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http://cms.math.ca/10.4153/CJM-2004-033-0	Canadian Mathematical Society www.cms.math.ca \| \| \| \| \| \|----------\|----\|-----------\|----\| \| \| \| \| \| \| \| \| Site map \| \| \| CMS store \| \| location: Publications → journals → CJM Abstract view # Fat Points in $\mathbb{P}^1 \times \mathbb{P}^1$ and Their Hilbert Functions Read article [PDF: 281KB] http://dx.doi.org/10.4153/CJM-2004-033-0 Canad. J. Math. 56(2004), 716-741 Published:2004-08-01 Printed: Aug 2004 • Elena Guardo • Adam Van Tuyl Features coming soon: Citations (via CrossRef) Tools: Search Google Scholar: Format: HTML LaTeX MathJax PDF PostScript ## Abstract We study the Hilbert functions of fat points in $\popo$. If $Z \subseteq \popo$ is an arbitrary fat point scheme, then it can be shown that for every $i$ and $j$ the values of the Hilbert function $_{Z}(l,j)$ and $H_{Z}(i,l)$ eventually become constant for $l \gg 0$. We show how to determine these eventual values by using only the multiplicities of the points, and the relative positions of the points in $\popo$. This enables us to compute all but a finite number values of $H_{Z}$ without using the coordinates of points. We also characterize the ACM fat point schemes sing our description of the eventual behaviour. In fact, n the case that $Z \subseteq \popo$ is ACM, then the entire Hilbert function and its minimal free resolution depend solely on knowing the eventual values of the Hilbert function. Keywords: Hilbert function, points, fat points, Cohen-Macaulay, multi-projective space MSC Classifications: 13D40 - Hilbert-Samuel and Hilbert-Kunz functions; Poincare series 13D02 - Syzygies, resolutions, complexes 13H10 - Special types (Cohen-Macaulay, Gorenstein, Buchsbaum, etc.) [See also 14M05] 14A15 - Schemes and morphisms	{"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": 0.18, "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": 0.6914311051368713, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/20713/can-a-photon-be-emitted-with-a-wavelength-299-792-458-meters-and-would-this-v	# Can a photon be emitted with a wavelength > 299,792,458 meters, and would this violate c? Just curious if the possibility exists (not necessarily spontaneously) for a photon with a wavelength greater than the distance component of c to be emitted, and would this inherently violate the scalar c? - 4 In addition to the answers you've been given, keep in mind that a speed does not have distance and time "components." It's just a speed. Given any time it allows you to find a distance, and vice versa, but just because the speed is special doesn't make any particular distance or time special. – David Zaslavsky♦ Feb 8 '12 at 18:01 1 The concept of a wavelength and frequency can be operationally defined only for classical electromagnetic waves, they cannot be defined for photons. – Revo Feb 8 '12 at 19:12 That's the number of meters light travels in one second. The length of "one second" was arbitrarily decided by humans, so passing that arbitrary boundary (causing the frequency to be <1Hz) shouldn't be anything special. – BlueRaja - Danny Pflughoeft Feb 8 '12 at 20:57 ## 3 Answers $$c=\nu\lambda$$ The waves will still travel at $c$. Changing $\lambda$ changes $\nu$, not $c$. If, in SI units, $c<\lambda$, then $\nu<1 \text{ Hz}$. These can exist, though we don't come across them often. Ultra-redshifted light coming from sources near a black hole have such frequencies (A source just entering the event horizon gets redshifted all the way to $\nu=0,\lambda=\infty$). Update: As leftaroundabout pointed out, such low frequency waves are possible when dealing with the transmission of electromagnetic fields. So shaking a charged balloon or any small current can give such photons. - 3 Actually, we come across them all the time; like when you shake a statically charged balloon, or when a lighning strikes. It's just not very useful to treat such kind of EMF as photons. – leftaroundabout Feb 8 '12 at 16:53 1 The expression $c<\lambda$ is meaningless since $c$ and $\lambda$ have different units. – Jan Feb 8 '12 at 20:39 Ive specified SI units to avoid exactly this discrepancy. The OPs question demands it anyways.. – Manishearth♦ Feb 9 '12 at 0:14 Even when SI units are specified, strictly speaking one cannot compare. I'm implying comparison of magnitudes. – Manishearth♦ Feb 9 '12 at 0:56 See http://en.wikipedia.org/wiki/Ultra_low_frequency EM frequencies below 1Hz, and therefore with a wavelength longer than c meters can be observed in nature. This does not violate relativity since those waves still propagate with velocity c (in vacuum). - Ah, ok. I get it. Regardless of the frequency or wavelength(cycle?) the entire quantum of EM will move at c. So even if the wavelength was greater than the distance component of c, the speed the quantum propagates at is still just c. I was confused. – DestressedData Feb 8 '12 at 16:51 4 @DestressedData: There would have to be something magical about 1 second for the wavelength you note to be special, but a second is just a convenient time scale for humans---not something that the universe gives special meaning to. – dmckee♦ Feb 8 '12 at 17:10 1 @dmckee: That seems like an answer to me! – Jefromi Feb 8 '12 at 17:35 It's probably best not to think of wavelength as anything to do with the size of the photon but more as a convenient way to think about frequency. Imagine a police car with a flashing light going at constant speed. If it's doing 36kph and the light flashes 1/second then the flash will appear every 10m along the road - you could think of this as a 10m wavelength. But really it's just a way of describing the speed and frequency. - Thank you, that helps me understand wavelength a little better. – DestressedData Feb 8 '12 at 16:51	{"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": 10, "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": 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": 0.9381968975067139, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/32110/classical-limit-of-the-path-integral-formulation-of-quantum-mechanics?answertab=active	# Classical limit of the path integral formulation of quantum mechanics It is well-known that if $S \gg \hbar$, then the classical path dominates the Feynman path integral. But is there some to show that if $S\gg\hbar$, then the particle's trajectory will approach the classical path? - 1 the argument $exp(iS/ \hbar )$ oscillates heavily whenever $\hbar \sim 0$ in this case only the trajectory which satisfy $\delta S =0$ contributes to the path integral.. the other trajectories 'interfere' each other – Jose Javier Garcia Jul 15 '12 at 21:58 1 This is only completely rigorously true if you add a small imaginary part to the time, so that high action paths are suppressed a little bit exponentially. This gives a cutoff which makes the Feynman integrals sensible mathematically, and it is always implicitly or explicitly used when you are doing a path integral. – Ron Maimon Jul 16 '12 at 6:22 1 The particle path will not approach the classical path exactly, rather the contributions of the important trajectories in the sum will be no different than just looking at the classical trajectory. This is subtly different, because you can still describe a probability distribution over phase space in the classical limit of quantum mechanics, there is still some left-over residual superposition principle, but it's without interference. – Ron Maimon Jul 16 '12 at 9:10 The question is not clear albeit the next related questions:19417,32112,32237, suggest what has been demonstrated in the article. – paritto Dec 8 '12 at 4:41 ## 1 Answer Actually there is a very interesting approach followed by E.Gozzi and his students to express the transition probability in Classical Mechanics in terms of a path integral. He is currently studying the relation between MQ and CM using this approach. You can take a look at his site : http://www-dft.ts.infn.it/~gozzi/, there you will find some interesting 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": 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": 0.9031707644462585, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/tagged/matrix-elements?sort=votes&pagesize=15	# Tagged Questions The matrix-elements tag has no wiki summary. 2answers 781 views ### Matrix Representations of Quantum States and Hamiltonians I am a high school student trying to teach himself quantum mechanics just for fun, and I am a bit confused. As a fun test of my programming/quantum mechanics skill, I decided to create a computer ... 2answers 185 views ### If the S-matrix has symmetry group G, must the fields be representations of G? If the fields in QFT are representations of the Poincare group (or generally speaking the symmetry group of interest), then I think it's a straight forward consequence that the matrix elements and ... 2answers 135 views ### Element of area in 4-dimensional space-time How would you proof that $$\mathrm {Tr} (\mathbf{S\cdot \bar S })=0$$ where $\mathbf S$ is an element of area delimited for the 4-vectors $\mathbf u$ and $\mathbf v$ given by S^{\alpha \beta}\equiv ... 4answers 384 views ### Is the momentum operator well-defined in the basis of standing waves? Suppose I want to describe an arbitrary state of a quantum particle in a box of side $L$. The relevant eigenmodes are those of standing waves, namely \left<x\|n\right>=\sqrt{\frac{2}{L}}\cdot ... 1answer 128 views ### Creation and Annialation Operators and Kinetic Energy Matrix Elements I'd like to write equations for $c_{ij}(t)$, With a hamiltonian of the form $$H=\sum_{kn}a^{\dagger}_k t_{kn}a_n + \frac{1}{2}\sum_{klmn}a^{\dagger}_k a^{\dagger}_l v_{klmn}a_m a_n$$ with $t_{kn}$ ... 2answers 139 views ### How to express continuous values as a matrix Usually a quantity of a matrix is defined as the eigenvalues of the matrix. If so, how can anyone express continuous values, as in Schrodinger picture, into a matrix? 3answers 125 views ### Vector space of $\mathbb{C}^4$ and its basis, the Pauli matrices How do I write an arbitrary $2\times 2$ matrix as a linear combination of the three Pauli Matrices and the $2\times 2$ unit matrix? Any example for the same might help ? 2answers 176 views ### Path integral on matrix model I was looking at a 0-dimensional matrix model, where the variables are $N\cdot N$ Hermitean matrices. It had a gauge symmetry, e.g. $U(N)$. And in the path integral, the Faddeev-Popov trick was used. ... 0answers 75 views ### Decay Amplitudes Notation This question is mostly about how to interpret notation used in Particle Physics. I am given that at lowest order the rate of $b\rightarrow s\gamma$ is proportional to \$\langle B_p\|b^\dagger ... 1answer 35 views ### Diagonal matrix in k-space I'm having some trouble with an integration I hope you guys can help me with. I have that: ${{\mathbf{v}}_{i}}\left( \mathbf{k} \right)=\frac{\hbar {{\mathbf{k}}_{i}}}{m}$ and ... 0answers 92 views ### Matrix element squared for up + antidown -> photon + W boson? [closed] I am in trouble… I am a master student (first year) in theoretical particle physics and I have been working for several months on the calculation of the total cross-section at leading order for the ...	{"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": 15, "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": 0.9060714244842529, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/77937?sort=votes	## Signed measure that is positive over convex sets ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I have a signed measure $\mu$ on a convex subset $C\subset \mathbb{R}^n$, and I want to prove that $\mu$ is a probability measure, most importantly that it is positive everywhere. I do know that $\int f(x)d\mu(x)\geq 0$ for any positive CONVEX function $f$. So if I could get this inequality for indicator functions I'd be done. Do you know if this suffices to get that the measure is positive, or maybe have a counterexample? - If $C=(a,b)$ is an interval, your assumption is that $\langle \mu,f\rangle\ge0$ for every $f$ such that $f''\ge0$ in distributional sense. Therefore an equivalent statement is that $\mu=N''$ where $N\ge0$ and $N(a)=N(b)=0$. If $n\ge2$, a characterization must be more involved. – Denis Serre Oct 13 2011 at 6:40 ## 1 Answer A counterexample is a signed measure on the interval $I:=[-1,1]$ concentrated in the points $\{-1\}$, $\{0\}$, $\{1\}$ with weights respectively $1/2$, $-1$, $1/2$. (Thus $\int_If d\mu= f(1)/2+f(-1)/2 - f(0)\ge0$ is just the convexity inequality). - 1 Thanks, that's a very simple example. I actually had the problem stated otherwise before, and I thought that the formulation I put here was simpler, but equivalent, but I see that was not really the case. Maybe I'll put out another question with the alternative formulation. – Henrique de Oliveira Oct 12 2011 at 18:35	{"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": 0.9608121514320374, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/98535/chebotarev-help	# Chebotarev help I'm new user here! I'm stuck on the following and I was wondering if some NT experts can prove some help. Question (from an entrance exam in my country) 5) What is the density of the rational primes $p$ that factor as $(p)=p_{1}p_{2}$ in $\mathbb{Q}[\sqrt[3]{2}]$. What about $\mathbb{Q}[\sqrt[5]{2}]$? I think Chebotarev is the answer but I kno only how to use it to get the density of the primes that split completely in those number fields. - 1 Ignore the ramified primes. If $p$ splits as $p_1p_2$ in $K=\mathbf{Q}(\root 3\of 2)$, then you must have $f(p_i\|p)=i$ for $i=1,2$. So when you go the normal closure $L=K(\omega), \omega=(-1+\sqrt{-3})/2$ you must get $f=2, g=3$. It shouldn't be too difficult to prove that all the primes with $f=2,g=3$ in $L/\mathbf{Q}$ behave in this way. Does that help? – Jyrki Lahtonen Jan 12 '12 at 19:19 1 Unless I made a mistake in $\mathbf{Q}(\root 5\of 2)$ $(p)=p_1p_2$ implies that the respective inertia degrees must be 1 and 4. As they must agree after extension to $\mathbf{Q}(\root 5\of 2,\zeta_5)$, we can deduce that $f=4$. So the decomposition groups must be cyclic subgroups of order 4 of the Galois group $$F_{20}=C_5\rtimes C_4=\langle s,t\mid s^5=1=t^4, tst^{-1}=s^2\rangle.$$ – Jyrki Lahtonen Jan 12 '12 at 19:29	{"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": 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": 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": 0.9518107771873474, "perplexity_flag": "head"}
http://mathoverflow.net/questions/103966?sort=newest	## General degree $d$ surface in $\mathbb{P}^3$ ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Let $H_{d_1,g_1}, H_{d_2,g_2}$ be two Hilbert schemes of curves in $\mathbb{P}^3$ with degrees $d_1, d_2$ and genus $g_1, g_2$. Denote by $H:=H_{d_1,g_1}\times H_{d_2,g_2}$ where an element in $H$ is a pair of curves $(C_1, C_2)$ with $C_i \in H_{d_i,g_i}$. As far as I understand, a generic element of $H$ consists of pairs $(C_1,C_2)$ such that $C_1 \cap C_2 = \emptyset$. Assume now that $g_i \not=\frac{1}{2}(d-1)(d-2)$ i.e., the Hilbert scheme do not parametrize plane curves. My question is whether we can say something similar for a degree $d$ surface in $\mathbb{P}^3$. More precisely, for a fixed $d \ge 5$, we denote by $H'$ the space parametrizing the family of all smooth degree $d$ surfaces $X$ in $\mathbb{P}^3$ such that $X$ contains at least one pair of curves $(C_1, C_2) \in H$. Then can we say that a general $X$ in $H'$ contains a non-intersection pair of curves $(C_1, C_2) \in H$. This is equivalent to saying that the dimension of an irreducible component of $H'$ is maximal (among the components of $H'$) if it parametrizes surfaces containing $(C_1, C_2) \in H$ with $C_1 \cap C_2 = \emptyset$. Note that we need to assume $C_i$ are not plane curves because as far as I recall it gives a counter-example to the statement. - ## 2 Answers By the Noether-Lefschetz theorem, a very general surface $X\subset \mathbb{P}^3$ of degree $d\geq 4$ has $\mathrm{Pic}(X) \cong \mathbb Z$, generated by a hyperplane section $H$. But then if $C \sim aH$ and $C'\sim bH$ on $X$, we have $C\cdot C' = abd^2 > 0$, and so $C,C'$ must intersect in at least one (and generically many) point. Thus every pair of curves on $X$ intersect each other. (This doesn't rule out that for some choices of pairs of curve classes that what you want could be true, but it does rule out that the most general possible result is true). - @Huizenga: However, the space of such surfaces in not dense in the space of degree $d$ surfaces in $\mathbb{P}^3$ (due to a result by Ciliberto, Miranda, et al.) Moreover, in most cases if the degree of the curve is somewhat lower than $d$ then the curves will not be complete intersection in the surface. – Naga Venkata Aug 4 at 22:24 1 Of course the space of such surfaces is dense. It doesn't contain an open dense subset (hence the quantifier "very general"), but it is dense. Basically, you can ask for exceptional cases where what you want to be true holds, but it certainly does not hold in general. – Jack Huizenga Aug 5 at 3:44 @Huizenga: I am a bit confused. The result of Ciliberto, Miranda, et. al. titled "General components of the Noether-Lefschetz locus" states that the space of degree $d$ surfaces with Picard number greater than $1$ is dense in the space of all degree $d$ surfaces (for $d \ge 4$). – Naga Venkata Aug 5 at 5:53 Certainly a set and its complement can both be dense. – Jack Huizenga Aug 5 at 8:14 ### 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. We can choose $d_1,g_1,d_2,g_2,d$ such that every pair of curves in $H$ contained in a single degree $d$ surface must intersect. The most well-behaved hypersurface that is not a plane is of course a quadric surface, so we choose $d=2$. A curve of degree $4$ and genus $1$ on a quadric surface must have bidegree $(2,2)$. Two such curves have intersection number $8$ and anyways must intersect. Clearly, they are not plane curves. - @Sawin: The question says $d \ge 5$. Is it obvious that this example can be extended to the higher degree case? – Naga Venkata Aug 4 at 22:19 No. We want to say that the only curves whose Hilbert polynomial is the same as the Hilbert polynomial of the divisor class $nH$ are the divisor class $nH$. Thanks to the intersection theory formula for genus and degree, we can compute $D \cdot D$ and $D \cdot H$ from the Hilbert polynomial, so we know that $(D - nH)\cdot (D-nH)=0$ and $(D-nH) \cdot H=0$ for any divisor of this type. So if we know that no nonzero divisors satisfy $X \cdot X=0$ and $X \cdot H=0$, we're good. Is that dense? – Will Sawin Aug 4 at 23:45 The question is primarily intended for curves that are not of complete intersection in a surface which is to say that the divisor class $[C_i]\not=kH$ for any $k$. This is because the space of degree d surfaces in $\mathbb{P}^3$ with picard number greater than $1$ is dense in the space of all degree $d$ surfaces. So in most cases the class of the curves are not of the form $kH$ for any $k$. – Naga Venkata Aug 5 at 0:09 My idea is that if the Picard number is small we can force the curves to be of the form $kH$ using their Hilbert polynomial. Do you know whether the surfaces with indefinite Neron lattices are dense? – Will Sawin Aug 5 at 2:44 @Sawin: I do not totally understand what you mean by "force to of the form". If you mean by adding it to a multiple of the hyperplane class, that might be possible. But this does not mean that the picard number becomes of the surface is $1$. By indefinite do you mean of infinite rank? – Naga Venkata Aug 5 at 6:04 show 1 more comment	{"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": 74, "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": 0.9318157434463501, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/16246/the-cardinality-of-a-countable-union-of-countable-sets-without-the-axiom-of-cho/16249	# The cardinality of a countable union of countable sets, without the axiom of choice One of my homework questions was to prove, from the axioms of ZF only, that a countable union of countable sets does not have cardinality $\aleph_2$. My solution shows that it does not have cardinality $\aleph_n$, where $n$ is any non-zero ordinal (not necessarily finite). I have a sneaking suspicion that my solution is actually invalid, but I can't find any reference which invalidates my conclusion. I have read that it is provable in ZF that there are no cardinals $\kappa$ such that $\aleph_0 < \kappa < \aleph_1$, but I believe the conclusion of my proof does not preclude the possibility that the cardinality is incomparable to $\aleph_1$ or some such. I think the weakest point in my solution is where I claim that the supremum of a countable set of countable ordinals is again countable. This is true, of course, but it sounds uncomfortably close to the claim "a countable union of countable sets is countable", which is well-known to be unprovable in ZF. Can anybody confirm that the ordinal version is provable in ZF though? If not, I think I can weaken the claim to "the supremum of any set of countable ordinals is at most $\omega_1$", and this establishes the weaker result that the cardinality of a countable union of countable sets is not $\aleph_n$ for any ordinal $n \ge 2$. - ## 1 Answer $\omega_1$ can be a countable union of countable ordinals, i.e., the sup of a countable set of countable ordinals. This consistency result was one of the first found with forcing. It was announced in S. Feferman-A. Levy, "Independence results in set theory by Cohen's method II", Notices Amer Math Soc., 10, (1963) 593. The result is that it is consistent with ZF that ${\mathbb R}$ is a countable union of countable sets. From this, it follows easily that $\omega_1$ also has this property. A proof can be found in Jech's book on the Axiom of Choice. The problem is that, as you suspect, "The supremum of a countable set of countable ordinals" cannot be proved without some choice to be countable. The issue is that although we know that each countable ordinal is in bijection with $\omega$, there is no uniform way of picking for each countable ordinal one such bijection. Now, you need this to run the usual proof that a countable union of countable sets is countable. In fact, things can be worse: Gitik showed that it is consistent with ZF that every infinite (well-ordered) cardinal has cofinality $\omega$. ("All uncountable cardinals can be singular", Israel J. Math, 35, (1980) 61-88.) On the other hand, one can check that a countable union of countable sets of ordinals must have size at most $\omega_1$ (which is essentially what your HW is asking to verify). So, in Gitik's model, $\omega_2$ is a countable union of countable unions of countable ordinals, but not a countable union of countable ordinals. Let me add two comments about other things you say in your question: You write "I have read that it is provable in ZF that there are no cardinals $\kappa$ such that $\aleph_0<\kappa<\aleph_1$". This is true, but it is stronger than that: By definition $\aleph_1$ is the first ordinal that is not countable, so of course there are no cardinals in between $\aleph_0$ and $\aleph_1$. Similarly, there are no cardinals between any (well-ordered) cardinal $\kappa$ and its successor $\kappa^+$, by definition. It is true, however, that a countable union of countable sets need not be comparable with $\aleph_1$ without choice. In fact, we can have a non-well-orderable set that can be written as a countable union of sets of size 2. - Thank you for the references and examples. I'm curious though — Can it be shown that the union of a countable family $S$ of countable unordered sets has cardinality at most $\aleph_1$? This seems to be a stronger claim than the conclusion of my proof (that $\|\bigcup S\| \ne \aleph_n$ for any $n \ge 2$). – Zhen Lin Jan 3 '11 at 17:24 1 This result of Gitik is terrifying, wonderful! – Asaf Karagila Jan 3 '11 at 17:43 @Zhen: No, this cannot be shown. As I mentioned, a countable union of sets of size 2 needs not be comparable with $\aleph_1$. – Andres Caicedo Jan 3 '11 at 17:50 @Asaf: I agree. And there are some very interesting questions still left. Schindler and Busche have recently looked at the consistency strength of Gitik's result, and Ioanna Fenderson Dimitriou, a student of Peter Koepke, has also been studying this model. – Andres Caicedo Jan 3 '11 at 17:53	{"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": 29, "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": 0.9600300788879395, "perplexity_flag": "head"}
http://mathhelpforum.com/advanced-algebra/169852-maximum-distance-point-sphere.html	# Thread: 1. ## maximum distance from point to sphere I have to calculate the maximum distance from the point $T(1, -1, 3)$ to the sphere $x^2+y^2+z^2-6x+4y-10z-62=0$. Here's what I did. The equation for the sphere is $(x-3)^2+(y+2)^2+(z-5)^2=100$. That means the center of the sphere is $C(3, -2, 5)$, and the radius is $r=10$. I think the maximum distance $d_m$ is the distance from the point $T$ to the center of the sphere plus the radius: $d_m=d(T, C)+r$. Is this okay? Thank you! 2. Everything looks good so far. What do you get? 3. I get $d((1, -1, 3), (3, -2, 5))=\sqrt((3-1)^2+(-2+1)^2+(5-2)^2 )=\sqrt(9)=3$ so $d_m=3+10=13$. It seems too simple, are you sure I'm not thinking something terribly wrong? 4. You're not computing the distance formula correctly. It should be more like $\sqrt{(3-1)^2+(-2+1)^2+(5-2)^2}=\sqrt{4+1+9}=\sqrt{14}.$ [EDIT]: Ignore this post. See below for correction. 5. Why do we have (5-2)^2 instead of (5-3)^2? I'm afraid I don't understand. 6. I'm sorry. You're correct. It should be 3, like you got. 7. Thank you so much! I'm quite insecure when I'm doing the exercises on my own, it really helps when someone checks your work. 8. If you feel insecure when you're doing exercises, I would come up with a system of checks you can run against your answers to justify them to yourself. It's good problem-solving strategy anyway. You're very welcome. Have a good 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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 11, "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": 0.9496886730194092, "perplexity_flag": "middle"}
http://en.wikipedia.org/wiki/Totally_disconnected	# Totally disconnected space (Redirected from Totally disconnected) In topology and related branches of mathematics, a totally disconnected space is a topological space that is maximally disconnected, in the sense that it has no non-trivial connected subsets. In every topological space the empty set and the one-point sets are connected; in a totally disconnected space these are the only connected subsets. An important example of a totally disconnected space is the Cantor set. Another example, playing a key role in algebraic number theory, is the field Qp of p-adic numbers. ## Definition A topological space X is totally disconnected if the connected components in X are the one-point sets. ## Examples The following are examples of totally disconnected spaces: • Discrete spaces • The rational numbers • The irrational numbers • The p-adic numbers; more generally, profinite groups are totally disconnected. • The Cantor set • The Baire space • The Sorgenfrey line • Zero dimensional T1 spaces • Extremally disconnected Hausdorff spaces • Stone spaces • The Knaster–Kuratowski fan provides an example of a connected space, such that the removal of a single point produces a totally disconnected space. • The Erdős space ℓp(Z)∩$\mathbb{Q}^{\omega}$ is a totally disconnected space that does not have dimension zero. ## Properties • Subspaces, products, and coproducts of totally disconnected spaces are totally disconnected. • Totally disconnected spaces are T1 spaces, since points are closed. • Continuous images of totally disconnected spaces are not necessarily totally disconnected, in fact, every compact metric space is a continuous image of the Cantor set. • A locally compact hausdorff space is zero-dimensional if and only if it is totally disconnected. • Every totally disconnected compact metric space is homeomorphic to a subset of a countable product of discrete spaces. • It is in general not true that every open set is also closed. • It is in general not true that the closure of every open set is open, i.e. not every totally disconnected Hausdorff space is extremally disconnected. ## Constructing a disconnected space Let $X$ be an arbitrary topological space. Let $x\sim y$ if and only if $y\in \mathrm{conn}(x)$ (where $\mathrm{conn}(x)$ denotes the largest connected subset containing $x$). This is obviously an equivalence relation. Endow $X/{\sim}$ with the quotient topology, i.e. the coarsest topology making the map $m:x\mapsto \mathrm{conn}(x)$ continuous. With a little bit of effort we can see that $X/{\sim}$ is totally disconnected. We also have the following universal property: if $f : X\rightarrow Y$ a continuous map to a totally disconnected space, then it uniquely factors into $f=\breve{f}\circ m$ where $\breve{f}:(X/\sim)\rightarrow Y$ is continuous. ## References • Willard, Stephen (2004), General topology, Dover Publications, ISBN 978-0-486-43479-7, MR2048350 (reprint of the 1970 original, MR0264581)	{"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": 12, "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": 0.8993088006973267, "perplexity_flag": "head"}
http://en.wikibooks.org/wiki/General_Chemistry/The_Quantum_Atom	# General Chemistry/The Quantum Atom \| \| \| \| \|-----------------------------------------------------------------------------------------------------------------------------------------------\|-------------------\|-------------------------\| \| ← The Quantum Model · \| General Chemistry \| · Shells and Orbitals → \| \| Book Cover · Introduction · v • d • e \| \| \| \| Units: Matter · Atomic Structure · Bonding · Reactions · Solutions · Phases of Matter · Equilibria · Kinetics · Thermodynamics · The Elements \| \| \| \| Appendices: Periodic Table · Units · Constants · Equations · Reduction Potentials · Elements and their Properties \| \| \| ## The Quantum Numbers These four numbers are used to describe the location of an electron in an atom. Number Symbol Possible Values Principal Quantum Number $n \,$ $\displaystyle 1, 2, 3, 4, \ldots$ Angular Momentum Quantum Number $\ell \,$ $\displaystyle 0, 1, 2, 3, \ldots , (n - 1)$ Magnetic Quantum Number $m_\text{l} \,$ $\displaystyle -\ell, \ldots, -1, 0, 1, \ldots ,\ell \,$ Spin Quantum Number $m_\text{s} \,$ $\displaystyle +1/2, -1/2$ ### Principal Quantum Number (n) Determines the shell the electron is in. The shell is the main component that determines the energy of the electron (higher n corresponds to higher energy), as well as nuclear distance (higher n means further from the nucleus). The row that an element is placed on the periodic table tells how many shells there will be. Helium (n = 1), neon (n = 2), argon (n = 3), etc. ### Angular Momentum Quantum Number (l) Also known as azimuthal quantum number. Determines the subshell the electron is in. Each subshell has a unique shape and a letter name. The s orbital is shaped like a sphere and occurs when l = 0. The p orbitals (there are three) are shaped like teardrops and occur when l = 1. The d orbitals (there are five) occur when l = 2. The f orbitals (there are seven) occur when l = 3. (By the way, when l = 4, the orbitals are "g orbitals", but they (and the l = 5 "h orbitals") can safely be ignored in general chemistry.) This number also gives information as to what the angular node of an orbital is. A node is defined as a point on a standing wave where the wave has minimal amplitude. When applied to chemistry this is the point of zero-displacement and thus where no electrons are found. In turn angular node means the planar or conical surface in which no electrons are found or where there is no electron density. Here are pictures of the orbitals. Keep in mind that they do not show the actual path of the electrons, due to the Heisenberg Uncertainty Principle. Instead, they show the area where the electron is most likely to occur (say, 90% of the probability). The two colors represent the two different spin numbers (the choice is arbitrary). m1 -3 -2 -1 0 1 2 3 S orbital → P orbitals → D orbitals → F orbitals → ### Magnetic Quantum Number (ml) Determines the orbital in which the electron lies. For example, there are three p orbitals in shell n = 2: the magnetic quantum number determines which one of these orbitals the electrons reside in. The different orbitals are oriented at different angles around the nucleus. See how each p orbital has the same general shape, but they point in different directions around the nucleus. ### Spin Quantum Number (ms) Determines the spin on the electron. ### Some Examples Let's examine the quantum numbers of electrons from a magnesium atom. Remember that each list of numbers corresponds to (n, l, ml, ms). \| \| \| \| \| \| \| \| \|------------------\|----------------\|----------------\|---------------\|---------------\|---------------\|---------------\| \| Two s electrons: \| (1, 0, 0, +½) \| (1, 0, 0, -½) \| \| \| \| \| \| Two s electrons: \| (2, 0, 0, +½) \| (2, 0, 0, -½) \| \| \| \| \| \| Six p electrons: \| (2, 1, -1, +½) \| (2, 1, -1, -½) \| (2, 1, 0, +½) \| (2, 1, 0, -½) \| (2, 1, 1, +½) \| (2, 1, 1, -½) \| \| Two s electrons: \| (3, 0, 0, +½) \| (3, 0, 0, -½) \| \| \| \| \| ### The Periodic Table Notice a pattern on the periodic table. Different areas, or blocks, have different types of electrons. The two columns on the left make the s-block. The six columns on the right make the p-block. The large area in the middle (transition metals) makes the d-block. The bottom portion makes the f-block. Each row introduces a new shell (aka energy level). Basically, the row tells you how many shells of electrons there will be, and the column tells you which subshells will occur (and which shells they occur in). The value of ml can be determined by some of the rules we will learn in the next chapter. The value of ms doesn't really matter as long as there are no repeating values in the same orbital. To see the pattern better, take a look at this periodic table.	{"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": 8, "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": 0.850818395614624, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/48809/relationships-between-the-roots-of-an-entire-function-and-the-roots-of-its-deriva	## Relationships between the roots of an entire function and the roots of its derivative ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Hey everyone, I would like to know if anybody could help me find references for the following. Take a suitably well defined entire function $f(x)$ and it's derivative $\tilde{f}(x)$ to which the roots $x_n$ and $\tilde{x}_n$ are associated. The function $f(x)$ may have infinitely many roots, though naturally one needs to be more careful in this case. Let's discount these subtleties for the time being. Define $$Z(s)=\sum_n\frac{1}{x_n^s} \hspace{8mm} \textrm{and} \hspace{8mm} \tilde{Z}(s)=\sum_n\frac{1}{\tilde{x}_n^s}.$$ One of the identities which I have proven which relates these is $$\tilde{Z}(3)=Z(3)-\left(\frac{Z(2)}{Z(1)}\right)^3+3\left(\frac{Z(3)Z(2)}{(Z(1))^2}-\frac{Z(4)}{Z(1)}\right).$$ I also have proven other identities (some much more simple) of this form. I have searched the internet, journals and every analysis book in my University library and found nothing of the sort. Also the main applications which I would expect to find curiously also do not appear in any of the aforementioned sources. Edit Just to say thanks to those who replied. The question has been answered :) - Sorry I have just realised that I completely misnamed the thread! – backstoreality Dec 9 2010 at 19:33 1 You can rename it if you edit the question. – Ryan Reich Dec 9 2010 at 19:53 Thank you Ryan :) – backstoreality Dec 9 2010 at 20:12 ## 2 Answers I don't know if there are any general results about these, but when $f$ is a polynomial, these must be in essence results about symmetric functions. If $f(z)=z^n+a_{n-1} z^{n-1}+\cdots+a_1z+a_0$ then $Z(1)=-a_1/a_0$, $Z(2)=Z(1)^2-2a_2/a_0$ etc. In this case your results surely specialize to polynomial identities in the $a_j$. - Right. The Z(s) here are the power symmetric functions, which are thoroughly studied in the literature. The \tilde{Z}(s) I am unsure about, but identities for the power symmetric functions will apply to them. – Qiaochu Yuan Dec 9 2010 at 19:46 Hey, Yes the results do specialise to give the results which you provide. However they also cover general entire functions, providing they obey certain qualities. Do you have any idea where I might find a list of references for the above polynomial identities? It might help Thanks very much! - Joe – backstoreality Dec 9 2010 at 19:46 @Joe: symmetric function identities are covered in Stanley's Enumerative Combinatorics (I can't remember which part at the moment), and they are also covered in Sagan's The Symmetric Group. – Qiaochu Yuan Dec 9 2010 at 19:54 @Qiaochu: Thank you, I found a copy on google books and I will have a read through. I guessed that my identities, when applied to polynomials, would surely be known. However it's seeming less likely that the results are known when applied to general entire functions – backstoreality Dec 9 2010 at 19:58 1 And also in Macdonald's Symmetric Functions and Hall Polynomials. I don't know off-hand if they have results like this though. – Robin Chapman Dec 9 2010 at 19:59 show 6 more comments ### 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. Let $f(x) = (1 - r_1 x)...(1 - r_n x)$ be a polynomial. Then $f(x) = 1 - e_1 x^1 + e_2 x^2 \mp ...$ where the $e_i$ are the elementary symmetric functions in the $r_i$. We define also $p_k = \sum_i r_i^k$, the power symmetric functions in the $r_i$. Then Newton's identities state that $$ke_k = \sum_{i=1}^{k} (-1)^{i-1} e_{k-i} p_i.$$ This identity is equivalent to the generating function identity $$\frac{f'(x)}{f(x)} = \sum_{i=1}^{n} \frac{r_i}{1 - r_i x} = \sum_{k \ge 0} p_{k+1} x^k$$ which follows from taking logarithmic derivatives on both sides. Now, you want to relate the functions $p_k$ to the functions $\tilde{p}_k$, the power symmetric functions of the reciprocals of the roots of the derivative $f'(x)$. Applying Newton's identities to $f'(x) = -e_1 + 2e_2 x - 3e_3 x^2 \pm ...$ gives $$k(k+1) \frac{e_{k+1}}{e_1} = \sum_{i=1}^k (-1)^{i-1} \frac{(k+1-i) e_{k+1-i}}{e_1} \tilde{p}_i.$$ This pair of identities should get you your results (at least formally, letting $n \to \infty$). You may or may not find it useful to write Newton's identities in the equivalent form $$e_n = \frac{1}{n!} \sum_{\sigma \in S_n} \text{sgn}(\sigma) p_{\sigma}$$ where $p_{\sigma} = p_{\lambda_1} ... p_{\lambda_i}$, where $\sigma$ has cycle type $(\lambda_1, ... \lambda_i)$. This gives $$\frac{(n+1) e_{n+1}}{e_1} = \frac{1}{n!} \sum_{\sigma \in S_n} \text{sgn}(\sigma) \tilde{p}_{\sigma}.$$ - Thank you again for your reply. This method seems similar to my method and I can see why it would give the same results, at least in the finite case :) – backstoreality Dec 9 2010 at 20:24 My method uses repeated derivatives of $\frac{f'(x)}{f(x)}$ evaluated at $x=0$ on infinite products. This is rigorous providing that the roots of the product grow at a sufficient rate for the product (in Hadamard form) to converge. – backstoreality Dec 9 2010 at 20:29 @Joe: the passage from the finite to the infinite case should be straightforward. – Qiaochu Yuan Dec 9 2010 at 20:33 @ Qiaochu: It is not completely straightforward as there is no assurance (that I can find) that the sum of the roots of $f'(x)$ should converge in the infinite case. In fact I only have on concrete example where I can prove that this is the case – backstoreality Dec 9 2010 at 20:59	{"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": 27, "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": 0.9445334076881409, "perplexity_flag": "head"}
http://mathhelpforum.com/trigonometry/171728-finding-height-angle-2.html	# Thread: 1. Originally Posted by scounged 12.5/24=(12.52)/(242)=25/48 I know it's rinsed by 2 but why do you times it by 2?? 2. Oh, that's just to get rid of the decimal in the numerator for simplicitys sake. 3. so would this mean the angle VAC is also 58.61? 4. Yes, I think so. 5. wouldn't you double it? because VCM if one triangle piece and VAC is double the size? 6. Originally Posted by andyboy179 wouldn't you double it? because VCM if one triangle piece and VAC is double the size? I think you're making this task more difficult than it has to be. Let's look at the pyramid from a side point of view: As you can see $\angle(A)=\angle(VAC)=\angle(VAM)$ Notice that the angles VAC and VAM are equal. This doesn't mean that $\triangle(VAC)=\triangle(VAM)$, which in fact is a false statement (oh, btw $\angle(VAC)=\angle(VCA) and \angle(VCM)$). Angles have nothing to do with the lengths of sides, it's a method of describing the difference in direction between two lines, regardless of the lengths. 7. VAM is 1 triangle part and VAC is 2 so why wouldn't you times 58.61 by 2?? i can't see how VAM and VAC are equal! VAM= VAC= 8. I'm not sure if I can explain this in a great way, but I'll try. We need to see the difference between what an angle is and what a triangle is. Therefore, I will strip my previous images from the sides h and VC (also the point C), leaving us with only two lines with an angle between them: In this picture we have two sides (VA and AM), and one angle(VAM). Let's see what happens if we extend the side AM a bit: In this picture the side AM has been extended. As we can see, the angle VAM has not been affected by this change of length in AM. Let's now examine what happens if we have a triangle. I'll create a triangle out of my first example by adding the side VM: Surprisingly enough, the angle VAM remains the same. Let's now take the triangle VAM and extend the side AM to see what happens: Let's see what has changed here. The length of the side AM has changed. The length of the side VM has changed. Even the area has changed. The angle VAM however, still remains the same. It seems like the angle VAM doesn't depend on the length of the side AM. Let's merge the triangles and name point M in the second triangle C, and call the distance of VM h: In this triangle we can see that the angle VAM is equal to the angle VAC (remember that C is the point M in the second triangle). Why is that? Well, it depends on the fact that the side AM has the same direction as the side AC, despite being of a different length, and an angle measures the difference of the direction of two lines, not the difference of lengths or anything else for that matter. I hope that this has cleared a few things up, and has helped you understand your problem a little better. 9. Oh I think I understand now, so VAC would be 58.61? 10. Also would the height = 20.49 11. Yes, you're absolutely right.	{"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": 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": 0.9668638110160828, "perplexity_flag": "middle"}
http://www.physicsforums.com/showthread.php?p=3805661	Physics Forums ## Help Understanding Stewart Chain Rule Proof [Picture Provided] Hi I've spent a long time trying to understand this chain rule proof but I just can't get it... I have attached 2 pictures: the second one is an intuitive chain rule proof that turns out to be bogus and the first is the correct proof. So I am trying to understand first of all what does the second proof actually mean, and secondly what did they do to fix the problem presented by the first invalid proof. Basically any and all help for me to gain insight into this proof would be appreciated. Don't be afraid to write too much or too little because I am very lost about what they are doing. What does the "property" that they lay out (resulting in equation 7) before they start the actual proof actually mean and how does it help their proof? Why introduce epsilon? I wish there was someone here in person to walk me through it but I guess I will have to try my best to get it over the internet. Thanks a lot! Attached Thumbnails 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 Here is the traditional proof of the chain rule: (Stewart's first proof made correct): We have a composite function y(u(x)), and assume both component functions y(u) and u(x) are differentiable, and we claim that also y(u(x)) is differentiable, and that its derivative equals y’(u).u’(x) = y’(u(x)).u’(x). All we have to do is show the limit of (∆y/∆x) as ∆x -->0, equals y’(u).u’(x). We are assuming that (∆u/∆x)-->u’(x) as ∆x -->0, and also that (∆y/∆u)-->y’(u) as ∆u -->0. Of course one needs to know the meaning of a limit. I.e. (∆y/∆u)-->y’(u) as ∆u -->0, means that the fraction (∆y/∆u) gets really close to the number y’(u) as long as ∆u is really small but not zero, and the same for the other limit. Unfortunately since the function u(x) is not assumed to be “one to one”, it can happen that two different values of x give the same value of u, and then we would have ∆u = 0 even though ∆x ≠ 0. So if we try to break up the fraction ∆y/∆x into a product (∆y/∆u)(∆u/∆x) and use the product rule for limits, we have a problem since this product may not really equal ∆y/∆x for all ∆x that is really small but not zero. I.e. if there is a small non zero ∆x such that ∆u = 0, then ∆y/∆x does not equal (∆y/∆u)( ∆u/∆x), since the fraction (∆y/∆u) does not make sense. Now we get to start from as small a ∆x as we want in this limit, so if there is ever a ∆x so small that ∆u ≠ 0 for that ∆x and also for all smaller ∆x, there is no problem. So the only case where we have not proved the chain rule is when there is a sequence of ∆x’s approaching zero, and for all of them we still have ∆u = 0. Now in that case, it follows that the fraction ∆u/∆x equals zero for all those ∆x’s, and since this fraction has a limit, the only possible limit is zero. I.e. in the only case where the proof does not work, we know that u’(x) = 0. Thus for the theorem to hold in that case, we only need to prove that y’(x) = y’(u).u’(x) = y’(u).0 = 0. I.e. all we have to do is prove that in this case the fraction ∆y/∆x still approaches zero even though we cannot always factor it into a product of fractions. The secret is to notice that we can still factor it as ∆y/∆x = (∆y/∆u)(∆u/∆x), as long as ∆u ≠ 0. I.e. there are two kinds of ∆x’s, those for which ∆u = 0, and those for which ∆u ≠ 0. But when ∆u = 0 we do not need to factor it, i.e. it is trivial then that the fraction ∆y/∆x = 0, since the top is the difference of the values of y at the same two values of u, so of course it equals zero. I.e. ∆u = 0 means the two values of u are the same, so y has the same value at bo0th of them so ∆y = 0, hence also ∆y/∆ = 0. And in the case where ∆u ≠ 0, we can still factor the fraction as ∆y/∆x = (∆y/∆u)(∆u/∆x), and use the other product argument. I.e. as long as ∆x is really small, if ∆u ≠ 0, then the fraction ∆y/∆x = (∆y/∆u)(∆u/∆x). And since u’(x) = 0 in this case, (∆u/∆x) is a small number, and (∆y/∆u) is close to the finite number y’(u), so the product (∆y/∆u)(∆u/∆x), is a small number. And in the case where ∆u = 0, things are actually even better. I.e. although we cannot factor the fraction, it does not matter because then ∆u= 0 implies also ∆y = 0, so the fraction ∆y/∆x is as close to zero as it can get, since it equals zero. Thus in the “bad” case where ∆x is small and non zero, but ∆u = 0, the chain rule still holds because both sides of the equaion equal zero. Thus Stewart’s second proof is unnecessary. It works because he has managed to take the denominators out of the argument. But he has also managed to make the argument less understandable. This result was traditionally proved correctly in turn of the century English language books, such as Pierpont's Theory of functions of a real variable, and in 19th century European books such as that of Tannery [see the article by Carslaw, in vol XXIX of B.A.M.S.], but unfortunately not in the first three editions of the influential book Pure Mathematics, by G.H.Hardy. Although Hardy reinstated the classical proof in later editions, modern books usually deal with the problem by giving the slightly more sophisticated linear approximation proof, or making what to me are somewhat artificial constructions. Summary: The point is simply that in proving a function has limit L, one only needs to prove it at points where the function does not already have value L. Thus to someone who says that the usual argument for the chain rule for y(u(x)), does not work for x's where ∆u = 0, one can simply reply that these points are irrelevant. Assume f is differentiable at g(a), g is differentiable at a, and on every neighborhood of a there are points x where g(x) = g(a). We claim the derivative of f(g(x)) at a equals f'(g(a))(g'(a)). Proof: 1) Clearly under these hypotheses, g'(a) = 0. Consequently, 2) the chain rule holds at a if and only if lim∆f/∆x = 0 as x approaches a. 3) Note that ∆f = ∆f/∆x = 0 at all x such that g(x) = g(a). 4) In general, to prove that lim h(x) = L, as x approaches a, it suffices to prove it for the restriction of h to those x such that h(x) ≠ L. 5) Thus in arguing that ∆f/∆x approaches 0, we may restrict to x such that g(x) ≠ g(a), where the usual argument applies. Recognitions: Homework Help Science Advisor The proof in Stewart is the linear approximation proof, that just says the same thing but takes out the division step. I.e. saying that deltay/deltax -->y'(a) as deltax -->0, is the same as saying that deltay/deltax - y'(a) -->0, i.e. that [deltay - y'(a)deltax]/deltax -->0. now if we multiply by deltax, we get that {[deltay - y'(a)deltax]/deltax}.delta x = = [deltay - y'(a)deltax] = e.deltax, where e-->0, i.e. e = [deltay - y'(a)deltax]/deltax. Thus we can state that y’(a) is the derivative of y without usin=g denominatiors, by saying that delta y = y’(a)deltax + e.deltax, where e-->0 as deltax does. so if y(x) = y(u(x)), in order to show that dy/dx (a) = y’(u(a)).(u’(a)), all we have to do is show that we can write y(u(x)) – y(u(a)) = [y’(u(a)).(u’(a))].deltax + E.deltax, where E is something that goes to zero as deltax does. This is a messy substitution using what we are given, i.e. y(u(x)) – y(u(a)) = y’(u(a)). delta u + e delta u, where e-->0, because y is differentiable wrt u. But u is also differentiable wrt x, so we can plug the same sort of thing in for delta u: y(u(x)) – y(u(a)) = y’(u(a)). {u’(a).delta x + e1.delta x} + e{u’(a).delta x + e1.delta x}, where e-->0 as delta u does, and e1-->0 as delta x does. Fortunately u is continuous in x, so delta u goes to zero when delta x does, hence e-->0 also when delta x does. Now just expand and collect terms getting: y(u(x)) – y(u(a)) = y’(u(a)). u’(a).delta x + {e1.y’(u(a)) + e.u’(a) + e1}.delta x} = y’(u(a)). u’(a).delta x + E. delta x, where E = {e1.y’(u(a)) + e.u’(a) + e1}, and that does go to zero as delta x does. Hence the multiplier of delta x must be the derivative dy/dx. i.e. dy/dx = y’(u(a)). u’(a). the one good thing about this more comp-licated proof is that this idea also works in several variables where you cannot just divide by the vector variable. also it reminds you that a derivative is really a linear approximation to the original function, which is important to know. But the idea that the original simpler proof does not work is just wrong, and may be evidence that a lot of calculus book writers just copy their stuff from other recent best sellers without thinking it through or doing historical research on the topic. Or to be fair, maybe they are aware of it but choose for pedagogical reasons not to mention it. ## Help Understanding Stewart Chain Rule Proof [Picture Provided] Quote by mathwonk Here is the traditional proof of the chain rule: (Stewart's first proof made correct): We have a composite function y(u(x)), and assume both component functions y(u) and u(x) are differentiable, and we claim that also y(u(x)) is differentiable, and that its derivative equals y’(u).u’(x) = y’(u(x)).u’(x). WANT TO SHOW: let dy = domain of y(u) du = domain of u(x) ru = range of u(x) $$\forall_{a\in d_y,\;b\in d_u,\;c\in r_u}\exists_{k_1, k_2, k_3\in\ \mathbb R}\;s.t.\;\left\{\lim_{x\to\ a}\frac{y(x)-y(a)}{y-a}=k_1 \wedge \lim_{x\to\ b}\frac{u(x)-u(b)}{x-b}=k_2 \wedge \lim_{x\to\ c}\frac{y(x)-y(c)}{y-c}=k_3\right\}\\\longrightarrow\;\forall_{a\in\ d_y,\;b\in\ d_u,\;c\in\ r_u}\left\{\frac{d}{dx}y(u(x)) = \left(\lim_{x\to\ a}\frac{y(x)-y(a)}{y-a}\cdot\lim_{x\to\ b}\frac{u(x)-u(b)}{x-b}\right) = \left(\frac{d}{dx}y(u(x))\cdot\lim_{x\to\ b}\frac{u(x)-u(b)}{x-b}\right)\right\}$$ ... did I interpret this first part correctly? I didn't know how to write the derivative of y(u(x)) in limit definition form. All we have to do is show the limit of (∆y/∆x) as ∆x -->0, equals y’(u).u’(x). ... do you say ∆y instead of ∆y(u(x)) because if we can show that lim ∆x->0(∆y/∆x) = y’(u).u’(x) for an arbitrary domain of y, then obviously the equality will hold true for the domain made up by the range of u(x)? We are assuming that (∆u/∆x)-->u’(x) as ∆x -->0, and also that (∆y/∆u)-->y’(u) as ∆u -->0. Of course one needs to know the meaning of a limit. I.e. (∆y/∆u)-->y’(u) as ∆u -->0, means that the fraction (∆y/∆u) gets really close to the number y’(u) as long as ∆u is really small but not zero, and the same for the other limit. ... okay I think i think this is consistent with what I typed in the "want to show" Unfortunately since the function u(x) is not assumed to be “one to one”, it can happen that two different values of x give the same value of u, and then we would have ∆u = 0 even though ∆x ≠ 0. ... okay because u(x) being one-to-one would imply each input x would give you a unique u(x) value, and we didn't assume this. So if we try to break up the fraction ∆y/∆x into a product (∆y/∆u)(∆u/∆x) and use the product rule for limits, we have a problem since this product may not really equal ∆y/∆x for all ∆x that is really small but not zero. I.e. if there is a small non zero ∆x such that ∆u = 0, then ∆y/∆x does not equal (∆y/∆u)( /∆x), since the fraction (∆y/∆u) does not make sense. [tex]\small\text{... okay because we have} \frac{∆y}{∆x}\text{and we want to multiply by} \frac{∆u}{∆u} \text{but} \frac{∆u}{∆u} \text{is undefined if ∆u=0}\\\small\text{and we can't guarantee ∆u≠0 since we didn't assume u(x) is 1-to-1}\\ \small\text{Also, if ∆u=0 for a nonzero ∆x that means there is at least one number p in the domain of u(x) where}\\\lim_{x\to\ p}\frac{u(x)-u(p)}{x-p} = \lim_{x\to\ p+∆x}\frac{u(x)-u(p+∆x)}{x-(p+∆x)}[/tex] i don't know if that's relevent or not (or even true for that matter) Now we get to start from as small a ∆x as we want in this limit, so if there is ever a ∆x so small that ∆u ≠ 0 for that ∆x and also for all smaller ∆x, there is no problem. So the only case where we have not proved the chain rule is when there is a sequence of ∆x’s approaching zero, and for all of them we still have ∆u = 0. gotcha Now in that case, it follows that the fraction ∆u/∆x equals zero for all those ∆x’s, and since this fraction has a limit, the only possible limit is zero. I.e. in the only case where the proof does not work, we know that u’(x) = 0. okay. Thus for the theorem to hold in that case, we only need to prove that y’(x) = y’(u).u’(x) = y’(u).0 = 0. I.e. all we have to do is prove that in this case the fraction ∆y/∆x still approaches zero even though we cannot always factor it into a product of fractions. is y'(x) the same as y'(u(x))? I thought we were trying to show y'(u(x)) = y'(u)*u'(x)? I was on board before because we were just doing ∆y/∆x which I assumed was considering u(x) to be the input into y. So I thought it was just saying "change in y(u(x)) resulting from a tiny change of x". But with y'(x) it seems like we're saying "change in y(u(x)) resulting from a tiny change in u(x)"? sorry; i'm stuck. edit: in the denominators of the limits in the "want to show" section i meant to have x-a and x-c not y-a and y-c can anyone help get me unstuck? I know it took me 5 months from the time I posted the initial question to when I posted my follow-up, but now I got stuck. :( I want to understand but I can't go anywhere with the proof at this point Thread Tools \| \| \| \| \|-------------------------------------------------------------------------------------\|-------------------------------\|---------\| \| Similar Threads for: Help Understanding Stewart Chain Rule Proof [Picture Provided] \| \| \| \| Thread \| Forum \| Replies \| \| \| Calculus & Beyond Homework \| 1 \| \| \| Calculus & Beyond Homework \| 5 \| \| \| Calculus & Beyond Homework \| 5 \| \| \| Calculus \| 15 \| \| \| Introductory Physics Homework \| 1 \|	{"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": 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": 0.9456312656402588, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/105088?sort=oldest	common dominating measure for a family of measures Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Given a family ${\mu }_{i\in I}$ on a Polish space (complete, separable metric space) $X$. When does there exist a measure $\lambda$ such that $$\mu_i=f_i \lambda$$ where the f_i are densities (Radon-Nikodym) of $mu_i$ wrt to $\lambda$. EDIT: What is a verifyable condition in the case I is uncountable. - What is $f_i$ supposed to be? – Aaron Tikuisis Aug 20 at 12:49 the densities/radon nikodym derivative – warsaga Aug 20 at 12:58 1 Answer A countable family of sigma-finite measures, yes. Can we drop a condition? Drop sigma-finite: consider two measures on $\mathbb R$: Lebesgue measure and counting measure. Drop countable: on $\mathbb R$ consider the family of measures, one of them is Lebesgue measure, and the rest are the unit point masses at all the points of $\mathbb R$. - I think that the OP has something about the dependence $i\mapsto \mu_i$ in mind. I could imagine positive answers for an interval $I$ and continuous dependence if the topology considered on all (finite) measures on $X$ is strong enough. – Jochen Wengenroth Aug 20 at 14:17	{"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": 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": 0.924249529838562, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/29883?sort=newest	## vanishing theorems ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I would be glad to know about possible generalizations of the following results: 1) (Grothendieck) Let $X$ be a noetherian topological space of dimension $n$. Then for all $i>n$ and all sheaves of abelian groups $\cal{F}$ on $X$, we have $H^i(X; \cal{F})=$ 0. [See Hartshorne, Algebraic Geometry, III.2.7.] 2) Let $X$ be an $n$-dimensional $C^0$-manifold. Then for all $i>n$ and all sheaves of abelian groups $\cal{F}$ on $X$, we have $H^i(X; \cal{F})=$ 0 . [See Kashiwara-Schapira, Sheaves on manifolds, III.3.2.2] More precisely, I'm interested in dropping the "abelian groups" hypothesis: could I take sheaves in any, say, AB5 abelian category? Apparently, in Grothendieck's theorem, the "abelian groups" hypothesis is necessary -at least in Hartshorne's proof-, because at the end you see a big constant sheaf $\mathbf{Z}$. But what happens if we talk about sheaves of $R$-modules, with $R$ any commutative ring with unit, for instance? Are those generalizations trivial ones? False for trivial reasons? Any hints or references will be welcome. - 2 For any sheaf of rings $O$, sheaf cohomology on the category of $O$-modules coincides with such cohomology on underlying abelian sheaves (due to acyclicity of flasques). So the generalizations are obvious. In (2) it isn't necessary to restrict to manifolds; any separable metric space (or disjoint union thereof) with dimension $n$ in the sense of topological dimension theory satisfies (2) (see Engelking's book "General Topology", especially the notion of "covering dimension"; recall that Cech = derived functor cohomology on paracompact Hausdorff spaces, and metric spaces are such spaces). – Boyarsky Jun 29 2010 at 10:44 @Boyarsky. Thanks. So the generalization to sheaves of O-modules is trivial. Do you know anything about possible generalizations to sheaves with values in an (AB5?) abelian category? – Agusti Roig Jun 29 2010 at 16:14 @Agusti: goodness, I can't even remember which one AB5 is...but is there some real reason for asking that kind of question? Like an example to motivate it? – Boyarsky Jun 29 2010 at 16:20 @Boyarsky. Thanks for your help again, Boyarsky. Well, I have a nasty spectral sequence with this kind of guys which I would like to converge strongly. For this, I need some zeros in it. As for AB5, is just a conjecture: exactness of filtered colimits seems to me, at first sight, the less you should ask -or the less I need- to work with sheaves with values in an abelian category – Agusti Roig Jun 29 2010 at 17:28 ## 1 Answer Well, I think I can answer my question, thanks to Boyarsky's remark. The point is that, since the theorem is also true for sheaves of $R$-modules, given a sheaf $\cal{F}$ with values in an abelian category $\cal{A}$, with the help of Mitchell's embedding theorem, http://en.wikipedia.org/wiki/Mitchell%27s_embedding_theorem, we can consider it as a sheaf of $R$-modules, for some ring $R$. Moreover, the embedding $V: {\cal A} \longrightarrow \mathbf{Mod}_R$ is full, faithful, and exact. That is to say, $V$ sends exact sequences to exact sequences. So $H^n(X;\cal{F})$ = $H^n(X;V(\cal{F}))$. Hence, both vanishing theorems are also (trivially) true for sheaves with values in any abelian category. - Unfortunately, the answer is wrong: see mathoverflow.net/questions/32173/… . – Agusti Roig Jul 16 2010 at 16:41	{"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": 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": 0.8940231800079346, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/53655?sort=votes	## Conformal-symplectic geometry ? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I think in priciple it's possible to consider a theory of "conformal-symplectic manifolds", in an analogous fashion as the usual conformal geometry. To spell out the spontaneous definitions: say that two symplectic forms $\omega_1$, $\omega_2$ on a smooth manifold $M$ are conformal to each other if there is a smooth positive function $\lambda \in \mathcal{C}^{\infty}(M,\mathbb{R}^{+})$ such that $\omega_1=\lambda\cdot \omega_2$ on $M$. Call a pair $(M,[\omega])$, with $[\omega]$ a conformal class of symplectic structures, a conformal-symplectic manifold. A smooth map $\varphi : M \to N$ between conformal-symplectic manifolds $(M,[\omega_1])$ and $(N,[\omega_2])$ is conformal-symplectic if $\varphi^(\omega_2)\in [\omega_1 ]$. Just out of curiosity, I would like to ask: Has such a theory been considered or studied? What can be said about these structures (provided it doesn't turn out to be somehow a "trivial" subject)? - ## 2 Answers If the manifold has dimension bigger than 2, I think the conformal class of $\omega$ is just $k\omega$ for constants $k$. Locally, by Darboux we can write $\omega = \sum_i dq^i \wedge dp_i$. If the dimension is greater than 2, the only way for $0 = d(f\omega) = df \wedge \omega$ is if $df = 0$. - Oh, so it was indeed a "trivial" subject, at least for symplectic manifolds (not almost symplectic)... – Qfwfq Jan 29 2011 at 1:25 ### 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. Yes, this has been considered (hasn't everything). See the following antique reference: http://archive.numdam.org/ARCHIVE/CTGDC/CTGDC_1978__19_3/CTGDC_1978__19_3_223_0/CTGDC_1978__19_3_223_0.pdf - It doesn't look like the same definition. The author there defines a "conformally symplectic* manifold" as a manifold that has a 2-form $\omega$ such that there's a 1-form $\rho$ such that $\mathrm{d} \omega = \rho \wedge \omega$. – Qfwfq Jan 28 2011 at 22:16 Ops, sorry, what I said above was what I just spotted on page 4... But it was not a definition but a theorem. – Qfwfq Jan 28 2011 at 22:21 Notice that this paper is talking about almost symplectic manifolds, so my comment does not apply. – Eric O. Korman Jan 28 2011 at 22:34	{"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": 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": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9298490881919861, "perplexity_flag": "middle"}
http://mathoverflow.net/revisions/16468/list	## Return to Question 6 discussed topological space structure Let X be an affine variety over ℂ. Consider X(ℂ) with the classical topology, and create the topologists loop space ΩX(ℂ) of maps from the circle into X(ℂ). One can also construct the ind-variety X((t)), whose R-points are given by X(R((t))) for any ℂ-algebra R. Take the ℂ-points of this ind-variety, and give them the usual topology. Is the topological space X((t))(ℂ) thus defined homotopy equivalent to ΩX(ℂ)? Edit: David Ben-Zvi's comment regarding using unbased loops instead of based loops is pertinent. We should be considering unbased loops (L not Ω). This checks out in the case where $X=\mathbb{G}_m$. The further commentary remark below applies to the affine Grassmannian case of based loopsalso provides positive evidence. Commentary (based on comments): Note that the space X((t)) is not the base change of X to ℂ((t)). It isn't the restriction of scalars either, since $R\otimes \mathbb{C}((t))\neq R((t))$ in general. I'll try to come back with more details, an example or a reference in due time, but if someone has one ready to go, let me know. Further commentary: Based on Regarding putting the comment classical topology of Stephen Griffeth, if loops on G is going to X((t))(ℂ), one should not be homotopic to the affine Grassmannian, and scared of the affine Grassmannian is ind-scheminess. ℂ((t)) has a homogenous space for the algebraic loop group, with the stabaliser natural structure of a point being topologically non-trivialtopological ring, then I guess and hence we topologise X(ℂ((t))) in the answer to usual manner, taking the original question as I asked it is nosubspace topology using a closed embedding into affine n-space for some n. [paragraph redacted] 5 added 280 characters in body Let X be an affine variety over ℂ. Consider X(ℂ) with the classical topology, and create the topologists loop space ΩX(ℂ) of maps from the circle into X(ℂ). One can also construct the ind-variety X((t)), whose R-points are given by X(R((t))) for any ℂ-algebra R. Take the ℂ-points of this ind-variety, and give them the usual topology. Is the topological space X((t))(ℂ) thus defined homotopy equivalent to ΩX(ℂ)? Edit: David Ben-Zvi's comment regarding using unbased loops instead of based loops is pertinent. We should be considering unbased loops (L not Ω). This checks out in the case where $X=\mathbb{G}_m$. The further commentary remark below applies to the case of based loops. Commentary (based on comments): Note that the space X((t)) is not the base change of X to ℂ((t)). It isn't the restriction of scalars either, since $R\otimes \mathbb{C}((t))\neq R((t))$ in general. I'll try to come back with more details, an example or a reference in due time, but if someone has one ready to go, let me know. Further commentary: Based on the comment of Stephen Griffeth, if loops on G is going to be homotopic to the affine Grassmannian, and the affine Grassmannian is a homogenous space for the algebraic loop group, with the stabaliser of a point being topologically non-trivial, then I guess the answer to the original question as I asked it is no. 4 further commentary added Let X be an affine variety over ℂ. Consider X(ℂ) with the classical topology, and create the topologists loop space ΩX(ℂ) of maps from the circle into X(ℂ). One can also construct the ind-variety X((t)), whose R-points are given by X(R((t))) for any ℂ-algebra R. Take the ℂ-points of this ind-variety, and give them the usual topology. Is the topological space X((t))(ℂ) thus defined homotopy equivalent to ΩX(ℂ)? Commentary (based on comments): Note that the space X((t)) is not the base change of X to ℂ((t)). It isn't the restriction of scalars either, since $R\otimes \mathbb{C}((t))\neq R((t))$ in general. I'll try to come back with more details, an example or a reference in due time, but if someone has one ready to go, let me know. Further commentary: Based on the comment of Stephen Griffeth, if loops on G is going to be homotopic to the affine Grassmannian, and the affine Grassmannian is a homogenous space for the algebraic loop group, with the stabaliser of a point being topologically non-trivial, then I guess the answer to the original question as I asked it is no. 3 it's not the restriction of scalars Let X be an affine variety over ℂ. Consider X(ℂ) with the classical topology, and create the topologists loop space ΩX(ℂ) of maps from the circle into X(ℂ). One can also construct the ind-variety X((t)), whose R-points are given by X(R((t))) for any ℂ-algebra R. Take the ℂ-points of this ind-variety, and give them the usual topology. Is the topological space X((t))(ℂ) thus defined homotopy equivalent to ΩX(ℂ)? Commentary (based on comments): Note that the space X((t)) is not the base change of X to ℂ((t)) but instead is ℂ((t)). It isn't the restriction of scalars either, since $R\otimes \mathbb{C}((t))\neq R((t))$ in general. I'll try to come back with more details, an example or a reference in due time, but if someone has one ready to go, let me know. 2 expanded definition of X((t)) Let X be an affine variety over ℂ. Consider X(ℂ) with the classical topology, and create the topologists loop space ΩX(ℂ) of maps from the circle into X(ℂ). One can also construct the ind-variety X((t)), whose R-points are given by X(R((t))) for any ℂ-algebra R. Take the ℂ-points of this ind-variety, and give them the usual topology. Is the topological space X((t))(ℂ) thus defined homotopy equivalent to ΩX(ℂ)? Commentary (based on comments): Note that the space X((t)) is not the base change of X to ℂ((t)) but instead is the restriction of scalars. I'll try to come back with more details, an example or a reference in due time, but if someone has one ready to go, let me know. 1 # Topologists loops versus algebraists loops Let X be an affine variety over ℂ. Consider X(ℂ) with the classical topology, and create the topologists loop space ΩX(ℂ) of maps from the circle into X(ℂ). One can also construct the ind-variety X((t)), whose R-points are given by X(R((t))) for any ℂ-algebra R. Take the ℂ-points of this ind-variety, and give them the usual topology. Is the topological space X((t))(ℂ) thus defined homotopy equivalent to ΩX(ℂ)?	{"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": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9147409796714783, "perplexity_flag": "middle"}
http://quant.stackexchange.com/questions/1320/vanilla-european-options-monte-carlo-vs-bs-formula?answertab=votes	# Vanilla European options: Monte carlo vs BS formula I have implemented a monte carlo simulation for a plain vanilla European Option and I am trying to compare it to the analytical result obtained from the BS formula. Assuming my monte carlo pricer is correctly implemented, am I supposed to get the very same result with both methods (Monte Carlo and BS analytical formula)? - with large amount of simulated paths (>200 000) - yes, you should have very close match. – Vytautas Jun 18 '11 at 17:19 @Vytautas. Thanks. What do you mean by close match? Up to how many decimals for instance? – balteo Jun 18 '11 at 17:25 You shouldn't be able to reject that the two solutions are different. – richardh♦ Jun 20 '11 at 3:07 The number of iterations is just a function of what size standard errors with which you are comfortable. – richardh♦ Jun 20 '11 at 3:08 Lets than 1 cent is considered good enough – Vytautas Jun 21 '11 at 15:34 ## 2 Answers Fundamentally this is no different from other simulation-based estimation---see this little experiment in R: ````R> set.seed(42) R> rowMeans(replicate(200,sapply(1:6, +> FUN=function(x) mean(rnorm(10^x)), simplify=TRUE))) [1] -2.47827e-02 -9.46800e-03 2.38226e-03 -1.08650e-03 9.41395e-05 1.06759e-05 R> ```` We are calculating the mean of a $N(0,1)$ vector for sample sizes from $10^1$ to $10^6$. That is then repeated 200 times, and we are calculating the mean of the 200 draws at the different sample sizes. We find that by and large, the mean gets closer to zero. But even at $10^6$, repeated 200 times, we are still pretty far from 'zero'. That is the way it goes with simulation, and it pays to get a feel for this. So while you have perfect benchmark with your analytical Black-Scholes result, you will be hard-pressed to get the difference to vanish completely. - Thanks! Just one last doubt remains: up to how many decimals do I need to get the difference to vanish? – balteo Jun 19 '11 at 12:35 As long as your simulations are independents, you can calculate some confidence interval thank's to the Central Limit Theorem and see if this interval is encompassing the true BS price. Regards - Thanks! – balteo Jun 19 '11 at 12:47	{"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": 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": 0.8762210607528687, "perplexity_flag": "middle"}
http://www.scholarpedia.org/article/Nambu-Jona-Lasinio_model	# Nambu-Jona-Lasinio model From Scholarpedia Giovanni Jona-Lasinio and Yoichiro Nambu (2010), Scholarpedia, 5(12):7487. Curator and Contributors 1.00 - Giovanni Jona-Lasinio The Nambu-Jona-Lasinio model is an effective chiral field theory realizing the spontaneous symmetry breaking mechanism. ## History Spontaneous breakdown of symmetry (SSB) is a concept that is applicable only to systems with infinitely many degrees of freedom. Although it pervaded the physics of condensed matter for a very long time, magnetism is a prominent example, its formalization and the recognition of its importance has been an achievement of the second half of the 20th century. Strangely enough the name was adopted only after its introduction in particle physics: it is due to Baker and Glashow (1962). What is SSB? In condensed matter physics it means that the lowest energy state of a system can have a lower symmetry than the forces acting among its constituents and on the system as a whole. As an example consider a long elastic bar on top of which we apply a compression force directed along its axis. Clearly there is rotational symmetry around the bar which is maintained as long as the force is not too strong: there is simply a shortening according to the Hooke's law. However when the force reaches a critical value the bar bends and we have an infinite number of equivalent lowest energy states which differ by a rotation. Heisenberg (1959, 1960) was probably the first to consider SSB as a possibly relevant concept in particle physics but his proposal was not the physically right one. The theory of superconductivity of Bardeen, Cooper and Schrieffer (1957) provided the key paradigm for the introduction of SSB in relativistic quantum field theory and particle physics on the basis of an analogy proposed by Nambu (1960a). To appreciate the innovative character of this concept in particle physics one should consider the strict dogmas which constituted the foundation of relativistic quantum field theory at the time. One of the dogmas stated that all the symmetries of the theory, implemented by unitary operators, must leave the lowest energy state, the vacuum, invariant. This property does not hold in presence of SSB and degenerate vacua. These vacua cannot be connected by local operations and are orthogonal to each other giving rise to different Hilbert spaces. If we live in one of them SSB will be manifested by its consequences, in particular the particle spectrum. The BCS theory of superconductivity, immediately after its appearance, was reformulated and developed by several authors including Bogolyubov, Valatin, Anderson, Ricayzen and Nambu. The following facts were emphasized 1. The ground state proposed by BCS is not invariant under gauge transformations. 2. The elementary fermionic excitations (quasi-particles) are not eigenstates of the charge as they appear as a superposition of an electron and a hole. 3. In order to restore charge conservation these excitations must be the source of bosonic excitations described by a long range (zero mass) field. In this way the original gauge invariance of the theory is restored. The peculiarity of the paper of Nambu (1960b), was that he used a language akin to quantum field theory, that is the Green's functions formalism, and the role of gauge invariance was discussed in terms of vertex functions and the associated Ward identities. The search for analogies in particle physics became quite natural. In particular, following the suggestion of Nambu (1960a), the study of chiral symmetry breaking was developed in detail in two papers by Nambu and Jona-Lasinio (1961a,1961b) which had a considerable influence on the evolution of elementary particle theories. ## The analogy with superconductivity Let us illustrate the elements of the analogy. Electrons near the Fermi surface are described by the following equation $\begin{align} E \psi_{p,+} &= \epsilon_p \psi_{p,+} + \phi \psi_{-p,-}^{\dagger} \\ E \psi_{-p,-}^{\dagger} &= -\epsilon_p \psi_{-p,-}^{\dagger} + \phi \psi_{p,+} , \end{align}$ with eigenvalues $E = \pm \sqrt{\epsilon_p^2 + \phi^2}.$ Here, $$\psi_{p,+}$$ and $$\psi_{-p,-}^{\dagger}$$ are the wavefunctions for an electron and a hole of momentum $$p$$ and spin $$+\ ;$$ $$\phi$$ is the gap. In the Weyl representation, the Dirac equation reads $\begin{align} E \psi_{1} &= \mathbf{\sigma} \cdot \mathbf{p} \psi_{1} + m \psi_{2} \\ E \psi_{2} &= -\mathbf{\sigma} \cdot \mathbf{p} \psi_{2} + m \psi_{1}, \end{align}$ with eigenvalues $E = \pm \sqrt{p^2 + m^2}.$ Here, $$\psi_{1}$$ and $$\psi_{2}$$ are the eigenstates of the chirality operator $$\gamma_5\ .$$ Particles with mass are superpositions of states of opposite chirality. The similarity is obvious. The bosonic excitations necessary to restore gauge invariance in a superconductor appear in the approximate expressions for the charge density and the current in a BCS superconductor (Nambu Y, 1960b), $\begin{align} \rho(x,t) &\simeq \rho_0 + \frac{1}{\alpha^2} \partial_t f \\ \mathbf{j}(x,t) &\simeq \mathbf{j}_0 - \mathbf{\nabla} f , \end{align}$ where $$\rho_0 = e \Psi^{\dagger} \sigma_3 Z \Psi$$ and $$\mathbf{j}_0 = e \Psi^{\dagger} (\mathbf{p}/m) Y \Psi$$ are the contributions of the quasi-particles, $$Y\ ,$$ $$Z\ ,$$ $$\alpha$$ are constants and $$f$$ satisfies the wave equation $\left( \nabla^2 - \frac{1}{\alpha^2} {\partial_t}^2 \right) f \simeq -2 e \Psi^{\dagger} \sigma_2 \phi \Psi.$ Here, $$\Psi^{\dagger} = (\psi^{\dagger}_1, \psi_2)\ .$$ In the elementary particle context the axial current $$\bar{\psi} \gamma_5 \gamma_{\mu} \psi$$ is the analog of the electromagnetic current in BCS theory. In the hypothesis of exact conservation, the matrix elements of the axial current between nucleon states of four-momentum $$p$$ and $$p'$$ have the form $\Gamma_{\mu}^A (p', p) = \left( i \gamma_5 \gamma_{\mu} - 2m \gamma_5 q_{\mu} / q^2 \right) F(q^2), \qquad q = p' - p .$ Exact conservation is compatible with a finite nucleon mass $$m$$ provided there exists a massless pseudoscalar particle. Assuming exact conservation of the chiral current, a picture of chiral SSB may consist in a vacuum of a massless Dirac field viewed as a sea of occupied negative energy states, and an attractive force between particles and antiparticles having the effect of producing a finite mass, the counterpart of the gap. The pseudoscalar massless particle, which may be interpreted as a forerunner of the pion, corresponds to the bosonic field associated to the fermionic quasi-particles in a superconductor. To implement this picture the construction of a relativistic field theoretic model was required. At that time Heisenberg and his collaborators had developed a comprehensive theory of elementary particles based on a non linear spinor interaction: the physical principle was that spin ½ fermions could provide the building blocks of all known elementary particles. Heisenberg was however very ambitious and wanted at the same time to solve in a consistent way the dynamical problem of a non renormalizable theory. This made their approach very complicated and not transparent. ## The Nambu-Jona-Lasinio model A Heisenberg type Lagrangian was adopted without pretending to solve the non-renormalizability problem and introducing a relativistic cut-off to cure the divergences. This model is known in the literature with the acronym NJL (Nambu-Jona-Lasinio model). The energy scale of interest was of the order of the nucleon mass and one hoped that higher energy effects would not change substantially the picture. The Lagrangian of the NJL model is $\mathcal {L} = -\bar{\psi} \gamma_{\mu} \partial_{\mu} \psi + g \left[ \left( \bar{\psi} \psi \right)^2 - \left( \bar{\psi} \gamma_5 \psi \right)^2 \right] .$ It is invariant under ordinary and chiral gauge transformations $\begin{align} \psi &\to e^{i \alpha} \psi, \qquad &\bar{\psi} &\to \bar{\psi} e^{-i \alpha} \\ \psi &\to e^{i \alpha \gamma_5} \psi, \qquad &\bar{\psi} &\to \bar{\psi} e^{i \alpha \gamma_5} . \end{align}$ To investigate the content of the model a simple mean field approximation for the mass was adopted $\begin{align} m &= -2g \left[ <\bar{\psi} \psi> - \gamma_5 <\bar{\psi} \gamma_5 \psi> \right] \\ &= 2g \left[ tr S^{(m)}(0) - tr \gamma_5 S^{(m)}(0) \right], \end{align}$ where $$S^{(m)}$$ is the propagator of the Dirac field of mass $$m\ ,$$ or more explicitly, $\frac{2 \pi^2}{g \Lambda^2} = 1 - \frac{m^2}{\Lambda^2} \ln \left(1 + \frac{\Lambda^2}{m^2} \right),$ where $$\Lambda$$ is the invariant cut-off. This equation is very similar to the gap equation in BCS theory. If $$\frac{2\pi^2}{g\Lambda^2} < 1$$ there exists a solution $$m>0\ .$$ From this relationship a rich spectrum of bound states follows, $\begin{array}{cccc} \hline \hline \text{nucleon} & \text{mass} \mu & \text{spin-parity} & \text{spectroscopic} \\ \text{number} & & & \text{notation} \\ \hline 0 & 0 & 0^- & ^1S_0 \\ 0 & 2m & 0^+ & ^3P_0 \\ 0 & \mu^2 > \frac{8}{3} m^2 & 1^- & ^3P_1 \\ \pm 2 & \mu^2 > 2 m^2 & 0^+ & ^1S_0 \\ \hline \hline \end{array}$ The bosonic field in the superconductor and the pseudoscalar particle in the NJL model are special cases of a general proposition formulated by Goldstone (1961). Whenever the original Lagrangian has a continuous symmetry group, which does not leave the ground state invariant, massless bosons appear in the spectrum of the theory. Other examples are: physical system broken symmetry massless bosons ferromagnets rotational invariance spin waves crystals translational and rotational invariance phonons These massless bosons are now known in the literature as Nambu-Goldstone bosons. In nature, however, the axial current is only approximately conserved. The model made contact with the real world under the hypothesis that the small violation of axial current conservation gives a mass to the massless boson, which is then identified with the $$\pi$$ meson. Actually the NJL model, reinterpreted in terms of quarks, has been the starting point of a successful effective theory of low energy QCD, see e.g. the review by Hatsuda and Kunihiro (1994). After the NJL model, SSB became a key concept in elementary particle physics, in particular electroweak unification (Weinberg S, 1967) requires SSB. ## References • Baker, M and Glashow, S L (1962). Spontaneous Breakdown of Elementary Particle Symmetries. Physical Review 128(5): 2462-2471. • Heisenberg, W; Dürr, H P; Mitter, H; Schlieder, S and Yamazaki, K (1959). Zur Theorie der Elementarteilchen. Zeitschrift für Naturforschung 14a: 441-485. • Heisenberg, W (1960). Recent research on the nonlinear spinor theory of elementary particles. Proc. 1960 Annual Intern. Conf. on High Energy Physics, Rochester : 851-858. • Bardeen, J; Cooper, L N and Schrieffer, J R (1957). Microscopic Theory of Superconductivity. Physical Review 106(1): 162-164. • Nambu, Y (1960a). Axial Vector Current Conservation in Weak Interactions. Physical Review Letters 4: 380-382. • Nambu, Y (1960b). Quasi-Particles and Gauge Invariance in the Theory of Superconductivity. Physical Review 117: 648-663. • Nambu, Y and Jona-Lasinio, G (1961a). Dynamical Model of Elementary Particles Based on an Analogy with Superconductivity. I. Physical Review 122(1): 345-358. • Nambu, Y and Jona-Lasinio, G (1961b). Dynamical Model of Elementary Particles Based on an Analogy with Superconductivity. II. Physical Review 124(1): 246-254. • Goldstone, J (1961). Field theories with Superconductor solutions. Nuovo Cimento 19: 154-164. • Hatsuda, T and Kunihiro, T (1994). QCD phenomenology based on a chiral effective Lagrangian. Physics Reports 247(5-6): 221-367. • Weinberg, S (1967). A Model of Leptons. Physical Review Letters 19(21): 1264-1266. ## See also Bardeen-Cooper-Schrieffer theory, Englert-Brout-Higgs-Guralnik-Hagen-Kibble mechanism, Gauge invariance, Gauge theories, Nambu-Jona-Lasinio model (Particle physics, Nuclear physics), Symmetry breaking in quantum systems	{"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": 25, "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": 0.9077751636505127, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/tagged/commutator+special-relativity	# Tagged Questions 2answers 205 views ### Causality and Quantum Field Theory I have a problem with proof of causality in Peskin & Schroeder, An Introduction to QFT, page 28. To avoid confusion I use three vectors notation, rewriting the Eq. (2.53) for $y=0$ as follows: ... 2answers 235 views ### In QFT, why does a vanishing commutator ensure causality? In relativistic quantum field theories (QFT), $$[\phi(x),\phi^\dagger(y)] = 0 \;\;\mathrm{if}\;\; (x-y)^2<0$$ On the other hand, even for space-like separation $$\phi(x)\phi^\dagger(y)\ne0.$$ ...	{"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": 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": 0.8337003588676453, "perplexity_flag": "middle"}
http://physics.aps.org/articles/large_image/f1/10.1103/Physics.2.99	Illustration: Alan Stonebraker Figure 1: (Top left) Many-particle systems exhibit complex, messy entanglement structures. (Top right) Entanglement combing straightens out the correlations among the parties so that ebits are created between Alice and each of the Bobs individually. A curly line on the right stands for one ebit, hence $n1=3$, $n2=1$, $n3=2$, and $n4=1$ in this example. (Bottom left) The initial entangled state of Alice, Bob, and Charlie is $\|ψ〉ABC$. If $S(B\|A)≤0$, then Bob can transfer (or merge) his state to Alice by sending classical messages only. (Bottom right) In addition to the successful transfer, Alice and Bob will share $-S(B\|A)$ ebits at the end of the protocol; in the displayed example we have $-S(B\|A)=2$.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 8, "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": 0.8946788907051086, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/278481/how-to-show-that-fracx2x-1-simplifies-to-x-frac1x-1-1/278488	# How to show that $\frac{x^2}{x-1}$ simplifies to $x + \frac{1}{x-1} +1$ How does $\frac{x^2}{(x-1)}$ simplify to $x + \frac{1}{x-1} +1$? The second expression would be much easier to work with, but I cant figure out how to get there. Thanks Alexander - $$\frac{x^2}{x-1}=\frac{x^2-1+1}{x-1}=\frac{(x-1)(x+1)+1}{x-1}=x+1+\frac1{x-1}$$ – lab bhattacharjee Jan 14 at 10:26 @Alexander I can see this is at least your second question on this site. When you get a satisfatory answer to your questions, you should accept it as an answer. – Git Gud Jan 14 at 10:59 This sort of problem has a universal solution called "the division algorithm" and "Bézout's identity ". – Luqing Ye Jan 14 at 12:12 ## 5 Answers A general $^1$ method is to perform the polynomial long division algorithm in the following form or another equivalent one: to get $$x^{2}=(x-1)(x+1)+1.$$ Divide both sides by $x-1$: $$\begin{eqnarray} \frac{x^{2}}{x-1} &=&\frac{(x-1)(x+1)+1}{x-1}=\frac{(x-1)(x+1)}{x-1}+\frac{1}{x-1} \\ &=&\frac{x+1}{1}+\frac{1}{x-1}=x+1+\frac{1}{x-1}=x+\frac{1}{x-1}+1 \end{eqnarray}$$ -- $^1$ The very same method can be applied to find (or show) that $$\frac{x^{4}-2x^{3}-6x^{2}+12x+15}{x^{3}+x^{2}-4x-4}=x-3+\frac{ x^{2}+4x+3}{x^{3}+x^{2}-4x-4};$$ In this case the polynomial long division algorithm corresponds to the following computation (or another equivalent one): - 1 Are the signs right in your long division? I've never seen it done like that before. – Peter Phipps Jan 14 at 11:34 @PeterPhipps The shown signs incorporate the minus sign of a difference. I posted this version because it is the output of the `\polylongdiv{x^2}{x-1}` LaTeX command (`\usepackage{polynom}`). – Américo Tavares Jan 14 at 11:53 1 Thanks for that. I've not seen the Polynom package. – Peter Phipps Jan 14 at 12:15 Very clever trick: If you have to show that two expressions are equivalent, you work backwards. $$\begin{align}=& x +\frac{1}{x-1} + 1 \\ \\ \\ =& \frac{x^2 - x}{x-1} +\frac{1}{x - 1} + \frac{x-1}{x-1} \\ \\ \\ =& \frac{x^2 - x + 1 + x - 1}{x - 1} \\ \\ \\ =&\frac{x^2 }{x - 1}\end{align}$$Now, write the steps backwards (if you're going to your grandmommy's place, you turn backwards and then you again turn backwards, you're on the right way!) and act like a know-it-all. $$\begin{align}=&\frac{x^2 }{x - 1} \\ \\ \\ =& \frac{x^2 - x}{x-1} +\frac{1}{x - 1} + \frac{x-1}{x-1} \\ \\ \\ =& x +\frac{1}{x-1} + 1 \end{align}$$ Q.E.Doodly dee! This trick works and you can impress your friends with such elegant proofs produced by this trick. - Instructive, but I $+1$ed it for the humor. – Git Gud Jan 14 at 10:45 2 +1 to emphasize the point that "show this is true" is a different problem than "figure out how to simplify $x^2/(x-1)$". Once you're in the "real world" rather than canned textbook problems, it is not uncommon that you can guess the answer through alternative means and then need to justify it. Eventually, you start solving some problems not by attacking it directly, but by trying to figure out ways to guess the answer! – Hurkyl Jan 14 at 11:32 +1 for the Q.E.Doodly dee. – dwarandae Jan 14 at 16:59 $\textbf{Hint:}$ Do you know about polynomial long divison? A simpler, though not universal way of dealing with this problem is noticing that: $$\frac{x^2}{x-1}=\frac{x^2-1+1}{x-1}=\frac{x^2-1}{x-1}+\frac{1}{x-1}=\frac{(x-1)(x+1)}{x-1}+\frac{1}{x-1}=x+1+\frac{1}{x-1}$$ It's not always this easy, though. So you should learn polynomial long divison. - We have that $$x^2=(x-1+1)^2=(x-1)^2+1+2(x-1)$$ and so for $x\neq 1$: $$\frac{x^2}{x-1}=\frac{(x-1)^2+1+2(x-1)}{x-1}=x-1+\frac{1}{x-1}+2=x+1+\frac{1}{x-1}.$$ - $$\frac{x^2}{x-1}=\frac{x^2-x+x-1+1}{x-1} = \frac{x(x-1) + (x-1)+1}{x-1}=\frac{x(x-1)}{x-1}+\frac{x-1}{x-1}+\frac{1}{x-1}$$ -	{"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": 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": 0.9289622902870178, "perplexity_flag": "middle"}
http://cs.stackexchange.com/questions/7577/time-complexity-for-synchronizer	# Time complexity for synchronizer In the field of Synchronizers on Distributed algorithms exists two type of time complexities of synchronizer $\xi$. Where $\xi$ denotes any synchronizer. $T_\text{pulse}(\xi) = \max \left \{ t_{\max}(p+1) -t_{\max}(p)\right \}$, where $t_{\max}(p)$ is the time when the slowest of the processors has progressed to pulse $p$. $T_\text{gap}(\xi) = \max \left \{ t(v,p+1) - t(v,p) \right \}$ among all nodes $v$ and all pulses $p$. $T_\text{gap}$ stands for the longest duration of some pulse among all nodes. The question is what is relationship between $T_\text{pulse}$ and $T_\text{gap}$. For me it seems intuitive that $T_\text{gap} \geq T_\text{pulse}$, because $T_\text{gap}$ shows the longest duration of the slowest node on the pulse $p$, and $T_\text{pulse}$ shows the duration of the longest pulse among all nodes, which is cannot be longer than the longest duration of the slowest node. In short, I don't have strong mathematical explanation, will appreciate for one. - Honestly I havent met $\xi$ synchronizer, I only know ABD, "Simple", $\alpha$, $\beta$ and $\gamma$. Could you provide some reference to $\xi$ synchronizer? – Bartek Dec 26 '12 at 22:05 Sorry, it's my mistake, $\xi$ denoting every synchronizer, the goal is to show relationship between $T_{gap}$ and $T_{pulse}$ for every synchronizer. – tam Dec 27 '12 at 5:34 Are there any additonal assumtpions about your distributed system model? Eg. how pulse messages are handled, are they prioritezed or handled after computation. Does pulse computation takes fixed time or are allowed to run for any time after pulse message. Because answer may vary depending on model assumptions. – Bartek Dec 28 '12 at 17:10 your question is not clear. I dont understand exactly what you mean by $T _{gap}$ and $T _{pulse}$. But usually, we dont study the time complexity of a "synchrnoizer" beceause there could be some random delays between pulses in asynchronous network. We usually study the message complexity of a synchrnoizer instead. – AJed Dec 28 '12 at 18:38 I don't agree. We can count time complexity for synchronizers. But under few additional assumptions. Like that message transimission time and message handling is bounded by some constant $\mu$ and $\bf pulse$ messages are handled as soon as they arrive to asynchronous process. This would allow us to see what is relation between $T_{\text{gap}}$ and $T_{\text{pulse}}$, relation would still be a little fuzzy but it will give us upper limit of relation. – Bartek Dec 29 '12 at 10:58 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": 29, "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": 0.9288378953933716, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/65642/symmetric-spaces-horocycle-spaces-and-intertwining-operators/92161	## Symmetric spaces, Horocycle spaces and intertwining operators ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Let $G=KAN$ be an Iwasawa decomposition of a connected semisimple Lie group with finite center. Let us assume for simplicity that the associated symmetric space $G/K$ has rank 1. Harish-Chandras plancherel theorem gives an explicit direct integral decomposition of the left-regular representation of $G$ on $L^2(G/K)$. $L^2(G/K) \cong \int_{i\mathbb{R}} \pi_\mu p(\mu)d\mu$. Here, $\pi_\mu$ is the spherical principle series representation with parameter $\mu$ and $p$ is an explicitly known density function. Now one has the normalized Knapp-Stein operators $A_\mu$ from $\pi_\mu$ to $\pi_{-\mu}$. These can be pieced together to give a unitary intertwining operator on the direct integral, and hence an operator $T$ on $L^2(G/K)$. Given the prominent role of the Knapp-Stein operators in representation theory I thought that the operator $T$ must have been studied extensively but I wasn't able to find anything on this subject. Specifically I'd like to know if the operator $T$ can be given in purely geometric terms without resorting to the direct integral decomposition. In a similiar vein, if $M$ is the centralizer of $A$ in $K$ then you can construct an analoguos operator on the space of $L^2$ functions on the horocycle space $G/MN$. Again, is there any way to define this operator in geometric terms without using the direct integral decomposition? I'd be very happy and grateful if a person more knowledgable then I could shed some light on this and could perhaps give a few references to artciles where these operators have been studied. - ## 2 Answers An explicit form is in my paper "A duality ... Advan. Math. 1970 and in my book Geometric Analysis on Symmetric Spaces AMS 2008, pp.554-556. Helgason. - @Sigurdur Helgason May I kindly ask you to look at my question mathoverflow.net/questions/92098/… , many thanks in advance... – Alexander Chervov Mar 25 2012 at 16:42 ### 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. I infer that the relevant normalizations of the intertwinings do not make them send the normalized spherical vector to itself, or else the Hilbert integral of these isomorphisms would be the identity map. A more interesting non-normalized version would be the integral over N whose meromorphic continuations gives the non-normalized intertwinings from pi(mu) to pi(-mu), whose effect on normalized spherical vectors is multiplication by an explicit constant (Gindikin-Karpelevich formula, even in general). This suggests that something like $Tf(g) = \int_N f(ng) dn$ is reasonable to consider on $L^2(G/K)$, insofar as it doesn't refer to the spectral decomposition. But I think it is not a bounded operator on $L^2(G/K)$. Certainly A. Knapp's book on Repn theory "by examples" discusses the individual intertwinings, but much less the spectral decomposition. Varadarajan's small book "introduction to harmonic analysis on semi-simple Lie groups" treats SL(2,R) in great detail. Hope this is relevant to the intent of your 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": 23, "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": 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": 0.9180399775505066, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/202556/bounding-the-number-of-monovalent-vertices-in-a-graph-satisfying-a-certain-condi	# Bounding the number of monovalent vertices in a graph satisfying a certain condition Show that a graph with exactly one vertex of degree $i$ : $2\le i \le m$ and $k$ other vertices (which are monovalent) satisfies $$\left\lfloor\frac{m+3}2\right\rfloor \le k\;.$$ Here is my current approach. I want to a construct a graph of minimal $k$ - to do this I need to make sure that all the all the other vertices of degree $i$ connect to the vertex of degree $m$, and that the other vertices of degree $m-1$, $m-2$, ... share the largest number of vertices that won't result in a double. Here is my current construction. Start with a vertex of degree $m$ $v_0$. choose some $v$ adjacent to that vertex, $v_1$. Connect it to $m-2$ of the other vertices giving it degree $m-1$ and creating $m-2$ vertices of degree $2$. At the next step, choose another of the vertices of degree $2$, call it $v_3$. connect it to $m-4$ vertices of degree $2$, leaving $1$ vertex of degree $2$, and creating $m-4$ vertices of degree $3$. This produces a sequence of graphs $G_n$, where each graph contains exactly one vertex of degree $1,\dots ,n+1$ and degree $m,\dots ,(m-n)+1$. Now at some point, we will reach an $i$ s.t $m-i = i+1$, at which point we will have two duplicate vertices. We then only have to move at most $i$ edges in order to get a graph with the "minimal number" of monovalent vertices satisfying this property. Problem is I can't prove this graph is minimal, and I can't figure out how to get the bound to arise from this construction. - ## 1 Answer The idea of this problem is to look at how many vertices we need to add such that the resulting degree sequence is actually realizable as a simple graph. You can draw a few examples yourself and you will quickly realize that the vertices of degree $2\le i \le m$ alone is not realizable as a graph. To produce a graph, we must add a number of monovalent vertices. Let us take a look at how many monovalent vertices we must add to produce a proper graph then. Suppose a graph with vertices of degree $i:\ 2\le i \le m$ along with $k$ monovalent vertices exist. Then the degree sequence of this graph $$(m, m-2, \cdots, 2, \underbrace{1, \cdots, 1}_k)$$ must satisfy the Erdős-Gallai Theorem, i.e. we must have $$\sum^{i}_{j=1}d_j\leq i(i-1)+ \sum^n_{j=i+1} \min(d_j,i)$$ for $1\le i \le n$ where $n$ is the number of vertices. The idea is to examine the associated inequalities at their "weakest" point. Consider the first $m-1$ vertices. Their associated numbering in the sequence is $$\begin{matrix}d_i=&m & m-1 & \cdots & 3 & 2\\i= & 1 & 2 & \cdots & m-2 & m-1 \end{matrix}$$ The point at which the latter sum begins to increase less rapidly is precisely when $i \ge d_{i+1}$ which first happens at $i=\lfloor\frac{m+1}{2}\rfloor$ and that is the point at which we want to take our inequality. To avoid tedious floor functions, I will work with even and odd cases separately (actually, I will do even and leave odd to you). Let $m=2\ell$. Then the inequality can be written as $$\sum_{j=1}^\ell d_j \le \ell(\ell-1) + \sum_{j=\ell+1}^{m-1}(d_j) + k$$ Evaluating the sums gives $$\sum_{j=1}^\ell d_j = m + \cdots + (m-\ell +1) = \frac{(2\ell)(2\ell+1)}{2} - \frac{(\ell + 1)(\ell)}{2}$$ $$\sum_{j=\ell+1}^{m-1}(d_j) = (m-\ell) + \cdots + 2 = \frac{(\ell + 1)(\ell)}{2} - 1$$ and putting it all together $$\frac{(2\ell)(2\ell+1)}{2} - \frac{(\ell + 1)(\ell)}{2} - \frac{(\ell + 1)(\ell)}{2} + 1 - \ell(\ell-1) \le k$$ after simplification, you yield $$\ell(2\ell + 1) - (\ell-1)\ell - (\ell + 1)\ell + 1 \le k$$ $$\ell + 1 = \frac{2\ell+2}{2} = \left\lfloor\frac{m + 3}{2}\right\rfloor\le k$$ The case for odd $m$ follows similarly and I will leave that proof to you. - Thank you very much, this was extremely helpful. – user42693 Sep 26 '12 at 4:17	{"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": 40, "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": 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": 0.9523604512214661, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/20307/free-boundary-conditions	# Free boundary conditions I am trying to simulate liquid film evaporation with free boundary conditions (in cartesian coordinates) and my boundary conditions are thus: $$\frac{\partial h}{\partial x} = 0, \qquad (1)$$ $$\frac{\partial^2 h}{\partial x^2} = 0, \qquad (2)$$ $$\frac{\partial^3 h}{\partial x^3}=0. \qquad (3)$$ However, I need only two of the above three conditions to satisfy my 4th order non-linear partial differential equation for film thickness, which looks something like. $$\frac{\partial h}{\partial t} + h^3\frac{\partial^3 h}{\partial x^3} + ... = 0$$ My question is: what does a combination of 1st and 2nd derivative conditions mean and what does a combination of 2nd and 3rd derivatives mean? If I apply (1) and (2), does it mean that slope and curvature are zero and if I apply (1) and (3), does it mean that slope and shear stress are zero (from analogies of bending beams etc.) - 1 @MaksimZholudev Thanks for making the edit. It looks much cleaner! – drN Jan 31 '12 at 16:48 ## 1 Answer Here is the answer that I gathered from months of looking at these boundary conditions: (1) and (2) would mean that the slope is zero and the bending moment / curvature at the ends is zero. (1) and (3) mean that the slope is zero and the shear stress at the end is zero. -	{"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": 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": 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": 0.9600231647491455, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/19170/generators-of-compact-lie-groups?answertab=votes	# Generators of compact Lie groups Suppose $G$ is a compact connected Lie group and let $\{X_i\}$ be a basis for its Lie algebra $\mathfrak g$. We know that the exponential $\exp:\mathfrak g \to G$ is surjective but when is it the case that $G$ is generated by $\{\exp(tX_i) : t\in \mathbb R\}$? - Ah, whoops... I forgot that the compact + connected implies surjectivity – Dylan Wilson Jan 27 '11 at 5:09 ## 1 Answer The map $\mathbb{R}^n \to G$ sending $(t_1, \dots, t_n)$ to $\mathrm{exp}(t_1 X_1) \dots \mathrm{exp}(t_n X_n)$ has nonsingular Jacobian at $0$, so its image contains a neighborhood of the origin. By a standard argument, a neighborhood of the origin in a connected topological group generates the full group. - Very nice! Thanks! – Eric O. Korman Jan 27 '11 at 3:04 Can you explain why for the compact connected Lie group the exponential map has to be surjective? (though there can be cases where the exponential map has a finite injectivity radius) – Anirbit Feb 17 '11 at 8:14 @Anirbit: One way to see this is to use the Hopf-Rinow theorem in Riemannian geometry (since the geodesics with respect to a bi-invariant metric are precisely $\exp(tX)$ for $X$ in the Lie algebra). Sorry for the very delayed response! – Akhil Mathew Feb 25 '12 at 16:00 @Anirbit Akhil's argument is great but if you do not wish to explicitly use Riemannian geometry, then there is an alternative proof if you assume Cartan's theorem on the maximal tori in a compact connected Lie group (Cartan's theorem can be proved using the Lefschetz fixed point theorem in algebraic topology). Indeed, each element of $G$ is an element of some maximal torus $T\subseteq G$ which means that we need only prove the surjectivity of the exponential map $\text{exp}:\mathfrak{t}\to T$ where $T$ is a torus. Of course, this is very easy to do. – Amitesh Datta Jun 27 '12 at 2:22	{"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": 16, "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": 0.8965293765068054, "perplexity_flag": "head"}
http://en.m.wikibooks.org/wiki/Trigonometry/Circles_and_Triangles/The_Excircles	# Trigonometry/Circles and Triangles/The Excircles An excircle of a triangle is a circle that has as tangents one side of the triangle and the other two sides extended. There are three such circles, one corresponding to each side of the triangle. The centre of each such circle, an excentre of the triangle, is at the intersection of the bisector of the angle opposite the side tangent to the circle and the external bisectors of the other two angles of the triangle. The proof is similar to that for the location of the incentre. Let the sides of a triangle be a, b and c and the angles opposite be A, B and C respectively. Denote the area of the triangle by Δ. Write s = 1⁄2(a+b+c). If the radius of the excircle touching side a is ra, then $r_a = \frac{\Delta}{s-a}$, with similar expressions for the other two excircles. If r is the inradius, we have $\frac{1}{r_a} + \frac{1}{r_b} + \frac{1}{r_c} = \frac{s-a}{\Delta} + \frac{s-b}{\Delta} + \frac{s-c}{\Delta} = \frac{s}{\Delta} = \frac{1}{r}$ If the vertices of a triangle are ABC and the excircle touching side BC does so at point D, then AB+BD = AC+CD. (This is easily proved using the theorem that the two tangents from a point to a circle are equal in length.) Both these expressions are s, the semiperimeter. The square of the distance of the excentre corresponding to side a from the circumcentre is R(R+2ra), with similar expressions for the other two excentres. Also, $\displaystyle r_a + r_b + r_c = r + 4R.$ The area of the triangle is $\displaystyle \sqrt {r_a r_b r_c r}.$ Three other expressions for ra are $r_a = a {{\cos(\frac{B}{2}) \cos(\frac{C}{2})} \over {\cos(\frac{A}{2})}}$ $r_a = 4R\sin(\frac{A}{2})\cos(\frac{B}{2})\cos(\frac{C}{2})$ $r_a = s \tan(\frac{A}{2})$ with similar expressions for rb and rc. It can readily be proved from the second of these expressions that if ra = r + rb + rc then A is a right angle. The distance of Ia from the three vertices A, B, C are $4R\cos\left(\frac{B}{2}\right)\cos\left(\frac{C}{2}\right) \text{, } 4R\sin\left(\frac{A}{2}\right)\cos\left(\frac{C}{2}\right) \text{, } 4R\sin\left(\frac{A}{2}\right)\cos\left(\frac{B}{2}\right)$ respectively, with similar expressions for the other excentres. $\displaystyle r.II_a.II_b.II_c = 4R.IA.IB.IC$	{"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": 9, "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": 0.8876017332077026, "perplexity_flag": "head"}
http://mathhelpforum.com/algebra/80441-equation-ellipse-standard-form.html	# Thread: 1. ## Equation of Ellipse in Standard Form I have to write the equation in standard form of the ellipse with foci (8,0) and (-8,0) if the minor axis has y-intercepts of 2 and -2. I know that the standard form of the equation if x^2/a^2 + y^2/b^2 =1. This is a horizontal elipse with center (0,0). The length of the minor axis is 4 units so I think c = 4 and c^2 = 16. I don't know where to go to find a^2 and b^2. Can you help? Thanks. Joanie 2. Originally Posted by Joanie I have to write the equation in standard form of the ellipse with foci (8,0) and (-8,0) if the minor axis has y-intercepts of 2 and -2. I know that the standard form of the equation if x^2/a^2 + y^2/b^2 =1. In any ellipse, the distance between the center of the ellipse and either focus is $c=\sqrt{a^2-b^2}.$ You know $b$ (it is half the length of the minor axis), and you know the distance between the center and the foci. So you can solve for $a^2.$ 3. ## Response to Equation of an Ellipse Thanks. I will do the problem and have you check my answer. Joanie 4. Originally Posted by Joanie I have to write the equation in standard form of the ellipse with foci (8,0) and (-8,0) if the minor axis has y-intercepts of 2 and -2. I know that the standard form of the equation if x^2/a^2 + y^2/b^2 =1. This is a horizontal elipse with center (0,0). The length of the minor axis is 4 units so I think c = 4 and c^2 = 16. I don't know where to go to find a^2 and b^2. Can you help? Thanks. Joanie Hi Joanie, $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ The two foci are (-c, 0) and (+c, 0). So c would have to be 8, wouldn't it? Not 4. The y-intercepts are at (0, b) and (0, -b) so b would have to be 2 and $b^2=4$. Using $c^2=a^2-b^2$, you can easily solve for $a^2$. Now just plug and chug.	{"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": 7, "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": 0.9403945803642273, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/206617/locally-convex-space-characterization-in-terms-of-duality?answertab=active	# Locally convex space characterization in terms of duality Let $(X,\tau)$ be a topological vector space (TVS for short). Denote by $X^$ the topological dual of $(X,\tau)$. If there exists a locally convex topology $\mu$ on $X$ compatible with the duality $(X,X^)$ (that is, $(X,\mu)^=X^$) and $\mu$ is finer than $\tau$ then is $(X,\tau)$ a locally convex space? An (equivalent) reformulation of the above question would be: - Is there an (infinite dimensional) locally convex space $(X,\mu)$, such that in between the weak topology, $w$ and $\mu$ there exists a linear topology $\tau$ on $X$ which is NOT locally convex? -	{"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": 15, "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": 0.9304940104484558, "perplexity_flag": "head"}
http://mathhelpforum.com/calculus/209354-consumers-surplus-calculus.html	2Thanks • 1 Post By richard1234 • 1 Post By skeeter Thread: 1. Consumers' Surplus for Calculus The demand function for a certain make of replacement cartridges for a water purifier is given by: p=-0.01x^2-0.1x+6 where p is the unit price in dollars and x is the quantity demanded each week, measured in units of a thousand. Determine the consumers' surplus if the market price is set at \$4/cartridge. I know the answer is \$11,667. How did they get this answer? Particularly, how would I find x for the bounds when I integrate? Please show steps in a thorough, understandable manner. (The problem involves integration.) Thank you. (If anyone asks for a supply function or any second equation -- there is none. This is how the question has been exactly written in the book.) 2. Re: Consumers' Surplus for Calculus Problem makes no sense. Why is the problem asking for a surplus (units: cartridges) when the answer is a dollar amount? 3. Re: Consumers' Surplus for Calculus Why would my book make me solve a problem that doesn't make sense? (That's also a question I'm asking myself since you say the problem makes no sense.) But that's exactly how the problem is written out in my book. I double-checked the answer and it is \$11,667. My book shows me how to solve problems like these (producers' and consumers' surplus) when they give me two equations (which I set them to each other, then set the resulting equation to 0 to find x - quantity demanded), but not what to do when I have only one equation. So, I took this equation, set it to 0 to find x, but gave me a weird number, which I used anyway. It didn't work. Or maybe it was the right number, but I did something else wrong. Someone has to know what do. There has to be someone who can help me. 4. Re: Consumers' Surplus for Calculus Originally Posted by AntoninaFinn So, I took this equation, set it to 0 to find x, but gave me a weird number, which I used anyway. It didn't work. Or maybe it was the right number, but I did something else wrong. You can't just set the RHS of that equation to zero. That would be like, solving for the demand when the price is \$0. If we let p = 4 and solve, we obtain x = 10 (10,000 cartridges). 5. Re: Consumers' Surplus for Calculus But that's not the answer in the book. Smiles So, are you saying that the book's answer is wrong? And so, if I set 4 to the equation, it would be: 4= -0.01x^2-0.1x+6 Subtract 4 to get to the other side, which would make: -.0.01x^2-0.1x+2 Is that what you mean? Thank you. 6. Re: Consumers' Surplus for Calculus Sorry, I might've mis-understood! I didn't know the difference between consumers' surplus and a surplus (they're two completely different things). According to Wikipedia, a consumers' surplus is "the monetary gain obtained by consumers because they are able to purchase a product for a price that is less than that they would be willing to pay." Using this definition, we find the highest price that consumers are willing to pay by maximizing $p$. To do this, we graph p and see that the highest p that produces a non-negative demand is p = 6. The consumers' surplus is equal to $S = \int_{4}^{6} x(p)\, dp$ where x(p) is the demand as a function of price. To find x(p) we must solve for x in the equation: $p = -0.01x^2 - 0.1x + 6 \Rightarrow 0.01x + 0.1x + (p-6) = 0$ $x = \frac{-0.1 + \sqrt{0.01 - 4(0.01)(p-6)}}{0.02} = -5 + 5\sqrt{25 - 4p}$ (taking positive root). However x is in thousands so we should multiply this expression by 1000 to get $1000(-5 + 5\sqrt{25 - 4p})$. Integrating from p = 4 to p = 6, $S = \int_{4}^{6} 1000(-5 + 5\sqrt{25-4p}) \, dp = \frac{35000}{3} = 11667$. Aha! Previously I didn't know the difference between "consumers' surplus" and "surplus." Sorry for the prior confusion. 7. Re: Consumers' Surplus for Calculus Aha! Previously I didn't know the difference between "consumers' surplus" and "surplus." Sorry for the prior confusion. No need for an apology ... it is incumbent on the original poster to clearly interpret all contextual terms used in posting a problem. 8. Re: Consumers' Surplus for Calculus I'm sorry, but I'm still confused. I've been going over the quadratic equation many times over and trying to figure out how you received -5+5(sq/25-4p) I've never solved a quadratic equation with an actual letter in it, so I was confused as to how to go about it. I thought to substitute 1 for a p so it would be (1-6), which within the square root of 0.1^2-4(0.01)(p-6) would equal square root 0.21 Many thanks if you would explain this part to me. And don't worry about confusing the "surplus" aspect, we all mess up sometimes. Good day. 9. Re: Consumers' Surplus for Calculus $p=-0.01x^2-0.1x+6$ $0.01x^2 + 0.1x + p - 6 = 0$ $a = 0.01$ $b = 0.1$ $c = p - 6$ $x = \frac{-0.1 + \sqrt{(0.1)^2 - 4(0.01)(p-6)}}{2(0.01)}$ $x = \frac{-0.1 + \sqrt{0.01 - 4(0.01)(p-6)}}{0.02}$ $x = \frac{-0.1 + \sqrt{0.01[1 - 4(p-6)]}}{0.02}$ $x = \frac{-0.1 + 0.1\sqrt{25 - 4p}}{0.02}$ $x = \frac{-0.1[1 - \sqrt{25 - 4p}]}{0.02}$ $x = -5 \left(1 - \sqrt{25-4p} \right)$ $x = 5 \left(\sqrt{25-4p} - 1 \right)$ 10. Re: Consumers' Surplus for Calculus Originally Posted by AntoninaFinn I'm sorry, but I'm still confused. I've been going over the quadratic equation many times over and trying to figure out how you received -5+5(sq/25-4p) Ah, good algebra skills are essential for a calculus course. Just use the quadratic formula and substitute a = 0.01, b = 0.1, c = p-6 as skeeter did.	{"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": 18, "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": 0.9640788435935974, "perplexity_flag": "middle"}
http://mathhelpforum.com/math-puzzles/167762-please-help-me-out.html	# Thread: 1. ## Please help me out on this For a 10x10x10 cube, 1000 cubes total (10^3), and 8^3 don't have paint on them. And what fraction of the cubes would have a painted side upward? 2. Originally Posted by abm For a 10x10x10 cube, 1000 cubes total (10^3), and 8^3 don't have paint on them. And what fraction of the cubes would have a painted side upward? I guess we assume the $8^3$ cubes are the "internal ones", since that is the number of cubes whose faces are not visible (think of an 8x8x8 cube covered by a new layer of cubes). If all of the cubes with "external faces" are painted, there are 2 ways to count them. The simpler way is "the number of cubes with external faces is the difference between the total amount of cubes and the amount of internal cubes". 3. Hello, abm! If I read the problem correctly, the question is rather silly. Besides, some introductory statements are missing: . . "An $n\!\times\!n\!\times\!n$ cube is painted on all sides, . . . . . then it is cut up into unit cubes." $\text{For a 10x10x10 cube, 1000 cubes total }(10^3),$ . . $\text{and }8^3\text{ don't have paint on them.}$ $\text{What fraction of the cubes would have a painted side upward?}$ Isn't that simply the top layer of cubes? . $\dfrac{100}{1000} \,=\,\dfrac{1}{10}$ 4. ## Thanks Thanks for your help. But can you tell me how to calculate internal cubes and external cubes. Suppose we have 3x3x3 cube what will be its internal and external cubes? 5. Originally Posted by abm Thanks for your help. But can you tell me how to calculate internal cubes and external cubes. Suppose we have 3x3x3 cube what will be its internal and external cubes? Let's say you meant to use the word "outward". You are in fact given the number of internal cubes. If the external cubes were removed, you'd have an 8x8x8 cube, since a layer of cubes would be removed from all 6 faces. Therefore the number of cubes with external faces are $10^3-8^3$ If you count them individually, there are 100 on the front and 100 on the back. There are 80 on the two sides that were not already counted (since cubes on the front and back have faces on the sides, top and bottom). Then there are 64 on the top and bottom not already counted. $200+160+128=488$ $10^3-8^3=1000-512=488$ Be sure that this isn't a trick question referring to the cubes on the top face only if placed flat on a table for instance. For a 3x3x3 cube, there is only 1 internal cube. The external cubes are 9+9+3+3+1+1=26 3x3x3=27.	{"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": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9298134446144104, "perplexity_flag": "middle"}
http://en.wikipedia.org/wiki/Euclidean_geometry	# Euclidean geometry "Plane geometry" redirects here. For other uses, see Plane geometry (disambiguation). A Greek mathematician (possilby Euclid or Archimedes) performing a geometric construction with a compass, from The School of Athens by Raphael Geometry Oxyrhynchus papyrus (P.Oxy. I 29) showing fragment of Euclid's Elements History of geometry Branches Research areas Important concepts Geometers Euclidean geometry is a mathematical system attributed to the Alexandrian Greek mathematician Euclid, which he described in his textbook on geometry: the Elements. Euclid's method consists in assuming a small set of intuitively appealing axioms, and deducing many other propositions (theorems) from these. Although many of Euclid's results had been stated by earlier mathematicians,[1] Euclid was the first to show how these propositions could fit into a comprehensive deductive and logical system.[2] The Elements begins with plane geometry, still taught in secondary school as the first axiomatic system and the first examples of formal proof. It goes on to the solid geometry of three dimensions. Much of the Elements states results of what are now called algebra and number theory, couched in geometrical language.[3] For over two thousand years, the adjective "Euclidean" was unnecessary because no other sort of geometry had been conceived. Euclid's axioms seemed so intuitively obvious (with the possible exception of the parallel postulate) that any theorem proved from them was deemed true in an absolute, often metaphysical, sense. Today, however, many other self-consistent non-Euclidean geometries are known, the first ones having been discovered in the early 19th century. An implication of Einstein's theory of general relativity is that physical space itself is not Euclidean, and Euclidean space is a good approximation for it only where the gravitational field is weak.[4] ## The Elements Main article: Euclid's Elements The Elements are mainly a systematization of earlier knowledge of geometry. Its superiority over earlier treatments was rapidly recognized, with the result that there was little interest in preserving the earlier ones, and they are now nearly all lost. Books I–IV and VI discuss plane geometry. Many results about plane figures are proved, e.g., If a triangle has two equal angles, then the sides subtended by the angles are equal. The Pythagorean theorem is proved.[5] Books V and VII–X deal with number theory, with numbers treated geometrically via their representation as line segments with various lengths. Notions such as prime numbers and rational and irrational numbers are introduced. The infinitude of prime numbers is proved. Books XI–XIII concern solid geometry. A typical result is the 1:3 ratio between the volume of a cone and a cylinder with the same height and base. The parallel postulate: If two lines intersect a third in such a way that the sum of the inner angles on one side is less than two right angles, then the two lines inevitably must intersect each other on that side if extended far enough. ### Axioms Euclidean geometry is an axiomatic system, in which all theorems ("true statements") are derived from a small number of axioms.[6] Near the beginning of the first book of the Elements, Euclid gives five postulates (axioms) for plane geometry, stated in terms of constructions (as translated by Thomas Heath):[7] "Let the following be postulated": 1. "To draw a straight line from any point to any point." 2. "To produce [extend] a finite straight line continuously in a straight line." 3. "To describe a circle with any centre and distance [radius]." 4. "That all right angles are equal to one another." 5. The parallel postulate: "That, if a straight line falling on two straight lines make the interior angles on the same side less than two right angles, the two straight lines, if produced indefinitely, meet on that side on which are the angles less than the two right angles." Although Euclid's statement of the postulates only explicitly asserts the existence of the constructions, they are also taken to be unique. The Elements also include the following five "common notions": 1. Things that are equal to the same thing are also equal to one another (Transitive property of equality). 2. If equals are added to equals, then the wholes are equal. 3. If equals are subtracted from equals, then the remainders are equal. 4. Things that coincide with one another equal one another (Reflexive Property). 5. The whole is greater than the part. ### Parallel postulate Main article: Parallel postulate To the ancients, the parallel postulate seemed less obvious than the others. They were concerned with creating a system which was absolutely rigorous and to them it seemed as if the parallel line postulate should have been able to be proven rather than simply accepted as a fact. It is now known that such a proof is impossible. Euclid himself seems to have considered it as being qualitatively different from the others, as evidenced by the organization of the Elements: the first 28 propositions he presents are those that can be proved without it. Many alternative axioms can be formulated that have the same logical consequences as the parallel postulate. For example Playfair's axiom states: In a plane, through a point not on a given straight line, at most one line can be drawn that never meets the given line. A proof from Euclid's elements that, given a line segment, an equilateral triangle exists that includes the segment as one of its sides. The proof is by construction: an equilateral triangle ΑΒΓ is made by drawing circles Δ and Ε centered on the points Α and Β, and taking one intersection of the circles as the third vertex of the triangle. ## Methods of proof Euclidean Geometry is constructive. Postulates 1, 2, 3, and 5 assert the existence and uniqueness of certain geometric figures, and these assertions are of a constructive nature: that is, we are not only told that certain things exist, but are also given methods for creating them with no more than a compass and an unmarked straightedge.[8] In this sense, Euclidean geometry is more concrete than many modern axiomatic systems such as set theory, which often assert the existence of objects without saying how to construct them, or even assert the existence of objects that cannot be constructed within the theory.[9] Strictly speaking, the lines on paper are models of the objects defined within the formal system, rather than instances of those objects. For example a Euclidean straight line has no width, but any real drawn line will. Though nearly all modern mathematicians consider nonconstructive methods just as sound as constructive ones, Euclid's constructive proofs often supplanted fallacious nonconstructive ones—e.g., some of the Pythagoreans' proofs that involved irrational numbers, which usually required a statement such as "Find the greatest common measure of ..."[10] Euclid often used proof by contradiction. Euclidean geometry also allows the method of superposition, in which a figure is transferred to another point in space. For example, proposition I.4, side-angle-side congruence of triangles, is proved by moving one of the two triangles so that one of its sides coincides with the other triangle's equal side, and then proving that the other sides coincide as well. Some modern treatments add a sixth postulate, the rigidity of the triangle, which can be used as an alternative to superposition.[11] ## System of measurement and arithmetic Euclidean geometry has two fundamental types of measurements: angle and distance. The angle scale is absolute, and Euclid uses the right angle as his basic unit, so that, e.g., a 45-degree angle would be referred to as half of a right angle. The distance scale is relative; one arbitrarily picks a line segment with a certain length as the unit, and other distances are expressed in relation to it. A line in Euclidean geometry is a model of the real number line. A line segment is a part of a line that is bounded by two end points, and contains every point on the line between its end points. Addition is represented by a construction in which one line segment is copied onto the end of another line segment to extend its length, and similarly for subtraction. Measurements of area and volume are derived from distances. For example, a rectangle with a width of 3 and a length of 4 has an area that represents the product, 12. Because this geometrical interpretation of multiplication was limited to three dimensions, there was no direct way of interpreting the product of four or more numbers, and Euclid avoided such products, although they are implied, e.g., in the proof of book IX, proposition 20. An example of congruence. The two figures on the left are congruent, while the third is similar to them. The last figure is neither. Note that congruences alter some properties, such as location and orientation, but leave others unchanged, like distance and angles. The latter sort of properties are called invariants and studying them is the essence of geometry. Euclid refers to a pair of lines, or a pair of planar or solid figures, as "equal" (ἴσος) if their lengths, areas, or volumes are equal, and similarly for angles. The stronger term "congruent" refers to the idea that an entire figure is the same size and shape as another figure. Alternatively, two figures are congruent if one can be moved on top of the other so that it matches up with it exactly. (Flipping it over is allowed.) Thus, for example, a 2x6 rectangle and a 3x4 rectangle are equal but not congruent, and the letter R is congruent to its mirror image. Figures that would be congruent except for their differing sizes are referred to as similar. Corresponding angles in a pair of similar shapes are congruent and corresponding sides are in proportion to each other. ## Notation and terminology ### Naming of points and figures Points are customarily named using capital letters of the alphabet. Other figures, such as lines, triangles, or circles, are named by listing a sufficient number of points to pick them out unambiguously from the relevant figure, e.g., triangle ABC would typically be a triangle with vertices at points A, B, and C. ### Complementary and supplementary angles Angles whose sum is a right angle are called complementary. Complementary angles are formed when one or more rays share the same vertex and are pointed in a direction that is in between the two original rays that form the right angle. The number of rays in between the two original rays are infinite. Those whose sum is a straight angle are supplementary. Supplementary angles are formed when one or more rays share the same vertex and are pointed in a direction that in between the two original rays that form the straight angle (180 degrees). The number of rays in between the two original rays are infinite like those possible in the complementary angle. ### Modern versions of Euclid's notation In modern terminology, angles would normally be measured in degrees or radians. Modern school textbooks often define separate figures called lines (infinite), rays (semi-infinite), and line segments (of finite length). Euclid, rather than discussing a ray as an object that extends to infinity in one direction, would normally use locutions such as "if the line is extended to a sufficient length," although he occasionally referred to "infinite lines." A "line" in Euclid could be either straight or curved, and he used the more specific term "straight line" when necessary. ## Some important or well known results • The bridge of asses theorem states that if A=B then C=D. • The sum of angles A, B, and C is equal to 180 degrees. • Pythagorean theorem: The sum of the areas of the two squares on the legs (a and b) of a right triangle equals the area of the square on the hypotenuse (c). • Thales' theorem: if AC is a diameter, then the angle at B is a right angle. ### Bridge of Asses The Bridge of Asses (Pons Asinorum) states that in isosceles triangles the angles at the base equal one another, and, if the equal straight lines are produced further, then the angles under the base equal one another.[12] Its name may be attributed to its frequent role as the first real test in the Elements of the intelligence of the reader and as a bridge to the harder propositions that followed. It might also be so named because of the geometrical figure's resemblance to a steep bridge that only a sure-footed donkey could cross.[13] ### Congruence of triangles Congruence of triangles is determined by specifying two sides and the angle between them (SAS), two angles and the side between them (ASA) or two angles and a corresponding adjacent side (AAS). Specifying two sides and an adjacent angle (SSA), however, can yield two distinct possible triangles. Triangles are congruent if they have all three sides equal (SSS), two sides and the angle between them equal (SAS), or two angles and a side equal (ASA) (Book I, propositions 4, 8, and 26). (Triangles with three equal angles (AAA) are similar, but not necessarily congruent. Also, triangles with two equal sides and an adjacent angle are not necessarily equal or congruent.) ### Sum of the angles of a triangle acute, obtuse, and right angle limits The sum of the angles of a triangle is equal to a straight angle (180 degrees).[14] This causes an equilateral triangle to have 3 interior angles of 60 degrees. Also, it causes every triangle to have at least 2 acute angles and up to 1 obtuse or right angle. ### Pythagorean theorem The celebrated Pythagorean theorem (book I, proposition 47) states that in any right triangle, the area of the square whose side is the hypotenuse (the side opposite the right angle) is equal to the sum of the areas of the squares whose sides are the two legs (the two sides that meet at a right angle). ### Thales' theorem Thales' theorem, named after Thales of Miletus states that if A, B, and C are points on a circle where the line AC is a diameter of the circle, then the angle ABC is a right angle. Cantor supposed that Thales proved his theorem by means of Euclid book I, prop 32 after the manner of Euclid book III, prop 31.[15] Tradition has it that Thales sacrificed an ox to celebrate this theorem.[16] ### Scaling of area and volume In modern terminology, the area of a plane figure is proportional to the square of any of its linear dimensions, $A \propto L^2$, and the volume of a solid to the cube, $V \propto L^3$. Euclid proved these results in various special cases such as the area of a circle[17] and the volume of a parallelepipedal solid.[18] Euclid determined some, but not all, of the relevant constants of proportionality. E.g., it was his successor Archimedes who proved that a sphere has 2/3 the volume of the circumscribing cylinder.[19] ## Applications This section requires expansion. (March 2009) Because of Euclidean geometry's fundamental status in mathematics, it would be impossible to give more than a representative sampling of applications here. • A surveyor uses a Level • Sphere packing applies to a stack of oranges. • A parabolic mirror brings parallel rays of light to a focus. As suggested by the etymology of the word, one of the earliest reasons for interest in geometry was surveying,[20] and certain practical results from Euclidean geometry, such as the right-angle property of the 3-4-5 triangle, were used long before they were proved formally.[21] The fundamental types of measurements in Euclidean geometry are distances and angles, and both of these quantities can be measured directly by a surveyor. Historically, distances were often measured by chains such as Gunter's chain, and angles using graduated circles and, later, the theodolite. An application of Euclidean solid geometry is the determination of packing arrangements, such as the problem of finding the most efficient packing of spheres in n dimensions. This problem has applications in error detection and correction. Geometric optics uses Euclidean geometry to analyze the focusing of light by lenses and mirrors. • Geometry is used in art and architecture. • The water tower consists of a cone, a cylinder, and a hemisphere. Its volume can be calculated using solid geometry. • Geometry can be used to design origami. Geometry is used extensively in architecture. Geometry can be used to design origami. Some classical construction problems of geometry are impossible using compass and straightedge, but can be solved using origami.[22] ## As a description of the structure of space Euclid believed that his axioms were self-evident statements about physical reality. Euclid's proofs depend upon assumptions perhaps not obvious in Euclid's fundamental axioms,[23] in particular that certain movements of figures do not change their geometrical properties such as the lengths of sides and interior angles, the so-called Euclidean motions, which include translations and rotations of figures.[24] Taken as a physical description of space, postulate 2 (extending a line) asserts that space does not have holes or boundaries (in other words, space is homogeneous and unbounded); postulate 4 (equality of right angles) says that space is isotropic and figures may be moved to any location while maintaining congruence; and postulate 5 (the parallel postulate) that space is flat (has no intrinsic curvature).[25] As discussed in more detail below, Einstein's theory of relativity significantly modifies this view. The ambiguous character of the axioms as originally formulated by Euclid makes it possible for different commentators to disagree about some of their other implications for the structure of space, such as whether or not it is infinite[26] (see below) and what its topology is. Modern, more rigorous reformulations of the system[27] typically aim for a cleaner separation of these issues. Interpreting Euclid's axioms in the spirit of this more modern approach, axioms 1-4 are consistent with either infinite or finite space (as in elliptic geometry), and all five axioms are consistent with a variety of topologies (e.g., a plane, a cylinder, or a torus for two-dimensional Euclidean geometry). ## Later work ### Archimedes and Apollonius A sphere has 2/3 the volume and surface area of its circumscribing cylinder. A sphere and cylinder were placed on the tomb of Archimedes at his request. Archimedes (ca. 287 BCE – ca. 212 BCE), a colorful figure about whom many historical anecdotes are recorded, is remembered along with Euclid as one of the greatest of ancient mathematicians. Although the foundations of his work were put in place by Euclid, his work, unlike Euclid's, is believed to have been entirely original.[28] He proved equations for the volumes and areas of various figures in two and three dimensions, and enunciated the Archimedean property of finite numbers. Apollonius of Perga (ca. 262 BCE–ca. 190 BCE) is mainly known for his investigation of conic sections. René Descartes. Portrait after Frans Hals, 1648. ### 17th century: Descartes René Descartes (1596–1650) developed analytic geometry, an alternative method for formalizing geometry which focused on turning geometry into algebra.[29] In this approach, a point is represented by its Cartesian (x, y) coordinates, a line is represented by its equation, and so on. In Euclid's original approach, the Pythagorean theorem follows from Euclid's axioms. In the Cartesian approach, the axioms are the axioms of algebra, and the equation expressing the Pythagorean theorem is then a definition of one of the terms in Euclid's axioms, which are now considered theorems. The equation $\|PQ\|=\sqrt{(p-r)^2+(q-s)^2} \,$ defining the distance between two points P = (p, q) and Q = (r, s) is then known as the Euclidean metric, and other metrics define non-Euclidean geometries. In terms of analytic geometry, the restriction of classical geometry to compass and straightedge constructions means a restriction to first- and second-order equations, e.g., y = 2x + 1 (a line), or x2 + y2 = 7 (a circle). Also in the 17th century, Girard Desargues, motivated by the theory of perspective, introduced the concept of idealized points, lines, and planes at infinity. The result can be considered as a type of generalized geometry, projective geometry, but it can also be used to produce proofs in ordinary Euclidean geometry in which the number of special cases is reduced.[30] Squaring the circle: the areas of this square and this circle are equal. In 1882, it was proven that this figure cannot be constructed in a finite number of steps with an idealized compass and straightedge. ### 18th century Geometers of the 18th century struggled to define the boundaries of the Euclidean system. Many tried in vain to prove the fifth postulate from the first four. By 1763 at least 28 different proofs had been published, but all were found incorrect.[31] Leading up to this period, geometers also tried to determine what constructions could be accomplished in Euclidean geometry. For example, the problem of trisecting an angle with a compass and straightedge is one that naturally occurs within the theory, since the axioms refer to constructive operations that can be carried out with those tools. However, centuries of efforts failed to find a solution to this problem, until Pierre Wantzel published a proof in 1837 that such a construction was impossible. Other constructions that were proved impossible include doubling the cube and squaring the circle. In the case of doubling the cube, the impossibility of the construction originates from the fact that the compass and straightedge method involve first- and second-order equations, while doubling a cube requires the solution of a third-order equation. Euler discussed a generalization of Euclidean geometry called affine geometry, which retains the fifth postulate unmodified while weakening postulates three and four in a way that eliminates the notions of angle (whence right triangles become meaningless) and of equality of length of line segments in general (whence circles become meaningless) while retaining the notions of parallelism as an equivalence relation between lines, and equality of length of parallel line segments (so line segments continue to have a midpoint). ### 19th century and non-Euclidean geometry In the early 19th century, Carnot and Möbius systematically developed the use of signed angles and line segments as a way of simplifying and unifying results.[32] The century's most significant development in geometry occurred when, around 1830, János Bolyai and Nikolai Ivanovich Lobachevsky separately published work on non-Euclidean geometry, in which the parallel postulate is not valid.[33] Since non-Euclidean geometry is provably relatively consistent with Euclidean geometry, the parallel postulate cannot be proved from the other postulates. In the 19th century, it was also realized that Euclid's ten axioms and common notions do not suffice to prove all of theorems stated in the Elements. For example, Euclid assumed implicitly that any line contains at least two points, but this assumption cannot be proved from the other axioms, and therefore must be an axiom itself. The very first geometric proof in the Elements, shown in the figure above, is that any line segment is part of a triangle; Euclid constructs this in the usual way, by drawing circles around both endpoints and taking their intersection as the third vertex. His axioms, however, do not guarantee that the circles actually intersect, because they do not assert the geometrical property of continuity, which in Cartesian terms is equivalent to the completeness property of the real numbers. Starting with Moritz Pasch in 1882, many improved axiomatic systems for geometry have been proposed, the best known being those of Hilbert,[34] George Birkhoff,[35] and Tarski.[36] ### 20th century and general relativity A disproof of Euclidean geometry as a description of physical space. In a 1919 test of the general theory of relativity, stars (marked with short horizontal lines) were photographed during a solar eclipse. The rays of starlight were bent by the Sun's gravity on their way to the earth. This is interpreted as evidence in favor of Einstein's prediction that gravity would cause deviations from Euclidean geometry. Einstein's theory of general relativity shows that the true geometry of spacetime is not Euclidean geometry.[37] For example, if a triangle is constructed out of three rays of light, then in general the interior angles do not add up to 180 degrees due to gravity. A relatively weak gravitational field, such as the Earth's or the sun's, is represented by a metric that is approximately, but not exactly, Euclidean. Until the 20th century, there was no technology capable of detecting the deviations from Euclidean geometry, but Einstein predicted that such deviations would exist. They were later verified by observations such as the slight bending of starlight by the Sun during a solar eclipse in 1919, and such considerations are now an integral part of the software that runs the GPS system.[38] It is possible to object to this interpretation of general relativity on the grounds that light rays might be improper physical models of Euclid's lines, or that relativity could be rephrased so as to avoid the geometrical interpretations. However, one of the consequences of Einstein's theory is that there is no possible physical test that can distinguish between a beam of light as a model of a geometrical line and any other physical model. Thus, the only logical possibilities are to accept non-Euclidean geometry as physically real, or to reject the entire notion of physical tests of the axioms of geometry, which can then be imagined as a formal system without any intrinsic real-world meaning. ## Treatment of infinity ### Infinite objects Euclid sometimes distinguished explicitly between "finite lines" (e.g., Postulate 2) and "infinite lines" (book I, proposition 12). However, he typically did not make such distinctions unless they were necessary. The postulates do not explicitly refer to infinite lines, although for example some commentators interpret postulate 3, existence of a circle with any radius, as implying that space is infinite.[26] The notion of infinitesimally small quantities had previously been discussed extensively by the Eleatic School, but nobody had been able to put them on a firm logical basis, with paradoxes such as Zeno's paradox occurring that had not been resolved to universal satisfaction. Euclid used the method of exhaustion rather than infinitesimals.[39] Later ancient commentators such as Proclus (410–485 CE) treated many questions about infinity as issues demanding proof and, e.g., Proclus claimed to prove the infinite divisibility of a line, based on a proof by contradiction in which he considered the cases of even and odd numbers of points constituting it.[40] At the turn of the 20th century, Otto Stolz, Paul du Bois-Reymond, Giuseppe Veronese, and others produced controversial work on non-Archimedean models of Euclidean geometry, in which the distance between two points may be infinite or infinitesimal, in the Newton–Leibniz sense.[41] Fifty years later, Abraham Robinson provided a rigorous logical foundation for Veronese's work.[42] ### Infinite processes One reason that the ancients treated the parallel postulate as less certain than the others is that verifying it physically would require us to inspect two lines to check that they never intersected, even at some very distant point, and this inspection could potentially take an infinite amount of time.[43] The modern formulation of proof by induction was not developed until the 17th century, but some later commentators consider it implicit in some of Euclid's proofs, e.g., the proof of the infinitude of primes.[44] Supposed paradoxes involving infinite series, such as Zeno's paradox, predated Euclid. Euclid avoided such discussions, giving, for example, the expression for the partial sums of the geometric series in IX.35 without commenting on the possibility of letting the number of terms become infinite. ## Logical basis This article needs attention from an expert in mathematics. Please add a reason or a talk parameter to this template to explain the issue with the article. WikiProject Mathematics (or its Portal) may be able to help recruit an expert. (December 2010) This section requires expansion. (June 2010) See also: Hilbert's axioms, Axiomatic system, and Real closed field ### Classical logic Euclid frequently used the method of proof by contradiction, and therefore the traditional presentation of Euclidean geometry assumes classical logic, in which every proposition is either true or false, i.e., for any proposition P, the proposition "P or not P" is automatically true. ### Modern standards of rigor Placing Euclidean geometry on a solid axiomatic basis was a preoccupation of mathematicians for centuries.[45] The role of primitive notions, or undefined concepts, was clearly put forward by Alessandro Padoa of the Peano delegation at the 1900 Paris conference:[45][46] ...when we begin to formulate the theory, we can imagine that the undefined symbols are completely devoid of meaning and that the unproved propositions are simply conditions imposed upon the undefined symbols. Then, the system of ideas that we have initially chosen is simply one interpretation of the undefined symbols; but..this interpretation can be ignored by the reader, who is free to replace it in his mind by another interpretation.. that satisfies the conditions... Logical questions thus become completely independent of empirical or psychological questions... The system of undefined symbols can then be regarded as the abstraction obtained from the specialized theories that result when...the system of undefined symbols is successively replaced by each of the interpretations... —Padoa, Essai d'une théorie algébrique des nombre entiers, avec une Introduction logique à une théorie déductive qulelconque That is, mathematics is context-independent knowledge within a hierarchical framework. As said by Bertrand Russell:[47] If our hypothesis is about anything, and not about some one or more particular things, then our deductions constitute mathematics. Thus, mathematics may be defined as the subject in which we never know what we are talking about, nor whether what we are saying is true. —Bertrand Russell, Mathematics and the metaphysicians Such foundational approaches range between foundationalism and formalism. ### Axiomatic formulations Geometry is the science of correct reasoning on incorrect figures. —George Polyá, How to Solve It, p. 208 • Euclid's axioms: In his dissertation to Trinity College, Cambridge, Bertrand Russell summarized the changing role of Euclid's geometry in the minds of philosophers up to that time.[48] It was a conflict between certain knowledge, independent of experiment, and empiricism, requiring experimental input. This issue became clear as it was discovered that the parallel postulate was not necessarily valid and its applicability was an empirical matter, deciding whether the applicable geometry was Euclidean or non-Euclidean. • Hilbert's axioms: Hilbert's axioms had the goal of identifying a simple and complete set of independent axioms from which the most important geometric theorems could be deduced. The outstanding objectives were to make Euclidean geometry rigorous (avoiding hidden assumptions) and to make clear the ramifications of the parallel postulate. • Birkhoff's axioms: Birkhoff proposed four postulates for Euclidean geometry that can be confirmed experimentally with scale and protractor.[49][50][51] The notions of angle and distance become primitive concepts.[52] • Tarski's axioms: Alfred Tarski (1902–1983) and his students defined elementary Euclidean geometry as the geometry that can be expressed in first-order logic and does not depend on set theory for its logical basis,[53] in contrast to Hilbert's axioms, which involve point sets.[54] Tarski proved that his axiomatic formulation of elementary Euclidean geometry is consistent and complete in a certain sense: there is an algorithm that, for every proposition, can be shown either true or false.[36] (This doesn't violate Gödel's theorem, because Euclidean geometry cannot describe a sufficient amount of arithmetic for the theorem to apply.[55]) This is equivalent to the decidability of real closed fields, of which elementary Euclidean geometry is a model. ### Constructive approaches and pedagogy The process of abstract axiomatization as exemplified by Hilbert's axioms reduces geometry to theorem proving or predicate logic. In contrast, the Greeks used construction postulates, and emphasized problem solving.[56] For the Greeks, constructions are more primitive than existence propositions, and can be used to prove existence propositions, but not vice versa. To describe problem solving adequately requires a richer system of logical concepts.[56] The contrast in approach may be summarized:[57] • Axiomatic proof: Proofs are deductive derivations of propositions from primitive premises that are ‘true’ in some sense. The aim is to justify the proposition. • Analytic proof: Proofs are non-deductive derivations of hypotheses from problems. The aim is to find hypotheses capable of giving a solution to the problem. One can argue that Euclid's axioms were arrived upon in this manner. In particular, it is thought that Euclid felt the parallel postulate was forced upon him, as indicated by his reluctance to make use of it,[58] and his arrival upon it by the method of contradiction.[59] Andrei Nicholaevich Kolmogorov proposed a problem solving basis for geometry.[60][61] This work was a precursor of a modern formulation in terms of constructive type theory.[62] This development has implications for pedagogy as well.[63] If proof simply follows conviction of truth rather than contributing to its construction and is only experienced as a demonstration of something already known to be true, it is likely to remain meaningless and purposeless in the eyes of students. —Celia Hoyles, The curricular shaping of students' approach to proof ## Notes 1. Eves, vol. 1., p. 19 2. Eves (1963), vol. 1, p. 10 3. Eves, p. 19 4. Misner, Thorne, and Wheeler (1973), p. 47 5. Euclid, book I, proposition 47 6. The assumptions of Euclid are discussed from a modern perspective in Harold E. Wolfe (2007). Introduction to Non-Euclidean Geometry. Mill Press. p. 9. ISBN 1-4067-1852-1. 7. tr. Heath, pp. 195–202. 8. Ball, p. 56 9. Within Euclid's assumptions, it is quite easy to give a formula for area of triangles and squares. However, in a more general context like set theory, it is not as easy to prove that the area of a square is the sum of areas of its pieces, for example. See Lebesgue measure and Banach–Tarski paradox. 10. Daniel Shanks (2002). Solved and Unsolved Problems in Number Theory. American Mathematical Society. 11. Coxeter, p. 5 12. Euclid, book I, proposition 5, tr. Heath, p. 251 13. Ignoring the alleged difficulty of Book I, Proposition 5, Sir Thomas L. Heath mentions another interpretation. This rests on the resemblance of the figure's lower straight lines to a steeply-inclined bridge that could be crossed by an ass but not by a horse: "But there is another view (as I have learnt lately) which is more complimentary to the ass. It is that, the figure of the proposition being like that of a trestle bridge, with a ramp at each end which is more practicable the flatter the figure is drawn, the bridge is such that, while a horse could not surmount the ramp, an ass could; in other words, the term is meant to refer to the surefootedness of the ass rather than to any want of intelligence on his part." (in "Excursis II," volume 1 of Heath's translation of The Thirteen Books of the Elements.) 14. Euclid, book I, proposition 32 15. Heath, p. 318 16. Euclid, book XII, proposition 2 17. Euclid, book XI, proposition 33 18. Ball, p. 66 19. Ball, p. 5 20. Eves, vol. 1, p. 5; Mlodinow, p. 7 21. 22. Richard J. Trudeau (2008). "Euclid's axioms". The Non-Euclidean Revolution. Birkhäuser. pp. 39 'ff. ISBN 0-8176-4782-1. 23. See, for example: Luciano da Fontoura Costa, Roberto Marcondes Cesar (2001). Shape analysis and classification: theory and practice. CRC Press. p. 314. ISBN 0-8493-3493-4. and Helmut Pottmann, Johannes Wallner (2010). Computational Line Geometry. Springer. p. 60. ISBN 3-642-04017-9. The group of motions underlie the metric notions of geometry. See Felix Klein (2004). Elementary Mathematics from an Advanced Standpoint: Geometry (Reprint of 1939 Macmillan Company ed.). Courier Dover. p. 167. ISBN 0-486-43481-8. 24. Roger Penrose (2007). The Road to Reality: A Complete Guide to the Laws of the Universe. Vintage Books. p. 29. ISBN 0-679-77631-1. 25. ^ a b Heath, p. 200 26. e.g., Tarski (1951) 27. Eves, p. 27 28. Ball, pp. 268ff 29. Eves (1963) 30. Hofstadter 1979, p. 91. 31. Eves (1963), p. 64 32. Ball, p. 485 33. * Howard Eves, 1997 (1958). Foundations and Fundamental Concepts of Mathematics. Dover. 34. Birkhoff, G. D., 1932, "A Set of Postulates for Plane Geometry (Based on Scale and Protractors)," Annals of Mathematics 33. 35. ^ a b Tarski (1951) 36. Misner, Thorne, and Wheeler (1973), p. 191 37. Rizos, Chris. University of New South Wales. GPS Satellite Signals. 1999. 38. Ball, p. 31 39. Heath, p. 268 40. Robinson, Abraham (1966). Non-standard analysis. 41. For the assertion that this was the historical reason for the ancients considering the parallel postulate less obvious than the others, see Nagel and Newman 1958, p. 9. 42. Cajori (1918), p. 197 43. ^ a b A detailed discussion can be found in James T. Smith (2000). "Chapter 2: Foundations". Methods of geometry. Wiley. pp. 19 ff. ISBN 0-471-25183-6. 44. 45. Bertrand Russell (2000). "Mathematics and the metaphysicians". In James Roy Newman. The world of mathematics 3 (Reprint of Simon and Schuster 1956 ed.). Courier Dover Publications. p. 1577. ISBN 0-486-41151-6. 46. Bertrand Russell (1897). "Introduction". An essay on the foundations of geometry. Cambridge University Press. 47. George David Birkhoff, Ralph Beatley (1999). "Chapter 2: The five fundamental principles". Basic Geometry (3rd ed.). AMS Bookstore. pp. 38 ff. ISBN 0-8218-2101-6. 48. James T. Smith. "Chapter 3: Elementary Euclidean Geometry". Cited work. pp. 84 ff. 49. Edwin E. Moise (1990). Elementary geometry from an advanced standpoint (3rd ed.). Addison–Wesley. ISBN 0-201-50867-2. 50. John R. Silvester (2001). "§1.4 Hilbert and Birkhoff". Geometry: ancient and modern. Oxford University Press. ISBN 0-19-850825-5. 51. Alfred Tarski (2007). "What is elementary geometry". In Leon Henkin, Patrick Suppes & Alfred Tarski. Studies in Logic and the Foundations of Mathematics – The Axiomatic Method with Special Reference to Geometry and Physics (Proceedings of International Symposium at Berkeley 1957–8; Reprint ed.). Brouwer Press. p. 16. ISBN 1-4067-5355-6. "We regard as elementary that part of Euclidean geometry which can be formulated and established without the help of any set-theoretical devices" 52. Keith Simmons (2009). "Tarski's logic". In Dov M. Gabbay, John Woods. Logic from Russell to Church. Elsevier. p. 574. ISBN 0-444-51620-4. 53. ^ a b Petri Mäenpää (1999). "From backward reduction to configurational analysis". In Michael Otte, Marco Panza. Analysis and synthesis in mathematics: history and philosophy. Springer. p. 210. ISBN 0-7923-4570-3. 54. Carlo Cellucci (2008). "Why proof? What is proof?". In Rossella Lupacchini, Giovanna Corsi. Deduction, Computation, Experiment: Exploring the Effectiveness of Proof. Springer. p. 1. ISBN 88-470-0783-6. 55. Eric W. Weisstein (2003). "Euclid's postulates". CRC concise encyclopedia of mathematics (2nd ed.). CRC Press. p. 942. ISBN 1-58488-347-2. 56. Deborah J. Bennett (2004). Logic made easy: how to know when language deceives you. W. W. Norton & Company. p. 34. ISBN 0-393-05748-8. 57. AN Kolmogorov, AF Semenovich, RS Cherkasov (1982). Geometry: A textbook for grades 6–8 of secondary school [Geometriya. Uchebnoe posobie dlya 6–8 klassov srednie shkoly] (3rd ed.). Moscow: "Prosveshchenie" Publishers. pp. 372–376. A description of the approach, which was based upon geometric transformations, can be found in Teaching geometry in the USSR Chernysheva, Firsov, and Teljakovskii 58. Viktor Vasilʹevich Prasolov, Vladimir Mikhaĭlovich Tikhomirov (2001). Geometry. AMS Bookstore. p. 198. ISBN 0-8218-2038-9. 59. Petri Mäenpää (1998). "Analytic program derivation in type theory". In Giovanni Sambin, Jan M. Smith. Twenty-five years of constructive type theory: proceedings of a congress held in Venice, October 1995. Oxford University Press. p. 113. ISBN 0-19-850127-7. 60. Celia Hoyles (Feb. 1997). "The curricular shaping of students' approach to proof". For the Learning of Mathematics (FLM Publishing Association) 17 (1): 7–16. JSTOR 40248217. `\|accessdate=` requires `\|url=` (help) ## References • Ball, W.W. Rouse (1960). A Short Account of the History of Mathematics (4th ed. [Reprint. Original publication: London: Macmillan & Co., 1908] ed.). New York: Dover Publications. pp. 50–62. ISBN 0-486-20630-0. • Coxeter, H.S.M. (1961). Introduction to Geometry. New York: Wiley. • Eves, Howard (1963). A Survey of Geometry. Allyn and Bacon. • Heath, Thomas L. (1956). The Thirteen Books of Euclid's Elements (3 vols.`\|format=` requires `\|url=` (help)) (2nd ed. [Facsimile. Original publication: Cambridge University Press, 1925] ed.). New York: Dover Publications. ISBN 0-486-60088-2 (vol. 1), ISBN 0-486-60089-0 (vol. 2), ISBN 0-486-60090-4 (vol. 3) Check `\|isbn=` value (help). Heath's authoritative translation of Euclid's Elements plus his extensive historical research and detailed commentary throughout the text. • Misner, Thorne, and Wheeler (1973). Gravitation. W.H. Freeman. • Mlodinow (2001). Euclid's Window. The Free Press. • Nagel, E. and Newman, J.R. (1958). Gödel's Proof. New York University Press. • Alfred Tarski (1951) A Decision Method for Elementary Algebra and Geometry. Univ. of California 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": 3, "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": 0.9219300150871277, "perplexity_flag": "middle"}
http://www.reference.com/browse/transcendental-number	Definitions # Transcendental number In mathematics, a transcendental number is a complex number that is not algebraic, that is, not a solution of a non-zero polynomial equation with rational coefficients. The most prominent examples of transcendental numbers are π and e. Only a few classes of transcendental numbers are known, indicating that it can be extremely difficult to show that a given number is transcendental. However, transcendental numbers are not rare: indeed, almost all real and complex numbers are transcendental, since the algebraic numbers are countable, but the sets of real and complex numbers are uncountable. All transcendental numbers are irrational, since all rational numbers are algebraic. (The converse is not true: not all irrational numbers are transcendental.) ## History Euler was probably the first person to define transcendental numbers in the modern sense. The name "transcendentals" comes from Leibniz in his 1682 paper where he proved sin x is not an algebraic function of x. Joseph Liouville first proved the existence of transcendental numbers in 1844, and in 1851 gave the first decimal examples such as the Liouville constant $sum_\left\{k=1\right\}^infty 10^\left\{-k!\right\} = 0.110001000000000000000001000ldots$ in which the nth digit after the decimal point is 1 if n is a factorial (i.e., 1, 2, 6, 24, 120, 720, ...., etc.) and 0 otherwise. Liouville showed that this number is what we now call a Liouville number; this essentially means that it can be particularly well approximated by rational numbers. Liouville showed that all Liouville numbers are transcendental. Johann Heinrich Lambert conjectured that e and π were both transcendental numbers in his 1761 paper proving the number π is irrational. The first number to be proven transcendental without having been specifically constructed for the purpose was e, by Charles Hermite in 1873. In 1874, Georg Cantor found the argument mentioned above establishing the ubiquity of transcendental numbers. In 1882, Ferdinand von Lindemann published a proof that the number π is transcendental. He first showed that e to any nonzero algebraic power is transcendental, and since eiπ = −1 is algebraic (see Euler's identity), iπ and therefore π must be transcendental. This approach was generalized by Karl Weierstrass to the Lindemann–Weierstrass theorem. The transcendence of π allowed the proof of the impossibility of several ancient geometric constructions involving compass and straightedge, including the most famous one, squaring the circle. In 1900, David Hilbert posed an influential question about transcendental numbers, Hilbert's seventh problem: If a is an algebraic number, that is not zero or one, and b is an irrational algebraic number, is ab necessarily transcendental? The affirmative answer was provided in 1934 by the Gelfond–Schneider theorem. This work was extended by Alan Baker in the 1960s in his work on lower bounds for linear forms in any number of logarithms (of algebraic numbers). ## Properties The set of transcendental numbers is uncountably infinite. The proof is simple: Since the polynomials with integer coefficients are countable, and since each such polynomial has a finite number of zeroes, the algebraic numbers must also be countable. But Cantor's diagonal argument proves that the real numbers (and therefore also the complex numbers) are uncountable; so the set of all transcendental numbers must also be uncountable. Transcendental numbers are never rational, but some irrational numbers are not transcendental. For example, the square root of 2 is irrational, but it is a solution of the polynomial x2 − 2 = 0, so it is algebraic, not transcendental. Any non-constant algebraic function of a single variable yields a transcendental value when applied to a transcendental argument. So, for example, from knowing that π is transcendental, we can immediately deduce that numbers such as 5π, (π − 3)/√2, (√π − √3)8 and (π5 + 7)1/7 are transcendental as well. However, an algebraic function of several variables may yield an algebraic number when applied to transcendental numbers if these numbers are not algebraically independent. For example, π and 1 − π are both transcendental, but π + (1 − π) = 1 is obviously not. It is unknown whether π + e, for example, is transcendental, though at least one of π + e and πe must be transcendental. More generally, for any two transcendental numbers a and b, at least one of a + b and ab must be transcendental. To see this, consider the polynomial (x − a) (x − b) = x2 − (a + b)x + ab. If (a + b) and ab were both algebraic, then this would be a polynomial with algebraic coefficients. Because algebraic numbers form an algebraically closed field, this would imply that the roots of the polynomial, a and b, must be algebraic. But this is a contradiction, and thus it must be the case that at least one of the coefficients is transcendental. The non–computable numbers are a strict subset of the transcendental numbers. All Liouville numbers are transcendental; however, not all transcendental numbers are Liouville numbers. Any Liouville number must have unbounded partial quotients in its continued fraction expansion. Using a counting argument one can show that there exist transcendental numbers which have bounded partial quotients and hence are not Liouville numbers. Using the explicit continued fraction expansion of e, one can show that e is not a Liouville number (although the partial quotients in its continued fraction expansion are unbounded). Kurt Mahler showed in 1953 that π is also not a Liouville number. It is conjectured that all infinite continued fractions with bounded terms that are not eventually periodic are transcendental (eventually periodic continued fractions correspond to quadratic irrationals). ## Known transcendental numbers and open problems Here is a list of some numbers known to be transcendental: • ea if a is algebraic and nonzero (by the Lindemann–Weierstrass theorem), and in particular, e itself. • π (by the Lindemann–Weierstrass theorem). • eπ, Gelfond's constant, as well as e-π/2=ii (by the Gelfond–Schneider theorem). • ab where a is non-zero algebraic and b is irrational algebraic (by the Gelfond–Schneider theorem), in particular: • $2^sqrt\left\{2\right\}$, the Gelfond–Schneider constant (Hilbert number), • sin(a), cos(a) and tan(a), and their multiplicative inverses csc(a), sec(a) and cot(a), for any nonzero algebraic number a (by the Lindemann–Weierstrass theorem). • ln(a) if a is algebraic and not equal to 0 or 1, for any branch of the logarithm function (by the Lindemann–Weierstrass theorem). • Γ(1/3), Γ(1/4), and Γ(1/6). • 0.12345678910111213141516..., the Champernowne constant. • Ω, Chaitin's constant (since it is a non-computable number). • Prouhet–Thue–Morse constant • $sum_\left\{k=0\right\}^infty 10^\left\{-leftlfloor beta^\left\{k\right\} rightrfloor\right\};$ where $beta > 1$ and $betamapstolfloor beta rfloor$ is the floor function. Numbers for which it is unknown whether they are transcendental or not: • Sums, products, powers, etc. (except for Gelfond's constant) of the number π and the number e: π + e, π − e, π·e, π/e, ππ, ee, πe • the Euler–Mascheroni constant γ (which has not even been proven to be irrational) • Catalan's constant, also not known to be irrational • Apéry's constant, ζ(3), and in fact, ζ(2n + 1) for any positive integer n (see Riemann zeta function). Conjectures: ## Proof sketch that e is transcendental The first proof that the base of the natural logarithms, e, is transcendental dates from 1873. We will now follow the strategy of David Hilbert (1862–1943) who gave a simplification of the original proof of Charles Hermite. The idea is the following: Assume, for purpose of finding a contradiction, that e is algebraic. Then there exists a finite set of integer coefficients $c_\left\{0\right\},c_\left\{1\right\},ldots,c_\left\{n\right\},$ satisfying the equation: $c_\left\{0\right\}+c_\left\{1\right\}e+c_\left\{2\right\}e^\left\{2\right\}+cdots+c_\left\{n\right\}e^\left\{n\right\}=0$ and such that $c_0$ and $c_n$ are both non-zero. Depending on the value of n, we specify a sufficiently large positive integer k (to meet our needs later), and multiply both sides of the above equation by $int^\left\{infty\right\}_\left\{0\right\}$, where the notation $int^\left\{b\right\}_\left\{a\right\}$ will be used in this proof as shorthand for the integral: $int^\left\{b\right\}_\left\{a\right\}:=int^\left\{b\right\}_\left\{a\right\}x^\left\{k\right\}\left[\left(x-1\right)\left(x-2\right)cdots\left(x-n\right)\right]^\left\{k+1\right\}e^\left\{-x\right\},dx.$ We have arrived at the equation: $c_\left\{0\right\}int^\left\{infty\right\}_\left\{0\right\}+c_\left\{1\right\}eint^\left\{infty\right\}_\left\{0\right\}+cdots+c_\left\{n\right\}e^\left\{n\right\}int^\left\{infty\right\}_\left\{0\right\} = 0$ which can now be written in the form $P_\left\{1\right\}+P_\left\{2\right\}=0;$ where $P_\left\{1\right\}=c_\left\{0\right\}int^\left\{infty\right\}_\left\{0\right\}+c_\left\{1\right\}eint^\left\{infty\right\}_\left\{1\right\}+c_\left\{2\right\}e^\left\{2\right\}int^\left\{infty\right\}_\left\{2\right\}+cdots+c_\left\{n\right\}e^\left\{n\right\}int^\left\{infty\right\}_\left\{n\right\}$ $P_\left\{2\right\}=c_\left\{1\right\}eint^\left\{1\right\}_\left\{0\right\}+c_\left\{2\right\}e^\left\{2\right\}int^\left\{2\right\}_\left\{0\right\}+cdots+c_\left\{n\right\}e^\left\{n\right\}int^\left\{n\right\}_\left\{0\right\}$ The plan of attack now is to show that for k sufficiently large, the above relations are impossible to satisfy because $frac\left\{P_\left\{1\right\}\right\}\left\{k!\right\}$ is a non-zero integer and $frac\left\{P_\left\{2\right\}\right\}\left\{k!\right\}$ is not. The fact that $frac\left\{P_\left\{1\right\}\right\}\left\{k!\right\}$ is a nonzero integer results from the relation $int^\left\{infty\right\}_\left\{0\right\}x^\left\{j\right\}e^\left\{-x\right\},dx=j!$ which is valid for any positive integer j and can be proved using integration by parts and mathematical induction. To show that $left\|frac\left\{P_\left\{2\right\}\right\}\left\{k!\right\}right\|<1$ for sufficiently large k we construct an auxiliary function $x^\left\{k\right\}\left[\left(x-1\right)\left(x-2\right)cdots\left(x-n\right)\right]^\left\{k+1\right\}e^\left\{-x\right\}$, noting that it is the product of the functions $\left[x\left(x-1\right)\left(x-2\right)cdots\left(x-n\right)\right]^\left\{k\right\}$ and $\left(x-1\right)\left(x-2\right)cdots\left(x-n\right)e^\left\{-x\right\}$. Using upper bounds for $\|x\left(x-1\right)\left(x-2\right)cdots\left(x-n\right)\|$ and $\|\left(x-1\right)\left(x-2\right)cdots\left(x-n\right)e^\left\{-x\right\}\|$ on the interval [0,n] and employing the fact $lim_\left\{ktoinfty\right\}frac\left\{G^k\right\}\left\{k!\right\}=0$ for every real number G is then sufficient to finish the proof. A similar strategy, different from Lindemann's original approach, can be used to show that the number π is transcendental. Besides the gamma-function and some estimates as in the proof for e, facts about symmetric polynomials play a vital role in the proof. For detailed information concerning the proofs of the transcendence of π and e see the references and external links. ## See also • Transcendence theory, the study of questions related to transcendental numbers ## References • David Hilbert, "Über die Transcendenz der Zahlen $e$ und $pi$", Mathematische Annalen 43:216–219 (1893). • Alan Baker, Transcendental Number Theory, Cambridge University Press, 1975, ISBN 0-521-39791-X. • Peter M Higgins, "Number Story" Copernicus Books, 2008, ISBN 978-84800-000-1.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 32, "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": 0.8886193633079529, "perplexity_flag": "head"}
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Problem: $\sum_{n=0}^\infty\sum_{m=n}^\infty\ a^nb^m$ I know ... 1answer 34 views ### Specific range of numbers is given, trying to get another number within same range I'm trying to calculate the width of an HTML element based on the window size. Here's what I have. These width values (first value) accurately match with the width the HTML element must be (second ... 0answers 147 views ### Equidistribution of roots of prime cyclotomic polynomials to prime moduli Here is a relevant - and longstanding, I'm told - conjecture. Let $f \in \mathbb{Z}[x]$ be irreducible and of degree > 1. 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For $k=1$, on MathWorld we have that ... 2answers 122 views ### Exponential formula to assign list items an equally distributed percentage I am going to explain this as best as I can: We have a number of list items, that dynamically changes (sometimes it has 5 items sometimes it has 7 items or 43 items, etc.) We are trying to find a ... 1answer 110 views ### summation of an arbitrary function multiplied by exponential I'm trying to find some function $g(k)$ such that $$\sum_{k=0}^{\infty} g(k) \frac{(n \lambda)^k}{k!} = 0$$ The textbook says that there is only one solution, that is $g(k)=0$ for all $k$. But I ... 1answer 214 views ### Non trivial upper bound for an exponential sum Suppose $h \in \mathbb{N}$, is there a known non trivial upper bound for $$\left\| \frac{1}{n} \sum_{m=1}^n e^{2 \pi i h (2 \pi m)} \right\|?$$	{"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": 77, "mathjax_display_tex": 19, "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": 0.9099400043487549, "perplexity_flag": "middle"}
http://mathoverflow.net/revisions/98730/list	## Return to Answer 2 added 27 characters in body So here is my comment again with slightly more details. Let $Y$ be a finite subset of $G$ such that its image generates $K$. As I was told by Yves, finite type means finitely generated. Thus, let $X$ be a finite subset of $N$ such that $N^{ab}$ is generated by its image as a $\mathbb{Z}_p[[K]]$-module. Then we claim that $H$ the subgroup generated (topologically) by $X \cup Y$ is equal $G$. Since $Y \subseteq H$ we have that $HN/N \cong$K\$. Therefore, $HN=G$. So it suffices to show that $N \leq H$. By definition, the action of $K$ on $N^{ab}$ is via the action of $G$. Since $N$ acts trivially on $N^{ab}$ we have that $H$ action on $N^{ab}$ is the same as $HN=G$ action. Hence, the image of $Y^{H}$ generates $N^{ab}$ as a $\mathbb{Z_p}$-module. So $Y^{H}$ generates $N/\Phi(N)$ as a group, where $\Phi(N)=[N,N]N^{p}$ is the Frattini subgroup of $N$. We deduce that $Y^{H}$ generates $N$ (topologically) and $N \leq H$ as we wanted. Note that it sufices to ask that $N/([N,K]N^{p})$ N/([N,N]N^{p})$(EDIT: see comments below) is finitely generatd as a$K\$-module. 1 So here is my comment again with slightly more details. Let $Y$ be a finite subset of $G$ such that its image generates $K$. As I was told by Yves, finite type means finitely generated. Thus, let $X$ be a finite subset of $N$ such that $N^{ab}$ is generated by its image as a $\mathbb{Z}_p[[K]]$-module. Then we claim that $H$ the subgroup generated (topologically) by $X \cup Y$ is equal $G$. Since $Y \subseteq H$ we have that $HN/N \cong$K\$. Therefore, $HN=G$. So it suffices to show that $N \leq H$. By definition, the action of $K$ on $N^{ab}$ is via the action of $G$. Since $N$ acts trivially on $N^{ab}$ we have that $H$ action on $N^{ab}$ is the same as $HN=G$ action. Hence, the image of $Y^{H}$ generates $N^{ab}$ as a $\mathbb{Z_p}$-module. So $Y^{H}$ generates $N/\Phi(N)$ as a group, where $\Phi(N)=[N,N]N^{p}$ is the Frattini subgroup of $N$. We deduce that $Y^{H}$ generates $N$ (topologically) and $N \leq H$ as we wanted. Note that it sufices to ask that $N/([N,K]N^{p})$ is finitely generatd as a $K$-module.	{"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": 68, "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": 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": 0.9695127606391907, "perplexity_flag": "head"}
http://cms.math.ca/10.4153/CJM-2003-001-0	Canadian Mathematical Society www.cms.math.ca \| \| \| \| \| \|----------\|----\|-----------\|----\| \| \| \| \| \| \| \| \| Site map \| \| \| CMS store \| \| location: Publications → journals → CJM Abstract view # An Exactly Solved Model for Mutation, Recombination and Selection Read article [PDF: 352KB] Published:2003-02-01 Printed: Feb 2003 • Michael Baake • Ellen Baake Features coming soon: Citations (via CrossRef) Tools: Search Google Scholar: Format: HTML LaTeX MathJax PDF PostScript ## Abstract It is well known that rather general mutation-recombination models can be solved algorithmically (though not in closed form) by means of Haldane linearization. The price to be paid is that one has to work with a multiple tensor product of the state space one started from. Here, we present a relevant subclass of such models, in continuous time, with independent mutation events at the sites, and crossover events between them. It admits a closed solution of the corresponding differential equation on the basis of the original state space, and also closed expressions for the linkage disequilibria, derived by means of M\"obius inversion. As an extra benefit, the approach can be extended to a model with selection of additive type across sites. We also derive a necessary and sufficient criterion for the mean fitness to be a Lyapunov function and determine the asymptotic behaviour of the solutions. Keywords: population genetics, recombination, nonlinear $\ODE$s, measure-valued dynamical systems, Möbius inversion MSC Classifications: 92D10 - Genetics {For genetic algebras, see 17D92} 34L30 - Nonlinear ordinary differential operators 37N30 - Dynamical systems in numerical analysis 06A07 - Combinatorics of partially ordered sets 60J25 - Continuous-time Markov processes on general state spaces	{"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": 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": 0.868360698223114, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/18988?sort=newest	## Motivation Suppose that $F\colon X\to A$ is left adjoint to $G\colon A\to X$, and let $\varepsilon\colon FG\stackrel{.}{\to}I_A$ be the counit of the adjunction. Suppose also that $A$ is $J$-complete (for some category $J$), so that $\operatorname{Lim}$ is a functor $C^J\to C$, where for an arrow $\alpha\colon T_1\stackrel{.}{\to} T_2$ of $C^J$, $\operatorname{Lim}(\alpha)$ is the unique arrow of $A$ for which the following diagram is commutative: ```$$ \begin{matrix} \operatorname{Lim}(T_1)& \stackrel{\text{limiting cone}}{\longrightarrow} & T_1\\ \| & & \|\\ \operatorname{Lim}(\alpha) & & \alpha\\ \downarrow & & \downarrow \\ \operatorname{Lim}(T_2)& \stackrel{\text{limiting cone}}{\longrightarrow} & T_2 \end{matrix} $$``` Let $T\colon J\to A$ be a functor. We have the natural transformation $\varepsilon T\colon FGT\stackrel{.}{\to} T$, and $\operatorname{Lim}(\varepsilon T)$ is the dotted line making the following diagram commutative: ```$$ \begin{matrix} \operatorname{Lim}(FGT)& \stackrel{\text{limiting cone}}{\longrightarrow} & FGT\\ \| & & \|\\ \operatorname{Lim}(\varepsilon T) & & \varepsilon T\\ \downarrow & & \downarrow \\ \operatorname{Lim}(T)& \stackrel{\text{limiting cone}}{\longrightarrow} & T \end{matrix} $$``` If $FG$ preserves $J$-limits, and $\tau\colon \operatorname{Lim}(T)\stackrel{.}{\to}T$ is the lower limiting cone, then $FG\tau\colon FG\operatorname{Lim}(T)\stackrel{.}{\to}FGT$ is the upper limiting cone, and the above diagram becomes ```$$ \begin{matrix} FG\operatorname{Lim}(T)& \stackrel{FG\tau}{\longrightarrow} & FGT\\ \| & & \|\\ \operatorname{Lim}(\varepsilon T) & & \varepsilon T\\ \downarrow & & \downarrow \\ \operatorname{Lim}(T)& \stackrel{\tau}{\longrightarrow} & T \end{matrix} $$``` Since the naturality of $\varepsilon$ implies that for all $j\in \operatorname{obj}(J)$ the diagram ```$$ \begin{matrix} FG\operatorname{Lim}(T)& \stackrel{FG\tau_j}{\longrightarrow} & FGT(j)\\ \| & & \|\\ \varepsilon_{\mathrm{Lim}T}& & \varepsilon_{T(j)}\\ \downarrow & & \downarrow \\ \operatorname{Lim}(T)& \stackrel{\tau_j}{\longrightarrow} & T(j) \end{matrix} $$``` is commutative, it follows that `$\varepsilon_{\mathrm{Lim}T}$` can replace $\operatorname{Lim}(\varepsilon T)$ in the last but one diagram while keeping it commutative. By uniqueness, we get the nice equation $$\varepsilon_{\mathrm{Lim}T} = \operatorname{Lim}(\varepsilon T).$$ Note that it seems that all depends on $FG$ preserving $J$ limits. ## Question If $F\colon X\to A$ is left adjoint to $G\colon A\to X$ and $A$ has $J$-limits, when does $FG$ preserve $J$-limits? This is obviously true when $F$ preserves limits (for example, when there is also a left adjoint to $F$), but are there other interesting situations? ## Background For solving an exercise from Mac Lane, I used some results from A. Gleason, ''Universally locally connected refinements,'' Illinois J. Math, vol. 7 (1963), pp. 521--531. In that paper, Gleason constructs a right adjoint to the inclusion functor $\mathbf{L\ conn}\subset \mathbf{Top}$ ($\mathbf{L\ conn}=$ locally connected spaces with continuous maps), and proves that the counit of the product of two topological spaces is the product of the counits (Theorem C). This made me curious when do counits and limits interchange. - 2 +1 for diagrams! – Theo Johnson-Freyd Mar 22 2010 at 16:40 I'm not a categorist, but that Gleason paper you link to surprises me a little, because I would have expected that if an inclusion or forgteful-like functor were to have an adjoint, it would have a left adjoint. (Think of CpctHff into RegularTop, or AbGp into Gp ...) So something atypical seems to be going on, unless I've misunderstood (which might admittedly very well be the case) – Yemon Choi Mar 22 2010 at 19:51 1 @Yemon Choi: Yes, I'm also more used to reflective subcategories than to coreflective ones. But Gleason does prove that Lconn is a coreflective subcategory (the inclusion functor has a right adjoint). Searching the web, I found some additional results on coreflective subcategories in general topology in H. Herrlich and G.E. Strecker, "Coreflective subcategories in general topology," Fund. Math. 73 (1972), pp. 199--218 (matwbn.icm.edu.pl/ksiazki/fm/fm73/fm73124.pdf). Mac Lane also gives an example from algebra: torsion abelian groups in abelian groups. – unknown (google) Mar 22 2010 at 20:19 @Yemon Choi: There are plenty of right adjoints to forgetful functors. A nice heuristic is that often a left adjoint freely adds the forgotten structure (destroying any that was already there) and a right adjoint kills off everything that doesn't have that structure. An example is the functor that sends cartesian categories to symmetric monoidal ones: the left adjoint freely makes every object a comonoid, while the right adjoint sends a category C to the category of comonoids in C. I think it was Lawvere who called these 'fascist' functors, because they're 'far right' adjoints. – Finn Lawler Jun 1 2010 at 19:56 Off topic, but we really need to have some way to use xypic for diagrams on this site. Though these are pretty impressive. – Daniel Litt Jun 29 2010 at 19:27 ## 1 Answer Have you looked at the paper by B. Eckmann and P. J. Hilton entitled "Commuting Limits with Colimits" in the "Journal of Algebra", 11, 116-144 (1969)? - Thanks for the reference, sounds interesting. Is there a free copy available somewhere? – unknown (google) Mar 23 2010 at 16:59 I can email you a copy off-line if you would like. I don't know of any places that have digitized the journal... – Glen M Wilson Mar 23 2010 at 17:07 Thanks a lot! Perhaps I will prepare an anonymous email address as a sink for such generous suggestions. Does this paper talk about the interchange of limits and colimits of a bifunctor from the product of a finite and a filtered category, as in Theorem IX.2.1 of Mac Lane (books.google.co.il/…)? If this is the case, I confess that I have no idea how to apply it for the current question. Hint? – unknown (google) Mar 23 2010 at 18:42 [deleted comment with full email address] – unknown (google) Apr 17 2010 at 18:00	{"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": 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": 0.9191001653671265, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/22189?sort=votes	## What is your favorite “strange” function? [closed] ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) There are many "strange" functions to choose from and the deeper you get involved with math the more you encounter. I consciously don't mention any for reasons of bias. I am just curious what you consider strange and especially like. Please also give a reason why you find this function strange and why you like it. Perhaps you could also give some kind of reference where to find further information. As usually: Please only mention one function per post - and let the votes decide :-) - 2 m.reddit.com/r/math/comments/9txzv/… – Regenbogen Apr 22 2010 at 14:21 2 Counterexamples in analysis has some nice ones: books.google.nl/… – skupers Aug 21 2010 at 18:40 show 3 more comments ## 37 Answers A Brownian motion sample path. These are about the most bizarrely behaved continuous functions on $\mathbb{R}^+$ that you can think of. They are nowhere differentiable, have unbounded variation, attain local maxima and minima in every interval... Many, many papers and books have been written about their strange properties. Edit: As commented, I should clarify the term "sample path". Brownian motion is a stochastic process $B_t$. We say a sample path of Brownian motion has some property if the function $t \mapsto B_t$ has that property almost surely. So, run a Brownian motion, and with probability 1 you will get a function with all these weird properties. - 3 Pedantic, but maybe worth mentioning that these functions have these properties almost surely rather than certainly. – Tom Smith Apr 22 2010 at 20:48 ### 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 empty function $\emptyset:\emptyset\to\emptyset$ is quite strange when you first meet it. - 49 For me, it makes a good argument that the "correct" value for $0^0$ is 1: it's the number of functions from a set with 0 elements to a set with 0 elements. – Nate Eldredge Apr 22 2010 at 19:13 39 And, of course, it gives combinatorial meaning to the fact that 0! = 1, since the empty function is a bijection! – Qiaochu Yuan Apr 23 2010 at 6:36 2 @Tom: …but it can always be very easily computed! – Peter LeFanu Lumsdaine Nov 13 2010 at 7:53 5 Elsewhere I have raised the question of whether this function should be considered a constant function. On the one hand, $f(x_1)=f(x_2)$ for every $x_1$ and $x_2$ in the domain; on the other hand, there is no $y$ in the codomain such that for every $x$ we have $f(x)=y$. I consider it non-constant. – Tom Goodwillie Nov 14 2010 at 3:29 5 A morphism in a category with terminal object $t$ may be called constant if it factors through $t$. According to this definition, $\emptyset \to S$ is constant iff $S$ is nonempty. – Martin Brandenburg Jul 16 at 16:50 show 4 more comments The Busy Beaver function Let Σ be a finite alphabet, for instance {0, 1}; let M be the set of Turing machines with alphabet Σ, and let H ⊆ M be the set of Turing machines that halt when given the empty string ε as input. For each M ∈ H, Let s(M) be the number of steps performed by M before halting (when given ε as input). Finally, let S : ℕ → ℕ be the function defined by S(n) = max {s(M) : M ∈ H and M has n states} Notice that S is well-defined, since only finitely many Turing machines with n states exist. In other words, S(n) is the maximum number of steps performed on ε among all halting Turing machines with n states. S is called the Busy Beaver function. It turns out that S is uncomputable because it grows faster than any computable function, that is, for all recursive functions f : ℕ → ℕ we have S(n) > f(n) for large enough n, and in particular f is o(S). - 6 In particular, it grows faster than the Ackermann function. – Joel David Hamkins Apr 22 2010 at 19:17 I like the Cantor function. A continuous, increasing function $f:[0,1]\rightarrow[0,1]$ with derivative $0$ almost everywhere. See wiki article here. - 18 The same is true of $f(x)=\sum_{0<p/q<x} \frac{1}{2^q-1}$, where the sum is over all irreducible fractions $p/q$. But this function is also strictly increasing! – Kevin O'Bryant Apr 23 2010 at 12:36 The Ackermann function $A(n,m)$ is defined on the natural numbers by a very simple recursion, but the values grow enormously, almost beyond conception. This function completely transcends any simple-minded system of rates-of-growth based on polynomial, exponential, double-exponential and so on. The first few values of the diagonal function $A(n) = A(n,n)$ are: • $A(0) = 1$ • $A(1) = 3$ • $A(2) = 7$ • $A(3) = 61$ • $A(4) = 2^{2^{2^{65536}}}-3$ • $A(5)$ is vast, and can be described in terms of exponential stacks of $2$s, whose height is a stack of $2$s, etc. 5 times. • $A(6)$ is so vast, it is best described using the Ackermann function itself. The levels of the Ackerman function $A_n(m)=A(n,m)$ stratify the primitive recursive functions, in the sense that they are each primitive recursive, but every primitive recursive function is bounded by such a level of the Ackermann function. Thus, the Ackermann function itself is not primitive recursive, although it is computable in the sense of computability theory. - 1 I never really thought of the Ackermann function as being strange, only big. But maybe that's just me. – Ketil Tveiten Apr 23 2010 at 12:13 1 Ketil, yes, perhaps I agree. But what is strangely wonderful about it is that the recursive definition A(n+1,m+1)=A(n,A(n+1,m)) is so simple, and yet leads immediately to such incomprehensible growth. – Joel David Hamkins Apr 23 2010 at 12:21 1 I can't remember where (probably tvtropes), but when reading something about the Ackermann numbers (1 ^ 1, 2 ^^ 2, 3 ^^^ 3, etc), which are related to the Ackermann function, the joke was "it's always weird when looking at a sequence of numbers that goes: 1, 4, too big to count." – Gabriel Benamy Jul 4 2010 at 15:39 show 1 more comment I suppose the strangest function in mainstream mathematics is Riemann zeta function http://en.wikipedia.org/wiki/Riemann_zeta_function $\zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s} = \frac{1}{1^s} + \frac{1}{2^s} + \frac{1}{3^s} + \cdots \;\;\;\;\;\;\;{Re}(s) >1.$ It is part of one of the most important hypothesis and is very influential in many branch of moder mathematics. It is actively used in many areas and is researched in many ways, it is not curiosity, or exotic example, but important mathematical being! And is mysterious and strange! Take a look: - 28 See also en.wikipedia.org/wiki/Zeta_function_universality . – Qiaochu Yuan Apr 23 2010 at 6:17 1 @Qiaochu Yuan: THAT IS SO COOL! – Vectornaut Apr 23 2010 at 18:27 I'm still quite impressed about $f(x)=\mathrm e^x$ … - 4 Very true, me too! Could you please elaborate on why you consider this function "strange" - thank you. – vonjd Apr 23 2010 at 9:29 2 Because of its stubborn nature: whether you differentiate it, or integrate it, it remains unmoved ;-) – S. Sra Nov 13 2010 at 17:39 1 There's a part of me which has never quite gotten over the Euler identity $e^{ix} = \cos(x) + i\sin(x)$, which was perhaps the biggest intellectual thrill of my early teenage years... – Todd Trimble May 29 2011 at 23:33 The Weierstrass function is particularly intriguing, as it's a function that's everywhere continuous, but nowhere differentiable. $f(x)= \sum_{n=0} ^\infty a^n \cos(b^n \pi x)$ where 0<a<1, and b is a positive odd integer such that $ab > 1 + \frac{3\pi}{2}$. It challenges the notion that, just because a function is continuous, it must also be differentiable in most places, which I think is pretty cool. - An unbounded operator with dense graph. In functional analysis, one deals with unbounded operators on Hilbert space, but usually ones that are closed, or are at least closable. At the opposite end of the spectrum, one can construct linear operators whose graph is dense: for any pair $(x,y)$, there is a sequence $x_n$ such that $x_n \to x$ and $A x_n \to y$ ! It's not so easy even on $\mathbb{R}$ to come up with a function whose graph is dense, and the examples I think of aren't measurable. But in infinite dimensions, you can find one that is linear! It's just an illustration that facts that are trivial in finite dimensions can be horribly, horribly false in infinite dimensions. A family of examples of this is constructed in MR0782615 (86i:47052) Lindsay, J. M. A family of operators with everywhere dense graphs. Exposition. Math. 2 (1984), no. 4, 375--378. Interestingly, as an appplication, Lindsay uses such operators to prove that a Brownian motion sample path is nowhere differentiable --- which is my other favorite strange function! - Please excuse me if I include two related functions in one answer. Any space filling curve is rather strange, at least for me. Let $\gamma\colon[0,1]\to[0,1]^2$ be such a curve, that is, $\gamma$ is continuous and surjective. Let $\gamma(t)=(x(t),y(t))$; then $x(t)$ (or $y(t)$) is my other candidate for strangest function: given any $z\in[0,1]$, $x^{-1}(z)$ has the cardinality of the continuum. - Thomae's function, also called the "popcorn function". It's continuous at all irrationals and discontinuous at all rationals. Here a picture: - The function defined by the power series $f(x)=x-x^2+x^4-x^8+x^{16}-\dots$ What is its limit as $x$ approaches $1$ from below? EDIT (This answer is a trick question.) - 1 SPOILER (albeit 1.5+ years late): math.harvard.edu/~elkies/Misc/sol8.html (and the PDF graph at math.harvard.edu/~elkies/Misc/gamma_pic0.pdf) – Noam D. Elkies Jun 8 at 23:06 The Conway base 13 function has to be the weirdest function I know. This function is continuous nowhere, yet it satisfies the intermediate value theorem. Only John Conway... - Minkowski's question mark function if only for the strange $?(\cdot)$ notation. - 1 en.wikipedia.org/wiki/… - it is closely related with Cantor function mentioned elsewhere. – kakaz Apr 22 2010 at 20:58 show 1 more comment It is pretty obvious after you've seen it, but I like the crinkled curve from Halmos's Hilbert Space Problem book: Let $f:\mathbb{R}\rightarrow(0,\infty)$ be an $L^2$ function, and define $t\mapsto g_t:\mathbb{R}\rightarrow L^2(\mathbb{R})$ by $$g_t(x)=\chi_{(-\infty,t)}(x) \times f(x).$$ Then $g_t$ has the property that for all $t_1 < t_2 < t_3$ the secants $g_{t_2}-g_{t_1}$ and $g_{t_3}-g_{t_2}$ are mutually orthogonal. (The curve turns a corner at every point.) - Dirac delta function seems strange to me since the first time I saw it. http://en.wikipedia.org/wiki/Dirac_delta_function - 5 Well, technically the Dirac delta function is a distribution, and not a function. – J.C. Ottem Apr 22 2010 at 14:57 16 Well, it is a function, just not on the space you might think... – Nate Eldredge Apr 22 2010 at 15:42 9 Haha, well everything is a function not on the space you might think. – Tom Ellis Apr 22 2010 at 22:08 11 I heard that after Schwartz got the Fields medal someone quipped "Now they're giving the Fields medal for integration by parts." – Jon Apr 23 2010 at 6:18 20 We used to say that Dirac's delta is the characteristic function of physicists: if you're a physicist, it's a function, otherwise, it isn't. – Federico Poloni Aug 21 2010 at 20:02 show 3 more comments Interpreting your questions a bit liberally, I suggest the Goodstein sequence: http://en.wikipedia.org/wiki/Goodstein%27s_theorem - f(x) = sin (1/x): (x not 0); f(x) = 0 (x equals 0) - 2 The functions one learns about early in studying mathematics are chosen to illustrate various "issues:' continuity, having a derivative, being periodic, etc. One of the functions one learns about in this way is y = sin(x). So while there are many functions that are "strange," the transition from y = sin (x) to y = sin (1/x) offers I feel lots of nice lessons about functions and their behavior. There are many web sites that use graphics to help one understand what is going on here. One such site is: math.washington.edu/~conroy/general/sin1overx – Joseph Malkevitch Apr 23 2010 at 15:32 show 2 more comments I like the Theta functions which are given by Fourier-type series. They show up in many areas in mathematics. For example: i)They are very important in the study of abelian varieties in algebraic geometry (for example, in the case of elliptic curves they are used in the proof of Abel's theorem and are related to Weierstrass $\mathcal{P}-$function). ii) They satisfy a number of interesting indentities. For example, in the one-dimensional case, they satisfy Jacobi's triple product identity which can be used to show Jacobi's four square theorem iii) They can be used to solve algebraic equations degree equations explicitly (see this link) iv) In the one-dimensional case, they solve the heat equation. - 2 But are they strange? – Gerry Myerson Apr 23 2010 at 4:41 1 Well, they are definitely unusual in the sense that they are not taught in school. Also, the fact that they appear in such a variety of mathematical disciplines is also rather surprising. But I see your point. – J.C. Ottem Apr 23 2010 at 7:31 show 1 more comment - These functions like the Cantor function and the continuous-but-not-differentiable function are all well and good, but contrived - the only place you ever see them is as counterexamples. Here is a function that has many uses in Number Theory, and still manages to have a strange property or two. Let $x=h/k$ with $h$ and $k$ integers, $k>0$. Define $$s(x)=\sum_{c=1}^{k-1}((c/k))((ch/k))$$ where $((y))=0$ if $y$ is an integer, $((y))=\lbrace y\rbrace-1/2$ otherwise. It is easily proved that the sum depends only on the ratio of $h$ and $k$, not on their individual values, so $s$ is a well-defined function from the rationals to the rationals. It is known as the Dedekind sum; it came up originally in Dedekind's study of the transformation formula of the Dedekind $\eta$-function. Now for the strange properties. Hickerson, Continued fractions and density results for Dedekind sums, J Reine Angew Math 290 (1977) 113-116, MR 55 #12611, proved that the graph of $s$ is dense in the plane. With Nick Phillips, I proved (Lines full of Dedekind sums, Bull London Math Soc 36 (2004) 547-552, MR 2005m:11075) that, with the exception of the line $y=x/12$, every line through the origin with rational slope passes through infinitely many points on the graph of $s$. We suspect that the points are dense on those lines, though we could only prove it for the line $y=x$. - The Banach limit assigns to every bounded sequence of real numbers a real number "limit" in a way that is linear, shift invariant, and agrees with the usual limit whenever it exists. Banach limits are among the mysterious examples of continuous linear functionals on $\ell^\infty$ that aren't represented by elements of $\ell^1$. Unfortunately, the Hahn-Banach theorem is used in the construction of the Banach limit, and the values aren't canonical. There's a precise definition at http://en.wikipedia.org/wiki/Banach_limit . - The Osgood curve ("A Jordan Curve of Positive Area") is an injective map from [0,1] into $\mathbb{R}^2$ which traces out an image of positive area. (This differs from standard space-filling curves, which are not injective.) - Among the "special" functions encountered in analysis in my view Ingrid Daubechies' waveletes with compact support are the strangest. - 1 Could you please give some more details? Thanks – vonjd Apr 22 2010 at 18:00 One can construct a natural 'metric' for the Riemann sphere which is equivalent to the spherical metric but which is singular on a dense set of points of the Riemann sphere though remains $L^1$ integrable. These are built from degree 2 rational maps (first constructed by Mary Rees) which have the whole Riemann sphere as their Julia sets, and have the orbits of their critical points also dense. The Carlesson-Jones-Yoccoz construction of a expanding metric for critically-finite rational maps actually extends to this case, and we get a metric in which this Julia set actually looks as if it was hyperbolic! [The details are worked out in my PhD thesis, never published as I decided that computer algebra suited me better than complex dynamics]. - Since Mariano took my favorite already, I'll go with the stopping time function for the 3x+1 problem: http://www.ieeta.pt/~tos/3x+1.html - Fix a probability $p < 1/2$ of winning an unfair coin toss. For $x \in [0,1]$ rational, let $f(x)$ be the probability that, if you started with $x$ dollars, you could make it to 1 dollar through optimal betting* on the outcome of these coin flips. This function $f(x)$ is obviously weakly increasing on $[0,1]$ (in fact strictly). Less obvious is that it extends to a continuous function on $[0,1]$, whose derivative exists almost everywhere, but that derivative is $0$. http://www.maa.org/joma/Volume8/Siegrist/RedBlack.pdf - I can't believe no one has mentioned the Dirichlet function: (I guess it's up to me to bring it up...) - 8 If you plant a post of unit height at every point in $\{(x, y) : x, y \in \mathbb{N}^+\}$ and stand at the origin, looking in the direction of $(1,1)$, you will see this picture. – Max Nov 13 2010 at 13:09 A canonical example from elementary real analysis - the Blancmange function. Consider $f$ defined piecewise by $f(x) = x - [x], \quad \text{if} \quad 0 \leq x- [x] \leq \frac{1}{2}$, and $f(x) = 1 - (x - [x]), \quad \text{if} \quad \frac{1}{2} < x - [x] < 1$, (where $[x]$ is the integer part of $x$). Then define the Blancmange function, $B$ $B(x) = \sum_{n=0}^{\infty}\dfrac{1}{2^n} f(2^{n}x)$. The series converges by the Comparison Test, since $\|f(2^{n}x)\| \leq \frac{1}{2}$, for all $x \in \mathbb{R}$, and it can be shown that $B$ is uniformly continuous but nowhere differentiable. Here a picture of the function: A tasty counterexample to the converse of "differentiability $\implies$ continuity". - 3 Closely related is the tavuk göğsü function, replete with shredded chicken. – Tom LaGatta Jul 1 2010 at 23:34 show 2 more comments Any of the isomorphisms $\mathbb{C}'\to S^{1}$, where $S^{1}$ is the unit circle and $\mathbb{C}'$ is the non-zero complex numbers, with the group operation for both being multiplication. -	{"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": 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": 0.3, "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": 0.9250450730323792, "perplexity_flag": "middle"}
http://www.physicsforums.com/showthread.php?t=134228	Physics Forums ## Mathematical intuition?? Where are students expected to learn all the little algebra tricks that can turn unsolvable looking diff EQ's, integrals, laplace and inverse laplace problems into cakewalks? Things like adding 5+(-5) to the numerator or multiplying by just the right x/x to nudge a nasty looking equation into something of the right form. Every time I encounter a problem that requires a trick like this, I usually spend hours researching on the internet or in other math books trying to figure it out, assuming that I'm making a mistake somewhere rather then just missing a trick. When I finally do figure it out, I feel like such an idiot because I didn't pick up that little trick along the way in my math classes. I'm Just curious, I've never considered the time I spent looking up a certain math subject and studying it in more detail than I normally would to be a waste of time. I've always gained a deeper understanding the material, but I've rarely actually found the trick on the web, I usually figured it out on my own, :LOL, usually in the shower. I have noticed though that a lot of my classmates are much better at spotting when these kinds of tricks are needed and seem to already have knowledge of them. Are their brains just geared more towards math then mine is? Did they have better algebra classes and calculus teachers? 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 Are their brains just geared more towards math then mine is? Did they have better algebra classes and calculus teachers? I would guess the latter. Recognitions: Gold Member Homework Help The answer just comes. Working at the problem consistently gets me nowhere. ## Mathematical intuition?? That's a good question. While taking differential equations, I usually found a trick quickly enough. It was just a "hmm, I think that doing this will simplify it", and it does! Maybe if you were to examine why multiplying by some form of x/x or something works, you may be able to over time learn to use tricks like these efficiently... If I absolutely can't solve a problem, after a little bit of not thinking about it, I figure out what would probably work. Here is a nasty little integral i found in Spivak, out of the 300 this was the only one that took longer than 30 mins to solve. Hope you guys have fun with it! $$\frac{dx}{dy}= x^4+1$$ kdinser, you just reason your way to the answer. Most tricks are really not that obscure. With enough practice, you can almost sense them immediately. For example let's look at SeReNiTy's problem. (Note: don't look further if you were gonna attempt to solve this on your own! ) dx/(x^4 + 1) = dy The right-hand side looks hopeless the way it is. So let's think about how we usually handle integrands of rational functions. First thing that pops into my mind is a trig substition. We know that 1+(tan(u))^2 = (sec(u))^2. Since we have an x^4, let's try x = sqrt(tan(u)). (Although you should be skeptic this would lead to anything useful, but you never know!) Then dx/du = (sec(u)tan(u))/(2 sqrt(tan(u))). The RHS now becomes (1/(sec(u))^2) * (sec(u)tan(u))/(2 sqrt(tan(u))) du Which simplifies to (tan(u))/(2 sec(u) sqrt(tan(u))) du This doesn't look too helpful. So let's move on. The second thing that popped into my mind was partial fractions. Since x^4 + 1 is irreducible (over R), I don't think this method will yield anything helpful -- but it might, so try it for yourself. The third thing that struck me was x^4 + 1 almost looks like (x^2 + 1)^2. In fact: x^4 + 1 = (x^2 + 1)^2 - 2x^2 Hey, that's the difference between two squares! Let's see where this goes... (x^2 + 1)^2 - 2x^2 = (x^2 + 1 - sqrt(2)x)(x^2 + 1 + sqrt(2)x) Split this into partial fractions, and you're off to the races. Recognitions: Homework Help Science Advisor Quote by devious_ Since x^4 + 1 is irreducible (over R), I don't think this method will yield anything helpful -- but it might, so try it for yourself. This isn't irreducible over R, you factored it a few lines later...no polynomial over R of degree 3 or higher is irreducible over R. You can always break them up into a product of linear and irreducible quadratic factors (then apply partial fractions). Quote by shmoe This isn't irreducible over R, you factored it a few lines later...no polynomial over R of degree 3 or higher is irreducible over R. You can always break them up into a product of linear and irreducible quadratic factors (then apply partial fractions). You're right of course. What I meant to say was something to the effect of "it isn't immediately obvious what this can reduced to"! I guess I got sidetracked and said that instead. Admin Quote by kdinser Where are students expected to learn all the little algebra tricks that can turn unsolvable looking diff EQ's, integrals, laplace and inverse laplace problems into cakewalks? I found myself asking the same question in my first few years of university. Quote by kdinser Are their brains just geared more towards math then mine is? Did they have better algebra classes and calculus teachers? I think the latter. It is often a matter of being exposed to the tricks and other formalities. Certainly, most high school math teachers are not necessarily exposed to these methods, unless they have degrees in mathematics. Recognitions: Gold Member Science Advisor Years of experience will help you in dcoding all the little short-cuts you need to know. I have to say, the 2 tricks that I mentioned no longer hold me at bay, so I guess I can say that I'm learning. My complaint is, that I had no chance to learn them before they showed up a calc 2 exam or a diff EQ final. Both times, they made the difference between a 3.2 and a 2.8. I don't worry to much about gpa because I know if I'm learning the material or not and I know that overall, I'll finish with a 3.4 to 3.6 gpa in my engineering classes. But it is frustrating to lose points on an exam because you were never exposed to a certain algebra trick or because you forgot some obscure trig identity. Hello kdinser, not too long ago I read this article (scientific american). In it, it details the possible thought proccess of a chess grandmaster. Aparently( and I agree with it) chess masters dont actually calculate all posible moves like a computer does. Instead, for a paticular arrangement of the pieces of the board they know based on experience what are usually the best paths to follow. Math is the same way, there are many paths that you can follow but only a few will get you to the end with relative ease. I think only experience and lots of practice will get you better at it. The more you practice the fundamentals, the deeper you are going to "see" the problem at a subconscious level. For example instead of readind (x^2-1) you read (x+1)(x-1). This is a very basic example but I bet it applies to much deeper processes. I would assume that the problem you are having is due to lack of practice of the fundamentals. I encounter the same problem that you do, but since I've become a pre-calc tutor I get more than enough practice and the paths to follow are becoming ever simpler. If I were you I would try becoming a homework helper for pre-calc level exersices. You ought to find more than enough practice there and you will be helping others. Just my sugestion. I pick it up algorithmic procedures/short cuts very fast. I think maths, like evrything else in life, is something which some people have an immediate aptitude for. Of course, you can train hard for it - read lots of books/do loads of exercises - but for some, it can also just come naturally. Admin Well, there is the "use it, or lose it" factor. If one learns the various 'tricks' or 'short cuts', one might forget them if one does not use them periodically. One option - when one learns a 'trick' or 'short cut', or comes across a handy piece of information, write it down in a personal notebook/handbook. I wish I had been more diligent as a student, particularly as a grad student. Lots of my notes have been boxed away, and periodically I see a problem on PF, particular ones involving PDE's and EM Theory, where I have done the solution of the problem or one similar, but I don't have those notes at hand. Somewhere I have an excellent set of notes on PDE's for various applications in heat transfer, wave and vibration problems, and stress analysis, not to mention diffusion problems, EM Theory and Schrödinger's equation. Thread Tools \| \| \| \| \|-----------------------------------------------\|------------------------------\|---------\| \| Similar Threads for: Mathematical intuition?? \| \| \| \| Thread \| Forum \| Replies \| \| \| Special & General Relativity \| 8 \| \| \| General Physics \| 2 \| \| \| General Discussion \| 10 \| \| \| General Discussion \| 20 \| \| \| General Discussion \| 12 \|	{"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": 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": 0.9630439281463623, "perplexity_flag": "middle"}
http://www.impan.pl/cgi-bin/dict?until	## until The algorithm compares $x$ to each entry in turn until a match is found or the list is exhausted. This process can be repeated until we obtain the promised triangulation. We postpone the proof until Section 2. We walk from $X$ to $Y$ until the first place where $G$ changes. We defer the proof of the `moreover' statement in Theorem 5 until after the proof of the lemma. There are, however, a few important papers of which we were unaware until fairly recently. Go to the list of words starting with: a b c d e f g h i j k l m n o p q r s t u v w y z	{"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": 0.9176663756370544, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/316062/affects-of-a-coordinate-transformation	Affects of a coordinate transformation I am attempting to solve a PDE of $f(r,t)$, where $r\in[0,g(t)]$ is a spacial coordinate and $t$ is time. The PDE is coupled to to an ODE for $g(t)$. I wish to simplify the problem by defining a new independent variable $$y \equiv \frac{r}{g(t)}.$$ Is it true that, since $y$ is an independent variable, $$\frac{dy}{dt} = 0 = -\frac{r}{\left(g(t)\right)^2}\frac{dg}{dt},$$ therefore $dg/dt = 0$ in this new coordinate system / reference frame? Since the PDE for $f$ involves $dg/dt$, this seems to make the problem too easy. Additionally, is there any recommended reading on this subject? - 1 Answer No, it is not true that $dy/dt = 0$. The independent variable must be restated in terms of the new variable, so the complete change of variables is described by $$\begin{align} q(y,t) &= f(r,t) \\ y &= r/R \end{align}$$ where by the chain rule $$\frac{\partial f}{\partial t} = \frac{\partial q}{\partial t} + \frac{\partial y}{\partial t}\frac{\partial q}{\partial y}.$$ Recommended reading on a formal treatment of change of variables is still unbeknownst to me. -	{"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": 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": 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": 0.9460219740867615, "perplexity_flag": "head"}
http://www.physicsforums.com/showthread.php?p=3517183	Physics Forums Page 4 of 47 < 1 2 3 4 5 6 7 14 > Last » Recognitions: Science Advisor ## CERN team claims measurement of neutrino speed >c Quote by DaleSpam Correct, they are claiming the the neutrinos travel at 299 799 893 m/s compared to the speed of light 299 792 458 m/s. So the massive-photon resolution would require that the invariant speed be something greater than 299 799 893 m/s, but that would have been detectable in other experiments. The bottomline is there is no way this can work theoretically. You could for instance look for departures from Lorentz invariance, but that's already been searched for ad naeusum through several different channels. No known violation of the SR dispersion relations have ever been discovered, and the bounds are already far in excess of the sensitivity of this experiment. It also contradicts well established neutrino measurements, like the Supernova ones. Trying to stay consistent with that, leads you into real absurdities (like modifying standard MSW physics in violent ways) Quote by DaleSpam Correct, they are claiming the the neutrinos travel at 299 799 893 m/s compared to the speed of light 299 792 458 m/s. So the massive-photon resolution would require that the invariant speed be something greater than 299 799 893 m/s, but that would have been detectable in other experiments. Thanks for clarifying that. So now that the paper clears that up, it appears the photon can resume it's original svelte, speedy status as "c". This sure is starting to look like an error in position measurement. Not as much fun, but still would be important, since they must have been closely studying that possibility all along. I suppose the little ones could be taking an extra-dimensional short cut or a convenient worm hole, but they'd all have to be taking the same short cut every time for years. I dunno... Thanks to those who are summarizing the paper's details for us non-physicists. Recognitions: Gold Member Science Advisor Agreed, the OPERA team is seeking confirmation [I agree with Pallen it appears unlikely]. Neutrino detection is tricky business and correlating capture with emission is no easy task. I can't help but wonder how many of the detected neutrinos were actually emitted by CERN and how that might skew the measurement. There was a paper about 10 years ago about neutrinos as tachyons by Chodos, IIRC. Quote by Chronos Agreed, the OPERA team is seeking confirmation [I agree with Pallen it appears unlikely]. Neutrino detection is tricky business and correlating capture with emission is no easy task. I can't help but wonder how many of the detected neutrinos were actually emitted by CERN and how that might skew the measurement. There was a paper about 10 years ago about neutrinos as tachyons by Chodos, IIRC. My background is engineering, not physics, but frankly, the method used to correlate the proton extractions with the v detections doesn't seem that bad to me so far, although at first blush, 16,111 detected events doesn't seem too great statistically. I'd like to see more expert comments on that however. Regarding potential contamination, would most contamination come from B-decay, which would be anti's? I think they accounted for anti's, counting about 2% unless I read it wrong. Could someone comment on that? I'm not sure about the potential sources of spurious neutrinos in significant numbers. I'm more struck by the timing aspects. There seem to be so many places in this system where inaccuracies can gang up on you. This is a pretty complex system with a lot of timing points, all with tolerances. I'd be the last person to second guess this work, but I think that's where I'd look. I think the question of clock synchronization may be tricky. In GR, there is no absolute definition of simultaneity. Due to differences in gravitational potential, as mentioned, clocks evolves differently at different points. So you must periodically resynchronize them, but how ? there is no unique choice, and the measured time of flight probably depends of how you define the timescale at each point. Quote by edgepflow There is one remote possibility I have not seen discussed in this thread. Is it possible in theory that a neutrino has zero mass and the test is showing tachyonic properties? This would not violate SR. An unlikely explanation but just wanted to see what an expert has to say. I'm afraid that's already ruled as a reasonable explanation for this by supernova 1987A. The problem is that the speed of a tachyon is given by $$v = c\sqrt{1+\frac{\|m^2\|c^4}{E^2}}$$. This means that a tachyon's speed increases as its energy decreases. As noted above, the OPERA neutrinos have higher energy than the 1987A neutrinos, meaning that, were they tachyonic, they should be slower, not faster, than the supernova neutrinos. But, in fact, the 1987A neutrinos have a discrepancy from c that is, at worst, something like 4 orders of magnitude smaller than the OPERA discrepancy. Quote by zadignose I'm sorry, but I don't quite buy a slight tweak to our definition of "c" as a complete answer. Anyone suggesting that simply adjusting the "c" constant will fix things needs to explain how 150 years of mathematical and physics equations didn't detect the discrepancy. Looking to an adjustment to c as the answer to this data, if correct, is... creative. It is not borne out of a dedication to science but a fear of change, as given data like this, that is certainly not the most likely cause, even within our CURRENT theories. What if neutrinos are very high energy tachions, so we never noticed that they are moving slightly faster than c? We can't detect low energy neutrinos, so usually dont see them moving much faster than c. Quote by Dmitry67 What if neutrinos are very high energy tachions, so we never noticed that they are moving slightly faster than c? We can't detect low energy neutrinos, so usually dont see them moving much faster than c. I believe someone else already asked this question. The answer that was given is that tachyons decrease in energy as they increase in speed. Using the neutrinos detected from the referenced supernova, were they tachyonic, the neutrinos should have been traveling even faster than the ones CERN is talking about. Instead we saw them arrive simultaneously with the photons. Blog Entries: 19 Recognitions: Science Advisor The idea that neutrinos are tachyons is not new. Many papers already exist: http://xxx.lanl.gov/find/hep-ph/1/ti.../0/1/0/all/0/1 It is reasonable that after this kind of announcements people starts getting nervous and all kind of silly things are said. Maybe it is not so normal that knowledgeable people first reaction to this apparently "FTL neutrinos" be that SR must be modified, or everything that was measured so far to a certain accuracy is now wrong. It is not. Let's listen to Vanadium 50 here. First thing to rule out is obviously some kind of error in the measurement, and this is explicit in most posts. Even if no measurement error is found, we must first look for explanations that are compatible with the accuracy level of thousands of previous experiments that can't just be ignored. So far little attention has been focused to the special nature of the subject particle, the neutrino and the way it is measured, I would say that this is the weakest link of the chain if no obvious claculational or silly error is found so I think the first serious theoretical searches must come from this side rather than question relativity. Recognitions: Homework Help Science Advisor Quote by Vanadium 50 This is a systematic effect. You can take that to the bank. They don't see a velocity dispersion. By itself, that's a huge problem. If you want to argue that not only are neutrinos faster than light, but they all travel at the same speed regardless of energy, you have to explain why the neutrinos from SN1987A arrived on the same day as the light did, instead of (as the Opera data would indicate) four years earlier. Thank you for posting this! I was pouring through all this info, with this same obvious fact in mind, wondering what I missed. The neutrino burst is part of the standard method for studying Type 2 supernovae in other galaxies, and they all arrive, exactly according to precise calculations, after the light gets here. So granted there's plenty I don't know or understand about the data, but place me in the camp that thinks a systematic error is to blame, rather than derailment of SR. But hey, I'm a good little scientist--I'll leave the door open. Quote by TrickyDicky Yes, it would be a big problem. The problem here theoreticians don't seem to make up their mind what speed neutrinos should travel at, when they were supposed to be massless they were expected to have light speed, and supernovae detections so seemed to verify, when agreement was reached that they had mass they obviously should be slower than c, but as Demystifier pointed out there were several people that hypothesized that they should be FTL. One has to wonder what they really are all measuring, is it really neutrinos? Is there a serious agreement about what its speed should be? The fact that particles arrive at the same moment from supernovae is a compelling argument it is a fluke, unfortunately. Success to you all. To be sure: it is not CERN who is claiming this, but a team outside of CERN, and all what CERN does is to provide a platform for today's press conference. Unfortunately so, and many colleagues strongly object this. Of course this is being mixed up all over in the media, as usual. Incidentally neither the General Director nor the research director will be present. Seems that I am only one here who bothered to read OPERA preprint: http://arxiv.org/ftp/arxiv/papers/1109/1109.4897.pdf Just some points after reading: 1. There is no information what reference frame they use for analysis and how they covered relativistic effects in their analysis: CERN? Gran-Sasso? Centre-of-Earth? Solar System? Please note, that SR time dilation between CERN and Gran-Sasso frames is 10 times stronger than the effect they report. How the clocks were corrected for dilation? There is also no information if GR effects were taken into account. 2. There is no discussion about systematic errors which may be caused by delays in readout electronics and scintillators itself (except of light propagation, which is the only one discussed). The systematic error caused by DAQ and detectors is estimated as for few ns each, which seems to be too optimistic. 3. Detailed experimental setup is delegated to other paper not available online. Quote by PAllen There would be a race to determine the mass of the photon. It would be a huge surprise, but I think it would be a bigger hit for QED than SR or GR - the latter rely only on the fact that there is a spacetime structure speed limit. Whether a particular particle reaches it is irrelevant. I would definitely take the bet against this being confirmed. My first thought was perhaps photons do no travel at "the speed of light", ie photons have (rest) mass. According to wikipedia http://en.wikipedia.org/wiki/Photon#...on_photon_mass the experimental limit is at least as good as m < 1e-14 eV/c^2 I could not find a formula to convert photon mass into speed, but I think I have worked it out: (v/c) = SQRT( (1+d^2)/(1+2d^2) ) where d = Lmc/h (L = wavelength, m = photon rest mass, c = "cosmic speed limit for which we need to find a new name", h = Planks constant). For small d this approximates to v/c = 1 - d^2/2 Using the mass given above and for a green photon of wavelength 500nm that comes out as one part in about 10^30, much smaller than the 20 parts in a million quoted for the neutrinos. To look at it the other way, for a photon to be travelling 6000m/s slower then true "c" would require it to have a rest mass of about 1.5e-2 ev/c^2 which would have been noticed. However my SR is a bit rusty so if anyone wants to check this I would be grateful. (AIUI it is not significant that light is observed to travel "at c" because since there is no evidence (as yet) that photons have mass, we have just taken "c" to be the speed of light). Page 4 of 47 < 1 2 3 4 5 6 7 14 > Last » Tags anisotropy, cern, ftl, gps, new math books Thread Tools \| \| \| \| \|------------------------------------------------------------------------\|----------------------------------------\|---------\| \| Similar Threads for: CERN team claims measurement of neutrino speed >c \| \| \| \| Thread \| Forum \| Replies \| \| \| Special & General Relativity \| 6 \| \| \| High Energy, Nuclear, Particle Physics \| 4 \| \| \| Introductory Physics Homework \| 1 \|	{"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": 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": 0.9673117995262146, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/84531/lower-semicontinuity-of-kullback-leibler-divergence	## Lower semicontinuity of Kullback-Leibler divergence ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) The Kullback-Leibler Divergence (KLD) of two PMF's $P(x)$ and $Q(x)$ is $D(P\|\|Q)=\sum_x P(x)\log(P(x)/Q(x))$, with the provisos that $0\cdot \log (0/p)=0$ and $p\cdot \log (p/0)=+\infty$ whenever $p>0$. It is known that KLD is continuous at $(P,Q)$ if $Q$ is strictly positive over all $x$'s. What can be said otherwise? To be more specific, assume we are given a sequence of PMF ${(P_n,Q_n)}_{n\geq 0}$ s.t. $(P_n,Q_n)\rightarrow (P,Q)$ in the simplex of PFM's (with the topology induced by, say, norm-1 distance). Is it correct to deduce that $\lim \inf_{n\rightarrow \infty} D(P_n\|\|Q_n) \geq D(P\|\|Q)$ ? This would follows if KLD is lower-semicontinuous, right? Many thanks. - ## 2 Answers In addition to the conventions you have mentioned, it is also assumed that $0\log(0/0)=0$. With these conventions, I think, in the finite case, it is always true that $$\lim_{n\to \infty} D(P_n\|\|Q_n)=D(P\|\|Q)$$ As you said, if $Q(x)>0$ for all $x$, its immediate from the Dominated Convergence theorem. The problem is only when for some $y$, $Q(y)=0$ whereas $P(y)>0$. In which case $P(y)\log(P(y)/Q(y))=\infty$ and $D(P\\|Q)=\infty$ But since $(P_n,Q_n)\to (P,Q)$, we have $P_n(y)\to P(y)$ and $Q_n(y)\to Q(y)$, whence $$P_n(y)\log(P_n(y)/Q_n(y))\to P(y)\log(P(y)/Q(y))=\infty.$$ Hence $D(P_n\|\|Q_n)\to \infty$ So in any case we have $\lim_{n\to \infty} D(P_n\|\|Q_n)=D(P\|\|Q)$. In a general measurable space (i.e., if $P_n, Q_n, P, Q$ are probability measures on some general measure space say $(\mathbb{X}, \mathcal{X})$), I think, we have only lower semicontinuity. Pardon me, if something is wrong. - Surely it is not always the case that if $(P_n,Q_n)\rightarrow (P,Q)$ then $\lim_{n\rightarrow \infty} D(P_n\|\|Q_n)= D(P\|\|Q)$, as you say. If $Q$ is the frontier of the PMF simplex (that is, $Q(x)=0$ for some $x$), consider a sequence $(P_n,Q_n)$ with $Q_n=Q$ and $P_n\rightarrow P$, where $P(x)=0$ whenever $Q(x)=0$. Moreover, assume all the $P_n$'s are in the interior of the simplex, that is they are all strictly positive on every $x$. Then, for each $n$, $D(P_n\|\|Q_n)=+\infty$, hence $\lim_{n\rightarrow \infty} D(P_n\|\|Q_n)=+\infty$, whereas $D(P\|\|Q)<+\infty$. – Michele Dec 30 2011 at 12:49 Yes Michele, you are right. My proof would not work in the case you have stated. But I am sure that its lower semi continuous. – Ashok Dec 30 2011 at 13:27 For example, you can find a proof in the paper (see section III) titled "Random coding strategies for minimum entropy" published in IT Transactions. – Ashok Dec 30 2011 at 13:45 Ashok, I found the paper, man thanks. Michele – Michele Dec 30 2011 at 13:55 ### 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. I believe that the lower semicontinuity of KBD is proved in Cover-Thomas Information Theory book. Also in Kullback's information theory book. - why the down-vote? – Anthony Quas Dec 29 2011 at 20:59	{"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": 43, "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": 0.9406031966209412, "perplexity_flag": "head"}
http://unapologetic.wordpress.com/2007/03/16/ams-sectional-day-1/?like=1&source=post_flair&_wpnonce=7b62a7b6bf	# The Unapologetic Mathematician ## AMS Sectional, Day 1 I don’t have wireless access here in the lecture hall, so I can’t “live blog”. I’m writing notes on the lectures I find noteworthy. David Radford spoke about something called the Hennings Invariant of a finite-dimensional Hopf algebra. I’ve always liked his style, since he manages to boil down a lot of complicated algebraic structure to what’s essential for the application at hand. He also describes it incredibly clearly. His lectures are very accessible to a grad student who has a basic background in algebra, which is more than I can say about many algebraists. I think it should be clear why I think this is a Good Thing. I’ll get to Hopf algebras in more depth eventually, but for now let me say this: they’re very much like groups, but using somewhat heavier machinery. In the long run, groups and Hopf algebras both work off of a very similar structure. Pat Gilmer gave a talk on “congruence” and “similarity” of 3-manifolds. A 3-manifold is a space that looks close-up like three-dimensional space, like the surface of the Earth looks flat since we’re so close to it. These two concepts he’s pushing are equivalence relations. Two 3-manifolds may be different, but might still be “similar” or “congruent” if they’re related by certain modifications, called “surgeries”. Congruence was evidently studied about ten years ago and Gilmer reinvented it himself along with similarity. He’s particularly interested in how certain well-known invariants of 3-manifolds change as you apply these surgeries. One interesting thing this brings to mind is the fact that we can get any 3-manifold from the 3-sphere (the surface of the Earth is a 2-sphere) by cutting out a bunch of bagel-shaped regions that might be knotted, twisting up the boundaries of the parts we cut out, and putting them back in. This means that there’s a connection between knot theory and 3-manifold theory. In fact, a very large portion of mathematicians calling themselves knot theorists are really more interested in 3-manifolds and just use knot theory as a tool. After lunch, Carmen Caprau gave her talk about an ${\mathfrak sl}_2$ tangle homology. It manages to fix a big problem I’ve had with Khovanov’s homology theory — it tends to screw up the signs. Knot homology theories are really big business these days. Most people I know who are on the job market and work directly with these sorts of things have jobs nailed down, and Carmen is no exception. Good luck to her. Scott Carter talked about cohomology in symmetric monoidal categories with products and coproducts. This extends the stuff he has done with Alissa Crans, Mohammed Elhamdadi, Pedro Lopes, and Masahico Saito. The last version of this talk I saw at last Spring’s Knots In Washington was some of the nicest theory I’ve seen. Now they’re taking this abstract setup describing Hoschschild homology theories (which try to capture the underlying essence of associativity), and “dualize” all the diagrams to get some sort of topological invariant. I always love mixing up notation and subject matter, and this is very much in that Kauffman-esque spirit. Hopefully there will be an updated version of their paper on the arXiv soon. Maciej Niebrzydowski spoke on homology of dihedral quandles, which he worked on with his advisor, Jozef Przytycki. I’ll leave this alone since I’m almost ready to talk about quandles in full detail. ### Like this: Posted by John Armstrong \| Uncategorized No comments yet. « 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": 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": 0.9507129788398743, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/32269/what-will-happen-if-a-plane-trys-to-take-off-whilst-on-a-treadmill/34752	# What will happen if a plane trys to take off whilst on a treadmill? So this has puzzled me for many a year... I still am no closer to coming to a conclusion, after many arguments that is. I don't think it can, others 100% think it will. If you have a plane trying to take off whilst on a tread mill which will run at the same speed as whatever the planes tyres rotation speed is will it take off? [edited to be more clear] The question is simple. Will a plane take off if you put this plane onto a treadmill that will equity whatever speed the plane wheels are moving at. So the plane should not be able to move. This is a hypothetical situation of course. But i am very interested. - my thoughts, you need a certain amount of thrust on the wings, its the hole basis of flight. – Jamie Hutber Jul 17 '12 at 22:11 4 – DJBunk Jul 17 '12 at 22:44 4 What matters to an airplane is relative wind, period. (This is not rocket science ;-) – Mike Dunlavey Jul 18 '12 at 14:41 1 – mwengler Jul 18 '12 at 15:11 mytherbusters version doesn't cut it for me though, as there is still friction between the floor and the tyres, this plus the fact we don't know how fast the treadmill is running at. – Jamie Hutber Jul 18 '12 at 16:07 show 2 more comments ## 6 Answers Idealizing the plane's wheels as frictionless, the thrust from the propeller accelerates the plane through the air regardless of the treadmill. The thrust comes from the prop, and the wheels, being frictionless, do not hold the plane back in any way. If the treadmill is too short, the plane just runs of the end of it and then continues rolling towards take off. If the treadmill is long enough for a normal takeoff roll, the plane accelerates through the air and rotates off of the treadmill. UPDATE: Don't take Alfred's word for it. Mythbusters has actually done the experiment. UPDATE 2: I've been thinking about how the problem is posed (for now as I'm typing this) and it occurred to me that the constraint "run at the same speed as whatever the planes tyres rotation speed" actually means run such that the plane doesn't move with respect to the ground. Consider a wheel of radius $R$ on a treadmill. The treadmill surface has a linear speed $v_T$ to the right. The center of the wheel has a linear speed $v_P$ to the left. The CCW angular speed of the wheel is: $\omega = \dfrac{v_T + v_P}{R}$ If "run at the same speed as whatever the planes tyres rotation speed" means: $\omega = \dfrac{v_T}{R}$ then the constraint requires $v_P = 0$. That is, the question, as posed, is: If the treadmill is run such that the plane doesn't move, will the plane take off? Obviously, the answer is no. The plane must move to take off. Looking at mwengler's long answer, we see what is happening. The rotational speed of the tires and treadmill are not the key, it is the acceleration of the treadmill that imparts a force on the wheel axles (ignoring friction for simplicity here). So, it is in fact the case that it is possible, in principle, ( don't think it is possible in practice though) to control the treadmill in such a way that it imparts a holding force on the plane, preventing it from moving. But, once again, this force is not proportional to the wheels rotational speed, but to the wheel's angular acceleration (note that in the idealized case of massless wheels, it isn't even possible in principle as the lower the moment of inertia of the wheels, the greater the required angular acceleration). - You make the most important point, which is the plane is thrusted through the air. – mwengler Jul 17 '12 at 22:25 1 @JamieHutber, the plane does move. The thrust from the propeller is far more significant that any nominal frictional force from the wheels. The plane moves forward. A plane isn't a car, the motor doesn't drive the wheels. – Alfred Centauri Jul 17 '12 at 22:44 1 @Jamie: I think the idea is that you can't match the propeller, the friction on the wheels will never be that high. The plane needs just a little bit of thrust to stay stationary: The treadmill will move and the wheels will rotate, but the plane itself will stay in place. Any more thrust and it will move forward. – Javier Badia Jul 17 '12 at 23:15 1 Another way to think about it is a plane taking off on ice - if the speed of the planes wheels mattered how would planes with skis manage? – Martin Beckett Jul 18 '12 at 4:17 1 Alfred Centauri is 100% correct. If you are still confused you should re-read his answer, re-read it again, and if need be get a pilots license to fully understand that plane speed is not at all influenced by wheel speed. – Benjamin Horowitz Jul 18 '12 at 16:12 show 7 more comments Simplify. Suppose the air is still - no wind. Suppose the wheels are truly frictionless - like greased skids. (After all, that's why they have ball bearings.) The aircraft starts from a standing position, and it accelerates to rotation airspeed, about 100 km/h. It does so by thrusting against the air, not against the surface it is standing on. As it accelerates, the pickup truck pulls the fabric under the plane (simulating a treadmill) in the opposite direction, up to 100 km/h. So, with respect to the fixed un-moving air, the aircraft is moving one way at 100, and the surface under the wheels is moving the opposite way at 100. The plane takes off, because of its airspeed. The wheels turn at 200 km/h, because somebody's dragging the runway backward. They don't care - they're frictionless. All the "treadmill" has done is make the wheels turn faster. - This is the correct answer. If the plane is stationary, the plane will obviously experience zero aerodynamic lift. – QuantumDot Aug 23 '12 at 3:29 Taking this as a logical question rather than physics based it is clearly playing on the mis-assumption that motive force is can only be aplied through contact with the floor. ie we walk forward by pushing on the floor, we drive by making car wheels push on the road. However a solution is to realise that a commercial jet will gain the force from pushing on the air as explained elsewhere, the floor contact is irrelevant to the problem. And so we say the plane takes off. Feel free to complicate the problem as you wish! - 1 Not just a commercial jet. All powered flying machines take there thrust from the air. – dmckee♦ Jul 18 '12 at 15:27 @dmckee: I was agreeing with you until I got a mental image of a leg-propelled hang-glider launching off a hillside :-) – Mike Dunlavey Jul 19 '12 at 18:04 @MikeDunlavey Uhm....er....I'm going to duck the issue by classifying that as unpowered. Yeah. That's it. – dmckee♦ Jul 20 '12 at 1:36 EDIT ADDED 7/18/12 Unfortunately, the original statement of the original question was totally different than the ACTUAL question the original poster intended to answer. That original question is simply asked and answerd by Mythbusters. If the original poster had simply referenced the source of his question, it would have been much clearer before I did my long answer below. The actual question the poster wanted to ask, and the one asked and answered by Mythbusters is this: an airplane is on a conveyor belt runway that can run backwards. The forward speed of the airplane is monitored and the conveyor belt is run backwards at that forward speed as the airplane tries to take off. The wheels on the airplane are free rolling (no brakes, no motors). Can the airplane take off? This is a WAY easier question than the one the poster originally asked in which the original question specified the conveyor belt would run at the speed of the WHEELS. So in the original question, the conveyor belt would run fast enough so that either the wheels were slipping on it (if the plane was moving forward) or the plane was forced to stand still (if the wheels were not slipping on it. That is the question I answered below. The Mythbusters question is much easier. First, we know a plane doesn't even need wheels to take off, water planes and planes that land on snow or ice on skis do it all the time. The wheels are just a convenient way to have a connection to the ground which is low friction in the forward-backward direction. All the conveyor belt causes is the free-turning wheels to turn twice as fast as they normally would on take-off. Does this cause the engine to put a little more (OK, 4X as much) rotational energy into the rotation of the wheels? Yes it does. It is even vaguely questionable that a plane with a margin-of-error extra power enough to take off by pulling itself through the air can spin its (rather small, relative to the airplane mass) wheels twice as fast? No, the wheel mass is way too small to be a big part of the equation of motion of an airplane being pulled through the air by a propeller. Watch the youtube video and watch the plane take off from the conveyor belt no problem. Below appears my answer to the original question, which was much more obscure, much more challenging to sort out from a physics perspective. What a wild question! The thing that determines take off is enough lift from the wings. The lift depends on the airspeed flowing over the wings. You might think on a windless day that the airspeed over the wings is zero if the airplane is not moving forward, but what if the airplane has a big propeller in front its wings? Then the propeller blows air over the wings. I don't know for sure, but maybe a very powerful acrobatic airplane can blow the wind across its wings with its propeller fast enough to create enough wing lift to take off, even when the airplane is not moving through the air itself. But certainly most front-propeller planes cannot do this, they need forward motion through the air to get enough airspeed across wings, and all jets and rear-propeller planes require forward motion to get airflow across wings. So the next question is: does the airplane develop any forward motion as you define the problem? Suppose it is a jet. The jet engine is sending a lot of mass of air very fast backwards behind the plane. To conserve momentum, that reverse momentum must be going somewhere. On a normal runway (or a treadmill that can't keep up with the tires) much of that momentum would go in to the forward motion of the airplane. Now we need to figure out something about what kind of force the treadmill can put on the airplane by running backwards. Suppose we had a tire (or a cylinder) on the treadmill, and the treadmill started running in such a direction as to start the tire spinning but not to translate the tire left or right. Would the tire move along the treadmill, or would the tire stay in place and simply turn as fast as the treadmill was moving? I feel as though I should stop here and let the students figure out their answer to this question. Instead I'll just continue. Actually lets look at a SLIGHTLY simpler question first. We have a post holding that tire down against the treadmill. If the treadmill is stationary and the tire is stationary, we know there is no force on the post holding the tire. The tire sits still, the post is not being pulled forward backward or sideways. Now what if the treadmill is running at a steady speed, then at steady state the tire is running at a steady rotational speed = to treadmill speed to stay in place as it will held in place by the post. But is there a forward or backward force on the post? If the bearing holding the wheel to its axle is frictionless, I am pretty sure there is no force. The tire is rotating at a contstant velocity, since the axle is frictionless, it doesn't need any force to keep it rotating at a constant speed. So in steady state, the tire rotating at a constant 100 kph on a treadmill running at a constant 100 kph puts no force one way or the other on the post holding it. Now how the heck can we couple translation motion of the treadmill into any translational force on the airplane? Assuming frictionless axles on the wheels? In steady state we can't. But what about as we accelerate? So we look at the problem where the wheel is on the treadmill stationary, and we speed the treadmill up to 100 kph. What happens 1. The wheel spins up slowly but does not forward or backward motion. 2. The wheel doesn't spin at all but does move in the direction of the treadmill 3. The wheel splits the difference, spinning up some as the treadmill accelerates, and picking up some forward motion as the treadmill accelerates. Now those of us who have been around the block a few times KNOW the answer must be number 3, that is, unless it isn't. But how do we show that? Consider a wheel in empty space, with its axle aligned with the x-axis, so it can spin freely through the y-z plane. At the lowest point (the most negative z point) we apply a force $+F\hat{y}$ for a time $t$, and then go back to applying zero force. $\hat{y}$ is a unit vector in the $y$ direction, that is the force we apply is along the surface of the wheel only. What does the wheel do? Well we are imparting linear "impulse" into the wheel of $Ft$ so we change its linear momentum by $Ft$ so we change its linear velocity by $v=Ft/m$ where $m$ is the mass of the wheel. But we are also putting torque around the axle of magnitude $Fr$ into the wheel where $r$ is the radius of the wheel. Thus we increase the angular momentum of the wheel by $Frt$. Which means we set the wheel to spinning with angular velocity $\omega = Frt/I$ where $I$ is the moment of inertia of the wheel about its axle. Seeing the linear dependence of $v$ and $omega$ on $Ft$ we can see that no matter what force at what time we put in, the ratio is fixed: $$v/\omega = I/mr$$ The point being, a force applied along the surface of the wheel imparts some linear momentum into the wheel (and whatever it is attached to) and some angular momentum into the wheel (which spins the wheel). So back to the airplane. We have this airplane with a powerful jet engine imparting a very large $-F\hat{y}$ into moving the airplane forward. If the treadmill is to keep the jet from accelerating forward, it will need to provide an equally large but opposite $F\hat{y}$ to the airplane. But as we saw above, whatever linear force the threadmill applies to the tire, it is applying a proportionally large torque to the wheel. We note the mass of the airplane $M$ is much more than the mass of the tire, $m$, so $I/r = m\ll M$. So to counteract the force of the jet engine, the treadmill is going to have to accelerate a lot. That is, $\omega=Ct$ to counteract the linear force of the jet engine on the airplane. So the wheel is going to have to spin up really really REALLY quickly, and keep spinning up faster and faster as long as we have the jet engine going. My intuition suggests that long before the wheel reaches relativistic speeds, it will be flung apart by centrifugal forces overcoming the molecular forces that usually keep solid matter solid. But until the wheel explodes (or the threadmill explodes), the jet is kept from having any linear acceleration, and so does not take off. - This particular problem is somewhat ill defined but, as usually posed, the idea is that the treadmill speed matches the speed of the plane as if it were accelerating on a runway. So, for example, let there be two identical planes, one on a runway, and one on a treadmill the length of the runway. Both planes apply the same power and now, let the treadmill speed match the speed of the plane on the runway. This is, I think, what most have in mind when they pose this problem. – Alfred Centauri Jul 17 '12 at 23:44 I'm sorry if I'm not understanding something (entirely possible), but consider this: The wheel has two forces in the horizontal direction acting on it: the friction with the treadmill and the thrust from the plane. The whole system will move forward if the thrust is higher than the friction. But the friction has a maximum: a certain coefficiente times the weight of the plane. Therefore, all the plane needs to do to take off is to create a thrust larger than this maximum. Is this right? – Javier Badia Jul 18 '12 at 0:24 @JavierBadia Not completely. It could be moving, but slowed down enough by the friction to not be able to achieve take off speed. So while it might get to 200 mph being pushed by the jet when it is rolling, it might get to only 50 mph if there was some excess friction going on. The FACT is that the wheels can hold a commercial jet against its jet engines at maximum thrust. A commercial jet does not produce enough thrust to overcome tire friction, the brakes must be released for the jet to move. – mwengler Jul 18 '12 at 0:54 @AlfredCentauri your comment here is entirely different than the question you ask in your original post, which i answer in my answer. Your comment means the jet on the treadmill must spin its tires twice as fast as the jet on the regular runway to achieve the take off speed. It seems likely to me the tires might burst or fail in some other way since they are not designed for 2x the speed, and centrifugal force is 2x higher in this case. – mwengler Jul 18 '12 at 0:57 1 It doesn't need to, unless the question assumes that the planes brakes are engaged, in which case this is a silly problem. Even with no treadmill, and a plane can't take off like that. – Colin K Jul 18 '12 at 1:51 show 4 more comments Quick answer: You better change your fitness center where you train or stop booze. Serious answer: The difference between the speed of air below and above the wing provides the lift. The velocity relative to the ground provides forward motion. The treadmill just zeros the latter. Case 1: The plane is still relative to the treadmill. But somewhat the air keeps flowing around the wings. Then the plane takes off vertically (if the lift is greater than the weight). Case 2: The plane is still relative to the treadmill. The air does not flow around the wings (the reactors are not good vaccum cleaners), or more likely, the lift is not greater than the weight, then you just burn fuel. - It will if its a harrier :) But if the treadmill is going fast enough so that the airplane has no velocity relative to the ground/air then it will not take off because there will be no lift from air flowing over the wings. This is of course not taking into the effects of propellers which shouldn't be able to lift the aircraft with 0 velocity relative to the ground/air - The speed of the treadmill affects the rotational speed of the tires, i.e., it changes the speed from what it would be if the treadmill were off. For example, let's say that the ground speed is zero. With the treadmill stopped, the tires aren't rotating. With the treadmill on, the tires rotate and, assuming insignificant friction, the ground speed is still zero. Essentially, the only thing the treadmill speed does is to add or subtract from the rotational speed of the tires for a given ground speed. – Alfred Centauri Jul 18 '12 at 14:57 2 My favorite (my only) airplane is a Cessna 172 Skyhawk. At takeoff power, it generates about 400lbs of thrust against the surrounding air. Since the plane weighs about 2000lbs, that's a pretty smart acceleration. It doesn't matter if somebody underneath is spinning up the wheels, forward or back. – Mike Dunlavey Jul 18 '12 at 18:13 @ Alfred, I don't know if the way I typed it was unclear but I totally agree with you and that was what I was trying to imply. – thebmags Jul 18 '12 at 19:23	{"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": 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": 0.9588834047317505, "perplexity_flag": "middle"}
http://www.haskell.org/haskellwiki/index.php?title=User:Michiexile/MATH198/Lecture_4&oldid=30809	# User:Michiexile/MATH198/Lecture 4 ### From HaskellWiki Revision as of 20:41, 14 October 2009 by Michiexile (Talk \| contribs) IMPORTANT NOTE: THESE NOTES ARE STILL UNDER DEVELOPMENT. PLEASE WAIT UNTIL AFTER THE LECTURE WITH HANDING ANYTHING IN, OR TREATING THE NOTES AS READY TO READ. ## Contents ### 1 Product Recall the construction of a cartesian product of two sets: $A\times B=\{(a,b) : a\in A, b\in B\}$. We have functions $p_A:A\times B\to A$ and $p_B:A\times B\to B$ extracting the two sets from the product, and we can take any two functions $f:A\to A'$ and $g:B\to B'$ and take them together to form a function $f\times g:A\times B\to A'\times B'$. Similarly, we can form the type of pairs of Haskell types: Pair s t = (s,t) . For the pair type, we have canonical functions fst :: (s,t) -> s and snd :: (s,t) -> t extracting the components. And given two functions f :: s -> s' and g :: t -> t' , there is a function f *** g :: (s,t) -> (s',t') . An element of the pair is completely determined by the two elements included in it. Hence, if we have a pair of generalized elements $q_1:V\to A$ and $q_2:V\to B$, we can find a unique generalized element $q:V\to A\times B$ such that the projection arrows on this gives us the original elements back. This argument indicates to us a possible definition that avoids talking about elements in sets in the first place, and we are lead to the Definition A product of two objects A,B in a category C is an object $A\times B$ equipped with arrows $A \leftarrow^{p_1} A\times B\rightarrow^{p_2} B$ such that for any other object V with arrows $A \leftarrow^{q_1} V \rightarrow^{q_2} B$, there is a unique arrow $V\to A\times B$ such that the diagram commutes. The diagram $A \leftarrow^{p_1} A\times B\rightarrow^{p_2} B$ is called a product cone if it is a diagram of a product with the projection arrows from its definition. In the category of sets, the unique map is given by q(v) = (q1(v),q2(v)). In the Haskell category, it is given by the combinator (&&&) :: (a -> b) -> (a -> c) -> a -> (b,c) . We tend to talk about the product. The justification for this lies in the first interesting Proposition If P and P' are both products for A,B, then they are isomorphic. Proof Consider the diagram Both vertical arrows are given by the product property of the two product cones involved. Their compositions are endo-arrows of P,P', such that in each case, we get a diagram like with $V=A\times B=P$ (or P'), and q1 = p1,q2 = p2. There is, by the product property, only one endoarrow that can make the diagram work - but both the composition of the two arrows, and the identity arrow itself, make the diagram commute. Therefore, the composition has to be the identity. QED. We can expand the binary product to higher order products easily - instead of pairs of arrows, we have families of arrows, and all the diagrams carry over to the larger case. #### 1.1 Binary functions Functions into a product help define the product in the first place, and function as elements of the product. Functions from a product, on the other hand, allow us to put a formalism around the idea of functions of several variables. So a function of two variables, of types A and B is a function f :: (A,B) -> C . The Haskell idiom for the same thing, A -> B -> C as a function taking one argument and returning a function of a single variable; as well as the curry / uncurry procedure is tightly connected to this viewpoint, and will reemerge below, as well as when we talk about adjunctions later on. ### 2 Coproduct The product came, in part, out of considering the pair construction. One alternative way to write the Pair a b type is: `data Pair a b = Pair a b` and the resulting type is isomorphic, in Hask, to the product type we discussed above. This is one of two basic things we can do in a data type declaration, and corresponds to the record types in Computer Science jargon. The other thing we can do is to form a union type, by something like `data Union a b = Left a \| Right b` which takes on either a value of type a or of type b , depending on what constructor we use. This type guarantees the existence of two functions ```Left :: a -> Union a b Right :: b -> Union a b``` Similarly, in the category of sets we have the disjoint union $S\coprod T = S\times 0 \cup T \times 1$, which also comes with functions $i_S: S\to S\coprod T, i_T: T\to S\coprod T$. We can use all this to mimic the product definition. The directions of the inclusions indicate that we may well want the dualization of the definition. Thus we define: Definition A coproduct A + B of objects A,B in a category C is an object equipped with arrows $A \rightarrow^{i_1} A+B \leftarrow^{i_2} B$ such that for any other object V with arrows $A\rightarrow^{q_1} V\leftarrow^{q_2} B$, there is a unique arrow $A+B\to V$ such that the diagram commutes. The diagram $A \rightarrow^{i_1} A+B \leftarrow^{i_2} B$ is called a coproduct cocone, and the arrows are inclusion arrows. For sets, we need to insist that instead of just any $S\times 0$ and $T\times 1$, we need the specific construction taking pairs for the coproduct to work out well. The issue here is that the categorical product is not defined as one single construction, but rather from how it behaves with respect to the arrows involved. With this caveat, however, the coproduct in Set really is the disjoint union sketched above. For Hask, the coproduct is the type construction of Union above - more usually written Either a b . And following closely in the dualization of the things we did for products, there is a first Proposition If C,C' are both coproducts for some pair A,B in a category D, then they are isomorphic. The proof follows the exact pattern of the corresponding proposition for products. ### 3 Cartesian Closed Categories and typed lambda-calculus A category is said to have pairwise products if for any objects A,B, there is a product object $A\times B$. A category is said to have pairwise coproducts if for any objects A,B, there is a coproduct object A + B. Recall when we talked about internal homs in Lecture 2. We can now define what we mean, formally, by the concept: Definition An object C in a category D is an internal hom object or an exponential object $[A\to B]$ or BA if it comes equipped with an arrow $ev: [A\to B] \times A \to B$, called the evaluation arrow, such that for any other arrow $f: C\times A\to B$, there is a unique arrow $\lambda f: C\to [A\to B]$ such that the composite $C\times A\to^{\lambda f\times 1_A} [A\to B]\times A\to^{ev} B$ is f. The idea here is that with something in an exponential object, and something in the source of the arrows we imagine live inside the exponential, we can produce the evaluation of the arrow at the source to produce something in the target. Using global elements, this reasoning comes through in a more natural manner: given $f: 1\to [A\to B]$ and $x: 1\to A$ we can produce the global element $f(x) = ev \circ f\times x: 1\to B$. Furthermore, we can always produce something in the exponential whenever we have something that looks as if it should be there. And with this we can define Definition A category C is a Cartesian Closed Category or a CCC if: 1. C has a terminal object 1 2. Each pair of objects $A, B\in C_0$ has a product $A\times B$ and projections $p_1:A\times B\to A$, $p_2:A\times B\to B$. 3. For every pair $A, B\in C_0$ of objects, there is an exponential object $[A\to B]$ with an evaluation map $[A\to B]\times A\to B$. #### 3.1 Currying Note that the exponential as described here is exactly what we need in order to discuss the Haskell concept of multi-parameter functions. If we consider the type of a binary function in Haskell: `binFunction :: a -> a -> a` This function really lives in the Haskell type a -> (a -> a) , and thus is an element in the repeated exponential object $[A \to [A\to A]]$. Evaluating once gives us a single-parameter function, the first parameter consumed by the first evaluation, and we can evaluate a second time, feeding in the second parameter to get an end result from the function. On the other hand, we can feed in both values at once, and get `binFunction' :: (a,a) -> a` which lives in the exponential object $[A\times A\to A]$. These are genuinely different objects, but they seem to do the same thing: consume two distinct values to produce a third value. The resolution of the difference lies, again, in a recognition from Set theory: there is an isomorphism $Hom(S, Hom(T, V)) = Hom(S\times T, V)$ which we can use as inspiration for an isomorphism $Hom(S,[T\to V]) = Hom(S\times T, V)$ valid in Cartesian Closed Categories. As it turns out, this is exactly what we need for λ-calculus. Any typed λ-calculus gives rise to a CCC in a natural manner, and any CCC has an internal language which satisfies, by the axioms for the CCC, all requirements to be a typed λ-calculus. More importantly, by stating λ-calculus in terms of a CCC instead of in terms of terms and rewriting rules is that you can escape worrying about variable clashes, alpha reductions and composability - the categorical translation ignores, at least superficially, the variables, reduces terms with morphisms that have equality built in, and provides associative composition for free. At this point, I'd reccomend reading more on Wikipedia [1] and [2], as well as in Lambek & Scott: Introduction to Higher Order Categorical Logic. The book by Lambek & Scott goes into great depth on these issues, but may be less than friendly to a novice. ### 4 Algebra of datatypes Recall from Lecture 3 that we can consider endofunctors as container datatypes. Some of the more obvious such container datatypes include: ```data 1 a = Empty data T a = T a``` These being the data type that has only one single element and the data type that has exactly one value contained. Using these, we can generate a whole slew of further datatypes. First off, we can generate a data type with any finite number of elements by $n = 1 + 1 + \dots + 1$ (n times). Remember that the coproduct construction for data types allows us to know which summand of the coproduct a given part is in, so the single elements in all the 1 s in the definition of n here are all distinguishable, thus giving the final type the required number of elements. Of note among these is the data type Bool = 2 - the Boolean data type, characterized by having exactly two elements. Furthermore, we can note that $1\times T = T$, with the isomorphism given by the maps ```f (Empty, T x) = T x g (T x) = (Empty, T x)``` Thus we have the capacity to add and multiply types with each other. We can verify, for any types A,B,C $A\times(B+C) = A\times B + A\times C$ We can thus make sense of types like T3 + 2T2 (either a triple of single values, or one out of two tagged pairs of single values). This allows us to start working out a calculus of data types with versatile expression power. We can produce recursive data type definitions by using equations to define data types, that then allow a direct translation back into Haskell data type definitions, such as: $List = 1 + T\times List$ $BinaryTree = T\times (1+BinaryTree\times BinaryTree)$ $TernaryTree = T\times (1+TernaryTree\times TernaryTree\times TernaryTree)$ $GenericTree = T\times (1+List\circ GenericTree)$ The real power of this way of rewriting types comes in the recognition that we can use algebraic methods to reason about our data types. For instance: ```List = 1 + T * List = 1 + T * (1 + T * List) = 1 + T * 1 + T * T* List = 1 + T + T * T * List``` so a list is either empty, contains one element, or contains at least two elements. Using, though, ideas from the theory of power series, or from continued fractions, we can start analyzing the data types using steps on the way that seem completely bizarre, but arriving at important property. Again, an easy example for illustration: ```List = 1 + T * List -- and thus List - T * List = 1 -- even though (-) doesn't make sense for data types (1 - T) * List = 1 -- still ignoring that (-)... List = 1 / (1 - T) -- even though (/) doesn't make sense for data types = 1 + T + TT + TT*T + ... -- by the geometric series identity``` and hence, we can conclude - using formally algebraic steps in between - that a list by the given definition consists of either an empty list, a single value, a pair of values, three values, et.c. At this point, I'd recommend anyone interested in more perspectives on this approach to data types, and thinks one may do with them, to read the following references: #### 4.1 Blog posts and Wikipages The ideas in this last section originate in a sequence of research papers from Conor McBride - however, these are research papers in logic, and thus come with all the quirks such research papers usually carry. Instead, the ideas have been described in several places by various blog authors from the Haskell community - which make for a more accessible but much less strict read. http://en.wikibooks.org/wiki/Haskell/Zippers -- On zippers, and differentiating types http://blog.lab49.com/archives/3011 -- On the polynomial data type calculus http://blog.lab49.com/archives/3027 -- On differentiating types and zippers http://comonad.com/reader/2008/generatingfunctorology/ -- Different recursive type constructions http://strictlypositive.org/slicing-jpgs/ -- Lecture slides for similar themes. http://blog.sigfpe.com/2009/09/finite-differences-of-types.html -- Finite differences of types - generalizing the differentiation approach. http://homepage.mac.com/sigfpe/Computing/fold.html -- Develops the underlying theory for our algebra of datatypes in some detail. ### 5 Homework 1. What are the products in the category C(P) of a poset P? What are the coproducts? 2. Prove that any two coproducts are isomorphic. 3. Prove that any two exponentials are isomorphic. 4. Prove that currying/uncurrying are isomorphisms in a CCC. Hint: the map $f\mapsto\lambda f$ is a map $Hom(C\times A, B)\to Hom(C,[A\to B])$. 5. Write down the type declaration for at least two of the example data types from the section of the algebra of datatypes, and write a Functor implementation for each.	{"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": 52, "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": 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": 0.9025548100471497, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/164340/directed-colimit-in-a-concrete-category?answertab=votes	# Directed colimit in a concrete category I recently found myself at a spot that I never believed I'll get (or at least not that soon in my career). I ran into a problem which seems to be best answered via categories. The situation is this, I have a directed system of structures and the maps are all the inclusion map, that is $X_i$ for $i\in I$ where $(I,\leq)$ is a directed set; and if $i\leq j$ then $X_i$ is a substructure of $X_j$. Suppose that the direct limit of the system exists. Can I be certain that this direct limit is actually the union? Namely, what sort of categories would ensure this, and what possible counterexamples are there? I asked several folks around the department today, some assured me that this is the case for concrete categories, while others assured me that a counterexample can be found (although it won't be organic, and would probably be manufactured for this case). The situation is such that the direct system is originating from forcing, so it's quite... wide and probably immune to some of the "thinkable" counterexamples (by arguments of [set theoretical] genericity from one angle or another), and so any counterexample which is essentially a linearly ordered system is not going to be useful as a counterexample. Another typical counterexample which is irrelevant here is finitely-generated things, for example we can take a direct system of f.g. vector spaces whose limit is not f.g. but this aspect is also irrelevant to me; although I am not sure how to accurately describe this sort of irrelevance. Last point (which came up with every person I discussed this question today), if we consider: $$\mathbb R\hookrightarrow\mathbb R^2\hookrightarrow\ldots$$ Then we consider those to be actually increasing sets in inclusion and not "natural identifications" as commonly done in categories. So the limit of the above would actually be $\mathbb R^{<\omega}$ (all finite sequences from $\mathbb R$). - I hope that I gave a proper title, tags and didn't abuse the language too much. Do feel free to correct things if you see mistakes. – Asaf Karagila Jun 28 '12 at 20:51 I should add that I did Google a bit, but I couldn't find anything that I figured was useful. Any particular pointing at a decent place to start (although I am mostly interested in the result and less in the theory of concrete categories) would be great. – Asaf Karagila Jun 28 '12 at 20:57 Haven't really thought this through, but I guess if there is a free functor from Set to your category then it must be the union (since 'adjoints preserve limits'). – wildildildlife Jun 28 '12 at 21:03 My category is not a specific one. I am sorta looking for some nice characterization of what sort of categories the direct limit would have to be union, and in what categories it won't be. – Asaf Karagila Jun 28 '12 at 21:05 1 @wildildildlife: Not relevant here, since a direct "limit" is a colimit. – Zhen Lin Jun 28 '12 at 21:23 show 3 more comments ## 2 Answers Your question essentially amounts to asking, "when does the forgetful functor $U : \mathcal{C} \to \textbf{Set}$ create directed colimits?" More generally, one could replace "directed colimit" by "filtered colimit". There is, as far as I know, no general answer. Here is one reasonably general class of categories $\mathcal{C}$ for which there is such a forgetful functor. Let us consider a finitary algebraic theory $\mathfrak{T}$, i.e. a one-sorted first-order theory with a set of operations of finite arity and whose axioms form a set of universally-quantified equations. For example, $\mathfrak{T}$ could be the theory of groups, or the theory of $R$-modules for any fixed $R$-module. Then, the category $\mathcal{C}$ of models of $\mathfrak{T}$ will be a category in which filtered colimits are, roughly speaking, obtained by taking the union of the underlying sets. This can be proven "by hand", by showing that the obvious construction has the required universal property: the key lemma is that filtered colimits commute with finite limits in $\textbf{Set}$ – so, for example, $\varinjlim_{\mathcal{J}} X \times \varinjlim_{\mathcal{J}} Y \cong \varinjlim_{\mathcal{J}} X \times Y$ if $\mathcal{J}$ is a filtered category. Mac Lane spells out the details in [CWM, Ch. IX, §3, Thm 1]. Addenda. Fix a one-sorted first-order signature $\Sigma$. Consider the directed colimit of the underlying sets of some $\Sigma$-structures: notice that the colimit inherits a $\Sigma$-structure if and only if the operations and predicates of $\Sigma$ are all finitary. Qiaochu's counterexample with $\{ 0 \} \subset \{ 0, 1 \} \subset \{ 0, 1, 2 \} \subset \cdots$ can be pressed into service here as well. So let us assume $\Sigma$ only has finitary operations and predicates. The problem is now to establish an analogue of Łoś's theorem for directed colimits. Let $\mathcal{J}$ be a directed set and let $X : \mathcal{J} \to \Sigma \text{-Str}$ be a directed system of $\Sigma$-structures. Let us say that a logical formula $\phi$ is "good" just if $X_j \vDash \phi$ for all $X_j$ implies $\varinjlim X \vDash \phi$ (where $\varinjlim X$ is computed in $\textbf{Set}$ and given the induced $\Sigma$-structure). 1. It is not hard to check that universally quantified equations and atomic predicates are good formulae. 2. The set of good formulae is closed under conjunction and disjunction. 3. The set of good formulae is closed under universal quantification. 4. The set of good formulae is not closed under existential quantification: the formula $\forall x . \, x \le m$ (with free variable $m$) is a good formula in the signature of posets, but $\exists m . \, \forall x . \, x \le m$ is clearly not preserved by direct limits. 5. However, a quantifier-free good formula is still a good formula when prefixed with any number of existential quantifiers. 6. In particular, the set of good formulae is not closed under negation: the property of being unbounded above can be expressed as a good formula in the signature of posets with inequality, but its negation is the property of being bounded above. Section 6.5 of [Hodges, Model theory] seems to have some relevant material, but I haven't read it yet. The point, I suppose, is that there are some fairly strong conditions that the theory in question must satisfy before the directed colimit in $\textbf{Set}$ is even a model of the theory, let alone be a directed colimit in the category of models of the theory. - One also remarks that some people define "finitary algebraic theory" to mean "a monad $T : \textbf{Set} \to \textbf{Set}$ that preserves filtered colimits", making this very tautologous... – Zhen Lin Jun 28 '12 at 22:05 I'm not sure if it's a good thing or a bad thing, but I didn't understand this answer due to an extreme chasm in form of [mathematical language. I'll get someone to translate it to me tomorrow (it's too late for understanding new category related topics anyway). Also, this makes me think that the model-theoretic approach would have been better for me... :-) – Asaf Karagila Jun 28 '12 at 22:50 I actually proved part of the Łos theorem analogue in my case. It needs a minor tweaking to avoid trivialities it it works. I hope to extend it to simple second-order formulae too. – Asaf Karagila Jun 29 '12 at 6:06 I don't understand whether you actually have a counterexample or not so let me supply one for the sake of completeness. Consider the category $\text{CHaus}$ of compact Hausdorff spaces. I can write down a filtered colimit of finite sets $\{ 1 \} \to \{ 1, 2 \} \to \{ 1, 2, 3, ... \}$ whose union is $\mathbb{N}$. The filtered colimit needs to be compact Hausdorff, so actually it's the Stone–Čech compactification $\beta \mathbb{N}$, which is much larger than the union. In general colimits in $\text{CHaus}$ are obtained by first taking the colimit in $\text{Top}$ and then taking the Stone–Čech compactification. Limits are as in $\text{Top}$ because the forgetful functor to $\text{Top}$ has a left adjoint, namely the Stone–Čech compactification! On the other hand, limits and colimits in $\text{Top}$ have underlying sets the expected thing because the forgetful functor to $\text{Set}$ has both a left and a right adjoint, so preserves both limits and colimits. If you just want to identify sufficient conditions, the forgetful functor having a right adjoint implies preserving colimits implies preserving filtered colimits so that seems like a good thing to try. (Note that this is not usually what is meant by being able to construct a free object; this is usually a left adjoint to the forgetful functor.) On the other hand this is far from necessary; the forgetful functor $\text{Ab} \to \text{Set}$ is far from preserving colimits but it preserves filtered colimits (I'm pretty sure). - I don't really have a counterexample. All those which were given to me were brushed aside as irrelevant to the case. I will have to sit and carefully verify this idea. It seems to have potential, but I wouldn't be surprised if the construction as a whole would cancel it out. As I said, linear systems are too "thin" to be of any real concern in this case. You have given me something to do for the time being, though. – Asaf Karagila Jun 28 '12 at 21:46 @Asaf: well there are "thick" counterexamples here too. You can consider very big filtered colimits of compact Hausdorff spaces. – Qiaochu Yuan Jun 28 '12 at 22:15 Well, as I said the directed systems rise naturally from forcing, and those tend to have quirks not always found in the "mathematical savanna". I have translated your suggestion into a forcing argument and I am about to finish the argument soon, although I cannot foresee the result (whether this counterexample is good, or does it falls into the list of interested-but-irrelevant). – Asaf Karagila Jun 28 '12 at 22:19 Hmmm. This is actually much harder than I expected it to be (and I have a grave suspicion why, too). I suppose that compact Hausdorff spaces provide a sufficient counterexample (although it may be possible to remedy the situation by a minor modification of the forcing, I will have to think about it). Thanks. – Asaf Karagila Jun 28 '12 at 22:36 I'm fairly certain how to fix things with the aid of an inaccessible cardinal, which would make this a non-issue. I was hoping large cardinals would come by eventually... Regardless to the above, I believe that there might be a way to get things working without the aid of large cardinals, but I'm not sure how (yet). – Asaf Karagila Jun 28 '12 at 22:52 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": 51, "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": 0.9544515013694763, "perplexity_flag": "head"}
http://mathoverflow.net/questions/81995/any-compactness-theorem-for-manifolds-which-has-ricci-lower-bound/81999	## any compactness theorem for manifolds which has Ricci lower bound? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) If manifolds have sectional curvature lower bound, then those manifolds has subsequence convergent to Alexandrov space. Is there similar results for manifolds with Ricci lower bound? - 1 Have you tried the Wikipedia page for Ricci curvature? en.wikipedia.org/wiki/Ricci_curvature – Ryan Budney Nov 27 2011 at 9:22 ## 2 Answers This is getting too long for a comment so I'm posting this as an answer... @guoyi xu No you can not claim in any sense that $\|\nabla_y dist_x\|> 1- \delta$ for any `$y \in B_{b(x)}(x)\backslash \{x\}$` for limits of spaces with just lower Ricci bounds. That is indeed different from Alexandrov spaces where you do have this property. That makes studying limits of manifolds with lower Ricci bounds quite a bit harder than Alexandrov spaces. Let me give a rough reason why you can not possibly hope for such a result. In Alexandrov spaces the above property is a fairly easy consequence of Toponogov comparison. Toponogov comparison is false with just lower Ricci bounds. There is an $L^2$ version of Toponogov for long thin triangles due to Colding (and later Colding and Cheeger) but that is only $L^2$ which means it's just a measured statement which can not give you full control on topology (except in some near extremal cases) as there is always a small set of points where you can say nothing about the gradient of the distance function but which might happen to contain a lot of topological information. One may hope that maybe Cheeger--Colding results are not optimal and maybe the result you want could be proved using some undiscovered tools but that is not so. If you could prove such a statement for lower Ricci then you'd be able to prove the Gromov betti number estimate for lower Ricci curvature bound instead of lower sectional curvature bound. However, that is known to be false. There are many counterexamples in collapsing situation ( examples by Sha- Yang etc) and it's not even possible in the noncollapsing case. I think the first example in the noncollapsing case is by Perelman Lastly, ( I should have probably started with that) there are some other pathalogical examples known which clearly violate the statement you are after. For example, Meguy constructed examples of limit spaces where some tangent cones are not polar meaning that not every geodesic starting at the origin can be extended to a ray. - @Vitali Kapovitch, I should be more specific here. I am focusing on 3-dimensional case, by Hamilton, Yau-schoen and recent Liu Gang's result, 3-dim manifolds with nonnegative Ricci curvature has finite topology type, so I guess my statement about limit space of 3-dim noncompact manifolds with nonnegative Ricci curvature could be true although I prefer more direct argument not using classification results. What do you think? – guoyi xu Nov 28 2011 at 5:49 1 Dimension 3 is certainly very special in a number of ways but I really don't expect the statement you are after to be true even there. Liu Gang's beautiful result (and the original paper by Schoen and Yau) is much more subtle and uses very different ideas which I don't think can be easily circumvented. – Vitali Kapovitch Nov 28 2011 at 13:46 ### 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. A well-known theorem of Gromov says that if $(M_i,g_i)$ is a sequence of manifolds with $Ric\geq -kg$ and fixed dimension, then it is a converging subsequence in the pointed Gromov-Hausdorff topology. References for this fact are Gromov's "Metric Structures for Riemannian and non-Riemannian spaces" and Burago-Burago-Ivanov's "A Course in Metric Geometry". A lot of investigation has been made on the possible limiting length space, see "On the structure of spaces with Ricci curvature bounded below." by Cheeger and Colding, and subsequent papers. On the other hand, one can have smoother convergence with stronger geometric requirements. For instance, one has $C^\alpha$ compactness when the injectivity radius is bounded from below, see "$C^\alpha$-compactness for manifolds with Ricci curvature and injectivity radius bounded below", by Anderson and Cheeger. - 2 In fact, precompactness of the class of $n$-manifolds with $Ric\ge k$ curvature bounds was proved by Gromov directly (sectional curvature is just a special case). It's an easy consequence of Bishop-Gromov volume comparison which implies a universal bound in terms of $\epsilon, k,n, R$ on the number of $\epsilon$-separated points in any $R$ ball. Such a bound for a class of proper inner metric spaces implies precompactness in pointed Gromov-Hausdorff topology. – Vitali Kapovitch Nov 27 2011 at 14:35 In fact, I am interested in whether the limit space's distance function has some C^1 property or critical point theory or not like Alexandrov space has. – guoyi xu Nov 27 2011 at 19:18 6 @guoyi xu No. there is no $C^1$ structure on limit spaces with lower Ricci curvature bounds - they can be extremely singular (there isn't one even for Alexandrov spaces btw) and there is no critical point theory either. Critical point theory for Alexandrov spaces is possible basically because distance functions on Alexandrov spaces are semiconcave. Distance functions on manifolds with lower Ricci curvature bounds are only semi- super harmonic. This is a much weaker condition which doesn't allow for much control of topology except on the level of $\pi_1$ and in near extremal cases. – Vitali Kapovitch Nov 27 2011 at 19:43 @Vitali Kapovitch, so for such limit space, can we claim that there exists a positive function b(x) on this space, such that \|\nabla_y dist_x\|> 1- \delta for any y \in B_{b(x)}(x)\{x}? In some sense, could we define spire on such limit space like in Alexandrov spaces? – guoyi xu Nov 27 2011 at 23:04	{"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": 14, "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": 0.9252812266349792, "perplexity_flag": "head"}
http://mathoverflow.net/revisions/87458/list	Return to Answer 2 Added word "with" which had been omitted. Brauer's proof that the number of similarity classes of irreducible representations of $G$ over an algebraically closed field of characteristic $p$ is equal to the number of $p$-regular conjugacy classes of $G$ is ring-theoretic in flavor, and rather tricky. There is also an easy character theoretic proof based on the following ideas. First, the set IBr$(G)$ of irreducible Brauer characters is in bijective correspondence with the irreducible representations, and this set of functions lives in the space $V$ of all complex-valued class functions defined on the set of p-regular elements. Since $\dim(V)$ equals the number of $p$-regular classes, it suffices to show that IBr($G$) is a basis for $V$. The linear independence of IBr$(G)$ is a standard result. To see that IBr$(G)$ spans, use the facts that Irr$(G)$ spans the space of all class functions and that on each $p$-regular class, the value of an ordinary character is a linear combination (and in fact, a sum) of values of Brauer characters. 1 Brauer's proof that the number of similarity classes of irreducible representations of $G$ over an algebraically closed field of characteristic $p$ is equal to the number of $p$-regular conjugacy classes of $G$ is ring-theoretic in flavor, and rather tricky. There is also an easy character theoretic proof based on the following ideas. First, the set IBr$(G)$ of irreducible Brauer characters is in bijective correspondence the irreducible representations, and this set of functions lives in the space $V$ of all complex-valued class functions defined on the set of p-regular elements. Since $\dim(V)$ equals the number of $p$-regular classes, it suffices to show that IBr($G$) is a basis for $V$. The linear independence of IBr$(G)$ is a standard result. To see that IBr$(G)$ spans, use the facts that Irr$(G)$ spans the space of all class functions and that on each $p$-regular class, the value of an ordinary character is a linear combination (and in fact, a sum) of values of Brauer characters.	{"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": 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": 0.9157941341400146, "perplexity_flag": "head"}
http://mathoverflow.net/questions/99319/asymptotic-behaviour-of-a-mean	## Asymptotic behaviour of a mean ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Fix $x>0$ and $c\in\mathbb{N}$. Let $f(x):=\frac{c}{4c-2+2x^2}$ and $$m_N(x):=\frac{1}{N} \sum_{i=0}^{f(x)N} \log(\frac{c N}{2}-i(2c-1))$$ I'm pretty sure $m_N(x)\to\infty$ as $N\to\infty$. I would like to know the asymptotic behaviour of $m_N(x)$, and I expect to find something like $$m_N(x)=f(x) \log{N} + g(x) + o(1)\ \ \text{ as }N\to\infty$$ Can you confirm this result? If this is the case, can you help me to compute the constant $g(x)$? - Unless you can give some background and context to this question, you are not likely to get any useful answers, and your question is likely to be closed. – Lee Mosher Jun 11 at 23:28 I'm trying to compute a limit in a Statistical Mechanics model. After some computations, I need to compute that $g(x)$. Just knowing it is true that $m_N(x)=f(x) logN + g(x) + o(1)$ for some $g(x)$ would be a good result. – xn--qwertyuiop-86a Jun 11 at 23:39 1 Rather than do the computation, I'll just tell you how I would do it: the sum looks a lot like a Riemann sum. So write the Riemann sum and the integral to which it is an approximation side by side and carefully estimate the differences. – Anthony Quas Jun 12 at 4:26 I have to guess that $i$ is an integer so, why is the upper limit of the sum the real number $f(x)N$? Is it a mistake or is it so? – Jon Jun 12 at 7:36 Jon you're right: the sum is over integer $i$. The correct upper bound of the sum would be $$[f(x) N + k(x)]+1$$ where $k(x)=\frac{-x^2}{2c-1+x^2}$ and $[]$ denotes the integer part. But I think it's not important for the asymptotic behaviour of the mean. – xn--qwertyuiop-86a Jun 12 at 9:52 show 1 more comment ## 1 Answer Let us consider the sum $$m_N(x)=\frac{1}{N}\sum_{i=0}^{[f(x)]N}\log\left(\frac{cN}{2}-i(2c-1)\right).$$ The first step to get an asymptotic approximation is to extract the leading term in $N$ to obtain $$m_N(x)=[f(x)]\log N+\frac{1}{N}\sum_{i=0}^{[f(x)]N}\log\left(\frac{c}{2}-\frac{i}{N}(2c-1)\right).$$ When $N$ is finite, we recognize a Riemann series and apply the average theorem. So, there exists a value of argument of the logarithm such that $$m_N(x)=[f(x)]\log N+[f(x)]\log[z(x)].$$ We can take $z(x)=\frac{c}{2}-t[f(x)](2c-1)$ being $t\in (0,1)$. Indeed, we can define a partition with $x_i=x_{i-1}+\frac{1}{[f(x)]N}$ and so $$\frac{1}{N}\sum_{i=0}^{[f(x)]N}\log\left(\frac{c}{2}-\frac{i}{N}(2c-1)\right)=[f(x)]\Delta x\sum_{i=0}^{[f(x)]N}\log\left(\frac{c}{2}-i[f(x)](2c-1)\Delta x\right)$$ being $\Delta x=\frac{1}{[f(x)]N}$. But this, in the given limit, is nothing else than $$\int_{\frac{c}{2}}^{\frac{c}{2}-[f(x)](2c-1)}\log(z)dz<\infty$$ as it should. - Yes in the end it's just Riemann sums – xn--qwertyuiop-86a Jun 14 at 15:20	{"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": 8, "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": 0.9370567202568054, "perplexity_flag": "head"}
http://chemed.chem.wisc.edu/chempaths/GenChem-Textbook/Atomic-Weights/chemprime/CoreChem3ATesting_the_Atomic_Theory-690.html	# Testing the Atomic Theory Two criteria are usually applied to any theory. First, does it agree with facts which are already known? Second, does it predict new relationships and stimulate additional observation and experimentation? Dalton’s atomic theory was able to do both of these things.It was especially useful in dealing with data regarding the masses of different elements which were involved in chemical compounds or chemical reactions. To test a theory, we first use it to make a prediction about the macroscopic world. If the prediction agrees with existing data, the theory passes the test. If it does not, the theory must be discarded or modified. If data are not available, then more research must be done. Eventually the results of new experiments can be compared with the predictions of the theory. Several examples of this process of testing a theory against the facts are afforded by Dalton’s work. For example, postulate 3 in Dalton’s Atomic Theory states that atoms are not created, destroyed, or changed in a chemical reaction. Postulate 2 says that atoms of a given element have a characteristic mass: By logical deduction, then, equal numbers of each type of atom must appear on left and right sides of chemical equations such as Hg(l) + Br2(l) → HgBr2(s) (1) and the total mass of reactants must equal the total mass of products. Dalton’s atomic theory predicts Lavoisier’s experimental law of conservation of mass. A second prediction of the atomic theory is a bit more complex. A compound a definite number of two or more types of atom. No matter how, when, or where a compound is made, it must always have the same ratios of different atoms. Thus mercuric bromide molecules always have the formula HgBr2 no matter how much we have or where the compound came from, there will always be twice as many bromine atoms as mercury atoms. Since each type of atom has a characteristic mass, the mass of one element which combines with a fixed mass of the other should always be the same. In mercuric bromide, for example, if each mercury atom is 2.510 times as heavy as a bromine atom, the ratio of masses would be No matter how many mercury (II) bromide molecules we have, each has the same proportion of mercury, and so any sample of mercury (II) bromide must have that same proportion of mercury. We have just derived the law of constant composition, sometimes called the law of definite proportions. When elements combine to form a compound, they always do so in exactly the same ratio of masses. This law had been postulated in 1799 by the French chemist Proust (1754 to 1826) 4 years before Dalton proposed the atomic theory, and its logical derivation from the theory contributed to the latter’s acceptance. The law of constant composition makes the important point that the composition and other properties of a pure compound are independent of who prepared it or where it came from. The carbon dioxide found on Mars, for instance, can be expected to have the same composition as that on Earth, while the natural vitamin C extracted and purified from rose hips has exactly the same composition as the synthetic vitamin C prepared by a drug company. Absolute purity is, however, an ideal limit which we can only approach, and the properties of many substances may be affected by the presence of very small quantities of impurities. The arrangement of atoms and molecules in a crystal of mercury (I) bromide, Hg2Br2. A third law of chemical composition may be deduced from the atomic theory. It involves the situation where two elements can combine in more than one way, forming more than one compound. For example, if mercury (II) bromide is ground and thoroughly mixed with liquid mercury, a new compound, mercury (I) bromide (mercurous bromide), is formed. Mercury (I) bromide is a white solid which is distinguishable from mercury (II) bromide because of its insolubility in hot or cold water. Mercury (I) bromide also changes directly from a solid to a gas at 345°C. From the microscopic view of solid mercury (I) bromide in figure, you can readily determine that its chemical formula is Hg2Br2. (Since there are two atoms of each kind in the molecule, it would be incorrect to write the formula as HgBr.) The chemical equation for synthesis of mercurous bromide is Hg(l) + HgBr2(s) → Hg2Br2(s) (2) From the formulas HgBr2 and Hg2Br2 we can see that mercury (II) bromide has only 1 mercury atom for every 2 bromines, while mercury (I) bromide has 2 mercury atoms for every 2 bromines. Thus, for a given number of bromine atoms, mercury (I) bromide will always have twice as many mercury atoms as mercury (II) bromide. Again using postulate 2 from Dalton’s Atomic Theory, the atoms have characteristic masses, and so a given number of bromine atoms corresponds to a fixed mass of bromine. Twice as many mercury atoms correspond to twice the mass of mercury. Therefore we can say that for a given mass of bromine, mercury (I) bromide will contain twice the mass of mercury that mercury (II) bromide will. [The doubled mass of mercury was provided by adding liquid mercury to mercuric bromide in Eq. (2).] EXAMPLE Given that the mass of a mercury atom is 2.510 times the mass of a bromine atom, calculate the mass ratio of mercury to bromine in mercury (I) bromide. Solution The formula Hg2Br2 tells us that there are 2 mercury atoms and 2 bromine atoms in each molecule. Thus the mass ratio is $\frac{\text{Mass of 2 mercury atoms}}{\text{Mass of 2 bromide atoms}} = \frac{\text{2}\times(\text{2.510}\times\text{Mass of 1 bromide atom})}{\text{2}\times\text{Mass of 1 bromide atom}}=\text{2.510}$ Note that the mass of mercury per unit mass of bromine is double that calculated earlier for mercury (II) bromide. The reasoning and calculations above illustrate the law of multiple proportions. When two elements form several compounds, the mass ratio in one compound will be a small whole-number multiple of the mass ratio in another. In the case of mercury (II) bromide and mercury (I) bromide the mass ratios of mercury to bromine are 1.255 and 2.510, respectively. The second value is a small whole-number multiple of (2 times) the first. Until the atomic theory was proposed, no one had expected any relationship to exist between mass ratios in two or more compounds containing the same elements. Because the theory predicted such relationships, Dalton and other chemists began to look for them. Before long, a great deal of experimental evidence was accumulated to show that the law of multiple proportions was valid. Thus the atomic theory was able to account for previously known facts and laws, and it also predicted a new law. In the process of verifying that prediction, Dalton and his contemporaries did many additional quantitative experiments. These led onward to more facts, more laws, and, eventually, new or modified theories. This characteristic of stimulating more research and thought put Dalton’s postulates in the distinguished company of other good scientific theories.	{"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": 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": 0.9382385015487671, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/101107/can-there-be-a-cubical-bubble?answertab=active	# Can there be a cubical bubble? Although not physically perfect, a reasonable mathematical model for a bubble's shape is that it minimizes surface area subject to fixed volume. A single floating bubble is usually a sphere, but bubbles only need to find local minima, not global minima. This makes more complicated bubble shapes possible. In a YouTube video, a performer discusses making a cubical compartment inside a complicated bubble. Is this possible in a bubble with finitely-many compartments and no wires or other framework to provide additional boundary conditions? - 7 +1 because you reminded me of a very old adventure of Donald Duck in which he was saved from captivity (after having been captured by an exotic tribe venerating cubes!) by his nephews' ability to make cubical bubbles with chewing-gum. – Georges Elencwajg Jan 21 '12 at 20:55 Once the bubble complex falls off his wire hoop, it looks like it remains (diffeomorphically) the same shape. Is that not an answer to your question? – Rahul Narain Jan 21 '12 at 21:51 @Rahul I don't understand, sorry. How does this address whether or not there can be a cubical bubble? – Mark Eichenlaub Jan 21 '12 at 21:54 Sorry, maybe I don't understand the question. I thought you were asking whether there can be a cubical compartment inside a free-floating "bubble complex" with no external boundary conditions. What am I misunderstanding? – Rahul Narain Jan 21 '12 at 22:01 @Rahul Yup, that's what I wanted to know. I just don't really understand how your comment addressed the question. I think there is something about it I'm missing. (I'm not a mathematician.) Will Jagy's answer was pretty much what I wanted. – Mark Eichenlaub Jan 21 '12 at 22:25 show 2 more comments ## 2 Answers Such a bubble is not actually cubical. See my answer on Quora. - Good point about the angles along the edges and at the vertices. – Will Jagy Nov 9 '12 at 21:34 EDIT, November 2012: another answer points out, quite correctly, that any three bubble compartments in a multiple bubble meet at $120^\circ$ around an edge, and four compartments meet around a vertex in such a way that the angles between pairs of edges are about $109^\circ,$ to be exact $\arccos \frac{-1}{3}.$ I actually did my dissertation in minimal surfaces and I should have thought of this, but all that was a long time ago. ORIGINAL: Yes, such a thing is possible. The real skill of the performer is in making the thing so quickly, real soap bubbles pop so quickly. Anyway, the rules for multiple bubbles are that any smooth part of bubble surface, either between one bubble and the outside or between two bubbles, be a surface of constant mean curvature. A flat square qualifies, it has mean curvature $0.$ A picture that is roughly what the performer creates: BRAKKE	{"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": 0.9492413401603699, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/188818/line-integral-of-sin-x/188826	# Line integral of $\sin x$. How to evaluate: $$\int \sqrt{1+\cos^2x} dx$$ Is a simple antiderivative known to exist? - Non-elementary, but traditional, in the family of elliptic integrals – André Nicolas Aug 30 '12 at 14:31 ## 2 Answers A simple antiderivative is given by $$\tag{1} \int_0^x\sqrt{1+\cos^2t}\,dt.$$ This is an example of what is called an elliptic integral and it cannot be expressed in terms of other well-known elementary functions (which is what you probably mean by "simple", and doesn't agree with what I would call "simple"). But let me stress my point: the expression in (1) can be used to approximate values of the function, and in that sense it is not necessarily worse than, say $\log x$. - This question is much like the other one. Using $\cos^2(x) = 1-\sin^2(x)$ $$\int \sqrt{1+\cos^2(x)} \mathrm{d} x = \int \sqrt{2-\sin^2(x)} \mathrm{d} x = \sqrt{2} \int \sqrt{1-\frac{1}{2} \sin^2(x)} \mathrm{d} x = \sqrt{2} \operatorname{E}\left(x, \frac{1}{2}\right) + C$$ where $\mathrm{E}(x, m)$ denotes the incomplete elliptic integral of the second kind: $$\operatorname{E}\left(\phi, m\right) = \int_0^\phi \sqrt{1-m \sin^2(\varphi)} \mathrm{d} \varphi$$ In particular, no elementary andi-derivative exists. -	{"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": 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": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9312829375267029, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/35107/why-doesnt-cantors-diagonal-argument-also-apply-to-natural-numbers	# Why Doesn't Cantor's Diagonal Argument Also Apply to Natural Numbers? In my understanding of Cantor's diagonal argument, we start by representing each of a set of real numbers as an infinite bit string. My question is, why can't we begin by representing each natural number as an infinite bit string? So that 0 = 00000000000..., 9 = 1001000000..., 255 = 111111110000000...., and so on. If we could, then the diagonal argument would imply that there is a natural number not in the natural numbers, which seems to be a bit of the contradiction.... Thanks! - If you try the diagonal argument on any ordering of the natural numbers, after every step of the process, your diagonal number (that's supposed to be not a natural number) is in fact a natural number. Also, the binary representation of the natural numbers terminates, whereas binary representations of real numbers do no. That's the basics for why the proof doesn't work. – Michael Chen Apr 26 '11 at 0:36 1 I don't think these arguments are sufficient though. For a) your diagonal number is a natural number, but is not in your set of rationals. For b), binary reps of the natural numbers do not terminate leftward, and diagonalization arguments work for real numbers between zero and one, which do terminate to the left. – bo1024 Apr 26 '11 at 1:11 ## 3 Answers If you represent a natural number as an infinite string, the string will become identically $0$ after a certain point. If you think it through, the "diagonal argument" in this case doesn't produce a natural number; it will produce a string with infinitely many $1$s. On the other hand, you can consider possibly infinite binary strings --- i.e. strings in which there can be infinitely many $1$; this is one way to think of the set of $2$-adic numbers, which is indeed an uncountable extension of the set of natural numbers (as one sees using the precise diagonal argument that you suggest). - 1 I was thinking about your first paragraph... what if you reordered the natural numbers such that the diagonal wasn't straight zeroes? – Michael Chen Apr 26 '11 at 0:45 1 You'd actually need the diagonal part to be identically 1 for digits large enough. If you have infinitely many zeros on the diagonal, then you still don't get a natural number. You'd need all but finitely many of the diagonals to be 1. But you can see that isn't possible, because if the $n$th binary digit of a number is 1, then that number is at least $2^n$. – Thomas Andrews Apr 26 '11 at 0:57 1 @Michael: I don't mean that all entries will necessarily be $1$, but there will necessarily be infinitely many $1$s. – Matt E Apr 26 '11 at 0:58 1 @b01024: but then since you have an enumerated list, you have to state where the numbers in between fall. The crux is that you can't have 1's everywhere on the diagonal, for then you have no place to put 1, 3, 5, 6, 7, etc. – Michael Chen Apr 26 '11 at 1:23 2 – joriki Apr 26 '11 at 4:49 show 4 more comments The reason is simply that natural numbers have a finite representation. (In set theory, Each one is a finite number of successions from 1, where the successor of n is n+1.) Your representation scheme essentially respects this, since for any natural number (which will be on the list), after some finite number of digits it becomes all zeros. Your diagonal element will either be one of these, and so on the list, or a sequence of ones and zeros which never 'zeros out'. This string is NOT a natural number; it corresponds to nothing in your representation scheme. The reason the diagonal argument works for real numbers is that they do not have a finite representation. In set theory they are represented as the limit of an infinite series. - Why can an infinite list of 0's and 1's not present a natural number? Because it does not suite us? Suppose we code a natural number n and from the kth position, only 0's will show up. Then n<2^k, so n is limited. But if you think that you can have an infinite set of natural number, than you allow code sequences with infinite length which do not end in only 0's (otherwise they were finite). - 1 Huh?${}{}{}{}{}$ – Asaf Karagila Apr 26 at 19:17	{"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": 0.9353590607643127, "perplexity_flag": "head"}
http://mathoverflow.net/questions/9396?sort=oldest	## The proper name for a kind of ordered space ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I'm trying to find the correct term for a specific kind of totally ordered space: Let $S$ be a totally ordered space with asymmetric relation <. Property: For any two `$s_{1}$` and `$s_{2}$` in $S$ where `$s_1 < s_2$`, there must exist some `$s_{3}$` such that `$s_{1} < s_{3}$` and `$s_{3} < s_{2}$`. What is the name of this property? Thank you! - ## 1 Answer Dense order is one name that concept goes by. - Thank you so much, Mariano! – chipuni Dec 20 2009 at 1:37	{"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": 0.3, "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": 0.8751001954078674, "perplexity_flag": "middle"}
http://mathhelpforum.com/advanced-statistics/168598-combinations-set.html	# Thread: 1. ## Combinations of a Set Hello Everyone! I just want to check something here... Say we have a set $E = {a_1, a_2, \cdots a_n}.$ And we want to get all the subsets of size $k$ and of size $p$ call this set of subsets $S$ So is $\|S\| = C_n ^k + C_n ^p$??	{"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": 0.8884273767471313, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/21177?sort=oldest	## classification of smooth involutions of torus ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Let $\mathbb{Z}_2={1,g},T^2={(e^{i\theta_1},e^{i\theta_2})}$ and place $T^2$ in $\mathbb{R}^3$ as the locus of the rotation of $2\pi$ rads of the circle${(y,z)\|(y-2)^2+z^2=1}$ around $z$ axis. It is known that there are 5 nonequivalent smooth involutions on torus,and they are: 1.$g(e^{i\theta_1},e^{i\theta_2})=(e^{i(\theta_1+\pi)},e^{i\theta_2})$ (rotation$\pi$ rads around $z$ axis) with null fixed point set and orbit space $T^2$ 2.$g(e^{i\theta_1},e^{i\theta_2})=(e^{-i\theta_1},e^{i\theta_2})$(reflection along $x=0$) with fixed point set $S^1\times S^0$ and orbit space an annulus 3.$g(e^{i\theta_1},e^{i\theta_2})=(e^{i\theta_2},e^{i\theta_1})$(switch the two coordinates) with fixed point set the diagonal circle and orbit space Mobius band 4.$g(e^{i\theta_1},e^{i\theta_2})=(e^{i(\theta_1 +\pi)},e^{-i\theta_2})$(restriction of the involution $(x,y,z,\mapsto (-x,-y,-z)$ of $\mathbb{R}^3$ to torus)with null fixed point set and orbit space klein bottle 5.$g(e^{i\theta_1},e^{i\theta_2})=(e^{-i\theta_1},e^{-i\theta_2})$(reflection along $x=0$ plus reflection along $z=0$) with fixed point set 4 points and orbit space $S^2$ i want to know how to derive the result above.for the free case it seems easy.since the action is free,the orbit space must be a manifold also,and has euler char 0,hence must be torus or klein bottle. for the nonfree case,the orbit is not manifold,but "orbifold". and we have Riemann-Hurwitz Formula: $\chi(O)=\chi(X_O)-\sum_{i=1}^n (1-\frac{1}{q_i})-\frac{1}{2}\sum_{j=1}^m (1-\frac{1}{r_j})$ here$\chi(O)$ is the orbifold euler char and $\chi(X_o)$ is the euler char of the underlying space associated to the orbifold $O$,and $q_i$and $r_j$ denote the angles for sigular points(cone points and reflector corners can we determine the remaining 3 involutions by using this formula?Thank you! - What equivalence are you talking about? Conjugacy under a diffeo? – Mariano Suárez-Alvarez Apr 13 2010 at 4:19 1 You're getting pretty close to duplicating this thread: mathoverflow.net/questions/7746/… – Ryan Budney Apr 13 2010 at 5:40 to Mariano Suárez-Alvarez:yes,two involutions are defined to be equivalent if they are conjugate in the group $Diff(T^2)$ – student Apr 13 2010 at 6:24 1 Do you really care how the torus is embedded in $R^3$? If not, perhaps edit the question to simplify the statement. – Sam Nead Apr 13 2010 at 14:54 ## 1 Answer Here is a sketch -- some of the details are a bit hazy: Suppose that $\iota$ is a smooth involution of $T^2$. Show that the fixed point set of $\iota$ is a submanifold. Show that the orbit space of $\iota$ is an orbifold with orbifold Euler characteristic zero. Using the orbifold Euler characteristic you can enumerate all 17 compact, connected, 2-dimensional orbifolds of orbifold Euler characteristic zero. Now rule out 12 of these for topological reasons. The second to last step is a nice exercise that everybody should do once, after learning about the orbifold Euler characteristic. The non-trivial part in the last step is eliminating $D(2,2;)$ and $P(2,2)$. Getting rid of the others is easy. -	{"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": 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": 0.8695812225341797, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/220996/can-any-nfa-be-converted-to-a-dfa	# Can _any_ NFA be converted to a DFA? I was wondering if for every NFA there exists an equivalent DFA? I think the answer is yes. How would one prove it? Since I'm just starting up learning about Automata I'm not confused about this and especially the proof of such a statement. - – Patrick87 Oct 25 '12 at 18:25 ## 2 Answers Indeed, every NFA can be converted to an equivalent DFA. In fact, DFAs, NFAs, and regular expressions are all equivalent. One approach would be to observe the NFA and if it simple enough, determine the regular expression that it recognizes, then convert the regular expression to a DFA. Yet, there are algorithms out there that can take a NFA and produce its equivalent DFA. For example, the powerset construction check out this link and google: http://en.wikipedia.org/wiki/Powerset_construction Furthermore, every DFA has a unique minimum state DFA that recognizes a regular expression using a minimal number of states. This DFA state minimization also has an algorithm. Good luck! - It can be done in two steps: 1) Use subset construction to construct DFA from NFA. 2) Then show for any w $\in$$\sum$$^*$, $\hat\delta$$_D$({q},a) = $\hat\delta$$_N$(q,a). That is for any string and for any set of states they both take you to same set of states. -	{"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": 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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9057732820510864, "perplexity_flag": "middle"}
http://mathhelpforum.com/geometry/106368-how-much-area-preserved-when-you-inscribe-circles-circles.html	# Thread: 1. ## How much of the area is preserved when you inscribe circles in circles What is the ratio of the area of the shaded portion in the figure on the right to the area of the shaded portion of the figure to the left? The circles are all tangent to each other even though the picture might not be exact. The inscribed circles in the left figure are congruent. The second figure is a result of inscribing 4 congruent circles in each of the 4 circles on the left. I have no idea how to approach this. Thanks for taking the time to read my question! 2. First, lets just look at the figure on the left with a single circle and four circles inscribed in it. Let $A$ be the area of the larger circle (with radius $R$) and $a$ be the area of each of the smaller circles (with radius $r$). You are looking for the following ratio: $\frac{A}{4a}$ which is $\frac{\pi R^2}{4(\pi r^2)}=\frac{R^2}{4r^2}$. So you essentially need to find the ratio of the radii. Put a coordinate system on your circle picture, with the center of the large, original circle at the origin. Look at the smaller circle in quadrant I of the coordinate system. It touches both the x-axis and the y-axis. Therefore, its center is at (r,r). Using this, you can calculate the distance from the origin to its center as $\sqrt{2}r$ and if you go from the center to the point of tangency (adding an r from this point) you will have gone R. This means $\sqrt{2}r+r=R$. Solving for r, you can see $r=\frac{R}{\sqrt{2}+1}$. Plug this in the original question and you will find $\frac{R}{4r}=\frac{R}{4\frac{R}{\sqrt{2}+1}}=\frac {\sqrt{2}+1}{4}$ Can you use this information to figure out the answer to your question? 3. Yes, thank you! That was very helpful.	{"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": 10, "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": 0.9514629244804382, "perplexity_flag": "head"}
http://en.wikipedia.org/wiki/Axiom_of_infinity	# Axiom of infinity In axiomatic set theory and the branches of logic, mathematics, philosophy, and computer science that use it, the axiom of infinity is one of the axioms of Zermelo–Fraenkel set theory. It guarantees the existence of at least one infinite set, namely a set containing the natural numbers. ## Formal statement In the formal language of the Zermelo–Fraenkel axioms, the axiom reads: $\exist \mathbf{I} \, ( \empty \in \mathbf{I} \, \and \, \forall x \in \mathbf{I} \, ( \, ( x \cup \{x\} ) \in \mathbf{I} ) ) .$ In words, there is a set I (the set which is postulated to be infinite), such that the empty set is in I and such that whenever any x is a member of I, the set formed by taking the union of x with its singleton {x} is also a member of I. Such a set is sometimes called an inductive set. ## Interpretation and consequences This axiom is closely related to the standard construction of the naturals in set theory, in which the successor of x is defined as x ∪ {x}. If x is a set, then it follows from the other axioms of set theory that this successor is also a uniquely defined set. Successors are used to define the usual set-theoretic encoding of the natural numbers. In this encoding, zero is the empty set: 0 = {}. The number 1 is the successor of 0: 1 = 0 ∪ {0} = {} ∪ {0} = {0}. Likewise, 2 is the successor of 1: 2 = 1 ∪ {1} = {0} ∪ {1} = {0,1}, and so on. A consequence of this definition is that every natural number is equal to the set of all preceding natural numbers. This construction forms the natural numbers. However, the other axioms are insufficient to prove the existence of the set of all natural numbers. Therefore its existence is taken as an axiom—the axiom of infinity. This axiom asserts that there is a set I that contains 0 and is closed under the operation of taking the successor; that is, for each element of I, the successor of that element is also in I. Thus the essence of the axiom is: There is a set, I, that includes all the natural numbers. The axiom of infinity is also one of the von Neumann–Bernays–Gödel axioms. ## Extracting the natural numbers from the infinite set The infinite set I is a superset of the natural numbers. To show that the natural numbers themselves constitute a set, the axiom schema of specification can be applied to remove unwanted elements, leaving the set N of all natural numbers. This set is unique by the axiom of extensionality. To extract the natural numbers, we need a definition of which sets are natural numbers. The natural numbers can be defined in a way which does not assume any axioms except the axiom of extensionality and the axiom of induction—a natural number is either zero or a successor and each of its elements is either zero or a successor of another of its elements. In formal language, the definition says: $\forall n (n \in \mathbf{N} \iff ([n = \empty \,\,\or\,\, \exist k ( n = k \cup \{k\} )] \,\,\and\,\, \forall m \in n[m = \empty \,\,\or\,\, \exist k \in n ( m = k \cup \{k\} )])).$ Or, even more formally: $\forall n (n \in \mathbf{N} \iff ([\forall k \in n(\bot) \or \exist k \in n( \forall j \in k(j \in n) \and \forall j \in n(j=k \lor j \in k))] \and$ $\forall m \in n[\forall k \in m(\bot) \or \exist k \in n(k \in m \and \forall j \in k(j \in m) \and \forall j \in m(j=k \lor j \in k))])).$ Here, $\bot$ denotes the logical constant "false", so $\forall k \in n(\bot)$ is a formula that holds only if n is the empty set. ### Alternative method An alternative method is the following. Let $\Phi(x)$ be the formula that says `x is inductive'; i.e. $\Phi(x) = (\varnothing \in x \wedge \forall y(y \in x \to (y \cup \{y\} \in x)))$. Informally, what we will do is take the intersection of all inductive sets. More formally, we wish to prove the existence of a unique set $W$ such that $\forall x(x \in W \leftrightarrow \forall I(\Phi(I) \to x \in I)).$ () For existence, we will use the Axiom of Infinity combined with the Axiom schema of specification. Let $I$ be an inductive set guaranteed by the Axiom of Infinity. Then we use the Axiom Schema of Specification to define our set $W = \{x \in I:\forall J(\Phi(J) \to x \in J)\}$ - i.e. $W$ is the set of all elements of $I$ which happen also to be elements of every other inductive set. This clearly satisfies the hypothesis of (), since if $x \in W$, then $x$ is in every inductive set, and if $x$ is in every inductive set, it is in particular in $I$, so it must also be in $W$. For uniqueness, first note that any set which satisfies () is itself inductive, since 0 is in all inductive sets, and if an element $x$ is in all inductive sets, then by the inductive property so is its successor. Thus if there were another set $W'$ which satisfied () we would have that $W' \subseteq W$ since $W$ is inductive, and $W \subseteq W'$ since $W'$ is inductive. Thus $W = W'$. Let $\omega$ denote this unique element. This definition is convenient because the principle of induction immediately follows: If $I \subseteq \omega$ is inductive, then also $\omega \subseteq I$, so that $I = \omega$. Both these methods produce systems which satisfy the axioms of second-order arithmetic, since the axiom of power set allows us to quantify over the power set of $\omega$, as in second-order logic. Thus they both completely determine isomorphic systems, and since they are isomorphic under the identity map, they must in fact be equal. ## Independence The axiom of infinity cannot be derived from the rest of the axioms of ZFC, if these other axioms are consistent. Nor can it be refuted, if all of ZFC is consistent. Indeed, using the Von Neumann universe, we can make a model of the axioms where the axiom of infinity is replaced by its negation. It is $V_\omega \!$, the class of hereditarily finite sets, with the inherited element relation. The cardinality of the set of natural numbers, aleph null ($\aleph_0$), has many of the properties of a large cardinal. Thus the axiom of infinity is sometimes regarded as the first large cardinal axiom, and conversely large cardinal axioms are sometimes called stronger axioms of infinity. ## References • Paul Halmos (1960) Naive Set Theory. Princeton, NJ: D. Van Nostrand Company. Reprinted 1974 by Springer-Verlag. ISBN 0-387-90092-6. • Thomas Jech (2003) Set Theory: The Third Millennium Edition, Revised and Expanded. Springer-Verlag. ISBN 3-540-44085-2. • Kenneth Kunen (1980) Set Theory: An Introduction to Independence Proofs. Elsevier. ISBN 0-444-86839-9. • Hrbacek, Karel; Jech, Thomas (1999). Introduction to Set Theory (3 ed.). Marcel Dekker. ISBN 0‐8247‐7915‐0.	{"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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 33, "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": 0.9235002994537354, "perplexity_flag": "head"}
http://en.wikibooks.org/wiki/Real_analysis/Sequences	# Real Analysis/Sequences (Redirected from Real analysis/Sequences) Real Analysis Sequences ## Definition Sequences occur frequently in analysis, and they appear in many contexts. While we are all familiar with sequences, it is useful to have a formal definition. Definition A sequence of real numbers is any function a : N→R. However, we usually write an for the image of n under a, rather than a(n). The values an are often called the elements of the sequence. To make a distinction between a sequence and one of its values it is often useful to denote the entire sequence by $(a_n)_{n=1}^\infty$, or just (an). When specifying a particular sequence, it may be written in the form (a1, a2, a3, …), where sufficiently many elements of the sequence are given so that the pattern is clear. For example, (1, 2, 3, 4, …), (1, -2, 3, -4, …), and (1, π, π2, π3, π4, …) are all sequences. Note, however, that there need not be any particular pattern to the elements of the sequence. For example, we may specify an to be the n-th digit of π. Often it is useful to specify a sequence recursively. That is, to specify some initial values of the sequence, and then to specify how to get the next element of the sequence from the previous elements. For example, consider the sequence x1=1, x2=1, and xn = xn−1 + xn−2 for n ≥ 3. This sequences is known as the Fibonacci sequence, and its first few terms are given by (1, 1, 2, 3, 5, 8, 13, …). Another familiar example of a recursive sequence is Newton's method. With an initial guess x0 for the zero of a function, Newton's method tells you how to construct the next guess. In this way you generate a sequence which (hopefully) converges to the zero of the function. Often sequences such as these are called real sequences, sequences of real numbers or sequences in R to make it clear that the elements of the sequence are real numbers. Analogous definitions can be given for sequences of natural numbers, integers, etc. Given a sequence (xn), a subsequence is a sequence $(x_{n_j})_{j=1}^\infty$, where (nj) is strictly increasing sequence of natural numbers. For example, taking nj=2j would the subsequence consisting of every other element of the original sequence, that is (x2, x4, x6, …). We can also perform algebraic operations on sequences. In other words, we can add, subtract, multiply, divide sequences. These operations are simply performed element by element, for completeness we give the definitions. Definition Given two sequences (xn) and (yn) and a real number c, we define the following operations: • (xn)+(yn)=(xn+yn); • (xn)−(yn)=(xn−yn); • (xn)·(yn)=(xn·yn); • if yn ≠ 0 for all n in N, (xn)/(yn)=(xn/yn); • c·(xn)=(c·xn) ## Classification of Sequences Some properties of sequence are so important that they are given special names. A sequence (an) is called: • strictly increasing if an < an+1 for all n in N; • non-decreasing if an ≤ an+1 for all n in N; • strictly decreasing if an > an+1 for all n in N; • non-increasing if an ≥ an+1 for all n in N; • monotone if it satisfies any of the above properties, that is, if it is either non-decreasing or non-increasing; • strictly monotone if it is either strictly increasing or strictly decreasing; • bounded above if there exists M in R such that an<M for all n in N. • bounded below if there exists M in R such that an>M for all n in N. • bounded if it is both bounded above and bounded below. • Cauchy if for all ε>0 there exists a natural number N so that, for all n, m > N, \|am-an\| < ε. The term increasing is used in some contexts with meaning either that of strictly increasing or of non-decreasing, and similarly decreasing can mean the same as either strictly decreasing, or non-increasing. As a result, these terms are ambiguous and we try to avoid their use here. ## Convergence A further important property of sequences (arguably the most important property from the perspective of analysis) is the property of convergence. Definition Let (xn) be a sequence of real numbers. The sequence (xn) is said to converge to a real number a. if for all ε>0, there exists N in N such that \|xn-a\|<ε for all n≥N. If (xn) converges to a then we say a is the limit of (xn) and write $\lim_{n\to\infty}x_n=a$ or $x_n\to a$ as $n\to\infty$. This is read xn approaches a as n approaches ∞. If it is clear which variable is playing the role of n then this may be abbreviated to simply xn→a or lim xn=a. If a sequence converges, then it is called convergent. It is also useful to extend this concept and allow sequences whose limits are either ∞ or −∞ Definition We say xn→∞ as n→∞ if for every M in R there is a natural number N so that xn≥ M for all n≥N. We say xn→−∞ as n→∞ if for every M in R there is a natural number N so that xn≤ M for all n≥N. Despite this, we do not refer to sequences such as these as convergent. The following theorems tells us that a convergent sequence converges to exactly one number. This may seem intuitively clear, mathematically we have to prove that limits behave the way we expect. After all, we gave the definition and nothing tells us ahead of time that it is correct. ### Theorem (Uniqueness of limits) A sequence can have at most one limit. In other words: if xn → a and xn → b then a = b. ##### Proof Suppose the sequence has two distinct limits, so a≠b. Let ε=\|a−b\|/3. Certainly ε>0, using the definition of convergence twice we can find natural numbers Na and Nb so that $\|x_n-a\|\leq\epsilon$ for all n > Na. and $\|x_n-b\|\leq\epsilon$ for all n > Nb. Taking k=max(Na,Nb) then both of these conditions hold for xk. Hence we deduce that \|xk−a\|≤ε and \|xk−b\|≤ε. Applying the triangle inequality, we see $\|a-b\|=\|(x_k-b)-(x_k-a)\|\leq 2\epsilon=(2/3)\|a-b\|,$ which is a contradiction. Thus, any sequence has at most one limit. $\Box$ ### Theorem (Convergent Sequences Bounded) If $(x_n)_{n=1}^{\infty}$ is a convergent sequence, then it is bounded. ##### Proof Let $a=\lim_{n\to\infty}x_n$, and let ε = 1. From the definition of convergence there exists a natural number N such that $\|x_n-a\|\leq 1$ for all n ≥ N. The sequence $(x_n)_{n=N+1}^{\infty}$ is bounded above by a+1 and below by a−1. Let M = max(\|x1\|,\|x2\|,\|x3\|,…,\|xN\|, \|a\|+1). It follows that −M ≤ xn ≤ M for all n in N. Hence the sequence is bounded. $\Box$ ### Theorem (Boundedness of Cauchy Sequences) If $(x_n)_{n=1}^{\infty}$ is a Cauchy sequence, then it is bounded. #### Proof Let (xn) be a Cauchy sequence. By the definition of a Cauchy sequence, there is a natural number N such that \|xn−xm\|<1 for all n,m > N. In particular, \|xN+1−xm\|<1 for all m > N. It follows by the reverse triangle inequality that \|xm\| < \|xN+1\| + 1. If we take M=max(\|x1\|, \|x2\|, …, \|xN\|, \|xN+1\| + 1), then \|xn\| ≤ M for all n in N. $\Box$ The following theorem tells us that algebraic operations on sequences commute with the taking limits. This simple theorem is a useful tool in computing limits. ### Theorem (Algebraic Operations) If (xn) and (yn) are convergent sequences and c ∈ R, the following properties hold: 1. $\lim_{n\to\infty}(x_n+y_n)=\lim_{n\to\infty}(x_n)+\lim_{n\to\infty}(y_n)$. 2. $\lim_{n\to\infty}(x_ny_n)=\left(\lim_{n\to\infty}x_n\right)\left(\lim_{n\to\infty}y_n\right)$. 3. $\lim_{n\to\infty}(ax_n)=a\lim_{n\to\infty}(x_n)$. 4. $\lim_{n\to\infty}\left(\frac{x_n}{y_n}\right)=\frac{\displaystyle \lim_{n\to\infty}x_n}{\displaystyle \lim_{n\to\infty}y_n}$ (assuming yn ≠ 0 for all n in N and lim y_n ≠ 0). 5. If xn ≤ yn for every n in N, then $\lim_{n\to\infty}x_n \leq \lim_{n\to\infty}y_n$. #### Proof 1. Let x=lim xn and y=lim yn. We need to show that for any ε>0 there is natural number N so that if n≥ N, then \|(xn + yn) − (x + y)\|≤ε. Given any ε>0 we have ε/3>0 so from the definition of convergence there is a natural number Nx so that \|xn−x\|≤ε/3 for all n>Nx, similarly we can choose Ny \|yn−y\|≤ε/3 for all n>Ny. Let N=max(Nx ,Ny). If n>N, then by the triangle inequality we have $\|(x_n+y_n)-(x+y)\|=\|(x_n-x)+(y_n-y)\| <= \epsilon /3 + \epsilon /3<\epsilon,$ which is what we needed to show. 2. Let x=lim xn and y=lim yn. Since these sequences are convergent they are bounded. Let Mx be a bound for (xn) and let My be a bound for (yn). By increasing these quantities of necessary we may also assume Mx > x and My > y. Given ε>0, there exists some Nx and Ny such that $\|x_n - x\| < {\epsilon \over 2M_y}$ for n > Nx and $\|y_n - y\| < {\epsilon \over 2M_x}$ for n > Ny. Then for every n > max(Nx, Ny), $\|x_n y_n - xy\|\,$ $= \|(x_n - x)y_n + x(y_n - y)\|\,$ $\leq \|x_n - x\|M_y + M_x \|y_n - y\|$ $\leq \frac{\epsilon}{2} + \frac{\epsilon}{2}\leq \epsilon.$ 3. Let yn = a for all n in N. The statement now follows from 2. 4. We can reduce this to showing that lim (1/yn) exists and equals 1/(lim yn). Then it follows by 2 that we have: $\lim_{n\to\infty}\left(\frac{x_n}{y_n}\right)= \left( \lim_{n \to \infty} {1 \over y_n} \right) \left( \lim_{n \to \infty} {x_n} \right)=\frac{\displaystyle \lim_{n\to\infty} x_n}{\displaystyle \lim_{n\to\infty} y_n}.$ Let y=lim yn. By the exercises, since y and yn are not 0, we can find δ > 0 so that \|y_n\| > δ and \|y\| > δ. It follows that 1/\|yny\|<1/δ2. Given ε > 0 choose n in N so that \|yn − y\| < δ2ε. We have $\left\| {1 \over y_n} - {1 \over y} \right\| = \|y - y_n\|{1 \over \|y_n y\|} \le \frac{\|y - y_n\|}{\delta^2} < \epsilon$. Hence, $\lim_{n \to \infty} {1 \over y_n} = {1 \over y}.$ 5. We first can reduce to the case when one sequence is identically 0. To see this let zn = xn − yn. Then zn < 0 for all n in N. Let z = lim zn. Suppose that z > 0 then we can then find an natural number N so that $\|z - z_N\| < z\,$. Since zN ≤ 0 < z, the absolute value equals z − zN. Subtracting z we find that −zN < 0. Hence zN is positive. Contradiction. Therefore we must have that z ≤ 0. Which means that by 1 we get: $\lim_{n \to \infty} x_n - \lim_{n \to \infty} y_n = \lim_{n \to \infty} z_n = z \le 0.$ Therefore lim xn ≤ lim yn $\square$ ### Theorem (Squeeze/Sandwich Limit Theorem) Given sequences (xn), (yn), and (wn), if (xn) and (yn) converge to a and xn ≤ wn ≤ yn, then wn converges to a. #### Proof Fix ε > 0. We need to find an N such that \|wn − a\| < ε if n > N. Since (xn) → a and (yn) → a the definition of convergence ensures that there exists integers Nx and Ny so that \|xn − a\| < ε for n > Nx and \|yn − a\| < ε for n > Ny. Let N=max(Nx, Ny). Then, for all n > N we have −ε < xn − a and yn − a < ε. Since xn < wn < yn, it follows that xn − a < wn − a < yn − a. Thus if n ≥ N, then −ε < xn − a < wn − a < yn − a < ε. In other words, \|wn − a\| < ε.$\Box$ ## Completeness The following results are closely related to the completeness of the real numbers. ### Theorem (Convergence of Monotone sequences) Any monotone, bounded sequence converges. If the sequence is non-decreasing, then the sequence converges to the least upper bound of the elements of the sequence. If the sequence is non-increasing, then the sequence converges to the greatest lower bound of the elements of the sequence #### Proof Let (xn) be any monotone sequence that is bounded by a real number M. Without loss of generality, assume (xn) is non-decreasing. Since (xn) is bounded above, it has a least upper bound by the least upper bound axiom. Let x = sup {xn \| n ∈ N}. We will now show that (xn) → x. Fix ε > 0. As was shown in the exercises, if s = sup(A), then for any ε > 0 there is an element a in A so that s − ε < a < s. Hence, it follows that there exists an N in N so that x − ε < xN < x. For any n > N, since xn is non-decreasing, we have that $x- \epsilon < x_N \leq x_n < x$. Thus \|x − xn\| < ε and by the definition of convergence, (xn) converges to x. $\Box$ ### Theorem (Nested intervals property) If there exists a sequence of closed intervals In = [an, bn] = {x \| an ≤ x ≤ bn} such that In+1 ⊆ In for all n, then ∩In is nonempty. #### Proof Since In+1 ⊆ In it follows that an ≤ an+1 and bn+1 ≤ bn. Since (an) and (bn) are monotonic sequences they converge by the previous theorem. Furthermore, since an < bn for all n, it follows that lim an ≤ lim bn . By the monotonicity of (an) and (bn) we have for every n $a_n \leq \lim_{n\to\infty}a_n \leq \lim_{n\to\infty}b_n \leq b_n.$ Therefore lim an ∈ [an, bn] for every n, which implies that $\lim_{n\to\infty}a_n \in \bigcap _{n=1}^{\infty} I_n.$ Thus the intersection is nonempty. $\Box$ ### Theorem (Bolzano—Weierstrass) Every bounded sequence of real numbers contains a convergent subsequence. #### Proof Let (xn) be a sequence of real numbers bounded by a real number M, that is \|xn\| < M for all n. We define the set A by A = {r \| \|r\| ≤ M and r < xn for infinitely many n}. We note that A is non-empty since it contains −M and A is bounded above by M. Let x = sup A. We claim that, for any ε > 0, there must be infinitely many points of xn in the interval (x − ε, x + ε). Suppose not and fix an ε > 0 so that there are only finitely many values of xn in the interval (x − ε, x + ε). Either x ≤ xn for infinitely many n or x ≤ xn for at most only finitely many n (possibly no n at all). Suppose x< xn for infinitely many n. Clearly in this case x ≠ M. If necessary restrict ε so that x + ε ≤ M. Set r = x + ε/2 we have that r < xn for infinitely many n because there are only finitely many xn in the set [x,r] and x must be less than infinitely many xn, furthermore \|r\| < M. Thus r is in A, which contradicts that x is an upper bound for A. Now suppose x< xn for at most finitely many n. Set y = x − ε/2. Then there are at most only finitely man n so that xn ≥ y. Thus, if r < xn for infinitely many n, we have that r ≤ y. This means that y is an upper bound for A that is less than x, contradicting that x wast the least upper bound of A. In either case we arrive at a contradiction, thus we must have that for any ε > 0, there must be infinitely many points of xn in the interval (x − ε, x + ε). Now we show there is a subsequence that converges to x. We define the subsequence inductively, choose any xn1 from the interval (x − 1, x + 1). Assuming we have chosen xn1, …, xnk−1, choose xnk to be an element in the interval (x − 1/k, x + 1/k) so that nk∉{n1, …, nk−1}, this is possible as there are infinitely many elements of (xn) in the interval. Notice that for this choice of xnk we have that \|x − xnk\|<1/k. Hence for any ε>0, if we take any k > 1/ε, then \|xnk-x\| < ε. That is the subsequence (xnk) → x. $\Box$ ### Theorem (Cauchy criterion) A sequence converges if and only if it is Cauchy. Although this seems like a weaker property than convergence, it is actually equivalent, as the following theorem shows: #### Proof First we show that if (xn) → x then xN is Cauchy. Now suppose that for a given ε > 0 we wish to find an N so that \|xn − xm\| < ε for all n, m > N. We will choose N so that for all n ≥ N we have that \|xn − x\| < ε/2. By the triangle inequality, for any n, m > N we have: $\|x_n - x_m\| \leq \|x_n - x\| + \|x_m - x\| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$. Thus (xn) is a Cauchy sequence. Now we show that if (xn) is a Cauchy sequence, then it converges to some x. Let (xn) be a Cauchy sequence, and let ε > 0. By the definition of a Cauchy Sequence, there exits a natural number L so that \|xn − xm\| < ε/2 whenever n, m > L. Since (xn) is a Cauchy sequence it is bounded. By the Bolzano—Weierstrass theorem, it has a convergent subsequence (xnk) that converges to some point x. Now we will show that the whole sequence converges to x Because (xnk) converges, we can choose a natural number M so that if nk > M, then \|xnk − x\| < ε/2. Let N = max(L, M), and fix any nk > N. For n > N we have that $\|x_n - x\| \leq \|x_n - x_{n_k}\| + \|x_{n_k} - x\| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$. Thus by definition of convergence (xn) → x. $\Box$ These theorems all describe different aspects of the completeness of the real numbers. The reader will notice that the least upper bound property was used heavily in this section, and it is the axiom that separates the real numbers from the rational numbers. While these theorems would be false for the rational numbers, not all of them can substitute for the least upper bound property. The Cauchy criterion and the nested intervals property are not strong enough to imply the least upper bound property without additional assumptions, while the Convergence of Monotone sequences theorem and the Bolzano—Weierstrass property do imply the least upper bound property. ## Limit superior and limit inferior Limits turn out to be a very useful tool in analysis, their primary draw back is that they may not always exist. Occasionally it is useful to have some notion of limit that makes sense for any sequence. To this end we introduce the limit superior (often just called the "lim sup") and the limit inferor (often called the "lim inf"). Definition For a sequence (xn) we define the limit superior, denoted lim sup by: $\limsup_{n\to\infty} x_n=\inf_{k}\sup_{n\geq k} x_n.$ Similarly we define the limit inferior, deonoted by lim inf by: $\liminf_{n\to\infty} x_n=\sup_{k}\inf_{n\geq k} x_n.$ If (xn) is not bounded above, we say that lim sup xn = ∞. If (xn) is not bounded we say that lim inf xn = −∞. Notice that for bounded sequences the lim sup and the lim inf always exist. As we know general bounded sequence the limit doesn't always exist. But in the case when the lim sup and lim inf are equal, life is nicer as the next theorem shows. ### Theorem (Limit Superior and Inferior) Let (xn) be a bounded sequence. Then (xn) → x if and only if lim sup xn = x = lim inf xn. #### Proof First suppose (xn) → x. Fix an ε > 0 choose a natural number N so that x − ε < xn < x + ε for any n > N. Hence for any k > N we have that $x-\epsilon\leq \sup_{n>k}x_n\leq x+\epsilon$ and hence x − ε < lim sup xn < x + ε. Since ε was arbitrary, this can only happen if lim sup xn = x. A similar argument shows that lim inf xn = x. Now suppose lim inf xn = x = lim sup xn, and we wish to show that lim xn = x. First recall that the x=lim sup xn is defined as: $x=\inf_k\sup_{n>k} x_n$ Given an ε > 0, since we can get arbitrarily close to the infimum, we can choose we will choose Nls so that $\sup_{n>N_{ls}} x_n - x < \epsilon$ Similarly recall that the x=lim inf xn is defined as: $x=\sup_k\inf_{n>k} x_n$ Since we can get arbitrarily close to the supremum, we can choose we will choose Nli so that $x-\sup_{n>N_{li}} x_n< \epsilon$ Let N = max(Nls, Nli). Now if n > N, then $\inf_{n> N_{li}}\leq \inf_{n\geq N}\leq x \leq \sup_{n\geq N}\leq \sup_{n> N_{ls}}$ Hence for any n > N $x-\sup_{n> N_{ls}} x_n \leq x-x_n\leq x-\inf_{n>N_{li}} x_n.$ By our choice of Nls and Nli this implies for any n > N $-\epsilon \leq x-x_n\leq \epsilon.\quad\Box$	{"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": 54, "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": 0.840153157711029, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/tagged/newtonian-mechanics	# Tagged Questions Newtonian mechanics covers the discussion of the movement of classical bodies under the influence of forces by making use of Newton’s three laws. For more general discussion of energy, momentum conservation etc., use classical-mechanics, for Newton’s description of gravity, use newtonian-gravity. 0answers 21 views ### How do determine the equation of motion of slinky? I'm trying to analyze the forces exerted to slinky. First, I've divided both ends. The red vector is the force exerted by slinky itself, and the green vector is gravitational force. Force exerted ... 1answer 27 views ### Problem with a rotating frame of reference on the South pole Consider this problem: A high-speed train is traveling at a constant 150 m/s (about 300 mph) on a straight horizontal track across the south pole. Find the angle between a plumb line suspended ... 2answers 63 views ### How can a car's engine move the car? Newton's First Law of Motion states that an object at rest or uniform motion tends to stay in that state of motion unless an unbalanced, external force acts on it. Say if I were in a car and I push ... 1answer 42 views ### Friction on a conveyor belt I was surfing through an Olympiad paper and I was caught with this question- A block of mass 1 kg is stationary with respect to a conveyor belt that is accelerating with $1\, \tfrac{m}{s^2}$ ... 4answers 147 views ### How can I determine whether the mass of an object is evenly distributed? How can I determine whether the mass of an object is evenly distributed without doing any permanent damage? Suppose I got all the typical lab equipment. I guess I can calculate its center of mass and ... 0answers 45 views ### Effective mass in Spring-with-mass/mass system Suppose you have a particle of mass $m$ fixed to a spring of mass $m_0$ that, in turn, is fixed to some wall. I'm trying to calculate the effective mass $m'$ that appears in the law of motion of the ... 1answer 27 views ### Static friction force on a block in a tunnel Linked to this Comparing Static Frictions Suppose there is a cuboidal vertical tunnel , and a cubical block in it such that all surface of the block are in contact with the four walls of the ... 1answer 32 views ### If An Object Explodes With A Force, What Force Are Fragments Given? So let's say for example I have an object with 5kg of mass. It explodes with a force of 500N. The object fragments into four fragments: a 0.5kg, a 1kg, a 1.5kg and a 2kg object. What force does ... 1answer 58 views ### Vector cross product of $\mathbf{r}$ and $\ddot{\mathbf{r}}$ in polar coordinates I'm struggling with the following question: Question 6 A planet of mass $m$ moves under the gravitational attraction of a central star of mass $M$. The equation of motion of the planet is ... 0answers 17 views ### newton's laws - involving block and conveyor belt [closed] i was surfing through a physics book when i encountered this problem- A block of mass 1 kg is stationary with respect to a conveyor belt that is accelerating with 1 m/s^2 upwards at an angle of ... 0answers 31 views ### Lagrangian with a general constraint [closed] Can any body help me out to solve this problem? I am familiar with mechanism of Lagrangian and I can solve some problems with constraints but this one is really hard to solve. 4answers 115 views ### Is it possible to whirl a point mass (attacted to a string) around in a horizontal circular motion above my hand? I'm studying circular motion and centripetal force in college currently and there is a very simple question but confuses me (our teacher doesn't know how to explain either :/), so I hope we can sort ... 1answer 22 views ### A rod of length $L$ & mass $M$ is rotating in a circle about one end then calculate tension in the rod at a distance $x$ from the support A rod of length L & mass M is rotating in a circle about one end then calculate tension in the rod at a distance 'x' from the support ? For its solution why should we take mass of L-x portion of ... 1answer 34 views ### Another Inclined plane question I did the FBD, and I found too many variables which are not eliminating...Moreover, I believe this question is based on kinetic and static friction. But, $\mu$ here is ambiguously defined...How Do I ... 3answers 29 views ### The motion of a spring I have a question about the force set by this spring, I saw many times that $\overrightarrow{F}=-Kx\overrightarrow{i}$. I'm asking why not using $\overrightarrow{F}=Kx\overrightarrow{i}$ without the ... 1answer 47 views ### Comparing Static Frictions In this figure, which of the static frictional forces will be more? My aim isn't to solve this particular problem but to learn how is static friction distributed . Since each of the rough-surfaces ... 3answers 96 views ### Moment of Inertia (triangular plate) I want to generalize the formula for the MOI of a triangular plate (sides $a,b,c$) about an axis passing through mid point of one sides and perpendicular to it's plane . The mass of plate $M$ is ... 2answers 55 views ### Massless string Paradox If we introduce the notion of a massless string to denote the fact that net force on a massless string will always be $0$, since it is massless . How can these massless strings ever accelerate when ... 1answer 55 views ### Work done by Static friction Here $v1$ is relative to the block on which sphere is pure rolling but static friction isn't $0$ as of now . In the following diagram, is work done by static friction $0$ ?, since the point of ... 1answer 46 views ### Boundary conditions on wave equation I am having trouble understanding the boundary conditions. From the solutions, the first is that $D_1(0, t) = D_2(0, t)$ because the rope can't break at the junction. The second is that ... 4answers 96 views ### Potential energy sign conventions Almost every book on physics that I read have some weird and non-clear explanations regarding the potential energy. Ok, I do understand that if we integrate a force over some path, we'll get a ... 2answers 65 views ### Centripetal Force Acceleration Suppose you want to perform a uniform circular motion . Then a body performing uniform circular motion horizontally needs an acceleration $= \frac{v^2}{r}$ at each point on the circular path with ... 1answer 51 views ### Physics of the point of contact for a spinning top I understand how spinning tops don't tip over, cf. e.g. this and this Phys.SE questions. What I'm more interested is in identifying the factors that determine the direction the spinning top moves to? ... 1answer 60 views ### In tennis, why does topspin serve bounce higher than flat serve? When receiving servers (while playing tennis), I've noticed that the tennis ball seems to bounce up higher on me when the server uses a topspin serve than when the server hits a flat serve. Why is ... 0answers 49 views ### A slender rod with a ball at the end A slender rod is attached to a block accelerating horizontally. The rod is free to rotate without friction. At the end of the rod is a ball. As the block accelerates, the slender rod will be deviated ... 0answers 45 views ### Object on an inclined plane I was solving a question about inclined planes, and it got me thinking: Imagine a scenario where, an object is on an inclined plane and the inclined plane is on a frictionless surface, and is itself ... 4answers 112 views ### What do people actually mean by “rolling without slipping”? I have never understood what's the meaning of the sentence "rolling without slipping". Let me explain. I'll give an example. Yesterday my mechanics professor introduced some concepts of rotational ... 1answer 145 views ### How reliant is the Solar System on being exactly the way it is? We know that all objects with mass exert forces on all other objects of mass such that $$F = \frac{GMm}{R^2}.$$ And as others have discussed the planets do interfere with each other ... 1answer 55 views ### Tension in a closed loop around an object, carrying stuff Sorry when this sounds like a first year physics homework question, but it's actually not. It arose during a discussion regarding safe practices for rock climbing anchors... So here we go: Imagine ... 0answers 46 views ### What force does the seat exert on the rider at the top and bottom of the ride? [closed] A 75 kg person rides a Ferris wheel which is rotating uniformly. The centripetal force acting on the person is 45 N. What force does the seat exert on the rider at the top and the bottom of the ride? 1answer 44 views ### A theoretical problem on Mechanics [closed] Two particles with masses $m_1$ and $m_2$ are moving in 3D space with some Cartesian coordinate system. There are known the laws of motion of these particles, i.e. the position vectors $\vec{r_1}(t)$ ... 1answer 96 views ### Chain of balls on an inclined plane Suppose we have some inclined plane, and there is some chain of balls of length $l$ and mass $m$ lying on it. No friction at all in the system. 1) What is $x_0$ (the vertical hanging part of the ... 1answer 60 views ### Centripetal force and circular motion I have a doubt, say when a body moves in a circular loop, there are basically two types of acceleration acting on the object. One is the linear acceleration which is basically tangent to the circle, ... 2answers 58 views ### Why is potential energy negative when orbiting in a gravitational field? I had to do a problem, and part of it was to find the mechanical energy of satellite orbiting around mars, and I had all of the information I needed. I thought the total mechanical energy would be the ... 1answer 79 views ### Initial position and velocity of rocket to escape earth's gravity I'm trying to numerically simulate a spacecraft trajectory between earth and mars. I already wrote the solar system model where the sun is at the origin of the x,y,z plane Earth and Mars are orbiting ... 5answers 1k views ### Why Won't a Tight Cable Ever Be Fully Straight? I posted this picture of someone on a zipline on Facebook. One of my friends saw it and asked this question, so he could try to calculate the speed at which someone on the zipline would be going ... 0answers 45 views ### Work And Energy Question [closed] $H = 3\text{ m}$,$m=2\text{ kg}$ The right side is rough. I want to figure: what is the coefficient of friction $\mu$? How high and exceed the maximum return on the plane right body? I know ... 1answer 51 views ### SHM of floating objects If we consider an object undergoing who has an acceleration proportional to the displacement of the object, it is going simple harmonic motion. In terms of Newton's second law, this is -\dfrac k ... 2answers 48 views ### Constant of gravity in earth fixed coordinate system I have this problem: If the constant of gravity is measured to be $g_0$ in an earth fixed coordinate system, what is the difference $g-g_0$ where $g$ is the real constant of gravity as ... 0answers 59 views +50 ### How multiple objects in contact are resolved in an inelastic collision, when edge normals don't “line up” In a case I understand, let's say I have an object A moving at velocity V toward 3 objects in contact B, C, and D: The momentum of A is the mass of A times its velocity. To figure out how the ... 2answers 43 views ### Simple conservation of momentum and frame of reference problem I'm making a very simple physics engine based on momentum, and I'm solving what response to use for a collision from each involved object's frame of reference. However, something about how I'm ... 2answers 46 views ### Center of mass problem According to the definition of potential energy, we use $U= mhg .$ In the figure below , A thin uniform rod of mass m and length h is positioned vertically above an anchored frictionless pivot ... 0answers 29 views ### Velocity versus time graph [closed] A stone is thrown at an angle of $45^0$ above the horizontal x-axis in the +x-direction. If air resistance is ignored, which of the velocity versus time graphs shown above best represents $v_x$ ... 2answers 54 views ### Angle between acceleration and velocity Problem: A particle is constrained to move in a circle with a 10-meter radius. At one instant, the particle's speed is 10 meters per second and is increasing at a rate of 10 meters per second ... 2answers 78 views ### Are there problems solvable with Newtonian physics, GR and QM? First I must let you know that I don't have much understanding of neither GR nor quantum mechanics, and therefore this question. I've mentally pictured Newtonian physics, GR and quantum mechanics all ... 2answers 52 views ### Time period for spring connected body Two identical springs with spring constant $k$ are connected to identical masses of mass $M$, as shown in the figures above. The ratio of the period for the springs connected in parallel (Figure 1) ... 0answers 38 views ### impulse problem [closed] The figure above shows a plot of the time-dependent force $F_x(t)$ acting on a particle in motion along the x-axis. What is the total impulse delivered to the particle? ... 1answer 56 views ### Why body starts moving when force is applied? [duplicate] The recent question by m.buettner regarding self-inductance and its resistance to EMF Faraday's law - does the induced current's magnetic field affect the change in flux?, recalled me the ... 0answers 36 views ### A space vehicle travelling at a velocity of 1296 (km/h) separates by a controlled explosion into two sections of mass 7276 (kg) and 236 (kg) [closed] A space vehicle traveling at a velocity of 1296 (km/h) separated by a controlled explosion into two sections of mass 7276 (kg) and 236 (kg). the two parts continue in the same direction with the ... 3answers 63 views ### Relation between the time, velocity and acceleration This is question from I.E. Iredov's General Physics: $1.22$ : The velocity of a particle moving in the positive direction of the $x-axis$ varies as $v = α \sqrt x$, where $α$ is a positive ...	{"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": 45, "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": 0.9309770464897156, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/tagged/rubiks-cube	# Tagged Questions Questions on the mathematics behind the famed toy invented by Ernő Rubik. 2answers 57 views ### combinations of Rubik cube A Rubick's Cube has owl heads on it, which can be misoriented. How many (times) MORE combinations are there of this cube vs. one that has blank stickers? Can someone give me some hints? Thanks 1answer 85 views ### Rubik's cube and counting Inspired by this question, I formulate the following: Suppose I have a $3\times3\times3$ Rubik's cube, call each small square on the surface a piece, there are then $336 = 54$ pieces. Enumerate ... 1answer 91 views ### Literature on group theory of Rubik's Cube While searching for literature on the group theory of Rubik's Cube, I mostly find introductions to group theory motivated by applications to Rubik's cube. I.e. the focus lies on elementary group ... 3answers 811 views ### Why should Rubik's cube get attention from mathematicians? I've seen a lot of math debate, calculations and other stuff related to Rubik's cube lately, but I don't really understand why is it important, why should anyone spend time asking and answering ... 2answers 175 views ### Rubik Cube finite non-abelian group I was reading the following paper from MIT: http://web.mit.edu/sp.268/www/rubik.pdf The paper is not difficult to understand, it is more or less a short introduction into group theory, taking the ... 1answer 87 views ### Rubik's cube as $2\times2\times2$ or $2\times2$? [closed] I'm surprised why people call a Rubik's cube as $2 \times2\times2$ or $3\times3\times3$? It's a CUBE! Isn't saying $2\times2$ or $3\times3$ cube (or rather just $2$ or $3$) sufficient? 2answers 77 views ### Lower bounds on numbers of arrangements of a Rubik's cube Last night, a friend of mine informed me that there were forty-three quintillion positions that a Rubik's Cube could be in and asked me how many there were for my Professor's Cube (5x5x5). So I gave ... 4answers 510 views ### Minimum number of steps needed to solve a rubik cube Long time ago I've seen a book on group theory and there was an appendix about rubik cube. I remember there were only three steps that enabled me to solve my cube (three strings with letters encoding ... 1answer 89 views ### The redundancy of Rubik's cube states [duplicate] Possible Duplicate: Rubik’s Cube Not a Group? I take a Rubik's cube in the solved state, and I secretly assign a unique integer label to each of the cubies. I then, via an arbitrarily long ... 2answers 170 views ### Rubik's Cube Combination Could anyone explain why the number of legal or reachable combinations of a $3\times 3\times 3$ Rubik's Cube is $1/12\mbox{th}$ of the total. I understood the logic behind the total number of ... 2answers 229 views ### Presentation of Rubik's Cube group The Rubik's Cube group is the group of permutations of the 20 cubes at the edges and vertices of a Rubik's group (taking into account their specific rotation) which are attainable by succesive ... 3answers 679 views ### Brute force method of solving the cube: How many moves would it take? Given that Rubik's cube has finitely many positions, one possible "brute force" method to solve it would be to determine once a sequence of moves which eventually reaches every possible position of ... 1answer 176 views ### Orders of moves in the Rubik's cube Playing with my Rubik's cube, I was thinking of facts about it that are immediate to mathematicians but novel to others. Here's one: Given a Rubik's cube in the starting position, any sequence of ... 1answer 102 views ### Rubik's cube notation [closed] Various Rubik cube guides use the notation F and R to describe the front and right faces of the cube. However, in the illustrations they show the cube in isometric projection. So which side is right, ... 3answers 573 views ### Is there a “most random” state in Rubik's cube? Is there a state in Rubik's cube which can be considered to have the highest degree of randomness (maximum entropy?) asssuming that the solved Rubik's cube has the lowest? 1answer 190 views ### Are actions in the $3\times 3\times 3$ rubik cube a group? Are actions in the $3\times 3\times 3$ rubik cube a group? You can see here Rubik's Cube Not a Group? that $4\times 4\times 4$ rubik cubes or higher arent groups. But what about \$3\times 3\times ... 1answer 636 views ### Rubik's Cube Not a Group? I read online that although the 3x3x3 is a great example of a mathematical group, larger cubes aren't groups at all. How can that be true? There is obviously an identity and it is closed, so ... 2answers 346 views ### Rubik's cube puzzle If we cut along the plane orthogonal to the largest diagonal of a Rubik's cube, what is the maximum number of small cubes can we cut? I thought this should be $9$, but apparently this is not the ... 2answers 2k views ### How to tell if a Rubik's cube is solvable How can I determine if a certain Rubik's cube, that is in a certain state, is solveable? By "certain state" it could mean that the cube has been dismantled and put together again. And in my experience ... 2answers 393 views ### All possible permutations on a Rubik cube ($3\times3\times 3$) can be reached from the initial state? If I were to represent a state in the Rubik cube as a permutation of the colors on the 9 tiles per side on all sides of the cube, could I reach all possible states (i.e. colorings) by the permutations ... 1answer 229 views ### Does solving a Rubik's cube imply alignment? Today, I got my hands on a Rubik's cube with text on it. It looks like this: Now, I would love to know whether solving the right cube will also always correctly align the text on the cube or ... 2answers 369 views ### Mathematics Behind the 4×4 and 5×5 Rubik's Cube A lot is known about the math behind the 3×3 Rubik's cube (symmetries, generators, group structure etc...). Is the same true for the 4×4 and 5×5 cubes? I haven't had much success finding this ... 1answer 325 views ### Solving a scrambled $3 \times 3 \times 3$ Rubik's Cube with at most 20 moves! I read somewhere that any scrambled form of $3 \times 3 \times 3$ Rubik's cube can be solved using at most $20$ moves, and I just said "wow"! I am wondering can we prove this by mathematical ways? Or ... 1answer 1k views ### What are the 2125922464947725402112000 symmetries of a Rubik's Cube? In a recent talk, Marcus du Sautoy says there are 2125922464947725402112000 (2.1*10^24) symmetries of a Rubik's cube, but doesn't explicitly identify what qualifies as a symmetry. What counts as a ... 1answer 529 views ### Rubik's cube interesting questions? The upper bound for the number of moves required to solve a regular Rubik's cube has been shown to be 20. Two questions come to mind: Does this result have more general significance? What are the ... 3answers 490 views ### How many ways can I make six moves on a Rubik's cube? I am writing a program to solve a Rubik's cube, and would like to know the answer to this question. There are 12 ways to make one move on a Rubik's cube. How many ways are there to make a sequence of ... 2answers 236 views ### How is done the calculation of the minimum number of movement to solve any configuration of Rubik's Cube? I have read a few weeks ago that some mathematical researchers have discover that the minimum number of movements to solve any initial configuration of a Rubik's cube has been downsized to 20. How do ... 1answer 462 views ### How many disconnected graphs of the Rubik's cube exist? Let us say that a Rubik's cube in a particular configuration is in a particular "state". All other configurations of this cube (other "states"), which can be achieved by rotations of the cube can be ...	{"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": 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": 0.9436934590339661, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/182926/a-question-about-integral-operator/182942	# A question about integral operator I have a question: Prove or disprove that: for every $f\in L^{1}\left(\mathbb{R}\right)$, $$\sup\left\{ { \int_{\mathbb{R}}\frac{\sqrt{n}}{\sqrt{\left\|x-y\right\|}\left(1+n^{2}\left(x-y\right)^{2}\right)}f\left(y\right)dy:n\in\mathbb{N}}\right\} <\infty,$$ for Lebesgue almost every $x\in\mathbb{R}$. I failed in my attempt to disprove this statement (I think so!!!). Can everybody help me? - ## 1 Answer Let $$f_n(x):=\int_{\Bbb R}\frac{\sqrt n}{\sqrt{\|x-y\|}(1+n^2\|x-y\|^2)}f(y)dy,$$ assuming WLOG that $f\geqslant 0$. We use the substitution $t=n(x-y)$, hence $dt=-ndy$ to get $$f_n(x)=\int_{\Bbb R}\frac 1{\sqrt{\|t\|}}\frac 1{1+t^2}f\left(x-\frac tn\right)dt.$$ By Fubini's theorem for non-negative functions, and since $f$ is integrable, we get that $f_n$ is integrable, and in particular almost everywhere finite. Approximate $f\in L^1$ by $g$, continuous with compact support, such that $\lVert f-g\rVert_{L^1}\leqslant 1$. Then, as the integral $\int_{\Bbb R}\frac 1{\sqrt{\|t\|}}\frac 1{1+t^2}dt<\infty$, to show the result when $f$ is continuous and bounded. With the latest formula, it's easier to see it. - 1 You are almost done, note however that one asks about $\sup\limits_{n\geqslant1}f_n(x)$, not $f_n(x)$ for some $n$. – Did Aug 15 '12 at 23:15 Supremum get on $n$ fixed $x$. I think the fact $f_{n}\in L^{1}\left(\mathbb{R}\right)$ is easy. Beside that, we also have \begin{eqnarray} T_{n}:L^{1}\left(\mathbb{R}\right) & \rightarrow & L^{1}\left(\mathbb{R}\right)\\ f & \rightarrow T_{n}\left(f\right)\left(x\right)= & \intop_{\mathbb{R}}\dfrac{\sqrt{n}}{\sqrt{\left\|x-y\right\|}\left(1+n^{2}\left(x-‌y\right)^{2}\right)}f\left(y\right)dy\end{eqnarray} is a continous linear operator and $\sup_{n\in\mathbb{N}}\left\Vert T_{n}\right\Vert <\infty.$ – user36548 Aug 16 '12 at 2:21	{"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": 19, "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": 0.8797354102134705, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/263413/zero-divisors-in-polynomial-ring	# zero divisors in polynomial ring The following is an exercise in Hungerford (Ch. III, ex. 5.6). Let $R$ be a commutative ring with identity. If $f=a_nx^n+\dots+a_0$ is a zero divisor in $R[x]$, then there exists a nonzero $b$ in $R$ such that $ba_n=ba_{n-1}=\dots=ba_0=0$. I can see for example that $\{g\in R[x]\|fg=0\}$ is a nonzero ideal, so it contains a nonzero element of smallest degree. But how to show that such an element is actually a constant? - Thanks rschwieb. The question in the other post is not an exact duplicate, but the answer from Gone proves the general case. – PatrickR Dec 21 '12 at 21:24 If you want you can delete it, but in this case you don't have to do anything. If someone really doesn't like it, they'll have to vote to delete. Please don't edit the other question's contents. – rschwieb Dec 21 '12 at 21:24 – Hurkyl Dec 21 '12 at 21:34 – YACP Dec 21 '12 at 21:58 ## 1 Answer This is McCoy's theorem, for which you can find a nice summary and information on in this solution. -	{"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": 0.9456015229225159, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/tagged/event-horizon	# Tagged Questions The event-horizon tag has no wiki summary. learn more… \| top users \| synonyms 1answer 13 views ### Can we build a synthetic event horizon? If we imagine ourselves to be a civilization capable of manipulating very heavy masses in arbitrary spatial and momentum configurations (because we have access to large amounts of motive force, for ... 1answer 55 views ### Relativistic Computation? Is it possible to employ relativity to develop computational technology? Here is a really basic example: Build a Computer and Feed it the Problem (say the problem is projected to take 10 years to ... 1answer 79 views ### Why does the Schwarzschild radius become excessively large after a certain point? Here's something that I've found difficult to wrap my head around. The relationship between the Schwarzschild radius and mass is linear. 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But does this not imply that a the geodesic of a particle crossing it must be spacelike at that point? But ... 1answer 82 views ### Gauss-Bonnet theorem in the Hawking/Ellis book At the page 336 of Hawking, Ellis: The Large Scale Structure of Space-Time, the Gauss-Bonnet theorem is stated as $$\int_H \hat{R}\ d\hat{S} = 2\pi \chi(H) \qquad (1)$$ with \hat{R} = R_{abcd} ... 3answers 120 views ### What actually happens to a light ray on a Schwarzschild black hole horizon? I know the Schwarzschild event horizon is a null surface generated by null geodesics. But what does that actually mean in terms of the path of a light ray that reaches it? Does that mean the geodesic ... 2answers 90 views ### Relaxation time for deviations from spherical shape of a black hole's event horizon (and waves) A different question about truly spherical objects in nature (Do spheres exist in nature?) made me think of a lecture I had been at where, as I recall, it was mentioned that the most perfectly ... 2answers 205 views ### How does the evaporation of a black hole look for a distant observer? Let's assume an observer looking at a distant black hole that is created by collapsing star. In observer frame of reference time near black hole horizon asymptotically slows down and he never see ... 1answer 101 views ### Would dense matter around a black hole event horizon eventually form a secondary black hole? [duplicate] Possible Duplicate: Black hole formation as seen by a distant observer Given that matter can never cross the event horizon of a black hole (from an external observer point of view), if a ... 0answers 36 views ### Can a black hole actually grow, from the point of view of a distant observer? [duplicate] Possible Duplicate: Black hole formation as seen by a distant observer I've read in several places that from the PoV of a distant observer it will take an infinite amount of time for new ... 2answers 183 views ### Fighting a black hole: Could a strong spherical shell inside an event horizon resist falling in to the singularity? As a thought experiment imagine an incredibly strong spherical shell with a diameter a bit smaller than the event horizon of a particular large black hole. The shell is split into two hemispheres, ... 3answers 179 views ### Black hole formation as seen by a distant observer [duplicate] Possible Duplicate: How can anything ever fall into a black hole as seen from an outside observer? Is black hole formation observable for a distant observer in finite amount of time? ... 2answers 170 views ### What is the current radius of cosmological event horizon? Doing some crude calculations (using the value of $H_0$ at this point of time only, since it is time dependent but not distance dependent thanks to Johannes answer) what is the radius of cosmological ... 1answer 84 views ### Are different frequencies of light lensed differently during gravitational lensing a bit like refraction? So I was wondering about the event horizon on a black hole. And wondering if the point of no return for radio waves vs gamma rays would be different. I guess the logic being, since gamma rays have ... 2answers 159 views ### What happens to the wavelength/frequency of a photon as it passes through an event horizon? I've asked a similar question about photons and black holes but wanted to rephrase it more specifically, so here goes... Ever since I learned how a photon's wavelength and frequency are indivisibly ... 2answers 398 views ### Can we have a black hole without a singularity? Assuming we have a sufficiently small and massive object such that it's escape velocity is greater than the speed of light, isn't this a black hole? It has an event horizon that light cannot escape, ... 1answer 116 views ### Way to escape from a black hole I’ve had a question on WHY a traveler couldn’t “escape” from a black hole under specific conditions (I have an image I'd like to send to clarify, but the website won't let me)> The key is for the ... 5answers 466 views ### Is it possible for one black hole to pull an object out of another black hole? Suppose we have a spacecraft just inside the event horizon of a black hole, struggling to escape, but slowly receding into it. Another (bigger) black hole expands until its event horizon includes the ... 2answers 79 views ### Causal reconnection Can causally disconnected regions join up again? For example the universe is expanding faster than light creating cosmological horizon, but what if something causes the expansion to slow down and ... 2answers 207 views ### What do you feel when crossing the event horizon? I have heard the claim over and over that you won't feel anything when crossing the event horizon as the curvature is not very large. But the fundamental fact remains that information cannot pass ... 4answers 538 views ### Gravity on supermassive black hole's event horizon M = black hole mass Gravitation is about 1/r^2 Schwarzschild radius r is ~ to M Greater BH -> weaker gravitation on its horizon. Lets take black hole so enormous that the gravitation on its ... 2answers 132 views ### event horizons are untraversable by observers far from the collapse? Consider this a followup question of this one In the classical schwarszchild solution with an eternal black hole, the user falls through the event horizon in finite local time, but this event does ... 1answer 193 views ### Falling into a black hole I've heard it mentioned many times that "nothing special" happens for an infalling observer who crosses the event horizon of a black hole, but I've never been completely satisfied with that statement. ... 1answer 109 views ### How can things at the event horizon slow down and appear to stop to a remote observer? So they say the remote observer will never see anything fallen to the black hole, because any object will slow down as it gets closer to the event horizon and eventually stop to stay there forever. Am ... 1answer 116 views ### Does cosmic expansion imply the possibility of the universe splitting in half; multiple big-crunches? There is an event horizon where cosmic expansion leads to superluminal recession speeds for sufficiently distant objects -- the Hubble Volume. 1) Does matter beyond the event horizon affect us ... 1answer 168 views ### Will acceleration rate of expansion of space become faster than speed of light? From watching cosmology lectures, it seems that the space between galaxies is expanding at an accelerating rate, my question is since it is the space that is (acceleratingly expanding), the special ... 1answer 251 views ### Can two particles remain entangled even if one is past the event horizon of a black hole? Can two particles remain entangled even if one is past the event horizon of a black hole? If both particles are in the black hole? What changes occur when the particle(s) crosses(cross) the event ... 8answers 2k views ### How can anything ever fall into a black hole as seen from an outside observer? The event horizon of a black hole is where gravity is such that not even light can escape. This is also the point I understand that according to Einstein time dilation will be infinite for a ... 4answers 861 views ### If an anti-matter singularity and a normal matter singularity, of equal masses, collided would we (outside the event horizon) see an explosion? If an anti-matter singularity and a normal matter singularity, of equal masses, collided would we (outside the event horizon) see an explosion? 1answer 175 views ### Could a ship equipped with Alcubierre drive theoretically escape from a black hole? Also could it reach parts of the universe that are receding faster than the speed of light from us? 3answers 222 views ### Solid objects inside the event horizon - can they remain “solid”? So, once something is inside a black hole's event horizon, it can only move towards the center. This is fine for a point-object. But 3D solid objects rely on molecular forces to stay in one piece. ... 2answers 168 views ### Do all event horizons emit radiation? So, the event horizon around a black hole emits radiation, and Rindler space is full of thermal energy. I guess I have two questions- does the Unruh effect have anything to do with radiation from the ... 2answers 210 views ### Is the observable region of the universe within the event horizon of a super-massive black hole? Observations: I have read that for a free-falling observer within the event horizon of a black hole that all lines of sight will end at the singularity which is black. I also look up and see that ... 2answers 183 views ### Direction of Time on Event Horizon Does the axis of Time point into a black hole or away from? Can you give a reference paper? 2answers 413 views ### Do apparent event horizons have Hawking radiation? As I understand it, black holes have an absolute event horizon and an apparent horizon specific an observer. In addition to black holes, an apparent horizon can come from any sustained acceleration. ... 3answers 397 views ### Horizon and Unruh radiation for a finite period of acceleration It's a well known fact that an observer that accelerates at a constant rate from $-c$ at past infinity to $+c$ at future infinity sees a horizon in flat Minkowski spacetime. This is easy to see from a ... 4answers 450 views ### de sitter cosmologic limit It has been said that our universe is going to eventually become a de sitter universe. Expansion will accelerate until their relative speed become higher than the speed of light. So i want to ...	{"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": 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": false}, "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": 0.9323810338973999, "perplexity_flag": "middle"}
http://www.reference.com/browse/sate+on+one+butt	Definitions Stone's theorem on one-parameter unitary groups In mathematics, Stone's theorem on one-parameter unitary groups is a basic theorem of functional analysis which establishes a one-to-one correspondence between self-adjoint operators on a Hilbert space H and one-parameter families of unitary operators $\left\{U_t\right\}_\left\{t in mathbb\left\{R\right\}\right\}$ which are strongly continuous, that is $lim_\left\{t rightarrow t_0\right\} U_t xi = U_\left\{t_0\right\} xi quad forall t_0 in mathbb\left\{R\right\}, xi in H$ and are homomorphisms: $U_\left\{t+s\right\} = U_t U_s. quad$ Such one-parameter families are ordinarily referred to as strongly continuous one-parameter unitary groups. The theorem is named after Marshall Stone who formulated and proved this theorem in 1932. Formal statement Let U be a strongly continuous 1-parameter unitary group, then there exists a unique self-adjoint operator A such that $U_t := e^\left\{i t A\right\} quad t in mathbb\left\{R\right\}.$ Conversely, let A be a self-adjoint operator on a Hilbert space H. Then $U_t := e^\left\{i t A\right\} quad t in mathbb\left\{R\right\}$ is a strongly continuous one-parameter family of unitary operators. The infinitesimal generator of {Ut}t is the operator $iA$. This mapping is a bijective correspondence. A will be a bounded operator iff the operator-valued function $t mapsto U_t$ is norm continuous. Example The family of translation operators $\left[T_t psi\right]\left(x\right) = psi\left(x + t\right) quad$ is a one-parameter unitary group of unitary operators; the infinitesimal generator of this family is an extension of the differential operator $frac\left\{d\right\}\left\{dx\right\} = i frac\left\{1\right\}\left\{i\right\} frac\left\{d\right\}\left\{dx\right\}$ defined on the space of complex-valued continuously differentiable functions of compact support on R. Thus $T_t = e^\left\{t , \left\{d\right\}/\left\{dx\right\}\right\}. quad$ Applications and generalizations Stone's theorem has numerous applications in quantum mechanics. For instance, given an isolated quantum mechanical system, with Hilbert space of states H, time evolution is a strongly continuous one-parameter unitary group on H. The infinitesimal generator of this group is the system Hamiltonian. The Hille–Yosida theorem generalizes Stone's theorem to strongly continuous one-parameter semigroups of contractions on Banach spaces. References • M. H. Stone, On one-parameter unitary groups in Hilbert Space, Annals of Mathematics 33, 643-648, (1932). • K. Yosida, Functional Analysis, Springer-Verlag, (1968) Last updated on Tuesday September 02, 2008 at 10:51:54 PDT (GMT -0700) Search another word or see sate on one button Dictionary \| Thesaurus \|Spanish	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 10, "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": 20, "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": 0.8661760091781616, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/251311/limits-trigonometry-tending-to-infinity	# Limits - trigonometry - tending to infinity How do we solve: $$\lim_{x\to \infty} 5^x \sin\left(\frac{a}{5^x}\right)$$ Thank You. - Gerry Myerson’s answer is the way to go, but you can easily see what the limit has to be if you remember that $\sin x\approx x$ when $\|x\|$ is small. Thus, $\sin\frac{a}{5^x}\approx\frac{a}{5^x}$ when $x$ is large, and ... . – Brian M. Scott Dec 5 '12 at 5:21 ## 1 Answer Convince yourself that it's the same as evaluating $\lim_{t\to0}{\sin at\over t}$, and then use other stuff you know to do that one. - Thanks. That works but I'm not sure if we can perform that operation on power functions... Can we? – Lavanya Dec 5 '12 at 5:20 Put $y=\frac{1}{5^x}$ and see what happens. – Mhenni Benghorbal Dec 5 '12 at 5:22 Thanks to all of you. I got it now! – Lavanya Dec 5 '12 at 5:31 1 Good. Then you can write it up and post it as an answer. Then, later, you can accept it. – Gerry Myerson Dec 5 '12 at 5:35 Well, now I have got another doubt.. Instead of doing it the above way, if I let it remain x -> infinity, then 5^infinity becomes infinity...so on solving I would get infinitysin(0) which is all too confusing! And if I can't directly put infinity in place of x, why is it so? Because I'll get an infinity0 form?? – Lavanya Dec 5 '12 at 9:20 show 1 more comment	{"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": 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": 0.9421858191490173, "perplexity_flag": "middle"}
http://en.wikipedia.org/wiki/Physical_cosmology	# Physical cosmology Physical cosmology Part of a series on Early universe Expanding universe Structure formation Future of universe Components History of cosmological theories Experiments Scientists Social impact Physical cosmology, as a branch of astronomy, is the study of the largest-scale structures and dynamics of the Universe and is concerned with fundamental questions about its formation and evolution.[1] For most of human history, it was a branch of metaphysics and religion. Cosmology as a science originated with the Copernican principle, which implies that celestial bodies obey identical physical laws to those on Earth, and Newtonian mechanics, which first allowed us to understand those laws. Physical cosmology, as it is now understood, began with the 20th century development of Albert Einstein's general theory of relativity, and better astronomical observations of extremely distant objects. These advances made it possible to speculate about the origin of the Universe, and allowed scientists to establish the Big Bang Theory as the leading cosmological model. Some researchers still advocate a handful of alternative cosmologies;[2] however, cosmologists generally agree that the Big Bang theory best explains observations. Cosmology draws heavily on the work of many disparate areas of research in theoretical and applied physics. Areas relevant to cosmology include particle physics experiments and theory, including astrophysics, general relativity, quantum mechanics, and plasma physics. Thus, cosmology unites the physics of the largest structures in the Universe with the physics of the smallest structures in the Universe. ## History of study See also: Timeline of cosmology and List of cosmologists Modern cosmology developed along tandem tracks of theory and observation. In 1916, Albert Einstein published his theory of general relativity, which provided a unified description of gravity as a geometric property of space and time.[3] At the time, Einstein believed in a static universe, but found that his original formulation of the theory did not permit it.[4] This is because masses distributed throughout the Universe gravitationally attract, and move toward each other over time.[5] However, he realized that his equations permitted the introduction of a constant term which could counteract the attractive force of gravity on the cosmic scale. Einstein published his first paper on relativistic cosmology in 1917, in which he added this cosmological constant to his field equations in order to force them to model a static universe.[6] However, this so-called Einstein model is unstable to small perturbations—it will eventually start to expand or contract.[4] The Einstein model describes a static universe; space is finite and unbounded (analogous to the surface of a sphere, which has a finite area but no edges). It was later realized that Einstein's model was just one of a larger set of possibilities, all of which were consistent with general relativity and the cosmological principle. The cosmological solutions of general relativity were found by Alexander Friedmann in the early 1920s.[7] His equations describe the Friedmann-Lemaître-Robertson-Walker universe, which may expand or contract, and whose geometry may be open, flat, or closed. In the 1910s, Vesto Slipher (and later Carl Wilhelm Wirtz) interpreted the red shift of spiral nebulae as a Doppler shift that indicated they were receding from Earth. However, it is difficult to determine the distance to astronomical objects. One way is to compare the physical size of an object to its angular size, but a physical size must be assumed to do this. Another method is to measure the brightness of an object and assume an intrinsic luminosity, from which the distance may be determined using the inverse square law. Due to the difficulty of using these methods, they did not realize that the nebulae were actually galaxies outside our own Milky Way, nor did they speculate about the cosmological implications. In 1927, the Belgian Roman Catholic priest Georges Lemaître independently derived the Friedmann-Lemaître-Robertson-Walker equations and proposed, on the basis of the recession of spiral nebulae, that the Universe began with the "explosion" of a "primeval atom"—which was later called the Big Bang. In 1929, Edwin Hubble provided an observational basis for Lemaître's theory. Hubble showed that the spiral nebulae were galaxies by determining their distances using measurements of the brightness of Cepheid variable stars. He discovered a relationship between the redshift of a galaxy and its distance. He interpreted this as evidence that the galaxies are receding from Earth in every direction at speeds proportional to their distance. This fact is now known as Hubble's law, though the numerical factor Hubble found relating recessional velocity and distance was off by a factor of ten, due to not knowing about the types of Cepheid variables. Given the cosmological principle, Hubble's law suggested that the Universe was expanding. Two primary explanations were proposed for the expansion. One was Lemaître's Big Bang theory, advocated and developed by George Gamow. The other explanation was Fred Hoyle's steady state model in which new matter is created as the galaxies move away from each other. In this model, the Universe is roughly the same at any point in time. For a number of years, support for these theories was evenly divided. However, the observational evidence began to support the idea that the Universe evolved from a hot dense state. The discovery of the cosmic microwave background in 1965 lent strong support to the Big Bang model, and since the precise measurements of the cosmic microwave background by the Cosmic Background Explorer in the early 1990s, few cosmologists have seriously proposed other theories of the origin and evolution of the cosmos. One consequence of this is that in standard general relativity, the Universe began with a singularity, as demonstrated by Stephen Hawking and Roger Penrose in the 1960s. ## Energy of the cosmos This section . (May 2011) Light chemical elements, primarily hydrogen and helium, were created in the Big Bang. These elements keep[clarification needed] spreading too[clarification needed] fast and too tightly[clarification needed] during the Big Bang process (see Nucleosynthesis). The small atomic nuclei combined into larger atomic nuclei to form heavier elements such as iron and nickel, which are more stable (see Nuclear fusion). This caused a later energy release. Such reactions of nuclear particles inside stars continue to contribute to sudden energy releases, such as in nova stars. Gravitational collapse of matter into black holes is also thought to power the most energetic processes, generally seen at the centers of galaxies (see Quasar and Active galaxy). Cosmologists cannot explain all cosmic phenomena exactly, such as those related to the accelerating expansion of the Universe, using conventional forms of energy. Instead, cosmologists propose a new form of energy called dark energy that permeates all space.[8] One hypothesis is that dark energy is the energy of virtual particles, which are believed to exist in a vacuum due to the uncertainty principle. There is no clear way to define the total energy in the Universe using the most widely accepted theory of gravity, general relativity. Therefore, it remains controversial whether the total energy is conserved in an expanding universe. For instance, each photon that travels through intergalactic space loses energy due to the redshift effect. This energy is not obviously transferred to any other system, so seems to be permanently lost. On the other hand, some cosmologists insist that energy is conserved in some sense; this follows the law of conservation of energy.[9] Thermodynamics of the Universe is a field of study that explores which form of energy dominates the cosmos - relativistic particles which are referred to as radiation, or non-relativistic particles referred to as matter. Relativistic particles are particles whose rest mass is zero or negligible compared to their kinetic energy, and so move at the speed of light or very close to it; non-relativistic particles have much higher rest mass than their energy and so move much slower than the speed of light. As the Universe expands, both matter and radiation in it become diluted. However, the Universe also cools down, meaning that the average energy per particle gets smaller. Therefore radiation becomes weaker, and dilutes faster[why?] than matter. Thus with the expansion of the Universe, radiation becomes less dominant than matter. The very early Universe is said to have been 'radiation dominated' and radiation controlled the deceleration of expansion. Later, as the average energy per photon becomes roughly 10 eV and lower, matter dictates the rate of deceleration and the Universe is said to be 'matter dominated'. The intermediate case is not treated well analytically. As the expansion of the universe continues, matter dilutes even further and the cosmological constant becomes dominant, leading to an acceleration in the universe's expansion. ## History of the universe See also: Timeline of the Big Bang The history of the Universe is a central issue in cosmology. The history of the Universe is divided into different periods called epochs, according to the dominant forces and processes in each period. The standard cosmological model is known as the Lambda-CDM model. ### Equations of motion Main article: Friedmann-Lemaître-Robertson-Walker metric The equations of motion governing the Universe as a whole are derived from general relativity with a small, positive cosmological constant.[10] The solution is an expanding universe; due to this expansion, the radiation and matter in the Universe cool down and become diluted. At first, the expansion is slowed down by gravitation attracting the radiation and matter in the Universe. However, as these become diluted, the cosmological constant becomes more dominant and the expansion of the Universe starts to accelerate rather than decelerate. In our universe this happened billions of years ago. ### Particle physics in cosmology Main article: Particle physics in cosmology Particle physics is important to the behavior of the early Universe, because the early Universe was so hot that the average energy density was very high. Because of this, scattering processes and decay of unstable particles are important in cosmology. As a rule of thumb, a scattering or a decay process is cosmologically important in a certain cosmological epoch if the time scale describing that process is smaller than, or comparable to, the time scale of the expansion of the Universe. The time scale that describes the expansion of the Universe is $1/H$ with $H$ being the Hubble constant, which itself actually varies with time. The expansion timescale $1/H$ is roughly equal to the age of the Universe at that time. ### Timeline of the Big Bang Main article: Timeline of the Big Bang Observations suggest that the Universe began around 13.8 billion years ago.[11] Since then, the evolution of the Universe has passed through three phases. The very early Universe, which is still poorly understood, was the split second in which the Universe was so hot that particles had energies higher than those currently accessible in particle accelerators on Earth. Therefore, while the basic features of this epoch have been worked out in the Big Bang theory, the details are largely based on educated guesses. Following this, in the early Universe, the evolution of the Universe proceeded according to known high energy physics. This is when the first protons, electrons and neutrons formed, then nuclei and finally atoms. With the formation of neutral hydrogen, the cosmic microwave background was emitted. Finally, the epoch of structure formation began, when matter started to aggregate into the first stars and quasars, and ultimately galaxies, clusters of galaxies and superclusters formed. The future of the Universe is not yet firmly known, but according to the ΛCDM model it will continue expanding forever. ## Areas of study Below, some of the most active areas of inquiry in cosmology are described, in roughly chronological order. This does not include all of the Big Bang cosmology, which is presented in Timeline of the Big Bang. ### Very early Universe The early, hot Universe appears to be well explained by the Big Bang from roughly 10−33 seconds onwards. But there are several problems. One is that there is no compelling reason, using current particle physics, for the Universe to be flat, homogeneous, and isotropic (see the cosmological principle). Moreover, grand unified theories of particle physics suggest that there should be magnetic monopoles in the Universe, which have not been found. These problems are resolved by a brief period of cosmic inflation, which drives the Universe to flatness, smooths out anisotropies and inhomogeneities to the observed level, and exponentially dilutes the monopoles. The physical model behind cosmic inflation is extremely simple, but it has not yet been confirmed by particle physics, and there are difficult problems reconciling inflation and quantum field theory. Some cosmologists think that string theory and brane cosmology will provide an alternative to inflation. Another major problem in cosmology is what caused the Universe to contain more particles than antiparticles. Cosmologists can observationally deduce that the Universe is not split into regions of matter and antimatter. If it were, there would be X-rays and gamma rays produced as a result of annihilation, but this is not observed. This problem is called the baryon asymmetry, and the theory to describe the resolution is called baryogenesis. The theory of baryogenesis was worked out by Andrei Sakharov in 1967, and requires a violation of the particle physics symmetry, called CP-symmetry, between matter and antimatter. However, particle accelerators measure too small a violation of CP-symmetry to account for the baryon asymmetry. Cosmologists and particle physicists look for additional violations of the CP-symmetry in the early Universe that might account for the baryon asymmetry. Both the problems of baryogenesis and cosmic inflation are very closely related to particle physics, and their resolution might come from high energy theory and experiment, rather than through observations of the Universe. ### Big bang nucleosynthesis Main article: Big bang nucleosynthesis Big Bang nucleosynthesis is the theory of the formation of the elements in the early Universe. It finished when the Universe was about three minutes old and its temperature dropped below that at which nuclear fusion could occur. Big Bang nucleosynthesis had a brief period during which it could operate, so only the very lightest elements were produced. Starting from hydrogen ions (protons), it principally produced deuterium, helium-4, and lithium. Other elements were produced in only trace abundances. The basic theory of nucleosynthesis was developed in 1948 by George Gamow, Ralph Asher Alpher, and Robert Herman. It was used for many years as a probe of physics at the time of the Big Bang, as the theory of Big Bang nucleosynthesis connects the abundances of primordial light elements with the features of the early Universe. Specifically, it can be used to test the equivalence principle, to probe dark matter, and test neutrino physics. Some cosmologists have proposed that Big Bang nucleosynthesis suggests there is a fourth "sterile" species of neutrino. ### Cosmic microwave background Main article: Cosmic microwave background The cosmic microwave background is radiation left over from decoupling after the epoch of recombination when neutral atoms first formed. At this point, radiation produced in the Big Bang stopped Thomson scattering from charged ions. The radiation, first observed in 1965 by Arno Penzias and Robert Woodrow Wilson, has a perfect thermal black-body spectrum. It has a temperature of 2.7 kelvins today and is isotropic to one part in 105. Cosmological perturbation theory, which describes the evolution of slight inhomogeneities in the early Universe, has allowed cosmologists to precisely calculate the angular power spectrum of the radiation, and it has been measured by the recent satellite experiments (COBE and WMAP) and many ground and balloon-based experiments (such as Degree Angular Scale Interferometer, Cosmic Background Imager, and Boomerang). One of the goals of these efforts is to measure the basic parameters of the Lambda-CDM model with increasing accuracy, as well as to test the predictions of the Big Bang model and look for new physics. The recent measurements made by WMAP, for example, have placed limits on the neutrino masses. Newer experiments, such as QUIET and the Atacama Cosmology Telescope, are trying to measure the polarization of the cosmic microwave background. These measurements are expected to provide further confirmation of the theory as well as information about cosmic inflation, and the so-called secondary anisotropies, such as the Sunyaev-Zel'dovich effect and Sachs-Wolfe effect, which are caused by interaction between galaxies and clusters with the cosmic microwave background. ### Formation and evolution of large-scale structure Understanding the formation and evolution of the largest and earliest structures (i.e., quasars, galaxies, clusters and superclusters) is one of the largest efforts in cosmology. Cosmologists study a model of hierarchical structure formation in which structures form from the bottom up, with smaller objects forming first, while the largest objects, such as superclusters, are still assembling. One way to study structure in the Universe is to survey the visible galaxies, in order to construct a three-dimensional picture of the galaxies in the Universe and measure the matter power spectrum. This is the approach of the Sloan Digital Sky Survey and the 2dF Galaxy Redshift Survey. Another tool for understanding structure formation is simulations, which cosmologists use to study the gravitational aggregation of matter in the Universe, as it clusters into filaments, superclusters and voids. Most simulations contain only non-baryonic cold dark matter, which should suffice to understand the Universe on the largest scales, as there is much more dark matter in the Universe than visible, baryonic matter. More advanced simulations are starting to include baryons and study the formation of individual galaxies. Cosmologists study these simulations to see if they agree with the galaxy surveys, and to understand any discrepancy. Other, complementary observations to measure the distribution of matter in the distant universe and to probe reionization include: • The Lyman alpha forest, which allows cosmologists to measure the distribution of neutral atomic hydrogen gas in the early Universe, by measuring the absorption of light from distant quasars by the gas. • The 21 centimeter absorption line of neutral atomic hydrogen also provides a sensitive test of cosmology • Weak lensing, the distortion of a distant image by gravitational lensing due to dark matter. These will help cosmologists settle the question of when and how structure formed in the Universe. ### Dark matter Main article: Dark matter Evidence from Big Bang nucleosynthesis, the cosmic microwave background and structure formation suggests that about 23% of the mass of the Universe consists of non-baryonic dark matter, whereas only 4% consists of visible, baryonic matter. The gravitational effects of dark matter are well understood, as it behaves like a cold, non-radiative fluid that forms haloes around galaxies. Dark matter has never been detected in the laboratory, and the particle physics nature of dark matter remains completely unknown. Without observational constraints, there are a number of candidates, such as a stable supersymmetric particle, a weakly interacting massive particle, an axion, and a massive compact halo object. Alternatives to the dark matter hypothesis include a modification of gravity at small accelerations (MOND) or an effect from brane cosmology. ### Dark energy Main article: Dark energy If the Universe is flat, there must be an additional component making up 73% (in addition to the 23% dark matter and 4% baryons) of the energy density of the Universe. This is called dark energy. In order not to interfere with Big Bang nucleosynthesis and the cosmic microwave background, it must not cluster in haloes like baryons and dark matter. There is strong observational evidence for dark energy, as the total energy density of the Universe is known through constraints on the flatness of the Universe, but the amount of clustering matter is tightly measured, and is much less than this. The case for dark energy was strengthened in 1999, when measurements demonstrated that the expansion of the Universe has begun to gradually accelerate. Apart from its density and its clustering properties, nothing is known about dark energy. Quantum field theory predicts a cosmological constant much like dark energy, but 120 orders of magnitude larger than that observed. Steven Weinberg and a number of string theorists (see string landscape) have invoked the 'weak anthropic principle': i.e. the reason that physicists observe a universe with such a small cosmological constant is that no physicists (or any life) could exist in a universe with a larger cosmological constant. Many cosmologists find this an unsatisfying explanation: perhaps because while the weak anthropic principle is self-evident (given that living observers exist, there must be at least one universe with a cosmological constant which allows for life to exist) it does not attempt to explain the context of that universe. For example, the weak anthropic principle alone does not distinguish between: • Only one universe will ever exist and there is some underlying priciple that constrains the CC to the value we observe. • Only one universe will ever exist and although there is no underlying principle fixing the CC, we got lucky. • Lots of universes exist (simultaneously or serially) with a range of CC values, and of course ours is one of the life-supporting ones. Other possible explanations for dark energy include quintessence or a modification of gravity on the largest scales. The effect on cosmology of the dark energy that these models describe is given by the dark energy's equation of state, which varies depending upon the theory. The nature of dark energy is one of the most challenging problems in cosmology. A better understanding of dark energy is likely to solve the problem of the ultimate fate of the Universe. In the current cosmological epoch, the accelerated expansion due to dark energy is preventing structures larger than superclusters from forming. It is not known whether the acceleration will continue indefinitely, perhaps even increasing until a big rip, or whether it will eventually reverse. ### Other areas of inquiry Cosmologists also study: • whether primordial black holes were formed in our universe, and what happened to them. • the GZK cutoff for high-energy cosmic rays, and whether it signals a failure of special relativity at high energies • the equivalence principle, whether or not Einstein's general theory of relativity is the correct theory of gravitation, and if the fundamental laws of physics are the same everywhere in the Universe. ## References 1. For an overview, see George FR Ellis (2006). "Issues in the Philosophy of Cosmology". In Jeremy Butterfield & John Earman. Philosophy of Physics (Handbook of the Philosophy of Science) 3 volume set. North Holland. pp. 1183ff. arXiv:astro-ph/0602280. ISBN 0-444-51560-7. 2. J. V. Narlikar, Introduction to Cosmology, Jones and Bartlett Publishers, Inc., Boston, MA, 1983. 3. "Nobel Prize Biography". Nobel Prize Biography. Nobel Prize. Retrieved 25 February 2011. 4. ^ a b Liddle, A. An Introduction to Modern Cosmology. Wiley. p. 51. ISBN 0-470-84835-9. 5. Vilenkin, Alex (2007). Many worlds in one : the search for other universes. New York: Hill and Wang, A division of Farrar, Straus and Giroux. p. 19. ISBN 978-0-8090-6722-0. 6. Jones, Mark; Lambourne, Robert (2004). An introduction to galaxies and cosmology. Milton Keynes Cambridge, UK New York: Open University Cambridge University Press. p. 228. ISBN 0-521-54623-0. 7. Jones, Mark; Lambourne, Robert (2004). An introduction to galaxies and cosmology. Milton Keynes Cambridge, UK New York: Open University Cambridge University Press. p. 232. ISBN 0-521-54623-0. 8. e.g. Liddle, A. An Introduction to Modern Cosmology. Wiley. ISBN 0-470-84835-9. This argues cogently "Energy is always, always, always conserved." 9. P. Ojeda and H. Rosu (Jun 2006). "Supersymmetry of FRW barotropic cosmologies". Internat. J. Theoret. Phys. (Springer) 45 (6): 1191–1196. arXiv:gr-qc/0510004. Bibcode:2006IJTP...45.1152R. doi:10.1007/s10773-006-9123-2. 10. "Cosmic Detectives". The European Space Agency (ESA). 2013-04-02. Retrieved 2013-04-25. ## Further reading ### Popular • Brian Greene (2005). . Penguin Books Ltd. ISBN 0-14-101111-4. • Alan Guth (1997). The Inflationary Universe: The Quest for a New Theory of Cosmic Origins. Random House. ISBN 0-224-04448-6. • Hawking, Stephen W. (1988). . Bantam Books, Inc. ISBN 0-553-38016-8. • Hawking, Stephen W. (2001). . Bantam Books, Inc. ISBN 0-553-80202-X. • Simon Singh (2005). Big Bang: the origins of the universe. Fourth Estate. ISBN 0-00-716221-9. • Steven Weinberg (1993; 1978). The First Three Minutes. Basic Books. ISBN 0-465-02437-8. ### Textbooks • Cheng, Ta-Pei (2005). Relativity, Gravitation and Cosmology: a Basic Introduction. Oxford and New York: Oxford University Press. ISBN 0-19-852957-0. Introductory cosmology and general relativity without the full tensor apparatus, deferred until the last part of the book. • Dodelson, Scott (2003). Modern Cosmology. Academic Press. ISBN 0-12-219141-2. An introductory text, released slightly before the WMAP results. • Grøn, Øyvind; Hervik, Sigbjørn (2007). Einstein's General Theory of Relativity with Modern Applications in Cosmology. New York: Springer. ISBN 978-0-387-69199-2. • Harrison, Edward (2000). Cosmology: the science of the universe. Cambridge University Press. ISBN 0-521-66148-X. For undergraduates; mathematically gentle with a strong historical focus. • Kutner, Marc (2003). Astronomy: A Physical Perspective. Cambridge University Press. ISBN 0-521-52927-1. An introductory astronomy text. • Kolb, Edward; Michael Turner (1988). The Early Universe. Addison-Wesley. ISBN 0-201-11604-9. The classic reference for researchers. • Liddle, Andrew (2003). An Introduction to Modern Cosmology. John Wiley. ISBN 0-470-84835-9. Cosmology without general relativity. • Liddle, Andrew; David Lyth (2000). Cosmological Inflation and Large-Scale Structure. Cambridge. ISBN 0-521-57598-2. An introduction to cosmology with a thorough discussion of inflation. • Mukhanov, Viatcheslav (2005). Physical Foundations of Cosmology. Cambridge University Press. ISBN 0-521-56398-4. • Padmanabhan, T. (1993). Structure formation in the universe. Cambridge University Press. ISBN 0-521-42486-0. Discusses the formation of large-scale structures in detail. • Peacock, John (1998). Cosmological Physics. Cambridge University Press. ISBN 0-521-42270-1. An introduction including more on general relativity and quantum field theory than most. • Peebles, P. J. E. (1993). Principles of Physical Cosmology. Princeton University Press. ISBN 0-691-01933-9. Strong historical focus. • Peebles, P. J. E. (1980). The Large-Scale Structure of the Universe. Princeton University Press. ISBN 0-691-08240-5. The classic work on large-scale structure and correlation functions. • Rees, Martin (2002). New Perspectives in Astrophysical Cosmology. Cambridge University Press. ISBN 0-521-64544-1. • Weinberg, Steven (1971). Gravitation and Cosmology. John Wiley. ISBN 0-471-92567-5. A standard reference for the mathematical formalism. • Weinberg, Steven (2008). Cosmology. Oxford University Press. ISBN 0-19-852682-2. • Benjamin Gal-Or, “Cosmology, Physics and Philosophy”, Springer Verlag, 1981, 1983, 1987, ISBN 0-387-90581-2, ISBN 0-387-96526-2.	{"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": 3, "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": 0.9117540121078491, "perplexity_flag": "middle"}
http://nrich.maths.org/2275/null	### GOT IT Now For this challenge, you'll need to play Got It! Can you explain the strategy for winning this game with any target? ### Is There a Theorem? Draw a square. A second square of the same size slides around the first always maintaining contact and keeping the same orientation. How far does the dot travel? ### Reverse to Order Take any two digit number, for example 58. What do you have to do to reverse the order of the digits? Can you find a rule for reversing the order of digits for any two digit number? # Picturing Square Numbers ##### Stage: 3 Challenge Level: The sum of consecutive odd numbers can be represented as square arrays. The diagram shows that 1 + 3 + 5 + 7 = 16. What is the sum of the first $30$ odd numbers? What is the sum of the first $60$ odd numbers? Have you a method for working this out quickly? Can you make $3249$ by adding odd numbers in this way? What is the sum of $1 + 3 + ... + 149 + 151 + 153$? What is the value of $83 + 81+ ... + 5 + 3 + 1$? What is the value of $51 + 53 + 55 + ... + 149 + 151 + 153$? What is the sum of $2 + 4 + ... + 150 + 152 + 154$? What is the sum of $2 + 6+ ... + 298+ 302+ 306$? Explain how you worked these out. 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": 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": 0.9117725491523743, "perplexity_flag": "middle"}
http://mathhelpforum.com/math-topics/182383-simple-harmonic-motion-string.html	# Thread: 1. ## simple harmonic motion on a string Hi I need some clarification on this question A light elastic string of natural length 50 cm and modulus 20 N has one end fastened to O on a smooth horizontal table. The other end has a particle of mass 0.4 kg attached to it and the particle is released from rest at a distance 60 cm from O. Find the time it takes to reach O. While the string is stretched T=kx/l =20x/0.5 =40x 40x=-.4a So,a = -100x w=10 T=2pi/10 =pi/5 I get stuck from here. Can anyone help please? ThanksHi I need some clarification on this question A light elastic string of natural length 50 cm and modulus 20 N has one end fastened to O on a smooth horizontal table. The other end has a particle of mass 0.4 kg attached to it and the particle is released from rest at a distance 60 cm from O. Find the time it takes to reach O. While the string is stretched T=kx/l =20x/0.5 =40x 40x=-.4a So,a = -100x w=10 T=2pi/10 =pi/5 I get stuck from here. Can anyone help please? Thanks 2. It would help if you said what "T", "w", etc. mean! I am going to guess that it is the period for the motion-the time required for the string to move from "all the way up" to "all the way down", then back to its starting point. The time required to go from "all the way up" to the central location is 1/4 of T. 3. Originally Posted by HallsofIvy It would help if you said what "T", "w", etc. mean! I am going to guess that it is the period for the motion-the time required for the string to move from "all the way up" to "all the way down", then back to its starting point. The time required to go from "all the way up" to the central location is 1/4 of T. Sincere apologies. 1st T is Tension and the second is the period. w is angular frequency. Thanks. Thats quite clear. How do I proceed from there please? It seems the total time would include the period and the time to the central location. Is that right. Thanks. 4. spring constant, $k = \frac{20 \, N}{0.5 \, m} = 40 \, N/m$ period, $T = 2\pi \sqrt{\frac{m}{k}} = \frac{\pi}{5}$ the time from maximum displacement back to equilibrium is 1/4 of the period ... $t = \frac{\pi}{20} \, s$ 5. Originally Posted by skeeter spring constant, $k = \frac{20 \, N}{0.5 \, m} = 40 \, N/m$ period, $T = 2\pi \sqrt{\frac{m}{k}} = \frac{\pi}{5}$ the time from maximum displacement back to equilibrium is 1/4 of the period ... $t = \frac{\pi}{20} \, s$ Thanks a lot! Sorry for late reply. #### Search Tags View Tag Cloud Copyright © 2005-2013 Math Help Forum. All rights reserved.	{"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": 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": 0.9333308339118958, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/39802/why-does-123-dots-1-over-12/39811	# Why does $1+2+3+\dots = {-1\over 12}$? $$\lim_{N\to\infty} \sum_{i=1}^N \, i = +\infty$$ $\displaystyle\sum_{n=1}^\infty n^{-s}$ only converges to $\zeta(s)$ if $\text{Re}(s) > 1$. Why should analytically continuing to $\zeta(-1)$ give the right answer? - 5 – joriki May 18 '11 at 4:40 16 What does the phrase "right answer" mean? If you define zeta(s) in the usual way for s with real part greater than 1, it has an analytic continuation to s = -1 and the value there is -1/12. That does not mean the series defining zeta(s) for Re(s) > 1 converges at s = -1 with value -1/12 except in a hand-waving Euler kind of way. This question doesn't really have a mathematical meaning as far as I can tell. – KCd May 18 '11 at 4:42 8 See: math.stackexchange.com/questions/37327/infinity-1-paradox, math.stackexchange.com/questions/27526/…, and terrytao.wordpress.com/2010/04/10/… for more information too. – JavaMan May 18 '11 at 4:54 ## 5 Answers Here is a variant on Lubos Motl's answer: Let $S = \sum_{n=1}^{\infty} n$. Then $S - 4 S = \sum_{n = 1}^{\infty} (-1)^{n-1} n.$ We will evaluate this latter expression with a regularization similar to Lubos Motl's. Namely, consider $$\sum_{n=1}^{\infty} (-1)^{n-1} n t^n = -t \dfrac{d}{dt} \sum_{n=1}^{\infty} (-t)^n = -t \dfrac{d}{dt} \dfrac{1}{1+t} = \dfrac{t}{(1+t)^2}.$$ Letting $t \to 1,$ we find that $-3 S = \dfrac{1}{4}$, and hence that $S = \dfrac{-1}{12}.$ To see the relationship between this approach and Lubos Motl's, note that if we write $t = e^{\epsilon},$ then $t\dfrac{d}{dt} = \dfrac{d}{d\epsilon},$ so in fact the arguments are essentially the same, except that Lubos doesn't perform the initial step of replacing $S$ by $S - 4S$, which means that he has the pole $\dfrac{1}{\epsilon^2}$ which he then subtracts away. As far as I know, this trick of replacing $\zeta(s)$ by $(1-2^{-s+1})^{-1}\zeta(s)$ is due to Euler, and it is a now standard method for replacing $\zeta(s)$ by a function which carries the same information, but does not have a pole at $s = 1$. The evaluation of $\zeta(s)$ at negative integers by passing to $(1-2^{-s+1})\zeta(s)$ and then performing Abelian regularization as above is also due to Euler, I believe. It is easy to see the Bernoulli numbers appearing in this way, for example. Of course, taken literally, the series $\sum_{n=1}^{\infty} n$ diverges to $+\infty$, so any attempt to assign it a finite value will involve some form of regularization. Analytic continuation of the $\zeta$-function is one form of regularization, and the Abelian regularization that Lubos Motl and I are making is another. I can't quote a precise theorem to this effect (although maybe others can), but with such a simple expression as $\sum_{n = 1}^{\infty} n,$ I'm reasonably confident that any sensible regularization will necessarily yield the same value of $-1/12$. (Lubos Motl makes the same assertion in his answer.) - 3 @Matt: My first example was flawed, but I think my general point still stands that a satisfactory theory of summation of divergent series would have to explain not only why several seemingly different methods yield the same result, but also why other methods don't. Here's a new example (hopefully less flawed): $S-2S=1+3+5+\dotso$ Now we can subtract $2S$ again, and depending on whether we start substracting the $2$ from the $3$ or the $1$, we get $1+1+\dotso$ or $-1-1-\dotso$ First, these should be different. Also, $1+1+\dotso=\zeta(0)=-1/2$, so that would yield $S-4S=-1/2$, $S=1/6$. – joriki May 18 '11 at 6:57 2 @Matt: (I'm not saying that this can't be explained, or that the results that do coincide do so by chance; to the contrary, I find it intriguing how many of them coincide; all I'm saying is that the ones that don't coincide are also part of the picture, and a satisfactory theory (one that would in my opinion allow us to speak of "the right answer") would have to have something to say about which of these methods are "valid", which aren't and why.) – joriki May 18 '11 at 7:31 17 @Luboš: Instead of insisting on using terms like "simple errors" "in reality" and "wrong interpretations", it would be more helpful if you could specifically point out in which sense you believe that your calculation of $S-2S-2S$ has more merit than mine. As I emphasized, it seems quite likely to me that there is such a difference; but I was looking for an explanation. How did I "randomly clump" terms in my calculation? In which framework does your combination of the terms appear more systematic than mine? (Not rhetorical questions.) – joriki May 18 '11 at 9:55 6 @joriki: Dear Joriki, I deleted my comment that was made in reference to your deleted comment. As for your revised example (obtaining $\zeta(0)$), there is a "graded" aspect to $S$, which is being disturbed in your computation. (This is what Lubos is referring to when he writes that you "randomly clump" terms.) In terms of Lubos's answer, this disturbing of the grading is not allowed because of dimensional considerations. Mathematically, all I can think to say right now is that both $\zeta$-regularization and Euler regularization involve the graded aspect of $S$, and preserve it. Regards, – Matt E May 18 '11 at 11:58 1 @joriki This is exactly why I think divergent series are interesting. I think people sense that the subject has to be founded on more rigorous foundations so they are more comfortable with approaching them. – Max Muller May 19 '11 at 12:53 show 3 more comments there are many ways to see that your result is the right one. What does the right one mean? It means that whenever such a sum appears anywhere in physics - I explicitly emphasize that not just in string theory, also in experimentally doable measurements of the Casimir force (between parallel metals resulting from quantized standing electromagnetic waves in between) - and one knows that the result is finite, the only possible finite part of the result that may be consistent with other symmetries of the problem (and that is actually confirmed experimentally whenever it is possible) is equal to $-1/12$. It's another widespread misconception (see all the incorrect comments right below your question) that the zeta-function regularization is the only way how to calculate the proper value. Let me show a completely different calculation - one that is a homework exercise in Joe Polchinski's "String Theory" textbook. Exponential regulator method Add an exponentially decreasing regulator to make the sum convergent - so that the sum becomes $$S = \sum_{n=1}^{\infty} n e^{-\epsilon n}$$ Note that this is not equivalent to generalizing the sum to the zeta-function. In the zeta-function, the $n$ is the base that is exponentiated to the $s$th power. Here, the regulator has $n$ in the exponent. Obviously, the original sum of integers is obtained in the $\epsilon\to 0$ limit of the formula for $S$. In physics, $\epsilon$ would be viewed as a kind of "minimum distance" that can be resolved. The sum above may be exactly evaluated and the result is (use Mathematica if you don't want to do it yourself, but you can do it yourself) $$S = \frac{e^\epsilon}{(e^\epsilon-1)^2}$$ We will only need some Laurent expansion around $\epsilon = 0$. $$S = \frac{1+\epsilon+\epsilon^2/2 + O(\epsilon^3)}{(\epsilon+\epsilon^2/2+\epsilon^3/6+O(\epsilon^4))^2}$$ We have $$S = \frac{1}{\epsilon^2} \frac{1+\epsilon+\epsilon^2/2+O(\epsilon^3)}{(1+\epsilon/2+\epsilon^2/6+O(\epsilon^3))^2}$$ You see that the $1/\epsilon^2$ leading divergence survives and the next subleading term cancels. The resulting expansion may be calculated with this Mathematica command `1/epsilon^2 * Series[epsilon^2 Sum[n Exp[-n epsilon], {n, 1, Infinity}], {epsilon, 0, 5}]` and the result is $$\frac{1}{\epsilon^2} - \frac{1}{12} + \frac{\epsilon^2}{240} + O(\epsilon^4)$$ In the $\epsilon\to 0$ limit we were interested in, the $\epsilon^2/240$ term as well as the smaller ones go to zero and may be erased. The leading divergence $1/\epsilon^2$ may be and must be canceled by a local counterterm - a vacuum energy term. This is true for the Casimir effect in electromagnetism (in this case, the cancelled pole may be interpreted as the sum of the zero-point energies in the case that no metals were bounding the region), zero-point energies in string theory, and everywhere else. The cancellation of the leading divergence is needed for physics to be finite - but one may guarantee that the counterterm won't affect the finite term, $-1/12$, which is the correct result of the sum. In physics applications, $\epsilon$ would be dimensionful and its different powers are sharply separated and may be treated individually. That's why the local counterterms may eliminate the leading divergence but don't affect the finite part. That's also why you couldn't have used a more complex regulator, like $\exp(-(\epsilon+\epsilon^2)n)$. There are many other, apparently inequivalent ways to compute the right value of the sum. It is not just the zeta function. Euler's method Let me present one more, slightly less modern, method that was used by Leonhard Euler to calculate that the sum of integers is $-1/12$. It's of course a bit more heuristic but his heuristic approach showed that he had a good intuition and the derivation could be turned into a modern physics derivation, too. We will work with two sums, $$S = 1+2+3+4+5+\dots, \quad T = 1-2+3-4+5-\dots$$ Extrapolating the geometric and similar sums to the divergent (and, in this case, marginally divergent) domain of values of $x$, the expression $T$ may be summed according to the Taylor expansion $$\frac{1}{(1+x)^2} = 1 - 2x + 3x^2 -4x^3 + \dots$$ Substitute $x=1$ to see that $T=+1/4$. The value of $S$ is easily calculated now: $$T = (1+2+3+\dots) - 2\times (2+4+6+\dots) = (1+2+3+\dots) (1 - 4) = -3S$$ so $S=-T/3=-1/12$. A zeta-function calculation A somewhat unusual calculation of $\zeta(-1)=-1/12$ of mine may be found here: http://www.kolej.mff.cuni.cz/~lmotm275/RUZE/09/node7.html The comments are in Czech but the equations represent bulk of the language that really matters, so the Czech comments shouldn't be a problem. A new argument (subscript) $s$ is added to the zeta function. The new function is the old zeta function for $s=0$ and for $s=1$, it only differs by one. We Taylor expand around $s=0$ to get to $s=1$ and we find out that only a finite number of terms survives if the main argument $x$ is a non-positive integer. The resulting recursive relations for the zeta function allow us to compute the values of the zeta-function at integers smaller than $1$, and prove that the function vanishes at negative even values of $x$. - 29 The fact that other methods also yield the result $-\frac{1}{12}$ doesn't make any of the comments under the question "incorrect". You interpreted "right answer" to mean the one that works in physical applications; that's fine (and I suggested in my comment that it was probably meant that way), but that doesn't make it the right answer in any mathematical sense of the term, and nor does the consistency of several summation methods. I'm not aware of any theory of summation of divergent series that provides a definition of what "the right answer" for a resummation is. – joriki May 18 '11 at 6:04 22 @Luboš: It seems we have a difference of opinion about the use of "right" in this context. I don't object to you using "right" in the sense that you're using it; what I object to is labelling other views as "incorrect". To call something "incorrect", you'd need a definition of what makes an answer the right answer. As long as you don't have one, it's a matter of taste that you call the consistent result of several different methods "the right answer" and I don't. – joriki May 18 '11 at 6:47 21 @Luboš: I object to your combative tone. None of the comments under the question "lead to incorrect conclusions in all of physics". Noone here has questioned the fact that $-1/12$ is the answer that works in physics. There is merely a difference of opinion about whether it should be called "the right answer". I accept that you call it that, and I'd appreciate if you would accept that I don't, instead of labelling that as "shallow". – joriki May 18 '11 at 7:26 29 @Luboš: I too object to your label of comments by others as "shallow". Look at the question: it asks specifically about analytically continuing and writes the value as $\zeta(-1)$. That is what I was responding to in my comment. I am well aware that there are other methods of "deriving" the value -1/12 and I had sketched Euler's technique a while ago on Mathoverflow (see my 2nd answer to mathoverflow.net/questions/13130/…). You answered a better question than the one that was asked. It's not a reason to say more focused answers are wrong. – KCd May 18 '11 at 7:59 16 @Luboš: OK, if that's how you see it, I don't think there's any point in exchanging further arguments. I would just suggest that you look at the upvotes for the comments both here and under the question. Truth isn't democratic, but the numbers might motivate you to consider the possibility that there is more than one valid viewpoint in this case. – joriki May 18 '11 at 10:09 show 5 more comments What a great method! I tried a few... $$\sum_{n = 1}^{\infty} \left(\frac{1}{(2 n)^{s}} - \frac{1}{(2 n - 1)^{s}}\right) = \zeta (s) \Bigl(2^{(1 - s)} - 1\Bigr)$$ put $s=-1$ to get $\sum_{n = 1}^{\infty} 1 = -\frac{1}{4}$ or $$\sum_{n = 1}^{\infty} \left(\frac{1}{(2 n + 1)^{s}} - \frac{1}{(2 n)^{s}}\right) = \zeta (s) - \frac{2 \zeta (s)}{2^{s}} - 1$$ put $s=-1$ to get $\sum_{n = 1}^{\infty} 1 = -\frac{3}{4}$ - You should read Terence Tao's article on this "oddities". – Peter Tamaroff Jun 11 '12 at 17:37 Take a look at Ramanujan summation article at wikipedia. - The notation "$1+2+3+\cdots$" is as meaningless as "$1/0$". If you treat such notation as though it defined a real number and conformed in its syntax to the rules of formation for genuine real numbers, you can easily "prove" it to equal any number you like, including $-1/12$. - 1 Dear John, Your claim that this notation is "meanlingless" seems unnecessarily absolute, in light of the answers explaining that it does in fact admit meaningful interpretations. Regards, – Matt E Sep 8 '12 at 19:11 Dear Matt: Sure, it admits meaningful interpretations, for some people. But they are not consistent. – John Bentin Sep 9 '12 at 12:03	{"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": 82, "mathjax_display_tex": 12, "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": 0.9583238959312439, "perplexity_flag": "head"}
http://mathhelpforum.com/calculus/170575-lines-planes.html	# Thread: 1. ## [SOLVED] Lines and Planes? It says, Find the equation of the plane in xyz-space through the point P = (3, 5, 3) and perpendicular to the vector n = (-5, -5, -4). Can someone show me, step by step how to work this problem out or explain to me what I'm doing? I'm so lost. 2. Originally Posted by Bracketology It says, Find the equation of the plane in xyz-space through the point P = (3, 5, 3) and perpendicular to the vector n = (-5, -5, -4). Can someone show me, step by step how to work this problem out or explain to me what I'm doing? I'm so lost. Let $R=<x,y,z>-<3.5,3>=<x-3,y-5,z-3>$ Then the plane is $n\cdot R=0.$ 3. Thanks, figured it out	{"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": 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": 0.9310818910598755, "perplexity_flag": "head"}
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http://physics.stackexchange.com/questions/38420/what-is-crystal-field-anisotropy-or-effect-it-forces-the-magnetic-moment-to-po?answertab=oldest	# What is crystal field anisotropy or effect ? It forces the magnetic moment to point in particular local direction.. Can you give a basic explanation of what is crystal field anisotropy ? What is the reason to arise ? In spin ice it forces the dipoles to point in the local 111 direction. For partially filled rare earth atoms hund rule requires S and L max. This leaves (2s +1)(2l+1) degeneracy which is partially lifted from the LS coupling. When inserted in anisotropic field inside crystal the expectation value , or mean value of L is 0 = 0 and the L is quenched leaving only S so is forced to point in some local direction, but this is if the field removes the deneracy ? Is it this or it is much more complicated ? - ## 1 Answer The $p$, $d$ and $f$ orbitals are only degenerate in a central potential. If you make the potential non-central, for example by putting the atom in an external magnetic field the energy levels are no longer degenerate. Putting the atom into a crystal is just another way of creating an non-central potential (though due to electostatic forces rather than magnetic). This is perhaps more clearly seen in the ligand field splitting you get when complexing your atom with suitable ligands. I'm not sure if that's all you were asking or if there's something else. If I've missed anything perhaps you could edit your question accordingly. -	{"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": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9341852068901062, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/285969/allowing-addition-of-ordinals-in-forcing	# Allowing addition of ordinals in forcing I'm reading Paul Cohen's "The Discovery of Forcing", this is a question related to my previous question: Why did Cohen require forcing to be such that no new ordinals are added in the process? Or, put differently: if we don't care and construct an extension $M[G]$ of a model $M$ of ZFC in such a way that there might be ordinals added, is the consequence that $M[G]$ does not satisfy ZFC? Thank you for your help. - For your second question, it would depend how you constructed the second model. If you "don't care" then I expect many bad things are possible :) – Trevor Wilson Jan 24 at 17:39 ## 1 Answer One of the key points of forcing is that you want to control "internally" the theory of the final model, so that starting from an (original) ground model $M$ you can predict what statements the resulting model $M'=M[G]$ will satisfy. A priori, it could well be that the resulting structure $M'$ we build is not even an end-extension of $M$, meaning that we could have, for example, new sets $c\in M'$ with $M'\models$"$c\in b$" for some $b\in M$. Or, it could be that $M'$ is no longer well-founded. Something like this would do the task of "predicting" $M'$'s theory unmanageable, since the key tool we use for this, absoluteness, would no longer be available. This suggests that we want $M'$ to be an end-extension, and (Cohen's key insight) we want it to be "standard", that is, transitive. (By the time forcing appeared, we had a fair amount of model theoretic tools to produce new models from old ones, but none of these tools allow us to control the theory of the final model in any meaningful way, which is why we did not have proofs of the consistency of $\lnot \mathsf{CH}$ prior to Cohen, for example.) But, if we were to add ordinals to $M'$, this would force $M'$ to be significantly larger than $M$: For example, if $\alpha=\mathsf{ORD}^M\in M'$, then $\beta=\alpha^{\alpha^{\alpha^{\dots}}}$ (ordinal exponentiation) would also have to be an element of $M'$. Actually, this ordinal, though immensely larger than $\alpha$, would dwarf when compared with $\mathsf{ORD}^{M'}$. For example, $M'$ would have a version of $\alpha^+$, which is much much larger than $\beta$ already, and it would have a version of fixed points of the aleph function above $\alpha^+$, and so on. In fact, $\mathsf{ORD}^{M'}$ would be unreachable from within $M$ in any practical sense, and we would have no means to predict the theory of $M'$ -- we wouldn't even have enough (definable) classes to use as names for the elements of $M'$! (There is a minor remark needed here: All this applies, even if the resulting $M'$ is still countable, because, from within $M$, there are no enumerations of $M$ itself, much less of $M'$.) An interesting direction in the model theory of set theory prior to Cohen was the study of chains of end-extensions; there are some classical nice results by Keisler, Morley, and others, and more recent work by Villaveces, and Hamkins, for example. Prior to Cohen, it was expected that an end-extension $M'$ of $M$ would in fact not add any sets of rank below $\alpha$, so it was expected $M'$ would be much larger than $M$, as explained above. Typical work on chains of end-extensions asks for $M'$ to be an elementary superstructure of $M$. This is certainly useless if our goal is to show relative consistency results, as $M'$ has the same theory as that of $M$. On the other hand, this setting is not completely superseded by forcing, since it is useful in the study of large cardinals of intermediate size, such as weakly compact cardinals. (A digression: It is an interesting question whether we can show the existence of "many" transitive models by means other than forcing, but I do not know of any methods to achieve this. See this MO question.) Let me close by saying that, if we take the idea that we want $M'$ to be transitive seriously, then we really ought to look for ways of achieving this while not adding any ordinals. For example, it could be that in our universe there is exactly one ordinal $\alpha$ such that there are transitive models $M$ of set theory of height $\alpha$. But then, short of making $M'$ have the size of the whole universe, there are not many other options. Of course, typically, we are in situations where we have plenty of ordinals $\alpha$ that are the height of a model of set theory, but we still want forcing to be a fairly local operation, so that if $M\in N$ are models, with $N\models$"$M$ is countable", then in $N'$ there are forcing extensions $M'$ of $M$. But then this will again force us to have $M'$ of the same height as $M$, as we could take $N$ to be of least possible height containing $M$ as an element. The other extreme is that we argue in the language of Boolean-valued models, and consider "virtual" forcing extensions of the whole universe, instead of talking of countable transitive models. Trying to formalize the required setting, so that we can in $V$ talk about virtual larger ordinals, is quite involved. I do not know of much serious work here. Reinhardt had some proposals, in terms of strong elementary embeddings, but I haven't seen any work that would make this setting feasible for a theory that would replace forcing. - Thank you! Actually, I had figured out that it was meant to be "end extension" after some googling. What I have not figured out and I'm currently thinking about is why we want the extension to be an end extension. What properties fail or cannot be guaranteed if it is not an end extension? – Matt N. Jan 25 at 20:22 (I don't yet see what happens if new sets as you mention in your answer are allowed) – Matt N. Jan 25 at 20:23 2 What happens is that absoluteness fails atrociously. For example, the empty set of $M$ may fail to be the empty set of $M'$. Absoluteness gives us an extensive library of basic terms that have the same meaning regardless of whether we are in $M$ or $M'$. Without it, we have no a priori tool to check the truth value of even the most basic of statements in $M'$. It may well be that $M'$ is still a nice model, but $M$ would not be able to predict much of it, or even if it could in some ad hoc situations, it would not be anything uniform that would apply every time. (Unmanageable in practice.) – Andres Caicedo Jan 25 at 20:29 Thank you. I still seem to have one piece missing to make a whole: I'm trying to work out why we want to control the outcome internally, that is, verify that the axioms of ZFC hold in $M[G]$ but verify them from within $M$. (Looking at "Generic Model Theorem" in Halbeisen on page 285 it looks to me as if this verification is done using $P$ names. If I read your sentence "Without it, we have no a priori tool to check the truth value of even the most basic of statements in M′." correctly this does not contradict you because $P$ names are a posteriori.) – Matt N. Jan 26 at 17:15	{"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": 72, "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": 0.9672276377677917, "perplexity_flag": "head"}
http://mathoverflow.net/revisions/7671/list	## Return to Answer 2 Included proof of the estimate on f.; added 4 characters in body; added 2 characters in body I have carried out the suggestion in the last paragraph of Yemon Choi's answer. Choose $\phi\in C^\infty(\mathbb{R})$, $\phi\ge0$ and $\int_{\mathbb{R}}\phi(x)dx=1$, and let $\phi_R(x)=R\phi(Rx)$. Define $$f=\phi_R\star\eta,\quad g=\eta-f.$$ Then it is easy to see that $$\\|f\\|{C^n}=O(R^{n-\alpha}),\quad \\|g\\|\|f\\|_{C^n}=O(R^{n-\alpha}),\quad \infty=O(R^{-\alpha}),$$ \|g\\|_\infty=O(R^{-\alpha}), but this is not what you are asking for. My feeling is that the constant $C$ must show some dependence on $n$. In response to your last comment, let me prove the estimate on $\\|f\\|_{C^n}$. We have $$f^{(n)}=(\phi_R)^{(n)}\star\eta=R^n(\phi^{(n)})_R\star\eta.$$ Since $(\phi^{(n)})_R$ has mean zero, for any $x\in\mathbb{R}$: $$\|f^{(n)}(x)\|\le R^n\int_{\mathbb{R}}\|\phi^{(n)}(y)\|\|\eta(x-\frac{y}{R})-\eta(x)\|dy\le HR^{n-\alpha}\int_{\mathbb{R}}\|\phi^{(n)}(y)\|\|y\|^\alpha dy,$$ where $H$ is $\eta$'s Hölder constant. 1 I have carried out the suggestion in the last paragraph of Yemon Choi's answer. Choose $\phi\in C^\infty(\mathbb{R})$, $\phi\ge0$ and $\int_{\mathbb{R}}\phi(x)dx=1$, and let $\phi_R(x)=R\phi(Rx)$. Define $$f=\phi_R\star\eta,\quad g=\eta-f.$$ Then it is easy to see that $$\\|f\\|{C^n}=O(R^{n-\alpha}),\quad \\|g\\|\infty=O(R^{-\alpha}),$$ but this is not what you are asking for. My feeling is that the constant $C$ must show some dependence on $n$.	{"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": 18, "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": 0.9413461685180664, "perplexity_flag": "head"}
http://mathoverflow.net/questions/46641/notation-for-bilinear-form-yt-m-z-where-m-is-a-matrix-and-y-z-are-vectors	## Notation for bilinear form $y^t M z$, where $M$ is a matrix and $y,z$ are vectors. ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I'm working on a problem where I need to consider a bilinear form of the form $y^t M z$ where $M$ is an $n$-by-$n$ real symmetric matrix and `$y,z \in \mathbb{R}^n$` are vectors. I also need to consider restricted forms of such a product, of the form $$\sum_{i,j=1}^n y_i m_{ij} z_j \mathbf{I}_{(y_i,z_j) \in E},$$ where $E$ is some subset of $\mathbb{R}^2$. We recover $y^t M z$ by taking $E=\mathbb{R}^2$. I want a common notation for $y^t M z$ and for these restricted sums, so I have been writing $y^t M z = \langle y,z\rangle_M$, and writing $\langle y,z\rangle_{M,E}$ for the restricted sum above. I see nothing wrong with the notation I'm using. However, if there's a standard notation for such things that I am unaware of, I would like to know about it. Is there? If you know of another notation, can you give me a reference? - I think it would be better (more standard) to call these bilinear forms rather than inner products. Also, did you really mean $R^2$? And what is the meaning of the bold-face I? I think that the pointed bracket notation may be misleading since it might make one expect it to be symmetric---or do you mean to restrict the matrix $M$ to be symmetric? – Dick Palais Nov 19 2010 at 17:52 Thanks for the suggestions, I will modify my question accordingly! Yes in my situation the matrix is in fact symmetric, I will mention this. – Louigi Addario-Berry Nov 19 2010 at 18:03 Also: the boldface I is meant to be the indicator function of the set E. – Louigi Addario-Berry Nov 19 2010 at 18:05 Note that a real inner product is usually defined to be non-degenerate, i.e. the matrix $M$ has to be non-singular. If that's not always the case in your situation, then the notation $\langle,\rangle_M$ is misleading. And certainly, the notation $\langle,\rangle_{M,E}$ seems misleading to me since these guys are degenerate. For the former, I would prefer "bilinear form $B(y,z)$", but for the latter, even that would be misleading, since it is not bi-linear. – Alex Bartel Nov 20 2010 at 5:01 Thanks Alex. The point of my question is to have a consistent notation for both the former and the latter, do you have a thought about what one might use in this case? – Louigi Addario-Berry Nov 20 2010 at 10:44 show 1 more comment	{"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": 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": 0.9572319984436035, "perplexity_flag": "head"}
http://en.wikipedia.org/wiki/Group_scheme	# Group scheme Group theory Group homomorphisms additive cyclic abelian dihedral • Integers (Z) • Lattice Modular groups • PSL(2,Z) • SL(2,Z) • General linear GL(n) • Special linear SL(n) • Orthogonal O(n) • Euclidean E(n) • Special orthogonal SO(n) • Unitary U(n) • Special unitary SU(n) • Symplectic Sp(n) Infinite dimensional Lie group • O(∞) • SU(∞) • Sp(∞) In mathematics, a group scheme is a type of algebro-geometric object equipped with a composition law. Group schemes arise naturally as symmetries of schemes, and they generalize algebraic groups, in the sense that all algebraic groups have group scheme structure, but group schemes are not necessarily connected, smooth, or defined over a field. This extra generality allows one to study richer infinitesimal structures, and this can help one to understand and answer questions of arithmetic significance. The category of group schemes is somewhat better behaved than that of group varieties, since all homomorphisms have kernels, and there is a well-behaved deformation theory. Group schemes that are not algebraic groups play a significant role in arithmetic geometry and algebraic topology, since they come up in contexts of Galois representations and moduli problems. The initial development of the theory of group schemes was due to Alexandre Grothendieck, Michel Raynaud and Michel Demazure in the early 1960s. ## Definition A group scheme is a group object in a category of schemes that has fiber products and some final object S. That is, it is an S-scheme G equipped with one of the equivalent sets of data • a triple of morphisms μ: G ×S G → G, e: S → G, and ι: G → G, satisfying the usual compatibilities of groups (namely associativity of μ, identity, and inverse axioms) • a functor from schemes over S to the category of groups, such that composition with the forgetful functor to sets is equivalent to the presheaf corresponding to G under the Yoneda embedding. A homomorphism of group schemes is a map of schemes that respects multiplication. This can be precisely phrased either by saying that a map f satisfies the equation fμ = μ(f × f), or by saying that f is a natural transformation of functors from schemes to groups (rather than just sets). A left action of a group scheme G on a scheme X is a morphism G ×S X→ X that induces a left action of the group G(T) on the set X(T) for any S-scheme T. Right actions are defined similarly. Any group scheme admits natural left and right actions on its underlying scheme by multiplication and conjugation. Conjugation is an action by automorphisms, i.e., it commutes with the group structure, and this induces linear actions on naturally derived objects, such as its Lie algebra, and the algebra of left-invariant differential operators. An S-group scheme G is commutative if the group G(T) is an abelian group for all S-schemes T. There are several other equivalent conditions, such as conjugation inducing a trivial action, or inversion map ι being a group scheme automorphism. ## Constructions • Given a group G, one can form the constant group scheme GS. As a scheme, it is a disjoint union of copies of S, and by choosing an identification of these copies with elements of G, one can define the multiplication, unit, and inverse maps by transport of structure. As a functor, it takes any S-scheme T to a product of copies of the group G, where the number of copies is equal to the number of connected components of T. GS is affine over S if and only if G is a finite group. However, one can take a projective limit of finite constant group schemes to get profinite group schemes, which appear in the study of fundamental groups and Galois representations or in the theory of the fundamental group scheme, and these are affine of infinite type. More generally, by taking a locally constant sheaf of groups on S, one obtains a locally constant group scheme, for which monodromy on the base can induce non-trivial automorphisms on the fibers. • The existence of fiber products of schemes allows one to make several constructions. Finite direct products of group schemes have a canonical group scheme structure. Given an action of one group scheme on another by automorphisms, one can form semidirect products by following the usual set-theoretic construction. Kernels of group scheme homomorphisms are group schemes, by taking a fiber product over the unit map from the base. Base change sends group schemes to group schemes. • Group schemes can be formed from smaller group schemes by taking restriction of scalars with respect to some morphism of base schemes, although one needs finiteness conditions to be satisfied to ensure representability of the resulting functor. When this morphism is along a finite extension of fields, it is known as Weil restriction. • For any abelian group A, one can form the corresponding diagonalizable group D(A), defined as a functor by setting D(A)(T) to be the set of abelian group homomorphisms from A to invertible global sections of OT for each S-scheme T. If S is affine, D(A) can be formed as the spectrum of a group ring. More generally, one can form groups of multiplicative type by letting A be a non-constant sheaf of abelian groups on S. • For a subgroup scheme H of a group scheme G, the functor that takes an S-scheme T to G(T)/H(T) is in general not a sheaf, and even its sheafification is in general not representable as a scheme. However, if H is finite, flat, and closed in G, then the quotient is representable, and admits a canonical left G-action by translation. If the restriction of this action to H is trivial, then H is said to be normal, and the quotient scheme admits a natural group law. Representability holds in many other cases, such as when H is closed in G and both are affine.[1] ## Examples • The multiplicative group Gm has the punctured affine line as its underlying scheme, and as a functor, it sends an S-scheme T to the multiplicative group of invertible global sections of the structure sheaf. It can be described as the diagonalizable group D(Z) associated to the integers. Over an affine base such as Spec A, it is the spectrum of the ring A[x,y]/(xy − 1), which is also written A[x, x−1]. The unit map is given by sending x to one, multiplication is given by sending x to x ⊗ x, and the inverse is given by sending x to x−1. Algebraic tori form an important class of commutative group schemes, defined either by the property of being locally on S a product of copies of Gm, or as groups of multiplicative type associated to finitely generated free abelian groups. • The general linear group GLn is an affine algebraic variety that can be viewed as the multiplicative group of the n by n matrix ring variety. As a functor, it sends an S-scheme T to the group of invertible n by n matrices whose entries are global sections of T. Over an affine base, one can construct it as a quotient of a polynomial ring in n2 + 1 variables by an ideal encoding the invertibility of the determinant. Alternatively, it can be constructed using 2n2 variables, with relations describing an ordered pair of mutually inverse matrices. • For any positive integer n, the group μn is the kernel of the nth power map from Gm to itself. As a functor, it sends any S-scheme T to the group of global sections f of T such that fn = 1. Over an affine base such as Spec A, it is the spectrum of A[x]/(xn−1). If n is not invertible in the base, then this scheme is not smooth. In particular, over a field of characteristic p, μp is not smooth. • The additive group Ga has the affine line A1 as its underlying scheme. As a functor, it sends any S-scheme T to the underlying additive group of global sections of the structure sheaf. Over an affine base such as Spec A, it is the spectrum of the polynomial ring A[x]. The unit map is given by sending x to zero, the multiplication is given by sending x to 1 ⊗ x + x ⊗ 1, and the inverse is given by sending x to −x. • If p = 0 in S for some prime number p, then the taking of pth powers induces an endomorphism of Ga, and the kernel is the group scheme αp. As a scheme, it is isomorphic to μp, but the group structures are different. Over an affine base such as Spec A, it is the spectrum of A[x]/(xp). • The automorphism group of the affine line is isomorphic to the semidirect product of Ga by Gm, where the additive group acts by translations, and the multiplicative group acts by dilations. The subgroup fixing a chosen basepoint is isomorphic to the multiplicative group, and taking the basepoint to be the identity of an additive group structure identifies Gm with the automorphism group of Ga. • A smooth genus one curve with a marked point (i.e., an elliptic curve) has a unique group scheme structure with that point as the identity. Unlike the previous positive-dimensional examples, elliptic curves are projective (in particular proper). ## Basic properties When working over a field, one often can analyze a group scheme by treating it as an extension of group schemes with distinguished properties. Any group scheme G of finite type is an extension of the connected component of the identity (i.e., the maximal connected subgroup scheme) by a constant group scheme. If G is connected, then it has a unique maximal reduced subscheme Gred, which is a smooth group variety that is a normal subgroup of G. The quotient group Ginf is the infinitesimal quotient, and is the spectrum of a local Hopf algebra of finite rank. Any affine group scheme is the spectrum of a commutative Hopf algebra (over a base S, this is given by the relative spectrum of an OS-algebra). The multiplication, unit, and inverse maps of the group scheme are given by the comultiplication, counit, and antipode structures in the Hopf algebra. The unit and multiplication structures in the Hopf algebra are intrinsic to the underlying scheme. For an arbitrary group scheme G, the ring of global sections also has a commutative Hopf algebra structure, and by taking its spectrum, one obtains the maximal affine quotient group. Affine group varieties are known as linear algebraic groups, since they can be embedded as subgroups of general linear groups. Complete connected group schemes are in some sense opposite to affine group schemes, since the completeness implies all global sections are exactly those pulled back from the base, and in particular, they have no nontrivial maps to affine schemes. Any complete group variety (variety here meaning reduced and geometrically irreducible separated scheme of finite type over a field) is automatically commutative, by an argument involving the action of conjugation on jet spaces of the identity. Complete group varieties are called abelian varieties. This generalizes to the notion of abelian scheme; a group scheme G over a base S is abelian if the structural morphism from G to S is proper and smooth with geometrically connected fibers They are automatically projective, and they have many applications, e.g., in geometric class field theory and throughout algebraic geometry. A complete group scheme over a field need not be commutative, however; for example, any finite group scheme is complete. ## Finite flat group schemes A group scheme G over a noetherian scheme S is finite and flat if and only if OG is a locally free OS-module of finite rank. The rank is a locally constant function on S, and is called the order of G. The order of a constant group scheme is equal to the order of the corresponding group, and in general, order behaves well with respect to base change and finite flat restriction of scalars. Among the finite flat group schemes, the constants (cf. example above) form a special class, and over an algebraically closed field of characteristic zero, the category of finite groups is equivalent to the category of constant finite group schemes. Over bases with positive characteristic or more arithmetic structure, additional isomorphism types exist. For example, if 2 is invertible over the base, all group schemes of order 2 are constant, but over the 2-adic integers, μ2 is non-constant, because the special fiber isn't smooth. There exist sequences of highly ramified 2-adic rings over which the number of isomorphism types of group schemes of order 2 grows arbitrarily large. More detailed analysis of commutative finite flat group schemes over p-adic rings can be found in Raynaud's work on prolongations. Commutative finite flat group schemes often occur in nature as subgroup schemes of abelian and semi-abelian varieties, and in positive or mixed characteristic, they can capture a lot of information about the ambient variety. For example, the p-torsion of an elliptic curve in characteristic zero is locally isomorphic to the constant elementary abelian group scheme of order p2, but over Fp, it is a finite flat group scheme of order p2 that has either p connected components (if the curve is ordinary) or one connected component (if the curve is supersingular). If we consider a family of elliptic curves, the p-torsion forms a finite flat group scheme over the parametrizing space, and the supersingular locus is where the fibers are connected. This merging of connected components can be studied in fine detail by passing from a modular scheme to a rigid analytic space, where supersingular points are replaced by discs of positive radius. ## Cartier duality Cartier duality is a scheme-theoretic analogue of Pontryagin duality. Given any finite flat commutative group scheme G over S, its Cartier dual is the group of characters, defined as the functor that takes any S-scheme T to the abelian group of group scheme homomorphisms from the base change GT to Gm,T and any map of S-schemes to the canonical map of character groups. This functor is representable by a finite flat S-group scheme, and Cartier duality forms an additive involutive antiequivalence from the category of finite flat commutative S-group schemes to itself. If G is a constant commutative group scheme, then its Cartier dual is the diagonalizable group D(G), and vice versa. If S is affine, then the duality functor is given by the duality of the Hopf algebras of functions. The definition of Cartier dual extends usefully to much more general situations where the resulting functor on schemes is no longer represented as a group scheme. Common cases include fppf sheaves of commutative groups over S, and complexes thereof. These more general geometric objects can be useful when one wants to work with categories that have good limit behavior. There are cases of intermediate abstraction, such as commutative algebraic groups over a field, where Cartier duality gives an antiequivalence with commutative affine formal groups, so if G is the additive group Ga, then its Cartier dual is the multiplicative formal group $\widehat{\mathbf{G}_m}$, and if G is a torus, then its Cartier dual is étale and torsion-free. For loop groups of tori, Cartier duality defines the tame symbol in local geometric class field theory. Laumon introduced a sheaf-theoretic Fourier transform for quasi-coherent modules over 1-motives that specializes to many of these equivalences. ## Dieudonné modules Main article: Dieudonné module Finite flat commutative group schemes over a perfect field k of positive characteristic p can be studied by transferring their geometric structure to a (semi-)linear-algebraic setting. The basic object is the Dieudonné ring D = W(k){F,V}/(FV − p), which is a quotient of the ring of noncommutative polynomials, with coefficients in Witt vectors of k. F and V are the Frobenius and Verschiebung operators, and they may act nontrivially on the Witt vectors. Dieudonne and Cartier constructed an antiequivalence of categories between finite commutative group schemes over k of order a power of "p" and modules over D with finite W(k)-length. The Dieudonné module functor in one direction is given by homomorphisms into the abelian sheaf CW of Witt co-vectors. This sheaf is more or less dual to the sheaf of Witt vectors (which is in fact representable by a group scheme), since it is constructed by taking a direct limit of finite length Witt vectors under successive Verschiebung maps V: Wn → Wn+1, and then completing. Many properties of commutative group schemes can be seen by examining the corresponding Dieudonné modules, e.g., connected p-group schemes correspond to D-modules for which F is nilpotent, and étale group schemes correspond to modules for which F is an isomorphism. Dieudonné theory exists in a somewhat more general setting than finite flat groups over a field. Oda's 1967 thesis gave a connection between Dieudonné modules and the first de Rham cohomology of abelian varieties, and at about the same time, Grothendieck suggested that there should be a crystalline version of the theory that could be used to analyze p-divisible groups. Galois actions on the group schemes transfer through the equivalences of categories, and the associated deformation theory of Galois representations was used in Wiles's work on the Shimura–Taniyama conjecture. ## References 1. Raynaud, Michel (1967), Passage au quotient par une relation d'équivalence plate, Berlin, New York: Springer-Verlag, MR0232781 • Demazure, Michel; Alexandre Grothendieck, eds. (1970). Séminaire de Géométrie Algébrique du Bois Marie – 1962–64 – Schémas en groupes – (SGA 3) – vol. 1 (Lecture notes in mathematics 151) (in French). Berlin; New York: Springer-Verlag. pp. xv+564. • Demazure, Michel; Alexandre Grothendieck, eds. (1970). Séminaire de Géométrie Algébrique du Bois Marie – 1962–64 – Schémas en groupes – (SGA 3) – vol. 2 (Lecture notes in mathematics 152) (in French). Berlin; New York: Springer-Verlag. pp. ix+654. • Demazure, Michel; Alexandre Grothendieck, eds. (1970). Séminaire de Géométrie Algébrique du Bois Marie – 1962–64 – Schémas en groupes – (SGA 3) – vol. 3 (Lecture notes in mathematics 153) (in French). Berlin; New York: Springer-Verlag. vii+529. • Gabriel, Peter; Demazure, Michel (1980). Introduction to algebraic geometry and algebraic groups. Amsterdam: North-Holland Pub. Co. ISBN 0-444-85443-6. • Berthelot, Breen, Messing Théorie de Dieudonné Crystalline II • Laumon, Transformation de Fourier généralisée • Shatz, Stephen S. (1986), "Group schemes, formal groups, and p-divisible groups", in Cornell, Gary; Silverman, Joseph H., Arithmetic geometry (Storrs, Conn., 1984), Berlin, New York: Springer-Verlag, pp. 29–78, ISBN 978-0-387-96311-2, MR861972 • Serre, Jean-Pierre (1984), Groupes algébriques et corps de classes, Publications de l'Institut Mathématique de l'Université de Nancago [Publications of the Mathematical Institute of the University of Nancago], 7, Paris: Hermann, ISBN 978-2-7056-1264-1, MR907288 • John Tate, Finite flat group schemes, from Modular Forms and Fermat's Last Theorem • Waterhouse, William (1979), Introduction to affine group schemes, Graduate Texts in Mathematics 66, Berlin, New York: Springer-Verlag, ISBN 978-0-387-90421-4	{"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": 1, "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": 0.880343496799469, "perplexity_flag": "head"}
http://mathhelpforum.com/calculus/154162-graphing-curve-both-tangents.html	# Thread: 1. ## Graphing the curve and both tangents... QUESTION: (a) Find the slope of tangent to the curve y=3+4x^2-2x^3 at the point where x=a. (b) Find equations of the tangent lines at the points (1,5) and (2,3). (c) Graph the curve and both tangents on a common screen. My answers: (a) m(TL) = dy/dx = 8x-6x^2 (i) at P(1,5); m(TL) = 2 (ii) at P(2,3); m(TL) = -8 (b) (i) y = 2x +3 (ii) y = -8x+19 (c) now in this part of graphing i don't know how to graph the curve since it is raised to cube..i don't know either what kind of curve is this.. ^_^ 2. A method that always works is to put in points for the x to solve for the y to plot on your graph. Eg. If x = -0.5, then y = 4.25 If x = 0, then y = 3 If x = 0.5, theny = 3.75 And then join up the dots Here's a common image of a cubic graph generated by Wolfram\|Alpha: Your graph should look something like this one, maybe moved up a bit. Then just add the tangent lines to your graph for the 2 tangent lines. 3. oh yeah!.. like test points.. thanks a lot ^_^ 4. Originally Posted by cutiemike1 QUESTION: (a) Find the slope of tangent to the curve y=3+4x^2-2x^3 at the point where x=a. (b) Find equations of the tangent lines at the points (1,5) and (2,3). (c) Graph the curve and both tangents on a common screen. My answers: (a) m(TL) = dy/dx = 8x-6x^2 Strictly speaking, this is wrong. The problem asked for the derivative " at the point where x=a." The correct answer would be $dy/dx= 8a- 6a^2$. (i) at P(1,5); m(TL) = 2 (ii) at P(2,3); m(TL) = -8 Except to solve part (b) there were no parts "i" and "ii". You were not asked for the slope of the tangent at x= 1 and x= 2 in part (a), (b) (i) y = 2x +3 (ii) y = -8x+19 (c) now in this part of graphing i don't know how to graph the curve since it is raised to cube..i don't know either what kind of curve is this.. ^_^ 5. 1. So does that mean, i cannot get the slopes given two points, it's because of "x=a"? 2. It is just a guide for me to know where the slope is leaning.	{"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": 0.9502978920936584, "perplexity_flag": "middle"}
http://mathhelpforum.com/discrete-math/173047-prove-equal-powers-probably-fairly-primitive.html	# Thread: 1. ## Prove equal powers (probably fairly primitive) Hello. My discrete mathematics course has just started, so I'm not very good even at probably fairly easy things: Prove if $\displaystyle \[G = A \times B\]$ is a bijection between finite sets $\displaystyle \[A\]$ and $\displaystyle \[B\]$ then $\displaystyle \[\left\| A \right\| = \left\| B \right\|\]$. It looks pretty obvious because of the nature of this bijection, but I just can't seem to figure out notations to prove it in a mathematical way. Thanks. 2. A bijection is a function, but if A is nonempty and \|B\| > 1, then A x B is not a function. If there is no mistake in the question, then the fact that A x B is a bijection implies that \|A\| = \|B\| = 1 or \|A\| = \|B\| = 0, so \|A\| = \|B\|.	{"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": 0.9549846053123474, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/9150/finding-the-fixed-points-of-a-contraction	# Finding the fixed points of a contraction Banach's fixed point theorem gives us a sufficient condition for a function in a complete metric space to have a fixed point, namely it needs be a contraction. I'm interested in how to calculate the limit of the sequence $x_0 = f(x), x_1 = f(x_0), \ldots, x_n = f(x_{n-1})$ for a fixed $x$. I couldn't figure out a way to do this limit with ordinary limits calculations. The only thing I have at my disposal is the proof of the theorem, from which we see that the sequence $x_n$ is a Cauchy sequence; from this, I'm able to say, for example, that $\left\|f(f(f(x))) - f(f(f(f(x))))\right\| \leq \left\|f(x_0)-f(x_1)\right\| ( \frac{k^3}{1-k})$, where $k$ is the contraction constant, but I can't get any further in the calculations. My question is: how should I procede to calculate this limit exactly? If there are non-numerical (read: analytical) way to do this. Remark: I'm interested in functions $\mathbb{R} \rightarrow \mathbb{R}$ (as it can be seen from my use of the euclidean metric in $\mathbb{R}$) - Your question is confusing. First of all, apply $\|f(x)-f(y)\| < k \|x-y\|$ to $y=f(x)$ to get $\|f(f(x))-f(x)\|<k \|f(x)-x\|$, and, by induction, $f^{(m+1)}(x)-f^{(m)}(x)\|< k^m \|f(x)-x\|$. The fixed point is the limit of $f^m(x)$. For any concrete computable $f$ you can get a good (exponentially fast) approximation to it just by repeatedly applying f. What exactly are you after? – Max Nov 6 '10 at 13:58 The problem I was encountering was calculating $\lim_{n \rightarrow \infty} x_n$. The question was about finding this limit without numerical approximations. – Andy Nov 6 '10 at 14:22 ## 3 Answers @Andy (in reply to your comment/question "Could you provide some example that has a closed form and explain if (and how) it is possible to find the fixed point without solving x = f(x) but trying to calculate the limit of x_n?": I believe that you would be hard-pressed to achieve this, since your function $f$ is a continuous function (being a contraction map in the first place); and if you then take limits of both sides of $x_n = f(x_{n-1})$, you will get: $$\lim_{n \rightarrow \infty} x_n = \lim_{n \rightarrow \infty} f(x_{n-1})$$ which (by continuity) leads to: $$\lim_{n \rightarrow \infty} x_n = f (\lim_{n \rightarrow \infty} x_{n-1})$$ or $$l = f(l)$$ with $l = \lim_{n \rightarrow \infty} x_n$ This means that you will have to solve $l = f(l)$, which was what you wanted to avoid in the first place! - Thank you for pointing the whole thing out.. :) It's all clear now. – Andy Nov 6 '10 at 18:36 You are welcome :-) – knightofmathematics Nov 6 '10 at 20:54 The limit is the fixpoint, so you "just" need to solve the equation $x=f(x)$. Whether this can be done in closed form or not depends on what $f$ looks like. - Alright, the solution of the equation is the fixed point, of course, but (I'm not sure I've been clear enough in the question) I wanted to know if there is a way to calculate $\lim_{n \rightarrow \infty} x_n$ in the "usual" way limits are calculated (like with big-O notation of asymptotics and such). – Andy Nov 6 '10 at 14:25 @Andy: without knowing the true nature of $f$, your question cannot have a better answer than Hans's suggestion. – J. M. Nov 6 '10 at 14:28 @J.M.: suppose $f(x) = \cos{x}$, then how could I do it the way I want to? i.e. not solving $x = \cos{x}$ – Andy Nov 6 '10 at 14:31 – J. M. Nov 6 '10 at 14:38 1 Andy, you are asking for something that is hopeless. In fact you've been misled by your experiences in calculus. Those are not the "usual" ways to find limits, because usually equations do not have closed form solutions. Anything that does is often a rigged example or a very elementary example. Heck, most infinite series like the power series for e^x or sin(x) can't be evaluated in closed form for generic choices of x. You have to settle for approximations with error estimate. That is real life, not a class. – KCd Nov 6 '10 at 17:58 show 3 more comments In addition to what Hans Lundmark has said about solving $x = f(x)$, you could also try writing a simple computer programme to read a number $n$ and a starting value $x_0$, and compute the result of applying f to $x_0$ $n$ times, thus delivering an approximation to the root that you are seeking. The value of $n$ may have to be fairly large in some cases. - Yes, that's what I wanted to avoid. But this is only because I don't like approximations much. :) – Andy Nov 6 '10 at 14:23 1 Probably a better idea is to set up a Newton-Raphson iteration for $x=f(x)$; the convergence might be better than doing vanilla fixed-point iteration. – J. M. Nov 6 '10 at 14:27	{"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": 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": 0.9538910388946533, "perplexity_flag": "head"}
http://crypto.stackexchange.com/questions/248/what-exactly-is-the-impact-of-the-hidden-subgroup-problem-on-cryptography/255	# What exactly is the impact of the hidden subgroup problem on cryptography? I understand my group theory (allegedly), so I can make partial sense of The Hidden Subgroup problem: Given a group $G$, a subgroup $H \leq G$, and a set $X$, we say a function $f : G \Rightarrow X$ separates cosets of $H$ if for all $g_1, g_2 \in G$, $f(g_1) = f(g_2)$ if and only if $g_1H = g_2H$. Hidden subgroup problem: Let $G$ be a group, $X$ a finite set, and $f : G \Rightarrow X$ a function such that there exists a subgroup $H \leq G$ for which $f$ separates cosets of $H$. The function $f$ is given via an oracle. Using information gained from evaluations of $f$ via its oracle, determine a generating set for $H$. In English as much as possible, $G$ is a group which means it satisfies certain conditions such as there being an identity and inverse element under an operation. Here, the operation isn't specified. So then we have some set $X$ which does not necessarily have to meet these criteria, and $f$ a function that maps $G$ the group onto $X$. Cosets e.g. $gH$ are the set of all $H$ operated on by a specific $g$, so $gH = {gh: h \in H, g}$. $Hg$ is the other ordering and exists because groups aren't required to be abelian. So this splitting function $f$ ensures each coset is mapped uniquely if when the function is applied to the group members it and derives the same result, when applied to the relevant cosets it must also derive the same result. This establishes that each coset does essentially map to a unique set of elements in $X$ if I understand this correctly. So, the HSP is essentially the task of finding $H$ given $G$, $f$ and $X$ if I follow that correctly. So, the obvious one-way-problem great-if-you-know-$H$ issue aside, how does the HSP affect cryptography? Specifically, aside from the discovery of the set $H$ in $X$ I see no particular direct use for the HSP, yet I have come across it frequently in discussions on cryptography, particularly with the odd link or reference towards this paper. Finally, am I missing anything, aside from the impacts of cryptography, in my summary of the HSP? - ## 3 Answers As far as I understand, the HSP is a hard problem such that: • some types of HSP (namely those operating in an abelian group) can (theoretically) be solved efficiently on a quantum computer (assuming one can be built); • many types of public key cryptosystems can be reduced to the HSP: if you can solve the HSP you can break the key. In particular, integer factorization and discrete logarithm (in any abelian group, which includes elliptic curves) can be reduced to HSP over abelian groups, and thus easily breakable on a quantum computer. This is not really new: Shor's algorithm already does that. It just happens that Shor's algorithm is a special case of HSP. The interesting part is that lattice reduction can be reduced to HSP over a non-abelian group, for which no efficient algorithm is known yet. So this formalism is actually giving reasons on why lattice-based algorithm may survive in the post-quantum world. In your summary of HSP, one might add two details: "finding `H`" really means "finding a set of elements of `G` which generate `H`" (i.e. `H` is the smallest subgroup of `G` which contains all those elements); and: "given `f`" means "a black box implementing `f` is given, and can be invoked repeatedly on any inputs". - In cryptography, you care not merely that some problem is hard but that hard instances are readily producible. Why don't people use NP-complete problems for cryptography, for example? An NP-complete problem would give you greater confidence asymptoticly speaking for two reasons : If any NP-complete problem were collapsed to P, then factoring becomes P too. Factoring is quantum polynomial time (BQP) but no relationship between BQP and NP is known. Instead, there is a more insidious problem that many NP-complete problems have too many easy cases to quickly find hard instances. There are many easy cases for the Hidden Subgroup Problem as well. In fact, there are interesting algorithms for identifying non-abelian finite groups that work by solving hidden subgroup problems for which polynomial time solutions exist thanks to the Classification of the Finite Simple Groups. - The hidden subgroup problem is very useful for developing an understanding of pairing-based non-interactive zero-knowledge proofs. You would need a suitable elliptic curve of large composite order in order to use the hidden subgroup problem securely so in practice, you probably wouldn't bother as the implementation would be very slow. However, the explanation and maths using the HSP is relatively understandable so it's useful for pedagogical purposes. See Jens Groth's talk http://research.microsoft.com/apps/video/default.aspx?id=103365 -	{"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": 36, "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": 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": 0.9404532313346863, "perplexity_flag": "head"}
http://amathew.wordpress.com/tag/simplicial-sets/	Climbing Mount Bourbaki Thoughts on mathematics January 9, 2012 Delooping and the bar construction Posted by Akhil Mathew under category theory, topology \| Tags: bar construction, delooping, iterated loop spaces, operads, simplicial sets \| [5] Comments The present post is motivated by the following problem: Problem: Given a pointed space ${X}$, when is ${X}$ of the homotopy type of a ${k}$-fold loop space ${\Omega^k Y}$ for some ${Y}$? One of the basic observations that one can make about a loop space ${\Omega Y}$ is that admits a homotopy associative multiplication map $\displaystyle m: \Omega Y \times \Omega Y \rightarrow \Omega Y.$ Having such an H structure imposes strong restrictions on the homotopy type of ${\Omega Y}$; for instance, it implies that the cohomology ring ${H^(\Omega Y; k)}$ with coefficients in a field is a graded Hopf algebra. There are strong structure theorems for Hopf algebras, though. For instance, in the finite-dimensional case and in characteristic zero, they are tensor products of exterior algebras, by a theorem of Milnor and Moore. Moreover, for a double loop space ${\Omega^2 Y}$, the H space structure is homotopy commutative. Nonetheless, it is not true that any homotopy associative H space has the homotopy type of a loop space. The problem with mere homotopy associativity is that it asserts that two maps are homotopic; one should instead require that the homotopies be part of the data, and that they satisfy coherence conditions. The machinery of operads was developed to codify these coherence conditions efficiently, and today it seems that one of the powers of higher (at least, ${(\infty, 1)}$) category theory is the ability to do this in a much more general context. For this post, I want to try to ignore all this operadic and higher categorical business and explain the essential idea of the delooping construction in May’s “The Geometry of Iterated Loop Spaces”; this relies on some category theory and a little homotopy theory, but the explicit operads play very little role. (more…) October 24, 2011 The commutative cochain problem and the polynomial de Rham functor Posted by Akhil Mathew under topology \| Tags: commutative cochain problem, de Rham functor, rational homotopy theory, simplicial sets \| Let ${k}$ be a field. The commutative cochain problem over ${k}$ is to assign (contravariantly) functorially, to every simplicial set ${K_\bullet}$, a commutative (in the graded sense) ${k}$-algebra ${A(K_\bullet)}$, which is naturally weakly equivalent to the algebra ${C^(K_\bullet, k)}$ of singular cochains (with ${k}$-coefficients). We also require that ${A(K_\bullet) \rightarrow A(L_\bullet)}$ is a surjection whenever ${L_\bullet \subset K_\bullet}$. Recall that ${C^(K_\bullet, k)}$ is an associative algebra, but it is not commutative; the commutativity only appears after one takes cohomology. The commutative cochain problem attempts to find an improvement to ${C^(K_\bullet, k)}$. If ${k}$ has finite characteristic, this problem cannot be solved, owing to the existence of nontrivial cohomology operations. (The answers at this MathOverflow question are relevant here.) However, there is a solution for ${k = \mathbb{Q}}$, given by the polynomial de Rham theory. In this post, I will explain this. (more…) October 22, 2011 The classifying space of a simplicial group Posted by Akhil Mathew under topology \| Tags: classifying space, model categories, simplicial groups, simplicial sets \| [2] Comments A simplicial group is, naturally enough, a simplicial object in the category of groups. It is equivalently a simplicial set ${G_\bullet}$ such that each ${G_n}$ has a group structure and all the face and degeneracy maps are group homomorphisms. By a well-known lemma, a simplicial group is automatically a Kan complex. We can thus think of a simplicial group in two ways. On the one hand, it is an ${\infty}$-groupoid, as a Kan complex. On the other hand, we can think of a simplicial group as a model for a “higher group,” or an ${\infty}$-group. This is the intuition that the nLab suggests. Now it is well-known that in ordinary category theory, a group is the same as a groupoid with one object. So an ${\infty}$-groupoid with one object should be the same as an ${\infty}$-group. This is in fact true with the above notation. In other words, if we say that an ${\infty}$-groupoid is a Kan complex (as usual), and decide that an ${\infty}$-group is a simplicial group, then the ${\infty}$-groups are the “same” as the ${\infty}$-groupoids with one object. Here the “same” means that the associated ${\infty}$-categories are equivalent. One way of expressing this is to say that there are natural model structures on ${\infty}$-groupoids with one object and simplicial groups, and that these are Quillen equivalent. This is a frequent way of expressing the idea that two ${\infty}$-categories (or at least, ${(\infty, 1)}$-categories) are equivalent. For instance, this is the way the ${\infty}$-categorical Grothendieck construction is stated in HTT, as a Quillen equivalence between model categories of “right fibrations” and simplicial presheaves. Let us express this formally. Theorem 1 (Kan) There are natural model structures on the category ${\mathbf{SGrp}}$ of simplicial groups and on the category ${\mathbf{SSet}_0}$ of reduced simplicial sets (i.e. those with one vertex), and the two are Quillen equivalent. To give this construction, we will first describe the classifying space of a simplicial group. One incarnation of this is going to give the functor from simplicial groups to reduced simplicial sets. This is a simplicial version of the usual topological classifying space. Recall that if ${G}$ is a topological group, then there is a classifying space ${BG}$ and a principal ${G}$-bundle ${EG \rightarrow BG}$ such that ${EG}$ is contractible. It follows from this that for any CW-complex ${X}$, the homotopy classes of maps ${X \rightarrow BG}$ are in bijection with the principal ${G}$-bundles on ${X}$ (in fact, ${EG \rightarrow BG}$ is a universal bundle). We will need the appropriate notion of a principal bundle in the simplicial setting. Let ${G_\bullet}$ be a simplicial group. Definition 2 A ${G_\bullet}$-simplicial set is a simplicial set ${X_\bullet}$ together with an action ${G_\bullet \times X_\bullet \rightarrow X_\bullet}$ satisfying the usual axioms. Thus, each ${X_n}$ is a ${G_n}$-set. There is a category ${\mathbf{SSet}_G}$ of ${G_\bullet}$-simplicial sets and ${G_\bullet}$-equivariant maps. We next give the definition of a principal bundle. Notice that because of the combinatorial nature of simplicial sets, we don’t make any local trivialty condition; that will fall out. Definition 3 Let ${E_\bullet \rightarrow B_\bullet}$ be a map of ${G_\bullet}$-simplicial sets. We say that ${E_\bullet \rightarrow B_\bullet}$ is a principal ${G_\bullet}$-bundle if ${B_\bullet}$ has trivial action, each ${E_n}$ is a free ${G_n}$-set, and if ${E_\bullet/G_\bullet \simeq B_\bullet}$ under the natural map. (more…) September 24, 2011 Theorem A via general formalism Posted by Akhil Mathew under category theory, topology \| Tags: algebraic K-theory, model categories, nerve, simplicial sets, Theorem A \| [3] Comments The following result is useful in algebraic K-theory. Theorem 1 Let ${F: \mathcal{C} \rightarrow \mathcal{D}}$ be a functor between categories. Suppose ${\mathcal{C}/d}$ is contractible for each ${d \in \mathcal{D}}$. Then ${F: N\mathcal{C} \rightarrow N \mathcal{D}}$ is a weak homotopy equivalence. I don’t really know enough to give a good justification for the usefulness, but in essence, what Quillen did in the 1970s was to show that the Grothendieck group of an “exact category” could be interpreted homotopically as the fundamental group of the nerve of the “Q-category” built from the exact category. As a result, Quillen was able to define higher K-groups as the higher homotopy groups of this space. He then proved a lot of results that were proved by ad hoc, homological means for the Grothendieck group of a category for the higher K-groups as well, by interpreting them in terms of homotopy theory. This result (together with the extension, “Theorem B”) is a key homotopical tool he used to analyze these nerves. Here ${N \mathcal{C}}$ denotes the nerve of the category ${\mathcal{C}}$: it is the simplicial set whose ${n}$-simplices consist of composable strings of ${n+1}$ morphisms of ${\mathcal{C}}$. The overcategory ${\mathcal{C}/d}$ has objects consisting of pairs ${(c, f)}$ for ${c \in \mathcal{C}}$, ${f: Fc \rightarrow d}$ a morphism in ${\mathcal{D}}$; morphisms in ${\mathcal{C}/d}$ are morphisms in ${\mathcal{C}}$ making the natural diagram commute. We say that a category is contractible if its nerve is weakly contractible as a simplicial set. There are other reasons to care. For instance, in higher category theory, the above condition on contractibility of over-categories is the analog of cofinality in ordinary category theory. Anyway, this result is pretty important. But what I want to explain in this post is that “Theorem A” (and Theorem B, but I’ll defer that) is really purely formal. That is, it can be deduced from some standard and not-too-difficult manipulations with model categories (which weren’t all around when Quillen wrote “Higher algebraic K-theory I”). To prove this, we shall obtain the following expression for a category: $\displaystyle N \mathcal{C} = \mathrm{colim}_d N (\mathcal{C}/d),$ where ${d}$ ranges over the objects of ${\mathcal{D}}$. This expresses the nerve of ${\mathcal{C}}$ as a colimit of simplicial sets arising as the nerves of ${\mathcal{C}/d}$. We will compare this with a similar expression for the nerve of $\mathcal{D}$, that is ${N \mathcal{D} = \mathrm{colim}_d N(\mathcal{D}/d)}$. Then, the point will be that ${N(\mathcal{C}/d) \rightarrow N(\mathcal{D}/d)}$ is a weak equivalence for each ${d}$; this by itself does not imply that the induced map on colimits is a weak equivalence, but it will in this case because both the colimits will in fact turn out to be homotopy colimits. I’ll start by explaining what those are. (more…) May 6, 2011 The Dold-Kan correspondence III Posted by Akhil Mathew under algebra, topology \| Tags: Dold-Kan correspondence, simplicial sets \| Recall that we were in the middle of establishing a crucial equivalence of categories between simplicial abelian groups and chain complexes. Last time, we had defined the two functors: on the one hand, we had the normalized chain complex of a simplicial abelian group; on the other hand, we had defined a functor $\sigma$ that amalgamated a chain complex into a simplicial abelian group. We were in the middle of proving that the two functors were quasi-inverse. With the same notation as before, we were trying to prove: Proposition 3 (One half of Dold-Kan) For a simplicial abelian group ${A_\bullet}$, we have for each ${n}$, an isomorphism of abelian groups $\displaystyle \bigoplus_{\phi: [n] \twoheadrightarrow [k]} NA_k \simeq A_n.$ Here the map is given by sending a summand ${NA_k}$ to ${A_n}$ via the pull-back by the term ${\phi: [n] \twoheadrightarrow [k]}$. Alternatively, the morphism of simplicial abelian groups $\displaystyle \sigma (N A_*)_\bullet \rightarrow A_\bullet$ is an isomorphism. May 6, 2011 The Dold-Kan correspondence II Posted by Akhil Mathew under algebra, topology \| Tags: Dold-Kan correspondence, simplicial sets \| 1 Comment Recall from last time that we were in the middle of proving the Dold-Kan correspondence, an important equivalence of categories between simplicial abelian groups and chain complexes. We defined three functors last time from simplicial abelian groups: the most obvious was the Moore complex, which just spliced all the components into one big chain complex with the differential the alternating sum of the face maps. But we noted that the functor one uses to construct this equivalence is ultimately either the normalized chain complex or the Moore complex modulo degeneracies. Today, I’ll show that the two functors from simplicial abelian groups to chain complexes are in fact the same, through a decomposition that next time will let us construct the inverse functor. I’ll also construct the functor in the reverse direction. A minor word of warning: the argument in Goerss-Jardine (which seems to be the main source nowadays for this kind of material) has a small mistake! See their errata. This confused me for quite a while. From chain complexes to simplicial groups A priori, the normalized chain complex of a simplicial abelian group ${A_\bullet}$ looks a lot different from ${A_\bullet}$, which a priori has much more structure. Nonetheless, we are going to see that it is possible to recover ${A_\bullet}$ entirely from this chain complex. A key step in the proof of the Dold-Kan correspondence will be the establishment of the functorial decomposition for any simplicial abelian group ${A_\bullet}$ $\displaystyle \bigoplus_{\phi: [n] \twoheadrightarrow [k]} NA_k \simeq A_n. \ \ \ \ \ (1)$ Here the map from a factor ${NA_k}$ corresponding to some ${\phi: [n] \twoheadrightarrow [k]}$ to ${A_n}$ is given by pulling back by ${\phi}$. We will establish this below. (more…) May 4, 2011 The Dold-Kan correspondence I Posted by Akhil Mathew under algebra, topology \| Tags: Dold-Kan correspondence, simplicial sets \| [5] Comments The next couple of posts will cover the Dold-Kan correspondence, which establishes an equivalence of categories between simplicial abelian groups and chain complexes. While this will not be strictly necessary for the introduction of the cotangent complex, it is a sufficiently important fact that it seems worth a digression. As far as I can tell, the Dold-Kan correspondence is a fairly technical result, and I’m not sure I have any good intuition for why one should expect it to work. But at least one can say the following: given a simplicial abelian group (that is, a contravariant functor from the simplex category to the category of abelian groups), one can form a chain complex in a fairly easy manner: just take the $n$-simplices of the simplicial group as the degree $n$ part of the complex, and define the boundary using the alternating sum of the simplicial boundary maps (defined below); this is the classical computation that one does in introductory algebraic topology, of showing that the singular chain complex is indeed a complex. So it’s natural that you would get a chain complex from a simplicial abelian group. Except, as it turns out, this is the wrong functor for the Dold-Kan correspondence; it is, however, close, being right up to (natural) homotopy. The other bit of intuition that I’ve heard is the following. Given a topological space $X$, there is a means of obtaining the homology of $X$ as the homotopy groups of the infinite symmetric product; this is the so-called Dold-Thom theorem. (See this, for instance.) The Dold-Kan correspondence is in some sense the simplicial analog of this. The infinite symmetric product is much like the abelianization functor from simplicial sets to simplicial abelian groups (that applies the free abelian group pointwise). Now, it will come out of the Dold-Kan correspondence that the so-called “simplicial homotopy groups” of a simplicial abelian group are going to be the same thing as the homology of the associated chain complex. This is a rather loose analogy, and my understanding is that one cannot derive Dold-Thom from Dold-Kan. (more…) May 3, 2011 Simplicial sets II: geometric realization Posted by Akhil Mathew under category theory, topology \| Tags: adjoint functors, geometric realization, simplicial sets \| [2] Comments The semester here finished recently, which means that I will (hopefully!) have more time to update this blog over the next couple of months. I intend to continue the brief series on simplicial methods, which should eventually lead into the cotangent complex soon. I have some other ideas for topics in the near future, but given my recent record at keeping promises, I shall perhaps refrain from divulging the information until I actually have the posts ready! Recall that last time, we introduced the notion of a simplicial set. As these were presheaves on the category of finite ordered sets (that is, the simplex category), we talked for a while about presheaves in general, on any small category. We did some abstract nonsense and showed in particular that any presheaf is (canonically, in fact!) the colimit of representable presheaves. In our case, that meant that the standard simplexes were enough to generate the entire category of simplicial sets. Today, using this formalism, we are going to see that functors can be defined solely on the standard simplices and thus extended canonically. Now you might be wondering why the simplex category itself is so special, especially since everything we’ve done so far has been for presheaves on any small category. In homotopy theory, which we haven’t gotten to, simplicial sets have the highly important property of admitting a closed model structure which is Quillen equivalent to the Serre model structure; you might thus wonder which categories of presheaves provide a model for classical homotopy theory in the above sense. I don’t have a complete answer; however, it seems worth mentioning that there is work by Cisinski that does. That is, there is a complete characterization of categories whose presheaf categories admit a model structure Quillen equivalent to spaces. But simplicial sets are (presumably) one of the simplest, and have the advantage of arising in many settings. 1.4. Adjunctions Let ${\mathcal{C}}$ be a category, and ${\mathcal{D}}$ a cocomplete category. We are interested in colimit-preserving functors $\displaystyle \overline{\mathbf{F}}: \hat{\mathcal{C}} \rightarrow \mathcal{D}.$ Here, as before, ${\hat{\mathcal{C}}}$ is the category of presheaves on ${\mathcal{C}}$. We shall, in this post, write functors out of a presheaf category with a line above them, and functors just defined out of ${\mathcal{C}}$ without the line. Functors will be in bold. (more…) April 16, 2011 Simplicial sets I Posted by Akhil Mathew under category theory, topology \| Tags: presheaves, simplicial sets \| 1 Comment As planned, I’m going to try to say a few words on the cotangent complex in the following posts. There is a bit of background I’d like to go through first, starting with the theory of simplicial sets. The reason is that the cotangent complex is most naturally thought of using the model structure on simplicial rings, on which it is a non-abelian version of a derived functor. 1.1. The simplex category Definition 1 Let ${\Delta}$ be the category of finite (nonempty) ordered sets and order-preserving morphisms. The object ${[n]}$ will denote the set ${\left\{0, 1, \dots, n\right\}}$ with the usual ordering. Thus ${\Delta}$ is equivalent to the subcategory consisting of the ${[n]}$. This is called the simplex category. There is a functor from ${\Delta}$ to the category ${\mathbf{Top}}$ of topological spaces. Given ${[n]}$, we send it to the standard topological ${n}$-simplex ${\Delta_n}$ that consists of points ${(t_0, \dots, t_{n}) \in \mathbb{R}^{n+1}}$ such that each ${t_i \in [0, 1]}$ and ${\sum t_i = 1}$. Given a morphism ${\phi: [m] \rightarrow [n]}$ of ordered sets, we define ${\Delta_m \rightarrow \Delta_n}$ by sending $\displaystyle (t_0, \dots, t_m) \mapsto (u_j), \quad u_j = \sum_{\phi(i) = j} t_i.$ Here the empty sum is to be regarded as zero. 1.2. Simplicial sets Definition 2 A simplicial set ${X_{\bullet}}$ is a contravariant functor from ${\Delta}$ to the category of sets. In other words, it is a presheaf on the simplex category. A morphism of simplicial sets is a natural transformation of functors. The class of simplicial sets thus becomes a category ${\mathbf{SSet}}$. Asimplicial object in a category ${\mathcal{C}}$ is a contravariant functor ${\Delta \rightarrow \mathcal{C}}$. We have just seen that the category ${\Delta}$ is equivalent to the subcategory consisting of the ${[n]}$. As a result, a simplicial set ${X_{\bullet}}$ is given by specifying sets ${X_n}$ for each ${n \in \mathbb{Z}_{\geq 0}}$, together with maps $\displaystyle X_n \rightarrow X_m$ for each map ${[m]\rightarrow [n]}$ in ${\Delta}$. The set ${X_n}$ is called the set of ${n}$-simplices of ${X_\bullet}$. 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http://en.wikipedia.org/wiki/Centrifugal_force_(fictitious)	# Centrifugal force (rotating reference frame) (Redirected from Centrifugal force (fictitious)) Classical mechanics Branches Formulations Fundamental concepts Core topics Scientists Centrifugal force (from Latin centrum "center" and fugere "to flee") can generally be any force directed outward relative to some origin. More particularly, in classical mechanics, the centrifugal force is an outward force which arises when describing the motion of objects in a rotating reference frame. Because a rotating frame is an example of a non-inertial reference frame, Newton's laws of motion do not accurately describe the dynamics within the rotating frame. However, a rotating frame can be treated as if it were an inertial frame so that Newton's laws can be used if so-called fictitious forces (also known as inertial or pseudo- forces) are included in the sum of external forces on an object. The centrifugal force is what is usually thought of as the cause for apparent outward movement like that of passengers in a vehicle turning a corner, of the weights in a centrifugal governor, and of particles in a centrifuge. From the standpoint of an observer in an inertial frame, the effects can be explained as results of inertia without invoking the centrifugal force. Centrifugal force should not be confused with centripetal force or the reactive centrifugal force, both of which are real forces independent of the frame of the observer. Analysis of motion within rotating frames can be greatly simplified by the use of the fictitious forces. By starting with an inertial frame, where Newton's laws of motion hold, and keeping track of how the time derivatives of a position vector change when transforming to a rotating reference frame, the various fictitious forces and their forms can be identified. Rotating frames and fictitious forces can often reduce the description of motion in two dimensions to a simpler description in one dimension (corresponding to a co-rotating frame). In this approach, circular motion in an inertial frame, which only requires the presence of a centripetal force, becomes the balance between the real centripetal force and the frame-determined centrifugal force in the rotating frame where the object appears stationary. If a rotating frame is chosen so that just the angular position of an object is held fixed, more complicated motion, such as elliptical and open orbits, appears because the centripetal and centrifugal forces will not balance. The general approach however is not limited to these co-rotating frames, but can be equally applied to objects at motion in any rotating frame. ## In classical Newtonian physics Although Newton's laws of motion hold exclusively in inertial frames, often it is far more convenient and more advantageous to describe the motion of objects within a rotating reference frame.[1][2] Sometimes the calculations are simpler (an example is inertial circles), and sometimes the intuitive picture coincides more closely with the rotational frame (an example is sedimentation in a centrifuge). By treating the extra acceleration terms due to the rotation of the frame as if they were forces, subtracting them from the physical forces, it's possible to treat the second time derivative of position (relative to the rotating frame) as absolute acceleration. Thus the analysis using Newton's laws of motion can proceed as if the reference frame was inertial, provided the fictitious force terms are included in the sum of external forces.[3] For example, centrifugal force is used in the FAA pilot's manual in describing turns.[4] Other examples are such systems as planets, centrifuges, carousels, turning cars, spinning buckets, and rotating space stations.[5][6][7] If objects are seen as moving within a rotating frame, this movement results in another fictitious force, the Coriolis force; and if the rate of rotation of the frame is changing, a third fictitious force, the Euler force is experienced.[8] Together, these three fictitious forces allow for the creation of correct equations of motion in a rotating reference frame.[9] ## Derivation Main article: Rotating reference frame For the following formalism, the rotating frame of reference is regarded as a special case of a non-inertial reference frame that is rotating relative to an inertial reference frame denoted the stationary frame. ### Velocity In a rotating frame of reference, the time derivatives of the position vector r, such as velocity and acceleration vectors, of an object will differ from the time derivatives in the stationary frame according to the frame's rotation. The first time derivative [dr/dt] evaluated within a reference frame with a coincident origin at $r=0$ but rotating with the absolute angular velocity ω is:[10] $\frac{\operatorname{d}\boldsymbol{r}}{\operatorname{d}t} = \left[\frac{\operatorname{d}\boldsymbol{r}}{\operatorname{d}t}\right] + \boldsymbol{\omega} \times \boldsymbol{r}\ ,$ where $\times$ denotes the vector cross product and square brackets [...] denote evaluation in the rotating frame of reference. In other words, the apparent velocity in the rotating frame is altered by the amount of the apparent rotation $\boldsymbol{\omega} \times \boldsymbol{r}$ at each point, which is perpendicular to both the vector from the origin r and the axis of rotation ω and directly proportional in magnitude to each of them. The vector ω has magnitude ω equal to the rate of rotation and is directed along the axis of rotation according to the right-hand rule. ### Acceleration Newton's law of motion for a particle of mass m written in vector form is: $\boldsymbol{F} = m\boldsymbol{a}\ ,$ where F is the vector sum of the physical forces applied to the particle and a is the absolute acceleration (that is, acceleration in an inertial frame) of the particle, given by: $\boldsymbol{a}=\frac{\operatorname{d}^2\boldsymbol{r}}{\operatorname{d}t^2} \ ,$ where r is the position vector of the particle. By twice applying the transformation above from the stationary to the rotating frame, the absolute acceleration of the particle can be written as: $\begin{align} \boldsymbol{a} &=\frac{\operatorname{d}^2\boldsymbol{r}}{\operatorname{d}t^2} = \frac{\operatorname{d}}{\operatorname{d}t}\frac{\operatorname{d}\boldsymbol{r}}{\operatorname{d}t} = \frac{\operatorname{d}}{\operatorname{d}t} \left( \left[\frac{\operatorname{d}\boldsymbol{r}}{\operatorname{d}t}\right] + \boldsymbol{\omega} \times \boldsymbol{r}\ \right) \\ &= \left[ \frac{\operatorname{d}^2 \boldsymbol{r}}{\operatorname{d}t^2} \right] + \frac{\operatorname{d} \boldsymbol{\omega}}{\operatorname{d}t}\times\boldsymbol{r} + 2 \boldsymbol{\omega}\times \left[ \frac{\operatorname{d} \boldsymbol{r}}{\operatorname{d}t} \right] + \boldsymbol{\omega}\times ( \boldsymbol{\omega} \times \boldsymbol{r}) \ . \end{align}$ ### Force The apparent acceleration in the rotating frame is [d2r/dt2]. An observer unaware of the rotation would expect this to be zero in the absence of outside forces. However Newton's laws of motion apply only in the stationary frame and describe dynamics in terms of the absolute acceleration d2r/dt2. Therefore the observer perceives the extra terms as contributions due to fictitious forces. These terms in the apparent acceleration are independent of mass; so it appears that each of these fictitious forces, like gravity, pulls on an object in proportion to its mass. When these forces are added, the equation of motion has the form:[11][12][13] $\boldsymbol{F} - m\frac{\operatorname{d} \boldsymbol{\omega}}{\operatorname{d}t}\times\boldsymbol{r} - 2m \boldsymbol{\omega}\times \left[ \frac{\operatorname{d} \mathbf{r}}{\operatorname{d}t} \right] - m\boldsymbol{\omega}\times (\boldsymbol{\omega}\times \boldsymbol{r})$ $= m\left[ \frac{\operatorname{d}^2 \boldsymbol{r}}{\operatorname{d}t^2} \right] \ .$ From the perspective of the rotating frame, the additional force terms are experienced just like the real external forces and contribute to the apparent acceleration.[14][15] The additional terms on the force side of the equation can be recognized as, reading from left to right, the Euler force $m \operatorname{d}\boldsymbol{\omega}/\operatorname{d}t \times\boldsymbol{r}$, the Coriolis force $2m \boldsymbol{\omega}\times \left[ \operatorname{d} \boldsymbol{r}/\operatorname{d}t \right]$, and the centrifugal force $m\boldsymbol{\omega}\times (\boldsymbol{\omega}\times \boldsymbol{r})$, respectively.[16] Unlike the other two fictitious forces, the centrifugal force always points radially outward from the axis of rotation of the rotating frame, with magnitude mω2r, and unlike the Coriolis force in particular, it is independent of the motion of the particle in the rotating frame. As expected, for a non-rotating inertial frame of reference $(\boldsymbol\omega=0)$ the centrifugal force and all other fictitious forces disappear.[17] ## Absolute rotation Main article: Absolute rotation The interface of two immiscible liquids rotating around a vertical axis is an upward-opening circular paraboloid. Centrifugal force causes rotating planets to assume the shape of an oblate spheroid Three scenarios were suggested by Newton to answer the question of whether the absolute rotation of a local frame can be detected; that is, if an observer can decide whether an observed object is rotating or if the observer is rotating.[18][19] • The shape of the surface of water rotating in a bucket. The shape of the surface becomes concave to balance the centrifugal force against the other forces upon the liquid. • The tension in a string joining two spheres rotating about their center of mass. The tension in the string will be proportional to the centrifugal force on each sphere as it rotates around the common center of mass. In these scenarios, the effects attributed to centrifugal force are only observed in the local frame (the frame in which the object is stationary) if the object is undergoing absolute rotation relative to an inertial frame. By contrast, in an inertial frame, the observed effects arise as a consequence of the inertia and the known forces without the need to introduce a centrifugal force. Based on this argument, the privileged frame, wherein the laws of physics take on the simplest form, is a stationary frame in which no fictitious forces need to be invoked. Within this view of physics, any other phenomenon that is usually attributed to centrifugal force can be used to identify absolute rotation. For example, the oblateness of a sphere of freely flowing material is often explained in terms of centrifugal force. The oblate spheroid shape reflects, following Clairaut's theorem, the balance between containment by gravitational attraction and dispersal by centrifugal force. That the Earth is itself an oblate spheroid, bulging at the equator where the radial distance and hence the centrifugal force is larger, is taken as one of the evidences for its absolute rotation.[20] ## Examples Below several examples illustrate both the stationary and rotating frames of reference, and the role of centrifugal force and its relation to Coriolis force in rotating frameworks. For more examples see Fictitious force, rotating bucket and rotating spheres. ### Dropping ball A ball moving vertically along the axis of rotation in a stationary frame appears to spiral downward in the rotating frame. The right panel shows a downward view in the rotating frame. The rate of rotation \|ω\| = ω is assumed constant in time An example of straight-line motion as seen in a stationary frame is a ball that steadily drops at a constant rate parallel to the axis of rotation. From a stationary frame of reference it moves in a straight line, but from the rotating frame it moves in a helix. The projection of the helical motion in a rotating horizontal plane is shown at the right of the figure. Because the projected horizontal motion in the rotating frame is a circular motion, the ball's motion requires an inward centripetal force, provided in this case by a fictitious force that produces the apparent helical motion. This force is the sum of an outward centrifugal force and an inward Coriolis force. The Coriolis force overcompensates the centrifugal force by exactly the required amount to provide the necessary centripetal force to achieve circular motion. ### Banked turn Main article: Banked turn Riding a car around a curve, we take a personal view that we are at rest in the car, and should be undisturbed in our seats. Nonetheless, we feel sideways force applied to us from the seats and doors and a need to lean to one side. To explain the situation, we propose a centrifugal force that is acting upon us and must be combated. Interestingly, we find this discomfort is reduced when the curve is banked, tipping the car inward toward the center of the curve. A different point of view is that of the highway designer. The designer views the car as executing curved motion and therefore requiring an inward centripetal force to impel the car around the turn. By banking the curve, the force exerted upon the car in a direction normal to the road surface has a horizontal component that provides this centripetal force. That means the car tires no longer need to apply a sideways force to the car, but only a force perpendicular to the road. By choosing the angle of bank to match the car's speed around the curve, the car seat transmits only a perpendicular force to the passengers, and the passengers no longer feel a need to lean nor feel a sideways push by the car seats or doors.[21] ### Earth A calculation for Earth at the equator ($\omega = 2\pi/86164$ seconds, $r = 6378100$ meters) shows that an object experiences a centrifugal force equal to approximately 1/289 of standard gravity.[22] Because centrifugal force increases according to the square of $\omega$, one would expect gravity to be cancelled for an object travelling 17 times faster than the Earth's rotation, and in fact satellites in low orbit at the equator complete 17 full orbits in one day.[23] Gravity diminishes according to the inverse square of distance, but centrifugal force increases in direct proportion to the distance. Thus a circular geosynchronous orbit has a radius of 42164 km; 42164/6378.1 = 6.61, the cube root of 289. ### Planetary motion See also: Orbit and Kepler's laws of planetary motion Centrifugal force arises in the analysis of orbital motion and, more generally, of motion in a central-force field: in the case of a two-body problem, it is easy to convert to an equivalent one-body problem with force directed to or from an origin, and motion in a plane,[24] so we consider only that. The symmetry of a central force lends itself to a description in polar coordinates. The dynamics of a mass, m, expressed using Newton's second law of motion (F = ma), becomes in polar coordinates:[25][26] $\boldsymbol{F} = m((\ddot r - r \dot \theta^2) \boldsymbol{\hat r} + (r \ddot\theta + 2 \dot r \dot\theta) \boldsymbol{\hat \theta})$ where $\boldsymbol{F}$ is the force accelerating the object and the "hat" variables are unit direction vectors ($\boldsymbol{\hat r}$ points in the centrifugal or outward direction, and $\boldsymbol{\hat \theta}$ is orthogonal to it). In the case of a central force, relative to the origin of the polar coordinate system, $\boldsymbol{F}$ can be replaced by $F(r) \boldsymbol{\hat r}$, meaning the entire force is the component in the radial direction. An inward force of gravity would therefore correspond to a negative-valued F(r). The components of F = ma along the radial direction therefore reduce to $F(r) = m(\ddot r - r \dot\theta^2)$ in which the term proportional to the square of the rate of rotation appears on the acceleration side as a "centripetal acceleration", that is, a negative acceleration term in the $\boldsymbol{\hat r}$ direction.[27] In the special case of a planet in circular orbit around its star, for example, where $\ddot r$ is zero, the centripetal acceleration alone is the entire acceleration of the planet, curving its path toward the sun under the force of gravity, the negative F(r). As pointed out by Taylor,[28] for example, it is sometimes convenient to work in a co-rotating frame, that is, one rotating with the object so that the angular rate of the frame, $\omega$, equals the $\dot\theta$ of the object in the stationary frame. In such a frame, the observed $\dot \theta$ is zero and $\ddot r$ alone is treated as the acceleration: so in the equation of motion, the $m r \dot\theta^2$ term is "reincarnated on the force side of the equation (with opposite signs, of course) as the centrifugal force mω2r in the radial equation":[28] The "reincarnation" on the force side of the equation is necessary because, without this force term, observers in the rotating frame would find they could not predict the motion correctly. They would have an incorrect radial equation: $F(r) + m r \dot\theta^2 = m\ddot r$ where the $m r \dot\theta^2$ term is known as the centrifugal force. The centrifugal force term in this equation is called a "fictitious force", "apparent force", or "pseudo force", as its value varies with the rate of rotation of the frame of reference. When the centrifugal force term is expressed in terms of parameters of the rotating frame, replacing $\dot\theta$ with $\omega$, it can be seen that it is the same centrifugal force previously derived for rotating reference frames. Because of the absence of a net force in the azimuthal direction, conservation of angular momentum allows the radial component of this equation to be expressed solely with respect to the radial coordinate, r, and the angular momentum $L=m \dot\theta r^2$, yielding the radial equation (a "fictitious one-dimensional problem"[24] with only an r dimension): $F(r) + \frac{L^2}{mr^3} = m \ddot r$. The $L^2/mr^3$ term is again the centrifugal force, a force component induced by the rotating frame of reference. The equations of motion for r that result from this equation for the rotating 2D frame are the same that would arise from a particle in a fictitious one-dimensional scenario under the influence of the force in the equation above.[24] If F(r) represents gravity, it is a negative term proportional to 1/r2, so the net acceleration in r in the rotating frame depends on a difference of reciprocal square and reciprocal cube terms, which are in balance in a circular orbit but otherwise typically not. This equation of motion is similar to one originally proposed by Leibniz.[29] Given r, the rate of rotation is easy to infer from the constant angular momentum L, so a 2D solution can be easily reconstructed from a 1D solution of this equation. When the angular velocity of this co-rotating frame is not constant, that is, for non-circular orbits, other fictitious forces—the Coriolis force and the Euler force—will arise, but can be ignored since they will cancel each other, yielding a net zero acceleration transverse to the moving radial vector, as required by the starting assumption that the $\hat r$ vector co-rotates with the planet.[30] In the special case of circular orbits, in order for the radial distance to remain constant the outward centrifugal force must cancel the inward force of gravity; for other orbit shapes, these forces will not cancel, so r will not be constant. ## History Main article: History of centrifugal and centripetal forces Concepts of centripetal and centrifugal force played a key early role in establishing the set of inertial frames of reference and the significance of fictitious forces, even aiding in the development of general relativity in which gravity itself becomes a fictitious force.[31] ## Applications The operations of numerous common rotating mechanical systems are most easily conceptualized in terms of centrifugal force. For example: • A centrifugal governor regulates the speed of an engine by using spinning masses that move radially, adjusting the throttle, as the engine changes speed. In the reference frame of the spinning masses, centrifugal force causes the radial movement. • A centrifugal clutch is used in small engine-powered devices such as chain saws, go-karts and model helicopters. It allows the engine to start and idle without driving the device but automatically and smoothly engages the drive as the engine speed rises. Inertial drum brake ascenders used in rock climbing and the inertia reels used in many automobile seat belts operate on the same principle. • Centrifugal forces can be used to generate artificial gravity, as in proposed designs for rotating space stations. The Mars Gravity Biosatellite will study the effects of Mars-level gravity on mice with gravity simulated in this way. • Spin casting and centrifugal casting are production methods that uses centrifugal force to disperse liquid metal or plastic throughout the negative space of a mold. • Centrifuges are used in science and industry to separate substances. In the reference frame spinning with the centrifuge, the centrifugal force induces a hydrostatic pressure gradient in fluid-filled tubes oriented perpendicular to the axis of rotation, giving rise to large buoyant forces which push low-density particles inward. Elements or particles denser than the fluid move outward under the influence of the centrifugal force. This is effectively Archimedes' principle as generated by centrifugal force as opposed to being generated by gravity. • Some amusement rides make use of centrifugal forces. For instance, a Gravitron's spin forces riders against a wall and allows riders to be elevated above the machine's floor in defiance of Earth's gravity.[32] Nevertheless, all of these systems can also be described without requiring the concept of centrifugal force, in terms of motions and forces in a stationary frame, at the cost of taking somewhat more care in the consideration of forces and motions within the system. ## Footnotes 1. Stephen T. Thornton & Jerry B. Marion (2004). Classical Dynamics of Particles and Systems (5th ed.). Belmont CA: Brook/Cole. Chapter 10. ISBN 0-534-40896-6. 2. John Robert Taylor (2004). Classical Mechanics. Sausalito CA: University Science Books. Chapter 9, pp. 327 ff. ISBN 1-891389-22-X. 3. Robert Resnick & David Halliday (1966). Physics. Wiley. p. 121. ISBN 0-471-34524-5. 4. Federal Aviation Administration (2007). Pilot's Encyclopedia of Aeronautical Knowledge. Oklahoma City OK: Skyhorse Publishing Inc. Figure 3–21. ISBN 1-60239-034-7. 5. Richard Hubbard (2000). Boater's Bowditch: The Small Craft American Practical Navigator. NY: McGraw-Hill Professional. p. 54. ISBN 0-07-136136-7. 6. Lawrence K. Wang & Norman C. Pereira (1979). Handbook of Environmental Engineering: Air and Noise Pollution Control. Humana Press. p. 63. ISBN 0-89603-001-6. 7. Lee M. Grenci & Jon M. Nese (2001). A World of Weather: Fundamentals of Meteorology. Kendall Hunt. p. 272. ISBN 0-7872-7716-9. 8. Jerrold E. Marsden & Tudor S. Ratiu (1999). Introduction to Mechanics and Symmetry: A Basic Exposition of Classical Mechanical Systems. Springer. p. 251. ISBN 0-387-98643-X. 9. Alexander L. Fetter & John Dirk Walecka (2003). Theoretical Mechanics of Particles and Continua. Courier Dover Publications. pp. 38–39. ISBN 0-486-43261-0. 10. John L. Synge (2007). Principles of Mechanics (Reprint of Second Edition of 1942 ed.). Read Books. p. 347. ISBN 1-4067-4670-3. 11. Taylor (2005). p. 342. 12. LD Landau and LM Lifshitz (1976). Mechanics (Third ed.). Oxford: Butterworth-Heinemann. p. 128. ISBN 978-0-7506-2896-9. 13. Louis N. Hand, Janet D. Finch (1998). Analytical Mechanics. Cambridge University Press. p. 267. ISBN 0-521-57572-9. 14. Mark P Silverman (2002). A universe of atoms, an atom in the universe (2 ed.). Springer. p. 249. ISBN 0-387-95437-6. 15. Taylor (2005). p. 329. 16. Cornelius Lanczos (1986). The Variational Principles of Mechanics (Reprint of Fourth Edition of 1970 ed.). Dover Publications. Chapter 4, §5. ISBN 0-486-65067-7. 17. Morton Tavel (2002). Contemporary Physics and the Limits of Knowledge. Rutgers University Press. p. 93. ISBN 0-8135-3077-6. "Noninertial forces, like centrifugal and Coriolis forces, can be eliminated by jumping into a reference frame that moves with constant velocity, the frame that Newton called inertial." 18. Louis N. Hand, Janet D. Finch (1998). Analytical Mechanics. Cambridge University Press. p. 324. ISBN 0-521-57572-9. 19. I. Bernard Cohen, George Edwin Smith (2002). The Cambridge companion to Newton. Cambridge University Press. p. 43. ISBN 0-521-65696-6. 20. Simon Newcomb (1878). Popular astronomy. Harper & Brothers. pp. 86–88. 21. Lawrence S. Lerner (1996). Physics for Scientists and Engineers. Jones & Bartlett Publishers. p. 129. ISBN 0-7637-0253-6. 22. Unknown parameter `\|name=` ignored (`\|author=` suggested) (help) 23. Robert and Gary Ehrlich (1998). What if you could unscramble an egg?. Rutgers University Press. ISBN 978-0-8135-2548-8. 24. ^ a b c Herbert Goldstein (1950). Classical Mechanics. Addison-Wesley. pp. 24–25, 61–64. ISBN 0-201-02918-9. 25. John Clayton Taylor (2001). Hidden unity in nature's laws. Cambridge University Press. p. 26. ISBN 0-521-65938-8. 26. Henry M. Stommel and Dennis W. Moore (1989). An introduction to the Coriolis force. Columbia University Press. pp. 28–40. ISBN 978-0-231-06636-5. 27. Taylor (2005). p. 358-9. 28. ^ a b Taylor (2005). p. 359. 29. Frank Swetz, John Fauvel, Otto Bekken, Bengt Johansson, and Victor Katz (1997). Learn from the masters!. Mathematical Association of America. pp. 268–269. ISBN 978-0-88385-703-8. 30. Whiting, J.S.S. (November 1983). "Motion in a central-force field". Physics Education 18 (6): pp. 256–257. Bibcode:1983PhyEd..18..256W. doi:10.1088/0031-9120/18/6/102. ISSN 0031-9120. Retrieved May 7, 2009. 31. Hans Christian Von Baeyer (2001). The Fermi Solution: Essays on science (Reprint of 1993 ed.). Courier Dover Publications. p. 78. ISBN 0-486-41707-7. 32. Myers, Rusty L. (2006). The basics of physics. Greenwood Publishing Group. p. 57. ISBN 0-313-32857-9. ## References • John Robert Taylor (2005). Classical Mechanics. University Science Books. ISBN 1-891389-22-X.	{"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": 38, "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": 0.8474919199943542, "perplexity_flag": "middle"}
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http://stats.stackexchange.com/questions/6601/what-is-the-hardest-statistical-concept-to-grasp	# What is the hardest statistical concept to grasp? This is a similar question to the one here, but different enough I think to be worthwhile asking. I thought I'd put as a starter, what I think one of the hardest to grasp is. Mine is the difference between probability and frequency. One is at the level of "knowledge of reality" (probability), while the other is at the level "reality itself" (frequency). This almost always makes me confused if I think about it too much. Edwin Jaynes Coined a term called the "mind projection fallacy" to describe getting these things mixed up. Any thoughts on any other tough concepts to grasp? - (I don't know enough to put this as an answer, hence adding a comment.) I always thought it was strange that PI crops up in statistical equations. I mean - what's PI got to do with statistics? :) – Wikis Jan 27 '11 at 9:16 2 – probabilityislogic Jan 27 '11 at 10:59 @Wiki If you accept that $\pi$ crops up when you go from measuring the length of a straigh piece of line to the length of a piece of circle, I don't see why it would not appear while going from measuring a probability to fall down on a segment to measuring the probability to fall down in a piece of circle ? – robin girard Jan 27 '11 at 12:19 @Wiki Whenever you have trigonometric funcions (sine, cosine, tangent etc.) you risk having $\pi$ pop up. And remember that whenever you derive a function you're actually finding a tangent. What is surprising is that $\pi$ doesn't appear more often. – Carlos Accioly Jan 28 '11 at 18:36 @Carlos I suspect the prevalence of $2\pi$ is mostly due to the use of the $\ell^2$ metric, leading to n-spheres. In the same vein, I would expect it's $e$ whose prevalence is due to analysis. – sesqu Jan 29 '11 at 22:00 show 1 more comment ## 12 Answers for some reason, people have difficulty grasping what a p-value really is. - 3 @shabbychef: Most of the people grasp it in the worst possible way i.e. probability of making Type I error. – suncoolsu Jan 27 '11 at 7:19 2 I think that's mostly related to how p-values are explained in classes (i.e.: just by giving a quick definition and without specifying what p-values are NOT) – nico Jan 27 '11 at 7:34 I think this is mainly to do with how it is introduced. For me, it was an "add-on" to the classical hypothesis test - so it appears as though its just another way to do a hypothesis test. The other problem is that it is usually only taught with respect to a normal distribution, where everything "works nice" (e.g. p-value is a measure of evidence in testing a normal mean). Generalising the p-value is not easy as there is no specific principles to guide the generalisation (e.g. there is no general agreement on how a p-value should vary with the sample size & multiple comparisons) – probabilityislogic Jan 27 '11 at 11:12 @shabbychef +1 though student often have difficulties with p-values (roughly because the concept in testing is a bit more subtle than a binary decision process and be cause "inverting a function" is not easy to aprehend). When you say "for some reason" do you mean it is unclear for you why people have difficulties ? PS: If I could, I would try to make statistics on this site about the relation between "being a top answer" and "talking about p-value" :) . I also even ask myself if the hardest statistical concept to grasp can have the most upvote (if it is difficult to grasp ... :) ) – robin girard Jan 27 '11 at 12:28 1 @eduardo - yes a small enough p-value is sufficient to cast doubt on the null hypothesis: but it is calculated in complete isolation to an alternative. Using p-values alone, you can never formally "reject" $H_0$, because no alternative has been specified. If you formally reject $H_0$, then you must also reject the calculations which was based on the assumption of $H_0$ being true, which means you must reject the calculation of the p-value that was derived under this assumption (it messes with your head, but it is the only way to reason consistently). – probabilityislogic Jan 30 '11 at 5:14 show 32 more comments Similar to shabbychef's answer, it is difficult to understand the meaning of a confidence interval in frequentist statistics. I think the biggest obstacle is that a confidence interval doesn't answer the question that we would like to answer. We'd like to know, "what's the chance that the true value is inside this particular interval?" Instead, we can only answer, "what's the chance that a randomly chosen interval created in this way contains the true parameter?" The latter is obviously less satisfying. - 1 The more I think about confidence intervals, the harder it is for me to think of what kind of question they can answer at a conceptual level that cannot be answered by asking for "the chance a true value is within an interval, given one's state of knowledge". If I were to ask "what is the chance (conditional on my information) that the average income in 2010 was between 10,000 and 50,000?" I don't think the theory of confidence intervals can give an answer to this question. – probabilityislogic Jan 27 '11 at 11:28 What is the meaning of "degrees of freedom"? How about df that are not whole numbers? - Conditional probability probably leads to most mistakes in everyday experience. There are many harder concepts to grasp, of course, but people usually don't have to worry about them--this one they can't get away from & is a source of rampant misadventure. - +1; could you add an example or two, favourite or current ? – Denis Jan 31 '11 at 14:53 For starters: P(you have the disease\|test is positive) != P(test is positive\|you have the disease). – xmjx Sep 23 '11 at 6:39 Tounge firmly in cheek: For frequentists, the Bayesian concept of probability; for Bayesians, the frequentist concept of probability. ;o) Both have merit of course, but it can be very difficult to understand why one framework is interesting/useful/valid if your grasp of the other is too firm. Cross-validated is a good remedy as asking questions and listening to answers is a good way to learn. - 2 I rule I use to remember: Use probabilities to predict frequencies. Once the frequencies have been observed, use them to evaluate the probabilities you assigned. The unfortunately confusing thing is that, often the probability you assign is equal to a frequency you have observed. One thing I have always found odd is why do frequentists even use the word probability? wouldn't it make their concepts easier to understand if the phrase "the frequency of an event" was used instead of "the probability of an event"? – probabilityislogic Jan 28 '11 at 16:58 Interestingly, cross validation can be seen as a Monte Carlo approximation to the integral of a loss function in Decision Theory. You have an integral $\int p(x) L(\textbf{x}_{n},x) dx$ and you approximate it by $\sum_{i=1}^{i=n} L(\textbf{x}_{[n-i]},x_i)$ Where $\textbf{x}_{n}$ is data vector, and $\textbf{x}_{[n-i]}$ is the data vector with the ith observation $x_i$ removed – probabilityislogic Jan 30 '11 at 1:30 I think that very few scientists understand this basic point: It is only possible to interpret results of statistical analyses at face value, if every step was planned in advance. Specifically: • Sample size has to be picked in advance. It is not ok to keep analyzing the data as more subjects are added, stopping when the results looks good. • Any methods used to normalize the data or exclude outliers must also be decided in advance. It isn't ok to analyze various subsets of the data until you find results you like. • And finally, of course, the statistical methods must be decided in advance. Is it not ok to analyze the data via parametric and nonparametric methods, and pick the results you like. Exploratory methods can be useful to, well, explore. But then you can't turn around and run regular statistical tests and interpret the results in the usual way. - 5 – Dikran Marsupial Jan 27 '11 at 17:32 3 I would partially disagree here. I think the caveat that people miss is that the appropriate conditioning operations are easy to ignore for these kinds of issues. Each of these operations change the conditions of the inference, and hence, they change the conditions of it applicability (and therefore to its generality). These is definitely only applicable to "confirmatory analysis", where a well defined model and question have been constructed. In exploratory phase, not looking to answer definite questions - more looking to build a model and come up with hypothesis for the data. – probabilityislogic Jan 27 '11 at 18:05 I edited my answer a bit to take into account the comments of Dikran and probabilityislogic. Thanks. – Harvey Motulsky Jan 28 '11 at 14:52 1 For me, the "excluding outliers" is not as clearly wrong as your answer implies. For example, you may only be interested in the relationships at a certain range of responses, and excluding outliers actually helps this kind of analysis. For example, if you want to model "middle class" income, then excluding the super rich and impoverished outliers is a good idea. It is only the outliers within your frame of inference (e.g. "strange" middle class observations) were your comments apply – probabilityislogic Jan 29 '11 at 6:10 1 Ultimately the real problem with the issues raised in the initial answer is that they (at least partially) invalidate p-values. If you are interested in quantifying an observed effect, one should be able to do any and all of the above with impunity. – Russell S. Pierce Jan 29 '11 at 19:24 show 3 more comments What do the different distributions really represent, besides than how they are used. - 2 This was the question I found most distracting after statistics 101. I would encounter many distributions with no motivation for them beyond "properties" that were relevant to topics at hand. It took unacceptably long to find out what any represented. – sesqu Jan 29 '11 at 22:12 1 Maximum entropy "thinking" is one method which helps understand what a distribution is, namely a state of knowledge (or a description of uncertainty about something). This is the only definition that has made sense to me in all situations – probabilityislogic Jan 30 '11 at 4:57 – David Sep 23 '11 at 5:04 Fiducial inference. Even Fisher admitted he didn't understand what it does, and he invented it. - From my personal experience the concept of likelihood can also cause quite a lot of stir, especially for non-statisticians. As wikipedia says, it is very often mixed up with the concept of probability, which is not exactly correct. - I think the question is interpretable in two ways, which will give very different answers: 1) For people studying statistics, particularly at a relatively advanced level, what is the hardest concept to grasp? 2) Which statistical concept is misunderstood by the most people? For 1) I don't know the answer at all. Something from measure theory, maybe? Some type of integration? I don't know. For 2) p-value, hands down. - Measure theory is neither a field of statistics nor hard. Some types of integration are hard, but, once again, that isn't statistics. – Eduardo León Jan 29 '11 at 22:12 Confidence interval in non-Bayesian tradition is a difficult one. - I think people miss the boat on pretty much everything the first time around. I think what most students don't understand is that they're usually estimating parameters based on samples. They don't know the difference between a sample statistic and a population parameter. If you beat these ideas into their head, the other stuff should follow a little bit easier. I'm sure most students don't understand the crux of the CLT either. -	{"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": 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": 0.9588578939437866, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/77628?sort=votes	## Question about Godel’s 2nd Theorem ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Let Con(PA) be the sentence of arithmetic which translates as "Peano Arithmetic is consistent." Then according to Godel's 2nd incompleteness theorem, assuming PA is consistent then PA can neither prove Con(PA) nor its negation. And in fact, if T contains PA and T is (omega-) consistent, then T can neither prove Con(T) nor its negation. In particular, if PA+Con(PA) is consistent then PA+Con(PA) can neither prove Con(PA+Con(PA)) nor its negation. But consider the following piece of reasoning: if PA is consistent, then Con(PA) is true, so PA+Con(PA) is consistent, so Con(PA+Con(PA)) is true. My question is, why can't this reasoning by formalized in PA, so that within PA you can prove Con(PA) implies Con(PA+Con(PA))? If you could prove that, then since you can obviously prove Con(PA) within PA+Con(PA), you would be able to prove Con(PA+Con(PA)) within PA+Con(PA), which is a contradiction. So where am I going wrong? We can even talk about this in terms of model theory. There are nonstandard models of PA in which Con(PA) does not hold: basically, you have infinitely large natural numbers, and infinitely long proofs of a contradiction in PA. This does not mean that PA is inconsistent, because there are no proofs of finite length of this contradiction. So are there also nonstandard models of PA+Con(PA) in which Con(PA+Con(PA)) does not hold? (That's a rhetorical question; clearly there must be, but what do they look like?) Any help would be greatly appreciated. Thank You in Advance. - Doesn't the statement "if T contains PA and T is (omega-) consistent, then T can neither prove Con(T) nor its negation." require that T be computable? – Quinn Culver Oct 14 2011 at 13:18 Quinn, we are of course talking about recursively axiomatized theories, so yes T is "computable." – Keshav Srinivasan Oct 15 2011 at 14:13 This is not about your question itself, but you should avoid editing the question too often for two reasons: (1) it bumps it up to the top of the question thread, which is not a desirable effect for a minor edit; (2) it will eventually become community-wiki. – Thierry Zell Oct 18 2011 at 1:58 ## 2 Answers Here's a somewhat more detailed version of Ricky Demer's answer. In the first sentence of your piece of reasoning, you say "if PA is consistent, then Con(PA) is true, so PA+Con(PA) is consistent". The most natural justification for the step from "PA consistent and Con(PA) true" to "PA+Con(PA) consistent" presupposes that PA is true. After all, a consistent but false theory could become inconsistent when some true statement is added to it. So formalizing your argument would require proving that PA is true; that can't be done in PA --- in fact, "PA is true" can't even be expressed in the language of PA. A more subtle justification for the step to "PA+Con(PA) is consistent" would use not the truth of PA but the weaker statement that all `$\Sigma^0_1$` sentences provable in PA are true. That can be expressed in the language of PA, but alas it's not provable in PA. So the argument still can't be formalized in PA. - All right, what if we replaced every occurrence of the word "consistent" in my original post with the phrase "sigma-1 sound", and we replaced Con(T) with Sig(T), a statement of arithmetic saying "T is sigma-1 sound." In that case what would be my error? – Keshav Srinivasan Oct 10 2011 at 0:01 1 Your error would be explained by doing the same replacements to my answer. $\hspace{1 in}$ – Ricky Demer Oct 10 2011 at 1:11 Ricky, your answer was "You can't carry that out in PA because it could be that PA is consistent and proves ¬Con(PA)." But is it possible for PA to be sigma-1 sound and prove ¬Sig(PA)? – Keshav Srinivasan Oct 10 2011 at 1:22 Yes. $\;$ (The only way I know to show that is from Godel's 2nd Thm.) $\;\;$ – Ricky Demer Oct 10 2011 at 3:03 1 @Keshav: No, it is $\Pi^0_2$ (and this is an optimal classification, as it is not contained in any consistent extension of PA by a set of $\Sigma^0_2$-sentences). By the way, the standard name of this statement is the uniform $\Sigma^0_1$-reflection principle, and it is denoted as $\mathrm{RFN}_T(\Sigma^0_1)$, $\mathrm{RFN}_{\Sigma^0_1}(T)$ or similar variations in the literature. – Emil Jeřábek Oct 12 2011 at 10:10 show 7 more comments ### 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. You can't carry that out in PA because it could be that PA is consistent and proves $\lnot$Con(PA). (in which case PA is not $\omega$-consistent) The best description of a model "of PA+Con(PA) in which Con(PA+Con(PA)) does not hold" would be "Take the standard model of PA, then 'insert' a nonstandard natural coding a proof of $\lnot$Con(PA)". -	{"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": 0.18, "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": 0.9464979767799377, "perplexity_flag": "middle"}
http://mathhelpforum.com/advanced-algebra/170513-find-basis.html	# Thread: 1. ## find the basis Let L be the line spanned by v_1=[1, 1, 1]. Find a basis {v_2, v_3} for the plane perpendicular to L, and verify that B= {v_1,, v_2, v_3} is a basis for R3. Let ProjL denote the projection onto the line L. Find the matrix B for ProjL with respect to the basis B. Can you just help me get started with the problem. I'll do the rest on my own. 2. Originally Posted by Taurus3 Let L be the line spanned by v_1=[1, 1, 1]. Find a basis {v_2, v_3} for the plane perpendicular to L, and verify that B= {v_1,, v_2, v_3} is a basis for R3. Let ProjL denote the projection onto the line L. Find the matrix B for ProjL with respect to the basis B. Can you just help me get started with the problem. I'll do the rest on my own. Since $v_1$ is perpendicular to the plane you want to span it is the planes normal vector. Since the equation of the plane must pass though the origin it will have the form $x+y+z=0$ Can you finish from here? 3. ummm......actually no. Sorry. I mean I know how to find the eigenvalues and then the basis. But this question is weird. 4. Originally Posted by Taurus3 ummm......actually no. Sorry. I mean I know how to find the eigenvalues and then the basis. But this question is weird. We don't need eigenvectors or eigenvalues. We need to find two vectors that span the above plane. e.g we need to find the basis for the null space of this matrix. $\begin{bmatrix} 1 & 1 & 1& 0\end{bmatrix}$ If it helps you can think of it as this matrix with added rows of zeros. $\begin{bmatrix} 1 & 1 & 1& 0 \\ 0 & 0 & 0& 0 \\ 0 & 0 & 0& 0 \\ \end{bmatrix}$ Since this is already in reduced row form We know that we have two free parameters. Let $z=t,y=s$ then $x+t+s=0 \iff x =-t-s$ So the basis of the nullspace is $\begin{pmatrix} x \\ y \\ z\end{pmatrix} = \begin{pmatrix} -t-s \\ s \\ t\end{pmatrix} = \begin{pmatrix} -t \\ 0 \\ t\end{pmatrix}+ \begin{pmatrix} -s \\ s \\ 0\end{pmatrix} = \begin{pmatrix} -1 \\ 0 \\ 1\end{pmatrix}t+\begin{pmatrix} 0\\ -1 \\ 1\end{pmatrix}s$ You can verify that the two above vectors are perpendicular to the first. Can you finish from here? 5. So for my 2nd question, would it just be B= [(1, 1, 1), (0, -1, 1), (-1, 0, 1)]? 6. So for my 2nd question, would it just be B= [(1, 1, 1), (0, -1, 1), (-1, 0, 1)]? No that is the matrix of the linear transformation from the standard basis to your new basis $B$. Hint: In your new basis $T(v_1)=v_1$ and $T(v_2)=T(v_2)=0$ So what is the matrix of this transformation? 7. this is what I don't get. How do you know T(v1)=v1 and that T(v2)=0? 8. Originally Posted by Taurus3 this is what I don't get. How do you know T(v1)=v1 and that T(v2)=0? Since $\displatstyle T(\vec{x})=\text{proj}_{v_1}\vec{x}=\frac{\vec{v_1 }\cdot \vec{x}}{\|\|v_1\|\|^2}\vec{v_1}$ This is the definition of projecting one vector onto another. This will verify the above "claim".	{"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": 11, "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": 0.9024828672409058, "perplexity_flag": "middle"}
http://nrich.maths.org/7048	### Ball Bearings If a is the radius of the axle, b the radius of each ball-bearing, and c the radius of the hub, why does the number of ball bearings n determine the ratio c/a? Find a formula for c/a in terms of n. ### Overarch 2 Bricks are 20cm long and 10cm high. How high could an arch be built without mortar on a flat horizontal surface, to overhang by 1 metre? How big an overhang is it possible to make like this? ### Cushion Ball The shortest path between any two points on a snooker table is the straight line between them but what if the ball must bounce off one wall, or 2 walls, or 3 walls? # Population Dynamics - Part 3 ### The Logistic Equation Unlimited growth is generally impossible, because birth and death rates are affected by factors such as lack of food, water and land. We now incorporate the effect of the environment on population size into the exponential model: $\frac {\mathrm{d}N}{\mathrm{d}t} =rN(t)$ Below are two derivations of this new model, called the logistic equation. It was originally devised by the mathematician Thomas Malthus. Notes on the discrete form of this equation can be found here. ### First Derivation The carrying capacity, K, is the largest population that can be supported indefinitely, given the resources available in the environment. When the population size is far below K, its growth is exponential. As the population approaches K, it begins to be affected by the reduced ability of the environment to provide necessary resources. In order to have this effect, we consider the term $\frac {K-N}{K}$ . Note that when $N< < K$ then $\frac {K-N}{K}\approx 1$ and when $N \rightarrow K$ then $\frac {K-N}{K} \rightarrow 0$. Including the above term, we now get the equation: $$\frac {\mathrm{d}N}{\mathrm{d}t}=r \frac{K-N}{K} N(t)= rN\left(1-\frac{N}{K}\right)$$Note that each individual added to the population reduces the rate of increase of the whole population. ### Second Derivation Consider a population of lions of size y, and begin by assuming the simple exponential equation $\frac {\mathrm{d}y}{\mathrm{d}t}=r y$ , where r is the intrinsic growth rate. If the probability of food being found by an individual lion is y, then the probability of some food being found by two lions is proportional to $y^2$. Fighting will occur within the species for these limited resources, so the death rate due to fighting can be represented by $\gamma y^2$ . Our equation then becomes: $$\frac {\mathrm{d}y}{\mathrm{d}t}=ry-\gamma y^2=ry\left(1-\frac{y}{Y}\right)$$where $Y=\frac{r}{\gamma}$. Again this is the logistic equation. Question: Graph the logistic equation to find the equilibrium points. Then solve the equation using standard integrals, showing that the solution is given by $N(t)=\frac {K N_0 e^{rt}}{K+N_0 (e^{rt}-1)}$ Graphically we describe this kind of population growth by a sigmoid, or S-shaped growth curve: By looking at the blue curve, we see the size of the population begins to level off and reaches a stable value, which must be less than the carrying capacity of the environment (marked in red). Mathematically, we can show this by looking at the above solution for $N(t)$ and showing $\lim_{t\to\infty} N(t)=K$ Question: A small lake has a carrying capacity of 100 geese. Starting with a pair of geese, how would the population change over 70 years if $r=0.1$? Draw a sigmoid graph of this change. Recall that r is the intrinsic growth rate. How will this affect the time taken for the number of geese to reach the carrying capacity of the lake? Perhaps start by drawing graphs with different r-values. 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": 14, "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": 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": 0.932403564453125, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/33242/must-all-symmetries-have-consequences?answertab=oldest	# Must all symmetries have consequences? Must all symmetries have consequences? We know that transnational invariance, for example, leads to momentum conservation, etc, cf. Noether's Theorem. Is it possible for a theory or a model to have a symmetry of some kind with no physical consequences at all for that symmetry? - 2 – Alex Nelson Aug 1 '12 at 4:31 ## 3 Answers Noether's theorem is overzealously applied--- it only applies to theories with a Lagrangian formulation, or to quantum mechanics. This is true of fundamental systems, but for non-fundamental systems, you can have classical equations which are symmetrical the symmetry does not imply a conservation law. The symmetry does not come with no consequence, however, it comes with the consequence of symmetrical solutions! If the equations are symmetrical under a transformation, the solutions must come in families that turn into each other under the symmetry. For classical systems, this is not a particularly profound consequence. So I will consider systems where this is the only consequence. For a stupid example, consider Newton's laws for an object free-falling in a gravitational field. The acceleration is uniform in the z direction, but z momentum is not conserved. The reason is that the Lagrangian is not invariant in the z direction. But you wouldn't know it from looking at the equations of motion. For a less stupid example, consider Newton's laws for a particle with constant force and a $v^3$ friction law, say: $${dv \over dt} = a + v^3$$ And there is a symmetry in time translation and for translations in x. But aside from telling you the trivial fact that solutions are translatable in x and in t, it doesn't tell you anything more. These problems are sort of silly, so I'll give the granddaddy of all examples--- the incompressible Navier stokes equations, with hyperviscosity (so that what I am saying is definitely true). Here you have a time-dependent diffeomorphism $X(x)$ from n-dimensional space whose general point is called x, to itself. The time derivative of X is the velocity field v, and v obeys the equation $${\partial_t} v = v\cdot \nabla v + \nabla P + \nu \nabla^2 v + \epsilon \nabla^4 v$$ Where the $\epsilon$ term is introduced to make sure that the equation has a unique and smooth initial value problem, so that the X diffeomorhism makes sense. This equation is translationally invariant to t translations, and completely diffeomorphism invariant--- composing X with a diffeomorphism takes a solution to a solution, but it doesn't have a conserved energy (although formally the limit $\nu=\epsilon=0$ does), nor does it have any conserved quantity corresponding to the diffeomorphism invariance. The diff invariance is a gauge redundancy in the X description. These symmetries still have the trivial consequence of symmetrical solutions--- so you can translate any solution in time, and do a diffeomorphism on the initial positions. It just doesn't mean anything to studying the equation. - What do you mean by "consequences"? Classically, topological numbers might fit that bill. In quantum field theory, that depends upon whether it is sensible to consider superpositions of topological sectors. If there is an S-duality, it might possibly be. - How are topological numbers a "symmetry"? Same for S-duality. This reads like nonsense superficially. – Ron Maimon Aug 1 '12 at 8:39 If there is a continuous symmetry of the action, it is necessary to take the quotient by the symmetry--gauge fix it--when quantizing. One way to see this is to just consider perturbation theory. The quadratic coupling in the Lagrangian will be constant along the flow of the symmetry, so it will be have degenerate derivative in the tangent space directions along that flow and hence not be invertible. Thus, we won't be able to find a propagator. Abstractly, perturbation theory works when the classical theory one is quantizing is integrable. When showing a 2n dimensional system is integrable, one must specify n Poisson-commuting integrals of motion. The continuous symmetry gives one by Noether's theorem, and n is the most possible in 2n dimensions, so one has to come up with n integrals of motion which combine linearly into the Noether charge of the symmetry. In other words, one eventually includes the symmetry in the solution regardless of whether he meant to. I feel like there might be weirder things that happen nonperturbatively, but I am not sure. Certainly when one wants all observables to also obey the symmetry, one can cook up a BRST operator. The existence of this operator implies that some states in the "big" Hilbert space that doesn't take account of the symmetry necessarily have zero norm. Can one do this in all cases? It is possible on the other hand to lose symmetries when one quantizes. Usually this is the fault of the regularization scheme, but in certain cases one can prove that no regularization scheme is invariant under the symmetry. An example is the 2+1 dimensional Chern Simons theory with Wilson loops. Even though the Lagrangian is fully topologically invariant, one needs to pick a framing of loops to regularize self-intersections. Similarly there are theories with manifestly conformal Lagrangians but which require breaking scale invariance in regularization. Discrete symmetries are of course of a much different flavor. - 1 This is false. You only need to take the quotient when doing the path integral over a locally infinite volume gauge invariance. You most definitely do not take the quotient for a single free particle which has a continuous symmetry which takes all points to the origin! Then you couldn't have free particle motion. – Ron Maimon Oct 31 '12 at 1:47	{"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": 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": 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": 0.9327443838119507, "perplexity_flag": "head"}
http://mathoverflow.net/questions/114337/representability-of-sheaf-of-ext1-of-a-neron-model-by-mathbbg-m	## Representability of sheaf of Ext^1 of a Néron model by $\mathbb{G}_m$ ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Let's work over a trait $S=\mathrm{Spec}R$, where $R$ is a dvr with fraction field $K$, residue field $k$. Given an abelian variety $A_K$ with semi-stable reduction, let $A$ over $S$ be its Néron model and $A^{\circ}$ the neutral component. We know the sheaf $\mathscr{E}xt^1(A^{\circ},\mathbb{G}_m)$ is represented by the Néron model of the dual of $A_K$ over the category of smooth scheme over $S$, see (Mazur and Messing's LNM Universal extensions and one dimensional crystalline cohomology, chapter I section 5) My question is: Is the sheaf $\mathscr{E}xt^1(A^{\circ},\mathbb{G}_m)$ also representable over the category of schemes over $S$? Also we know the Poincaré biextension $W_K$ of $A_K$ and $A_K^'$ by $\mathbb{G}_m$ extends to a biextension $W$ of certain open subgroups of $A$ and $A'$ (i.e. the subgroups making the component pairing vanish), my second question is if $W$ is represented by a scheme over $S$? - ## 1 Answer First question: no. Assume, to fix ideas, that $R$ is complete with uniformizer $\pi$, $k$ is algebraically closed, and $A$ is an elliptic curve with multiplicative reduction. Denote by $\mathscr{E}$ the Ext sheaf in question. Then the restriction of $A^\circ$ to `$S_n:=\mathrm{Spec\,}(R/(\pi^{n+1}))$` is isomorphic to $\mathbb{G}_{m}$, so $\mathscr{E}(S_n)$ is zero for all $n$. If $\mathscr{E}$ were a scheme, this would imply $\mathscr{E}(R)=0$ (the functor of points of a scheme "commutes with completion" for local rings), a contradiction. Second question: yes, because $W$ is a $\mathbb{G}_m$-torsor over the product, and torsors under affine group schemes are schemes. - Thanks a lot! Could you please tell me the reference for 1. "the functor of points of a scheme "commutes with completion" for local rings"; 2. "torsors under affine group schemes are schemes" – Heer Nov 25 at 16:53 @ Laurent Moret-Bailly: Is $A^{'\circ}$ like an "open" subsheaf of $\mathscr{E}xt^1(A,\mathbb{G}_m)$ over the category of schemes over S? How to understand the relation between the two? – Heer Nov 25 at 17:48 Concerning completions: I don't know a reference but this is an exercise. First observe that for an affine scheme, the functor of points commutes with arbitrary inverse limits of rings (this is trivial). For an arbitrary scheme $X$, a filtered inverse system of local rings $R_i$ (with local transition maps) and compatible elements $f_i$ of $X(R_i)$, observe that all closed points of the spectra must map to the same point $x$ of $X$, hence all $f_i$'s must factor through any affine neighborhood of $x$, so we are reduced to the affine case. – Laurent Moret-Bailly Nov 26 at 18:52 Concerning representability of torsors: this is just flat descent of affine morphisms (SGA1, VIII, th. 2.1). For the special case of torsors, I am sure this is stated in Demazure-Gabriel. – Laurent Moret-Bailly Nov 26 at 19:01	{"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": 46, "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": 0.9120337963104248, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/258665/what-does-the-localization-of-a-finite-ring-look-like?answertab=active	# What does the localization of a finite ring look like? Let $A$ be the ring $\mathbb{Z}/60$. What does it look like under the localization at the prime (2)? Everything that is not divisible by 2 should be a unit, so one might thing this looks like $\mathbb{Z}/4$ - but on the other hand, rings always inject into their localizations... I'm asking because Vakil in his notes says that the stalk at (2) of this ring is $\mathbb{Z}/4$. But this doesn't make sense to me because of the above confusion. - 2 Dear @Wouter, It is not true that rings always inject into their localizations. If $R$ is a ring and $S$ a multiplicative set, then the kernel of $R\rightarrow S^{-1}R$ is the set of $r\in R$ for which there exists some $s\in S$ with $sr=0$. So if $S$ contains zero divisors, $R$ will not inject into $S^{-1}R$ (unless $R=0$). If there are no zero divisors, that is, if $R$ is a domain, then it is true that $R$ injects into all of its (non-zero) localizations. – Keenan Kidwell Dec 14 '12 at 13:26 A nice exercise says that any ring of fractions of $\mathbb Z_n$ is isomorphic to some $\mathbb Z_d$ with $d\mid n$. – YACP Dec 14 '12 at 15:23 ## 1 Answer but on the other hand, rings always inject into their localizations... This is false: you've been spoiled by the case of integral domains! Actually this can even fail for integral domains: $0^{-1} \mathbb{Z}$ is the zero ring, since this forces $1 = 0 \cdot 0^{-1} = 0$. In the more general case, localizations have the effect of discarding any "components" of your ring that are away from your ideal (this sentence makes more sense in the geometric picture). In this case, the components are $\mathbb{Z} / 4$, $\mathbb{Z} / 3$, and $\mathbb{Z} / 5$. So, localization at $(2)$ has the effect of focusing only on the $\mathbb{Z} / 4$ case, and discarding the other two components. If you understand the example above of inverting zero, this might be more clear if you write it as $$\left(\mathbb{Z}/60 \mathbb{Z}\right)_{(2)} = \{3^{-1}, 5^{-1}\} \mathbb{Z} / 60 \mathbb{Z}$$ - I see my error, but how does your definition correspond to the formal definition of a localization? Isn't 3/2 an element of the localization, for instance? – Wouter Zeldenthuis Dec 14 '12 at 13:45 No. $2$ cannot appear in the denominator of a fraction in the localisation at $(2)$, because we are not inverting the elements in $(2)$! – Zhen Lin Dec 14 '12 at 13:58 Sorry, I meant 2/3... – Wouter Zeldenthuis Dec 14 '12 at 14:09 Yes, $2/3$ is an element of the localization; it equals $2$. – Hurkyl Dec 14 '12 at 14:16	{"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": 31, "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": 0.9376362562179565, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/tagged/time?page=2&sort=newest&pagesize=50	# Tagged Questions Time is defined operationally to be that which is measured by clocks. The SI unit of time is the second, which is defined to be 7answers 732 views ### What grounds the difference between space and time? We experience space and time very differently. From the point of view of physics, what fundamentally grounds this difference? Dimensionality (the fact that there are three spatial dimensions but only ... 3answers 266 views ### Is time dilation an illusion? It is said that we can verify time dilation by flying a very accurate clock on a fast jet or spaceship and prove that it registers less time than the clocks on earth. However, the clocks on earth ... 1answer 74 views ### Degree of Time Dilation At a Distance From the Sun where acceleration = g? At higher altitudes above a body, clocks tick more slowly, and gravitational field is weaker. But what is the relationship? 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Is there (or can there be) a device that could measure the speed and acceleration of time?	{"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": 15, "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": 0.9480692148208618, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/39693/calculating-torque-adjustment-for-offset-of-pivot-point?answertab=votes	# Calculating torque adjustment for offset of pivot point I'm trying to figure out how much to adjust torque, on a torque wrench when using an extension that offsets the pivot point. See this calculator for a diagram of what I'm talking about: The trouble is this calculator can't be correct for the problem I have in mind. It adjusts the torque based on the torque wrench's handle length (i.e. the adjustment factor is simply L/(L+E)), but on a clicker type torque wrench, the measured torque is independent of the handle length. I have several different torque wrenches, all with different handle lengths; if configured to apply the same torque to the extension, it makes no sense that the resultant torque would be different. Another way to think about it is using a torque screwdriver instead of a torque wrench. A torque screwdriver has no lever (beyond the rubber on the grip). Of course the motion is slightly more complicated; as the screwdriver rotates, the screwdriver itself would need to orbit around the offset pivot point. Do I have this all wrong? Is the naive and straightforward calculation actually correct? - ## 2 Answers This can be solved as a statics problem... Consider the wrench in the setting with A=0. You press down on the wrench at a distance $L$ from the pivot point with a force $F$, which is compensated by the nut with an equal but opposite force $F$ at a distance $E$ from the pivot point. The total torque generated by this two forces is $F(E+l)$ CW, which is compensated by an equal but CCW torque from the nut. Now consider what happens at the pivot point, removing either the left or the right part of the wrench. • If you keep the left side, to have equilibrium, at the pivot point you will have to have a force of magnitude $F$ acting down, plus a CW torque of value $F(E+L) - FE = FL$, because the reaction force is also generating torque. • If you keep the right side it is clear that you only have to resist the force on the handle, and the torque it generates, $FL$ So for the $A=0$ case the torque is augmented by a factor of $\frac{E+L}{L}$. For different angles, instead of $E$ you have to use $E\cos A$. This formulas hold if you are delivering the torque by exerting a force which is resisted by the nut. If instead you apply a pair couple that generates torque without a resultant total force, there is no conversion to be done, as the torque you apply will be what is delivered to the nut. That's how you normally apply torque to a screwdriver, as you suggest, and you could try and do something similar with a wrench, by holding it with both hands and pushing in opposing directions, although it would be hard to be sure that there is no net resulting force. But there is nothing wrong with the calculator either. EDIT Another way of looking at the same problem... The whole wrench arrangement behaves as an end loaded cantilever beam: The torque applied at the support is $PL$, but since we don't have our torque measuring pivot at the support, but at a distance $E$ from it, we need to consider what the bending moment at a distance $E$ from the support. Bending moment varies linearly from a maximum at the support, to 0 where the force is applied: So unless you are measuring your torque at the very nut, i.e. without an extension, the total length of the lever has an effect. - OK; now think about this adjustment. Suppose I have a very high torque requirement, and I slip a cheater bar onto the torque wrench. If I'm not using an extension that changes the pivot point, I don't need to modify the torque setting on the wrench; but if I am, then I do. Why is this? – Barry Kelly Oct 12 '12 at 22:00 I mean, I knew the mathematics beforehand - it's straightforward - but it seems counter-intuitive, and that's always been a red flag for me, hinting something may be wrong with the model. Holding F constant but moving it further away from the pivot point certainly increases torque, but we don't lengthen the handle to increase torque. The target torque is always the same. We lengthen the handle to reduce F. Same thing, you say, but the point is that FL is constant no matter what L is. Why does changing the pivot make it variable? – Barry Kelly Oct 12 '12 at 22:20 Actually, it is F(E+L) that stays constant, which means that F*L, which is what your pivot measures, changes... – Jaime Oct 12 '12 at 22:25 OK, I think I'm just about convinced. Thanks! – Barry Kelly Oct 12 '12 at 22:31 Sorry, but I still think there is something missing. At what point is the equilibrium resolved? answer: at the point where the wrench clicks. If the wrench clicks always at the calibrated torque, then F will be just the resulting force to achieve that torque. A good way to visualize this is to imagine a very long extension and a short and tiny clicker wrench calibrated to a very low value with a tiny handle. One would imagine that the torque will be amplified, but in the end we will keep trying to move the nut and wrench will keep clicking... I argue that the nut will see a torque of whatever the wrench is calibrated to times Cos(A). The free body diagrams or energy conservation calculations will show that. Therefore there is no augmentation when wrench is aligned, which is contrary to the first answer. -	{"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": 15, "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": 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": 0.9462507963180542, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/3773/shape-of-the-higgs-branching-ratio-to-zz	# Shape of the Higgs branching ratio to ZZ I've been looking at the, now very popular, graph of the SM Higgs decay branching ratios: You see that the ZZ branching ratio has a funny dip around the $170\, GeV$, very different from the WW counterpart. It's true that graphs with Z will always have the $\gamma$ interference and, perhaps, this is the cause for this funny shape. However, the value of the mass for which this happens is not something that I can intuitively explain. Given that I have no idea of the right answer, let me ask: What causes this dip in the $H\rightarrow ZZ$ branching ratio around $M_H=170\, GeV$? - ## 2 Answers The dip of the ZZ branching ratio - and all other branching ratios except for the WW branching ratio - near 170 GeV is caused by the increase of the total width (decay rate) at those masses. And the total width (decay rate) around 170 GeV increases exactly because the Higgs decays to the WW final states start to become possible. Because the total width goes up, the ratio of a (non-WW) partial width and the total width goes down - and this ratio is what we call the branching ratio. Kostya wrote almost the same thing. But I want to emphasize a subtle point: note that in your graph, the branching ratio to WW is nonzero already from Higgs masses at 80 GeV or so. Similarly, the branching ratio to ZZ is nonzero from 90 GeV. How can a 90 GeV Higgs decay to two Z's, each of which has mass close to 90 GeV? Doesn't it violate energy conservation? The answer is that the graph shows the decays to off-shell particles, not the final states. A Higgs boson may decay to one virtual and one real Z-boson. The virtual particle continues in its decay. To check this hypothesis, note that all decay channels in the graph are composed out of two particles. The actual final states of the decay will often include (many) more than two particles. When the Higgs mass exceeds two times the mass of the W-bosons, the total width genuinely goes up because there's suddenly a lot of new "phase space" of the final states. - Thanks for the answer, but could you please write my name with f? I appreciate. – Rafael Jan 24 '11 at 19:41 Fixed, @Rafael, apologies. I would always write it with a "f" but ten years in the U.S. and interactions with the likes of Raphael Bousso changes some detailed things. – Luboš Motl Jan 26 '11 at 9:25 Nice question. I did notice this "dip" too, but didn't think too much about it. While the explanation seems to be pretty simple: $M_W \simeq 80\,\mbox{GeV},\quad M_Z \simeq 91\,\mbox{GeV}$ Therefore, around 170 GeV you are below the threshold for ZZ, but above the threshold for WW. But I also think that this dip tends to be overemphasised on such a graph. First because of the logarithmic scale, and also since the branching ratio is the relative to the full width. Edit: In response to Luboš Motl's response I'll add two more points: 1. The reference to the original of the plot: hep-ph/9704448 2. In fact, below the threshold it is safe (with 1% accuracy) to assume that one boson is on shell. -	{"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": 0.3, "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": 0.9502456784248352, "perplexity_flag": "middle"}
http://mathhelpforum.com/math-topics/163209-heat-transfer-problem.html	# Thread: 1. ## Heat transfer problem, A heat exchanger produces steam by boiling water within vertical tubes, as shown in Figure 3.1 i)Water is supplied at a rate of 2 kg s–1 and an initial temperature of 80°C. Calculate the rate of heat transfer required to produce saturated steam at 140°C. What would the pressure in the top of the heat exchanger be (in kPa gauge)? ii)The water/steam passes through a bundle of 50 vertical tubes, each of 3 cm internal diameter, 4 cm outer diameter, and 4 m in length, made of stainless steel with thermal conductivity 19 W m–1 K–1. Calculate the overall resistance to heat transfer across the tube walls. Hence calculate the required temperature driving force to deliver the heat transfer rate calculated in part (i). I am really stuck, don't know what to do. Any help appreciated. Attached Thumbnails 2. Okay I asked my teacher and he said I have to use this equation for the second part, thermal resistance for a single pipe = $\frac{ln(\frac{r_{o}}{r_{i}})}{2 \pi H \lambda}$ $H = 4m$ so $r_{o} = 2 cm = 0.02 m$ $r_{i} = 1.5cm = 0.015 m$ $\lambda = 19 W m^{-1} K^{-1}$ pluggiing these numbers in I get $6.02\times 10^{-4}$ so the overall resistance, I am not sure but Can I just use the same formula as in electrical circuits, e.g $\frac{1}{R} = \frac{1}{R} + \frac{1}{R} ...$ and times that by 50, since there are 50 tubes so R = $\frac{1}{ 6.02 x 10^{-4}} \times 50$ Is this correct? I really need help, with the first part, since we are turning water into steam will I have to use Q = ml , for latent heat? or is it $Q = mc \delta T$. 3. 1. For the first part, you'll need to need to use: $Q = mc_{water}\theta$ That'll give you the energy required to raise the temperature of water from 80 to 100 C Then, use $Q = mL_{vaporisation}$ That'll give you the energy required to boil the water. $Q = mc_{steam}\theta$ Again to get the heat required to heat the steam at 100 to 140 C. Use m = 2 kg, this will give the heat required in 1 second. 4. Originally Posted by Tweety Okay I asked my teacher and he said I have to use this equation for the second part, thermal resistance for a single pipe = $\frac{ln(\frac{r_{o}}{r_{i}})}{2 \pi H \lambda}$ $H = 4m$ so $r_{o} = 2 cm = 0.02 m$ $r_{i} = 1.5cm = 0.015 m$ $\lambda = 19 W m^{-1} K^{-1}$ pluggiing these numbers in I get $6.02\times 10^{-4}$ so the overall resistance, I am not sure but Can I just use the same formula as in electrical circuits, e.g $\frac{1}{R} = \frac{1}{R} + \frac{1}{R} ...$ and times that by 50, since there are 50 tubes so R = $\frac{1}{ 6.02 x 10^{-4}} \times 50$ Is this correct? I really need help, with the first part, since we are turning water into steam will I have to use Q = ml , for latent heat? or is it $Q = mc \delta T$. I don't know how to do this... but if what you wrote is correct, then this is not correct: $R = \dfrac{1}{ 6.02 \times 10^{-4}} \times 50$ You should be having: $\dfrac{1}{R} = \dfrac{1}{ 6.02 \times 10^{-4}} \times 50$ So, $R = \dfrac{6.02 \times 10^{-4}}{50}$ 5. Originally Posted by Unknown008 1. For the first part, you'll need to need to use: $Q = mc_{water}\theta$ That'll give you the energy required to raise the temperature of water from 80 to 100 C Then, use $Q = mL_{vaporisation}$ That'll give you the energy required to boil the water. $Q = mc_{steam}\theta$ Again to get the heat required to heat the steam at 100 to 140 C. Use m = 2 kg, this will give the heat required in 1 second. Thank you very much, so for the first one $Q = 2 kg \times 4.198 KJ Kg^{-1} K^{-1} \times 100-80 C^{o} = 167.92$ for the next one I looked up the latent heat of vaporization for water, in my steam table, at 100C and found it to be 2257 KJ/Kg $Q = 2kg \times 2257 = 4514 KJ$ $Q = 2kg \times 2.034 KJ Kg^{-1} K^{-1} \times 140-100C^{o} = 162.72$ Is this correct? So now do I just add all of my values up to get the rate of heat required to produce saturated steam at 140? Thanks so much 6. Yes, you're right For the pressure, I'm sorry but I don't know what you are supposed to be using. Personally, I would have use the ideal gas equation, but this might be seen mostly as chemistry... 2 kg of water -> 111.1 moles of water. Assuming that steam behaves as an ideal gas, the pressure is given by: $PV = nRT$ I would say that the volume is kept constant from liquid to gas, hence a volume of 2 litres, or 2 dm^3 = 0.002 m^3 R = 8.31, T = 273.15 + 140 = 412.15 K n = 111.11 So, we can get P. However, this gives an incredibly large value... 190275916.7 Pa = 190 MPa 7. Thank you, On my steam table, it shows saturated vapour pressure, at 140C to be 0.3614 MPa. Also I know that $pressure_ {absolute} = Pressure_{gauge} + pressure_{barometric}$ $1 MPa = 10^{6} Nm^{-2} = 10bar$ thanks, you have been a great help. 8. ## Heat Transfer Problem Hi Tweety, Part 1 is an enthalpy balance as shown by others. The pressure of saturated steam at 140 C is 53 psi absolute Part 2 requires the heating media ,hot oil . high pressure steam, or other fluids bjh	{"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": 33, "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": 0.948215901851654, "perplexity_flag": "middle"}
http://nrich.maths.org/public/leg.php?code=12&cl=3&cldcmpid=7141	nrich enriching mathematicsSkip over navigation # Search by Topic #### Resources tagged with Factors and multiples similar to Weekly Problem 47 - 2010: Filter by: Content type: Stage: Challenge level: ##### Other tags that relate to Weekly Problem 47 - 2010 Equivalent fractions, decimals and percentages. Fractions. Creating expressions/formulae. Continued fractions. Calculating with fractions. History of number systems. Factors and multiples. Generalising. smartphone. Ratio. ### There are 91 results Broad Topics > Numbers and the Number System > Factors and multiples ### Ben's Game ##### Stage: 3 Challenge Level: Ben passed a third of his counters to Jack, Jack passed a quarter of his counters to Emma and Emma passed a fifth of her counters to Ben. 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Given a target number, say 23, and a range of numbers to choose from, say 1-4, players take it in turns to add to the running total to hit their target. ### GOT IT Now ##### Stage: 2 and 3 Challenge Level: For this challenge, you'll need to play Got It! Can you explain the strategy for winning this game with any target? ### Reverse to Order ##### Stage: 3 Challenge Level: Take any two digit number, for example 58. What do you have to do to reverse the order of the digits? Can you find a rule for reversing the order of digits for any two digit number? ### Repeaters ##### Stage: 3 Challenge Level: Choose any 3 digits and make a 6 digit number by repeating the 3 digits in the same order (e.g. 594594). Explain why whatever digits you choose the number will always be divisible by 7, 11 and 13. ### Cuboids ##### Stage: 3 Challenge Level: Find a cuboid (with edges of integer values) that has a surface area of exactly 100 square units. Is there more than one? Can you find them all? ### Shopping Basket ##### Stage: 3 Challenge Level: A mathematician goes into a supermarket and buys four items. Using a calculator she multiplies the cost instead of adding them. How can her answer be the same as the total at the till? ### Take Three from Five ##### Stage: 3 and 4 Challenge Level: Caroline and James pick sets of five numbers. Charlie chooses three of them that add together to make a multiple of three. Can they stop him? ### Hidden Squares ##### Stage: 3 Challenge Level: Rectangles are considered different if they vary in size or have different locations. How many different rectangles can be drawn on a chessboard? ### Eminit ##### Stage: 3 Challenge Level: The number 8888...88M9999...99 is divisible by 7 and it starts with the digit 8 repeated 50 times and ends with the digit 9 repeated 50 times. What is the value of the digit M? ### Adding All Nine ##### Stage: 3 Challenge Level: Make a set of numbers that use all the digits from 1 to 9, once and once only. Add them up. The result is divisible by 9. Add each of the digits in the new number. What is their sum? Now try some. . . . ### Mathematical Swimmer ##### Stage: 3 Challenge Level: Twice a week I go swimming and swim the same number of lengths of the pool each time. As I swim, I count the lengths I've done so far, and make it into a fraction of the whole number of lengths. . . . ### Remainders ##### Stage: 3 Challenge Level: I'm thinking of a number. When my number is divided by 5 the remainder is 4. When my number is divided by 3 the remainder is 2. Can you find my number? ### Product Sudoku ##### Stage: 3, 4 and 5 Challenge Level: The clues for this Sudoku are the product of the numbers in adjacent squares. ### AB Search ##### Stage: 3 Challenge Level: The five digit number A679B, in base ten, is divisible by 72. What are the values of A and B? ### Factors and Multiple Challenges ##### Stage: 3 Challenge Level: This package contains a collection of problems from the NRICH website that could be suitable for students who have a good understanding of Factors and Multiples and who feel ready to take on some. . . . ### Stars ##### Stage: 3 Challenge Level: Can you find a relationship between the number of dots on the circle and the number of steps that will ensure that all points are hit? ### Can You Find a Perfect Number? ##### Stage: 2 and 3 Can you find any perfect numbers? Read this article to find out more... ### What a Joke ##### Stage: 4 Challenge Level: Each letter represents a different positive digit AHHAAH / JOKE = HA What are the values of each of the letters? ### Factoring Factorials ##### Stage: 3 Challenge Level: Find the highest power of 11 that will divide into 1000! exactly. ### Inclusion Exclusion ##### Stage: 3 Challenge Level: How many integers between 1 and 1200 are NOT multiples of any of the numbers 2, 3 or 5? ### Hot Pursuit ##### Stage: 3 Challenge Level: The sum of the first 'n' natural numbers is a 3 digit number in which all the digits are the same. How many numbers have been summed? ### A First Product Sudoku ##### Stage: 3 Challenge Level: Given the products of adjacent cells, can you complete this Sudoku? ### Remainder ##### Stage: 3 Challenge Level: What is the remainder when 2^2002 is divided by 7? What happens with different powers of 2? ### Times Right ##### Stage: 3 Challenge Level: Using the digits 1, 2, 3, 4, 5, 6, 7 and 8, mulitply a two two digit numbers are multiplied to give a four digit number, so that the expression is correct. How many different solutions can you find? ### Shifting Times Tables Game ##### Stage: 2 and 3 Challenge Level: In this activity, the computer chooses a times table and shifts it. Can you work out the table and the shift each time? ### Diggits ##### Stage: 3 Challenge Level: Can you find what the last two digits of the number $4^{1999}$ are? ### Digat ##### Stage: 3 Challenge Level: What is the value of the digit A in the sum below: [3(230 + A)]^2 = 49280A ### Ewa's Eggs ##### Stage: 3 Challenge Level: I put eggs into a basket in groups of 7 and noticed that I could easily have divided them into piles of 2, 3, 4, 5 or 6 and always have one left over. How many eggs were in the basket? ### Gaxinta ##### Stage: 3 Challenge Level: A number N is divisible by 10, 90, 98 and 882 but it is NOT divisible by 50 or 270 or 686 or 1764. It is also known that N is a factor of 9261000. What is N? ### Expenses ##### Stage: 4 Challenge Level: What is the largest number which, when divided into 1905, 2587, 3951, 7020 and 8725 in turn, leaves the same remainder each time? ### Oh! Hidden Inside? ##### Stage: 3 Challenge Level: Find the number which has 8 divisors, such that the product of the divisors is 331776. ### What Numbers Can We Make Now? ##### Stage: 3 and 4 Challenge Level: Imagine we have four bags containing numbers from a sequence. What numbers can we make now? ### A Biggy ##### Stage: 4 Challenge Level: Find the smallest positive integer N such that N/2 is a perfect cube, N/3 is a perfect fifth power and N/5 is a perfect seventh power. ### Sixational ##### Stage: 4 and 5 Challenge Level: The nth term of a sequence is given by the formula n^3 + 11n . Find the first four terms of the sequence given by this formula and the first term of the sequence which is bigger than one million. . . . ### Sieve of Eratosthenes ##### Stage: 3 Challenge Level: Follow this recipe for sieving numbers and see what interesting patterns emerge. ### Counting Cogs ##### Stage: 2 and 3 Challenge Level: Which pairs of cogs let the coloured tooth touch every tooth on the other cog? Which pairs do not let this happen? Why? ### How Old Are the Children? ##### Stage: 3 Challenge Level: A student in a maths class was trying to get some information from her teacher. She was given some clues and then the teacher ended by saying, "Well, how old are they?" ### Pebbles ##### Stage: 2 and 3 Challenge Level: Place four pebbles on the sand in the form of a square. Keep adding as few pebbles as necessary to double the area. How many extra pebbles are added each time? ### Divisively So ##### Stage: 3 Challenge Level: How many numbers less than 1000 are NOT divisible by either: a) 2 or 5; or b) 2, 5 or 7? ### Number Rules - OK ##### Stage: 4 Challenge Level: Can you convince me of each of the following: If a square number is multiplied by a square number the product is ALWAYS a square number... 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": 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": 0.3, "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": 0.8787553310394287, "perplexity_flag": "middle"}

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