sense? Go back to the diagrammatics of braided monoidal categories with duals. There we could draw knots and links to represent morphisms from the monoidal identity object to itself. And topologically deformed versions of the same knot encoded the same morphism. This is the basic idea of the category $\mathcal{T}ang$ of tangles. But if we shift our perspective a bit, we consider the 2-category of tangles. Instead of saying that deformations are “the same” tangle, we consider explicit 2-isomorphisms between tangles. We’ve got basic 2-isomorphisms for each of the Reidemeister moves, and a couple to create or cancel caps and cups in pairs (duality) and to pull crossings past caps or cups (naturality). Just like we can write out any link diagram in terms of a small finite collection of basic tangles, we can write out any link diagram isotopy in terms of a small finite collection of basic moves. What does a link diagram isotopy describe? Links (in our picture) are described by collections of points moving around in the plane. As we stack up pictures of these planes the points trace out a link. So now we’ve got links moving around in space. As we stack up pictures of these spaces, the links trace out linked surfaces in four-dimensional space. And we can describe any such surface in terms of a small collection of basic 2-morphisms in the braided monoidal 2-category of 2-tangles. These are analogous to the basic cups, caps, and crossings for tangles. Of course the natural next step is to consider how to deform 2-tangles into each other. And we again have a small collection of basic 3-morphisms that can be used to describe any morphisms of 2-tangles. These are analogous to the Reidemeister moves. Any deformation of a surface (which is written in terms of the basic 2-morphisms) can be written out in terms of these basic 3-morphisms. We can simplify our picture a bit. Instead of knotting surfaces in four-dimensional space, let’s just let them intersect each other in three-dimensional space. To do this, we need to use a symmetric monoidal 3-category with duals, since there’s no distinction between two types of crossings. And now we come back to eversions. We write the sphere as a 2-dimensional cup followed by a 2-dimensional cap. Since we have duals, we can consider one side to be “painted red” and one side “painted blue”. One way of writing the sphere has the outside painted red and the other side is painted blue. An eversion in our language will be an explicit list of 3-morphisms that run from one of these spheres to the other. Scott Carter and Sarah Gelsinger have now created just such an explicit list of directions to evert a sphere. And, what’s more, they’ve rendered pictures of it! Here, for the first time in public, is a 50MB PDF file showing the Carter-Gelsinger eversion. First they illustrate the basic pieces of a diagram of knotted surfaces (pp. 1-4). Then they illustrate the basic 2-morphisms that build up surfaces (pp. 5-6), and write out a torus as an example (p. 7). Then come a few more basic 2-morphisms that involve self-intersections (pp. 8-9) and a more complicated immersed sphere (pp. 10-11). Each of these is written out also as a “movie” of self-intersecting loops in the plane. Next come the “movie moves” — the 3-morphisms connecting the 2-morphism “movies” (pp. 12-17). These are the basic pieces that let us move from one immersed surface to another. Finally, the eversion itself, consisting of the next 79 pages. Each one consists of an immersed sphere, rendered in a number of different ways. On the left is a movie of immersed plane curves. On the top are three views of the sphere as a whole — a “solid” view on the right, a sketch of the double-point curves in the middle, and a “see-through” view on the left. The largest picture on each page is a more schematic view I don’t want to say too much about. The important thing to see here is that between each two frames of this movie is exactly one movie move. Everything here is rendered into pictures, but we could write out the movie on each page as a sequence of 2-morphisms form the top of the page to the bottom. Then moving from one page to the next we trace out a sequence of 3-morphisms, writing out the eversion explicitly in terms of the basic 3-morphisms. As an added bonus, there’s only ever one quadruple point — where we pass from Red 26 to Blue 53 — and no higher degree points. I’d like to thank Scott for not only finishing off this rendering he’s been promising for ages, but for allowing me to host its premiere weblog appearance. I, for one, am looking forward to the book, although I’m not sure this one will be better than the movie. [UPDATE] Some people have been having trouble with the whole 50MB PDF (and more people might as the Carnival comes to see this page. Scott Carter broke the file up into five pieces, and I’ve put them up here in a new post. There’s a glitch in part 4, but I’ll have that one up as soon as I can. Like this: Posted by John Armstrong | Category theory, Knot theory, Topology 7 Comments » 1. Just like the enemy’s gate, the link is down. Comment by | July 6, 2008 | Reply 2. bah! Comment by | July 6, 2008 | Reply 3. better now Comment by | July 6, 2008 | Reply 4. John, I’m really excited about the overview you wrote describing the logic leading to the tools used for the Carter-Gelsinger eversion. I’m not a mathematician, so I truly appreciate your top-down “systems engineering” (hope you don’t find that offensive!)overview showing how how to extend known concepts into the problem-solution domain. To me, this is an amazing example of how an understanding of n-categories helps to derive the solution of a specific problem. It is even more understandable because the basic tools being extended don’t require an advanced mathematics degree to understand (at least the elementary versions). You have sketched out all the steps required to have a preliminary understanding of what is really going on here, a road map of how to proceed. Outstanding! Now I can more fully appreciate the accomplishment of Carter and Gelsinger (two smart people whose work I never thought I’d be able to understand). There’s no way I could have appreciated the basis for their achievement without your contribution. Thank you for this excellent overview! Charlie C. Comment by Charlie C | July 6, 2008 | Reply 5. [...] on the C-G Eversion Some people had trouble grabbing the whole 50MB file that I posted, so Scott Carter broke it into pieces. He also included these comments: The red, blue, and purple [...] Pingback by | July 10, 2008 | Reply 6. [...] this, me droogs Scott Carter recently took a break from turning spheres inside out and made a bunch of videos for YouTube. Since I’m stuck all day tomorrow learning which [...] Pingback by | August 14, 2008 | Reply 7. this is pretty cool Comment by prashant sharma | November 1, 2008 | Reply « Previous | Next » About this weblog This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”). I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.

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http://mathoverflow.net/questions/110505/is-there-a-simple-test-to-determine-whether-a-polytope-is-integral

## Is there a simple test to determine whether a polytope is integral? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) It is known that any rational convex polytope expressed as $\{ x\in\mathbb{R}^d : Ax \ge b \}$, where $A\in\mathbb{Z}^{k\times d}$ and $b\in\mathbb{Z}^k$, can be written as the convex hull of finitely many points. My question is, given the above representation in terms of hyperplanes, one can determine (easily) whether the polytope is integral ---that is, whether the polytope can be written as the convex hull of points in the integer lattice. - Are you looking for a theoretical solution for a class of examples, or a computational solution for specific instances? – gordon-royle Oct 24 at 5:28 I am currently interested in figuring out whether, under some reasonable conditions, certain polytopes I am studying are integral. The class of polytopes I am looking at must satisfy some specific symmetry conditions. If for certain parameters I could argue that the polytope must be totally unimodular (as someone pointed out below), then that would be nice ... – Steven Collazos Oct 27 at 18:46 ## 2 Answers An incomplete solution: There is a polynomial-time test for total unimodularity, and if $A$ is totally unimodular, then the polytope is integral. - What if $A$ is not unimodular? E.g., $A=\begin{pmatrix}1 & 0\\ 0 & 1\\ 2 & 1 \end{pmatrix}$ and $b=\begin{pmatrix}0 \\ 0 \\ 2\end{pmatrix}$? – J.C. Ottem Oct 24 at 4:52 Obviously he doesn't (yet) have an answer for that case. $\:$ – Ricky Demer Oct 24 at 4:53 (the example should have been $A=\begin{pmatrix}1 & 0\\ 0 &1\\ 2 & -1\end{pmatrix}$ and $b=\begin{pmatrix}0 \\ 0 \\ -2\end{pmatrix}$.) – J.C. Ottem Oct 24 at 4:56 Right, if $A$ is not totally unimodular then this test is inconclusive. I'm not aware of an efficient test that works for all $A,b$, though I'd certainly be interested to see one! – Michael Biro Oct 24 at 5:29 ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. This paper shows that testing integrality is coNP-complete. -

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http://mathoverflow.net/questions/16888/when-do-divisors-pull-back/59694

## When do divisors pull back? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) If we have a nonconstant map of nonsingular curves $\varphi:X\rightarrow Y$, then Hartshorne defines a map $\varphi^* Div(Y)\rightarrow Div(X)$ using the fact that codimension one irreducibles are just points, and looking at `$\mathcal{O}_{Y,f(p)} \rightarrow \mathcal{O}_{X,p}$`. My question is if we don't have a nice map of curves, what conditions can we put on the morphism so that we may pull divisors back? Clearly it's not true in general, since we can take a constant map and then topologically the inverse image doesn't even have the right codim. Thinking about this in terms of Cartier divisors (and assuming the schemes are integral), it seems like we just need a way to transport functions in $K(Y)$ to functions in $K(X)$. If $\varphi$ is dominant, then we'll get such a map. Is this sufficient? Also is there something we can say when $\varphi$ is not dominant? Something like we have a way to map divisors with support on $\overline{\varphi(X)}$ to divisors on $X$? - ## 5 Answers If you want to pull back a Cartier divisor $D$, you can do that provided the image of $f$ is not contained in the support of $D$: just pull back the local equations for $D$. If this does not happen, on an integral scheme, you can just pass to the associated line bundle $\mathcal{O}_X(D)$ and pull back that, obtaining `$f^{*} \mathcal{O}_X(D)$`; of course you lose some information because a line bundle determines a Cartier divisor only up to linear equivalence. Fulton invented a nice way to avoid this distinction. Define a pseudodivisor on $X$ to be a triple $(Z, L, s)$ where $Z$ is a closed subset of $X$, $L$ a line bundle and $s$ a nowhere vanishing section on $X \setminus Z$, hence a trivialization on that open set. Then you can simply define the pullback of this triple as `$(f^{-1}(Z), f^{*} L, f^{*} s)$`, so you can always pull back pseudo divisors, whatever $f$ is. The relation with Cartier divisors is the following: to a Cartier divisor $D$ you can associate a pseudodivisor $(|D|, \mathcal{O}_X(D), s)$, where $s$ is the section of $\mathcal{O}_X(D)$ which gives a local equation for $D$. This correspondence is not bijective. First, a pseudodivisor $(Z, L, s)$ determines a Cartier divisor if $Z \subsetneq X$; note that in this case enlarging $Z$ will not change the associated Cartier divisor, so to obtain a bijective correspondence with Cartier divisors you have to factor out pseudodivisors by an equivalence relation, which I leave to you to formulate. But if $Z = X$, you only obtain a line bundle on $X$, and you have no way to get back a Cartier divisor. - ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. If you have a morphism $f:X\rightarrow Y$ of schemes and $D$ is a Cartier divisor on $Y$, then you can pull back the invertible sheaf $\mathcal{O}(D)$ to $X$ and it will be an invertible subsheaf of the sheaf of total quotients $\mathcal{K}_X$. This sheaf defines a linear equivalence class of Cartier divisors but there's no clear choice of a specific divisor. If however you furthermore assume that $X$ is reduced, and $D$ is given by sections $f_i$ on an open covering $U_i$ of $Y$ whose support doesn't contain the image of any of the irreducible components of $X$, then you can define a Cartier divisor $f^*D$ on $X$ as $(f^{-1}(U_i), f_i\circ f)$ and this lies in the class of $f^*\mathcal{O}(D)$. This kind of "support-avoidance" is quite common with these arguments. - Did not ask X to be integral？ Why should the pull back be a subsheaf of the sheaf of total quotients？ – MZWang May 7 2012 at 1:46 It seems to me that you are approaching this in the wrong way. First of all you have to decide whether you want to pull-back divisors or divisor classes Pulling back divisors means pulling back cycles and you run into all sorts of problems. If your map is flat, then this is OK. See Fulton's Intersection theory book for more on this. (There is a section called "flat pull-back of cycles" somewhere early). This is actually why Hartshorne did not have to worry about pulling back divisors or divisor classes. Any non-constant morphism between non-singular curves is flat. Of course, I am not saying this is the most general condition under which you can pull-back divisors, just that this is the "natural" one. Then again, this is a little too much, because it works for any dimensional cycles, not just divisors. Pulling back divisor classes is somewhat easier (and harder at the same time). It is easier, because then you don't have to worry about the support. As already pointed out by Frank, you just "convert" your divisor class to a line bundle, pull that back and "convert" it back to a Cartier divisor. Actually for this you need $X$ to be integral, but you seem to be OK with assuming that. Anyway, another, perhaps more interesting question is how to pull-back Weil divisors. The main problem there is that they are not defined locally by a single equation, so their associated sheaf is not an line bundle, only a reflexive sheaf of rank $1$. Unfortunately pulling back those does not necessarily give you a reflexive sheaf. One solution is to still do that and then take the reflexive hull, but this will likely not give a group homomorphism. Another partial way to handle Weil divisors is to restrict to $\mathbb Q$-Cartier divisors, i.e., a divisor that itself may not be Cartier, but a multiple of which is Cartier. Then you just take that multiple, pull that back and then divide by the same number you multiplied with. Interestingly this can result in fractional coefficients, so it will only preserve linear equivalence mod $\mathbb Q$. The simplest example of a $\mathbb Q$-Cartier, but not Cartier divisor is a ruling of a quadratic cone (look at intersection numbers to prove this). A little more interesting example is included in this MO answer. For an example of a non-$\mathbb Q$-Cartier divisor consider the cone over a quadric surface in $\mathbb P^3$ and let the divisor be the cone over one line on the surface. - Assume that both $X$ and $Y$ are locally noetherian and take a Cartier divisor $D$ on Y. I claim that $D$ pullbacks to a Cartier divisor on $X$ if and only if $f(Ass\ X) \cap |D|=\emptyset$. Clearly this doesn't make sense unless I define what is a pullback of $D$: a Cartier divisor $E$ on $X$ is a pullback of $D$, written $E=f^*D$, if 1) $|E|\subseteq f^{-1}(|D|)$. 2) $U=X-f^{-1}(|D|)$ is schematically dense, i.e. $Ass\ X \subseteq U$, and there exists an isomorphism $\phi : f^*\mathcal{O}_Y(D)\longrightarrow \mathcal{O}_X(E)$ such that $\phi(f^*1)=1$ on $U$. Condition $1)$ relates the supports $|D|$ and $|f^* D|$ while condition $2)$ states that $f^*\mathcal{O}_Y(D)\simeq \mathcal{O}_X(f^* D)$ and assures that $f^* D$ is uniquely determined. The idea behind the above definition is that if $\mathcal{L}$ is an invertible sheaf on a scheme $X$ then $\mathcal{L}\simeq \mathcal{O}_X(E)$ for a Cartier divisor $E$ if and only if there exists a schematically dense open subset $U\subseteq X$ and a nowhere zero section $s\in \mathcal{L}(U)$. Moreover given $(U,s)$ there exists a unique $E$ s.t. $|E|\cap U = \emptyset$ and there exists an isomorphism $\mathcal{L}\simeq \mathcal{O}_X(E)$ sending $s$ to $1$ on $U$. Note that in general if $C$ is a divisor $X-|C|$ is schematically dense and $1$ is a nowhere zero section of $\mathcal{O}_X(C)$ on it. In our case $f(Ass\ X) \cap |D|=\emptyset$ means that $U=X-f^{-1}(|D|)$ is schematically dense. Since $f^* 1$ is a nowhere zero section of $f^*\mathcal{O}_Y(D)$ over $U$ we obtain a divisor $f^*D$ satisfying conditions $1),2)$. For example if $f(Ass\ X) \subseteq Ass Y$ you can pullback any Cartier divisor on $Y$. This happens if $f$ is flat or a dominant map between integral schemes. Finally if $D$ is effective $f^*D$ exists, i.e. $f(Ass\ X) \cap |D|=\emptyset$, if and only if $f^{-1}(D)$ is an effective divisor and in this case they coincide. Moreover the exact sequence defining $D$ on $Y$ pullbacks to the exact sequence defining $f^{-1}(D)$ on $X$, which is very useful. (For the effective case see [Stacks project, 22.3]). - You shouldn't need to worry about whether or not the support intersects the closure $\overline{\varphi(X)}$. If it isn't, then your divisor simply pulls back to the zero divisor on X. - The problem is that the support may contain $\varphi(X)$ (or the image of a component). In that case, the thing you want the pullback to be ends up having the wrong codimension. – Anton Geraschenko♦ Nov 12 2010 at 3:20 Ah, that's fair. For example, given the inclusion $i :D \hookrightarrow X$, then if we do this naively, then the "pullback" $i*D = D$ which is certainly not a divisor. – Simon Rose Nov 12 2010 at 15:54

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http://physics.stackexchange.com/questions/54413/what-conservation-law-corresponds-to-this-local-u1-symmetry-of-the-ccr?answertab=oldest

What conservation law corresponds to this local $U(1)$ symmetry of the CCR? It is known that canonical commutation relations do not fix the form of momentum operator. That means that if canonical commutation relations (CCR) are given by $$[\hat{x}^i,\hat{p}_j]~=~i\hbar~\delta^i_j~ {\bf 1}$$ they can be satisfied by the following choice of momentum operators: $p_x = -ih\frac{∂}{∂x}+\frac{∂f}{∂x}$ $p_y = -ih\frac{∂}{∂y}+\frac{∂f}{∂y}$ $p_z = -ih\frac{∂}{∂z}+\frac{∂f}{∂z}$ where $f(x,y,z)$ - arbitrary function. On the other hand, for any choice of $f(x,y,z)$ momentum operators can be transformed to their most frequently used form $(-ih\frac{∂}{∂x})$ (etc for $y$ and $z$) by the following transformation of the wave function $\psi$ and operators $p$: $\psi'=e^{-\frac{i}{h}f(x,y,z)}\psi$ $p^{'}_x =e^{-\frac{i}{h}f(x,y,z)}p_x e^{+\frac{i}{h}f(x,y,z)}=-ih\frac{∂}{∂x}$ Hence, we obtain $U(1)$ gauge transformation using only canonical commutation relations for momentum and position operators. Does this mean that $U(1)$ gauge invariance corresponds to conservation of momentum rather than to conservation of electric charge? - 1 Answer 1) If we interpret OP's transformation$^1$ as a passive transformation, i.e. a mere change of coordinates/description that doesn't alter the system, then there is no conservation law. 2) So in the following, let us interpret OP's transformation as an active transformation. For a system to have a symmetry, its action $S$ (or more precisely, in this quantum mechanical context, its Hamiltonian operator $\hat{H}$) must respect this symmetry. The corresponding conservation law would depend on the specific form of Hamiltonian $\hat{H}$. This is all we have to say about QM. 3) Finally, in a field theoretic context, we can interpret OP's transformation $$\tag{1} \frac{\hbar}{i} {\bf \nabla} ~\longrightarrow~ e^{-i\Lambda({\rm r})}\frac{\hbar}{i} {\bf\nabla} e^{i\Lambda({\rm r})}$$ as a pure gauge transformation in an electromagnetic theory with zero electromagnetic field strength. When interpreted as an EM theory, on one hand, global gauge symmetry leads to electric charge conservation, cf. Noether's first Theorem. See also this Phys.SE question. On the other hand, there is no conserved quantity associated with local gauge symmetry per se, cf. Noether's second Theorem. (Its off-shell Noether identity is a triviality. See also this Phys.SE question.) -- $^1$ OP's transformation is related to the choice of phase factors in overlaps between position and momentum eigenstates in QM, cf. this Phys.SE post. -

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http://stats.stackexchange.com/tags/moments/info

# Tag info ## About moments For questions about the moments of a random variable, that is, the expectation of $X^k$ where $X$ is a random variable and $k$ an integer (or a real number with $X$ non-negative).

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http://math.stackexchange.com/questions/193872/prove-1-2-4-8-dots-1

# Prove $1 + 2 + 4 + 8 + \dots = -1$ [duplicate] Possible Duplicate: Infinity = -1 paradox I was told by a friend that $1 + 2 + 4 + 8 + \dots$ equaled negative one. When I asked for an explanation, he said: Do I have to? Okay so, Let $x = 1+2+4+8+\dots$ $2x-x=x$ $2(1+2+4+8+\dots) - (1+2+4+8+\dots) = (1+2+4+8+\dots)$ Therefore, $(2+4+8+16+\dots) + (-1-2-4-8+\dots) = (1+2+4+8+\dots)$. Now $-2$ and $2$, $-4$ and $4$, $-8$ and $8$ and so on, cancel out, and the only thing left is $-1$. Therefore, $1+2+4+8+\dots = -1$. I feel that this conclusion is not right, but I cannot express it. Can anyone tell if this proof is wrong, and if it is, how it is wrong? - 9 The algebraic operations are not valid if the sum doesn't exist (which it doesn't). Also, some infinite series (conditionally convergent) are very sensitive to rearrangement, so the rearrangements must be valid on the level of partial sums in order to be valid for the infinite sums. However, these very same algebraic manipulations can be used to give analytic continuations which give the regularized value of $-1$ for the divergent sum. (Alternatively you can switch out the Euclidean topology for the $2$-adic topology, in which case this is completely valid.) – anon Sep 11 '12 at 1:00 1 If your friend had told you that this sum was $-\frac{1}{12}$.. now that would've been much more interesting! – student Sep 11 '12 at 1:04 1 (Yes, there is another divergent series $1+2+3+\cdots$ whose zeta-regularized value is $-1/12$.) – anon Sep 11 '12 at 1:05 5 – MJD Sep 11 '12 at 1:48 3 +1 MJD. This series converges (to -1) in the 2-adic metric. – GEdgar Sep 11 '12 at 1:51 show 2 more comments ## marked as duplicate by Argon, Austin Mohr, William, Jyrki Lahtonen, mt_Sep 11 '12 at 10:15 This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question. ## 5 Answers The first mistake is right at the beginning, in writing "Let $x=1+2+4+\cdots$." This builds in the assumption that there is such an object as $1+2+4+\cdots$. The second mistake lies in treating this supposed object as if it were a finite but maybe very long sum, to which the sensible rules for manipulating finite sums apply. Remark: Think about the "sum" $1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots$. If we call this $x$, and use a manipulation analogous to the one that you made, we end up with the conclusion that the "sum" is $2$. When infinite series are formally defined, it turns out that the answer is indeed $2$. So some "natural" manipulations yield nonsense, and some yield correct results. Of course that is not a tolerable state of affairs: we cannot use manipulational techniques that sometimes yield a correct result, and sometimes don't. This sort of issue, at a more sophisticated level, led mathematicians in the second half of the $19$th century to look for very careful definitions of the fundamental objects of mathematics, and rigorous proofs of their basic properties. - And if the sum is finite, say length $n$, the proof is flawed anyway, as the multiplication of the LHS sum by two yields a term of the sum that doesn't exist in the RHS, namely $2^n$ - so the two sums do not fully cancel out and the net result is $2^n - 1$ which is the expected result. – Thomas Sep 11 '12 at 2:41 While true, I consider this answer incomplete, since there are generalized summations that allow these kinds of manipulations even for divergent series. And even when those do not work, one has further methods like zeta-regularization and p-adic interpretation. Or in other words, there are many definitions of summation (and arguably much more useful than the standard one in many situations). – Marek Sep 11 '12 at 9:51 @Marek: Yes, it is incomplete. My aim was to present a short fairly vigorously expressed answer that would be locally useful. – André Nicolas Sep 11 '12 at 14:39 The infinite series $\displaystyle 1 + 2 + 4 + 8 + \ldots$ diverges. However, the sum $f(z) = 1 + z + z^2 + z^3 + \ldots$, which converges to $1/(1-z)$ for $|z| < 1$, has an analytic continuation to the complex plane with the point $1$ removed, and indeed $f(2) = -1$. So in that sense you could regard $-1$ as the value of the divergent series. For more on this, see http://en.wikipedia.org/wiki/1_%2B_2_%2B_4_%2B_8_%2B_%C2%B7_%C2%B7_%C2%B7 and references there. - Look at Calculus textbook, undergraduate level. When we treat infinite sum, we cannot change the order to compute. Example. $1-1+1-\cdots$ $$(1-1)+(1-1)+\cdots=0+0+\cdots=0$$ $$1+(-1+1)+(-1+1)+\cdots=1.$$ - That is conditional convergence, not relevant in this case, for example $1-\frac{1}{2^2}+\frac{1}{3^2}-\frac{1}{4^2}+..$ you can change the order. – Arjang Sep 11 '12 at 5:47 @Arjang: True, but the naive mistake is the same - that you can do whatever to an infinite sum and everything is fine. – user1729 Sep 11 '12 at 9:15 @user1729 : everything is fine, what is not fine is the naive use of convergence where the more complete forms of convergence need to be considered, cesaro summibility etc.. And this example converges using extended version of convergence. – Arjang Sep 11 '12 at 11:24 1 @Arjang: But surely that is the point - that naive convergence doesn't work. – user1729 Sep 11 '12 at 12:36 Thanks user1729. – Tom Sep 19 '12 at 13:28 show 1 more comment It is similar to "proving" that 1 = 2 by saying that 1+infinity = 2+infinity. - What is wrong is that $1+2+4+8+\cdots$'' is not a number, and so you cannot treat it like one. - try saying that to Euler. – Arjang Sep 11 '12 at 5:48 2 @Arjang: I think Euler is dead for quite some time now. – Asaf Karagila Sep 11 '12 at 9:20 1 @AsafKaragila : His works make him more alive than many alive people. – Arjang Sep 11 '12 at 11:26 2 @Arjang: I think you and modern science have very different definition of "alive". – Asaf Karagila Sep 11 '12 at 12:04 1 @Arjang: the fact that he was genius doesn't preclude that he did make mistakes. – Martin Argerami Sep 11 '12 at 20:24

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http://nrich.maths.org/2370/solution

### Instant Insanity Given the nets of 4 cubes with the faces coloured in 4 colours, build a tower so that on each vertical wall no colour is repeated, that is all 4 colours appear. ### Magic Caterpillars Label the joints and legs of these graph theory caterpillars so that the vertex sums are all equal. ### Plum Tree Label this plum tree graph to make it totally magic! # Limiting Probabilities ##### Stage: 5 Challenge Level: Congratulations to Angus from Bexhill College; Richard from Bournemouth School and Andrei from Tudor Vianu National College, Bucharest, Romania for their clearly explained and excellent solutions to this problem. The numbers on the edges of this graph give the probabilities of a particle travelling along those edges in the direction given by the arrow. The same information is given by the entry $a_{ij}$ in the following matrix which gives the probability of travelling from vertex $i$ to vertex $j$. $$A=\left (\begin{array}{cccc} 0 &1 &0 &0\\ 0 &0 &0.5 &0.5\\ 1 &0 &0 &0\\ 0 &0 &0 &1 \end{array} \right)$$ This is Angus's solution. Firstly, we see that when we multiply matrices : $$\left ( \begin{array}{ccc} A &B &C \\ D &E &F \\ G &H &I \end{array} \right) \left ( \begin{array}{ccc} a &b &c \\ d &e &f \\ g &h &i \end{array} \right) = \left ( \begin{array}{ccc} Aa+Bd+Cg & Ab+Be+Ch &Ac+Bf+Ci\\ Da+Ed+Fg &Db+Ee+Fh &Dc+Ef+Fi \\ Ga+Hd+Ig &Gb+He+Ih &Gc+Hf+Ii \end{array} \right)$$ and similarly for larger matrices. Hence when the matrix gives probabilities: $$A= \left( \begin{array}{cccc} P(1\to 1) &P(1\to 2) &P(1\to 3) &P(1\to 4)\\ P(2\to 1) &P(2\to 2) &P(2\to 3) &P(2\to 4)\\ P(3\to 1) &P(3\to 2) &P(3\to 3) &P(3\to 4)\\ P(4\to 1) &P(4\to 2) &P(4\to 3) &P(4\to 4) \end{array} \right)$$ where $P(1\to 2)$ is the probability of travelling to $2$ when at $1$. Then the first column of $A^2$ is: $$\left ( \begin{array}{c} P(1\to 1)P (1\to 1) + P (1\to 2) P(2\to 1) + P(1\to 3)P(3\to 1) + P(1\to 4) P(4\to 1)\\ P(2\to 1)P (1\to 1) + P (2\to 2) P(2\to 1) + P(2\to 3)P(3\to 1) + P(2\to 4) P(4\to 1)\\ P(3\to 1)P (1\to 1) + P (3\to 2) P(2\to 1) + P(3\to 3)P(3\to 1) + P(3\to 4) P(4\to 1) \\ P(4\to 1)P (1\to 1) + P(4\to 2) P(2\to 1) + P(4\to 3)P(3\to 1) + P(4\to 4) P(4\to 1) \end{array} \right )$$ and so on for the four columns. Now, the probability of getting from $1$ to $2$ in $x$ steps equals the sum of the probabilities of all the ways of getting from 1 to 2 in x steps : $P(1\to 2 \text{ (2 steps)}) = P(1\to 1)P(1\to 2)+ P(1\to 2)P (2\to 2)+ P(1\to 3)P(3\to 2)+\ldots$. Thus we have in $A^2$ the probability of getting from one point to another in two steps, as each entry in the matrix is the sum of probabilities of getting between the two points in two steps. Similarly, we find that if we cube the matrix, we are adding the probabilities of three-stage routes between two points, and so on for higher powers. Now, we look at what happens for higher powers of $A$. $$A^{20}= \left( \begin{array}{cccc} 0 &0 &0.008 &0.992 \\ 0.008 &0 &0 &0.992 \\ 0 &0016 &0 &0.984 \\ 0 &0 &0 &1 \end{array} \right)$$ $$A^{21}= \left( \begin{array}{cccc} 0.008 &0 &0 &0.992 \\ 0 &0.008 &0 &0.992 \\ 0 &0 &0.008 &0.992 \\ 0 &0 &0 &1 \end{array} \right)$$ $$A^{22}= \left( \begin{array}{cccc} 0 &0.008 &0 &0.992 \\ 0 &0 &0.004 &0.996 \\ 0.008 &0 &0 &0.992 \\ 0 &0 &0 &1 \end{array} \right)$$ First (and easiest) we explain the fourth row. In the network, an object at vertex (4) has a probability of 1 of returning to (4) at the next journey. It therefore always returns to (4) and never reaches any other vertex. The fourth row will remain at $0$ $0$ $0$ $1$ as an object at (4) will remain at (4). Next we come to the cycling of the three values in the loop at the left. At vertices (3) and (1), an object may only travel to one other vertex ((1) and (2) respectively), and at vertex (2) it travels to (3) (with a probability of $0.5$ which we ignore to explain the cycling). Therefore, an object at vertex (1) can travel to (2), then (3) then (1) again and so on. However, it moves on one vertex at a time so, if it cycles through the vertices, after n journeys the object will be at vertex $n (\text{mod } 3)$. Similarly, if it starts at (2), it will be at vertex $(n + 1)(\text{mod } 3)$ and, from (3), at $(n + 2)(\text{mod } 3)$. Lastly, we explain the numerical values of the probabilities. Every time an object is at vertex (2), it travels to (4) with a probability p where p = 0.5 . Therefore, it will only avoid going to (4) with a probability of $(1 - p)^m$ , where $m$ is the number of times the object is at vertex (2) and $m$ is roughly $n/3$. Therefore, the probability of an object starting at vertices (1), (2) or (3) arriving at (4) is $1 - (1 - p)^m$ . As once the object is at (4) it cannot travel to any other vertex, the probability of being at (4) after n journeys can only increase. With this information, we can predict the behaviour of $A^n$ as we let $n$ increase. All values in the fourth column will tend to one as $n$ tends to infinity, as they are given (roughly) by $1 - 0.5^{n/3}$ and $0.5^{n/3}$ tends to 0. The other nine values will cycle as shown above, in the manner : $$\left( \begin{array}{ccc} 0 &a &0 \\ 0 &0 &b \\ c &0 &0 \end{array} \right) \to \left( \begin{array}{ccc} 0 &0 &b \\ c/2 &0 &0 \\ 0 &a &0 \end{array} \right) \to \left( \begin{array}{ccc} c/2 &0 &0 \\ 0 &a/2 &0 \\ 0 &0 &b \end{array} \right) \to \left( \begin{array}{ccc} 0 &a/2 &0 \\ 0 &0 &b/2 \\ c/2 &0 &0 \end{array} \right)$$ and so on for $n=3x+1,\ 3x+2,\ 3x+3,\ 3(x+1)+1$ respectively where $x$ is an integer. The values of $a,\ b$ and $c$ will tend to zero as they are given by $0.5^m$ for $n$ stages where $m$ is of the order of $n/3$. Hence $$A^{100}= \left ( \begin{array}{cccc} 1 &1\times 10^{-10} &0 &0.9999999999 \\ 0 &0 &5\times 10^{-11} &0.99999999995\\ 1\times 10^{-10} &0 &0 &0.9999999999 \\ 0 &0 &0 &1 \end{array} \right)$$ The NRICH Project aims to enrich the mathematical experiences of all learners. To support this aim, members of the NRICH team work in a wide range of capacities, including providing professional development for teachers wishing to embed rich mathematical tasks into everyday classroom practice. More information on many of our other activities can be found here.

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http://math.stackexchange.com/questions/30330/fast-algorithm-for-solving-system-of-linear-equations/30802

# fast algorithm for solving system of linear equations I have a system of linear equations, $Ax=b$, with $N$ equations and $N$ unknowns ($N$ large) If I am interested in the solution for only one of the unknowns, what are the best approaches? for example, Assume N=50,000 and we want the solution for $x_1$ through $x_{100}$ only. is there any trick that does not require $O(n^{3})$ (or O(matrix inversion) ) for that ? - 1 Is $A$ full or sparse? Are you doing this once or many times with the same $A$? – joriki Apr 1 '11 at 16:56 A is not sparse and has many nonzero elements, however, the coefficients themselves are derived from a smaller set of variables. – mghandi Apr 2 '11 at 3:37 About your second question, yes I'm doing this many times. I am familiar with the LU decomposition trick to speed up when the coeffs matrix is unchanged. Is there any other tricks to do that even more efficiently? – mghandi Apr 2 '11 at 3:51 – Dirk Dec 15 '12 at 17:43 ## 2 Answers Unless your matrix is sparse or structured (e.g. Vandermonde, Hankel, or those other named matrix families that admit a fast solution method), there is not much hope of doing things better than $O(n^3)$ effort. Even if one were to restrict himself to solving for just one of the 50,000 variables, Cramer will demand computing two determinants for your answer, and the effort for computing a determinant is at least as much as decomposing/inverting a matrix to begin with. - [this should be a comment to @Juan Joder's answer or to the OP's question, but not enough repo] strang gilbert - linear algebra and its applications 3rd edition page 16 at the footer mentions that complexity had fallen already fallen below $O(Cn^{2.376})$ at the date of writing 1988, altough $C$ is so large that makes the algorithm impractical for most matrix sizes found in practice today. It does not mention the name of the algorithm though. personal guess: not sure about solving for single variables, but I don't think you can reduce complexity like that since the entire system is coupled. -

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http://mathoverflow.net/questions/63187?sort=votes

## prime powers between n and 2n ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Is there a result in the spirit of Bertrand-Chebyshev which talks about the existence of prime powers between n and 2n (or 3n or something like that) for n large? - ## 3 Answers For fixed $k$, the existence of a prime power $p^k$ between $n$ and $2n$ is (asymptotically) equivalent to the existence of a prime $p$ between $m$ and $\sqrt[k]{2} m$ where $m = \sqrt[k]{n}$. Bachraoui gives an elementary proof here that there exists a prime between $m$ and $\frac{3}{2} m$ for sufficiently large $m$. I remember reading on MO that it is known that similar elementary proofs exist for showing the existence of primes between $m$ and $(1 + \epsilon) m$ for any $\epsilon > 0$ and, furthermore, that the existence of these proofs itself depends on the PNT. However, I can't track down where I read this. - 3 Possibly it is this question you have in mind: mathoverflow.net/questions/53498/… – quid Apr 27 2011 at 19:00 Yes, that's the one. – Qiaochu Yuan Apr 27 2011 at 20:11 ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. Actually there is a power of 2. It goes to show the power of binary arithmetic ... : write 2n in binary and write zeroes after the initial one. - 1 In other words, if $2^{k - 1}$ is the largest power of 2 less than n, then $n \leq 2^k < 2n$? – Ryan Reich Apr 27 2011 at 18:14 Yes, you said it. – Charles Matthews Apr 27 2011 at 18:43 It follows from the prime number theorem that for fixed $k$, provided $n$ is sufficiently large, there is a prime $p$ such that $p^k$ is between $n$ and $2n$. Does this answer your question? - Thank you very much for your quick answer. Yes, it answers the question, and it goes to show the power of the PNT. – Chebolu Apr 27 2011 at 16:01 1 I wonder if there is an elementary proof of this without using PNT, like Erdős' proof of Bertrand's Postulate? – Tony Huynh Apr 27 2011 at 17:24

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http://mathhelpforum.com/pre-calculus/150765-equation-line-parallel.html

# Thread: 1. ## Equation of a line which is parallel I know it sounds funny but please help i got headache from this equation even thou i know its easy. Write down the equation of a line which is a) parallel to y=4x+2, and passes through 4 on the y- axis b) parallel to y=10x -5, passes through -1 on the y-axis Thanks For The Help In Advance 2. Write down the equation of a line which is a) parallel to y=4x+2, and passes through 4 on the y- axis b) parallel to y=10x -5, passes through -1 on the y-axis[/QUOTE] the equation of a line which is a) parallel to y=mx+b, and passes through c on the y-axis is y=mx+c. 3. Originally Posted by MrGenuouse a) parallel to y=4x+2, and passes through 4 on the y- axis Headaches are bad so i'll help you out with the first question and you can apply the same method to the second. We seek a line in the form of $y=mx+c$ . We need to find m (the gradient) and c (the y-intercept) . The gradient in the given equation is $m= 4$ and as the line we are after is parallel we can simply substitute this value into the new line giving $y=4x+c$ Now passing through 4 on the y-axis means the y-intercept is 4 so we can substitute that in for c and we are finished $y=4x+4$ How easy was that? 4. Thank You Very much for your help! My headache is over so I am really thankful about that too! i hope i will be able to gain more knowledge from this community as i am 16 years old boy who is really interested in maths.

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http://ams.org/bookstore?fn=20&arg1=chelsealist&ikey=CHEL-131-H

New Titles | FAQ | Keep Informed | Review Cart | Contact Us Quick Search (Advanced Search ) Browse by Subject General Interest Logic & Foundations Number Theory Algebra & Algebraic Geometry Discrete Math & Combinatorics Analysis Differential Equations Geometry & Topology Probability & Statistics Applications Mathematical Physics Math Education Return to List The Theory of Matrices, Volume 1 SEARCH THIS BOOK: AMS Chelsea Publishing 1959; 374 pp; hardcover Volume: 131 Reprint/Revision History: second AMS printing 2000 ISBN-10: 0-8218-1376-5 ISBN-13: 978-0-8218-1376-8 List Price: US\$55 Member Price: US\$49.50 Order Code: CHEL/131.H This item is also sold as part of the following set: CHELGANTSET This treatise, by one of Russia's leading mathematicians, gives in easily accessible form a coherent account of matrix theory with a view to applications in mathematics, theoretical physics, statistics, electrical engineering, etc. The individual chapters have been kept as far as possible independent of each other, so that the reader acquainted with the contents of Chapter 1 can proceed immediately to the chapters of special interest. Much of the material has been available until now only in the periodical literature. Reviews "This is an excellent textbook." -- Zentralblatt MATH From a review of the original Russian edition ... "The first part (10 chapters; "General theory") gives in satisfactory detail, and with more than customary completeness, the topics which belong to the main body of the ... subjects ... The point of view is broad and includes much abstract treatment ... "The number of subjects which the book treats well is great ... would appeal to a wide audience." -- Mathematical Reviews From a review of the English translation ... "The work is an outstanding contribution to matrix theory and contains much material not to be found in any other text." -- Mathematical Reviews Volume 1 • I. Matrices and operations on matrices: 1. Matrices. Basic notation; 2. Addition and multiplication of rectangular matrices; 3. Square matrices; 4. Compound matrices. Minors of the inverse matrix • II. The algorithm of Gauss and some of its applications: 1. Gauss's elimination method; 2. Mechanical interpretation of Gauss's algorithm; 3. Sylvester's determinant identity; 4. The decomposition of a square matrix into triangular factors; 5. The partition of a matrix into blocks. The technique of operating with partitioned matrices. The generalized algorithm of Gauss • III. Linear operators in an $$n$$-dimensional vector space: 1. Vector spaces; 2. A linear operator mapping an $$n$$-dimensional space into an $$m$$-dimensional space; 3. Addition and multiplication of linear operators; 4. Transformation of coordinates; 5. Equivalent matrices. The rank of an operator. Sylvester's inequality; 6. Linear operators mapping an $$n$$-dimensional space into itself; 7. Characteristic values and characteristic vectors of a linear operator; 8. Linear operators of simple structure • IV. The characteristic polynomial and the minimal polynomial of a matrix: 1. Addition and multiplication of matrix polynomials; 2. Right and left division of matrix polynomials; 3. The generalized Bézout theorem; 4. The characteristic polynomial of a matrix. The adjoint matrix; 5. The method of Faddeev for the simultaneous computation of the coefficients of the characteristic polynomial and of the adjoint matrix; 6. The minimal polynomial of a matrix • V. Functions of matrices: 1. Definition of a function of a matrix; 2. The Lagrange-Sylvester interpolation polynomial; 3. Other forms of the definition of $$f(A)$$. The components of the matrix $$A$$; 4. Representation of functions of matrices by means of series; 5. Application of a function of a matrix to the integration of a system of linear differential equations with constant coefficients; 6. Stability of motion in the case of a linear system • VI. Equivalent transformations of polynomial matrices. Analytic theory of elementary divisors: 1. Elementary transformations of a polynomial matrix; 2. Canonical form of a $$\lambda$$-matrix; 3. Invariant polynomials and elementary divisors of a polynomial matrix; 4. Equivalence of linear binomials; 5. A criterion for similarity of matrices; 6. The normal forms of a matrix; 7. The elementary divisors of the matrix $$f(A)$$; 8. A general method of constructing the transforming matrix; 9. Another method of constructing a transforming matrix • VII. The structure of a linear operator in an $$n$$-dimensional space: 1. The minimal polynomial of a vector and a space (with respect to a given linear operator); 2. Decomposition into invariant subspaces with co-prime minimal polynomials; 3. Congruence. Factor space; 4. Decomposition of a space into cyclic invariant subspaces; 5. The normal form of a matrix; 6. Invariant polynomials. Elementary divisors; 7. The Jordan normal form of a matrix; 8. Krylov's method of transforming the secular equation • VIII. Matrix equations: 1. The equation $$AX=XB$$; 2. The special case $$A=B$$. Commuting matrices; 3. The equation $$AX-XB=C$$; 4. The scalar equation $$f(X)=O$$; 5. Matrix polynomial equations; 6. The extraction of $$m$$-th roots of a non-singular matrix; 7. The extraction of $$m$$-th roots of a singular matrix; 8. The logarithm of a matrix • IX. Linear operators in a unitary space: 1. General considerations; 2. Metrization of a space; 3. Gram's criterion for linear dependence of vectors; 4. Orthogonal projection; 5. The geometrical meaning of the Gramian and some inequalities; 6. Orthogonalization of a sequence of vectors; 7. Orthonormal bases; 8. The adjoint operator; 9. Normal operators in a unitary space; 10. The spectra of normal, hermitian, and unitary operators; 11. Positive-semidefinite and positive-definite hermitian operators; 12. Polar decomposition of a linear operator in a unitary space. Cayley's formulas; 13. Linear operators in a euclidean space; 14. Polar decomposition of an operator and the Cayley formulas in a euclidean space; 15. Commuting normal operators • X. Quadratic and hermitian forms: 1. Transformation of the variables in a quadratic form; 2. Reduction of a quadratic form to a sum of squares. The law of inertia; 3. The methods of Lagrange and Jacobi of reducing a quadratic form to a sum of squares; 4. Positive quadratic forms; 5. Reduction of a quadratic form to principal axes; 6. Pencils of quadratic forms; 7. Extremal properties of the characteristic values of a regular pencil of forms; 8. Small oscillations of a system with $$n$$ degrees of freedom; 9. Hermitian forms; 10. Hankel forms • Bibliography • Index AMS Home | Comments: webmaster@ams.org © Copyright 2012, American Mathematical Society Privacy Statement

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http://physics.stackexchange.com/questions/tagged/classical-mechanics+action

# Tagged Questions 1answer 152 views ### What's the motivation behind the action principle? [closed] What's the motivation behind the action principle? Why does the action principle lead to Newtonian law? If Newton's law of motion is more fundamental so why doesn't one derive Lagrangians and ... 4answers 454 views ### Is the principle of least action a boundary value or initial condition problem? Here is a question that's been bothering me since I was a sophomore in university, and should have probably asked before graduating: In analytic (Lagrangian) mechanics, the derivation of the ... 2answers 206 views ### What is the significance of action? What is the physical interpretation of $$\int_{t_1}^{t_2} (T -V) dt$$ where, $T$ is Kinetic Energy and $V$ is potential energy. How does it give trajectory? 2answers 615 views ### Hamilton-Jacobi Equation In the Hamilton-Jacobi equation, we take the partial time derivative of the action. But the action comes from integrating the Lagrangian over time, so time seems to just be a dummy variable here and ... 2answers 281 views ### Is it circular reasoning to derive Newton's laws from action minimization? Usually, a typical example of the use of the action principle that I've read a lot is the derivation of Newton's equation (generalized to coordinate $q(t)$). However, in the classical mechanics ... 2answers 1k views ### Deriving the Lagrangian for a free particle I'm a newbie in physics. Sorry, if the following questions are dumb. I began reading "Mechanics" by Landau and Lifshitz recently and hit a few roadblocks right away. Proving that a free particle ... 4answers 725 views ### Why can't any term which is added to the Lagrangian be written as a total derivative (or divergence)? All right, I know there must be an elementary proof of this, but I am not sure why I never came across it before. Adding a total time derivative to the Lagrangian (or a 4D divergence of some 4 ... 2answers 197 views ### How do I show that there exists variational/action principle for a given classical system? We see variational principles coming into play in different places such as Classical Mechanics (Hamilton's principle which gives rise to the Euler-Lagrange equations), Optics (in the form of Fermat's ... 6answers 2k views ### Why the Principle of Least Action? I'll be generous and say it might be reasonable to assume that nature would tend to minimize, or maybe even maximize, the integral over time of $T-V$. Okay, fine. You write down the action ... 1answer 203 views ### What variables does the action $S$ depend on? Action is defined as, $$S ~=~ \int L(q, q', t) dt,$$ but my question is what variables does $S$ depend on? Is $S = S(q, t)$ or $S = S(q, q', t)$ where $q' := \frac{dq}{dt}$? In ... 1answer 484 views ### Does Action in Classical Mechanics have a Interpretation? [duplicate] Possible Duplicate: Hamilton's Principle The Lagrangian formulation of Classical Mechanics seem to suggest strongly that "action" is more than a mathematical trick. I suspect strongly ...

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http://mathoverflow.net/revisions/68599/list

## Return to Answer 2 edited body; deleted 1 characters in body This is now an old thread, but I just came across it. The question you are asking was asked by Behrends a couple of years ago, and he knew about the counter example on the plane that Gjergji points out. Behrends was specifically asking for an incomplete metric space that is a subset of the line that and has the fixed point property. I came up with a partition of $\mathbb R$ into two dense sets $X_0$ and $X_1$ such that for all $i\in{0,1}$, every map $f:X_i\to X_i$ that satisfies $$\forall x,y\in X_i(x\not=y\rightarrow|f(x)-f(y)|<|x-y|)$$ is actually constant (and in particular has a fixed point). The $X_i$ can be constructed by an easy transfinite recursion and there is no control of how complicated the sets are. This is the huge advantage of Elekes' examples which are Borel. The notes with my proof are here: http://www.hausdorff-center.uni-bonn.de/people/geschke/papers/Behrends.dvihttp://www.hausdorff-center.uni-bonn.de/people/geschke/papers/Behrends.pdf 1 This is now an old thread, but I just came across it. The question you are asking was asked by Behrends a couple of years ago, and he knew about the counter example on the plane that Gjergji points out. Behrends was specifically asking for an incomplete metric space that is a subset of the line that has the fixed point property. I came up with a partition of $\mathbb R$ into two dense sets $X_0$ and $X_1$ such that for all $i\in{0,1}$, every map $f:X_i\to X_i$ that satisfies $$\forall x,y\in X_i(x\not=y\rightarrow|f(x)-f(y)|<|x-y|)$$ is actually constant (and in particular has a fixed point). The $X_i$ can be constructed by an easy transfinite recursion and there is no control of how complicated the sets are. This is the huge advantage of Elekes' examples which are Borel. The notes with my proof are here: http://www.hausdorff-center.uni-bonn.de/people/geschke/papers/Behrends.dvi

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http://mathhelpforum.com/calculus/36478-need-help-few-important-homework-questions.html

# Thread: 1. ## Need help with a few important homework questions Graph y^2=2xy The thing with this is that I just don't know how to solve for Y. Apparently I'm not supposed to divide Y from both sides? Also I have no idea how to do this next one. A machine is rolling a metal cylinder under pressure. The radius of the cylinder is decreasing at a constant rate of 0.05 inches per second and the volume is a constant 128(pi) cubic inches. At what rate is the length h changing when the radius r is 1.8 inches? And I'll just throw in this third one cuz I'm not sure if I did it right. What Values of c are guaranteed by the Mean Value Theorem for f(x)=(x-2)^3 on the interval [-1,2]? I got 2-(the square root of)3. Thanks 2. Originally Posted by sean1058 Graph y^2=2xy The thing with this is that I just don't know how to solve for Y. Apparently I'm not supposed to divide Y from both sides? Also I have no idea how to do this next one. A machine is rolling a metal cylinder under pressure. The radius of the cylinder is decreasing at a constant rate of 0.05 inches per second and the volume is a constant 128(pi) cubic inches. At what rate is the length h changing when the radius r is 1.8 inches? And I'll just throw in this third one cuz I'm not sure if I did it right. What Values of c are guaranteed by the Mean Value Theorem for f(x)=(x-2)^3 on the interval [-1,2]? I got 2-(the square root of)3. Thanks For the first one just plot points..if you have to for the mean value one did you solve for c in the following $3(c-2)^2=\frac{0+9}{2+1}$? 3. No I have this $3(c-2)^2=\frac{0+27}{2+1}$ cuz the 3 is to the third power and I can't just plot points for that one, I have to solve it first. 4. Originally Posted by sean1058 Graph y^2=2xy The thing with this is that I just don't know how to solve for Y. Apparently I'm not supposed to divide Y from both sides? Also I have no idea how to do this next one. A machine is rolling a metal cylinder under pressure. The radius of the cylinder is decreasing at a constant rate of 0.05 inches per second and the volume is a constant 128(pi) cubic inches. At what rate is the length h changing when the radius r is 1.8 inches? And I'll just throw in this third one cuz I'm not sure if I did it right. What Values of c are guaranteed by the Mean Value Theorem for f(x)=(x-2)^3 on the interval [-1,2]? I got 2-(the square root of)3. Thanks $y^2-2xy=0\Rightarrow{(y-x)^2-x^2=0}$ so $(y-x)^2=x^2\Rightarrow{|y-x|=|x|}\Rightarrow{y=x\pm|x|}$ 5. Originally Posted by sean1058 Graph y^2=2xy The thing with this is that I just don't know how to solve for Y. Apparently I'm not supposed to divide Y from both sides? Thanks Well, you can't divide each side by y because it's $y^2$, so to undo that you would have to take the square root to get plain old y. Do you know polar coordinates? That could help possibly. Otherwise, the comment above this one is the best you will get. 6. thanks if anybody can help me with that middle one it would be most appreciated 7. nevermind i got it

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http://physics.stackexchange.com/questions/52821/for-an-isolated-system-can-the-entropy-decrease-or-increase/52823

# For an isolated system, can the entropy decrease or increase? In any sizable system, the number of equilibrium states are much, much greater then the number of non-equilibrium states. Since each accessible micro state is equally probably, it is overwhelmingly likely that we will find the system in an equilibrium state. However, for a closed system, one that does not interact with any external system, the number of micro states is fixed. Therefore the entropy is fixed. At any given time, the system will be in one of its micro states. Even if the system is in a micro state that does not look like equilibrium from a macroscopic point of view, its entropy will remain the same, since entropy is a property of all the micro states, not just a given micro state. So, are the number of accessible micro states (and hence entropy) of an isolated system constant? - ## 5 Answers Isolated system: Since the matter, energy, and momentum is fixed, the total number of microstates available that satisfy these constraints is fixed/constant. So is the entropy constant? Yes, if the system is in equilibrium. No, if the system is not in equilibrium. What does that mean in terms of microstates? If the system is in equilibrium, all these microstates are equally probable and the system visits each of these microstates over the course of time (also known as ergodic hypothesis). Therefore the each micostate has equal probability and $S=k ln\Omega$. In such a state there is no more increase in entropy possible. If the system is in non-equilibrium the system doesn't have equal probability of being in every microstate. In fact if the system is stuck in non-equilibrium ( e.g., a hot part and cold part in the box separated by a thermally insulating wall) it cannot access some microstates at all. Hence the system is restricted to fewer microstates. Technically, there is no unique global thermodyanmic state for the whole system and you cannot define entropy. But you can calculate entropy by summing up entropy of different local equilibrium parts (e.g., entropy of the hold part and cold part separately). Since the total number of microstates you can access is small, entropy is lesser than what it could be if you remove that thermally insulating wall letting the system equilibrate. Thus, when you equilibrate more microstates become accessible. Think of the system spreading in the phase space. Thus entropy increases. Once the system has reached complete thermodynamic equilibrium no more entropy increase in possible. All allowable microstates have been made accessible and equally probable! - Your first sentence, answers, I think, the question. The rest is a subject for another question. – yalis Feb 2 at 2:20 +1 I just had a long conversation with a grad student friend of mine, and we came to essentially the same conclusion as this answer. This is a much better answer than mine. – joshphysics Feb 2 at 2:23 Glad to be of help. – Sankaran Feb 2 at 16:15 Since entropy is $$S=k_b \ln \Omega,$$ it is synonymous to a certain phase space volume $\Omega$ which we have to fix somehow. Phase space volume is something like a counter of possible states of the system (in the most simple case, the positions and momenta of all particles). This is (normally) done by setting some macroscopic parameters, like temperature, pressure, number of particles etc. So, if you have a closed system (i.e. a box where nothing can get in or out), you can e.g. fix volume, energy and the number of particles. You end up with $S(E,V,N)=k_b \ln \Omega(E,V,N)$. This can be understood as the number of all possible configurations inside the box which have the total energy $E$, the total possible volume $V$ and consist of $N$ particles - including all the stuff in the box on the left side, on the right side, evenly distributed, half of it moving and the other half standing still etc. $S$ is a constant number because $E$, $V$ and $N$ are fixed in a closed system. The problem is that you can make no statement about the system whatsoever. But - instead of $S(E,V,N)$, you can look at other dependencies of $S$. Let's say you divide up the box into two equally sized parts, a left part and a right part. $N_1$ is the number of particles in the left part, and $N_2$ is the number of particles in the right part (also $N_1+N_2=N$). Now, you can calculate $S(E,V,N_1,N_2)$ and choose $N_1$ and $N_2$. That way, $S$ can change because you can choose $N_1$ and $N_2$ (particles may go from the left to the right part and vice versa). The important point is that $S$ is much greater for $N_1\approx N_2$ than for any other values. This means that most of the possible configurations have approximately the same number of particles on the left and on the right side. Because every configuration is equally probable, most of the time, the system will look like this. But not all the time, though - even when $S(E,V,N_1,N_2)=0$, there is exactly one microscopic configuration which fulfils the macroscopic parameters, and it will occur at some time (if the system is ergodic, but that is a different story). In the given example, the entropy will increase and decrease all the time when particles go from the left half to the right half. But by definition, the percentage of time when the system is in a specific configuration $N_1$,$N_2$ is proportional to $\Omega(E,V,N_1,N_2)=\exp\left( \frac {S(E,V,N_1,N_2)} {k_b}\right)$ (I just thought of this last formula, please correct me if I'm wrong). In conclusion: The term entropy is meaningless without the parameters one wants to fix/vary. Funny example: Let's calculate $S(N_{\textrm{people alive on earth}})$ for the whole universe which describes the number of possible configurations of the universe for a given number of living people on earth. $S(0)$ will probably be much bigger than for any other $N_{paoe}$, which means that, most of the time, there will be no people on earth and the entropy will be huge (this is my interpretation of the the heat death of the universe). Then, they might come back and the entropy will be very small. - ok, but in your example, you are dividing the isolated system into two non-isolated, interacting subsystems (left and right). The entropy of either system can increase or decrease but on the whole we are just sampling different micro states. Entropy should not be dependent on the particular micro state that the system is currently sampling. It only depends on the total number of micro states. In a way that seems to contradict intuition. The universe is closed (?) but it seems that its entropy will increase until heat death is reached. Yet, the number of micro states of the universe is fixed? – yalis Feb 2 at 1:17 The division is necessary for the entropy to be able to change. If the entropy only depends on constant parameters (e.g. E, V, N), it is constant. This also applies to a closed universe. For other parameters (e.g. $N_1$, rotation speed of mars, number of posts on physics.SE), the entropy changes (see my last example). – Rafael Reiter Feb 2 at 10:01 Yes the entropy of an isolated system can increase. Here's an example. Take a system consisting of an ideal gas in a thermally insulated box. Let the initial condition of the system be such that the gas occupies the left half of the box with the right side being vacuum. After the initial time, the gas will expand freely to fill the box. Thermal insulation and free expansion guarantee that $Q = 0$ and $W = 0$, so by the first law $\Delta U = 0$. The entropy of the system will be larger in this final state because of this equation (which gives ideal gas entropy) and the fact that $\Delta U = 0$ and $\Delta N$ = 0 while $\Delta V > 0$. Hope that helps! Cheers! - 2 I actually wrote up this example in the initial draft of the question, but deleted it. The thing here is this: Are we not considering the left half of the system as a sub-system of the closed the system. This subsystem interacts with the whole system by expanding. You are "adding volume" to the subsystem and thus increasing the number of micro states. Thus, I would think that a truly isolated system cannot expand. – yalis Feb 2 at 0:49 1 Another angle. Say gas particles uniformly distributed at some initial time. It's quite possible, though extremely unlikely, that at at some point in the future, all the particles were on one side of the box. Has the entropy of the system decreased? Not sure how it can, since entropy is associated with the number of micro states, which has not changed. – yalis Feb 2 at 0:50 Yea you make a very good point that I was actually thinking about as well when I was writing up this example. To put it another way, viewed as a quantum system, the Von-Neumann entropy cannot change since isolated systems undergo unitary time evolution and the Von-Neumann entropy remains invariant under such an evolution of the density matrix of the system. I've heard that a resolution of such issues has to do with quantum "decoherence," but I'm by no means an expert on such things. – joshphysics Feb 2 at 1:26 I'm almost tempted to start another thread asking about this specific example of the gas in the box and how it can be reconciled with the quantum/microscopic considerations we are both concerned about. I'm not entirely sure such questions have even been resolved to be honest. – joshphysics Feb 2 at 1:36 Yes, if we say the entropy is just the log of the number of accessible microstates, then the entropy cannot change for an isolated system as it evolves. But this contradicts common sense. For an isolated system, we must introduce a notion of coarse graining for entropy to be a useful concept. Coarse graining can be done by dividing the system into subsystems, and then adding the entropies if the subsystems. This coarse grained entropy can and will change over time, as energy is exchanged between the different subsystems. The exact value of the coarse grained entropy will in general depend on the details of how you form the subsystems. The coarse grained entropy should eventually approach the original fine grained entropy as the system reaches equilibrium. - Yes, the idea of breaking an isolated system into subsystems. A given subsystem can have its entropy increase or decrease, until such time as all subsystems have reached maximum entropy. This is in line with the the following statement attributed to Rudolf Clausius on wikipedia: "The entropy of an isolated system not in equilibrium will tend to increase over time, approaching a maximum value at equilibrium" Suggesting that an isolated system can be away from equilibrium. Yet the information content in an isolated system cannot change, so there's still a bit of fuzziness here. – yalis Feb 2 at 2:32 the entropy can decrease as well. On average the entropy will increases (answer above), but it is possible that the entropy decreases. - You are right, but if it is a macroscopic system, don't hold your breath waiting for it to happen. – Paul J. Gans Feb 2 at 1:16

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http://www.nag.com/numeric/CL/nagdoc_cl23/html/E05/e05sac.html

NAG Library Function Documentnag_glopt_bnd_pso (e05sac) Note: this function uses optional arguments to define choices in the problem specification and in the details of the algorithm. If you wish to use default settings for all of the optional arguments, you need only read Sections 1 to 9 of this document. If, however, you wish to reset some or all of the settings please refer to Section 10 for a detailed description of the algorithm and to Section 11 for a detailed description of the specification of the optional arguments. 1 Purpose nag_glopt_bnd_pso (e05sac) is designed to search for the global minimum or maximum of an arbitrary function, using Particle Swarm Optimization (PSO). Derivatives are not required, although these may be used by an accompanying local minimization function if desired. nag_glopt_bnd_pso (e05sac) is essentially identical to nag_glopt_nlp_pso (e05sbc), but with a simpler interface and with various optional arguments removed; otherwise most arguments are identical. In particular, nag_glopt_bnd_pso (e05sac) does not handle general constraints. 2 Specification #include <nag.h> #include <nage05.h> void nag_glopt_bnd_pso (Integer ndim, Integer npar, double xb[], double *fb, const double bl[], const double bu[], void (*objfun)(Integer *mode, Integer ndim, const double x[], double *objf, double vecout[], Integer nstate, Nag_Comm *comm), void (*monmod)(Integer ndim, Integer npar, double x[], const double xb[], double fb, const double xbest[], const double fbest[], const Integer itt[], Nag_Comm *comm, Integer *inform), Integer iopts[], double opts[], Nag_Comm *comm, Integer itt[], Integer *inform, NagError *fail) Before calling nag_glopt_bnd_pso (e05sac), nag_glopt_opt_set (e05zkc) must be called with optstr set to ‘Initialize = e05sac’. Optional arguments may also be specified by calling nag_glopt_opt_set (e05zkc) before the call to nag_glopt_bnd_pso (e05sac). 3 Description nag_glopt_bnd_pso (e05sac) uses a stochastic method based on Particle Swarm Optimization (PSO) to search for the global optimum of a nonlinear function $F$, subject to a set of bound constraints on the variables. In the PSO algorithm (see Section 10), a set of particles is generated in the search space, and advances each iteration to (hopefully) better positions using a heuristic velocity based upon inertia, cognitive memory and global memory. The inertia is provided by a decreasingly weighted contribution from a particles current velocity, the cognitive memory refers to the best candidate found by an individual particle and the global memory refers to the best candidate found by all the particles. This allows for a global search of the domain in question. Further, this may be coupled with a selection of local minimization functions, which may be called during the iterations of the heuristic algorithm, the interior phase, to hasten the discovery of locally optimal points, and after the heuristic phase has completed to attempt to refine the final solution, the exterior phase. Different options may be set for the local optimizer in each phase. Without loss of generality, the problem is assumed to be stated in the following form: $minimize x∈Rndim Fx subject to ℓ ≤ x ≤ u ,$ where the objective $F\left(\mathbf{x}\right)$ is a scalar function, $\mathbf{x}$ is a vector in ${R}^{\mathit{ndim}}$ and the vectors $\mathbf{\ell }\le \mathbf{u}$ are lower and upper bounds respectively for the $\mathit{ndim}$ variables. The objective function may be nonlinear. Continuity of $F$ is not essential. For functions which are smooth and primarily unimodal, faster solutions will almost certainly be achieved by using Chapter e04 functions directly. For functions which are smooth and multi-modal, gradient dependent local minimization functions may be coupled with nag_glopt_bnd_pso (e05sac). For multi-modal functions for which derivatives cannot be provided, particularly functions with a significant level of noise in their evaluation, nag_glopt_bnd_pso (e05sac) should be used either alone, or coupled with nag_opt_simplex_easy (e04cbc). The $\mathit{ndim}$ lower and upper box bounds on the variable $\mathbf{x}$ are included to initialize the particle swarm into a finite hypervolume, although their subsequent influence on the algorithm is user determinable (see the option ${\mathbf{Boundary}}$ in Section 11). It is strongly recommended that sensible bounds are provided for all variables. nag_glopt_bnd_pso (e05sac) may also be used to maximize the objective function (see the option ${\mathbf{Optimize}}$). Due to the nature of global optimization, unless a predefined target is provided, there is no definitive way of knowing when to end a computation. As such several stopping heuristics have been implemented into the algorithm. If any of these is achieved, nag_glopt_bnd_pso (e05sac) will exit with NW_SOLUTION_NOT_GUARANTEED, and the parameter inform will indicate which criteria was reached. See inform for more information. In addition, you may provide your own stopping criteria through monmod and objfun. nag_glopt_nlp_pso (e05sbc) provides a comprehensive interface, allowing for the inclusion of general nonlinear constraints. 4 References Gill P E, Murray W and Wright M H (1981) Practical Optimization Academic Press Kennedy J and Eberhart R C (1995) Particle Swarm Optimization Proceedings of the 1995 IEEE International Conference on Neural Networks 1942–1948 Koh B, George A D, Haftka R T and Fregly B J (2006) Parallel Asynchronous Particle Swarm Optimization International Journal for Numerical Methods in Engineering 67(4) 578–595 Vaz A I and Vicente L N (2007) A Particle Swarm Pattern Search Method for Bound Constrained Global Optimization Journal of Global Optimization 39(2) 197–219 Kluwer Academic Publishers 5 Arguments Note: for descriptions of the symbolic variables, see Section 10. 1: ndim – IntegerInput On entry: $\mathit{ndim}$, the number of dimensions. Constraint: ${\mathbf{ndim}}\ge 1$. 2: npar – IntegerInput On entry: $\mathit{npar}$, the number of particles to be used in the swarm. Assuming all particles remain within bounds, each complete iteration will perform at least npar function evaluations. Otherwise, significantly fewer objective function evaluations may be performed. Suggested value: ${\mathbf{npar}}=10×{\mathbf{ndim}}$. Constraint: ${\mathbf{npar}}\ge 5$. 3: xb[ndim] – doubleOutput On exit: the location of the best solution found, $\stackrel{~}{\mathbf{x}}$, in ${R}^{\mathit{ndim}}$. 4: fb – double *Output On exit: the objective value of the best solution, $\stackrel{~}{f}=F\left(\stackrel{~}{\mathbf{x}}\right)$. 5: bl[ndim] – const doubleInput 6: bu[ndim] – const doubleInput On entry: ${\mathbf{bl}}$ is $\mathbf{\ell }$, the array of lower bounds, bu is $\mathbf{u}$, the array of upper bounds. The ndim entries in bl and bu must contain the lower and upper simple (box) bounds of the variables respectively. These must be provided to initialize the sample population into a finite hypervolume, although their subsequent influence on the algorithm is user determinable (see the option ${\mathbf{Boundary}}$ in Section 11). If ${\mathbf{bl}}\left[i-1\right]={\mathbf{bu}}\left[i-1\right]$ for any $i\in \left\{1,\dots ,{\mathbf{ndim}}\right\}$, variable $i$ will remain locked to ${\mathbf{bl}}\left[i-1\right]$ regardless of the ${\mathbf{Boundary}}$ option selected. It is strongly advised that you place sensible lower and upper bounds on all variables, even if your model allows for variables to be unbounded (using the option ${\mathbf{Boundary}}=\mathrm{ignore}$) since these define the initial search space. Constraints: • ${\mathbf{bl}}\left[\mathit{i}-1\right]\le {\mathbf{bu}}\left[\mathit{i}-1\right]$, for $\mathit{i}=1,2,\dots ,{\mathbf{ndim}}$; • ${\mathbf{bl}}\left[i-1\right]\ne {\mathbf{bu}}\left[i-1\right]$ for at least one $i\in \left\{1,\dots ,{\mathbf{ndim}}\right\}$. 7: objfun – function, supplied by the userExternal Function objfun must, depending on the value of mode, calculate the objective function and/or calculate the gradient of the objective function for a $\mathit{ndim}$-variable vector $\mathbf{x}$. Gradients are only required if a local minimizer has been chosen which requires gradients. See the option ${\mathbf{Local Minimizer}}$ for more information. The specification of objfun is: void objfun (Integer *mode, Integer ndim, const double x[], double *objf, double vecout[], Integer nstate, Nag_Comm *comm) 1: mode – Integer *Input/Output On entry: indicates which functionality is required. ${\mathbf{mode}}=0$ $F\left(\mathbf{x}\right)$ should be returned in objf. The value of objf on entry may be used as an upper bound for the calculation. Any expected value of $F\left(\mathbf{x}\right)$ that is greater than objf may be approximated by this upper bound; that is objf can remain unaltered. ${\mathbf{mode}}=1$ ${\mathbf{Local Minimizer}}='\mathrm{e04ucc}'$ only First derivatives can be evaluated and returned in vecout. Any unaltered elements of vecout will be approximated using finite differences. ${\mathbf{mode}}=2$ ${\mathbf{Local Minimizer}}='\mathrm{e04ucc}'$ only $F\left(\mathbf{x}\right)$ must be calculated and returned in objf, and available first derivatives can be evaluated and returned in vecout. Any unaltered elements of vecout will be approximated using finite differences. ${\mathbf{mode}}=5$ $F\left(\mathbf{x}\right)$ must be calculated and returned in objf. The value of objf on entry may not be used as an upper bound. ${\mathbf{mode}}=6$ ${\mathbf{Local Minimizer}}='\mathrm{e04dgc}'$ only All first derivatives must be evaluated and returned in vecout. ${\mathbf{mode}}=7$ ${\mathbf{Local Minimizer}}='\mathrm{e04dgc}'$ only $F\left(\mathbf{x}\right)$ must be calculated and returned in objf, and all first derivatives must be evaluated and returned in vecout. On exit: if the value of mode is set to be negative, then nag_glopt_bnd_pso (e05sac) will exit as soon as possible with NE_USER_STOP and ${\mathbf{inform}}={\mathbf{mode}}$. 2: ndim – IntegerInput On entry: the number of dimensions. 3: x[ndim] – const doubleInput On entry: $\mathbf{x}$, the point at which the objective function and/or its gradient are to be evaluated. 4: objf – double *Input/Output On entry: the value of objf passed to objfun varies with the argument mode. ${\mathbf{mode}}=0$ objf is an upper bound for the value of $F\left(\mathbf{x}\right)$, often equal to the best value of $F\left(\mathbf{x}\right)$ found so far by a given particle. Only objective function values less than the value of objf on entry will be used further. As such this upper bound may be used to stop further evaluation when this will only increase the objective function value above the upper bound. ${\mathbf{mode}}=1$, $2$, $5$, $6$ or $7$ objf is meaningless on entry. On exit: the value of objf returned varies with the argument mode. ${\mathbf{mode}}=0$ objf must be the value of $F\left(\mathbf{x}\right)$. Only values of $F\left(\mathbf{x}\right)$ strictly less than objf on entry need be accurate. ${\mathbf{mode}}=1$ or $6$ Need not be set. ${\mathbf{mode}}=2$, $5$ or $7$ $F\left(\mathbf{x}\right)$ must be calculated and returned in objf. The entry value of objf may not be used as an upper bound. 5: vecout[ndim] – doubleInput/Output On entry: if ${\mathbf{Local Minimizer}}=\mathrm{e04ucc}$, the values of vecout are used internally to indicate whether a finite difference approximation is required. See nag_opt_nlp (e04ucc). On exit: the required values of vecout returned to the calling function depend on the value of mode. ${\mathbf{mode}}=0$ or $5$ The value of vecout need not be set. ${\mathbf{mode}}=1$ or $2$ vecout can contain components of the gradient of the objective function $\frac{\partial F}{\partial {x}_{i}}$ for some $i=1,2,\dots {\mathbf{ndim}}$, or acceptable approximations. Any unaltered elements of vecout will be approximated using finite differences. ${\mathbf{mode}}=6$ or $7$ vecout must contain the gradient of the objective function $\frac{\partial F}{\partial {x}_{i}}$ for all $i=1,2,\dots {\mathbf{ndim}}$. Approximation of the gradient is strongly discouraged, and no finite difference approximations will be performed internally (see nag_opt_conj_grad (e04dgc)). 6: nstate – IntegerInput On entry: nstate indicates various stages of initialization throughout the function. This allows for permanent global arguments to be initialized the least number of times. For example, you may initialize a random number generator seed. ${\mathbf{nstate}}=2$ objfun is called for the very first time. You may save computational time if certain data must be read or calculated only once. ${\mathbf{nstate}}=1$ objfun is called for the first time by a NAG local minimization function. You may save computational time if certain data required for the local minimizer need only be calculated at the initial point of the local minimization. ${\mathbf{nstate}}=0$ Used in all other cases. 7: comm – Nag_Comm * Pointer to structure of type Nag_Comm; the following members are relevant to objfun. user – double * iuser – Integer * p – Pointer The type Pointer will be void *. Before calling nag_glopt_bnd_pso (e05sac) you may allocate memory and initialize these pointers with various quantities for use by objfun when called from nag_glopt_bnd_pso (e05sac) (see Section 3.2.1 in the Essential Introduction). 8: monmod – function, supplied by the userExternal Function A user-specified monitoring and modification function. monmod is called once every complete iteration after a finalization check. It may be used to modify the particle locations that will be evaluated at the next iteration. This permits the incorporation of algorithmic modifications such as including additional advection heuristics and genetic mutations. monmod is only called during the main loop of the algorithm, and as such will be unaware of any further improvement from the final local minimization. If no monitoring and/or modification is required, monmod may be NULLFN. The specification of monmod is: void monmod (Integer ndim, Integer npar, double x[], const double xb[], double fb, const double xbest[], const double fbest[], const Integer itt[], Nag_Comm *comm, Integer *inform) 1: ndim – IntegerInput On entry: the number of dimensions. 2: npar – IntegerInput On entry: the number of particles. 3: x[${\mathbf{ndim}}×{\mathbf{npar}}$] – doubleInput/Output Note: the $i$th component of the $j$th particle, ${x}_{j}\left(i\right)$, is stored in ${\mathbf{x}}\left[\left(j-1\right)×{\mathbf{ndim}}+i-1\right]$. On entry: the npar particle locations, ${\mathbf{x}}_{j}$, which will currently be used during the next iteration unless altered in monmod. On exit: the particle locations to be used during the next iteration. 4: xb[ndim] – const doubleInput On entry: the location, $\stackrel{~}{\mathbf{x}}$, of the best solution yet found. 5: fb – doubleInput On entry: the objective value, $\stackrel{~}{f}=F\left(\stackrel{~}{\mathbf{x}}\right)$, of the best solution yet found. 6: xbest[${\mathbf{ndim}}×{\mathbf{npar}}$] – const doubleInput Note: the $i$th component of the position of the $j$th particle's cognitive memory, ${\stackrel{^}{x}}_{j}\left(i\right)$, is stored in ${\mathbf{xbest}}\left[\left(j-1\right)×{\mathbf{ndim}}+i-1\right]$. On entry: the locations currently in the cognitive memory, ${\stackrel{^}{\mathbf{x}}}_{\mathit{j}}$, for $\mathit{j}=1,2,\dots ,{\mathbf{npar}}$ (see Section 10). 7: fbest[npar] – const doubleInput On entry: the objective values currently in the cognitive memory, $F\left({\stackrel{^}{\mathbf{x}}}_{\mathit{j}}\right)$, for $\mathit{j}=1,2,\dots ,{\mathbf{npar}}$. 8: itt[$6$] – const IntegerInput On entry: iteration and function evaluation counters (see description of itt below). 9: comm – Nag_Comm * Pointer to structure of type Nag_Comm; the following members are relevant to monmod. user – double * iuser – Integer * p – Pointer The type Pointer will be void *. Before calling nag_glopt_bnd_pso (e05sac) you may allocate memory and initialize these pointers with various quantities for use by monmod when called from nag_glopt_bnd_pso (e05sac) (see Section 3.2.1 in the Essential Introduction). 10: inform – Integer *Input/Output On entry: ${\mathbf{inform}}=0$ On exit: setting ${\mathbf{inform}}<0$ will cause near immediate exit from nag_glopt_bnd_pso (e05sac). This value will be returned as inform with NE_USER_STOP. You need not set inform unless you wish to force an exit. 9: iopts[$\mathit{dim}$] – IntegerCommunication Array Note: the dimension of the array iopts, corresponding to the array length liopts (see e05zkc), must be at least $\mathrm{100}$. On entry: optional parameter array as generated and possibly modified by calls to nag_glopt_opt_set (e05zkc). The contents of iopts MUST NOT be modified directly between calls to nag_glopt_bnd_pso (e05sac), nag_glopt_opt_set (e05zkc) or nag_glopt_opt_get (e05zlc). 10: opts[$\mathit{dim}$] – doubleCommunication Array Note: the dimension of the array opts, corresponding to the array length lopts (see e05zkc), must be at least $\mathrm{100}$. On entry: optional parameter array as generated and possibly modified by calls to nag_glopt_opt_set (e05zkc). The contents of opts MUST NOT be modified directly between calls to nag_glopt_bnd_pso (e05sac), nag_glopt_opt_set (e05zkc) or nag_glopt_opt_get (e05zlc). 11: comm – Nag_Comm *Communication Structure The NAG communication argument (see Section 3.2.1.1 in the Essential Introduction). 12: itt[$6$] – IntegerOutput On exit: integer iteration counters for nag_glopt_bnd_pso (e05sac). ${\mathbf{itt}}\left[0\right]$ Number of complete iterations. ${\mathbf{itt}}\left[1\right]$ Number of complete iterations without improvement to the current optimum. ${\mathbf{itt}}\left[2\right]$ Number of particles converged to the current optimum. ${\mathbf{itt}}\left[3\right]$ Number of improvements to the optimum. ${\mathbf{itt}}\left[4\right]$ Number of function evaluations performed. ${\mathbf{itt}}\left[5\right]$ Number of particles reset. 13: inform – Integer *Output On exit: indicates which finalization criterion was reached. The possible values of inform are: inform Meaning $<0$ Exit from a user-supplied subroutine. 0 nag_glopt_bnd_pso (e05sac) has detected an error and terminated. 1 The provided objective target has been achieved. (${\mathbf{Target Objective Value}}$). 2 The standard deviation of the location of all the particles is below the set threshold (${\mathbf{Swarm Standard Deviation}}$). If the solution returned is not satisfactory, you may try setting a smaller value of ${\mathbf{Swarm Standard Deviation}}$, or try adjusting the options governing the repulsive phase (${\mathbf{Repulsion Initialize}}$, ${\mathbf{Repulsion Finalize}}$). 3 The total number of particles converged (${\mathbf{Maximum Particles Converged}}$) to the current global optimum has reached the set limit. This is the number of particles which have moved to a distance less than ${\mathbf{Distance Tolerance}}$ from the optimum with regard to the ${L}^{2}$ norm. If the solution is not satisfactory, you may consider lowering the ${\mathbf{Distance Tolerance}}$. However, this may hinder the global search capability of the algorithm. 4 The maximum number of iterations without improvement (${\mathbf{Maximum Iterations Static}}$) has been reached, and the required number of particles (${\mathbf{Maximum Iterations Static Particles}}$) have converged to the current optimum. Increasing either of these options will allow the algorithm to continue searching for longer. Alternatively if the solution is not satisfactory, re-starting the application several times with ${\mathbf{Repeatability}}=\mathrm{OFF}$ may lead to an improved solution. 5 The maximum number of iterations (${\mathbf{Maximum Iterations Completed}}$) has been reached. If the number of iterations since improvement is small, then a better solution may be found by increasing this limit, or by using the option ${\mathbf{Local Minimizer}}$ with corresponding exterior options. Otherwise if the solution is not satisfactory, you may try re-running the application several times with ${\mathbf{Repeatability}}=\mathrm{OFF}$ and a lower iteration limit, or adjusting the options governing the repulsive phase (${\mathbf{Repulsion Initialize}}$, ${\mathbf{Repulsion Finalize}}$). 6 The maximum allowed number of function evaluations (${\mathbf{Maximum Function Evaluations}}$) has been reached. As with ${\mathbf{inform}}=5$, increasing this limit if the number of iterations without improvement is small, or decreasing this limit and running the algorithm multiple times with ${\mathbf{Repeatability}}=\mathrm{OFF}$, may provide a superior result. 14: fail – NagError *Input/Output The NAG error argument (see Section 3.6 in the Essential Introduction). nag_glopt_bnd_pso (e05sac) will return NW_SOLUTION_NOT_GUARANTEED if no error has been detected, and a finalization criterion has been achieved which cannot guarantee success. This does not indicate that the function has failed, merely that the returned solution cannot be guaranteed to be the true global optimum. The value of inform should be examined to determine which finalization criterion was reached. 6 Error Indicators and Warnings NE_ALLOC_FAIL Dynamic memory allocation failed. NE_BAD_PARAM On entry, argument $⟨\mathit{\text{value}}⟩$ had an illegal value. NE_BOUND On entry, ${\mathbf{bl}}\left[i\right]={\mathbf{bu}}\left[i\right]$ for all $i$. Constraint: ${\mathbf{bu}}\left[i\right]>{\mathbf{bl}}\left[i\right]$ for at least one $i$. On entry, ${\mathbf{bl}}\left[⟨\mathit{\text{value}}⟩\right]=⟨\mathit{\text{value}}⟩$ and ${\mathbf{bu}}\left[⟨\mathit{\text{value}}⟩\right]=⟨\mathit{\text{value}}⟩$. Constraint: ${\mathbf{bu}}\left[i\right]\ge {\mathbf{bl}}\left[i\right]$ for all $i$. NE_DERIV_ERRORS Derivative checks indicate possible errors in the supplied derivatives. NE_INT On entry, ${\mathbf{ndim}}=⟨\mathit{\text{value}}⟩$. Constraint: ${\mathbf{ndim}}\ge 1$. On entry, ${\mathbf{npar}}=⟨\mathit{\text{value}}⟩$. Constraint: ${\mathbf{npar}}\ge 5$. NE_INTERNAL_ERROR An internal error has occurred in this function. Check the function call and any array sizes. If the call is correct then please contact NAG for assistance. NE_INVALID_OPTION Either the option arrays have not been initialized for nag_glopt_bnd_pso (e05sac), or they have become corrupted. The option arrays are not consistent. The option arrays have not been initialized for nag_glopt_bnd_pso (e05sac). NE_USER_STOP User requested exit $⟨\mathit{\text{value}}⟩$ during call to monmod. User requested exit $⟨\mathit{\text{value}}⟩$ during call to objfun. NW_FAST_SOLUTION Target $⟨\mathit{\text{value}}⟩$ achieved after the first iteration. ${\mathbf{fb}}=⟨\mathit{\text{value}}⟩$. NW_SOLUTION_NOT_GUARANTEED A finalization criterion was reached that cannot guarantee success. On exit, ${\mathbf{inform}}=⟨\mathit{\text{value}}⟩$. 7 Accuracy If NE_NOERROR (or NW_FAST_SOLUTION) or NW_SOLUTION_NOT_GUARANTEED on exit, either a ${\mathbf{Target Objective Value}}$ or finalization criterion has been reached, depending on user selected options. As with all global optimization software, the solution achieved may not be the true global optimum. Various options allow for either greater search diversity or faster convergence to a (local) optimum (See Sections 10 and 11). Provided the objective function and constraints are sufficiently well behaved, if a local minimizer is used in conjunction with nag_glopt_bnd_pso (e05sac), then it is more likely that the final result will at least be in the near vicinity of a local optimum, and due to the global search characteristics of the particle swarm, this solution should be superior to many other local optima. Caution should be used in accelerating the rate of convergence, as with faster convergence, less of the domain will remain searchable by the swarm, making it increasingly difficult for the algorithm to detect the basin of attraction of superior local optima. Using the options ${\mathbf{Repulsion Initialize}}$ and ${\mathbf{Repulsion Finalize}}$ described in Section 11 will help to overcome this, by causing the swarm to diverge away from the current optimum once no more local improvement is likely. On successful exit with guaranteed success, NE_NOERROR. This may only happen if a ${\mathbf{Target Objective Value}}$ is assigned and is reached by the algorithm. On successful exit without guaranteed success, NW_SOLUTION_NOT_GUARANTEED is returned. This will happen if another finalization criterion is achieved without the detection of an error. In both cases, the value of inform provides further information as to the cause of the exit. 8 Further Comments The memory used by nag_glopt_bnd_pso (e05sac) is relatively static throughout. As such, nag_glopt_bnd_pso (e05sac) may be used in problems with high dimension number (${\mathbf{ndim}}>100$) without the concern of computational resource exhaustion, although the probability of successfully locating the global optimum will decrease dramatically with the increase in dimensionality. Due to the stochastic nature of the algorithm, the result will vary over multiple runs. This is particularly true if arguments and options are chosen to accelerate convergence at the expense of the global search. However, the option ${\mathbf{Repeatability}}=\mathrm{ON}$ may be set to initialize the internal random number generator using a preset seed, which will result in identical solutions being obtained. 9 Example Note: a modified example is supplied with the NAG Library for SMP & Multicore and links have been supplied in the following subsections. This example uses a particle swarm to find the global minimum of the Schwefel function: $minimize x∈Rndim f = ∑ i=1 ndim xi sinxi$ $xi ∈ -500,500 , for i=1,2,…,ndim .$ In two dimensions the optimum is ${f}_{\mathrm{min}}=-837.966$, located at $\mathbf{x}=\left(-420.97,-420.97\right)$. The example demonstrates how to initialize and set the options arrays using nag_glopt_opt_set (e05zkc), how to query options using nag_glopt_opt_get (e05zlc), and finally how to search for the global optimum using nag_glopt_bnd_pso (e05sac). The function is minimized several times to demonstrate using nag_glopt_bnd_pso (e05sac) alone, and coupled with local minimizers. This program uses the nondefault option ${\mathbf{Repeatability}}=\mathrm{ON}$ to produce repeatable solutions. 9.1 Program Text Program Text (e05sace.c) None. 9.3 Program Results Program Results (e05sace.r) 10 Algorithmic Details The following pseudo-code describes the algorithm used with the repulsion mechanism. $INITIALIZE for j=1, npar xj = R ∈ Uℓbox,ubox x^ j = R∈ Uℓbox,ubox vj = R ∈ U -V max ,Vmax f^j = F x^ j initialize wj wj = Wmax Weight Initialize=MAXIMUM Wini Weight Initialize=INITIAL R ∈ U W min , W max Weight Initialize=RANDOMIZED end for x~ = 12 ℓbox + ubox f~ = F x~ Ic = Is = 0 SWARM while (not finalized), Ic = Ic + 1 for j = 1 , npar xj = BOUNDARYxj,ℓbox,ubox fj = F xj if fj < f^j f^j = fj , x^j = xj if fj < f~ f~ = fj , x~ = xj end for if new f~ LOCMINx~,f~,Oi , Is=0 [see note on repulsion below for code insertion] else Is = Is + 1 for j = 1 , npar vj = wj vj + Cs D1 x^j - xj + Cg D2 x~ - xj xj = xj + vj if xj - x~ < dtol reset xj, vj, wj; x^j = xj else update wj end for if (target achieved or termination criterion satisfied) finalized=true monmod xj end LOCMINx~,f~,Oe$ The definition of terms used in the above pseudo-code are as follows. $\mathit{npar}$ the number of particles, npar ${\mathbf{\ell }}_{\mathrm{box}}$ array of ndim lower box bounds ${\mathbf{u}}_{\mathrm{box}}$ array of ndim upper box bounds ${\mathbf{x}}_{j}$ position of particle $j$ ${\stackrel{^}{\mathbf{x}}}_{j}$ best position found by particle $j$ $\stackrel{~}{\mathbf{x}}$ best position found by any particle ${f}_{j}$ $F\left({\mathbf{x}}_{j}\right)$ ${\stackrel{^}{f}}_{j}$ $F\left({\stackrel{^}{\mathbf{x}}}_{j}\right)$, best value found by particle $j$ $\stackrel{~}{f}$ $F\left(\stackrel{~}{\mathbf{x}}\right)$, best value found by any particle ${\mathbf{v}}_{j}$ velocity of particle $j$ ${w}_{j}$ weight on ${\mathbf{v}}_{j}$ for velocity update, decreasing according to ${\mathbf{Weight Decrease}}$ ${V}_{\mathrm{max}}$ maximum absolute velocity, dependent upon ${\mathbf{Maximum Variable Velocity}}$ ${I}_{c}$ swarm iteration counter ${I}_{s}$ iterations since $\stackrel{~}{\mathbf{x}}$ was updated ${\mathbf{D}}_{1}$,${\mathbf{D}}_{2}$ diagonal matrices with random elements in range $\left(0,1\right)$ ${C}_{s}$ the cognitive advance coefficient which weights velocity towards ${\stackrel{^}{\mathbf{x}}}_{j}$, adjusted using ${\mathbf{Advance Cognitive}}$ ${C}_{g}$ the global advance coefficient which weights velocity towards $\stackrel{~}{\mathbf{x}}$, adjusted using ${\mathbf{Advance Global}}$ $\mathit{dtol}$ the ${\mathbf{Distance Tolerance}}$ for resetting a converged particle $\mathbf{R}\in U\left({\mathbf{\ell }}_{\mathrm{box}},{\mathbf{u}}_{\mathrm{box}}\right)$ an array of random numbers whose $i$-th element is drawn from a uniform distribution in the range $\left({{\mathbf{\ell }}_{\mathrm{box}}}_{\mathit{i}},{{\mathbf{u}}_{\mathrm{box}}}_{\mathit{i}}\right)$, for $\mathit{i}=1,2,\dots ,{\mathbf{ndim}}$ ${O}_{i}$ local optimizer interior options ${O}_{e}$ local optimizer exterior options $\mathrm{LOCMIN}\left(\mathbf{x},f,O\right)$ apply local optimizer using the set of options $O$ using the solution $\left(\mathbf{x},f\right)$ as the starting point, if used (not default) monmod monitor progress and possibly modify ${\mathbf{x}}_{j}$ BOUNDARY apply required behaviour for ${\mathbf{x}}_{j}$ outside bounding box, (see ${\mathbf{Boundary}}$) new ($\stackrel{~}{f}$) true if $\stackrel{~}{\mathbf{x}}$, $\stackrel{~}{\mathbf{c}}$, $\stackrel{~}{f}$ were updated at this iteration Additionally a repulsion phase can be introduced by changing from the default values of options ${\mathbf{Repulsion Finalize}}$ (${r}_{f}$), ${\mathbf{Repulsion Initialize}}$ (${r}_{i}$) and ${\mathbf{Repulsion Particles}}$ (${r}_{p}$). If the number of static particles is denoted ${n}_{s}$ then the following can be inserted after the new($\stackrel{~}{f}$) check in the pseudo-code above. $else if ( ns ≥ rp and ri ≤ Is ≤ ri + rf ) LOCMINx~,f~,Oi use -Cg instead of Cg in velocity updates if Is = ri + rf Is = 0$ 11 Optional Arguments This section can be skipped if you wish to use the default values for all optional arguments, otherwise, the following is a list of the optional arguments available and a full description of each optional argument is provided in Section 11.1. 11.1 Description of the Optional Arguments For each option, we give a summary line, a description of the optional argument and details of constraints. The summary line contains: • the keywords; • a parameter value, where the letters $a$, $i\text{ and }r$ denote options that take character, integer and real values respectively; • the default value, where the symbol $\epsilon $ is a generic notation for machine precision (see nag_machine_precision (X02AJC)), and $\mathit{Imax}$ represents the largest representable integer value (see nag_max_integer (X02BBC)). All options accept the value ‘DEFAULT’ in order to return single options to their default states. Keywords and character values are case insensitive, however they must be separated by at least one space. For nag_glopt_bnd_pso (e05sac) the maximum length of the argument cvalue used by nag_glopt_opt_get (e05zlc) is $15$. Advance Cognitive $r$ Default$\text{}=2.0$ The cognitive advance coefficient, ${C}_{s}$. When larger than the global advance coefficient, this will cause particles to be attracted toward their previous best positions. Setting $r=0.0$ will cause nag_glopt_bnd_pso (e05sac) to act predominantly as a local optimizer. Setting $r>2.0$ may cause the swarm to diverge, and is generally inadvisable. At least one of the global and cognitive coefficients must be nonzero. Advance Global $r$ Default$\text{}=2.0$ The global advance coefficient, ${C}_{g}$. When larger than the cognitive coefficient this will encourage convergence toward the best solution yet found. Values $r\in \left(0,1\right)$ will inhibit particles overshooting the optimum. Values $r\in \left[1,2\right)$ cause particles to fly over the optimum some of the time. Larger values can prohibit convergence. Setting $r=0.0$ will remove any attraction to the current optimum, effectively generating a Monte–Carlo multi-start optimization algorithm. At least one of the global and cognitive coefficients must be nonzero. Boundary $a$ Default$\text{}=\text{FLOATING}$ Determines the behaviour if particles leave the domain described by the box bounds. This only affects the general PSO algorithm, and will not pass down to any NAG local minimizers chosen. This option is only effective in those dimensions for which ${\mathbf{bl}}\left[i-1\right]\ne {\mathbf{bu}}\left[i-1\right]$, $i=1,2,\dots ,{\mathbf{ndim}}$. IGNORE The box bounds are ignored. The objective function is still evaluated at the new particle position. RESET The particle is re-initialized inside the domain. ${\stackrel{^}{\mathbf{x}}}_{j}$ and ${\stackrel{^}{f}}_{j}$ are not affected. FLOATING The particle position remains the same, however the objective function will not be evaluated at the next iteration. The particle will probably be advected back into the domain at the next advance due to attraction by the cognitive and global memory. HYPERSPHERICAL The box bounds are wrapped around an $\mathit{ndim}$-dimensional hypersphere. As such a particle leaving through a lower bound will immediately re-enter through the corresponding upper bound and vice versa. The standard distance between particles is also modified accordingly. FIXED The particle rests on the boundary, with the corresponding dimensional velocity set to $0.0$. Distance Scaling $a$ Default$\text{}=\mathrm{ON}$ Determines whether distances should be scaled by box widths. ON When a distance is calculated between $\mathbf{x}$ and $\mathbf{y}$, a scaled ${L}^{2}$ norm is used. $L2 x,y = ∑ i | ui ≠ ℓi , i≤ndim xi - yi ui - ℓi 2 1 2 .$ OFF Distances are calculated as the standard ${L}^{2}$ norm without any rescaling. $L2 x,y = ∑ i=1 ndim xi - yi 2 1 2 .$ Distance Tolerance $r$ Default$\text{}={10}^{-4}$ This is the distance, $\mathit{dtol}$ between particles and the global optimum which must be reached for the particle to be considered converged, i.e., that any subsequent movement of such a particle cannot significantly alter the global optimum. Once achieved the particle is reset into the box bounds to continue searching. Constraint: $r>0.0$. Function Precision $r$ Default$\text{}={\epsilon }^{0.9}$ The argument defines ${\epsilon }_{r}$, which is intended to be a measure of the accuracy with which the problem function $F\left(\mathbf{x}\right)$ can be computed. If $r<\epsilon $ or $r\ge 1$, the default value is used. The value of ${\epsilon }_{r}$ should reflect the relative precision of $1+\left|F\left(\mathbf{x}\right)\right|$; i.e., ${\epsilon }_{r}$ acts as a relative precision when $\left|F\right|$ is large, and as an absolute precision when $\left|F\right|$ is small. For example, if $F\left(\mathbf{x}\right)$ is typically of order $1000$ and the first six significant digits are known to be correct, an appropriate value for ${\epsilon }_{r}$ would be ${10}^{-6}$. In contrast, if $F\left(\mathbf{x}\right)$ is typically of order ${10}^{-4}$ and the first six significant digits are known to be correct, an appropriate value for ${\epsilon }_{r}$ would be ${10}^{-10}$. The choice of ${\epsilon }_{r}$ can be quite complicated for badly scaled problems; see Chapter 8 of Gill et al. (1981) for a discussion of scaling techniques. The default value is appropriate for most simple functions that are computed with full accuracy. However when the accuracy of the computed function values is known to be significantly worse than full precision, the value of ${\epsilon }_{r}$ should be large enough so that no attempt will be made to distinguish between function values that differ by less than the error inherent in the calculation. Local Boundary Restriction $r$ Default$\text{}=0.5$ Contracts the box boundaries used by a box constrained local minimizer to, $\left[{\beta }_{l},{\beta }_{u}\right]$, containing the start point $x$, where $∂i = r × ui - ℓi βli = maxℓi, xi - ∂i2 βui = minui, xi + ∂i2 , i=1,…,ndim .$ Smaller values of $r$ thereby restrict the size of the domain exposed to the local minimizer, possibly reducing the amount of work done by the local minimizer. Constraint: $0.0\le r\le 1.0$. Local Interior Iterations ${i}_{1}$ Local Interior Major Iterations ${i}_{1}$ Local Exterior Iterations ${i}_{2}$ Local Exterior Major Iterations ${i}_{2}$ The maximum number of iterations or function evaluations the chosen local minimizer will perform inside (outside) the main loop if applicable. For the NAG minimizers these correspond to: | | | | | |-------------------------------|------------------------------------------------------------------------------|------------------------------------------------------------------------------|------------------------------------------------------------------------------| | Minimizer | Argument/option | Default Interior | Default Exterior | | nag_opt_simplex_easy (e04cbc) | maxcal | ${\mathbf{ndim}}+10$ | $2×{\mathbf{ndim}}+15$ | | nag_opt_conj_grad (e04dgc) | ${\mathbf{max_iter}}$ | $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(30,3×{\mathbf{ndim}}\right)$ | $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(50,5×{\mathbf{ndim}}\right)$ | | nag_opt_nlp (e04ucc) | ${\mathbf{max_iter}}$ | $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(10,2×{\mathbf{ndim}}\right)$ | $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(30,3×{\mathbf{ndim}}\right)$ | Unless set, these are functions of the arguments passed to nag_glopt_bnd_pso (e05sac). Setting $i=0$ will disable the local minimizer in the corresponding algorithmic region. For example, setting ${\mathbf{Local Interior Iterations}}=0$ and ${\mathbf{Local Exterior Iterations}}=30$ will cause the algorithm to perform no local minimizations inside the main loop of the algorithm, and a local minimization with upto $30$ iterations after the main loop has been exited. Constraint: ${i}_{1}\ge 0$, ${i}_{2}\ge 0$. Local Interior Tolerance ${r}_{1}$ Default$\text{}={10}^{-4}$ Local Exterior Tolerance ${r}_{2}$ Default$\text{}={10}^{-4}$ This is the tolerance provided to a local minimizer in the interior (exterior) of the main loop of the algorithm. Constraint: ${r}_{1}>0.0$, ${r}_{2}>0.0$. Local Interior Minor Iterations ${i}_{1}$ Local Exterior Minor Iterations ${i}_{2}$ Where applicable, the secondary number of iterations the chosen local minimizer will use inside (outside) the main loop. Currently the relevant default values are: | | | | | |----------------------|------------------------------------------------------------------------------|------------------------------------------------------------------------------|------------------------------------------------------------------------------| | Minimizer | Argument/option | Default Interior | Default Exterior | | nag_opt_nlp (e04ucc) | ${\mathbf{minor_max_iter}}$ | $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(10,2×{\mathbf{ndim}}\right)$ | $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(30,3×{\mathbf{ndim}}\right)$ | Constraint: ${i}_{1}\ge 0$, ${i}_{2}\ge 0$. Local Minimizer $a$ Default$\text{}=\mathrm{OFF}$ Allows for a choice of Chapter e04 functions to be used as a coupled, dedicated local minimizer. OFF No local minimization will be performed in either the INTERIOR or EXTERIOR sections of the algorithm. e04cbc Use nag_opt_simplex_easy (e04cbc) as the local minimizer. This does not require the calculation of derivatives. On a call to objfun during a local minimization, ${\mathbf{mode}}=5$. e04dgc Use nag_opt_conj_grad (e04dgc) as the local minimizer. Accurate derivatives must be provided, and will not be approximated internally. Additionally, each call to objfun during a local minimization will require either the objective to be evaluated alone, or both the objective and its gradient to be evaluated. Hence on a call to objfun, ${\mathbf{mode}}=5$ or $7$. e04ucc Use nag_opt_nlp (e04ucc) as the local minimizer. This operates such that any derivatives of the objective function that you cannot supply, will be approximated internally using finite differences. Either, the objective, objective gradient, or both may be requested during a local minimization, and as such on a call to objfun, ${\mathbf{mode}}=1$, $2$ or $5$. The box bounds forwarded to this function from nag_glopt_bnd_pso (e05sac) will have been acted upon by ${\mathbf{Local Boundary Restriction}}$. As such, the domain exposed may be greatly smaller than that provided to nag_glopt_bnd_pso (e05sac). Maximum Function Evaluations $i$ Default $=\mathit{Imax}$ The maximum number of evaluations of the objective function. When reached this will return NW_SOLUTION_NOT_GUARANTEED and ${\mathbf{inform}}=6$. Constraint: $i>0$. Maximum Iterations Completed $i$ Default$\text{}=1000×{\mathbf{ndim}}$ The maximum number of complete iterations that may be performed. Once exceeded nag_glopt_bnd_pso (e05sac) will exit with NW_SOLUTION_NOT_GUARANTEED and ${\mathbf{inform}}=5$. Unless set, this adapts to the parameters passed to nag_glopt_bnd_pso (e05sac). Constraint: $i\ge 1$. Maximum Iterations Static $i$ Default$\text{}=100$ The maximum number of iterations without any improvement to the current global optimum. If exceeded nag_glopt_bnd_pso (e05sac) will exit with NW_SOLUTION_NOT_GUARANTEED and ${\mathbf{inform}}=4$. This exit will be hindered by setting ${\mathbf{Maximum Iterations Static Particles}}$ to larger values. Constraint: $i\ge 1$. Maximum Iterations Static Particles $i$ Default$\text{}=0$ The minimum number of particles that must have converged to the current optimum before the function may exit due to ${\mathbf{Maximum Iterations Static}}$ with NW_SOLUTION_NOT_GUARANTEED and ${\mathbf{inform}}=4$. Constraint: $i\ge 0$. Maximum Particles Converged $i$ Default $=\mathit{Imax}$ The maximum number of particles that may converge to the current optimum. When achieved, nag_glopt_bnd_pso (e05sac) will exit with NW_SOLUTION_NOT_GUARANTEED and ${\mathbf{inform}}=3$. This exit will be hindered by setting ‘Repulsion’ options, as these cause the swarm to re-expand. Constraint: $i>0$. Maximum Particles Reset $i$ Default $=\mathit{Imax}$ The maximum number of particles that may be reset after converging to the current optimum. Once achieved no further particles will be reset, and any particles within ${\mathbf{Distance Tolerance}}$ of the global optimum will continue to evolve as normal. Constraint: $i>0$. Maximum Variable Velocity $r$ Default$\text{}=0.25$ Along any dimension $j$, the absolute velocity is bounded above by $\left|{\mathbf{v}}_{j}\right|\le r×\left({\mathbf{u}}_{j}-{\mathbf{\ell }}_{j}\right)={\mathbf{V}}_{\mathrm{max}}$. Very low values will greatly increase convergence time. There is no upper limit, although larger values will allow more particles to be advected out of the box bounds, and values greater than $4.0$ may cause significant and potentially unrecoverable swarm divergence. Constraint: $r>0.0$. Optimize $a$ Default$\text{}=\mathrm{MINIMIZE}$ Determines whether to maximize or minimize the objective function. MINIMIZE The objective function will be minimized. MAXIMIZE The objective function will be maximized. This is accomplished by minimizing the negative of the objective. Repeatability $a$ Default$\text{}=\mathrm{OFF}$ Allows for the same random number generator seed to be used for every call to nag_glopt_bnd_pso (e05sac). ${\mathbf{Repeatability}}=\mathrm{OFF}$ is recommended in general. OFF The internal generation of random numbers will be nonrepeatable. ON The same seed will be used. Repulsion Finalize $i$ Default $=\mathit{Imax}$ The number of iterations performed in a repulsive phase before re-contraction. This allows a re-diversified swarm to contract back toward the current optimum, allowing for a finer search of the near optimum space. Constraint: $i\ge 2$. Repulsion Initialize $i$ Default $=\mathit{Imax}$ The number of iterations without any improvement to the global optimum before the algorithm begins a repulsive phase. This phase allows the particle swarm to re-expand away from the current optimum, allowing more of the domain to be investigated. The repulsive phase is automatically ended if a superior optimum is found. Constraint: $i\ge 2$. Repulsion Particles $i$ Default$\text{}=0$ The number of particles required to have converged to the current optimum before any repulsive phase may be initialized. This will prevent repulsion before a satisfactory search of the near optimum area has been performed, which may happen for large dimensional problems. Constraint: $i\ge 0$. Swarm Standard Deviation $r$ Default$\text{}=0.1$ The target standard deviation of the particle distances from the current optimum. Once the standard deviation is below this level, nag_glopt_bnd_pso (e05sac) will exit with NW_SOLUTION_NOT_GUARANTEED and ${\mathbf{inform}}=2$. This criterion will be penalized by the use of ‘Repulsion’ options, as these cause the swarm to re-expand, increasing the standard deviation of the particle distances from the best point. Constraint: $r\ge 0.0$. Target Objective $a$ Default$\text{}=\mathrm{OFF}$ Target Objective Value $r$ Default$\text{}=0.0$ Activate or deactivate the use of a target value as a finalization criterion. If active, then once the supplied target value for the objective function is found (beyond the first iteration if ${\mathbf{Target Warning}}$ is active) nag_glopt_bnd_pso (e05sac) will exit with NE_NOERROR and ${\mathbf{inform}}=1$. Other than checking for feasibility only (${\mathbf{Optimize}}=\mathrm{CONSTRAINTS}$), this is the only finalization criterion that guarantees that the algorithm has been successful. If the target value was achieved at the initialization phase or first iteration and ${\mathbf{Target Warning}}$ is active, nag_glopt_bnd_pso (e05sac) will exit with NW_FAST_SOLUTION. This option may take any real value $r$, or the character ON/OFF as well as DEFAULT. If this option is queried using nag_glopt_opt_get (e05zlc), the current value of $r$ will be returned in rvalue, and cvalue will indicate whether this option is ON or OFF. The behaviour of the option is as follows: $r$ Once a point is found with an objective value within the ${\mathbf{Target Objective Tolerance}}$ of $r$, nag_glopt_bnd_pso (e05sac) will exit successfully with NE_NOERROR and ${\mathbf{inform}}=1$. OFF The current value of $r$ will remain stored, however it will not be used as a finalization criterion. ON The current value of $r$ stored will be used as a finalization criterion. DEFAULT The stored value of $r$ will be reset to its default value ($0.0$), and this finalization criterion will be deactivated. Target Objective Safeguard $r$ Default$\text{}=100.0\epsilon $ If the magnitude of ${\mathbf{Target Objective Value}}$ is below $r$, then the absolute error between the optimum and the target will be used. Otherwise the relative error will be used. If a value entered is below $100.0\epsilon $, the default value will be used. Target Objective Tolerance $r$ Default$\text{}=0.0$ The optional tolerance to a user-specified target value. Constraint: $r\ge 0.0$. Target Warning $a$ Default$\text{}=\mathrm{OFF}$ Activates or deactivates the error exit associated with the target value being achieved before entry into the main loop of the algorithm, NW_FAST_SOLUTION. OFF No error will be returned, and the function will exit normally. ON An error will be returned if the target objective is reached prematurely, and the function will exit with NW_FAST_SOLUTION. Verify Gradients $a$ Default$\text{}=\mathrm{ON}$ Adjusts the level of gradient checking performed when gradients are required. Gradient checks are only performed on the first call to the chosen local minimizer if it requires gradients. There is no guarantee that the gradient check will be correct, as the finite differences used in the gradient check are themselves subject to inaccuracies. OFF No gradient checking will be performed. ON A cheap gradient check will be performed on both the gradients corresponding to the objective through objfun. OBJECTIVE FULL A more expensive gradient check will be performed on the gradients corresponding to the objective objfun. Weight Decrease $a$ Default$\text{}=\mathrm{INTEREST}$ Determines how particle weights decrease. OFF Weights do not decrease. INTEREST Weights decrease through compound interest as ${w}_{\mathit{IT}+1}={w}_{\mathit{IT}}\left(1-{W}_{\mathit{val}}\right)$, where ${W}_{\mathit{val}}$ is the ${\mathbf{Weight Value}}$ and $\mathit{IT}$ is the current number of iterations. LINEAR Weights decrease linearly following ${w}_{\mathit{IT}+1}={w}_{\mathit{IT}}-\mathit{IT}×\left({W}_{\mathit{max}}-{W}_{\mathit{min}}\right)/{\mathit{IT}}_{\mathit{max}}$, where $\mathit{IT}$ is the iteration number and ${\mathit{IT}}_{\mathit{max}}$ is the maximum number of iterations as set by ${\mathbf{Maximum Iterations Completed}}$. Weight Initial $r$ Default$\text{}={W}_{\mathit{max}}$ The initial value of any particle's inertial weight, ${W}_{\mathit{ini}}$, or the minimum possible initial value if initial weights are randomized. When set, this will override ${\mathbf{Weight Initialize}}=\mathrm{RANDOMIZED}$ or $\mathrm{MAXIMUM}$, and as such these must be set afterwards if so desired. Constraint: ${W}_{\mathit{min}}\le r\le {W}_{\mathit{max}}$. Weight Initialize $a$ Default$\text{}=\mathrm{MAXIMUM}$ Determines how the initial weights are distributed. INITIAL All weights are initialized at the initial weight, ${W}_{\mathit{ini}}$, if set. If ${\mathbf{Weight Initial}}$ has not been set, this will be the maximum weight, ${W}_{\mathit{max}}$. MAXIMUM All weights are initialized at the maximum weight, ${W}_{\mathit{max}}$. RANDOMIZED Weights are uniformly distributed in $\left({W}_{\mathit{min}},{W}_{\mathit{max}}\right)$ or $\left({W}_{\mathit{ini}},{W}_{\mathit{max}}\right)$ if ${\mathbf{Weight Initial}}$ has been set. Weight Maximum $r$ Default$\text{}=1.0$ The maximum particle weight, ${W}_{\mathit{max}}$. Constraint: $1.0\ge r\ge {W}_{\mathit{min}}$ (If ${W}_{\mathit{ini}}$ has been set then $1.0\ge r\ge {W}_{\mathit{ini}}$.) Weight Minimum $r$ Default$\text{}=0.1$ The minimum achievable weight of any particle, ${W}_{\mathit{min}}$. Once achieved, no further weight reduction is possible. Constraint: $0.0\le r\le {W}_{\mathit{max}}$ (If ${W}_{\mathit{ini}}$ has been set then $0.0\le r\le {W}_{\mathit{ini}}$.) Weight Reset $a$ Default$\text{}=\mathrm{MAXIMUM}$ Determines how particle weights are re-initialized. INITIAL Weights are re-initialized at the initial weight if set. If ${\mathbf{Weight Initial}}$ has not been set, this will be the maximum weight. MAXIMUM Weights are re-initialized at the maximum weight. RANDOMIZED Weights are uniformly distributed in $\left({W}_{\mathit{min}},{W}_{\mathit{max}}\right)$ or $\left({W}_{\mathit{ini}},{W}_{\mathit{max}}\right)$ if ${\mathbf{Weight Initial}}$ has been set. Weight Value $r$ Default$\text{}=0.01$ The constant ${W}_{\mathit{val}}$ used with ${\mathbf{Weight Decrease}}=\mathrm{INTEREST}$. Constraint: $0.0\le r\le \frac{1}{3}$.

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http://math.stackexchange.com/questions/252629/generalization-of-bernoullis-inequality

# Generalization of Bernoulli's Inequality. When you're first taking an Introduction to Proofs class, you eventually learn about induction. One of the first induction proofs I did was to prove for $n \geq 1$ that $(1+x)^n \geq 1+nx$. You first prove the base case, $n=1$, which yields $(1+x)^1 \geq 1+1x$, which is true. Then you assume that the inequality holds for for the arbitrary $n$ case. Then you try to prove that the inequality holds for $n+1$, i.e. $(1+x)^(n+1) \geq 1+nx$ as well. Using our assumption that $(1+x)^n \geq 1+nx$, we can multiply both sides by $1+x$, which yields $(1+x)*(1+x)^n \geq (1+nx)*(1+x).$ Since $1+nx^2+nx +x \geq 1+(n+1)x$, the inequality is proven. Now what I’m trying to do is prove a generalization of Bernoulli’s inequality, i.e. the following. “Find the least possible $\alpha < 0$ so that $\forall x \geq \alpha$ and $n \in \mathbb{N}$, that $(1+x)^n \geq 1+nx$. Prove the inequality as well as prove that your $\alpha$ is best possible.” What I’m trying to figure out is what is more special about this question rather than the typical inductive Bernoulli’s inequality proof? What is the purpose of introducing this alpha restricted to less than zero? There must be a reason for it. I read a few different textbooks and they all say that it (it referring to Bernoulli's inequality) works for $x > -1$, so I’m assuming that the alpha would also be -1 because then for any $x \in \mathbb{R}$ such that $x \geq \alpha$, the normal Bernoulli’s inequality holds. Am I completely going off the deep end? Any input would be very much appreciated. - Does the inequality hold when $x=-1$ ? – Ju'x Dec 6 '12 at 22:55 ## 2 Answers We find the answer, using basic calculus. The magic $\alpha$ is not $-1$. Consider the function $f(x)=(1+x)^n -(1+nx)$ for $n \gt 1$. Then $$f'(x)=n((1+x)^{n-1}-1).$$ For $n\gt 1$ and even, the derivative vanishes only at $x=0$. For $n\gt 1$ and odd, the derivative vanishes at $x=-2$ and $x=0$. The only interesting case is $n$ odd. By the usual increasing/dcreasing analysis, we find that $f(x)\ge 0$ when $x\ge -2$. For any $w\lt -2$, by taking a suitably large $n$, we can make $f(w)\lt 0$. The part about $f(x)\ge 0$ for $x\ge -2$ can be pushed through by induction. - So the critical points of f(x) occur at x = 0 for all n and x = -2 for n odd. So x = -2 is not a critical point for n even; therefore, how can x = -2 be a minimum for all n? -

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http://math.stackexchange.com/questions/87104/shouldnt-the-coaddition-on-mathbb-zx-be-a-morphism-of-rings

# Shouldn't the coaddition on $\mathbb Z[x]$ be a morphism of rings? $\operatorname{Spec} \mathbb Z[x]$ is a group scheme. Then there is a morphism of schemes: $\operatorname{Spec} \mathbb Z[x]\times\operatorname{Spec} \mathbb Z[x]\rightarrow\operatorname{Spec} \mathbb Z[x]$ induced by the coaddition $a:\mathbb Z[x]\rightarrow \mathbb Z[x] \otimes\mathbb Z[x]$ mapping $x$ to $1\otimes x+x\otimes 1$. Now, the coaddition is not a morphism of rings, since it does not respect multiplication: $a(x^2)=(1\otimes x^2+x^2\otimes 1)\neq(1\otimes x + x\otimes 1)^2=a(x)\cdot a(x)$. But $a$ should be a morphism of rings, since the addition on $\operatorname{Spec} \mathbb Z[x]$ is a morphism of affine schemes. - ## 1 Answer I think you are being confused by the notation. There is a canonical isomorphism $\mathbb{Z}[t] \otimes_{\mathbb{Z}} \mathbb{Z}[t] \cong \mathbb{Z}[x, y]$ such that $t \otimes 1 \mapsto x$ and $1 \otimes t \mapsto y$. In that light, coaddition is the unique ring homomorphism $\alpha : \mathbb{Z}[t] \to \mathbb{Z}[x, y]$ such that $t \mapsto x + y$, since $\mathbb{Z}[t]$ is freely generated by $t$. In particular, $\alpha(t^2) = x^2 + 2 x y + y^2 = \alpha(t)^2$. - Thank you very much! – Felix Wellen Dec 1 '11 at 15:04

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http://mathhelpforum.com/calculus/62706-integrate.html

# Thread: 1. ## Integrate I have to integrate K(1-t)^4 dt between 0 and y I get: K(1/5)(1-t)^5 the book has k(1-t)^5/(-5) i dont know how they got -5 then when i evaluate at 0 and y i get: (k/5)[(1-y)^5 -1] and at this spot the book has (k/5)[1-(1-y)^5] thanks 2. You're integrating $k(1-t)^4$, right? Be sure to account for the coefficient of t, which is negative one. That's the only difference between your answer and the correct one. And whenever in doubt, just try differentiating the answer you determined along with the answer given by the textbook. You'll probably notice the difference. 3. Yes, that is what I am integrating. I guess I was just thinking that I could raise the power and divide by the new power. I just ignored what was in the parenthesis. Thanks.

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http://physics.stackexchange.com/questions/31143/counting-degrees-of-freedom-of-gauge-bosons

# Counting degrees of freedom of gauge bosons Gauge bosons are represented by $A_{\mu}$, where $\mu = 0,1,2,3$. So in general there are 4 degrees of freedom. But in reality, a photon (gauge boson) has two degrees of freedom (two polarization states). So, when someone asks about on-shell and off-shell degrees of freedom, I thought they are 2 and 4. But I read that the off-shell d.of. are 3. And my question is how to see this? - ## 2 Answers On-shell dof: Consider the constraints of "Equation of Motion" plus "Gauge Condition"(Lorentz gauge, Coulomb gauge,...), #dof=4-2. Off-Shell dof: Consider the constraints of "Gauge Condition" (Lorentz gauge, Coulomb gauge,...) only, #dof=4-1. - One way to see this is the fact that if you write out the Lagragian in terms of the $A^0$ and $A^i$ explicitly, you will find there is no time derivatives for the $A^0$. The only candidate term $F^{00} = \partial^0 A^0 - \partial^0 A^0$ vanishes. This means we shouldn't count $A^0$ as a degree of freedom, and in fact, we should integrate it out. Now the resulting Lagrangian explicitly only has 3 DOF. If you now take this Lagrangian and vary it you will get the EOM which puts one constraint on the 3 DOF which knocks you down to the 2 DOF on-shell. - No $\partial_0 A_0$ term may be interpreted that $A_0$ is not a dynamical dof rather than dof. Three dimensional gravity for example, the gravity has a dof which is not dynamical. – Craig Thone Jul 2 '12 at 16:12 Craig - sorry, I don't understand your comment. Did I make an error in my statement? – DJBunk Jul 2 '12 at 17:25 No. I just want to emphasize $A_0$ is not a dynamical degree of freedom. In Lorentz gauge, I am not sure whether $A_0$ is a degree of freedom. – Craig Thone Jul 3 '12 at 1:42

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http://mathoverflow.net/questions/15836/oneupsmanship-and-publishing-etiquette/15837

## Oneupsmanship and Publishing Etiquette ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Apologies for the anonymous (and, in hindsight, mildly longish) question. I have an etiquette question on publishing an improvement of another author's work, but I figured a thread on the matter to address generalities might be nice to have as well. Another paper in my field reports to have solved an open problem by finding a necessary and sufficient condition for deciding if an object $X$ has property $P$. As it turns out, there are cases in which this theorem reduces to a tautology, and not in the good "every theorem is a tautology" way -- it reduces to the statement $X$ has property $P$ if and only if it has property $P$. This happens frequently enough that it seems at the very least a little disingenuous to have called it a necessary and sufficient condition. In writing up a stronger version of the result, I have to decide how to address the previous work, leading to some fairly general questions about publishing etiquette to which the MO audience at large might have some insightful responses: 1) Usually when improving on someone's work, there are a variety of diplomatic words to use -- "We generalize soandso's work" being a standard one, providing both a nod to their efforts while also emphasizing your improvement. Or, if a result is plain wrong, no amount of diplomacy will salvage their result (though could possibly soften the blow), so you can just provide the counter-example and move on. In my specific example, it feels a little harder to address -- their result isn't incorrect, just not as complete as they seem to take credit for. I know I've had both colleagues who positively light up with joy at the thought of besting someone else's paper, and those who I think would likely contact the original author to give them a chance to contribute to the discussion in the paper. Ignoring it altogether doesn't seem like an option given that they imply that the problem is now closed. Let's take the point of view that I am a recent graduate and thus the prestige of improving a published result presumably conveys a non-trivial benefit to me -- any heuristics on how to make decisions of this form? 2) On a logistical level, their techniques are completely different from ours. What's the extent to which we should re-introduce all of their (particularly burdensome, in my opinion) notation and terminology in order to prove that our result is an objective improvement? It's a fairly simple matter to give an example which our result provides an answer to that theirs does not, but it's rather technical (maybe longer than the proof of the result itself) to show that our result handles every case that theirs does (non-tautologically). In addition to the extra clutter in the paper, the prolonged emphasis on provably improving their work seems to work against the diplomacy hoped for in the previous question. - 2 One question is, how important this diplomacy is to you? For example, you are worried about spending time in the paper to show that you provably improve their work. I don't think that mathematical community in general will find this "undiplomatic". They may or may not find your improvements interesting, but just because you prove that you improved somebody else's work will not bother anyone (again, assuming the improvements are actually interesting and you are not dwelling on them just to show how much smarter you are). The only person who could be bothered by this is the original author. – sheldon-cooper Feb 19 2010 at 20:21 So the question is, do you really care about the feelings of that one person enough to modify your paper significantly and maybe even withhold some interesting results? – sheldon-cooper Feb 19 2010 at 20:23 3 @sheldon-cooper: You make a valid point, but I can certainly imagine contexts where this would antagonize more than just one person. – François G. Dorais♦ Feb 19 2010 at 20:29 2 I'm not going to submit an "answer," per se, given the response to Andrew's below. But I personally have grown more and more disillusioned over the years with the lack "humanness" (for lack of a better term) in how we do things. I go out of my way to at the very least acknowledge people somewhere. A recent paper of mine was the result of a pub discussion with 7 or 8 people - I acknowledged them all (and the paper won an award, so it obviously wasn't considered bad form). – Ian Durham Feb 20 2010 at 3:30 1 @ID: I don't quite take your point. It is always good form to acknowledge help given by others, however indirect. (And it doesn't hurt you to do so.) So having given acknowledgments in your situation seems unquestionably appropriate. But it doesn't make sense to acknowledge people with whom you have not had any relevant interactions. OP doesn't need to acknowledge the author whose result he is improving: he needs to cite that person's work. And I don't see anything inhumane about improving on another person's work. That is most of what we do, after all. – Pete L. Clark Feb 20 2010 at 5:23 show 1 more comment ## 6 Answers I am not sure I see what exactly is the ethical issue. You need to be very honest and very clear - you are writing a paper not for yourself but for other people to read and understand, so that should be your first priority. Write something like this. "We prove property P. This result is strongly related to the result in [..] which proves property P' and extends it in the following sense ... Let us note that in all cases our property P is at least as strong as P' (see Thm~?), while in some special cases our property P favorably compares to property P'. Below we give some examples which emphasize the connection." If the matter is easy - work out carefully some examples right in the introduction. If this is more delicate as it seems to be the case, make a new section where you can work out several examples so the reader can see that P' is saying in some cases P' implies P', while your P is saying something stronger. If you stick to math, no one I can think of will get upset over this, and there is no need to make short broad characterizations in the introduction which might upset somebody. At the same time you "besting somebody" will come across in exactly the way you desire. The only downside of this approach I can see is that the paper might get a little longer due to this examples section. But so what - the reader will appreciate the clarity, and an extra couple of pages cannot possibly affect the publication chances. - ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. Igor Pak gave good advice on what to write in your paper. Additionally, I would advise you to contact the author of the original paper and tell them about your work before it is published. This can be a good opportunity to get to know someone who shares common interests with you and it opens possibility for future joint work. I'll bet if you talk to the other author the two of you together are going to have a new idea which is even better than his or yours alone. And another thing. If you write to the author and start talking to him, you will have a chance to find out about the background of his work and you won't have to imagine useless things (such as "this other guy is so stupid"). Perhaps he knows that his theorem sometimes reduces to a tautology and will readily agree with you. Perhaps he thinks it only happens occasionally and will be grateful to you for explaining that this is not so. In any case, just talk to the other guy. - It's hard to judge without details, but this is perhaps a question of metatheory. (This is what item 2 sounds like to me, but this is just an impression.) It is not a requirement in mathematical papers to carefully explain aspects of the metatheory, but it is not prohibited to do so. In your case, it seems that their result is trivial in your metatheory while yours isn't. By clarifying your metatheory, you can carefully explain that in your paper. In many cases, you really don't need to say much about their result since the issue is likely to be with their assumptions or methodology, not with their conclusions. Such a critique should sound much tamer than an attack on their result. Clarifying your metatheory will also separate your result from theirs in other ways. It is possible that their metatheory (presumably not carefully explained in their paper) does not see their result as trivial. Maybe it even sees your result as trivial! A reader that shares their point of view and not yours will immediately see the difference when reading your paper and will be able to appropriately interpret your work regardless of their personal stance. This could avoid further confusion of the same kind. PS: Being a logician, I prefer the term "metatheory" for what many other mathematicians would call "context," "framework," "perspective," "methodology," or whatever. Don't read too much technicality into the term, the metatheory is always informal. (Only logicians occasionally need to make the metatheory formal in order to formally prove metatheorems.) If this is indeed the kind of scenario you're looking at, it should be relatively clear to you exactly what and how much you need to clarify about your own metatheory. - I largely agree with Igor Pak's response, and I think he has the right perspective on the matter. Two comments: [X is the author of the first paper, Y is the OP!] 1) I'm not sure it's critical to show that your result completely subsumes the result of X if that verification is going to, say, double the length of your paper and require you to introduce a lot of terminology and notation that is far away from your approach to the problem. It is not clear what value you are adding to the community by doing this: is your proof going to be any more pleasant to read or insightful than X's? On the other hand, I find it slightly odd that it should be as hard or harder to demonstrate that your results imply X's than it is to prove your (in fact more general) result. I'm having trouble thinking of an example of this phenomenon from my own experience. Is it possible that there's more insight to gained here, and perhaps a(n obvious!) common generalization of your two approaches? 2) A good rule about describing your own work -- borrowed from the creative writing community -- is: show, don't tell. Specifically, you should seek to minimize the number of times in which you tell the reader how she should feel about your work. Rather, you want to present the mathematical information which brings the reader to this conclusion. In this case, I would recommend putting a lot of effort into the writing of the example(s) of the novelty of your approach. The more clear and detailed your writing is, the smaller your sales pitch needs to be. Some direct comparison may be necessary, but if you stick to comparing one mathematical statement to another, then "stronger than" has an objective meaning. It seems that there is little in such a practice that could reasonably cause offense. Unfortunately 1) and 2) are somewhat at cross-purposes to each other. Regarding 1), in lieu of writing out all the gory details it is tempting to write something like "It can be shown that in every case where X's theorem applies, so does mine". If this takes pages and pages to show, then the author might well not see / believe it at first, and it could be annoying to have to work for hours to verify someone's claim that they have trumped you. Maybe the best solution -- although not the easiest -- is for you to carefully write up the implication that your result implies X's result in a separate document, which you submit along with the paper itself and post a copy on your webpage. This also places some of the judgment of whether this gory extra part should be included in the paper in the hands of the referee and the editor, which is perhaps as it should be. - This suggestion may only work in certain situations, but is it possible to split your paper into two? One could be for your new results in the most elegant form, and the other could be primarily for comparing your results to the previous work. You can sent the former to a good journal and throw the latter in a lesser journal or just post on the arXiv. - See if the other author wants to be a co-author. This is playing nice, and also, it eliminates the chance that they will be a referee. - 21 I'd advise against that. First, you will be sharing credit with other people for something you already did all by yourself. More importantly, I suspect your contribution may go completely unnoticed in such case. If a person X is already known for working on this problem, them most of the credit will go to them. It will be just another paper in a series "X with this student", "X with another student", "X with third student", and most of the credit goes to X. – sheldon-cooper Feb 19 2010 at 19:53 4 I see by my draining reputation that this one is not too popular. But consider the original questioner has pointed out that his result is not completely new - he is simply patching up the lack of generality of previous work. Drawing lines in the sand about credit for that seems a bit of a stretch. – Andrew Mullhaupt Feb 19 2010 at 22:04 6 Downvotes on community wiki answers won't actually drain your reputation. – Noah Snyder Feb 19 2010 at 23:27 23 "On appropriate occasions, it may be desirable to offer or accept joint authorship when independent researchers find that they have produced identical results. All the authors listed for a paper, however, must have made a significant contribution to its content, and all who have made such a contribution must be offered the opportunity to be listed as an author." That is from ams.org/secretary/ethics.html and should be taken seriously. If the other person's contribution is in an already published paper, having his or her name on the new paper looks to me to be a violation. – Bill Johnson Feb 20 2010 at 0:08 10 In that case, "See if the other author wants to collaborate" seems to be a better way to phrase your suggestion. – stankewicz Feb 20 2010 at 17:05 show 5 more comments

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http://math.stackexchange.com/questions/279838/is-the-point-at-0-0-of-the-graph-z-x3-y3-considered-a-saddle-point

# Is the point at $(0,0)$ of the graph $z=x^3 + y^3$ considered a saddle point? Is the point at $(0,0)$ of the graph $z=x^3 + y^3$ considered a saddle point? I was given that function, and I used the second-order derivative test only to find that the $Hf(0,0) = 0$, with the critical point being at 0,0. Is anything that is not a local minimum or a local maximum considered a saddle point? What the graph looks like - ## 1 Answer 1. In your case, the Hessian is a null matrix. This does not guarantee that the point is a saddle point. Although I cannot come up with an example but I think you can have a null matrix as hessian at the minima. But don't quote me for that. 2. Such a point, AFAIK, is given the name "Degenerate Critical Point". Whether it is considered saddle or not, I am not sure. 3. If I am standing at (0,0,0) and you give me an $\epsilon$, I can step $(\epsilon,0)$ and make the function positive or $(-\epsilon,0)$ and make the function negative. So, no matter how small epsilon you give me, I can always move to points that increase the function value and reduce it. This makes the (0,0,0) a saddle point. 4. Does this question help : Classifying singular points as local min, max or saddle points? - Standard example of a local min with vanishing Hessian is $x^4 + y^4$. The common definition of saddle point is that it is neither a local min nor local max, and this is defined without any reference to derivatives or Hessian. – Erick Wong Jan 16 at 5:31 @ErickWong, Fair enough. But given that the gradient and hessian is both 0 (in their own dimensions) at that point, how do you know that it is not maxima or minima? (One can say so by observation or by computing the function at other points but that's not robust). Is there any other way I am missing? – Inquest Jan 16 at 5:34 @Inquest You're not missing anything. Things get harder when the second derivative vanishes, except in simple cases such as these. This homework problem is meant to point out a limitation of the method. – Michael E2 Jan 17 at 12:43

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http://mathoverflow.net/questions/21031/ultrafilters-vs-well-orderings/21034

## Ultrafilters vs Well-orderings ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) This question was actually asked by John Stillwell in a comment to an answer to this question. I thought I would advertise it as a separate question since no one has yet answered and I am also curious about it. Question: Is the existence of a non-principal ultra-filter on $\omega$ a weaker assumption than the existence of a well-ordering of $\mathbb{R}$? - ## 2 Answers It is consistent that there exists a non-principal ultrafilter over $\omega$ while $\mathbb{R}$ is not well-ordered. To see this, suppose that the partition relation $\omega \to (\omega)^{\omega}$ holds in $L(\mathbb{R})$. Then forcing with $\mathbb{P}= [\omega]^{\omega}$ adjoins a selective ultrafilter $\mathcal{U}$ over $\omega$ and $\mathcal{P}(\omega)$ cannot be well-ordered in $L(\mathbb{R})[\mathcal{U}]$. (See Eisworth's paper: Selective ultrafilters and $\omega \to (\omega)^{\omega}$.) Thus $L(\mathbb{R})[\mathcal{U}]$ is a model of $ZF$ which contains the nonprincipal ultrafilter $\mathcal{U}$ and yet $\mathbb{R}$ cannot be well-ordered in $L(\mathbb{R})[\mathcal{U}]$. An update: it is perhaps also interesting to note that in $L(\mathbb{R})[\mathcal{U}]$, the ultraproduct $\prod_{\mathcal{U}} \bar{\mathbb{F}}_{p}$ of the algebraic closures of the fields of prime order $p$ is not isomorphic to $\mathbb{C}$. - How about relative consistency rather than direct implication? – Harry Gindi Apr 11 2010 at 21:55 Thanks a lot for the answer, Simon. – John Stillwell Apr 12 2010 at 10:52 1 It was a very interesting question. While it is well-known that the existence of non-principal ultrafilters is in general strictly weaker than the axiom of choice, this special case is probably the one which would interest most mathematicians. Plus I now understand a conversation that I recently had with a much more knowledgeable set theorist ... – Simon Thomas Apr 12 2010 at 12:45 ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. Historically it was proved that the ultrafilter lemma is independent from the axiom of choice by showing that there is a model in which there is an infinite Dedekind-finite set of real numbers, but every filter can be extended to an ultrafilter. Where an infinite Dedekind-finite set is an infinite set which does not have a countably infinite subset. The existence of infinite Dedekind-finite sets negates not only the axiom of choice, but also the [much] weaker axiom of countable choice. These sets cannot be well-ordered, and since the real numbers have such subset they cannot be well-ordered themselves in such model. The proof was given by Halpern and Levy in 1964. -

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http://www.scholarpedia.org/article/Numerical_analysis

# Numerical analysis From Scholarpedia Kendall E. Atkinson (2007), Scholarpedia, 2(8):3163. Curator and Contributors 1.00 - Kendall E. Atkinson Numerical analysis is the area of mathematics and computer science that creates, analyzes, and implements algorithms for solving numerically the problems of continuous mathematics. Such problems originate generally from real-world applications of algebra, geometry, and calculus, and they involve variables which vary continuously. These problems occur throughout the natural sciences, social sciences, medicine, engineering, and business. Beginning in the 1940's, the growth in power and availability of digital computers has led to an increasing use of realistic mathematical models in science, medicine, engineering, and business; and numerical analysis of increasing sophistication has been needed to solve these more accurate and complex mathematical models of the world. The formal academic area of numerical analysis varies from highly theoretical mathematical studies to computer science issues involving the effects of computer hardware and software on the implementation of specific algorithms. ## Areas of numerical analysis A rough categorization of the principal areas of numerical analysis is given below, keeping in mind that there is often a great deal of overlap between the listed areas. In addition, the numerical solution of many mathematical problems involves some combination of some of these areas, possibly all of them. There are also a few problems which do not fit neatly into any of the following categories. ### Systems of linear and nonlinear equations • Numerical solution of systems of linear equations. This refers to solving for $$x$$ in the equation $$Ax=b$$ with given matrix $$A$$ and column vector $$b\ .$$ The most important case has $$A$$ a square matrix. There are both direct methods of solution (requiring only a finite number of arithmetic operations) and iterative methods (giving increased accuracy with each new iteration). This topic also includes the matrix eigenvalue problem for a square matrix $$A\ ,$$ solving for $$x$$ and $$\lambda$$ in the equation $$Ax=\lambda x\ .$$ • Numerical solution of systems of nonlinear equations. This refers to rootfinding problems which are usually written as $$f(x)=0$$ with $$x$$ a vector with $$n$$ components and $$f(x)$$ a vector with $$m$$ components. The most important case has $$n=m\ .$$ • Optimization. This refers to minimizing or maximizing a real-valued function $$f(x)\ .$$ The permitted values for $$x=(x_{1},\dots,x_{n})$$ can be either constrained or unconstrained. The 'linear programming problem' is a well-known and important case; $$f(x)$$ is linear, and there are linear equality and/or inequality constraints on $$x\ .$$ ### Approximation theory Use computable functions $$p(x)$$ to approximate the values of functions $$f(x)$$ that are not easily computable or use approximations to simplify dealing with such functions. The most popular types of computable functions $$p(x)$$ are polynomials, rational functions, and piecewise versions of them, for example spline functions. Trigonometric polynomials are also a very useful choice. • Best approximations. Here a given function $$f(x)$$ is approximated within a given finite-dimensional family of computable functions. The quality of the approximation is expressed by a functional, usually the maximum absolute value of the approximation error or an integral involving the error. Least squares approximations and minimax approximations are the most popular choices. • Interpolation. A computable function $$p(x)$$ is to be chosen to agree with a given $$f(x)$$ at a given finite set of points $$x\ .$$ The study of determining and analyzing such interpolation functions is still an active area of research, particularly when $$p(x)$$ is a multivariate polynomial. • Fourier series. A function $$f(x)$$ is decomposed into orthogonal components based on a given orthogonal basis $$\left\{ \varphi _{1},\varphi_{2},\dots\right\} \ ,$$ and then $$f(x)$$ is approximated by using only the largest of such components. The convergence of Fourier series is a classical area of mathematics, and it is very important in many fields of application. The development of the Fast Fourier Transform in 1965 spawned a rapid progress in digital technology. In the 1990s wavelets became an important tool in this area. • Numerical integration and differentiation. Most integrals cannot be evaluated directly in terms of elementary functions, and instead they must be approximated numerically. Most functions can be differentiated analytically, but there is still a need for numerical differentiation, both to approximate the derivative of numerical data and to obtain approximations for discretizing differential equations. ### Numerical solution of differential and integral equations These equations occur widely as mathematical models for the physical world, and their numerical solution is important throughout the sciences and engineering. • Ordinary differential equations. This refers to systems of differential equations in which the unknown solutions are functions of only a single variable. The most important cases are initial value problems and boundary value problems, and these are the subjects of a number of textbooks. Of more recent interest are 'differential-algebraic equations', which are mixed systems of algebraic equations and ordinary differential equations. Also of recent interest are 'delay differential equations', in which the rate of change of the solution depends on the state of the system at past times. • Partial differential equations. This refers to differential equations in which the unknown solution is a function of more than one variable. These equations occur in almost all areas of engineering, and many basic models of the physical sciences are given as partial differential equations. Thus such equations are a very important topic for numerical analysis. For example, the Navier-Stokes equations are the main theoretical model for the motion of fluids, and the very large area of 'computational fluid mechanics' is concerned with solving numerically these and other equations of fluid dynamics • Integral equations. These equations involve the integration of an unknown function, and linear equations probably occur most frequently. Some mathematical models lead directly to integral equations; for example, the radiosity equation is a model for radiative heat transfer. Another important source of such problems is the reformulation of partial differential equations, and such reformulations are often called 'boundary integral equations'. ## Some common viewpoints and concerns in numerical analysis Most numerical analysts specialize in small sub-areas of the areas listed above, but they share some common concerns and perspectives. These include the following. • The mathematical aspects of numerical analysis make use of the language and results of linear algebra, real analysis, and functional analysis. • If you cannot solve a problem directly, then replace it with a 'nearby problem' which can be solved more easily. This is an important perspective which cuts across all types of mathematical problems. For example, to evaluate a definite integral numerically, begin by approximating its integrand using polynomial interpolation or a Taylor series, and then integrate exactly the polynomial approximation. • All numerical calculations are carried out using finite precision arithmetic, usually in a framework of floating-point representation of numbers. What are the effects of using such finite precision computer arithmetic? How are arithmetic calculations to be carried out? Using finite precision arithmetic will affect how we compute solutions to all types of problems, and it forces us to think about the limits on the accuracy with which a problem can be solved numerically. Even when solving finite systems of linear equations by direct numerical methods, infinite precision arithmetic is needed in order to find a particular exact solution. • There is concern with 'stability', a concept referring to the sensitivity of the solution of a given problem to small changes in the data or the given parameters of the problem. There are two aspects to this. First, how sensitive is the original problem to small changes in the data of the problem? Problems that are very sensitive are often referred to as 'ill-conditioned' or 'ill-posed', depending on the degree of sensitivity. Second, the numerical method should not introduce additional sensitivity that is not present in the original mathematical problem being solved. In developing a numerical method to solve a problem, the method should be no more sensitive to changes in the data than is true of the original mathematical problem. • There is a fundamental concern with error, its size, and its analytic form. When approximating a problem, a numerical analyst would want to understand the behaviour of the error in the computed solution. Understanding the form of the error may allow one to minimize or estimate it. A 'forward error analysis' looks at the effect of errors made in the solution process. This is the standard way of understanding the consequences of the approximation errors that occur in setting up a numerical method of solution, e.g. in numerical integration and in the numerical solution of differential and integral equations. A 'backward error analysis' works backward in a numerical algorithm, showing that the approximating numerical solution is the exact solution to a perturbed version of the original mathematical problem. In this way the stability of the original problem can be used to explain possible difficulties in a numerical method. Backward error analysis has been especially important in understanding the behaviour of numerical methods for solving linear algebra problems. • In order to develop efficient means of calculating a numerical solution, it is important to understand the characteristics of the computer being used. For example, the structure of the computer memory is often very important in devising efficient algorithms for large linear algebra problems. Also, parallel computer architectures lead to efficient algorithms only if the algorithm is designed to take advantage of the parallelism. • Numerical analysts are generally interested in measuring the efficiency of algorithms. What is the cost of a particular algorithm? For example, the use of Gaussian elimination to solve a linear system $$Ax=b$$ containing $$n$$ equations will require approximately $$2n^{3}/3$$ arithmetic operations. How does this compare with other numerical methods for solving this problem? This topic is a part of the larger area of 'computational complexity'. • Use information gained in solving a problem to improve the solution procedure for that problem. Often we do not fully understand the characteristics of a problem, especially very complicated and large ones. Such a solution process is sometimes referred to as being an 'adaptive procedure', and it can also be viewed as a feedback process. ## Development of numerical methods Numerical analysts and applied mathematicians have a variety of tools which they use in developing numerical methods for solving mathematical problems. An important perspective, one mentioned earlier, which cuts across all types of mathematical problems is that of replacing the given problem with a 'nearby problem' which can be solved more easily. There are other perspectives which vary with the type of mathematical problem being solved. ### Numerical solution of systems of linear equations Linear systems arise in many of the problems of numerical analysis, a reflection of the approximation of mathematical problems using linearization. This leads to diversity in the characteristics of linear systems, and for this reason there are numerous approaches to solving linear systems. As an example, numerical methods for solving partial differential equations often lead to very large 'sparse' linear systems in which most coefficients are zero. Solving such sparse systems requires methods that are quite different from those used to solve more moderate sized 'dense' linear systems in which most coefficients are non-zero. There are 'direct methods' and 'iterative methods' for solving all types of linear systems, and the method of choice depends on the characteristics of both the linear system and on the computer hardware being used. For example, some sparse systems can be solved by direct methods, whereas others are better solved using iteration. With iteration methods, the linear system is sometimes transformed to an equivalent form that is more amenable to being solved by iteration; this is often called 'pre-conditioning' of the linear system. With the matrix eigenvalue problem $$Ax=\lambda x\ ,$$ it is standard to transform the matrix $$A$$ to a simpler form, one for which the eigenvalue problem can be solved more easily and/or cheaply. A favorite choice are 'orthogonal transformations' because they are a simple and stable way to convert the given matrix $$A\ .$$ Orthogonal transformations are also very useful in transforming other problems in numerical linear algebra. Of particular importance in this regard is the least squares solution of over-determined linear systems. The linear programming problem was solved principally by the 'simplex method' until new approaches were developed in the 1980s, and it remains an important method of solution. The simplex method is a direct method that uses tools from the numerical solution of linear systems. ### Numerical solution of systems of nonlinear equations With a single equation $$f(x)=0\ ,$$ and having an initial estimate $$x_{0}$$ of the root $$\alpha\ ,$$ approximate $$f(x)$$ by its tangent line at the point $$\left(x_{0},f(x_{0})\right) \ .$$ Find the root of this tangent line as an approximation to the root $$\alpha$$ of the original equation $$f(x)=0\ .$$ This leads to 'Newton's iteration method', $x_{n+1}=x_{n}-\frac{f(x_{n})}{f^{\prime}(x_{n})},\quad\quad n=0,1,\dots$ Other linear and higher degree approximations can be used, and these lead to alternative iteration methods. An important derivative-free approximation of Newton’s method is the 'secant method'. For a system of $$m$$ nonlinear equations for a solution vector $$x$$ in $$R^m\ ,$$ we approximate $$f(x)$$ by its linear Taylor approximation about the initial estimate $$x_{0}\ .$$ This leads to Newton's method for nonlinear systems, $x_{n+1}=x_{n}-[f^{\prime}(x_{n})]^{-1}f(x_{n}),\quad\quad n=0,1,\dots$ in which $$f^{\prime}(x)$$ denotes the Jacobian matrix, of order $$m\times m$$ for $$f(x)\ .$$ In practice, the Jacobian matrix for $$f(x)$$ is often too complicated to compute directly; instead the partial derivatives in the Jacobian matrix are approximated using 'finite differences'. This leads to a 'finite difference Newton method'. As an alternative strategy and in analogy with the development of the secant method for the single variable problem, there is a similar rootfinding iteration method for solving nonlinear systems. It is called 'Broyden’s method' and it uses finite difference approximations of the derivatives in the Jacobian matrix, avoiding the evaluation of the partial derivatives of $$f(x)\ .$$ ### Numerical methods for solving differential and integral equations With such equations, there are usually at least two general steps involved in obtaining a nearby problem from which a numerical approximation can be computed; this is often referred to as 'discretization' of the original problem. The given equation will have a domain on which the unknown function is defined, perhaps an interval in one dimension and maybe a rectangle, ellipse, or other simply connected bounded region in two dimensions. Many numerical methods begin by introducing a mesh or grid on this domain, and the solution is to be approximated using this grid. Following this, there are several common approaches. One approach approximates the equation with a simpler equation defined on the mesh. For example, consider approximating the boundary value problem $u^{\prime\prime}(s)=f\left( s,u(s)\right) ,\quad0\leq s\leq1$ $u(0)=u(1)=0.$ Introduce a set of mesh points $$s_{j}=jh\ ,$$ $$j=0,1,\dots,n\ ,$$ with $$h=1/n$$ for some given $$n\geq2\ .$$ Approximate the boundary value problem by $\frac{1}{h^{2}}\left[ \tilde{u}_{n}(s_{j+1})-2\tilde{u}_{n}(s_{j})+\tilde {u}_{n}(s_{j-1})\right] =f\left( s_{j},\tilde{u}_{n}\left( s_{j}\right) \right) ,\quad j=1,\dots,n-1$ $\tilde{u}_{n}(0)=\tilde{u}_{n}(1)=0$ The second derivative in the original problem has been replaced by a numerical approximation to the second derivative. The new problem is a finite system of nonlinear equations, presumably amenable to solution by known techniques. The solution to this new problem is $$\tilde{u}_{n}\ ,$$ and it is defined on only the mesh points $$\left\{ s_{j}\right\} \ .$$ A second approach to discretizing differential and integral equations is as follows. Choose a finite-dimensional family of functions, denoted here by $$\mathcal{F}\ ,$$ with which to approximate the unknown solution function $$u\ .$$ Write the given differential or integral equation as $$L\left( u\right) =0\ ,$$ with $$L(v)$$ a function for any function $$v\ ,$$ perhaps over a restricted class of functions $$v\ .$$ The numerical method consists of selecting a function $$\tilde{u}\in\mathcal{F}$$ such that $$L(\tilde{u})$$ is a small function in some sense. The various ways of doing this lead to 'Galerkin methods', 'collocation methods', and 'least square methods'. Yet another approach is to reformulate the equation $$L\left( u\right) =0$$ as an optimization problem. Such reformulations are a part of the classical area of mathematics known as the 'calculus of variations', a subject that reflects the importance in physics of minimization principles. The well-known 'finite element method' for solving elliptic partial differential equations is obtained in this way, although it often coincides with a Galerkin method. The approximating functions in $$\mathcal{F}$$ are often chosen as piecewise polynomial functions which are polynomial over the elements of the mesh chosen earlier. Such methods are sometimes called 'local methods'. When the approximating functions $$p\in\mathcal{F}$$ are defined without reference to a grid, then the methods are sometimes called 'global methods' or 'spectral methods'. Examples of such $$\mathcal{F}$$ are sets of polynomials or trigonometric functions of some finite degree or less. With all three approaches to solving a differential or integral equations, the intent is that the resulting solution $$\tilde{u}$$ be close to the desired solution $$u\ .$$ The business of theoretical numerical analysis is to analyze such an algorithm and investigate the size of $$u-\tilde{u}\ .$$ ## References For an historical account of early numerical analysis, see • Herman Goldstine. A History of Numerical Analysis From the 16th Through the19th Century, Springer-Verlag, New York, 1977. For a current view of numerical analysis as taught at the undergraduate level, see • Cleve Moler. Numerical Computing with MATLAB, SIAM Pub., Philadelphia, 2004. For a current view of numerical analysis as taught at the advanced undergraduate or beginning graduate level, see • Alfio Quarteroni, Riccardo Sacco, and Fausto Saleri. Numerical Mathematics, Springer-Verlag, New York, 2000. • Christoph W. Ueberhuber. Numerical Computation: Vol. 1: Methods, Software, and Analysis, Vol. 2: Methods, Software, and Analysis, Springer-Verlag, New York, 1997. For one perspective on a theoretical framework using functional analysis for studying many problems in numerical analysis, see • Kendall Atkinson and Weimin Han. Theoretical Numerical Analysis: A Functional Analysis Framework, 2nd ed., Springer-Verlag, New York, 2005. As references for numerical linear algebra, see • Gene Golub and Charles Van Loan. Matrix Computations, 3rd ed., Johns Hopkins University Press, 1996. • Nicholas Higham. Accuracy and Stability of Numerical Algorithms, SIAM Pub., Philadelphia, 1996. For an introduction to practical numerical analysis for solving ordinary differential equations, see • Lawrence Shampine, Ian Gladwell, Skip Thompson. Solving ODEs with Matlab, Cambridge University Press, Cambridge, 2003. For information on computing aspects of numerical analysis, see • Michael Overton. Numerical computing with IEEE floating point arithmetic, SIAM Pub., Philadelphia, 2001. • Suely Oliveira and David Stewart. Writing Scientific Software: A Guide to Good Style, Cambridge University Press, Cambridge, 2006. Internal references • Olaf Sporns (2007) Complexity. Scholarpedia, 2(10):1623. • Skip Thompson (2007) Delay-differential equations. Scholarpedia, 2(3):2367. • James Meiss (2007) Dynamical systems. Scholarpedia, 2(2):1629. • Eugene M. Izhikevich (2007) Equilibrium. Scholarpedia, 2(10):2014. • Lawrence F. Shampine and Skip Thompson (2007) Initial value problems. Scholarpedia, 2(3):2861. • Mark Aronoff (2007) Language. Scholarpedia, 2(5):3175. • Philip Holmes and Eric T. Shea-Brown (2006) Stability. Scholarpedia, 1(10):1838.

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http://stats.stackexchange.com/questions/45174/voting-system-that-uses-confidence-for-each-voter

# Voting system that uses confidence for each voter Let's say, we have simple "yes/no" question that we want to know answer to. And there are N people "voting" for correct answer. Every voter has a history - list of 1's and 0's, showing whether they were right or wrong about this kind of questions in the past. If we assume history as a binomial distribution, we can find voters' mean performance on such questions, their variation, CI and any other kind of confidence metrics. Basically, my question is: how to incorporate confidence information into voting system? For example, if we consider only mean performance of each voter, then we can construct simple weighted voting system: $$result = sign(\sum_{v \in voters}\mu_v \times (-1)^{1-vote})$$ That is, we can just sum voters' weights multiplied either by $+1$ (for "yes") or by $-1$ (for "no"). It makes sense: if voter 1 has average of correct answers equal to $.9$, and voter 2 has only $.8$, than, probably, 1st person's vote should be considered as more important. On other hand, if 1st person have answered only 10 questions of this kind, and 2nd person have answered 1000 such questions, we are much more confident about 2nd person's skill level than about those of the 1st - it's just possible that 1st person was lucky, and after 10 relatively successful answers he will continue with much worse results. So, more precise question may sound like this: is there statistical metric that incorporates both - strength and confidence about some parameter? - ## 1 Answer You should consider the expertise of a voter as a latent variable of your system. You may then be able to solve your problem with bayesian inference. A representation as graphical model could be like this : Let's denote the variables $A$ for the true answer, $V_i$ for the vote of the voter $i$ and $H_i$ for its history. Say that you also have an "expertise" parameter $\mu_i$ such that $\Pr(A=V_i) = \mu_i$. If you put some prior on these $\mu_i$ -for example a Beta prior- you should be able to use the Bayes theorem to infer $\Pr(\mu_i \mid H_i)$, and then integrate over $\mu_i$ to compute $$\Pr(A \mid V_i, H_i) = \int_{\mu_i} \Pr(A, \mu_i \mid A_i, H_i)~ \mathrm{d}\mu_i$$ These systems are difficult to solve. You can use the EM algorithm as an approximation, or use complete likelihood maximisation scheme to perform exact Bayesian inference. Take a look on this paper Variational Inference for Crowdsourcing, Liu, Peng and Ihler 2012 (presented yesterday at NIPS !) for detailed algorithms for solving this task. - 1 Thanks for your answer, but could you please be a bit more specific about it? In particular, what you mean by expertise? If it's just probability that the person will answer correctly, then we already have its estimate as an average of previous answers, so it's not latent. If you mean than expertise incorporates both average and confidence about our estimate, then how can we propagate probabilities to get expertise and result? – ffriend Dec 6 '12 at 8:21 Yes, you can represent both average and confidence with this "expertise" variable and Bayesian inference. I have added a few explanations and a reference to my answer. Hope that helps ! – Emile Dec 7 '12 at 1:04 Great! This is exactly what I need. Thank you! – ffriend Dec 10 '12 at 8:46

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http://math.stackexchange.com/questions/255048/suppose-that-we-have-a-p-1-cdot-p-2-cdot-p-3-q-1-cdot-q-2-cdots-q-s-where-p

# Suppose that we have $a=p_1 \cdot p_2\cdot p_3=q_1\cdot q_2\cdots q_s$ where $p_1,p_2,p_3$ and $q_1,q_2,…,q_s$ are primes. Explain why $s=3$. Suppose that we have $a=p_1 \cdot p_2\cdot p_3=q_1\cdot q_2\cdots q_s$ where $p_1,p_2,p_3$ and $q_1,q_2,...,q_s$ are primes. Explain why $s=3$. This seems like a pointless question, but is it $3$ because $p_n$ doesn't go past $p_3$? I'm still having a hard time with this question. I'm mainly not understanding what it is asking. - Please check to see that I didn't alter the intended meaning with my edit. – Cameron Buie Dec 10 '12 at 2:10 Yes, there was an edit that changed the meaning. – Dmitri.Mendeleev Dec 10 '12 at 2:13 Aha! Now it makes sense. – MJD Dec 10 '12 at 2:14 ## 3 Answers I don't think this question is posed correctly, because it's definitely not true: $$30=2\cdot3\cdot5=13+17\Rightarrow s=2$$ and $$42=2\cdot3\cdot7=5+7+13+17\Rightarrow s=4$$ are just two counterexamples. For the corrected post: We want to show that if $a=p_1p_2p_3=q_1q_2\dots q_s$, then $s=3$. Clearly, $a$ divides $p_1$, so by the theorem on prime division, one of the $q_i$ divides $p_1$. Since $q_i$ is prime, then, $q_i=p_1$ and we can cancel it to get $p_2p_3=r_1r_2\dots r_{s-1}$ with $r_i$ prime. Repeating this process (assuming $s\geq 3$), we get $1=r_1r_2\dots r_{s-3}$, and all of the $r_i$ are integers, so either $s-3=0$ or $|r_i|=1$. But $r_i\geq 2$, so $s=3$. If $s<3$, then we will instead end up as $p_2p_3=1$ or similar. In this case, $p_i>1$ implies a contradiction. Thus $s=3$ is the only possible conclusion. - 2 Please see corrected post. – amWhy Dec 10 '12 at 2:15 I might be wrong, but isn't this just an application of the Fundamental Theorem of Arithmetic (FTA)? This theorem states that for any given positive integer, there is only one way to express it as a product of primes. The solution of the problem is then obvious. Of course, for a more complete solution you would have to first prove the FTA, but I won't go into that here.. - If you're worried that the question seems pointless, notice that the corresponding claim for addition is not true, since: $$p_1 + p_2 + p_3 = q_1 + \cdots + q_s$$ does not require that $s=3$—for example, $$10 + 11 + 12 = 1 + 2 + 3 + 4 + 5 + 18.$$ Even for multiplication the claim is not true unless you require the $p$s and $q$s to be prime: $$20 \cdot 25 \cdot 30 = 2 \cdot 10\cdot 15 \cdot 50$$ So this claim depends in a crucial way on special properties of multiplication and of prime numbers, and that is why it is not a silly question. -

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http://physics.stackexchange.com/questions/tagged/magnetic-fields?page=2&sort=newest&pagesize=15

# Tagged Questions The magnetic-fields tag has no wiki summary. 1answer 52 views ### What geological properties of the earth could we deduce by measuring magnetic field strength and direction? I wonder if it is of any use for a geophysicist, if we measure the magnetic field strength and direction of the earth. Could we make valid statements about the composition of the the earth, earth ... 1answer 186 views ### If the Moon had gravity as good as the Earth and a magnetic field could it have supported life? If the Moon had gravity as good as Earth and a magnetic field could it have supported life? Because if the Moon had gravity, it could have retained water more than is present today on the surface. ... 0answers 72 views ### Is there an equation for the magnetic field of a conductor attached to a magnet? Let's say I have a hollow conductive rod, 10mm diameter (O.D.), and I place a magnet of known strength 50mm up the shaft. 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http://mathforum.org/mathimages/index.php?title=Markus-Lyapunov_Fractals&diff=25118&oldid=18868

# Markus-Lyapunov Fractals ### From Math Images (Difference between revisions) | | | | | |----------------------------------------|-----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------|------------------------------------------------------|--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------| | | | Current revision (16:17, 14 July 2011) (edit) (undo) | | | (20 intermediate revisions not shown.) | | | | | Line 1: | | Line 1: | | | - | {{Image Description | + | {{Image Description Ready | | | |ImageName=Markus-Lyapunov Fractal | | |ImageName=Markus-Lyapunov Fractal | | | |Image=Markus-Lyapunov1.gif | | |Image=Markus-Lyapunov1.gif | | - | |ImageIntro=A representation of the regions of chaos and stability over the space of two population growth rates. | + | |ImageIntro=Markus-Lyapunov fractals are representations of the regions of chaos and stability over the space of two population growth rates. | | - | |ImageDescElem=The Markus-Lyapunov fractal is much more than a pretty picture; it is a map. The curving bodies and sweeping arms of the image are a color-coded plot that shows us how a population changes as its rate of growth moves between two values. All the rich variations of color in the fractal come from the different levels of stability and chaos possible in such change. | + | |ImageDescElem=The Markus-Lyapunov fractal is much more than a pretty picture – it is a chart. The curving bodies and sweeping arms of the image are a color-coded plot that shows us how a population changes as its rate of growth moves between two values. All the rich variations of color in the fractal come from the different levels of stability and chaos possible in such change. | | | | | | | - | The [[Logistic Bifurcation|logistic map]] is one of the simplest mathematical representations of population growth. Depending on the rate of fecundity used in the map, it will generate either a neutral system, a stable oscillating system, or a [[chaos|chaotic]] system. To help determine which of these outcomes would occur, the mathematician Aleksandr Lyapunov developed a method for comparing changes in growth and time in order to calculate what has been dubbed the {{EasyBalloon|Link=Lyapunov exponent|Balloon=The Lyapunov exponent represents the overall rate of change of a system over many iterations, expressed logarithmically.}}. This is a useful indicator, and here's why: | + | The [[Logistic Bifurcation#Jump2|logistic map]] is one of the most concise mathematical representations of population growth. Depending on the rate of fecundity used in the map, it will generate either a neutral system, a stable oscillating system, or a [[chaos|chaotic]] system. To help determine which of these outcomes will occur, the mathematician Aleksandr Lyapunov developed a method for comparing changes in growth rate in order to calculate a value called the {{EasyBalloon|Link=Lyapunov exponent|Balloon=The Lyapunov exponent represents the overall rate of change of a system over many iterations, expressed logarithmically.}}. This is a useful indicator because, for the logistic map, | | - | *If it is zero, the population change is neutral; at some point in time, it reaches a fixed point and remains there. | + | | | - | *If it is less than zero, the population will become {{EasyBalloon|Link=stable|Balloon=Stability is different from a fixed point; A system that oscillates between two values is stable, and a system that oscillates between sixteen values is still stable.}}. The lower the number, the faster and more thoroughly the population will stabilize. | + | *If the Lyapunov exponent is zero, the population change is neutral; the population size begins and remains at a constant, fixed point. | | - | *If it is positive, the population will become chaotic. | + | *If the Lyapunov exponent is less than zero, the population will become {{EasyBalloon|Link=stable|Balloon=Stability is different from a fixed point. A system that oscillates between two values is stable, and a system that oscillates between sixteen values is still stable.}}. The more negative the number, the faster and more thoroughly the population will stabilize. | | | | + | *If the Lyapaunov exponent is positive, the population will become chaotic. | | | | | | | | [[Image:Markus-Lyapunov2.jpg|left|frame|Another example of a Markus-Lyapunov fractal, this one with chaos in black and stability in gold.]] | | [[Image:Markus-Lyapunov2.jpg|left|frame|Another example of a Markus-Lyapunov fractal, this one with chaos in black and stability in gold.]] | | - | What does all this have to do with the fantastical shapes of the Markus-Lyapunov fractal? The scientist Mario Markus wanted a way to visualize the potential represented by the Lyapunov exponent as a population moved between ''two'' different rates of growth. So he created a graphical space with one rate of growth measured along the ''x''-axis and the other along the ''y''. Thus for any point, (''x'',''y''), there is one specific Lyapunov exponent that predicts how a population with those rates of change will behave. Markus then created a color scheme to represent different Lyapunov exponents – one color represents positive numbers, and another represents negative numbers and zero. This second color he placed on a gradient from light to dark, so that lower negative numbers are lighter and those closer to zero are darker. The bands of black that appear in many fractals therefore show where the Lyapunov exponent is exactly zero, and bands of white indicate {{EasyBalloon|Link=superstable|Balloon=<math>\lambda=-\infty</math>, as the lowest possible Lyapunov exponent, indicates the fastest possible approach to stability.}} points. By this code, Markus could color every point on his graph space based on its Lyapunov exponent. | + | What does all this have to do with the fantastical shapes of the Markus-Lyapunov fractal? The scientist Mario Markus wanted a way to visualize the potential represented by the Lyapunov exponent as a population moved between ''two'' different rates of growth. So he created a graphical space with one rate of growth measured along the ''x''-axis and the other along the ''y''. Thus for any point (''x'',''y'') there is one specific Lyapunov exponent that predicts how a population with those rates of change will behave. | | | | + | | | | | + | Markus then created a color scheme to represent different Lyapunov exponents – one color represents positive numbers, and another represents negative numbers and zero. This second color he placed on a gradient from light to dark, so that lower negative numbers are lighter and those closer to zero are darker. The bands of black that appear in many fractals therefore show where the Lyapunov exponent is exactly zero, and bands of white indicate {{EasyBalloon|Link=superstable|Balloon=<math>\lambda=-\infty</math>, as the lowest possible Lyapunov exponent, indicates the fastest possible approach to stability.}} points. By this code, Markus could color every point on his graph space based on its Lyapunov exponent. | | | | | | | | Consider the main image on this page. The blue "background" shows all the points where the combination of the rates of change on the ''x'' and ''y'' axes will result in chaotic population growth. The "floating" yellow shapes show where the population will move toward stability. The lighter the yellow, the more stable the population. | | Consider the main image on this page. The blue "background" shows all the points where the combination of the rates of change on the ''x'' and ''y'' axes will result in chaotic population growth. The "floating" yellow shapes show where the population will move toward stability. The lighter the yellow, the more stable the population. | | | | + | | | | | + | Based on this color assignment, if a logistic system with rates of change ''x'' = 2 and ''y'' = 1.6 has a Lyapunov exponent of -0.3, there will be a dark yellow pixel at the graph location (2, 1.6), showing that the system moves slowly toward stability. If, instead, that logistic system had a Lyapunov exponent of 1, that same pixel would be blue, showing chaos. | | | | | | | | |ImageDesc= | | |ImageDesc= | | | ===The Lyapunov Exponent=== | | ===The Lyapunov Exponent=== | | - | The discrete form of the Lyapunov exponent is | + | [[Image:Dynamical1.gif|right|frame|As a logistic system progresses through ''n'' iterations, the distance, d''x''0, between two arbitrarily close points entered into the system will become a new distance, d''x''''n''.]] | | - | {{EquationRef2|4}}<math>\lambda=\lim_{N \to \infty}\frac{1}{N}\sum_{n=1}^N \log_2 \frac{dx_{(n+1)}}{dx_n}</math> | + | [[Image:LambdaGraphs.gif|right|frame|In both graphs, we see the evolution of <math>{\operatorname{d}x_n\over\operatorname{d}x_0}</math> from ''n'' = 0 to ''n'' = 10. The upper graph shows this evolution for λ = -1, while the lower shows this for λ = 1. Notice how quickly we observe convergence in the upper graph and divergence in the lower.]] | | - | In other words, the Lyapunov exponent <math>\lambda</math> represents the [[Limit|limit]] of the {{EasyBalloon|Link=mean|Balloon=The presence of <math>\frac{1}{N}\sum_{n=1}^N ...</math> shows that this is a mean.}} of the {{EasyBalloon|Link=exponential|Balloon=The presence of <math>\log_2</math> makes the relationship between <math>\lambda</math> and the original equation exponential.}} {{EasyBalloon|Link=rates of change|Balloon=<math>\frac{dx_{(n+1)}}{dx_n}</math> defines the rate of change.}} that occur in each transition, ''x''''n'' <math>\rightarrow</math> ''x''(''n''+1), as the number of transitions approaches infinity. | + | The Lyapunov exponent is a measure of the rate of divergence of two infinitesimally close points in a dynamic system. For a single-variable system such as the logistic map, we can consider two points at an arbitrarily small distance, d''x''0 from each other. After ''n'' iterations of the system, they will be at distance d''x''''n'' from each other. The Lyapunov exponent λ represents this change with the approximation: | | | | + | {{EquationRef2|Eq. 1}}<math>{\operatorname{d}x_n\over\operatorname{d}x_0}\approx 2^{\lambda n}</math> | | | | + | Or, isolating the Lyapunov exponent λ: | | | | + | :::<math>\frac{1}{n}\log_2{\operatorname{d}x_n\over\operatorname{d}x_0}\approx \lambda</math> | | | | + | Generalizing this for all ''n'', we use [[Summation Notation|summation notation]] to consider every step of iteration: | | | | + | {{EquationRef2|Eq. 2}}<math>\frac{1}{n}\sum_{1}^n \log_2 {\operatorname{d}x_{(n+1)}\over\operatorname{d}x_n} \approx \lambda</math> | | | | + | In order for this to be accurate, however, it must measure the system's divergence over an infinite period of time, so we define λ as the limit of {{EquationNote|Eq. 2}} as ''n'' approaches infinity: | | | | + | {{EquationRef2|Eq. 3}}<math>\lambda=\lim_{N \to \infty}\frac{1}{N}\sum_{n=1}^N \log_2 {\operatorname{d}x_{(n+1)}\over\operatorname{d}x_n}</math> | | | | + | This is the discrete form of the Lyapunov exponent. | | | | | | | - | What does this have to do with stability? The key is the log2 component, which renders numbers under 1 negative and those over 1 positive. This is what yields the properties of Lyapunov exponents laid out in the "Basic Explanation" – those mean overall rates of change that make the system {{EasyBalloon|Link=finite|Balloon=A geometric series or sequence is finite if is multiplied by a factor, '''r''' < 1, that makes it converge to a discrete value or set of values.}} must be less than 1, giving us a negative Lyapunov exponent, while those rates of change that expand the system to the point of chaos must be greater than one, giving us a positive exponent. When the mean overall rate of change is zero, the logarithmic component no longer exists, showing exactly what happens in a superstable system; the rate of change ceases to exist. | + | To consider the implications of λ, let us return to {{EquationNote|Eq. 1}}: | | | | + | :::<math>{\operatorname{d}x_n\over\operatorname{d}x_0}\approx 2^{\lambda n}</math> | | | | + | We can see that, for λ < 0, the difference between d''x''0 to d''x''''n'' will disappear as ''n'' grows. (Indeed, the lower the value of λ, the more quickly this change will disappear.) Similarly, for λ = 0, there will be no difference at all. For λ > 0, however, the difference between the initial distance between the points and the final distance between the points will expand exponentially as ''n'' grows. In other words, a positive Lyapunov exponent indicates a system in which an infinitesimal change in initial conditions can result in massively different final conditions. A negative Lyapunov exponent, on the other hand, indicates a system in which the effects of such an initial change will fade over time. | | | | | | | - | In other words, the Lyapunov exponent is a method for examining the rate of change of a system considered over infinite iterations, then taking that rate of change and making it easily identifiable as a value that induces either chaos or stability. | + | Recall that the phenomenon captured by a positive Lyapunov exponent – wide variation resulting from infinitely small initial changes – is one of the conditions for a system to be chaotic. Thus a positive Lyapunov, in the presence of other conditions of chaos, implies a chaotic system. | | | | | | | - | ====In the Logistic Formula==== | + | ====In the Logistic Map==== | | - | Basic differentiation shows us that, for the logistic formula ({{EquationNote|3}}): | + | Recall that the [[Logistic Bifurcation#Jump3|logistic map]] is: | | - | {{EquationRef2|5}}<math>\frac{dx_{(n+1)}}{dx_n}=\mathbf{r}-2\mathbf{r}x_0^2</math> | + | {{EquationRef2|Eq. 4}}<math>x_{(n+1)}=\mathbf{r}x_n(1-x_n)</math> | | - | Using this and a sufficiently large ''N'' number of iterations, we can approximate the Lyapunov exponent for the logistic formula to be: | + | From which we find | | - | {{EquationRef2|6}}<math>\lambda \approx \frac{1}{N}\sum_{n=1}^N \log_2 \mathbf{r}-2\mathbf{r}x_0^2</math> | + | :::<math>{\operatorname{d}x_{(n+1)}\over\operatorname{d}x_n}=\mathbf{r}-2\mathbf{r}x_n</math> | | - | Here we can see much more clearly a property that we have been assuming -- the variable that has the greatest impact on the stability of the logistic equation is '''r''', not ''x''''n''. This is clear here because, as we differentiate the logistic formula in order to examine its overall rate of change, ''x''''n'' is reduced to the constant value ''x''0 (the starting volume of the population). It is therefore no longer a variable when we look at the formula on the level of its Lyapunov exponent, and so changing the value of ''x''''n'' does not change whether the logistic function yields chaos or stability. | + | Inserting this into the Lyapunov exponent, {{EquationNote|Eq. 3}}, we have the Lyapunov exponent for the logistic map: | | | | + | {{EquationRef2|Eq. 5}}<math>\lambda=\lim_{N \to \infty}\frac{1}{N}\sum_{n=1}^N \log_2 \mathbf{r}-2\mathbf{r}x_n</math> | | | | + | As noted above, negative values of λ here indicate stability in the logistic map. Also, in the case of the logistic map, any system with a Lyapunov exponent greater than zero is a chaotic system. We can see this when we compare the [[Logistic Bifurcation|logistic bifurcation]] diagram with a graph of the Lyapunov exponents for the logistic maps of changing '''r''' values: | | | | + | [[Image:Lyap Bifurc1.gif|center|frame|Here a graph of the Lyapunov exponents for logistic maps with 0 < '''r''' < 4 is overlaid in red on the [[Logistic Bifurcation|logistic bifurcation]] diagram. The points of the red graph were calculated directly in Matlab using {{EquationNote|Eq. 5}}.]] | | | | | | | | ===Forcing the Rates of Change=== | | ===Forcing the Rates of Change=== | | - | Mathematically, the important part of Markus's contribution to understanding this type of system was not his method for generating fractals, but his use of periodic rate-of-change forcing. We have been discussing the great impact of the '''r''' value in determining the output of the logistic formula, but this value can have still greater impact if we do not choose to keep it constant. Anyone who has studied biology, as Markus has, knows that the rates of change of a populations size do not simply fluctuate with changing supplies of food and space, but also often alternate between two or more specific potential rates of change depending on such things as weather and mating seasons.[[Image:AB.gif|frame|A Markus-Lyapunov fractal with rate-of-change pattern ''ab'']] | + | Mathematically, the important part of Markus's contribution to understanding this type of system was not his method for generating fractals, but his use of periodic rate-of-change forcing. The value of '''r''' has a great impact on the output of the logistic map, but this value can have still greater impact if we do not choose to keep it constant. | | | | + | [[Image:AB.gif|frame|A Markus-Lyapunov fractal with rate-of-change pattern ''ab'']] | | | | | | | - | In terms of the logistic formula, this means we choose a set of rates of change, '''r'''1, '''r'''2, '''r'''3,..., '''r'''''p'', where ''p'' is the period over which the rates of change loop. When we force the rates of change to follow such a loop, we have a new, [[Modular arithmetic|modular]] logistic equation ({{EquationNote|3}}): | + | In terms of the logistic map, this means we choose a set of rates of change, '''r'''1, '''r'''2, '''r'''3,..., '''r'''''p'', where ''p'' is the period over which the rates of change loop. When we force the rates of change to follow such a loop, we have a new, [[Modular arithmetic|modular]] logistic map ({{EquationNote|Eq. 4}}): | | | | | | | - | {{EquationRef2|7}}<math>x_{(n+1)}=\mathbf{r}_{n \text{mod} p}x_n(1-x_n)</math> | + | {{EquationRef2|Eq. 6}}<math>x_{(n+1)}=\mathbf{r}_{n \text{mod} p}x_n(1-x_n)</math> | | | | | | | - | It is in these forced alterations in rates of change that the fascinating shapes of the Markus-Lyapunov fractal come out. Each of the fractals is formed from some pattern of two rates of change, ''a'' and ''b''. So a pattern ''aba'' would mean each point on the fractal is colored based on the Lyapunov exponent of the logistic formula {{EquationNote|7}}, where '''r'''1 = ''a'', '''r'''2 = ''b'', and '''r'''3 = ''a''. That is, the '''r''' values would cycle ''a,b,a,a,b,a,a,b,a...''. | + | It is in these forced alterations in rates of change that the fascinating shapes of the Markus-Lyapunov fractal come out. Each of the fractals is formed from some pattern of two rates of change, ''a'' and ''b''. So a pattern ''aba'' would mean each point on the fractal is colored based on the Lyapunov exponent of the logistic map {{EquationNote|Eq. 5}}, where '''r'''1 = ''a'', '''r'''2 = ''b'', and '''r'''3 = ''a''. That is, the '''r''' values would cycle ''a,b,a,a,b,a,a,b,a...''. | | | | | | | - | Because the axes used to map these fractals are measurements of changes in ''a'' and ''b'', the pattern ''a'' would simply yield a set of vertical bars, just as the pattern ''b'' would yield horizontal bars. However, once the patterns start to become mixed, more interesting results come out. The image to the right shows an ''ab'' pattern. Note that it is much simpler than other images shown on this page; the main image, for instance, is a ''bbbbbbaaaaaa'' pattern. | + | Because the axes used to map these fractals are measurements of changes in ''a'' and ''b'', the pattern ''a'' would simply yield a set of vertical bars, just as the pattern ''b'' would yield horizontal bars. However, once the patterns start to become mixed, more interesting results come out. The image to the right shows an ''ab'' pattern. Note that it is much simpler, in the quantity of spires and crossing arms, than other images shown on this page; the main image, for instance, is a ''bbbbbbaaaaaa'' pattern. | | | | | | | - | |AuthorName=BernardH, using Mathematica 5 | + | |other=understanding [[Logistic Bifurcation]] | | | | + | |AuthorName=BernardH | | | |SiteURL=http://en.wikipedia.org/wiki/File:Lyapunov-fractal.png | | |SiteURL=http://en.wikipedia.org/wiki/File:Lyapunov-fractal.png | | | |Field=Dynamic Systems | | |Field=Dynamic Systems | | | |Field2=Fractals | | |Field2=Fractals | | - | |WhyInteresting=[[Image:MarkusLyapunovZoom.gif|200px|frame|An enlargement of a section of "Zircon Zity," showing self-similarity.]] | + | |WhyInteresting= | | | | + | | | | | + | [[Image:MLBlowup.gif|frame|An enlargement of a section of "Zircon Zity," showing self-similarity.]] | | | ===Fractal Properties=== | | ===Fractal Properties=== | | - | The movements from light to dark and the dramatic curves of the boundaries between stability and chaos here create an astonishing 3D effect. But the image is striking not only for its beauty but also for its self-similarity. Self-similarity is that trait that makes [[fractals]] what they are – zooming in on the image reveals smaller and smaller parts that resemble the whole. Consider the image to the right, enlarged from a section of the main image above. Here we see several shapes that repeat in smaller and smaller iterations. Perhaps ironically, this type of pattern is a common property of chaos. | + | The movements from light to dark and the dramatic curves of the boundaries between stability and chaos here create an astonishing 3D effect. But the image is striking not only for its beauty but also for its '''self-similarity'''. Self-similarity is that trait that makes [[Field:Fractals|fractals]] what they are – zooming in on the image reveals smaller and smaller parts that resemble the whole. Consider the image to the right, showing an enlarged section of the main image above. Here we see several shapes that repeat in smaller and smaller iterations. Perhaps ironically, this type of pattern is a common property of chaos. | | | | | | | - | For more images of the fractal properties of chaotic systems, see the [[Henon Attractor]], the [[Harter-Heighway Dragon]] Curve, and [[Julia Sets]]. | + | For more images of the fractal properties of chaotic systems, see the [[Blue Fern]], the [[Henon Attractor]], the [[Harter-Heighway Dragon]] Curve, and [[Julia Sets]]. | | | | | | | | | | | | Line 59: | | Line 81: | | | | |References=<references /> | | |References=<references /> | | | | | | | - | ;Other Sources Consulted: | + | |InProgress=No | | - | : Elert, G. (2007). ''The Chaos Hypertextbook''. http://hypertextbook.com/chaos/ | + | | | - | |InProgress=Yes | + | | | | }} | | }} | ## Current revision Markus-Lyapunov Fractal Fields: Dynamic Systems and Fractals Image Created By: BernardH Website: [1] Markus-Lyapunov Fractal Markus-Lyapunov fractals are representations of the regions of chaos and stability over the space of two population growth rates. # Basic Description The Markus-Lyapunov fractal is much more than a pretty picture – it is a chart. The curving bodies and sweeping arms of the image are a color-coded plot that shows us how a population changes as its rate of growth moves between two values. All the rich variations of color in the fractal come from the different levels of stability and chaos possible in such change. The logistic map is one of the most concise mathematical representations of population growth. Depending on the rate of fecundity used in the map, it will generate either a neutral system, a stable oscillating system, or a chaotic system. To help determine which of these outcomes will occur, the mathematician Aleksandr Lyapunov developed a method for comparing changes in growth rate in order to calculate a value called the Lyapunov exponentThe Lyapunov exponent represents the overall rate of change of a system over many iterations, expressed logarithmically.. This is a useful indicator because, for the logistic map, • If the Lyapunov exponent is zero, the population change is neutral; the population size begins and remains at a constant, fixed point. • If the Lyapunov exponent is less than zero, the population will become stableStability is different from a fixed point. A system that oscillates between two values is stable, and a system that oscillates between sixteen values is still stable.. The more negative the number, the faster and more thoroughly the population will stabilize. • If the Lyapaunov exponent is positive, the population will become chaotic. Another example of a Markus-Lyapunov fractal, this one with chaos in black and stability in gold. What does all this have to do with the fantastical shapes of the Markus-Lyapunov fractal? The scientist Mario Markus wanted a way to visualize the potential represented by the Lyapunov exponent as a population moved between two different rates of growth. So he created a graphical space with one rate of growth measured along the x-axis and the other along the y. Thus for any point (x,y) there is one specific Lyapunov exponent that predicts how a population with those rates of change will behave. Markus then created a color scheme to represent different Lyapunov exponents – one color represents positive numbers, and another represents negative numbers and zero. This second color he placed on a gradient from light to dark, so that lower negative numbers are lighter and those closer to zero are darker. The bands of black that appear in many fractals therefore show where the Lyapunov exponent is exactly zero, and bands of white indicate superstable$\lambda=-\infty$, as the lowest possible Lyapunov exponent, indicates the fastest possible approach to stability. points. By this code, Markus could color every point on his graph space based on its Lyapunov exponent. Consider the main image on this page. The blue "background" shows all the points where the combination of the rates of change on the x and y axes will result in chaotic population growth. The "floating" yellow shapes show where the population will move toward stability. The lighter the yellow, the more stable the population. Based on this color assignment, if a logistic system with rates of change x = 2 and y = 1.6 has a Lyapunov exponent of -0.3, there will be a dark yellow pixel at the graph location (2, 1.6), showing that the system moves slowly toward stability. If, instead, that logistic system had a Lyapunov exponent of 1, that same pixel would be blue, showing chaos. # A More Mathematical Explanation Note: understanding of this explanation requires: * [Click to view A More Mathematical Explanation] ### The Lyapunov Exponent [[Image:Dynamical1.gif|right|frame|As a logistic system progresses throug [...] [Click to hide A More Mathematical Explanation] ### The Lyapunov Exponent As a logistic system progresses through n iterations, the distance, dx0, between two arbitrarily close points entered into the system will become a new distance, dxn. In both graphs, we see the evolution of ${\operatorname{d}x_n\over\operatorname{d}x_0}$ from n = 0 to n = 10. The upper graph shows this evolution for λ = -1, while the lower shows this for λ = 1. Notice how quickly we observe convergence in the upper graph and divergence in the lower. The Lyapunov exponent is a measure of the rate of divergence of two infinitesimally close points in a dynamic system. For a single-variable system such as the logistic map, we can consider two points at an arbitrarily small distance, dx0 from each other. After n iterations of the system, they will be at distance dxn from each other. The Lyapunov exponent λ represents this change with the approximation: ${\operatorname{d}x_n\over\operatorname{d}x_0}\approx 2^{\lambda n}$ Or, isolating the Lyapunov exponent λ: $\frac{1}{n}\log_2{\operatorname{d}x_n\over\operatorname{d}x_0}\approx \lambda$ Generalizing this for all n, we use summation notation to consider every step of iteration: $\frac{1}{n}\sum_{1}^n \log_2 {\operatorname{d}x_{(n+1)}\over\operatorname{d}x_n} \approx \lambda$ In order for this to be accurate, however, it must measure the system's divergence over an infinite period of time, so we define λ as the limit of Eq. 2 as n approaches infinity: $\lambda=\lim_{N \to \infty}\frac{1}{N}\sum_{n=1}^N \log_2 {\operatorname{d}x_{(n+1)}\over\operatorname{d}x_n}$ This is the discrete form of the Lyapunov exponent. To consider the implications of λ, let us return to Eq. 1: ${\operatorname{d}x_n\over\operatorname{d}x_0}\approx 2^{\lambda n}$ We can see that, for λ < 0, the difference between dx0 to dxn will disappear as n grows. (Indeed, the lower the value of λ, the more quickly this change will disappear.) Similarly, for λ = 0, there will be no difference at all. For λ > 0, however, the difference between the initial distance between the points and the final distance between the points will expand exponentially as n grows. In other words, a positive Lyapunov exponent indicates a system in which an infinitesimal change in initial conditions can result in massively different final conditions. A negative Lyapunov exponent, on the other hand, indicates a system in which the effects of such an initial change will fade over time. Recall that the phenomenon captured by a positive Lyapunov exponent – wide variation resulting from infinitely small initial changes – is one of the conditions for a system to be chaotic. Thus a positive Lyapunov, in the presence of other conditions of chaos, implies a chaotic system. #### In the Logistic Map Recall that the logistic map is: $x_{(n+1)}=\mathbf{r}x_n(1-x_n)$ From which we find ${\operatorname{d}x_{(n+1)}\over\operatorname{d}x_n}=\mathbf{r}-2\mathbf{r}x_n$ Inserting this into the Lyapunov exponent, Eq. 3, we have the Lyapunov exponent for the logistic map: $\lambda=\lim_{N \to \infty}\frac{1}{N}\sum_{n=1}^N \log_2 \mathbf{r}-2\mathbf{r}x_n$ As noted above, negative values of λ here indicate stability in the logistic map. Also, in the case of the logistic map, any system with a Lyapunov exponent greater than zero is a chaotic system. We can see this when we compare the logistic bifurcation diagram with a graph of the Lyapunov exponents for the logistic maps of changing r values: Here a graph of the Lyapunov exponents for logistic maps with 0 < r < 4 is overlaid in red on the logistic bifurcation diagram. The points of the red graph were calculated directly in Matlab using Eq. 5. ### Forcing the Rates of Change Mathematically, the important part of Markus's contribution to understanding this type of system was not his method for generating fractals, but his use of periodic rate-of-change forcing. The value of r has a great impact on the output of the logistic map, but this value can have still greater impact if we do not choose to keep it constant. A Markus-Lyapunov fractal with rate-of-change pattern ab In terms of the logistic map, this means we choose a set of rates of change, r1, r2, r3,..., rp, where p is the period over which the rates of change loop. When we force the rates of change to follow such a loop, we have a new, modular logistic map (Eq. 4): $x_{(n+1)}=\mathbf{r}_{n \text{mod} p}x_n(1-x_n)$ It is in these forced alterations in rates of change that the fascinating shapes of the Markus-Lyapunov fractal come out. Each of the fractals is formed from some pattern of two rates of change, a and b. So a pattern aba would mean each point on the fractal is colored based on the Lyapunov exponent of the logistic map Eq. 5, where r1 = a, r2 = b, and r3 = a. That is, the r values would cycle a,b,a,a,b,a,a,b,a.... Because the axes used to map these fractals are measurements of changes in a and b, the pattern a would simply yield a set of vertical bars, just as the pattern b would yield horizontal bars. However, once the patterns start to become mixed, more interesting results come out. The image to the right shows an ab pattern. Note that it is much simpler, in the quantity of spires and crossing arms, than other images shown on this page; the main image, for instance, is a bbbbbbaaaaaa pattern. # Why It's Interesting An enlargement of a section of "Zircon Zity," showing self-similarity. ### Fractal Properties The movements from light to dark and the dramatic curves of the boundaries between stability and chaos here create an astonishing 3D effect. But the image is striking not only for its beauty but also for its self-similarity. Self-similarity is that trait that makes fractals what they are – zooming in on the image reveals smaller and smaller parts that resemble the whole. Consider the image to the right, showing an enlarged section of the main image above. Here we see several shapes that repeat in smaller and smaller iterations. Perhaps ironically, this type of pattern is a common property of chaos. For more images of the fractal properties of chaotic systems, see the Blue Fern, the Henon Attractor, the Harter-Heighway Dragon Curve, and Julia Sets. One artist superimposed and edited several real Markus-Lyapunov fractals to create this piece of art. ### Artistic Extensions After Markus saw the incredible beauty and intriguing three-dimensionality of the images generated by his plotting system, he immediately sent the images to a gallery in the hopes that it would display his images in an exhibition.[1] It's easy to see why he did so, and in fact, pictures based on these fractals have become a large part of what is called "fractalist" art. As with all domains of fractalist art, there is a great deal of debate in the art community over whether these images are truly "art" given their intrinsic reliance on a purely scientific, algorithmically-generated chart. One could say that such a process is devoid of creativity, but it is equally valid to say that the identification and presentation of the beauty in the science is an art in itself – a concept that is critical in modern art. Either way, there has been an undeniable artistic fascination with Markus-Lyapunov fractals; if the image seems familiar, you have likely seen it on posters, t-shirts, or any other canvas for graphic design. # Teaching Materials There are currently no teaching materials for this page. Add teaching materials. # References 1. ↑ Dewdney, A. K. (1991). Leaping into Lyapunov Space. Scientific American, (130-132). Leave a message on the discussion page by clicking the 'discussion' tab at the top of this image page.

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http://en.wikipedia.org/wiki/Optimal_control

# Optimal control Optimal control theory, an extension of the calculus of variations, is a mathematical optimization method for deriving control policies. The method is largely due to the work of Lev Pontryagin and his collaborators in the Soviet Union[1] and Richard Bellman in the United States. ## General method Optimal control deals with the problem of finding a control law for a given system such that a certain optimality criterion is achieved. A control problem includes a cost functional that is a function of state and control variables. An optimal control is a set of differential equations describing the paths of the control variables that minimize the cost functional. The optimal control can be derived using Pontryagin's maximum principle (a necessary condition also known as Pontryagin's minimum principle or simply Pontryagin's Principle[2]), or by solving the Hamilton–Jacobi–Bellman equation (a sufficient condition). We begin with a simple example. Consider a car traveling on a straight line through a hilly road. The question is, how should the driver press the accelerator pedal in order to minimize the total traveling time? Clearly in this example, the term control law refers specifically to the way in which the driver presses the accelerator and shifts the gears. The "system" consists of both the car and the road, and the optimality criterion is the minimization of the total traveling time. Control problems usually include ancillary constraints. For example the amount of available fuel might be limited, the accelerator pedal cannot be pushed through the floor of the car, speed limits, etc. A proper cost functional is a mathematical expression giving the traveling time as a function of the speed, geometrical considerations, and initial conditions of the system. It is often the case that the constraints are interchangeable with the cost functional. Another optimal control problem is to find the way to drive the car so as to minimize its fuel consumption, given that it must complete a given course in a time not exceeding some amount. Yet another control problem is to minimize the total monetary cost of completing the trip, given assumed monetary prices for time and fuel. A more abstract framework goes as follows. Minimize the continuous-time cost functional $J=\Phi\,[\,\textbf{x}(t_0),t_0,\textbf{x}(t_f),t_f\,] + \int_{t_0}^{t_f} \mathcal{L}\,[\,\textbf{x}(t),\textbf{u}(t),t\,] \,\operatorname{d}t$ subject to the first-order dynamic constraints $\dot{\textbf{x}}(t) = \textbf{a}\,[\,\textbf{x}(t),\textbf{u}(t),t\,],$ the algebraic path constraints $\textbf{b}\,[\,\textbf{x}(t),\textbf{u}(t),t\,] \leq \textbf{0},$ and the boundary conditions $\boldsymbol{\phi}\,[\,\textbf{x}(t_0),t_0,\textbf{x}(t_f),t_f\,] = 0$ where $\textbf{x}(t)$ is the state, $\textbf{u}(t)$ is the control, $t$ is the independent variable (generally speaking, time), $t_0$ is the initial time, and $t_f$ is the terminal time. The terms $\Phi$ and $\mathcal{L}$ are called the endpoint cost and Lagrangian, respectively. Furthermore, it is noted that the path constraints are in general inequality constraints and thus may not be active (i.e., equal to zero) at the optimal solution. It is also noted that the optimal control problem as stated above may have multiple solutions (i.e., the solution may not be unique). Thus, it is most often the case that any solution $[\textbf{x}^*(t^*),\textbf{u}^*(t^*),t^*]$ to the optimal control problem is locally minimizing. ## Linear quadratic control A special case of the general nonlinear optimal control problem given in the previous section is the linear quadratic (LQ) optimal control problem. The LQ problem is stated as follows. Minimize the quadratic continuous-time cost functional $J=\tfrac{1}{2} \textbf{x}^{\text{T}}(t_f)\textbf{S}_f\textbf{x}(t_f) + \tfrac{1}{2} \int_{t_0}^{t_f} [\,\textbf{x}^{\text{T}}(t)\textbf{Q}(t)\textbf{x}(t) + \textbf{u}^{\text{T}}(t)\textbf{R}(t)\textbf{u}(t)\,]\, \operatorname{d}t$ Subject to the linear first-order dynamic constraints $\dot{\textbf{x}}(t)=\textbf{A}(t) \textbf{x}(t) + \textbf{B}(t) \textbf{u}(t),$ and the initial condition $\textbf{x}(t_0) = \textbf{x}_0$ A particular form of the LQ problem that arises in many control system problems is that of the linear quadratic regulator (LQR) where all of the matrices (i.e., $\textbf{A}$, $\textbf{B}$, $, \textbf{Q}$, and $\textbf{R}$) are constant, the initial time is arbitrarily set to zero, and the terminal time is taken in the limit $t_f\rightarrow\infty$ (this last assumption is what is known as infinite horizon). The LQR problem is stated as follows. Minimize the infinite horizon quadratic continuous-time cost functional $J=\tfrac{1}{2} \int_{0}^{\infty}[\,\textbf{x}^{\text{T}}(t)\textbf{Q}\textbf{x}(t) + \textbf{u}^{\text{T}}(t)\textbf{R}\textbf{u}(t)\,]\, \operatorname{d}t$ Subject to the linear time-invariant first-order dynamic constraints $\dot{\textbf{x}}(t)=\textbf{A} \textbf{x}(t) + \textbf{B} \textbf{u}(t),$ and the initial condition $\textbf{x}(t_0) = \textbf{x}_0$ In the finite-horizon case the matrices are restricted in that $\textbf{Q}$ and $\textbf{R}$ are positive semi-definite and positive definite, respectively. In the infinite-horizon case, however, the matrices $\textbf{Q}$ and $\textbf{R}$ are not only positive-semidefinite and positive-definite, respectively, but are also constant. These additional restrictions on $\textbf{Q}$ and $\textbf{R}$ in the infinite-horizon case are enforced to ensure that the cost functional remains positive. Furthermore, in order to ensure that the cost function is bounded, the additional restriction is imposed that the pair $(\textbf{A},\textbf{B})$ is controllable. Note that the LQ or LQR cost functional can be thought of physically as attempting to minimize the control energy (measured as a quadratic form). The infinite horizon problem (i.e., LQR) may seem overly restrictive and essentially useless because it assumes that the operator is driving the system to zero-state and hence driving the output of the system to zero. This is indeed correct. However the problem of driving the output to a desired nonzero level can be solved after the zero output one is. In fact, it can be proved that this secondary LQR problem can be solved in a very straightforward manner. It has been shown in classical optimal control theory that the LQ (or LQR) optimal control has the feedback form $\textbf{u}(t)=-\textbf{K}(t)\textbf{x}(t)$ where $\textbf{K}(t)$ is a properly dimensioned matrix, given as $\textbf{K}(t)=\textbf{R}^{-1}\textbf{B}^{\text{T}}\textbf{S}(t),$ and $\textbf{S}(t)$ is the solution of the differential Riccati equation. The differential Riccati equation is given as $\dot{\textbf{S}}(t) = -\textbf{S}(t)\textbf{A}-\textbf{A}^{\text{T}}\textbf{S}(t)+\textbf{S}(t)\textbf{B}\textbf{R}^{-1}\textbf{B}^{\text{T}}\textbf{S}(t)-\textbf{Q}$ For the finite horizon LQ problem, the Riccati equation is integrated backward in time using the terminal boundary condition $\textbf{S}(t_f) = \textbf{S}_f$ For the infinite horizon LQR problem, the differential Riccati equation is replaced with the algebraic Riccati equation (ARE) given as $\textbf{0} = -\textbf{S}\textbf{A}-\textbf{A}^{\text{T}}\textbf{S}+\textbf{S}\textbf{B}\textbf{R}^{-1}\textbf{B}^{\text{T}}\textbf{S}-\textbf{Q}$ Understanding that the ARE arises from infinite horizon problem, the matrices $\textbf{A}$, $\textbf{B}$, $\textbf{Q}$, and $\textbf{R}$ are all constant. It is noted that there are in general multiple solutions to the algebraic Riccati equation and the positive definite (or positive semi-definite) solution is the one that is used to compute the feedback gain. The LQ (LQR) problem was elegantly solved by Rudolf Kalman.[3] ## Numerical methods for optimal control Optimal control problems are generally nonlinear and therefore, generally do not have analytic solutions (e.g., like the linear-quadratic optimal control problem). As a result, it is necessary to employ numerical methods to solve optimal control problems. In the early years of optimal control (circa 1950s to 1980s) the favored approach for solving optimal control problems was that of indirect methods. In an indirect method, the calculus of variations is employed to obtain the first-order optimality conditions. These conditions result in a two-point (or, in the case of a complex problem, a multi-point) boundary-value problem. This boundary-value problem actually has a special structure because it arises from taking the derivative of a Hamiltonian. Thus, the resulting dynamical system is a Hamiltonian system of the form $\begin{array}{lcl} \dot{\textbf{x}} & = & \partial H/\partial\boldsymbol{\lambda} \\ \dot{\boldsymbol{\lambda}} & = & -\partial H/\partial\textbf{x} \end{array}$ where $H=\mathcal{L}+\boldsymbol{\lambda}^{\text{T}}\textbf{a}-\boldsymbol{\mu}^{\text{T}}\textbf{b}$ is the augmented Hamiltonian and in an indirect method, the boundary-value problem is solved (using the appropriate boundary or transversality conditions). The beauty of using an indirect method is that the state and adjoint (i.e., $\boldsymbol{\lambda}$) are solved for and the resulting solution is readily verified to be an extremal trajectory. The disadvantage of indirect methods is that the boundary-value problem is often extremely difficult to solve (particularly for problems that span large time intervals or problems with interior point constraints). A well-known software program that implements indirect methods is BNDSCO.[4] The approach that has risen to prominence in numerical optimal control over the past two decades (i.e., from the 1980s to the present) is that of so-called direct methods. In a direct method, the state and/or control are approximated using an appropriate function approximation (e.g., polynomial approximation or piecewise constant parameterization). Simultaneously, the cost functional is approximated as a cost function. Then, the coefficients of the function approximations are treated as optimization variables and the problem is "transcribed" to a nonlinear optimization problem of the form: Minimize $F(\textbf{z})\,$ subject to the algebraic constraints $\begin{array}{lcl} \textbf{g}(\textbf{z}) & = & \textbf{0} \\ \textbf{h}(\textbf{z}) & \leq & \textbf{0} \end{array}$ Depending upon the type of direct method employed, the size of the nonlinear optimization problem can be quite small (e.g., as in a direct shooting or quasilinearization method) or may be quite large (e.g., a direct collocation method[5]). In the latter case (i.e., a collocation method), the nonlinear optimization problem may be literally thousands to tens of thousands of variables and constraints. Given the size of many NLPs arising from a direct method, it may appear somewhat counter-intuitive that solving the nonlinear optimization problem is easier than solving the boundary-value problem. It is, however, the fact that the NLP is easier to solve than the boundary-value problem. The reason for the relative ease of computation, particularly of a direct collocation method, is that the NLP is sparse and many well-known software programs exist (e.g., SNOPT[6]) to solve large sparse NLPs. As a result, the range of problems that can be solved via direct methods (particularly direct collocation methods which are very popular these days) is significantly larger than the range of problems that can be solved via indirect methods. In fact, direct methods have become so popular these days that many people have written elaborate software programs that employ these methods. In particular, many such programs written in FORTRAN include DIRCOL,[7] SOCS,[8] OTIS,[9] GESOP/ASTOS[10] and DITAN.[11] In recent years, due to the advent of the MATLAB programming language, optimal control software in MATLAB has become more common. Examples of academically developed MATLAB software tools implementing direct methods include RIOTS,[12]DIDO,[13] DIRECT,[14] and GPOPS,[15] while an example of an industry developed MATLAB tool is PROPT.[16] These software tools have increased significantly the opportunity for people to explore complex optimal control problems both for academic research and industrial-strength problems. Finally, it is noted that general-purpose MATLAB optimization environments such as TOMLAB have made coding complex optimal control problems significantly easier than was previously possible in languages such as C and FORTRAN. ## Discrete-time optimal control The examples thus far have shown continuous time systems and control solutions. In fact, as optimal control solutions are now often implemented digitally, contemporary control theory is now primarily concerned with discrete time systems and solutions. The Theory of Consistent Approximations[17] provides conditions under which solutions to a series of increasingly accurate discretized optimal control problem converge to the solution of the original, continuous-time problem. Not all discretization methods have this property, even seemingly obvious ones. For instance, using a variable step-size routine to integrate the problem's dynamic equations may generate a gradient which does not converge to zero (or point in the right direction) as the solution is approached. The direct method RIOTS is based on the Theory of Consistent Approximation. ## Examples A common solution strategy in many optimal control problems is to solve for the costate (sometimes called the shadow price) $\lambda(t)$. The costate summarizes in one number the marginal value of expanding or contracting the state variable next turn. The marginal value is not only the gains accruing to it next turn but associated with the duration of the program. It is nice when $\lambda(t)$ can be solved analytically, but usually the most one can do is describe it sufficiently well that the intuition can grasp the character of the solution and an equation solver can solve numerically for the values. Having obtained $\lambda(t)$, the turn-t optimal value for the control can usually be solved as a differential equation conditional on knowledge of $\lambda(t)$. Again it is infrequent, especially in continuous-time problems, that one obtains the value of the control or the state explicitly. Usually the strategy is to solve for thresholds and regions that characterize the optimal control and use a numerical solver to isolate the actual choice values in time. ### Finite time Consider the problem of a mine owner who must decide at what rate to extract ore from his mine. He owns rights to the ore from date $0$ to date $T$. At date $0$ there is $x_0$ ore in the ground, and the instantaneous stock of ore $x(t)$ declines at the rate the mine owner extracts it u(t). The mine owner extracts ore at cost $u(t)^2/x(t)$ and sells ore at a constant price $p$. He does not value the ore remaining in the ground at time $T$ (there is no "scrap value"). He chooses the rate of extraction in time u(t) to maximize profits over the period of ownership with no time discounting. 1. Discrete-time version The manager maximizes profit $\Pi$: $\Pi = \sum_{t=0}^{T-1} \left[ pu_t - \frac{u_t^2}{x_t} \right]$ subject to the law of evolution for the state variable $x_t$ $x_{t+1} - x_t = - u_t\!$ Form the Hamiltonian and differentiate: $H = pu_t - \frac{u_t^2}{x_t} - \lambda_{t+1} u_t$ $\frac{\partial H}{\partial u_t} = p - \lambda_{t+1} - 2\frac{u_t}{x_t} = 0$ $\lambda_{t+1} - \lambda_t = -\frac{\partial H}{\partial x_t} = -\left( \frac{u_t}{x_t} \right)^2$ As the mine owner does not value the ore remaining at time $T$, $\lambda_T = 0\!$ Using the above equations, it is easy to solve for the $x_t$ and $\lambda_t$ series $\lambda_t = \lambda_{t+1} + \frac{(p-\lambda_{t+1})^2}{4}$ $x_{t+1} = x_t \frac{2 - p + \lambda_{t+1}}{2}$ and using the initial and turn-T conditions, the $x_t$ series can be solved explicitly, giving $u_t$. 2. Continuous-time version The manager maximizes profit $\Pi$: $\Pi = \int_0^T \left[ pu(t) - \frac{u(t)^2}{x(t)} \right] dt$ subject to the law of evolution for the state variable $x(t)$ $\dot x(t) = - u(t)$ Form the Hamiltonian and differentiate: $H = pu(t) - \frac{u(t)^2}{x(t)} - \lambda(t) u(t)$ $\frac{\partial H}{\partial u} = p - \lambda(t) - 2\frac{u(t)}{x(t)} = 0$ $\dot\lambda(t) = -\frac{\partial H}{\partial x} = -\left( \frac{u(t)}{x(t)} \right)^2$ As the mine owner does not value the ore remaining at time $T$, $\lambda(T) = 0$ Using the above equations, it is easy to solve for the differential equations governing $u(t)$ and $\lambda(t)$ $\dot\lambda(t) = -\frac{(p-\lambda(t))^2}{4}$ $u(t) = x(t) \frac{p- \lambda(t)}{2}$ and using the initial and turn-T conditions, the functions can be solved numerically. ## References 1. L. S. Pontryagin, 1962. The Mathematical Theory of Optimal Processes. 2. I. M. Ross, 2009. A Primer on Pontryagin's Principle in Optimal Control, Collegiate Publishers. ISBN 978-0-9843571-0-9. 3. Kalman, Rudolf. A new approach to linear filtering and prediction problems. Transactions of the ASME, Journal of Basic Engineering, 82:34–45, 1960 4. Oberle, H. J. and Grimm, W., "BNDSCO-A Program for the Numerical Solution of Optimal Control Problems," Institute for Flight Systems Dynamics, DLR, Oberpfaffenhofen, 1989 5. Betts, J. T., Practical Methods for Optimal Control Using Nonlinear Programming, SIAM Press, Philadelphia, Pennsylvania, 2001 6. Gill, P. E., Murray, W. M., and Saunders, M. A., User's Manual for SNOPT Version 7: Software for Large-Scale Nonlinear Programming, University of California, San Diego Report, 24 April 2007 7. von Stryk, O., User's Guide for DIRCOL (version 2.1): A Direct Collocation Method for the Numerical Solution of Optimal Control Problems, Fachgebiet Simulation und Systemoptimierung (SIM), Technische Universität Darmstadt (2000, Version of November 1999). 8. Betts, J.T. and Huffman, W. P., Sparse Optimal Control Software, SOCS, Boeing Information and Support Services, Seattle, Washington, July 1997 9. Hargraves, C. R. and Paris, S. W., "Direct Trajectory Optimization Using Nonlinear Programming and Collocation", Journal of Guidance, Control, and Dynamics, Vol. 10, No. 4., 1987, pp. 338–342 10. Gath, P.F., Well, K.H., "Trajectory Optimization Using a Combination of Direct Multiple Shooting and Collocation", AIAA 2001–4047, AIAA Guidance, Navigation, and Control Conference, Montréal, Québec, Canada, 6–9 August 2001 11. Vasile M., Bernelli-Zazzera F., Fornasari N., Masarati P., "Design of Interplanetary and Lunar Missions Combining Low-Thrust and Gravity Assists", Final Report of the ESA/ESOC Study Contract No. 14126/00/D/CS, September 2002 12. Schwartz, Adam, Theory and Implementation of Methods based on Runge–Kutta Integration for Solving Optimal Control Problems, University of California at Berkeley, PhD Dissertation, 1996. 13. Ross, I. M. and Fahroo, F., User's Manual for DIDO: A MATLAB Package for Dynamic Optimization, Dept. of Aeronautics and Astronautics, Naval Postgraduate School Technical Report, 2002 14. Williams, P., User's Guide to DIRECT, Version 2.00, Melbourne, Australia, 2008 15. Rao, A. V., Benson, D. A., Huntington, G. T., Francolin, C., Darby, C. L., and Patterson, M. A., , University of Florida Report, August 2008. 16. Rutquist, P. and Edvall, M. M, PROPT – MATLAB Optimal Control Software," 1260 S.E. Bishop Blvd Ste E, Pullman, WA 99163, USA: Tomlab Optimization, Inc. 17. E. Polak, On the use of consistent approximations in the solution of semi-infinite optimization and optimal control problems Math. Prog. 62 pp. 385–415 (1993). ## Further reading Books • Athans, M. A. and Falb, P. L., Optimal Control, McGraw–Hill, New York, 1966. • Becerra, V.M., 2008, Optimal control. Scholarpedia, 3(1):5354 • Bryson, A. E., 1969. Applied Optimal Control: Optimization, Estimation, & Control. • Bryson, A.E and Ho, Y., "Applied Optimal Control: Optimization, Estimation and Control (Revised Printing)", John Wiley and Sons, New York, 1975. • Evans, L.C., An Introduction to Optimal Control Theory (available free online) • Ross, I. M. A Primer on Pontryagin's Principle in Optimal Control, Collegiate Publishers, 2009. ISBN 978-0-9843571-0-9. (http://www.ElissarGlobal.com free chapter available online) • H. O. Fattorini and S. S. Sritharan, "Existence of Optimal Controls for Viscous Flow Problems," Proceedings of the Royal Society of London Series A, Vol. 439, 1992, pp. 81–102. • H. O. Fattorini and S. S. Sritharan, "Necessary and Sufficient Conditions for Optimal Controls in Viscous Flow," Proceedings of the Royal Society of Edinburgh, Series A, Vol. 124A, 1994, pp. 211–251. • H. O. Fattorini and S. S. Sritharan, "Optimal chattering controls for viscous flow," Nonlinear analysis, Theory, Methods and Applications, Vol. 25, No. 8, pp. 763–797, 1995. • H. O. Fattorini and S. S. Sritharan,"Optimal control problems with state constraints in fluid mechanics and combustion," Applied Mathematics and Optimization, Vol. 38(2), 1998, pp. 159–192. • Kirk, D. E., 2004. Optimal Control Theory: An Introduction. • Lebedev, L. P., and Cloud, M. J., 2003. The Calculus of Variations and Functional Analysis with Optimal Control and Applications in Mechanics. World Scientific. Especially chpt. 2. • Lewis, F. L., and Syrmos, V. L., 19nn. Optimal Control, 2nd ed. John Wiley & Sons. • S. S. Sritharan, "Optimal Control of Viscous Flow", SIAM, 1998. ( http://www.nps.edu/academics/schools/gseas/sri/Sritharan-Optimal_Control_of_Viscous_Flow.pdf ) • S. S. Sritharan, "An Optimal Control Problem in Exterior Hydrodynamics," Proceedings of the Royal Society of Edinburgh, Series 121A, 1992, pp. 5–32. • S. S. Sritharan,"Deterministic and stochastic control of viscous flow with linear, monotone and hyper viscosities," Applied Mathematics and Optimization, Vol. 41(2), pp. 255–308, 2000. • Stengel, R. F., 1994. Optimal Control and Estimation. Dover. • Sethi, S. P., and Thompson, G. L., 2000. Optimal Control Theory: Applications to Management Science and Economics, 2nd edition, Springer (ISBN 0387280928 and ISBN 0-7923-8608-6). Slides are available at http://www.utdallas.edu/~sethi/OPRE7320presentation.html • Sontag, Eduardo D. Mathematical Control Theory: Deterministic Finite Dimensional Systems. Second Edition. Springer. (ISBN 0-387-984895) (available free online) • Brogan, William L. 1990. Modern Control Theory. ISBN 0-13-589763-7 • Bryson, A.E.; Ho, Y.C. (1975). Applied optimal control. Washington, DC: Hemisphere. Journals • Optimal Control Applications and Methods. John Wiley & Sons, Inc. • SIAM Journal of Control and Optimization.

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http://math.stackexchange.com/questions/241735/the-roots-of-the-cubic-x3qxr-0-are-a-b-c-how-can-i-find-the-equation-who

# The roots of the cubic $x^3+qx+r=0$ are $a,b,c$. How can I find the equation whose roots are $la+mbc,lb+mca,lc+mab$ The roots of the cubic $x^3+qx+r=0$ are $a,b,c$. How can I find the equation whose roots are $la+mbc,lb+mca,lc+mab$? Can anyone help me to solve this problem? - ## 2 Answers The long, brute-force way: Expand out $\left(x-(la+mbc)\right)\left(x-(lb+mca)\right)\left(x-(lc+mab)\right)$ to get its full form as a cubic; you should find that all of the coefficients are symmetric functions of $(a,b,c)$. Then use the Fundamental Theorem Of Symmetric Polynomials to express those coefficients in terms of the basic symmetric polynomials $S_1(a,b,c) = a+b+c$, $S_2(a,b,c) = ab+bc+ca$, and $S_3(a,b,c)=abc$. Finally, use the traditional theorems on expressing the coefficients of a polynomial in terms of its roots (e.g., $r=-abc$) to express the coefficients in terms of $q$ and $r$. - $la+mbc,lb+mca,lc+mab$ as our roots means that From the given equation we have $a+b+c = 0$ since coefficient of $x^2$ is 0. and $ab+ba+bc =q$ $x^3 + px^2+ qx + r$ (independent of the equation in your question) $p = l(a+b+c) + m(bc+ca+ab)$ $p = l(0) + m(q) = mq$ where q is the coefficient of x in your expression Similarly (I haven't really expressed it in terms of the coefficients of your given equation), $q = (la+mbc)(lb+mca)+(la+mbc)(lc+mba)+lb(mca)(lc+mba)$ $r = (la+mbc)(lb+mca)(lc+mba)$ The cubic polynomial with these roots is of the form $x^3 - (l(a+b+c) + m(bc+ca+ab)))x^2 + ((la+mbc)(lb+mca)+(la+mbc)(lc+mba)+lb(mca)(lc+mba)) x - (la+mbc)(lb+mca)(lc+mba)$ -

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http://mathhelpforum.com/advanced-math-topics/33063-solar-funnel-trig-problem.html

# Thread: 1. ## solar funnel trig problem I posted this in a high school forum yesterday, but didn't get any response, so sorry if you see this twice. I'm trying to work out a formula that solves theta given x and y. I'll attach a diagram. What I'm trying to work out is the smallest theta (widest funnel) where light hitting the top of the funnel just enters the oven, and doesn't hit the funnel on the other side. any help? Attached Thumbnails 2. Originally Posted by nervousnelly I posted this in a high school forum yesterday, but didn't get any response, so sorry if you see this twice. I'm trying to work out a formula that solves theta given x and y. I'll attach a diagram. What I'm trying to work out is the smallest theta (widest funnel) where light hitting the top of the funnel just enters the oven, and doesn't hit the funnel on the other side. any help? You need to solve: $\tan(180-2\theta)=\frac{x+y \cos(\theta)}{y \sin(\theta)}$ use trig identities to reduce to something containing $\sin$s and $\cos$s of $\theta$ only. then see what you can do RonL 3. Thanks CaptainBlack, I appreciated the time you took to getting me closer to the solution. The trouble is high school trig was 16 years ago, and even if I learnt trig identities, i forgot them 2 hours after the last exam. This is for a personal project I'm working on, a solar funnel. I'm not in any classes, it's a one off problem I googled trig identities, and it may as well have been written in greek. (sight pun there) 4. Originally Posted by nervousnelly Thanks CaptainBlack, I appreciated the time you took to getting me closer to the solution. The trouble is high school trig was 16 years ago, and even if I learnt trig identities, i forgot them 2 hours after the last exam. This is for a personal project I'm working on, a solar funnel. I'm not in any classes, it's a one off problem I googled trig identities, and it may as well have been written in greek. (sight pun there) Don't bother, now you will learn the dreadfull secret of maths, when we have a practical problem we don't usually bother with all that. Just post the values of x and y and we will solve this numerically for you (without tidying up the trig) If you need to find theta for many x's and y's what we can do is observe that theta depends on z=x/y, so we compute a table of theta against z. RonL 5. thanks, what I'm hoping for is a little web page with javascript or php function to calculate theta. so x might be 40 and y might be 50, but could be anything. i think if x = y, theta is 90 - 45 / 2 6. Originally Posted by nervousnelly thanks, what I'm hoping for is a little web page with javascript or php function to calculate theta. so x might be 40 and y might be 50, but could be anything. i think if x = y, theta is 90 - 45 / 2 I think the formula you are looking for is $\boxed{\cos\theta = \frac{\sqrt{y^2+8x^2}-y}{4x}}$. When x = y, this gives θ = 60°. This is certainly correct, because if x=y then the triangle formed by the sides labelled x and y, together with the long dashed red line of incident light, is isosceles (another long-forgotten word from high school?), and it's then easy to verify geometrically that its angles must be 30°, 30° and 120°. 7. that's awesome opalg, thank you very much. i made 50 = x = y and got cos t = 1/2 not sure what to do from there? (you're right, no idea what an isosceles is ) edit: i worked it out (cos^-1), thank you very much captainblack and opalg. I appreciate it. 8. I think something is a miss. I seem to be getting numbers between 45 and 90, rather than 0 and 90. If I make y=2 and x=200, I should get a figure close to 0, but I get 45.something. any ideas? 9. Originally Posted by nervousnelly I think something is amiss. I seem to be getting numbers between 45 and 90, rather than 0 and 90. If I make y=2 and x=200, I should get a figure close to 0, but I get 45.something. any ideas? The angle θ can never be less than 45°. If you look at your diagram, you'll see that if θ=45° then the vertical ray of incident light gets reflected horizontally across to the opposite rim of the funnel. If θ is less than 45° then the ray will be reflected back upwards and certainly cannot enter the funnel. Attached Thumbnails 10. you're absolutely right, i see it now. thank you very much, i appreciate it. 11. I might as well show this done numerically: Code: ```>function bi(z1) $ global z $ ll=length(z1);rv=[]; $ ss="tan(pi-2*x)-(z+cos(x))/sin(x)"; $ for idx=1 to ll $ z=z1(idx); $ rv=rv_[z,bisect(ss,pi/4,pi/2)]; $ end $ return rv $endfunction > >bi([0.1:0.1:3]) 0.1 1.47256 0.2 1.38356 0.3 1.30822 0.4 1.24641 0.5 1.19606 0.6 1.15483 0.7 1.12072 0.8 1.09215 0.9 1.06794 1 1.0472 1.1 1.02925 1.2 1.01358 1.3 0.999785 1.4 0.987552 1.5 0.976632 1.6 0.966827 1.7 0.957974 1.8 0.949944 1.9 0.942625 2 0.935929 2.1 0.92978 2.2 0.924112 2.3 0.918873 2.4 0.914014 2.5 0.909497 2.6 0.905286 2.7 0.901352 2.8 0.897667 2.9 0.894209 3 0.890958 >``` RonL

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http://nrich.maths.org/792/note

### Prompt Cards These two group activities use mathematical reasoning - one is numerical, one geometric. ### Consecutive Numbers An investigation involving adding and subtracting sets of consecutive numbers. Lots to find out, lots to explore. ### Exploring Wild & Wonderful Number Patterns EWWNP means Exploring Wild and Wonderful Number Patterns Created by Yourself! Investigate what happens if we create number patterns using some simple rules. # Dodecamagic ### Why do this problem? Visualising is a very important mathematical skill. Representing 3d shapes in a 2d form is a sophisticated form of visualising. This problem offers opportunities to visualise and to use deductive reasoning with small numbers. ### Possible approach If your pupils are used to visualising, you may wish to go straight into the problem and see how far they can get. Display the pictures and ensure that everyone understands the problem, then after a little time draw the pupils together and ask for any useful statements or observations they can make. At this stage you may want to suggest that they could choose to use a 3d model if that would help, but you might also want to challenge them to do it 'in their heads'. If your pupils are not used to visualising, you may want to begin by organising them into pairs to make a 3-d model from a net, or from other plastic shapes you have in the classroom, such as Polydron. The main challenge is to match the information from one diagram onto the other. Once the children realise how the two diagrams are connected, the arithmetic is relatively trivial. Labelling the vertices and then opening up the net of the solid can help make the connections too. ### Key questions There are $9$s on both diagrams. Does that help? How? What about the $25$? What number does the F represent? How does the name of the shape help you to know? ### Possible extension You may wish to offer some other model-making activities which serve to consolidate 2d representation of 3d objects. There is a paper folding example of a dodecahedron here, which dextrous children may be able to do, perhaps with a little support. You might want to try making one yourself first! ### Possible support Support children who find visualising difficult with the latest in digital technology - use a digital camera to take photographs of the front and back of different 3d shapes including polyhedra and ask them to match them up, identifying common edges or vertices. The NRICH Project aims to enrich the mathematical experiences of all learners. To support this aim, members of the NRICH team work in a wide range of capacities, including providing professional development for teachers wishing to embed rich mathematical tasks into everyday classroom practice. More information on many of our other activities can be found here.

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http://mathhelpforum.com/calculus/42264-limits-continuity.html

# Thread: 1. ## Limits and continuity... Hello... Q1) Find the limit of the following using L'Hospital rule.... a) lim x-> 0+ [sin(x) - x] / [e^x - 1] b) lim x-> pi/2 [pi/2 - x] . tan(x) Q2) Prove using the Mean value theorem: (x-1)/x < ln(x) < x-1, when x>1 Thanks! 2. Originally Posted by Vedicmaths Hello... Q1) Find the limit of the following using L'Hospital rule.... a) lim x-> 0+ [sin(x) - x] / [e^x - 1] b) lim x-> pi/2 [pi/2 - x] . tan(x) Q2) Prove using the Mean value theorem: (x-1)/x < ln(x) < x-1, when x>1 Thanks! Must you use L'hopital's? Ok for the first one seeing that if we let $sin(x)-x=f(x)$ and $g(x)=e^x-1$ and $g(x)=f(x)=0$ Then $\lim_{x\to{0}}\frac{f(x)}{g(x)}=\lim_{x\to{0}}\fra c{f'(x)}{g'(x)}$ 3. Edit: duplicated mathstuds fine work, sorry L'Hopital's rule states that $\lim_{x\to x_0}\frac{f(x)}{g(x)} = \lim_{x\to x_0}\frac{f'(x)}{g'(x)}$ if $f(x_0)=0$ and $g(x_0) = 0$ or $f(x_0) = \pm \infty$ and $g(x_0) = \pm \infty$ Go for it. Question 1 is easy. For question 2 start by taking the derivative of all parts of the inequality. 4. Originally Posted by Vedicmaths Hello... Q1) Find the limit of the following using L'Hospital rule.... a) lim x-> 0+ [sin(x) - x] / [e^x - 1] b) lim x-> pi/2 [pi/2 - x] . tan(x) Q2) Prove using the Mean value theorem: (x-1)/x < ln(x) < x-1, when x>1 Thanks! The first one is of the form $\frac{0}{0}$ so now using L'hospitals rule we take a derivative of the numerator and the denominator to get $\lim_{x \to 0^+}\frac{\cos(x)-1}{e^x}=\frac{\cos(0)-1}{e^{0}}=\frac{1-1}{1}=0$ For the 2nd rewrite it as $\frac{[x-\pi/2]\sin(x)}{\cos(x)}$ now as $x\to \frac{\pi}{2}$ this is of the form $\frac{0}{0}$ Using L.H again we get $\lim_{x \to \frac{\pi}{2}}\frac{[x-\pi/2]\cos(x)+\sin(x)}{-\sin(x)}=\frac{[0]0+1}{-1}=-1$ For the 2nd question consider the function $f(x)=\ln(x)$ f is continous on $[1,\infty)$ and differentiable on $(1,\infty)$ so it satisfies the hypothesis of the MVT by the MVT $f(b)-f(a)=f'(c)(b-a)$ where $c \in (a,b)$ Let a=1 and b=x and we get $\ln(x)-\ln(1)=\frac{1}{c}\left( x-1 \right)$ simplifying we get $\ln(x) =\frac{x-1}{c} \iff c\ln(x)=x-1$ Since $c \in (1,x) \mbox{ or } 1< c < x$ $\ln(x) < x-1$ Also by the same reasoning $c < x \iff \frac{1}{x} < \frac{1}{c}$ so $\frac{x-1}{x}< \frac{x-1}{c}=\ln(x)$ so finally we get $\frac{x-1}{x} < \ln(x) < x-1$ Yeah!!! 5. Is it necessary L'Hôpital? Q2 can also be killed as follows: Since $\ln x=\int_{1}^{x}{\frac{dy}{y}},$ we have $\frac{1}{x}\le \frac{1}{y}\le 1\implies \frac{x-1}{x}\le \ln x\le x-1.\quad\blacksquare$ (This holds for $x>0.$) 6. Thanks a lot! That really helps alot! 7. Dear Empty Set! In the second problem you suggested to rewrite as ....but how can we say that...If we use (pi/2) form we will end up getting 1. Ans you are getting -1. Will that make any difference? Thanks! 8. Originally Posted by Vedicmaths Dear Empty Set! In the second problem you suggested to rewrite as ....but how can we say that...If we use (pi/2) form we will end up getting 1. Ans you are getting -1. Will that make any difference? Thanks! $\lim_{x \to \frac{\pi}{2}}\frac{[x-\frac{\pi}{2}]\sin(x)}{\cos(x)}=\frac{[\frac{\pi}{2}-\frac{\pi}{2}]\sin(\frac{\pi}{2})}{\cos(\frac{\pi}{2})}=\frac{[0]\cdot 1}{0}=\frac{0}{0}$ So now we can use L'hospitals rule. Then we can take the derivatives to get what I did above. I hope this helps. Good luck. 9. ## confusion! Yes Sir! I got what you were trying to do after using the L'hospital Rule but I was asking that why are you manipulating the term? Since, in the problem it was given as ( pi/2 - x) sinx / cosx...but you have mentioned that we can rewrite the function as So why are you changing the sign here? And I guess that will change the answer. I got 1 from the function given in the original question and you are getting -1 from this one. I am sorry to bother you. But I am just a little bit confused and want to learn if something I do not know. Thanks so much once again! Regards, Originally Posted by Vedicmaths b) lim x-> pi/2 [pi/2 - x] . tan(x) If we take the limit as $x \to \frac{\pi}{2}$ we get $0\cdot \pm \infty$ This doesnt quite fit the form to use L'hospitals rule. We need to manipulate the equation into the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$ Hence I changed $[\frac{\pi}{2}-x]\tan(x)=\frac{[\frac{\pi}{2}-x]\sin(x)}{\cos(x)}$ Now this is of the form $\frac{0}{0}$ So from here we take the derivative of the numerator and the denomiator as per L.H's rule. I guess I don't understand what you mean changing the sign? 11. Originally Posted by TheEmptySet If we take the limit as $x \to \frac{\pi}{2}$ we get $0\cdot \pm \infty$ This doesnt quite fit the form to use L'hospitals rule. We need to manipulate the equation into the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$ Hence I changed $[\frac{\pi}{2}-x]\tan(x)=\frac{[\frac{\pi}{2}-x]\sin(x)}{\cos(x)}$ Now this is of the form $\frac{0}{0}$ So from here we take the derivative of the numerator and the denomiator as per L.H's rule. I guess I don't understand what you mean changing the sign? You did Originally Posted by TheEmptySet For the 2nd rewrite it as $\frac{[x-\pi/2]\sin(x)}{\cos(x)}$ Which is the negative of the original problem. -Dan 12. Sorry, I didn't even notice that I switched them. Thanks

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http://www.physicsforums.com/showthread.php?t=100606

Physics Forums ## Complex integration I have the function $$f(z)=\frac{e^z}{1+e^{4z}}$$. and the loop $$\gamma_{r_1,r_2}=I_{r_1,r_2}+II_{r_2}+III_{r_1,r_2}+IV_{r_1},\quad r_1,r_2>0,$$ which bounds the domain $$A_{r_1,r_2}=\{z\in\mathbb{C}\mid -r_1<\Re(z)< r_2\wedge0<\Im(z)<\pi\}.$$ Now I have to show that $$\int_{\gamma_{r_1,r_2}}f(z)\,dz=\tfrac{\sqrt2}{2}\pi.$$ I know that the integral along $\gamma_{r_1,r_2}$ is the sum of the integrals along the four countours $I_{r1,r_2},\dots,IV_{r_1}$. I also know that if $f$ is continuous in a domain D and has an antiderivative $F$ throughout D, then the integral along a contour lying in D is $F(z_T)-F(z_I)$, where $z_T$ is the terminal point, and $z_I$ the initial point, of the countour. However, using this, I am not able to show the aforementioned result, i.e, I get $\int_{\gamma_{r_1,r_2}}f(z)\,dz=0$. Could anyone try to calculate the integral, and then report back what they've found. PhysOrg.com science news on PhysOrg.com >> Front-row seats to climate change>> Attacking MRSA with metals from antibacterial clays>> New formula invented for microscope viewing, substitutes for federally controlled drug Recognitions: Homework Help Science Advisor Quote by sigmund I have the function $$f(z)=\frac{e^z}{1+e^{4z}}$$. and the loop $$\gamma_{r_1,r_2}=I_{r_1,r_2}+II_{r_2}+III_{r_1,r_2}+IV_{r_1},\quad r_1,r_2>0,$$ which bounds the domain $$A_{r_1,r_2}=\{z\in\mathbb{C}\mid -r_1<\Re(z)< r_2\wedge0<\Im(z)<\pi\}.$$ Now I have to show that $$\int_{\gamma_{r_1,r_2}}f(z)\,dz=\tfrac{\sqrt2}{2}\pi.$$ I know that the integral along $\gamma_{r_1,r_2}$ is the sum of the integrals along the four countours $I_{r1,r_2},\dots,IV_{r_1}$. I also know that if $f$ is continuous in a domain D and has an antiderivative $F$ throughout D, then the integral along a contour lying in D is $F(z_T)-F(z_I)$, where $z_T$ is the terminal point, and $z_I$ the initial point, of the countour. However, using this, I am not able to show the aforementioned result, i.e, I get $\int_{\gamma_{r_1,r_2}}f(z)\,dz=0$. Could anyone try to calculate the integral, and then report back what they've found. Ok, your contour as stated is tough for me to follow but looks like you're going around the origin in such a was as to encompass poles of the integrand. Right? When is the denominator zero in the complex plane? Thus can't rely on Cauchy's Theorem but rather the Residue Theorem. Edit: Alright, suppose Cauchy's Theorem is a special case of the Residue Theorem but you know what I mean. Quote by saltydog Ok, your contour as stated is tough for me to follow, but looks like you're going around the origin in such a was as to encompass poles of the integrand. Right? When is the denominator zero in the complex plane? Thus can't rely on Cauchy's Theorem but rather the Residue Theorem. Edit: Alright, suppose Cauchy's Theorem is a special case of the Residue Theorem but you know what I mean. The function f has singularities in $z=\frac{i}{4}\left(\pi+k2\pi\right),~k=0,1,2,\dots$, and the contour is a rectangle with vertices at $z=-r_1$, $z=r_2$, $z=r_2+i\pi$, and $z=-r_1+i\pi$, where $r_1,r_2>0$. It is also oriented counter-clockwise. Moreover, in the first question of the exercise, I am asked to calculate the residues at the singularities. Comparing Cauchy's Integral Theorem and Cauchy's Residue Theorem, I clearly see that the Residue Theorem can be used here, because the Integral Theorem requires the domain, which contains the contour, to be simply connected, while the Residue Theorem allows singularities inside the contour. Thus the Residue Theorem would be the right one to use here. I have not tried to apply this yet, but thanks for pointing me towards this. EDIT: I have just used Maple to quickly calculate the integral, using the Residue Theorem, and I got the right result. Recognitions: Homework Help Science Advisor ## Complex integration Quote by sigmund EDIT: I have just used Maple to quickly calculate the integral, using the Residue Theorem, and I got the right result. Hey Sigmund, you do know how to calculate the residues by hand right? That is, how to show: $$\mathop\lim\limits_{z\to \pi i/4} (z-\pi i/4) \frac{e^z}{1+e^{4z}}=-\frac{1/ \sqrt{2}+i/ \sqrt{2}}{4}$$ Quote by saltydog Hey Sigmund, you do know how to calculate the residues by hand right? That is, how to show: $$\mathop\lim\limits_{z\to \pi i/4} (z-\pi i/4) \frac{e^z}{1+e^{4z}}=-\frac{1/ \sqrt{2}+i/ \sqrt{2}}{4}$$ Just factor the denominator by determining the poles : $$z = \frac{i( \pi + 2k \pi)}{4}$$ and k = 0,1,2,3 So you get 4 factors, one for each k. Then you will see that the $$(z-\pi i/4)$$ term will also appear in the denominator. Be sure that this only goes for first order poles marlon Quote by marlon Just factor the denominator by determining the poles : $$z = \frac{i( \pi + 2k \pi)}{4}$$ and k = 0,1,2,3 So you get 4 factors, one for each k. Then you will see that the $$(z-\pi i/4)$$ term will also appear in the denominator. Be sure that this only goes for first order poles marlon I am not sure how to do this. The denominator of f has a zero of order 1 at $z=i\pi/4$, whence it can be written as $(z-i\pi/4)g(z)$, where g(z) is analytic at $i\pi/4$ and $g(i\pi/4)\neq0$. I do not know how to do this. Couldn't you give me a hint to how to solve this? Recognitions: Homework Help Science Advisor Quote by sigmund I am not sure how to do this. The denominator of f has a zero of order 1 at $z=i\pi/4$, whence it can be written as $(z-i\pi/4)g(z)$, where g(z) is analytic at $i\pi/4$ and $g(i\pi/4)\neq0$. I do not know how to do this. Couldn't you give me a hint to how to solve this? Yea, I'd like to know too. I was referring to the relation: $$\text{Res}_{z=z_0} \left(\frac{p(z)}{q(z)}\right)=\frac{p(z_0)}{q^{'}(z_0)}$$ Note: Only for a simple pole. Quote by saltydog Yea, I'd like to know too. I was referring to the relation: $$\text{Res}_{z=z_0} \left(\frac{p(z)}{q(z)}\right)=\frac{p(z_0)}{q^{'}(z_0)}$$ Note: Only for a simple pole. The function f has a simple pole at $z=z_0$ (z_0 has been stated several times in this thread, so I omit it here). Hence the residue at that point is $\text{Res}(f;z_0)=\lim_{z\to z_0}(z-z_0)f(z)$. If we let $f(z)=p(z)/q(z)$, where p and q are both analytic at z_0, and q has a simple zero at z_0, while $p(z_0)\neq0$, we can do the following simplification: $$\text{Res}(f;z_0)=\lim_{z\to z_0}(z-z_0)f(z)=\lim_{z\to z_0}(z-z_0)\frac{p(z)}{q(z)}=\lim_{z\to z_0}\frac{p(z)}{\frac{q(z)-q(z_0)}{z-z_0}}=\frac{\lim_{z\to z_0}p(z)}{\lim_{z\to z_0}\frac{q(z)-q(z_0)}{z-z_0}}=\frac{p(z_0)}{q'(z_0)}$$ This can be used in the actual problem, because the requirements about analyticity at z_0, and that p(z) is nonzero at z_0, are met. Using this, it is easy to calculate the residue at z_0, while factoring the denominator undoubtedly could be done, although difficult, i presume. As Albert Einstein said: "Everything should be made as simple as possible, but no simpler." Thread Tools | | | | |------------------------------------------|-------------------------------|---------| | Similar Threads for: Complex integration | | | | Thread | Forum | Replies | | | Introductory Physics Homework | 6 | | | Calculus | 1 | | | Calculus & Beyond Homework | 1 | | | Calculus & Beyond Homework | 19 | | | Calculus | 6 |

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http://math.stackexchange.com/questions/100518/closed-form-for-sum-over-orbit-of-roots-of-unity

# Closed form for sum over orbit of roots of unity Let $\sigma$ act on the $m$th roots of unity by $\sigma \zeta = \zeta^r$ for $1\le r <m, (r,m)=1.$ I'm looking to evaluate $\displaystyle\sum_{i=0}^{t(\zeta )-1} \zeta^{r^i},$ where $t(\zeta )$ is the size of the orbit of $\zeta$ under $\sigma .$ I keep running into this pesky little sum and was wondering if anyone on this forum might be able to point me in the direction of converting it to something simple (or at least more amenable for further computation.) It wouldn't surprise me if I'm overlooking something elementary, but it's already frustrated me to the point of posting here, so any guidance whatsoever would be greatly appreciated. - Where does $\sigma$ come from? Isn't $r$ fixed, but arbitrary? – user21436 Jan 19 '12 at 18:17 $\sigma$ is just a name I gave to the map $\zeta \mapsto \zeta^r.$ $r$ is fixed but arbitrary. – Tim Duff Jan 19 '12 at 18:22 ## 1 Answer You're looking at a sum consisting of values of an additive character on ${\mathbb Z}_m$, where the inputs run over a multiplicative subgroup of ${\mathbb Z}_m^\times$. You can use a linear combination of Dirichlet characters to detect that multiplicative subgroup, and this converts your problem to a linear combination of Gauss sums. The relevant Dirichlet characters will be those $\chi$ such that $\chi(r)=1$. - 2 PS: Welcome back Wikipedia :) – Greg Martin Jan 19 '12 at 18:25 Looks like I'll be busy, thank you! – Tim Duff Jan 19 '12 at 18:36

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http://physics.stackexchange.com/questions/9551/applications-of-the-spectral-theorem-to-quantum-mechanics?answertab=active

# Applications of the Spectral Theorem to Quantum Mechanics I'm currently learning some basic functional analysis. Yesterday I arrived at the spectral theorem of self-adjoint operators. I've heard that this theorem has lots of applications in Quantum Mechanics. But let me first state the formulation of the theorem that I'm using: Let $H$ be a Hilbert space. There's a 1-to-1-correspondence between self-adjoint operators $A$ on $H$ and spectral measures $P^{A}$ given by $$A~=~\int_{\mathbb{R}} \lambda ~dP^{A}.$$ ($\lambda$ denotes a constant, $\mathbb{R}$ denotes the real numbers.) A corollary is: Let $g:\mathbb{R}\to\mathbb{R}$ be a function. (Again: $\mathbb{R}$ denotes the set of real numbers.) Then: $$g(A)~:=~\int_{\mathbb{R}} g(\lambda)~ dP^{g(A)}$$ $$P^{g(A)}(\Delta) ~=~ P^{A}(g^{-1}(\Delta))$$ where $\Delta$ denotes a set in the $\sigma$-algebra of $\mathbb{R}$. Okay. Now this is the theorem. First I don't really the application of the corollary in Quantum mechanics. I've heard that suppose you're given an operator $A$ this means that it's easy for you to define operators like $\exp(A)$, especially on infinite dimensional Hilbert spaces. This indeed could be useful in quantum mechanics. Especially when thinking about the "time-evolution operator" of a system. However then I say: Why do you make things so complicated? Suppose you want to calculate $\exp(A)$. Why don't you define $$\exp(A)~:=~1+A+1/2 A^2 + \ldots$$ and require convergence with respect to the operator norm. An example: Consider the vectorspace spanned by the monomials $1,x,x^2,\ldots$ and let $A=d/dx$. Then you can perfectly define $$\exp(d/dx)~:=~1+ d/dx + 1/2 d^2/dx^2 + \ldots$$ and require convergence with respect to the operator norm. In addition to that I've heard that the spectral theorem gives a full description of all self-adjoint operators. Now why is that the case? I mean okay..there's a one to one correspondence between self-adjoint operators and spectral measures...but why does this give me any information about "the inner structure of the operator"? (And why is there this $\lambda$ in the integral? Looks somehow like an eigenvalue of $A$? But I'm just guessing) I'd me more than happy, if you could provide me with some intuition and ideas of how the theorem can be used. - 1 Dear @Matt_Quantum. Two comments: 1) Would you be willing to restrict to separable Hilbert spaces only? 2) The two examples $A=\frac{d}{dx}$ and $\exp(\frac{d}{dx})$ are not self-adjoint operators, if $x$ is supposed to be a real variable. – Qmechanic♦ May 7 '11 at 8:52 1) Okay 2) Okay. Sorry for that. But assume A is some arbitrary self-adjoint operator, one could still define exp(A)=1+A+1/2 A^2 +... and require convergence with respect to the operator norm. – Matt_Quantum May 7 '11 at 12:36 Dear @Matt_Quantum. But the operator norm $||A||$ of an arbitrary self-adjoint operator $A$ could be $\infty$. In the question formulation (v1) you never restricted yourself to only bounded operators, which is actually good, because in quantum mechanics there are lots of unbounded operators. For instance, $|| \frac{d}{dx} ||=\infty$. – Qmechanic♦ May 7 '11 at 14:56 I may try to work this up into an answer, but for now, I'll just comment: in a way, the spectral measure determines the operator up to a change of variables. Every operator whose spectrum is the real line can be transformed to $id\over dx$ acting on some subspace of the functions on the real line, after a change of variable, for example. It's almost true that the spectrum determines the operator (up to a change of basis) but you do actually need a little more info: multiplicites of eigenvavlues, whether they are part of a continuous set of eigenvalues, etc., spectral measure type thingies. – joseph f. johnson Jan 7 '12 at 22:09 ## 2 Answers Meta: It is true that a lot of quantum mechanics can be taught and understood without much knowledge of the mathematical foundations, and usually it is. Since QM is a mandatory class at many faculties that future experimental physicists have to attend, too, this also makes sense. But for future theoretical and mathematical physicists, it may pay off to learn a little bit about the math, too. A little anecdote: John von Neumann once said to Werner Heisenberg that mathematicians should be grateful for QM, because it led to the invention of a lot of beautiful mathematics, but that mathematicians repaid this by clarifying, e.g., the difference between a selfadjoint and a symmetric operator. Heisenberg asked: "What is the difference?" Suppose you want to calculte exp(A). Why don't you define exp(A):=1+A+1/2 A^2 + ... and require convergence with respect to the operator norm. That's correct. The benefit of the spectral theorem is that you can define f(A) for any selfadjoint (or more generally, normal) operator for any bounded Borel function. This comes in handy in many proofs in operator theory. In addition to that I've heard that the spectral theorem gives a full description of all self-adjoint operators. Now why is that the case? I mean okay..there's a one to one correspondence between self-adjoint operators and spectral measures.. That's correct, too. Spectral measures are much much simpler objects than selfadoint operators, that's why. Futhermore, you can use the spectral theorem to prove that every selfadjoint operator is unitarily equivalent to a multiplication operator (multiply f(x) by x). From an abstract viewpoint, this is a very satisfactory characterization. It does not help much for concrete calcuations in QM, though. BTW: On a more advanced level, you'll need to understand the spectral theorem to understand what a mass gap is in Yang-Mills Theory (millenium problem). Hint: In QFT in Minkowski-Spacetime, one usually assumes that there is a continuous representation of the Poincaré group, especially of the commutative subgroup of translations, on the Hilbert space that contains all physical states. The operators that form the representation have a common spectral measure, this is an application of the SNAG-theorem. The support of this spectral measure is bounded away from zero, that's the definition of the mass gap. - Dear Constantin, the $A=\int_R \lambda\,dP^A$ is just a continuous version of spectral decomposition. Here $dP^A$ is a differential version of the projection operators that define the Hermitian operator. For a discrete spectrum, the corresponding equation would be $$A = \sum_{i} \lambda_i P_{\lambda_{i}}$$ where the sum goes over the eigenvalues $\lambda_i$ and $P_{\lambda_i}$ are the projection operators on the subspace of the Hilbert space that contains the eigenvectors with the eigenvalue $\lambda_i$. Indeed, $\lambda$ is always meant to be a possible eigenvalue of the operator. And indeed, a Hermitian operator is fully determined by its spectrum and the corresponding eigenvectors (and multiplicities) for each eigenvalue, which is why the formula above is an equivalent way to rewrite a Hermitian operator. When the spectrum of $A$ is continuous, the summation over $i$ has to be replaced by an integral, and the corresponding differential $d$ is added in front of $dP_{\lambda}$: it's really the differential of the projection operator on the space of eigenstates with eigenvalues in the interval $[-\infty, \lambda]$; $dP_\lambda = dP_\lambda / d\lambda \cdot d\lambda$, if you wish. But it's really morally the same thing as in the case of the discrete spectrum (which produces delta functions in $dP_\lambda / d\lambda$ if we adopt this terminology). Also, some of your additional proofs involving $\Delta$ are just trivial substitutions under the integral sign. One would need to know lots of details of your mathematical axioms - lots of the particular "math culture" you're coming from - to figure out what could exactly be difficult for a mathematician about the substitutions under the integral sign. There are no difficulties from a physicist's perspective - it's high school maths. http://en.wikipedia.org/wiki/Spectral_theorem#Hermitian_matrices Mathematicians may worry about boundedness and well-definedness of all these things for most of their careers but these things are totally vacuous from the viewpoint of physics. If a physicist finds out that the answer to a physics question requires him to calculate $f(A)$, a function of an operator - an observable - he just has to calculate it whether or not it looks hard or well-defined. In particular, the Taylor expansion for functions such as the exponential is always assumed to hold. You discuss the function $g(A)$ of the operator as an example. The procedures you outline physically mean that one diagonalizes $A$ - which brings the projection operators to a simple form (only one number $1$ on the diagonal) - and then he simply applies the function $g$ to the eigenvalues. In other words, $$A = U D U^{-1} \quad \Rightarrow \quad g(A) = U g(D) U^{-1}$$ where $g(D)$ is simply a diagonal matrix with entries $g(D_{ii})$ on the diagonal. The formula above works because $U^{-1} U$ cancel everywhere in the middle if we write $g(A)$ e.g. as a Taylor expansion - and by generalization, we just declare the formula above to be right even if the Taylor expansion is not appropriate for a mathematician. The Taylor expansion for the exponential is always OK from a physicist's viewpoint. All these objects - Hilbert spaces, operators, their functions (especially exponentials), spectra, eigenvalues - and all these operations - exponentiation, search for projection operators etc. - are important in physics, indeed. And that's what mathematicians often declare when they want their students to listen. But it's equally true that all the points that mathematicians actually focus on most of their lives are totally uninteresting from a scientific viewpoint. This is why the comments that the "material is important in physics" is morally wrong. Mathematicians shouldn't try to support the attractiveness of their teachings by physical applications - especially because their real goal (and the real goal of pure maths) is to make them as independent of natural science as possible. One can't have it both ways. Doing physics or science means that one is allowed to "prove" all these things - such as $g(A)=g(A)$ which is really the content of the "difficult" proof you sketched) - much more elegantly and arguably naively than in maths. One is only worried about the lack of rigor if he can actually find a contradiction with the experiment or other calculations. If it doesn't exist, the science is just perfectly OK. On the other hand, mathematicians are usually looking for problems even if no problems from a scientist's viewpoint exist. This implies a totally different set of priorities and it is unlikely that a student is going to be excited about both. One either prefers to measure the truth according to nitpicking based on predetermined sets of axioms - which is the mathematician's viewpoint - or one is ready to adjust his methods, axioms, and precise definitions of objects (and invent or learn totally new branches of physics) as needed to agree with the empirical data and other accurate calculations - which is the physicist's viewpoint. They're different attitudes and this is why I think that your question should have been posted on a math forum because it's not really following physicist's way of thinking and it is not really motivated by the desire to understand Nature. -

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http://math.stackexchange.com/questions/41646/golden-ratio-powers-tend-to-integer-values

# Golden ratio powers tend to integer values If $G$ is the golden ratio, then $\lim_{n \to \infty}G^n$ tends ever nearer to integer values that approach $\infty$. Can it therefore be proved that $\infty$ is itself an integer? If not, why not? - 3 – Qiaochu Yuan May 27 '11 at 12:48 1 That is more to do with whether $\infty$ is a number. I am certain it is because, given any really big number n, we know that n+1 is one step closer to our goal of $\infty$ and all the steps we take to get there (albeit an infinite number of steps), are all considered numbers. Even if we never reach our goal of $\infty$ we can certainly say that we have only ever encountered numbers on the way. Surely our destination will ultimately be a number even if we never reach it! – Graphic Equaliser May 27 '11 at 13:02 4 Why are you assuming that "integerness" respects limits? (What does this discussion even mean? Mathematics is nothing without definitions and you have not provided a definition of anything.) – Qiaochu Yuan May 27 '11 at 13:30 2 Let's take a look at your argument that $\infty$ is a number. In your argument, I believe your use of "number" is equivalent to "real number" and the real numbers can be identified 1-1 with points on a Euclidean line. The process you describe is moving in some direction along that line. For every point on that line, there is at least one point on each side of it, so for every real number, there is at least one real number greater than it (and one less than it). If the "goal" is $\infty$, then there cannot be any real number past $\infty$, so $\infty$ cannot be a number. – Isaac May 27 '11 at 13:56 1 @lhf We're attempting to get rid of the [number] tag. I am forced to revive some dead questions as a part of this. :-/ – Srivatsan Dec 1 '11 at 0:04 show 2 more comments ## 3 Answers At least your observation that the powers of the golden ratio $G$ seem to approach large integers is true and interesting. The fundamental equation defining $G$ is $G^2=G+1$. From it you can deduce that $G^n = G \cdot F_n + F_{n-1}$, where $F_n$ is the $n$-th Fibonacci number. Since for large $n$ we have $G \approx F_{n+1}/F_n$, we get $G^n \approx F_{n+1}+F_{n-1}$, an integer. A more precise statement of this approximation is $G^n+H^n = (G+H)F_n +2F_{n-1}=F_{n+1}+F_{n-1}$, where $H=-1/G=1-G$ is the other root of the equation defining $G$. Since $|H|<1$, its powers go to zero. Wikipedia mentions this in almost integer. The golden ratio is a Pisot–Vijayaraghavan number, whose characteristic property is that their powers approach integers at an exponential rate. - What does the golden ratio have to do with it? The integer values 1, 2, 3, ..., approach infinity, too, but that doesn't prove infinity is an integer. - Because the numbers 1.5, 2.5, 3.5, ... also approach $\infty$ but they are not integer. However, $G^\infty$ is definitely $\infty$ and also approaches an integer value the nearer it gets to $\infty$ so you can see my argument that $\infty$ is an integer. – Graphic Equaliser May 27 '11 at 12:57 1 @Graphic: $\infty$ is not a number. You cannot say that it is integer, rational, real, imaginary. – Beni Bogosel May 27 '11 at 13:04 6 @Graphic: I don't understand your argument. What do you make of $\lim_{n\to \infty} (G^n+.5)$? It tends to $\infty$ too, but is not integral. – Ross Millikan May 27 '11 at 13:04 $\infty$ can't be a natural number. The natural numbers satisfy: 1. There is exactly one element called '$0$' with no predecessor($-1$ isn't a natural number). 2. Every element has a unique successor, different from itself. You could argue that we excluded by definition in (1) by using 'exactly one' and change it, but you don't get the natural numbers - you get your own custom-made structure. The integers can be defined very conservatively using the natural numbers, so you would have to change this if you wanted "\infty" to be a number. But if you change (1), you lose out on the Principle of Induction which is needed to prove anything useful about the natural numbers. And if you change (2) to remove "different from itself", then {0,1,2,3,4,5,6} with 6+1 = 6 could be considered "the natural numbers" because it satisfies our defining properties. It's not easy to change the definitions to allow $\infty$, and it doesn't seem to give much benefit. It's perfectly possible to define your own structure where $\infty$ exists, but you're asking whether it meets the criteria for a specific already-existing structures and it doesn't. 3,3.1,3.14,3.141,3.1415, etc. is a sequence of rational numbers tending to $\pi$, but $\pi$ is not rational. You can't always argue that because $P(n)$ holds for every $n$, that it must also hold for the limit.

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http://physics.stackexchange.com/questions/45737/directionality-of-angular-momentum?answertab=votes

# Directionality of angular momentum I was told that the sum of linear and angular momentum is conserved. Given that angular momentum's direction as a vector is completely arbitrary (I believe there is no physical reason for choosing the cross product going the way it does (that is, perpendicular up or down, it just matters that it's perpendicular), correct me if I'm wrong), and linear momentum is often interchanged into angular momentum, surely the vector sum cannot be conserved? Or am I wrong, and the meaning is that the magnitude is conserved? As a sidenote, I suppose this would be a good place to ask why such an odd system was devised for calculating the direction (not magnitude) of angular momentum. Why not have the arrow pointing in the direction of motion, with its base at the point that we are considering angular momentum about? Is it useful simply to compute the conservation of it about a point, as it just involves addition? - 1 Did you considered that your system should be closed to have them conserved? in sense no outer forces are applied? – TMS Dec 2 '12 at 22:19 1 It was an implicit assumption, as with any momentum conservation questions. – Alyosha Dec 2 '12 at 22:21 The conserved quantities is the vectors, and thus there magnitude too, and there sum is conserved, and if interchanging happens it will always happen in away that will keep the sum intact-ed, and of course there is a reasons to take the momentum as a cross product, one of them is to make the sum conservative! – TMS Dec 2 '12 at 22:28 The problem I was having is that he seemed to imply that the value [sum of all linear momentum] could grow or shrink to [sum of all angular momentum]'s loss or gain (losses/gains are equal in magnitude). On reflection, I think he may have just been wrong, but I wanted to check here that I wasn't being an ignoramus. – Alyosha Dec 2 '12 at 22:31 You should be more specific, who is he? if that a book you should mention it in your question to make it clear, anyway hope you got it. – TMS Dec 2 '12 at 22:33 show 2 more comments ## 2 Answers Linear momentum and angular momentum are each conserved separately. It is not meaningful to talk about their sum. To talk about angular momentum using a concrete example, let's consider a wheel spinning about its axis (whose center we will call the origin). What really matters in this case is the plane in which the rotation is taking place; let's call this the x-y plane. In three dimensions, it is possible to uniquely specify a plane by giving a vector normal (i.e. perpendicular) to the plane (in the z direction, in this case). (In four dimensions, though, a plane cannot be uniquely specified using a vector; one would need to use a wedge product instead.) The magnitude of the vector is the magnitude of the angular momentum, but there are still two directions in which to choose the vector normal to the plane of rotation. By convention, we choose this using the right-hand rule. You could use a left-hand rule instead if you really wished, but nobody does it this way, and it is crucial to be consistent in the definition of the cross product (or you will find contradictions). As an aside, any "vector" which results from the cross product of two physical vectors is actually a pseudovector. This means that if you look at the system in a mirror, the "vector" will actually point the other way. For instance, if a wheel is spinning counterclockwise (angular momentum pointing toward the observer), then a mirror placed beside the wheel would show it spinning clockwise (i.e. with the angular momentum pointing away from the observer). Physical vectors do not act this way under a mirror ("parity") transformation. - If $x$ and $y$ are two conserved quantities, then $Ax + By$ is also a conserved quantity, for any nonzero real constants $A$ and $B$. This includes both $x+y$ and $x-y$, so it is true that their sum is conserved, no matter which way around you define the direction of angular momentum. However, it is not a very interesting or fundamental result that the sum of linear and angular momentum is conserved, because they have different units, so their sum isn't a physically meaningful quantity. It certainly isn't true that one can be converted into the other. -

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http://math.stackexchange.com/questions/23305/derivation-of-fourier-series

# Derivation of Fourier Series? Can someone point me to the full derivation of the Fourier Series? I'm having problems understanding how the a's and b's coeffients are worked out. - 1 This isn't really a physics question. I'll flag for the mods to migrate it to math.stackexchange. – Mark Eichenlaub Feb 17 '11 at 20:54 I don't quite understand the question: the Fourier Series is a definition: for any integrable function on a closed interval, its coefficients are given by a particular formula: see e.g. en.wikipedia.org/wiki/Fourier_series#Definition. So what exactly is it that you want to work out? – Pete L. Clark Feb 23 '11 at 2:16 I think the OP wants to see a proof that if $f(x) = \sum a_n\cos nx + b_n\sin nx$, where the convergence is (at least) pointwise, then the coefficients $a_n$ and $b_n$ are given by those standard formulas. – Jesse Madnick Feb 23 '11 at 2:43 ## 2 Answers If you already believe that any function can be written as an infinite series of sines and cosines with appropriate coefficients, and all you are wondering about is how we go about actually computing the coefficients, then you're most of the way there. The tricky part is proving that the Fourier series works at all; the derivation of the coefficients is pretty straightforward. It's still not exactly intuitive and involves some steps that you would never think of, but it's easy enough to follow. A good derivation is here: http://cnx.org/content/m10733/latest/ - Imagine that $f(x) = \sum_n a_n \cos n x + \sum_n b_n \sin n x$ (for $x \in [0,2\pi]$, say). The idea for computing the $a_n$s and $b_n$s is that when you write down integrals of the form $\int_0^{2\pi} \cos m x \cos n x,$ or $\int_0^{2\pi} \sin m x \sin n x$, or $\int_0^{2 \pi} \cos m x \sin n x$, then the integrals vanish (just compute them!) unless $m = n$ and the functions coincide; and in the cases when they don't vanish, their values are easily computed. So taking $f$, and then computing $\int_0^{2\pi} f(x) \cos n x$ or $\int_0^{2 \pi} f(x) \sin n x$, one exactly reads off $a_n$ or $b_n$ (for the value of $n$ you chose). This is where the formulas come from. The way people normally think about this is as a kind of orthogonal projection: the functions $\cos n x$ and $\sin n x$ are like orthogonal basis vectors in a vector space, and the integral is like an inner product. So to find the coefficient of a given basis vector (i.e. an $a_n$ or a $b_n$) one takes the inner product against that particular basis vector. -

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http://math.stackexchange.com/questions/135056/radius-of-convergence-of-power-series-or-geometric-series

# Radius of convergence of power series or geometric series To find radius of convergence of geometric series $$\sum_{n=1}^\infty a_n$$ I need to use ratio/root test to find $|L|<1$ To find radius of convergence of power series $$\sum_{n=1}^\infty c_n (x-a)^n$$ I am supposed to find the limit $L$ of just the constant term $c_n$? $$\sum_{n=1}^\infty \frac{(2x-5)^n}{n^2}, \qquad c_n = \frac{2^n}{n^2}, \qquad R = 2^{-1}=1/2$$ How do I know what is a constant? Free of $n$? Or I should focus on getting it into the form $\sum_{n=1}^\infty c_n (x-a)^n$. Thats what I did below $$\sum_{n=1}^\infty \frac{(x-1)^{n-3} + (x-1)^{n-1}}{4^n + 2^{2n-1}}$$ I first factorized what I can $$= \sum_{n=1}^\infty (x-1)^n \cdot \frac{(x-1)^{-3} + (x-1)^{-1}}{4^n(1 + 2^{-1})}$$ So I guess $$c_n=\frac{(x-1)^{-3} + (x-1)^{-1}}{4^n(1 + 2^{-1})}$$ Then $$L=|\frac{c_{n+1}}{c_n}|$$ But what do I do with the $x$? Correct method is supposed to be knowing its a geometric series of common ratio $\frac{x-1}{4}$. Then radius is |x-1|<4. But here, don't I need to get the radius of $x$ alone, without the $-1$? Anyways, I think the main thing I am confused about is when to use which method. For geometric series, I am doing the ratio/root test for the whole $a_n$ while in power series, I am "separating" the $a_n$ into $c_n (x-a)^n$ form? And doing test on the constant terms ($c_n$)? - ## 2 Answers The ratio test is a more general approach than the "recognize it as a geometric sum test" - obviously because the latter won't work if the sum is not, in fact, a geometric series - but they are both valid. One thing to note is that if you compute $|c_{n+1}/c_n|$ using your definitions of $c_n$ in the example you give, the numerator $(x-1)^{-3}+(x-1)^{-1}$ does not depend on $n$ and so it will disappear when we compute the ratio of the two coefficients. Same with $(1+2^{-1})$ in the denominator. Define $f_n:= b_n(x-a)^n$ and consider the power series $$f(x)=\sum_{n=0}^\infty f_n=\sum_{n=0}^\infty b_n(x-a)^n.$$ If $\lim\limits_{n\to\infty} |b_{n+1}/b_n|=L$ then our "second" test says the radius of convergence is $R=1/L$. Why? Well, $$\lim_{n\to\infty}\left|\frac{f_{n+1}}{f_n}\right| =\lim_{n\to\infty}\left|\frac{b_{n+1}(x-a)^{n+1}}{b_n(x-a)^n}\right|=L|x-a|,$$ so the series $\sum\limits_{n=0}^\infty f_n$ converges if $\big|L|x-a|\big|<1$ by our "original" ratio test, equivalently $|x-a|<R$, which defines a circle of radius $R$ (note: $a$ is the center of the circle and tells us nothing about the radius $R$). This exhibits how the "secondary" test, of just comparing the coefficients of $(x-a)^n$, is actually a shortcut to applying the original ratio test, where we test the sequence $f_n$, so to speak. - Without using ratio test u can also find radius of convergence in this example, $\sum_{n=1}^\infty c_n$ $\frac{(2x-5)^n} {n^2}$ Comparing the given series with power series, $\sum_{n=1}^{\infty}c_n(x-x_0)^n$ We have, $c_n=\frac{1}{2^n n^2}$ and $x_0=(\frac{5}{2})^n$ $R=\lim_{n\to \infty}|c_n|^\frac{1}{n}$ $\ \ \ =\lim_{n\to \infty}|\frac{1}{2[(n)^\frac{1}{n}]^2}|$ $\ \ \ R=\frac{1}{2}$ $\ \ \ R=\frac{1}{r}$ $\ \ \ r=2$ $\therefore$ radius of convergence is 2 Moreover,interval of convergence $(x_0-r,x_0+r)$ In this example root test is very useful because here constant having power n -

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http://physics.stackexchange.com/questions/43545/disappearance-of-moduli-for-condensate-of-open-strings?answertab=active

# Disappearance of moduli for condensate of open strings Consider a Dp-brane. Compactify $d$ spatial dimensions over a torus $T^d$. Suppose $d\geqslant p$, and that the Dp-brane is completely wrapped around the compactified dimensions. Look at the open string modes ending on this wrapped D-brane. There is a zero energy open string mode associated with each spatial dimension. That corresponds to the orientation of the worldsheet field excitation. If the direction is along an uncompactified spatial dimension, that corresponds to quanta of brane displacements along that direction. These cases needn't concern us. If the direction is normal to the brane but lies along a compactified dimension, that corresponds to quanta of brane displacements along that direction. If it's tangent to the brane, it corresponds to quanta of the Wilson line of the brane gauge field along that wrapped direction. Here's the question. Suppose the wrapped D-brane has a total mass of $M$. Suppose there is a compactified spatial dimension of radius $R$ along which the brane isn't wrapped, i.e. $p<d$. The brane has Kaluza-Klein momenta along that direction of value $n/R$ where n is integral. The energy spectrum is given by $$\sqrt {M^2 +n^2/R^2} ~\approx~ M + \frac{n^2}{2MR^2} +\mathcal{O}(M^{-3}).$$ A condensate of zero energy open strings with orientation along that direction ought to give a continuous moduli? Why is there no moduli then, and why is the energy spectrum discretized? Or consider a direction in which the brane is wrapped. We ought to have a continuous modulus of Wilson line of the brane gauge field along that dimension? Once again, we have discretization. Why? Anyway, how can a completely wrapped D-brane have KK momentum? In the string worldsheet picture, we have open strings ending at the D-brane background at a fixed position. Along those compactified dimensions along which the brane isn't wrapped, we have Dirichlet boundary conditions. Such open strings can only have winding numbers, but no KK momentum either. Even then, we're dealing with a condensate of open strings with zero winding number. This discretization doesn't occur if the brane remains unwrapped under at least two uncompactified spatial dimensions because the brane now has infinite mass. If it remains unwrapped only along one uncompactified spatial dimension, there's the Mermin-Wagner theorem, meaning there’s no fixed brane position or Wilson line. PS: Maybe this question can be rephrased in terms of BPS. We have a wrapped BPS brane. But nonperturbatively, somehow, the BPS state has to be delocalized along the compactified dimension or its Wilson line in a superposition over all possible values? Open strings with no energy are also BPS. So, we can have any condensate of them and still remain BPS? This clearly isn't the case with a lifted modulus, and a discretization of the energy spectrum. PPS: How do you even express the KK modes of the D-brane, or the dual to its Wilson lines, in terms of a condensate of open strings? Suppose you have the lowest energy state with zero KK momentum. Then, disregarding transverse displacements along the uncompactified spatial dimensions, there is an energy gap to the next energy state. However, a condensate of open strings with internal mode excitations along the compactified dimensions would naively give no energy gap. PPPS: Is the number of open string modes corresponding to these "disappearing moduli" even a well-defined operator? This is precisely because at the perturbative level, such open string modes have no energy. If it's not a well defined operator, just how do you even express this in terms of open string worldsheets? - ## 1 Answer By T-dualities along the finite directions on which the D-brane is wrapped, your first problem is equivalent to the problem of a D0-brane moving on a circle. The massless open-string scalar corresponding to the direction of the circle produces the field $X(\tau)$ living on the world line of the D0-brane. And indeed, by assumption, the global topology of the configuration space for this field $X(\tau)$ includes the periodic identification $X(\tau)\sim X(\tau)+2\pi R$. This is equivalent to the quantization for the complementary momentum, $P=N/R$. But none of these two equivalent signs of the periodicity/quantization of the transverse location/momentum is visible at the level of the spectrum of the open strings. In the perturbative treatment of the D-brane dynamics, we effectively expand around a classical shape of the "heavy" D-brane, i.e. around a classical solution. Imagine $X(\tau)=0$. The open-string excitation corresponding to the transverse field is a quantum of a "wave" that only deviates "infinitesimally" from $X(\tau)=0$. The quantitative distance scale at which $X$ deviates is parameterically shorter than then string scale. So the open-strings know about all the potential of the D-brane to change its shape and vibrate but a finite number of open-string excitations still leaves you near $X(\tau)=0$ so it isn't capable of probing the global properties of the $X$ configuration space. To reach the point $X=2\pi R$, you would need an infinite number of open strings – the number would scale like $1/g$ or something like that – and the perturbative expansion would break down or would become insufficient at that point because it's an expansion around the point $g=0$ at which point the required number of open strings is strictly infinite. So at the leading order, the open-string calculations will be totally insensitive to the global properties of the D-brane configuration space. This is also reflected by your $n^2/2MR^2$ kinetic term from the D-brane. Because $M\sim 1/g$, this kinetic energy scales like $g$ and vanishes in the $n=0$ limit. But yes, once you go to the subleading order in $g_s$ when you're calculating the energy or evolution, you will be able to see that the field on the D0-brane is periodic and the coefficient of this $1/2M$-like kinetic term is a quantized $n^2/R^2$. Analogously, you may solve the other universal problem with the Wilson line which is just the T-dual of the D0-brane above performed along the $X$ axis. You get a D1-brane with a Wilson line that plays exactly the role of the periodic $X$ for the D0-brane. The dual quantity to this periodic Wilson line is the electric flux, $\int d \tilde X\,F_{01}$, along the D-string, which is quantized much like the momentum of the D0-brane – they're dual descriptions of each other. Again, the open-string excitations only produce "infinitesimal variations" (or Planckian, substringy variations) of the Wilson line that are too small to see the periodicity of the Wilson line or the quantization of the complementary electric flux. It's helpful, at least for me, to look at this problem via Matrix theory which picks some low-energy limit of the D-brane dynamics. Indeed, for D0-branes on a circle, one may also describe the situation by D1-branes on the dual circle. The matrix model, matrix string theory, will automatically give you the right gauge theory with the right periodicity conditions of the Wilson line and the quantization for the electric flux. You may be asking how one derives that the gauge theory has the right identifications of the configuration space. One may be more or less rigorous but it's clear that it has to have this identification. This identification boils down to the fact that e.g. the D0-brane at places $X$ and $X+2\pi R$ is really the same state – it's the same state from the perturbative string theory viewpoint as well because the D0-brane locations define two totally identical (i.e. one) boundary conditions for the world sheet, the same boundary CFT. The configuration space for D-branes is the space of different boundary conditions and because $X$ and $X+2\pi R w$ give the same boundary conditions, they're identified. - I know this is a nonperturbative effect. You are trying to reexpress it in terms of matrix theory, but how do you even deal with this question at the level of a condensate of open strings? Or is the description in terms of open strings incomplete, with matrix theory being the complete theory? – Hecles Nov 6 '12 at 14:36 If path integrals over string worldsheets is fundamental, surely there ought to be a way to answer my question purely at the level of open string worldsheets, shouldn't there? A failure to do so is an indictment of the completeness of the string worldsheet description. – Hecles Nov 6 '12 at 14:38 @Hecles: Yes, the perturbative string picture is incomplete. – user1504 Nov 6 '12 at 14:57 Dear @Hecles, yes, I confirm user1504, perturbative string theory is an incomplete description of string theory. It only describes things that exist at $g=0$, and what they're doing at a general $g$ is expressed as a Taylor expansion - which isn't the most general function even among functions that are smooth near $g=0$. Things that are singular or nonperturbative near $g=0$ are invisible in perturbative string theory. Moduli spaces whose volumes go like $1/g$ look infinite in perturbative expansion, and so on... D-branes give you a way to "extend" perturbative expansions if done right. – Luboš Motl Nov 16 '12 at 10:38

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http://mathoverflow.net/questions/43538?sort=newest

## Wonderful applications of the Vandermonde determinant ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) This semester I am assisting my mentor teaching a first-year undergraduate course on linear algebra in Peking University, China. And now we have come to the famous Vandermonde determinant, which has many useful applications. I wonder if there are some applications of the Vandermonde determinant that are suitable for students without much math background. For example, using the Vandermonde determinant, we can prove that a vector space $V$ over a field $F$ of characteristic 0 cannot be expressed as a finite union of its nontrivial subspaces, i.e., there do not exist subspaces $V_1,\ldots,V_m$ that satisfy $$V_1\cup \cdots\cup V_m=V,$$ where $V_i\ne {0}$ and $V_i\ne V$ for all $i=1,2,\ldots,m$. This can be proved as follows: choose $v_1,\ldots,v_n$ as a basis of $V$, and consider the infinite series $$\alpha_i = v_1 + iv_2+\cdots+i^{n-1}v_n.$$ Using our knowledge of the Vandermonde determinant, one can show that every subset of the $\alpha$'s having $n$ vectors in it consists of a basis of $V$, hence each of the $V_i$'s can contain at most $n-1$ of the $\alpha$'s in it, so there must be infinitely many $\alpha$'s not contained in any of the $V_i$'s. - 2 You should probably add the 'big-list' and 'example' tags, and set this as Community Wiki. – Mariano Suárez-Alvarez Oct 25 2010 at 16:20 The basic proof seems however more simple : take $x\in V_1 \setminus \bigcup_{i>1}V_i$ and $y\in V_2\setminus V_1$; then for some $\lambda \neq \mu$ in $F$, we have $x+\lambda y, x+\mu y \in V_j$ for some $j$, absurd. – Henri Oct 25 2010 at 16:27 yes, but here the set has an extra property: each subset with n elements is a base of $V$, which is a bit more interesting. – zhaoliang Oct 25 2010 at 16:35 2 I think the Vandermonde can also be used to compute the discriminant of a cyclotomic number field. Can an algebraic NTist step forward and elaborate on this? – darij grinberg Oct 26 2010 at 0:12 ## 17 Answers I found some remarkable property of the Vandermonde determinant. Based on this property we are able to introduce a notion "difference between n>2 quantities". Below I give the abstract of this result. The notion of difference between two quantities plays a basic role in mathematics, consequently in all branches of human activity where the mathematics is applied. However the long stand question is: what is {\it the difference between three (or more) quantities}? The binary operation $[a,b]=(a-b)$ possesses the following principal feature: with respect to the third quantity $c$ this operation is decomposed into a sum of the same operations between $a$ and $c$, and $c$ and $b$, i.e., $$[a,b]=[a,c]+[c,b].$$ Denote by $[a,b,c]$ difference between three quantities $a,b,c$. With respect to additional quantity $d$ this definition of the difference has to possess with the following property $$[a,b,c]=[d,b,c]+[a,d,c]+[a,b,d].$$ We prove that this property of difference between three (or $n\geq 2$) quantities is satisfied by one of the features of Vandermonde determinant. - ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. Especially for students with just very basic background, it might be a fun fact that companion matrices are diagonalized by the vandermonde matrix corresponding to the zeros of the characteristic polynomial they encode - provided, of course, that the roots are all distinct. - The Vandermonde determinant is used to prove that cyclic polytopes maximize the number of $i$-dimensional faces for each $i$ among all triangulations of a $(d-1)$-dimensional sphere having exactly n vertices. The cyclic polytope $C(n,d)$ is the convex hull of any $n$ distinct points on the moment curve {$(t,t^2,\dots ,t^d) | t\in {\mathbf R}$} $\subseteq {\mathbf R}^d$. I like the discussion of cyclic polytopes in G"unter Ziegler's book "Lectures on Polytopes". The wikipedia article en.wikipedia.org/wiki/Cyclic_polytope also seems to give a nice quick summary. - Probably not an application for just any audience, but I thought I'd share... The Vandermonde determinant shows up in matrix models of quantum field theories. Roughly speaking, in these we consider integrals of the form $\int dM f(M)$, over the space of Hermitian matrices, where $f$ is invariant under conjugation by unitary matrices, and $dM$ is the (also conjugation invariant) measure $dM = (\prod_i dM_{ii})(\prod_{i\lt j} dM_{ij})$. We want to calculate the integral by gauge fixing. In other words, to integrate over a judiciously chosen set of representatives of each $U(N)$ orbit. Usually the best set of representatives to take are the diagonal matrices. The procedure for evaluating the integral in that case is exactly that of the Weyl integration formula. In doing so we get what physicists call the Faddeev-Popov determinant, which turns out to be the Vandermonde determinant! In other words, we get an equivalent integral $\int d\lambda_1 ... d\lambda_N \Delta(\lambda_1, ...,\lambda_N) f(diag(\lambda_1,...,\lambda_N))$, where $\Delta$ is the Vandermonde determinant. - The vectors corresponding to any $n$ distinct points on the rational normal curve $x\to (x,x^2,\dots,x^n) \subset \mathbb{A}^n$ span $\mathbb{A}^n$. - The Vandermonde determinant plays a role in the proof by Kempf and Kleiman-Laksov of the existence portion of the famous Brill-Noether theorem (formulated, but not proved, by Brill and Noether): a general, genus $g$ projective curve has an algebraic line bundle of degree $d$ and $(r+1)$-dimensional space of global sections if and only if the "naive parameter count" for the dimension of such, $\rho(g,r,d) = g-(r+1)(g-r+d)$, is nonnegative. The point is that they set up the enumerative formula to count the number of such line bundles (satisfying some appropriate additional conditions). Miraculously the formula comes out to a Vandermonde determinant which can be explicitly evaluated as being nonzero (as opposed to many similar enumerative problems in algebraic geometry which have no such closed formula). - Maybe this should be a comment under Darij Grinberg's or Terry Tao's answer, but anyway: the discriminant of a monic polynomial is the square of the Vandermonde determinant evaluated at roots of the polynomial. Undergraduate students who advanced to a linear algebra course must have encountered at least the discriminant of a quadratic--although in this case the relation between the Vandermondian and the discriminant is not so wonderful... This relation between Vandermondian and discriminant also determines the relation between Euler class and Pontryagin class: http://en.wikipedia.org/wiki/Splitting_principle - The Discrete Fourier Transform, which sends a vector $x=\left(x_j\right)_{j=0}^{N-1}$ to $y=\mathrm{DFT}(x)$ such that $$y_k=\frac{1}{\sqrt{N}}\sum_{j=0}^{N-1}e^{2\pi i \times jk/N}x_j$$ has a matrix representation $$\mathrm{DFT}_{jk}=e^{2\pi i \times jk/N}=\left(e^{2\pi i /N}\right)^{j\times k},$$ which is in fact a doubly Vandermonde matrix: both it and its transpose are Vandermonde matrices. With this you can use the Vandermonde determinant to prove that $\mathrm{DFT}$ is nonsingular, and if you prove using other means that it is unitary (rather easy) then you will get, I think, a nontrivial expression for 1 as a product of differences of roots of unity. - There is an elegant (and short!) application of the (generalized) Vandermonde determinant to the famous problem of D. H. Lehmer in the article [D.C. Cantor and E.G. Straus, Acta Arith. 42 (1982/83), no. 1, 97-–100]. I put here a scan of the article together with corrections given by the authors later. - Maybe it is not really suitable to undergrads (unless they are really problem solving-oriented), but there is a nice proof that $$\prod_{1\leq i \lt j\leq n} \frac{x_j-x_i}{j-i}$$ is integer for all integer sequences $(x_k)_{k=1}^n$ using Vandermonde determinants. The idea is reducing the thesis to the fact that the matrix $$\begin{bmatrix}1 & 1 & \dots \newline \binom{x_1}{1} & \binom{x_2}{1} & \dots \newline \binom{x_1}{2} & \binom{x_2}{2} & \dots \newline \vdots & \vdots & \ddots \end{bmatrix}$$ has integer entries, and thus integer determinant. After clearing the denominators (which give the factor $\prod \frac{1}{j-i}$), one can transform the resulting determinant to a Vandermonde with basic row operations. - The Vandermonde determinant formula implies that Vandermonde matrices are maximum distance separable and can therefore be used to construct error-correcting codes with good properties (BCH codes in particular). - As for applications in symmetric function theory... well, open a book on symmetric function theory at a random page and stare at the formulas. Usually something looking like a Vandermonde determinant will stare back at you. For example: Crawley-Boevey, Lectures on Representation Theory uses it on page 18. Oh, and of course he uses the Cauchy determinant too. While he derives it from a geometric argument, it can also be proven purely algebraically, and the proof uses Vandermonde. (Again, I'm not claiming the proof is new. In fact I believe I have seen it somewhere, but I couldn't find it again...) - I wanted to put this online long ago but somehow never came to actually doing it. Now is a good occasion: http://mit.edu/~darij/www/index.html#hyperfactorial or, directly, the PDF file: http://mit.edu/~darij/www/hyperfactorialBRIEF.pdf This is about a theorem by MacMahon stating that for any three nonnegative integers $a$, $b$, $c$, the number $\frac{H\left(a\right)H\left(b\right)H\left(c\right)H\left(a+b+c\right)}{H\left(b+c\right)H\left(c+a\right)H\left(a+b\right)}$ is an integer, where $H\left(m\right)$ means $0!\cdot 1!\cdot ...\cdot \left(m-1\right)!$. There are various proofs of this now, some of them combinatorial (see the references in the note), but the simplest one is probably the one I give using the Vandermonde determinant (I don't think it's new...). The note is a bit long (10 pages), but the proof ends at page 6. Also note that I prove Vandermonde itself, which takes up some space as well. Another application of Vandermonde appears on page 9: If $a_1$, $a_2$, ..., $a_m$ are $m$ integers, then $\prod\limits_{1\leq i < j\leq m}\left(a_i-a_j\right)$ is divisible by $H\left(m\right)$. This is very well-known (and so is the proof). Finally, a little question - slightly offtopic, I know. Back to the $\frac{H\left(a\right)H\left(b\right)H\left(c\right)H\left(a+b+c\right)}{H\left(b+c\right)H\left(c+a\right)H\left(a+b\right)}$ problem, one might try proving that this is an integer by showing that every prime $p$ divides $H\left(a\right)H\left(b\right)H\left(c\right)H\left(a+b+c\right)$ at least as often as it divides $H\left(b+c\right)H\left(c+a\right)H\left(a+b\right)$. This can be easily shown equivalent to the following: Any nonnegative integers $a$, $b$, $c$ satisfy $\sum\limits_{k=0}^{a-1} \lfloor \frac{k}{p} \rfloor + \sum\limits_{k=0}^{b-1} \lfloor \frac{k}{p} \rfloor + \sum\limits_{k=0}^{c-1} \lfloor \frac{k}{p} \rfloor + \sum\limits_{k=0}^{a+b+c-1} \lfloor \frac{k}{p} \rfloor$ $\geq \sum\limits_{k=0}^{b+c-1} \lfloor \frac{k}{p} \rfloor + \sum\limits_{k=0}^{c+a-1} \lfloor \frac{k}{p} \rfloor + \sum\limits_{k=0}^{a+b-1} \lfloor \frac{k}{p} \rfloor$. (Yes, it can be shown that the $p^2$, $p^3$, ... terms can be ignored.) Is there an easy way to see this? Or any way at all, without going back to the Vandermonde determinant? - (1) Not really an application, but the paper by I. Gessel, Tournaments and Vandermonde's determinant, J. Graph Theory 3 (1979), 305-308, gives a nice connection with tournaments. See also Exercise 2.16 of my book Enumerative Combinatorics, vol. 1 (equivalent to Exercise 2.35 at http://math.mit.edu/~rstan/ec/ec1.pdf). (2) Probably not suitable for an undergraduate course, but if $s_{(n-1,n-2,\dots,1)}$ denotes the Schur function of the staircase shape $(n-1,n-2,\dots,1)$, then the evaluation $$s_{(n-1,n-2,\dots,1)}(x_1,\dots,x_n)=\prod_{1\leq i \lt j\leq n} (x_i+x_j)$$ follows immediately from the bialternant formula for Schur functions, since it reduces to the quotient of two Vandermonde's: $\prod (x_i^2-x_j^2)/\prod(x_i-x_j)$. See Exercise 7.30 of Enumerative Combinatorics, vol. 2. - Vandermonde determinants + Cramer's rule = Lagrange interpolation. (EDIT: Also, there is a qualitative version of the above identity: just from knowing that the Vandermonde determinant is non-vanishing when the $x_i$ are distinct, one can already deduce that polynomial interpolation is theoretically possible, though to get the precise formula one still needs to go through the above identity.) - When I was typing this item, I had Lagrange interpolation in my mind :P I think there should be some applications in combinatorics or symmetric function theory – zhaoliang Oct 25 2010 at 17:54 Of course, what is "wonderful" is quite subjective. One simple application that I like is showing that the functions $e^{cx}$ are linearly independent over ${\mathbb R}$. Another is that if $tr (A^n)=0$ for all $n$, then $A$ is nilpotent. - this reminds me the Linderman-Weierstrass theorem on the transcendence of e. – zhaoliang Oct 25 2010 at 17:50 1 @zhaoliang: In fact, one uses the Vandermonde determinant in the standard proof of Baker's theorem. – Daniel Litt Oct 25 2010 at 23:02 The independence of $e^{c x}$ is a rather trivial fact. To prove it using Vandermond determinant, just differentiate any linear dependence multiple times and pack it into a matrix equality. – Keivan Karai Oct 26 2010 at 13:43 1 @Keivan. A beautiful application of this characterization of nilpotent matrices is Amitsur-Levitski's polynomial identity in $M_n(k)$. – Denis Serre Oct 26 2010 at 16:52 Another application is to prove that for any $0<\alpha_1<\cdots<\alpha_n$ the family $$\sin \alpha_1x,\cdots,\sin\alpha_nx$$ is linearly independent in $C^{4(n-1)}(\mathbb R)$. – Hany Oct 26 2010 at 19:20 I loved encountering the following as a student: The Vandermonde determinant plays a role in the proof of Hilbert's Theorem 90 in Section 9.6 of Schilling and Piper's Basic Abstract Algebra. I certainly should take the time to type this argument in a bit. I just wanted to get the reference out there. - gigapedia.com/… – zhaoliang Oct 25 2010 at 17:18

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http://unapologetic.wordpress.com/2009/12/29/the-mean-value-theorem-for-multiple-integrals/?like=1&source=post_flair&_wpnonce=a832ba55c5

# The Unapologetic Mathematician ## The Mean Value Theorem for Multiple Integrals As in the single variable case, multiple integrals satisfy a mean value property. First of all, we should note that, like one-dimensional Riemann-Stieltjes integrals with increasing integrators, integration preserves order. That is, if $f$ and $g$ are both integrable over a Jordan-measurable set $S$, and if $f(x)\leq g(x)$ at each point $x\in S$, then we have $\displaystyle\int\limits_Sf(x)\,dx\leq\int\limits_Sg(x)\,dx$ This is a simple consequence of the definition of a multiple integral as the limit of Riemann sums, since every Riemann sum for $f$ will be smaller than the corresponding sum for $g$. Now if $f$ and $g$ are integrable on $S$ and $g(x)\geq0$ for every $x\in S$, then we set $m=\inf f(S)$ and $M=\sup f(S)$ — the infimum and supremum of the values attained by $f$ on $S$. I assert that there is some $\lambda$ in the interval $[m,M]$ so that $\displaystyle\int\limits_Sf(x)g(x)\,dx=\lambda\int\limits_Sg(x)\,dx$ In particular, we can set $g(x)=1$ and find $\displaystyle mc(S)\leq\int\limits_Sf(x)\,dx\leq Mc(S)$ giving bounds on the integral in terms of the Jordan content of $S$. Incidentally, $g(x)\,dx$ here is serving a similar role to the integrator $d\alpha$ in the integral mean value theorem for Riemann-Stieltjes integrals. Okay, so since $g(x)\geq0$ we have $mg(x)\leq f(x)g(x)\leq Mg(x)$ for every $x\in S$. Since integration preserves order, this yields $\displaystyle m\int\limits_Sg(x)\,dx\leq\int\limits_Sf(x)g(x)\,dx\leq M\int\limits_Sg(x)\,dx$ If the integral of $g$ is zero, then our result automatically holds for any value of $\lambda$. Otherwise we can divide through by this integral and set $\displaystyle\lambda=\frac{\displaystyle\int\limits_Sf(x)g(x)\,dx}{\displaystyle\int\limits_Sg(x)\,dx}$ which will be between $m$ and $M$. One particularly useful case is when $S$ has Jordan content zero. In this case, we find that any integral over $S$ is itself automatically zero. ### Like this: Posted by John Armstrong | Analysis, Calculus ## 3 Comments » 1. [...] in for integrals are Jordan measurable, and their boundaries have zero Jordan content, so we know changing things along these boundaries in an integral will make no [...] Pingback by | December 30, 2009 | Reply 2. [...] infinitesimal pieces of -dimensional volume. Now, with the change of variables formula and the mean value theorem in hand, we can pull out a macroscopic [...] Pingback by | January 8, 2010 | Reply 3. [...] analogue of the integral mean value theorem that holds not just for single integrals, not just for multiple integrals, but for integrals over any measure [...] Pingback by | June 14, 2010 | Reply « Previous | Next » ## About this weblog This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”). I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.

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http://physics.stackexchange.com/questions/15907/pressure-inside-an-ideal-gas

# Pressure Inside an Ideal Gas A non-relativistic ideal gas exerts a pressure at the surface of its container $p = \frac13 \rho \langle v^2 \rangle$ where $\rho$ is the mass density of the gas and $\langle v^2 \rangle$ is the average square of the Maxwell velocity distribution. This is the relation for the pressure at the boundary of the vessel containing the ideal gas. However, if one were to place an infinitesimal "test" area inside the boundary of the vessel the momentum flux across that area would be 0 since the distribution of velocities is symmetric. That is, as many particles would cross one way across the area as cross the opposite way. This suggests that the pressure inside an ideal gas is 0. In general relativity an ideal gas is usually presented as an example of a perfect fluid, that is one with stress energy tensor equal to $T_{\mu \nu} = \left( \begin{array}{cccc} \rho & 0 & 0 & 0 \\ 0 & p & 0 & 0 \\ 0 & 0 & p &0 \\ 0& 0 & 0 & p \end{array} \right)$ Since the stress energy tensor should be a local function it seems strange to assign to $p$ (in the frame in which the velocity distribution is isotropic) in the above equation the value of the pressure at the boundary of the vessel. I can see why this is done: one wants the stress energy tensor to be a smooth function, and also it should be that a gas at finite temperature should gravitate more than a gas at zero temperature (dust)... however this is point is not articulated, in for instance MTW. Perhaps I am missing something elementary? - 1 Why does a net momentum flux of zero for some area suggest a pressure of zero? You completely lost me at that point. – AlanSE Oct 19 '11 at 5:30 Pressure is defined to be the force per unit area. If when the particles collide with the test area they do so elastically, the force (integrated over a small time window) exerted will be the momentum flux (integrated over that time window). Basically as many particles bounce off on one side as they do on the other... the net force is zero... the pressure is zero – Ian Oct 19 '11 at 5:42 1 Your wording is proposing a closed area integral - a Gaussian surface. The integral over any such surface for an idealized gas model should give zero, of course. If I put any object into a gas, I do not expect a net force (aside from buoyancy, drag, etc). In order to define "pressure" you need to take a once-sided integral, and only count those particles going in one direction, like the case of a boundary surface. If I have a jar in a vacuum with a gas in it, the basic idea of pressure is that I count the collisions on one side of the wall only. – AlanSE Oct 19 '11 at 6:19 I think Z is missing the more general definition of pressure: perpendicular momentum transported across an area. In the case where the area sits on the wall of the container, that definition can be restated as the usual "particles bouncing off of the wall" argument. But his question still stands in the case that the area is within the gas. An interstellar gas has no containing wall. Is pressure a sensible concept in this case? Of course it is! Is the pressure zero? Certainly not! I'll give an answer below... – garyp Oct 20 '11 at 1:43 The momentum flux is not zero. If it were, there would be no stress. @Zassounotsukushi: this is the definition of stress. – Ron Maimon Oct 20 '11 at 6:01 ## 2 Answers The stress is the current of the conserved momentum. If momentum is going from one place to another, you have a stress, because the momentum has to go through all the places inbetween. If your intuition comes from action-at-a-distance physics, this can be confusing, because in Newtonian gravity, momentum can jump across empty space between two gravitating objects, leading to a nonlocal force. But this is no longer true in relativity. When you have a gas compressed by a container, the x-momentum flows from one x-end of the container, where a positive x-force is applied (x-momentum pumped into the gas) to the other x-end, going in the x-direction. This means that there is a flow of x-momentum in the x-direction, a pressure. If you place an x-wall somewhere, the force exerted by the left half of the gas on the right half is equal and opposite to the force exerted by the right half on the left half. But this does not mean that there is zero stress. The rightward force is a flow of x momentum in the positive x direction, and the leftward force is also a flow of x momentum in the positive x direction. This is a bit confusing because of the vector nature of momentum, and the tensor nature of stress, so consider a scalar conserved charge. Suppose you have a wire with a nonzero flow of electric charge, a nonzero current, and you cut the wire somewhere with a plane. The current through the plane is the flow of charge from left to right, and if 8 coulombs per second are flowing from left to right, then it is obvious that -8 coulombs per second are flowing from right to left, in order for charge to be conserved. But does this mean that the current is always zero? Obviously not--- the current is defined by one or the other of these quantities, not the sum. Similarly, in a gas, the momentum currents are defined by the force one part of the gas exerts on another, both through particles leaving the region carrying a certain amount of momentum with them, and collisions right by the boundary transferring momentum from particles inside to particles outside. The net momentum flow is defined by the force of one half on the other, or by the negative of the force the other way, not by the sum of the two. - Somebody gets a kick from downvoting all my answers. This answer is the correct resolution of the confusion, regardless. – Ron Maimon Oct 20 '11 at 9:33 1 Well, I up-voted it. – garyp Oct 20 '11 at 12:17 [see my correction in the comments. Sorry, Ian.] Ian is missing the fact that the area segment that he is considering is a directed area. In our imagination we think of the area segment as a tiny transparent, infinitely thin rectangle, and if we flip it 180 degrees it looks the same. We can't be imposing that limitation on our area. We can only reasonably insist that it returns to its original state after a rotation of 360 degrees. Imagine that the little area has an arrow pointing perpendicularly from the surface. Now the momentum that crosses the surface has to be reckoned according to whether or not your atom is crossing in the positive direction or the negative direction. When two atoms cross, one from the left the other from the right, the momentum that crosses the directed area is 2mv. Suddenly the pressure is no longer zero. A directed area is something we don't ordinarily think about, but we do use them without naming them when we use the integral form of Gauss' Law. On the other hand, a directed line segment is quite familiar; we call it a vector instead of "directed line segment". The directed area is more naturally described mathematically by an even more unfamiliar concept: the bivector, which naturally accounts for the directionality of an area segment without having to glue perpendicular arrows on to it. - I don't believe I was missing that fact.... the test area is of course directed. Consider the test area to have unit normal $\hat{n} = (1,0,0)$. A sensible definition of the momentum flux would be $\vec{p} \cdot \hat{n}$. A particle traveling from the "left" would have momentum $\vec{p} = (mv ,0 ,0)$ and would contribute a flux $mv$. A particle traveling from the "right" would have momentum $\vec{p} = ( -mv ,0,0)$ and would contribute a flux of $-mv$. These cancel. – Ian Oct 20 '11 at 2:40 The source of my confusion was in how the Stress tensor (or Stress energy tensor) is defined. It turns out that it is defined such that given a test area one only considers the force that would be exerted on the area from the side opposite the direction of the unit normal. This means that the pressure inside the gas is identical to the pressure at the boundary. – Ian Oct 20 '11 at 3:50 Yes. (lesson learned: I posted an answer because I struggled with the same question some years ago, and came up with a somewhat satisfying answer. I forgot to make sure I correctly remembered the answer.) What I neglected was that in considering the momentum added by the two particles ... one from the right, the other from the left ... one has to ask how much momentum the two particles carry to the right (for instance). They both carry $mv$ to the right, giving $2mv$. I'll go out on another limb and suggest that the factor of two is important. But Ron's answer is better anyway. – garyp Oct 20 '11 at 12:45

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http://mathforum.org/mathimages/index.php/Hyperbolic_Tilings

# Hyperbolic Tilings ### From Math Images Hyperbolic Tilings Field: [[Field:| ]] Image Created By: [[Author:| ]] Hyperbolic Tilings This image is a hyperbolic tiling made from alternating two shapes: heptagons and triangles. # Basic Description In addition to tessellations found in the traditional Euclidean geometry, there are also tilings that can be created in non-tradition or non-Euclidean geometry. Hyperbolic tiling is a type of tessellation that is visualized in a non-Euclidean geometry called hyperbolic geometry. This particular tessellation is viewed in the Poincaré disk model, where straight lines are arcs that form right angles at the boundary of the disk. A hyperbolic tiling The polygons in a hyperbolic tiling grow smaller as they reach the edge of the disk. This is due to the representation of the tiling in the Poincaré disk model that we use to visualize hyperbolic tilings. The Poincaré disk model does not show traditional distances, so that as we get closer and closer to the edge of the disk, the distances increase and the boundary of the disk is in fact infinite. # A More Mathematical Explanation [Click to view A More Mathematical Explanation] ## Schläfli symbol The Schläfli symbol is a notation that can be used to describe the properties o [...] [Click to hide A More Mathematical Explanation] ## Schläfli symbol The Schläfli symbol is a notation that can be used to describe the properties of polygons in a tessellation. The symbol is in the form {n , k}, where n refers to the number of edges of a regular polygon and k refers to the number of regular polygons that meet at each vertex. Thus, if the summation of all the angles at a vertex is 360 degrees: the angle at each vertex is $\frac{360}{k}$ and the polygon has a summation of angles $\frac{n360}{k}$. Regarding the image shown to the right, the tessellation is made up solely of triangles, so that n = 3. At each vertex in the tessellation, there is always a meeting of six triangles, giving us k = 8. Thus, the Schläfli symbol for this tessellation is {3,8}. Also, the angle at each vertex is 35 degrees, and the summation of angles of the polygon is 135 degrees (which is less than the traditional 180 degrees, a characteristic of hyperbolic tilings). ## Types of Tilings Most tessellations are made from regular polygons, which are polygons with edges of equal side length and vertices of equal angle. This allows the tessellation to have the same number of polygons meeting at each vertex. ### Quasi-regular Tilings You will notice that some hyperbolic tilings utilize two polygons instead of one polygon. These types of tilings are called quasi-regular tilings. The two polygons will alternate meeting at each vertex and are usually colored differently for emphasis. To describe a quasi-regular tiling, the Schläfli symbol is still used with an prefix quasi and with n being the number of edges of one polygon and k being the number of edges of one polygon: quasi-{n,k}. It's interesting to note that every regular tessellation can be made into a quasi-regular tessellation. You simply connect the midpoints of all the edges of the polygons in the tessellation. In the image gallery below, the first tessellation shown is a regular tessellation {5,4}. The second image shows how to connect the midpoints of the edges to create a quasi-{5,4} tessellation like the one shown in the third image of the gallery. The n in the quasi-regular tessellation now refers to the the four-sided squares, while the k now represents the five-sided pentagons alternating within the tessellation. ### Dual Tessellations There is another type of tessellation that occurs when the Schläfli symbols {n,k} are switched to {k,n}. This is called a dual tessellation, and an example can be found in the final image in the gallery: {4,5}. (Note, although the Schläfli symbols for the quasi- and duel tessellations are the same for this particular set of examples, they are very different tessellations). Hyperbolic Tiling {5,4} Making Quasi-{5,4} Quasi-{5,4} Dual {4,5} ## Tilings in Different Geometries There are various geometries that can be used to draw tilings, including in the Euclidean, hyperbolic, and elliptic planes. It is interesting to note that Euclidean geometry only allows for three regular tilings (using the regular polygons squares, triangles, and hexagons), while non-Euclidean geometry allows for an infinite amount of regular tilings. A simple way to determine which geometry a tiling can be draw in is by using its Schläfli symbol. Every regular tessellation can be completely broken up into triangles, because every regular polygon is made up of n-2 triangles, where n is the number of sides of the polygon. The Three Regular Euclidean Tessellations In Euclidean tessellations, the summation of angles in all triangles is 180 degrees, so the summation of angles of a n-gon is $(n-2)180\,$. and $\frac{n360}{k}$ is the summation of the angles of the polygon in a tilling then the following must be true: $\frac{1}{n} + \frac{1}{k} = \frac{1}{2}$ [Show Proof][Hide Proof] $\frac{n360}{k} = (n-2)180$ $\frac{n360}{k} = n180 - 360$ $\left( \frac{n360}{k} = n180 - 360 \right)\frac{1}{n360}$ $\frac{n360}{k (n360)} = \frac{n180}{(n360)} - \frac{360}{(n360)}$ $\frac{1}{k} = \frac{1}{2} - \frac{1}{n}$ $\frac{1}{n} + \frac{1}{k} = \frac{1}{2}$ Elliptic Tilings In hyperbolic tessellations, triangles always have a summation of angles less than 180 degrees, so that: $\frac{1}{n} + \frac{1}{k} < \frac{1}{2}$ In elliptic tessellations (also called spherical tessellations), triangles always have a summation of angles greater than 180 degrees, so that: $\frac{1}{n} + \frac{1}{k} > \frac{1}{2}$ # Teaching Materials There are currently no teaching materials for this page. Add teaching materials. Leave a message on the discussion page by clicking the 'discussion' tab at the top of this image page. [[Category:]] [[Category:]] |other=Hyperbolic Geometry |AuthorName=Jos Leys |AuthorDesc=Jos Leys creates images from mathematics using programs such as Ultrafractal and Povray. He has created numerous tessellations, fractals, and other images. |SiteName=Mathematical Imagery |SiteURL=http://www.josleys.com/show_gallery.php?galid=262 |Field=Geometry |Field2=Topology |FieldLinks=:Jos Ley's hyperbolic tilings gallery at http://www.josleys.com/show_gallery.php?galid=262 Also, Jos Ley's page on M.C. Echer's artistic works http://www.josleys.com/show_gallery.php?galid=325. Make Your Own Hyperbolic Tilings Applet at http://aleph0.clarku.edu/~djoyce/poincare/PoincareApplet.html |References= David E. Joyce, Hyperbolic Tessellations Wilhelm Magnus, Non-Euclidean Tessellations and their Groups (www.cims.nyu.edu/vigrenew/ug_research/JohnAdamski05.pdf) Wolfram MathWorld, Wolfram MathWorld Wikipedia, Uniform Tilings in Hyperbolic Plane |ImageRelates=This image is a quasi-regular hyperbolic tiling with Schläfli symbol of quasi-{7,3}, created using heptagons and triangles. This tessellation was created in the hyperbolic plane, so it must be visualized using the Poincaré disk model. |ToDo=*The Tilings in Different Geometries section should be expanded so that separate section for each geometry should be created and perhaps merged with the Tessellations page. • An applet for the allowing users to make their own tilings by specifying the Schläfli symbol (similar to this applet) |InProgress=No |HideMME=No }}

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http://math.stackexchange.com/questions/102959/predicate-logic-notation-where-to-put-the-parentheses-etc

# Predicate logic notation: where to put the parentheses, etc. My math professor tends to write $\exists x\in\mathbf{X} \ni P(x)$. Is this a correct use of the such that symbol $\ni$? If not, what is the use of that symbol? Isn't it better to write $\exists x\in\mathbf{X} (P(x))$ assuming $P(x)$ is a complex proposition? Also, while we're at it, is $\forall x\in\mathbf{X} : P(x)$ an acceptable notation or should it really be $\forall x\in\mathbf{X} (P(x))$ for precision? - 2 I'm guessing he means it as "such that." Often we just use a colon: $\exists x\in\mathbf{X}: P(x)$. You could put parentheses around it, as you do. Incidentally, even that is shorthand, since what it really means is: $\exists x(x\in\mathbf{X} \text{ and } P(x))$. – Thomas Andrews Jan 27 '12 at 13:54 This is semi-formal notation, at some distance from the standard ways of expressing the idea in a formal language. – André Nicolas Jan 27 '12 at 15:15 I would describe $\exists x\in\mathbf{X}\ni P(x)$, $\exists x\in\mathbf{X}: P(x)$, and $\exists x\in\mathbf{X}$ s.t. $P(x)$ as shorthand for an English sentence, There is an x in ... . $\exists x\in\mathbf{X}\big(P(x)\big)$ and $(\exists x\in\mathbf{X}) P(x)$, on the other hand, are expressions in slightly different formal languages, abbreviating the even more formal $\exists x\big(x\in\mathbf{X}\land P(x)\big)$ or $(\exists x)\big(x\in\mathbf{X}\land P(x)\big)$. – Brian M. Scott Jan 27 '12 at 15:29 ## 1 Answer The backwards epsilon means "such that", but in this context it's slightly bizarre since the usual set membership symbol appears symmetrically before the $\mathbf{X}$ and of course means something quite different. After all, if $P$ were a function symbol rather than a relation symbol then we might parse the statement quite differently, as $\exists{x} (x \in \mathbf{X} \wedge P(x) \in \mathbf{X})$. We can't see inside your professor's mind, but with usual usage in mind, your expansion is correct: it abbreviates $\exists{x \in \mathbf{X}} (P(x))$, which is in turn an abbreviation for $\exists{x} (x \in \mathbf{X} \wedge P(x))$. Universal quantifiers expand similarly: $\forall{y \in Y} \; \varphi(y)$ is short for $\forall{y} (y \in Y \rightarrow \varphi(y))$. Both colons and full stops (periods) are acceptable separators: $\exists{y \in Y} : \varphi(y)$ and $\forall{z \in Z} \; . \; \psi(z)$ are reasonably common. The key to using abbreviations effectively is to employ them to increase clarity. They increase brevity, but if using one might be ambiguous, go for the expanded version instead. I recommend reading things like Knuth on mathematical writing for further explanations of good practice in mathematical writing. -

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http://crypto.stackexchange.com/questions/tagged/group-theory+pairings

# Tagged Questions 2answers 156 views ### when do we need composite order groups for bilinear maps and when prime order? Why we need bilinear groups of composite order? What's the special security property of the composite order group in comparison with one of prime order?To put it in another way when do we need ... 1answer 126 views ### Must the order of the groups in a bilinear map be the same? I've been reading up on bilinear maps and their application to cryptography and one thing I keep seeing hasn't yet clicked. If $e:G_1\times G_2\to G_n$ is a bilinear map, $G_1,G_2,G_n$ are always ...

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http://stats.stackexchange.com/questions/4768/amoeba-interview-question/4773

# Amoeba Interview Question I was asked this question during an interview for a trading position with a proprietary trading firm. I would very much like to know the answer to this question and the intuition behind it. Thank you! Amoeba Question: A population of amoebas starts with 1. After 1 period that amoeba can divide into 1, 2, 3, or 0 (it can die) with equal probability. What is the probability that the entire population dies out eventually? - are we to suppose it does each of these with probability $1/4$? – shabbychef Nov 21 '10 at 5:40 yes, each outcome has a 1/4 probability. – AME Nov 21 '10 at 6:10 8 from a biological point of view, that chance is 1. The environment is bound to change to a point that no population can survive, given that in x billion years the sun is to explode. But I guess that's not really the answer he was looking for. ;-) The question doesn't make sense either. An amoebe can only divide into 2 or 0. Moral: traders shouldn't ask questions about biology. – Joris Meys Nov 21 '10 at 12:34 5 – mbq♦ Nov 21 '10 at 14:02 This is a cute question as Mike mentions. The intuition here is that the eventual survival/extinction probability is the same between two generations. A more creative version could be thought of when the survival probability itself varies as a function of the number amoeba present. I've added it to my site blog. – broccoli Aug 23 '12 at 2:33 ## 3 Answers Cute problem. This is the kind of stuff that probabilists do in their heads for fun. The technique is to assume that there is such a probability of extinction, call it $P$. Then, looking at a one-deep decision tree for the possible outcomes we see--using the Law of Total Probability--that $P=\frac{1}{4} + \frac{1}{4}P + \frac{1}{4}P^2 + \frac{1}{4}P^3$ assuming that, in the cases of 2 or 3 "offspring" their extinction probabilities are IID. This equation has two feasible roots, $1$ and $\sqrt{2}-1$. Someone smarter than me might be able to explain why the $1$ isn't plausible. Jobs must be getting tight -- what kind of interviewer expects you to solve cubic equations in your head? - 1 The reason 1 is not a root is easily seen by considering the expected number of Amoeba after $k$ steps, call it $E_k$. One can easily show that $E_k = E_1^k$. Because the probability of each outcome is $1/4,$ we have $E_1 = 3/2$, and so $E_k$ grows without bound in $k$. This clearly does not gibe with $P = 1$. – shabbychef Nov 21 '10 at 18:03 3 @shabbychef It's not so obvious to me. You can have the expectation grow exponentially (or even faster) while the probability of dying out still approaches unity. (For example, consider a stochastic process in which the population either quadruples in each generation or dies out entirely, each with equal chances. The expectation at generation n is 2^n but the probability of extinction is 1.) Thus there is no inherent contradiction; your argument needs something additional. – whuber♦ Nov 21 '10 at 21:02 @shabbychef -- thanks for the edit. I didn't realize we could use embedded TeX for math! @whuber -- shabbychef's statement $E_k = E_1^k$ is just a variation on my statement about the extinction probability, just add expectations instead of multiplying probabilities. Nice work, shab. – Mike Anderson Nov 21 '10 at 23:04 That's clear, Mike, but what's your point? Aren't we talking about how to rule out 1 as a solution? By the way, it's obvious (by inspection and/or by understanding the problem) that 1 will be a solution. This reduces it to a quadratic equation which one can easily solve on the spot. That's not usually the point of an interview question, though. The asker is probably probing to see what the applicant actively knows about stochastic processes, Brownian motion, the Ito calculus, etc., and how they go about solving problems, not whether they can solve this particular question. – whuber♦ Nov 22 '10 at 15:18 2 @shabbychef: One way to rule out P=1 is to study the evolution of the probability generating function. The pgf is obtained by starting with t (representing an initial population of 1) and iteratively replacing t by (1+t+t^2+t^3)/4. For any starting value of t less than 1, a graphic easily shows the iterates converge to Sqrt(2)-1. In particular, the pgf is staying away from 1, showing it cannot converge to 1 everywhere, which would represent complete extinction. This is why "the 1 isn't plausible." – whuber♦ Nov 22 '10 at 21:53 Some back of the envelope calculation (litterally - I had an envelope lying around on my desk) gives me a probability of 42/111 (38%) of never reaching a population of 3. I ran a quick Python simulation, seeing how many populations had died off by 20 generations (at which point they usually either died out or are in the thousands), and got 4164 dead out of 10000 runs. So the answer is 42%. - 3 $\sqrt{2}-1$ is 0.4142, so it is in agreement with Mike's analytical result. And +1, because I like simulations ;-) – mbq♦ Nov 21 '10 at 14:04 Also +1 because I like simulations. Which would have been my answer ;). – EpiGrad Aug 23 '12 at 4:22 This sounds related to the Galton Watson process, originally formulated to study the survival of surnames. The probability depends on the expected number of sub-amoebas after a single division. In this case that expected number is $3/2,$ which is greater than the critical value of $1$, and thus the probability of extinction is less than $1$. By considering the expected number of amoeba after $k$ divisions, one can easily show that if the expected number after one division is less than $1$, the probability of extinction is $1$. The other half of the problem, I am not so sure about. - ## protected by whuber♦Aug 23 '12 at 13:45 This question is protected to prevent "thanks!", "me too!", or spam answers by new users. To answer it, you must have earned at least 10 reputation on this site.

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http://www.physicsforums.com/showthread.php?t=106556

Physics Forums Thread Closed Page 1 of 2 1 2 > ## Universe Expansion If our universe is expanding and light travels at a finite speed, then, will there be a time in future, where we won't see any light from stars and our whole sky will be black, since all the lights have not reach us? PhysOrg.com science news on PhysOrg.com >> Front-row seats to climate change>> Attacking MRSA with metals from antibacterial clays>> New formula invented for microscope viewing, substitutes for federally controlled drug We should always see the current and possibly new stars in the future. I'm not sure but I beleive that near the beginning (when ever that was) there might have been no stars in the night sky. I think this because the universe has expanded faster then the speed of light. (Which I do not fully understand) The way I understand it; Almost all galaxies are recceding from us (very nearby galaxies excluded). The galaxies themselves are not moving as such, the space between the galaxies is increasing, and appears to be so at an ever faster rate. So in theory, one day we wont be able to see any distant galaxies. The stars in our own galaxy are gravitationally bound to the galaxy, and hence these stars are not receeding from us. So we will still be able to see the stars in our galaxy, at a time when we cannot see any other distant gallaxies. ## Universe Expansion If this were the case, wouldn't that mean that only "space" is expanding (as opposed to galactic regions)? If gravity can eventually overcome spacetime expansion, then doesn't that mean that the expansion is not a spacetime one but an addition of "outer space"? Conversely, if the entirety of spacetime is expanding, wouldn't it be correct to say that, at some point in the future (provided a long enough period has elapsed) my big toe will have moved further away from my pinky toe? Recognitions: Gold Member Science Advisor Note: It is the space-like hyper-surface slice through space-time that expands as the cosmological parameter t increases. Space-time does not 'expand' as it is static, time having been already accounted for within the continuum. The question is; "If space is expanding then what expands with it?" It is generally accepted that gravitational attraction of the local gravitational fields, of the Earth, Sun, Milky Way galaxy, and possibly the Local Group, overwhelm the cosmological expansion and these bodies do not expand with the universe. Einstein himself wrote a paper in the 1940's to prove that the solar system was not co-expanding with the universe. He did so by cutting out a spherical volume from the cosmological model and replacing it with a void with a spherical mass inserted in the middle; thus embedding a Schwarzschild solution inside a cosmological one. The question is how do you take the limit of the Schwarzschild metric as $r \rightarrow \infty$ It could be argued that the Pioneer Anomaly indicates that something might be wrong with this understanding. Garth Blog Entries: 9 Recognitions: Science Advisor Quote by touqra If our universe is expanding and light travels at a finite speed, then, will there be a time in future, where we won't see any light from stars and our whole sky will be black, since all the lights have not reach us? Let’s consider we are located in a spatially infinite universe which contains a homogeneous distribution of light sources in space and let’s assume that no new light sources are created nor destroyed as time passes. Consider a spherical volume with a given radius. Due to expansion of space, it is true that this volume will contain a decreasing number of light sources as time increases and that it will be void sometime. However, the light we are receiving is not determined by the number of light sources contained in this volume as $t \rightarrow \infty$, but by the shape of our past lightcone as $t \rightarrow \infty$. In my opinion the technical answer to your question is based on the fact that our past light cone will always intersect all past spatial hypersurfaces. This means that we are going to receive always light from every epoch in the universe (in our model). Consider a light source A from which we are receiving light emitted N years ago. As $t \rightarrow \infty$, we may not receive the light this source is currently emitting, if this source is located outside our current cosmological event horizon. However, we will receive light from some source located within our cosmological event horizon today (in the same spatial hypersurface than A). The problem is that, as $t \rightarrow \infty$, this source will be located very far in our past lightcone and therefore its light will be very redshifted. Summary: some radiation will always reach us. However, this radiation will not be visible light, due to the increasing redshift of all sources. So if I (being able to live forever without aging) were standing a meter away from a lamp (which also could go unchanging forever) - PRESUMING LOCAL GRAVITATION DID NOT INTERFERE WITH UNIVERSE EXPANSION - would the light from the lamp appear dimmer to me, say, a million or five hundred million years later? If it does, is this because the lamp is further away from me or because the strength of the light "dims" (which I guess would mean redshifts) over the course of millions of years? And would it thus be correct to say that light "dims" (in a non-relativistic or subjective sense) over time (again, observed time - not the actual light itself)? More over, the universe can only expand if it also redshifts, because to expand without redshifting (or contract without blueshifting, I guess) would be a violation of the total conservation of a finite universe's energy? Recognitions: Gold Member Science Advisor I totally agree with hellfire. In a casually connected universe, all light cones, once connected, will forever remain connected [albeit they may redshift beyond detectability over time]. And all light cones too remote to make initial causal contact with our universe prior to inflation will forever remain disconnected. Quote by jhe1984 So if I (being able to live forever without aging) were standing a meter away from a lamp (which also could go unchanging forever) - PRESUMING LOCAL GRAVITATION DID NOT INTERFERE WITH UNIVERSE EXPANSION - would the light from the lamp appear dimmer to me, say, a million or five hundred million years later? If it does, is this because the lamp is further away from me or because the strength of the light "dims" (which I guess would mean redshifts) over the course of millions of years? I don't know whether this have been implicitly answered anywhere, but I sure want to know the opinion to this. Recognitions: Gold Member Science Advisor Staff Emeritus Quote by jhe1984 So if I (being able to live forever without aging) were standing a meter away from a lamp (which also could go unchanging forever) - PRESUMING LOCAL GRAVITATION DID NOT INTERFERE WITH UNIVERSE EXPANSION - would the light from the lamp appear dimmer to me, say, a million or five hundred million years later? That question can't really be answered because it depends primarily on the local distribution of matter instead of the Hubble flow. If you were instead to ask about a "lamp" one hundred million light years away, then yes, it would appear dimmer. If we only consider cosmology, its apparent brightness will depend on its redshift. The more redshifted its light is: 1) The less total energy it carries. This is just because red photons are less energetic than blue ones. 2) The slower the photons arrive at our telescope. This is a GR effect that leads to the lamp's clock ticking slower than ours. 3) The further it will have travelled after one hundred million years. All of these things add up to a dimmer source. Remember, the faster the universe is expanding, the more the object is redshifted. Thus, the fact that the universe is accelerating means that the lamp will appear dimmer in a hundred million years. More over, the universe can only expand if it also redshifts, because to expand without redshifting (or contract without blueshifting, I guess) would be a violation of the total conservation of a finite universe's energy? Actually, energy is not always conserved in GR. The fact that light redshifts to begin with could be considered a violation of the energy conservation law, depending on your definition of "energy". See here for more details: Is Energy Conserved in General Relativity? Recognitions: Gold Member Science Advisor I feel compelled to vote in that quasi-poll, ST! I unwaveringly vote yes, energy is conserved when all is said and done . . . albeit with difficulty. Quote by SpaceTiger That question can't really be answered because it depends primarily on the local distribution of matter instead of the Hubble flow. If you were instead to ask about a "lamp" one hundred million light years away, then yes, it would appear dimmer. If we only consider cosmology, its apparent brightness will depend on its redshift. Actually, energy is not always conserved in GR. The fact that light redshifts to begin with could be considered a violation of the energy conservation law, depending on your definition of "energy". See here for more details: Is Energy Conserved in General Relativity? According to SR, the [Energy,momentum] pair could be thought as another way to describe the physical world by the other pair [displacement,velocity] "Energy" should be regarded as an operator, which should be conserved, if we are describing the physical model correctly. A physical model is "correct" because it has its own invariant, and the term "energy" could be defined as one of those invariants Recognitions: Science Advisor Staff Emeritus The FAQ and Space Tiger are correct on the issue of energy conservation in GR being problematic. Aside from the FAQ, there is a discussion of the history of energy conservation and Emily Noether's contribution in http://arxiv.org/PS_cache/physics/pdf/9807/9807044.pdf In general relativity, on the other hand, it has no meaning to speak of a definite localization of energy. One may define a quantity which is divergence free analogous to the energy-momentum density tensor of special relativity , but it is gauge dependent: i.e., it is not covariant under general coordinate transformations. Consequently the fact that it is divergence free does not yield a meaningful law of local energy conservation. Thus one has, as Hilbert saw it, in such theories ‘improper energy theorems’. A key feature for physics of Noether’s I.V. paper is the clarity her theorems brought to our understanding of the principle of energy conservation. As Feza Gursey wrote [18]: “Before Noether’s Theorem the principle of conservation of energy was shrouded in mystery, leading to the obscure physical systems of Mach and Ostwald. Noether’s simple and profound mathematical formulation did much to demystify physics.” Noether showed in her theorem I that the principle of energy conservation follows from symmetry under time translations. This applies to theories having a finite continuous symmetry group; theories that are Galilean or Poincare invariant, for example. In general relativity, on the other hand, energy conservation takes a different form as will be shown below. Noether’s theorem II applies in the case of general relativity and one sees that she has proved Hilbert’s assertion that in this case one has ‘improper energy theorems’, and that this is a characteristic feature of the theory. It is owing to the fact that the theory is a gauge theory; i.e., that it has an infinite continuous group of symmetries of which time translations are a subgroup. Indeed generally she defines as “improper” divergence relationships, which vanish when the field equations are satisfied, which correspond to a finite continuous subgroup of an infinite continuous group. Generally they do not have the required invariance or covariance properties under the larger group. For example, in general relativity a divergence free energy-momentum (pseudo) tensor can be constructed but it is gauge dependent (see below). Because it is not covariant under general coordinate transformations, it is more properly called a pseudotensor. Such pseudotensors are covariant with respect to the linear transformations of the Poincare group and may be used in asymptotic spacetime regions far from gravitating sources to derive a principle of energy conservation. The bottom line is that in GR we have theories of energy conservation that apply to static space-times, and theories of energy conservation that apply to asymptotically flat space-times, but there is no truly _general_ theorem of energy conservation in GR for arbitrary space-times. The FRW metric is neither asymptotically flat nor is it static, as a consequence there is no conserved "energy of the universe". This point is also made in most GR textbooks, including MTW's "Gravitation". Recognitions: Gold Member Science Advisor Quote by pervect In general relativity, on the other hand, it has no meaning to speak of a definite localization of energy. One may define a quantity which is divergence free analogous to the energy-momentum density tensor of special relativity , but it is gauge dependent: i.e., it is not covariant under general coordinate transformations. Therein lies the problem of a Quantum Gravity, may energy be represented by an operator or not? Does this not require a preferred foliation of space-time in contradiction to the principle of relativity? In order to be able to define such a locally conserved energy one has to select out a preferred foliation, slice, of space-time. SR is formulated for an empty, i.e. curvature free, universe, and there is nothing 'to hang' a selected frame on. However, once one has introduced curvature, i.e. proceeded from the SR limit into the more general GR situation, then matter or energy has been introduced into the system. Such matter/energy may be used to define a specific frame of reference - the Centre of Mass/momentum of the system or the frame in which radiation is globally isotropic. Thus Mach's Principle ought to be important in the local conservation of energy. This is the basis of Self Creation Cosmology. Garth Hi! Do scientists know why universe expands and even accelerates? Something is pushing or pulling? Recognitions: Gold Member Science Advisor hejs In the standard General Relativity (GR) cosmological theory the universe either has to expand or contract, it cannot 'stay still'. The Hubble red shift indicates that distant galaxies are all isotropically moving away from us. Either we live in a very special position in space or the universe as a whole is expanding. It is generally accepted that it is the universe as a whole that is expanding and that agrees with the prediction of GR. But what has it expanded from? If you follow GR all the way back you come to the singularity of the Big Bang in which the volume of the whole universe reduces to zero. Although before you reach that point quantum effects must come into play and they may well prevent the initial singularity from happening, instead the universe expanded from a very small and ultra dense volume, or 'bounced' from the collapse of a previous universe. What made it expand in the first place? The energy of the Big Bang was such that it had to expand, even in the face of a fierce cosmological gravitational field. Interestingly, in GR the kind of energy you need is provided by a negative pressure. This negative pressure shows up at two other places in the mainstream theory. 1. At an early stage of the Big Bang, 10-35 sec after BB, it is thought the universe underwent a burst or very powerful expansion called Inflation. 2. At a later stage of its expansion (at z ~ 1) it is thought the universe underwent an extended period of accelerated expansion called cosmic acceleration. Both would have been caused by two different kinds of negative pressure. Garth What do you mean by negative pressure? Could you give an analogy perhaps? Thread Closed Page 1 of 2 1 2 > Thread Tools | | | | |-----------------------------------------|------------------------------|---------| | Similar Threads for: Universe Expansion | | | | Thread | Forum | Replies | | | Astrophysics | 8 | | | Special & General Relativity | 12 | | | General Astronomy | 13 | | | General Astronomy | 6 | | | General Astronomy | 6 |

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http://mathematica.stackexchange.com/questions/tagged/polynomials

# Tagged Questions Questions on the functionality operating on polynomials learn more… | top users | synonyms (1) 3answers 102 views ### List of Tribonacci Polynomials with Mathematica? [duplicate] I want to list top ten of Tribonacci polynomials. I have following algorithm, but it doesnt work. ... 3answers 99 views ### The plot of roots of polynomials I have polynomial equation like Tribonacci Polynomials for example: $T_3(x)=x^4+x$. After finding the roots of this polynomial, I want to show these roots in the complex plane. I have tried lots of ... 1answer 48 views ### Table of Polynomials [closed] I made a function makePolynomial that creates a random Polynomial of a certain degree, e.g. 2+T+3T^2+8T^3-5T^4... now I ... 2answers 88 views ### What are Root objects with multiple polynomials? In Mathematica 9 a new flavor of Root object with multiple polynomials was introduced. For example, ... 2answers 76 views ### How to define a polynomial/function from an array of coefficients? I have the coefficients of my desired polynomial in an array CoefArr (I'm new to mathematica, so I think of everything as arrays, it is actually a list I believe) starting with the constant at index ... 3answers 142 views ### Symbolic manipulation of functional form I have a functional polynomial expression of the form: ... 2answers 82 views ### Evaluating Polynomials at Grid Points I am continuing my quest on B-splines. The code below builds a 5x5 matrix out of B-splines, using the BSplineBasis[] routine. I now want to evaluate the polynomials that are stored in each matrix ... 0answers 57 views ### Apart may use Padé method: what's that? How does Apart work? The page tutorial/SomeNotesOnInternalImplementation#7441 says, "Apart ... 5answers 284 views ### Series expansion in terms of Hermite polynomials I am trying to expand a polynomial in terms of orthogonal polynomials (in my case, Hermite). Maple has a nice built-in function for this, ChangeBasis. Is there a ... 2answers 165 views ### Calculating Taylor polynomial of an implicit function given by an equation I'd like to write a procedure that will take an equation: F(x,y,z) = 0 chosen variable: x a point: ... 4answers 133 views ### How to collect terms with positive powers in polynomial I am trying to collect all terms with non-negative powers of $x$ in polynomials like $\frac{1}{x^2}\left(a x^2+x^{\pi }+x+z\right)^2$ First expand the polynomial ... 7answers 215 views ### Defining a function that completes the square given a quadratic polynomial expression How can I write a function that would complete the square in a quadratic polynomial expression such that, for example, CompleteTheSquare[5 x^2 + 27 x - 5, x] ... 1answer 92 views ### Is there any way to force Mathematica to collect a symbol in a polynomial? Let's say that I have a polynomial like this: a + b + c Is there any way that I can get Mathematica to transform it to: ... 0answers 102 views ### Negative power instead of fraction Solve returns a solution in the form {{x->y/a^2 + y^2/a^7}}. Since I want to process the input (with another program) in terms of Laurent polynomials, I would ... 1answer 46 views ### Factorize and find the null space of a polynomial in several variables [duplicate] I've been asked to factor the following polynomial: poly = 6 x^3 + x^2 y - 11 xy^2 - 6 y^3 - 5 x^2 z + 11 xyz + 11 y^2 z - 2 xz^2 - 6 yz^2 + z^3 And to solve for z so that poly = 0 Can anyone help ... 3answers 118 views ### Convert coefficients of polynomials into a matrix I have several sets of 5 polynomials of the form: ... 0answers 90 views ### Computing Ehrhart's polynomial for a convex polytope Is there a Mathematica implementation for computing the Ehrhart polynomial of a convex polytope which is specified either by its vertices or by a set of inequalities? I am interested in knowing this ... 1answer 72 views ### Integrating polynomial functions over polytopes with an add-on package There is a Mathematica package to evaluate integrals over polytopes: http://library.wolfram.com/infocenter/Books/3652/ In the documentation (Functions.nb file) I ... 3answers 158 views ### How to evaluate all the essentially distinct polynomials in 4 variables over $F_2$ on points of $F_2 ^ 4$ I am a beginner with Mathematica. For my research purpose I would like to get a list of all the polynomials in $F_2[x,y,z,w]$ and for each polynomial I would like to know the result that it gives then ... 1answer 55 views ### FindFit and Integration errors First off, appologies for what may sound like a newbie question, as I am very new to using Mathematica. I am trying to find a way to get Mathematica to give me an expression that would describe the ... 1answer 175 views ### How do I get a two-term polynomial with a leading negative sign to display in the correct (i.e. textbook) order? The first three expressions evaluate as expected and the polynomial is displayed in what I would call "textbook" form. The last expression, however, switches the order of terms. Mathematica employs ... 1answer 93 views ### How to transform an expression using algebraical instead of pattern rules [duplicate] I would like to transform rules algebraically. A very simple example would be: - k^2 - 2 k x + x^2 /. {2*k -> 1} This transforms to: - $$k^2-2 k x+x^2$$ ... 1answer 124 views ### Expanding a polynomial with fractional powers Given an expression like a + b*y + c*y^2 + d*Sqrt[f + g*y + h*y^2] How can I programatically, expand this to a quartic without any fractional powers? Right ... 1answer 89 views ### How can I prevent a polynomial from being simplified? I'm having a problem with polynomials. Let's say I have a polynomial "2x^2 - 5x + 6 - 3x^2" .. How can I check that this expression is not simplified ? Additionally, I would like to locate the ... 1answer 90 views ### How can I get an exponent vector from monomials? I am trying to get an exponent vector from a list of monomials. I am using the CoefficientRules command; however, it is returning a list that includes the ... 3answers 436 views ### Solving cubic equation for real roots I'm looking to solve the following cubic equation for x: $\beta\, x^3 - \gamma \,x = c$. I have plugged in some sample values ($\beta = 2$, $\gamma = 5$ and $c = 2$). When I try to solve this ... 1answer 117 views ### Implementation of Decompose I'm curious as to how Decompose works so I decided to use Trace with the option ... 2answers 190 views ### How can I make the output from Solve look nice? I have a problem with presenting solutions. Roots of 4th order polynomials are big expressions. Is there a way to present the roots, s2 and s3, in normal form with some substitutions? Maybe a way to ... 0answers 25 views ### Handling “Solve was unable to solve the system with inexact coefficient” errors [duplicate] Possible Duplicate: How to get rid of warnings when using Solve on an equation with inexact coefficients? I've been trying to calculate the following in Mathematica: ... 4answers 511 views ### How to get exact roots of this polynomial? The equation $$64x^7 -112x^5 -8x^4 +56x^3 +8x^2 -7x - 1 = 0$$ has seven solutions $x = 1$, $x = -\dfrac{1}{2}$ and $x = \cos \dfrac{2n\pi}{11}$, where $n$ runs from $1$ to $5$. With ... 3answers 375 views ### First positive root Simple question but problem with NSolve. I need help how to extract first positive root? For example eq=-70.5 + 450.33 x^2 - 25 x^4; NSolve[eq== 0, x] If I ... 0answers 78 views ### Know the degree of the equation with the radicals expanded I have an equation with radicals. I would like to know what would be the degree of the polynomials if I'd move the terms and square the equation a number of times sufficient to remove all the ... 1answer 155 views ### How to do the polynomial stuff over finite fields extensions fast? This question is raised from the problem of package FiniteFields being very slow (please, see the corresponding question): I have had an evidence that Mathematica ... 2answers 196 views Equations ... 3answers 257 views ### How to keep Collect[] result in order? For example, Collect[(1 + x + Cos[s] x^2)^3, x] gives the result ... 1answer 187 views ### Generating lots of Examples in Polynomials Rings I'm studying polynomial rings and i would like to know some tricks for generating lots of examples. For instance, suppose i'm interested in polynomials over the integers mod (2,x^3 + 1). To get a ... 4answers 315 views ### How do I find the degree of a multivariable polynomial automatically? I have a very simple question which appears not to have already been answered on this forum. Is there built-in functionality that returns the degree of a multivariable polynomial? For example if the ... 1answer 92 views ### Factorize Parametric Polynomials Is there a possibility to factorize a parametric polynomial expression - meaning that the coefficients are defined as parameters, and not as specific numbers? My example - a polynomial in ... 1answer 157 views ### Small Issue with Chebyshev Derivative Appoximation I am trying to get approximate the derivative of a function from its Chebyshev expansion. I start out with the following random function ... 3answers 977 views ### Get polynomial interpolation formula I'm attempting to get a polynomial interpolation formula out of Mathematica but I am absolutely lost. I stared out using ... 2answers 265 views ### How to deduce a generator formula for a polynomial sequence? Consider a polynomial sequence $\{p_n\}$ generated by some (simple) rule: \begin{array}{l} p_1(x)=x \\ p_2(x)=2 x-x^2 \\ p_3(x)= x^3-3 x^2+3 x \\ p_4(x)=-x^4+4 x^3-6 x^2+4 x \\ p_5(x)= x^5-5 ... 1answer 324 views ### Polynomial Approximation from Chebyshev coefficients I would like to expand a function $f(r)$ in the domain $[0,R]$, around the points $r =0$, and $r = R$ in the following manner $f(r = 0) = \Sigma_{i=0,i = even}^{imax} f_i (r/R)^i$ and \$f(r = R) = ... 2answers 242 views ### 3D Plot: Number of Roots in x of a polynomial in x, a, b and c I have a polynomial in four variables x,a,b and c. The number of roots of the polynomial in x depends of the choice of a, b and c. I would like to have a 3D-Plot with a, b and c on the axes, while the ... 2answers 188 views ### How can I compute the representation matrices of a point group under given basis functions? Take the $C_{3v}$ point group for example: ... 3answers 211 views ### Is it possible to use Composition for polynomial composition? I want to do this: $P = (x^3+x)$ $Q = (x^2+1)$ $P \circ Q = P \circ (x^2+1) = (x^2+1)^3+(x^2+1) = x^6+3x^4+4x^2+2$ I used Composition for testing if that could ... 5answers 367 views ### How do I reassign canonical ordering of symbols? I have a big polynomial that evaluates to: $$A^2 e^2 \phi ^- \phi ^++A e \phi ^- \phi ^+ c_{2 w} g_Z+\frac{1}{2} A e g h W^- \phi ^+ +\ll13\gg,$$ which is supposed to represent some terms in the ... 2answers 139 views ### Strange integration result In Mathematica 8, the Integrate command sometimes strangely integrates polynomials yielding unsimplified (and unexpected) fractional results. As an example, the line: ... 1answer 131 views ### How to express the original ideal elements in the Groebner basis? Suppose I call GroebnerBasis[{f1, f2, ...}, {x1,x2, ...}] The output is a list {g1,g2,...} For each $g_j$, there should be ... 4answers 299 views ### “Evaluating” polynomials of functions (Symbols) I want to implement the following type evaluation symbolically $$(f^2g + fg + g)(x) \to f(x)^2 g(x) + f(x) g(x) + g(x)$$ In general, on left hand side there is a polynomial in an arbitrary number of ... 3answers 460 views ### What function can I use to evaluate $(x+y)^2$ to $x^2 + 2xy + y^2$? What function can I use to evaluate $(x+y)^2$ to $x^2 + 2xy + y^2$? I want to evaluate It and I've tried to use the most obvious way: simply typing and evaluating $(x+y)^2$, But it gives me only ...

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http://unapologetic.wordpress.com/2011/08/05/chains/?like=1&source=post_flair&_wpnonce=420e8d64d7

# The Unapologetic Mathematician ## Chains We can integrate on the standard cube, and on singular cubes in arbitrary manifolds. What if it’s not very easy to describe a region as the image of a singular cube? This is where chains come in. So a chain is actually pretty simple; it’s just a formal linear combination of singular $k$-cubes. That is, for each $k$ we build the free abelian group $C_k(M)$ generated by the singular $k$-cubes in $M$. If we have a formal sum $c=a_1c_1+\dots+a_lc_l$ — the $c_i$ are all singular $k$-cubes and the $a_i$ are all integers — then we define integrals over the chain by linearity: $\displaystyle\int\limits_c\omega=a_1\int\limits_{c_1}\omega+\dots+a_l\int\limits_{c_l}\omega$ And that’s all there is to it; just cover the $k$-dimensional region you’re interested in with singular $k$-cubes. If there are some overlaps, those areas will get counted twice, so you’ll have to cover them with their own singular $k$-cubes with negative multipliers to cancel them out. Take all the integrals — by translating each one back to the standard $k$-cube — and add (or subtract) them up to get the result! ### Like this: Posted by John Armstrong | Differential Topology, Topology ## 4 Comments » 1. [...] that we’re armed with chains — formal sums — of singular cubes we can use them to come up with a homology theory. [...] Pingback by | August 9, 2011 | Reply 2. [...] anyway, on to the theorem! We know how to integrate a differential -form over a -chain . We also have a differential operator on differential forms [...] Pingback by | August 17, 2011 | Reply 3. [...] defined how to integrate forms over chains made up of singular cubes, but we still haven’t really defined integration on manifolds. [...] Pingback by | September 5, 2011 | Reply 4. [...] sensible to identify an orientation-preserving singular cube with its image. When we write out a chain, a positive multiplier has the sense of counting a point in the domain more than once, while a [...] Pingback by | September 8, 2011 | Reply « Previous | Next » ## About this weblog This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”). I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.

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http://mathoverflow.net/questions/63129/how-do-you-calculate-the-solid-angle-of-a-rectangular-axis-aligned-section-of-a

## How do you calculate the solid angle of a rectangular, axis aligned section of a surface defined by a two dimensional function? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I have $f(x,y) = \frac{1}{2} (1 - x^2 - y^2)$, which is a paraboloid centered around the origin (plot). Now I want to calculate the solid angle (with the origin as the viewpoint) of the surface area defined by f(x,y) with a rectangular, axis aligned section of the xy plane as its input, e.g. $-1 \leq x \leq 1$ and $-1 \leq y \leq 1$. So each point of the surface area has the coordinates `$\vec f(x,y) = ( x, y, f(x, y)) = ( x, y, \frac{1 - x^2 - y^2}{2} )$`. The problem is that I don't know how I convert the Integral over the Surface $\iint_S dS$ in `$\Omega = \iint_S \frac { \vec{r} \cdot \hat{n} \,dS }{r^3}$` into an Integral over $x$ and $y$: $\int_X \int_Y dx dy$. ### If I... ...simply replace it with $\int_{-1}^{1} \int_{-1}^{1} dx dy$, replacing the other values accordingly with • each point on the surface: `$\vec r = \vec f(x, y) = (x, y, \frac{1 - x^2 - y^2}{2})$`, • normal at each point on the surface: `$\hat{n} = \frac{\vec f_x \times \vec f_y} {|\vec f_x \times \vec f_y|} = (x, y, 1) $` with `$\vec f_x(x, y) = (1, 0, -x)$` and `$\vec f_y(x,y) = (0, 1, -y)$`, I receive `$\Leftrightarrow \int_x \int_y \frac{ \vec f(x, y) \cdot (x,y,1) } { |\vec f(x, y)|^3 \cdot |(x, y, 1)|} $` `$\Leftrightarrow \int_x \int_y \frac{ (x, y, \frac{1 - x^2 - y^2}{2}) \cdot (x, y 1)} {|(x, y, \frac{1 - x^2 - y^2}{2})|^3 \cdot |(x, y, 1)|}$` `$\Leftrightarrow \int_x \int_y \frac{ x^2 + y^2 + \frac{1 - x^2 - y^2}{2} } { ( x^2 + y^2 + (\frac{1 - x^2 - y^2}{2})^2)^{\frac{3}{2}} \cdot (x^2 + y^2 + 1)^{\frac{1}{2}} }$` If I let wolframalpha calculate[1] that, the result is $5.87$. This is clearly wrong though, because the paraboloid covers more than the hemisphere, and thus needs to have a solid angle of more than $2\pi$. So what do I need to change? ### Background I'm using this paraboloid as a mapping to project geometry into a texture, and for the next step I need to find out what portion of the hemisphere each pixel covers. So ideally I need a way to calculate this as fast as possible - I might need to search for a similar, faster function for actual usage. [1] http://www.wolframalpha.com/input/?i=Integrate+(x%5e2+%2b+y%5e2+%2b+(1-x%5e2-y%5e2)%2f2)+%2f+((x%5e2+%2b+y%5e2+%2b+((1-x%5e2-y%5e2)%2f2)%5e2)%5e(3%2f2)+*+(x%5e2+%2b+y%5e2+%2b+1)%5e(1%2f2))+from+x%3d-1+to+1+y%3d-1+to+1&incParTime=true - I edited the tags - this doesn't fall under topology or math. phys. as far as I can see. – David Roberts Apr 27 2011 at 11:34 ## 3 Answers You need to include the differential surface area in your parametrized version of the integral. In effect, you replace the $\hat{n} dS$ term with $$\frac{\vec{f}_x\times\vec{f}_y}{\|\vec{f}_x\times\vec{f}_y\|} {\|\vec{f}_x\times\vec{f}_y\|} dxdy.$$ Although, really the two norms just cancel, so you needn't calculate them. Your integral then becomes $$\int_x \int_y \frac{( x,y,\frac{1-x^2-y^2}{2})}{(x^2+y^2+(\frac{1-x^2-y^2}{2})^2)^{(3/2)}}\cdot( x,y,1) dx dy.$$ This simplifies to $$\int_x \int_y \frac{4}{(1+x^2+y^2)^2} dx dy.$$ WolframAlpha gives the value of this as about 6.96336. - ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. This is not an answer, just another way of viewing the calculation. You need only compute the area of the roughly one-eighth of the sphere that falls below the equator (green below) and beneath the line formed by the origin and the curve $(x,1,-x^2 /2)$ for $x\in[-1,1]$, tracing out one boundary curve. The portion outside your solid angle is composed of four of these regions. - I'm afraid I need to calculate the solid angle for arbitrary sections/intervals. – hrehf Apr 27 2011 at 16:39 The situation differs little with arbitrary intervals. Still there are curves on the sphere bounding the area you want to measure. If you need an exact answer, you will need to explicitly compute those curves. If you are content with a numerical approximation, you could estimate that area in many ways. – Joseph O'Rourke Apr 27 2011 at 18:03 The normal to the paraboloid plays no role in this. A "surface element" ${\rm d}(x,y)$ at the point $(x,y)$ in the $(x,y)$-parameter plane produces via $\vec f$ (or rather $\vec f_*$) a surface element $dS$ at the point $\vec f(x,y)$ on your paraboloid $S$, and then this surface element $dS$ casts a shadow $d\omega$ on the unit sphere $S^2$ through central projection from $O$, i.e., via normalization of $\vec f$. Since $$|\vec f(x,y)|^2=x^2+y^2+{1\over4}(1-x^2-y^2)^2={1\over4}(1+x^2+y^2)^2$$ it follows that the shadow on $S^2$ is produced by the map $$\vec g: \quad (x,y) \mapsto {2\over 1+x^2+y^2} \bigl(x,y,{1\over2}(1-x^2-y^2)\bigr)\ .$$ This $\vec g$ is nothing else but an (unusual) parametric representation of $S^2$. In order to compute the area of the shadowed part of $S^2$ one has to compute $d\omega=|g_x\times g_y|{\rm d}(x,y)$ and to integrate this over the intended rectangle in the $(x,y)$-plane. The computation gives, as already remarked by Ben, $$d\omega={4\over(1+x^2+y^2)^2}{\rm d}(x,y)\ .$$ Transforming to polar coordinates one finds for the $[-1,1]^2$-rectangle the exact value $8\sqrt 2\ \arctan(1/\sqrt 2)\doteq 6.96366$. - The shadowing makes sense to me and fits my view of the problem. I see that an alternate (?) view of the problem is that we need to integrate over the area of the projection of $S$ onto the unit sphere. However this confuses me even more, because I'm not sure what we need to do in order to represent the projection/transformation/etc. in the first place. Also, I do not see how you arrived at $\frac{2}{1 + x^2 + y^2} \cdot \vec f$, can you explain some more? Is this the complete integral that gives us the solid angle? – hrehf Apr 27 2011 at 16:37 @href: You were lucky: x^2+y^2+{1/over4}(1-x^2-y^2)^2={1/over4}(1+x^2+y^2)^2\$. – Christian Blatter Apr 27 2011 at 22:33

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http://www.oilfieldwiki.com/wiki/Archimedes'_principle

# Archimedes' principle From Oilfield Wiki - the Oilfield Encyclopedia Archimedes (287 BC - 212 BC), the discoverer of this principle Archimedes' principle is a law of physics stating that the upward force (buoyancy) exerted on a body immersed in a fluid is equal to the weight of the amount of fluid the body displaces. In other words, an immersed object is buoyed up by a force equal to the weight of the fluid it actually displaces. Archimedes' principle is an important and underlying concept in the field of fluid mechanics. This principle is named after its discoverer, Archimedes of Syracuse.[1] ## Principle Archimedes' treatise "On Floating Bodies" states that: Any object, wholly or partially immersed in a fluid, is buoyed up by a force equal to the weight of the fluid displaced by the object. — Archimedes of Syracuse with the clarifications that for a sunken object the volume of displaced fluid is the volume of the object. Thus, in short, buoyancy = weight of displaced fluid. Archimedes' principle is true of liquids and gases, both of which are fluids. If an immersed object displaces 1 kilogram of fluid, the buoyant force acting on it is equal to the weight of 1 kilogram (technically, as a kilogram is unit of mass and not of force, the buoyant force is the weight of 1 kg, which is 9.8 Newtons.) It is important to note that the term immersed refers to an object that is either completely or partially submerged. If a sealed 1-liter container is immersed halfway into the water, it will displace a half-liter of water and be buoyed up by a force equal to the weight of a half-liter of water, no matter what is in the container. If such an object is completely immersed (submerged), it will be buoyed up by a force equivalent to the weight of a full liter of water (1 kilogram of mass). If the container is completely submerged and does not compress, the buoyant force will equal to the weight of 1 kilogram of water at any depth. This is due to the fact that at any depth, the container can displace no greater volume of water than its own volume. The weight of this displaced water (not the weight of the submerged object) equals the buoyant force. Thus, for objects floating or sunken, Archimedes' principle may be stated in terms of forces: ### Floatation Archimedes' principle relates buoyant force and displacement of fluid. However, the concept of Archimedes' principle can be applied when considering why objects float. Proposition 5 of Archimedes' treatise On Floating Bodies states that: Any floating object displaces its own weight of fluid. — Archimedes of Syracuse[2] In other words, for a floating object on a liquid, the weight of the displaced liquid is the weight of the object. Thus, only in the special case of floating does the buoyant force acting on an object equal the objects weight. Consider a 1-ton block of solid iron. As iron is nearly eight times denser than water, it displaces only 1/8 ton of water when submerged, which is not enough to keep it afloat. Suppose the same iron block is reshaped into a bowl. It still weighs 1 ton, but when it is put in water, it displaces a greater volume of water than when it was a block. The deeper the iron bowl is immersed, the more water it displaces, and the greater the buoyant force acting on it. When the buoyant force equals 1 ton, it will sink no farther. When any boat displaces a weight of water equal to its own weight, it floats. This is often called the "principle of floatation": A floating object displaces a weight of fluid equal to its own weight. Every ship, submarine, and dirigible must be designed to displace a weight of fluid equal to its own weight. A 10,000-ton ship must be built wide enough to displace 10,000 tons of water before it sinks too deep in the water. The same is true for vessels in air (as air is a fluid): a dirigible that weighs 100 tons displaces at least 100 tons of air. If it displaces more, it rises; if it displaces less, it falls. If the dirigible displaces exactly its weight, it hovers at a constant altitude. ## Refinements Archimedes' principle does not consider the surface tension (capillarity) acting on the body.[3] ## Formula The weight of the displaced fluid is directly proportional to the volume of the displaced fluid (if the surrounding fluid is of uniform density). In simple terms, the principle states that the buoyant force on an object is going to be equal to the weight of the fluid displaced by the object, or the density of the fluid multiplied by the submerged volume times the gravitational constant, g. Thus, among completely submerged objects with equal masses, objects with greater volume have greater buoyancy. Suppose a rock's weight is measured as 10 newtons when suspended by a string in a vacuum with gravity acting upon it. Suppose that when the rock is lowered into water, it displaces water of weight 3 newtons. The force it then exerts on the string from which it hangs would be 10 newtons minus the 3 newtons of buoyant force: 10 − 3 = 7 newtons. Buoyancy reduces the apparent weight of objects that have sunk completely to the sea floor. It is generally easier to lift an object up through the water than it is to pull it out of the water. Assuming Archimedes' principle to be reformulated as follows, $\text{apparent immersed weight} = \text{weight} - \text{weight of displaced fluid}\,$ then inserted into the quotient of weights, which has been expanded by the mutual volume $\frac { \text{density}} { \text{density of fluid} } = \frac { \text{weight}} { \text{weight of displaced fluid} }$ yields the formula below. The density of the immersed object relative to the density of the fluid can easily be calculated without measuring any volumes: $\frac { \text {density of object}} { \text{density of fluid} } = \frac { \text{weight}} { \text{weight} - \text{apparent immersed weight}}\,$ (This formula is used for example in describing the measuring principle of a dasymeter and of hydrostatic weighing.) Example: If you drop wood into water, buoyancy will keep it afloat. Example: A helium balloon in a moving car. In increasing speed or driving a curve, the air moves in the opposite direction of the car's acceleration. The balloon however, is pushed due to buoyancy "out of the way" by the air, and will actually drift in the same direction as the car's acceleration. When an object is immersed in a liquid the liquid exerts an upward force which is known as buoyant force and it is proportional to the weight of displaced liquid. The sum force acting on the object, then, is proportional to the difference between the weight of the object ('down' force) and the weight of displaced liquid ('up' force), hence equilibrium buoyancy is achieved when these two weights (and thus forces) are equal. ## See also • Buoyancy • "Eureka", reportedly exclaimed by Archimedes upon discovery that the volume of displaced fluid is equal to the volume of the submerged object ## References 1. {{#invoke:Citation/CS1|citation |CitationClass=journal }} 2. "The works of Archimedes". p. 257. Retrieved 11 March 2010. "Any solid lighter than a fluid will, if placed in the fluid, be so far immersed that the weight of the solid will be equal to the weight of the fluid displaced." ar:مبدأ أرخميدس ca:Principi d'Arquimedes cs:Archimédův zákon de:Archimedisches Prinzip et:Archimedese seadus el:Αρχή του Αρχιμήδη es:Principio de Arquímedes eu:Arkimedesen printzipioa fr:Poussée d'Archimède hy:Արքիմեդի օրենք hr:Arhimedov zakon it:Principio di Archimede he:חוק ארכימדס ka:არქიმედეს კანონი kk:Архимед заңы sw:Kanuni ya Archimedes lv:Arhimēda spēks lt:Archimedo dėsnis hu:Arkhimédész törvénye nl:Wet van Archimedes ja:アルキメデスの原理 no:Arkimedes' prinsipp nn:Arkimedes-lova pms:Prinsipi d'Archimede pl:Prawo Archimedesa ro:Forță arhimedică ru:Закон Архимеда sq:Ligji i Arkimedit si:ආකිමිඩීස් නියමය simple:Archimedes' principle sk:Archimedov zákon sl:Arhimedov zakon sr:Архимедов закон sv:Arkimedes princip ta:ஆர்க்கிமிடீசு தத்துவம் th:หลักการของอาร์คิมิดีส tr:Arşimet prensibi uk:Закон Архімеда vi:Lực đẩy Archimedes wo:Dalu Archiede zh:阿基米德浮體原理

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http://mathoverflow.net/questions/47523?sort=oldest

Example of: K-regular graph with girth K, for a given K Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I've been reading some papers about (n,g)cages, but it doesn't seem trivial to me that there is, indeed, for example, a K-regular graph with girth K, for large Ks. So: Can you give me an example of a K-regular graph (it may be a graph with minimum degree K, if it's easier) and girth K (it may be >= K, if it's easier) for a given integer K. Thanks in advance. - 3 Answers In this paper http://onlinelibrary.wiley.com/doi/10.1002/jgt.3190070210/abstract even more strict theorem is proved. If I remember well, the example you are asking about belongs to Tutte. - I did not look closely, but it seems that that article starts with a given graph have a certain degree and minimal girth and uses it to make something meeting more conditions (on even and odd girth) – Aaron Meyerowitz Dec 19 2010 at 0:26 You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. It is one of the problems for which Random Graph theory seems amazingly powerful. Bollobas gives a very short (non-deterministic) construction of graphs with "minimum degree at least K" and "girth at least g". He's actually creating a large random graph with few short cycles, then removing an edge from each of them to break them all. As the average degree of his graph is more than 2K, he then removes vertices of degree less than K until the minimum degree is large enough. You can read all about that in "Extremal Graph Theory (Bollobas)", chapter 3.1 : "Graphs with large minimum degree and large girth" (he also gives a construction of K-regular graphs of girth "at least something") Nathann - I am not sure this construction originates with Bollobas; a closely related construction (high chromatic number, high girth) goes back to Erdos. – Qiaochu Yuan Dec 18 2010 at 3:39 The easiest way to do this is as said the probabilistic method. However, for those who prefer non-random constructions, here is a greedy method. Choose some large $n$ (you can calculate what is needed easily enough). Start with the empty $n$-vertex graph. Add edges greedily between vertices subject to two conditions. First, the vertices you join should always have distance at least $K$ in the current graph (if not connected, then assume distance is infinite). Second, both vertices should have degree at most $K-1$. When this procedure is forced to terminate for lack of such pairs, you have a graph with maximum degree $K$ and girth at least $K$. Now take any vertex $v$ of degree less than $K$. Look at all the vertices at distance less than $K$ from $v$ (including $v$). This set must include all the vertices of degree less than $K$, or you would not have terminated. But the set has at most $1+K+K(K-1)+K(K-1)^2+..+K(K-1)^{K-1}=C$ vertices, since the maximum degree is $K$. Similarly, the set of vertices at distance at most $2K$ from $v$ is bounded, and we can presume there are at least $KC^2$ vertices at distance more than $2K$ from $v$. Now joining greedily vertices of degree less than $K$ to these far-away vertices greedily without creating short cycles must succeed (each edge added blocks at most $C$ vertices, and there are certainly not more than $CK$ edges required). To make this have girth exactly $K$, start with a $K$-cycle. To make it regular is a little harder: one option is to run the first procedure (starting with a $K$-cycle which we insist on preserving forever, to fix the girth) with a much higher distance requirement to join two edges (say $3K$), then after termination, identify a low-degree vertex $u$ and adding an edge to some far-away $v$ (as before) then removing some edge $vw$. Now $w$ cannot have been (before edge removal) a low degree vertex (it is too far away from the first low degree vertex) and furthermore it cannot be within distance $K$ of any remaining low degree vertex, so you can join it to a remaining low degree vertex. Rinse, repeat, assuming $n$ satisfies the parity condition for a $K$-regular graph to exist (and is large enough) you will succeed. -

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http://physics.stackexchange.com/questions/tagged/temperature

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I'm sorry if this was asked before but with all the stars and dark matter and all the other stuff, curently in the Universe, what's the avarage temperature of the Universe? Is it like extremely high ... 5answers 3k views ### How long can you survive 1 million degrees? I asked my Dad this once when I was about 14, and he said that no matter how short the amount of time you were exposed to such a great temperature, you would surely die. The conversation went ... 0answers 50 views ### Temperature of an Object in Space [closed] Rotating cylinder in space Hi all I've been having problems trying to calculate the temperature of an object in space, and was hoping I could get some help. Say we have a cylinder in space rotating ... 0answers 43 views ### How to solve state parameters using these givens for an ideal gas? 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In general, the energy is $kT$ times the ... 1answer 97 views ### Does the Kelvin have a rigorous definition? From Wikipedia: The kelvin is defined as the fraction 1⁄273.16 of the thermodynamic temperature of the triple point of water. That presupposes that we can take a fraction of temperature. Now, ... 2answers 136 views ### What's the basic difference between heat and temperature? Temperature is usually seen as a calibrated representation of heat but what about latent heat? Eg. Ice and water have different amounts of heat at 0 degree c. 2answers 112 views ### Room temperature and fan orientation So I'm in a tiny dorm room and I normally point my fan blowing outside the window to cool my room off. I've been in some debates on blowing air out or in is more effective, so I'm hoping to get some ... 1answer 99 views ### Does increasing the density of a solution decrease the rate of temperature change? I did an experiment to compare whether salt water (5% concentration of salt) or fresh water of the same volume took longer to heat up to a certain temperature. We found that salt water took longer to ... 1answer 57 views ### Predict final temperature by taking temperature samples? Is it possible to predict what the final temperature will be by taking temperature samples. For example, an object is 0ºC and moved to a room above 0ºC. I'm taking temperature of the object using a ... 3answers 191 views ### Is rate of temperature change constant? Is the rate of change in temperature for an object constant? For example, from 0ºC to 25ºC, or from 25ºC to -10ºC? Does it take the same amount of time to increase temperature from 1º to 2º as 24º to ... 2answers 172 views ### Ideal gas concentration under temperature gradient I'm trying to calculate the concentration of an ideal gas in an adiabatic container as a function of position where the top and bottom plates of the container are fixed at temperatures $T_1$ and ... 3answers 198 views ### Does sound propagate further in freezing weather? A few days ago I went for a walk in the evening. We're having winter with a little snow and freezing temperatures. We're in a quiet, shallow valley with a train station about 1km from us. I heard a ... 3answers 122 views ### Temperature in space Temperature is a measure of kinetic energy transferred to particles, henceforth, space being vacuum, temperature cannot be measured. But then, there is cosmic background radiation. It is the leftover ... 1answer 100 views ### Does electric potential have a temperature? When I took my first thermo class a tucked away chapter introduced Exergy in terms of electrical energy, meaning that the amount of electrical energy you could get from something is functionally its ... 2answers 135 views ### With ideal gases, varying quantity of moles, and having a constant volume how do temperature and pressure behave? I'm trying to build a simulation of gases so I ended-up trying to use law of ideal gases ($PV = nRT$). In my scenario: volume is constant ($V=1\rm{m}^3$); a known quantity of moles are being added ... 1answer 128 views ### Why aren't two systems in thermal equilibrium the same as one system? I am reading Molecular Driving Forces, 2nd ed., by Dill & Bromberg. On page 53, example 3.9, we consider why energy exchanges between two systems from the point of view of the 2nd law. We ... 1answer 149 views ### How to simulate temperature change of oven? I am trying to write a software, which will model the oven temperature change when turning on/off. The data I can get is graph, by taking temperature reading each second from T0 time up to some ... 0answers 55 views ### How can I read density fluctuation from microwaves? The Cosmic Microwave Background Radiation shows temperature differences. The red and yellow areas are warmer. The green and blue areas are cooler. For example consider this picture of CMBR ... 2answers 92 views ### How to get “real-time” temperature from sensor? The following is LM35 Thermal response time in air The following is temperature reading from LM35 sensor. Horizontal axis is time in sec. So this is not "real-time" temperature graph. The ... 3answers 296 views ### Less than absolute zero possible? [duplicate] Possible Duplicate: Temperature below absolute zero? According to this article http://www.sciencemag.org/content/339/6115/52 (preprint: http://arxiv.org/abs/1211.0545) it is. What do you ... 3answers 260 views ### Could temperature have been defined as $-\partial S/\partial U$? When coming up with a definition of temperature, it's typical to start with an empirical definition that a system with a hotter temperature tends to lose heat to a system with a colder temperature. ... 1answer 115 views ### How can gas from compressed air can “take” heat from surrounding environment? I have recently been reading about why a can of compressed air gets cold when the air it contains is discharged. From what I understand the change from a liquid to a gas requires energy and therefore ...

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http://stats.stackexchange.com/questions/20870/is-there-a-fast-algorithm-to-check-for-arp-stationarity/26822

# Is there a fast algorithm to check for AR(p) stationarity? It is well-known that an AR(p) process $$x_t=\sum_{i=1}^p \varrho_i x_{t-i} + \epsilon_t \,,$$ is causal and stationary if and only if the roots of the polynomial $$\mathcal{P}(u) = 1 - \sum_{i=1}^p \varrho_i u^i$$ are all outside the unit circle in the complex plane. (Here is a Cross Validated discussion on the topic.) My question is whether or not there exists a faster algorithm than the naïve one that consists in (a) finding the roots of $\mathcal{P}$ and (b) checking none one them is inside the unit circle. - This is a mathematical question. – vinux Jan 10 '12 at 12:53 1 @vinux: It is more of a computational question in that I am looking for the smallest power $d$ such that the answer can be produced in a time $\text{O}(p^d)$... – Xi'an Jan 10 '12 at 14:23 1 The process, as it's written is always causal. The condition about the roots tells if the system is stable - which implies stationarity. But it's more common to speak about stability conditions, rather than stationarity conditions, IMO. – leonbloy Apr 20 '12 at 18:39 ## 2 Answers The Schur-Cohn algorithm has $d=2$; this is what I learned in a computational statistics class at Berkeley some years ago. - 1 Thank you, this is exactly what I was looking for. – Xi'an Jan 10 '12 at 15:43 For comparison purposes, do you also know of the order of an efficient root finder? – Xi'an Jan 10 '12 at 19:35 1 I think they are also $d=2$, using the following thought process. Evaluation of a p-degree polynomial is $O(p)$. Finding a single root should be independent of the order of the polynomial. After the root is found, the polynomial can be reduced to a polynomial of degree $p-1$, but this doesn't affect the order. So each of the $p$ roots takes $O(p)$ to find, leading to an $O(p^2)$ algorithm. – jbowman Jan 10 '12 at 20:49 1 – jbowman Jan 10 '12 at 20:55 2 – jbowman Jan 11 '12 at 15:22 show 1 more comment It is unnecessary to find the $p$ complex roots as far as these are not to be used for themselves. Moreover, most (if not all) root finding processes can fail for large $p$. Another solution is as follows. The $\mathrm{AR}(p)$ model can be reparametrised thanks to its $p$ partial autocorrelations (PACs) $\zeta_k$ for $1 \le k \le p$. The PACs are often denoted as $\phi_{k,k}$ because their computation involves an array $\phi_{k,\ell}$. There is a one-to-one correspondence between the vector $\boldsymbol{\rho}$ of the $p$ coefficients in the stationarity region and the vector $\boldsymbol{\zeta}$ in the region defined by the $p$ conditions $|\zeta_k | < 1$ for $1 \le k \le p$. The transformation "AR to PAC" $\boldsymbol{\rho} \mapsto \boldsymbol{\zeta}$ is quite simple and is given as a pretty recursion formula due to Barndorff-Nielsen and Schou. Testing stationarity from the vector of coefficients boils down to computing the $\zeta_k$ and stop as soon as one condition $| \zeta_k | \ge 1$ is found. The less well-known inverse transform "PAC to AR" $\boldsymbol{\zeta} \mapsto \boldsymbol{\rho}$ is available explicitly (due to Monahan), and is as simple as is the direct one. It might also be useful in some cases. The two transformations are implemented (in R) in a CRAN R package named FitAR which provides as well an efficient `invertibleQ` function to test stationarity. The package is described in the following article where list of references is provided. McLeod, A.I. and Zhang Y., "Improved Subset Autoregression: With R Package" Journal of Statistical Software, vol. 28, Issue 2, Oct 2008. http://www.jstatsoft.org/v28/i02 - 2 – leonbloy Apr 20 '12 at 18:34

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http://mathoverflow.net/revisions/122725/list

## Return to Question 2 added 102 characters in body I believe that the following questions are very basic, but I don't know how to get a reference. Consider a curve in the plane $C\in \mathbb C^2$ with a singularity at $0$ and suppose it is unibranch at zero (i.e. analytically irreducible). Then I guess one should be able to define "arithmetic genus defect" of the curve at $0$. Namely if one smooths analytically $C$, its geometric genus will grow by a positive number (in case of the cusp $x^2=y^3$ it will grow by one), and let us call this number the defect. Question 1. Is this defect well defined (independent of a smoothing)? How is it called and how one should calculate it (say it terms of the local ring of $C$ at $0$)? Question 2. Suppose we have an explicit local parametrisation of $C$ at $0$, say by two holomorphic functions $f(t), g(t)$ (polynomials if you wish). Is it possible to find this "defect" as a certain invariant of this pair of functions at $t=0$? Question 1 is settled in the answer of unknown and Question 2 in comments to it by Roy and Vivek 1 # "Arithmetic genus" of a plane curve singularity. I believe that the following questions are very basic, but I don't know how to get a reference. Consider a curve in the plane $C\in \mathbb C^2$ with a singularity at $0$ and suppose it is unibranch at zero (i.e. analytically irreducible). Then I guess one should be able to define "arithmetic genus defect" of the curve at $0$. Namely if one smooths analytically $C$, its geometric genus will grow by a positive number (in case of the cusp $x^2=y^3$ it will grow by one), and let us call this number the defect. Question 1. Is this defect well defined (independent of a smoothing)? How is it called and how one should calculate it (say it terms of the local ring of $C$ at $0$)? Question 2. Suppose we have an explicit local parametrisation of $C$ at $0$, say by two holomorphic functions $f(t), g(t)$ (polynomials if you wish). Is it possible to find this "defect" as a certain invariant of this pair of functions at $t=0$?

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http://math.stackexchange.com/questions/248770/simplify-frac5-sqrt206-sqrt20-to-frac5-sqrt58/248771

# Simplify $\frac{5+ \sqrt{20}}{6+ \sqrt{20}}$ to $\frac{5+ \sqrt{5}}{8}$ I would like to know how the following expression was simplified? From this $$\frac{5+ \sqrt{20}}{6+ \sqrt{20}}$$ to this $$\frac{5+ \sqrt{5}}{8}$$ - 1 Multiply the numerator and denominator by $6-\sqrt{20}$ and simplify – Simon Hayward Dec 1 '12 at 20:26 Thank you. I totally forgot this trick! – Mohammed Saleh Alorayed Dec 4 '12 at 3:27 ## 2 Answers You could simplify it a step prematurely for kicks and giggles. $\sqrt{20}=\sqrt{4\cdot 5}=2\sqrt{5}$. Now, $\frac{5+\sqrt{20}}{6+\sqrt{20}}=\frac{5+2\sqrt{5}}{6+2\sqrt{5}}$. Then, of course, multiply by the fraction by $\frac{6-2\sqrt{5}}{6-2\sqrt{5}}$. Multiplying by the conjugate to remove the radical is a common practice. It comes from the fact that $(a-b)(a+b)=a^2+ab-ba-b^2=a^2-b^2.$ So, in the case of $\frac{\text{something}}{u+\sqrt{v}}$, we multiply by $\frac{u-\sqrt{v}}{u-\sqrt{v}}$ since we know that $(u+\sqrt{v})(u-\sqrt{v})=u^2-v^2$. Similarly, if it is $u-\sqrt{v}$ in the denominator, we simply multiply by $\frac{u+\sqrt{v}}{u+\sqrt{v}}$. - You can multiply by the "conjugate": $$\frac{(5 +\sqrt{20})}{(6+\sqrt{20})}\frac{(6 - \sqrt{20})}{(6 - \sqrt{20})} = \dots$$ -

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http://mathhelpforum.com/advanced-algebra/207579-graph-matrices.html

# Thread: 1. ## As graph matrices? I would like to please guide me on this question: Let $S_+$ denote the set of semi positive definite matrices in $\mathbb{R}^{2\times 2}$ is known that $S_+\subseteq Sym \simeq\mathbb{R}^{3}$,wherein $Sym$ are the matrices symmetric. But Is it possible to give a geometric interpretation of $S_+$ in $\mathbb{R}^{3}$? Can be graphed? Thank you very much for your attention, any suggestions are welcome.

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http://en.wikipedia.org/wiki/Universal_algebra

# Universal algebra Universal algebra (sometimes called general algebra) is the field of mathematics that studies algebraic structures themselves, not examples ("models") of algebraic structures. For instance, rather than take particular groups as the object of study, in universal algebra one takes "the theory of groups" as an object of study. ## Basic idea From the point of view of universal algebra, an algebra (or algebraic structure) is a set A together with a collection of operations on A. An n-ary operation on A is a function that takes n elements of A and returns a single element of A. Thus, a 0-ary operation (or nullary operation) can be represented simply as an element of A, or a constant, often denoted by a letter like a. A 1-ary operation (or unary operation) is simply a function from A to A, often denoted by a symbol placed in front of its argument, like ~x. A 2-ary operation (or binary operation) is often denoted by a symbol placed between its arguments, like x * y. Operations of higher or unspecified arity are usually denoted by function symbols, with the arguments placed in parentheses and separated by commas, like f(x,y,z) or f(x1,...,xn). Some researchers allow infinitary operations, such as $\textstyle\bigwedge_{\alpha\in J} x_\alpha$ where J is an infinite index set, thus leading into the algebraic theory of complete lattices. One way of talking about an algebra, then, is by referring to it as an algebra of a certain type $\Omega$, where $\Omega$ is an ordered sequence of natural numbers representing the arity of the operations of the algebra. ### Equations After the operations have been specified, the nature of the algebra can be further limited by axioms, which in universal algebra often take the form of identities, or equational laws. An example is the associative axiom for a binary operation, which is given by the equation x * (y * z) = (x * y) * z. The axiom is intended to hold for all elements x, y, and z of the set A. ## Varieties Main article: Variety (universal algebra) An algebraic structure that can be defined by identities is called a variety, and these are sufficiently important that some authors consider varieties the only object of study in universal algebra, while others consider them an object.[citation needed] Restricting one's study to varieties rules out: • Predicate logic, notably quantification, including existential quantification ( $\exists$ ) and universal quantification ($\forall$) • Relations, including inequalities, both $a \neq b$ and order relations In this narrower definition, universal algebra can be seen as a special branch of model theory, in which we are typically dealing with structures having operations only (i.e. the type can have symbols for functions but not for relations other than equality), and in which the language used to talk about these structures uses equations only. Not all algebraic structures in a wider sense fall into this scope. For example ordered groups are not studied in mainstream universal algebra because they involve an ordering relation. A more fundamental restriction is that universal algebra cannot study the class of fields, because there is no type in which all field laws can be written as equations (inverses of elements are defined for all non-zero elements in a field, so inversion cannot simply be added to the type). One advantage of this restriction is that the structures studied in universal algebra can be defined in any category which has finite products. For example, a topological group is just a group in the category of topological spaces. ### Examples Most of the usual algebraic systems of mathematics are examples of varieties, but not always in an obvious way – the usual definitions often involve quantification or inequalities. #### Groups To see how this works, let's consider the definition of a group. Normally a group is defined in terms of a single binary operation *, subject to these axioms: • Associativity (as in the previous section): x * (y * z) = (x * y) * z. • Identity element: There exists an element e such that for each element x, e * x = x = x * e. • Inverse element: It can easily be seen that the identity element is unique. If we denote this unique identity element by e then for each x, there exists an element i such that x * i = e = i * x. (Sometimes you will also see an axiom called "closure", stating that x * y belongs to the set A whenever x and y do. But from a universal algebraist's point of view, that is already implied when you call * a binary operation.) Now, this definition of a group is problematic from the point of view of universal algebra. The reason is that the axioms of the identity element and inversion are not stated purely in terms of equational laws but also have clauses involving the phrase "there exists ... such that ...". This is inconvenient; the list of group properties can be simplified to universally quantified equations if we add a nullary operation e and a unary operation ~ in addition to the binary operation *, then list the axioms for these three operations as follows: • Associativity: x * (y * z) = (x * y) * z. • Identity element: e * x = x = x * e. • Inverse element: x * (~x) = e = (~x) * x. (Of course, we usually write "x −1" instead of "~x", which shows that the notation for operations of low arity is not always as given in the second paragraph.) What has changed is that in the usual definition there are: • a single binary operation (signature (2)) • 1 equational law (associativity) • 2 quantified laws (identity and inverse) ...while in the universal algebra definition there are • 3 operations: one binary, one unary, and one nullary (signature (2,1,0)) • 3 equational laws (associativity, identity, and inverse) • no quantified laws It is important to check that this really does capture the definition of a group. The reason that it might not is that specifying one of these universal groups might give more information than specifying one of the usual kind of group. After all, nothing in the usual definition said that the identity element e was unique; if there is another identity element e', then it is ambiguous which one should be the value of the nullary operator e. Proving that it is unique is a common beginning exercise in classical group theory textbooks. The same thing is true of inverse elements. So, the universal algebraist's definition of a group is equivalent to the usual definition. At first glance this is simply a technical difference, replacing quantified laws with equational laws. However, it has immediate practical consequences – when defining a group object in category theory, where the object in question may not be a set, one must use equational laws (which make sense in general categories), and cannot use quantified laws (which do not, as objects in general categories do not have elements). Further, the perspective of universal algebra insists not only that the inverse and identity exist, but that they be maps in the category. The basic example is of a topological group – not only must the inverse exist element-wise, but the inverse map must be continuous (some authors also require the identity map to be a closed inclusion, hence cofibration, again referring to properties of the map). ## Basic constructions We assume that the type, $\Omega$, has been fixed. Then there are three basic constructions in universal algebra: homomorphic image, subalgebra, and product. A homomorphism between two algebras A and B is a function h: A → B from the set A to the set B such that, for every operation fA of A and corresponding fB of B (of arity, say, n), h(fA(x1,...,xn)) = fB(h(x1),...,h(xn)). (Sometimes the subscripts on f are taken off when it is clear from context which algebra your function is from) For example, if e is a constant (nullary operation), then h(eA) = eB. If ~ is a unary operation, then h(~x) = ~h(x). If * is a binary operation, then h(x * y) = h(x) * h(y). And so on. A few of the things that can be done with homomorphisms, as well as definitions of certain special kinds of homomorphisms, are listed under the entry Homomorphism. In particular, we can take the homomorphic image of an algebra, h(A). A subalgebra of A is a subset of A that is closed under all the operations of A. A product of some set of algebraic structures is the cartesian product of the sets with the operations defined coordinatewise. ## Some basic theorems • The Isomorphism theorems, which encompass the isomorphism theorems of groups, rings, modules, etc. • Birkhoff's HSP Theorem, which states that a class of algebras is a variety if and only if it is closed under homomorphic images, subalgebras, and arbitrary direct products. ## Motivations and applications This section does not cite any references or sources. Please help improve this section by adding citations to reliable sources. Unsourced material may be challenged and removed. (April 2010) In addition to its unifying approach, universal algebra also gives deep theorems and important examples and counterexamples. It provides a useful framework for those who intend to start the study of new classes of algebras. It can enable the use of methods invented for some particular classes of algebras to other classes of algebras, by recasting the methods in terms of universal algebra (if possible), and then interpreting these as applied to other classes. It has also provided conceptual clarification; as J.D.H. Smith puts it, "What looks messy and complicated in a particular framework may turn out to be simple and obvious in the proper general one." In particular, universal algebra can be applied to the study of monoids, rings, and lattices. Before universal algebra came along, many theorems (most notably the isomorphism theorems) were proved separately in all of these fields, but with universal algebra, they can be proven once and for all for every kind of algebraic system. The 1956 paper by Higgins referenced below has been well followed up for its framework for a range of particular algebraic systems, while his 1963 paper is notable for its discussion of algebras with operations which are only partially defined, typical examples for this being categories and groupoids. This leads on to the subject of higher dimensional algebra which can be defined as the study of algebraic theories with partial operations whose domains are defined under geometric conditions. Notable examples of these are various forms of higher dimensional categories and groupoids. ### Category theory and operads Further information: Category theory and Operad theory A more generalised programme along these lines is carried out by category theory. Given a list of operations and axioms in universal algebra, the corresponding algebras and homomorphisms are the objects and morphisms of a category. Category theory applies to many situations where universal algebra does not, extending the reach of the theorems. Conversely, many theorems that hold in universal algebra do not generalise all the way to category theory. Thus both fields of study are useful. A more recent development in category theory that generalizes operations is operad theory – an operad is a set of operations, similar to a universal algebra. ## History In Alfred North Whitehead's book A Treatise on Universal Algebra, published in 1898, the term universal algebra had essentially the same meaning that it has today. Whitehead credits William Rowan Hamilton and Augustus De Morgan as originators of the subject matter, and James Joseph Sylvester with coining the term itself.[1] At the time structures such as Lie algebras and hyperbolic quaternions drew attention to the need to expand algebraic structures beyond the associatively multiplicative class. In a review Alexander Macfarlane wrote: "The main idea of the work is not unification of the several methods, nor generalization of ordinary algebra so as to include them, but rather the comparative study of their several structures." At the time George Boole's algebra of logic made a strong counterpoint to ordinary number algebra, so the term "universal" served to calm strained sensibilities. Whitehead's early work sought to unify quaternions (due to Hamilton), Grassmann's Ausdehnungslehre, and Boole's algebra of logic. Whitehead wrote in his book: "Such algebras have an intrinsic value for separate detailed study; also they are worthy of comparative study, for the sake of the light thereby thrown on the general theory of symbolic reasoning, and on algebraic symbolism in particular. The comparative study necessarily presupposes some previous separate study, comparison being impossible without knowledge."[2] Whitehead, however, had no results of a general nature. Work on the subject was minimal until the early 1930s, when Garrett Birkhoff and Øystein Ore began publishing on universal algebras. Developments in metamathematics and category theory in the 1940s and 1950s furthered the field, particularly the work of Abraham Robinson, Alfred Tarski, Andrzej Mostowski, and their students (Brainerd 1967). In the period between 1935 and 1950, most papers were written along the lines suggested by Birkhoff's papers, dealing with free algebras, congruence and subalgebra lattices, and homomorphism theorems. Although the development of mathematical logic had made applications to algebra possible, they came about slowly; results published by Anatoly Maltsev in the 1940s went unnoticed because of the war. Tarski's lecture at the 1950 International Congress of Mathematicians in Cambridge ushered in a new period in which model-theoretic aspects were developed, mainly by Tarski himself, as well as C.C. Chang, Leon Henkin, Bjarni Jónsson, Roger Lyndon, and others. In the late 1950s, Edward Marczewski[3] emphasized the importance of free algebras, leading to the publication of more than 50 papers on the algebraic theory of free algebras by Marczewski himself, together with Jan Mycielski, Władysław Narkiewicz, Witold Nitka, J. Płonka, S. Świerczkowski, K. Urbanik, and others. ## Footnotes 1. Grätzer, George. Universal Algebra, Van Nostrand Co., Inc., 1968, p. v. 2. Quoted in Grätzer, George. Universal Algebra, Van Nostrand Co., Inc., 1968. 3. Marczewski, E. "A general scheme of the notions of independence in mathematics." Bull. Acad. Polon. Sci. Ser. Sci. Math. Astronom. Phys. 6 (1958), 731–736. ## References • Bergman, George M., 1998. An Invitation to General Algebra and Universal Constructions (pub. Henry Helson, 15 the Crescent, Berkeley CA, 94708) 398 pp. ISBN 0-9655211-4-1. • Birkhoff, Garrett, 1946. Universal algebra. Comptes Rendus du Premier Congrès Canadien de Mathématiques, University of Toronto Press, Toronto, pp. 310–326. • Brainerd, Barron, Aug–Sep 1967. Review of Universal Algebra by P. M. Cohn. American Mathematical Monthly, 74(7): 878–880. • Burris, Stanley N., and H.P. Sankappanavar, 1981. A Course in Universal Algebra Springer-Verlag. ISBN 3-540-90578-2 Free online edition. • Cohn, Paul Moritz, 1981. Universal Algebra. Dordrecht, Netherlands: D.Reidel Publishing. ISBN 90-277-1213-1 (First published in 1965 by Harper & Row) • Freese, Ralph, and Ralph McKenzie, 1987. Commutator Theory for Congruence Modular Varieties, 1st ed. London Mathematical Society Lecture Note Series, 125. Cambridge Univ. Press. ISBN 0-521-34832-3. Free online second edition. • Grätzer, George, 1968. Universal Algebra D. Van Nostrand Company, Inc. • Higgins, P. J. Groups with multiple operators. Proc. London Math. Soc. (3) 6 (1956), 366–416. • Higgins, P.J., Algebras with a scheme of operators. Mathematische Nachrichten (27) (1963) 115–132. • Hobby, David, and Ralph McKenzie, 1988. The Structure of Finite Algebras American Mathematical Society. ISBN 0-8218-3400-2. Free online edition. • Jipsen, Peter, and Henry Rose, 1992. Varieties of Lattices, Lecture Notes in Mathematics 1533. Springer Verlag. ISBN 0-387-56314-8. Free online edition. • Pigozzi, Don. General Theory of Algebras. • Smith, J.D.H., 1976. Mal'cev Varieties, Springer-Verlag. • Whitehead, Alfred North, 1898. A Treatise on Universal Algebra, Cambridge. (Mainly of historical interest.)

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http://cameroncounts.wordpress.com/2011/07/20/the-field-with-one-element/

always busy counting, doubting every figured guess . . . ## The field with one element Posted on 20/07/2011 by Two weeks ago I was at the British Combinatorial Conference in Exeter. It was smaller than many recent BCCs, but a really good conference: the invited talks were uniformly excellent. I want to focus here on one talk, that of Koen Thas. He covered a lot of ground, mentioning for example the conjecture that finite projective planes necessarily have prime power order, and the conjecture (due to Jacques Tits, on which I inadvertently made a contribution) that there is no generalised quadrangle with a finite number (greater than 2) of points on a line and an infinite number of lines through a point. (I settled the first case, where there are three points on a line; the case of four points on a line was done by Kantor and independently Brouwer, and five points on a line by Cherlin, and that is the limit of our knowledge.) I will focus even more narrowly, on the last part of his talk, and the last of seventeen sections in his paper, concerning “the field with one element”. Some explanation for non-mathematicians is in order. A field is an algebraic system in which addition and multiplication are defined, and “the usual rules apply”: addition and multiplication are commutative and associative, there are identity elements and inverses for both operations, and the distributive law (a.k.a. expanding brackets) holds. In a field, we can’t divide by zero. An absolutely standard part of the definition is that the additive identity element 0 and the multiplicative identity 1 are unequal: 0≠1. So, by definition, there is no “field with one element”: every field has at least two elements. Thus, the field with one element is essentially a poetic concept (I am not sure which figure of speech best describes it). But, undaunted, I will try to explain. ### A tale of three qs The Gaussian, or q-binomial, coefficient, which I will write here as Gauss(n,k)q, is given by the formula (qn−1)(qn-1−1)…(qn-k+1−1)/(qk−1)(qk-1−1)…(q−1). Despite appearances, it is a polynomial in q, of degree k(n−k). This can be seen by showing, from the definition, that the Gaussian coefficients can also be specified by Gauss(n,0)q = Gauss(n,n)q = 1 and the “Pascal-like” recurrence Gauss(n,k)q = Gauss(n−1,k−1)q+qkGauss(n−1,k)q. For example, Gauss(4,2)q = q4+q3+2q2+q+1. A simple property is that, if we substitute q=1 in the Gaussian coefficient Gauss(n,k)q, we obtain the corresponding binomial coefficient Bin(n,k). The easiest way to see this is from the recurrence; alternatively, use the fact that the limit of (qn−1)/(qk−1) as q→1 is n/k (either by l’Hôpital’s rule, or by simply dividing top and bottom by q−1). #### Fixed sets for powers of a cycle So are the Gaussian coefficients nothing more than polynomials which happen to take interesting values when q=1? To see that this is not so, turn to another of the invited speakers at the BCC, Bruce Sagan, who talked about a recent discovery which has produced a lot of activity, the cyclic sieving phenomenon. Incidentally, Bruce used remarkably well the opportunity given by the publication of the main speakers’ talks in advance: his talk was slow and beautifully clear, while his paper is densely packed with interesting mathematics. The binomial coefficient Bin(n,k) counts the k-element subsets of an n-element set. Now suppose we have a cyclic permutation σ of the set, and we wish to count its orbits on k-subsets. By the Orbit-counting Lemma, we have to count the subsets fixed by any power of σ. Now it turns out that the number of sets fixed by a power of σ of order d is the result of substituting a primitive dth root of unity for q in Gauss(n,k)q. This fact is only the tip of a very large iceberg, for which see Bruce’s paper. But I want to speak of two other, completely different, occurrences of the Gaussian coefficients, discovered much earlier. In both cases, the letter q is traditionally used: is this just coincidence? (A question for a historian, maybe.) #### Finite vector spaces A finite field necessarily has a prime power number of elements; Galois showed that for any prime power q there is a unique field with q elements, up to isomorphism. This is now called a Galois field, and denoted by GF(q), or Fq. Now let V be an n-dimensional vector space over GF(q). Then V can be identified with the set of all n-tuples of coordinates (relative to a basis), so |V|=qn. The number of k-dimensional subspaces of V is the Gaussian coefficient Gauss(n,k)q. The standard proof of this fact involves noticing that the number of linearly independent k-tuples of vectors in V is (qn−1)(qn−q)…(qn−qk-1), while each k-dimensional subspace contains (qk−1)(qk−q)…(qk−qk-1) such tuples; dividing these numbers gives the Gaussian coefficient. There is an alternative counting proof. Take V=Fn, where F=GF(q), the space of all n-tuples. Now every k-dimensional subspace has a unique basis consisting of k vectors in reduced echelon form with no zero vectors: this means that, if the vectors are the rows of a k×n matrix, then • the first non-zero entry in each row is a 1; • these “leading 1s” occur further to the right as we go down the matrix; • all other entries in the column of a “leading 1″ are zero. Now it is easy to verify the Pascal-like recurrence for the number of matrices in reduced echelon form: the two terms on the right correspond to matrices where the leading 1 in the last row is in the last position (so deleting the last row and column gives a reduced echelon matrix) or in an earlier position (so the last column is arbitrary, and deleting it gives a reduced echelon matrix). Indeed the definition of “reduced echelon form” makes sense over any alphabet containing elements called 0 and 1; and the argument shows that the number of k×n matrices in reduced echelon form is the Gaussian coefficient, where q is the alphabet size (not necessarily a prime power). #### Area under lattice paths Consider lattice points (with integer coordinates) in the plane. You are standing at the origin, and have to move to the point (n−k,k); you are allowed to step from any point to its neighbour to the right or above. Clearly you must take n steps, of which k should be vertical. The vertical steps can be chosen anywhere in the sequence of moves. So the number of possible walks is Bin(n,k). Now suppose that we are interested in the area under such a path. This can take any value between 0 (for the path that takes all the horizontal steps first) to k(n−k) (for the path that takes the vertical steps first). A solution to this problem could be a generating function, a polynomial in an indeterminate q where the coefficient of qm is the number of paths with area m. It turns out that this generating function is Gauss(n,k)q. This can be seen most easily from the recurrence relation. The first term accounts for paths with the last step vertical (this step adds nothing to the area); the second counts paths with the last step horizontal (the last step adds k to the area, so multiplies the generating function by qk. Remarkably, we have seen three occurrences of the Gaussian coefficient: in the first, q is typically a root of unity; in the second, a prime power; and in the third, an indeterminate. ### Geometry over the one-element field, 1 In 1957, Jacques Tits suggested an analogy between the combinatorics of sets and subsets and the geometry of vector spaces and subspaces: roughly speaking, the combinatorics was the case q=1 of the geometry, where q is the order of the field. The (n−1)-dimensional projective space over a field F is the geometry whose points, lines, planes, … are the subspaces of dimension 1, 2, 3, … of an n-dimensional vector space over F. (Caution: dimension shift!) Axiomatically, the geometries are very similar. According to the Veblen–Young axioms, the projective space is characterised by the properties • Two points lie on a unique line. • If a line meets two sides of a triangle, not at their intersection, then it also meets the third. • A subspace is a set of points containing the line through any two of its points. We have to make some low-dimensional exceptions: a 1-dimensional projective space (according to these axioms) is just an arbitrary set of points forming a line, with no algebraic structure; and there are non-Desarguesian projective planes which are not obtained from vector spaces. But the correspondence is exact for higher dimensions, assuming that any line has at least three points. (If the lines are finite, then the number of points on a line is q+1, where q is the size of the coordinatising field.) By contrast, if every line has just two points, then an arbitrary set trivially satisfies the axioms, with the lines being all the 2-element subsets and the subspaces arbitrary subsets. Thus, the subsets of an arbitrary set “are” the projective spaces over the “field with one element”. What is an affine space? The number qn of points in an n-dimensional affine space reduces to 1 when q=1; so the affine space is a single point; but it has n ghostly directions through the point (lines with only one point). Recall that we obtain a projective space from an affine space by adding a point at infinity on each parallel class of lines; thus to get from the n-dimensional affine space to the projective space, we add n points, obtaining n+1 altogether, in agreement with our previous description. If this analogy is to be fruitful, it should suggest new insights; and indeed this one does. The “general linear group” in n dimensions over the field of one element is the symmetric group Sn; this reflects the BN-pair structure of the general linear group. Indeed, this extends to the other groups of Lie type: the group over the field of one element is the Weyl group of the BN-pair. There is interesting combinatorics too. In fact, one of my earliest papers exploited this analogy. A theorem of Fisher (for t=1), Petrenjuk (for t=2), and Ray-Chaudhuri and Wilson (in general) says that a 2t-design on v points with block size at most v−t has at least Bin(v,t) blocks. (Such a design is a collection of k-subsets, with 2t≤k≤v−t, such that any 2t-set is contained in a constant number of these sets.) I extended the result by replacing “set” and “subset” by “vector space over finite field” and “subspace”; the binomial lower bound is replaced by Gauss(v,t)q, where q is the field order. This result was almost immediately superseded by the much more general results of Philippe Delsarte. Several similar results exist. Probably the most famous is the vector space Ramsey theorem of Graham, Leeb and Rothschild, an exact translation of the classical Ramsey theorem. ### Geometry over the field of one element, 2 In recent years, this enterprise has become much more serious. In 2009, Javier Lopez talked about this in our Pure Mathematics seminar: I wish I had got more out of it than I did. (The abstract is here.) Enough to say that, if a geometer wants to make sense of the field with one element, it is necessary to be able to define varieties, schemes, etc. over it. For some particular varieties, we know how to proceed: we have already dealt with projective and affine spaces over the 1-element field, and for algebraic groups we just get the Weyl group. The situation for other varieties is more complicated. Soulé proposed a definition, which led to some results disagreeing with the naive interpretation. A revised version due to Connes and Consani seems to have fared better. Manin took things even further, proposing a particularly simple form for the zeta function of a variety over the 1-element field F1. (This presupposes that we have a notion of an extension of degree n of F1; even though this is another 1-element field, it is not the same as F1 itself!) Soulé proposed a zeta-function associated with any suitable counting function: for the Gaussian coefficient, the result is in agreement with Manin’s suggestion. ### So what is the one-element field? It seems that the 1-element field should have multiplication but not addition. But addition has been restored with the concept of a hyperfield. An abelian hypergroup is a set H with a “hyperoperation” +: this means that u+v is a non-empty subset, rather than an element, of H, and an appropriate collection of axioms is satisfied. A typical hypergroup is the set of orbits of an automorphism group of an abelian group, the hyperoperation giving the set of all orbits hit by sums of elements in the chosen orbits. Now a hyperfield is a set with operations of addition and multiplication, so that addition is a hypergroup, multiplication is a group on the non-zero elements, and the distributive laws hold. The Krasner hyperfield has elements 0 and 1: hyperaddition is given by 1+1 = {0,1} (all other instances follow from the axioms), and multiplication is obvious. This is the algebraic object which might play the role of the 1-element field. There is more to the story: the adèle ring of a global field modulo multiplication by non-zero field elements forms a hyperfield which is an extension of the Krasner hyperfield. Moreover, one can do geometry over the Krasner hyperfield, in a way which seems more satisfying than the hints and guesses we have had hitherto. Connes and Consani show that finite commutative hyperfield extensions of the Krasner hyperfield corresond to “incidence groups” of finite projective spaces, together with some low-dimensional extensions (including projective planes with sharply point-transitive groups, a notorious problem). Clearly a very productive analogy, when we have made sense of it! See Thas’ paper for more insights. ## About Peter Cameron I count all the things that need to be counted. This entry was posted in events, exposition and tagged affine geometry, algebraic geometry, binomial coefficient, Bruce Sagan, cyclic sieving, Gaussian coefficient, hyperfield, hypergroup, Koen Thas, projective geometry, q-analogue, scheme. Bookmark the permalink. ### 8 Responses to The field with one element 1. Peter Cameron says: Read more about the field with one element here: http://matrix.cmi.ua.ac.be/XTRA/ncg.pdf 2. Pingback: Do We Need Mysticism In Theory? « Gödel’s Lost Letter and P=NP 3. Jon Awbrey says: Hendiadys? — or maybe the converse of that? 4. Jon Awbrey says: This reminds me of some things I used to think about — I always loved working playing with generating functions and I can remember a time in the 80s when I was very pleased with myself for working out the q-analogue of integration by parts — but it will probably take me a while to warm up those old gray cells today. Let me first see if I can get LaTex to work in these comment boxes … The Gaussian coefficient, also know as the q-binomial coefficient, is notated as Gauss(n, k)q and given by the following formula: (qn−1)(qn−1−1) … (qn−k+1−1) / (qk−1)(qk−1−1) … (q−1). $\text{Gauss}(n, k)_q = \frac{(q^n - 1)(q^{n-1} - 1) \ldots (q^{n-k+1} - 1)}{(q^k - 1)(q^{k-1} - 1) \ldots (q - 1)}$ • Jon Awbrey says: Not sure what happened with the HTML version … • Jon Awbrey says: I tried a test copy on my blog and it came out like this. • Peter Cameron says: I haven’t figured out what the WordPress rules actually are. Even in posts, it is not proper HTML: for example, it obeys line ends in the input file. But rules for titles are different, and comments are different again. • Jon Awbrey says: Yes, it even seems to vary with the same theme used on different sides of the Atlantic. Maybe my use of strikeouts on “working” interfered with the later sub-&-superscript formatting? Feel free to delete or edit any of this discussion that is too junky or might mislead readers.

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http://physics.stackexchange.com/questions/29209/can-every-particle-be-regarded-as-being-a-combination-of-black-holes-and-white-h

# Can every particle be regarded as being a combination of Black holes and White holes? Can the statement be regarded as true? That every particle, or element in the universe can be regarded as a combination of black hole and white hole in variable proportion. - 4 Except for colorful objects that also include red holes, green holes, blue holes. Huh!? There aren't any white holes in the Universe because they're the time-reversed histories to black holes that would violate the second law of thermodynamics. Each microstate one could call a "white hole microstate" is actually a black hole microstate. And black holes only deserve to be called in this way if they're large enough - heavier than the Planck mass and compact objects. Normal objects in the Universe are composed of elementary particles which are much lighter and they're not (usefully) black holes. – Luboš Motl May 30 '12 at 7:06 The Compton wavelength of elementary particles is too large to allow them to be black holes (or white holes for that matter). Also, white holes don't exist as Lubos says correctly. Which you could have found out easily by Googling. – WIMP May 30 '12 at 7:32 ## 3 Answers You first have to understand what a "white hole" is. It's the time reverse of a black hole. It was rightly pointed out in previous answers that white holes violate the second law of thermodynamics. Now, like anything in thermodynamics, this makes them unlikely but not impossible (unlikely here usually means unlikely even in an astronomical number of universes ...). But if you keep that aside, there is a more beautiful picture of what a white hole is: it is a quantum superposition of all possible black holes of about the same size. There are so many ways to make a black hole, and they come in so many microstates, that the quantum superposition corresponding to a white hole is astronomically unlikely, ... but not impossible. Such considerations could be important if you ask what the relation is between elementary particles and black holes. This has been answered already: if you try to make such a comparison you would have to realise first that all known particles are so light that those black holes would be tiny, in fact billions of times smaller than the smallest distance conceivable in physics: the Planck length. Because of this, comparing known particles with black holes, or imagining them as being built from black holes, black, white or otherwise, is not considered to be a fruitful exercise in theoretical physics. - so, from the perspective of observers that cross a Kerr black hole and exit after a while through the corresponding white hole; how do they explain away to the observers they might find outside, the astronomical improbability of having found a white hole? will the outside observers agree that this is just a fortituous fluctuation of an otherwise thermal membrane? – user56771 Aug 21 '12 at 2:11 @user56771: Your comment is why the extended Kerr is not taken seriously--- it can't possibly describe travel to another universe. Penrose said that you just get cooked at the Cauchy horizon, but I think you come out in this universe. – Ron Maimon Aug 21 '12 at 6:12 1 @RonMaimon, good, i want to believe that too, but suppose we say that in this universe we have an excess of kerr black holes relative to kerr white holes given by some astronomical statistical number of $10^{10}$ against 1. Where all those extra kerr black holes are connecting to? if the white hole is an analytic continuation of the black hole then there can be no excess: the mapping needs to be one-to-one – user56771 Aug 21 '12 at 12:53 @user56771: It is impossible to have an "excess" of Kerr black holes--- the two concepts of Kerr black hole and Kerr white hole are the same. You can't come out of another black hole, you just come out of the same one you fell into. This is strongly suggested by AdS/CFT, but the gluing is tricky, as it is easy to get traversable CTC's in a wrong gluing, and you get no hints from the classical solution as to how it's supposed to be done. – Ron Maimon Aug 21 '12 at 15:48 ok @RonMaimon, but if all these observers/travelers can really exit in the same universe, they might end up afterwards and compare notes, and they'll have to agree that white holes are more abundant than what quantum gravity theory predicts. How do they reconcile this? – user56771 Aug 22 '12 at 0:29 show 3 more comments Over the years there have been suggestions that elementary particles may be black holes. However no-one has ever been able to make this quantitative and I doubt anyone believes it these days. There was some discussion of this in what is the difference between a blackhole and a point particle, and Googling will find you lots of hits on this subject. I've never heard of the idea that particles may be a combination of a black hole and a white hole. It's hard to see how such an arrangement could be possible. The term "white hole" is widely used in science fiction, but SF has the wrong idea about them. The equations of general relativity are time symmetric, and if you reverse the direction of time the black hole solution turns into a white hole. However there is no evidence that the time reversed solution has any physical significance, and as the comments above suggest, no-one (in the mainstream physics community) believes that white holes exist. The nearest we come go to a white hole existing are suggestions that the Big Bang was a white hole. Have a look at John Baez's article on this for more info. - A "White hole" is the time reverse of a black hole, as discovered by Hawking and explained by 'tHooft. The reason is that black holes can be in thermal equilibrium with radiation, and the time reverse of a thermal equilibrium state is still thermal equilibrium. This intuition is confirmed in AdS/CFT, where a thermal black hole in AdS can be described by a time-reversal invariant thermal boundary state, and so one is safe to conclude Hawking's argument is valid in quantum generality. The question of whether particles are black holes is to my mind entirely answered by string theory. The first point is that black holes are classical, and you are talking about quantum particles, so the question is only whether the particles can be interpreted as black holes in an appropriate limit. The F-strings from which the particles in the standard model are made are dual to branes, either by going to strong coupling if you are in a heterotic string theory, where the strings are revealed to be 2-branes wrapping a dimension, or else by going to strong coupling in IIB string theory, where you find that the F-string swaps with a D-brane. In either case, M2 branes or D1 branes are both classically intepretable as black holes, because when you stack them up, their supergravity solution is exactly a black hole. So within string theory, it is not wrong to say that every particle is a particularly highly quantum black hole, in the smallest quantum in which it comes. This doesn't mean the classical GR picture works for these objects, it doesn't. The other answers explain that it fails completely. But the quantum picture that all matter is extremal black holes (excluding orbifolds and other exotica) is the basic picture of modern string theory. It unifies particle physics and black hole physics in a fundamental way. -

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