keirp/tiny_math · Datasets at Hugging Face

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http://physics.stackexchange.com/questions/49807/what-information-is-stored-on-gramaphones-tape-recorders-cds-dvds/49810	# What information is stored on gramaphones/tape recorders/CDs/DVDs I'm a Software Developer by profession and my physics knowledge is limited what I had learned at high school level. Please excuse me if the question is trivial. Question: From what I know, a sound wave is set of different amplitudes spread across a time line. The amplitudes vary greatly. Hence, sound wave is largely aperiodic (amplitudes don't repeat often). Now, where does frequency come into picture here? If the wave is periodic like a sine wave or a deterministic mathematical function of time, then frequency can be measured as the number of cycles (wave reaching same amplitude) in one second. How can we define frequency of highly aperiodic sounds like human speech. If all that is recorded on a gramophone disc is varying amplitudes across timeline, where is frequency accounted for? Does the frequency remain constant in a typical human speech? - ## 3 Answers The sound that reaches your ear is just air pressure fluctuating over time. You can use a transducer of some sort to convert the value of air pressure to some other form - for example: • to the depth of a groove being cut into a helical track on a layer of wax on a rotating drum • to the depth of a groove being cut into a spiral track on a circular disc of metal from which other plastic disks are pressed. • to a strength of magnetisation of a magnetic layer on a plastic tape being wound onto a spool • to a series of numbers representing the pressure at regular tiny intervals of time. The idea that the variations in pressure over time are due to, or consist of, a collection of frequencies is just a mathematically equivalent description but it does not represent extra information, it's just a different way of describing the same information. Here's some diagrams from a synthesizer manual Above are three very different sounds with apparently the same frequency (say 440 Hz) Above is shown how you can add sine waves of two frequencies to produce a more complex waveform Above is shown how you can continue adding sine waves of differing frequencies to construct an arbitrary waveform (a sawtooth). The sawtooth waverform can be recorded directly as depths in a groove on a record. But you could "record" the same thing as a set of numbers that represent the frequencies of a dozen sine waves you could add together to produce a single pressure wave that varies over time in the same way. See Fast Fourier Transform - @RedGrittyBrick..thanks for the answer! The first part of the answer pretty much explains what I want. Is it like at every point in time (say at every ms) we record all the harmonics that contribute to the sound wave at that instant of time? Also, I can understand that recording this information in a digital form is easy. How is stored on mechanical device like Gramophone? Do we just record grooves with varying depths over time? – Gopal Jan 22 at 10:21 @Gopal: at every interval in time we record the single value of air pressure that results from the sum of all the harmonics (+ other sounds/noise). Yes, this information can be stored as the depth of a groove cut in some surface. The depth varies over distance along the groove, a needle moving along the groove experiences changes in depth over time. – RedGrittyBrick Jan 22 at 10:36 @RedGrittyBrick..as I accept the answer I would like to ask you a closing question, Essentially a gramophone doesn't actually do any fourier transforms. All it records is the displacement of microphone at every instant of time and reproduces same using a diaphragm. – Gopal Jan 22 at 14:33 @Gopal: That is true. – RedGrittyBrick Jan 22 at 15:10 You can make an arbitrary sound (or any waveform) by adding together a bunch of pure tones at different frequencies. So a sound, unless it happens to be a pure tone, does not contain a single frequency component, rather a range of frequencies. The mathematics behind this is called Fourier analysis and you can see many examples on Wikipedia or by searching the web. - ..>>You can make an arbitrary sound (or any waveform) by adding together a bunch of pure tones at different frequencies << bang on! Liked this lucid explanation. – Gopal Jan 23 at 7:18 Your question is specifically about how the concept of frequency can be applied to aperiodic signals. Simplest example is a finite width rectangular pulse - the signal is zero outside the pulse. This is certainly aperiodic. Now for periodic signals f(t), you can, for each frequency which is a multiple of the period, compute the n'th Fourier coefficient as $$c_n = \int^{\pi}_{-\pi}f(t)e^{-int}dt$$ This represents "how much" of frquency n is present in the periodic signal, as explained in the other answers. Correspondingly, for our aperiodic isolated rectangular pulse f(t), we can choose an interval of length T which is bigger than the pulse width, and again calculate the Fourier coefficients over this interval$$c_n=\int^{\frac{T}{2}}_{-\frac{T}{2}}f(t)e^{-2\pi i (\frac{n}{T})t}dt$$ Now if we let $T\rightarrow \infty$, the discrete values $\frac{n}{T}$ can be replaced by a continuous variable $\xi$, and the set of Fourier coefficients is replaced by a function $\hat{f}(\xi)$: $$\hat{f}(\xi) = \int^{\infty}_{-\infty}f(t)e^{2\pi i \xi t}dt$$ This is the Fourier transform of f. So for aperiodic signals (or truncated periodic signals), this is what you want. You can calculate the Fourier transform of anything from a rectangular pulse to the latest David Bowie track. The "frequencies" that are present in the signal are just the Fourier transformed variables $\xi$. - @twistor..thanks for detailed explanation. I'll try to understand fourier transform (had a bit of it in my Engineering course) – Gopal Jan 23 at 7:17	{"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": 5, "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": 0.9169406294822693, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/67366?sort=votes	## Non-measurable sets and Determinacy… ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Assume AC. Suppose $X$ is a subset of the irrationals (Baire Space) for which neither player has a winning strategy (i.e. the game $G(\omega, X)$ is not determined). Is $X$ non-measurable in the Lebesgue sense as a subset of $\mathbb{R}$? - ## 2 Answers (My argument is somewhat easier if you consider games where the players play $0$s and $1$s, so that the payoff set is in Cantor space $2^\omega$, and we use the usual coin-flipping probability measure; but an essentially similar idea works in Baire space.) For any game with payoff set $A$, where player I wins if the play is in $A$, consider the following slightly modified game $A^\ast$, which is just like $A$, except we insert a pair of dummy moves between each pair of actual moves, and insist that player I play a $0$ in this dummy round, while player II can play anything. Thus, a sequence or play is in the payoff set $A^\ast$ if indeed that sequence shows that player I did play a $0$ in all the dummy rounds (so every fourth digit is $0$), and furthermore, if we omit the dummy rounds entirely from the sequence, we get a sequence in $A$. Thus, playing the game $A^*$ is just like playing $A$, except that the play is interrupted for these silly dummy rounds. Note that player I has no incentive not to play a $0$ on those rounds, and player II's plays in the dummy rounds are ignored entirely. Thus, it is clear that a player has a winning strategy for $A$ if and only if he or she has a winning strategy for $A^\ast$, since we can translate the strategies from $A$ to $A^\ast$ and back again. The dummy rounds really don't change the difficulty of winning the game. But the point now is that because every fourth digit of $A^\ast$ is $0$, it follows that $A^\ast$ has measure $0$. (Every time you insist that a particular digit is $0$, it cuts the measure in half again.) The conclusion, therefore, which does not use the axiom of choice, is that if there is a non-determined set, then there is a non-determined set with measure $0$. In particular, there is a non-determined set that is measurable. - I'm not convinced... Cantor Space is homeomorphic to the Cantor Set which has measure zero... Thus any payoff set in it will have measure zero when considered as a subset of $\mathbb{R}$ by virtue of being a subset of the Cantor Set... I'm not convinced that your argument about halving the measure carries through to non-measurable sets... further my guess (I should probably ask this as a question) is that the existence of a non-determined game in Cantor Space is stronger than AC... – George Lazou Jun 9 2011 at 23:51 I am not using the Cantor set as a subset of $\mathbb{R}$ and the Lebesgue measure on that, but rather, using the natural probability measure on $2^\omega$, for which $2^\omega$ is the whole space and has measure $1$; the measure of the basic open set determined by a finite binary sequence of length $n$ has measure $\frac{1}{2^n}$, like flipping a coin. A subset of $2^\omega$ which is $0$ in every fourth digit has measure $0$ (an easy calculation). The same idea works in Baire space, but you seem to be mapping spaces by homeomorphisms that may not be measure-preserving... – Joel David Hamkins Jun 10 2011 at 0:24 1 As for your final question, I think you mean to ask whether the existence of a non-determined game is weaker than AC, rather than stronger, since ZFC proves the existence of such non-determined games for Cantor space in exactly the same way that it does for Baire space. I don't expect the existence of such a non-determined set to imply full AC, but I'd have to think a bit more to give a precise model showing this. If this is right, then the existence of non-determined sets would be a weak choice principle. – Joel David Hamkins Jun 10 2011 at 0:29 1 But I would encourage you to ask that as a question, since perhaps someone knows a good model, and I would be interested to see it. – Joel David Hamkins Jun 10 2011 at 2:16 1 There is an old beautiful argument of Sierpinski that shows that the axiom of choice for PAIRS $AC_2$ implies the existence of a nonmeasurable set; which in turn shows that $AD$ fails as soon as $AC_2$ is true. So a model in which $AC$ fails but $AC_2$ holds is yet another way of seeing that the negation of $AD$ does not imply $AC$. Cohen's so-called "first (symmetric) model" in which $AC$ fails is such a model (since every set has can be linearly ordered in that model, but the reals cannot be well-ordered there). – Ali Enayat Jun 10 2011 at 19:14 show 3 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. I have two possible answers. The first is short and possibly not the one you want. The second is probably the right one. First: take any (co)analytic subset $X$ of the Baire space $\omega^{\omega}$, which is not Borel. Than it is consistent with ZFC that the game $G(\omega,X)$ is not determined (ZFC just proves the determinacy of Borel games), but $X$ is Lebesgue-measurable (in fact universally measurable). I guess this is a consistent proof of "no", to your question. Second Assume AC. Then there exists a universally-null set (hence Lebesgue null, since the Lebesgue measure is atomless) $X$ of cardinality $\geq\aleph_{1}$. Then one can show that such a set $X$ can not be a Perfect set. Now let us consider the so-called Perfect-set game $PSG(\omega, X)$, which is technically a Gale -Stewart game $G(\omega, Y)$, with $Y$ of about the same complexity of $X$ (in particular $Y$ is universally-null non-perfect set of cardinality $\geq\aleph_{1}$). This game is not determined since $Y$ is not Perfect nor countable by construction, and it is knonw that such a game is determined only if $Y$ is Perfect or countable. Yet $Y$ is universally null, hence Lebesgue measurable. This second example is mentioned in Martin's "Blackwell's determinacy", where it is credited to Greg Hjorth. This is a proof in ZFC of "no", to your quesiton. -	{"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": 63, "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": 0.9644228219985962, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/262284/is-there-a-functional-definition-of-a-limit	# Is there a “functional” definition of a limit? Say we have a convergent sequence $(x_n)$ where $x_n \in E$ for all $n \in \mathbb{N}$ and $E$ is a subset of a metric space $(X,d)$. With this setup, we usually define it's limit as a point $x \in X$ such that for every $\epsilon > 0$, there exists $N \in \mathbb{N}$ such that $n > N$ implies $d(x_n,x) < \epsilon$. In some sense, I think that the above definition of the limit effectively mean that given any "$\epsilon > 0$, there is some kind of $N$ that we can use to get the sequence within $\epsilon$ of the limit $x$. I am wondering whether this can be reformulated as follows: there is a function $f: \mathbb{R} \rightarrow \mathbb{N}$, so that $f(\epsilon) = N$ and $n > N$ implies $d(x_n,x) < \epsilon$. If so, the function $f$ would have some nice properties (it would be onto, and monotonically decreasing in $\epsilon$ for instance). Is there any use to thinking about functions in this way / has it been introduced in this way? - 1 Note that your $f$ need not be onto. – Hagen von Eitzen Dec 19 '12 at 21:14 ## 2 Answers This concept is often known as the modulus of convergence, though for recursion-theoretic purposes it's often given as a function $g(n): \mathbb{N}\to\mathbb{N}$, where (in terms of your definition for $f()$) $g(n)$ would be defined as $g(n) = f(\frac{1}{n})$ or $g(n) = f(2^{-n})$. The concept is important in recursion theory because, for instance, we can compute the digits of the limit recursively in $g$ (i.e., using a Turing machine for $g$). Note that the function $g$ doesn't have to be computable even if the series converges; see, for instance, http://mathoverflow.net/questions/51794/simple-example-of-a-sequence-without-computable-modulus-of-convergence . - To obtain a well-defined function you might define $f(\epsilon)$ to be the smallest $N$ such that $d(x_n,x)<\epsilon$ for all $n>N$. Convergence of the sequence $x_n$ guarantees existence of such $N$'s and one of them has to be the smallest. Conversely, if there exists a function $f$ such that, for every $\epsilon$, $d(x,x_n)<\epsilon$ for all $n>f(\epsilon)$, then $x_n$ is convergent with limit $x$, because you may take $N=f(\epsilon)$. These are nothing but quite tautological reformulations of each other, though. -	{"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": 42, "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": 0.927402675151825, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/297954/embedding-ordered-number-fields-into-bbb-r	# Embedding ordered number fields into $\Bbb R$ Consider a number field $K$ equipped with a total order $\le$. Is there always a field homomorphism $\phi:K \rightarrow \Bbb R$ which respects the total order, i. e. with $\phi(x)>\phi(y)$ for $x>y$? - If the number field is contained in $\Bbb{R}$, shouldn't such a $\phi$ always exist? – BenjaLim Feb 8 at 11:14 @BenjaLim: The question is, whether the only possible orderings of $K$ are those inherited from an embedding in $\Bbb R$. I see no appareant reason why their couldn't exist other "exotic" orderings of $K$. – Dominik Feb 8 at 11:18 Ok, then use the fact that a field can be totally ordered iff $a_1^2+a_2^2+\dots+a_n^2=0 \implies a_1=a_2=\dots=0$. – Berci Feb 8 at 11:47 1 @Berci: I don't see how this fact relates to the question how to extend the total order of $K$. – Dominik Feb 8 at 13:13 1 – JSchlather Feb 8 at 19:39 ## 2 Answers Here is Theorem 11.6 from N. Jacobson's Basic Algebra II: Let $F$ be an algebraic number field, and let $\mathbb{R}_0$ be the field of real algebraic numbers [i.e., those real numbers which are algebraic over $\mathbb{Q}$]. Then we have a $1-1$ correspondence between the set of orderings of $F$ and the set of monomorphisms of $F$ into $\mathbb{R}_0$. The ordering determined by the monomorphism $\sigma$ is that in which $a > 0$ for $a \in F$ if $\sigma a > 0$ in $\mathbb{R}_0$. It follows from this that the orderings on a number field $F$ are in bijection with the number of roots of any minimal polynomial $P$ for $F$ in $\mathbb{Q}_0$. One knows -- e.g. by model completeness of the theory of real-closed fields -- that this is the same as the number of roots of $P$ in $\mathbb{R}$ and thus the same as the number of embeddings of $F$ into $\mathbb{R}$. Thus the number of (total, compatible with the field structure) orderings of a number field $F$ is equal to the number of embeddings of $F$ into $\mathbb{R}$. Added: If you don't like the model completeness bit, you can replace it with the fact that any homomorphism of real-closed fields must be order-preserving. Indeed, if not then by restriction one would get a different ordering on the smaller real-closed field, but the ordering on a real-closed field is unique: the positive elements are precisely the squares. If you are interested in the proof of this theorem (and don't have ready access to Jacobson's book), please let me know. - Take $L$ the Dedekind completion of $K$. Because the rational nmbrrs are dense in $K$ we have that $L$ must be isomorphic to the reals (as an ordered field). Now restrict the isomorphism to $K$ and we are done. Of course one has to make a severe distinction between orderable and ordered, one only requires that an order exists, and the other specifies such order. Pete L. Clark has answered regarding to how many orders are compatible with a number field. - How do you know the rationals are dense in $K$? (It's true, but I'm wondering whether it's clear.) – Pete L. Clark Feb 10 at 3:18 @Pete Well, number fields are Archemedian. I don't see an immediate argument better than "obviously", but I am already in bed. I'll give it some thought during the first two dreams, and the last one. I hope it turns up nicely in the morning. :-) – Asaf Karagila Feb 10 at 3:22 $@$Asaf: the natural orderings on number fields are Archimedean, which, because the answer to this question is yes, are all the orderings. But if you contemplate an arbitrary ordering of a number field? I feel confident that if $L/K$ is a finite degree field extension and $\sigma$ is an ordering of $L$, then $(L,\sigma)$ is Archimedean iff $(K,\sigma\|K)$ is Archimedean, but this too seems to require proof. – Pete L. Clark Feb 10 at 3:25 2 Hmm, I think I see how to prove that: the real closure of an ordered field $K$ is unique up to $K$-isomorphism, so if $L/K$ is a finite degree ordered extension it must be realizable as an ordered subfield of the real closure. Since $\mathbb{R}_0$ is a subfield of the Archimedean ordered field $\mathbb{R}$, it must itself be Archimedean. – Pete L. Clark Feb 10 at 3:30	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 53, "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": 0.9339014291763306, "perplexity_flag": "head"}
http://mathhelpforum.com/algebra/94722-unknown-roots-two-quadratics.html	Thread: 1. Unknown roots of two quadratics Let $(a, c)$ and $(b, d)$ be the roots of the equation $x^2 + ax - b = 0$ and $x^2 + cx + d = 0$ respectively. Find all possible real values for $a, b, c, d$ I'm not sure how to do this question and not sure how to approach it. 2. Hello, In a quadratic equation $x^2+mx+n=0$, m is the sum of the roots, and n is the product of the roots. So from the second equation, we have $bd=d$ (1). And $c=b+d$ (2) And from the first one, we have $ac=-b$ (3) and $a+c=a$ (4) From (4), we get c=0 Substituting in (3), we get b=0 Substituting in (1), we get d=0 Then, a can take any values you want...	{"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": 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": 0.8537924885749817, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/69544/create-polynomial-coefficients-from-its-roots?answertab=votes	# Create polynomial coefficients from its roots Given some roots : $r_1,r_2,\ldots,r_n$, how can we reconstruct polynomial coefficients? I know the Horner scheme and that we can just go backwards receiving those coefficients. But I'm curious if there's some other nice solution to this problem. - 4 I'm not sure I'm answering your question, but to get the coefficients from the roots, just expand $(X-r_1) (X-r_2) \ldots (X-r_n)$. You get $$(X-r_1) (X-r_2) \ldots (X-r_n) = X^n - (r_1 + r_2 + \cdots + r_n) X^{n-1}+ \cdots + (-1)^n r_1 r_2 \ldots r_n X^0$$ The expressions you find are called symmetric polynomials. – Joel Cohen Oct 3 '11 at 15:21 2 – Srivatsan Oct 3 '11 at 15:22 1 It's not clear to me how you can take the Horner scheme "backwards" to get the coefficients. – Thomas Andrews Oct 3 '11 at 15:27 Joel could you elaborate hows this equation goes further? I can't get it – Chris Oct 3 '11 at 15:52 1 @Chris : The $X^{n-2}$ coefficient is $\sum_{1 \le i, j \le n} r_i r_j$ (sum of products of $2$ roots), then the $X^{n-3}$ coefficient $-\sum_{1 \le i, j, k \le n} r_i r_j r_k$ (sum of products of $3$ roots), the $X^{n-4}$ coefficient $\sum_{1 \le i, j, k, l \le n} r_i r_j r_k r_k$ (sum of products of $4$ roots) and so on. The $X^{n-k}$ coefficient is $(-1)^k\sum_{1 \le i_1, i_2, \ldots, i_k \le n} r_{i_1} r_{i_2} \ldots r_{i_k}$ (sum of products of $k$ roots). – Joel Cohen Oct 3 '11 at 16:37 show 2 more comments ## 3 Answers Knowing the location of roots really only pins down a family of polynomials, which vary by multiplicative constants. If you know the roots of $P(x)$ are $r_1, r_2 \cdots r_n$ then you can write the polynomial as $$P(x) = a (x-r_1)(x-r_2) \cdots (x-r_n)$$ where $a$ is some non-zero constant. However, we can not be more precise than that. We can pin down $a$ if we know the value of $P$ are any other point. If we knew $a$, then simply expanding the right hand side would produce the polynomial in a form where you could read off the constants. - As a tiny note: the requirement of an $a$ along with the $n$ roots is very much consistent with the fact that one requires $n+1$ conditions to uniquely determine a polynomial. – J. M. Oct 4 '11 at 0:42 You might want to read this well written article about your problem: http://stochastix.wordpress.com/2008/11/22/building-a-polynomial-from-its-roots/ http://stochastix.wordpress.com/2008/12/10/building-a-polynomial-from-its-roots-ii/ - The polynomial is $P(z) = a_0 (z-r_1)\times (z-r_2) \times \cdots \times (z-r_n)$. Now labeling coefficients as $P(z) = a_0 z^n + a_1 z^{n-1} + \ldots + a_n$ we have $$a_1 = -a_0 ( r_1 + r_2 + \ldots + r_n) \qquad a_n = (-1)^n a_0 r_1 \times r_2 \times \cdots \times r_n$$ In general: $$a_k = \frac{(-1)^k}{k!} a_0 \sum_\pi r_{\pi(1)} \times \cdots \times r_{\pi(k)}$$ where the sum is over permutations $\pi \in \mathfrak{S}_n$. -	{"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": 26, "mathjax_display_tex": 4, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9152019619941711, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/75789/what-is-step-by-step-logic-of-pinv-pseudoinverse/75793	What is step by step logic of pinv (pseudoinverse)? So we have a matrix $A$ size of $M \times N$ with elements $a_{i,j}$. What is a step by step algorithm that returns the Moore-Penrose inverse $A^+$ for a given $A$ (on level of manipulations/operations with $a_{i,j}$ elements, not vectors)? - 3 Answers 1. Perform a singular value decomposition $\mathbf A=\mathbf U\mathbf \Sigma\mathbf V^\top$ 2. Check for "tiny" singular values in $\mathbf \Sigma$. A common criterion is that anything less than $\\|\mathbf A\\|\epsilon$ is "tiny". Set those singular values to zero. 3. Form $\mathbf \Sigma^+=\mathrm{diag}(1/\sigma_1,1/\sigma_2,\dots, 0)$. That is, reciprocate the nonzero singular values, and leave the zero ones untouched. 4. $\mathbf A^+=\mathbf V\mathbf \Sigma^+\mathbf U^\top$ - Did you find this question interesting? Try our newsletter Sign up for our newsletter and get our top new questions delivered to your inbox (see an example). email address As far as I know and from what I have read, there is no direct formula (or algorithm) for the (left) psuedo-inverse of $A$, other than the "famous formula" $$(A^\top A)^{-1}A^\top$$ which is computationally expensive. I will describe here an algorithm that does indeed calculate it efficiently. As I said, I believe this may be previously un-published, so cite me appropriately, or if you know of a reference please state it. Here is my algorithm. Set $B_0 = A$ and for each iteration step, take a column of $B_i$ and orthogonalize against the columns of $A$. Here is the algorithm ($A$ has $n$ independent columns): 1. Initialize $B_0=A$. 2. For $j \in 1\ldots n$ do \begin{align} \mathbf{t} &= B_i^\top A_{\star j} \\ R\mathbf{t} &= e_j & \text{Find such $R$, an elementary row operation matrix}\\ B_{i+1}^\top &= R B_i^\top \\ \end{align} Notice that each step does one vector/matrix multiplication and one elementary matrix row operation. This will require much less computation than the direct evaluation of $(A^\top A)^{-1}A^\top$. Notice also that there is a nice opportunity here for parallel computation. (One more thing to notice--this may calculate a regular inverse, starting with $B_0=A$, or with $B_0=I$.) The step that calculates $R$ may partially be orthogonal rather than elementary (orthonormal/unitary -thus better in the sense of error propagation) if desired, but should not make use of the previously orthonormalized rows of $B$, since they must remain orthogonalized in the process. If this is done, the process becomes a "cousin" to the QR algorithm, whereas before it would be considered a "cousin" to the LU factorization. "Cousin" because the matrix operations of those are self referential, and this algorithm references the matrix $A$. The following is a hopefully enlightening description. See first edits for a more wordy description. But I think the following is more concise and to the point. Consider the conformable matrix $B^\top$ such that $B$ has the same dimension as $A$ and \begin{align} B^\top A &= I \tag{1}\\ B &= A B_A \tag{2}\\ \end{align} Here $B_A$ is arbitrary and exists only to ensure that the matix $B$ shares the same column space as $A$. (1) states that $B^\top$ is a (not necessarily unique) left inverse for $A$. (2) states that $B$ shares the same column space as $A$. Claim: $B^\top$ is the Moore-Penrose pseudoinverse for $A$. Proof: The Moore-Penrose pseudoinverse for $A$ is $$A^\dagger = \left(A^\top A\right)^{-1}A^\top$$ Given (1) and substituting (2) we have \begin{align} \left(AB_A\right)^\top A &= I \\ B_A^\top A^\top A &= I \\ B_A^\top &= \left( A^\top A\right)^{-1} \\ B_A &= \left( A^\top A\right)^{-1} \\ \end{align} Now solving for $B$ from (2): \begin{align} B &= A\left(B_A\right) \\ B &= A\left( A^\top A\right)^{-1} \\ \Rightarrow B^\top &=\left( A^\top A\right)^{-1}A^\top \\ \end{align} Q.E.D. - But the question is about the Moore-Penrose pseudoinverse, not the left inverse. If left inverses exist, the Moore-Penrose pseudoinverse is one of them, isn't it? And it can be computed using the algorithm in J.M.'s answer. So I'm not sure what you mean by "there is no direct formula (or algorithm)". – Rahul Narain Feb 28 at 20:17 @Rahul The algorithm in J.M.'s answer is iterative. What I am saying is that this algorithm computes the particular left inverse which is indeed the Moore-Penrose pseudoinverse, and does so in exact arithmetic, not by convergence. – adam W Feb 28 at 20:29 – Gottfried Helms Feb 28 at 21:45 @Gottfried Well, I do get the same numbers up to accuracy. For instance, $-0.012390122873089741$ is what I get for the lower right element. The way you describe "putting A into upper triangular shape" seems like you are doing regular LU or QR though...so not too certain. – adam W Feb 28 at 23:02 Is this faster than SVD? It's $\Omega(mn^2)$, right? – user7530 Mar 1 at 2:05 show 4 more comments First step is to buy Matlab. Second step is to type `pinv`. Edit: Here is an implementation. - 5 Funny ^^ but hardly an answer to his question. – Olivier Bégassat Oct 25 '11 at 17:07 or get octave for free=) – Kabumbus Oct 25 '11 at 17:38	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 42, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "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": 0.8833175301551819, "perplexity_flag": "middle"}
http://www.physicsforums.com/showthread.php?t=7454	Physics Forums Toe must include ourselves I know we have discussed that a theory for everything needs to include our spirituality. This is true but also it must include our skin cells, our hearts, our brains for these too are a part of our "Physical" universe. A theory of everything must include our ability to imagine and think, our ability to articulate and our ability to express our selves with free will. These are all physical manifestations and certainly not delusional. Mankind continously endeavours to create and recreate it self and this too has many physical implications. WE are very good at thinking that a theory of everything is only about stars and planets and phantom particles, The nature of thought it-self is missing from any theories so far. WE tend to think that we may be close to a TOE but I have to suggest that we are so far away from it that we have no hope of even getting close. For if one wants to quantify some thing like the human brain (a physical reality) into a theory that can be valued and determined by a formula then I am sorry they must be very deluded. Even Einstiens theories espouse that the perception of the observer is to be considered. Could this not be also a theory of subjectivity. A simple truth that we have always known. I think if science forgets the human or life element it is badly mistaken and I am not even talking about religion or spirituality or souls or life after death or anything. So are we taking on an impossible challenge as surely the closer one gets to this Holy grail of the theory for everything the more complex the mathematics must get, the harder it is to quantify and prove, and the less constancy there is. PhysOrg.com physics news on PhysOrg.com >> Promising doped zirconia>> New X-ray method shows how frog embryos could help thwart disease>> Bringing life into focus Recognitions: Gold Member Staff Emeritus The skin cells and all the construction of our bodies is referred to molecular chemistry, where it is being better and better understood. Changeing from the present theories to a TOE won't change that, and the genome and proteomic revolutions now ongoing are going to show us everything about our own biology before 2050. Consciousness is the last refuge of the "more than chemistry" faith. However much people like Dennet insist there is nothing there, the believers will insist that qualia, or free will, or something must transcend the dance of the molecules. so, From a mathematical perpsective how can you write a mathematical equation that not only includes black hole and gravity etc but also a single skin cell? Toe must include ourselves Originally posted by selfAdjoint The skin cells and all the construction of our bodies is referred to molecular chemistry, where it is being better and better understood. Changeing from the present theories to a TOE won't change that, and the genome and proteomic revolutions now ongoing are going to show us everything about our own biology before 2050. Consciousness is the last refuge of the "more than chemistry" faith. However much people like Dennet insist there is nothing there, the believers will insist that qualia, or free will, or something must transcend the dance of the molecules. Yes, well, the "believers" (as you put it) always insisted that we were the center of the Universe, and that the Earth was created in 7 days of 24 hours each, and that the Universe was static, and that God doesn't play dice... I hope that the belief that consciousness should remain mysterious is destroyed as these others were. Dennett has taken a powerful first step, and I hope he and his fellow materialists are not ignored. Originally posted by scott_sieger so, From a mathematical perpsective how can you write a mathematical equation that not only includes black hole and gravity etc but also a single skin cell? You shouldn't have to. As selfAdjoint pointed out, molecular biology will continue to be the field that deals with skin cells, long after the TOE is discovered. The term "Theory of Everything" is often misleading, since it implies that this theory will actually explain all phenomena in one equation. This is not the case. The TOE equation will be one that explains what the nature of spacetime and particles are. That's pretty much it. From this equation, we will be able to derive explanations of such phenomena as blackholes and the big bang, and so on, but we will never have every answer - and the ToE is not looking for every answer. I see your point,which is well taken. I for one understand the need for the human brain/mind to search out new knowledge and go where we have not yet been. Thank you for sharing it. if spirituality b a part of this...it must b "Flawless" i.e, it must not have any mistakes in it! it is not. because of a few things, such as: 1) it is made by humans. 2) it is like a pair of hand-cuffs holdin u back from a lot of things(remember the wars against the churches in Europe, which led to the industrial revolution). 3) it must b common 2 all races,species etc. living on the earth at-least... else there will b difference of opinion(and we all know what tht cud lead to) therefore we can conclude tht "SPIRITUALITY" can not b a part of ToE [;)] Call me a materialist, but I agree with anyone who says that spirituality, free will, and things of the sort are nothing but illusions\delusions. The bottom line is that the concept of the thinking mind is a consequence of chemical\sub-atomic interactions in that lumpy mass of grey matter known as the human brain. There is no free will. The concept is a consequence of vibrating strings. I realize that my login name is a misnomer, and that I should be more like "free brain". Originally posted by Mentat You shouldn't have to. As selfAdjoint pointed out, molecular biology will continue to be the field that deals with skin cells, long after the TOE is discovered. The term "Theory of Everything" is often misleading, since it implies that this theory will actually explain all phenomena in one equation. This is not the case. The TOE equation will be one that explains what the nature of spacetime and particles are. That's pretty much it. From this equation, we will be able to derive explanations of such phenomena as blackholes and the big bang, and so on, but we will never have every answer - and the ToE is not looking for every answer. Who says ToE has to be only one equation though? Maybe the simplest answer is the most complex system to understand. Originally posted by Jeebus Who says ToE has to be only one equation though? Maybe the simplest answer is the most complex system to understand. Well, Michio Kaku and the like have often expressed a desire for their ToE to be expressable in one small equation. What part of the word 'everything' do some posters here not understand? It also has to define itself. Therefore it must be the most simple equation. There can be no more fundamental than itself. And... the search is for a law of everything, not a theory. A theory is a way to 'get there'. True, a law of everything would be the ultimate achievement of science, but, if you ask me, that is impossible. Why? Because in order to have a law of everything, the writer would then have to know everything, be able to have absolute knowledge at one time, and would then most likely (Pardon the sound) become one with existence and thus have ascended beyond this realm to at least have absolute contact with what is higher. Such a person would cease to be human, and probably wouldn't see anything positive in giving us the law of everything, for then everything would change and the law would have to be rewritten. A theory of everything is much more beneficial, because we will be able to understand all that there is to understand without absolutely knowing from observation. We could predict everything, most likely, and any inconsistencies could be plugged in and used to modify the theory for accuracy. If it was a law, we'd have to know everything, and thus everything would be different. Those that know everything would probably not have any desire to share it with us. That may be difficult for some to grasp... but think about it. A law is proven by nature. A theory is supported by observations and calculations. We could quite possibly come up with a theory of everything, but a law of everything, though desirable, is beyond our grasp. To know everything is similar to becoming all powerful. Originally posted by scott_sieger Even Einstiens theories espouse that the perception of the observer is to be considered. Could this not be also a theory of subjectivity. A simple truth that we have always known. My favorite misperception of relativity (and QM). Both have nothing to do with perception, and by observer, all that is meant is point of reference (in QM, it's particle interaction). Neither theory in any way cares about a conscious observer. You sound like lifegazer... The term "spirituality" could generate confusion. I prefer to say that TOE must explain how is it possible from inside the universe to perceive the rules that govern the universe itself (i.e. TOE). It seems that TOE is to be a tautology. Or merely it cannot be stated. My think is that TOE should bring us to a possible new dimensionless situation, in which the observer is outside space-time frame and therefore at an actually infinite distance from any experiment. To achieve this result, TOE should start with an antinomic statement that would grant a precise initial formula for a bootstrap of Universe and would grant freedom as well. Like this: $$\Rightarrow \emptyset$$ That reads: Nothingness (i.e. the missing term on the left of implication symbol) implicates the "Empty Set" (i.e. a concept that is something else than Nothingness). Since Nothingness itself is an antinomic concept, the statement is stated by Nothingness itself in an originary situation of no-space, no-time. From this situation an endless oscillation is generated without any external agent in each possible point of Nothingness. Imagine that the "modes" of implication leads to "space" and the "occurrencies" of implication leads to "time". If the series of infinite ratios modes/occurrencies converges to a finite value (say c) and the modes present some symmetry, we could build some set of points that synchronically could interact among themselves and self-define a space-time frame. This could be the seed of any knowable universe (included our's). Thread Tools Similar Threads for: Toe must include ourselves Thread Forum Replies Programming & Comp Sci 20 General Math 29 Forum Feedback & Announcements 5 General Math 0 Forum Feedback & Announcements 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": 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": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.958895206451416, "perplexity_flag": "middle"}
http://mathhelpforum.com/calculus/187093-find-maximum-value-function-two-variables.html	# Thread: 1. ## Find maximum value of a function in two variables Hi, I have a problem. Given a+b=10. Find maximum value of a^3b^2 How do i go about doing this? Also i heard, a^xb^y*c^z is maximum when a/x=b/y=c/z, provided a+b+c=constant? Can someone give me the proof? Also i am appearing for an exam that tests me in such applications of math, can someone post a link to the most common axioms and proofs? Thanks. 2. ## Re: Find maximum value of a function in two variables There are various ways to go about it. The simplest might be a quick substitution. a = 10-b might lead somewhere useful. 3. ## Re: Find maximum value of a function in two variables Originally Posted by TKHunny There are various ways to go about it. The simples might be a quick substitution. a = 10-b might lead somewhere useful. The substituion is a valid process. But i wanted to know if i could apply the concept i described earlier. The reason is this problem needs to be solved in 1 minute. If the concept(theorem/axiom) described earlier is accurate, could anybody explain to me the proof or a high level reasoning. Thanks 4. ## Re: Find maximum value of a function in two variables By the concept i meant --> The max value of a^x.b^y is obtained when a/x=b/y. 5. ## Re: Find maximum value of a function in two variables Ummm...That may be right, but I'm not sure it's useful. It has very limited scope (solves very few problems) and is unlikely to appear in that form ever again - unless you already know what will be on the exam. 6. ## Re: Find maximum value of a function in two variables True about the scope. But can you help me deduce or prove the theorem. I would want to know how they arrived at the concept. That helps me understand the direction in which to approach the problem. And some variations of the problem can be expected. Also any links to pages explaining some useful theorems? WHat i am looking for is a site that just lists down some salient and useful theorems. Eg 1+2+3+...+n = [n(n+1)]/2 Thanks. 7. ## Re: Find maximum value of a function in two variables How is your experience with Lagrangian Multipliers? If it can be proven, that should work. 8. ## Re: Find maximum value of a function in two variables Originally Posted by TKHunny How is your experience with Lagrangian Multipliers? If it can be proven, that should work. I can't recollect that. Are you suggesting Lagrange's to deduce the theorem or to solve the problem? Either way could you explain the same? THanks 9. ## Re: Find maximum value of a function in two variables To maximize $f(a,b,c)= a^xb^yc^z$ subject to the condition that g(a, b, c)= a+ b+ c= constant, make the gradients parallel: $\nabla f= xa^{x-1}b^yc^z\vec{i}+ ya^xb^{y-1}c^z\vec{j}+ za^xb^yz^{c-1}\vec{k}= \lambda\nabla g= \lambda\vec{i}+ \lambda\vec{j}+ \lambda\vec{k}$ where $\lambda$ is the Lagrange multiplier. Then we must have $xa^{x-1}b^y c^z= \lambda$, $ya^xb^{y-1}c^z= \lambda$, and $za^xb^yc^{z-1}= \lambda$. We can eliminate $\lambda$ from those equations by dividing one by another: Dividing the first equation by the second, $\frac{xa^{x-1}b^yc^z}{ya^xb^{y-1}c^z}= \frac{xb}{ya}= 1$ or $\frac{x}{a}= \frac{y}{b}$. Dividing the second equation by the third gives $\frac{ya^xb^{y-1}c^z}{za^xb^yc^{z-1}}= 1$ or $\frac{y}{b}= \frac{z}{c}$. If you are unfamiliar with "Lagrange Multipliers", here's a way to think about them- if you want to maximize f(x,y,z), calculate $\nabla f$, which always points in the direction of fastest increase of f, and follow it until you can't follow it any more- until it is 0. If you are constrained to stay on surface g(x,y,z)= constant, you may not be able to move in that direction ( $\nabla f$ may point off the surface), but you could move in the direction of the projection of $\nabla f$ onto the surface. You can do that, getting higher values of f, until there is NO projection on the surface- that is, until $\nabla f$ is perpendicular to the surface. But that means that it will be parallel to the normal vector to g at that point: $\nabla f$ is parallel to $\nabla g$ or $\nabla f= \lambda\nabla g$ for some number $\lambda$. 11. ## Re: Find maximum value of a function in two variables Originally Posted by TKHunny	{"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": 19, "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": 0.915532112121582, "perplexity_flag": "middle"}
http://michaelnielsen.org/polymath1/index.php?title=The_Erd%C5%91s_discrepancy_problem&diff=3206&oldid=2749	# The Erdős discrepancy problem ### From Polymath1Wiki (Difference between revisions) () Current revision (18:29, 20 July 2010) (view source)m () (40 intermediate revisions not shown.) Line 3: Line 3: ==Introduction and statement of problem== ==Introduction and statement of problem== - Is there a <math>\scriptstyle \pm 1</math> sequence <math>(x_n)</math> and constant <math>C</math>, such that for all positive integers <math> d,k </math> + Let <math>(x_n)</math> be a <math>\scriptstyle \pm 1</math> sequence and let <math>C</math> be a constant. Must there exist positive integers <math> d,k </math> such that - :<math> \left\| \sum_{i=1}^k x_{id} \right\| \leq C </math>? + :<math> \left\| \sum_{i=1}^k x_{id} \right\| > C </math>? - Known colloquially as "The Erdős discrepancy problem", this question has remained unanswered since the 1930s [citation needed]. It was also asked by Chudakov [again a citation needed]. It is extremely easy to state, and a very appealing problem, so perhaps Polymath can succeed where individual mathematicians have failed. If so, then, given the notoriety of the problem, it would be a significant coup for the multiply collaborative approach to mathematics. But even if a full solution is not reached, preliminary investigations have already thrown up some interesting ideas which could perhaps be developed into publishable results. + Known colloquially as "The Erdős discrepancy problem", this question has remained unanswered since the 1930s (Erdős, 1957) and Erdős offered \$500 for an answer. It was also asked by Chudakov (1956). It is extremely easy to state, and a very appealing problem, so perhaps Polymath can succeed where individual mathematicians have failed. If so, then, given the notoriety of the problem, it would be a significant coup for the multiply collaborative approach to mathematics. But even if a full solution is not reached, preliminary investigations have already thrown up some interesting ideas which could perhaps be developed into publishable results. - Anticipating a negative answer to Erdős's question, we now quantify the problem. For any finite set <math>A</math> of integers, we define the error <math>A</math> by + It seems likely that the answer to Erdős's question is yes. If it is, then an easy compactness argument tells us that for every <math>C</math> there exists <math>n</math> such that for every <math>\scriptstyle \pm 1</math> sequence of length <math>n</math> there exist <math>d,k</math> with <math>dk\leq n</math> such that <math> \left\| \sum_{i=1}^k x_{id} \right\| > C </math>. In view of this, we make some definitions that allow one to talk about the dependence between <math>C</math> and <math>n</math>. For any finite set <math>A</math> of integers, we define the <em>error</em> on <math>A</math> by :<math> E(A) := \sum_{a\in A} x_a </math>, :<math> E(A) := \sum_{a\in A} x_a </math>, - and for a set <math>\mathcal{A}</math> of finite sets of integers, we define the discrepancy + and for a set <math>\mathcal{A}</math> of finite sets of integers, we define the <em>discrepancy</em> :<math> \delta(\mathcal{A},x) := \sup_{A\in \mathcal{A}} \|E(A)\|. </math> :<math> \delta(\mathcal{A},x) := \sup_{A\in \mathcal{A}} \|E(A)\|. </math> We can think of the values taken by the sequence <math>(x_n)</math> as a red/blue colouring of the integers that tries to make the number of reds and blues in each <math>\scriptstyle A\in\mathcal{A}</math> as equal as possible. The discrepancy measures the extent to which the sequence fails in this attempt. Taking <math>\scriptstyle \mathcal{HAP}(N)</math> to be the set of <em>homogeneous arithmetic progressions</em> <math>\{d, 2d, 3d, ..., nd\}</math> contained in <math>\{1, 2, ..., N\}</math>, we can restate the question as whether <math>\scriptstyle \delta(\mathcal{HAP}(N),x) \to \infty</math> for every sequence <math>x</math>. We can think of the values taken by the sequence <math>(x_n)</math> as a red/blue colouring of the integers that tries to make the number of reds and blues in each <math>\scriptstyle A\in\mathcal{A}</math> as equal as possible. The discrepancy measures the extent to which the sequence fails in this attempt. Taking <math>\scriptstyle \mathcal{HAP}(N)</math> to be the set of <em>homogeneous arithmetic progressions</em> <math>\{d, 2d, 3d, ..., nd\}</math> contained in <math>\{1, 2, ..., N\}</math>, we can restate the question as whether <math>\scriptstyle \delta(\mathcal{HAP}(N),x) \to \infty</math> for every sequence <math>x</math>. - Two related questions have already been solved in the literature. Letting <math>\scriptstyle \mathcal{AP}(N) </math> be the collection of all (not necessarily homogeneous) arithmetic progressions in <math>\{1, 2, ..., N\}</math>, Roth [citation needed] proved that <math> \scriptstyle \delta(\mathcal{AP}(N),x) \geq c n^{1/4}</math>, independent of the sequence <math>x</math>. Letting <math> \scriptstyle \mathcal{HQAP}(N) </math> be the collection of all homogeneous quasi-arithmetic progressions <math>\scriptstyle\{\lfloor \alpha \rfloor,\lfloor 2\alpha \rfloor,\dots,\lfloor k\alpha \rfloor\}</math> contained in <math>\{1, 2, ..., N\}</math>, Vijay [citation needed] proved that <math>\scriptstyle \delta(\mathcal{HQAP}(N),x) \geq 0.02 n^{1/6}</math>, independent of the sequence <math>x</math>. + Two related questions have already been solved in the literature. Letting <math>\scriptstyle \mathcal{AP}(N) </math> be the collection of all (not necessarily homogeneous) arithmetic progressions in <math>\{1, 2, ..., N\}</math>, Roth (1964) proved that <math> \scriptstyle \delta(\mathcal{AP}(N),x) \geq c n^{1/4}</math>, independent of the sequence <math>x</math>. Letting <math> \scriptstyle \mathcal{HQAP}(N) </math> be the collection of all homogeneous quasi-arithmetic progressions <math>\scriptstyle\{\lfloor \alpha \rfloor,\lfloor 2\alpha \rfloor,\dots,\lfloor k\alpha \rfloor\}</math> contained in <math>\{1, 2, ..., N\}</math>, Vijay (2008) proved that <math>\scriptstyle \delta(\mathcal{HQAP}(N),x) \geq 0.02 n^{1/6}</math>, independent of the sequence <math>x</math>. ==Some definitions and notational conventions== ==Some definitions and notational conventions== Line 29: Line 29: The EDP is the Erdős discrepancy problem. (This may conceivably be changed if enough people don't like it.) The EDP is the Erdős discrepancy problem. (This may conceivably be changed if enough people don't like it.) - A sequence <math>(x_n)</math> is <em>completely multiplicative</em> if <math>x_{mn}=x_mx_n</math> for any two positive integers m and n. We shall sometimes abbreviate this to "multiplicative", but the reader should be aware that the word "multiplicative" normally refers to the more general class of sequences such that <math>x_{mn}=x_mx_n</math> whenever m and n are coprime. A completely multiplicative function is determined by the values it takes at primes. The <em>Liouville function</em> λ is the unique completely multiplicative function that takes the value -1 at every prime: if the prime factorization of n is <math>\prod p_i^{a_i}</math> then λ(n) equals <math>(-1)^{\sum a_i}</math>. + A sequence <math>(x_n)</math> is <em>completely multiplicative</em> if <math>x_{mn}=x_mx_n</math> for any two positive integers m and n. We shall sometimes abbreviate this to "multiplicative", but the reader should be aware that the word "multiplicative" normally refers to the more general class of sequences such that <math>x_{mn}=x_mx_n</math> whenever m and n are coprime. A completely multiplicative function is determined by the values it takes at primes. The <em>Liouville function</em> λ is the unique completely multiplicative function that takes the value -1 at every prime: if the prime factorization of n is <math>\prod p_i^{a_i}</math> then λ(n) equals <math>(-1)^{\sum a_i}</math>. Another important multiplicative function is <math>\mu_3</math>, the multiplicative function that’s -1 at 3, 1 at numbers that are 1 mod 3 and -1 at numbers that are 2 mod 3. This function has mean-square partial sums which grow logarithmically, [http://gowers.wordpress.com/2010/01/26/edp3-a-very-brief-report-on-where-we-are/#comment-5585 see this calculation]. A <em>HAP-subsequence</em> of a sequence <math>(x_n)</math> is a subsequence of the form <math>x_d,x_{2d},x_{3d},...</math> for some d. If <math>(x_n)</math> is a multiplicative <math>\pm 1</math> sequence, then every HAP-subsequence is equal to either <math>(x_n)</math> or <math>(-x_n)</math>. We shall call a sequence <em>weakly multiplicative</em> if it has only finitely many distinct HAP-subsequences. Let us also call a <math>\pm 1</math> sequence <em>quasi-multiplicative</em> if it is a composition of a completely multiplicative function from <math>\mathbb{N}</math> to an Abelian group G with a function from G to {-1,1} (This definition is too general. See [http://gowers.wordpress.com/2010/01/09/erds-discrepancy-problem-continued/#comment-4843 this and the next comment]). [http://gowers.wordpress.com/2009/12/17/erdoss-discrepancy-problem/#comment-4705 It can be shown] that if <math>(x_n)</math> is a weakly multiplicative sequence, then it has an HAP-subsequence that is quasi-multiplicative. A <em>HAP-subsequence</em> of a sequence <math>(x_n)</math> is a subsequence of the form <math>x_d,x_{2d},x_{3d},...</math> for some d. If <math>(x_n)</math> is a multiplicative <math>\pm 1</math> sequence, then every HAP-subsequence is equal to either <math>(x_n)</math> or <math>(-x_n)</math>. We shall call a sequence <em>weakly multiplicative</em> if it has only finitely many distinct HAP-subsequences. Let us also call a <math>\pm 1</math> sequence <em>quasi-multiplicative</em> if it is a composition of a completely multiplicative function from <math>\mathbb{N}</math> to an Abelian group G with a function from G to {-1,1} (This definition is too general. See [http://gowers.wordpress.com/2010/01/09/erds-discrepancy-problem-continued/#comment-4843 this and the next comment]). [http://gowers.wordpress.com/2009/12/17/erdoss-discrepancy-problem/#comment-4705 It can be shown] that if <math>(x_n)</math> is a weakly multiplicative sequence, then it has an HAP-subsequence that is quasi-multiplicative. Line 39: Line 39: To answer the problem negatively, it is sufficient to find a completely multiplicative sequence (that is, a sequence <math>(x_n)</math> such that <math>x_{mn}=x_mx_n</math> for every m,n) such that its partial sums <math>x_1+\dots+x_n</math> are bounded. First mentioned in the proof of the theorem in [http://gowers.wordpress.com/2009/12/17/erdoss-discrepancy-problem/ this] post. To answer the problem negatively, it is sufficient to find a completely multiplicative sequence (that is, a sequence <math>(x_n)</math> such that <math>x_{mn}=x_mx_n</math> for every m,n) such that its partial sums <math>x_1+\dots+x_n</math> are bounded. First mentioned in the proof of the theorem in [http://gowers.wordpress.com/2009/12/17/erdoss-discrepancy-problem/ this] post. - The function that takes n to 1 if the last non-zero digit of n in its ternary representation is 1 and -1 if the last non-zero digit is 2 is completely multiplicative and the partial sum up to n is easily shown to be at most <math>\log_3n +1</math>. Therefore, the rate at which the worst discrepancy grows, as a function of the length of the homogeneous progression, can be as slow as logarithmic. [http://gowers.wordpress.com/2009/12/17/erdoss-discrepancy-problem/] <s>[http://gowers.wordpress.com/2009/12/17/erdoss-discrepancy-problem/#comment-4553 It turns out] that the base can be made significantly higher than 3, so this example is not best possible.</s> [http://gowers.wordpress.com/2010/01/06/erdss-discrepancy-problem-as-a-forthcoming-polymath-project/#comment-4757 Correction] + The function that takes n to 1 if the last non-zero digit of n in its ternary representation is 1 and -1 if the last non-zero digit is 2 is completely multiplicative and the partial sum up to n is easily shown to be at most <math>\log_3n +1</math>. Therefore, the rate at which the worst discrepancy grows, as a function of the length of the homogeneous progression, can be as slow as logarithmic. [[Matryoshka_Sequences\|More examples of this sort are here.]] [http://gowers.wordpress.com/2009/12/17/erdoss-discrepancy-problem/] <s>[http://gowers.wordpress.com/2009/12/17/erdoss-discrepancy-problem/#comment-4553 It turns out] that the base can be made significantly higher than 3, so this example is not best possible.</s> [http://gowers.wordpress.com/2010/01/06/erdss-discrepancy-problem-as-a-forthcoming-polymath-project/#comment-4757 Correction] To answer the problem negatively, it is also sufficient to find a completely multiplicative function f from <math>\mathbb{N}</math> to a finite Abelian group G (meaning that f(mn)=f(m)+f(n) for every m,n) and a function h from G to {-1,1} such that for each g in G there exists d with f(d)=g and with the partial sums hf(d)+hf(2d)+...+hf(nd) bounded. In that case, one can set <math>x_n=hf(n)</math>. (Though this observation is simple with the benefit of hindsight, it might not have been made had it not been for the fact that this sort of structure was identified in a long sequence that had been produced experimentally). First mentioned [http://gowers.wordpress.com/2009/12/17/erdoss-discrepancy-problem/#comment-4696 here]. Se also [http://gowers.wordpress.com/2010/01/06/erdss-discrepancy-problem-as-a-forthcoming-polymath-project/ this post] and [http://gowers.wordpress.com/2010/01/06/erdss-discrepancy-problem-as-a-forthcoming-polymath-project/#comment-4717 this comment]. To answer the problem negatively, it is also sufficient to find a completely multiplicative function f from <math>\mathbb{N}</math> to a finite Abelian group G (meaning that f(mn)=f(m)+f(n) for every m,n) and a function h from G to {-1,1} such that for each g in G there exists d with f(d)=g and with the partial sums hf(d)+hf(2d)+...+hf(nd) bounded. In that case, one can set <math>x_n=hf(n)</math>. (Though this observation is simple with the benefit of hindsight, it might not have been made had it not been for the fact that this sort of structure was identified in a long sequence that had been produced experimentally). First mentioned [http://gowers.wordpress.com/2009/12/17/erdoss-discrepancy-problem/#comment-4696 here]. Se also [http://gowers.wordpress.com/2010/01/06/erdss-discrepancy-problem-as-a-forthcoming-polymath-project/ this post] and [http://gowers.wordpress.com/2010/01/06/erdss-discrepancy-problem-as-a-forthcoming-polymath-project/#comment-4717 this comment]. Line 46: Line 46: The problem for the positive integers is equivalent to the problem for the positive rationals. [http://gowers.wordpress.com/2009/12/17/erdoss-discrepancy-problem/#comment-4676](Sketch proof: if you have a function f from the positive integers to {-1,1} that works, then define <math>f_q(x)</math> to be f(q!x) whenever q!x is an integer. Then take a pointwise limit of a subsequence of the functions <math>f_q</math>. [[Limits with better properties\|Click here for a more detailed discussion of this construction and what it is good for]]. The problem for the positive integers is equivalent to the problem for the positive rationals. [http://gowers.wordpress.com/2009/12/17/erdoss-discrepancy-problem/#comment-4676](Sketch proof: if you have a function f from the positive integers to {-1,1} that works, then define <math>f_q(x)</math> to be f(q!x) whenever q!x is an integer. Then take a pointwise limit of a subsequence of the functions <math>f_q</math>. [[Limits with better properties\|Click here for a more detailed discussion of this construction and what it is good for]]. + + Define the [[drift]] of a sequence to be the maximal value of \|f(md)+...+f(nd)\|. The discrepancy problem is equivalent to showing that the drift is always infinite. It is obvious that it is at least 2 (because \|f(2)+f(4)\|, \|f(2)+f(3)\|, \|f(3)+f(4)\| cannot all be at most 1); one can show that [[drift\|it is at least 3]] (which implies as a corollary that the discrepancy is at least 2). One can also show that the [[upper and lower discrepancy]] are each at least 2. + + One can [[topological dynamics formulation\|formulate the problem using topological dynamics]]. + + Using Fourier analysis, one can [[Fourier reduction\|reduce the problem to one about completely multiplicative functions]]. + + Here is an [[algorithm for finding multiplicative sequences with bounded discrepancy]]. + + Here is a page whose aim is to record a [[human proof that completely multiplicative sequences have discrepancy at least 2]]. + + Here is an argument that shows that [[bounded discrepancy multiplicative functions do not correlate with characters]]. + + * The answer to a [[function field version]] of the problem seems to be negative. + + * The problem can be generalized to [[pseudointegers]]. Here we have found [[EDP on pseudointegers\|problems similar to EDP]] with a negative answer. ==Experimental evidence== ==Experimental evidence== + + ''Main page: [[Experimental results]]'' A nice feature of the problem is that it lends itself well to investigation on a computer. Although the number of <math>\pm 1</math> sequences grows exponentially, the constraints on these sequences are such that depth-first searches can lead quite quickly to interesting examples. For instance, at the time of writing they have yielded several examples of sequences of length 1124 with discrepancy 2. (This is the current record.) See [[Experimental results\|this page]] for more details and further links. A nice feature of the problem is that it lends itself well to investigation on a computer. Although the number of <math>\pm 1</math> sequences grows exponentially, the constraints on these sequences are such that depth-first searches can lead quite quickly to interesting examples. For instance, at the time of writing they have yielded several examples of sequences of length 1124 with discrepancy 2. (This is the current record.) See [[Experimental results\|this page]] for more details and further links. Line 57: Line 75: for sufficiently large <math> N </math>. Currently, this is the record-holder for slow growing discrepancy. for sufficiently large <math> N </math>. Currently, this is the record-holder for slow growing discrepancy. - The Thue-Morse sequence has discrepancy <math> \delta(N) \gg \sqrt{N} </math>. (See the discussion following [http://gowers.wordpress.com/2010/01/06/erdss-discrepancy-problem-as-a-forthcoming-polymath-project/#comment-4775 this comment] and the next one.) + The [[Thue-Morse-Hedlund Sequence\|Thue-Morse sequence]] has discrepancy <math> \delta(N) \gg N^{\log_4(3)} </math>. A random sequence (each <math> x_i </math> chosen independently) has discrepancy at least (asymptotically) <math> \sqrt{2N \log\log N} </math> by the law of the iterated logarithm. A random sequence (each <math> x_i </math> chosen independently) has discrepancy at least (asymptotically) <math> \sqrt{2N \log\log N} </math> by the law of the iterated logarithm. Line 80: Line 98: Does there exist a <math>\pm 1</math> sequence such that the corresponding discrepancy function grows at a rate that is slower than <math>c\log n</math> for any positive c? Does there exist a <math>\pm 1</math> sequence such that the corresponding discrepancy function grows at a rate that is slower than <math>c\log n</math> for any positive c? + + Does the number of sequences of length n and discrepancy at most C grow exponentially with n or more slowly than exponentially? (Obviously if the conjecture is true then it must be zero for large enough n, but the hope is that this question is a more realistic initial target.) + + ==General proof strategies== + + This section contains links to other pages in which potential approaches to solving the problem are described. + + [[First obtain multiplicative structure and then obtain a contradiction]] + + [[Find a different parameter, show that it tends to infinity, and show that that implies that the discrepancy tends to infinity]] + + [[Prove the result for shifted HAPs instead of HAPs]] + + [[Find a good configuration of HAPs]] + + [[Generalize to a graph-theoretic formulation]] + + [[Representation of the diagonal]] + + [[Bounding the discrepancy in terms of the common difference]] ==Annotated Bibliography== ==Annotated Bibliography== Line 88: Line 126: [http://gowers.wordpress.com/2010/01/09/erds-discrepancy-problem-continued/ Second Post] (Jan 9 - Jan 11). [http://gowers.wordpress.com/2010/01/09/erds-discrepancy-problem-continued/ Second Post] (Jan 9 - Jan 11). [http://gowers.wordpress.com/2010/01/11/the-erds-discrepancy-problem-iii/ Third Post] (Jan 11 - Jan 14). [http://gowers.wordpress.com/2010/01/11/the-erds-discrepancy-problem-iii/ Third Post] (Jan 11 - Jan 14). - [http://gowers.wordpress.com/2010/01/14/the-erds-discrepancy-problem-iv/ Fourth Post] (Jan 14 - ?). + [http://gowers.wordpress.com/2010/01/14/the-erds-discrepancy-problem-iv/ Fourth Post] (Jan 14 - 16). + [http://gowers.wordpress.com/2010/01/16/the-erds-discrepancy-problem-v/ Fifth Post] (Jan 16-19). + [http://gowers.wordpress.com/2010/01/19/edp1-the-official-start-of-polymath5/ First Theoretical Post] (Jan 19-21) + [http://gowers.wordpress.com/2010/01/21/edp2-a-few-lessons-from-edp1/ Second Theoretical Post] (Jan 21-26) + [http://gowers.wordpress.com/2010/01/26/edp3-a-very-brief-report-on-where-we-are/ Third Theoretical Post] (Jan 26 -?) + [http://gowers.wordpress.com/2010/01/30/edp4-focusing-on-multiplicative-functions/ Fourth Theoretical Post] (Jan 30- Feb 2) + [http://gowers.wordpress.com/2010/02/02/edp5-another-very-brief-summary/ Fifth Theoretical Post] (Feb 2 - Feb 5) + [http://gowers.wordpress.com/2010/02/05/edp6-what-are-the-chances-of-success/ Sixth Theoretical Post] (Feb 5- Feb 8) + [http://gowers.wordpress.com/2010/02/08/edp7-emergency-post/ Seventh Theoretical Post] (Feb 8 - 19) + [http://gowers.wordpress.com/2010/02/19/edp8-what-next/ Eighth Theoretical Post] (Feb 19-Feb 24) + [http://gowers.wordpress.com/2010/02/24/edp9-a-change-of-focus/ Ninth Theoretical Most] (Feb 24-Mar 2) + [http://gowers.wordpress.com/2010/03/02/edp10-a-new-and-very-promising-approach/ Tenth Theoretical Post] (Mar 2-Mar 7) + [http://gowers.wordpress.com/2010/03/07/edp11-the-search-continues/ Eleventh Theoretical Post] (Mar 7-Mar 12) + [http://gowers.wordpress.com/2010/03/13/edp12-representing-diagonal-maps/ Twelfth Theoretical Post] (Mar 13-Mar 22) + [http://gowers.wordpress.com/2010/03/23/edp13-quick-summary/ Thirteenth Theoretical Post] (Mar 23-Apr 24) + [http://gowers.wordpress.com/2010/04/25/edp14-strategic-questions/ Fourteenth Theoretical Post] (Apr 25-?) + + [http://mathoverflow.net/questions/tagged/polymath5 Several questions on MathOverflow have been given the `polymath5' tag.] Mathias, A. R. D. [http://www.dpmms.cam.ac.uk/~ardm/erdoschu.pdf On a conjecture of Erdős and Čudakov]. Combinatorics, geometry and probability (Cambridge, 1993). Mathias, A. R. D. [http://www.dpmms.cam.ac.uk/~ardm/erdoschu.pdf On a conjecture of Erdős and Čudakov]. Combinatorics, geometry and probability (Cambridge, 1993). - This is a short paper establishing the maximal length of sequence for the case where C=1 is 11, and is the starting point for our experimental studies. + This one page paper establishes that the maximal length of sequence for the case where C=1 is 11, and is the starting point for our experimental studies. + + Tchudakoff, N. G. Theory of the characters of number semigroups. J. Indian Math. Soc. (N.S.) 20 (1956), 11--15. MR0083515 (18,719e) + + The Mathias paper states that one of Erdős's problem lists states that this paper "studies related questions". The Math Review for this paper states that it summarizes results from seven other papers that are in Russian. The Math Review leaves the impression that the topic concerns characters of modulus 1. Borwein, Peter, and Choi, Stephen. [http://www.cecm.sfu.ca/personal/pborwein/PAPERS/P204.pdf A variant of Liouville's lambda function: some surprizing formulae]. Borwein, Peter, and Choi, Stephen. [http://www.cecm.sfu.ca/personal/pborwein/PAPERS/P204.pdf A variant of Liouville's lambda function: some surprizing formulae]. - Explicit formulas for the discrepancy of some completely multiplicative functions whose discrepancy is logarithmic. A typical example is: Let <math>\lambda_3(n) = (-1)^{\omega_3(n)}</math> where <math>\omega_3(n)</math> is the number of distinct prime factors congruent to <math>-1 \bmod 3</math> in <math>n</math> (with multiple factors counted multiply). Then the discrepancy of the <math>\lambda_3</math> sequence up to <math>N</math> is exactly the number of 1's in the base three expansion of <math>N</math>. + Explicit formulas for the discrepancy of some completely multiplicative functions whose discrepancy is logarithmic. A typical example is: Let <math>\lambda_3(n) = (-1)^{\omega_3(n)}</math> where <math>\omega_3(n)</math> is the number of distinct prime factors congruent to <math>-1 \bmod 3</math> in <math>n</math> (with multiple factors counted multiply). Then the discrepancy of the <math>\lambda_3</math> sequence up to <math>N</math> is exactly the number of 1's in the base three expansion of <math>N</math>. (This turns out not to be so surprising after all, since <math>\lambda_3(n)</math> is precisely the same as the ternary function defined in the second item of the Simple Observations section.) Borwein, Peter, Choi, Stephen and Coons, Michael. [http://www.cecm.sfu.ca/personal/pborwein/PAPERS/P225.pdf Completely multiplicative functions taking values in {-1,1}]. The published version of the above paper, which explains the results and proofs more clearly than the preprint, and is more explicit about the relationship with Erdős's question. Borwein, Peter, Choi, Stephen and Coons, Michael. [http://www.cecm.sfu.ca/personal/pborwein/PAPERS/P225.pdf Completely multiplicative functions taking values in {-1,1}]. The published version of the above paper, which explains the results and proofs more clearly than the preprint, and is more explicit about the relationship with Erdős's question. - - Vijay, Sujith. [http://www.combinatorics.org/Volume_15/PDF/v15i1r104.pdf On the discrepancy of quasi-progressions]. Electronic Journal of Combinatorics, R104 of Volume 15(1), 2008. - - A quasi-progression is a sequence of the form <math>\lfloor k \alpha \rfloor </math>, with <math>1\leq k \leq K</math> for some K. The main theorem of interest is: If the integers from 0 to n are 2-coloured, there exists a quasi-progression that has discrepancy at least <math> (1/50)n^{1/6}</math>. Granville and Soundararajan. [http://arxiv.org/abs/math/0702389 Multiplicative functions in arithmetic progressions] Granville and Soundararajan. [http://arxiv.org/abs/math/0702389 Multiplicative functions in arithmetic progressions] Line 108: Line 163: In this [http://gowers.wordpress.com/2010/01/06/erdss-discrepancy-problem-as-a-forthcoming-polymath-project/#comment-4729 blog post] Terry Tao wrote: "Granville and Soundararajan have made some study of the discrepancy of bounded multiplicative functions. The situation is remarkably delicate and number-theoretical (and is closely tied with the Granville-Soundararajan theory of pretentious characters)." In this [http://gowers.wordpress.com/2010/01/06/erdss-discrepancy-problem-as-a-forthcoming-polymath-project/#comment-4729 blog post] Terry Tao wrote: "Granville and Soundararajan have made some study of the discrepancy of bounded multiplicative functions. The situation is remarkably delicate and number-theoretical (and is closely tied with the Granville-Soundararajan theory of pretentious characters)." - [http://www.math.niu.edu/~rusin/known-math/93_back/prizes.erd As this web page reveals], the Erdős discrepancy problem was a \$500-dollar problem of Erdős, so it is clear that he regarded it as pretty hard. + Wirsing, E. Das asymptotische Verhalten von Summen über multiplikative Funktionen. II. (German) [The asymptotic behavior of sums of multiplicative functions. II.] Acta Math. Acad. Sci. Hungar. 18 1967 411--467. MR0223318 (36 #6366) [http://michaelnielsen.org/polymath1/index.php?title=Wirsing_translation (partial) English translation] - P. Erdős. [http://www.renyi.hu/~p_erdos/1957-13.pdf Some unsolved problems], Michigan Math. J. 4 (1957), 291--300 MR20 #5157; Zentralblatt 81,1. + Newman, D. J. [http://www.jstor.org/stable/2036455 On the number of binary digits in a multiple of three]. Proc Amer Math Soc. 21(1969): 719--721. + Proves that the Thue-Morse sequence has discrepancy <math> \gg N^{\log_4(3)}</math> - Problem 9 of this paper is ours. Erdős dates the problem to around 1932, and notes what we know about Liouville's function (lower and upper bound). + Halász, G. [http://gowers.wordpress.com/2010/02/02/edp5-another-very-brief-summary/#comment-5806 On random multiplicative functions]. Hubert Delange colloquium (Orsay, 1982), 74–96, Publ. Math. Orsay, 83-4, Univ. Paris XI, Orsay, 1983. + + The discrepancy of a random multiplicative function is close to \$\sqrt{N}\$. + + === Problem lists on which this problem appears === Erdős, P. and Graham, R. [http://www.math.ucsd.edu/~fan/ron/papers/79_09_combinatorial_number_theory.pdf Old and New Problems and Results in Combinatorial Number Theory: van der Waerden's Theorem and Related Topics], L'Enseignement Math. 25 (1979), 325-344. Erdős, P. and Graham, R. [http://www.math.ucsd.edu/~fan/ron/papers/79_09_combinatorial_number_theory.pdf Old and New Problems and Results in Combinatorial Number Theory: van der Waerden's Theorem and Related Topics], L'Enseignement Math. 25 (1979), 325-344. Line 118: Line 178: Our problem is mentioned at the bottom of page 331, where they indicate knowledge of a coloring with logarithmic discrepancy. On pages 330-1, they review work on the non-homogeneous problem. Our problem is mentioned at the bottom of page 331, where they indicate knowledge of a coloring with logarithmic discrepancy. On pages 330-1, they review work on the non-homogeneous problem. - Hochberg, Robert. [http://www.springerlink.com/content/q02424284373qw45/ Large Discrepancy In Homogenous Quasi-Arithmetic Progressions]. Combinatorica, Volume 26 (2006), Number 1. + Erdős, Paul, Some of my favourite unsolved problems. A tribute to Paul Erdős, 467--478, Cambridge Univ. Press, Cambridge, 1990. MR1117038 (92f:11003) - Roth's method for the <math> n^{1/4}</math> lower bound on nonhomogeneous AP discrepancy is adapted to give a lower bound for homogeneous quasi-AP discrepancy. This result is weaker than the Vijay result, but uses a different method. + "Let <math>f(n)=\pm1</math>. Is it true that for every <math>c</math> there is a <math>d</math> and an <math>m</math> so that + :<math> \left\| \sum_{k=1}^m f(kd) \right\| > c</math>? + I will pay \$500 for an answer. Let <math>f(n)=\pm1</math> and <math>f(ab)=f(a)f(b)</math>. Is it then true that <math>\left\| \sum_{k=1}^m f(kd) \right\|</math> cannot be bounded?" + + P. Erdős. [http://www.renyi.hu/~p_erdos/1957-13.pdf Some unsolved problems], Michigan Math. J. 4 (1957), 291--300 MR20 #5157; Zentralblatt 81,1. + + Problem 9 of this paper is ours. Erdős dates the problem to around 1932, and notes what we know about Liouville's function (lower and upper bound). Finch, S. [http://algo.inria.fr/csolve/ec.pdf Two-colorings of positive integers]. Dated May 27, 2008. Finch, S. [http://algo.inria.fr/csolve/ec.pdf Two-colorings of positive integers]. Dated May 27, 2008. Summarizes knowledge of discrepancy of two colorings when the discrepancy is restricted to homogeneous progressions, nonhomogeneous progressions, and homogeneous quasi-progressions. Contains bibliography with 17 entries, including most of those above. Summarizes knowledge of discrepancy of two colorings when the discrepancy is restricted to homogeneous progressions, nonhomogeneous progressions, and homogeneous quasi-progressions. Contains bibliography with 17 entries, including most of those above. + + === non-homogeneous AP, quasi-AP, and other related discrepancy papers === + + Hochberg, Robert. [http://www.springerlink.com/content/q02424284373qw45/ Large Discrepancy In Homogenous Quasi-Arithmetic Progressions]. Combinatorica, Volume 26 (2006), Number 1. + + Roth's method for the <math> n^{1/4}</math> lower bound on nonhomogeneous AP discrepancy is adapted to give a lower bound for homogeneous quasi-AP discrepancy. This result is weaker than the Vijay result, but uses a different method. + + Vijay, Sujith. [http://www.combinatorics.org/Volume_15/PDF/v15i1r104.pdf On the discrepancy of quasi-progressions]. Electronic Journal of Combinatorics, R104 of Volume 15(1), 2008. + + A quasi-progression is a sequence of the form <math>\lfloor k \alpha \rfloor </math>, with <math>1\leq k \leq K</math> for some K. The main theorem of interest is: If the integers from 0 to n are 2-coloured, there exists a quasi-progression that has discrepancy at least <math> (1/50)n^{1/6}</math>. + + * Doerr, Benjamin; Srivastav, Anand; Wehr, Petra. [http://www.combinatorics.org/Volume_11/Abstracts/v11i1r5.html Discrepancy of Cartesian products of arithmetic progressions]. Electron. J. Combin. 11 (2004), no. 1, Research Paper 5, 16 pp. (electronic). + + Abstract: We determine the combinatorial discrepancy of the hypergraph H of cartesian + products of d arithmetic progressions in the <math> [N]^d</math> –lattice. + The study of such higher dimensional arithmetic progressions is motivated by a + multi-dimensional version of van derWaerden’s theorem, namely the Gallai-theorem + (1933). We solve the discrepancy problem for d–dimensional arithmetic progressions + by proving <math>disc(H) = \Theta�(N^{d/4})</math> for every fixed positive integer d. This extends the famous + lower bound of <math> \Omega(N^{1/4})</math> of Roth (1964) and the matching upper bound <math> O(N^{1/4})</math> + of Matoušek and Spencer (1996) from d = 1 to arbitrary, fixed d. To establish + the lower bound we use harmonic analysis on locally compact abelian groups. For + the upper bound a product coloring arising from the theorem of Matoušek and + Spencer is sufficient. We also regard some special cases, e.g., symmetric arithmetic + progressions and infinite arithmetic progressions. + + Cilleruelo, Javier; Hebbinghaus, Nils [http://www.ams.org/mathscinet-getitem?mr=2547932 Discrepancy in generalized arithmetic progressions]. European J. Combin. 30 (2009), no. 7, 1607--1611. MR 2547932 + + Valkó, Benedek. [Discrepancy of arithmetic progressions in higher dimensions]. (English summary) + J. Number Theory 92 (2002), no. 1, 117--130. MR1880588 (2003b:11071) + + Abstract: + K. F. Roth (1964, Acta. Arith.9, 257–260) proved that the discrepancy of arithmetic progressions contained in [N] is at least <math> cN^{1/4} </math>, and later it was proved that this result is sharp. We consider the d-dimensional version of this problem. We give a lower estimate for the discrepancy of arithmetic progressions on <math> [N]^d </math> and prove that this result is nearly sharp. We use our results to give an upper estimate for the discrepancy of lines on an N×N lattice, and we also give an estimate for the discrepancy of a related random hypergraph. + + *Roth, K. F., Remark concerning integer sequences. Acta Arith. 9 1964 257--260. MR0168545 (29 #5806) + + Shows that the discrepancy on APs is at least <math>cN^{1/4}</math>. ## Current revision If you want to you can jump straight to the main experimental results page. ## Introduction and statement of problem Let (xn) be a $\scriptstyle \pm 1$ sequence and let C be a constant. Must there exist positive integers d,k such that $\left\| \sum_{i=1}^k x_{id} \right\| > C$? Known colloquially as "The Erdős discrepancy problem", this question has remained unanswered since the 1930s (Erdős, 1957) and Erdős offered \$500 for an answer. It was also asked by Chudakov (1956). It is extremely easy to state, and a very appealing problem, so perhaps Polymath can succeed where individual mathematicians have failed. If so, then, given the notoriety of the problem, it would be a significant coup for the multiply collaborative approach to mathematics. But even if a full solution is not reached, preliminary investigations have already thrown up some interesting ideas which could perhaps be developed into publishable results. It seems likely that the answer to Erdős's question is yes. If it is, then an easy compactness argument tells us that for every C there exists n such that for every $\scriptstyle \pm 1$ sequence of length n there exist d,k with $dk\leq n$ such that $\left\| \sum_{i=1}^k x_{id} \right\| > C$. In view of this, we make some definitions that allow one to talk about the dependence between C and n. For any finite set A of integers, we define the error on A by $E(A) := \sum_{a\in A} x_a$, and for a set $\mathcal{A}$ of finite sets of integers, we define the discrepancy $\delta(\mathcal{A},x) := \sup_{A\in \mathcal{A}} \|E(A)\|.$ We can think of the values taken by the sequence (xn) as a red/blue colouring of the integers that tries to make the number of reds and blues in each $\scriptstyle A\in\mathcal{A}$ as equal as possible. The discrepancy measures the extent to which the sequence fails in this attempt. Taking $\scriptstyle \mathcal{HAP}(N)$ to be the set of homogeneous arithmetic progressions {d,2d,3d,...,nd} contained in {1,2,...,N}, we can restate the question as whether $\scriptstyle \delta(\mathcal{HAP}(N),x) \to \infty$ for every sequence x. Two related questions have already been solved in the literature. Letting $\scriptstyle \mathcal{AP}(N)$ be the collection of all (not necessarily homogeneous) arithmetic progressions in {1,2,...,N}, Roth (1964) proved that $\scriptstyle \delta(\mathcal{AP}(N),x) \geq c n^{1/4}$, independent of the sequence x. Letting $\scriptstyle \mathcal{HQAP}(N)$ be the collection of all homogeneous quasi-arithmetic progressions $\scriptstyle\{\lfloor \alpha \rfloor,\lfloor 2\alpha \rfloor,\dots,\lfloor k\alpha \rfloor\}$ contained in {1,2,...,N}, Vijay (2008) proved that $\scriptstyle \delta(\mathcal{HQAP}(N),x) \geq 0.02 n^{1/6}$, independent of the sequence x. ## Some definitions and notational conventions A homogeneous arithmetic progression, or HAP, is an arithmetic progression of the form {d,2d,3d,...,nd}. When x is clear from context, we write δ(N) in place of $\delta(\mathcal{HAP}(N),x)$. The discrepancy of a sequence (xn) is the supremum of $\|\sum_{n\in P}x_n\|$ over all homogeneous arithmetic progressions P. (Strictly speaking, we should call this something like the HAP-discrepancy, but since we will almost always be talking about HAPs, we adopt the convention that "discrepancy" always means "HAP-discrepancy" unless it is stated otherwise.) Insertformulahere A sequence (xn) has discrepancy at most φ(n) if for every natural number N the maximum value of $\|\sum_{n\in P}x_n\|$ over all homogeneous arithmetic progressions P that are subsets of {1,2,...,N} is at most φ(N). The EDP is the Erdős discrepancy problem. (This may conceivably be changed if enough people don't like it.) A sequence (xn) is completely multiplicative if xmn = xmxn for any two positive integers m and n. We shall sometimes abbreviate this to "multiplicative", but the reader should be aware that the word "multiplicative" normally refers to the more general class of sequences such that xmn = xmxn whenever m and n are coprime. A completely multiplicative function is determined by the values it takes at primes. The Liouville function λ is the unique completely multiplicative function that takes the value -1 at every prime: if the prime factorization of n is $\prod p_i^{a_i}$ then λ(n) equals $(-1)^{\sum a_i}$. Another important multiplicative function is μ3, the multiplicative function that’s -1 at 3, 1 at numbers that are 1 mod 3 and -1 at numbers that are 2 mod 3. This function has mean-square partial sums which grow logarithmically, see this calculation. A HAP-subsequence of a sequence (xn) is a subsequence of the form xd,x2d,x3d,... for some d. If (xn) is a multiplicative $\pm 1$ sequence, then every HAP-subsequence is equal to either (xn) or ( − xn). We shall call a sequence weakly multiplicative if it has only finitely many distinct HAP-subsequences. Let us also call a $\pm 1$ sequence quasi-multiplicative if it is a composition of a completely multiplicative function from $\mathbb{N}$ to an Abelian group G with a function from G to {-1,1} (This definition is too general. See this and the next comment). It can be shown that if (xn) is a weakly multiplicative sequence, then it has an HAP-subsequence that is quasi-multiplicative. It is convenient to define the maps Tk for $k \geq 1$ by Tk(x)n = xkn, where x = (x1,x2,...) is a sequence. ## Simple observations • To answer the problem negatively, it is sufficient to find a completely multiplicative sequence (that is, a sequence (xn) such that xmn = xmxn for every m,n) such that its partial sums $x_1+\dots+x_n$ are bounded. First mentioned in the proof of the theorem in this post. • The function that takes n to 1 if the last non-zero digit of n in its ternary representation is 1 and -1 if the last non-zero digit is 2 is completely multiplicative and the partial sum up to n is easily shown to be at most log3n + 1. Therefore, the rate at which the worst discrepancy grows, as a function of the length of the homogeneous progression, can be as slow as logarithmic. More examples of this sort are here. [1] It turns out that the base can be made significantly higher than 3, so this example is not best possible. Correction • To answer the problem negatively, it is also sufficient to find a completely multiplicative function f from $\mathbb{N}$ to a finite Abelian group G (meaning that f(mn)=f(m)+f(n) for every m,n) and a function h from G to {-1,1} such that for each g in G there exists d with f(d)=g and with the partial sums hf(d)+hf(2d)+...+hf(nd) bounded. In that case, one can set xn = hf(n). (Though this observation is simple with the benefit of hindsight, it might not have been made had it not been for the fact that this sort of structure was identified in a long sequence that had been produced experimentally). First mentioned here. Se also this post and this comment. • It can be shown that if a $\pm 1$ sequence (xn) starts with 1 and has only finitely many distinct subsequences of the form $(x_d,x_{2d},x_{3d},\dots)$, then it must have a subsequence that arises in the above way. (This was mentioned just above in the terminology section.) • The problem for the positive integers is equivalent to the problem for the positive rationals. [2](Sketch proof: if you have a function f from the positive integers to {-1,1} that works, then define fq(x) to be f(q!x) whenever q!x is an integer. Then take a pointwise limit of a subsequence of the functions fq. Click here for a more detailed discussion of this construction and what it is good for. • Define the drift of a sequence to be the maximal value of \|f(md)+...+f(nd)\|. The discrepancy problem is equivalent to showing that the drift is always infinite. It is obvious that it is at least 2 (because \|f(2)+f(4)\|, \|f(2)+f(3)\|, \|f(3)+f(4)\| cannot all be at most 1); one can show that it is at least 3 (which implies as a corollary that the discrepancy is at least 2). One can also show that the upper and lower discrepancy are each at least 2. • Using Fourier analysis, one can reduce the problem to one about completely multiplicative functions. • Here is a page whose aim is to record a human proof that completely multiplicative sequences have discrepancy at least 2. • Here is an argument that shows that bounded discrepancy multiplicative functions do not correlate with characters. • The answer to a function field version of the problem seems to be negative. • The problem can be generalized to pseudointegers. Here we have found problems similar to EDP with a negative answer. ## Experimental evidence Main page: Experimental results A nice feature of the problem is that it lends itself well to investigation on a computer. Although the number of $\pm 1$ sequences grows exponentially, the constraints on these sequences are such that depth-first searches can lead quite quickly to interesting examples. For instance, at the time of writing they have yielded several examples of sequences of length 1124 with discrepancy 2. (This is the current record.) See this page for more details and further links. Another sort of evidence is to bound the discrepancy for specific sequences. For example, setting x1 = − x2 = 1 and xn = − xn / 3 or xn = xn − 3 according to whether n is a multiple of 3 or not, yields a sequence with $\delta(N) \leq \log_9(N)+1$ for sufficiently large N. Currently, this is the record-holder for slow growing discrepancy. The Thue-Morse sequence has discrepancy $\delta(N) \gg N^{\log_4(3)}$. A random sequence (each xi chosen independently) has discrepancy at least (asymptotically) $\sqrt{2N \log\log N}$ by the law of the iterated logarithm. Determining if the discrepancy of Liouville's lambda function is O(n1 / 2 + ε) is equivalent to solving the Riemann hypothesis. However, this growth cannot be bounded above by n1 / 2 − ε for any positive ε. ## Interesting subquestions Given the length of time that the Erdős discrepancy problem has been open, the chances that it will be solved by Polymath5 are necessarily small. However, there are a number of interesting questions that we do not know the answers to, several of which have arisen naturally from the experimental evidence. Good answers to some of these would certainly constitute publishable results. Here is a partial list -- further additions are very welcome. (When the more theoretical part of the project starts, this section will probably grow substantially and become a separate page.) • Is there an infinite sequence of discrepancy 2? (Given how long a finite sequence can be, it seems unlikely that we could answer this question just by a clever search of all possibilities on a computer.) • The long sequences of low discrepancy discovered by computer all have some kind of approximate weak multiplicativity. If we take a hypothetical counterexample (xn) (which we could even assume has discrepancy 2), can we prove that it, or some other counterexample derived from it by passing to HAP-subsequences and taking pointwise limits, has some kind of interesting multiplicative structure? • Closely related to the previous question. If (xn) is a hypothetical counterexample, must it satisfy a compactness property of the following kind: for every positive c there exists M such that if you take any M HAP-subsequences, then there must be two of them, (yn) and (zn), such that $\lim\inf N^{-1}\sum_{n=1}^Ny_nz_n\geq 1-c$? • An even weaker question. If (xn) is a hypothetical counterexample, must it have two HAP-subsequences (yn) and (zn) such that the lim inf of $N^{-1}\sum_{n=1}^N y_nz_n$ is greater than 0? • A similar question, perhaps equivalent to the previous one (this should be fairly easy to check). Given a sequence (xn) define f(N) to be the average of xaxbxcxd over all quadruples (a,b,c,d) such that ab=cd and $a,b,c,d\leq N$. If (xn) is a counterexample, does that imply that $\lim\inf f(N)$ is greater than 0? See here for a computation of this average for the first 1124 sequence. • Is there any completely multiplicative counterexample? (This may turn out to be a very hard question. If so, then answering the previous questions could turn out to be the best we can hope for without making a substantial breakthrough in analytic number theory.) • Does there exist a $\pm 1$ sequence such that the corresponding discrepancy function grows at a rate that is slower than clogn for any positive c? • Does the number of sequences of length n and discrepancy at most C grow exponentially with n or more slowly than exponentially? (Obviously if the conjecture is true then it must be zero for large enough n, but the hope is that this question is a more realistic initial target.) ## General proof strategies This section contains links to other pages in which potential approaches to solving the problem are described. First obtain multiplicative structure and then obtain a contradiction Find a different parameter, show that it tends to infinity, and show that that implies that the discrepancy tends to infinity Prove the result for shifted HAPs instead of HAPs Find a good configuration of HAPs Generalize to a graph-theoretic formulation Representation of the diagonal Bounding the discrepancy in terms of the common difference ## Annotated Bibliography • Blog Discussion on Gowers's Weblog • Zeroth Post (Dec 17 - Jan 6). • First Post (Jan 6 - Jan 12). • Second Post (Jan 9 - Jan 11). • Third Post (Jan 11 - Jan 14). • Fourth Post (Jan 14 - 16). • Fifth Post (Jan 16-19). • First Theoretical Post (Jan 19-21) • Second Theoretical Post (Jan 21-26) • Third Theoretical Post (Jan 26 -?) • Fourth Theoretical Post (Jan 30- Feb 2) • Fifth Theoretical Post (Feb 2 - Feb 5) • Sixth Theoretical Post (Feb 5- Feb 8) • Seventh Theoretical Post (Feb 8 - 19) • Eighth Theoretical Post (Feb 19-Feb 24) • Ninth Theoretical Most (Feb 24-Mar 2) • Tenth Theoretical Post (Mar 2-Mar 7) • Eleventh Theoretical Post (Mar 7-Mar 12) • Twelfth Theoretical Post (Mar 13-Mar 22) • Thirteenth Theoretical Post (Mar 23-Apr 24) • Fourteenth Theoretical Post (Apr 25-?) • Mathias, A. R. D. On a conjecture of Erdős and Čudakov. Combinatorics, geometry and probability (Cambridge, 1993). This one page paper establishes that the maximal length of sequence for the case where C=1 is 11, and is the starting point for our experimental studies. • Tchudakoff, N. G. Theory of the characters of number semigroups. J. Indian Math. Soc. (N.S.) 20 (1956), 11--15. MR0083515 (18,719e) The Mathias paper states that one of Erdős's problem lists states that this paper "studies related questions". The Math Review for this paper states that it summarizes results from seven other papers that are in Russian. The Math Review leaves the impression that the topic concerns characters of modulus 1. • Borwein, Peter, and Choi, Stephen. A variant of Liouville's lambda function: some surprizing formulae. Explicit formulas for the discrepancy of some completely multiplicative functions whose discrepancy is logarithmic. A typical example is: Let $\lambda_3(n) = (-1)^{\omega_3(n)}$ where ω3(n) is the number of distinct prime factors congruent to − 1mod 3 in n (with multiple factors counted multiply). Then the discrepancy of the λ3 sequence up to N is exactly the number of 1's in the base three expansion of N. (This turns out not to be so surprising after all, since λ3(n) is precisely the same as the ternary function defined in the second item of the Simple Observations section.) • Borwein, Peter, Choi, Stephen and Coons, Michael. Completely multiplicative functions taking values in {-1,1}. The published version of the above paper, which explains the results and proofs more clearly than the preprint, and is more explicit about the relationship with Erdős's question. • Granville and Soundararajan. Multiplicative functions in arithmetic progressions In this blog post Terry Tao wrote: "Granville and Soundararajan have made some study of the discrepancy of bounded multiplicative functions. The situation is remarkably delicate and number-theoretical (and is closely tied with the Granville-Soundararajan theory of pretentious characters)." • Wirsing, E. Das asymptotische Verhalten von Summen über multiplikative Funktionen. II. (German) [The asymptotic behavior of sums of multiplicative functions. II.] Acta Math. Acad. Sci. Hungar. 18 1967 411--467. MR0223318 (36 #6366) (partial) English translation • Newman, D. J. On the number of binary digits in a multiple of three. Proc Amer Math Soc. 21(1969): 719--721. Proves that the Thue-Morse sequence has discrepancy $\gg N^{\log_4(3)}$ • Halász, G. On random multiplicative functions. Hubert Delange colloquium (Orsay, 1982), 74–96, Publ. Math. Orsay, 83-4, Univ. Paris XI, Orsay, 1983. The discrepancy of a random multiplicative function is close to \$\sqrt{N}\$. ### Problem lists on which this problem appears • Erdős, P. and Graham, R. Old and New Problems and Results in Combinatorial Number Theory: van der Waerden's Theorem and Related Topics, L'Enseignement Math. 25 (1979), 325-344. Our problem is mentioned at the bottom of page 331, where they indicate knowledge of a coloring with logarithmic discrepancy. On pages 330-1, they review work on the non-homogeneous problem. • Erdős, Paul, Some of my favourite unsolved problems. A tribute to Paul Erdős, 467--478, Cambridge Univ. Press, Cambridge, 1990. MR1117038 (92f:11003) "Let $f(n)=\pm1$. Is it true that for every c there is a d and an m so that $\left\| \sum_{k=1}^m f(kd) \right\| > c$? I will pay \$500 for an answer. Let $f(n)=\pm1$ and f(ab) = f(a)f(b). Is it then true that $\left\| \sum_{k=1}^m f(kd) \right\|$ cannot be bounded?" • P. Erdős. Some unsolved problems, Michigan Math. J. 4 (1957), 291--300 MR20 #5157; Zentralblatt 81,1. Problem 9 of this paper is ours. Erdős dates the problem to around 1932, and notes what we know about Liouville's function (lower and upper bound). • Finch, S. Two-colorings of positive integers. Dated May 27, 2008. Summarizes knowledge of discrepancy of two colorings when the discrepancy is restricted to homogeneous progressions, nonhomogeneous progressions, and homogeneous quasi-progressions. Contains bibliography with 17 entries, including most of those above. ### non-homogeneous AP, quasi-AP, and other related discrepancy papers • Hochberg, Robert. Large Discrepancy In Homogenous Quasi-Arithmetic Progressions. Combinatorica, Volume 26 (2006), Number 1. Roth's method for the n1 / 4 lower bound on nonhomogeneous AP discrepancy is adapted to give a lower bound for homogeneous quasi-AP discrepancy. This result is weaker than the Vijay result, but uses a different method. • Vijay, Sujith. On the discrepancy of quasi-progressions. Electronic Journal of Combinatorics, R104 of Volume 15(1), 2008. A quasi-progression is a sequence of the form $\lfloor k \alpha \rfloor$, with $1\leq k \leq K$ for some K. The main theorem of interest is: If the integers from 0 to n are 2-coloured, there exists a quasi-progression that has discrepancy at least (1 / 50)n1 / 6. • Doerr, Benjamin; Srivastav, Anand; Wehr, Petra. Discrepancy of Cartesian products of arithmetic progressions. Electron. J. Combin. 11 (2004), no. 1, Research Paper 5, 16 pp. (electronic). Abstract: We determine the combinatorial discrepancy of the hypergraph H of cartesian products of d arithmetic progressions in the [N]d –lattice. The study of such higher dimensional arithmetic progressions is motivated by a multi-dimensional version of van derWaerden’s theorem, namely the Gallai-theorem (1933). We solve the discrepancy problem for d–dimensional arithmetic progressions by proving disc(H) = Θ(Nd / 4) for every fixed positive integer d. This extends the famous lower bound of Ω(N1 / 4) of Roth (1964) and the matching upper bound O(N1 / 4) of Matoušek and Spencer (1996) from d = 1 to arbitrary, fixed d. To establish the lower bound we use harmonic analysis on locally compact abelian groups. For the upper bound a product coloring arising from the theorem of Matoušek and Spencer is sufficient. We also regard some special cases, e.g., symmetric arithmetic progressions and infinite arithmetic progressions. • Cilleruelo, Javier; Hebbinghaus, Nils Discrepancy in generalized arithmetic progressions. European J. Combin. 30 (2009), no. 7, 1607--1611. MR 2547932 • Valkó, Benedek. [Discrepancy of arithmetic progressions in higher dimensions]. (English summary) J. Number Theory 92 (2002), no. 1, 117--130. MR1880588 (2003b:11071) Abstract: K. F. Roth (1964, Acta. Arith.9, 257–260) proved that the discrepancy of arithmetic progressions contained in [N] is at least cN1 / 4, and later it was proved that this result is sharp. We consider the d-dimensional version of this problem. We give a lower estimate for the discrepancy of arithmetic progressions on [N]d and prove that this result is nearly sharp. We use our results to give an upper estimate for the discrepancy of lines on an N×N lattice, and we also give an estimate for the discrepancy of a related random hypergraph. • Roth, K. F., Remark concerning integer sequences. Acta Arith. 9 1964 257--260. MR0168545 (29 #5806) Shows that the discrepancy on APs is at least cN1 / 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": 46, "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": 0.8485578894615173, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/26079?sort=votes	## Algorithm for determining if a path exists in a graph or if not, the closest edit distance. ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Given a directed acyclic graph `G` and a path made up from its set of nodes `N`, what is the closest approximate match to N, equipped with an intuitive notion of distance? A path in a directed acyclic graph is essentially a string (that uses the set of nodes as the alphabet), so when comparing paths one can make use of edit distances developed for approximate string matching: There are many algorithms for approximate string matching: http://en.wikipedia.org/wiki/Approximate_string_matching This string matching answers the question when the `G` itself is also a path.. then we're merely asking to compare two strings. Asking if an arbitrary graph `A` built from the same set of nodes is a sub-graph of `G` is the general case of the problem, but I'm only interested in the case where `A` is a path and `G` is directed & acyclic. Any general pointers are also welcome. -- Edit: I ended up using a dynamic programming algorithm, independently also suggested in the accepted answer. Good call! It is probably the most accurate option as well, and barely more "complex" than the string-to-string case when the average # of edges per node is low. - 3 for q1, I'm not seeing the problem. given the sequence of nodes in the path, why can't you just verify that directed edges exist between each adjacent pair of nodes ? – Suresh Venkat May 27 2010 at 1:09 good point, I guess q2 is what I'd like know more about. – Deniz May 27 2010 at 3:53 ## 1 Answer If graph is acyclic you can use some sort of dynamic programming. Let $a_{u,k}$ be the best distance you can get if you start from vertex $u \in G$ and consider only $k$ last vertices of your given path. It's quite straightforward how to calculate all $a_{u,k}$ based on all values of $a_{u',k'}$ with $u'$ "after" $u$ and $k' < k$: you just iterate over all edges going from vertex $u$. This approach works in $O(\|G\| \times strlen)$ 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": 10, "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": 0.9380460381507874, "perplexity_flag": "head"}
http://mathoverflow.net/revisions/68849/list	## Return to Answer 2 added 29 characters in body Oh sure, this is quite well-known. The closure of an element is the smallest closed element which is greater than or equal to the given element. Dually for the interior operator. In general, a closure operator $\phi: P \to P$ on a poset $P$ is an order-preserving, inflationary ($x \leq \phi(x)$), idempotent ($\phi(\phi(x)) = \phi(x)$) operation, and if you want a topological closure operator, then you demand $\phi(x \vee y) = \phi(x) \vee \phi(y)$ as well. Alternatively, a closure operator can be specified by an inclusion $i: C \hookrightarrow P$ such that every $p \in P$ has a least upper bound $c \in C$. The assignment $p \mapsto c$ gives an order-preserving mapping $j: P \to C$ with the property $$j(q) \leq d \Leftrightarrow q \leq i(d)$$ (where the left side is to be read in $C$ and the right in $P$), and the closure operator $\phi$ is the composite $i \circ j$; it is easy to check the order-preserving inflationary idempotent properties. In the context of Boolean algebras $P$, as in Sikorski, you demand that $C$ be closed under joins as well (and that $i$ preserve them), to make $\phi$ a topological closure. I am writing this answer to bear out a far wider connection with category theory. The posets here are special cases of categories, where there is at most one arrow in any hom-set $\hom(x, y)$, which we write as $x \leq y$. Functors between poset categories amount to order-preserving maps. A closure operator amounts to a monad on a poset. The subcollection $i: C \subseteq P$ of closed elements for that operator is the category of algebras for that monad. The corresponding mapping $j: P \to C$ is the left adjoint to $i$. Any adjoint pair (known under another name as a Galois connection) between posets always induces a closure operator. And so on. There is an embarrassment of riches of references (i.e., so many that it's hard to think of one that stands out). But a trade secret among category theorists is to understand what general concepts mean in the simplified case of poset categories, and this point of view is explicitly declared in Paul Taylor's Practical Foundations of Mathematics. I think just about any introductory book on category theory will refer to this, though, and you will also find it used heavily in more specialized treatises like Johnstone's Stone Spaces that involve looking at lattices from a categorical point of view. 1 Oh sure, this is quite well-known. The closure of an element is the smallest closed element which is greater than or equal to the given element. Dually for the interior operator. In general, a closure operator $\phi: P \to P$ on a poset $P$ is an order-preserving, inflationary ($x \leq \phi(x)$), idempotent ($\phi(\phi(x)) = \phi(x)$) operation, and if you want a topological closure operator, then you demand $\phi(x \vee y) = \phi(x) \vee \phi(y)$ as well. Alternatively, a closure operator can be specified by an inclusion $i: C \hookrightarrow P$ such that every $p \in P$ has a least upper bound $c \in C$. The assignment $p \mapsto c$ gives an order-preserving mapping $j: P \to C$ with the property $$j(q) \leq d \Leftrightarrow q \leq i(d)$$ (where the left side is to be read in $C$ and the right in $P$), and the closure operator $\phi$ is the composite $i \circ j$; it is easy to check the order-preserving inflationary idempotent properties. In the context of Boolean algebras $P$, as in Sikorski, you demand that $C$ be closed under joins as well, to make $\phi$ a topological closure. I am writing this answer to bear out a far wider connection with category theory. The posets here are special cases of categories, where there is at most one arrow in any hom-set $\hom(x, y)$, which we write as $x \leq y$. Functors between poset categories amount to order-preserving maps. A closure operator amounts to a monad on a poset. The subcollection $i: C \subseteq P$ of closed elements for that operator is the category of algebras for that monad. The corresponding mapping $j: P \to C$ is the left adjoint to $i$. Any adjoint pair (known under another name as a Galois connection) between posets always induces a closure operator. And so on. There is an embarrassment of riches of references (i.e., so many that it's hard to think of one that stands out). But a trade secret among category theorists is to understand what general concepts mean in the simplified case of poset categories, and this point of view is explicitly declared in Paul Taylor's Practical Foundations of Mathematics. I think just about any introductory book on category theory will refer to this, though, and you will also find it used heavily in more specialized treatises like Johnstone's Stone Spaces that involve looking at lattices from a categorical point of view.	{"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": 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": 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": 0.9430294036865234, "perplexity_flag": "head"}
http://pediaview.com/openpedia/Quadrilateral	# Quadrilateral Quadrilateral Six different types of quadrilaterals Edges and vertices 4 Schläfli symbol {4} (for square) Area various methods; see below Internal angle (degrees) 90° (for square) In Euclidean plane geometry, a quadrilateral is a polygon with four sides (or edges) and four vertices or corners. Sometimes, the term quadrangle is used, by analogy with triangle, and sometimes tetragon for consistency with pentagon (5-sided), hexagon (6-sided) and so on. The origin of the word "quadrilateral" is the two Latin words quadri, a variant of four, and latus, meaning "side." Quadrilaterals are simple (not self-intersecting) or complex (self-intersecting), also called crossed. Simple quadrilaterals are either convex or concave. The interior angles of a simple (and planar) quadrilateral ABCD add up to 360 degrees of arc, that is $\angle A+\angle B+\angle C+\angle D=360^{\circ}.$ This is a special case of the n-gon interior angle sum formula (n − 2) × 180°. In a crossed quadrilateral, the four interior angles on either side of the crossing add up to 720°.[1] All convex quadrilaterals tile the plane by repeated rotation around the midpoints of their edges. ## Convex quadrilaterals – parallelograms[] Euler diagram of some types of quadrilaterals. (UK) denotes British English and (US) denotes American English. A parallelogram is a quadrilateral with two pairs of parallel sides. Equivalent conditions are that opposite sides are of equal length; that opposite angles are equal; or that the diagonals bisect each other. Parallelograms also include the square, rectangle, rhombus and rhomboid. • Rhombus or rhomb: all four sides are of equal length. An equivalent condition is that the diagonals perpendicularly bisect each other. An informal description is "a pushed-over square" (including a square). • Rhomboid: a parallelogram in which adjacent sides are of unequal lengths and angles are oblique (not right angles). Informally: "a pushed-over rectangle with no right angles." • Rectangle: all four angles are right angles. An equivalent condition is that the diagonals bisect each other and are equal in length. Informally: "a box or oblong" (including a square). • Square (regular quadrilateral): all four sides are of equal length (equilateral), and all four angles are right angles. An equivalent condition is that opposite sides are parallel (a square is a parallelogram), that the diagonals perpendicularly bisect each other, and are of equal length. A quadrilateral is a square if and only if it is both a rhombus and a rectangle (four equal sides and four equal angles). • Oblong: a term sometimes used to denote a rectangle which has unequal adjacent sides (i.e. a rectangle that is not a square). ## Convex quadrilaterals – other[] • Kite: two pairs of adjacent sides are of equal length. This implies that one diagonal divides the kite into congruent triangles, and so the angles between the two pairs of equal sides are equal in measure. It also implies that the diagonals are perpendicular. • Right kite: a kite with two opposite right angles. • Trapezoid (North American English) or Trapezium (British English): at least one pair of opposite sides are parallel. • Trapezium (NAm.): no sides are parallel. (In British English this would be called an irregular quadrilateral, and was once called a trapezoid.) • Isosceles trapezoid (NAm.) or isosceles trapezium (Brit.): one pair of opposite sides are parallel and the base angles are equal in measure. Alternative definitions are a quadrilateral with an axis of symmetry bisecting one pair of opposite sides, or a trapezoid with diagonals of equal length. • Tangential trapezoid: a trapezoid where the four sides are tangents to an inscribed circle. • Tangential quadrilateral: the four sides are tangents to an inscribed circle. A convex quadrilateral is tangential if and only if opposite sides have equal sums. • Cyclic quadrilateral: the four vertices lie on a circumscribed circle. A convex quadrilateral is cyclic if and only if opposite angles sum to 180°. • Bicentric quadrilateral: it is both tangential and cyclic. • Orthodiagonal quadrilateral: the diagonals cross at right angles. • Equidiagonal quadrilateral: the diagonals are of equal length. • Ex-tangential quadrilateral: the four extensions of the sides are tangent to an excircle. ## More quadrilaterals[] • An equilic quadrilateral has two opposite equal sides that, when extended, meet at 60°. • A Watt quadrilateral is a quadrilateral with a pair of opposite sides of equal length.[2] • A quadric quadrilateral is a convex quadrilateral whose four vertices all lie on the perimeter of a square.[3] • A geometric chevron (dart or arrowhead) is a concave quadrilateral with bilateral symmetry like a kite, but one interior angle is reflex. • A self-intersecting quadrilateral is called variously a cross-quadrilateral, crossed quadrilateral, butterfly quadrilateral or bow-tie quadrilateral. A special case of crossed quadrilaterals are the antiparallelograms, crossed quadrilaterals in which (like a parallelogram) each pair of nonadjacent sides has equal length. The diagonals of a crossed or concave quadrilateral do not intersect inside the shape. • A non-planar quadrilateral is called a skew quadrilateral. Formulas to compute its dihedral angles from the edge lengths and the angle between two adjacent edges were derived for work on the properties of molecules such as cyclobutane that contain a "puckered" ring of four atoms.[4] See skew polygon for more. Historically the term gauche quadrilateral was also used to mean a skew quadrilateral.[5] A skew quadrilateral together with its diagonals form a (possibly non-regular) tetrahedron, and conversely every skew quadrilateral comes from a tetrahedron where a pair of opposite edges is removed. ## Special line segments[] The two diagonals of a convex quadrilateral are the line segments that connect opposite vertices. The two bimedians of a convex quadrilateral are the line segments that connect the midpoints of opposite sides.[6] They intersect at the "vertex centroid" of the quadrilateral (see Remarkable points below). The four maltitudes of a convex quadrilateral are the perpendiculars to a side through the midpoint of the opposite side.[7] ## Notations in metric formulas[] In the metric formulas below, the following notations are used. A convex quadrilateral ABCD has the sides a = AB, b = BC, c = CD, and d = DA. The diagonals are p = AC and q = BD, and the angle between them is θ. The semiperimeter s is defined as $s=\tfrac{1}{2}(a+b+c+d)$. ## Area of a convex quadrilateral[] There are various general formulas for the area K of a convex quadrilateral. ### Trigonometric formulas[] The area can be expressed in trigonometric terms as $K = \tfrac{1}{2} pq \cdot \sin \theta,$ where the lengths of the diagonals are p and q and the angle between them is θ.[8] In the case of an orthodiagonal quadrilateral (e.g. rhombus, square, and kite), this formula reduces to $K=\tfrac{1}{2}pq$ since θ is 90°. Bretschneider's formula[9] expresses the area in terms of the sides and two opposite angles: $\begin{align} K &= \sqrt{(s-a)(s-b)(s-c)(s-d) - \tfrac{1}{2} abcd \; [ 1 + \cos (A + C) ]} \\ &= \sqrt{(s-a)(s-b)(s-c)(s-d) - abcd \left[ \cos^2 \left( \tfrac{A + C}{2} \right) \right]} \\ \end{align}$ where the sides in sequence are a, b, c, d, where s is the semiperimeter, and A and C are two (in fact, any two) opposite angles. This reduces to Brahmagupta's formula for the area of a cyclic quadrilateral when A+C = 180°. Another area formula in terms of the sides and angles, with angle C being between sides b and c, and A being between sides a and d, is $K = \tfrac{1}{2}ad \cdot \sin{A} + \tfrac{1}{2}bc \cdot \sin{C}.$ In the case of a cyclic quadrilateral, the latter formula becomes $K = \tfrac{1}{2}(ad+bc)\sin{A}.$ In a parallelogram, where both pairs of opposite sides and angles are equal, this formula reduces to $K=ab \cdot \sin{A}.$ Alternatively, we can write the area in terms of the sides and the intersection angle θ of the diagonals, so long as this angle is not 90°:[10] $K = \frac{\|\tan \theta\|}{4} \cdot \left\| a^2 + c^2 - b^2 - d^2 \right\|.$ In the case of a parallelogram, the latter formula becomes $K = \tfrac{1}{2}\|\tan \theta\|\cdot \left\| a^2 - b^2 \right\|.$ Another area formula including the sides a, b, c, d is[11] $K=\tfrac{1}{4}\sqrt{(2(a^2+c^2)-4x^2)(2(b^2+d^2)-4x^2)}\sin{\varphi}$ where x is the distance between the midpoints of the diagonals and φ is the angle between the bimedians. ### Non-trigonometric formulas[] The following two formulas expresses the area in terms of the sides a, b, c, d, the semiperimeter s, and the diagonals p, q: $K = \sqrt{(s-a)(s-b)(s-c)(s-d) - \tfrac{1}{4}(ac+bd+pq)(ac+bd-pq)},$ [12] $K = \frac{1}{4} \sqrt{4p^{2}q^{2}- \left( a^{2}+c^{2}-b^{2}-d^{2} \right) ^{2}}.$ [13] The first reduces to Brahmagupta's formula in the cyclic quadrilateral case, since then pq = ac + bd. The area can also be expressed in terms of the bimedians m, n and the diagonals p, q: $K=\tfrac{1}{2}\sqrt{(m+n+p)(m+n-p)(m+n+q)(m+n-q)},$ [14] $K=\tfrac{1}{2}\sqrt{p^2q^2-(m^2-n^2)^2}.$ [15] ### Vector formulas[] The area of a quadrilateral ABCD can be calculated using vectors. Let vectors AC and BD form the diagonals from A to C and from B to D. The area of the quadrilateral is then $K = \tfrac{1}{2} \|\mathbf{AC}\times\mathbf{BD}\|,$ which is half the magnitude of the cross product of vectors AC and BD. In two-dimensional Euclidean space, expressing vector AC as a free vector in Cartesian space equal to (x1,y1) and BD as (x2,y2), this can be rewritten as: $K = \tfrac{1}{2} \|x_1 y_2 - x_2 y_1\|.$ ### Area inequalities[] If a convex quadrilateral has the consecutive sides a, b, c, d and the diagonals p, q, then its area K satisfies[16] $K\le \tfrac{1}{4}(a+c)(b+d)$ with equality only for a rectangle. $K\le \tfrac{1}{4}(a^2+b^2+c^2+d^2)$ with equality only for a square. $K\le \tfrac{1}{4}(p^2+q^2)$ with equality only if the diagonals are perpendicular and equal. $K\le \tfrac{1}{2}\sqrt{(a^2+c^2)(b^2+d^2)}$ with equality only for a rectangle.[11] From Bretschneider's formula it directly follows that the area of a quadrilateral satisfies $K \le \sqrt{(s-a)(s-b)(s-c)(s-d)}$ with equality if and only if the quadrilateral is cyclic or degenerate such that one side is equal to the sum of the other three (it has collapsed into a line segment, so the area is zero). The area of any quadrilateral also satisfies the inequality[17] $\displaystyle K\le \tfrac{1}{2}\sqrt[3]{(ab+cd)(ac+bd)(ad+bc)}.$ ## Diagonals[] ### Properties of the diagonals in some quadrilaterals[] In the following table it is listed if the diagonals in some of the most basic quadrilaterals bisect each other, if their diagonals are perpendicular, and if their diagonals have equal length.[18] The list applies to the most general cases, and excludes named subsets. Quadrilateral Bisecting diagonals Perpendicular diagonals Equal diagonals Trapezoid No See note 1 No Isosceles trapezoid No See note 1 Yes Parallelogram Yes No No Kite See note 2 Yes See note 2 Rectangle Yes No Yes Rhombus Yes Yes No Square Yes Yes Yes Note 1: The most general trapezoids and isosceles trapezoids do not have perpendicular diagonals, but there are infinite numbers of (non-similar) trapezoids and isosceles trapezoids that do have perpendicular diagonals and are not any other named quadrilateral. Note 2: In a kite, one diagonal bisects the other. The most general kite has unequal diagonals, but there is an infinite number of (non-similar) kites in which the diagonals are equal in length (and the kites are not any other named quadrilateral). ### Length of the diagonals[] The length of the diagonals in a convex quadrilateral ABCD can be calculated using the law of cosines. Thus $p=\sqrt{a^2+b^2-2ab\cos{B}}=\sqrt{c^2+d^2-2cd\cos{D}}$ and $q=\sqrt{a^2+d^2-2ad\cos{A}}=\sqrt{b^2+c^2-2bc\cos{C}}.$ Other, more symmetric formulas for the length of the diagonals, are[19] $p=\sqrt{\frac{(ac+bd)(ad+bc)-2abcd(\cos{B}+\cos{D})}{ab+cd}}$ and $q=\sqrt{\frac{(ab+cd)(ac+bd)-2abcd(\cos{A}+\cos{C})}{ad+bc}}.$ ### Generalizations of the parallelogram law and Ptolemy's theorem[] In any convex quadrilateral ABCD, the sum of the squares of the four sides is equal to the sum of the squares of the two diagonals plus four times the square of the line segment connecting the midpoints of the diagonals. Thus $a^2 + b^2 + c^2 + d^2 = p^2 + q^2 + 4x^2$ where x is the distance between the midpoints of the diagonals.[20]:p.126 This is sometimes known as Euler's quadrilateral theorem and is a generalization of the parallelogram law. A corollary is the inequality $a^2 + b^2 + c^2 + d^2 \ge p^2 + q^2$ where equality holds if and only if the quadrilateral is a parallelogram. Euler also generalized Ptolemy's theorem, which is an equality in a cyclic quadrilateral, into an inequality for a convex quadrilateral. It states that $pq \le ac + bd$ where there is equality if and only if the quadrilateral is cyclic.[20]:p.128–129 This is often called Ptolemy's inequality. The German mathematician Carl Anton Bretschneider derived in 1842 the following generalization of Ptolemy's theorem, regarding the product of the diagonals in a convex quadrilateral[21] $p^2q^2=a^2c^2+b^2d^2-2abcd\cos{(A+C)}.$ This relation can be considered to be a law of cosines for a quadrilateral. In a cyclic quadrilateral, where A + C = 180°, it reduces to pq = ac + bd. Since cos (A + C) ≥ -1, it also gives a proof of Ptolemy's inequality. ### Other metric relations[] If X and Y are the feet of the normals from B and D to the diagonal AC = p in a convex quadrilateral ABCD with sides a = AB, b = BC, c = CD, d = DA, then[22]:p.14 $XY=\frac{\|a^2+c^2-b^2-d^2\|}{2p}.$ In a convex quadrilateral ABCD with sides a = AB, b = BC, c = CD, d = DA, and where the diagonals intersect at E, $efgh(a+c+b+d)(a+c-b-d) = (agh+cef+beh+dfg)(agh+cef-beh-dfg)$ where e = AE, f = BE, g = CE, and h = DE.[23] The shape of a convex quadrilateral is fully determined by the lengths of its sides in sequence and of one diagonal between two specified vertices. The two diagonals p, q and the four side lengths a, b, c, d of a quadrilateral are related[24] by the Cayley-Menger determinant, as follows: $\det \begin{bmatrix} 0 & a^2 & p^2 & d^2 & 1 \\ a^2 & 0 & b^2 & q^2 & 1 \\ p^2 & b^2 & 0 & c^2 & 1 \\ d^2 & q^2 & c^2 & 0 & 1 \\ 1 & 1 & 1 & 1 & 0 \end{bmatrix} = 0.$ ## Bimedians[] The Varignon parallelogram EFGH The midpoints of the sides of any quadrilateral (convex, concave or crossed) are the vertices of a parallelogram called the Varignon parallelogram. It has the following properties: • Each pair of opposite sides of the Varignon parallelogram are parallel to a diagonal in the original quadrilateral. • The length of a side in the Varignon parallelogram is half as long as the diagonal in the original quadrilateral it is parallel to. • The area of the Varignon parallelogram equals half the area of the original quadrilateral. This is true in convex, concave and crossed quadrilaterals provided the area of the latter is defined to be the difference of the areas of the two triangles it is composed of.[25] • The perimeter of the Varignon parallelogram equals the sum of the diagonals of the original quadrilateral. The diagonals of the Varignon parallelogram are the bimedians of the original quadrilateral. The two bimedians in a quadrilateral and the line segment joining the midpoints of the diagonals in that quadrilateral are concurrent and are all bisected by their point of intersection.[20]:p.125 In a convex quadrilateral with sides a, b, c and d, the length of the bimedian that connects the midpoints of the sides a and c is $m=\tfrac{1}{2}\sqrt{-a^2+b^2-c^2+d^2+p^2+q^2}$ where p and q are the length of the diagonals.[26] The length of the bimedian that connects the midpoints of the sides b and d is $n=\tfrac{1}{2}\sqrt{a^2-b^2+c^2-d^2+p^2+q^2}.$ Hence[20]:p.126 $\displaystyle p^2+q^2=2(m^2+n^2).$ This is also a corollary to the parallelogram law applied in the Varignon parallelogram. The length of the bimedians can also be expressed in terms of two opposite sides and the distance x between the midpoints of the diagonals. This is possible when using Euler's quadrilateral theorem in the above formulas. Whence[15] $m=\tfrac{1}{2}\sqrt{2(b^2+d^2)-4x^2}$ and $n=\tfrac{1}{2}\sqrt{2(a^2+c^2)-4x^2}.$ Note that the two opposite sides in these formulas are not the two that the bimedian connects. In a convex quadrilateral, there are the following dual connection between the bimedians and the diagonals:[22] • The two bimedians have equal length if and only if the two diagonals are perpendicular • The two bimedians are perpendicular if and only if the two diagonals have equal length ## Trigonometric identities[] The four angles of a simple quadrilateral ABCD satisfy the following identities:[27] $\sin{A}+\sin{B}+\sin{C}+\sin{D}=4\sin{\frac{A+B}{2}}\sin{\frac{A+C}{2}}\sin{\frac{A+D}{2}}$ and $\frac{\tan{A}\tan{B}-\tan{C}\tan{D}}{\tan{A}\tan{C}-\tan{B}\tan{D}}=\frac{\tan{(A+C)}}{\tan{(A+B)}}.$ Also,[28] $\frac{\tan{A}+\tan{B}+\tan{C}+\tan{D}}{\cot{A}+\cot{B}+\cot{C}+\cot{D}}=\tan{A}\tan{B}\tan{C}\tan{D}.$ In the last two formulas, no angle is allowed to be a right angle, since then the tangent functions are not defined. ## Maximum and minimum properties[] Among all quadrilaterals with a given perimeter, the one with the largest area is the square. This is called the isoperimetric theorem for quadrilaterals. It is a direct consequence of the area inequality[17]:p.114 $K\le \tfrac{1}{16}L^2$ where K is the area of a convex quadrilateral with perimeter L. Equality holds if and only if the quadrilateral is a square. The dual theorem states that of all quadrilaterals with a given area, the square has the shortest perimeter. The quadrilateral with given side lengths that has the maximum area is the cyclic quadrilateral.[29] Of all convex quadrilaterals with given diagonals, the orthodiagonal quadrilateral has the largest area.[17]:p.119 This is a direct consequence of the fact that the area of a convex quadrilateral satisfies $K=\tfrac{1}{2}pq\sin{\theta}\le \tfrac{1}{2}pq,$ where θ is the angle between the diagonals p and q. Equality holds if and only if θ = 90°. If P is an interior point in a convex quadrilateral ABCD, then $AP+BP+CP+DP\ge AC+BD.$ From this inequality it follows that the point inside a quadrilateral that minimizes the sum of distances to the vertices is the intersection of the diagonals. Hence that point is the Fermat point of a convex quadrilateral.[30]:p.120 ## Remarkable points and lines in a convex quadrilateral[] The centre of a quadrilateral can be defined in several different ways. The "vertex centroid" comes from considering the quadrilateral as being empty but having equal masses at its vertices. The "side centroid" comes from considering the sides to have constant mass per unit length. The usual centre, called just centroid (centre of area) comes from considering the surface of the quadrilateral as having constant density. These three points are in general not all the same point.[31] The "vertex centroid" is the intersection of the two bimedians.[32] As with any polygon, the x and y coordinates of the vertex centroid are the arithmetic means of the x and y coordinates of the vertices. The "area centroid" of quadrilateral ABCD can be constructed in the following way. Let Ga, Gb, Gc, Gd be the centroids of triangles BCD, ACD, ABD, ABC respectively. Then the "area centroid" is the intersection of the lines GaGc and GbGd.[33] In a general convex quadrilateral ABCD, there are no natural analogies to the circumcenter and orthocenter of a triangle. But two such points can be constructed in the following way. Let Oa, Ob, Oc, Od be the circumcenters of triangles BCD, ACD, ABD, ABC respectively; and denote by Ha, Hb, Hc, Hd the orthocenters in the same triangles. Then the intersection of the lines OaOc and ObOd is called the quasicircumcenter; and the intersection of the lines HaHc and HbHd is called the quasiorthocenter of the convex quadrilateral.[33] These points can be used to define an Euler line of a quadrilateral. In a convex quadrilateral, the quasiorthocenter H, the "area centroid" G, and the quasicircumcenter O are collinear in this order, and HG = 2GO.[33] There can also be defined a quasinine-point center E as the intersection of the lines EaEc and EbEd, where Ea, Eb, Ec, Ed are the nine-point centers of triangles BCD, ACD, ABD, ABC respectively. Then E is the midpoint of OH.[33] Another remarkable line in a convex quadrilateral is the Newton line. ## Other properties of convex quadrilaterals[] • Let exterior squares be drawn on all sides of a quadrilateral. The segments connecting the centers of opposite squares are (a) equal in length, and (b) perpendicular. Thus these centers are the vertices of an orthodiagonal quadrilateral. This is called Van Aubel's theorem. • The internal angle bisectors of a convex quadrilateral either form a cyclic quadrilateral[20]:p.127 or they are concurrent. In the latter case the quadrilateral is a tangential quadrilateral. • For any simple quadrilateral with given edge lengths, there is a cyclic quadrilateral with the same edge lengths.[29] • The four smaller triangles formed by the diagonals and sides of a convex quadrilateral have the property that the product of the areas of two opposite triangles equals the product of the areas of the other two triangles.[34] ## Taxonomy[] A taxonomy of quadrilaterals is illustrated by the following graph. Lower forms are special cases of higher forms. Note that "trapezium" here is referring to the British definition (the North American equivalent is a trapezoid), and "kite" excludes the concave kite (arrowhead or dart). Inclusive definitions are used throughout. ## References[] 1. G. Keady, P. Scales and S. Z. Németh, "Watt Linkages and Quadrilaterals", The Mathematical Gazette Vol. 88, No. 513 (Nov., 2004), pp. 475–492. 2. A. K. Jobbings, "Quadric Quadrilaterals", The Mathematical Gazette Vol. 81, No. 491 (Jul., 1997), pp. 220–224. 3. M.P. Barnett and J.F. Capitani, Modular chemical geometry and symbolic calculation, International Journal of Quantum Chemistry, 106 (1) 215–227, 2006. 4. William Rowan Hamilton, , Proceedings of the Royal Irish Academy, 4 (1850), pp. 380–387. 5. E.W. Weisstein. "Bimedian". MathWorld - A Wolfram Web Resource. 6. E.W. Weisstein. "Maltitude". MathWorld - A Wolfram Web Resource. 7. Harries, J. "Area of a quadrilateral," Mathematical Gazette 86, July 2002, 310–311. 8. R. A. Johnson, Advanced Euclidean Geometry, 2007, Dover Publ., p. 82. 9. Mitchell, Douglas W., "The area of a quadrilateral," Mathematical Gazette 93, July 2009, 306–309. 10. ^ a b Josefsson, Martin (2013), "Five Proofs of an Area Characterization of Rectangles", Forum Geometricorum 13: 17–21. 11. J. L. Coolidge, "A historically interesting formula for the area of a quadrilateral", American Mathematical Monthly, 46 (1939) 345–347. 12. E.W. Weisstein. "Bretschneider's formula". MathWorld - A Wolfram Web Resource. 13. Archibald, R. C., "The Area of a Quadrilateral", American Mathematical Monthly, 29 (1922) pp. 29–36. 14. ^ a b Josefsson, Martin (2011), "The Area of a Bicentric Quadrilateral", Forum Geometricorum 11: 155–164. 15. O. Bottema, Geometric Inequalities, Wolters-Noordhoff Publishing, The Netherlands, 1969, pp. 129, 132. 16. ^ a b c Alsina, Claudi; Nelsen, Roger (2009), When Less is More: Visualizing Basic Inequalities, Mathematical Association of America, p. 68. 17. Rashid, M. A. & Ajibade, A. O., "Two conditions for a quadrilateral to be cyclic expressed in terms of the lengths of its sides", Int. J. Math. Educ. Sci. Technol., vol. 34 (2003) no. 5, pp. 739–799. 18. Altshiller-Court, Nathan, College Geometry, Dover Publ., 2007. 19. Andreescu, Titu & Andrica, Dorian, Complex Numbers from A to...Z, Birkhäuser, 2006, pp. 207–209. 20. ^ a b Josefsson, Martin (2012), "Characterizations of Orthodiagonal Quadrilaterals", Forum Geometricorum 12: 13–25. 21. Hoehn, Larry (2011), "A New Formula Concerning the Diagonals and Sides of a Quadrilateral", Forum Geometricorum 11: 211–212. 22. E.W. Weisstein. "Quadrilateral". MathWorld - A Wolfram Web Resource. 23. H. S. M. Coxeter and S. L. Greitzer, Geometry Revisited, MAA, 1967, pp. 52-53. 24. C. V. Durell & A. Robson, Advanced Trigonometry, Dover, 2003, p. 267. 25. MathPro Press, "Original Problems Proposed by Stanley Rabinowitz 1963–2005", p. 23, [2] 26. ^ a b Thomas Peter, "Maximizing the Area of a Quadrilateral", The College Mathematics Journal, Vol. 34, No. 4 (September 2003), pp. 315–316. 27. Alsina, Claudi and Nelsen, Roger, Charming Proofs. A Journey Into Elegant Mathematics, Mathematical Association of America, 2010, pp. 114, 119, 120, 261. 28. King, James, Two Centers of Mass of a Quadrilateral, [3], Accessed 2012-04-15. 29. Honsberger, Ross, Episodes in Nineteenth and Twentieth Century Euclidean Geometry, Math. Assoc. Amer., 1995, pp. 35–41. 30. ^ a b c d Myakishev, Alexei (2006), "On Two Remarkable Lines Related to a Quadrilateral", Forum Geometricorum 6: 289–295. 31. Martin Josefsson, "Characterizations of Trapezoids", Forum Geometricorum 13 (2013) 23–35. ## Source Content is authored by an open community of volunteers and is not produced by or in any way affiliated with ore reviewed by PediaView.com. Licensed under the Creative Commons Attribution-ShareAlike 3.0 Unported License, using material from the Wikipedia article on "Quadrilateral", which is available in its original form here: http://en.wikipedia.org/w/index.php?title=Quadrilateral • ## Finding More You are currently browsing the the PediaView.com open source encyclopedia. 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Wikipedia® itself is a registered trademark of the Wikimedia Foundation, Inc.	{"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": 45, "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": 0.861242949962616, "perplexity_flag": "middle"}
http://www.nag.com/numeric/CL/nagdoc_cl23/html/S/s21bcc.html	NAG Library Function Documentnag_elliptic_integral_rd (s21bcc) 1 Purpose nag_elliptic_integral_rd (s21bcc) returns a value of the symmetrised elliptic integral of the second kind. 2 Specification #include <nag.h> #include <nags.h> double nag_elliptic_integral_rd (double x, double y, double z, NagError fail) 3 Description nag_elliptic_integral_rd (s21bcc) calculates an approximate value for the integral $R D x,y,z = 3 2 ∫ 0 ∞ dt t+x t+y t+z 3$ where $x$, $y\ge 0$, at most one of $x$ and $y$ is zero, and $z>0$. The basic algorithm, which is due to Carlson (1979) and Carlson (1988), is to reduce the arguments recursively towards their mean by the rule: $x 0 = x , y 0 = y , z 0 = z μ n = x n + y n + 3 z n / 5 X n = 1 - x n / μ n Y n = 1 - y n / μ n Z n = 1 - z n / μ n λ n = x n y n + y n z n + z n x n x n+1 = x n + λ n / 4 y n+1 = y n + λ n / 4 z n+1 = z n + λ n / 4$ For $n$ sufficiently large, $ε n = max X n , Y n , Z n ∼ 1 / 4 n$ and the function may be approximated adequately by a 5th-order power series $R D x,y,z = 3 ∑ m=0 n-1 4 -m z m + λ n z m + 4 -n μ n 3 1 + 3 7 S n 2 + 1 3 S n 3 + 3 22 S n 2 2 + 3 11 S n 4 + 3 13 S n 2 S n 3 + 3 13 S n 5 ,$ where ${S}_{n}^{\left(m\right)}=\left({X}_{n}^{m}+{Y}_{n}^{m}+3{Z}_{n}^{m}\right)/2m$. The truncation error in this expansion is bounded by $3{\epsilon }_{n}^{6}/\sqrt{{\left(1-{\epsilon }_{n}\right)}^{3}}$ and the recursive process is terminated when this quantity is negligible compared with the machine precision. The function may fail either because it has been called with arguments outside the domain of definition, or with arguments so extreme that there is an unavoidable danger of setting underflow or overflow. Note: ${R}_{D}\left(x,x,x\right)={x}^{-3/2}\text{,}$ so there exists a region of extreme arguments for which the function value is not representable. . 4 References Abramowitz M and Stegun I A (1972) Handbook of Mathematical Functions (3rd Edition) Dover Publications Carlson B C (1979) Computing elliptic integrals by duplication Numerische Mathematik 33 1–16 Carlson B C (1988) A table of elliptic integrals of the third kind Math. Comput. 51 267–280 5 Arguments 1: x – doubleInput 2: y – doubleInput 3: z – doubleInput On entry: the arguments $x$, $y$ and $z$ of the function. Constraint: x, ${\mathbf{y}}\ge 0.0$, ${\mathbf{z}}>0.0$ and only one of x and y may be zero. 4: fail – NagError Input/Output The NAG error argument (see Section 3.6 in the Essential Introduction). 6 Error Indicators and Warnings NE_REAL_ARG_EQ On entry, ${\mathbf{x}}+{\mathbf{y}}$ must not be equal to 0.0: ${\mathbf{x}}+{\mathbf{y}}=〈\mathit{\text{value}}〉$. Both x and y are zero and the function is undefined. NE_REAL_ARG_GE On entry, x must not be greater than or equal to $〈\mathit{\text{value}}〉$: ${\mathbf{x}}=〈\mathit{\text{value}}〉$. On entry, y must not be greater than or equal to $〈\mathit{\text{value}}〉$: ${\mathbf{y}}=〈\mathit{\text{value}}〉$. On entry, z must not be greater than or equal to $〈\mathit{\text{value}}〉$: ${\mathbf{z}}=〈\mathit{\text{value}}〉$. There is a danger of setting underflow and the function returns zero. NE_REAL_ARG_LE On entry, z must not be less than or equal to 0.0: ${\mathbf{z}}=〈\mathit{\text{value}}〉$. The function is undefined. NE_REAL_ARG_LT On entry, either z is too close to zero or both x and y are too close to zero: there is a danger of setting overflow. On entry, $〈\mathit{parameters}〉$ must not be less than $〈\mathit{\text{value}}〉$: $〈\mathit{parameters}〉=〈\mathit{\text{value}}〉$. On entry, x must not be less than 0.0: ${\mathbf{x}}=〈\mathit{\text{value}}〉$. On entry, y must not be less than 0.0: ${\mathbf{y}}=〈\mathit{\text{value}}〉$. The function is undefined. 7 Accuracy In principle the function is capable of producing full machine precision. However, round-off errors in internal arithmetic will result in slight loss of accuracy. This loss should never be excessive as the algorithm does not involve any significant amplification of round-off error. It is reasonable to assume that the result is accurate to within a small multiple of the machine precision. 8 Further Comments Symmetrised elliptic integrals returned by functions nag_elliptic_integral_rd (s21bcc), nag_elliptic_integral_rc (s21bac), nag_elliptic_integral_rf (s21bbc) and nag_elliptic_integral_rj (s21bdc) can be related to the more traditional canonical forms (see Abramowitz and Stegun (1972)), as described in the s Chapter Introduction. 9 Example This example program simply generates a small set of nonextreme arguments which are used with the function to produce the table of low accuracy results. 9.1 Program Text Program Text (s21bcce.c) None. 9.3 Program Results Program Results (s21bcce.r)	{"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": 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": 0.7396653890609741, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/44209/ground-state-energy-e-0-and-evaluation-of-physical-energies/44215	# Ground state energy $E_{0}$ and evaluation of physical energies Given the lowest eigenvalue $E_0$ of an Schrödinguer operator, do the other energies $E_{n}$ for $n >0$ depend strongly on the lowest eigenvalue of the system? I mean, if we somehow fixed the eigenvalue $E_{0}$, could we get more or at least better approximations to the other eigenenergies of the system? thanks - ## 1 Answer No. The differences between the eigenvalues (giving the spectral frequencies) are highly specific for each chemical substance, so knowledge of $E_0$ tells very little about the remainder of the spectrum. -	{"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": 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": 0.8559458255767822, "perplexity_flag": "head"}
http://www.sciforums.com/showthread.php?113574-Has-James-Gates-Discovered-Computer-Code-in-String-Theory-Equations&p=2936937	• Forum • New Posts • FAQ • Calendar • Ban List • Community • Forum Actions • Encyclopedia • What's New? 1. If this is your first visit, be sure to check out the FAQ by clicking the link above. You need to register and post an introductory thread before you can post to all subforums: click the register link above to proceed. To start viewing messages, select the forum that you want to visit from the selection below. # Thread: 1. ## Has James Gates Discovered Computer Code in String Theory Equations? Is physical reality ultimately constructed of computer code? http://www.transcend.ws/?p=3020 Dr. S. James Gates, Jr., a theoretical physicist, the John S. Toll Professor of Physics at the University of Maryland, and the Director of The Center for String & Particle Theory, is reporting that certain string theory, super-symmetrical equations, which describe the fundamental nature of the Universe and reality, contain embedded computer codes. These codes are digital data in the form of 1′s and 0′s. Not only that, these codes are the same as what make web browsers work and are error-correction codes! Gates says, “We have no idea what these ‘things’ are doing there”. http://being.publicradio.org/program...sofpower.shtml Physicists have long sought to describe the universe in terms of equations. Now, James Gates explains how research on a class of geometric symbols known as adinkras could lead to fresh insights into the theory of supersymmetry — and perhaps even the very nature of reality. [...] However, with the observation that structures from information theory — codes — control the structure of equations with the SUSY property, we may be crossing a barrier. I know of no other example of this particular intermingling occurring at such a deep level. Could it be that codes, in some deep and fundamental way, control the structure of our reality? In asking this question, we may be ending our "treasure hunt" in a place that was anticipated previously by at least one pioneering physicist: John Archibald Wheeler. [...] As for my own collaboration on adinkras, the path my colleagues and I have trod since the early 2000s has led me to conclude that codes play a previously unsuspected role in equations that possess the property of supersymmetry. This unsuspected connection suggests that these codes may be ubiquitous in nature, and could even be embedded in the essence of reality. If this is the case, we might have something in common with the Matrix science-fiction films, which depict a world where everything human beings experience is the product of a virtual-reality-generating computer network. http://arxiv.org/abs/0806.0051 Which leads to the spooky idea that our universe is a computer simulation... http://www.simulation-argument.com/simulation.html http://rationalwiki.org/wiki/Simulation_argument 2. Originally Posted by khan Which leads to the spooky idea that our universe is a computer simulation... Most Cosmologists acknowledge the plausible (yet not probable) possibility that we exist in a computer-generated universe. 3. Originally Posted by Xotica Most Cosmologists acknowledge the plausible (yet not probable) possibility that we exist in a computer-generated universe. I don't know why you're singling out cosmologists because all scientists would admit we cannot disprove that notion but that's a long way from saying it's plausible. Atheists acknowledge they (we) cannot prove there is no deity or deities but that doesn't mean we consider the notion to be plausible. As for the work in question he's dealing with spinor structures, specifically spin 1/2 structures so you get binary representations naturally. This $\pm \frac{1}{2}$ structure can be written as 0,1 structure instead, so it's not like binary representations are unknown things in physics. In this case it happens the particular spinor constructs he's considering also can be put in a graph structure. Again, this is not a new concept, using graphs to represent decompositions of algebras is a centuries old procedure, known to anyone studying theoretical physics, particularly gauge theory. In this case the graph structure happens to be equivalent to error correcting code. It's important not to fall into the trap of confirmation bias. As I said, there's many many years of physics involving graph structures, spinors, decompositions and symmetries. It's not terribly shocking that at some point someone comes across a structure which has been seen elsewhere. Saying "Wow, this is clear justification for the Matrix view of the universe!" is a bit like me asking someone to think of a number between 1 and 1000, trying to guess it and when I finally guess right after many failed tries I declare it evidence I'm telepathic. Part of being a good mathematician is noticing structures which link seemingly different concepts. Realising "Oh, if I just write this like that then it becomes a well known expression whose solutions are known" is something every mathematician will do at some point in their research. In this case it's connected error correcting code with string theory and so people have started reading all sorts of things from it. If this is reason to consider the universe as just a computer simulation can we therefore use the same reasoning to say the thousands of things not related to computer programming in string theory are pieces of evidence against that notion? Because if we can't then you're making a logical fallacy. 4. Originally Posted by AlphaNumeric I don't know why you're singling out cosmologists because all scientists would admit we cannot disprove that notion but that's a long way from saying it's plausible. plau·si·ble adj \ˈplȯ-zə-bəl\ Definition of PLAUSIBLE 1 :superficially fair, reasonable, or valuable but often specious The notion is indeed plausible, albeit fantastical. 5. Math and reality ...what is the connection? http://arxiv.org/abs/0704.0646 I explore physics implications of the External Reality Hypothesis (ERH) that there exists an external physical reality completely independent of us humans. I argue that with a sufficiently broad definition of mathematics, it implies the Mathematical Universe Hypothesis (MUH) that our physical world is an abstract mathematical structure. I discuss various implications of the ERH and MUH, ranging from standard physics topics like symmetries, irreducible representations, units, free parameters, randomness and initial conditions to broader issues like consciousness, parallel universes and Godel incompleteness. I hypothesize that only computable and decidable (in Godel's sense) structures exist, which alleviates the cosmological measure problem and help explain why our physical laws appear so simple. I also comment on the intimate relation between mathematical structures, computations, simulations and physical systems. 6. How many scientists have claimed the universe computes? What does it mean? Does it just underline that any information we record (or observe) is the result of some physical process which can be argued is like an algorithm? Why do scientists balk at the idea of the universe (and all the information in it) being algorithmic? Or why do they mostly take the view that such a claim is somehow "lowering" the status of physical theories, to "just a computer simulation", isn't that like saying "the universe is just a universe"? Well, yeah, I guess that's what it is, just a universe. Maybe only just . . . It seems a bit contrary: every physical equation I've ever seen is arguably algorithmic, it describes what happens to a given input, or what some physical system does to an input--it either gives an output or perhaps goes into a nonterminating loop of some kind. 7. It depends on how far you take it. Someone like Wolfram (of Mathematica fame) believes the universe is, at it's most basic, a cellular automaton which is computing the passage of time. This is a somewhat extreme view lacking justification. On the other hand the rules of the universe do seem to be consistent and allow for systems to be set up and then just allowed to evolve under said rules to then spit out some 'answer' to a question encoded within the initial conditions. This is precisely what a computer does, we set up all the electrons and silicon to compute a new state from which we extract an answer. Using this we can even compute the answer to some set of equations which we model the universe (or some part of it) by, so you could argue that the universe itself computes the exact answer to such questions as it evolves in time. The reason a lot of scientists don't like phrasing it that way is that it seems to imply there's some kind of intent behind the universe doing what it does, ie what or who is asking the questions or that the universe isn't just doing whatever it is it does but actually 'computing' something. For example, when I jump in the air I don't have to compute the acceleration imparted on me by my leg muscles so I can compute how high to rise, it just happens. I can indeed make a machine or some algorithm for me to do in my head which will compute those answers but the universe doesn't compute "You will rise 0.442 metres in the air", things just bounce off one another, interact, exchange energy and momentum etc and that's it. So while anyone doing physics is obviously implementing an algorithm to allow them to determine the outcome of some set of dynamics based on initial conditions saying the universe computes is saying something else and it's that scientists don't like the connotations of. 8. Originally Posted by AlphaNumeric For example, when I jump in the air I don't have to compute the acceleration imparted on me by my leg muscles so I can compute how high to rise, it just happens. I can indeed make a machine or some algorithm for me to do in my head which will compute those answers but the universe doesn't compute "You will rise 0.442 metres in the air", things just bounce off one another, interact, exchange energy and momentum etc and that's it. You're missing something here, I think. You're implying that "compute" how high you rise when you jump in the air means "return an answer in metres" say, but despite that your height above the ground still changes. The change (whatever you want to call it, or however you measure it) is the output. That is, the height your body rises above the ground is the output "computed" by the energy you impart to it, via leg muscles, a bungy cord or whatever. You know your height changes because you experience the change, so you therefore "measure" the output, you don't need to specify it in metres or any other standard. So the universe does compute that you will rise in the air; that it doesn't "tell" you what that is in some arbitrary measurement standard is clearly irrelevant. But: Criticism The critics of digital physics—including physicists[citation needed] who work in quantum mechanics—object to it on several grounds. Physical symmetries are continuous One objection is that extant models of digital physics are incompatible[citation needed] with the existence of several continuous characters of physical symmetries, e.g., rotational symmetry, translational symmetry, Lorentz symmetry, and electroweak symmetry, all central to current physical theory. Proponents of digital physics claim that such continuous symmetries are only convenient (and very good) approximations of a discrete reality. For example, the reasoning leading to systems of natural units and the conclusion that the Planck length is a minimum meaningful unit of distance suggests that at some level space itself is quantized.[29] Locality Some argue[citation needed] that extant models of digital physics violate various postulates of quantum physics. For example, if these models are not grounded in Hilbert spaces and probabilities, they belong to the class of theories with local hidden variables that some deem ruled out experimentally using Bell's theorem. This criticism has two possible answers. First, any notion of locality in the digital model does not necessarily have to correspond to locality formulated in the usual way in the emergent spacetime. A concrete example of this case was recently given by Lee Smolin.[30] Another possibility is a well-known loophole in Bell's theorem known as superdeterminism (sometimes referred to as predeterminism).[31] In a completely deterministic model, the experimenter's decision to measure certain components of the spins is predetermined. Thus, the assumption that the experimenter could have decided to measure different components of the spins than he actually did is, strictly speaking, not true. Physical theory requires the continuum It has been argued[weasel words] that digital physics, grounded in the theory of finite state machines and hence discrete mathematics, cannot do justice to a physical theory whose mathematics requires the real numbers, which is the case for all physical theories having any credibility. But computers can manipulate and solve formulas describing real numbers using symbolic computation, thus avoiding the need to approximate real numbers by using an infinite number of digits. Before symbolic computation, a number—in particular a real number, one with an infinite number of digits—was said to be computable if a Turing machine will continue to spit out digits endlessly. In other words, there is no "last digit". But this sits uncomfortably with any proposal that the universe is the output of a virtual-reality exercise carried out in real time (or any plausible kind of time). Known physical laws (including quantum mechanics and its continuous spectra) are very much infused with real numbers and the mathematics of the continuum. "So ordinary computational descriptions do not have a cardinality of states and state space trajectories that is sufficient for them to map onto ordinary mathematical descriptions of natural systems. Thus, from the point of view of strict mathematical description, the thesis that everything is a computing system in this second sense cannot be supported".[32] For his part, David Deutsch generally takes a "multiverse" view to the question of continuous vs. discrete. In short, he thinks that “within each universe all observable quantities are discrete, but the multiverse as a whole is a continuum. When the equations of quantum theory describe a continuous but not-directly-observable transition between two values of a discrete quantity, what they are telling us is that the transition does not take place entirely within one universe. So perhaps the price of continuous motion is not an infinity of consecutive actions, but an infinity of concurrent actions taking place across the multiverse.” January, 2001 The Discrete and the Continuous, an abridged version of which appeared in The Times Higher Education Supplement. --http://en.wikipedia.org/wiki/Digital_physics 9. I agree with AlphaNumeric, I think that it is a coincidence. Sort of like the Bible codes found in the Bible, that it correlates to known codes doesn't mean that it was intentionally put there or that they even function the way they function in computers. If they do function as error-correcting methods in reality then you might be on to something though but I don't see from the quote that this is the case. Another explanation could be that the theory is, in fact, man-made and structures could have been put there that are borrowed from other fields (such as the technology of computers), that a theory works to describe reality doesn't mean that all the structures within the theory has to apply to structures in reality. In fact, if the theory auto-corrects itself through those structures then it could be a indication that it is false. 10. mathematics it can prove anything it wants so is string theory the of everything ? 11. Originally Posted by AlphaNumeric I don't know why you're singling out cosmologists because all scientists would admit we cannot disprove that notion but that's a long way from saying it's plausible. Atheists acknowledge they (we) cannot prove there is no deity or deities but that doesn't mean we consider the notion to be plausible. As for the work in question he's dealing with spinor structures, specifically spin 1/2 structures so you get binary representations naturally. This $\pm \frac{1}{2}$ structure can be written as 0,1 structure instead, so it's not like binary representations are unknown things in physics. In this case it happens the particular spinor constructs he's considering also can be put in a graph structure. Again, this is not a new concept, using graphs to represent decompositions of algebras is a centuries old procedure, known to anyone studying theoretical physics, particularly gauge theory. In this case the graph structure happens to be equivalent to error correcting code. It's important not to fall into the trap of confirmation bias. As I said, there's many many years of physics involving graph structures, spinors, decompositions and symmetries. It's not terribly shocking that at some point someone comes across a structure which has been seen elsewhere. Saying "Wow, this is clear justification for the Matrix view of the universe!" is a bit like me asking someone to think of a number between 1 and 1000, trying to guess it and when I finally guess right after many failed tries I declare it evidence I'm telepathic. Part of being a good mathematician is noticing structures which link seemingly different concepts. Realising "Oh, if I just write this like that then it becomes a well known expression whose solutions are known" is something every mathematician will do at some point in their research. In this case it's connected error correcting code with string theory and so people have started reading all sorts of things from it. If this is reason to consider the universe as just a computer simulation can we therefore use the same reasoning to say the thousands of things not related to computer programming in string theory are pieces of evidence against that notion? Because if we can't then you're making a logical fallacy. Well said, Alphanumeric. It still is fundamentally remarkable that our mathematical observation mimic the microcosms we create. The question which is more genuinely remarkable (in my view). Is this observer bias? Or is our fundamental human experience merely a replica of our physical reality? Both solution's are remarkable, but for greatly different reasons. #### Posting Permissions • You may not post new threads • You may not post replies • You may not post attachments • You may not edit your posts • • BB code is On • Smilies are On • [IMG] code is On • [VIDEO] code is On • HTML code is Off Forum Rules All times are GMT -5. The time now is 06:54 PM. sciforums.com	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 2, "/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": 0.9451698660850525, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/29285/rotational-speed-of-a-discus/29287	# Rotational speed of a discus I was wondering whether the rotational speed of a discus has any influence on the flight of the discus. Would slowing the rotation or speeding it up change the trajectory in any way or would the flight simply become unstable when slowing down? - ## 3 Answers Allegedly the rotation has two effects. I say "allegedly" because although I was told this in physics lectures at university I'm not sure if anyone has ever rigorously proved it. Anyhow, with the disclaimer behind me, the first effect is that the angular momentum stabilises the angle of the discus as it travels through the air. That allows the angle of attack to be maintained at the optimum value and hence increases the lift and therefore the range. The second effect is that a high rotational speed makes the boundary layer turbulent, and this decreases aerodynamic drag and once again increases the range. Given this I guess higher rotational speed is better, though presumably an athlete is limited to how high a rotational speed they can generate without compromising the speed they can throw at. - + This reminds me of a related issue - how does a frisbee work - but in that case there's more of an airfoil effect, I would think. – Mike Dunlavey May 31 '12 at 20:21 I take it then that accelerating the angular moment influences only the distance and stability of the flight, but I cannot "steer" the discus to the left or the right by accelerating or slowing down the angular moment, right? – cdecker May 31 '12 at 20:46 @cdecker: don't forget, it's a gyroscope, and gyroscopes precess. So if there is a force to cause it to "pitch up", that will make it roll right or left, and vice versa. Which way it goes depends on the direction and rate of spin. – Mike Dunlavey Jun 1 '12 at 0:36 The faster it spins, the greater the aerodynamic side force on it; see Magnus effect. Also, higher rotation increases the $\mu$ (ratio of edge speed relative to body to airspeed of the body) of the disc; the higher airspeed of the advancing edge relative to the retreating edge creates asymmetric lift & drag. The former would impart a rolling moment, while the latter would impart a moment opposing the in-plane rotation of the discus. All that said, I doubt either are particularly significant effects. - 1 Simply on one side the airspeed is higher than the other side, affecting the lift and drag on each side. This causes the discus to drift or slice. – ja72 Jun 1 '12 at 1:55 Given the physical conditions, this seems like an appropriate explanation: The faster the discus rotates, the more violently and quickly it displaces the air around it. Now the absence or scarcity of air causes a reduction in air friction or viscosity around the discus and this allows it to move onward in the direction of propulsion; now that depends on what angle the athlete projects it. After a certain distance there begins a constant deceleration of rotational speed because at some point, the air friction starts overpowering the rotation and this results in the discus entering the second half of its trajectory, i.e., moves downward along a curved path. I hope that answers your question. - Maybe I'm not used to the lingo, but "violently and quickly it displaces the air" and "absence or scarcity of air" doesn't sound like the kind of aerodynamics I'm used to. Want to give it another shot? – Mike Dunlavey May 31 '12 at 20:25 What I was trying to convey by the use of the quoted phrases was that Bernoulli's Principle comes into play in this situation. – Graviton Jun 1 '12 at 6:53 Let me suggest alternate wording, then you do what you want: The disk is an airfoil whose orientation is gyroscopically stabilized. As it follows its (nominally parabolic) arc, as it goes into the descending portion of the arc, its angle of attack increases. The rotational speed could have an effect of delaying the onset of aerodynamic stall, thus lengthening the trajectory. It also could have increased lift on the forward-moving side, causing precession that could also affect the angle of attack. – Mike Dunlavey Jun 1 '12 at 13:22	{"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": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9386345744132996, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/45644/can-bosons-that-are-composed-of-several-fermions-occupy-the-same-state/45645	Can bosons that are composed of several fermions occupy the same state? It is generally assumed that there is no limit on how many bosons are allowed to occupy the same quantum mechanical state. However, almost every boson encountered in every-day physics is not a fundamental particle (with the photon being the most prominent exception). They are instead composed of a number of fermions, which can not occupy the same state. Is it possible for more than one of these composite bosons to be in the same state even though their constituents are not allowed to be in the same state? If the answer is "yes", how does this not contradict the more fundamental viewpoint considering fermions? - 2 Answers This is a nice puzzle--- but the answer is simple: the composite bosons can occupy the same state when the state is spatially delocalized on a scale larger than the scale of the wavefunction of the fermions inside, but they feel a repulsive force which prevents them from being at the same spatial point, so that they cannot sit at the same point at the same time. The potential energy of this force is always greater than the excitation energy of the composite system, so if you force the bosons to sit at the same point, you will excite one of them, so that the composing fermions are no longer in the same state, and the two particles become distinguishable. The scale for this effective repulsion is the decay-length of the wavefunction of the composing fermions, and this repulsion is what leads matter to feel hard. The reason you haven't heard this is somewhat political--- there are people who say that the exclusion principle is not the cause of the repulsive contact forces in ordinary matter, that this force is electrostatic, and despite this being ridiculously false, nobody wants to get into the mud and argue with them. So people don't explain the fermionic exclusion principle forces properly. If you have a two-fermion composite which is net bosonic, like a H atom with a proton nucleus and spin-polarized electron, when you bring the H-atoms close, the energy of the electronic ground state is the effective Hamiltonian potential energy for the nuclei. When the nuclei are close enough so that the electronic wavefunctions have appreciable overlap, you get a strong repulsion. You can see that this repulsion is pure Pauli, because if the electrons have opposite spins, you don't get repulsion at short distances, you get attraction, and the result is that you form an H2 molecule of the two H atoms. You can see this exclusion force emerge in an exactly solvable toy model. Consider a 1d line with two attractive unit delta function pontetials at positions a and -a, each with a fermion attached in the ground state. Each one has an independent ground state wavefunction that has the shape $exp(-\|x\|)$, but when the two are together at separation 2a, the two states are deformed, and the ground state energy for the fermions goes up. The effect is quadratic in the separation, because the ground state (one fermion) goes down in energy, and the first excited state goes up in energy, and to leading order in perturbations, the two are cancelling when both states are occupied. To next leading order, the effect is positive potential energy, a repulsion. This potential is the effective potnetial of the two delta functions when you make them dynamical instead of fixed. The maximum value of the repulsive potential in this model is exactly where the model breaks down, which is at a=1. At this point, the ground state is exp(-2x) to the left of -1, constat between the two delta functions, then exp(2x) to the right, with energy -2, and the first excited state is constant to the left of -1, a straight line from -1 to 1, and constant past 1, with energy 0. The result is a net energy of -1 unit. This is half the binding energy of the two separated delta functions, which is -2. This effect is the exclusion repulsion, and it reconciles the fermionic substructure with the net bosonic behavior of the particle. You can only see the substructure when the wavefunction of the boson is concentrated enough to have appreciable overlap on the scale of the composing fermion wavefunctions, and this is why you need high energies to probe the compositeness of the Higgs (or for that matter, the alpha particle). To get the wavefunctions to sit at the same point to this accuracy, you need to localize them at high energy. - You state that "if you force the bosons to sit at the same point, you will excite one of them". There is the problem I am having. When the two bosons are in the same state, they are on the same point, in the sense that their wavefunctions are the same and therefore overlap entirely. It all comes down to this: If the wavefunctions of two bosons are the same, the wavefunctions of their constituents should pairwise be the same, which is impossible. What am I missing? – Friedrich Dec 3 '12 at 0:10 @Friedrich: Not exactly--- the wavefunction is only exactly the same for two bosons if they are non-interacting. If they have a repulsion at ultra-short distances, the wavefunction is entangled so that it is very close to the product of the wavefunction with itself at most points, but the entanglement zeros out the diagonal (so they aren't ever at the same point). This is a property of higher dimensional multi-particle wavefunctions. The best definition for the condensate wavefunction by giving the effective field for the boson a VEV, and this only produces a product state for a free field. – Ron Maimon Dec 3 '12 at 3:26 Ron, how can the volume of the wavefunction of a composite boson be larger than the volume of the wavefunction of its fermions? – lurscher Dec 4 '12 at 19:44 @lurscher: The volume of the wavefunction of the fermions is in the dimension of the relative coordinate, while the volume of the composite is in the center of mass coordinate. – Ron Maimon Dec 4 '12 at 23:37 @Ron: I hope you will return safely in four days. If two interacting bosons can not be in a product state, is it even meaningful to speak about them being in "the same state"? – Friedrich Dec 31 '12 at 1:39 show 2 more comments Yes, they can, an experimental example of that is Bose-Einstein Condensate of fermions. And that is possible because actually they will have the same wave function, in sense that nature no more capable of making any distinguish between them. Regarding everyday life, actually saying that it is bosonic is just a formal statement, in sense that Pauli exclusion principle not working here, but that not because the composite things are bosons, but because in everyday life there is almost nothing in the same state, because accomplishing that is very hard, and if done, it will reproduce Bose-Einstein condensate. -	{"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": false, "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": 0.9388255476951599, "perplexity_flag": "head"}
http://mathhelpforum.com/differential-geometry/86622-integrable-functions.html	Thread: 1. Integrable Functions Let f be a function on the interval [a, b] such that for every epsilon > 0 there exist integrable functions g, h on [a, b] with h(x) <= f(x) <= g(x) and ((integral from a to b g(x)dx) - (integral from a to b h(x)dx)) < epsilon. Show that f is integrable on [a, b]. Any ideas on where to start and/or how I would go about showing this? Thanks. 2. Riemann integrable? Assuming that, take some partition of [a, b] and look at the Riemann sums for g, f, and h. You should be able to show that [itex]R(g)\le R(f)\le r(h)[/tex] (R(f) is the Riemann sum of f for the particular partition). "((integral from a to b g(x)dx) - (integral from a to b h(x)dx)) < epsilon" is a bit peculiar since the two integrals are specific numbers. That really just says the integrals of g and h are the same. That is that the lim, as the number of intervals in the partition goes to infinity, of the Riemann sums, R(g) and R(h), are the same so the limit of R(f) exists and is that joint value. 3. Originally Posted by kjwill1776 Let f be a function on the interval [a, b] such that for every epsilon > 0 there exist integrable functions g, h on [a, b] with h(x) <= f(x) <= g(x) and ((integral from a to b g(x)dx) - (integral from a to b h(x)dx)) < epsilon. Show that f is integrable on [a, b]. Any ideas on where to start and/or how I would go about showing this? Thanks. $\int h \leq \int_{-} f \leq \int g$ and $\int h \leq \int^{-} f \leq \int g.$ thus $0 \leq \int^{-}f - \int_{-}f \leq \int g - \int h < \epsilon.$ since $\epsilon > 0$ is arbitrary, we must have $\int^{-}f - \int_{-}f=0.$	{"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": 0, "codecogs_latex": 5, "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": 0.910717248916626, "perplexity_flag": "middle"}
http://crypto.stackexchange.com/questions/1767/proving-knowledge-of-a-preimage-of-a-hash-without-disclosing-it/1771	Proving knowledge of a preimage of a hash without disclosing it? We consider a public hash function $H$, assumed collision-resistant and preimage-resistant (for both first and second preimage), similar in construction to SHA-1 or SHA-256. Alice discloses a value $h$, claiming that she (or/and parties she can communicate with or/and devices they have access to) knows a message $m$ such that $H(m)=h$. Can some protocol convince Bob of this claim without help of a third party/device that Bob trusts, nor allowing Bob to find $m$? At Crypto 98 rump session, Hal Finney made a 7-minutes presentation A zero-knowledge proof of possession of a pre-image of a SHA-1 hash which seems to be intended for that. This remarkable result is occasionally stated as fact, including recently here and next door. But I do not get how it is supposed to work. Update: This talk mentions using the protocol in the Crypto'98 paper of Ronald Cramer and Ivan B. Damgård: Zero-Knowledge Proofs for Finite Field Arithmetic or: Can Zero-Knowledge be for Free? (this freely downloadable version is very similar, or there is this earlier, longer version). - I could not find any article that supported his talk in the rump session. Do you have an access to any such article. I would be very much interested in that because it sounds like a challenging problem especially when one sees hash function in the random oracle model. Maybe, then I could be of some help. – Jalaj Jan 28 '12 at 17:58 1 I went beyond what was in the talk. I think that considering a RO-model and solving this problem will be a theoretically great result. – Jalaj Jan 29 '12 at 2:51 My current intuition is that no practical protocol can rightly convince Bob of Alice's claim, for that very protocol would be able to distinguish $H$ from a random oracle, something that was never done for SHA-1 (restricted to messages of constant length). I reason that in the random oracle model, the only way for Bob to be convinced that the oracle outputs $h$ for message $m$ is to submit it himself, which implies knowledge of $m$. – fgrieu Jan 29 '12 at 9:38 1 The protocol can allow Bob to distinguish $H$ from a random oracle, but it requires Alice's input. I don't think that contradicts indistinguishability. For example, Elgamal is IND-CPA-secure but also admits proofs of plaintext knowledge. Alice could allow Bob to distinguish ciphertexts he couldn't on his own. In any case, no implemented hash function can act as a random oracle in the real world. – PulpSpy Jan 30 '12 at 2:02 2 The hash design needs to be known so both parties can follow the progress through the circuit. I don't think we can always have something real-world practical (it isn't claimed in the talk that we can; only that we can for SHA-1). It depends on how well the function lends itself to certain tricks for speeding up the proof. What he did was as much art as science in converting SHA-1 into a blend of boolean/arithmetic circuitry. – PulpSpy Jan 30 '12 at 15:12 show 2 more comments 2 Answers I'm not sure what I can add that wasn't covered in the talk. The approach is that Alice commits to the preimage and sends the commitment to Bob. The commitment has homomorphic properties, meaning it is possible to do computation on the value. For example, if Alice commits to $x$ and $y$, Bob may be able to compute a commitment to $z$ where $z=f(x,y)$ for some function $f$. For example, if the commitments are additively homomorphic, Bob can add two commitments or multiply by a constant for free. Alternatively, $f$ may not be directly computable. In this case, Alice who knows the actual values computes $z$, sends a commitment to $z$ to Bob asserting that it is correct. Bob holds commitments to $x$, $y$, and $z$; as well as knowing $f$. Alice can then interactively prove that for $f$, the commitment containing $z$ does contain the correct output for the inputs contained in the commitments to $x$ and $y$. The Cramer-Damgaard paper shows how to do these proofs for a simple Turning-complete set of both boolean and arithmetic gates (for example, NAND and modular addition/multiplication). The size of the circuit implementing SHA-1, expressed with only NAND gates for example, will be very large and infeasible. The art to doing the proof in practice is breaking it into subprotocols that are best represented by either boolean or arithmetic circuits and switching between the proof systems as appropriate. For certain SHA operations, he works at the bit level with boolean operations and for other operations, he works with integers in a finite field. - Do you feel confident that the answer to the question is a clear "yes, using the techniques in the Cramer-Damgård paper"? Or something weaker? I get that there is no evidence of impossibility. – fgrieu Jan 30 '12 at 12:54 2 An unbounded Alice can always prove. A bounded Alice can only prove when the circuit complexity of the hash has the same bound. A real-life Alice can only prove if she can generate a relatively simple description of the circuit with operations that have efficient zero-knowledge proofs. The latter is the case with SHA-1, but necessarily other hash functions. – PulpSpy Jan 30 '12 at 15:08 I guess the above is intended to read "but not necessarily other hash functions", and otherwise is supportive that a practical protocol can be devised in the case of SHA-1 as claimed in the talk. I'm still a tad skeptical, on these (admittedly weak) arguments: 1) such a protocol is something we can't devise in general for a perfect hash, yet we are computationally unable to distinguish SHA-1 (with fixed message size) from a perfect hash, other than by the fact that it is SHA-1; 2) details on the talk's method are scarce; 3) this remarkable result does not appear to have been reproduced. – fgrieu Jan 30 '12 at 20:10 @fgrieu: 3) Do you have use cases where this construction would be useful (and more desirable than a ZKPoK of a plaintext)? – bob Oct 13 '12 at 19:49 @bob: to be honest, no I do not have a use case. My interest at that point is purely theoretical. – fgrieu Oct 15 '12 at 11:14 I wanted to comment on PulpSpy's answer, but my comment turned out to be too large! I have an intuitive understand why they does not stick to the idea of using Boolean circuits alone for all the proofs which I am writing it down here. I might be wrong and I would like to be cross examined on this. There will be an issue with the boolean and arithmetic circuits. I view this from a linear algebra point of view. We can write the boolean circuit in form of a corresponding matrix that performs the transformation. Since every matrix will be linear transformation, there are known lower bounds which are of form $\Omega(n^2/r^c)$, where $n$ is the input size, $c$ is an arbitrary constant, and $r$ is the size of the biggest sub-matrix that is linear dependent. Since SHA-1 does mixing very well, I am sure we won't be able to reject even a constant fraction of input for any small (in matrix form, a submatrix of large size, comparable to $n$). This further implies that computing the circuit will require a lot of computation one that we are not willing to take. I feel this argument works well for most of the candidate hash functions that uses linear arithmetic circuit to do the mixing. -	{"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": 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": 0.9400736689567566, "perplexity_flag": "middle"}
http://mathhelpforum.com/advanced-algebra/192784-irreducible-polynomials-over-ring-integers.html	Thread: 1. Irreducible polynomials over ring of integers ? Is it true that polynomials of the form : $f_n= x^n+x^{n-1}+\cdots+x^{k+1}+ax^k+ax^{k-1}+\cdots+a$ where $\gcd(n+1,k+1)=1$ , $a\in \mathbb{Z^{+}}$ , $a$ is odd number , $a>1$, and $a_1\neq 1$ are irreducible over the ring of integers $\mathbb{Z}$? Note that general form of $f_n$ is : $f_n=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$ , so condition $a_1 \neq 1$ is equivalent to the condition $k \geq 1$ . Also polynomial can be rewritten into form : $f_n=\frac{x^{n+1}+(a-1)x^{k+1}-a}{x-1}$ Eisenstein's criterion , Cohn's criterion , and Perron's criterion cannot be applied to the polynomials of this form. Example : The polynomial $x^4+x^3+x^2+3x+3$ is irreducible over the integers but none of the criteria above can be applied on this polynomial. 2. Re: Irreducible polynomials over ring of integers ? Originally Posted by princeps Is it true that polynomials of the form : $f_n= x^n+x^{n-1}+\cdots+x^{k+1}+ax^k+ax^{k-1}+\cdots+a$ where $\gcd(n+1,k+1)=1$ , $a\in \mathbb{Z^{+}}$ , $a$ is odd number , $a>1$, and $a_1\neq 1$ are irreducible over the ring of integers $\mathbb{Z}$? Note that general form of $f_n$ is : $f_n=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$ , so condition $a_1 \neq 1$ is equivalent to the condition $k \geq 1$ . Also polynomial can be rewritten into form : $f_n=\frac{x^{n+1}+(a-1)x^{k+1}-a}{x-1}$ Eisenstein's criterion , Cohn's criterion , and Perron's criterion cannot be applied to the polynomials of this form. Example : The polynomial $x^4+x^3+x^2+3x+3$ is irreducible over the integers but none of the criteria above can be applied on this polynomial. i don't know the answer to the general case but it's easy to prove that if $n = 2^m - 1, \ m \geq 2, \ 2 \mid k$ and $a \equiv 3 \mod 4,$ then $f$ is irreducible. to see this, note that the numbers $\binom{2^m}{i}, \ 0 < i < 2^m,$ and $a - 1$ are even and hence $f(x+1) = \frac{(x+1)^{n+1}+(a-1)(x+1)^{k+1}-a}{x}=x^{n} + a_{n-1}x^{n-1}+ \ldots +$ $a_1x + a_0,$ where all $a_j$ are even and $a_0 = 2^m + (k+1)(a-1) \not \equiv 0 \mod 4.$ thus $f(x+1),$ and so $f(x),$ is irreducible by the Eisenstein's criterion. 3. Re: Irreducible polynomials over ring of integers ? Originally Posted by NonCommAlg $f(x+1) = \frac{(x+1)^{n+1}+(a-1)(x+1)^{k+1}-a}{x}=x^{n} + a_{n-1}x^{n-1}+ \ldots +$ $a_1x + a_0,$ where all $a_j$ are even and $a_0 = 2^m + (k+1)(a-1) \not \equiv 0 \mod 4.$ There is condition in the text of the question $a_k=a_{k-1}=\ldots =a_1=a_0=a$ , where $a$ is an odd number so $a_j$ cannot be even number. 4. Re: Irreducible polynomials over ring of integers ? Originally Posted by princeps There is condition in the text of the question $a_k=a_{k-1}=\ldots =a_1=a_0=a$ , where $a$ is an odd number so $a_j$ cannot be even number. in my solution, $a_j$ are the coefficients of $f(x+1)$ not $f(x)$. 5. Re: Irreducible polynomials over ring of integers ? Can you give me an example of f(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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 50, "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": 0.8640118837356567, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/216049/lim-x-to-infty-fracxex-1x-1ex	# $\lim_{x \to \infty} \frac{xe^{x-1}}{(x-1)e^x}$ $$\lim_{x \to \infty} \frac{xe^{x-1}}{(x-1)e^x}$$ I don't know what to do. At all. I've read the explanations in my book at least a thousand times, but they're over my head. Oh, and I'm not allowed to use L'Hospital's rule. (I'm guessing it isn't needed for limits of this kind anyway. This one is supposedly simple - a beginners problem.) Most of the answers I've seen on the Internet simply says "use L'Hospital's rule". Any help really appreciated. I'm so frustrated right now... - What is $e^{x-1}/e^x$? – wj32 Oct 18 '12 at 2:50 Do you know any of the limit laws? Such as $\lim cf = c\lim f$? They are quite helpful in this case. – Raymond Cheng Oct 18 '12 at 2:53 ## 4 Answers Multiply the fraction by $1$ in the carefully chosen disguise $\dfrac{e^{-x}}{e^{-x}}$ and do a bit of algebra: $$\begin{align} \lim_{x \to \infty} \frac{xe^{x-1}}{(x-1)e^x}&=\lim_{x\to\infty}\left(\frac{xe^{x-1}}{(x-1)e^x}\cdot\frac{e^{-x}}{e^{-x}}\right)\\ &=\lim_{x\to\infty}\frac{xe^{-1}}{x-1}\\ &=\lim_{x\to\infty}\frac{x}{e(x-1)}\\ &=\frac1e\lim_{x\to\infty}\frac{x}{x-1}\\ &=\frac1e\lim_{x\to\infty}\frac{x-1+1}{x-1}\\ &=\frac1e\lim_{x\to\infty}\left(1+\frac1{x-1}\right)\;. \end{align}$$ That last limit really is easy. You may wonder how I came up with some of the steps. The very first one was simply recognizing that if I divided numerator and denominator by $e^x$, the resulting fraction would be a lot simpler, in that all of the exponentials would be gone. Pulling constant factors outside the limit is usually useful and almost never hurts. The simplification of $\frac{x}{x-1}$ could also have been achieved by doing a straightforward polynomial long division of $x-1$ into $x$, but the trick of subtracting and adding the same amount (here $1$) in order to get an expression that can be split in some nice way is a pretty common one that’s worth remembering. - Clearly, $$\frac{{x{e^{x - 1}}}}{{(x - 1){e^x}}} \sim \frac{{x{e^{x - 1}}}}{{x{e^x}}} = \frac{{{e^{x - 1}}}}{{{e^x}}}.$$ So ... - $$\lim_{x\to\infty}\frac{xe^{x-1}}{(x-1)e^x}=\lim_{x\to\infty}\frac{x}{x-1}\cdot\lim_{x\to\infty}\frac{e^{x-1}}{e^x}=1\cdot\frac{1}{e}=\frac{1}{e}$$ the first equality being justified by the fact that each of the right hand side limits exists finitely. - first downvote didn't come from me. I usually don't downvote neither questions, nor answers. – Chris's wise sister Oct 20 '12 at 5:42 THere seems to be a serial downvoter around here. I could almost have sworn it was you but, of course, I didn't since I'm not sure. I don't care either, to be honest, as in any case the serial downvotes are reversed back by the system and also because 2,4, or 40 points less or more make no real difference. – DonAntonio Oct 20 '12 at 17:20 honestly, if I wanted to downvote you then I'd have no problem telling you this direclty. I didn't downvote you this time. I downvote you ONLY when you downvote me. I do not like to downvote people (neither do I, nor my sister). – Chris's wise sister Oct 20 '12 at 17:25 then after I saw that you downvoted me I downvoted you. I mean that the first downvote didn't come from me. – Chris's wise sister Oct 20 '12 at 17:27 As before, I won't get into long discussions with you and I'll end this one after this message: I've no idea what you're talking about. I didn't downvote you, and I really don't care whether you downvoted me because you thought I did with you or because any other reason. I'm not interested in this kids game and please do feel as free as you want to downvote me: I don't care. Have a good day. – DonAntonio Oct 20 '12 at 18:25 show 1 more comment This is straightforward $$\lim_{x \to \infty} \frac{xe^{x-1}}{(x-1)e^x}=\lim_{x \to \infty} \frac{x}{(x-1)e}=\frac{1}{e}$$ (Chris) -	{"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": 5, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": 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": 0.9765657782554626, "perplexity_flag": "middle"}
http://sumidiot.wordpress.com/tag/approximation/	∑idiot's Blog The math fork of sumidiot.blogspot.com Posts Tagged ‘approximation’ Hardy and Wright, Chapter 11 (part 2) June 16, 2009 Today we finished off Chapter 11. We worked through some of the proofs, and discussing the meaning of some of theorems, and a good time was had by all. In 11.10 it is mentioned that there are some notable constant multiples besides $\sqrt{5}$ and $2\sqrt{2}$ in the bounding inequalities on approximating irrationals by rationals. However, the text doesn’t mention what they are, which I thought was unfortunate. I also wondered what sort of numbers are the troublesome examples for the constant $2\sqrt{2}$. That is, the troublesome number for $\sqrt{5}$ is the golden ratio (or anybody whose continued fraction ends in a string of 1s), so what numbers do it for $2\sqrt{2}$. I think we decided that probably it was not a single number, but more like… any number whose continued fraction expansion is just lots of 1s and 2s. The more ones, the worse the number, in some sense. But as long as there are infinitely many twos, maybe you start running into, or getting close to, this $2\sqrt{2}$ bound. We talked a little bit of our way through the proof that almost all numbers have arbitrarily large “quotients” (the $a_n$ in the continued fraction). I tried to dig up some memories from my reading of Khinchin’s book about how to picture some of the intervals and things in the proof. I have this pictures in my head of rectangles over the interval $[1/(n+1),1/n]$ of height the length of the interval (I guess that makes them squares, huh?). So the biggest rectangle is the one between 1/2 and 1, and they get smaller as you move left. Then each rectangle is split up again, this time with the rectangles getting smaller as you move to the right (within one of the first-stage rectangles). The first set of rectangles correspond, somehow, to the first term of continued fractions, and the second (smaller) rectangles correspond to the second term. Probably I should dig out that book and try to figure out what this picture actually says, but for now… that’s the picture I have in my head. We were all a little bit slow in understanding some of the later proofs about things like the discussion in 11.11: “Further theorems concerning approximations”. But we also didn’t seem interested enough to really dive in to it. In the section on simultaneous approximations, Eric mentioned that similar things are done in other contexts (like, perhaps, $p$-adics). When you have valuations, you prove a weak (single) and strong (simultaneous) theorem about approximations. While we were talking about it, I wondered if there was some analogy to the distinction between continuous (at each point in an interval) and uniformly continuous (on that interval). It seems like there maybe should be. Finally, we spent a while digging through the proof that $e$ is transcendental. Mostly because I was stubbornly refusing to believe I wasn’t being lied to throughout the proof. Setting $h^r=r!$ and then “plugging $h$ into” polynomials really made me uncomfortable. As we went, I joked about things not having any actual meaning. Eventually Chris and Eric pointed out that they do, actually, have meaning. This “plugging $h$ in” thing is actually giving you an integer (if your polynomial has integer coefficients). That calmed me down a bit. I still feel like I don’t understand the proof at all, and certainly couldn’t explain even an outline of it. Eric said similar things, but asked if we should have expected that somehow. Eric also mentioned that these sorts of formal manipulations with things that look wrong can sometimes be ok, and that it was something related to umbral calculus. He showed us an identity (Vandermonde’s) associated with binomial coefficients that does similar sorts of symbolic trickery. Which apparently I should now go read some more about. I had printed out a paper about the continued fraction expansion of $e$ (which maybe was pointed out to me in this comment), which talked about Pade approximations. Some of the things looked somewhat similar to what was going on in the proof that $e$ is transcendental (which the paper said was where they came from), but I couldn’t explain the paper well during out meeting (since I don’t understand it well enough), and we ran out of time. Tags:approximation, continued fraction, hardy and wright, number theory, transcendental, umbral calculus	{"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": 15, "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": 0.9735337495803833, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/5815/does-string-theory-disagree-with-general-relativity	# Does String Theory disagree with General Relativity? I would like to expand on what I mean by the title of this question to focus the answers. Normally whenever a theory (e.g. General Relativity) replaces another (e.g. Newtonian Gravity) there is a correspondence requirement in some limit. However there is also normally some experimental area where the new larger theory makes predictions which are different from the older theory which made predictions of the same phenomena. This is ultimately because the newer theory has a deeper view of physics with its own structures which come into play in certain situations that the old theory didnt cover well. Additionally the newer theory will make predictions based on its novel aspects which the older theory did not consider. I know that String Theory is quite rich in this regard, but am not interested in that here. Nor am I concerned as to whether experiment has caught up, as I know that ST (and Quantum Gravity in general) is not easy to test. So for the GR to Newtonian example an answer to this question would be: bending of light rays; Mercury perhelion movement - GR had a different results to Newton. What would not count as an answer would be new structures which GR introduces like Black Holes or even gravitational curvature per se. So does ST have anything like Mercury perhelion movement waiting to be experimentally verified, and thus "improving" on GR within GR's own back yard? - ## 2 Answers String theory implies new physics in - and only in - the quantum regime. In particular, at distances that are very short - comparable to the string scale or Planck scale - there are new effects. The black holes decay (while they preserve the information), effective actions have higher-derivative terms, e.g. $R^2$, there are strings, branes, fluxes, extra dimensions, and so on. The new physics at the Planck scale - which is very far - is almost certainly not testable by the naive 19th century observations such as the Mercury perihelion's precession. This is not a problem of string theory in any sense: it is a tautological consequence of the questions that string theory addresses - namely the behavior of the Universe at the most fundamental scale - and any other theory that addresses the same questions inevitably shares the inaccessibility of the phenomena by direct tests. When you only look at the classical limit or classical physics, string theory exactly agrees with general relativity. In some sense, this is true even in the quantum regime: string theory is the only consistent quantum completion of general relativity. - 1 I think the higher derivative terms are a good answer to the question - if you had sensitive enough probes, even without using any new "stringy" physics, you'd be able to detect them. Also - they are a result of classical string theory ($\alpha'$ corrections). – user566 Feb 24 '11 at 19:36 1 Motl: So this is saying that String Theory doesnt replace General Relativity - in its predictive and testable aspects - even in principle, at the larger-than-Planck scale? – Roy Simpson Feb 24 '11 at 19:45 @Moshe: I dont understand what the higher derivative terms are doing in this answer, but in other theories higher derivative terms correspond to non-local effects. If "local" means Planck scale, then "non-local" could mean "greater than Planck scale" - perhaps much greater. Thus these effects would be measurable - was that your point? – Roy Simpson Feb 24 '11 at 20:26 1 @Roy: Higher derivative terms are not "non-local", they correspond to additional terms beyond the Einstein-Hilbert action, which modify in a small and specific way any solution of GR. One example is $R^2$ terms, which by dimensional analysis are suppressed by a power of high mass scale (string scale in this case). Those occur with calculable coefficients in string theory. Generally, GR occurs automatically at distances larger than Planck length in string theory, with corrections that are small in that regime or become more dramatic and qualitative at short distances. – user566 Feb 24 '11 at 20:32 1 @Raindrop yes, is the short answer. – anna v Feb 28 at 7:28 show 1 more comment The higher derivative terms are there just like in semiclassical gravity simply because when you introduce quantum mechanics, any term that is not forbidden by some symmetry principle or other physical criteria must be present, albeit Planck suppressed. So it's not s that urprising that string theory predicts them. The interesting thing is that there are additional objects beyond the usual spectrum of GR. Things like the dilaton, supersymmetric objects (gravitino et al) and so on. -	{"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": 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": 0.942741334438324, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/31882/why-does-string-theory-require-9-dimensions-of-space-and-one-dimension-of-time?answertab=votes	# Why does string theory require 9 dimensions of space and one dimension of time? String theorists say that there are many more dimensions out there, but they are too small to be detected. 1. However, I do not understand why there are ten dimensions and not just any other number? 2. Also, if all the other dimensions are so coiled up in such a tiny space, how do we distinguish one dimension from the other? 3. If so, how do we define dimension? - 1 (1) It's all in the mathematics. (2) Can you distinguish your everyday 3 dimensions? Nope. So then there's no problem if the curled up ones are indistinguishable (not saying they are, though). [Atleast, I think it's this.] – Manishearth♦ Jul 12 '12 at 12:27 – Dilaton Jul 12 '12 at 13:53 – Dilaton Jul 12 '12 at 13:53 1 – Dilaton Jul 12 '12 at 13:54 – Rody Oldenhuis Jul 12 '12 at 18:58 ## 2 Answers One can posit mathematical string theories in any dimensions of any kind. However, I do not understand why there are ten dimensions and not just any other number? The specific dimensions arise from the requirements of the known physics encapsulated in the Standard Model and other data coming from particle physics, plus the requirement of General Relativity and its quantization. The Special Unitary groups whose representations accommodate the SM need at least these dimensions. There are models with more dimensions than this. Also, if all the other dimensions are so coiled up in such a tiny space, how do we distinguish one dimension from the other? We cannot move into the coiled ones, only in x.y,z. We do not need to distinguish them, as we do not distinguish the molecules in the air. The predictions from this type of theory on the behavior of particles is the only way of checking for their existence: consistency of theory with data. If so, how do we define dimension? A space variable ( centimeters) or time one ( seconds) that is continuous and maps the real numbers, each dimension at 90degrees to the rest, an extension of how we define normal x,y,z.That some are curled should not bother one. The coordinates over the earth are curled over the sphere's surface, for example, though the 90degree does not hold there. It would hold on the surface of a cylinder , z from -infinity to infinity, x from 0 to 2rpi. - 2 This answer is not accurate, you cannot formulate any sort of string theory in 60 dimensions, if you have too many spatial dimensions, there are too many degrees of freedom on the horizon. Regarding spherical coordinates, these are orthogonal. – Ron Maimon Jul 13 '12 at 1:45 – anna v Jul 13 '12 at 4:35 1 I see what you mean. But if you have ghosts in the theory, and divergent loop integrals, what does the mathematics mean? I agree that in all dimensions less than 10 and in all dimensions less than 26 you can formulate a fermionic/bosonic string theory (if you use a Polyakov noncritical string or linear dilaton), but in general I don't like to say this for dimensions higher than 26, because ghosts are not the same kind of problem qualitatively. BTW, only the poles are bad in spherical coordinates. – Ron Maimon Jul 13 '12 at 4:52 (1) String Theory is a very mathematical theory based on some natural assumptions, and this ends up relating Quantum Mechanics and General Relativity, as we want. Some of the equations in String Theory, however, have a proportionality constant $c$ in it, called the central charge. And when we manipulate these equations and set them equal to each other, we see that they ONLY make sense if $c=26$. This $c$ is the dimension of space that String Theory is a priori defined over, so now we see that we need 26 dimensions to not have absurdities... BUT that only made use of the bosonic particles in the world -- we forgot about fermions!! This is where Supersymmetry comes into play, and it throws in the fermions, and the equations are perturbed and leads to a new dimension of 10 for everything to make sense. (2) Just because we can't see it, doesn't mean it's not there... we can't see atoms with the eye, but we can use tools to see them... same thing happens here, our current technology can't see them, but we hope to change this in the future. EVEN BETTER though, is that the formula for gravitational force should actually be different because of these extra small dimensions -- thus we plan to figure these extra dimensions out by testing the gravitational force at small distances and seeing a perturbation to the standard inverse-square law of Newton. These extra dimensions are what is supposed to make gravity so weak compared to the other forces of nature. (3) a dimension is just a coordinate axis... so time is a dimension too. And just like your clock, this axis can repeat itself and not stretch to infinity. - I don't know how to make sense of compactified time, especially in string theory. Even if you compactify euclidean time, thermal string theory is hard to make sense of too, because gravity doesn't allow thermal ensembles of infinite extent. Regarding the central charge argument, it's fine, but it doesn't require 10 dimensions per-se, just an equivalent central charge, so you can have a non-geometric compactification. – Ron Maimon Jul 13 '12 at 1:42	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 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": 0.9138901829719543, "perplexity_flag": "middle"}
http://openwetware.org/index.php?title=User:Pranav_Rathi/Notebook/OT/2011/03/01/Laser_Shutter_.2&diff=next&oldid=657478	# User:Pranav Rathi/Notebook/OT/2011/03/01/Laser Shutter .2 ### From OpenWetWare (Difference between revisions) \| \| \| \| \| \|----------\|-----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------\|----------\|-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------\| \| () \| \| () \| \| \| Line 42: \| \| Line 42: \| \| \| \| \| \| \| \| \| === Construction=== \| \| === Construction=== \| \| - \| The buildup is divided is the following categories. In the start make sure that you have a good set of screw driver, players, wire cutter & stripper and solder. \| + \| The buildup is divided is the following categories. In the start make sure that you have a good set of screw driver, pliers, wire cutter, stripper and solder. \| \| \| ==== Shutter==== \| \| ==== Shutter==== \| \| - \| I will start with the choice of the motor. A 12V DC motor is a good choice because it can provide a wide rang of torques. The motor and the spring works against each other so it important to have the right set. So choose a 12V DC motor with a shaft length of at least 15mm. Mount the motor on the cage plate as shown in the picture, you will need 4/40 screws to tight it. once this is done, unscrew and take the motor out. \| + \| I will start with the choice of the motor. A 12V DC motor is a good choice because it can provide a wide rang of torques. The motor and the spring works against each other so it important to have the right set. So choose a 12V DC motor with a shaft length of at least 15mm. Mount the motor on the cage plate or platform as shown in the picture, you will need 4/40 screws to tight it (in present setup). once this is done, unscrew and take the motor out. \| \| \| \| \| \| \| - \| Now next task is to choose the right spring. The spring should be half the length of the shaft and the shaft should easily fit through the spring. I want to keep all this really simple, so there is a very easy way to choose the right spring. To choose the spring first need to know the torque of the motor. This is really simple, it can be mathematically calculate if the voltage and the current is known. But I like the experimental way; the setup is shown in the slide2. In the slide I have a DC motor with a leverage pivoted at the shaft and a free weight hanging with a string on the other side. In this, all I am doing is balancing the weigh at that point by giving the motor just enough power. \| + \| Now next task is to choose the right spring. The spring should be half the length of the shaft and the shaft should easily fit through the spring. I want to keep the spring-choice procedure really simple and easy. To choose the spring first need to know the torque of the motor. This is really simple, it can be mathematically calculate if the voltage and the current is known. But I like the experimental way; the setup is shown in the slide2. In the slide I have a DC motor with a leverage pivoted at the shaft and a free weight hanging with a string on the other side. In this, all I am doing is balancing the weigh at that point by giving the motor just enough power. \| \| \| so the torque is equal to the weight at the distance from the shaft: \| \| so the torque is equal to the weight at the distance from the shaft: \| \| \| \| + \| \| \| \| :: <math> \mathbf{\tau}=r \times F = r \times m .g </math> \| \| :: <math> \mathbf{\tau}=r \times F = r \times m .g </math> \| \| \| \| \| \| \| - \| Once this is know we can choose the spring with less torque than this. The restoring torque applied by the spring can also be measure in the same fashion (as motor) as shown in slide1, with hanging known weight. The spring stiffness and the torque can be related through: \| + \| Once this is know we can choose the spring with less torque. The restoring torque applied by the spring can also be measure in the same fashion (as motor) as shown in slide1, with hanging known weight. The spring stiffness and the torque can be related through: \| \| \| \| + \| \| \| \| :: <math> \mathbf{\tau}=2{\pi}n.K.r^2 </math> \| \| :: <math> \mathbf{\tau}=2{\pi}n.K.r^2 </math> \| \| \| \| \| \| ## Motivation Motivation behind designing this shutter is speed, accuracy and variability (activation-time). In our optical tweezers we need to center the trap over the tethered bead, to get the geometry right (this affects the force measurement) and make our feedback program run. To center the tether we need to turn the laser intensity (trap) on/off quickly for various time intervals. To do this we used to use AOM (acoustooptic module), because it’s extremely quick (nano second on/off time). But AOM has some inherent oscillation problem which makes it use as a switch, unsuccessful. So I needed something else, (a shutter) which can replace this AOM function. This shutter does it exact and it is fast (activation-time) with an opening time of 4 and closing time of 2 μsec. The shutter runs on a +5 volts voltage. It moves to an on-position with an active voltage (controlled by the toggle foot switch) working against a restore-spring and remains on when the voltage is applied and turns to off when the voltage is removed. This gives a freedom of choice for the duration shutter is active with very simple electronics used to control the speed (activation time.) In the design the laser passes through an aperture on the only moving part a cylinder. No gears and no electronics in the design make it very simple, stable and accurate, even under the heat produced by a laser beam. The shutter needs no special power supply and can be run on a cell phone charge with an output of roughly 300mA/5V. Cost and construction time is also important. With this design, a shutter can be prepared under \$40 with 10 hours of construction time (10 x 25 (hourly wage of a technician) =\$250+40=\$290), still better than many commercially available shutter systems with same performance. ## Design & Construction ### Components There are three major parts of the system. • Shutter. • Control box. • Power supply. The components used: #### Shutter • 12V DC motor • Wood rotation stage • Spring with torque of .1 N m • Rubber padding • Pillar Post Extension, Length=1" from Thorlabs (shutter cylinder) • 30mm Cage Plate Optic Mount from Thorlabs • Post-holder, base-plate ext... #### Control Box • 1 power jack M&F • 1 1/4" mono Panel-Mount Audio Jack M&F • 1 Foot Paddle • 1 100Ω pot with, 1 220Ω resistor • 1 on/off toggle switch • 1 LED • 1 box enclosure • some connection wires, solder gun and solder wire #### Power Supply Any power-supply which can provide 300mA at 5V and above. The motor torque is power dependent and the shutter speed is resorting spring's stiffness (torque) dependent. So choose the spring carefully before decide on the power supply. I would recommend a variable power-supply which can be bought easily from any where. ### Construction The buildup is divided is the following categories. In the start make sure that you have a good set of screw driver, pliers, wire cutter, stripper and solder. #### Shutter I will start with the choice of the motor. A 12V DC motor is a good choice because it can provide a wide rang of torques. The motor and the spring works against each other so it important to have the right set. So choose a 12V DC motor with a shaft length of at least 15mm. Mount the motor on the cage plate or platform as shown in the picture, you will need 4/40 screws to tight it (in present setup). once this is done, unscrew and take the motor out. Now next task is to choose the right spring. The spring should be half the length of the shaft and the shaft should easily fit through the spring. I want to keep the spring-choice procedure really simple and easy. To choose the spring first need to know the torque of the motor. This is really simple, it can be mathematically calculate if the voltage and the current is known. But I like the experimental way; the setup is shown in the slide2. In the slide I have a DC motor with a leverage pivoted at the shaft and a free weight hanging with a string on the other side. In this, all I am doing is balancing the weigh at that point by giving the motor just enough power. so the torque is equal to the weight at the distance from the shaft: • $\mathbf{\tau}=r \times F = r \times m .g$ Once this is know we can choose the spring with less torque. The restoring torque applied by the spring can also be measure in the same fashion (as motor) as shown in slide1, with hanging known weight. The spring stiffness and the torque can be related through: • $\mathbf{\tau}=2{\pi}n.K.r^2$ Where K is the stiffness, r distance from the shaft center and n number of times spring twisted one complete 360 degrees circle before it was hacked. Since the torque is n dependent, it is a matter of great convenience, because now it is really easy to pick the right spring. All you have to do, is twist the required number of times before hacked to get the right restoring torque. Now we are ready to put the the shutter together. Start with the motor, most of the motors have little holes for the screws to hold slide3. Choose a screw which can fit into that but still sticking out few millimeters slide4. Next is the rotation stage which joins the cylinder to the motor and holds the spring assembly. I choose the wood for this function, because its easy to machine and a good thermal-insulator slide5. I drilled two hole on both sides; to fit the shaft and 1/4 screw on either sides. To drill for 1/4 I used 15/64 size drill-bit slide6,7,8 & 9. Now the motor can be attached to one side of the stage and cylinder to other slide10 & 11. Now we can machine the spring assembly. Slide 12 shows the parts of spring assembly. We need two metal strips of .75 and 1 cm long with a hole in the end for the screws. This strips with the screw on the motor will define the boundaries of the movement. Cylinder will rotate end to end between these strips. To screw the strips directly on the stage we need to drill two holes for the screws. The holes are 135 degree apart over the face of the stage as shown in the slide13. We also drill a small hole for the spring attachment near the edge. Now all parts are ready to get together slide 14. This shutter is little different of what is shown in the video but follows the same concept. Put the shorter end of the spring in the hole on the stage like slide 15. Now the motor shaft will go through the spring into the stage. The other side of the spring is hold against the screw. You can twist the required number of turns before it holds to get the right restoring torque. Now the spring wants to move the stage clockwise but the strip against the screw does not let it (this is position 1). As the voltage is applied the motor turns counterclockwise working against the spring until the second strip stops it. When the voltage is applied the stage remains in the second position. As soon as the voltage is turned down the spring restores the stage into the previous position. These two positions can be chosen for the shutter to be on/off slide 16,17 & 18. A fully assembled shutter is ready. Now the next task is to prepare a control box. #### Control Box I used a 2X3 inch aluminium box to house the electronics slide20. The electronics is really simple. The circuit is shown in the slide21. The foot switch is used to connect the motor ground to the power-supply hence activates the shutter. Since the shutter works on active voltage, the voltage is always on, when the foot paddle is active. So it is very important to give the right voltage to the motor to avoid damage to the motor over the long duration of active voltage. To do this i use a 100 ohm POT. POT let me choose the right voltage and current. This can be done once the shutter is ready. Now connect the shutter to the shutter input and foot-paddle to its input. #### Power Supply The good thing about the system is that it can use any power supply which can give enough power to operate the motor. The POT inside the control box let choose the appropriate voltage and current and hence makes easy to choose a power-supply. ## Comments Motor shutter buildup View more presentations from pranavrathi. Full Shutter System Shutter Control box Shutter different view Control box different view	{"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": 2, "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": 0.9220695495605469, "perplexity_flag": "middle"}
http://mathoverflow.net/revisions/78891/list	## Return to Answer 5 Added property P_d A simple example is obtained by taking $P$ to mean "has positive dimension". Every local domain of positive dimension $(A,\mathfrak m)$ has $P$ at all maximal ideals (i.e. just at $\mathfrak m$ !) since $A_{\mathfrak m}=A$ , but $P$ fails at the generic point $\eta=(0)$ since $A_\eta=Frac(A)$ has dimension zero, being a field. Edit In order to address Chuck's comment, let me emphasize that the answer above is very easily adapted to non local non-local rings. For example any finitely generated domain $A$ of positive dimension $d$ over a field has property $P$ when localized at a maximal ideal $\frak m$ but not at the zero ideal. More precisely, $dim A_{(0)}=0$ and $dimA_{\frak m}=d$ for any maximal ideal ${\frak m}$ : this equidimensionality result follows from Noether's normalization theorem. This shows that if property $P_d$ is " has dimension $d$ ", then $P_d$ holds for $A_{\frak m}$ if ${\frak m}$ is maximal and does not hold for $A_{\frak p}$ if the prime $\frak p$ is not maximal. 4 added non local example. A simple example is obtained by taking $P$ to mean "has positive dimension". Every local domain of positive dimension $(A,\mathfrak m)$ has $P$ at all maximal ideals (i.e. just at $\mathfrak m$ !) since $A_{\mathfrak m}=A$ , but $P$ fails at the generic point $\eta=(0)$ since $A_\eta=Frac(A)$ has dimension zero, being a field. Edit In order to address Chuck's comment, let me emphasize that the answer above is very easily adapted to non local rings. For example any finitely generated domain $A$ of positive dimension $d$ over a field has property $P$ when localized at a maximal ideal $\frak m$ but not at the zero ideal. More precisely, $dim A_{(0)}=0$ and $dimA_{\frak m}=d$ for any maximal ideal ${\frak m}$ : this equidimensionality result follows from Noether's normalization theorem. 3 added 8 characters in body A simple example is obtained by taking $P$ to mean "has positive dimension". Every local domain of positive dimension $(A,\mathfrak m)$ has $P$ at all maximal ideals (i.e. just at $\mathfrak m$ !) since $A_{\mathfrak m}=A$ , but $P$ fails at the generic point $\eta=(0)$ since $A_\eta$ A_\eta=Frac(A)\$ has dimension zero, being a field. 2 Replaced "integral ring" by "domain" A simple example is obtained by taking $P$ to mean "has positive dimension". Every integral local ring domain of positive dimension $(A,\mathfrak m)$ has $P$ at all maximal ideals (i.e. just at $\mathfrak m$ !) since $A_{\mathfrak m}=A$ , but $P$ fails at the generic point $\eta=(0)$ since $A_\eta$ has dimension zero, being a field. 1 A simple example is obtained by taking $P$ to mean "has positive dimension". Every integral local ring of positive dimension $(A,\mathfrak m)$ has $P$ at all maximal ideals (i.e. just at $\mathfrak m$ !) since $A_{\mathfrak m}=A$ , but $P$ fails at the generic point $\eta=(0)$ since $A_\eta$ has dimension zero, being a field.	{"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": 61, "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": 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": 0.9184466004371643, "perplexity_flag": "head"}
http://www.physicsforums.com/showthread.php?p=3814788	Physics Forums ## Conservation of momentum and energy contradiction? Suppose we have two objects and we're only talking about rectilinear motion. Initially, one object has mass m and is moving at velocity V. The other has mass M and is standing still. Then they hit each other and suppose that all kinetic energy is conserved and they stick together and move at velocity v. Then this follows from the conservation of energy: 1/2mV$^{2}$ = 1/2(m+M)v$^{2}$ And this follows from the conservation of momentum: mV = (m+M)v But when we divide the first equation by the second one we get: 1/2V = 1/2v V = v Which implies that M is zero but I haven't stated that anywhere, at least not directly. What did I do wrong? PhysOrg.com physics news on PhysOrg.com >> Promising doped zirconia>> New X-ray method shows how frog embryos could help thwart disease>> Bringing life into focus Mentor If they stick together, then the collision must be inelastic ie kinetic energy cannot be conserved. Remember that momentum is always conserved in a collision (provided there are no external forces). So, to see that what I'm saying is true, first apply conservation of momentum to figure out the final velocity in terms of the initial one. Then use these velocities to compare the final kinetic energy to the initial. Quote by cepheid If they stick together, then the collision must be inelastic ie kinetic energy cannot be conserved. Can you elaborate on this? This is kind of too abstract for me. Does this mean that for these masses and velocity always the same amount of kinetic energy is lost? I'd imagine it would also depend on the materials or contact area etc. Mentor Blog Entries: 1 ## Conservation of momentum and energy contradiction? Quote by Fyreth Can you elaborate on this? This is kind of too abstract for me. Just do the calculation. Momentum is always conserved. And since they stick together, it's easy to calculate the final velocity. Then you can directly compare the final KE with the initial. Does this mean that for these masses and velocity always the same amount of kinetic energy is lost? Right. Since they stick together, the maximum amount of KE is lost. I'd imagine it would also depend on the materials or contact area etc. All you need to know is that they stick together. Quote by Doc Al Just do the calculation. Momentum is always conserved. And since they stick together, it's easy to calculate the final velocity. Then you can directly compare the final KE with the initial. Right. Since they stick together, the maximum amount of KE is lost. All you need to know is that they stick together. I understand it mathematically but not physically. It's my first time looking into these collisions. How does this work? Which process turns the uniform motion into other forms of energy? Why is it not dependent on other properties? Why does sticking together leads to KE loss? Quote by Fyreth I understand it mathematically but not physically. It's my first time looking into these collisions. How does this work? Which process turns the uniform motion into other forms of energy? Why is it not dependent on other properties? Why does sticking together leads to KE loss? Hey, Try this. Do NOT assume that the balls stick together afterwards. Assume the initial conditions you gave. Conserve BOTH momentum and energy, and see what you expect the system to do. Now, ask yourself: Why do the balls not naturally stick together. The answer is, you need extra energy to force the balls to stick together. This energy comes from the kinetic energy of the ball. Thus, some of the energy is lost there and we cannot account for it. Of course, total energy = KE of both balls afterwards + Energy lost in forcing balls together is still conserved. But since we cannot calculate the latter (atleast directly), we cannot use energy conservation. You may ask of course, Why is MOMENTUM not lost in forcing the balls to stick together. The answer is that momentum is ALWAYS conserved if there are no net force. (Energy conservation is a more subtle issue). In this case, the balls experience forces that make it stick together, but all these forces are internal and effectively cancel out leaving zero net force. Thus, you may use momentum conservation, but not energy conservation here. Quote by Fyreth I understand it mathematically but not physically. It's my first time looking into these collisions. How does this work? Which process turns the uniform motion into other forms of energy? Why is it not dependent on other properties? Why does sticking together leads to KE loss? When the balls stick together, some energy is converted to heat (assuming no other losses, like sound waves, etc.) Suppose you had two balls one coming from the left with a speed of v, one coming from the right at the same speed. Suppose they have the same mass, and they stick together. Then when they stick together, they will stop dead. The total momentum is zero before and after, because their velocities were equal and opposite, no problem, but their kinetic energy has gone to zero. It all turned into heat, the balls are warmer after the collision. This usually happens because the two balls deform when they collide, and the different parts of the ball moving past each other creates friction, which creates the heat. You could also have the collision create sound waves inside the balls, and the sound waves would bounce around inside the balls, eventually getting absorbed and turned into heat. If there is a sound when they collide, and that sound gets radiated away, and carries some of the kinetic energy with it. What I'm saying below is parasitic on Rap's post. As Doc Al suggested, use momentum conservation to find v. As you wrote: mV = (m+M)v. So v = mV/(m+M). Now imagine that you view the collision from a car which is moving with this just this velocity. In your new frame of reference, m and M collide and each becomes stationary. That's what makes sticking together so special: there is a frame of reference in which both bodies lose all their KE. This won't happen for all colliding bodies, but there are quite a few materials (the obvious one is putty) for which this does happen, owing to the nature of the forces between the particles making up the material. But losing all KE in this special frame equates to sticking together in other frames of reference, and the loss of some of the original KE. The fact that only a certain fraction of the KE is lost in these frames, and that this fraction can be calculated with the aid of momentum conservation, makes it seem mysterious – as if the properties of the material aren't involved. But they are – as we saw by looking at the collision from the frame of reference (called the 'centre of mass frame') in which both bodies finish up stationary. Mentor Quote by Fyreth I understand it mathematically but not physically. It's my first time looking into these collisions. How does this work? Which process turns the uniform motion into other forms of energy? Why is it not dependent on other properties? Why does sticking together leads to KE loss? Think of different kinds of collisions with the ground. If something collides with the ground and maintains its KE then it bounces back up. If something collides with the ground and loses its KE then it splats on the ground. So, just by experience you know that "bouncy" collisions maintain KE and "sticky" collisions do not. Now, think about the result of a bouncy or a sticky collision on the bouncy or sticky object. A rubber ball is largely unaffected by a collision with the ground. It temporarily deforms, and then bounces right back to its original shape. A ball of mud or clay on the other hand deforms permanently. After the sticky collision it is no longer a ball. So, what happened to the KE of the clay? "1/2mV2 = 1/2(m+M)v2" This equation is wrong,because when they hit each other,some energy chenge into heat Q,which is about the movement of molecule in the body.So he have: So, conservation of energy is also true,but conservation of kinetic energy is not satisfied. With the equation which contain the heat ,you won't get the contradiction which make m=0.So conservation of momentum is also not broken. Tags conservation laws, contradiction Thread Tools \| \| \| \| \|-------------------------------------------------------------------------\|-------------------------------\|---------\| \| Similar Threads for: Conservation of momentum and energy contradiction? \| \| \| \| Thread \| Forum \| Replies \| \| \| Introductory Physics Homework \| 18 \| \| \| Classical Physics \| 12 \| \| \| General Physics \| 1 \| \| \| Engineering Systems & Design \| 9 \| \| \| Introductory Physics Homework \| 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": 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": false, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9502913951873779, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/68220/what-makes-simple-groups-so-special	# What makes simple groups so special? The classification of finite simple groups was one of the most important problem in group theory. But what makes simple groups so interesting and special? - 5 – user1729 Sep 28 '11 at 13:31 1 What makes prime numbers so interesting and special? Since I have class starting in a minute, I'll just post a quote from the Wikipedia page for simple groups: "The finite simple groups are important because in a certain sense they are the "basic building blocks" of all finite groups, somewhat similar to the way prime numbers are the basic building blocks of the integers. This is expressed by the Jordan–Hölder theorem which states that any two composition series of a given group have the same length and the same factors, up to permutation and isomorphism." – Alex Sep 28 '11 at 13:31 2 – Bruno Stonek Sep 28 '11 at 13:32 Simple groups, and their relatives the quasi-simple and almost simple groups, tend to come up a lot. Volume and form preserving matrix groups tend to be quasi-simple, and automorphisms of highly symmetric geometries tend to be almost simple. While simple groups are building blocks in general, the non-abelian ones tend to build very interesting things with only one (non-abelian) block. – Jack Schmidt Sep 28 '11 at 14:26 ## 1 Answer Simple groups are like prime numbers........ One must not let the account of the matter end with a full stop after the last word above. If one could say that all finite groups are products of finite simple groups, then the analogy would be simpler. But one can say that simple groups are as far as you can take the process of taking quotient groups without going to the very smallest quotient group: the group with only one element. The complication is that just taking Cartesian products of simple groups doesn't give all finite groups, and in fact, I think the understanding the ways in which other finite groups are built out of finite simple groups is quite a substantial problem in itself. However. Say you have a composite number like 299. You can form quotients: $$299/13 = 23.$$ $$299/23 = 13.$$ But from the prime number 23, you can't form any quotients except 23 and 1. Similarly there are no quotient groups of a simple group except itself and the group with only one element. -	{"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": 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": 0.9419145584106445, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/76566/what-is-transcendental-algebraic-geometry?answertab=active	# What is 'Transcendental algebraic geometry'? What is 'Transcendental algebraic geometry'? Could you give me some good references in this field? Thanks. - ## 3 Answers Transcendental Algebraic Geometry is the study of the algebraic geometry of a variety defined over the complex numbers $\mathbb C$ by concentrating on its undelying structure as a holomorphic manifold or variety. This allows one to study the variety through the powerful tools of topology, analysis and differential geometry like : characteristic classes, elliptic partial differential equations, Kähler structure, ... Serre wrote a remarkable article (always quoted as GAGA!) giving a very precise functorial correspondence between a projective algebraic variety $X$ over $\mathbb C$ and the corresponding complex analytic variety $X^{an}$: here it is. Voisin's book mentioned by sweetjazz is indeed masterful but rather advanced. Two good introductory, fairly reader friendly, books to that area are Taylor's textbook and Wells's classic Differential Analysis on Complex Manifolds . The latter book culminates in the profound solution by Kodaira of Hodge's problem of characterizing the Kähler complex manifolds underlying a projective algebraic variety. - Griffith & Harris - "Principles of algebraic geometry" is also very good reference. - Transcendental Algebraic Geometry generally refers to algebraic geometry studied using techniques from the theory of complex variables, so that the results generally only apply to varieties defined over $\mathbb{C}$. One of the basic references for this area is Claire Voisin's two volume series Hodge Theory and Complex Algebraic Geometry. -	{"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": 0.8985762000083923, "perplexity_flag": "middle"}
http://mathhelpforum.com/calculus/59466-real-valued-function.html	# Thread: 1. ## Real-valued function Hi ! Do you know how to solve this problem: Find a real-valued function on R possessing derivatives of all orders whose Taylor series at a certain point converges to the function only at that point. 2. Originally Posted by Milus Hi ! Do you know how to solve this problem: Find a real-valued function on R possessing derivatives of all orders whose Taylor series at a certain point converges to the function only at that point. Consider $x^{x^x}$ at $x=0$ 3. Thank,but still I do not know what means that up side down L inside the log parentasice. Also, do you have any idea how to prove it. 4. Originally Posted by Milus Thank,but still I do not know what means that up side down L inside the log parentasice. Also, do you have any idea how to prove it. That is my signature, that is not part of the post. And for the proof try finding the general form for the n-th derivative of x. 5. Can you help me a little more? I am still strugling with general form of the derivative 6. Originally Posted by Milus Hi ! Do you know how to solve this problem: Find a real-valued function on R possessing derivatives of all orders whose Taylor series at a certain point converges to the function only at that point. I didn't know MathStud28's example. The example that I would have called the "classical" one is: $f(x)=e^{-\frac{1}{\|x\|}}$ for $x\neq 0$ and $f(0)=0$. I can't remember which famous mathematician introduced it, perhaps Cauchy, anyway it has a long history. And it seems simpler to deal with. For $x>0$, $f(x)=e^{-\frac{1}{x}}$. First notice that $f(x)\to_{x\to 0^+} 0$. Notice as well $\frac{e^{-\frac{1}{x}}}{x^n}\to_{x\to 0^+}0$ for any integer $n$. To convince yourself of this, substitute $u=\frac{1}{x}$, so that the limit is $\lim_{u\to\infty} u^n e^{-u}$. Now, the $n$-th derivative of $f$ is (easily) seen to be a rational function of $x$ (a quotient of two polynomials) times $e^{-\frac{1}{x}}$. This is proved by induction (suppose there is a rational function $R$ such that the $n$-th derivative is, for $x>0$, $R(x)e^{-\frac{1}{n}}$, and prove that the next derivative is again of the same kind). Because of the previously mentioned limit, you obtain that the n-th derivative converges to 0 as $x$ tends to 0 from the right. By symmetry, the same holds from the left. Finally, there is a theorem (that you probably know if you're asked this problem) telling you that, as a consequence, $f$ is indefinitely differentiable at 0 and the derivatives at 0 are the limits of the derivatives at $x$ when $x$ tends to 0. That is to say, all derivatives at 0 exist and are 0. Yet, the function is only zero at 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": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 24, "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": 0.9561836123466492, "perplexity_flag": "head"}
http://blogs.ubc.ca/math101sec210portal/2013/02/05/lecture-10/	Home › Essential › News › Lecture › Lecture #10 # Lecture #10 Posted on February 5, 2013 by Today we covered some tough examples of trigonometric substitution, for easier examples, read examples 1-7 in p.479 – p.482 of the textbook. The examples I gave are unusual not in the techniques of trig sub itself, because the first step of identifying of the correct substitution is not really so hard. However, they are unusual second step, where we need to evaluate the weird trigonometric integrals that come out, and by now I have shown pretty much all special cases of $$\int \tan^m x \sec^n x dx$$. Note that the examples I gave doesn’t fall into the following cases • m is not odd (otherwise, substitute $$u = \sec \theta$$ will work) • n is not even (otherwise, substitute $$u = \tan \theta$$ will work) Situations when m and n are super large numbers AND does not conform to the rules will MOST LIKELY NOT APPEAR again in MATH101, because it requires reduction formulas to be solved with a reasonable amount of work. Similarly, for integrals of sines and cosines, when the powers of sine and cosine are odd, we can substitute cosine and sine respectively. But the case when both powers are even and large will most likely not appear again in this course. (for small even powers, though, we will use half angle formulas) So before it confuses you finally, we isolate these really evil special cases: • $$\int \sec x \ dx$$ • $$\int \sec^3 x dx$$ • $$\int \tan^2 x \sec x dx$$ The lecture note is here: wk5day10, and the promised extra discussion on \int\sec x dx[\latex] is here: wk5day10_spec_int. Posted in Lecture Search the Weblog	{"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": 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": 0.9067398309707642, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/27987/are-all-probabilities-conditional-probabilities/27993	## Are all probabilities conditional probabilities? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) We know that $P(A\|B) = \frac{P(A \cap B)}{P(B)}$. So $P(B) = P(A\|B)P(A \cap B)$. Thus are all probabilities conditional probabilities? Can one make a probability more accurate by introducing a conditional component? For example, the probability of rolling a six on a fair die is $1/6$. But can we make this more accurate by assuming prior events? - The definition of the words "fair die" is that the probability that each face comes up is $\frac{1}{6}$. What would it mean to make this more accurate? (Of course, in the physical world it is likely that there is no such thing as a fair die.) – JBL Jun 13 2010 at 2:53 @JBL, that's the kind of accuracy the OP is referring to: introducing a bias based on prior knowledge about the die etc. – Suresh Venkat Jun 13 2010 at 3:27 ## 5 Answers There's an obvious algebra error in Tony's question. In one trivial sense all probabilities are conditional: just condition on the whole "sample space". In (another?) sense, probabilities are necessarily conditional on what is known. - ### 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. In a sense this is precisely the Bayesian persective on what probabilities mean. Rather than being the limit fraction of a number of times an event occurs, it is a strength of belief conditioned on a prior. Bayesian approaches to statistical reasoning always start with a prior belief and condition based on that. - It is possible to develop probability theory taking conditional probability as one of the basic definitions; see section 3.2 in this book and the references mention there. Renyi was one of the first mathematicians to favor this approach, which is described in his book Probability Theory. Part of the motivation for this approach is to directly build in the ability to condition on measure-zero events without having to make a limiting argument. Another key word related to this idea is disintegration. So, as Kjetil mentions, it all depends on what one takes as axioms. But certainly it is possible to develop theories that take conditional probability as the centerpiece. - I think that's the main difference between the Bayesian approach vs. the frequentist approach, having prior knowledge is supposed to make your predictions more accurate. See [LessWrong][1] Bailey was trained in statistics, and when he joined an insurance company he was horrified to see them using Bayesian techniques developed in 1918. They asked not "What should the new rates be?" but instead "How much should the present rates be changed?" But after a year of trying different things, he realized that the Bayesian actuarial methods worked better than frequentist methods. Bailey "realized that the hard-shelled underwriters were recognizing certain facts of life neglected by the statistical theorists." For example, Fisher's method of maximum likelihood assigned a zero probability to nonevents. But since many businesses don't file insurance claims, Fisher's method produced premiums that were too low to cover future costs.cover future costs. [1]: http://lesswrong.com/lw/774/a_history_of_bayes_theorem/cover future costs. - I think this is a tricky question! and the answer depends on the exact meaning of the question. In the usual standard probability theory, based on Kolmogoro'vs axioms, in the axioms itself there are no conditional probabilities, and only later conditional probability are separately axiomatized. This seems to me a most strange procedure, and I am not sure why it is done that way. But in the lighjt of this theory, there are in fact nn-conditional probabilities. From an applied point of view, that of course do not make sense. Others can maybe comment on alternative axiomatizations. -	{"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": 0.9599866271018982, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/23741?sort=votes	## How well does a truncated fourier expansion of a stepfunction perform near the expansionpoint ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Hey, i'am currently trying to make a stability analysis of a binary fluid at its phase border. As the governing equations in this double-diffusive problem are rather complicated i have to do a series expansion of my density profile. My problem is, that my PDEs contain a hevyside step function right at the border and i want to do a fourier expansion (i can then use symmetries of my problem and do some horrible algebra to actually get a "solution" which can be computed for up to 5 or 6 modes). Now i know that fourier expansion performs quite horribly at a discontinous jump if only the first few modes are kept. Is it nervertheless possible to obtain a "margin of error"? Numerical precision is not of utmost concern, but the physics behind my problem should not all be thrown out by such an expansion. Do you know of any criteria or "better" expansion where its possible to make heavy use of symmetry/antisymmetry? all the best, jan - SFAIK the standard way of handling the "ringing" that comes up in Fourier approximants of discontinuous functions requires multiplication by a sine cardinal (sinc(x)=sin(x)/x), as recommended by Lanczos. Have you tried this already? – J. M. Aug 12 2010 at 8:02 ## 3 Answers Too lazy to think about the precise answer, but using something like a Cesaro sum should improve matters. There are some useful references in the Wikipedia pages for Gibbs phenomenon, Fejer's theorem, etc. - ### 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. As Nate pointed out, Fejer's summation should be better than the standard Fourier sum. The standard Fourier sum will converge pointwise like 1/n near the discontinuity. - Isn't it just the Gibbs phenomenom you are asking for? If there is a discontinuity (as for the Heaviside function) the discontinuity in the partial Fourier series does not die out for $n \to \infty$ , but will be about 18% larger than the discontinuity in the original function. This does not give you the precise value for 5 or 6 modes, but tells you that for Heaviside function the overshot will be 0,18 for $n \to \infty$, which will give you the order of magnitude also for 5 or 6 modes. (Please see also http://en.wikipedia.org/wiki/Gibbs_phenomenon.) - Is this an expansion of Nate Eldridge's answer? – S. Carnahan♦ Jun 18 2010 at 19:16	{"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": 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": 0.9076010584831238, "perplexity_flag": "middle"}
http://mathhelpforum.com/advanced-applied-math/28887-heat-problems.html	# Thread: 1. ## Heat problems 1) Determine the amount of heat needed to turn 500g of ice in a 100g aluminum pan into steam. 2) What will be the phase of 1 liter of water in a 200g copper pot if it is placed in a coolant that absorbs 100,000J of heat? 3) How much heat is required to turn 50g of ice with temperature of -50 C to steam at 150 C? Specific Heat: Water = 4180 J/kgK or 1.000 cal/gC Copper = 392 J/kgK or 0.093 cal/gC Aluminum = 895 J/kgK or 0.214 cal/gC Latent Heat: = Heat of fusion Water = $3.34 * 10^5 J/kg$ or $80 cal/g$ Copper = $2.05 * 10^5 J/kg$ or $48.9 cal/g$ Aluminum = $3.76 * 10^5 J/kg$ or $89.9 cal/g$ = Heat of vaporization Water = $22.6 * 10^5 J/kg$ or $540 cal/g$ Copper = $48 * 10^5 J/kg$ or $1150 cal/g$ Aluminum = $113.71 * 10^5 J/kg$ or $2717.7 cal/g$ 2. Originally Posted by ihmth 1) Determine the amount of heat needed to turn 500g of ice in a 100g aluminum pan into steam. My answer is 360,000 cal If you are going to specify that we have an aluminum pan then we need to know what temperature the pan is starting at. Additionally we need to know what temperature the ice starts at as well. There is not enough information present to do this problem. -Dan 3. Originally Posted by ihmth 2) What will be the phase of 1 liter of water in a 200g copper pot if it is placed in a coolant that absorbs 100,000J of heat? I dont have an answer Specific Heat: Water = 4180 J/kgK or 1.000 cal/gC Copper = 392 J/kgK or 0.093 cal/gC Latent Heat: = Heat of fusion Water = $3.34 * 10^5 J/kg$ or $80 cal/g$ The maximum amount of heat energy the water can have is when its initial temperature is at the boiling point, 373 K. (Otherwise it is steam, not water.) Assume the copper pot also has this temperature at the beginning. Let's find out how much heat needs to be absorbed from the system to turn the water at the boiling point into ice: $c_wm_w(373 - 273) + c_{Cu}m_{Cu}(373 - 273) + L_wm_w = 7.5984 \times 10^5~J$ (The copper pot must be a maximum of 273 K for the water to freeze, so I'm assuming that it drops at the same temperature rate as the water. This is not quite physically true, but will do as an approximation.) The first two terms take care of the heat needed to be absorbed to take the water and copper from 100 C to 0 C, the third term is the heat to be absorbed from the water at 0 C to turn it into ice. The coolant absorbs $1 \times 10^5~J$ of heat energy. This is less than the heat we can absorb from the water and copper at 100 C to freeze the ice, so we would need a starting temperature for the water-ice system in order to give a definitive answer to this question. (Now if the coolant absorbed, say, $1 \times 10^6~J$ then we would know for sure that the water would freeze.) -Dan 4. Originally Posted by ihmth 3) How much heat is required to turn 50g of ice with temperature of -50 C to steam at 150 C? My answer is 41,000 cal Specific Heat: Water = 4180 J/kgK or 1.000 cal/gC Latent Heat of fusion: Water = $3.34 * 10^5 J/kg$ or $80 cal/g$ = Heat of vaporization Water = $22.6 * 10^5 J/kg$ or $540 cal/g$ This one we can do: Heat needed to turn the 50 g of ice at -50 C to ice at 0 C: $c_im_i(0 - -50) = 5125~J$ ( $c_i = 2.050~J/g ^o C$. You didn't give this, I had to look it up. It's Wikipedia so I'm only assuming the numbers are correct.) Heat needed to turn 50 g of ice at 0 C to water at 0 C: $L_fm_i = 16700~J$ Heat needed to turn 50 g water at 0 C to water at 100 C: $c_wm_w(100 - 0) = 20900~J$ Heat needed to turn 50 g water at 100 C to steam at 100 C: $L_vm_w = 113000~J$ Finally, heat need to turn 50 g of steam at 100 C to steam at 150 C: $c_sm_s(150 - 100) = 5200~J$ ( $c_s = 2.080~J/g ^oC$. Again, you didn't give us this information.) Adding these up gives a total of 160925 J of heat required. In the future, please provide us with all the information you are given. -Dan 5. Originally Posted by topsquark If you are going to specify that we have an aluminum pan then we need to know what temperature the pan is starting at. Additionally we need to know what temperature the ice starts at as well. There is not enough information present to do this problem. -Dan ice and the aluminum pan is at 0 C 6. Originally Posted by ihmth ice and the aluminum pan is at 0 C Originally Posted by ihmth 1) Determine the amount of heat needed to turn 500g of ice in a 100g aluminum pan into steam. My answer is 360,000 cal Specific Heat: Water = 4180 J/kgK or 1.000 cal/gC Aluminum = 895 J/kgK or 0.214 cal/gC Latent Heat: = Heat of fusion Water = $3.34 * 10^5 J/kg$ or $80 cal/g$ = Heat of vaporization Water = $22.6 * 10^5 J/kg$ or $540 cal/g$ So $L_f m_i + c_wm_i(100 - 0) + c_{Al}m_{Al}(100 - 0) + L_vm_i$ -Dan	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 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": 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": 0.9272366166114807, "perplexity_flag": "middle"}
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http://mathoverflow.net/questions/114245/i-know-that-you-know/114265	## I know that you know… ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) A bit unsure if the following vague question has enough mathematical content to be suitable upon here. In the case, please feel free to close it. In several circumstances of competition, a particular situation of partial information occurs, usually described as "I know that you know that I know... something". We may distinguish a whole hierarchy of more and more complicate situations closer and closer to a complete information. E.g. : • $I_0$: I know $X$, but you don't know that I know. • $I_1$: I know $X$, you know that I know, but I don't know that you know that I know. • $I_1$: I know $X$, you know that I know, I know that you know that I know, but you don't know that I know that you know that I know. • .... &c. For small values of $k$, I can imagine simple situations where passing from $I_k$ to $I_ {k+1}$ really makes a difference (for instance: you are Grandma Duck, and $X$ is : "you left a cherry pie to cool on the window ledge". Clearly, $I_0$ is quite agreeable position; $I_1$ may lead to an unpleasant end (for me); $I_2$ leaves me some hope, if I behave well, and so on). But, I can't imagine how passing from $I_6$ to $I_7$ may affect my strategy, or Grandma's. Are there situations, real or factitious, concrete or abstract, where $I_k$ implies a different strategy than $I_{k+1}$ for the competitors? What about $I_{\omega}$ and, more generally, $I_\alpha$ for an ordinal $\alpha$ (suitably defined by induction)? How these situations are modeled mathematically? - 9 I assume you have heard of the blue eyed islander problem? This is a classic example of where arbitrarily high orders of knowledge of this form become relevant. – Steven Gubkin Nov 23 at 15:01 2 There is a (not serious) mention of this in "A canticle for Leibowitz", but sadly I don't remember where exactly. Can somebody find the exact quotation? – Laurent Berger Nov 23 at 15:35 3 A whole load of fictitious examples can be found at tvtropes.org/pmwiki/pmwiki.php/Main/… – Terry Tao Nov 23 at 18:03 There are some papers on "reflexive game theory" by Alexander Chkhartishvili, see e.g. link.springer.com/article/…, where this hierarchy of reflexive levels is formalized and he shows how this affects the Nash equilibrium etc. Chkhartishvili apparently did not put any of his work on arXiv, so unfortunately one has to resort to English translations of some Russian journals where this is published. – ansobol Nov 23 at 19:49 ## 6 Answers My wife and I have a standing agreement where I pick up our son Horatio from school and she picks up our daughter Hypatia. One day, because I knew I would be near Hypatia's school, it was convenient to swap duties. I emailed her a message, "I'll pick up Hypatia today, and you get Horatio. Please confirm; otherwise it is as usual." She texted me back, "Let's do it. Let me know if you get this message, so that I know we're really on." I left her a voicemail, "OK, we're definitely on for the swap! ....as long as I know you get this message." She emailed me back, "Got the message. We're on! But let me know that you get this message so I can count on you." You see, without confirmation she couldn't be sure that I knew she had gotten my earlier confirmation of her acknowledgement of my first message, and she may have worried that the plan to swap was consequently off. And so on ad infinitum...... How truly frustrating it was for us that at no stage of our conversation could we seem to know for certain that the other person had all the necessary information to ensure that the plan would be implemented! The result, of course, since we had time to exchange only at most a finite number of messages, was that the only rational course of action was for us each to abandon the plan to swap: we both independently decided just to pick up the usual child. To see that this was rational, observe that clearly the first message needed to have been confirmed in order for the plan to be implemented properly. Furthermore, if the $n$-th message need not have been confirmed, then it wasn't important to know that it had been received and the algorithm should have worked whether or not it was received, meaning that it didn't actually need to have been sent. So by induction, no number of confirmations suffices to implement the common knowledge that we both needed, namely, that we had each agreed to make the swap. See also the two generals problem. - 1 Incidentally, it seems to me that the same problem still arises even if we were to speak in person, because one would have to make sure that the previous comment was received and understood, before its acknowledgement would rise to the level of common knowledge. I am left to wonder how it is possible ever for us to attain common knowledge... – Joel David Hamkins Nov 23 at 21:25 Tweetie-bird, of course my intention here was to provide a comparatively concrete example exhibiting the principle issues of common knowledge. I am actually fascinated by the logic and mathematics of this kind of situation, which I view as a genunine, sophisticated mathematical topic. – Joel David Hamkins Nov 23 at 22:50 I deleted my previous comment so as not to confuse others. This logic puzzle is already confusing enough. – tweetie-bird Nov 23 at 23:18 .....principal. – Joel David Hamkins Nov 24 at 1:55 5 As someone who didn't have the easiest time in school, I am glad that I was not named horatio or hypatia... – Bern Oay Nov 24 at 5:15 show 4 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. Here is another puzzle I like which is in the spirit of the question. 100 people play a game as follows. Each person secretly writes a number between 1 and 1000000. The numbers are then all revealed and the person who is closest to 2/3 of the average wins a prize. If there is a tie, the prize is shared between the winners. What number should one write down? Now, it is clearly foolish to write down any number greater than 666667, since 2/3 of the average cannot be more than 666667. But now we can view the game as being played on the interval from 1 to 666667 instead of from 1 to 1000000. Now we can iterate again and conclude that it is foolish to choose any number greater than 444444. Ultimately (but this requires many iterations of knowledge), the only rational choice is for all players to choose 1 and to split the prize. - 1 A agree that choosing $1$ is the only rational choice, but still I would like to actually do this experiment with a bunch of non-mathematicians and see what $2/3$ of the average turns out to be. I suspect it changes after the first go, or if you give them much time to think, but such empirical information could have been useful in a pub quiz I used to go to. – Paul Reynolds Nov 23 at 19:43 1 IIRC, I’ve read that in human experiments the 2/3 of the average tends to be around 20% of the maximum. – Emil Jeřábek Nov 23 at 19:58 Interesting! I wonder how this varies with the fraction $r = 2/3$... – Paul Reynolds Nov 23 at 20:02 3 Paul & Emil: such "beauty contest" experiments reveal baffling play: for r = 0, not everyone plays 0. Some do not play increasing functions of r. Many people seems to play roughly linearly in r relative to their play at r=1. Here is my analysis of some data I recently collected from web surveys: faculty.chicagobooth.edu/richard.hahn/…. Plots of observed strategies: faculty.chicagobooth.edu/richard.hahn/… – R Hahn Nov 23 at 21:09 1 ---Now, it is clearly foolish to write down any number greater than 666667, since 2/3 of the average cannot be more than 666667.--- Not if you playing with a partner. Of course, you won't take the prize yourself, but you can raise your partner's chances quite a bit by writing a large number if you agree upon a decent strategy. Now, how do I know that there are no coalitions out there? And if I don't, the second step of the induction fails... – fedja Nov 26 at 0:38 show 6 more comments Check out the wikipedia page on common knowlege, which includes some mathematical formalizations of the concept as well as the famous blue-eyed islander problem mentioned by Steven Gubkin. This problem was also discussed by Terrence Tao on his blog. - The classic model of these things is due to Robert Aumann and was introduced in his 1976 Agreeing to Disagree. There is a famous example due to Ariel Rubinstein, the electronic mail game, in which the behavior of $I_\omega$ radically differs from $I_n$ for any $n$. Here is a way to show that one might have to apply this reasoning up to an arbitrarily large ordinal. Let $\alpha$ be a successor ordinal. Ann and Bob play the game of picking ordinals in $\alpha$ simultaneously. The one who chooses the highest number wins. One can never win by choosing $0$, so rationality rules out choosing $0$. It is however possible to win by choosing $1$ if the other player plays $0$. But if both Ann and Bob know that they are both rational, they have to choose at least $2$. It is clear that one has to iterate this reasoning up to the predecessor of $\alpha$. - I don’t get the example. Whatever the other player knows, thinks, or plays, I’m always better off playing the predecessor of $\alpha$ than any other ordinal, hence rationality dictates I do just that, without any iteration. – Emil Jeřábek Nov 23 at 16:03 @Emil It is correct that playing the predecessor is always compatible with common knowledge of rationality up to some ordinal $\beta<\alpha$. But there are many more choices that are compatible with common rationality up to a lower level. If you want to role them out, you have to go all the way. The notion of rationality being used is not using a choice you know is not a best response. – Michael Greinecker Nov 23 at 16:15 3 Playing an ordinal $\beta$ smaller than the predecessor of $\alpha$ is never rational, because there is the possibility that the opponent may play something larger than $\beta$, which would make me lose, whereas if I play the predecessor, it is either a win for me or a draw. I know this a priori, without considering what the opponent knows. – Emil Jeřábek Nov 23 at 16:43 @Emil: No using weakly dominated strategies is stronger requirement than rationality. And having common knowledge of not playing weakly dominated strategies might be impossible. There is a well known paper by L. Samuelson, in which he show that there are games in which common knowledge of not-playing weakly dominated strategies is inconsistent (alturl.com/d2s5m). The notion I used coincides with point-rationalizability in the sense of Bernheim (alturl.com/fc9oe). – Michael Greinecker Nov 23 at 17:55 But I am not assuming any common knowledge, I am in fact not assuming anything whatsoever about the opponent. You defined rationality as “not using a choice you know is not a best response”, and I know choosing an ordinal smaller than the predecessor is not the best response, because choosing the predecessor is a better response. I have no idea whether this conforms to one formal definition or another, but if not, then IMHO the example merely illustrates that the definition is contrived rather than the phenomena mentioned by the OP. – Emil Jeřábek Nov 23 at 18:39 show 3 more comments A good source for this sort of reasoning is: J.-J. Meyer and W. van der Hoek, Epistemic Logic for AI and Computer Science, Number 41 in Cambridge Tracts in Theoretical Computer Science, Cambridge University Press, 1995. The question of common knowledge, and of the algorithmics of epistemic modal logics, provides a lot of good material on this sort of problem. (It is good fun to try the mathematical side of this in Popularisation activities. I used Muddy Children with various groups with a lot of laughter and appreciation of the mathematical models.) - The article http://plato.stanford.edu/entries/common-knowledge/ is more extensive than wikipedia's. However, it doesn't deal with the issue of infinite information. -	{"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": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9605563879013062, "perplexity_flag": "middle"}
http://mathhelpforum.com/advanced-statistics/79415-joint-probability-question.html	# Thread: 1. ## Joint Probability Question Hi, I am having trouble with the following question: Five balls are drawn without replacement from an urn containing 7 blue balls, 4 green balls, and 8 yellow balls. Let B, G, and Y be the total number of blue, green and yellow balls drawn, respectively. Find a) The joint probability mass function of B, G, and Y. b) The joint probability mass function of Y and B c) THe marginal probability mass function for B, G, and Y separately. Do these marginal probabilities look familiar? ID them. For a) I tried modelling the joint probability mass function using a combinatorial approach. c(7 choose B)c(4 choose G)c(8 choose Y)/c(19 choose 5). And B+G+Y=5 (because you can only choose 5 balls). However, after this I'm stuck. 2. these are all hypergeometrics joint is $P(B=b,G=g,Y=y)={ {7 \choose b} {4 \choose g} {8 \choose y}\over {19 \choose 5}}$. b is really the same... $P(B=b,G=g)={ {7 \choose b} {4 \choose g} {8 \choose 5-b-g}\over {19 \choose 5}}$. I'll only do B's distribution. It's B vs the world so... $P(B=b)={ {7 \choose b} {12 \choose 5-b} \over {19 \choose 5}}$. you do the other two. 3. I understand how you got part c, but I don't really understand how you got part b. Is it because you are focusing on B and G, so Y is just what's left over? 4. I was in a rush. I wrote the joint distribution of B and G. Here's the joint of Y and B... $P(B=b,Y=y)={ {7 \choose b} {8 \choose y} {4 \choose 5-b-y}\over {19 \choose 5}}$. This is because b+y+g=5. So, if you know two of them, you know the third 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": 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": 0.18, "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": 0.9279760718345642, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/44694/understanding-algebraic-manipulation	Understanding algebraic manipulation If $a+b+c \neq 0$ where $a,b$ and $c$ are three non-zero distinct integers, then find the value of: $$\frac{ab+ca}{a^2+ab+ca} + \frac{ab+cb}{b^2+ab+bc} + \frac{ac+cb}{c^2+ac+bc}$$ What confusing me here, is the not so obvious hint which is given with the problem,which says that that form could be written as: $$3- \frac{a^2}{a^2+ab+ca} - \frac{b^2}{b^2+ab+bc} - \frac{c^2}{c^2+ac+bc}$$ But how is this possible? - What have you tried? Where are you stuck? Have you tried seeing what you get when you plug in numbers (say $a = b = c = 1$)? – JavaMan Jun 11 '11 at 3:19 @DJC:Please read my question carefully. – Quixotic Jun 11 '11 at 3:20 I have read your question carefully. So where are you stuck? What have you tried? – JavaMan Jun 11 '11 at 3:21 $\quad$@DJC:Edited. – Quixotic Jun 11 '11 at 3:25 3 Answers All you need to do is "simplify" the fractions, for example by dividing "top" and "bottom" of the first one by $a$, of the second by $b$, of the third by $c$. We get $$\frac{b+c}{a+b+c}+\frac{a+c}{a+b+c}+\frac{a+b}{a+b+c}$$ We have a common denominator $a+b+c$. The numerators add up to $2(a+b+c)$. Cancel. We get $2$. Comment: If you find this not obvious, let's look at the first term, that is, at $$\frac{ab+ca}{a^2+ab+ca}$$ The "top" is $a(b+c)$. The "bottom" is $a(a+b+c)$. So the fraction is $$\frac{a(b+c)}{a(a+b+c)}$$ Divide top and bottom by $a$, or equivalently, "cancel" the $a$'s. We are using the "algebra" version of the familiar fact that $$\frac{2\cdot 3}{2 \cdot 5}=\frac{3}{5}$$ Additional comment The hint is strange. True, we can rewrite the expression as $$\left(1-\frac{a^2}{a^2+ab+ca}\right) +\left(1-\frac{b^2}{b^2+ab+bc}\right)+\left(1-\frac{c^2}{c^2+bc+ca}\right)$$ which is equivalent to what was given. And then we could divide top and bottom by $a$ in the first fraction, by $b$ in the second, by $c$ in the third, and end up with $$3-\frac{a+b+c}{a+b+c}$$ But it sure seems like a lot of work when we can cancel immediately! You are sure that you quoted the problem correctly? - Thanks this is a really easy approach the hint given confused me. – Quixotic Jun 11 '11 at 3:27 I can understand your answer,but I don't understand the hint given in my module (my edited question),Of-course the same thing could be done after that step given $3-1=2$. – Quixotic Jun 11 '11 at 3:32 Yes,I am sure I have quoted the problem correctly. – Quixotic Jun 12 '11 at 23:55 @Debanjan: The reason I wondered is that the hint seemed unreasonable in view of the obvious cancellations. – André Nicolas Jun 13 '11 at 0:06 André's approach is pretty simple. In case you really must follow the hint, here's how to get it: note, for example, that in the first fraction, $$\frac{ab+ca}{a^2+ab+ca}$$ the only thing that is missing from having the numerator exactly equal to the denominator is an $a^2$; so adding it and taking it away again we get $$\frac{ab+ca}{a^2+ba+ca} = \frac{ab+ca+a^2-a^2}{a^2+ba+ca} = \frac{a^2+ab+ca}{a^2+ab+ca} - \frac{a^2}{a^2+ba+ca} = 1-\frac{a^2}{a^2+ba+ca}.$$ The same thing can be done with $$\frac{ab+cb}{b^2+ab+bc}$$ by adding and subtracting $b^2$ to the numerator. And also with $$\frac{ac+cb}{c^2+ac+bc}$$ by adding and subtracting $c^2$ to the numerator. Doing all three, we have: $$\begin{align} \frac{ab+ca}{a^2+ba+ca} &+ \frac{ab+cb}{b^2+ab+bc} + \frac{ac+cb}{c^2+ac+bc} \\&= \left(1 - \frac{a^2}{a^2+ba+ca}a\right)\\ &\qquad\quad\mathop{+}\left(1 - \frac{b^2}{b^2+ab+bc}\right)\\ &\qquad\quad\mathop{+}\left(1 - \frac{c^2}{c^2+ac+bc}\right)\\ &= 3 - \frac{a^2}{a^2+ba+ca}a - \frac{b^2}{b^2+ab+bc} - \frac{c^2}{c^2+ac+bc}, \end{align}$$ which is exactly the formula your hint gives. As to how the hint gives the answer any better than the straightforward method suggested by André (which would be my own approach, to simplify first), I couldn't tell you, but that is how the equation can be rewritten in that form. - It is obvious that we have to use the cancellation after the hint step,thanks for the explanation and I agree this hint is not necessary and lengthy to some extent. – Quixotic Jun 11 '11 at 3:38 Or just add and subtract $a^2$, $b^2$, $c^2$ to the numerators of the three fractions respectively. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 25, "mathjax_display_tex": 13, "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": 0.9533253908157349, "perplexity_flag": "head"}
http://mathhelpforum.com/calculus/152044-determining-if-series-convergent-divergent.html	# Thread: 1. ## Determining if a series is convergent or divergent Hi I was wondering if anyone could explain this to me, I have Series (-1)^n n/ln(n) sum : infinite n:2 I took the derivative and got : 1/ln(x) - 1/(ln(x)^2) How do I determine if this is a decreasing or increasing function. From what I could tell 2 is the only value to make it go <0 , so is this neither increasing or decreasing? 2. Originally Posted by illidari Hi I was wondering if anyone could explain this to me, I have Series (-1)^n n/ln(n) sum : infinite n:2 I took the derivative and got : 1/ln(x) - 1/(ln(x)^2) How do I determine if this is a decreasing or increasing function. From what I could tell 2 is the only value to make it go <0 , so is this neither increasing or decreasing? You wrote $\sum\limits^\infty_{n=2}(-1)^n\frac{n}{\ln n}$ . This series diverges since the general term' sequence doesn't converge to zero. I can't understand what you're trying to derivate and wrt what... Tonio 3. Yeah that is the series I wrote, I was trying to apply the alternating series test. The 2nd part of the equation, excluding (-1)^n, needs to be decreasing I took the derivative of it and need to determine if that derivative is <0 . That is the part I got stuck :/ 4. Originally Posted by illidari Yeah that is the series I wrote, I was trying to apply the alternating series test. The 2nd part of the equation, excluding (-1)^n, needs to be decreasing I took the derivative of it and need to determine if that derivative is <0 . That is the part I got stuck :/ Are you trying to derivate wrt n, which is a discrete variable?! It can't be done, of course. Besides this the series does not converge at all, Leibnitz or not. Tonio	{"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": 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": 0.9644781947135925, "perplexity_flag": "middle"}
http://cogsci.stackexchange.com/questions/tagged/experimental-psychology+reading	# Tagged Questions 5answers 296 views ### How long does it take to read X number of characters? How does the time needed to read a sentence scale with the number of characters? Or does this time scaling depend on something more than just character count? For example, let $X$ be the number of ...	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9300304651260376, "perplexity_flag": "middle"}
http://www.mathplanet.com/education/algebra-1/quadratic-equations/completing-the-square	# Completing the square If you've got a quadratic equation on the form of $\\ ax^{2}+c=0 \\$ Then you can solve the equation by using the square root of $\\ x=\pm \sqrt{\frac{-c}{a}} \\$ Example: $\\ 3x^{2}-243=0 \\\\3x^{2}=243 \\\\x^{2}=\frac{243}{3} \\\\x^{2}=81 \\x=\pm \sqrt{81} \\x=9\: \: or\: \: x=-9 \\$ This method can only be used if b = 0. If we instead have an equation on the form of $\\ x^{2}+bx=0 \\$ we can't use the square root initially since we do not have c-value. But we can add a constant d to both sides of the equation to get a new equivalent equation that is a perfect square trinomial. Remember that a perfect square trinomial can be written as $\\ x^{2}+bx + d=\left ( x+d \right )^{2}=0 \\$ This process is called completing the square and the constant d we're adding is $\\ d=\left (\frac{b}{2} \right )^{2} \\$ Example: $\\ x^{2}+12x=0 \\$ We begin by finding the constant d that can be used to complete the square. $\\ d=\left (\frac{b}{2} \right )^{2}=\left ( \frac{12}{2} \right )^{2}=6^{2}=36 \\\\\\ x^{2}+12x+d=0+d\Rightarrow \\x^{2}+12x+36=0+36\Rightarrow \\\\\begin{pmatrix}x+6 \end{pmatrix}^{2}=36 \\\\\\\sqrt{\begin{pmatrix} x+6 \end{pmatrix}^{2}}=\pm \sqrt{36} \\\\\\\begin{matrix} x+6=6\: \: &or\: \: & x+6=-6\\ x=0 & & x=-12 \end{matrix} \\$ The completing the square method could of course be used to solve quadratic equations on the form of $\\ ax^{2}+bx+c=0 \\$ In this case you will add a constant d that satisfy the formula $\\ d=\left ( \frac{b}{2} \right )^{2}-c \\$ Video lesson: Solve the equation by completing the squares $\\x^{2} - 3x - 10=0 \\$ Next Class: Quadratic equations, The quadratic formula • Pre-Algebra • Algebra 1 • Algebra 2 • Geometry • Sat • Act	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 11, "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": 0.8883190751075745, "perplexity_flag": "head"}
http://www.encyclopediaofmath.org/index.php/Analytic_function	# Analytic function From Encyclopedia of Mathematics A function that can be locally represented by power series. The exceptional importance of the class of analytic functions is due to the following reasons. First, the class is sufficiently large; it includes the majority of functions which are encountered in the principal problems of mathematics and its applications to science and technology. Secondly, the class of analytic functions is closed with respect to the fundamental operations of arithmetic, algebra and analysis. Finally, an important property of an analytic function is its uniqueness: Each analytic function is an "organically connected whole" , which represents a "unique" function throughout its natural domain of existence. This property, which in the 18th century was considered as inseparable from the very notion of a function, became of fundamental significance after a function had come to be regarded, in the first half of the 19th century, as an arbitrary correspondence. The theory of analytic functions originated in the 19th century, mainly due to the work of A.L. Cauchy, B. Riemann and K. Weierstrass. The "transition to the complex domain" had a decisive effect on this theory. The theory of analytic functions was constructed as the theory of functions of a complex variable; at present (the 1970's) the theory of analytic functions forms the main subject of the general theory of functions of a complex variable. There are different approaches to the concept of analyticity. One definition, which was originally proposed by Cauchy, and was considerably advanced by Riemann, is based on a structural property of the function — the existence of a derivative with respect to the complex variable, i.e. its complex differentiability. This approach is closely connected with geometric ideas. Another approach, which was systematically developed by Weierstrass, is based on the possibility of representing functions by power series; it is thus connected with the analytic apparatus by means of which a function can be expressed. A basic fact of the theory of analytic functions is the identity of the corresponding classes of functions in an arbitrary domain of the complex plane. Exact definitions are given below. Let $D$ be a domain in the complex plane $\mathbb C$. If to each point $z\in D$ there has been assigned some complex number $w$, then one says that on $D$ a (single-valued) function $f$ of the complex variable $z$ has been defined and one writes: $w=f(z), z\in D$ (or $f:D\to\mathbb C$). The function $w=f(z)=f(x+iy)$ may be regarded as a complex function of two real variables $x$ and $y$, defined in the domain $D\subset\mathbb R^2$ (where $\mathbb R^2$ is the Euclidean plane). To define such a function is tantamount to defining two real functions \begin{equation} u=\phi(x,y),\quad v=\psi(x,y),\quad (x,y)\in D\quad (w = u+iv). \end{equation} Having fixed a point $z\in D$, one gives $z$ the increment $\Delta z = \Delta x+ i\Delta y$ (such that $z+\Delta z \in D$) and considers the corresponding increment of the function $f$: \begin{equation} \Delta f(z) = f(z+\Delta z) - f(z). \end{equation} If \begin{equation} \Delta f(z) = A\Delta z + o(\Delta z) \end{equation} as $\Delta z\to 0$, or in other words, if \begin{equation} \lim_{\Delta z\to 0}\frac{\Delta f(z)}{\Delta z} = A \end{equation} exists, the function $f$ is said to be differentiable (in the sense of complex analysis or in the sense of $\mathbb C$) at $z$; $A = f'(z)$ is the derivative of $f$ at $z$, and \begin{equation} A\Delta z = f'(z)dz = df(z) \end{equation} is its differential at that point. A function $f$ which is differentiable at every point of $D$ is called differentiable in the domain $D$. One may compare the concepts of differentiability of $f$ considered as a function of two variables (in the sense of $\mathbb R^n$) and in the sense of $\mathbb C$. In the former case the differential $df$ has the form \begin{equation} \frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y}dy, \end{equation} where \begin{equation} \frac{\partial f}{\partial x} = \frac{\partial \phi}{\partial x} + i\frac{\partial \psi}{\partial x},\quad \frac{\partial f}{\partial y} = \frac{\partial \phi}{\partial y} + i\frac{\partial \psi}{\partial y}, \end{equation} are the partial derivatives of $f$. Passing from the independent variables $x, y$ to the variables $z, \overline{z}$, which may formally be considered as new independent variables, related to the old ones by the equations $z = x+iy$, $\overline{z}=x-iy$ (from this point of view, the function $f$ may also be written as $f(z,\overline{z})$) and expressing $dx$ and $dy$ in terms of $dz$ and d\overline{z} according to the usual rules of differential calculus, one can write $df$ in its complex form: \begin{equation} df = \frac{\partial f}{\partial z}dz + \frac{\partial f}{\partial \overline{z}}z\overline{z} \end{equation} where \begin{equation} \frac{\partial f}{\partial z} = \frac{1}{2}\left(\frac{\partial f}{\partial x} - i\frac{\partial f}{\partial y}\right), \quad \frac{\partial f}{\partial \overline{z}} = \frac{1}{2}\left(\frac{\partial f}{\partial x} + i\frac{\partial f}{\partial y}\right), \end{equation} are the (formal) derivatives of $f$ with respect to $z$ and $\overline{z}$, respectively. It is seen, accordingly, that $f$ is differentiable in the sense of $\mathbb C$ if and only if it is differentiable in the sense of $\mathbb R^2$ and if the equation $\partial f /\partial\overline{z}=0$ is satisfied, which in expanded form may be written as If is differentiable in the sense of in , the latter relations are satisfied at all point of the domain; they are called the Cauchy–Riemann equations. These equations occurred already in the 18th century in J.L. d'Alembert's and L. Euler's studies on functions of a complex variable. The initial definition may now be rendered more precise as follows. A function , defined in a domain , is said to be holomorphic (analytic) at a point if there exists a neighbourhood of this point in which may be represented by a power series: If this requirement is satisfied at every point of , the function is said to be holomorphic (analytic) in the domain . A function which is holomorphic at a point is differentiable at that point. In addition, the sum of a convergent power series has derivatives of all orders (is infinitely differentiable) with respect to the complex variable ; the coefficients of the series may be expressed in terms of the derivatives of at by the formulas . The power series, written in the form is known as the Taylor series of at . Thus, holomorphy of a function in a domain means that it is infinitely differentiable at any point in and that its Taylor series converges to it in some neighbourhood of this point. On the other hand, the following noteworthy fact is established in the theory of analytic functions: If a function is differentiable in a domain , it is holomorphic in this domain (for a single point this statement is not true: is differentiable at , but is nowhere holomorphic). Thus, the concepts of complex differentiability and holomorphy of a function in a domain are identical; each one of the following properties of a function in a domain — differentiability in the sense of , differentiability in the sense of together with satisfaction of the Cauchy–Riemann equations, holomorphy — may serve as definition of analyticity of in this domain. One other characteristic of an analytic function is connected with the notion of an integral. The integral of a function along an (oriented rectifiable) curve : , , may be defined by the formula: or by means of a curvilinear integral: A key result in the theory of analytic functions is Cauchy's integral theorem: If is an analytic function in a domain , then for any closed curve bounding a domain inside . The converse result (Morera's theorem) is also true: If is continuous in a domain and if for any such curve , then is an analytic function in . In particular, in a simply-connected domain, those and only those continuous functions are analytic, whose integral along any closed curve is zero (or, which is the same thing, the integral along any curve connecting two arbitrary points does depend only on the points and themselves and not on the shape of the curve). This characterization of analytic functions forms the basis of many of their applications. Cauchy's integral theorem yields Cauchy's integral formula, which expresses the values of an analytic function inside a domain in terms of its values on the boundary: Here, is a domain whose boundary consists of a finite number of non-intersecting rectifiable curves (the orientation of is assumed to be positive with respect to ), and is a function which is analytic in some domain . This formula makes it possible, in particular, to reduce the study of many problems connected with analytic functions to the corresponding problems for a very simple function — the Cauchy kernel , , . For more details see Integral representation of an analytic function. A very important property of analytic functions is expressed by the following uniqueness theorem: Two functions which are analytic in a domain and which coincide on some set with a limit point in , coincide throughout (are identical). In particular, an analytic function , , which is not identically zero can only have isolated zeros in . If, in addition, is a zero of , then one has, in some neighbourhood of , , where is a natural number (called the multiplicity of the zero of at ), while is a analytic function in . An important role in the theory of analytic functions is played by the points at which the function is not analytic — the so-called singular points of the analytic function. Here, only isolated singular points of (single-valued) analytic functions are considered; for more details cf. Singular point. If is an analytic function in an annulus of the form , it may be expanded there in a Laurent series which contains, as a rule, not only positive but also negative powers of . If there are no terms with negative powers in the series ( for ), is called a regular point of (a removable singular point). At a regular point there also exists a finite limit Putting , one obtains an analytic function in the whole disc . If the Laurent series of the function contains only a finite number of terms with negative powers of : the point is called a pole of (of multiplicity ); a pole is characterized by The function has a pole of multiplicity at the point if and only if the function has a zero of multiplicity at this point. If the Laurent series contains an infinite number of negative powers of ( for an infinite set of negative indices ), then is called an essential singular point; at such points there is no finite and no infinite limit for . The coefficient in the Laurent series for with centre at the isolated singular point is called the residue of at : It can be defined by the formula where and is sufficiently small (so that the disc does not contain singular points of other than ). The important role of residues is made clear by the following theorem: If is an analytic function in a domain , except for some set of isolated singular points, if is a contour bounding a domain and not passing through any singular points of , and if are all the singular points of inside , then This theorem is an effective tool in calculating integrals. See also Residue of an analytic function. The sum of the terms of the Laurent series for at corresponding to the negative indices , is known as the principal part of the Laurent series (or of the function ) at the point . This principal part determines the nature of the singularity of at . Functions which are representable as a quotient of two functions that are holomorphic in a domain are called meromorphic in the domain . A function which is meromorphic in a domain is holomorphic in that domain, except possibly at a finite or countable set of poles; at the poles the values of a meromorphic function are considered to be infinite. If such values are allowed, then meromorphic functions in a domain may be defined as functions that in a neighbourhood of each point can be represented by a Laurent series in with a finite number of terms involving negative powers of (depending on ) in a neighbourhood of each point . Both holomorphic and meromorphic functions in a domain are often designated as analytic in the domain . In this a case holomorphic functions are also said to be regular analytic or simply regular functions. The simplest class of analytic functions consists of the functions which are holomorphic in the whole plane; such functions are called entire functions. Entire functions are represented by series which are convergent in the whole plane. This class includes the polynomials in , the functions etc. Weierstrass' theorem states that, for any sequence of complex numbers , without limit points in , there exists an entire function that vanishes at the points and only at these points (among the there may be coincident points, to which correspond zeros of of corresponding multiplicity). Here, the function may be represented as a (generally infinite) product of entire functions each one of which has only one zero. Functions that are meromorphic in the plane (i.e. that may be represented as quotients of entire functions) are called meromorphic functions. These include rational functions, , , elliptic functions, etc. According to Mittag-Leffler's theorem, for any sequence , without limit points in , there exists a meromorphic function with poles at the points and only at those points, such that its principal parts at the points coincide with pre-assigned polynomials in . The function may be represented as a (usually infinite) sum of meromorphic functions, each one with a pole at a single point only. Theorems on the existence of a holomorphic function with pre-assigned zeros and of meromorphic functions with pre-assigned poles and principal parts are also valid for an arbitrary domain . In the study of analytic functions the related geometric notions are also of importance. If is an analytic function, the image of the domain is also a domain (principle of preservation of domains). , If the mapping preserves angles at both in value and in sign, i.e. it is conformal. Thus, there exists a close connection between analyticity and the important geometric notion of conformal mapping. If is an analytic function in and for (such functions are called univalent), then in and defines a one-to-one and conformal mapping of the domain onto the domain . Riemann's theorem, which is the fundamental theorem in the theory of conformal mappings, says that on any simply-connected domain whose boundary contains more than one point there exist univalent analytic functions which conformally map this domain onto a disc or a half-plane (cf. Conformal mapping; Univalent function). The real and imaginary parts of a function which is holomorphic in a domain satisfy the Laplace equation in that domain: i.e. they are harmonic functions (cf. Harmonic function). Two harmonic functions which are connected by the Cauchy–Riemann equations are called conjugate. In a simply-connected domain any harmonic function has a conjugate function and is thus the real part of some holomorphic function in . The connections with conformal mappings and harmonic functions form the basis of many applications of the theory of analytic functions. A function ( being an arbitrary set), is called analytic at a point if there exists a neighbourhood of this point such that may be represented by a convergent power series on the intersection of this neighbourhood with . The function is called analytic on the set if it is analytic on some open set which contains (or, more exactly, if there exist both an open set containing and an analytic function on this set which coincides with on ). For open sets the notion to analyticity coincides with the notion of differentiability with respect to the set. However, this is not the case in general; in particular, on the real line there exist functions which not only have a derivative, but which are infinitely differentiable at every point and are not analytic even at a single point of this line. The property of connectedness of the set is necessary in order that the uniqueness theorem for analytic functions holds. This is why analytic functions are usually considered in domains, i.e. on connected open sets. All the preceding refers to single-valued analytic functions , considered in a given domain (or on a given set ) of the complex plane. In considering the possible extension of a function , as an analytic function, to a larger domain, one arrives at the concept of the analytic function considered as a whole, i.e. throughout its whole natural domain of existence. If the function is thus extended, its domain of analyticity becomes larger, and may overlap itself, supplying new values of the function at points in the plane where it already was defined. Accordingly, an analytic function considered as a whole is generally multi-valued. Many problems in analysis (inversion of a function, the determination of a primitive and the construction of an analytic function with a given real part in multiply-connected domains (cf. Multiply-connected domain), the solution of algebraic equations with analytic coefficients, etc.) require the study of multi-valued functions; such functions include , , , , algebraic functions, etc. (cf. Algebraic function). A regular process which yields the complete analytic function, considered throughout its natural domain of existence, was proposed by Weierstrass; it is known as Weierstrass' method of analytic continuation. The initial concept is that of an element of an analytic function, viz. a power series with a non-zero radius of convergence. Such an element : defines a certain analytic function on its disc of convergence . Let be a point of different from . Expanding in a series with centre at , one obtains a new element : whose disc of convergence will be denoted by . In the intersection of and the series converges to the same function as the series . If extends beyond the boundary of , the series defines the function determined by on some set outside (where the series is divergent). In such a case the element is called a direct analytic continuation of the element . Let be a chain of elements in which is a direct analytic continuation of (); the element is then said to be an analytic continuation of the element (by means of the given chain of elements). When the centre of the disc belongs to it may happen that the element is not a direct analytic continuation of the element . In such a case the sums of the series and will have different values in the intersection of and ; thus analytic continuation may lead to new values of the function inside . The totality of all elements which may be obtained by analytic continuation of an element forms the complete analytic function (in the sense of Weierstrass) generated by ; the union of their discs of convergence represents the (Weierstrass) domain of existence of this function. It follows from the uniqueness theorem for analytic functions that an analytic function in the sense of Weierstrass is completely determined by the given element . The initial element may be any other element belonging to this function; the complete analytic function will not be affected. A complete analytic function , considered as a function of the points in the plane belonging to its domain of existence , is generally multi-valued. In order to eliminate this feature, is considered not as a function of the points in the plane domain , but rather as a function of the points on some multi-sheeted surface (lying above ) such that to each point of there correspond as many points of the surface (projecting onto the given point of ) as there are different elements with centre at this point in the complete analytic function ; on the surface the function becomes single-valued. The idea of passing to such surfaces is due to Riemann, and the surfaces themselves are known as Riemann surfaces. The abstract definition of the notion of a Riemann surface has made it possible to replace the theory of multi-valued analytic functions by the theory of single-valued analytic functions on Riemann surfaces (cf. Riemann surface). Now fix a domain belonging to the domain of existence of the complete analytic function , and fix some element of with centre at a point in . The totality of all elements which may be obtained by analytic continuation of by means of chains with centres belonging to is called a branch of the analytic function . A branch of a multi-valued analytic function may turn out to be a single-valued analytic function in the domain . Thus, arbitrary branches of the functions and which correspond to an arbitrary simply-connected domain not containing the point 0, are single-valued functions. The function has exactly different branches in such a domain, while has an infinite set of such branches. The selection of single-valued branches (using some cuts in the domain of existence) and their study by the theory of single-valued analytic functions constitute one of the principal methods of studying specific multi-valued analytic functions. A.A. Gonchar ## Analytic functions of several complex variables. The complex space , consisting of the points , , is a vector space over the field of complex numbers with the Euclidean metric It differs from the -dimensional Euclidean space by a certain asymmetry: on passing from to (i.e. on introducing a complex structure in ), the coordinates are subdivided into pairs which appear in the complex combinations . If a complex function is defined in a domain and is differentiable at all points in the sense of (i.e. as a function of the real variables and ), its differential may be represented in the form Here while the symbols and are defined as in the case of the plane. If is of the form i.e. is a complex linear function in , the function is said to differentiable in the sense of or holomorphic or analytic in the domain . Thus, the condition of holomorphy of in a domain consists of the condition that it be differentiable in the sense of and that it satisfy the system of complex equations (), which is equivalent to the system of first-order partial differential equations (Cauchy–Riemann system). In the case of space () as distinct from that of the plane () this system is overdetermined, since the number of equations is larger than that of the unknown functions. It remains overdetermined on passing to the (geometrically more natural) spatial analogue of a holomorphic function of one complex variable — a holomorphic mapping , which is realized by a system of functions which are holomorphic in a domain . The mapping is called biholomorphic if it is one-to-one and if it is holomorphic together with its inverse . The conditions for holomorphy of a mapping are expressed by a system of real equations involving real functions. The overdeterminacy of the conditions of holomorphy for is the cause of a number of effects typical of the spatial case — such as the absence of a spatial analogue of the Riemann theorem on the existence of conformal mappings. According to Riemann's theorem, if , any two simply-connected domains whose boundaries do not reduce to a single point are isomorphic. However, if , even such simple simply-connected domains as the ball and the product of discs (polydisc) are non-isomorphic. The non-isomorphism is brought to light on comparing the groups of automorphisms of these domains (i.e. their biholomorphic mappings onto themselves, cf. Biholomorphic mapping) — the groups prove to be algebraically non-isomorphic, whereas a biholomorphic mapping of one domain onto another, if it existed, would establish an isomorphism of these groups. Owing to this, the theory of biholomorphic mappings of domains in complex space is substantially different from the theory of conformal mappings in the plane. A function is called holomorphic at a point if it is holomorphic in some neighbourhood of this point. According to the Cauchy–Riemann criterion, a function of several variables which is holomorphic at a point is holomorphic with respect to each variable (if the values of the other variables are fixed). The converse proposition is also true: If, in a neighbourhood of some point, a function is holomorphic with respect to each variable separately, then it is holomorphic at this point (Hartogs' fundamental theorem). In analogy with the case of the plane the holomorphy of a function at a point is equivalent to its expandability in a multiple power series in a neighbourhood of this point or, in abbreviated notation, where is a multi-index of integers , and A holomorphic function is infinitely differentiable, and the above series is its Taylor series, i.e. the derivatives being taken at the point . The fundamental facts of the theory of holomorphic functions of one variable extend to holomorphic functions of several variables, sometimes in an altered form. An instance of this is the Weierstrass preparation theorem (cf. Weierstrass theorem), which extends the property of holomorphic functions of one variable to become zero as an integral power of to the spatial case. The theorem is formulated as follows: If a function which is holomorphic at a point is equal to zero at that point, then it may be represented, in a certain neighbourhood (possibly after a non-degenerate linear transformation of the independent variables) in the form where is an integer, are functions of which are holomorphic in a neighbourhood of the point (a "prime" preceding a letter denotes the projection on the space of the first coordinates) and which are equal to zero at , while is holomorphic and zero-free in . This theorem is of fundamental importance in the study of analytic sets (cf. Analytic set), which are described locally, in a neighbourhood of each one of their points, as sets of common zeros of a certain number of functions which are holomorphic at this point. By Weierstrass' preparation theorem such sets may be locally described as sets of common zeros of polynomials in the variable , with coefficients from the ring of holomorphic functions in the other variables . This circumstance permits extensive use of algebraic methods in the local study of analytic sets. Cauchy's integral theorem must also be slightly modified in the spatial case, and is then known as the Cauchy–Poincaré theorem: Let a function be holomorphic in a domain ; then, for any -dimensional surface compactly imbedded in , with piecewise-smooth boundary , As in the planar case, this integral is defined by a parametric representation of the given set: If has the equation , where the parameter varies over an -dimensional cell , then, by definition, The difference between the spatial and the planar cases consists in the fact that in the former case the dimension of the surface is less than that of the domain , while in the planar case the dimensions are the same . The spatial analogue of the Cauchy integral formula can be written in a particularly simple form for polycylinder domains, i.e. for products of plane domains. Let be a domain in which is a domain in the complex plane with a piecewise-smooth boundary (), while the function is holomorphic in a domain which compactly contains . Successive application of Cauchy's formula for one variable then yields, for any point , where is an -dimensional surface in the boundary , , and However, polycylinder domains are only a very special class, and in general domains such a separation of variables is not possible. The role of Cauchy's integral for arbitrary domains with a piecewise-smooth boundary is played by the Martinelli–Bochner integral formula: For any function which is holomorphic in a domain containing , and for any point , where , and This is Green's formula for a pair of functions, one of which is holomorphic in , while the other is a fundamental solution of the Laplace equation in the space with singular point . If , this is the ordinary Cauchy integral. If , the formula differs from Cauchy's multiple integral for a product of plane domains in that, first, the integration is not over an -dimensional part of the boundary, but over the whole -dimensional boundary of the domain, and, secondly, its kernel (the factor multiplying under the integral sign) does not depend analytically on the parameter . An analytic kernel, however, is essential in a number of problems, and it is therefore desirable to construct an integral formula with such a kernel for as large a class of domains as possible. An ample supply of integral formulas, including formulas with an analytic kernel for many domains, is contained in the general Leray formula. This formula is where is a smooth vector function depending also on , and are defined above, and ; it is assumed that for any fixed and running over . The value of the integral in this formula does not depend on the choice of the vector function (provided that for all , ), and if , this integral coincides with the Martinelli–Bochner integral. By varying the choice of for different classes of domains, the Leray formula will yield various integral formulas. In the theory of analytic functions of several variables other integral representations, which are valid only for certain classes of domains, are also considered. An important class of this kind consists of the so-called Weil domains, which are a generalization of the product of plane domains. For such domains one has the Bergman–Weil representation with a kernel which also depends analytically on the parameter. As in the planar case, the study of the singularities of analytic functions is of fundamental interest; the main difference between the planar and the spatial cases is expressed by the Osgood–Brown theorem on the removability of compact singularities, according to which any function which is holomorphic in , where is a domain in () and is compact subset of which does not subdivide , extends holomorphically to the whole domain . By this theorem, holomorphic functions of several variables cannot have isolated singular points. These are replaced in () by singular sets which are analytic if their dimension is lower than . This fact is essential in the theory of multi-dimensional residues. This theory deals with the problem of computing the integral of a function , which is holomorphic everywhere in a domain except for an analytic set , over a closed -dimensional surface not intersecting . Since the dimension of the singular set is lower than the dimension of by at least two, does not subdivide . If the surface is not linked with , i.e. bounds an -dimensional surface compactly belonging to , then by the Cauchy–Poincaré theorem . In order to calculate this integral in the general case, it is necessary to clarify how is linked with the singular set , and to calculate the integrals over special -dimensional surfaces associated with separate portions of the set (residues). The solution of this problem involves considerable topological and analytic difficulties. These may often be overcome by the methods proposed by E. Martinelli and J. Leray. The Martinelli method is based on the use of the topological Aleksander–Pontryagin duality principle, and reduces the study of the -dimensional homologies of the set to the study of the -dimensional homologies of the singular set . The Leray method is more general: it is based on the examination of special homology classes and on the calculation of certain differential forms (residue forms). The multi-dimensional theory of residues has also found applications in theoretical physics (cf. Feynman integral). The Osgood–Brown theorem reveals an important fundamental difference between the spatial and the planar theories. In the plane one can, for any domain , construct a function which is holomorphic in but which cannot be extended analytically beyond its boundary, i.e. is the natural domain of existence. In space the situation is different: thus, the spherical shell cannot be the domain of existence of any holomorphic function, since by the Osgood–Brown theorem, any function which is holomorphic in it will certainly extend analytically to the entire ball . Thus arises the problem of the characterization of the natural domains of existence of holomorphic functions — the so-called domains of holomorphy. A simple sufficient condition may be formulated with the aid of the concept of a barrier at a boundary point of the domain, i.e. a function which is holomorphic in this domain and which increases without limit as tends to . will be a domain of holomorphy if it is possible to construct a barrier for an everywhere-dense set of points in its boundary. This condition is satisfied, in particular, by any convex domain; for any point it is sufficient to select, in the -dimensional supporting plane to at the point , a -dimensional plane of the form the function will then be a barrier. Consequently, every convex domain in is a domain of holomorphy. However, convexity is not a necessary condition for holomorphy: A product of plane domains is always a domain of holomorphy, and such a product need not be convex. Nevertheless, if the notion of convexity is suitably generalized, it is possible to arrive at a necessary and sufficient condition. One such generalization is based on the observation that the convex hull of a set may be described as the set of points at which the value of any linear function does not exceed the supremum of the values of this function on . By analogy, the holomorphically convex hull of a set is defined by where denotes the set of all functions which are holomorphic in . A domain is called holomorphically convex if, for every compact subset of , the hull also is a compact subset of . Holomorphic convexity is a necessary and sufficient condition for a domain of holomorphy. However, this criterion is not very effective, since holomorphic convexity is difficult to verify. Another generalization is connected with the notion of a plurisubharmonic function, which is the complex analogue of a convex function. A convex function in a domain of may be defined as a function for which the restrictions to the segments in of the straight lines (where and is a real parameter) are convex functions of . A real function , defined and upper semi-continuous in a domain , is said to be plurisubharmonic in if for each complex line (, is a complex parameter), its restriction to the parts of this line in is a subharmonic function of . If is twice continuously differentiable, then the condition of plurisubharmonicity, in accordance with the rules of differentiation of composite functions, is that the Hermitian form — the so-called Levi form — be non-negative. A domain is called pseudo-convex if the function , where denotes the Euclidean distance from the point to the boundary , is plurisubharmonic in this domain. Pseudo-convexity is also a necessary and sufficient condition for a domain to be a domain of holomorphy. In some cases it is possible to verify effectively the pseudo-convexity of a domain. As regards domains which are not domains of holomorphy, there arises the problem of describing their envelope of holomorphy, i.e. the smallest domain of holomorphy to which any function holomorphic in extends analytically. For domains of the simplest types envelopes of holomorphy can be effectively constructed, but in the general case the problem is unsolvable within the class of single-sheeted domains. Under analytic continuation of functions beyond the boundary of a given domain , multi-valuedness may result, which can be avoided by introducing multi-sheeted covering domains over , analogous to Riemann surfaces (cf. Riemann surface). In the class of covering domains, the problem of constructing envelopes of holomorphy is always solvable. This problem also has applications in theoretical physics, to wit in quantum field theory. The transition from the plane to a complex space substantially increases the variety of geometrical problems related to holomorphic functions. In particular, such functions are naturally considered not only in domains, but also on complex manifolds — smooth manifolds of even real dimension, the neighbourhood relations of which are biholomorphic. Among these, Stein manifolds (cf. Stein manifold) — natural generalizations of domains of holomorphy — play a special role Several problems in analysis may be reduced to the problem of constructing, in a given domain, a holomorphic function with given zeros or a meromorphic function with given poles and principal parts of the Laurent series. In the plane case, these problems have been solved for arbitrary domains by the theorems of Weierstrass and Mittag-Leffler and their generalizations. The spatial case is different — the solvability of the corresponding problems, the so-called Cousin problems, depends on certain topological and analytic properties of the complex manifolds considered. The key step in the solution of the Cousin problems is to construct — starting from locally-defined functions with given properties — a global function, defined on the whole manifold under consideration and having the same local properties. Such kinds of constructions are very conveniently effected using the theory of sheaves, which arose from the algebraic-topological treatment of the concept of an analytic function, and which has found important applications in various branches of mathematics. The solution of the Cousin problems by methods of the theory of sheaves was realized by H. Cartan and J.-P. Serre. B.V. Shabat ## The contemporary theory of analytic functions and their generalizations. This is one of the most important branches of analysis, it is closely connected with quite diverse branches of mathematics and it has numerous applications in theoretical physics, mechanics and technology. Fundamental investigations on the theory of analytic functions have been carried out by Soviet mathematicians. Extensive interest in the theory of functions of a complex variable emerged in the Soviet Union at the beginning of the 20th century. This was in connection with noteworthy investigations by Soviet scientists on applications of the theory of analytic functions to various problems in the mechanics of continuous media. N.E. Zhukovskii and S.A. Chaplygin solved very important problems in hydrodynamics and aerodynamics by using methods of the theory of analytic functions. In the works of G.V. Kolosov and N.I. Muskhelishvili these methods were applied to fundamental problems in the theory of elasticity. In subsequent years the theory of functions of a complex variable underwent extensive development. The development of various aspects of the theory of analytic functions was determined by the fundamental research of, among others, V.V. Golubev, N.N. Luzin, I.I. Privalov and V.I. Smirnov (boundary properties), M.A. Lavrent'ev (geometric theory, quasi-conformal mappings and their applications to gas dynamics), M.V. Keldysh, M.A. Lavrent'ev and L.I. Sedov (applications to problems in the mechanics of continuous media), D.E. Men'shov (theory of monogeneity), M.V. Keldysh, M.A. Lavrent'ev and S.N. Mergelyan (approximation theory), I.N. Vekua (theory of generalized analytic functions and their applications), A.O. Gel'fond (theory of interpolation), N.N. Bogolyubov and V.S. Vladimirov (theory of analytic functions of several variables and its application to quantum field theory). The development of the theory of analytic functions of one and several complex variables and their generalizations is continuing. See Boundary properties of analytic functions; Quasi-conformal mapping; Boundary value problems of analytic function theory; Approximation of functions of a complex variable. #### References [1] I.I. [I.I. Privalov] Priwalow, "Einführung in die Funktionentheorie" , 1–3 , Teubner (1958–1959) (Translated from Russian) MR0342680 MR0264037 MR0264036 MR0264038 MR0123686 MR0123685 MR0098843 Zbl 0177.33401 Zbl 0141.26003 Zbl 0141.26002 Zbl 0082.28802 [2] V.I. Smirnov, "A course of higher mathematics" , III-2 , Addison-Wesley (1964) (Translated from Russian) MR0182690 MR0182688 MR0182687 MR0177069 MR0168707 Zbl 0122.29703 Zbl 0121.25904 Zbl 0118.28402 Zbl 0117.03404 [3] A.I. Markushevich, "Theory of functions of a complex variable" , 1–3 , Chelsea (1977) (Translated from Russian) MR0444912 Zbl 0357.30002 [4] M.A. Lavrent'ev, B.V. Shabat, "Methoden der komplexen Funktionentheorie" , Deutsch. Verlag Wissenschaft. (1967) (Translated from Russian) [5] G.M. Goluzin, "Geometric theory of functions of a complex variable" , Transl. Math. Monogr. , 26 , Amer. Math. Soc. (1969) (Translated from Russian) MR0247039 Zbl 0183.07502 [6] M.A. Evgrafov, "Analytic functions" , Saunders , Philadelphia (1966) (Translated from Russian) MR0197686 Zbl 0147.32605 [7] A.G. Sveshnikov, A.N. Tikhonov, "The theory of functions of a complex variables" , Moscow (1967) (In Russian) MR0722295 MR0349962 MR0349961 [8] B.A. Fuks, "Theory of analytic functions of several complex variables" , 1–2 , Amer. Math. Soc. (1963–1965) (Translated from Russian) MR0188477 MR0174786 MR0168793 MR0155003 MR0037915 MR0027069 Zbl 0146.30802 Zbl 0138.30902 Zbl 0040.19002 [9] V.S. Vladimirov, "Methods of the theory of functions of several complex variables" , M.I.T. (1966) (Translated from Russian) [10] A.I. Markushevich, "Notes on the history of the theory of analytic functions" , Moscow-Leningrad (1951) (In Russian) [11] , Mathematics in the USSR during thirty years: 1917–1947 , Moscow-Leningrad (1948) pp. 319–414 (In Russian) [12] , Mathematics in the USSR during 40 years: 1917–1957 , 1 , Moscow (1959) pp. 381–510 (In Russian) [13] B.V. Shabat, "Introduction of complex analysis" , 1–2 , Moscow (1976) (In Russian) Zbl 0799.32001 Zbl 0732.32001 Zbl 0732.30001 Zbl 0578.32001 Zbl 0574.30001 [14] A.V. Bitsadze, "Fundamentals of the theory of analytic functions of a complex variable" , Moscow (1972) (In Russian) A.A. GoncharB.V. Shabat #### Comments Complex analysis, the theory of analytic functions, has been an active field of research up to the present time also in the West. Some of the history may be found in the extensive article in the Encyclopaedia Britannica entitled Analysis, complex. There is a large number of textbooks on complex analysis besides those listed above. Some good current books are listed below: [a1]–[a10] for the case of one variable, [a11]–[a18] for the case of several variables. Some final comments on complex analysis of several variables are in order. The "Osgood–Brown theorem on the removability of compact singularities" is usually called the Hartogs extension theorem in the West. The modern proof, due to L. Ehrenpreis, makes use of the inhomogeneous Cauchy–Riemann or -equations. The so-called -method, extensively developed by J.J. Kohn and L. Hörmander, has become one of the three most powerful techniques available in complex analysis today. Among other things, it can be used to obtain solutions of the Levi problem (are domains of holomorphy the same as pseudo-convex domains?) and the Cousin problems, cf. [a15]. These fundamental problems had been settled previously by methods now belonging to sheaf theory, and which go back to K. Oka (1936–1954), and have culminated in the sheaf-cohomology theory of H. Cartan, J.-P. Serre, H. Grauert and others, cf. the elegant treatment in [a12]. The third and most recent technique consists in the use of suitable integral representations as developed by G.M. Henkin [G.M. Khenkin] and E. Ramirez de Arellano, cf. [a14]. #### References [a1] L.V. Ahlfors, "Complex analysis" , McGraw-Hill (1966) MR0188405 Zbl 0154.31904 [a2] J.B. Conway, "Functions of one complex variable" , Springer (1973) MR0447532 Zbl 0277.30001 [a3] A. Dinghas, "Vorlesungen über Funktionentheorie" , Springer (1961) MR0179329 Zbl 0102.29301 [a4] P. Henrici, "Applied and computational complex analysis" , 1–3 , Wiley (1974–1986) MR1164865 MR1541308 MR1008928 MR0822470 MR0453984 MR0372162 Zbl 1107.30300 Zbl 0925.30003 Zbl 0635.30001 Zbl 0578.30001 Zbl 0363.30001 Zbl 0313.30001 [a5] E. Hille, "Analytic function theory" , 1–2 , Chelsea, reprint (1974) MR1532177 MR0201608 MR1530816 MR0107692 Zbl 0273.30002 Zbl 0102.29401 Zbl 0088.05204 [a6] Z. Nehari, "Conformal mapping" , Dover, reprint (1975) pp. 2 MR0377031 Zbl 0071.07301 Zbl 0052.08201 Zbl 0048.31503 Zbl 0041.41201 [a7] R. Nevanilinna, "Analytic functions" , Springer (1970) (Translated from German) [a8] R. Nevanlinna, V. Paatero, "Introduction to complex analysis" , Addison-Wesley (1969) (Translated from German) MR0239056 Zbl 0169.09001 [a9] W. Rudin, "Real and complex analysis" , McGraw-Hill (1974) pp. 24 MR0344043 Zbl 0278.26001 [a10] E.C. Titchmarsh, "The theory of functions" , Oxford Univ. Press (1939) MR0593142 MR0197687 MR1523319 Zbl 65.0302.01 [a11] H. Grauert, K. Fritzsche, "Several complex variables" , Springer (1976) (Translated from German) MR0414912 Zbl 0381.32001 [a12] H. Grauert, R. Remmert, "Theory of Stein spaces" , Springer (1979) (Translated from German) MR0580152 Zbl 0433.32007 [a13] R.C. Gunning, H. Rossi, "Analytic functions of several complex variables" , Prentice-Hall (1965) MR0180696 Zbl 0141.08601 [a14] G.M. [G.M. Khenkin] Henkin, J. Leiterer, "Theory of functions on complex manifolds" , Birkhäuser (1984) MR0795028 MR0774049 [a15] L. Hörmander, "An introduction to complex analysis in several variables" , North-Holland (1973) MR0344507 Zbl 0271.32001 [a16] S.G. Krantz, "Function theory of several complex variables" , Wiley (Interscience) (1982) MR0635928 Zbl 0471.32008 [a17] R. Narasimhan, "Several complex variables" , Univ. Chicago Press (1971) MR0342725 Zbl 0223.32001 [a18] W. Rudin, "Function theory in the unit ball in " , Springer (1980) MR601594 Zbl 0495.32001 [a19] R.M. Range, "Holomorphic functions and integral representation in several complex variables" , Springer (1986) MR0847923 How to Cite This Entry: Analytic function. Encyclopedia of Mathematics. URL: http://www.encyclopediaofmath.org/index.php?title=Analytic_function&oldid=29387 This article was adapted from an original article by A.A. Gonchar, B.V. Shabat (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article	{"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": 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": 0.9175985455513, "perplexity_flag": "head"}
http://mathhelpforum.com/calculus/64839-solved-triple-integral-issue.html	# Thread: 1. ## [SOLVED] triple integral issue I need some help with a triple integral problem that I keep getting wrong in the last step somehow. the problem is: Evaluate $\int\int\int$ on E xyz dV where E lies between the spheres $\rho$=2 and $\rho$=4 and above the cone $\phi$= $\pi$/3. There work I've done so far is set up the integral integrated it and got: 1/4 sin^4 $\phi$ from 0 to $\pi$ * 1/2sin^2 $\theta$ from 0 to 2 $\pi$ * 1/6 $\rho$^6 from 2 to 4. I get that the middle term should be zero, but for some reason my prof gives the answer to be (1562 $\pi$)/15 2. That is a large value your professor got. I am not so sure he/she is correct. 3. That's what I was thinking as I am pretty sure the answer should be zero, but I thought that it wouldn't hurt to check. 4. Consider symmetry. The region of integration is rotationally symmetric about the z axis, and so if we rotate the function being integrated about the z axis, the integral should be unchanged. Now, let us rotate the integrand by 90 degrees about the z axis: we transform x and y to x' and y' via x'=-y and y'=x. Then the integrand is f(x',y',z)=xyz=(y')(-x')z=-x'y'z. Thus, we have $\iiint_{E}xyz\,dx\,dy\,dz=\iiint_{E'}-x'y'z\,dx'\,dy'\,dz$. But, if we rename the variables x' and y' as x and y, (and as E'=E), we have $\iiint_{E}xyz\,dx\,dy\,dz=-\iiint_{E}xyz\,dx\,dy\,dz$ which means $\iiint_{E}xyz\,dx\,dy\,dz=0$. Thus, due to the symmetry, the integral is zero. [We can also look at this as the fact that xyz's dependence on the azimuthal angle $\theta$ is of the form $\sin(2\theta)$, and the region has no dependence on the azimuthal angle (the rotational symmetry), so the integral over $\theta$ will be 0] --Kevin C. 5. Thanks everyone. I got an e-mail back from my professor and he said that it was a typo. Thanks again!	{"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": 17, "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": 0.9471530914306641, "perplexity_flag": "head"}
http://crypto.stackexchange.com/questions/6417/why-do-we-assume-un-security-of-communication-channel-on-every-cryptography-syst	# Why do we assume un-security of communication channel on every cryptography system While reading about a few cryptographic systems, I noticed that we always assume the communication channel is not secured. Why is this assumption made? And, why the effort is being put into designing cryptographic systems rather than working on making these communication channels more secure? I know the question is basic, but really cannot find a complete answer. - 3 I suppose because you need cryptography to make those channels secure. And you can't just create a mega-channel and say "this is secure, use this for encrypted communication", because a "secure channel" requires authentication of both parties, which is inherently an individual process. – Thomas Feb 21 at 2:56 Additionally, if we can find ways of solving problems without having to rely on a secure channel, that's one fewer point of failure. What happens when our "secure" channel turns out not to be? – Stephen Touset Feb 21 at 3:54 Mathematics in general and cryptology in particular is about getting strong conclusions from weak assumptions. Public key cryptography for example can convert an insecure channel from A to B into a secret one assuming there was an authentic channel from B to A before. – jug Feb 21 at 8:43 ## 3 Answers The goal of cryptography is to create these secure communication channels. However, keep in mind that a secure channel is more than just an encrypted channel. A secure channel should be able to provide confidentiality (using encryption), data integrity (using something like signatures), and data authentication (using something like certificates, MACs). The goal of cryptography is to create these properties to make a "secure channel" out of an unsecured channel. Note that some cryptosystems build on top of other types of channels. For instance, in public-key cryptography, you generally need to assume that you have access to a authenticated channel for key exchanges (which is a stronger assumption than a completely unsecured channel). - For public key cryptography you need at least a trusted certificate, and this has to be secured through a secure channel. In this case the secured channel could however be e.g. a browser installation. After that you don't need secure channels anymore. – owlstead Feb 21 at 22:56 The point of the trusted certificate is to ensure that you are talking to who you think you're talking to. In other words, the point of this CA is to create an authentication channel to exchange keys. After the keys are exchange, the rest of the communication can occur over the fully unsecured channel. – Oleksi Feb 22 at 0:31 +1 although the second paragraph is still a bit vague to me - which is understandable given the question – owlstead Feb 22 at 13:09 If you already had a secure channel, you wouldn't need any cryptography. So if you're reading about cryptographic systems, you're reading about how to make things secure. Now, if you want to make a secure communications channel, the first thing you need is a communications channel. - 1 There would still be digital signatures, multi-party computation, and string commitment. $\hspace{0.95 in}$ – Ricky Demer Feb 21 at 5:02 1 Don't forget fair coin tosses. – Stephen Touset Feb 21 at 18:12 Any time you transmit information you have have to send it through some kind of medium ( air, wire, optical fiber, ect.). The medium can be tapped into at any point by an adversary and is thus considered insecure. -	{"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": 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": 0.9486883282661438, "perplexity_flag": "middle"}
http://unapologetic.wordpress.com/2009/08/10/the-complex-spectral-theorem/	# The Unapologetic Mathematician ## The Complex Spectral Theorem We’re now ready to characterize those transformations on complex vector spaces which have a diagonal matrix with respect to some orthonormal basis. First of all, such a transformation must be normal. If we have a diagonal matrix we can find the matrix of the adjoint by taking its conjugate transpose, and this will again be diagonal. Since any two diagonal matrices commute, the transformation must commute with its adjoint, and is therefore normal. On the other hand, let’s start with a normal transformation $N$ and see what happens as we try to diagonalize it. First, since we’re working over $\mathbb{C}$ here, we can pick an orthonormal basis that gives us an upper-triangular matrix and call the basis $\left\{e_i\right\}_{i=1}^n$. Now, I assert that this matrix already is diagonal when $N$ is normal. Let’s write out the matrices for $N$ $\displaystyle\begin{pmatrix}a_{1,1}&\cdots&a_{1,n}\\&\ddots&\vdots\\{0}&&a_{n,n}\end{pmatrix}$ and $N^$ $\displaystyle\begin{pmatrix}\overline{a_{1,1}}&&0\\\vdots&\ddots&\\\overline{a_{1,n}}&\cdots&\overline{a_{n,n}}\end{pmatrix}$ Now we can see that $N(e_1)=a_{1,1}e_1$, while $N^(e_1)=\overline{a_{1,1}}e_1+\dots+\overline{a_{1,n}}e_n$. Since these bases are orthonormal, it’s easy to calculate the squared-lengths of these two: $\displaystyle\begin{aligned}\lVert N(e_1)\rVert^2&=\lvert a_{1,1}\rvert^2\\\lVert N^(e_1)\rVert^2&=\lvert a_{1,1}\rvert^2+\dots+\lvert a_{1,n}\rvert^2\end{aligned}$ But since $N$ is normal, these two must be the same. And so all the entries other than maybe $a_{1,1}$ in the first row of our matrix must be zero. We can then repeat this reasoning with the basis vector $e_2$, and reach a similar conclusion about the second row, and so on until we see that all the entries above the diagonal must be zero. That is, not only is it necessary that a transformation be normal in order to diagonalize it, it’s also sufficient. Any normal transformation on a complex vector space has an orthonormal basis of eigenvectors. Now if we have an arbitrary orthonormal basis — say $N$ is a transformation on $\mathbb{C}^n$ with the standard basis already floating around — we may want to work with the matrix of $N$ with respect to this basis. If this were our basis of eigenvectors, $N$ would have the diagonal matrix $\Lambda=\Bigl(\lambda_i\delta_{ij}\Bigr)$. But we may not be so lucky. Still, we can perform a change of basis using the basis of eigenvectors to fill in the columns of the change-of-basis matrix. And since we’re going from one orthonormal basis to another, this will be unitary! Thus a normal transformation is not only equivalent to a diagonal transformation, it is unitarily equivalent. That is, the matrix of any normal transformation can be written as $U\Lambda U^{-1}$ for a diagonal matrix $\Lambda$ and a unitary matrix $U$. And any matrix which is unitarily equivalent to a diagonal matrix is normal. That is, if you take the subspace of diagonal matrices within the space of all matrices, then use the unitary group to act by conjugation on this subspace, the result is the subspace of all normal matrices, which represent normal transformations. Often, you’ll see this written as $U\Lambda U^$, which is really the same thing of course, but there’s an interesting semantic difference. Writing it using the inverse is a similarity, which is our notion of equivalence for transformations. So if we’re thinking of our matrix as acting on a vector space, this is the “right way” to think of the spectral theorem. On the other hand, using the conjugate transpose is a congruence, which is our notion of equivalence for bilinear forms. So if we’re thinking of our matrix as representing a bilinear form, this is the “right way” to think of the spectral theorem. But of course since we’re using unitary transformations here, it doesn’t matter! Unitary equivalence of endomorphisms and of bilinear forms is exactly the same thing. ### Like this: Posted by John Armstrong \| Algebra, Linear Algebra ## 9 Comments » 1. Should your first sentence end “with respect to some orthonormal basis”? Comment by Å \| August 11, 2009 \| Reply 2. Yeah, probably… Comment by \| August 11, 2009 \| Reply 3. I’m really REALLY enjoying these. I’m going nuts over it being 2 weeks until I theoretically am teaching full-time High School Math someplace in Los Angeles County. But no interviews, as school districts are paralyzed by the legislature having rejected several Billion dollars (yes, \$10^9) of Federal Funds because they don’t like the strings attached. Really. The legislators are lobbied by the teachers unions, who would rather their members be laid off than accept money tainted with the willingness to allow ANY use of student test score in evaluating teacher merit. Tactically, it’s excruciating for me to do on-line applications with district-by-district tweaks of a State-based “EdJoin” online system, that requires endless attachments to everything, but won’t take .doc files or TFF files, so that I physically had to scan 50 pages of hardcopy to blurry GIF files and attach them. It’s easy of newly graduated teachers, but I have 1/3 century of teaching and professorship and research and letters of recommendation to drag around, and they don’t imagine that this makes length-limit problems for online applications. Great. Chase away the over-qualified such as me and thee, and then complain that it’s hard to find good Math profs and secondary school teachers… Comment by \| August 11, 2009 \| Reply 4. It’s intruiging how all the anglo societies are setting themselves up ASAP for their own Century of Humiliation – I guess the only bright spot is that with the speed at which things move these days, it will only last a few decades Comment by Avery Andrews \| August 12, 2009 \| Reply 5. [...] the last couple lemmas we’ve proven and throw them together to prove the real analogue of the complex spectral theorem. We start with a self-adjoint transformation on a finite-dimensional real inner-product space [...] Pingback by \| August 14, 2009 \| Reply 6. [...] Singular Value Decomposition Now the real and complex spectral theorems give nice decompositions of self-adjoint and normal transformations, [...] Pingback by \| August 17, 2009 \| Reply 7. [...] Here’s a neat thing we can do with the spectral theorems: we can take square roots of positive-semidefinite transformations. And this makes sense, [...] Pingback by \| August 20, 2009 \| Reply 8. [...] we work with a normal transformation on a complex inner product space we can pick orthonormal basis of eigenvectors. That is, we can find a unitary transformation and a diagonal transformation so that [...] Pingback by \| August 24, 2009 \| Reply 9. [...] We're now ready to characterize those transformations on complex vector spaces which have a diagonal matrix with respect to some orthonormal basis. First of all, such a transformation must be normal. If we have a diagonal matrix we can find the matrix of the adjoint by taking its conjugate transpose, and this will again be diagonal. Since any two diagonal matrices commute, the transformation must commute with its adjoint, and is therefore normal. … Read More [...] Pingback by \| November 18, 2010 \| Reply « Previous \| Next » ## About this weblog This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”). I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 23, "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": 0.9120751023292542, "perplexity_flag": "head"}
http://mathoverflow.net/questions/120121?sort=newest	Mean value theorems for the Haar integral? Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Let $G$ be a compact topological group (feel free to add hypotheses if necessary). Is there any mean value theorem for its (normalized to 1) Haar integral? In general, are there mean value theorems for abstract spaces with measures? (Or at least for Borel measures?) Later edit: After reading the first two comments, let me be more precise; I'm looking for a theorem giving something like: for any continuous $f$ on $G$, there exist $x \in G$ such that $\int_G f(g) \mathrm{d}g = f(x)$. Does such an $x$ really exist? Can anything else be said about it (the integral being so special, maybe this $x$ can be made more precise)? - What is your substitute for the notion of an interval when you move outside the setting of R? Already for domains in R^2 one has to be a bit careful in finding a correct generalization of R^2? What notions have you tried thus far? – Yemon Choi Jan 28 at 16:20 Maybe it also helps if you indicate what do you need such a result for. – András Bátkai Jan 28 at 16:45 1 Answer Say $\mu$ is a Borel probability measure on a connected set $A$ in a topological space. Let $f : A \to \mathbb R$ be continuous. Then the mean value $\int_A f\;d\mu$ is equal to $f(a)$ for some $a \in A$. Proof: the mean value is between the sup of all values and the inf of all values, so (by connectedness) it is a value of the function. Of course the desired result fails for non-connected sets. Even in the two-point group we get a counterexample. - Excellent answer, thank you. But please note that your proof works in general, while I require more: if there is an underlying group structure and the integral is translation-invariant, do I get any extra information about the points where the mean value is attained? One minor remark: one must have $\mu (A)<\infty$ in order for the above proof to work. – Alex M Jan 28 at 20:40 One seems to need $\mu(A)=1$ in fact! Either that or Gerald forgot to divide by $\mu(A)$; both are trivial ways to fix things. – wccanard Jan 28 at 22:10 3 $\mu(A)=1$ is implied: $\mu$ was a probability measure on $A$. – Vince Jan 29 at 1:32 How would one extend the proof presented by Gerald Edgar above to the case of complex-valued functions? Separating into the real and imaginary part is a wrong approach, since this would give two different points. – Alex M Feb 7 at 17:16	{"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": 19, "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": 0.9243728518486023, "perplexity_flag": "head"}
http://unapologetic.wordpress.com/2009/01/28/the-characteristic-polynomial/?like=1&_wpnonce=76a2d9cd23	# The Unapologetic Mathematician ## The Characteristic Polynomial Given a linear endomorphism $T:V\rightarrow V$ on a vector space $V$ of dimension $d$, we’ve defined a function — $\det(T-\lambda1_V)$ — whose zeroes give exactly those field elements $\lambda$ so that the $\lambda$-eigenspace of $T$ is nontrivial. Actually, we’ll switch it up a bit and use the function $\det(\lambda1_V-T)$, which has the same useful property. Now let’s consider this function a little more deeply. First off, if we choose a basis for $V$ we have matrix representations of endomorphisms, and thus a formula for their determinants. For instance, if $T$ is represented by the matrix with entries $t_i^j$, then its determinant is given by $\displaystyle\sum\limits_{\pi\in S_d}\mathrm{sgn}(\pi)\prod\limits_{k=1}^dt_k^{\pi(k)}$ which is a sum of products of matrix entries. Now, the matrix entries for the transformation $\lambda1_V-T$ are given by $t_i^j-\lambda\delta_i^j$. Each of these new entries is a polynomial (either constant or linear) in the variable $\lambda$. Any sum of products of polynomials is again a polynomial, and so our function is actually a polynomial in $\lambda$. We call it the “characteristic polynomial” of the transformation $T$. In terms of the matrix entries of $T$ itself, we get $\displaystyle\sum\limits_{\pi\in S_d}\mathrm{sgn}(\pi)\prod\limits_{k=1}^d(\lambda\delta_k^{\pi(k)}-t_k^{\pi(k)})$ What’s the degree of this polynomial? Well, first let’s consider the degree of each term in the sum. Given a permutation $\pi\in S_d$ the term is the product of $d$ factors. The $k$th of these factors will be a field element if $k\neq\pi(k)$, and will be a linear polynomial if $k=\pi(k)$. Since multiplying polynomials adds their degrees, the degree of the $\pi$ term will be the number of $k$ such that $k=\pi(k)$. Thus the highest possible degree happens if $k=\pi(k)$ for all index values $k$. This only happens for one permutation — the identity — so there can’t be another term of the same degree to cancel the highest-degree monomial when we add them up. And so the characteristic polynomial has degree $d$, equal to the dimension of the vector space $V$. What’s the leading coefficient? Again, the degree-$d$ monomial can only show up once, in the term corresponding to the identity permutation. Specifically, this term is $\prod\limits_{k=1}^d(\lambda-t_k^k)$ Each factor gives $\lambda$ a coefficient of ${1}$, and so the coefficient of the $\lambda^d$ term is also ${1}$. Thus the leading coefficient of the characteristic polynomial is ${1}$ — a fact which turns out to be useful. ### Like this: Posted by John Armstrong \| Algebra, Linear Algebra ## 13 Comments » 1. [...] from the Characteristic Polynomial This one’s a pretty easy entry. If we know the characteristic polynomial of an endomorphism on a vector space of finite dimension , then we can get its determinant from [...] Pingback by \| January 29, 2009 \| Reply 2. [...] let’s take a linear endomorphism on a vector space of finite dimension . We know that its characteristic polynomial can be defined without reference to a basis of , and so each of the coefficients of is independent [...] Pingback by \| January 30, 2009 \| Reply 3. [...] does this assumption buy us? It says that the characteristic polynomial of a linear transformation is — like any polynomial over an algebraically closed field [...] Pingback by \| February 2, 2009 \| Reply 4. [...] remember that we can always calculate the characteristic polynomial of . This will be a polynomial of degree — the dimension of . Further, we know that a field [...] Pingback by \| February 10, 2009 \| Reply 5. [...] So we’ve got a linear transformation on a vector space of finite dimension . We take its characteristic polynomial to find the eigenvalues. If all of its roots are distinct (and there are of them, as there must be [...] Pingback by \| February 11, 2009 \| Reply 6. please send me the coefficient in the characteristic polynomial of signed graph of order n. Comment by \| February 13, 2009 \| Reply 7. A little demanding, aren’t we? Comment by \| February 13, 2009 \| Reply 8. [...] we saw that if the entries along the diagonal of an upper-triangular matrix are , then the characteristic polynomial [...] Pingback by \| February 19, 2009 \| Reply 9. [...] we can calculate the characteristic polynomial of , whose roots are the eigenvalues of . For each eigenvalue , we can define the generalized [...] Pingback by \| March 4, 2009 \| Reply 10. [...] of an Eigenpair An eigenvalue of a linear transformation is the same thing as a root of the characteristic polynomial of . That is, the characteristic polynomial has a factor . We can evaluate this polynomial at to [...] Pingback by \| April 3, 2009 \| Reply 11. [...] we found this in the complex case we saw that the characteristic polynomial had to have a root, since is algebraically closed. It’s the fact that isn’t [...] Pingback by \| August 12, 2009 \| Reply 12. [...] on the remaining -dimensional space. This tells us that all of our eigenvalues are , and the characteristic polynomial is , where . We can evaluate this on the transformation to find [...] Pingback by \| January 19, 2010 \| Reply 13. [...] the Jordan-Chevalley decomposition. We let have distinct eigenvalues with multiplicities , so the characteristic polynomial of [...] Pingback by \| August 28, 2012 \| Reply « Previous \| Next » ## About this weblog This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”). I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 38, "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": 0.9117995500564575, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/7185/simple-solution-to-diffusion-equation	# Simple Solution to Diffusion equation A very simple solution to the diffusion equation is $u(x,t)=x^2+2 t$ My question: How can this be a solution to the diffusion equation when nothing really diffuses, but just stays the same - see plot: Here I can see that it is a solution but can't believe my eyes... Is there some kind of intuition other than: "it satisfies the PDE"? Are there other examples of solutions to well-known equations that behave completely un-intuitive in some situations? - ## 3 Answers Here are a few pointers, from the more obtuse to the more direct. 1. "Meeting a friend in a corridor, Wittgenstein said: 'Tell me, why do people always say it was natural for men to assume that the sun went round the earth, rather than that the earth was rotating?' His friend said, 'Well, obviously, because it looks as if the sun is going round the earth.' To which the philosopher replied, 'Well, what would it have looked like if it had looked as if the earth was rotating?'" Apparently from Tom Stoppard's play Jumpers. More to the point: what does your intuition think the solution to the diffusion equation with initial conditions $u(x,0) = x^2$ should have looked like? 2. $u_1(x,t) = \text{const}$, and $u_2(x,t) = x$ are also solutions of the diffusion equation. Do you find them more or less surprising than the $x^2 + t$ solution? 3. The standard intuitive interpretation of the diffusion equation is that it "smooths out" variations in the solution. The way it finds these variations is by looking at the second derivative. But the second derivative of $x^2$ is constant everywhere, so what can the diffusion equation do? 4. Locally, every point "wants to be" equal to the average of the value in its neighbourhood. If the second derivative at a point is positive, the function looks like a valley, so the point increases its value over time; if the second derivative is negative, the point is on a peak, and tries to decrease. With $x^2$, all of $\mathbb{R}$ is a valley. - Well, concerning 1. I would have thought that some flattening out takes place ending in a flat line, but for 2.: This really dropped my jaw to the floor! Esp. with $u_2$ there is such an disequilibrium (and it could be multiplied by any const. to make it even worse) and the whole situation doesn't change over time - now, where is the intuition here?!? – vonjd Oct 20 '10 at 9:54 @vonjd: Well, diffusion is certainly taking place, in that heat is flowing at a constant rate from the right to the left. But the temperature $u$ cannot change on any region, because exactly as much heat flows out from the left as flows in from the right. A more intuitive variant may be to think of diffusion on an interval $[-a,a]$ with the boundaries fixed at temperatures $-a$ and $a$. Then $u(x,t) = x$ on $x \in [-a,a]$ is a solution: heat comes in from the right, goes out the left. My example is what you get when you let $a \rightarrow \infty$. – Rahul Narain Oct 20 '10 at 18:58 But there is diffusion! Heat – or whatever quantity $u$ represents – flows from the far left and right ($x\to\pm\infty$, where there's lots of it!) towards the middle ($x=0$). For every fixed point $x$ there is always more heat flowing towards it from the outside than what it emits inwards, and the result is that the heat keeps increasing at each point. Now this might seem unphysical, and that's because it is. For this particular solution, the total amount of heat, $\int_{-\infty}^{\infty} u(x,t) dx$, is infinite for each $t$. - This is a very intuitive explanation - the fascinating thing is that it keeps it form which is lifted upwards over time. – vonjd Oct 20 '10 at 9:56 It is difficult to interpret a PDE without knowing its boundary conditions and/or initial values. But, at first guess, I'd say you should look at individual time slices of the solution on the same graph and see if that helps your interpretation. The 3D image is nice for showing everything at once, but that can obscure an interpretation. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 22, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9569216966629028, "perplexity_flag": "middle"}
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Let the angle between the sphere's velocity and the surface be 90 ...	{"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": 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": 0.93470698595047, "perplexity_flag": "middle"}
http://en.wikipedia.org/wiki/Divergent_series	# Divergent series In mathematics, a divergent series is an infinite series that is not convergent, meaning that the infinite sequence of the partial sums of the series does not have a limit. If a series converges, the individual terms of the series must approach zero. Thus any series in which the individual terms do not approach zero diverges. However, convergence is a stronger condition: not all series whose terms approach zero converge. The simplest counterexample is the harmonic series $1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \cdots =\sum_{n=1}^\infty\frac{1}{n}.$ The divergence of the harmonic series was proven by the medieval mathematician Nicole Oresme. In specialized mathematical contexts, values can be usefully assigned to certain series whose sequence of partial sums diverges. A summability method or summation method is a partial function from the set of sequences of partial sums of series to values. For example, Cesàro summation assigns Grandi's divergent series $1 - 1 + 1 - 1 + \cdots$ the value 1/2. Cesàro summation is an averaging method, in that it relies on the arithmetic mean of the sequence of partial sums. Other methods involve analytic continuations of related series. In physics, there are a wide variety of summability methods; these are discussed in greater detail in the article on regularization. ## Theorems on methods for summing divergent series A summability method M is regular if it agrees with the actual limit on all convergent series. Such a result is called an abelian theorem for M, from the prototypical Abel's theorem. More interesting and in general more subtle are partial converse results, called tauberian theorems, from a prototype proved by Alfred Tauber. Here partial converse means that if M sums the series Σ, and some side-condition holds, then Σ was convergent in the first place; without any side condition such a result would say that M only summed convergent series (making it useless as a summation method for divergent series). The operator giving the sum of a convergent series is linear, and it follows from the Hahn–Banach theorem that it may be extended to a summation method summing any series with bounded partial sums. This fact is not very useful in practice since there are many such extensions, inconsistent with each other, and also since proving such operators exist requires invoking the axiom of choice or its equivalents, such as Zorn's lemma. They are therefore nonconstructive. The subject of divergent series, as a domain of mathematical analysis, is primarily concerned with explicit and natural techniques such as Abel summation, Cesàro summation and Borel summation, and their relationships. The advent of Wiener's tauberian theorem marked an epoch in the subject, introducing unexpected connections to Banach algebra methods in Fourier analysis. Summation of divergent series is also related to extrapolation methods and sequence transformations as numerical techniques. Examples for such techniques are Padé approximants, Levin-type sequence transformations, and order-dependent mappings related to renormalization techniques for large-order perturbation theory in quantum mechanics. ## Properties of summation methods Summation methods usually concentrate on the sequence of partial sums of the series. While this sequence does not converge, we may often find that when we take an average of larger and larger initial terms of the sequence, the average converges, and we can use this average instead of a limit to evaluate the sum of the series. So in evaluating a = a0 + a1 + a2 + ..., we work with the sequence s, where s0 = a0 and sn+1 = sn + an+1. In the convergent case, the sequence s approaches the limit a. A summation method can be seen as a function from a set of sequences of partial sums to values. If A is any summation method assigning values to a set of sequences, we may mechanically translate this to a series-summation method AΣ that assigns the same values to the corresponding series. There are certain properties it is desirable for these methods to possess if they are to arrive at values corresponding to limits and sums, respectively. 1. Regularity. A summation method is regular if, whenever the sequence s converges to x, A(s) = x. Equivalently, the corresponding series-summation method evaluates AΣ(a) = x. 2. Linearity. A is linear if it is a linear functional on the sequences where it is defined, so that A(k r + s) = k A(r) + A(s) for sequences r, s and a real or complex scalar k. Since the terms an = sn+1 − sn of the series a are linear functionals on the sequence s and vice versa, this is equivalent to AΣ being a linear functional on the terms of the series. 3. Stability. If s is a sequence starting from s0 and s′ is the sequence obtained by omitting the first value and subtracting it from the rest, so that s′n = sn+1 − s0, then A(s) is defined if and only if A(s′) is defined, and A(s) = s0 + A(s′). Equivalently, whenever a′n = an+1 for all n, then AΣ(a) = a0 + AΣ(a′). The third condition is less important, and some significant methods, such as Borel summation, do not possess it. One can also give a weaker alternative to the last condition. 1. Finite Re-indexability. If s and s′ are two sequences such that there exists a bijection $f: \mathbb{N} \rightarrow \mathbb{N}$ such that si = s′f(i) for all i, and if there exists some $N \in \mathbb{N}$ such that si = s′i for all i > N, then A(s) = A(s′). (In other words, s′ is the same sequence as s, with only finitely many terms re-indexed.) Note that this is a weaker condition than Stability, because any summation method that exhibits Stability also exhibits Finite Re-indexability, but the converse is not true. A desirable property for two distinct summation methods A and B to share is consistency: A and B are consistent if for every sequence s to which both assign a value, A(s) = B(s). If two methods are consistent, and one sums more series than the other, the one summing more series is stronger. There are powerful numerical summation methods that are neither regular nor linear, for instance nonlinear sequence transformations like Levin-type sequence transformations and Padé approximants, as well as the order-dependent mappings of perturbative series based on renormalization techniques. ## Axiomatic methods Taking regularity, linearity and stability as axioms, it is possible to sum many divergent series by elementary algebraic manipulations. For instance, whenever r ≠ 1, the geometric series $\begin{align} G(r,c) & = \sum_{k=0}^\infty cr^k & & \\ & = c + \sum_{k=0}^\infty cr^{k+1} & & \mbox{ (stability) } \\ & = c + r \sum_{k=0}^\infty cr^k & & \mbox{ (linearity) } \\ & = c + r \, G(r,c), & & \mbox{ whence } \\ G(r,c) & = \frac{c}{1-r} , & & \\ \end{align}$ can be evaluated regardless of convergence. More rigorously, any summation method that possesses these properties and which assigns a finite value to the geometric series must assign this value. However, when r is a real number larger than 1, the partial sums increase without bound, and averaging methods assign a limit of ∞. ## Nörlund means Suppose pn is a sequence of positive terms, starting from p0. Suppose also that $\frac{p_n}{p_0+p_1 + \cdots + p_n} \rightarrow 0.$ If now we transform a sequence s by using p to give weighted means, setting $t_m = \frac{p_m s_0 + p_{m-1}s_1 + \cdots + p_0 s_m}{p_0+p_1+\cdots+p_m}$ then the limit of tn as n goes to infinity is an average called the Nörlund mean Np(s). The Nörlund mean is regular, linear, and stable. Moreover, any two Nörlund means are consistent. The most significant of the Nörlund means are the Cesàro sums. Here, if we define the sequence pk by $p_n^k = {n+k-1 \choose k-1}$ then the Cesàro sum Ck is defined by Ck(s) = N(pk)(s). Cesàro sums are Nörlund means if k ≥ 0, and hence are regular, linear, stable, and consistent. C0 is ordinary summation, and C1 is ordinary Cesàro summation. Cesàro sums have the property that if h > k, then Ch is stronger than Ck. ## Abelian means Suppose λ = {λ0, λ1, λ2, ...} is a strictly increasing sequence tending towards infinity, and that λ0 ≥ 0. Recall that an = λn+1 − λn is the associated series whose partial sums form the sequence λ. Suppose $f(x) = \sum_{n=0}^\infty a_n \exp(-\lambda_n x)$ converges for all real numbers x>0. Then the Abelian mean Aλ is defined as $A_\lambda(s) = \lim_{x \rightarrow 0^{+}} f(x).$ A series of this type is known as a generalized Dirichlet series; in applications to physics, this is known as the method of heat-kernel regularization. Abelian means are regular, linear, and stable, but not always consistent between different choices of λ. However, some special cases are very important summation methods. ### Abel summation See also: Abel's theorem If λn = n, then we obtain the method of Abel summation. Here $f(x) = \sum_{n=0}^\infty a_n \exp(-nx) = \sum_{n=0}^\infty a_n z^n,$ where z = exp(−x). Then the limit of ƒ(x) as x approaches 0 through positive reals is the limit of the power series for ƒ(z) as z approaches 1 from below through positive reals, and the Abel sum A(s) is defined as $A(s) = \lim_{z \rightarrow 1^{-}} \sum_{n=0}^\infty a_n z^n$ Abel summation is interesting in part because it is consistent with but more powerful than Cesàro summation: A(s) = Ck(s) whenever the latter is defined. The Abel sum is therefore regular, linear, stable, and consistent with Cesàro summation. ### Lindelöf summation If λn = n ln(n), then (indexing from one) we have $f(x) = a_1 + a_2 2^{-2x} + a_3 3^{-3x} + \cdots .$ Then L(s), the Lindelöf sum (Volkov 2001), is the limit of ƒ(x) as x goes to zero. The Lindelöf sum is a powerful method when applied to power series among other applications, summing power series in the Mittag-Leffler star. If g(z) is analytic in a disk around zero, and hence has a Maclaurin series G(z) with a positive radius of convergence, then L(G(z)) = g(z) in the Mittag-Leffler star. Moreover, convergence to g(z) is uniform on compact subsets of the star. ## References • Arteca, G.A.; Fernández, F.M.; Castro, E.A. (1990), Large-Order Perturbation Theory and Summation Methods in Quantum Mechanics, Berlin: Springer-Verlag . • Baker, Jr., G. A.; Graves-Morris, P. (1996), Padé Approximants, Cambridge University Press . • Brezinski, C.; Zaglia, M. Redivo (1991), Extrapolation Methods. Theory and Practice, North-Holland . • Hardy, G. H. (1949), Divergent Series, Oxford: Clarendon Press . • LeGuillou, J.-C.; Zinn-Justin, J. (1990), Large-Order Behaviour of Perturbation Theory, Amsterdam: North-Holland . • Volkov, I.I. (2001), "Lindelöf summation method", in Hazewinkel, Michiel, , Springer, ISBN 978-1-55608-010-4 . • Zakharov, A.A. (2001), "Abel summation method", in Hazewinkel, Michiel, , Springer, ISBN 978-1-55608-010-4 .	{"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": 13, "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": 0.9060743451118469, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/36128/laplacian-of-a-delta-function-as-an-interaction-potential-for-laughlin-state/36130	# Laplacian of a delta function as an interaction potential for Laughlin state I am reading Xiao-Gang Wen's paper "Pattern-of-zeros approach to Fractional quantum Hall states and a classification of symmetric polynomial of infinite variables", on page 8, he gives three interaction potentials for various Laughlin state and claim that these Laughlin state are exactly the zero-energy ground state of the given interaction potential. I do have some questions about these interaction potentials. 1. I do believe that the second term in the interaction potential for $\nu=1/4$ is from Kivelson-Trugman's delta function expansion ("Exact results for the fractional quantum Hall effect with general interactions"), but I think in their paper, the Laplacian is with respect to the relative coordinate which is $z_{rel}=z_1-z_2$, but why in Xiao-Gang's paper that Laplacian is only with respect to $z_1$? 2. Why the delta function only involves $z$, but no conjugate $z^$? Isn't a two dimensional delta function usually defined as $\delta(z,z^)$? - – Qmechanic♦ Sep 11 '12 at 7:45 ## 1 Answer I haven't read your paper but is Laplacian not invariant by translation ? Z=z1-z2 For the notation for δ(z,z∗), you are right to say that to recover x, y you need z and z*, but it might just be a shorthening of the writing. - The situation I am concerned is that when you treat the delta function as a limiting case of a Gaussian function(say if we want to investigate the interaction range for this Laplacian of the delta function), then if z is the only variable wouldn't make sense. – huyichen Sep 11 '12 at 6:16 – Shaktyai Sep 11 '12 at 8:21 I am aware of that definition, but I don't think that is quite equivalent to a two dimensional delta function, maybe a delta function for analytic complex functions. – huyichen Sep 11 '12 at 18:36	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 6, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.920626699924469, "perplexity_flag": "middle"}
http://www.physicsforums.com/showthread.php?t=53270	Physics Forums ## Fluid Mechanics - Pressure measurements I'm having difficulty with one problem and was hoping someone could help me out. Here's the problem: Mercury is poured into a U-tube as in Figure P14.18a. The left arm of the tube has a cross-sectional area A1 of 107.0 cm2, and the right arm has a cross-sectional area A2 of 4.10 cm2. One hundred grams of water are then poured into the right arm, as in Figure P14.18b. (a) Determine the length of the water column in the right arm of the U-tube (b) Given that the density of mercury is 13.6 g/cm3, what distance h does the mercury rise in the left arm? I've attached the picture of the figure to this post For part (a), given the radius of A2 and the volume of water, I easily got the length of the water column: 24.5cm Part (b) is giving me a problem. How do I equate the change in height of the mercury due to the pressure of the water on it? I can determine the pressure of the water (P=F/A) on the mercury and thereby determine the upward pressure on the left arm (A2/A1=F2/F1). Once I get the change in height (dy) of the right arm I can use it to figure out the subsequent change on the left arm (A1/dy1=A2/dy2) but getting to that point is difficult for me. I would greatly appreciate any suggestions or a point in the right direction. Thanks! 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: Gold Member Science Advisor Open the archive attached, I have modified the yours one a bit. Pay attention to the new references z, P_1, and L. P_1 is the pressure just at the interface water-mercury, and L is the lenght obtained in a). Pressure at the interface (Hydrostatics): $$P_1=P_{at}+\rho_w g L$$ that pressure is the same at the same height inside the mercury. So that, we can relate the jump of pressures at the left side: $$P_1=P_{at}+\rho_{Hg} g (h+z)$$ So that: $$\rho_w g L=\rho_{Hg} g (h+z)$$ (1) In addition and due to the mass conservation: $$\rho_{Hg} z A_2=\rho_{Hg} h A_1$$ (2) With (1) and (2) you can solve for h. Attached Images untitled.bmp (47.2 KB, 167 views) Thank you very much! The solution you provided worked perfectly. I don't think I would have ever come to the same conclusion on my own but with your explanation I understand the physics involved and how you arrived at the answer. Thanks again! Thread Tools Similar Threads for: Fluid Mechanics - Pressure measurements Thread Forum Replies Introductory Physics Homework 1 Classical Physics 8 Introductory Physics Homework 1 Introductory Physics Homework 12 Introductory Physics Homework 14	{"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": 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": 0.9122685194015503, "perplexity_flag": "middle"}
http://cs.stackexchange.com/questions/439/which-combinations-of-pre-post-and-in-order-sequentialisation-are-unique	# Which combinations of pre-, post- and in-order sequentialisation are unique? We know post-order, ````post L(x) => [x] post N(x,l,r) => (post l) ++ (post r) ++ [x] ```` and pre-order ````pre L(x) => [x] pre N(x,l,r) => [x] ++ (pre l) ++ (pre r) ```` and in-order traversal resp. sequentialisation. ````in L(x) => [x] in N(x,l,r) => (in l) ++ [x] ++ (in r) ```` One can easily see that neither describes a given tree uniquely, even if we assume pairwise distinct keys/labels. Which combinations of the three can be used to that end and which can not? Positive answers should include an (efficient) algorithm to reconstruct the tree and a proof (idea) why it is correct. Negative answers should provide counter examples, i.e. different trees that have the same representation. - ## 1 Answer First, I'll assume that all elements are distinct. No amount of sequentialisations is going to tell you the shape of a tree with elements `[3,3,3,3,3]`. It is possible to reconstruct some trees with duplicate elements, of course; I don't know what nice sufficient conditions exist. Continuing on the negative results, you can't fully rebuild a binary tree from its pre-order and post-order sequentializations alone. `[1,2]` preorder, `[2,1]` post-order has to have `1` at the root, but `2` can be either the left child or the right child. If you don't care about this ambiguity, you can reconstruct the tree with the following algorithm: • Let $[x_1,\dots,x_n]$ be the pre-order traversal and $[y_n,\ldots,y_1]$ be the post-order traversal. We must have $x_1=y_1$, and this is the root of the tree. • $x_2$ is the leftmost child of the root, and $y_2$ is the rightmost child. If $x_2 = y_2$, the root node is unary; recurse over $[x_2,\ldots,x_n]$ and $[y_n,\ldots,y_2]$ to build the single subtree. • Otherwise, let $i$ and $j$ be the indices such that $x_2 = y_i$ and $y_2 = x_j$. $[x_2,\ldots,x_{j-1}]$ is the pre-order traversal of the left subtree, $[x_j,\ldots,x_n]$ that of the right subtree, and similarly for the post-order traversals. The left subtree has $j-2=n-i+1$ elements, and the right subtree has $i-2=n-j+1$ elements. Recurse once for each subtree. By the way, this method generalizes to trees with arbitrary branching. With arbitrary branching, find out the extent of the left subtree and cut off its $j-2$ elements from both lists, then repeat to cut off the second subtree from the left, and so on. As stated, the running time is $O(n^2)$ with $\Theta(n^2)$ worst case (in the case with two children, we search each list lineraly). You can turn that into $O(n\,\mathrm{lg}(n))$ if you preprocess the lists to build an $n\,\mathrm{lg}(n)$ finite map structure from element values to positions in the input lists. Also use an array or finite map to go from indices to values; stick to global indices, so that recursive calls will receive the whole maps and take a range as argument to know what to act on. With the pre-order traversal $[x_1,\ldots,x_n]$ and the in-order traversal $[z_1,\ldots,z_n]$, you can rebuild the tree as follows: • The root is the head of the pre-order traversal $x_1$. • Let $k$ be the index such that $z_k = x_1$. Then $[z_1,\ldots,z_{k-1}]$ is the in-order traversal of the left child and $[z_{k+1},\ldots,z_n]$ is the in-order traversal of the right child. Going by the number of elements, $[x_2,\ldots,x_k]$ is the pre-order traversal of the left child and $[x_{k+1},\ldots,x_n]$ that of the right child. Recurse to build the left and right subtrees. Again, this algorithm is $O(n^2)$ as stated, and can be performed in $O(n\,\mathrm{lg}(n))$ if the list is preprocessed into a finite map from values to positions. Post-order plus in-order is of course symmetric. - Is there a typo here: "[1,2] preorder, [1,2] post-order has to have 1 at the root, but 2 can be either the left child or the right child. " The post order of such a tree would be [2,1] not [1,2] whether 2 is a left or right child. Also, do you mean if both preorder and postorder are given, we cannot reconstruct the tree, or do you mean if we are given only one of them then we cannot reconstruct the tree? – CEGRD Jun 20 '12 at 14:55 @CEGRD Indeed, the postorder was a typo. The example shows that you cannot fully reconstruct the tree in this case: you can't know whether `2` is a left child or a right child. This corresponds to the “single subtree” case of the reconstruction algorithm. – Gilles♦ Jun 20 '12 at 19:55	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 32, "mathjax_display_tex": 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": 0.9048203825950623, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/159700/complex-analysis-exercises?answertab=oldest	# Complex analysis exercises These two questions are driving me mad as I need to help my daughter but I can't remember all this stuff. $\,(1)\,\,$ Let $\,p(z)\,,\,q(z)\,$ be two non-constant complex polynomials of the same degree s.t. $$\text{whenever \|z\|=1}\,\,,\,\|p(z)\|=\|q(z)\|$$ If all the zeros of both $\,p(z)\,,\,q(z)\,$ are within the open unit disk $\,\|z\|<1\,$ , prove that $$\forall z\in\mathbb{C}\,\,,\,q(z)=\lambda\, p(z)\,,\,\lambda\in\mathbb{C}\,\,\text{a constant}$$ What've I thought: since the polynomials are of the same degree, I know that $$\lim_{\|z\|\to\infty}\frac{q(z)}{p(z)}$$ exists finitely, so we can bound $\,\displaystyle{\frac{q(z)}{p(z)}}\,$ say in $\,\|z\|>1\,$ . Unfortunately, I can't use Liouville's Theorem to get an overall bound as the rational function is not entire within the unit disk... $\,(2)\,\,$ Let $\,f(z)\,$ be analytic in the punctured disk $\,\{z\in\mathbb{C}\;\|\;0<\|z-a\|<r\,,\,\text{for some}\,\,0<r\in\mathbb R\}\,$. Prove that if $\,\displaystyle{\lim_{z\to a}f'(z)}\,$ exists and finite, then $\,a\,$ is a removable singularity of $\,f(z)\,$ . My thoughts: We have a Laurent expansion in the above disk$$f(z)=\ldots +\frac{a_{-n}}{(x-a)^n}+\ldots +\frac{a_{-1}}{z-a}+a_0+a_1(z-a)+\ldots$$ so taking the derivative term-term (is there any special condition that must be fulfilled in this particular case to do so?) we get$$f'(z)=\ldots -\frac{na_{-n}}{(z-a)^{n+1}}-\ldots -\frac{a_{-1}}{(z-a)^2}+a_1+2a_2(z-a)+\ldots$$Now, as the limit of the above when $\,z\to a\,$ exists finitely, it must be that all the terms with negative power of $\,z-a\,$ vanish, thus $$\ldots =\,a_{-n}=a_{-n+1}=\ldots =a_{-1}=0$$ and we get that the above Laurent series for $\,f(z)\,$ is, in fact, a Taylor one and, thus, the function's limit (not the derivative's!) exists and finite when $\,z\to a\,$ and $\,a\,$ is then a removable singularity. Any help in (1) if I got right (2), or in both if there's some problem with the latter will be much appreciated. - 1 +1 for your carefully formulated question, your solution of (2) and above all for being such a great daddy! – Georges Elencwajg Jun 18 '12 at 11:16 Hehe...thanks, @George! But this girl's exercises in complex functions are making me realize I should have taught that course or, at least, follow it more closely. Most of the time it was linear, abstract algebra, calculus...I need now to get into shape in order to be a good dad... – DonAntonio Jun 18 '12 at 12:33 ## 1 Answer The function $g(z)= \frac{p(\frac{1}{z})}{q(\frac{1}{z})}$ has a removable singularity at $0$ since the $\lim_{z \rightarrow \infty }\frac{p(z)}{q(z)}$ exists. Furthermore since $p(z),q(z)$ have all their zeros inside $\|z\|=1\,$ , the functions $g(z)$ and $\frac{1}{g(z)}$ are analytic in $\|z\|<1+\epsilon\,$ , for some $\,\epsilon >0\,$ , so by the maximum principle both functions $\left(\|g(z)\|, \left\|\frac{1}{g(z)}\right\|\right)$ are bounded by $1$ for $\|z\| \leq 1$ since the condition $\|p(z)\|=\|q(z)\|$ for $\|z\|=1$ that means $\|g(z)\|=1$ for $\|z\| \leq 1$ and you are done. - you needed that $\epsilon >0$ in order to apply the Maxixmum principle, also you can find it because the zeros inside the circe are finite so they have a positive distance from the circe. – clark Jun 18 '12 at 0:13 thanks for the answer but I still don't follow. At the end of your answer, did you mean $\,\displaystyle{\|f(z)\|:=\left\|\frac{p(z)}{q(z)}\right\|=1\,\,\,for\,\,\,\|z\|\leq 1}\,$? Or what is that function $\,f(z)\,$? And how exactly does any of the above end problem (1)? – DonAntonio Jun 18 '12 at 2:05 I fixed it, it was a typo mistake I meant $g(z)$ instead of $f(z)$ and you conclude $\|g(z)\|=1$ for $\|z\| \leq 1$ because it is analytic you can coclude that $g(z)$ is constant you can go there using the cauchy riemann equations for instance. And then $\frac{p(z)}{q(z)} = e^{i \theta} = g(\frac{1}{z}) \forall z \geq 1$ and since they are polynomials and they take the same value for infinite $z$ they must coincide – clark Jun 18 '12 at 10:42	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 43, "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": 0.9506472945213318, "perplexity_flag": "head"}
http://en.wikipedia.org/wiki/TCP_Tuning	TCP tuning From Wikipedia, the free encyclopedia (Redirected from TCP Tuning) Jump to: navigation, search Internet protocol suite Application layer Transport layer Internet layer Link layer TCP tuning techniques adjust the network congestion avoidance parameters of TCP connections over high-bandwidth, high-latency networks. Well-tuned networks can perform up to 10 times faster in some cases.[1] However, blindly following instructions without understanding their real consequences can hurt the performance as well. Network and system characteristics Bandwidth-delay product (BDP) Bandwidth-delay product (BDP) is a term primarily used in conjunction with the TCP to refer to the number of bytes necessary to fill a TCP "path", i.e. it is equal to the maximum number of simultaneous bits in transit between the transmitter and the receiver. High performance networks have very large BDPs. To give a practical example, two nodes communicating over a geostationary satellite link with a round trip delay of 0.5 seconds and a bandwidth of 10 Gbit/s can have up to 0.5×1010 bits, i.e., 5 Gbit = 625 MB of unacknowledged data in flight. Despite having much lower latencies than satellite links, even terrestrial fiber links can have very high BDPs because their link capacity is so large. Operating systems and protocols designed as recently as a few years ago when networks were slower were tuned for BDPs of orders of magnitude smaller, with implications for limited achievable performance. Buffers The original TCP configurations supported buffers of up to 65,535 Bytes (64 KiB-1) TCP receive window size, which was adequate for slow links or links with small round trip times (RTTs). Larger buffers are required by the high performance options described below. Buffering is used throughout high performance network systems to handle delays in the system. In general, buffer size will need to be scaled proportionally to the amount of data "in flight" at any time. For very high performance applications that are not sensitive to network delays, it is possible to interpose large end to end buffering delays by putting in intermediate data storage points in an end to end system, and then to use automated and scheduled non-real-time data transfers to get the data to their final endpoints. TCP speed limits Maximum achievable throughput for a single TCP connection is determined by different factors. One trivial limitation is the maximum bandwidth of the slowest link in the path. But there are also other, less obvious limits for TCP throughput. Bit errors can create a limitation for the connection as well as round-trip time. Window size See also: TCP window scale option In computer networking, RWIN (TCP Receive Window) is the amount of data that a computer can accept without acknowledging the sender. If the sender has not received acknowledgement for the first packet it sent, it will stop and wait and if this wait exceeds a certain limit, it may even retransmit. This is how TCP achieves reliable data transmission. Even if there is no packet loss in the network, windowing can limit throughput. Because TCP transmits data up to the window size before waiting for the acknowledgements, the full bandwidth of the network may not always get used. The limitation caused by window size can be calculated as follows: $\mathrm{Throughput} \le \frac {\mathrm{RWIN}} {\mathrm{RTT}} \,\!$ where RWIN is the TCP Receive Window and RTT is the round-trip time for the path. At any given time, the window advertised by the receive side of TCP corresponds to the amount of free receive memory it has allocated for this connection. Otherwise it would risk dropping received packets due to lack of space. The sending side should also allocate the same amount of memory as the receive side for good performance. That is because, even after data has been sent on the network, the sending side must hold it in memory until it has been acknowledged as successfully received, just in case it would have to be retransmitted. If the receiver is far away, acknowledgments will take a long time to arrive. If the send memory is small, it can saturate and block emission. A simple computation gives the same optimal send memory size as for the receive memory size given above. Packet loss When packet loss occurs in the network, an additional limit is imposed on the connection.[2] In the case of light to moderate packet loss when the TCP rate is limited by the congestion avoidance algorithm, the limit can be calculated according to the formula (Mathis et al.): $\mathrm{Throughput} \le \frac {\mathrm{MSS}} {\mathrm{RTT} \sqrt{ P_{\mathrm{loss} }}}$ where MSS is the maximum segment size and Ploss is the probability of packet loss.[3] If packet loss is so rare that the TCP window becomes regularly fully extended, this formula doesn't apply. TCP Options for High Performance A number of extensions have been made to TCP over the years to increase its performance over fast high-RTT links ("long fat networks", or LFNs for short). TCP timestamps (RFC 1323) play a double role: they avoid ambiguities due to the 32-bit sequence number field wrapping around, and they allow more precise RTT estimation in the presence of multiple losses per RTT. With those improvements, it becomes reasonable to increase the TCP window beyond 64 kB, which can be done using the window scaling option (RFC 1323). The TCP selective acknowledgment options (SACK, RFC 2018) allows a TCP receiver to precisely inform the TCP server about which segments have been lost. This increases performance on high-RTT links, when multiple losses per window are possible. Path MTU discovery avoids the need for in-network fragmentation, which increases performance in the presence of losses. References Retrieved from "http://en.wikipedia.org/w/index.php?title=TCP_tuning&oldid=531135675"	{"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": 2, "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": 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": 0.9340457916259766, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/151081/gcd-of-rationals/151090	# GCD of rationals Disclaimer: I'm an engineer, not a mathematician Somebody claimed that $\gcd$ only is applicable for integers, but it seems I'm perfectly able to apply it to rationals also: $$\gcd\left(\frac{13}{6}, \frac{3}{4} \right) = \frac{1}{12}$$ I can do this for a number of cases on sight, but I need a method. I tried Euclid's algorithm, but I'm not sure about the end condition: a remainder of 0 doesn't seem to work here. So I tried the following: $$\gcd\left(\frac{a}{b}, \frac{c}{d} \right) = \frac{\gcd(a\cdot d, c \cdot b)}{b \cdot d}$$ This seems to work, but I'm just following my intuition, and I would like to know if this is a valid equation, maybe the correct method. (It is in any case consistent with $\gcd$ over the natural numbers.) I'm not a mathematician, so please type slowly :-) - @WillieWong - Why did you remove the [GCD] tag? Because it's a new one? But it is about GCD, and the [GCD] tag may help people who're looking for information. – stevenvh May 29 '12 at 11:25 the gcd tag is too specific and too broad at the same time (it is about a particular operation which has interpretation in many different levels/specialties of mathematics). And any question about gcd will have the phrase "gcd" or "greatest common denominator" in the question body, which can be easily found using the search function. – Willie Wong♦ May 29 '12 at 11:44 To illustrate: math.stackexchange.com/search?q=gcd returns over 1000 questions about greatest common denominators in different contexts. It is much more useful to use the tags to help specify in which context your question about gcd is asked. – Willie Wong♦ May 29 '12 at 11:46 Since you are an engineer: Think about the numbers as representing some quantities. Divide your unit in smaller units so that both quantities become integers. Apply the gcd, and then switch back to the original unit....This is exactly what Greek's understanding mentioned in Andre's answer means.... – N. S. May 31 '12 at 8:27 ## 3 Answers In the ancient Greek sense, if we have two quantities $x$ and $y$, then the quantity $z$ is a common measure of $x$ and $y$ if each of $x$ and $y$ is an integer multiple of $z$. The quantities involved were not thought of as numbers, but of course they were what we think of as positive. So for example the diagonal of a square, and the diagonal of a square whose sides are $50\%$ bigger, have a common measure. But some pairs of quantities are incommensurable, like the side and diagonal of a square. Any two rationals, unless they are both $0$, have a greatest common measure. You are therefore perfectly correct. The two rationals also have a least number that they both measure. So if you had further argued that two rationals, not both $0$, have a least positive common multiple, you would also be right. Your calculation method is also correct. One brings the two rationals to a common denominator $q$, say as $\frac{x}{q}$ and $\frac{y}{q}$. Then the $\gcd$ is $\frac{\gcd(x,y)}{q}$. You chose the common denominator $q=bd$. Any common denominator will do, and will produce the same $\gcd$. Most of the standard theorems about $\gcd$ and lcm for integers extend in a straightforward way to results about the $\gcd$ and lcm for rationals. For example, a mild variant of what we nowadays call the Euclidean algorithm will compute the $\gcd$ of two positive rationals. Actually, this variant is the original Euclidean algorithm! Remark: A number of programs, including Mathematica, accept rationals as inputs to their $\gcd$ function. The same appears to be true of WolframAlpha, which has the great advantage of being freely accessible. One way, perhaps, to (sort of) settle the argument in your favour is to type $\gcd(3/7, 12/22)$ into Alpha. It will, probably, return $\frac{3}{77}$ (it did when I used it). With other rationals, test first, Alpha is not bug-free. - Thanks for the Wolfram Alpha reference too. Interesting site, though I guess you guys would use Mathematica. Also gives additional information like series representation and prime factorization. I took the liberty of adding a link to it. – stevenvh May 31 '12 at 7:56 Euclid's algorithm (originally formulated using repeated subtractions rather than divisions) can be applied to any pair of comparable quantities (two lengths, two masses, two frequencies): just keep subtracting the smaller of a pair from the larger and then replace the larger one by the difference found; stop when the difference becomes $0$, returning the smaller (which is now also the larger) retained quantity as GCD. Although I haven't actually read Euclid, I would be surprised if the original formulation wasn't actually in terms of geometric lengths, given that Greek mathematicians were more at ease with geometry than with numbers. The only caveat is that with lengths it is not an algorithm in the sense that it is guaranteed to terminate; two lengths are by definition "commensurable" if the procedure does terminate, and it then returns the largest unit of length for which both intitial lengths are integer multiples. One can easily show (without using any arithmetic) that the procedure does indeed never terminate (because one gets back to the original ratio of lengths, but at a smaller scale) when starting with the lengths of a side and a chord of a regular pentagon (giving the golden ratio) or for a side and a diagonal of a square (ratio $1:\sqrt2$); this is a nice exercise in geometry. $\qquad$ Now if you apply this to rational numbers, you see there is no problem, since they can be compared, subtracted, and any pair of them is necessarily commensurable (one over the product of the denominators gives common measure, though not necessarily the largest one). Bringing both rational numbers to this common denominator, one sees that the formula you guessed is indeed the correct one. If you have prime factorizations of $a,b,c,d$ available, you can also find the $\gcd(\frac ab,\frac cd)$ by taking every prime number $p$ to a power that is the minimum of the valuations of $p$ in $\frac ab$ and in $\frac cd$, where the valuation in $\frac ab$ is its multiplicity in $a$ minus its multiplicity in $b$ (a possibly negative integer). However your formula combined with Euclid's algorithm for integers (or directly Euclid's algorithm for rational numbers) is a more efficient method of computing. Added in reply to comment by @Michael Hardy: When I say that the Greek mathematicians were more at ease with geometry than with numbers, I did not mean to say the did not know about numbers, but that they would prefer stating and treating a statement about numbers in geometric terms, quite opposite to our modern tendency to translate geometric problems into numerical terms. A case in point is Euclid's proof of what we would formulate as the infiniteness of the set of prime numbers (he never uses terms "finite" or "infinite", and I don't think these notions were even used at the time, but that is not my point here). I think most people would agree this is a purely number theoretic statement with absolutely no geometric content, yet Euclid states and proves it geometrically (essentially by constructing a least common multiple of a given (finite) set of prime multiples of a given unit length). And for this reason of geometric preference I think it is probably historically inaccurate to use the term "Euclidean algorithm" for a procedure restricted to the case of a pair of positive integers (and therefore assured to terminate) rather than a pair of lengths (which could fail to terminate); however I'll repeat my ignorance of the precise sources here. Added later Thank you @Robert Isreal for the reference to the translation of Elements. So Euclid operates in his description of his algorithm, in Book VII Propositions 2,3, on numbers, which are integer multiples some unit veiled in mystery ("that by virtue of which each of the things that exist is called one"). Yet the language suggests that numbers are thought of as lengths; if not, why denote individual numbers by $AB$, $CD$, talk about one number measuring another, etc.? Moreover, in book X, Proposition 2 one finds the same procedure, but with numbers replaced by "magnitudes", treated exactly like numbers are (also denoted $AB$ etc.) but with the difference that they need not have a common measure. So rather than the relatively prime/relatively composite distinction, one has a (rather different) commensurable/incommensurable distinction for magnitudes. And in the incommensurable case, during the procedure "that which is left never measures the one before it", implying that the procedure never terminates. Maybe I haven't looked well enough, but it would seem that Euclid isn't very strong in proving the existence of incommensurable magnitudes. The example given (the same one I gave above) doesn't seem to be Euclid's, and Book X, Proposition 10, apart from being non genuine, fails to prove the existence of (integer) ratios that are not the ratio of two square numbers. For instance he does not show (at least not at that proposition) why $1:2$ cannot also be the ratio of two square numbers, even though the proof would be easy. - Right: The Greeks didn't think of A mod B; they thought of "the line segment left over when you start with a line segment of length A and subtract as many line segments of length B as possible". – Dan May 29 '12 at 14:09 "Greek mathematicians were more at ease with geometry than with numbers" isn't a very precise statement of the matter to say the least. Numbers, i.e. 2,3,4,5,6,... were something Euclid was quite comfortable with; he used the word "arithmoi" frequently. The Greeks had no concept at all of what we now call "real numbers". That's not just "discomfort" with them. They had a concept of congruence of line segments; they had a concept of laying a specified number of segments end to end; the had a concept of the ratio of two lengths being the same as.... – Michael Hardy May 29 '12 at 17:48 ....another ratio of two lengths even when the ratios were of incommensuable pairs of segments, etc. They could multiply two lengths and get an area. But you couldn't say the area was greater than or less than or equal to any particular length: these were not real numbers. – Michael Hardy May 29 '12 at 17:49 – Robert Israel May 30 '12 at 7:51 He is talking about numbers here, not lengths. He needs two propositions, one for the case of relatively prime numbers and one for the greatest common measure of two numbers that are not relatively prime, because for the Greeks $1$ was not a "number". – Robert Israel May 30 '12 at 7:54 Yes, your formula yields the unique extension of $\rm\:gcd\:$ from integers to rationals (fractions), presuming the natural extension of the divisibility relation from integers to rationals, i.e. for rationals $\rm\:r,s,\:$ we define $\rm\:r\:$ divides $\rm\:s,\:$ if $\rm\ s/r\:$ is an integer, in symbols $\rm\:r\:\|\:s\:$ $\!\iff\!$ $\rm\:s/r\in\mathbb Z.\:$ [Such divisibility relations induced by subrings are discussed further here] Essentially your formula for the gcd of rationals works by scaling the gcd arguments by a factor that yields a known gcd (of integers), then performing the inverse scaling back to rationals. Even in more general number systems (integral domains), where gcds need not always exist, this scaling method still works to compute gcds from the value of a known scaled gcd, namely $\rm{\bf Lemma}\ \ \ gcd(a,b)\ =\ gcd(ac,bc)/c\ \ \ if \ \ \ gcd(ac,bc)\$ exists $\rm\quad$ Therefore $\rm\ \ gcd(a,b)\, c = gcd(ac,bc) \ \ \ \ \ if\ \ \ \ gcd(ac,bc)\$ exists $\quad$ (GCD distributive law) The reverse direction fails, i.e. $\rm\:gcd(a,b)\:$ exists does not generally imply that $\rm\:gcd(ac,bc)\:$ exists. $\$ For a counterexample see my post here, which includes further discussion and references. More generally, as proved here, we have these dual formulas for reduced fractions $$\rm\ gcd\left(\frac{a}b,\frac{c}d\right) = \frac{gcd(a,c)}{lcm(b,d)}\ \ \ if\ \ \ \gcd(a,b) = 1 = \gcd(c,d)$$ $$\rm\ lcm\left(\frac{a}b,\frac{c}d\right) = \frac{lcm(a,c)}{gcd(b,d)}\ \ \ if\ \ \ \gcd(a,b) = 1 = \gcd(c,d)$$ Some of these ideas date to Euclid, who computed the greatest common measure of line segments, by anthyphairesis (continually subtract the smaller from the larger), i.e. the subtractive form of the Euclidean algorithm. The above methods work much more generally since they do not require the existence of a Euclidean (division) algorithm but, rather, only the existence of (certain) gcds. - I don't follow- $\mathbb Q$ is a field. What is the meaning of $gcd(x,y)$ where $x,y$ are members of a field ? – users31526 May 30 '12 at 4:10 @Kuashik As mentioned in the first paragraph, divisibility is w.r.t. to a specified subdomain Z of "integers". In an integral domain, $\rm\:d = gcd(a,b)\:$ is a greatest common divisor of $\rm\:a,b\:$ if it is a common divisor $\rm\:d\:\|\:a,b\:$ that is (divisibly) greatest, i.e. divisible by every common divisor, i.e. $\rm\: c\:\|\:a,b\Rightarrow c\:\|\:d.\:$ Said more universally $\rm\:c\:\|\:a,b\iff c\:\|\:gcd(a,b),\:$ analogous to $\rm\:c\le a,b\iff c\le min(a,b).\:$ See here for more on such universal definitions of GCD and LCM. – Gone May 30 '12 at 4:31 @ Bill Dubuque: There should be a sense of ordering in a Integral Domain otherwise no one can say which one is greater among $x,y$ where $x,y$ are members of an Integral Domain. In a field except additive identity (usually we say $0$) "everyone divisible by everyone". So in that sense, I can say $gcd(\frac{13}{2},\frac{3}{4})=100$ or $1000$.... I have doubt that's why I am asking don't get angry – users31526 May 30 '12 at 4:47 @Kuashik The (partial) order is that given by divisibility, as I said. This is the standard definition of gcd in a general integral domain. It specializes to the well-known definition in $\mathbb Z$ (or any Euclidean domain). Again, the divisibility relation used in $\mathbb Q$ is the natural extension of integer divisibility - see the first paragraph of my answer. – Gone May 30 '12 at 4:53 @ Bill Dubuque: Yes, any Integral domain $R$ has natural partial order (antisymmetric, antireflexive, transitive) say $x\leq y\iff \exists$ non unit $c\in R$ such that $y=cx$. Then one can talk about chain and one can talk about $gcd(x,y)$ where $x,y\in R$ – users31526 May 30 '12 at 5:19 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": 81, "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": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9595109224319458, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/tagged/taylor-expansion+trigonometry	# Tagged Questions 2answers 40 views ### Taylor Series of $\sin 2x$ finding $f^{(n)} (a)$ where $a = 0$ ok so i get; f (x) = sin 2x f ' = 2cos 2x f '' = -4sin 2x f ''' = -8cos 2x f '''' = 16sin 2x f ''''' = 32cos 2x f (0) = 0 f '(0) = 2 f ''(0) = 0 f '''(0) = -8 f ''''(0) = 0 f '''''(0) = 32 ... 2answers 165 views ### Finding the 9th derivative of $\frac{\cos(5 x^2)-1}{x^3}$ How do you find the 9th derivative of $(\cos(5 x^2)-1)/x^3$ and evaluate at $x=0$ without differentiating it straightforwardly with the quotient rule? The teacher's hint is to use Maclaurin Series, ... 2answers 82 views ### Infinite (Taylor) Series What is the general term for a Taylor series of $\sin(x)$ centered at $\pi/4$? It should be $(-1)^{[??]} \times \sqrt{2}/2 \times \frac{(x-\pi/4)^n}{n!}$ What power is $(-1)$ supposed to be raised ... 1answer 253 views ### Approximating $\arctan x$ for large $\|x\|$ I would like to know if there is reasonably fast converging method for computing large arguments of arctan. Until now I've came across Taylor series that converges only on interval $(-1,1)$ and for ... 2answers 304 views ### How do we know Taylor's Series works with complex numbers? Euler famously used the Taylor's Series of $\exp$: $$\exp (x)=\sum_{n=0}^\infty \frac{x^n}{n!}$$ and made the substitution $x=i\theta$ to find $$\exp(i\theta) = \cos (\theta)+i\sin (\theta)$$ How ... 2answers 416 views ### Is the Maclaurin series expansion of $\sin x$ related to the inclusion-exclusion principle? When I see the alternating signs in the infinite series expansion of $\sin x$, I'm reminded of the inclusion-exclusion principle. Could there be any way to visualize it in such a way? Also, is there ... 1answer 192 views ### Taylor series expansion of $\sec(x +y^2)$ We have $f(x,y) = \sec(x+y^2)$ I want to find the first two non-zero terms of $f$ at $(0,0)$ starting by Taking the first few terms of $\cos x$ centered at zero, $1 - \frac{x^2}{2!}$ Using this ...	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 23, "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": 0.9124918580055237, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/49849/decomposition-of-an-abelian-von-neumann-algebra	## Decomposition of an abelian von Neumann algebra ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Hi, I came across the statement below and I couldn't figure out why it is true. I was hoping someone could explain it or give me a good reference. Thank you in advance. "Let $\pi$ be a non-degenerate representation of a separable C-algebra $A$ on a separable Hilbert space $H$. Suppose $M$ is an abelian von Neumann algebra contained in the commutant of $\pi(A)$. Then there exists a unique sequence of mutually orthogonal projections $(e_{n})$ in $M$ such that $\sum e_{n}=I$ and $M'e_{n}$ is spatially isomorphic to $L^{\infty}(E_{n})\otimes M_{n}(C)$. Moreover, each $e_{n}$ is invariant under spatial automorphisms of $M$. - ## 2 Answers Apparently, the data of π and A is irrelevant here. We might as well start with an arbitrary representation of an abelian von Neumann algebra M on a Hilbert space H. Geometrically, commutative von Neumann algebras are just measurable spaces (the opposite category of the category of commutative von Neumann algebras is equivalent to the category of (localizable) measurable spaces). Likewise, representations of von Neumann algebras on Hilbert spaces can be thought of geometrically as bundles of Hilbert spaces over measurable spaces, i.e., there is a similar equivalence of categories here. Spatial automorphisms of representations of von Neumann algebras are precisely automorphisms f of the underlying measurable space together with an isomorphism fV→V, where V is the bundle of Hilbert spaces corresponding to the representation. More generally, a morphism of von Neumann algebras M→N, where M is abelian, can be thought of as a bundle of von Neumann algebras over the spectrum (i.e., the corresponding measurable space) of M. Again, this can be stated as an equivalence of categories. In our case we have a morphism M→M' and M is abelian. The corresponding bundle of von Neumann algebras is precisely the bundle of (fiberwise) endomorphisms of the bundle V of Hilbert spaces corresponding to the representation of M on H. Denote by E_n the measurable subset of Spec M consisting of points with the dimension of the fiber of V being equal to n. Denote by e_n the corresponding projection in M. Observe that E_n are disjoint and ⋃_n E_n=Spec M, i.e., e_n are mutually orthogonal and ∑_n e_n=1. Moreover, M'e_n is the algebra of endomorphisms of the restriction of V to E_n. The latter bundle is trivial and has dimension n, hence its algebra of endormophisms is just L^∞(E_n)⊗M_n(C) as required. From the geometric interpretation of spatial isomorphisms it follows that every e_n is invariant under any spatial isomorphism. Finally, the sequence e is unique because the property M'f_n=L^∞(E_n)⊗M_n(C) forces every F_n to be a subset of E_n, and because ⋃_n F_n=Spec M we have F_n=E_n. - ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. It's a standard result in von Neumann algebras, called the homogeneous decomposition of a type I von Neumann algebra. To quote results from the literature, I will use Takesaki's book (volume I). Since your $M$ is an abelian von Neumann algebra, it is type I. Then it's commutant $M'$ is also type I (V.1.30). Then Theorem V.1.27 expresses exactly the result you are asking about. -	{"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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8771220445632935, "perplexity_flag": "head"}
http://mathhelpforum.com/calculus/204600-problems-spivak-calculus-chapter-6-problem-6-a.html	1Thanks # Thread: 1. ## Problems with Spivak Calculus Chapter 6 Problem 6 I have the solutions book and see the answers, but they make no sense to me. Any/All help is appreciated. a) Find a function that is discontinuous at 1, 1/2, 1/4, 1/4... but continuous at all other points. b) Find a function that is discontinuous at 1, 1/2, 1/4, 1/4... and at 0 but continuous at all other points. Problem 7 doesn't seem to have enough explanation in the solution either. How is f being continuous 0 used? 7) Suppose f satisfies f(x+y) = f(x) + f(y) and f is continuous at 0. Prove it is continuous at A for all values of A. 2. ## Re: Problems with Spivak Calculus Chapter 6 Problem 6 Attached Thumbnails 3. ## Re: Problems with Spivak Calculus Chapter 6 Problem 6 Well that certainly didn't help at all. Thanks, anyway. 4. ## Re: Problems with Spivak Calculus Chapter 6 Problem 6 For 6a, you can also consider $f(x)=\begin{cases}\frac{1}{k}& \frac{1}{2^k}<x\le \frac{1}{2^{k-1}}, k=1,2,\dots\\0&\text{otherwise}\end{cases}$ 5. ## Re: Problems with Spivak Calculus Chapter 6 Problem 6 Problem #7: Ignoring issues of the domain, $f$ continuous at $a$ means: $\lim_{x\to a} f(x) = f(a)$. Now observe that, always, $\lim_{x\to a} f(x) = \lim_{h\to 0} f(a + h)$. 6. ## Re: Problems with Spivak Calculus Chapter 6 Problem 6 Originally Posted by johnsomeone Problem #7: Ignoring issues of the domain, $f$ continuous at $a$ means: $\lim_{x\to a} f(x) = f(a)$. Now observe that, always, $\lim_{x\to a} f(x) = \lim_{h\to 0} f(a + h)$. I think that one also has to prove that f(0) = 0. 7. ## Re: Problems with Spivak Calculus Chapter 6 Problem 6 Thanks, I managed something with 7, but I still don't understand 6. Is there any way to show how you construct an answer? 8. ## Re: Problems with Spivak Calculus Chapter 6 Problem 6 Originally Posted by patrickmanning I still don't understand 6. Is there any way to show how you construct an answer? Can you construct a function that is discontinuous at x = 0? For example, f(x) = 0 for x <= 0 and f(x) = 1 for x > 0. If this is clear, then what is difficult about problem 6? I suggested a function whose graph looks like a staircase. 9. ## Re: Problems with Spivak Calculus Chapter 6 Problem 6 Originally Posted by emakarov Can you construct a function that is discontinuous at x = 0? For example, f(x) = 0 for x <= 0 and f(x) = 1 for x > 0. If this is clear, then what is difficult about problem 6? I suggested a function whose graph looks like a staircase. In your example in this post, f(x) is really discontinuous at x = 0? It seems I don't even understand continuity properly. At the point where there is a jump, that point is discontinuous? Because it is different when approached from left and right sides, right? Wouldn't that mean in your first example, f(x) was also discontinuous at x 0? 10. ## Re: Problems with Spivak Calculus Chapter 6 Problem 6 Originally Posted by patrickmanning At the point where there is a jump, that point is discontinuous? Because it is different when approached from left and right sides, right? Yes. Would you agree that a person's spacial position as a function of time is a continuous function? If f(t) = 0 for t <= 0 and f(1) = 1 for t > 0 described my spacial position, then I would be at position 0 before time 0 and then suddenly move to position 1 (say, 1 mile from position 0). If you know how to construct such transportation, I would certainly like to hear! Formally, f(t) is not continuous at 0 because any interval containing t = 0 is mapped by f to a set that contains some $y_1$ and $y_2$ such that $\|y_1 - y_2\| = 1$. In other words, however small you make an interval around 0, the image of that interval contains numbers that are sufficiently far apart, i.e., not arbitrarily close. It does not matter how precisely you approximate t = 0: if you change t just a little bit, f(t) can change by 1. 11. ## Re: Problems with Spivak Calculus Chapter 6 Problem 6 Thank you. I hate to be a bother, but can you continue helping by answering my question about your first example? Isn't f(x) discontinuous at x = 0 in your example with f(x) = 1/k? 12. ## Re: Problems with Spivak Calculus Chapter 6 Problem 6 Originally Posted by patrickmanning sn't f(x) discontinuous at x = 0 in your example with f(x) = 1/k? As defined in post #4, f(x) = 0 for $x\le 0$. As far as x > 0, f(x) decreases as x decreases. In fact, f(x) tends to 0 as x tends to 0 from the right. Since $\lim_{x\to0}f(x)=f(0)$, f is continuous at 0. 13. ## Re: Problems with Spivak Calculus Chapter 6 Problem 6 Thank you. I understand well now. Your example actually is not discontinuous at 1/3 then. Changing it a bit to: f(x) = 1/b , 1/a < x <= 1/b where a = b+1 and b = 1, 2, 3... will fix that. 14. ## Re: Problems with Spivak Calculus Chapter 6 Problem 6 Originally Posted by patrickmanning Your example actually is not discontinuous at 1/3 then. Changing it a bit to: f(x) = 1/b , 1/a < x <= 1/b where a = b+1 and b = 1, 2, 3... will fix that. I though that "1, 1/2, 1/4, 1/4..." from post #1 was supposed to mean "1, 1/2, 1/4, 1/8...," but you are right if it should be "1, 1/2, 1/3, 1/4..." 15. ## Re: Problems with Spivak Calculus Chapter 6 Problem 6 Haha, you are right. I didn't realise I typed 1/4 twice. Thank you so much for everything. I understand this all now. For 6b, you can just change it to "otherwise, x = -1" and now x = 0 will be discontinuous as from the left, f(x) approaches -1 but from the right, f(x) still approaches 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": 0, "codecogs_latex": 14, "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": 0.9464513063430786, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/46541/how-to-introduce-notions-of-flat-projective-and-free-modules/46627	## How to introduce notions of flat, projective and free modules? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) In the coming spring semester I will be teaching for the first time an introductory (graduate) course in Commutative Algebra. As many people know, I have been plugging away for a while at this subject and keeping my own lecture notes. So I feel relatively prepared to teach the course in the sense that I know more than enough theorems and proofs to cover. However, there is more to teaching a graduate course than just theorems and proofs. Commutative algebra has a reputation for being somewhat dry and unmotivated. After something like 10 years of hard work (over a period of about 15 years!), I am at a point where I find the subject both interesting in and of itself and useful. But how to communicate this to students? Looking through my notes, the technicalities begin with the introduction of various fundamental classes of modules, especially free, projective and flat modules (but also including other things like finite generation and finite presentation). I remember well that when I first learned this material, projective and (especially) flat modules were a tough sell: for instance, the first time I picked up Bourbaki's Commutative Algebra out of curiosity, I saw that the very first chapter was on flat modules, and I put it down in horror. How would you introduce these concepts to an early career graduate student? What are the fundamental differences between these classes of modules, and why do we care? Here are some of my preliminary thoughts: 1) Free modules are of course an easy sell: for such things the usual notions of linear algebra work well, including that of dimension (called "rank" in this case: note that my rings are commutative!). One can easily prove that for a commutative ring R, all R-modules are free iff R is a field, so the need to move beyond free modules is easily linked to the ideal theory of R. 2) Projective modules are important for at least the following reasons. a) Geometric: A finitely generated module over a ring R is projective iff it is locally free (in the stronger sense of an open cover of $\operatorname{Spec} R$). In other words, projective modules are the way to express vector bundles in algebraic language. I plan to drive this point home by discussing Swan's theorem on modules over $C^{\infty}(M)$. b) Homological: projective modules play a distinguished role in homological algebra, i.e., in the construction of left-derived functors. c) K-theoretic: It is of interest to know to what extent finitely generated projective modules must be free. This leads to the construction of $K_0(R)$ -- i.e., stable isomorphism classes of projective modules -- and in the rank one case, to the Picard group $\operatorname{Pic}(R)$. (After writing the above, I feel that in some sense it is a good enough answer -- certainly these are three important reasons for studying projective modules. But I'm not sure how to explain them to beginning students. Let's be clear that this is part of the question.) 3) Flat modules: this is harder to explain! a) Geometric: Flatness is the "right" condition for things to vary nicely in families, but this is more of a mantra than an explanation. I think that most people hear this at one point and come to believe and understand it slowly over time. b) Homological: Flat modules are those which are acyclic for the Tor functors. But this is not a homological algebra course: I would be happiest not to mention Tor at all. c) Near equivalence with projective modules in the finitely generated case: the difference between finitely generated flat and finitely generated projective modules is very subtle. Recall for instance: Theorem: For a finitely generated flat module $M$ over a commutative ring $R$, TFAE: (i) $M$ is projective. (ii) $M$ is finitely presented. (iii) The associated rank function is locally constant. Thus in almost all of the cases of importance to a beginning algebra student -- e.g. if $R$ is Noetherian or is an integral domain -- finitely generated flat modules are projective. d) However infinitely generated flat modules are a much bigger class, since flatness is preserved under direct limits. In particular, there are large classes of domains $R$ for which a module is flat iff it is torsionfree (namely this holds iff $R$ is a Prufer domain; even the case $R = \mathbb{Z}$ is useful). Perhaps this is significant? Your insight will be much appreciated. Update: the prevailing sentiment of the answers thus far seems to be that a very little bit of homological algebra will go a long way in presenting the basic definitions in a unified and useful way. Duly noted. - 3 In my opinion, the very basic projective non-free modules are (fractional) ideals in number rings (and Dedekind domains); this particular case of Pic is very intriguing for those who are interested in algebraic number theory. By the way, do you want to introduce the defintion of a Dedekind domain (early in your course)? – Mikhail Bondarko Nov 18 2010 at 21:55 2 For projective modules, you might start with the observation that many of the nice properties of free modules follow only from the lifting property. – Kevin Ventullo Nov 18 2010 at 22:51 1 This may or may not be helpful to you, but there are many different characterizations of flatness in Lam's book Lectures on Modules and Rings, Ch. 4. Flatness can be viewed purely in terms of linear equations as in Thm. 4.24, the "equational criteria for flatness." This may not add much intuition, but it at least avoids any mention of homological algebra. – Manny Reyes Nov 18 2010 at 23:10 1 Why would you avoid mentioning "homological algebra"(whereby one means, in this context "preservation of exactness properties") when introducing a property which is essentially of homological nature? – Mariano Suárez-Alvarez Nov 19 2010 at 0:13 2 IMHO, the motivation of flatness based on the single case in which one considers the problem of when extension of scalars along a map of rings $A\to B$ preserves exactness is quite good already. In fact, I think that extension of scalars is a great way to motivate tensor products, because it is very very concrete and natural. – Mariano Suárez-Alvarez Nov 19 2010 at 0:39 show 3 more comments ## 11 Answers Hi Pete, this sounds like a lot of fun! I wish I could be there (-: Here is a concrete and useful property of flatness, you can explain it without using Tor. Suppose $R\to S$ is a flat extension. Then if $I$ is an ideal of $R$, tensoring the exact sequence: $$0 \to I \to R \to R/I \to 0$$ with $S$ gives that $I\otimes_RS = IS$. The left hand side is somewhat abstract object, but the right hand side is very concrete. There are very natural extensions which are flat but not projective. For example, if $R$ is Noetherian and $\dim R>0$, then $S=R[[X]]$ is flat but never projective over $R$. - 3 @Hailong: you can be there as far as I'm concerned. Please feel free to drop by, especially if you don't mind being handed the chalk. – Pete L. Clark Nov 19 2010 at 3:23 Dear Long, Thanks for the interesting reference! – Emerton Dec 3 2010 at 8:15 This is indeed a very nice motivation of flatness. If you wanted even less homological algebra and your students already believed free and projective modules were interesting then I suppose you could introduce flat modules as direct limits of f.g. free modules. After talking about direct summands of free modules, I can't imagine people would be shocked when you asked about direct limits. – David White Jun 22 2011 at 14:35 ### 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. All of this is coming from the point of view of being a student. I took a commutative algebra course last winter and am currently taking a homological algebra course, so maybe I can touch on your differentiation between those two perspectives. It sounds like you want the motivating ideas, and I think those make the most sense homologically. I completely understand your not wanting to introduce Tor and/or Ext, but you don't have to. $Hom_R (P,-)$ is exact iff $P$ is projective, $Hom_R (-,I)$ is exact iff $I$ is injective, and $F \otimes_R -$ is exact iff $F$ is flat. These can even be taken as definitions (these are equivalent to $Tor_1$ and $Ext^1$ vanishing which is each equivalent to all the higher derived functors vanishing). Edit: What if you don't know why you should care about exact sequences as Pete asked in a comment? Well, you should still care about when functors preserve injections (tensor-ing with flat modules) and surjections (Hom-ing out of projective modules), and this is precisely what flat and projective do for you. This is the only place where exactness might fail, so just skip it and talk about injections and surjections and when they are preserved. This is how I feel most comfortable thinking about these concepts. It is pretty concrete and not far away from Commutative Algebra (I think, meaning that this is how we talked about it in my commutative algebra class). I should also mention I appreciate you posting all of those course notes, it is a great resource for us students out there. - 1 @Sean: thanks. I agree that these are nice, clean definitions. But I think more motivation would be helpful: why do we care whether some functor is exact? – Pete L. Clark Nov 18 2010 at 22:41 4 @Pete: In my mind, the exactness of the tensor product is relatively easy to motivate through change of rings. Say you've got your sequence over $R$ and you've got an ideal $I$. Is it still exact when we think of everything as $R/I$ modules? Well, that's what the exactness of the tensor product will tell you. Of course this becomes very useful once you have localization and completions to play with. For $Hom$, it seems easiest to motivate $Hom(-,R)$ by analogy with vector space duals, and you can build up from there to $Hom(-,M)$ and $Hom(M,-)$. – Justin DeVries Nov 18 2010 at 23:42 Please forgive these very naïve remarks. I am enjoying the chance to learn something about flatness in trying to contribute to this question. First of all, since as pointed out here, the definition of flatness is that the object behaves as simply as possible under tensor product, it follows that the primary use of the concept is in applications of the tensor product. Now there are three of these that come to mind, (after some review of the literature), namely 1) forming fibers, 2) localizing, (and changing rings) 3) completions. These are all local operations, looking at the inverse image of one point, restricting to a Zariski neighborhood of a point, and restriction to an analytic or formal neighborhood of one point. Thus one wants to compare the geometry at a point with the geometry near that point. E.g. given an algebraic subvariety through a point, one can take its algebraic germ, formal germ, or analytic germ, and then ask whether one can recover the original germ from these. This is equivalent to asking whether one recovers the original ideal after extending to the localization or completion, and then restricting back to the original ring. Flatness is the property that tells us yes to all these questions. I.e. both the localization and the completion are flat over the original ring. Moreover, the algebraic and analytic local rings form a “flat pair”, slightly stronger than one being flat over the other, and this apparently implies they have the same completions. This lets us compare analytic and algebraic local rings, by comparing both with their completions. It follows e.g. that the algebraic dimension of an algebraic variety equals the analytic dimension of the associated analytic variety. Reasoning of this sort allows Serre to prove geometric results such as those mentioned above as well as homological ones. Homological results desired are of the sort that compare the analytic cohomology to the algebraic cohomology. The simplest way to do this is to show that the analytic sheaves and their analytic cohomology is obtained from the algebraic ones by tensoring with flat objects, i.e. changing rings in the simplest way. Then the desired results say that homomorphisms of sheaves, and cohomology of sheaves, commutes with this process of tensoring, i.e. of applying the functor of making these objects analytic. Flatness is the key to all these results. Thus flatness of one object over another seems to imply a relation between their local geometric structures. This is my take on Serre’s lovely paper GAGA, available free online at NUMDAM, after a brief perusal. It certainly looks worth a careful read. We have mentioned before the result that a surjective morphism of smooth varieties is flat if and only if the fibers have constant dimension. From perusing Matsumura, other geometric properties of flat maps seem to be: the dimension of (non empty) fibers is always the expected one, i.e. dim(source) - dim(target); the going down theorem holds, hence every subvariety through the point f(p) is the image of a subvariety through p, in particular a flat map cannot separate branches at a point; indeed flat maps cannot remove singularities in any fashion, i.e. f is flat and if f(p) is a regular point, so is p; and a birational flat map is an open immersion. Another remark building on those above is that flatness is a natural weakening of the property of projectiveness, and hence for local rings, of freeness. If finite map behaves well locally when it defines a locally free module, what about a map that lowers dimension? What is the closest thing to locally free that still holds when the fibers vary nicely but not smoothly? I.e. a flat map is a slight weakening of a smooth map. I am struggling here from ignorance. Thanks for the question! - 6 Dear Roy: Last spring I asked Serre about the history. He viewed it as a purely algebraic notion, inspired by usefulness for the back-and-forth passage between analytic and algebraic local rings through their common completion (noetherian property of the former being pretty serious, of course) and clarifying localization. It came as a total surprise to him later that flatness is an important concept in algebraic geometry, and he said that all credit for its geometric significance belongs to Grothendieck; he (Serre) did not anticipate it would be of interest beyond pure algebra. – BCnrd Nov 19 2010 at 4:43 Thanks. Do you have a favorite reference for where Grothendieck exploits it, analogous to this one I just perused by Serre? It seems to me that even in GAGA Serre shows how it relates local geometric concepts like dimension, but maybe I am stretching, if he says otherwise! – roy smith Nov 19 2010 at 4:53 BC, I agree there is no visible occurrence in GAGA of the geometry of a general flat map. I was just to imagine how the idea extends. So that was Grothendieck's contribution, that's nice to know. – roy smith Nov 19 2010 at 5:02 3 Dear Roy: My impression is that Serre viewed local analytic dimension as part of algebra. Usefulness of flatness for geometric problems (Hilbert schemes, deformation theory, etc.) is what I think surprised him. There are too many amazing geometric uses of flatness by Grothendieck, so no "favorite": fpqc descent (to work intuitively with $G/H$ and etale topology, etc.), fibral flatness criteria, openness results for loci defined by fibral properties relative to proper fppf maps, flatness of formally etale lfp maps,... Some of your examples are really about excellence (another miracle!), btw. – BCnrd Nov 19 2010 at 5:22 Gee, this is nice. In line with the idea that flat maps are a slight weakening of smooth ones, one can just read the titles of the first few sections of SGA 1: i.e. etale maps, smooth maps, flat maps,..... Thanks for prompting me to look at Grothendieck. – roy smith Nov 19 2010 at 5:24 show 1 more comment For me, the most intuitive characterization of flatness of an $R$-module $M$ is the "principle of sufficient reason": Whenever elements `$x_i$` of $M$ satisfy a linear equation with coefficients in $R$, say `$\sum_{i=1}^n r_ix_i=0$`, then this is "because" of linear equations holding in $R$; that is, the $x_i$ can be expressed as linear combinations of other elements `$y_j\in M$`, say `$x_i=\sum_j s_{ij}y_j$` (where `$s_{ij}\in R$`), such that `$\sum_i r_is_{ij}=0$` for all $j$. Thus the original linear relation among the $x$'s follows from the expressions in terms of the $y$'s plus properties in $R$ of the coefficients `$r_i$` and `$s_{ij}$`. Admittedly, this description of flatness doesn't help much with geometry, but it clarifies the algebraic situation. The same intuition also applies, for example, to the notion of "flat functor" that occurs in the study of classifying topoi. - 1 @Andreas: right, this is the "equational criterion of flatness" that Manny Reyes referred to above. I do plan on covering it (it's in my private copy of my notes, although not yet the one posted on my webpage) because it's used to prove that a finitely generated flat module over a local ring is projective. I admit though that I don't have a lot of intuition for what this criterion "really means". Any further hints would be appreciated. – Pete L. Clark Nov 19 2010 at 3:19 2 The equational criterion more or less amounts to the statement: "$M$ is flat IFF every homomorphism from a finitely presented module factors through a free module". (With the instant consequence that flat & finitely presented imply projective.) – Charles Rezk Nov 19 2010 at 3:39 (every homomorphism from a finitely presented module to $M$, of course) – Charles Rezk Nov 19 2010 at 3:41 Pete I agree that locally free is a good geometric intuition for projective. In my algebra class I gave an exercise for the students to prove that the module of tangent vector fields on a 2-sphere is locally free over the ring of smooth functions, and challenged them to show it is not free. If you introduce direct limits, maybe the fact you mentioned about flatness being preserved is a way to motivate it. I.e. you want to take direct limits of finite generated projectives, but they lose the property of projectivity. So what property do they keep? But that may be artificial. it is hard to get away from the points made by Sean Tilson. I guess the difficulty there is that Hom is for many of us more intuitive than tensor. So lifting and extension problems are more intuitive than the distinction between IM and ItensorM. You might review your reasons why flat modules are in the course. If you have an application for them in mind, maybe the application can motivate the concept. In the spirit of "flat families", I used to cling to the geometric fact that a surjective morphism of smooth algebraic varieties is flat if and only if the fiber dimension is constant, but I don't know if you can use that. Those "local criteria for flatness" are usually among the more advanced flatness topics. E.g. that is treated in the books of Matsumura near the end. But this suggests that in some contexts flatness is the algebraic version of locally constant geometry. I have heard that Serre focused on the concept in some of his work relating algebraic and analytic varieties. Maybe the properties he used are illuminating. Great question. - 6 @Roy: +1 for the audacious suggestion that I track down why I am discussing flat modules in the first place. (If I found out that I had no good application, my natural reaction would be to find one, rather than cut this important concept.) – Pete L. Clark Nov 18 2010 at 22:50 This was one of the aspects of algebra that I enjoyed the most while first learning it. This is the way I would develop the subject: 1) Introduction to exact sequences with an emphasis on short exact sequences. We can use these to illustrate how the properties of any module in such a sequence are related to those of others. Relevant here would be discussions of theorems such as, length of a module is additive, a module has ACC/DCC if and only if a submodule and quotient modulo the submodule have ACC/DCC. We can also develop direct sums via short exact sequences, the short five lemma, the snake lemma here. 2) Projective modules: Given a short exact sequence, $0\to L\to M\to N\to 0$, of $R$-modules, and some other $R$-module $T$ and a homomorphism $T\to L$ there exists a homomorphism $T\to M$ via composition. We can ask the opposite question. When does giving a homomorphism $T\to M$ give a homomorphism $T\to L$. This can be used to motivate $Hom_R(T,-)$. Then we can go on to show this functor is left exact. Then we define projective modules as those which make this functor exact. 3) At this point I would introduce Free modules and motivating them via vector spaces as you suggested. I would also discuss free resolutions since this is a very elegant machinery. Then, I would show that projective modules are just direct summands of free modules. 4) Injective modules: A similar development as for projective modules, this time via the functor $Hom_R(-,T)$. I would also develop the characterizations of injective modules via Baer's theorem and divisible modules. 5) Flat modules: I shall then introduce the functors $T\otimes -$ and $- \otimes T$ and show these are right exact. Then define flat modules as those which make the above functors exact. I shall also discuss that projective modules are flat. To tie all of this together, I shall then discuss the adjointness of Hom and $\otimes$ followed by Homological algebra if that is part of the course. - @Timothy: thanks for your response. Interestingly (to me!), in my notes I currently do not discuss injective modules at all. Obviously in some sense I should, but I think there is something to ponder here in the notion that in commutative algebra itself, projective modules are much more important than injective modules. (Or do people disagree with this? Let me know!) – Pete L. Clark Nov 18 2010 at 22:52 @Timothy: thinking out loud...whether a projective module is free is one of the great questions of commutative algebra. Asking whether an injective module is cofree does not have the same allure. In my experience, although it is very useful to know that an abelian category has enough injectives, this does not necessarily make injective objects worthy of study. I think that I have never met an injective sheaf in person, for instance. – Pete L. Clark Nov 18 2010 at 22:57 2 @Pere: As someone who studies algebra, I find injective modules are quite ubiquitous. They show up as injective hulls which is a central notion in local cohomology. – Timothy Wagner Nov 18 2010 at 22:59 @Timothy: fair enough. I don't know anything about local cohomology. I do remember injective envelopes coming up in Part III of Serre's book on representations of finite groups, albeit for modules over non-commutative rings. I would be happy to hear of a reference, so as to broaden my horizons... – Pete L. Clark Nov 19 2010 at 3:59 1 @Timothy: thanks very much for the link, and please don't worry about calling me "Pere": I knew you were talking to me and not to your father. – Pete L. Clark Nov 19 2010 at 4:56 show 1 more comment Another student answer: Considering 2b, it seems a bit strange that you want to mention derived functors and not to go into $\mathbf{Tor}$ and $\mathbf{Ext}$. The problem with derived functors is that first of all, categories usually have enough injective objects. Secondly, projectives are just a suitable collection of objects to use, but not always. However, the universal property is really important here. Lifting morphisms is something extremely natural to ask (which is the exactness of $\mathbf{Hom}$). This is also a way to mention that projective modules are direct summands of free (which fits perfectly into you sequence of presentation). Speaking of flat modules, I'd avoid saying about "varying nicely". It's not very convincing the first time you see it (like telling a child that to have a baby one should kiss his partner). I'd also be nice to somehow sneak in the adjointness of $\otimes$ and $\mathbf{Hom}$. - @Anton: sorry, maybe I wasn't clear. When I say that I would ideally like to avoid derived functors, that includes Tor and Ext. I am however willing to reconsider that -- a lot of people are saying that it's very important. – Pete L. Clark Nov 19 2010 at 3:22 +1 for "like telling a child that to have a baby one should kiss his partner." – Qiaochu Yuan Nov 20 2010 at 0:15 You might perhaps mention Lazard's theorem. - @Greg: Thanks. The Govorov-Lazard Theorem (a module is flat iff it can be expressed as the direct limit of finitely generated free modules) is in the latest copy of my notes. – Pete L. Clark Nov 19 2010 at 20:58 In the question there have been already mentioned many deep applications of these notions for modules (flat, projective, free). However, I think that an introductory course should not focus on these aspects. It would be very nice (at least for the ones who can follow...) if some applications are at least mentioned and are partially worked out as exercises. The students will learn the "big story" later, perhaps even in the same course of commutative algebra. But as for the introduction of these notions, I would just use the standard definitions (see Sean Tilson's answer). They later also apply in abelian categories (whereas, "free" generalizes rather to left adjoint functors). I think an introductory course should show the students how to work with the new objects and what can be done with them. For example, it is pretty useful that a short exact sequence $0 \to I \to M \to P \to 0$ splits if $P$ is projective, and that the converse is also true: If $P$ has the property that every short exact sequence of the above type splits, then $P$ is projective. By duality, there is a dual statement characterizing the injectivity of $I$. A propos, it should be partially pointed out that the notions of projective and injective are dual to each other, and the reason that they differ so horribly in $R$-Mod comes from the fact that this category has no autoequivalence. Concerning flatness, it should be pointed out that they are stable under directed colimits and have various other nice local properties which fails for the notions of projective and free. Here is a beautiful theorem which fascinated me when I started to learn commutative algebra and algebraic geometry. It basically says that freeness of a module "extends from points to open neighborhoods". Let $M$ be a finitely presented $A$-module and $\mathfrak{p} \subseteq A$ a prime ideal such that $M_\mathfrak{p}$ is free. Then there is some $f \in A \backslash \mathfrak{p}$, such that $M_f$ is free. I would certainly include this in the course. The proof is also pretty illuminating, using some of the devissage arguments which are important in commutative algebra and algebraic geometry. - The question reminds me of the time when I was studying mathematics. I had attended a course on algebra (some basic theory of group, rings, categories, loads of Galois theory and some valuation theory, which was the main area of research of the lecturer). Commutative algebra was the first subject I studied on my own, because I wanted to attend a course on algebraic curves, and because I needed some commutative algebra for my diploma thesis. This is not a direct answer to your question, but maybe some thoughts from that time are of use... Literature: As a student I found Bourbaki, Nagata (local rings) and Matsumura (Comm. rings) too difficult as a starting point. I liked the book of Atiyah-MacDonald and I loved the book "Kommutative Algebra" of Brüske, Ischebeck, Vogel. The Brüske-Ischebeck-Vogel book is out of print and available online http://wwwmath.uni-muenster.de/u/ischebeck/SkriptBrskeIschebeckVogel.pdf. If I had to teach a course on commutative algebra, then I would surely have this book on my desk again. Unfortunately it is written in German, but nevertheless it might be worth to have a look... Free, projective and flat modules: I remember that I needed them for my thesis project and my thesis contained a section summarizing these things. I think at that time I was fine with the account in the BIV book. They define M projective iff $Hom(M, -)$ is exact, flat iff $M\otimes -$ is exact and injective if $Hom(-, M)$ is exact. I found this natural at a first reading, and later on the step to $Ext$ and $Tor$ was quite natural as well. (Maybe I was influenced a bit by the fact that categories and functors were always in the air in Munich at that time and were mentioned in my algebra course.) Having these definitions at hand, one can directly go into the proof of theorems comparing these classes of modules... - Coincidentally, I was on the verge of posing a flatness question. But it concerned flat homomorphisms of commutative rings, rather than flat modules. I finally slogged through Durov's proof of the "affine base change theorem" for his generalized rings. As far as I can tell, the only use he makes of flatness is the fact that it forces base change to preserve finite products. No one has mentioned that motivation so far. Is that a typical use? - Yes flatness is a typical ingredient for base change theorems in algebraic geometry. – Martin Brandenburg Jan 31 2012 at 10:22	{"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": 85, "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": 0.9526337385177612, "perplexity_flag": "head"}
http://en.wikipedia.org/wiki/Repeat-accumulate_code	# Repeat-accumulate code In computer science, repeat-accumulate codes (RA codes) are a low complexity class of error-correcting codes. They were devised so that their ensemble weight distributions are easy to derive. RA codes were introduced by Divsalar et al. In an RA code, an information block of length ${N}$ is repeated ${q}$ times, scrambled by an interleaver of size ${qN}$, and then encoded by a rate 1 accumulator. The accumulator can be viewed as a truncated rate 1 recursive convolutional encoder with transfer function ${1/(1 + D)}$, but Divsalar et al. prefer to think of it as a block code whose input block ${(z_1, \ldots , z_n)}$ and output block ${(x_1, \ldots , x_n)}$ are related by the formula ${x_1 = z_1}$ and $x_i = x_{i-1}+z_i$ for $i > 1$. The encoding time for RA codes is linear and their rate is $1/q$. They are nonsystematic. ## References • D. Divsalar, H. Jin, and R. J. McEliece. "Coding theorems for ‘turbo-like’ codes." Proc. 36th Allerton Conf. on Communication, Control and Computing, Allerton, Illinois, Sept. 1998, pp. 201–210.	{"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": 10, "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": 0.9333856701850891, "perplexity_flag": "middle"}
http://mathoverflow.net/revisions/96102/list	## Return to Question 4 added 7 characters in body Suppose you are given a single unit square, and you are permitted to cut it into $k$ (connected) pieces (where $k=1$ means just the square). Your task is to construct the largest volume convex body in $\mathbb{R}^3$ by pasting the $k$ pieces together as its complete surface; so its surface area is $\le 1$. Denote its volume by $V_\max(k)$. (This is a more general version of my earlier question, Covering a Cube with a Square.) The best one could hope for is to create a sphere with unit surface area, when the radius satisfies $4 \pi r^2 =1$ and so $r=\frac{1}{2\sqrt{\pi}} \approx 0.28$, and the volume is $V_\max(\infty)= V_\max = \frac{4}{3} \pi r^3 = \frac{1}{6 \pi} sqrt{\pi}} \approx 0.094$. In response to a question of Joe Malkevitch, with students I computed the optimum for $k=1$, with no cuts and no overlap, and found a maximum volume of $V_\max(1)\approx 0.056$, which is about 60% of the sphere volume. [p.418 in Geometric Folding Algorithms: Linkages, Origami, Polyhedra, Demaine, O'Rourke, Cambridge, 2007.]: The seeming randomness of this shape discouraged my further pursuit. The case $k=2$ resembles the famous tea-bag problem, but here I am restricting attention to convex bodies. As $k \to \infty$, one should be able to approach $V_\max$ by, for example, cutting the square into many nearly equilateral triangles and constructing a geodesic dome. I am wondering if anyone has heard of this problem in any guise before? And in any case, can anyone see any clear hypothesis, for any particular $k$? Presumably one could establish that $V_\max(k+1) > V_\max(k)$, but beyond that, I do not see how to make inroads. Is there an obvious candidate for $k=2$? Update. Here is my interpretation of Gerhard's suggestion for $k=2$: 3 added 233 characters in body Suppose you are given a single unit square, and you are permitted to cut it into $k$ (connected) pieces (where $k=1$ means just the square). Your task is to construct the largest volume convex body in $\mathbb{R}^3$ by pasting the $k$ pieces together as its complete surface; so its surface area is $\le 1$. Denote its volume by $V_\max(k)$. (This is a more general version of my earlier question, Covering a Cube with a Square.) The best one could hope for is to create a sphere with unit surface area, when the radius satisfies $4 \pi r^2 =1$ and so $r=\frac{1}{2\sqrt{\pi}} \approx 0.28$, and the volume is $V_\max(\infty)= V_\max = \frac{4}{3} \pi r^3 = \frac{1}{6 \pi} \approx 0.094$. In response to a question of Joe Malkevitch, with students I computed the optimum for $k=1$, with no cuts and no overlap, and found a maximum volume of $V_\max(1)\approx 0.056$, which is about 60% of the sphere volume. [p.418 in Geometric Folding Algorithms: Linkages, Origami, Polyhedra, Demaine, O'Rourke, Cambridge, 2007.]: The seeming randomness of this shape discouraged my further pursuit. The case $k=2$ resembles the famous tea-bag problem, but here I am restricting attention to convex bodies. As $k \to \infty$, one should be able to approach $V_\max$ by, for example, cutting the square into many nearly equilateral triangles and constructing a geodesic dome. I am wondering if anyone has heard of this problem in any guise before? And in any case, can anyone see any clear hypothesis, for any particular $k$? Presumably one could establish that $V_\max(k+1) > V_\max(k)$, but beyond that, I do not see how to make inroads. Is there an obvious candidate for $k=2$? Update. Here is my interpretation of Gerhard's suggestion for $k=2$: 2 added 80 characters in body Suppose you are given a single unit square, and you are permitted to cut it into $k$ (connected) pieces (where $k=1$ means no cuts)just the square). Your task is to construct the largest volume convex body in $\mathbb{R}^3$ by pasting the $k$ pieces together as its complete surface; so its surface area is $\le 1$. Denote its volume by $V_\max(k)$. (This is a more general version of my earlier question, Covering a Cube with a Square.) The best one could hope for is to create a sphere with unit surface area, when the radius satisfies $4 \pi r^2 =1$ and so $r=\frac{1}{2\sqrt{\pi}} \approx 0.28$, and the volume is $V_\max(\infty)= V_\max = \frac{4}{3} \pi r^3 = \frac{1}{6 \pi} \approx 0.094$. In response to a question of Joe Malkevitch, with students I computed the optimum for $k=1$, with no cuts and no overlap, and found a maximum volume of $V_\max(1)\approx 0.056$, which is about 60% of the sphere volume. [p.418 in Geometric Folding Algorithms: Linkages, Origami, Polyhedra, Demaine, O'Rourke, Cambridge, 2007.]: The seeming randomness of this shape discouraged my further pursuit. The case $k=2$ resembles the famous tea-bag problem, but here I am restricting attention to convex bodies. As $k \to \infty$, one should be able to approach $V_\max$ by, for example, cutting the square into many nearly equilateral triangles and constructing a geodesic dome. I am wondering if anyone has heard of this problem in any guise before? And in any case, can anyone see any clear hypothesis, for any particular $k$? Presumably one could establish that $V_\max(k+1) > V_\max(k)$, but beyond that, I do not see how to make inroads. Is there an obvious candidate for $k=2$? 1 # Maximum volume convex body coverable by a unit square Suppose you are given a single unit square, and you are permitted to cut it into $k$ (connected) pieces (where $k=1$ means no cuts). Your task is to construct the largest volume convex body in $\mathbb{R}^3$ by pasting the $k$ pieces together as its surface. Denote its volume by $V_\max(k)$. (This is a more general version of my earlier question, Covering a Cube with a Square.) The best one could hope for is to create a sphere with unit surface area, when the radius satisfies $4 \pi r^2 =1$ and so $r=\frac{1}{2\sqrt{\pi}} \approx 0.28$, and the volume is $V_\max(\infty)= V_\max = \frac{4}{3} \pi r^3 = \frac{1}{6 \pi} \approx 0.094$. In response to a question of Joe Malkevitch, with students I computed the optimum for $k=1$, and found a maximum volume of $V_\max(1)\approx 0.056$, which is about 60% of the sphere volume. [p.418 in Geometric Folding Algorithms: Linkages, Origami, Polyhedra, Demaine, O'Rourke, Cambridge, 2007.]: The seeming randomness of this shape discouraged my further pursuit. The case $k=2$ resembles the famous tea-bag problem, but here I am restricting attention to convex bodies. As $k \to \infty$, one should be able to approach $V_\max$ by, for example, cutting the square into many nearly equilateral triangles and constructing a geodesic dome. I am wondering if anyone has heard of this problem in any guise before? And in any case, can anyone see any clear hypothesis, for any particular $k$? Presumably one could establish that $V_\max(k+1) > V_\max(k)$, but beyond that, I do not see how to make inroads. Is there an obvious candidate for $k=2$?	{"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": 69, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9429001808166504, "perplexity_flag": "head"}
http://unapologetic.wordpress.com/2008/03/05/functions-of-bounded-variation-i/?like=1&source=post_flair&_wpnonce=bbf4ba6a50	# The Unapologetic Mathematician ## Functions of Bounded Variation I In our coverage of the Riemann-Stieltjes integral, we have to talk about Riemann-Stieltjes sums, which are of the form $\displaystyle\sum\limits_{i=1}^nf(t_i)(\alpha(x_i)-\alpha(x_{i-1}))$ How can we choose a function $f$ to make sums like this get really big? Well, we could make the function $f$ get big, but that’s sort of a cheap trick. Let’s put a cap that $\|f(x)\|\leq1$. Now how can we make a Riemann-Stieltjes sum big? Well, sometimes $\alpha(x_i)>\alpha(x_{i-1})$ and sometimes $\alpha(x_i)<\alpha(x_{i-1})$. In the first case, we can try to make $f(t_i)=1$, and in the second we can try to make $f(t_i)=-1$. In either case, we’re adding as much as we possibly can, subject to our restriction. We’re saying a lot of words here, but what it boils down to is this: given a partition we can get up to $\sum\limits_{i=1}^n\|\alpha(x_i)-\alpha(x_{i-1})\|$ in our sum. As we choose finer and finer partitions, it’s not hard to see that this number can only go up. So, if there’s an upper bound $M$ for our $\alpha$, then we’re never going to get a Riemann-Stieltjes sum bigger than $\max\limits_{x\in\left[a,b\right]}(f(x))M$. That will certainly make it easier to get nets of sums to converge. So let’s make this definition: a function $\alpha$ is said to be of “bounded variation” on the interval $\left[a,b\right]$ if there is some $M$ so that for any partition $a=x_0<...<x_n=b$ we have $\displaystyle\sum\limits_{i=1}^n\|\alpha(x_i)-\alpha(x_{i-1})\|\leq M$ That is, $M$ is an upper bound for the set of variations as we look at different partitions of $\left[a,b\right]$. Then, by the Dedekind completeness of the real numbers, we will have a least upper bound $V_\alpha(a,b)$, which we call the “total variation” of $\alpha$ on $\left[a,b\right]$ Let’s make sure that we’ve got some interesting examples of these things. If $\alpha$ is monotonic increasing — if $c<d$ implies that $\alpha(c)\leq\alpha(d)$ — then we can just drop the absolute values here. The sum collapses, and we’re just left with $\alpha(b)-\alpha(a)$ for every partition. Similarly, if $\alpha$ is monotonic decreasing then we always get $\alpha(a)-\alpha(b)$. Either way, $\alpha$ is clearly of bounded variation. An easy consequence of this condition is that functions of bounded variation are bounded! In fact, if $\alpha$ has variation $V_\alpha(a,b)$ on $\left[a,b\right]$, then $\|\alpha(x)\|\leq\|\alpha(a)\|+V_\alpha(a,b)$ for all $x\in\left[a,b\right]$. Indeed, we can just use the partition $a<x<b$ and find that $\displaystyle\|\alpha(x)-\alpha(a)\|\leq\|\alpha(x)-\alpha(a)\|+\|\alpha(b)-\alpha(x)\|\leq V_\alpha(a,b)$ from which the bound on $\|\alpha(x)\|$ easily follows. ### Like this: Posted by John Armstrong \| Analysis ## 12 Comments » 1. The strict inequality \|f\|<1 later turns out to be not so strict in your post. Comment by Johan Richter \| March 5, 2008 \| Reply 2. Thanks for catching that. Comment by \| March 5, 2008 \| Reply 3. [...] of Bounded Variation II Let’s consider the collection of functions of bounded variation on a little more deeply. It turns out that they form a subring of the ring of all real-valued [...] Pingback by \| March 6, 2008 \| Reply 4. [...] continue our study of functions of bounded variation by showing that total variation is “additive” over its interval. That is, if is of [...] Pingback by \| March 9, 2008 \| Reply 5. [...] over subintervals As I noted when I first motivated bounded variation, we’re often trying to hold down Riemann-Stieltjes sums to help them converge. In a sense, [...] Pingback by \| March 21, 2008 \| Reply 6. [...] Oscillation in a function is sort of a local and non-directional version of variation. If is a bounded function on some region , and if is a nonempty subset of , then we define the [...] Pingback by \| December 7, 2009 \| Reply 7. [...] explicitly, let be a continuous function on the rectangle , and let be of bounded variation on . Define the function on [...] Pingback by \| January 12, 2010 \| Reply 8. Why is the upper bound M on alpha, and not on the set of all possible “distances” (i.e. absolute values of differences) between successive alpha values? Comment by Kizi \| November 14, 2012 \| Reply 9. It may not be completely clear; if there’s an upper bound on the $\alpha$ values then there’s an upper bound on their differences. That’s in the motivation, but the actual definition that follows is written in terms of an upper bound on the sums of absolute differences. Comment by \| November 14, 2012 \| Reply • Thank you for the prompt response and clarification. In the motivation, do the alpha values need to be bounded below as well (or is there some property of the alpha values that I am missing, like non-negativity)? I do apologise if I am missing something obvious. Thanks. Comment by Kizi \| November 14, 2012 \| Reply 10. Yeah, they’d probably need to be bounded above and below; the motivation is meant to give the idea, but the definition is precise. Comment by \| November 14, 2012 \| Reply • Thank you Comment by Kizi \| November 14, 2012 \| Reply « Previous \| Next » ## About this weblog This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”). I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 38, "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": 0.9166198372840881, "perplexity_flag": "head"}
http://mathoverflow.net/questions/28470/infinite-system-of-stochastic-ordinary-differential-equations-coupled-by-infinite	## Infinite System of Stochastic Ordinary Differential Equations Coupled by Infinite Graphs ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) $\ \ \$ In Time Evolution of Infinite Anharmonic Systems, Lanford,Lebowitz and Lieb, roughly speaking, proved that for some families of functions $F_v$ $(v\in\mathbb Z^d)$ and a large set of initial condition, there exist a unique solution defined on a non-degenerated interval $[0,T]$ for the infinite system ```$$ \left\{ \begin{array}{rcl} \frac{d}{dt}q_v(t)&=&p_v(t)\\ \frac{d}{dt}p_v(t)&=&F_v(X(t))\\ X(0)&=&X_0, \end{array} \right. $$``` where $X_0$ and $X(t)$ belongs to some Banach space contained in $(\mathbb R^n\times\mathbb R^n)^{\mathbb Z^d}$ and as mentioned before $v\in\mathbb Z^d$. $\ \ \$ Although the authors do not mention, their results can be generalized for any infinite uniformly bounded degree graph. But a more interesting generalization, would be made by considering the stochastic case. Because it seems to be physically relevant to consider that the forces in each vertex of the graph are not completely known, due some random perturbations. $\ \ \$ Could you help me to find a reference where this kind of model is considered ? -	{"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": 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": 0.8814870119094849, "perplexity_flag": "head"}
http://mathhelpforum.com/calculus/198661-related-rates-cylinder.html	# Thread: 1. ## Related Rates of a Cylinder Water is being pumped into a vertical cylinder of radius 4 meters and height 15 meters at a rate of 2 meters3/min. How fast is the water level rising when the cylinder is half full? 2. ## Re: Related Rates of a Cylinder Originally Posted by titansfreak93 Water is being pumped into a vertical cylinder of radius 4 meters and height 15 meters at a rate of 2 meters3/min. How fast is the water level rising when the cylinder is half full? $V = \pi r^2 h$ since radius is constant ... $V = 16\pi h$ you are given $\frac{dV}{dt}$ , take the time derivative of the volume equation and determine $\frac{dh}{dt}$ ... you'll find that the amount in the cylinder doesn't matter.	{"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": 0.18, "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": 0.9405333995819092, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/59588/where-can-i-find-information-on-the-calculations-of-limits	Where can I find information on the calculations of limits? I'm doing some research for an algorithm, and I have a lot of curiosity, too. I'd like to learn how limits are calculated by mathematicians and mathematical software. I understand calculus somewhat, but I'm a litty rusty on it and I sometimes get perplexed by calculations involving limits. What's my best bet for learning about how limits are calculated? Should I review my calculus, or is there more science to limits (like numerical methods?!?)? - 1 – anon Aug 25 '11 at 2:17 – JavaMan Aug 25 '11 at 2:25 On the numerical front, things get a little hazy... you'll want to look into the use of sequence transformation methods. Briefly, you construct an appropriate sequence for your limit; e.g. for something like $\lim\limits_{x\to 0} f(x)$, you might consider the sequence $f(x_k)$ where $x_k$ is a sequence that tends to 0 like $k^-n,\quad n>0$ or $c^k,\quad 0 < c < 1$, and then apply a transformation that accelerates the convergence. If that's what you're interested in, I'll write up something. – J. M. Aug 25 '11 at 2:51 @J.M. Yes-I'm very interested in sequence transformation methods. Anything relating to limits is good, and this seems prime territory to explore! I'd like to thank you in advance for bringing this up. :-) – Matt Groff Aug 25 '11 at 3:27 – J. M. Aug 25 '11 at 3:29 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": 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": false, "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": 0.9402629733085632, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/42065?sort=oldest	## Posets of finite sequences are highly connected ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I need the following result for an example in a paper I'm writing. It's easy enough to prove, but I'd prefer to just give a reference. Does anyone know one? Fix $1 \leq k \leq n$. Define $X_{n,k}$ to be the following poset. The elements of $X_{n,k}$ are ordered sequences $\omega = (x_1,\ldots,x_m)$, where the $x_i$ are distinct elements of the $n$-element set `$\{1,\ldots,n\}$` and $m \geq k$. The order relation is that $\omega_1 \leq \omega_2$ if $\omega_1$ is a subsequence of $\omega_2$. The theorem then is that the geometric realization of $X_{n,k}$ is $(n-1-k)$-connected. - what's the geometric realization of a poset ? – Suresh Venkat Oct 13 2010 at 20:47 1 The set of chains is an abstract simplicial complex, so the geometric realization is just the realization of this complex. – Richard Kent Oct 13 2010 at 20:54 ah ok. i was wondering if that is what it was. – Suresh Venkat Oct 13 2010 at 21:44 ## 1 Answer If I understand your question correctly, an answer should appear in the paper "On lexicographically shellable posets" of Anders Bj\"orner and Michelle Wachs, in Transactions of the AMS 277, pp. 323-341. - Yes, this paper gives me exactly what I want. Thanks! – Andy Putman Oct 14 2010 at 1:39	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 12, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9114333391189575, "perplexity_flag": "head"}
http://mathoverflow.net/questions/8874/what-are-some-slogans-that-express-mathematical-tricks/40743	## What are some slogans that express mathematical tricks? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Many "tricks" that we use to solve mathematical problems don't correspond nicely to theorems or lemmas, or anything close to that rigorous. Instead they take the form of analogies, or general methods of proof more specialized than "induction" or "reductio ad absurdum" but applicable to a number of problems. These can often be summed up in a "slogan" of a couple of sentences or less, that's not fully precise but still manages to convey information. What are some of your favorite such tricks, expressed as slogans? (Note: By "slogan" I don't necessarily mean that it has to be a well-known statement, like Hadamard's "the shortest path..." quote. Just that it's fairly short and reasonably catchy.) Justifying blather: Yes, I'm aware of the Tricki, but I still think this is a useful question for the following reasons: 1. Right now, MO is considerably more active than the Tricki, which still posts new articles occasionally but not at anything like the rate at which people contribute to MO. 2. Perhaps causally related to (1), writing a Tricki article requires a fairly solid investment of time and effort. The point of slogans is that they can be communicated without much of either. If you want, you can think of this question as "Possible titles for Tricki articles," although that's by no means its only or even main purpose. - 2 I realize this question's borderline (and I wasn't sure whether I should ask it), but I don't think it's any less MO-appropriate than "Fundamental examples" or the mathematical joke thread. People who downvoted, care to explain why you disagree? – Harrison Brown Dec 14 2009 at 15:31 5 I wasn't the downvoter, but I think some people are getting a little annoyed at the number of these "produce a ginormous list of answers" soft questions. You're entirely right that there are worse offenders on the site, but I think the issue is in part the density of them, not any particular ones. – Ben Webster♦ Dec 14 2009 at 16:07 1 @Ben: That makes sense, although a better solution might be to ignore the "soft-question" tag. Or make a new tag for "ginormous list" questions. But this would be more appropriate on meta, so I'll start a topic there. – Harrison Brown Dec 14 2009 at 16:11 5 Ben: I really like the ginormous list of answer questions. It's nice to have some big picture responses from a variety of mathematicians. Peter: If you don't like soft answer questions, you don't have to read them. There's even a box on the front page for ignoring tags you don't like. – Tom LaGatta Jan 13 2010 at 9:51 2 @Tom: soft-answer is not the same as big-list; the meta discussion that came from this thread actually inspired the "big-list" tag, for exactly the reason that it can be ignored. But if you have more to say, feel free to contribute on meta.MO. – Harrison Brown Jan 13 2010 at 11:01 show 2 more comments ## 21 Answers If something does not hold, make it true! Examples: - Sobolev spaces (not necessarily differentiable functions satisfy differential equations) - distribution theory (think of identities involving the delta "function") - no converging? take the closure of your vector space (analysis) or compactify your space (geometry) - 8 Related to that, "Look in a bigger bag." For example, to find an integer solution, first find a rational or even complex solution then try to show it's an integer. Or find a solution in a Sobolev space then prove the solution is actually a classical solution. – John D. Cook Dec 14 2009 at 21:15 3 Another famous example of this is the notion of stacks: quotients of schemes by group actions do not necessarily exist? We make them exist by sheer brute force! (And it's funny that the definition of a stack also mimics the definition of a distribution somehow -- in both situations, the idea is to forget the object itself and only remember how it acts on some class of "test objects".) – Dan Petersen Dec 15 2009 at 8:19 ### 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 analyst's toolbox consists of three things: 1. The Cauchy-Schwarz inequality 2. Changing the order of integration/summation 3. Integration by parts (I'm not saying I believe that; it's just a very common saying.) - 17 You forgot 4. Adding and substracting something. – Mariano Suárez-Alvarez Jan 13 2010 at 7:37 8 Would you allow me the principle that if the average of some numbers is at least C then one of the numbers must be at least C? – gowers Sep 23 2010 at 21:42 3 An analyst once told me: "When in doubt, integrate by parts." – Micah Milinovich Sep 23 2010 at 22:42 5 I've also heard the following cited as a tool: if $a \leq b + \epsilon$ for all $\epsilon > 0$, then $a \leq b$. – S. Carnahan♦ Sep 24 2010 at 3:02 2 When Peter Lax went to receive the national medal of science, he was asked by the other recipients about his merits. His answer was (apocryph) I integrated by parts. – Denis Serre Apr 7 2011 at 9:33 show 3 more comments Try to replace a structure on an object with a map to a clasifying object. E.g., replace a cohomology class of a space with a map to an Eilenberg-MacLane space. Replace a vector/general bundle on a manifold with a map to the Grassmanian/other classifying space. There must also be plenty of examples outside algebraic topology, though this technique seems to be most popular there... - Along these lines, replace a subset of a finite set by an element of a vector space over GF(2). – Harrison Brown Dec 15 2009 at 2:50 Devissage is a useful tool when proving something holds for a general class of objects, at least in algebraic geometry, like all schemes/stacks/morphisms. - 3 What is devissage? – Ilya Grigoriev Dec 15 2009 at 18:13 8 I don't know if there is a formal definition. For me it roughly means that, when proving something in general, one reduces step by step to special cases. Some people say that it's an important feature in Grothendieck's style proofs. Here's an "example". Suppose we want to prove something for any morphism $f:X\to Y$ of schemes of finite type over a field k. One can check if this property is local in the sense that, if it's true for all fibers of $f,$ then it's true for $f.$ If it is local then we reduce to the case where Y is a point Spec k. (too long to fit in one comment; to be continued) – shenghao Dec 15 2009 at 20:37 3 Let U be any open subset of X with complement Z. Suppose that if the property holds for both U and Z, then it holds for X. Then we may shrink X to any open subset and use noetherian induction, so we can assume X is affine of equidimension d. There exists a map $X\to X'$ with fibers of dimension <=1, and dim X'=d-1, so by induction on d we reduce to two cases: X is a curve, or X is a point. Finally we prove these special cases. – shenghao Dec 15 2009 at 20:39 I forget who this is attributed to, but someone said something like "A technique is a trick used twice." - Weil's "three columns": Number fields over $\mathbb{Q}$ behave like function fields of curves over finite fields which are related to the field of algebraic functions over $\mathbb{C}$. (This is far removed from my comfort zone, so please fix it if I'm off the mark.) - "Think homologically, prove cohomologically!" definitely sounds like a slogan. One argument for this is that homology has a nice explanation in terms of geometry, think singular simplices or cells, so you can think about a space in terms of its cellular homology. When proving things you might want to have more structure around, like a product, and this is where cohomology comes in. - If you have to chose some auxiliary object and that object is not unique, it's better to make all choices simultaneously. I think there are many examples of this, but for me it first hit home when I learned about crystalline cohomology. There you want to lift varieties in positive characteistic to characteristic zero. Locally there are many nonisomorphic lifts, and rather than picking one, it's better to work with the category of all of them. I've absorbed this lesson pretty fully, to the point where I don't need to remind myself of it, but at first it seemed revolutionary. - You must exchange the order of summation in order to prove any identity involving multiple sums. - The best way to solve a problem is to define it out of existence. Typical example: Weil constructed abelian varieties over finite fields, and at first he did not know if these were varieties because it was not clear that they were projective. Weil defined this problem out of existence by changing the definition of variety and inventing abstract varieties. - There are two interesting tricks in K-theory / operator algebras / homotopy theory - one attached to an amusing slogan and the other with an amusing name - that I think foot the bill. The first is "uniqueness is a relative form of existence", due apparently to Shmuel Weinberger. This slogan seems to appear frequently in operator theory. Take, for example the problem of proving that K-theory commutes with direct limits (say, of C* algebras $A_1 \subseteq A_2 \subseteq \ldots \subseteq A$). There are two components to the proof: surjectivity (the "existence" part) which amounts to showing that every element of $K_0(A)$ lies in the image of some $K_0(A_j) \to K_0(A)$, and injectivity (the "uniqueness" part) which involves proving that if two elements of $K_0(A_j)$ are equivalent in $K_0(A)$ then they are equivalent in $K_0(A_j)$. Once you have proven existence you can verify uniqueness by joining representatives of your chosen $K_0(A_j)$ classes by a homotopy in the space of generators for $K_0(A)$ and then use your existence argument to lift to a homotopy in $A_j$. In other words, prove uniqueness by applying your existence argument to a pair. The second is the (in)famous "Eilenberg Swindle" which seems to come up everywhere. I first encountered it in K-theory, but I think the canonical example is the argument which proves that the $n$-sphere is prime with respect to connected sum (which I will denote +). Suppose that $M$ and $N$ are manifolds such that $M + N = S^n$. We have that $(M + N) + (M + N) + (M + N) + \ldots$ is homeomorphic to $\mathbb{R}^n$ (it is a cylinder with the left opening glued shut), and similarly so is $(N + M) + (N + M) + \ldots$. Since $M + (N + M) + \ldots = (M + N) + (M + N) + \ldots$, we have shown that $M + \mathbb{R}^n = \mathbb{R}^n$ which forces $M$ to be homeomorphic to $S^n$. - 1 This is definitely older than Shmuel, but i guess you are referring specifically to the quote. The main point is that uniqueness in homotopy theory is just saying everything you want to construct is connected by a homotopy. So you need that homotopy to exist, but you have an existence result that you can usually apply to construct your homotopy. Also think of the example provided by whitney's embedding theorem and how it shows any two embeddings are homotopic. – Sean Tilson Sep 26 2010 at 6:02 Look at flabbier objects. This seems to be especially useful in complex algebraic geometry. Hard to prove something for varieties? See if there's a version that's true for schemes. Or maybe Kahler manifolds. Or worse: stacks. Vector bundles giving you trouble? Try coherent sheaves. Try quasi-coherent sheaves. In fact, try complexes of them. This is really just a special case of "Generalize the question as far as you can" but in this specific case, it's rather clarifying, here are some examples in algebraic geometry: 1. It's hard to say anything about fundamental groups of complex projective varieties that isn't also true about compact Kahler manifolds. Perhaps the proof should focus on using the Kahler structure, when you're working on these. 2. Want to parameterize subvarieties of a projective variety? Tough, it doesn't work. SubSCHEMES, however, gives the Hilbert Scheme. 3. Proving things about ideals is often easier to do with modules in general - can you elaborate on 3? – Ho Chung Siu Jan 13 2010 at 17:04 Well, that's pretty much one of the main themes of commutative algebra books: that looking at ideals is the wrong point of view, really they're just submodules of the free module on one generator, and many (most, perhaps) results about ideals are actually true about modules, and how to prove them is easier to see when you realize just what the right way to look at things is. – Charles Siegel Jan 13 2010 at 17:10 "If you count something two different ways, you get the same result." This is related to the trick of changing the order of integration (or summation) discussed above, but discrete and more general. This method is used all the time in combinatorics. I think it has also been phrased differently, but I don't remember the exact phrasing. - If you want to show that a graph has few edges, prove that not too many vertices can have large degree. (The complementary statement is the main trick in the solution to this MO question, by way of example. It's also used in the proof of the Stanley-Wilf conjecture.) - is there away of enlarging this to other fields? or rather, how should it be phrased? – Sean Tilson Sep 26 2010 at 6:04 Something like "If the average is small, and you have controlled the large outliers, then everything must be small" but that's not very snappy. – Matthew Daws Oct 1 2010 at 14:04 I'm not sure if this is a bit too general, but it is a slogan/heuristic that I find very useful and that I think most people will be able to come up with plenty of examples of: "Extremalities always arise from symmetry." - 1 I would personally change 'always' to 'often'. Life for geometric analysts will be somewhat easier if this were actually a theorem. In the theoreical physics literature this is related to the so-called Coleman's Principle. However, it was shown by Kapitanski and Ladyzhenskaya that unless one makes additional assumptions, this principle is generally incorrect. ams.org/mathscinet-getitem?mr=711846 – Willie Wong Oct 1 2010 at 15:14 "When in doubt, differentiate." I've heard this attributed to Chern. - A perfect example are the twelve heuristics listed on page 1 of L. Larson, "Problem solving through problems": http://books.google.com/books?id=qFNZIUQ_MYUC&lpg=PP1&dq=larson%20problem%20solving&pg=PA1#v=onepage&q&f=false - One of the slogans in T. W. Körner's book Fourier Analysis that is definitely in the harmonic analyst's toolbox: The function $f*g$ has the good properties both of f and g. An example of its use is in approximating functions by trigonometric polynomials: convolving the function with any trigonometric polynomial gives you a trigonometric polynomial, and if you pick the polynomial carefully the resulting function will have similar properties to the original one. - 2 With one exception: support in physical space. If $f$ has compact support and $g$ does not, the convolution does not have compact support. Of course, another way of looking at it is that having compact support is such an unstable property that it is actually not good in harmonic analysis. – Willie Wong Oct 1 2010 at 15:06 ### "It is easy to prove existence when there is only one, or when there are many" explanation: If there is only one object with a certain property, you can sometimes use it to define it. For example, in geometric situations, you can sometimes define it locally and glue the patches since uniqueness guarantees compatibility on overlaps. It suggests that you should try proving uniqueness before proving existence and if uniqueness fails, maybe you should add constraints (thus, paradoxically, adding constrains can help in proving existence). On the other hand, sometimes it is easier to prove that there are many than to point out one specific example (transcendental numbers, continues nowhere differentiable functions,...). Therefore, you may want to seek for the right notion of "many" in your universe (cardinality, measure, "topological bigness" like the baire property,...) and try to prove that actually there are "few" objects that don't have the required property. comment: This relates to the answer saying that when you can't avoid making a choice, make all of them simultaneously. This happens when there are more than one, but not many... - 1. Pick a random example. 2. If you add lots of small and reasonably independent things together then the result will be highly concentrated about its mean. - Jacobi's famous quote that "one must always invert." He had elliptic integrals in mind. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 32, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9480271935462952, "perplexity_flag": "middle"}
http://mathoverflow.net/revisions/25973/list	## Return to Answer 2 added 1 characters in body; deleted 47 characters in body I would also go for $\Pi(t)$ or $t!$, but a possible reason to prefer the shifted version, $\Gamma(t)$, is the following. The gamma densities $\gamma_t$, $t\in \mathbb{R}$ defined as $\gamma_t(x):=\frac{x_+^{t-1}e^{-x}}{\Gamma(t)},$ are a convolution semigroup, so that $\Gamma(t)$ appears naturally as the normalization factor of $\gamma_t$. (And, of course, the semigroup relation $\gamma_t\gamma_s=\gamma_{t+s}$ would be destroyed shifting from t-1 to t in the definition of $\gamma_t$) Also note that the expression of the Beta function $B(t,s):=\int_0^1 x^{t-1}(1-x)^{s-1}dx$ in terms of the $\Gamma$ function, if shifted, would also loose the useful form $B(t,s)=\frac{\Gamma(t)\Gamma(s)}{\Gamma(t+s)}.$ (incidentally note that this relation follows plainly from the semigroup property since as a general fact, the integral of a convolution of two functions is the product of their integrals, it you feel like tryin the short computation)integrals). 1 I would also go for $\Pi(t)$ or $t!$, but a possible reason to prefer the shifted version, $\Gamma(t)$, is the following. The gamma densities $\gamma_t$, $t\in \mathbb{R}$ defined as $\gamma_t(x):=\frac{x_+^{t-1}e^{-x}}{\Gamma(t)},$ are a convolution semigroup, so that $\Gamma(t)$ appears naturally as the normalization factor of $\gamma_t$. (And, of course, the semigroup relation $\gamma_t\gamma_s=\gamma_{t+s}$ would be destroyed shifting from t-1 to t in the definition of $\gamma_t$) Also note that the expression of the Beta function $B(t,s):=\int_0^1 x^{t-1}(1-x)^{s-1}dx$ in terms of the $\Gamma$ function, if shifted, would also loose the useful form $B(t,s)=\frac{\Gamma(t)\Gamma(s)}{\Gamma(t+s)}.$ (incidentally note that this relation follows plainly from the semigroup property since as a general fact, the integral of a convolution of two functions is the product of their integrals, it you feel like tryin the short computation).	{"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": 26, "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": 0.9527852535247803, "perplexity_flag": "head"}
http://mathhelpforum.com/advanced-algebra/178902-normalizing-4x4-matrix-unknown-functions-elements.html	Thread: 1. Normalizing a 4x4 matrix with unknown functions as elements Hi All! This is my first time putting up my own thread on MF. I can usually find what I'm looking for, but this time: no go. As the title says, I'm trying to find the normalization constant of this 4x4 matrix (g and f are functions): (1-g^2) 0 0 0 0 (1+f^2) (-g^2-f^2) 0 0 (-g^2-f^2) (1+f^2) 0 0 0 0 (1-g^2) It's a matrix that's in a research paper which gives the normalization constant (they use the word 'factor'. I'm not sure if that means something diff.) as: N=4-2g^2+2f^2. 1]I've been looking up online and found that N can be found with: N=\sqrt{\sum{X^2}} where X represents the elements of the matrix. 2]I also found somewhere which said that I need to find the determinant. I'm not sure who's right, but I'm not getting what's in the paper. For method [1] I'm getting as far as: N^2 = 4(1+f^4+f^2g^2+f^2) and got stuck trying to find the square root (it's been a while since I've done multinomial theorem). So I backtracked to see if their N^2 matches my N^2. But their N^2=16+4g^4+4f^4+16g^2-8g^2f^2+16f^2. and method [2] is giving me something so long, with so many variables of (g^2), (f^2), (g^4), (f^4),(g^2f^4) (and it keeps going for about 3tysomething variables) that I've given up. So I'm wrong all over the place. Can someone help me out? booklist05 confused 2. I don't know what is meant by the normalisation constant of a matrix. But it looks from your suggested definitions 1] and 2] as though it is meant to be some function of the eigenvalues of a (positive?) matrix. The definition 1] gives the root mean square sum of the eigenvalues. The suggestion in 2] is to find the determinant (product of the eigenvalues). It looks to me as though the formula for N given here ( $N=4-2g^2+2f^2$) is simply that N is the sum of the eigenvalues. This is equal to the trace of the matrix, which is the sum of its diagonal elements. That is easily seen to be $4-2g^2+2f^2$. Or is that too simple-minded a suggestion? 3. Omg! can it really be that simple? Later on in the paper, they take each of the elements and divide them each by N. This is why I think they call it "normalization constant". I'm still in shock. so simple? Well since it works, that's what I'm going with. No one else I've asked has been able to give me an answer. Thanks!	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 2, "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": 0.9537609815597534, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/25802/why-does-a-black-hole-have-a-finite-mass/25803	# Why does a black hole have a finite mass? I mean besides the obvious "it has to have finite mass or it would suck up the universe." A singularity is a dimensionless point in space with infinite density, if I'm not mistaken. If something is infinitely dense, must it not also be infinitely massive? How does a black hole grow if everything that falls into it merges into the same singularity, which is already infinitely dense? - 3 Note that not all theories explain a black hole as a singularity. There are some exotic ideas out there, but in some cases the black hole is simply extremely dense, but not necessarily infinitely dense (but still dense enough to have an event horizon). The singularity arises from relativity. – voithos Jun 14 '11 at 6:33 ## 4 Answers If something is infinitely dense, must it not also be infinitely massive? Nope. The singularity is a point where volume goes to zero, not where mass goes to infinity. It is a point with zero volume, but which still holds mass, due to the extreme stretching of space by gravity. The density is $\frac{mass}{volume}$, so we say that in the limit $volume\rightarrow 0$, the density goes to infinity, but that doesn't mean mass goes to infinity. The reason that the volume is zero rather than the mass is infinite is easy to see in an intuitive sense from the creation of a black hole. You might think of a volume of space with some mass which is compressed due to gravity. Normal matter is no longer compressible at a certain point due to Coulomb repulsion between atoms, but if the gravity is strong enough, you might get past that. You can continue compressing it infinitely (though you'll probably have to overcome some other force barriers along the way) - until it has zero volume. But it still contains mass! The mass can't just disappear through this process. The density is infinite, but the mass is still finite. - 2 Right. I guess the division by zero means there are infinite solutions, not necessarily that the parameters are infinite. I blurred that distinction. – Carson Myers Jun 14 '11 at 2:34 1 Also, black holes are constructed from a finite amount of mass-energy. – Andrew Jun 14 '11 at 17:20 A black hole has an infinite density; since its volume is zero, it is compressed to the very limit. So it also has infinite gravity, and sucks anything which is near it! Not everything there is!! Now above all when it sucks things it adds up to its mass, which remains finite and it always will, even if it did suck in the whole universe!! It's all for the formula: density in a black hole is mass divided by volume (0) so the density is infinite, not the mass!!! So thus a black hole has mass which is finite and will always be finite. - if it has infinite density, that would mean it has infinite gravity (or infinite space curvature in special relativity, which i believe in). so, it obviously doesn't have infinite density because that would suck the whole entire universe into itself. that would mean it would have to have an unmeasurably small amount of volume. and even if it did have infinite density, space curvature should be more powerful nearer to the singularity, but how can we add more to an infinite amount of gravity? if it had infinite density, its curvature should be equal everywhere, which is also impossible. What do you think of this theory? - I'm not going to claim any real knowledge on the topic, or most of what I bring up, so feel free to correct me for any extremely rash or impossible aspects, but here it goes. A black hole to me, seems like it's just the epitomy of nothing. There's something, then general space, and then a black hole. It appears to me that something to nothing is relatively possible, as at one point nothing came to be something, however I feel that the universe itself wouldn't allow such a flaw, as the universe seems quite perfect in its design. One theory I've stumbled upon is a theory that all atoms have a negative and positive form of which they can swap between. Kind of a matter and anti-matter deal, but all atoms experience this, and that we live in the positive. This kind of interacts with the idea of a black hole, a massive explosion of something that created energy to powers beyond numbers we have words for (googlplex's), for a length of years we barely use the number for (billions). And it eventually explodes, and in the wake of such energy, a black hole is formed (If that's still what people believe). So a rediculous mass of positive energy goes out from a point in the universe, and in this point, it sucks in mass, everything, a perfect unescapable gravity pit. Could it possibly be that after so much positive energy radiates from one point, it explodes, then begins sucking in all the positive energy that comes into it, to rebalance the rediculous amount of negative energy that would then be in the wake of such a thing? I can kind of imagine it, more like an explosion under water (an implosion really). First it explodes, and then everything gets sucked in until it's at its natural point. Except this is on a massive (much bigger than any bomb we could ever make or imagine), yet scale so small (fills itself in to a sub-atomic level, if not to an electron level) it takes an unbelievably long time to go back to normal. And even then that doesn't factor in to say that all this energy its taking in, isn't being vortex'd to another space or even time. So, as a summary, I suppose what I'm concluding black holes to be, is an enormous sphere of negative. And as all things want to be neutral, it sucks in as much positive (the reality we live in) as possible, practically, indefinitely, all positive matter that goes near it to be sucked in without flaw, no possible escape. To anwser whether it has mass, I really don't know for sure, but from what I can tell, it has a Finite mass. But the time required for it to reach a neutral point is absolutely rediculous, seeing as how it would have to fill all that space flawlessly, to the absolute with the little bit of energy that it gets coming in from light, or small fragments of atoms. But again, I have no real knowledge on the topic, its just my educated guess on what I've read and learned over a little while. - This isn't really coherent. There's also no science behind it. – Daniel Blay Dec 11 '12 at 22:55	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9657410383224487, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/68746/riemannian-length-of-an-element-of-the-fundamental-group-of-a-manifold/68750	## riemannian length of an element of the fundamental group of a manifold ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) It is a stupid question i guess but like they say if you ask you are stupid for 5 minutes and if you don't ask you are stupid forever . here is the question given a closed manifold $(M,g)$ and $\alpha$ in $\pi_1(M,p)$ we can define the length of $\alpha$ as the minimum riemannian length of a representative now obviousle length $\alpha^2$ is less or equal then $2\mathrm{length}(\alpha)$ but it seems it is always equal $2\mathrm{length}(\alpha)$ i want to know why ? - 4 The statement is false. Think projective plane. – Richard Kent Jun 24 2011 at 16:01 under what conditions the statement is true ? and can the length of \alpha^2 be less the length of \alpha ? – unkown Jun 24 2011 at 16:34 3 When $\pi_1$ has an element $\alpha$ of order 2, $\alpha^2$ can have length 0 (that's what R.Kent was suggesting). There are intermediate possibilities too, such as $\alpha$ of order 3 implies $\alpha$ and $\alpha^2$ have the same length; this doesn't happen in two dimensions but does in three (e.g. lens space). – Noam D. Elkies Jun 24 2011 at 17:02 8 Generically, a geodesic loop realizing the minimum length does not close up smoothly at $p$, and then the double loop is not a geodesic and hence not a minimizer. So the statement is almost never true unless you minimize in a free homotopy class. – Sergei Ivanov Jun 24 2011 at 17:40 (sorry, I see I have unwillingly down-voted this question) – Pietro Majer Jun 27 2011 at 16:16 ## 1 Answer There are geometric hypotheses that ensure the property you want. For example, suppose that $(M,g)$ has negative curvature. Then every $\alpha \in \pi_1(M)$ is freely homotopic to a unique geodesic representative $\alpha^$. Usually people write $\ell_g(\alpha)$ for the length of $\alpha^$. Finally, uniqueness of geodesic representatives implies that $\ell_g(\alpha^k) = k \cdot \ell_g(\alpha)$. This is just the beginning of an important area in Riemannian geometry. (When are geodesic representatives unique? What is the interaction between the metric and the variational properties of geodesics? And in a different direction: How does the fundamental group act on the universal cover? What does the metric tell us about the algebraic topology of $M$ and the universal cover? Etc.) It is amusing to contemplate all the ways in which the real projective plane, or more generally any closed manifold with finite fundamental group, differs from a negatively curved manifold. - is there is some algebraic properties of the fundamental group that can garuentee the property without talking about curvature say for example if the fundamental group is torsion free – unkown Jun 24 2011 at 17:22 From another hand thanks for your answer but it doesn't answer the question because the length i am talking about is for \alpha=[c] in pi_1(M,p) length(\alpha) is the infimum length of loops that are homotpic to c and not freely homotopic – unkown Jun 24 2011 at 17:26 I personally seriously doubt that there exists such an algebraic condition, as there are many manifolds with a given fundamental group and you can perturb the metric on a Riemannian manifold in several ways. I would be very surprised if the property you require was stable under those perturbations (except in the case of `$S^1$`), see the comment by Sergei Ivanov. Btw, the only examples of manifolds with that property I can think of are tori (with flat metric). And simply connected manifolds. – Alessandro Sisto Jun 24 2011 at 18:47 @unknown - I assumed that free homotopy classes were meant and I explicitly wrote free homotopy in my answer. In the setting of based homotopy classes (as others have pointed out) length is basically never multiplicative with respect to powers... Somethings can still be shown even in the based case. For example, if you look at $\ell_g(\alpha^k)/k$ using based length (where $g$ is negatively curved) then this approaches the free length. – Sam Nead Jun 24 2011 at 22:24 @unknown - Regarding your first comment. The Klein bottle has torsion free fundamental group. At the same time there is a based loop whose (based) square has length less than twice the original. @Alessandro - and products of spheres and tori, I guess. Are there other examples? – Sam Nead Jun 24 2011 at 22:34 show 3 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": 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": 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": 0.9344087243080139, "perplexity_flag": "head"}
http://nrich.maths.org/4935	### Area L Draw the graph of a continuous increasing function in the first quadrant and horizontal and vertical lines through two points. The areas in your sketch lead to a useful formula for finding integrals. ### Integral Equation Solve this integral equation. ### Integral Sandwich Generalise this inequality involving integrals. # Integral Inequality ##### Stage: 5 Challenge Level: (i)Suppose that $a$, $b$ and $t$ are positive. Which of the following two expressions is the larger $$P=\left(\int_0^t x^{a+b}dx\right)^2, \qquad Q=\left(\int_0^t x^{2a}dx \right) \left(\int_0^t x^{2b}dx\right)\ ?$$ (ii)By considering the inequality $$\int_0^t [f(x)+\lambda g(x)]^2 dx \geq 0,$$ prove that, for all functions $f(x)$ and $g(x)$, $$\left(\int_0^t f(x)g(x)dx\right)^2 \leq \left(\int_0^t f(x)^2 dx\right) \left(\int_0^t g(x)^2 dx\right).$$ The NRICH Project aims to enrich the mathematical experiences of all learners. To support this aim, members of the NRICH team work in a wide range of capacities, including providing professional development for teachers wishing to embed rich mathematical tasks into everyday classroom practice. More information on many of our other activities can be found here.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 5, "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": 0.9046816229820251, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/106604/min-number-of-vertices-in-graph-as-function-of-kappag-and-diamg	# Min. number of vertices in graph as function of $\kappa(G)$ and $diam(G)$ Question I found in "Introduction to Graph Theory" by Douglas B. West: Let $G$ graph on $n$ vertices with connectivity $\kappa(G)=k \geq 1$. Prove that $n \geq k(diam(G)-1)+2$, when $diam(G)$ is the the maximal (edge) distance between a pair of vertices in $G$ Seems easy and I've seen the answer, too, which goes the same as thought: by taking the vertices the end vertices of the path of length $diam(G)$, and applying Menger's Theorem. I'm having a hard time understanding one part - it seems like all the paths need to be of the same length for the $n \geq k(diam(G)-1)+2$ to apply. Am I missing some part here? Thanks in advance. - ## 1 Answer Let's write $d$ for the diameter of $G$. Then there is some pair of vertices such that one path between them has exactly $d$ edges, and each path between them has at least $d$ edges. Menger says there are $k$ independent paths. So the number of vertices is at least $k(d-1)+2$. I'm not sure why you think the paths all have to be the same length. They just have to have length at least $d$ - and, they do. - I don't think that all the paths have the same length, but that the maximal path length is $d$ (or am I misunderstanding the notion of $diam(G)$?). – Pavel Feb 8 '12 at 19:25 My understanding is that the diameter of a graph is the maximum over all pairs of the minimum distance between the pair. Once you have that maximizing pair, all paths joining those two vertices have length at least the maximum (and at least one path has length exactly the maximum). Maybe I'm wrong - find a definition of diameter in some textbook or on the web, and show me. – Gerry Myerson Feb 8 '12 at 23:10 Oh, now I see that I was wrong about the definition. Thanks for the answer! – Pavel Feb 9 '12 at 11:19	{"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": 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": 0.9709724187850952, "perplexity_flag": "head"}
http://torus.math.uiuc.edu/cal/math/cal?year=2010&month=10&day=28&interval=day	Seminar Calendar for events the day of Thursday, October 28, 2010. . events for the events containing More information on this calendar program is available. Questions regarding events or the calendar should be directed to Tori Corkery. ``` September 2010 October 2010 November 2010 Su Mo Tu We Th Fr Sa Su Mo Tu We Th Fr Sa Su Mo Tu We Th Fr Sa 1 2 3 4 1 2 1 2 3 4 5 6 5 6 7 8 9 10 11 3 4 5 6 7 8 9 7 8 9 10 11 12 13 12 13 14 15 16 17 18 10 11 12 13 14 15 16 14 15 16 17 18 19 20 19 20 21 22 23 24 25 17 18 19 20 21 22 23 21 22 23 24 25 26 27 26 27 28 29 30 24 25 26 27 28 29 30 28 29 30 31 ``` Thursday, October 28, 2010 Group Theory Seminar 1:00 pm in 347 Altgeld Hall, Thursday, October 28, 2010 Del Edit Copy Submitted by clein. Enric Ventura (Universitat Politecnica de Catalunya and CRM-Montreal)The conjugacy problem for some extensions of F_n, Z^m, B_n, and FAbstract: We will review the main idea in the solution of the conjugacy problem (CP) for free-by-cyclic groups given by Bogopolski-Martino-Maslavova-Ventura in 2006. A close analysis of this argument having the classical Miller's groups in mind (which are free-by-free and have unsolvable conjugacy problem) gave rise to a subsequent and stronger result by the same authors, giving an explicit characterization of the solvability of the conjugacy problem within the family of free-by-free groups. It turns out that the freeness of the base group is irrelevant in the whole proof, and the only crucial property is the solvability of the so-called twisted conjugacy problem (TCP). This way, we shall give characterizations of the solvability of the conjugacy problem for certain families of extensions of groups with solvable TCP. We will discuss the particular cases of extensions of free groups $F_n$, free abelian groups $Z^m$, Braid groups $B_n$, and Thomson's group $F$. Number Theory 1:00 pm in 241 Altgeld Hall, Thursday, October 28, 2010 Del Edit Copy Submitted by berndt. Kevin Ford (Illinois)R. I. P.Abstract: Recently, Candes and Tao showed that a k-sparse vector x in R^N (think of x as a signal) can be effectively recovered from a set of n-dimensional linear measurements with n much smaller than N, if the matrix of the vectors v_i satisfies the Restricted Isometry Property (RIP for short) of order k. All known explicit constructions of RIP matrices make use of number theory, and all have order k=O(sqrt(n)) (we will review these). We give a new explicit construction of RIP matrices of order n^{1/2+c} for some positive c, based on additive combinatorics. This is joint work with J. Bourgain, S. J. Dilworth, S. Konyagin and D. Kutzarova. Mathematics Colloquium 4:00 pm in 245 Altgeld Hall, Thursday, October 28, 2010 Del Edit Copy Submitted by clein. David Dumas (University of Illinois at Chicago)Complex projective structures and character varietiesAbstract: A complex projective structure is a way to build a surface from pieces of the Riemann sphere glued together using Mobius transformations. After giving a more precise definition and some examples, we will describe the moduli space of such structures and discuss how this space is related to the character variety of maps from a surface group in PSL(2,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": 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": 0.3, "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": 0.8717068433761597, "perplexity_flag": "middle"}
http://www.physicsforums.com/showthread.php?t=517665	Physics Forums ## Finding Moment of Interia of a 'Loop' when given Density & Cross sectional area... 1. The problem statement, all variables and given/known data I eventually have to solve for maximum angular acceleration of the loop in a magnetic field, and I have gotten everything with the exception of the moment of inertia, so I won't include the emf and B known variables. Known: a copper wire with a density of $\rho$ = 8960 kg/m3 is formed into a circular loop of radius 0.50 m. Cross sectional area of the wire is 1.00 x 10-5m2. 2. Relevant equations I=MR2 (and eventually) $\tau$ = $\alpha$I 3. The attempt at a solution I know since mass isn't given, I need to integrate something so I can use the density. However, it's been a really long time since I've integrated, so I'm not very familar with it. I've been unable to find an equation to find the volume of the 'loop,' so I know integration is the only way. 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 No integration required. A loop is just a cylinder bent into a circle. Imagine bending the loop back into a "normal" cylinder, and find the volume of that object. (If you don't remember the formula for the volume if a cylinder, it is easy to find). You can approximate the loop as a circle (line) because the cross section is much smaller than the radius. ## Finding Moment of Interia of a 'Loop' when given Density & Cross sectional area... Yes, I forgot to add that part. But first, OP needs to find the mass, which requires finding the volume. Once that is done you would forget about the finite width and simply use I = MR2 Tags angular acceleration, density, magnetic field, moment of interia Thread Tools \| \| \| \| \|---------------------------------------------------------------------------------------------------------\|----------------------------------\|---------\| \| Similar Threads for: Finding Moment of Interia of a 'Loop' when given Density & Cross sectional area... \| \| \| \| Thread \| Forum \| Replies \| \| \| Precalculus Mathematics Homework \| 1 \| \| \| General Math \| 11 \| \| \| Classical Physics \| 0 \| \| \| Introductory Physics Homework \| 2 \| \| \| Introductory Physics Homework \| 3 \|	{"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": 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": 0.9016082882881165, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/127964/about-probability/127967	# About probability Is a probability function in $\Omega{(a1, a2, a3)}$, find $P(a1)$ whether $P({a2, a3}) = 2P(a1)$. I know that $1 = a1 + a2 + a3$. from where I have to start ? - Do you mean to say that the sample space is a three element set? That is $\Omega=\{a_1,a_2,a_3\}$? – user21436 Apr 4 '12 at 12:48 This is exactly, the sample space is Ω={a1,a2,a3}. – mastergoo Apr 4 '12 at 12:50 You have to start with a comprehensible question. What does $P(a_2,a_3)$ mean? – Gerry Myerson Apr 4 '12 at 12:52 $a2 \Cup a3$, this is ? – mastergoo Apr 4 '12 at 12:53 ## 1 Answer Start by using additivity: $P(\{a_2, a_3\} \cup \{a_1\}) = P(\{a_2,a_3\})+P(\{a_1\})$. Since $\{a_2, a_3\} \cup \{a_1\} = \Omega$, and $P(\Omega) = 1$, we have found that $$P(\{a_2,a_3\})+P(\{a_1\}) = 1.$$ Now substitute $P(\{a_2,a_3\}) = 2P(\{a_1\})$ into this equation, and you'll find $P(\{a_1\}) = \frac{1}{3}$. - Oh yes, I've find the following result: $1 = a1 + a2 + a3$, so $1 = a1 + (2a1)$, so $a1 = \frac{1}{3}$ – mastergoo Apr 4 '12 at 12:59 THANK YOU SO MUCH. Very good explanation. – mastergoo Apr 4 '12 at 13:02 1 You're welcome, but you really need to be more precise with your notation: $a_1$ is the event (or rather $\{a_1\}$ is), which is very different from its probability $P(\{a_1\})$. – lazyhaze Apr 4 '12 at 13:10 By the way, don't put "Solved" in the title, but instead accept my answer (if you find it acceptable, of course). – lazyhaze Apr 4 '12 at 13:11 Ok ok, I undertood.. – mastergoo Apr 4 '12 at 13:23	{"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": 18, "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": 0.9103912711143494, "perplexity_flag": "middle"}
http://chemistry.stackexchange.com/questions/151/why-do-elements-in-columns-6-and-11-assume-abnormal-electron-configurations	# Why do elements in columns 6 and 11 assume 'abnormal' electron configurations? When I look around for why copper and chromium only have one electron in their outermost s orbital and 5/10 in their outermost d orbital, I'm bombarded with the fact that they are more stable with a half or completely filled d orbital, so the final electron enters that orbital instead of the 'usual' s orbital. What I'm really looking for is why the d orbital is more stable this way. I assume it has to do with distributing the negative charge of the electrons as evenly as possible around the nucleus since each subshell of the d orbital is in a slightly different location, leading to a more positive charge in the last empty or half-filled d orbital. Putting the final electron in the s orbital would create a more negative charge around the atom as a whole, but still leave that positive spot empty. Why does this not happen with the other columns as well? Does this extra stability work with all half or completely filled orbitals, except columns 6 and 11 are the only cases where the difference is strong enough to 'pull' an electron from the s orbital? It seems like flourine would have a tendency to do do this as well, so I suppose the positive gap left in the unfilled p orbital isn't strong enough to remove an electron from the lower 2s orbital. - ## 5 Answers As I understand this, there are basically two effects at work here. When you populate an s orbital, you add a significant amount of electron density close to the nucleus. This screens the attractive charge of the nucleus from the d orbitals, making them higher in energy (and more radially diffuse). The difference in energy between putting all the electrons in d orbitals and putting one in an s orbital increases as you fill the d orbitals. Additionally, pairing electrons in one orbital (so, adding the second s electron) carries a significant energy cost in terms of Coulombic repulsion because you're adding an electron essentially in exactly the same space as there's already an electron. I'm assuming that the effect isn't strong enough to avert fluorine having a $2s^2$ occupation, and if you look at gadolinium, the effect there isn't strong enough to stop the s from filling (large nuclear charge and orbital extent at the nucleus is a good combination energy-wise), it does manage to make it more favourable to add the electron into the 5d instead of the 4f orbitals. Also, if you take a look at tungsten vs gold, there the effect isn't strong enough for tungsten to avoid a $6s^2$ occupation, but is for gold - more d electrons making the screening effect overcome the strong nuclear charge and enhanced nuclear penetration of an s orbital. - ## Did you find this question interesting? Try our newsletter email address This is just a confirmation to Aesin's answer... Say, we take copper. The expected electronic configuration (as we blindly fill the d-orbitals along the period) is $[Ar]\ 3d^9\ 4s^2$, whereas the real configuration is $[Ar]\ 3d^{10}\ 4s^1$. There is a famous interpretation for this, that d-orbitals are more stable when half-filled and completely-filled. That's a complete myth. There are very few pages explaining this myth. As we fill the electrons starting from $3d^1$, we'd be stuck at Chromium and also at Copper. While filling Chromium and Copper, it has been observed that the energies of $4s$ and $3d$ orbitals are fairly close to each other. The increasing nuclear charge (as we go along the period) and the size & shape of d-orbital should be a reason. This similarity makes the energy for pairing up electrons in d-orbital very less than that of pairing up in s-orbital (i.e.) the energy difference between these orbitals is much less than the pairing energy required to fill the electrons in 4s orbital. Moreover, the energy for the configuration $3d^5\ 4s^1$ is much less than that of $3d^4\ 4s^2$. Since we usually fill electrons in the order of increasing energy, the next electron (in case of Mn) goes into the 4s-orbital. The same reason for effective nuclear charge makes the 3d-orbitals somewhat lower in energy than 4s-orbitals and hence, the unusual configuration of Cr and Cu. Here's a paper which supports that this is quite untrue... In the case of chromium, this means that $4s^1\ 3d^5$ will be lower in energy than $4s^2\ 3d^4$, because in the second case you have to "pay" the electron pairing energy. Since this pairing energy is larger than any difference in the energies of the 4s and 3d orbitals, the lowest energy electron configuration will be the one which has one electron in each of the six orbitals that are available. Effectively this is Hund's rule applying not just to strictly degenerate orbitals (orbitals with the same energy), but to all orbitals that are (significantly) closer in energy than the electron pairing energy. In the case of copper, the 3d orbital has dropped in energy below that of the 4s, so that it is better to have the paired electrons in the d and the unpaired one in the s. The reason why the 3d is lower than 4s is tied to the high effective nuclear charge. The high effective nuclear charge gives rise to the small size of Cu compared with the earlier transition metals, and also means that orbitals in inner shells are more stabilised with respect to those further out for copper than for earlier elements. - We know that the electronic configuration of chromium is $\ce{[Ar] 3d^5 4s^1}$. It is because promotion occurs in case of $3d$ and $4s$ orbitals -- in other words, the electron is shifted from a lower energy level to a higher one (also known as excitation). Promotional energy and pairing energy both are endergonic so the process which requires less energy would preferably take place. - 1 Could you explain a bit more? For example, why does that process have less energy? Elaborate a bit and your post will be improved greatly :) – ManishEarth♦ Nov 19 '12 at 18:51 The experimental data has 6-unpaired e's for Cr and 1 for Cu, therefore, the spdf model must "adapt". It is easy to see Cu (s1d10), but less for Cr (s1d5). The MCAS orbital model presents a different explanation as to why these are different. The MCAS atomic orbital model provides a more logical sstem for the periodic table, too. See the Cu, Cr issue and the periodic table according to this orbital model at this site You can see more about Modeling the MCAS Way at this other site which is the sharper version of the original that is in the arxiv.org database as html/physics/9902046v2. Of course, the physics and chemistry communities stand entrenched behind the spdf model that is heavily steeped in math. - 1 Welcome to Chemistry.SE. Your answer would be improved if you could summarize the content of the links you present rather than just putting the url into the text. – Ben Norris Apr 17 at 1:32 Thanks for the advise. I wrote much as it was. The two links have abstracts. It would be difficult to "summarize" something when a reader would undoubtedly have little knowledge of the model that is different that the spdf one. The URL links let the reader know where to look for what the model is and, as I pointed out, addresses the Cu, Cr electron issue explicitedly, which is why I chose to answer the question. – Joel M Williams Apr 17 at 2:21 Perhaps this shouldn't be counted as an answer, but since this topic has been resurrected, I'd like to point to Cann (2000). He explains the apparent stability of half-filled and filled subshells by invoking exchange energy (actually more of a decrease in destabilization due to smaller-than-expected electron-electron repulsions). According to him, there is a purely quantum mechanical energy term proportional to $$\frac{n_{↑}(n_{↑}-1)}{2} + \frac{n_{↓}(n_{↓}-1)}{2}$$, where $n_{↑}$ and $n_{↓}$ represent the number of spin-up and spin-down electrons in a subshell. This term decreases the potential energy of the atom, and it can be shown that the difference in exchange energy between two consecutive subshell populations (such as $p^{3}/p^{4}$, $f^{9}/f^{10}$, etc) has local maxima at half-filled and filled subshell configurations. This can be used to defend the favourability of $s^{1}d^{5}$ and $s^{1}d^{10}$ configurations relative to $s^{2}d^{4}$ and $s^{2}d^{9}$, even though there would be a slight increase of electron density in the more compact $d$ subshells. However, this clashes with Crazy Buddy's reference, which seems to deny any stabilization effect. So, which is (more) true? Or is neither? -	{"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": 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": 0.944912314414978, "perplexity_flag": "middle"}
http://mathhelpforum.com/calculus/114574-intersecting-points-polar-coordinates.html	# Thread: 1. ## intersecting points - polar coordinates Hi C1 r= 3cos x C2 r= 1 + cos x First question is find the polar coordinates of the intersection points of the two curves. I have found (3/2, pi/3) and (3/2, -pi/3) now what I am wondering is the point that we would normally consider (0,0) also an intersecting point and how would I find the coordinates for this point? Thanks Calculus beginner 2. Originally Posted by calcbeg Hi C1 r= 3cos x C2 r= 1 + cos x First question is find the polar coordinates of the intersection points of the two curves. I have found (3/2, pi/3) and (3/2, -pi/3) now what I am wondering is the point that we would normally consider (0,0) also an intersecting point and how would I find the coordinates for this point? Thanks Calculus beginner Your question seems to be a bit unclear. (0, 0) is not in the solution set. In fact neither of your given curves passes through this point. As to the rest we have $r = 3~cos( \theta)$ and $r = 1 + cos( \theta)$ Since r = r this means $3~cos( \theta) = 1 + cos( \theta)$ so $2~cos( \theta) = 1$ $cos( \theta ) = \frac{1}{2}$ You have mentioned the only two correct solutions. ( pi / 3 is the reference angle so there is another solution 4 pi / 3, or -pi / 3 which you already gave. -Dan	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 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": 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": 0.9263052344322205, "perplexity_flag": "middle"}
http://mathematica.stackexchange.com/questions/11542/numerical-comparisons-of-matrices	# Numerical comparisons of matrices I have a matrix which should be equal to a null matrix. However due to the numerical precision, a brutal equality test with a matrix initialized with zeros does not work. How should I perform the numerical equality test (with a given threshold for the precision) ? - ## 1 Answer A simpler way than to adjust the threshold for `Equal` is to use `Chop` which replaces approximate real numbers in expr that are close to zero by the exact integer 0. Adding the suggestions from the comments you have the following possibilities: • Use `Chop` as it is. Here, you may only chop the `Norm` at the end by `Chop[Norm[mat, 1]] == 0`. • Look at the second argument to `Chop` when you want to adjust the default tolerance. Ideally, it should correspond to the "sizes" of the matrices from which the putative null was constructed. For instance, if those matrices involve numbers in the millions, then any matrix whose coordinates are all less than about 0.0001 would likely need to be considered null. Typically, the second argument will be somewhere between the smallest and largest absolute values of the eigenvalues of those matrices. (These eigenvalues can reliably be found with `SingularValueDecomposition`; in many applications, they are already available from previous computations.) • Look at the `Internal`$EqualTolerance` (which is probably not the best idea in your case). - 1 Note, also, that one can control the tolerance used by `Chop[]` through its second argument. See the docs for details. – J. M.♦ Oct 4 '12 at 10:03 1 Another possible route to check if the matrix `mat` is a null matrix: `Chop[Norm[mat, 1]] == 0`; only a null matrix has zero norm. – J. M.♦ Oct 4 '12 at 10:14 1 It might be good to not compare against zero but test if the norm is smaller than an epsilon. – ruebenko Oct 4 '12 at 11:44 1 Be careful. There does not exist any absolute universal test like this. The comparison of a matrix to zero needs to account for the matrices used to create it. As an example, emulate the `SingularValueDecomposition` help by creating a random matrix `m` with entries on the order of, say, $10^{12}$, reconstruct `m` via its SVD, and subtract the reconstruction from `m` to see whether the two are equal. They're not--due to imprecision--but `chop` won't help. How much imprecision should one expect? Use the sizes of the eigenvalues of `m` (the diagonal of the SVD) to estimate the tolerance. – whuber Oct 4 '12 at 14:44 1 @whuber I gave you comment +1 and I would be happy if you insert your concerns into the answer. – halirutan Oct 4 '12 at 15:01 show 3 more comments lang-mma	{"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": 0.9100081920623779, "perplexity_flag": "middle"}
http://stats.stackexchange.com/questions/16349/how-to-compute-the-confidence-interval-of-the-ratio-of-two-normal-means	# How to compute the confidence interval of the ratio of two normal means I want to derive the limits for the $100(1-\alpha)\%$ confidence interval for the ratio of two means. Suppose, $X_1 \sim N(\theta_1, \sigma^2)$ and $X_2 \sim N(\theta_2, \sigma^2)$ being independent, the mean ratio $\Gamma = \theta_1/\theta_2$. I tried to solve: $$\text{Pr}(-z(\alpha/2)) \leq X_1 - \Gamma X_2 / \sigma \sqrt {1 + \gamma^2} \leq z(\alpha/2)) = 1 - \alpha$$ but that equation couldn't be solved for many cases (no roots). Am I doing something wrong? Is there a better approach? Thanks - – mbq♦ Oct 2 '11 at 8:58 1 @mbq - the Cauchy distribution presents no problems for confidence intervals, as the CDF is the inverse tangent function. Variance need not be defined for CIs to work. And the ratio of two normal RVs with zero mean is Cauchy, but not necessarily two normal RVs with non-zero mean. – probabilityislogic Oct 2 '11 at 10:58 @probabilityislogic Sure, I must stop trying to think on Sunday mornings. – mbq♦ Oct 2 '11 at 12:13 ## 2 Answers Fieller's method does what you want -- compute a confidence interval for the quotient of two means, both assumed to be sampled from Gaussian distributions. The original citation is: Fieller EC: The biological standardization of Insulin. Suppl to J R Statist Soc 1940, 7:1-64. The Wikipedia article does a good job of summarizing, and an appendix to the first edition of my Intutitive Biostatistics is even more concise. I've created an online calculator that does the computation. - It's very good references, I also like that you actually made a calculator for it (+1). As expected though, in your calculator you clearly state that when the confidence interval of the denominator includes zero, it is not possible to compute the CI of the quotient. I think it's the same that happens when I try solving the quadratic equation. suppose variance is 1, mu1 = 0 and mu2=1, N=10000. It's unsolvable. – francogrex Oct 2 '11 at 16:03 1 thanks for the online calculator Harvey, I'm a typical biologist with insufficient background in statistics and your calculator was exactly what I needed. – Timtico Jul 5 '12 at 8:29 R has the package `mratios` with the function `t.test.ratio`. Gemechis Dilba Djira, Mario Hasler, Daniel Gerhard and Frank Schaarschmidt (2011). mratios: Inferences for ratios of coefficients in the general linear model. R package version 1.3.15. http://CRAN.R-project.org/package=mratios See also http://www.r-project.org/user-2006/Slides/DilbaEtAl.pdf -	{"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": 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": 0.9081093072891235, "perplexity_flag": "middle"}
http://mathhelpforum.com/algebra/86465-compounded-intrest-continuously.html	# Thread: 1. ## Compounded intrest continuously Suppose an investor deposits \$4,000 in an account bearing 2.5% intrest compounded continuously. Find the total amount in the account for the following time period. 29 years The total amount of money is=? I am not asking for the answer, I just want to know if I need to know how to get the answer....do I multiply the 2.5 by 29 then divide? I can't figure out how to set it up???? Please help. Thanks 2. You should be studying the formulas for compound interest if you are given this sort of problem. It isn't something students usually can derive on their own. When the interest in compounded continuously, than the number "e" comes into play and the formula for calculating the total value after t-time is: $P=P_{0}e^{rt}$ Look familiar? 3. ## compounded It does, and i understand the formula. What I did not know was that I was able to plug that into my calculator. Unfortunatly I do not have the e on the calculator I am using. So that brings me to my next question. Am I screwed because I do not have that capability??? 4. Originally Posted by tpoma It does, and i understand the formula. What I did not know was that I was able to plug that into my calculator. Unfortunatly I do not have the e on the calculator I am using. So that brings me to my next question. Am I screwed because I do not have that capability??? Really? Have you tried looking near the ln key - it might be shift+ln for example. On my Casio fx-85ES is it Code: ``` img.top {vertical-align:15%;} $Alpha \rightarrow \text { x }10^x$``` Alternatively google has a calculator function: below is my search for e.	{"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": 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": 0.9736664891242981, "perplexity_flag": "middle"}
http://mathhelpforum.com/advanced-algebra/108926-change-basis.html	# Thread: 1. ## change of basis find the change of basis matrix from the basis B = (1, t-3, (t-3)^2) to the standard basis U = (1, t, t^2) 2. Originally Posted by noles2188 find the change of basis matrix from the basis B = (1, t-3, (t-3)^2) to the standard basis U = (1, t, t^2) The "change of basis matrix" maps the coefficients of vectors written in terms of the first basis to the coefficients of the same vectors written in the second basis. Further, a simple way of finding the matrix corresponding to a linear transformation in terms of bases for the "domain" and "range" spaces is: Apply the linear transformation to each of the basis vectors for the domain in turn. Write the result in terms of the range basis. The coefficients for that linear combination form the column of the matrix. For example, 1 is just mapped into 1 since the first "basis vector" in each basis is the same: 1= 1(1)+ 0(t)+ 0(t^2) so the first column of the matrix is $\begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix}$. t- 3 is mapped into -3(1)+ 1(t)+ 0(t^2) so the second column of the matrix is $\begin{bmatrix}-2 \\ 1 \\ 0\end{bmatrix}$. Can you do the third column? 3. slight typo: it's $-3.$ 4. I got 9, -6, and 1 for the third column	{"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": 0.8840592503547668, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/276522/how-to-solve-this-nonlinear-equation-system-by-changing-this-system-to-linear-sy	# how to solve this nonlinear equation system by changing this system to linear system consider following nonlinear equation system how solve it? $$x'=\|y\|$$ $$y'=x$$ and whats the matrix that associated to this system - I haven't worked out the details, but here's a thought: at first glance, your system appears to be nonlinear (in fact, the vector field isn't even C^1, which makes me nervous) but I suspect (and this will be easy to check) that an initial condition in a given quadrant (e.g. x>0, y>0) will remain in that quadrant for all times. So my gut now tells me that the fundamental solution matrix is going to have a piecewise definition based on the quadrant your initial conditions lie in (this makes sense, because your solutions should not be C^1). – A Blumenthal Jan 12 at 20:35 all none linear system change to this form AX+G(t)=0 by Jacobian matrix i mean whats this matrix and this question is exercise of book and doesn't define initial condition – Maisam Hedyelloo Jan 12 at 20:56 What's the book? On what pages does it discuss your $AX+G(t)=0$? On what page is the exercise? – Gerry Myerson Jan 25 at 5:49 ## 1 Answer It seems like you would have to be given some ICs that keep you in a particular quadrant or that you would require that with your solution. In order to solve, we would have to break this up into two systems as: $$x'=y$$ $$y'=x$$ This would give us eigenvalues $\lambda_{1,2} = \pm 1$, which yields a solution: $x(t) = c_1 e^{-t} (e^{2 t}+1)+c_2 e^{-t} (e^{2 t}-1)$, and $y(t) = c_1 e^{-t} (e^{2 t}-1)+c_2 e^{-t} (e^{2 t}+1)$ $$x'=-y$$ $$y'=x$$ This would give us eigenvalues $\lambda_{1,2} = \pm i$, which yields a solution: $x(t) = c_2 \sin(t)+c_1 \cos(t)$, and $y(t) = c_1 \sin(t)+c_2 \cos(t).$ Of course, there would have to be restrictions placed on the ICs and I suppose you can just plot them with different values of x and y and see if they are consistent for all time. It is possible that you would have to glue together the solutions for both systems based on the the values of $x$ and $y$. Regards - $+1\;$You look good in green $\;\ddot\smile\;$ – amWhy Apr 22 at 0:11 @amWhy: Thank you, I never feel like I get enough of those and some of you seem to be in the stratosphere with the number of answers and accepts (green)! :-) Yourself included! regards – Amzoti Apr 22 at 0:16	{"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": 11, "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": 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": 0.9591816067695618, "perplexity_flag": "middle"}
http://nrich.maths.org/5876	### Circles Ad Infinitum A circle is inscribed in an equilateral triangle. Smaller circles touch it and the sides of the triangle, the process continuing indefinitely. What is the sum of the areas of all the circles? ### Climbing Powers $2\wedge 3\wedge 4$ could be $(2^3)^4$ or $2^{(3^4)}$. Does it make any difference? For both definitions, which is bigger: $r\wedge r\wedge r\wedge r\dots$ where the powers of $r$ go on for ever, or $(r^r)^r$, where $r$ is $\sqrt{2}$? ### Flexi Quads A quadrilateral changes shape with the edge lengths constant. Show the scalar product of the diagonals is constant. If the diagonals are perpendicular in one position are they always perpendicular? # Root Hunter ##### Stage: 5 Challenge Level: . This graph is positive for $x = 5$ and negative for $x = 3$. This means that the graph must cut the $x$ axis somewhere between $3$ and $5$. Although in this case the result is obvious (because we have the whole graph to look at!), we can also use this idea to show that more tricky functions also have roots. Use this idea to show that these functions possess at least one solution $f(x) = 0$: $$f(x)=\frac{1}{x-2}+\frac{1}{x-3}$$ $$f(x)= x^x - 1.5 x$$ $$f(x)= x^{1000000}+{1000000}^x - 17$$ $$f(x)=\cos(\sin(\cos x)) - \sin(\cos(\sin x))$$ Optional extension activity: Can you make a spreadsheet that helps you find the numerical values of the roots to, say, four decimal places? 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": 4, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9159531593322754, "perplexity_flag": "head"}
http://nrich.maths.org/6276	### I'm Eight Find a great variety of ways of asking questions which make 8. ### More Mods What is the units digit for the number 123^(456) ? ### Expenses What is the largest number which, when divided into 1905, 2587, 3951, 7020 and 8725 in turn, leaves the same remainder each time? # Galley Division ##### Stage: 4 Challenge Level: The 'Galley' method of division was probably the most widely used method of division prior to the 1600s. It is called 'Galley' because, when completed, a division looks a bit like a galley (boat). The Galley method is thought to be Hindu in origin and an early version of this method was known to be used in the ninth century by a Persian called Al-Khowarizmi. It is still taught in parts of North Africa and the Middle East today. This is how you could work out $65284$ divided by $594$ using the Galley method: Can you see the two numbers I started with? The calculation tells me that $625284 \div 594 = 109$ remainder $538$. Can you see the answer and the remainder? Here's another division done using the Galley method: What division do you think is being done here? What is the answer to this division? Here are two video clips of divisions using Galley arithmetic: This text is usually replaced by the Flash movie. This text is usually replaced by the Flash movie. If you are struggling to see what is happening, try the division using your own method and see if you can recognise numbers that are appearing in the galley method that are also in your own recording. If you need further guidance, there is a division, showing each of the main steps and some explanation, in the hints section of this problem. Send us answers to the following divisions showing that you have used the same method. $$2986 \div 47$$ $$76254 \div 235$$ Why does the method work? How does this method compare with our modern methods of division? 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": 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": 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": 0.950797438621521, "perplexity_flag": "middle"}
http://en.wikipedia.org/wiki/Menger_sponge	# Menger sponge An illustration of M4, the fourth iteration of the construction process. In mathematics, the Menger sponge is a fractal curve. It is a universal curve, in that it has topological dimension one, and any other curve (more precisely: any compact metric space of topological dimension 1) is homeomorphic to some subset of it. It is sometimes called the Menger-Sierpinski sponge or the Sierpinski sponge. It is a three-dimensional extension of the Cantor set and Sierpinski carpet. It was first described by Karl Menger (1926) while exploring the concept of topological dimension. The Menger sponge simultaneously exhibits an infinite surface area and encloses zero volume. ## Construction The construction of a Menger sponge can be described as follows: 1. Begin with a cube (first image). 2. Divide every face of the cube into 9 squares, like a Rubik's Cube. This will sub-divide the cube into 27 smaller cubes. 3. Remove the smaller cube in the middle of each face, and remove the smaller cube in the very center of the larger cube, leaving 20 smaller cubes (second image). This is a level-1 Menger sponge (resembling a Void Cube). 4. Repeat steps 2 and 3 for each of the remaining smaller cubes, and continue to iterate ad infinitum. The second iteration will give you a level-2 sponge (third image), the third iteration gives a level-3 sponge (fourth image), and so on. The Menger sponge itself is the limit of this process after an infinite number of iterations. An illustration of the iterative construction of a Menger sponge up to M3, the third iteration. The number of cubes is 20n, with n being the number of iterations performed on the first cube. A sculptural representation of the previous illustration. ## Properties Each face of the Menger sponge is a Sierpinski carpet; furthermore, any intersection of the Menger sponge with a diagonal or medium of the initial cube M0 is a Cantor set. The Menger sponge is a closed set; since it is also bounded, the Heine–Borel theorem implies that it is compact. It has Lebesgue measure 0. It is an uncountable set. The topological dimension of the Menger sponge is one, the same as any curve. Menger showed, in the 1926 construction, that the sponge is a universal curve, in that any possible one-dimensional curve is homeomorphic to a subset of the Menger sponge, where here a curve means any compact metric space of Lebesgue covering dimension one; this includes trees and graphs with an arbitrary countable number of edges, vertices and closed loops, connected in arbitrary ways. In a similar way, the Sierpinski carpet is a universal curve for all curves that can be drawn on the two-dimensional plane. The Menger sponge constructed in three dimensions extends this idea to graphs that are not planar, and might be embedded in any number of dimensions. The Menger sponge simultaneously exhibits an infinite surface area and encloses zero volume. In spite of this, there exists a homeomorphism of the cube having finite distortion that "squeezes the sponge" in the sense that the holes in the sponge go to a Cantor set of zero measure (Iwaniec & Martin 2001, §6.5.6). The sponge has a Hausdorff dimension of (log 20) / (log 3) (approximately 2.726833). ## Formal definition Formally, a Menger sponge can be defined as follows: $M := \bigcap_{n\in\mathbb{N}} M_n$ where M0 is the unit cube and $M_{n+1} := \left\{\begin{matrix} (x,y,z)\in\mathbb{R}^3: & \begin{matrix}\exists i,j,k\in\{0,1,2\}: (3x-i,3y-j,3z-k)\in M_n \\ \mbox{and at most one of }i,j,k\mbox{ is equal to 1}\end{matrix} \end{matrix}\right\}.$ ## References • Iwaniec, Tadeusz; Martin, Gaven (2001), Geometric function theory and non-linear analysis, Oxford Mathematical Monographs, The Clarendon Press Oxford University Press, ISBN 978-0-19-850929-5, MR1859913 . • Menger, Karl (1926), "Allgemeine Räume und Cartesische Räume. I.", Communications to the Amsterdam Academy of Sciences English translation reprinted in Edgar, Gerald A., ed. (2004), Classics on fractals, Studies in Nonlinearity, Westview Press. Advanced Book Program, Boulder, CO, ISBN 978-0-8133-4153-8, MR2049443 • Karl Menger, Dimensionstheorie, (1928) B.G Teubner Publishers, Leipzig. • Zhou, Li (2007), "Problem 11208: Chromatic numbers of the Menger sponges", 114 (9): 842	{"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": 2, "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": 0.844073474407196, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/203640/ratio-between-two-sets-of-numbers	# Ratio between two sets of numbers? I have two sets of numbers. The first set is 1 to 4. The second set is 0 to 190. What is the calculation to get the proportionate number in the second set if a number in the first set is, for example, 1.2? Math is not my forte! - ## 1 Answer It is $\displaystyle\frac{1.2-1}{4-1}\cdot (190-0) +0$. - Can you express that equation without the fraction? For a programming equation? – user1342133 Sep 27 '12 at 23:19 If your input is $x$, then to go from $[1,4]$ to $[0,190]$, compute $(x-1)*190/3$. – Théophile Sep 27 '12 at 23:32 That's it, thanks! – user1342133 Sep 27 '12 at 23:37	{"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": 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": 0.9148833751678467, "perplexity_flag": "middle"}
http://mathhelpforum.com/advanced-algebra/95583-algebra-problems-fun-35-a.html	# Thread: 1. ## Algebra, Problems For Fun (35) Characterize all subgroups of $\mathbb{Q}^{\times}.$ 2. here's an idea: let $G=\{1,-1 \},$ considered as a multiplicative group. show that $\mathbb{Q}^{\times}$ and $G \oplus \mathbb{Z} \oplus \mathbb{Z} \oplus \cdots$ are isomorphic. 3. My guess would be that $~ \mathbb{Q}^{\times}_{>0} = \displaystyle\bigoplus_{p ~ prime} \langle p \rangle ~$. Clearly each $\langle p \rangle \simeq \mathbb{Z}$ Then $\mathbb{Q}^{\times} \simeq G \oplus \mathbb{Q}^{\times}_{>0}$ as we can define the trivial isomorphism $\theta$ : $(g,q) \rightarrow gq$ Just the first step needs some justification... 4. Originally Posted by pomp My guess would be that $~ \mathbb{Q}^{\times}_{>0} = \oplus_p \langle p \rangle ~$ for p prime. Clearly each $\langle p \rangle \simeq \mathbb{Z}$ Then $\mathbb{Q}^{\times} \simeq G \oplus \mathbb{Q}^{\times}_{>0}$ as we can define the trivial isomorphism $\theta$ : $(g,q) \rightarrow gq$ Just the first step needs some justification... well, every $r \in \mathbb{Q}^{\times}$ can be written uniquely as $r=\pm p_1^{n_1}p_2^{n_2} \cdots p_k^{n_k},$ where $p_j$ is the j-th prime number and $n_j \in \mathbb{Z}.$ now define $f(r)=(\text{sgn}(r),n_1,n_2, \cdots, n_k, 0, 0, \cdots ).$ 5. EDIT: Aw man, ninja'd >< Original post: As a consequence of uniqueness of prime factorization, every nonzero rational number can be expressed uniquely as $q=(-1)^{r_0}p_1^{r_1}p_2^{r_2}\ldots p_s^{r_s}$ where $r_i \in \mathbb{Z}^\times$. Then given a particular enumeration of the primes $P=\{p_1, p_2, \ldots\}$, there are only a finite number of prime divisors, so with respect to our enumeration there is a unique such representation with $q=(-1)^{r_0}p_1^{r_1}p_2^{r_2}\ldots =(-1)^{r_0}\prod_{i=1}^{\infty} p_i^{r_i}$ where $r_i \in \mathbb{Z}$ such that only a finite number of $r_i$ are nonzero This would suggest an isomorphism $Q^\times \rightarrow G\times\displaystyle\bigoplus_{i=1}^\infty \mathbb{Z}$ given by $(-1)^{r_0}\prod_{i=1}^{\infty} p_i^{r_i} \mapsto ((-1)^{r_0}, r_1, r_2, \ldots)$. Uniqueness of factorization assures this is a well defined and 1-1 function. The finite number of prime factors assures this indeed maps into and onto the direct sum. Finally the morphism property follows from the properties of exponents. Hand wavy but conveys the ideas I think. 6. so, to answer the question you only need to find the subgroups of $G \oplus \mathbb{Z} \oplus \mathbb{Z} \oplus ...$. it shouldn't be hard! 7. Either the subgroups are finitely generated or not. If they are finitely generated, then the fundamental theorem of finitely generated abelian groups tells us that up to isomorphisms the subgroups are the direct product of copies of the integers with the 2 element group. If a subgroup is not finitely generated, then intuition says we should expect an isomorphic copy of the group, or an isomorphic copy of the infinite direct sum of integers. 8. let $N$ be a subgroup of $\mathbb{Q}^{\times}.$ then either $\{1,-1 \}=G \subseteq N$ or $N \cap G = \{1 \}.$ if $G \subseteq N,$ then $N/G$ would be a subgroup of the free abelian group $\mathbb{Q}^{\times}/G \cong \mathbb{Z} \oplus \mathbb{Z} \oplus \cdots.$ we know that any (non-trivial} subgroup of a free abelian group is free. so $N$ is isomorphic to $G \oplus K,$ where $K$ is the direct sum of finitely or infinitely (countable) many copies of $\mathbb{Z}.$ if $N \cap G = \{1 \},$ then $NG/G \cong N/(N \cap G) \cong N.$ thus $NG/G \cong N$ is a subgroup of the free abelian group $\mathbb{Q}^{\times}/G \cong \mathbb{Z} \oplus \mathbb{Z} \oplus \cdots.$ so $N$ is just a direct sum of finitely or infinitely (countable) many of copies of $\mathbb{Z}.$ briefly subgroups of $\mathbb{Q}^{\times}$ are in one of these forms: $\{1 \}, \ \{1,-1 \}, \ K, \ \{1,-1 \} \oplus K,$ where $K$ is any free abelian group of at most countable rank.	{"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": 48, "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": 0.9175580739974976, "perplexity_flag": "head"}
http://mathoverflow.net/questions/114881/a-consequence-of-convexity/114885	## A consequence of convexity ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Let $f:\mathbb{R}\to\mathbb{R}$ a convex decreasing function. Let $x_0 < x_1 < x_2$. Studying the behaviour of the difference quotient, it is clear that $$f(x_0)-f(x_2) \leq M (f(x_0)-f(x_1))$$ with $M=\frac{x_2-x_0}{x_1-x_0}>0$. Now take $F:\mathbb{R}^2\to\mathbb{R}$ convex and decreasing with respect to each variable. Let $x_0 < x_1 < x_2$ and $y_0 < y_1 < y_2$. I ask if a similar condition holds, say for example $$F(x_0,y_0)-F(x_2,y_2) \leq M (F(x_0,y_0)-F(x_1,y_1))$$ with $M=\max ( \frac{x_2-x_0}{x_1-x_0}, \frac{y_2-y_0}{y_1-y_0} )$ or $M=\frac{x_2-x_0}{x_1-x_0}+\frac{y_2-y_0}{y_1-y_0}$. Edit after Brian's answer: you may add the hypothesis that $F$ is $C^2$ and also the mixed derivative $\frac{\partial^2F}{\partial x\partial y}$ is non-negative. Edit: under the additional assumption the answer is yes with $M=\max(\frac{x_2-x_0}{x_1-x_0},\ \frac{x_2-x_0}{x_1-x_0})$. But is it possible to reach the same conclusion without the additional assumption on the mixed derivatives? - Do I understand correctly that you want your inequality for $f$ to hold with SOME $M$ that depends only on $x_j,y_j$ but not of $f$ ? Or you only want one of the two expressions you proposed for $M$ ? – Alexandre Eremenko Nov 29 at 13:34 Yes I want an inequality with some $M$ that depends only on the points $x_,y_j$, not necessarely one of those I proposed. – xn--qwertyuiop-86a Nov 29 at 17:01 ## 3 Answers You can use the triangle inequality to solve this by looking at each coordinate separately. $F(x_0,y_0)-F(x_2,y_2)=F(x_0,y_0)-F(x_2,y_0)+F(x_2,y_0)-F(x_2,y_2)$ $\leq M_1(F(x_0,y_0)-F(x_1,y_0))+M_2(F(x_2,y_0)-F(x_2,y_1))$. Replacing $F(x_1,y_0)$ with $F(x_1,y_1)$ only increases the right side, and replacing both $x_2$'s on the far right side with $x_0$ only makes the right hand side larger by convexity. Finally, replace the last $x_0$ that we just added with $x_1$ without ncreasing the right hand side. This gives us $F(x_0,y_0)-F(x_2,y_2)\leq M(F(x_0,y_0)-F(x_1,y_1))$, where $M$ is $2\max(\frac{x_2-x_0}{x_1-x_0},\frac{y_2-y_0}{y_1-y_0})$. $M$ can also be the sum instead of the max, in which case we can drop the 2. Edit: This argument doesn't work without additional assumptions; see the comments. - Thank you Brian. I think the statement after the triangular inequality should be "Repalcing $F(x_1,y_0)$ with $F(x_1,y_1)$". Please could you explain how do you use convexity to prove that $F(x_2,y_0)-F(x_2,y_1)\leq F(x_0,y_0)-F(x_0,y_1)$ ? – xn--qwertyuiop-86a Nov 29 at 16:59 1 @xn I can't explain it because it's not true! I had a false picture in my head. I was essentially claiming (in the smooth case) that having positive `unmixed' second partial derivatives implies that the mixed partial derivative is positive, which is not true. I apologize! – Brian Rushton Nov 29 at 17:25 Thank you as well, maybe I can extend my assumptions. If I assume all the second partial derivatives (mixed and pures) of F are $\geq0$, all should work right? – xn--qwertyuiop-86a Nov 29 at 17:49 1 As long as the derivatives exist, it should work. If the derivatives do not exist, then you would have to assume that $F(x,y)-F(x,y+\Delta Y)$ is a decreasing function of $x$ (you could also interchange $x$ and $y$ and have this work). This is analagous to regular convexity where $F(x)-F(x+\Delta x)$ is a decreasing function of $x$, and it probably has a name. But yes, it should work if all second derivatives are positive and $F$ is decreasing in each variable separately. – Brian Rushton Nov 29 at 18:43 1 To improve the bound one may use: 1) triangular inequality; 2) convexity in $x$ and $y$; 3) $\frac{\partial F}{\partial x\partial y}\geq0$. And to sum up one obtains the bound $M=\max(\frac{x_2-x_0}{x_1-x_0},\ \frac{y_2-y_0}{y_1-y_0})$. In this way there's no $2$, so that $M$ does not depend on the dimension of the space. – xn--qwertyuiop-86a Nov 30 at 9:48 show 1 more comment ### 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. Without loss of generality, we can assume that $$\frac{y_2-y_0}{y_1-y_0}\ge\frac{x_2-x_0}{x_1-x_0}.$$ Let $x_2'$ be the real number number such that $$\frac{y_2-y_0}{y_1-y_0}=\frac{x'_2-x_0}{x_1-x_0}.$$ Clearly $x_2'\ge x_2$; therefore $$F(x_2',y_2)\le F(x_2,y_2).$$ From the one-dimensional case, you get $$F(x_0,y_0)-F(x_2',y_2) \leq M\cdot (F(x_0,y_0)-F(x_1,y_1))$$ with $$M=\frac{y_2-y_0}{y_1-y_0}=\max\left\{\frac{y_2-y_0}{y_1-y_0},\frac{x_2-x_0}{x_1-x_0}\right\}.$$ Hence the result follows. It seems to be the optimal bound. P.S. It turned out that the term "convex" was used in a nonstandard way; namely, xm wants to consider $F$ such that the restriction to any line parallel to the $x$-axis or to the $y$-axis is convex. In particular $F(x,y)= -x\cdot y$ is "xm"-convex in positive quadrant. Note that in this example $F(t,t)=- t^{2}$, so $M$ has to be much bigger; maybe $$M=\left(\max\left\{\frac{y_2-y_0}{y_1-y_0},\frac{x_2-x_0}{x_1-x_0}\right\}\right)^2$$ will do. - Thank you Anton. Please could you explain how you use the one dimensional case to get the inequality? I found the same bound but adding the hypothesis $\frac{\partial F}{\partial x\partial y}\geq0$ (see the last comment to Brian's answer). Instead you don't use this hypothesis, right? – xn--qwertyuiop-86a Nov 30 at 10:01 @xn--qwertyuiop-86a, consider function $$f(t)=F(x_0+(x_1-x_0)\cdot t, y_0+(y_1-y_0)\cdot t).$$ – Anton Petrunin Nov 30 at 23:46 @Anton Petrunin, let me see if I've understood. The inequality $F(x_0,y_0)-F(x_2',y_2)\leq M\cdot(F(x_0,y_0)-F(x_1,y_1))$ is equivalent to $f(0)-f(\frac{y_2-y_0}{y_1-y_0}\leq M\cdot(f(0)-f(1)))$ and this is true if $f$ is convex. But to have $f$ convex one need also $\frac{\partial^2F}{\partial x\partial y}\geq0$: it isn't enough to assume $F$ convex w.r.t. each variable separately. Correct? – xn--qwertyuiop-86a Dec 1 at 10:26 I understood the question as "convex" and "decreasing with respect to each variable". Convex usually means convex on any line. It seems that you wanted to say "restriction to any line t↦(x,t) and t↦(t,y) is convex". Let me know if this is the case. – Anton Petrunin Dec 1 at 19:16 Yes, I wanted to say $F$ restricted to any line parallel to the $x$-axis or to the $y$-axis is convex. In other words (assuming $F$ regular enough) my hypotesis is that $\frac{\partial^2 F}{\partial x^2}\geq0$ and $\frac{\partial^2 F}{\partial y^2}\geq0$ . – xn--qwertyuiop-86a Dec 2 at 12:09 show 1 more comment In general, whenever you have a (separately) convex function $f :\mathbb{R}^N \to \mathbb{R}$, which means it is convex in each variable, this implies the function is locally Lipschitz. The fact that it is decreasing allows you to make some explicit calculations of the constant, as Brian Rushton does, but in general the inequality looks like this: $\|f(x)-f(y)\| \leq \frac{N}{r} (\sup_{B_{2r}} f - \inf_{B_{2r}} f) \|x-y\|$ for all $x,y \in B_{r}$. -	{"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": 9, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9439836144447327, "perplexity_flag": "head"}
http://stats.stackexchange.com/questions/16998/accuracy-of-a-random-classifier	# Accuracy of a random classifier I was wondering how to compare accuracy of my classifier to a random one. I'm going to elaborate further. Let's say we have a binary classification problem. We have $n^+$ positive examples and $n^-$ negative examples in the test set. I say that and record is positive with a probability of $p$. I can estimate that, on average, I get: $$TP = pn^+ \ TN=(1-p)n^- \ FN = pn^- \ FP = (1-p)n^+$$ thus $$\mbox{acc} = \frac{TP+TN}{TP+TN+FP+FN} = \frac{pn^+ + (1-p)n^-}{pn^+(1-p)n^-pn^- + (1-p)n^+}$$ that is $$= \frac{pn^+ + (1-p)n^-}{n^+ + n^-}$$ For example if we have $n^+=n^-$ accuracy is always $1/2$ for any $p$. This can be extended in multiclass classification: $$\mbox{acc} = \frac{\sum_{i=1}^c p_i n_i}{\sum n_i}$$ where $p_i$ is the probability to say "it is in the $i$th class", and $n_i$ is the count of records of class $i$. Also in this case, if $n_i = n/c \ \forall i$ then $$\mbox{acc} = 1/c$$ But, how can I compare the accuracy of my classifier without citing a test set? For example if I said: my classifier accuracy is 70% (estimated somehow, e.g. Cross-Validation), is it good or bad compared to a random classifier? - ## 1 Answer I am not sure I understand the last part of your question But, how can I compare the accuracy of my classifier without citing a test set? but I think I understand your concern. A given binary classifier's accuracy of 90% may be misleading if the natural frequency of one case vs the other is 90/100. If the classifier simply always chooses the most common case then it will, on average, be correct 90% of the time. A useful score to account for this issue is the Information score. A paper describing the score and its rationale can be found here. I learned about this score because it is part of the cross-validation suite in the excellent Orange data mining tools (you can use no-coding-needed visual programming or call libraries from Python). - Thank you. You made me realize that you have always to take into account prior probabilities. Even if the accuracy is estimated by Cross-Validation, the classes do have prior probabilities. And, for example, you can exstimate them with $n_i$ that I wrote in my question. – Simone Oct 16 '11 at 10:01 2 – Simone Oct 16 '11 at 10:19 Last comment. Classifying the positive class with probability $p$ means that your classifier lies along the main diagonal of the ROC curve (if you are able to compute it, because not all the classifier gives probabilities for output, as state in Kononenko and Bratko paper). Because $TPR = \frac{pn^+}{pn^+ + (1-p)n^+} = p$ and $FPR = \frac{pn^-}{pn^- + (1-p)n^-} = p$. With a diagonal ROC curve you can understand your performance is similar to a random one. Instead, changing $p$ changes $acc$, and it is somehow misleading. Thus, there should be better performance measures. – Simone Oct 16 '11 at 10:51 @Simone: +1 for finishing your thoughts and the useful link, but mostly for exposing me to a new word (maieutics). – Josh Hemann Oct 17 '11 at 4:53	{"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": 17, "mathjax_display_tex": 5, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9190309643745422, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/182704/a-question-on-iterated-sums?answertab=active	# A question on iterated sums We are to prove that the set $$\Bigg\{ \sum^m_{i=1}\sum^n_{j=1}\|a_ib_j\| :m,n\in \mathbb N \Bigg\}$$ is bounded. Also, We are to use this to show that the iterated sum $\sum^\infty_{i=1} \sum^\infty_{j=1}\|a_ib_j\|$ converges. I could prove that the set is bounded, however, I can't solve the second part. I become very confused when I have to deal with iterated sums. Could anyone please help me with this problem. - 2 There are two hints: (1) The infinite summation of a non-negative sequence is equal to the supremum of the set of partial sums: $$\sum_{i=1}^{\infty} \|a_i\| = \sup_{n\in\mathbb{N}} \sum_{i=1}^{n} \|a_i\|,$$ and (2) supremum can be distributed: $$\sup_{(x,y)\in A\times B}f(x, y) = \sup_{x\in A}\sup_{y\in B} f(x,y).$$ – sos440 Aug 15 '12 at 5:16 ## 2 Answers Although by first proving that the sum is bounded, it can be proved that the sum converges. But, there is a way to directly prove that it converges. For that, let $$\sum^\infty_{i=1}\|a_i\|=L~~~~~\text{and}~~~~~\sum^\infty_{j=1}\|b_j\|=M$$ For each fixed $i \in \mathbb N$, the Algebraic Limit Theorem will allow us to write $\sum^\infty_{j=1}\|a_ib_j\|$=$\|a_i\|\sum^\infty_{j=1}\|b_j\|$. If one continues the process, we will see that: $$\sum^\infty_{i=1}\sum^\infty_{j=1}\|a_ib_j\|=\sum^\infty_{i=1}\|a_i\|\sum^\infty_{j=1}\|b_j\|=ML$$ Hence,$\sum^\infty_{i=1}\sum^\infty_{j=1}\|a_ib_j\|$ converges (to $ML$). - In order to show that $\sum_{i=1}^\infty\sum_{j=1}^\infty\|a_ib_j\|$ converges, you must show that there is some real number $a$ such that for every $\epsilon>0$ there is some $n_0\in\Bbb Z^+$ such that $$\left\|a-\sum_{i=1}^m\sum_{j=1}^n\|a_ib_j\|\right\|<\epsilon$$ whenever $m,n\ge n_0$. For $m,n\in\Bbb Z^+$ let $$s_{mn}=\sum_{i=1}^m\sum_{j=1}^n\|a_ib_j\|\;.$$ You know that the set $S=\{s_{mn}:m,n\in\Bbb Z^+\}$ is bounded. You also know that $s_{k\ell}\le s_{mn}$ whenever $k\le m$ and $\ell\le n$, since the terms $\|a_ib_j\|$ are non-negative. For $n\in\Bbb Z^+$ let $d_n=s_{nn}$; then $s_{k\ell}\le d_n$ whenever $k,\ell\le n$. In other words, the sums $d_n$ are cofinal in $S$: for each $s\in S$ there is an $n\in\Bbb Z^+$ such that $s\le d_n$. Consider the sequence $\langle d_n:n\in\Bbb Z^+\rangle$: it’s bounded and non-decreasing, so it converges to some limit $a$; I’ll leave to you the easy verification that $s\le a$ for all $s\in S$. Fix $\epsilon>0$. There is some $n_0\in\Bbb Z^+$ such that $\|a-d_n\|<\epsilon$ whenever $n\ge n_0$. Suppose that $m,n\ge n_0$; then $a-\epsilon<d_{n_0}\le s_{mn}\le a$, so $$\left\|a-\sum_{i=1}^m\sum_{j=1}^n\|a_ib_j\|\right\|<\epsilon$$ whenever $m,n\ge n_0$, exactly as we wanted. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 37, "mathjax_display_tex": 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": 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": 0.9593819379806519, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/22928/limit-of-algebraic-function-avoiding-lhopitals-rule	# Limit of algebraic function avoiding l'Hôpital's rule Find $$\lim_{x \to \infty} \sqrt[3]{1 + x^{2} + x^{3}} -x$$ - 1 @white: I have edited the question, please if this is what you wanted to post or not. – anonymous Feb 20 '11 at 13:39 Hi ? Please ? Homework ? Any thoughts ? – Sam Feb 20 '11 at 13:39 1 @white: As Arturo, always keeps mentioning, please pose the question in a more polite form. This is like asking a homework question. We would like to know what you have tried and where you are finding difficulty. – anonymous Feb 20 '11 at 13:40 lol..ok i am sorry if i am rude.. – user7284 Feb 20 '11 at 13:43 3 In general, $\lim_{x\rightarrow\infty} \sqrt[n]{x^n+ax^{n-1}+O(x^{n-2})}-x=\frac{a}{n}$. – Eric♦ Feb 20 '11 at 23:59 show 6 more comments ## 5 Answers HINT $\$ It's simply a first derivative: $\:$ changing variables $\rm\ x\to 1/x\$ transforms it to $$\rm\displaystyle\ \lim_{x\to\ 0^{+}}\ \frac{f(x)-f(0)}x\ =\ f\:'(0) \ \ \ for\ \ \ f(x) = (1+x+x^3)^{1/3}$$ Now it is easy to calculate $\rm\ f\:'(0)\ =\ 1/3\$ by direct evaluation (it's not indeterminate). Namely $$\rm f\:'(x)\ =\ \frac{d}{dx}\ (1+x+x^3)^{1/3}\ =\ \frac{1+3\ x^2}{3\ (1+x+x^3)^{2/3}}\ \ \Rightarrow\ \ f\:'(0)\ =\ \frac{1}3$$ Note that this method employs only knowledge of the definition of the derivative and basic rules for calculating derivatives of polynomial and powers. It does not employ more advanced techniques such as L'Hospital's rule, or (binomial) power series expansions, etc. - is this lopital?? – user7284 Feb 20 '11 at 19:29 @white: The above easy calculation of $\rm\ f\:'(0)\$ doesn't use L'Hopital's rule (it's not indeterminate). While there is indeed a close relationship between L'Hospital's rule, derivatives, Taylor's formula, the mean-value-theorem, etc., that doesn't imply that use of the latter implies use of L'Hospital's rule. The above answer uses only the definition of the derivative and the basic rules for calculating derivatives. – Gone Feb 20 '11 at 19:53 I like this answer, +1. To explain something white, this actually is L'hopitals rule, but not quite. I say not quite because Bill's argument actually shows that L'Hopitals Rule follows immediately from the definition of the derivative when the denominator is $x$. – Eric♦ Feb 20 '11 at 22:41 Here is a more elementary way using the difference of cubes identity. (It is not as elegant as the Taylor series presented by Willie Wong, but requires less background) Since $x=\sqrt[3]{x^{3}}$, we are looking at $\lim_{x\rightarrow\infty}\sqrt[3]{1+x^{2}+x^{3}}-\sqrt[3]{x^{3}}$.Recall that the cubic identity $a^{3}-b^{3}=\left(a-b\right)\left(a^{2}+ab+b^{2}\right)$, which tells us that $$\left(\sqrt[3]{1+x^{2}+x^{3}}\right)^{3}-\left(\sqrt[3]{x^{3}}\right)^{3}$$ $$=\left(\sqrt[3]{1+x^{2}+x^{3}}-\sqrt[3]{x^{3}}\right)\cdot \left(\left(1+x^{2}+x^{3}\right)^{\frac{2}{3}}+\sqrt[3]{x^{3}}\cdot\left(1+x^{2}+x^{3}\right)^{\frac{1}{3}}+\left(x^{3}\right)^{\frac{2}{3}}\right)$$ and hence $$\frac{1+x^{2}}{\left(1+x^{2}+x^{3}\right)^{\frac{2}{3}}+x\cdot\left(1+x^{2}+x^{3}\right)^{\frac{1}{3}}+x^{2}}=\sqrt[3]{1+x^{2}+x^{3}}-\sqrt[3]{x^{3}}.$$ Divide the top and bottom of the left hand side by $x^{2}$ to find $$\frac{1+\frac{1}{x^{2}}}{\left(1+\frac{1}{x^{2}}+\frac{1}{x^{3}}\right)^{\frac{2}{3}}+\left(1+\frac{1}{x^{2}}+\frac{1}{x^{3}}\right)^{\frac{1}{3}}+1}=\sqrt[3]{1+x^{2}+x^{3}}-\sqrt[3]{x^{3}}.$$ Since $$\lim_{x\rightarrow\infty}1+\frac{1}{x^{2}}=1$$ and $$\lim_{x\rightarrow\infty}\left(1+\frac{1}{x^{2}}+\frac{1}{x^{3}}\right)^{\frac{2}{3}}+\left(1+\frac{1}{x^{2}}+\frac{1}{x^{3}}\right)^{\frac{1}{3}}+1=3$$ we see by the quotient rule for limits $$\lim_{x\rightarrow\infty}\sqrt[3]{1+x^{2}+x^{3}}-\sqrt[3]{x^{3}}=\frac{1}{3}.$$ Hope that helps, Edit: This faq question: Limits: How to evaluate $\lim\limits_{x\rightarrow \infty}\sqrt[n]{x^{n}+a_{n-1}x^{n-1}+\cdots+a_{0}}-x$ was made shortly after this post to help answer it more generally. - hey..thx for the info..but unfortunately i'm not so understand about this method..abit confusing as from the 3rd part of ur solution,the x has been forfited n i duno where to cancel it off..sry bro i not so good in maths..XD – user7284 Feb 20 '11 at 17:19 Which part exactly? I might be able to explain better. – Eric♦ Feb 20 '11 at 22:43 hint: you can rewrite the terms inside the limit as $$x \cdot \left( \sqrt[3]{\frac{1}{x^3} + \frac{1}{x} + 1} - 1\right)$$ for the term underneath the cube-root, use the Taylor expansion of the cube-root function near the value 1: $$\sqrt[3]{1 + y} = 1 + \frac{1}{3}y - \frac{1}{9} y^2 + \ldots$$ - bionomial???ok..i try it nw..thx for ur help.. – user7284 Feb 20 '11 at 14:32 ermm,@Willie Wong♦ ,i am not so sure about the taylor expansion as u mention..can u explain more on hw to apply it on the cube roots?? – user7284 Feb 20 '11 at 14:41 1 The Taylor expansion of (1+x)^k starts with 1+kx, even if k is not an integer. For this problem you only need the $y/3$ term. – Ross Millikan Feb 20 '11 at 16:39 – Willie Wong♦ Feb 20 '11 at 17:45 A very similar question was asked recently here... The limit follows immediately upon showing, using the mean value theorem, that $$\sqrt[3]{{\bigg(x + \frac{1}{3}\bigg)^3 }} - \sqrt[3]{{\bigg(x + \frac{1}{3}\bigg)^3 - \bigg(\frac{x}{3} - \frac{{26}}{{27}}\bigg)}} \to 0$$ as $x \to \infty$. - I hope you understand where the term $(x + \frac{1}{3})^3 - (\frac{x}{3} - \frac{{26}}{{27}})$ comes from. – Shai Covo Feb 20 '11 at 19:02 nope..i am noob maths..lol – user7284 Feb 20 '11 at 19:28 @white: It is equal to $x^3 + x^2 +1$. – Shai Covo Feb 20 '11 at 19:37 For such problems, quite often it pays off to use the identity $a^b = \exp[ b * \log(a)]$ because then one can apply all his/her knowledge about exp and log. - Can you explain? I don't see how you use this for this example unless you expand $e^{b\log a}=1+b\log a+\frac{1}{2}(b\log a)^2+\cdots$ and then expand the $\log$'s into their power series as well. Am I missing something? – Eric♦ Feb 20 '11 at 16:38 The comment was for an earlier version of the question which looked quite a bit different [$\lim_{x\to \infty} (1 + x^2 + x^3)^{1/3 -x}$]. – Fabian Feb 20 '11 at 20:27	{"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": 20, "mathjax_display_tex": 13, "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": 0.917233407497406, "perplexity_flag": "middle"}
http://new.wikipedia.org/wiki/%E0%A4%97%E0%A5%8D%E0%A4%B0%E0%A5%81%E0%A4%AA_%E0%A4%B8%E0%A4%BF%E0%A4%A6%E0%A5%8D%E0%A4%A7%E0%A4%BE%E0%A4%A8%E0%A5%8D%E0%A4%A4	# ग्रुप सिद्धान्त विकिपिडिया नं ग्रुप सिद्धान्त गणितयागु छगु ख्यः ख। थुकिगु छ्येलेज्या भौतिकशास्त्र व रसायनशास्त्रय् जुइ। पुचः यागु छ्येलेज्या गणितय् यक्व थासे अप्वयाना इन्टर्नल सिमेट्रि क्याप्चर यायेत अटोमर्फिज ग्रुप यागु रुपे जुइ। An internal symmetry of a structure is usually associated with an invariant property; the set of transformations that preserve this invariant property, together with the operation of composition of transformations, form a group called a symmetry group. In Galois theory, which is the historical origin of the group concept, one uses groups to describe the symmetries of the equations satisfied by the solutions to a polynomial equation. The solvable groups are so-named because of their prominent role in this theory. Abelian groups underlie several other structures that are studied in abstract algebra, such as rings, fields, and modules. In algebraic topology, groups are used to describe invariants of topological spaces (the name of the torsion subgroup of an infinite group shows the legacy of this field of endeavor). They are called "invariants" because they are defined in such a way that they do not change if the space is subjected to some deformation. Examples include the fundamental group, homology groups and cohomology groups. The concept of the Lie group (named after mathematician Sophus Lie) is important in the study of differential equations and manifolds; they combine analysis and group theory and are therefore the proper objects for describing symmetries of analytical structures. Analysis on these and other groups is called harmonic analysis. In combinatorics, the notion of permutation group and the concept of group action are often used to simplify the counting of a set of objects; see in particular Burnside's lemma. An understanding of group theory is also important in physics and chemistry and material science. In chemistry, groups are used to classify crystal structures, regular polyhedra, and the symmetries of molecules. In physics, groups are important because they describe the symmetries which the laws of physics seem to obey. Physicists are very interested in group representations, especially of Lie groups, since these representations often point the way to the "possible" physical theories. Physics examples: Standard Model, Gauge theory ## इतिहास There are three historical roots of group theory: the theory of algebraic equations, number theory and geometry. Euler, Gauss, Lagrange, Abel and French mathematician Galois were early researchers in the field of group theory. Galois is honored as the first mathematician linking group theory and field theory, with the theory that is now called Galois theory.[१] An early source occurs in the problem of forming an $m$th-degree equation having as its roots m of the roots of a given $n$th-degree equation ($m < n$). For simple cases the problem goes back to Hudde (1659). Saunderson (1740) noted that the determination of the quadratic factors of a biquadratic expression necessarily leads to a sextic equation, and Le Sœur (1748) and Waring (1762 to 1782) still further elaborated the idea.[१] A common foundation for the theory of equations on the basis of the group of permutations was found by mathematician Lagrange (1770, 1771), and on this was built the theory of substitutions. He discovered that the roots of all resolvents (résolvantes, réduites) which he examined are rational functions of the roots of the respective equations. To study the properties of these functions he invented a Calcul des Combinaisons. The contemporary work of Vandermonde (1770) also foreshadowed the coming theory.[१] Ruffini (1799) attempted a proof of the impossibility of solving the quintic and higher equations. Ruffini distinguished what are now called intransitive and transitive, and imprimitive and primitive groups, and (1801) uses the group of an equation under the name l'assieme delle permutazioni. He also published a letter from Abbati to himself, in which the group idea is prominent.[१] Galois found that if $r_1, r_2, \ldots, r_n$ are the $n$ roots of an equation, there is always a group of permutations of the $r$'s such that (1) every function of the roots invariable by the substitutions of the group is rationally known, and (2), conversely, every rationally determinable function of the roots is invariant under the substitutions of the group. Galois also contributed to the theory of modular equations and to that of elliptic functions. His first publication on group theory was made at the age of eighteen (1829), but his contributions attracted little attention until the publication of his collected papers in 1846 (Liouville, Vol. XI).[१] Arthur Cayley and Augustin Louis Cauchy were among the first to appreciate the importance of the theory, and to the latter especially are due a number of important theorems. The subject was popularised by Serret, who devoted section IV of his algebra to the theory; by Camille Jordan, whose Traité des Substitutions is a classic; and to Eugen Netto (1882), whose Theory of Substitutions and its Applications to Algebra was translated into English by Cole (1892). Other group theorists of the nineteenth century were Bertrand, Charles Hermite, Frobenius, Leopold Kronecker, and Emile Mathieu.[१] It was Walther von Dyck who, in 1882, gave the modern definition of a group. The study of what are now called Lie groups, and their discrete subgroups, as transformation groups, started systematically in 1884 with Sophus Lie; followed by work of Killing, Study, Schur, Maurer, and Cartan. The discontinuous (discrete group) theory was built up by Felix Klein, Lie, Poincaré, and Charles Émile Picard, in connection in particular with modular forms and monodromy. The classification of finite simple groups is a vast body of work from the mid 20th century, which is thought to classify all the finite simple groups. Other important mathematicians in this subject area include Emil Artin, Emmy Noether, Sylow, and many others. ## ग्रुप सिद्धान्त विचातः ### ग्रुपयागु अर्थ मू पौ: पुचः (गणित) A group (G, ) is a set G with a binary operation : G × G → G (one that assigns each ordered pair (a,b) in G an element in G denoted by ab) that satisfies the following 3 axioms: • Associativity: For all a, b and c in G, (a b) * c = a * (b * c). • Identity element: There is an element e in G such that for all a in G, e * a = a * e = a. • Inverse element: For each a in G, there is an element b in G such that a * b = b * a = e, where e is an identity element. ### Order of groups and elements The order of a group G is the number of elements in the set G. If the order is not finite, then the group is an infinite group. The order of an element a in a group G is the least positive integer n such that an=e, where an is multiplication of a by itself n times. For finite groups, it can be shown that the order of every element in the group must divide the order of the group. ### Subgroups A set H is a subgroup of a group G if it is a subset of G and is a group using the operation defined on G. In other words, H is a subgroup of (G, ) if the restriction of to H is a group operation on H. If G is a finite group, the order of H divides the order of G. A subgroup H is a normal subgroup of G if for all h in H and g in G, ghg-1 is also in H. An alternate definition is that a subgroup is normal if its left and right cosets coincide. Normal subgroups are useful because they can be used to create quotient groups. ### Special classes of groups A group is abelian (or commutative) if the operation is commutative (that is, for all a, b in G, a * b = b * a). A non-abelian group is a group that is not abelian. The term "abelian" is named after the mathematician Niels Abel. A cyclic group is a group that is generated by a single element. A simple group is a group that has no nontrivial normal subgroups. A solvable group , or soluble group, is a group that has a normal series whose quotient groups are all abelian. The fact that S5, the symmetric group in 5 elements, is not solvable proves that some quintic polynomials cannot be solved by radicals. ### Operations on groups A homomorphism is a map between two groups that preserves the structure imposed by the operator. If the map is bijective, then it is an isomorphism. An isomorphism from a group to itself is an automorphism. The set of all automorphisms of a group is a group called the automorphism group. The kernel of a homomorphism is a normal subgroup of the group. A group action is a map involving a group and a set, where each element in the group defines a bijective map on a set. Group actions are used to prove the Sylow theorems and to prove that the center of a p-group is nontrivial. ## Some useful theorems • Some basic results in elementary group theory • Lagrange's theorem: if G is a finite group and H is a subgroup of G, then the order (that is, the number of elements) of H divides the order of G. • Cayley's Theorem: every group G is isomorphic to a subgroup of the symmetric group on G. • Sylow theorems: perhaps the most useful of the group theorems. Among them, that if pn (and p prime) divides the order of a finite group G, then there exists a subgroup of order pn. • The Butterfly lemma is a technical result on the lattice of subgroups of a group. • The Fundamental theorem on homomorphisms relates the structure of two objects between which a homomorphism is given, and of the kernel and image of the homomorphism. • Jordan-Hölder theorem: any two composition series of a given group are equivalent. • Krull-Schmidt theorem: a group G, subjected to certain finiteness conditions of chains of subgroups, can be uniquely written as a finite product of indecomposable subgroups. • Burnside's lemma: the number of orbits of a group action on a set equals the average number of points fixed by each element of the group ## Miscellany James Newman summarized group theory as follows: The theory of groups is a branch of mathematics in which one does something to something and then compares the results with the result of doing the same thing to something else, or something else to the same thing. One application of group theory is in musical set theory. Group theory is also very important to the field of chemistry, where it is used to assign symmetries to molecules. The assigned point groups can then be used to determine physical properties (such as polarity and chirality), spectroscopic properties (particularly useful for Raman spectroscopy and Infrared spectroscopy), and to construct molecular orbitals. In Philosophy, Ernst Cassirer related the theory of group to the theory of perception as described by Gestalt Psychology; Perceptual Constancy is taken to be analogous to the invariants of group theory. ## लिधंसा 1. ↑ Smith, D. E., History of Modern Mathematics, Project Gutenberg, 1906. • Rotman, Joseph (1994). An introduction to the theory of groups. New York: Springer-Verlag. ISBN 0-387-94285-8. A standard modern reference. • Scott, W. R. [1964] (1987). Group Theory. New York: Dover. ISBN 0-486-65377-3. An inexpensive and fairly readable textbook (somewhat outdated in emphasis, style, and notation). • Livio, M. (2005). The Equation That Couldn't Be Solved: How Mathematical Genius Discovered the Language of Symmetry. Simon & Schuster. ISBN 0-7432-5820-7. Pleasant to read book that explains the importance of group theory and how its symmetries lead to parallel symmetries in the world of physics and other sciences. Really helps congeal the importance of group theory as a practical tool). पुच:	{"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": 6, "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": 0.905517041683197, "perplexity_flag": "head"}
http://mathematica.stackexchange.com/questions/19699/badly-conditioned-matrix-generalluc	# badly conditioned matrix (General::luc) With some matrices I am receiving the following message Inverse::luc Result for Inverse of badly conditioned matrix (M) may contain significant numerical errors. How can I tell to Mathematica to isolate(or detect) that result (or badly conditioned matrix)? For instance, using the "If" statement, which condition should I put into the argument of If statement when my matrix is badly conditioned? p.s. Some advice to tackle with these kind of matrices would be good. - 3 You can use `LinearAlgebra`MatrixConditionNumber` to find the condition number of your matrix. What value you choose as the threshold depends on your particular application... In general, a condition number that is $\mathcal{O}(10^n)$ can make you lose up to $n$ digits of accuracy (in addition to FP errors). – rm -rf♦ Feb 16 at 19:28 3 Numerical inversion of a matrix is both slow and poorly conditioned, a fact mentioned in most books on numerical analysis. Typically, a solution to an applied problem can be formulated in such a way that matrix inversion is expressed in terms of the solution of a system of equations. Naturally, it would be much easier to provide details if you provided more details of your problem. – Mark McClure Feb 16 at 19:36 1 – whuber Feb 16 at 19:44 ## 2 Answers This might be as good a time as any to distill the collective wisdom of Messrs. Huber, McClure, and Toad R. M. As already mentioned, there is this quantity of great interest to people in the business of solving simultaneous linear equations, called the condition number, and conventionally denoted by the symbol $\kappa$. This is usually associated with a matrix $\mathbf A$ that figures as the matrix of coefficients in the linear system, and thus, we have the symbol $\kappa(\mathbf A)$. Further, there is not just one condition number, but a number of them, all dependent on the underlying matrix norm used in their definition. One then speaks of the "2-norm condition number", $\kappa_2(\mathbf A)$, the "$\infty$-norm condition number", $\kappa_\infty(\mathbf A)$... and so on. Now, why should we be interested in this condition number? One for instance cannot depend on the determinant, as it is not a reliable measure of how badly a coefficient matrix will behave in a linear system (see also this answer). In inexact arithmetic, the condition number is pretty much the only nice diagnostic you have. For applications, usually the choice of condition number does not matter much, since if some matrix $\mathbf A$ has a large/small $p$-norm condition number, the $q$-norm condition number of the same matrix will be of comparable magnitude. In any event, one usually wants matrices whose condition numbers are as near to unity as can be, as this is the case of well-conditioning. Conversely, if your matrix's condition number is "huge" (for some application-dependent definition of "huge"), then your matrix might as well be singular. (A matrix that is truly singular has a condition number of $\infty$.) The two-norm condition number is easily computed in Mathematica: ````cond2[mat_?MatrixQ] := Divide @@ Through[{Max, Min} SingularValueList[mat, Tolerance -> 0]] ```` As you might ascertain from this routine, this condition number requires the computation of the singular value decomposition (SVD). This is quite expensive, and sometimes one instead uses condition number estimators, which require much less computational effort. (One thing it can do that the next method I am about to describe can't is that it can be applied to non-square matrices as well.) Mathematica has a condition number estimator built-in, in the form of the function `LinearAlgebra`MatrixConditionNumber[]`. This function can be set to estimate either the $1$-norm or $\infty$-norm condition number, depending on the setting of its `Norm` option. The Hager-Higham condition estimator, which is the underlying algorithm, almost always gives a result that is equal to or near the exact condition number, though there are matrices that can defeat it. (Luckily, these counterexamples are rather contrived and do not seem to crop up in practice.) So, what to do in Mathematica? As R. M. notes, this is application dependent, but you will want to use the heuristic that when solving a linear system with coefficient matrix $\mathbf A$, you stand to lose $\approx\log_b(\kappa(\mathbf A))$ base-$b$ digits in the solution of your linear system. You will then want to do something like `If[LinearAlgebra`MatrixConditionNumber[A] < 1/$MachineEpsilon, (* code for well-conditioned case ), ( code for ill-conditioned case *)]`, to use a typical example. Mark's advice is only slightly related to the matter at hand, but he is of course right: one often does not need to compute inverses, and what an inverse can do, an appropriate matrix decomposition can do as well or better. (There are instances where inverses are genuinely needed, like variance-covariance matrices, but they are few and far between.) For instance, if you are doing something like ````x = Inverse[A].b ```` this is better done using the functionality of `LinearSolve[]`, which internally stores a matrix decomposition: ````lf = LinearSolve[A]; x = lf[b] ```` There are still a number of details to say (or I have forgotten), but this post is getting too long already, and I think it best to stop here. - ## Did you find this question interesting? Try our newsletter Use `Check` with 3 arguments, possibly in conjunction with `Quiet`. For example: ````Quiet[Check[Inverse[N[HilbertMatrix[20]]], "Badly conditioned", Inverse::luc], Inverse::luc] ````	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 17, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 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": 0.8958756923675537, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/99549/how-does-my-radio-work/100249	## How Does My Radio Work? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Bear with me for a moment while I invoke the real world; the main question at the end is purely mathematical. I live in an area with $n$ AM radio stations and $m$ FM radio stations. AM station number $j$ wants to send me the signal $\phi_j(t)$. FM station number $k$ wants to send me the signal $\psi_k(t)$. Of course if they just sent those signals, my radio would recieve their sum and have no idea how to disentangle them. Therefore, the signals are first encoded. My (possibly ill-informed) understanding is that (modulo a gazillion bells and whistles), AM station $j$ sends the signal $\phi_j(t)\sin(\omega_j t)$. where $\omega_j$ is some constant, and FM station $k$ sends the signal $A_k \sin(\psi_k(t))$ where $A_k$ is some constant. My radio then receives the signal $$\sum_{j=1}^n\phi_j(t)\sin(\omega_j t)+\sum_{k=1}^mA_k\sin(\psi_k(t))\quad\quad(1)$$ Having received this signal, and knowing the values of the $\omega_j$ and the $A_k$, my radio is then somehow able to compute any one of the signals $\phi_j(t)$ or $\psi_k(t)$ and play it for me on request. (In fact, I'm pretty sure it can recover $\phi_j$ on the basis of $\omega_j$ alone, without knowing the values of the other $\omega$'s.) It's not obvious to me that this is mathematically possible, though my radio seems to have no problem doing it. Question 1 (Pure Mathematics). For what values of $\omega_1,\ldots,\omega_n,A_1,\ldots,A_m$ is it possible to recover the functions $\phi_1,\ldots,\phi_n,\psi_1,\ldots\psi_m$ from expression (1) alone? And what assumptions are being made on the class of allowable functions from which the $\phi_j$ and $\psi_k$ are drawn? Question 2 (Part Engineering, part Pure Mathematics). If (as is not impossible), AM and/or FM works entirely differently than I think it does, thus rendering Question 1 entirely unmotivated, then how do AM and FM work, what is the correct analogue of expression (1), and what is the right answer to the corresponding new version of Question 1? Edited to add: I'm aware that there are all sorts of issues with distorted transmissions, error-correcting, etc. I want to abstract away from all of these and understand the basics. - 9 Follow up question: "... and can you guys fix it?" – Vidit Nanda Jun 14 at 4:42 23 Have you tried turning it off and on again? – François G. Dorais♦ Jun 14 at 4:58 5 ...or just take a very simple circuit diagram for a crystal diode radio and write down the ODE for the response in the earphones from the earphones with the given signal. This should extract out frequencies close to the one to which you are tuned and perform the (amplitude) demodulation. – George Lowther Jun 14 at 8:57 6 @David Roberts' points out that you did not quite express FM correctly – you did not incorporate the base frequency, whereas I care whether I'm listening to 88.5 or 89.1. According to Wikipedia, FM radio transmits the data $\psi_k(t)$ via an encoding of the form $A_k\sin(\lambda_k t + \Psi_k(t))$, where $\lambda_k$ is a frequency much higher (three orders of magnitude, or so) than any of the Fourier modes in $\psi_k(t)$, and $\Psi_k(t)$ is (a multiple of – the units would be off otherwise) an antiderivative of $\psi_k(t)$. – Theo Johnson-Freyd Jun 15 at 2:35 4 Also, again according to Wikipedia, AM is more accurately described via a signal of the form $\Phi_j(t) \sin(\omega_j t)$, where $\Phi_j(t) = A_j(1 + M_j\phi_j(t))$. Here $A$ and $M$ are unit-full quantities, usually set so that $1 + M\phi_j(t)$ is always positive (otherwise the signal is distorted), but has minimum value $1/c$ for some single-digit number $c$. – Theo Johnson-Freyd Jun 15 at 2:49 show 10 more comments ## 4 Answers Modulation is used to reduce antenna height, noise distortion, to avoid interference... The low frequency signal (e.g., human voice) is superimposed to a high frequency signal (the carrier) and transmited. Every radio station have its own high frequency (carrier frequency) of transmission. When you tune your radio (choosing the carrier frequency of the radio station), you're also indicating the electronic circuit to be used to demodulate (extract information from the carrier) AM or FM signals to recover the (low frequency) human audible signal. See any book from B.P. Lathi (e.g. Modern Digital and Analog Communication System) or from Oppenheim (e.g Signals and Systems). These are classical books used in Electric Enginnering courses. According to Lathi, the AM signal can be demodulated coherently (demodulation synchronous) or noncoherently (demodulation assynchronous). In practice, two demodulation noncoherent methods are used: (1) rectifier detection and (2) envelope detection. (1) Rectifier Detector: If an AM signal is applied to a diode and resistor circuit, the negative part of the AM wave will be supressed. The rectified output, $v_r(t)$, is: $$v_r(t) = [A+m(t)]\cos \omega_ct [1/2 + 2/\pi(\cos\omega_ct- 1/3 \cos3\omega_ct + \ldots)] = 1/\pi[A + m(t)] + hft$$ where $hft$ are high frequency terms, $m(t) = B \cos\omega_mt$ is the information (low frequency signal), $\omega_m$, information frequency, $\omega_c$, carrier frequency, $A$ and $B$, amplitude of carrier and low frequency signals, respectively. When $v_r(t)$ is applied to a low-pass filter of cutoff $B$ Hz, the output is $[A +m(t)]/\pi$, and all the other terms in $v_r$ of frequencies higher than $B$ Hz are supressed. The dc term $a/\pi$ may be blocked by a capacitor to give the desired output $m(t)/\pi$. (2) Envelope Detector: the output of the detector follows the envelop of the modulated signal. The circuit is a diode followed by a RC-filter. Mathematical details in Lathi. The information (low frequency signal) in FM resides in the instantaneous frequency $\omega_i = \omega_c + k_f m(t)$, $k_f$ is a modulation index. A frequency-selective network with a transfer function $$\|H(\omega)\| = a\omega + b$$ over the FM band would yield an output proportional to the instantaneous frequency, $\omega_i$. There are several possible networks with such characteristics, the simplest is an ideal differentiator with transfer function $j\omega$. The mathematical details can be seen in Lathi. There is a tutorial here or here. BTW, this is a very beautiful (real) application of Fourier Theory. ADDED: In relation to your question, if we have $$\phi_{AM}(t) = A \cos\omega_ct + f(t)\cos\omega_ct$$ representing the AM signal and $$\phi_{FM}(t) = A \cos\omega_ct - A k_f g(t)\sin\omega_ct$$ representing the FM signal, then $$\phi_{AM}(t) \leftrightarrow \frac{1}{2}[F(\omega + \omega_c) + F(\omega - \omega_c)] + \pi A [\delta(\omega + \omega_c) + \delta(\omega - \omega_c)]$$ and $$\phi_{FM}(t) \leftrightarrow \pi A [\delta(\omega + \omega_c) + \delta(\omega - \omega_c)] + j\frac{Ak_f}{2}[G(\omega - \omega_c) - G(\omega + \omega_c)]$$ Considering $$v(t) = \phi_{FM}(t) + \phi_{AM}(t)$$ we have the Fourier transform: $$\frac{1}{2}[F(\omega + \omega_c) + F(\omega - \omega_c)] + 2 \pi A [\delta(\omega + \omega_c) + \delta(\omega - \omega_c)] + j\frac{Ak_f}{2}[G(\omega - \omega_c) - G(\omega + \omega_c)]$$ The next step is to adapt your equation to these equations. Then, procedures (1) or (2) should be used to recover the low frequency signal in the receiver side. - Contrary to your second sentence, the low-frequency signal is not superimposed on the high-frequency signal. Superimposing means addition. Modulation maintains the signal within a small neighborhood of the carrier in frequency space. – S. Carnahan♦ Jun 20 at 8:47 Sorry I disagree with you. The "superimposing" word was used in the sentence with the meaning of this definition: Modulation is the addition of information (or the signal) to an electronic or optical signal carrier as you can see at searchnetworking.techtarget.com/definition/…. Furthermore, in any part of my answer can be found the word "modulation" meaning "addition". This also applies to the equations I wrote. Maybe I used the wrong word. My apologies, English is not my first language. Thank you for your comments. – PaPiro Jun 20 at 14:38 ### 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. Entirely engineering: I believe "different frequency bands" means that the Fourier transform of $\sin (\phi_k(t))$ is small in some interval containing the $\omega_j$. Note that I am completely unqualified to speak about the engineering here. Both engineering and mathematics: I believe that the radio recovers, approximately, the functions $\phi_j$ by convolving the signal with $e^{(-\beta+\omega_j i)t}$. Mathematics: One cannot recover the signal exactly without strong assumptions on the $\phi_j$. Take the signal $\sin(\omega_1 t )\sin(\omega_2 t)$. This could represent $\phi_1=\sin(\omega_2 t)$, $\phi_2=0$, or $\phi_1=0$, $\phi_2=\sin(\omega_1 t)$ or any linear combination. Assume that the $\phi_i$ have Fourier transforms supported on the interval $[-\delta,\delta]$ and the $\omega_i$ are separated by gaps of size at least $2\delta$. Assume no FM stations. Then we can recover each AM station by taking the Fourier transform and setting everything outside $[\omega_i -\delta,\omega_i+\delta]$ to $0$. I don't know the right solution for FM. - I've been trained as an engineer, and I can tell you that engineers have a somewhat simplified view of the matter. (But, not only a simplified view, of course.) The other answers fill in some detail, but I think a higher-level view is useful. There is no such thing as perfect recovery of the transmitted signal. The best you can hope is to bound the error. For most modulation techniques the basic idea is that the spectrum $X$ of a signal $x$ is nearly 0 outside a narrow band: $X(f)\approx0$ when $\|f \pm f_0\| < B$. Both AM and FM are essentially means of transforming a spectrum centered around $0$ into one centered around $f_0$. So, in order to recover a signal, the main concern is to make sure that the spectrums $X_1$, $X_2$, …, $X_n$ do not overlap. This is achieved in a rather uninteresting way: regulation. Then you can extract one signal by shifting $f_0$ to $0$ (convolution with a Dirac impulse in frequency domain, meaning multiplication with a harmonic signal in the time domain), and then applying a low pass filter (multiplication with a rectangular function in the frequency domain, meaning convolution with a sinc in the time domain). See also this related question. There are broad-spectrum modulation techniques, which are used for example in fourth generation mobile-phone networks, that do not rely on the assumption that the signal covers a narrow band. The two main ones are frequency hoping (use some narrow-band modulation technique but change $f_0$ often in some pseudorandom sequence) and spread spectrum (multiply the signal with a pseudorandom sequence before using a narrow-band modulation technique). The signals obtained thru such methods have a wide band, but are bounded $\|X(f)\| < c$ for some $c$ for all $f$. This way they behave as background noise as far as demodulating any narrow-band signal is concerned. - Your expression for FM transmissions is not quite right - it's missing the radio frequency! The simple model that captures the essentials of what FM station $k$ is sending you is the function $$B_k\sin\left((\omega_k+\gamma_k\psi_k(t))t\right),$$ where $\gamma_k\psi_k$ never gets close to $\omega_k$ (so you're only modulating the frequency and not completely disrupting it). If the interesting signal $\psi_k$ is a pure note at frequency $\omega$, then the spectrum of the actual radio signal can be found in terms of Bessel functions and consists of sidebands separated from the carrier by spacing $\omega$. (The number of sidebands is controlled by how large $\gamma_k$ is.) The real radio signal your device is getting, then, is $$F(t)=\sum_{j=1}^n\phi_j(t)\sin(\omega_j t)+\sum_{k=1}^m B_k\sin\left((\omega_k+\gamma_k\psi_k(t))t\right).$$ Because of the modulation, none of the stations' radio signals are single peaks; instead they are spread over a bandwidth roughly given by the frequency content of the audio signals they encode. (For comparison, human hearing can detect 16 Hz to roughly 20,000 Hz, AM frequencies are medium-wave radio at 520 kHz to 1,610 kHz, and FM stations run at 87.5 to 108 MHz. Thus in reality the peaks are quite narrow!) To detect a signal, your device uses a combination of antennas, loops of wire, parallel plates, and the like, which contrive to give to the decoding device (the one that takes a radio signal and gives you an audio one) a voltage $f$ that's controlled by a damped harmonic oscillator equation of the form $$\frac{d^2}{dt^2}f-2\gamma\frac{d}{dt}f+\omega_0^2f=F,$$ where the resonance frequency $\omega_0$ is controlled by a knob on the device. The spectral response of this dynamical system is routinely evaluated in college ODE courses, and comes out as a Lorentzian bell-shaped curve centred at $\omega_0$ and of width $\gamma$. Choose $\gamma$ to match the spectral width of the typical radio station, and you've got a fantastic filter! EDIT: After doing some looking up, I find that the $\psi_k$ here is not exactly the audio signal the station is trying to encode, but rather something like its average over the interval $[0,t]$, so it is equivalent to it up to simple mathematical operations performed at the decoder. -	{"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": 99, "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": 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": 0.9340865612030029, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/44983/how-do-i-combine-multiple-sets-of-ratios-in-order-to-meet-an-overall-target-ra?answertab=votes	# How do I combine multiple sets of ratios in order to meet an overall “target” ratio? (word problem supplied to illustrate) (I had no idea how to tag this, any help would be great) :) I am a programmer with little math experience beyond high school - this will be demonstrated by my explanation. I am developing an application that needs a formula. The formula involves calculating what amounts of food to eat based on factors like calories and concentrations of macronutrients (carbs, protein, fats). Since there are other calculations need that aren't relevant to main problem, I have abstracted it into the following word problem. Here it is: Your job is to keep your store, Colorful Widgets, Inc., supplied with widgets. There are four suppliers that you can order from every month. There is a set ratio of colors that must be met as closely as possible to match sales expectations. The ratio is: Target Color Ratios: 40% Red widgets 40% Blue widgets 20% Yellow widgets The catch is that each supplier offers shipments of mixed colors in different ratios, with each company specializing in one color. Here are the ratios available from the suppliers: Mostly Red Widgets, Inc. 86% Red 11% Blue 03% Yellow Acme Widgets, Inc (Limit 3 boxes) 79% Red 09% Blue 20% Yellow Mostly Blue Widgets, Inc. 15% Red 77% Blue 08% Yellow Mostly Yellow Widgets, Inc. 10% Red 00% Blue 90% Yellow All four companies sell their widgets in boxes containing 100 widgets each. Boxes cannot be split into smaller amounts. You need to order 24,000 widgets, with the total ratio of colors matching the targets laid out above. Restrictions: You have a contract with the first three companies - they must be ordered from every month, no exceptions. The fourth (Mostly Yellow Widgets, Inc.) is a stop gap - if your total ratio is low on yellow widgets, you can order from them. Additionally, Acme Widget, Inc. makes the best widgets, but only in limited quantity. You must order 3 boxes from them each ordering cycle. I understand that this problem probably will not have a neat and tidy catch-all formula. Even if there is no formula for this problem with it's restrictions, there should be a formula or something that will help with the problem of combining three different ratios in such a way to come as close as possible to the target ratio. I also understand that if the target ratios or ratios of suppliers change too much, a solution may not be possible. - I don't really understand the restrictions. Are you saying that you must order at least one box from each of the first three companies? What is meant by "best"? – Qiaochu Yuan Jun 12 '11 at 18:09 The numbers for Acme don't add up to 100%. – Robert Israel Jun 12 '11 at 18:53 @Qiaochu: This word problem was my poor attempt to abstract what is actually a food calculation. The "Acme Widget" actually represents green veggies in my real problem - something that cannot be skipped. :) – Matt Miller Jun 13 '11 at 18:00 @Robert: That's what I get for making a last minute change to my question. – Matt Miller Jun 13 '11 at 18:00 ## 2 Answers For Acme widgets, you may order either $0$ or $1$ or $2$ or $3$ boxes; this integer will be called $M$. It's just four possibilities and all of them are small numbers of boxes and it is easy to manually check which of these four values will you the best results. Most of the adjustment is done by and most of the boxes will come from the mostly red, mostly blue, and mostly yellow companies. You must order $R, B, Y$ boxes from each, respectively. The conditions, neglecting $M$ for a while, are $$R + B + Y = 240,$$ $$0.86 R + 0.15 B + 0.10 Y = 96,$$ $$0.11 R + 0.77 B + 0 Y = 96,$$ $$0.03 R + 0.08 B + 0.90 = 48.$$ The four equations are not independent: the sum of the last three gives the first. So eliminate the first. We have a set of three linear equations for three unknowns $R,B,Y$. Define matrix $A$ as the $3\times 3$ matrix of the entries on the left hand side of the three equations. Then the solution is $$(R,B,Y) = A^{-1} (96,96,48)^T \approx (87.36,112.20, 40.45)$$ That's the number of boxes you would ideally want to order if $M=0$: you would have to round it to $(87,112,41)$. Now, solve the same equations with the extra terms for $M=1,2,3$ and decide which of them is the best according to your best criteria (which, I guess, is either the sum of the deviation of the three percentages from $.4,.4,.2$ or the sum of the squares of these numbers). At any rate, the number of boxes $R,B,Y$ won't deviate by more than 3 from the numbers written above. - – Matt Miller Jun 13 '11 at 18:05 It's basically a linear programming problem: if $x_1$, $x_2$, $x_3$, $x_4$ are the numbers of boxes ordered from the four companies, you would like $$\eqalign{ x_1 + x_2 + x_3 + x4 &= 240 \cr 86 x_1 + 79 x_2 + 15 x_3 + 10 x_4 &= 40 \times 240\cr 11 x_1 + 9 x_2 + 77 x_3 &= 40 \times 240\cr 3 x_1 + 20 x_2 + 8 x_3 + 90 x_4 &= 20 \times 240\cr x_2 \le 3,\ {\rm all}\ x_i \ge 0}$$ However, this may not be possible to attain precisely, so you allow some slack. One way is to minimize $r$ subject to $$\eqalign{ x_1 + x_2 + x_3 + x_4 &= 240 \cr 86 x_1 + 79 x_2 + 15 x_3 + 10 x_4 &\le 40 \times 240 + r \cr 86 x_1 + 79 x_2 + 15 x_3 + 10 x_4 &\ge 40 \times 240 - r \cr 11 x_1 + 9 x_2 + 77 x_3 &\le 40 \times 240 + r\cr 11 x_1 + 9 x_2 + 77 x_3 &\ge 40 \times 240 - r\cr 3 x_1 + 20 x_2 + 8 x_3 + 90 x_4 &\le 20 \times 240 + r\cr 3 x_1 + 20 x_2 + 8 x_3 + 90 x_4 &\ge 20 \times 240 - r\cr x_2 \le 3,\ {\rm all}\ x_i \ge 0}$$ Now usually the solution values will not be integers. If you want integer values, this becomes an integer linear programming problem. Many software packages will solve these. In this case the optimal solution, as given by Maple's Optimization package, is $x_1 = 85$, $x_2 = 3$, $x_3 = 112$, $x_4 = 40$, $r = 27$. -	{"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": 25, "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": 0.9420636296272278, "perplexity_flag": "middle"}
http://mathhelpforum.com/advanced-math-topics/28696-solved-can-anyone-help.html	# Thread: 1. ## [SOLVED] Can anyone help? I was wondering if there is any kind of formula that will allow me to do this: I have about 20 numbers. I want a combination of some of the numbers to add up to be (let's say) \$25,768.45. I don't have to use all the numbers to make them add up to that number. I can use 2 or I can use all 20. Is there any kind of formula I can use to help me do this? The numbers are all big and the final number I need them to add up to is big too so I've just been plugging in numbers and guessing but it is taking forever AND I cannot make it work. I realize it may not work, but I thought maybe there is a trick or a shortcut to this. Thanks. Sorry if this makes no sense! 2. Given a set of twenty numbers, there are a maximum of $2^{20} - 1 = 1048575$ possible sums. (There may be several equal sums, but that is the number of different sums.) Given the power of today’s computers, it should be possible to actually check out each possible sum. But there is no mathematical formula to do this for an arbitrary set.	{"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": 0.9462314248085022, "perplexity_flag": "head"}
http://mathoverflow.net/questions/17295/results-about-the-order-of-a-group-forcing-a-particular-property/17304	## Results about the order of a group forcing a particular property. ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Given a group of order $n$ where $n$ is either a specific number, or a number of a particular form, e.g. square-free, when does $n$ completely determine a particular group property among all groups of that order? Vipul's group theory wiki has several stubs on this topic, and in the language of his wiki, I will call this a $P$-forcing number, where $P$ is a particular group theoretic property. We already have quite a few easy examples, for instance, orders $pq$, $pqr$, and $p^2q$ force solvability, and $p^2$ forces abelian. Then there are more specific results like 99 is an abelian-forcing number. I am interested in general, in any results of this flavor beyond what would be considered a common result in a standard graduate-level group theory book. - 1 See Pete Clark's answer at: mathoverflow.net/questions/11001/… – S. Carnahan♦ Mar 8 2010 at 5:20 ## 4 Answers The numbers n such that every group of order n is cyclic, abelian, nilpotent, supersolvable, or solvable are known. Most are described in an easy to read survey: Pakianathan, Jonathan; Shankar, Krishnan. "Nilpotent numbers." Amer. Math. Monthly 107 (2000), no. 7, 631-634. MR 1786236 DOI: 10.2307/2589118 If you want to go beyond results like this, you may have better luck looking at a slightly more refined version of the order: the isomorphism type of the Sylow subgroups. Sometimes a p-group P has the property that every group G containing it as a Sylow p-subgroup has a normal subgroup Q of order coprime to p such that G is the semi-direct product of P and Q. An easy version of this that does appear in many group theory texts is that if n=4k+2, then in each group G of order n there is a normal subgroup Q of order 2k+1 so that G is the semi-direct product of any of its Sylow 2-subgroups and Q. Groups all of whose Sylow p-subgroups are cyclic have very nice properties, subsuming those of groups of square-free order. Groups all of whose Sylows are abelian have more flexibility, but are still basically under control. - ### 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 haven't had time to read it yet, but The influence of conjugacy-class sizes on the structure of finite groups seems tailor made to answer these questions. It's a survey article, too. - Very famous one is the Feit-Thompson theorem: if n is odd, then G is solvable. Though I suppose this is stated (but not proved) in most modern algebra texts. - $(n,\phi(n)) = 1$ forces $G$ cyclic. $p^nq^m$ forces $G$ solvable. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 13, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 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": 0.9367812275886536, "perplexity_flag": "middle"}

http://physics.stackexchange.com/questions/49807/what-information-is-stored-on-gramaphones-tape-recorders-cds-dvds/49810

# What information is stored on gramaphones/tape recorders/CDs/DVDs I'm a Software Developer by profession and my physics knowledge is limited what I had learned at high school level. Please excuse me if the question is trivial. Question: From what I know, a sound wave is set of different amplitudes spread across a time line. The amplitudes vary greatly. Hence, sound wave is largely aperiodic (amplitudes don't repeat often). Now, where does frequency come into picture here? If the wave is periodic like a sine wave or a deterministic mathematical function of time, then frequency can be measured as the number of cycles (wave reaching same amplitude) in one second. How can we define frequency of highly aperiodic sounds like human speech. If all that is recorded on a gramophone disc is varying amplitudes across timeline, where is frequency accounted for? Does the frequency remain constant in a typical human speech? - ## 3 Answers The sound that reaches your ear is just air pressure fluctuating over time. You can use a transducer of some sort to convert the value of air pressure to some other form - for example: • to the depth of a groove being cut into a helical track on a layer of wax on a rotating drum • to the depth of a groove being cut into a spiral track on a circular disc of metal from which other plastic disks are pressed. • to a strength of magnetisation of a magnetic layer on a plastic tape being wound onto a spool • to a series of numbers representing the pressure at regular tiny intervals of time. The idea that the variations in pressure over time are due to, or consist of, a collection of frequencies is just a mathematically equivalent description but it does not represent extra information, it's just a different way of describing the same information. Here's some diagrams from a synthesizer manual Above are three very different sounds with apparently the same frequency (say 440 Hz) Above is shown how you can add sine waves of two frequencies to produce a more complex waveform Above is shown how you can continue adding sine waves of differing frequencies to construct an arbitrary waveform (a sawtooth). The sawtooth waverform can be recorded directly as depths in a groove on a record. But you could "record" the same thing as a set of numbers that represent the frequencies of a dozen sine waves you could add together to produce a single pressure wave that varies over time in the same way. See Fast Fourier Transform - @RedGrittyBrick..thanks for the answer! The first part of the answer pretty much explains what I want. Is it like at every point in time (say at every ms) we record all the harmonics that contribute to the sound wave at that instant of time? Also, I can understand that recording this information in a digital form is easy. How is stored on mechanical device like Gramophone? Do we just record grooves with varying depths over time? – Gopal Jan 22 at 10:21 @Gopal: at every interval in time we record the single value of air pressure that results from the sum of all the harmonics (+ other sounds/noise). Yes, this information can be stored as the depth of a groove cut in some surface. The depth varies over distance along the groove, a needle moving along the groove experiences changes in depth over time. – RedGrittyBrick Jan 22 at 10:36 @RedGrittyBrick..as I accept the answer I would like to ask you a closing question, Essentially a gramophone doesn't actually do any fourier transforms. All it records is the displacement of microphone at every instant of time and reproduces same using a diaphragm. – Gopal Jan 22 at 14:33 @Gopal: That is true. – RedGrittyBrick Jan 22 at 15:10 You can make an arbitrary sound (or any waveform) by adding together a bunch of pure tones at different frequencies. So a sound, unless it happens to be a pure tone, does not contain a single frequency component, rather a range of frequencies. The mathematics behind this is called Fourier analysis and you can see many examples on Wikipedia or by searching the web. - ..>>You can make an arbitrary sound (or any waveform) by adding together a bunch of pure tones at different frequencies << bang on! Liked this lucid explanation. – Gopal Jan 23 at 7:18 Your question is specifically about how the concept of frequency can be applied to aperiodic signals. Simplest example is a finite width rectangular pulse - the signal is zero outside the pulse. This is certainly aperiodic. Now for periodic signals f(t), you can, for each frequency which is a multiple of the period, compute the n'th Fourier coefficient as $$c_n = \int^{\pi}_{-\pi}f(t)e^{-int}dt$$ This represents "how much" of frquency n is present in the periodic signal, as explained in the other answers. Correspondingly, for our aperiodic isolated rectangular pulse f(t), we can choose an interval of length T which is bigger than the pulse width, and again calculate the Fourier coefficients over this interval$$c_n=\int^{\frac{T}{2}}_{-\frac{T}{2}}f(t)e^{-2\pi i (\frac{n}{T})t}dt$$ Now if we let $T\rightarrow \infty$, the discrete values $\frac{n}{T}$ can be replaced by a continuous variable $\xi$, and the set of Fourier coefficients is replaced by a function $\hat{f}(\xi)$: $$\hat{f}(\xi) = \int^{\infty}_{-\infty}f(t)e^{2\pi i \xi t}dt$$ This is the Fourier transform of f. So for aperiodic signals (or truncated periodic signals), this is what you want. You can calculate the Fourier transform of anything from a rectangular pulse to the latest David Bowie track. The "frequencies" that are present in the signal are just the Fourier transformed variables $\xi$. - @twistor..thanks for detailed explanation. I'll try to understand fourier transform (had a bit of it in my Engineering course) – Gopal Jan 23 at 7:17

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http://mathoverflow.net/questions/67366?sort=votes

## Non-measurable sets and Determinacy… ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Assume AC. Suppose $X$ is a subset of the irrationals (Baire Space) for which neither player has a winning strategy (i.e. the game $G(\omega, X)$ is not determined). Is $X$ non-measurable in the Lebesgue sense as a subset of $\mathbb{R}$? - ## 2 Answers (My argument is somewhat easier if you consider games where the players play $0$s and $1$s, so that the payoff set is in Cantor space $2^\omega$, and we use the usual coin-flipping probability measure; but an essentially similar idea works in Baire space.) For any game with payoff set $A$, where player I wins if the play is in $A$, consider the following slightly modified game $A^\ast$, which is just like $A$, except we insert a pair of dummy moves between each pair of actual moves, and insist that player I play a $0$ in this dummy round, while player II can play anything. Thus, a sequence or play is in the payoff set $A^\ast$ if indeed that sequence shows that player I did play a $0$ in all the dummy rounds (so every fourth digit is $0$), and furthermore, if we omit the dummy rounds entirely from the sequence, we get a sequence in $A$. Thus, playing the game $A^*$ is just like playing $A$, except that the play is interrupted for these silly dummy rounds. Note that player I has no incentive not to play a $0$ on those rounds, and player II's plays in the dummy rounds are ignored entirely. Thus, it is clear that a player has a winning strategy for $A$ if and only if he or she has a winning strategy for $A^\ast$, since we can translate the strategies from $A$ to $A^\ast$ and back again. The dummy rounds really don't change the difficulty of winning the game. But the point now is that because every fourth digit of $A^\ast$ is $0$, it follows that $A^\ast$ has measure $0$. (Every time you insist that a particular digit is $0$, it cuts the measure in half again.) The conclusion, therefore, which does not use the axiom of choice, is that if there is a non-determined set, then there is a non-determined set with measure $0$. In particular, there is a non-determined set that is measurable. - I'm not convinced... Cantor Space is homeomorphic to the Cantor Set which has measure zero... Thus any payoff set in it will have measure zero when considered as a subset of $\mathbb{R}$ by virtue of being a subset of the Cantor Set... I'm not convinced that your argument about halving the measure carries through to non-measurable sets... further my guess (I should probably ask this as a question) is that the existence of a non-determined game in Cantor Space is stronger than AC... – George Lazou Jun 9 2011 at 23:51 I am not using the Cantor set as a subset of $\mathbb{R}$ and the Lebesgue measure on that, but rather, using the natural probability measure on $2^\omega$, for which $2^\omega$ is the whole space and has measure $1$; the measure of the basic open set determined by a finite binary sequence of length $n$ has measure $\frac{1}{2^n}$, like flipping a coin. A subset of $2^\omega$ which is $0$ in every fourth digit has measure $0$ (an easy calculation). The same idea works in Baire space, but you seem to be mapping spaces by homeomorphisms that may not be measure-preserving... – Joel David Hamkins Jun 10 2011 at 0:24 1 As for your final question, I think you mean to ask whether the existence of a non-determined game is weaker than AC, rather than stronger, since ZFC proves the existence of such non-determined games for Cantor space in exactly the same way that it does for Baire space. I don't expect the existence of such a non-determined set to imply full AC, but I'd have to think a bit more to give a precise model showing this. If this is right, then the existence of non-determined sets would be a weak choice principle. – Joel David Hamkins Jun 10 2011 at 0:29 1 But I would encourage you to ask that as a question, since perhaps someone knows a good model, and I would be interested to see it. – Joel David Hamkins Jun 10 2011 at 2:16 1 There is an old beautiful argument of Sierpinski that shows that the axiom of choice for PAIRS $AC_2$ implies the existence of a nonmeasurable set; which in turn shows that $AD$ fails as soon as $AC_2$ is true. So a model in which $AC$ fails but $AC_2$ holds is yet another way of seeing that the negation of $AD$ does not imply $AC$. Cohen's so-called "first (symmetric) model" in which $AC$ fails is such a model (since every set has can be linearly ordered in that model, but the reals cannot be well-ordered there). – Ali Enayat Jun 10 2011 at 19:14 show 3 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. I have two possible answers. The first is short and possibly not the one you want. The second is probably the right one. First: take any (co)analytic subset $X$ of the Baire space $\omega^{\omega}$, which is not Borel. Than it is consistent with ZFC that the game $G(\omega,X)$ is not determined (ZFC just proves the determinacy of Borel games), but $X$ is Lebesgue-measurable (in fact universally measurable). I guess this is a consistent proof of "no", to your question. Second Assume AC. Then there exists a universally-null set (hence Lebesgue null, since the Lebesgue measure is atomless) $X$ of cardinality $\geq\aleph_{1}$. Then one can show that such a set $X$ can not be a Perfect set. Now let us consider the so-called Perfect-set game $PSG(\omega, X)$, which is technically a Gale -Stewart game $G(\omega, Y)$, with $Y$ of about the same complexity of $X$ (in particular $Y$ is universally-null non-perfect set of cardinality $\geq\aleph_{1}$). This game is not determined since $Y$ is not Perfect nor countable by construction, and it is knonw that such a game is determined only if $Y$ is Perfect or countable. Yet $Y$ is universally null, hence Lebesgue measurable. This second example is mentioned in Martin's "Blackwell's determinacy", where it is credited to Greg Hjorth. This is a proof in ZFC of "no", to your quesiton. -

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http://math.stackexchange.com/questions/262284/is-there-a-functional-definition-of-a-limit

# Is there a “functional” definition of a limit? Say we have a convergent sequence $(x_n)$ where $x_n \in E$ for all $n \in \mathbb{N}$ and $E$ is a subset of a metric space $(X,d)$. With this setup, we usually define it's limit as a point $x \in X$ such that for every $\epsilon > 0$, there exists $N \in \mathbb{N}$ such that $n > N$ implies $d(x_n,x) < \epsilon$. In some sense, I think that the above definition of the limit effectively mean that given any "$\epsilon > 0$, there is some kind of $N$ that we can use to get the sequence within $\epsilon$ of the limit $x$. I am wondering whether this can be reformulated as follows: there is a function $f: \mathbb{R} \rightarrow \mathbb{N}$, so that $f(\epsilon) = N$ and $n > N$ implies $d(x_n,x) < \epsilon$. If so, the function $f$ would have some nice properties (it would be onto, and monotonically decreasing in $\epsilon$ for instance). Is there any use to thinking about functions in this way / has it been introduced in this way? - 1 Note that your $f$ need not be onto. – Hagen von Eitzen Dec 19 '12 at 21:14 ## 2 Answers This concept is often known as the modulus of convergence, though for recursion-theoretic purposes it's often given as a function $g(n): \mathbb{N}\to\mathbb{N}$, where (in terms of your definition for $f()$) $g(n)$ would be defined as $g(n) = f(\frac{1}{n})$ or $g(n) = f(2^{-n})$. The concept is important in recursion theory because, for instance, we can compute the digits of the limit recursively in $g$ (i.e., using a Turing machine for $g$). Note that the function $g$ doesn't have to be computable even if the series converges; see, for instance, http://mathoverflow.net/questions/51794/simple-example-of-a-sequence-without-computable-modulus-of-convergence . - To obtain a well-defined function you might define $f(\epsilon)$ to be the smallest $N$ such that $d(x_n,x)<\epsilon$ for all $n>N$. Convergence of the sequence $x_n$ guarantees existence of such $N$'s and one of them has to be the smallest. Conversely, if there exists a function $f$ such that, for every $\epsilon$, $d(x,x_n)<\epsilon$ for all $n>f(\epsilon)$, then $x_n$ is convergent with limit $x$, because you may take $N=f(\epsilon)$. These are nothing but quite tautological reformulations of each other, though. -

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http://math.stackexchange.com/questions/297954/embedding-ordered-number-fields-into-bbb-r

# Embedding ordered number fields into $\Bbb R$ Consider a number field $K$ equipped with a total order $\le$. Is there always a field homomorphism $\phi:K \rightarrow \Bbb R$ which respects the total order, i. e. with $\phi(x)>\phi(y)$ for $x>y$? - If the number field is contained in $\Bbb{R}$, shouldn't such a $\phi$ always exist? – BenjaLim Feb 8 at 11:14 @BenjaLim: The question is, whether the only possible orderings of $K$ are those inherited from an embedding in $\Bbb R$. I see no appareant reason why their couldn't exist other "exotic" orderings of $K$. – Dominik Feb 8 at 11:18 Ok, then use the fact that a field can be totally ordered iff $a_1^2+a_2^2+\dots+a_n^2=0 \implies a_1=a_2=\dots=0$. – Berci Feb 8 at 11:47 1 @Berci: I don't see how this fact relates to the question how to extend the total order of $K$. – Dominik Feb 8 at 13:13 1 – JSchlather Feb 8 at 19:39 ## 2 Answers Here is Theorem 11.6 from N. Jacobson's Basic Algebra II: Let $F$ be an algebraic number field, and let $\mathbb{R}_0$ be the field of real algebraic numbers [i.e., those real numbers which are algebraic over $\mathbb{Q}$]. Then we have a $1-1$ correspondence between the set of orderings of $F$ and the set of monomorphisms of $F$ into $\mathbb{R}_0$. The ordering determined by the monomorphism $\sigma$ is that in which $a > 0$ for $a \in F$ if $\sigma a > 0$ in $\mathbb{R}_0$. It follows from this that the orderings on a number field $F$ are in bijection with the number of roots of any minimal polynomial $P$ for $F$ in $\mathbb{Q}_0$. One knows -- e.g. by model completeness of the theory of real-closed fields -- that this is the same as the number of roots of $P$ in $\mathbb{R}$ and thus the same as the number of embeddings of $F$ into $\mathbb{R}$. Thus the number of (total, compatible with the field structure) orderings of a number field $F$ is equal to the number of embeddings of $F$ into $\mathbb{R}$. Added: If you don't like the model completeness bit, you can replace it with the fact that any homomorphism of real-closed fields must be order-preserving. Indeed, if not then by restriction one would get a different ordering on the smaller real-closed field, but the ordering on a real-closed field is unique: the positive elements are precisely the squares. If you are interested in the proof of this theorem (and don't have ready access to Jacobson's book), please let me know. - Take $L$ the Dedekind completion of $K$. Because the rational nmbrrs are dense in $K$ we have that $L$ must be isomorphic to the reals (as an ordered field). Now restrict the isomorphism to $K$ and we are done. Of course one has to make a severe distinction between orderable and ordered, one only requires that an order exists, and the other specifies such order. Pete L. Clark has answered regarding to how many orders are compatible with a number field. - How do you know the rationals are dense in $K$? (It's true, but I'm wondering whether it's clear.) – Pete L. Clark Feb 10 at 3:18 @Pete Well, number fields are Archemedian. I don't see an immediate argument better than "obviously", but I am already in bed. I'll give it some thought during the first two dreams, and the last one. I hope it turns up nicely in the morning. :-) – Asaf Karagila Feb 10 at 3:22 $@$Asaf: the natural orderings on number fields are Archimedean, which, because the answer to this question is yes, are all the orderings. But if you contemplate an arbitrary ordering of a number field? I feel confident that if $L/K$ is a finite degree field extension and $\sigma$ is an ordering of $L$, then $(L,\sigma)$ is Archimedean iff $(K,\sigma|K)$ is Archimedean, but this too seems to require proof. – Pete L. Clark Feb 10 at 3:25 2 Hmm, I think I see how to prove that: the real closure of an ordered field $K$ is unique up to $K$-isomorphism, so if $L/K$ is a finite degree ordered extension it must be realizable as an ordered subfield of the real closure. Since $\mathbb{R}_0$ is a subfield of the Archimedean ordered field $\mathbb{R}$, it must itself be Archimedean. – Pete L. Clark Feb 10 at 3:30

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http://mathhelpforum.com/algebra/94722-unknown-roots-two-quadratics.html

Thread: 1. Unknown roots of two quadratics Let $(a, c)$ and $(b, d)$ be the roots of the equation $x^2 + ax - b = 0$ and $x^2 + cx + d = 0$ respectively. Find all possible real values for $a, b, c, d$ I'm not sure how to do this question and not sure how to approach it. 2. Hello, In a quadratic equation $x^2+mx+n=0$, m is the sum of the roots, and n is the product of the roots. So from the second equation, we have $bd=d$ (1). And $c=b+d$ (2) And from the first one, we have $ac=-b$ (3) and $a+c=a$ (4) From (4), we get c=0 Substituting in (3), we get b=0 Substituting in (1), we get d=0 Then, a can take any values you want...

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http://math.stackexchange.com/questions/69544/create-polynomial-coefficients-from-its-roots?answertab=votes

# Create polynomial coefficients from its roots Given some roots : $r_1,r_2,\ldots,r_n$, how can we reconstruct polynomial coefficients? I know the Horner scheme and that we can just go backwards receiving those coefficients. But I'm curious if there's some other nice solution to this problem. - 4 I'm not sure I'm answering your question, but to get the coefficients from the roots, just expand $(X-r_1) (X-r_2) \ldots (X-r_n)$. You get $$(X-r_1) (X-r_2) \ldots (X-r_n) = X^n - (r_1 + r_2 + \cdots + r_n) X^{n-1}+ \cdots + (-1)^n r_1 r_2 \ldots r_n X^0$$ The expressions you find are called symmetric polynomials. – Joel Cohen Oct 3 '11 at 15:21 2 – Srivatsan Oct 3 '11 at 15:22 1 It's not clear to me how you can take the Horner scheme "backwards" to get the coefficients. – Thomas Andrews Oct 3 '11 at 15:27 Joel could you elaborate hows this equation goes further? I can't get it – Chris Oct 3 '11 at 15:52 1 @Chris : The $X^{n-2}$ coefficient is $\sum_{1 \le i, j \le n} r_i r_j$ (sum of products of $2$ roots), then the $X^{n-3}$ coefficient $-\sum_{1 \le i, j, k \le n} r_i r_j r_k$ (sum of products of $3$ roots), the $X^{n-4}$ coefficient $\sum_{1 \le i, j, k, l \le n} r_i r_j r_k r_k$ (sum of products of $4$ roots) and so on. The $X^{n-k}$ coefficient is $(-1)^k\sum_{1 \le i_1, i_2, \ldots, i_k \le n} r_{i_1} r_{i_2} \ldots r_{i_k}$ (sum of products of $k$ roots). – Joel Cohen Oct 3 '11 at 16:37 show 2 more comments ## 3 Answers Knowing the location of roots really only pins down a family of polynomials, which vary by multiplicative constants. If you know the roots of $P(x)$ are $r_1, r_2 \cdots r_n$ then you can write the polynomial as $$P(x) = a (x-r_1)(x-r_2) \cdots (x-r_n)$$ where $a$ is some non-zero constant. However, we can not be more precise than that. We can pin down $a$ if we know the value of $P$ are any other point. If we knew $a$, then simply expanding the right hand side would produce the polynomial in a form where you could read off the constants. - As a tiny note: the requirement of an $a$ along with the $n$ roots is very much consistent with the fact that one requires $n+1$ conditions to uniquely determine a polynomial. – J. M. Oct 4 '11 at 0:42 You might want to read this well written article about your problem: http://stochastix.wordpress.com/2008/11/22/building-a-polynomial-from-its-roots/ http://stochastix.wordpress.com/2008/12/10/building-a-polynomial-from-its-roots-ii/ - The polynomial is $P(z) = a_0 (z-r_1)\times (z-r_2) \times \cdots \times (z-r_n)$. Now labeling coefficients as $P(z) = a_0 z^n + a_1 z^{n-1} + \ldots + a_n$ we have $$a_1 = -a_0 ( r_1 + r_2 + \ldots + r_n) \qquad a_n = (-1)^n a_0 r_1 \times r_2 \times \cdots \times r_n$$ In general: $$a_k = \frac{(-1)^k}{k!} a_0 \sum_\pi r_{\pi(1)} \times \cdots \times r_{\pi(k)}$$ where the sum is over permutations $\pi \in \mathfrak{S}_n$. -

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http://math.stackexchange.com/questions/75789/what-is-step-by-step-logic-of-pinv-pseudoinverse/75793

What is step by step logic of pinv (pseudoinverse)? So we have a matrix $A$ size of $M \times N$ with elements $a_{i,j}$. What is a step by step algorithm that returns the Moore-Penrose inverse $A^+$ for a given $A$ (on level of manipulations/operations with $a_{i,j}$ elements, not vectors)? - 3 Answers 1. Perform a singular value decomposition $\mathbf A=\mathbf U\mathbf \Sigma\mathbf V^\top$ 2. Check for "tiny" singular values in $\mathbf \Sigma$. A common criterion is that anything less than $\|\mathbf A\|\epsilon$ is "tiny". Set those singular values to zero. 3. Form $\mathbf \Sigma^+=\mathrm{diag}(1/\sigma_1,1/\sigma_2,\dots, 0)$. That is, reciprocate the nonzero singular values, and leave the zero ones untouched. 4. $\mathbf A^+=\mathbf V\mathbf \Sigma^+\mathbf U^\top$ - Did you find this question interesting? Try our newsletter Sign up for our newsletter and get our top new questions delivered to your inbox (see an example). email address As far as I know and from what I have read, there is no direct formula (or algorithm) for the (left) psuedo-inverse of $A$, other than the "famous formula" $$(A^\top A)^{-1}A^\top$$ which is computationally expensive. I will describe here an algorithm that does indeed calculate it efficiently. As I said, I believe this may be previously un-published, so cite me appropriately, or if you know of a reference please state it. Here is my algorithm. Set $B_0 = A$ and for each iteration step, take a column of $B_i$ and orthogonalize against the columns of $A$. Here is the algorithm ($A$ has $n$ independent columns): 1. Initialize $B_0=A$. 2. For $j \in 1\ldots n$ do \begin{align} \mathbf{t} &= B_i^\top A_{\star j} \\ R\mathbf{t} &= e_j & \text{Find such $R$, an elementary row operation matrix}\\ B_{i+1}^\top &= R B_i^\top \\ \end{align} Notice that each step does one vector/matrix multiplication and one elementary matrix row operation. This will require much less computation than the direct evaluation of $(A^\top A)^{-1}A^\top$. Notice also that there is a nice opportunity here for parallel computation. (One more thing to notice--this may calculate a regular inverse, starting with $B_0=A$, or with $B_0=I$.) The step that calculates $R$ may partially be orthogonal rather than elementary (orthonormal/unitary -thus better in the sense of error propagation) if desired, but should not make use of the previously orthonormalized rows of $B$, since they must remain orthogonalized in the process. If this is done, the process becomes a "cousin" to the QR algorithm, whereas before it would be considered a "cousin" to the LU factorization. "Cousin" because the matrix operations of those are self referential, and this algorithm references the matrix $A$. The following is a hopefully enlightening description. See first edits for a more wordy description. But I think the following is more concise and to the point. Consider the conformable matrix $B^\top$ such that $B$ has the same dimension as $A$ and \begin{align} B^\top A &= I \tag{1}\\ B &= A B_A \tag{2}\\ \end{align} Here $B_A$ is arbitrary and exists only to ensure that the matix $B$ shares the same column space as $A$. (1) states that $B^\top$ is a (not necessarily unique) left inverse for $A$. (2) states that $B$ shares the same column space as $A$. Claim: $B^\top$ is the Moore-Penrose pseudoinverse for $A$. Proof: The Moore-Penrose pseudoinverse for $A$ is $$A^\dagger = \left(A^\top A\right)^{-1}A^\top$$ Given (1) and substituting (2) we have \begin{align} \left(AB_A\right)^\top A &= I \\ B_A^\top A^\top A &= I \\ B_A^\top &= \left( A^\top A\right)^{-1} \\ B_A &= \left( A^\top A\right)^{-1} \\ \end{align} Now solving for $B$ from (2): \begin{align} B &= A\left(B_A\right) \\ B &= A\left( A^\top A\right)^{-1} \\ \Rightarrow B^\top &=\left( A^\top A\right)^{-1}A^\top \\ \end{align} Q.E.D. - But the question is about the Moore-Penrose pseudoinverse, not the left inverse. If left inverses exist, the Moore-Penrose pseudoinverse is one of them, isn't it? And it can be computed using the algorithm in J.M.'s answer. So I'm not sure what you mean by "there is no direct formula (or algorithm)". – Rahul Narain Feb 28 at 20:17 @Rahul The algorithm in J.M.'s answer is iterative. What I am saying is that this algorithm computes the particular left inverse which is indeed the Moore-Penrose pseudoinverse, and does so in exact arithmetic, not by convergence. – adam W Feb 28 at 20:29 – Gottfried Helms Feb 28 at 21:45 @Gottfried Well, I do get the same numbers up to accuracy. For instance, $-0.012390122873089741$ is what I get for the lower right element. The way you describe "putting A into upper triangular shape" seems like you are doing regular LU or QR though...so not too certain. – adam W Feb 28 at 23:02 Is this faster than SVD? It's $\Omega(mn^2)$, right? – user7530 Mar 1 at 2:05 show 4 more comments First step is to buy Matlab. Second step is to type `pinv`. Edit: Here is an implementation. - 5 Funny ^^ but hardly an answer to his question. – Olivier Bégassat Oct 25 '11 at 17:07 or get octave for free=) – Kabumbus Oct 25 '11 at 17:38

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http://www.physicsforums.com/showthread.php?t=7454

Physics Forums Toe must include ourselves I know we have discussed that a theory for everything needs to include our spirituality. This is true but also it must include our skin cells, our hearts, our brains for these too are a part of our "Physical" universe. A theory of everything must include our ability to imagine and think, our ability to articulate and our ability to express our selves with free will. These are all physical manifestations and certainly not delusional. Mankind continously endeavours to create and recreate it self and this too has many physical implications. WE are very good at thinking that a theory of everything is only about stars and planets and phantom particles, The nature of thought it-self is missing from any theories so far. WE tend to think that we may be close to a TOE but I have to suggest that we are so far away from it that we have no hope of even getting close. For if one wants to quantify some thing like the human brain (a physical reality) into a theory that can be valued and determined by a formula then I am sorry they must be very deluded. Even Einstiens theories espouse that the perception of the observer is to be considered. Could this not be also a theory of subjectivity. A simple truth that we have always known. I think if science forgets the human or life element it is badly mistaken and I am not even talking about religion or spirituality or souls or life after death or anything. So are we taking on an impossible challenge as surely the closer one gets to this Holy grail of the theory for everything the more complex the mathematics must get, the harder it is to quantify and prove, and the less constancy there is. PhysOrg.com physics news on PhysOrg.com >> Promising doped zirconia>> New X-ray method shows how frog embryos could help thwart disease>> Bringing life into focus Recognitions: Gold Member Staff Emeritus The skin cells and all the construction of our bodies is referred to molecular chemistry, where it is being better and better understood. Changeing from the present theories to a TOE won't change that, and the genome and proteomic revolutions now ongoing are going to show us everything about our own biology before 2050. Consciousness is the last refuge of the "more than chemistry" faith. However much people like Dennet insist there is nothing there, the believers will insist that qualia, or free will, or something must transcend the dance of the molecules. so, From a mathematical perpsective how can you write a mathematical equation that not only includes black hole and gravity etc but also a single skin cell? Toe must include ourselves Originally posted by selfAdjoint The skin cells and all the construction of our bodies is referred to molecular chemistry, where it is being better and better understood. Changeing from the present theories to a TOE won't change that, and the genome and proteomic revolutions now ongoing are going to show us everything about our own biology before 2050. Consciousness is the last refuge of the "more than chemistry" faith. However much people like Dennet insist there is nothing there, the believers will insist that qualia, or free will, or something must transcend the dance of the molecules. Yes, well, the "believers" (as you put it) always insisted that we were the center of the Universe, and that the Earth was created in 7 days of 24 hours each, and that the Universe was static, and that God doesn't play dice... I hope that the belief that consciousness should remain mysterious is destroyed as these others were. Dennett has taken a powerful first step, and I hope he and his fellow materialists are not ignored. Originally posted by scott_sieger so, From a mathematical perpsective how can you write a mathematical equation that not only includes black hole and gravity etc but also a single skin cell? You shouldn't have to. As selfAdjoint pointed out, molecular biology will continue to be the field that deals with skin cells, long after the TOE is discovered. The term "Theory of Everything" is often misleading, since it implies that this theory will actually explain all phenomena in one equation. This is not the case. The TOE equation will be one that explains what the nature of spacetime and particles are. That's pretty much it. From this equation, we will be able to derive explanations of such phenomena as blackholes and the big bang, and so on, but we will never have every answer - and the ToE is not looking for every answer. I see your point,which is well taken. I for one understand the need for the human brain/mind to search out new knowledge and go where we have not yet been. Thank you for sharing it. if spirituality b a part of this...it must b "Flawless" i.e, it must not have any mistakes in it! it is not. because of a few things, such as: 1) it is made by humans. 2) it is like a pair of hand-cuffs holdin u back from a lot of things(remember the wars against the churches in Europe, which led to the industrial revolution). 3) it must b common 2 all races,species etc. living on the earth at-least... else there will b difference of opinion(and we all know what tht cud lead to) therefore we can conclude tht "SPIRITUALITY" can not b a part of ToE [;)] Call me a materialist, but I agree with anyone who says that spirituality, free will, and things of the sort are nothing but illusions\delusions. The bottom line is that the concept of the thinking mind is a consequence of chemical\sub-atomic interactions in that lumpy mass of grey matter known as the human brain. There is no free will. The concept is a consequence of vibrating strings. I realize that my login name is a misnomer, and that I should be more like "free brain". Originally posted by Mentat You shouldn't have to. As selfAdjoint pointed out, molecular biology will continue to be the field that deals with skin cells, long after the TOE is discovered. The term "Theory of Everything" is often misleading, since it implies that this theory will actually explain all phenomena in one equation. This is not the case. The TOE equation will be one that explains what the nature of spacetime and particles are. That's pretty much it. From this equation, we will be able to derive explanations of such phenomena as blackholes and the big bang, and so on, but we will never have every answer - and the ToE is not looking for every answer. Who says ToE has to be only one equation though? Maybe the simplest answer is the most complex system to understand. Originally posted by Jeebus Who says ToE has to be only one equation though? Maybe the simplest answer is the most complex system to understand. Well, Michio Kaku and the like have often expressed a desire for their ToE to be expressable in one small equation. What part of the word 'everything' do some posters here not understand? It also has to define itself. Therefore it must be the most simple equation. There can be no more fundamental than itself. And... the search is for a law of everything, not a theory. A theory is a way to 'get there'. True, a law of everything would be the ultimate achievement of science, but, if you ask me, that is impossible. Why? Because in order to have a law of everything, the writer would then have to know everything, be able to have absolute knowledge at one time, and would then most likely (Pardon the sound) become one with existence and thus have ascended beyond this realm to at least have absolute contact with what is higher. Such a person would cease to be human, and probably wouldn't see anything positive in giving us the law of everything, for then everything would change and the law would have to be rewritten. A theory of everything is much more beneficial, because we will be able to understand all that there is to understand without absolutely knowing from observation. We could predict everything, most likely, and any inconsistencies could be plugged in and used to modify the theory for accuracy. If it was a law, we'd have to know everything, and thus everything would be different. Those that know everything would probably not have any desire to share it with us. That may be difficult for some to grasp... but think about it. A law is proven by nature. A theory is supported by observations and calculations. We could quite possibly come up with a theory of everything, but a law of everything, though desirable, is beyond our grasp. To know everything is similar to becoming all powerful. Originally posted by scott_sieger Even Einstiens theories espouse that the perception of the observer is to be considered. Could this not be also a theory of subjectivity. A simple truth that we have always known. My favorite misperception of relativity (and QM). Both have nothing to do with perception, and by observer, all that is meant is point of reference (in QM, it's particle interaction). Neither theory in any way cares about a conscious observer. You sound like lifegazer... The term "spirituality" could generate confusion. I prefer to say that TOE must explain how is it possible from inside the universe to perceive the rules that govern the universe itself (i.e. TOE). It seems that TOE is to be a tautology. Or merely it cannot be stated. My think is that TOE should bring us to a possible new dimensionless situation, in which the observer is outside space-time frame and therefore at an actually infinite distance from any experiment. To achieve this result, TOE should start with an antinomic statement that would grant a precise initial formula for a bootstrap of Universe and would grant freedom as well. Like this: $$\Rightarrow \emptyset$$ That reads: Nothingness (i.e. the missing term on the left of implication symbol) implicates the "Empty Set" (i.e. a concept that is something else than Nothingness). Since Nothingness itself is an antinomic concept, the statement is stated by Nothingness itself in an originary situation of no-space, no-time. From this situation an endless oscillation is generated without any external agent in each possible point of Nothingness. Imagine that the "modes" of implication leads to "space" and the "occurrencies" of implication leads to "time". If the series of infinite ratios modes/occurrencies converges to a finite value (say c) and the modes present some symmetry, we could build some set of points that synchronically could interact among themselves and self-define a space-time frame. This could be the seed of any knowable universe (included our's). Thread Tools Similar Threads for: Toe must include ourselves Thread Forum Replies Programming & Comp Sci 20 General Math 29 Forum Feedback & Announcements 5 General Math 0 Forum Feedback & Announcements 4

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http://mathhelpforum.com/calculus/187093-find-maximum-value-function-two-variables.html

# Thread: 1. ## Find maximum value of a function in two variables Hi, I have a problem. Given a+b=10. Find maximum value of a^3*b^2 How do i go about doing this? Also i heard, a^x*b^y*c^z is maximum when a/x=b/y=c/z, provided a+b+c=constant? Can someone give me the proof? Also i am appearing for an exam that tests me in such applications of math, can someone post a link to the most common axioms and proofs? Thanks. 2. ## Re: Find maximum value of a function in two variables There are various ways to go about it. The simplest might be a quick substitution. a = 10-b might lead somewhere useful. 3. ## Re: Find maximum value of a function in two variables Originally Posted by TKHunny There are various ways to go about it. The simples might be a quick substitution. a = 10-b might lead somewhere useful. The substituion is a valid process. But i wanted to know if i could apply the concept i described earlier. The reason is this problem needs to be solved in 1 minute. If the concept(theorem/axiom) described earlier is accurate, could anybody explain to me the proof or a high level reasoning. Thanks 4. ## Re: Find maximum value of a function in two variables By the concept i meant --> The max value of a^x.b^y is obtained when a/x=b/y. 5. ## Re: Find maximum value of a function in two variables Ummm...That may be right, but I'm not sure it's useful. It has very limited scope (solves very few problems) and is unlikely to appear in that form ever again - unless you already know what will be on the exam. 6. ## Re: Find maximum value of a function in two variables True about the scope. But can you help me deduce or prove the theorem. I would want to know how they arrived at the concept. That helps me understand the direction in which to approach the problem. And some variations of the problem can be expected. Also any links to pages explaining some useful theorems? WHat i am looking for is a site that just lists down some salient and useful theorems. Eg 1+2+3+...+n = [n(n+1)]/2 Thanks. 7. ## Re: Find maximum value of a function in two variables How is your experience with Lagrangian Multipliers? If it can be proven, that should work. 8. ## Re: Find maximum value of a function in two variables Originally Posted by TKHunny How is your experience with Lagrangian Multipliers? If it can be proven, that should work. I can't recollect that. Are you suggesting Lagrange's to deduce the theorem or to solve the problem? Either way could you explain the same? THanks 9. ## Re: Find maximum value of a function in two variables To maximize $f(a,b,c)= a^xb^yc^z$ subject to the condition that g(a, b, c)= a+ b+ c= constant, make the gradients parallel: $\nabla f= xa^{x-1}b^yc^z\vec{i}+ ya^xb^{y-1}c^z\vec{j}+ za^xb^yz^{c-1}\vec{k}= \lambda\nabla g= \lambda\vec{i}+ \lambda\vec{j}+ \lambda\vec{k}$ where $\lambda$ is the Lagrange multiplier. Then we must have $xa^{x-1}b^y c^z= \lambda$, $ya^xb^{y-1}c^z= \lambda$, and $za^xb^yc^{z-1}= \lambda$. We can eliminate $\lambda$ from those equations by dividing one by another: Dividing the first equation by the second, $\frac{xa^{x-1}b^yc^z}{ya^xb^{y-1}c^z}= \frac{xb}{ya}= 1$ or $\frac{x}{a}= \frac{y}{b}$. Dividing the second equation by the third gives $\frac{ya^xb^{y-1}c^z}{za^xb^yc^{z-1}}= 1$ or $\frac{y}{b}= \frac{z}{c}$. If you are unfamiliar with "Lagrange Multipliers", here's a way to think about them- if you want to maximize f(x,y,z), calculate $\nabla f$, which always points in the direction of fastest increase of f, and follow it until you can't follow it any more- until it is 0. If you are constrained to stay on surface g(x,y,z)= constant, you may not be able to move in that direction ( $\nabla f$ may point off the surface), but you could move in the direction of the projection of $\nabla f$ onto the surface. You can do that, getting higher values of f, until there is NO projection on the surface- that is, until $\nabla f$ is perpendicular to the surface. But that means that it will be parallel to the normal vector to g at that point: $\nabla f$ is parallel to $\nabla g$ or $\nabla f= \lambda\nabla g$ for some number $\lambda$. 11. ## Re: Find maximum value of a function in two variables Originally Posted by TKHunny

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http://michaelnielsen.org/polymath1/index.php?title=The_Erd%C5%91s_discrepancy_problem&diff=3206&oldid=2749

# The Erdős discrepancy problem ### From Polymath1Wiki (Difference between revisions) () Current revision (18:29, 20 July 2010) (view source)m () (40 intermediate revisions not shown.) Line 3: Line 3: ==Introduction and statement of problem== ==Introduction and statement of problem== - Is there a <math>\scriptstyle \pm 1</math> sequence <math>(x_n)</math> and constant <math>C</math>, such that for all positive integers <math> d,k </math> + Let <math>(x_n)</math> be a <math>\scriptstyle \pm 1</math> sequence and let <math>C</math> be a constant. Must there exist positive integers <math> d,k </math> such that - :<math> \left| \sum_{i=1}^k x_{id} \right| \leq C </math>? + :<math> \left| \sum_{i=1}^k x_{id} \right| > C </math>? - Known colloquially as "The Erdős discrepancy problem", this question has remained unanswered since the 1930s [citation needed]. It was also asked by Chudakov [again a citation needed]. It is extremely easy to state, and a very appealing problem, so perhaps Polymath can succeed where individual mathematicians have failed. If so, then, given the notoriety of the problem, it would be a significant coup for the multiply collaborative approach to mathematics. But even if a full solution is not reached, preliminary investigations have already thrown up some interesting ideas which could perhaps be developed into publishable results. + Known colloquially as "The Erdős discrepancy problem", this question has remained unanswered since the 1930s (Erdős, 1957) and Erdős offered \$500 for an answer. It was also asked by Chudakov (1956). It is extremely easy to state, and a very appealing problem, so perhaps Polymath can succeed where individual mathematicians have failed. If so, then, given the notoriety of the problem, it would be a significant coup for the multiply collaborative approach to mathematics. But even if a full solution is not reached, preliminary investigations have already thrown up some interesting ideas which could perhaps be developed into publishable results. - Anticipating a negative answer to Erdős's question, we now quantify the problem. For any finite set <math>A</math> of integers, we define the error <math>A</math> by + It seems likely that the answer to Erdős's question is yes. If it is, then an easy compactness argument tells us that for every <math>C</math> there exists <math>n</math> such that for every <math>\scriptstyle \pm 1</math> sequence of length <math>n</math> there exist <math>d,k</math> with <math>dk\leq n</math> such that <math> \left| \sum_{i=1}^k x_{id} \right| > C </math>. In view of this, we make some definitions that allow one to talk about the dependence between <math>C</math> and <math>n</math>. For any finite set <math>A</math> of integers, we define the error on <math>A</math> by :<math> E(A) := \sum_{a\in A} x_a </math>, :<math> E(A) := \sum_{a\in A} x_a </math>, - and for a set <math>\mathcal{A}</math> of finite sets of integers, we define the discrepancy + and for a set <math>\mathcal{A}</math> of finite sets of integers, we define the discrepancy :<math> \delta(\mathcal{A},x) := \sup_{A\in \mathcal{A}} |E(A)|. </math> :<math> \delta(\mathcal{A},x) := \sup_{A\in \mathcal{A}} |E(A)|. </math> We can think of the values taken by the sequence <math>(x_n)</math> as a red/blue colouring of the integers that tries to make the number of reds and blues in each <math>\scriptstyle A\in\mathcal{A}</math> as equal as possible. The discrepancy measures the extent to which the sequence fails in this attempt. Taking <math>\scriptstyle \mathcal{HAP}(N)</math> to be the set of homogeneous arithmetic progressions <math>\{d, 2d, 3d, ..., nd\}</math> contained in <math>\{1, 2, ..., N\}</math>, we can restate the question as whether <math>\scriptstyle \delta(\mathcal{HAP}(N),x) \to \infty</math> for every sequence <math>x</math>. We can think of the values taken by the sequence <math>(x_n)</math> as a red/blue colouring of the integers that tries to make the number of reds and blues in each <math>\scriptstyle A\in\mathcal{A}</math> as equal as possible. The discrepancy measures the extent to which the sequence fails in this attempt. Taking <math>\scriptstyle \mathcal{HAP}(N)</math> to be the set of homogeneous arithmetic progressions <math>\{d, 2d, 3d, ..., nd\}</math> contained in <math>\{1, 2, ..., N\}</math>, we can restate the question as whether <math>\scriptstyle \delta(\mathcal{HAP}(N),x) \to \infty</math> for every sequence <math>x</math>. - Two related questions have already been solved in the literature. Letting <math>\scriptstyle \mathcal{AP}(N) </math> be the collection of all (not necessarily homogeneous) arithmetic progressions in <math>\{1, 2, ..., N\}</math>, Roth [citation needed] proved that <math> \scriptstyle \delta(\mathcal{AP}(N),x) \geq c n^{1/4}</math>, independent of the sequence <math>x</math>. Letting <math> \scriptstyle \mathcal{HQAP}(N) </math> be the collection of all homogeneous quasi-arithmetic progressions <math>\scriptstyle\{\lfloor \alpha \rfloor,\lfloor 2\alpha \rfloor,\dots,\lfloor k\alpha \rfloor\}</math> contained in <math>\{1, 2, ..., N\}</math>, Vijay [citation needed] proved that <math>\scriptstyle \delta(\mathcal{HQAP}(N),x) \geq 0.02 n^{1/6}</math>, independent of the sequence <math>x</math>. + Two related questions have already been solved in the literature. Letting <math>\scriptstyle \mathcal{AP}(N) </math> be the collection of all (not necessarily homogeneous) arithmetic progressions in <math>\{1, 2, ..., N\}</math>, Roth (1964) proved that <math> \scriptstyle \delta(\mathcal{AP}(N),x) \geq c n^{1/4}</math>, independent of the sequence <math>x</math>. Letting <math> \scriptstyle \mathcal{HQAP}(N) </math> be the collection of all homogeneous quasi-arithmetic progressions <math>\scriptstyle\{\lfloor \alpha \rfloor,\lfloor 2\alpha \rfloor,\dots,\lfloor k\alpha \rfloor\}</math> contained in <math>\{1, 2, ..., N\}</math>, Vijay (2008) proved that <math>\scriptstyle \delta(\mathcal{HQAP}(N),x) \geq 0.02 n^{1/6}</math>, independent of the sequence <math>x</math>. ==Some definitions and notational conventions== ==Some definitions and notational conventions== Line 29: Line 29: The EDP is the Erdős discrepancy problem. (This may conceivably be changed if enough people don't like it.) The EDP is the Erdős discrepancy problem. (This may conceivably be changed if enough people don't like it.) - A sequence <math>(x_n)</math> is completely multiplicative if <math>x_{mn}=x_mx_n</math> for any two positive integers m and n. We shall sometimes abbreviate this to "multiplicative", but the reader should be aware that the word "multiplicative" normally refers to the more general class of sequences such that <math>x_{mn}=x_mx_n</math> whenever m and n are coprime. A completely multiplicative function is determined by the values it takes at primes. The Liouville function λ is the unique completely multiplicative function that takes the value -1 at every prime: if the prime factorization of n is <math>\prod p_i^{a_i}</math> then λ(n) equals <math>(-1)^{\sum a_i}</math>. + A sequence <math>(x_n)</math> is completely multiplicative if <math>x_{mn}=x_mx_n</math> for any two positive integers m and n. We shall sometimes abbreviate this to "multiplicative", but the reader should be aware that the word "multiplicative" normally refers to the more general class of sequences such that <math>x_{mn}=x_mx_n</math> whenever m and n are coprime. A completely multiplicative function is determined by the values it takes at primes. The Liouville function λ is the unique completely multiplicative function that takes the value -1 at every prime: if the prime factorization of n is <math>\prod p_i^{a_i}</math> then λ(n) equals <math>(-1)^{\sum a_i}</math>. Another important multiplicative function is <math>\mu_3</math>, the multiplicative function that’s -1 at 3, 1 at numbers that are 1 mod 3 and -1 at numbers that are 2 mod 3. This function has mean-square partial sums which grow logarithmically, [http://gowers.wordpress.com/2010/01/26/edp3-a-very-brief-report-on-where-we-are/#comment-5585 see this calculation]. A HAP-subsequence of a sequence <math>(x_n)</math> is a subsequence of the form <math>x_d,x_{2d},x_{3d},...</math> for some d. If <math>(x_n)</math> is a multiplicative <math>\pm 1</math> sequence, then every HAP-subsequence is equal to either <math>(x_n)</math> or <math>(-x_n)</math>. We shall call a sequence weakly multiplicative if it has only finitely many distinct HAP-subsequences. Let us also call a <math>\pm 1</math> sequence quasi-multiplicative if it is a composition of a completely multiplicative function from <math>\mathbb{N}</math> to an Abelian group G with a function from G to {-1,1} (This definition is too general. See [http://gowers.wordpress.com/2010/01/09/erds-discrepancy-problem-continued/#comment-4843 this and the next comment]). [http://gowers.wordpress.com/2009/12/17/erdoss-discrepancy-problem/#comment-4705 It can be shown] that if <math>(x_n)</math> is a weakly multiplicative sequence, then it has an HAP-subsequence that is quasi-multiplicative. A HAP-subsequence of a sequence <math>(x_n)</math> is a subsequence of the form <math>x_d,x_{2d},x_{3d},...</math> for some d. If <math>(x_n)</math> is a multiplicative <math>\pm 1</math> sequence, then every HAP-subsequence is equal to either <math>(x_n)</math> or <math>(-x_n)</math>. We shall call a sequence weakly multiplicative if it has only finitely many distinct HAP-subsequences. Let us also call a <math>\pm 1</math> sequence quasi-multiplicative if it is a composition of a completely multiplicative function from <math>\mathbb{N}</math> to an Abelian group G with a function from G to {-1,1} (This definition is too general. See [http://gowers.wordpress.com/2010/01/09/erds-discrepancy-problem-continued/#comment-4843 this and the next comment]). [http://gowers.wordpress.com/2009/12/17/erdoss-discrepancy-problem/#comment-4705 It can be shown] that if <math>(x_n)</math> is a weakly multiplicative sequence, then it has an HAP-subsequence that is quasi-multiplicative. Line 39: Line 39: *To answer the problem negatively, it is sufficient to find a completely multiplicative sequence (that is, a sequence <math>(x_n)</math> such that <math>x_{mn}=x_mx_n</math> for every m,n) such that its partial sums <math>x_1+\dots+x_n</math> are bounded. First mentioned in the proof of the theorem in [http://gowers.wordpress.com/2009/12/17/erdoss-discrepancy-problem/ this] post. *To answer the problem negatively, it is sufficient to find a completely multiplicative sequence (that is, a sequence <math>(x_n)</math> such that <math>x_{mn}=x_mx_n</math> for every m,n) such that its partial sums <math>x_1+\dots+x_n</math> are bounded. First mentioned in the proof of the theorem in [http://gowers.wordpress.com/2009/12/17/erdoss-discrepancy-problem/ this] post. - *The function that takes n to 1 if the last non-zero digit of n in its ternary representation is 1 and -1 if the last non-zero digit is 2 is completely multiplicative and the partial sum up to n is easily shown to be at most <math>\log_3n +1</math>. Therefore, the rate at which the worst discrepancy grows, as a function of the length of the homogeneous progression, can be as slow as logarithmic. [http://gowers.wordpress.com/2009/12/17/erdoss-discrepancy-problem/] <s>[http://gowers.wordpress.com/2009/12/17/erdoss-discrepancy-problem/#comment-4553 It turns out] that the base can be made significantly higher than 3, so this example is not best possible.</s> [http://gowers.wordpress.com/2010/01/06/erdss-discrepancy-problem-as-a-forthcoming-polymath-project/#comment-4757 Correction] + *The function that takes n to 1 if the last non-zero digit of n in its ternary representation is 1 and -1 if the last non-zero digit is 2 is completely multiplicative and the partial sum up to n is easily shown to be at most <math>\log_3n +1</math>. Therefore, the rate at which the worst discrepancy grows, as a function of the length of the homogeneous progression, can be as slow as logarithmic. [[Matryoshka_Sequences|More examples of this sort are here.]] [http://gowers.wordpress.com/2009/12/17/erdoss-discrepancy-problem/] <s>[http://gowers.wordpress.com/2009/12/17/erdoss-discrepancy-problem/#comment-4553 It turns out] that the base can be made significantly higher than 3, so this example is not best possible.</s> [http://gowers.wordpress.com/2010/01/06/erdss-discrepancy-problem-as-a-forthcoming-polymath-project/#comment-4757 Correction] *To answer the problem negatively, it is also sufficient to find a completely multiplicative function f from <math>\mathbb{N}</math> to a finite Abelian group G (meaning that f(mn)=f(m)+f(n) for every m,n) and a function h from G to {-1,1} such that for each g in G there exists d with f(d)=g and with the partial sums hf(d)+hf(2d)+...+hf(nd) bounded. In that case, one can set <math>x_n=hf(n)</math>. (Though this observation is simple with the benefit of hindsight, it might not have been made had it not been for the fact that this sort of structure was identified in a long sequence that had been produced experimentally). First mentioned [http://gowers.wordpress.com/2009/12/17/erdoss-discrepancy-problem/#comment-4696 here]. Se also [http://gowers.wordpress.com/2010/01/06/erdss-discrepancy-problem-as-a-forthcoming-polymath-project/ this post] and [http://gowers.wordpress.com/2010/01/06/erdss-discrepancy-problem-as-a-forthcoming-polymath-project/#comment-4717 this comment]. *To answer the problem negatively, it is also sufficient to find a completely multiplicative function f from <math>\mathbb{N}</math> to a finite Abelian group G (meaning that f(mn)=f(m)+f(n) for every m,n) and a function h from G to {-1,1} such that for each g in G there exists d with f(d)=g and with the partial sums hf(d)+hf(2d)+...+hf(nd) bounded. In that case, one can set <math>x_n=hf(n)</math>. (Though this observation is simple with the benefit of hindsight, it might not have been made had it not been for the fact that this sort of structure was identified in a long sequence that had been produced experimentally). First mentioned [http://gowers.wordpress.com/2009/12/17/erdoss-discrepancy-problem/#comment-4696 here]. Se also [http://gowers.wordpress.com/2010/01/06/erdss-discrepancy-problem-as-a-forthcoming-polymath-project/ this post] and [http://gowers.wordpress.com/2010/01/06/erdss-discrepancy-problem-as-a-forthcoming-polymath-project/#comment-4717 this comment]. Line 46: Line 46: *The problem for the positive integers is equivalent to the problem for the positive rationals. [http://gowers.wordpress.com/2009/12/17/erdoss-discrepancy-problem/#comment-4676](Sketch proof: if you have a function f from the positive integers to {-1,1} that works, then define <math>f_q(x)</math> to be f(q!x) whenever q!x is an integer. Then take a pointwise limit of a subsequence of the functions <math>f_q</math>. [[Limits with better properties|Click here for a more detailed discussion of this construction and what it is good for]]. *The problem for the positive integers is equivalent to the problem for the positive rationals. [http://gowers.wordpress.com/2009/12/17/erdoss-discrepancy-problem/#comment-4676](Sketch proof: if you have a function f from the positive integers to {-1,1} that works, then define <math>f_q(x)</math> to be f(q!x) whenever q!x is an integer. Then take a pointwise limit of a subsequence of the functions <math>f_q</math>. [[Limits with better properties|Click here for a more detailed discussion of this construction and what it is good for]]. + + *Define the [[drift]] of a sequence to be the maximal value of |f(md)+...+f(nd)|. The discrepancy problem is equivalent to showing that the drift is always infinite. It is obvious that it is at least 2 (because |f(2)+f(4)|, |f(2)+f(3)|, |f(3)+f(4)| cannot all be at most 1); one can show that [[drift|it is at least 3]] (which implies as a corollary that the discrepancy is at least 2). One can also show that the [[upper and lower discrepancy]] are each at least 2. + + *One can [[topological dynamics formulation|formulate the problem using topological dynamics]]. + + *Using Fourier analysis, one can [[Fourier reduction|reduce the problem to one about completely multiplicative functions]]. + + *Here is an [[algorithm for finding multiplicative sequences with bounded discrepancy]]. + + *Here is a page whose aim is to record a [[human proof that completely multiplicative sequences have discrepancy at least 2]]. + + *Here is an argument that shows that [[bounded discrepancy multiplicative functions do not correlate with characters]]. + + * The answer to a [[function field version]] of the problem seems to be negative. + + * The problem can be generalized to [[pseudointegers]]. Here we have found [[EDP on pseudointegers|problems similar to EDP]] with a negative answer. ==Experimental evidence== ==Experimental evidence== + + ''Main page: [[Experimental results]]'' A nice feature of the problem is that it lends itself well to investigation on a computer. Although the number of <math>\pm 1</math> sequences grows exponentially, the constraints on these sequences are such that depth-first searches can lead quite quickly to interesting examples. For instance, at the time of writing they have yielded several examples of sequences of length 1124 with discrepancy 2. (This is the current record.) See [[Experimental results|this page]] for more details and further links. A nice feature of the problem is that it lends itself well to investigation on a computer. Although the number of <math>\pm 1</math> sequences grows exponentially, the constraints on these sequences are such that depth-first searches can lead quite quickly to interesting examples. For instance, at the time of writing they have yielded several examples of sequences of length 1124 with discrepancy 2. (This is the current record.) See [[Experimental results|this page]] for more details and further links. Line 57: Line 75: for sufficiently large <math> N </math>. Currently, this is the record-holder for slow growing discrepancy. for sufficiently large <math> N </math>. Currently, this is the record-holder for slow growing discrepancy. - The Thue-Morse sequence has discrepancy <math> \delta(N) \gg \sqrt{N} </math>. (See the discussion following [http://gowers.wordpress.com/2010/01/06/erdss-discrepancy-problem-as-a-forthcoming-polymath-project/#comment-4775 this comment] and the next one.) + The [[Thue-Morse-Hedlund Sequence|Thue-Morse sequence]] has discrepancy <math> \delta(N) \gg N^{\log_4(3)} </math>. A random sequence (each <math> x_i </math> chosen independently) has discrepancy at least (asymptotically) <math> \sqrt{2N \log\log N} </math> by the law of the iterated logarithm. A random sequence (each <math> x_i </math> chosen independently) has discrepancy at least (asymptotically) <math> \sqrt{2N \log\log N} </math> by the law of the iterated logarithm. Line 80: Line 98: *Does there exist a <math>\pm 1</math> sequence such that the corresponding discrepancy function grows at a rate that is slower than <math>c\log n</math> for any positive c? *Does there exist a <math>\pm 1</math> sequence such that the corresponding discrepancy function grows at a rate that is slower than <math>c\log n</math> for any positive c? + + *Does the number of sequences of length n and discrepancy at most C grow exponentially with n or more slowly than exponentially? (Obviously if the conjecture is true then it must be zero for large enough n, but the hope is that this question is a more realistic initial target.) + + ==General proof strategies== + + This section contains links to other pages in which potential approaches to solving the problem are described. + + [[First obtain multiplicative structure and then obtain a contradiction]] + + [[Find a different parameter, show that it tends to infinity, and show that that implies that the discrepancy tends to infinity]] + + [[Prove the result for shifted HAPs instead of HAPs]] + + [[Find a good configuration of HAPs]] + + [[Generalize to a graph-theoretic formulation]] + + [[Representation of the diagonal]] + + [[Bounding the discrepancy in terms of the common difference]] ==Annotated Bibliography== ==Annotated Bibliography== Line 88: Line 126: **[http://gowers.wordpress.com/2010/01/09/erds-discrepancy-problem-continued/ Second Post] (Jan 9 - Jan 11). **[http://gowers.wordpress.com/2010/01/09/erds-discrepancy-problem-continued/ Second Post] (Jan 9 - Jan 11). **[http://gowers.wordpress.com/2010/01/11/the-erds-discrepancy-problem-iii/ Third Post] (Jan 11 - Jan 14). **[http://gowers.wordpress.com/2010/01/11/the-erds-discrepancy-problem-iii/ Third Post] (Jan 11 - Jan 14). - **[http://gowers.wordpress.com/2010/01/14/the-erds-discrepancy-problem-iv/ Fourth Post] (Jan 14 - ?). + **[http://gowers.wordpress.com/2010/01/14/the-erds-discrepancy-problem-iv/ Fourth Post] (Jan 14 - 16). + **[http://gowers.wordpress.com/2010/01/16/the-erds-discrepancy-problem-v/ Fifth Post] (Jan 16-19). + **[http://gowers.wordpress.com/2010/01/19/edp1-the-official-start-of-polymath5/ First Theoretical Post] (Jan 19-21) + **[http://gowers.wordpress.com/2010/01/21/edp2-a-few-lessons-from-edp1/ Second Theoretical Post] (Jan 21-26) + **[http://gowers.wordpress.com/2010/01/26/edp3-a-very-brief-report-on-where-we-are/ Third Theoretical Post] (Jan 26 -?) + **[http://gowers.wordpress.com/2010/01/30/edp4-focusing-on-multiplicative-functions/ Fourth Theoretical Post] (Jan 30- Feb 2) + **[http://gowers.wordpress.com/2010/02/02/edp5-another-very-brief-summary/ Fifth Theoretical Post] (Feb 2 - Feb 5) + **[http://gowers.wordpress.com/2010/02/05/edp6-what-are-the-chances-of-success/ Sixth Theoretical Post] (Feb 5- Feb 8) + **[http://gowers.wordpress.com/2010/02/08/edp7-emergency-post/ Seventh Theoretical Post] (Feb 8 - 19) + **[http://gowers.wordpress.com/2010/02/19/edp8-what-next/ Eighth Theoretical Post] (Feb 19-Feb 24) + **[http://gowers.wordpress.com/2010/02/24/edp9-a-change-of-focus/ Ninth Theoretical Most] (Feb 24-Mar 2) + **[http://gowers.wordpress.com/2010/03/02/edp10-a-new-and-very-promising-approach/ Tenth Theoretical Post] (Mar 2-Mar 7) + **[http://gowers.wordpress.com/2010/03/07/edp11-the-search-continues/ Eleventh Theoretical Post] (Mar 7-Mar 12) + **[http://gowers.wordpress.com/2010/03/13/edp12-representing-diagonal-maps/ Twelfth Theoretical Post] (Mar 13-Mar 22) + **[http://gowers.wordpress.com/2010/03/23/edp13-quick-summary/ Thirteenth Theoretical Post] (Mar 23-Apr 24) + **[http://gowers.wordpress.com/2010/04/25/edp14-strategic-questions/ Fourteenth Theoretical Post] (Apr 25-?) + + *[http://mathoverflow.net/questions/tagged/polymath5 Several questions on MathOverflow have been given the `polymath5' tag.] *Mathias, A. R. D. [http://www.dpmms.cam.ac.uk/~ardm/erdoschu.pdf On a conjecture of Erdős and Čudakov]. Combinatorics, geometry and probability (Cambridge, 1993). *Mathias, A. R. D. [http://www.dpmms.cam.ac.uk/~ardm/erdoschu.pdf On a conjecture of Erdős and Čudakov]. Combinatorics, geometry and probability (Cambridge, 1993). - This is a short paper establishing the maximal length of sequence for the case where C=1 is 11, and is the starting point for our experimental studies. + This one page paper establishes that the maximal length of sequence for the case where C=1 is 11, and is the starting point for our experimental studies. + + *Tchudakoff, N. G. Theory of the characters of number semigroups. J. Indian Math. Soc. (N.S.) 20 (1956), 11--15. MR0083515 (18,719e) + + The Mathias paper states that one of Erdős's problem lists states that this paper "studies related questions". The Math Review for this paper states that it summarizes results from seven other papers that are in Russian. The Math Review leaves the impression that the topic concerns characters of modulus 1. *Borwein, Peter, and Choi, Stephen. [http://www.cecm.sfu.ca/personal/pborwein/PAPERS/P204.pdf A variant of Liouville's lambda function: some surprizing formulae]. *Borwein, Peter, and Choi, Stephen. [http://www.cecm.sfu.ca/personal/pborwein/PAPERS/P204.pdf A variant of Liouville's lambda function: some surprizing formulae]. - Explicit formulas for the discrepancy of some completely multiplicative functions whose discrepancy is logarithmic. A typical example is: Let <math>\lambda_3(n) = (-1)^{\omega_3(n)}</math> where <math>\omega_3(n)</math> is the number of distinct prime factors congruent to <math>-1 \bmod 3</math> in <math>n</math> (with multiple factors counted multiply). Then the discrepancy of the <math>\lambda_3</math> sequence up to <math>N</math> is exactly the number of 1's in the base three expansion of <math>N</math>. + Explicit formulas for the discrepancy of some completely multiplicative functions whose discrepancy is logarithmic. A typical example is: Let <math>\lambda_3(n) = (-1)^{\omega_3(n)}</math> where <math>\omega_3(n)</math> is the number of distinct prime factors congruent to <math>-1 \bmod 3</math> in <math>n</math> (with multiple factors counted multiply). Then the discrepancy of the <math>\lambda_3</math> sequence up to <math>N</math> is exactly the number of 1's in the base three expansion of <math>N</math>. (This turns out not to be so surprising after all, since <math>\lambda_3(n)</math> is precisely the same as the ternary function defined in the second item of the Simple Observations section.) *Borwein, Peter, Choi, Stephen and Coons, Michael. [http://www.cecm.sfu.ca/personal/pborwein/PAPERS/P225.pdf Completely multiplicative functions taking values in {-1,1}]. The published version of the above paper, which explains the results and proofs more clearly than the preprint, and is more explicit about the relationship with Erdős's question. *Borwein, Peter, Choi, Stephen and Coons, Michael. [http://www.cecm.sfu.ca/personal/pborwein/PAPERS/P225.pdf Completely multiplicative functions taking values in {-1,1}]. The published version of the above paper, which explains the results and proofs more clearly than the preprint, and is more explicit about the relationship with Erdős's question. - - *Vijay, Sujith. [http://www.combinatorics.org/Volume_15/PDF/v15i1r104.pdf On the discrepancy of quasi-progressions]. Electronic Journal of Combinatorics, R104 of Volume 15(1), 2008. - - A quasi-progression is a sequence of the form <math>\lfloor k \alpha \rfloor </math>, with <math>1\leq k \leq K</math> for some K. The main theorem of interest is: If the integers from 0 to n are 2-coloured, there exists a quasi-progression that has discrepancy at least <math> (1/50)n^{1/6}</math>. *Granville and Soundararajan. [http://arxiv.org/abs/math/0702389 Multiplicative functions in arithmetic progressions] *Granville and Soundararajan. [http://arxiv.org/abs/math/0702389 Multiplicative functions in arithmetic progressions] Line 108: Line 163: In this [http://gowers.wordpress.com/2010/01/06/erdss-discrepancy-problem-as-a-forthcoming-polymath-project/#comment-4729 blog post] Terry Tao wrote: "Granville and Soundararajan have made some study of the discrepancy of bounded multiplicative functions. The situation is remarkably delicate and number-theoretical (and is closely tied with the Granville-Soundararajan theory of pretentious characters)." In this [http://gowers.wordpress.com/2010/01/06/erdss-discrepancy-problem-as-a-forthcoming-polymath-project/#comment-4729 blog post] Terry Tao wrote: "Granville and Soundararajan have made some study of the discrepancy of bounded multiplicative functions. The situation is remarkably delicate and number-theoretical (and is closely tied with the Granville-Soundararajan theory of pretentious characters)." - *[http://www.math.niu.edu/~rusin/known-math/93_back/prizes.erd As this web page reveals], the Erdős discrepancy problem was a \$500-dollar problem of Erdős, so it is clear that he regarded it as pretty hard. + *Wirsing, E. Das asymptotische Verhalten von Summen über multiplikative Funktionen. II. (German) [The asymptotic behavior of sums of multiplicative functions. II.] Acta Math. Acad. Sci. Hungar. 18 1967 411--467. MR0223318 (36 #6366) [http://michaelnielsen.org/polymath1/index.php?title=Wirsing_translation (partial) English translation] - *P. Erdős. [http://www.renyi.hu/~p_erdos/1957-13.pdf Some unsolved problems], Michigan Math. J. 4 (1957), 291--300 MR20 #5157; Zentralblatt 81,1. + *Newman, D. J. [http://www.jstor.org/stable/2036455 On the number of binary digits in a multiple of three]. Proc Amer Math Soc. 21(1969): 719--721. + Proves that the Thue-Morse sequence has discrepancy <math> \gg N^{\log_4(3)}</math> - Problem 9 of this paper is ours. Erdős dates the problem to around 1932, and notes what we know about Liouville's function (lower and upper bound). + *Halász, G. [http://gowers.wordpress.com/2010/02/02/edp5-another-very-brief-summary/#comment-5806 On random multiplicative functions]. Hubert Delange colloquium (Orsay, 1982), 74–96, Publ. Math. Orsay, 83-4, Univ. Paris XI, Orsay, 1983. + + The discrepancy of a random multiplicative function is close to \$\sqrt{N}\$. + + === Problem lists on which this problem appears === *Erdős, P. and Graham, R. [http://www.math.ucsd.edu/~fan/ron/papers/79_09_combinatorial_number_theory.pdf Old and New Problems and Results in Combinatorial Number Theory: van der Waerden's Theorem and Related Topics], L'Enseignement Math. 25 (1979), 325-344. *Erdős, P. and Graham, R. [http://www.math.ucsd.edu/~fan/ron/papers/79_09_combinatorial_number_theory.pdf Old and New Problems and Results in Combinatorial Number Theory: van der Waerden's Theorem and Related Topics], L'Enseignement Math. 25 (1979), 325-344. Line 118: Line 178: Our problem is mentioned at the bottom of page 331, where they indicate knowledge of a coloring with logarithmic discrepancy. On pages 330-1, they review work on the non-homogeneous problem. Our problem is mentioned at the bottom of page 331, where they indicate knowledge of a coloring with logarithmic discrepancy. On pages 330-1, they review work on the non-homogeneous problem. - *Hochberg, Robert. [http://www.springerlink.com/content/q02424284373qw45/ Large Discrepancy In Homogenous Quasi-Arithmetic Progressions]. Combinatorica, Volume 26 (2006), Number 1. + *Erdős, Paul, Some of my favourite unsolved problems. A tribute to Paul Erdős, 467--478, Cambridge Univ. Press, Cambridge, 1990. MR1117038 (92f:11003) - Roth's method for the <math> n^{1/4}</math> lower bound on nonhomogeneous AP discrepancy is adapted to give a lower bound for homogeneous quasi-AP discrepancy. This result is weaker than the Vijay result, but uses a different method. + "Let <math>f(n)=\pm1</math>. Is it true that for every <math>c</math> there is a <math>d</math> and an <math>m</math> so that + :<math> \left| \sum_{k=1}^m f(kd) \right| > c</math>? + I will pay \$500 for an answer. Let <math>f(n)=\pm1</math> and <math>f(ab)=f(a)f(b)</math>. Is it then true that <math>\left| \sum_{k=1}^m f(kd) \right|</math> cannot be bounded?" + + *P. Erdős. [http://www.renyi.hu/~p_erdos/1957-13.pdf Some unsolved problems], Michigan Math. J. 4 (1957), 291--300 MR20 #5157; Zentralblatt 81,1. + + Problem 9 of this paper is ours. Erdős dates the problem to around 1932, and notes what we know about Liouville's function (lower and upper bound). *Finch, S. [http://algo.inria.fr/csolve/ec.pdf Two-colorings of positive integers]. Dated May 27, 2008. *Finch, S. [http://algo.inria.fr/csolve/ec.pdf Two-colorings of positive integers]. Dated May 27, 2008. Summarizes knowledge of discrepancy of two colorings when the discrepancy is restricted to homogeneous progressions, nonhomogeneous progressions, and homogeneous quasi-progressions. Contains bibliography with 17 entries, including most of those above. Summarizes knowledge of discrepancy of two colorings when the discrepancy is restricted to homogeneous progressions, nonhomogeneous progressions, and homogeneous quasi-progressions. Contains bibliography with 17 entries, including most of those above. + + === non-homogeneous AP, quasi-AP, and other related discrepancy papers === + + *Hochberg, Robert. [http://www.springerlink.com/content/q02424284373qw45/ Large Discrepancy In Homogenous Quasi-Arithmetic Progressions]. Combinatorica, Volume 26 (2006), Number 1. + + Roth's method for the <math> n^{1/4}</math> lower bound on nonhomogeneous AP discrepancy is adapted to give a lower bound for homogeneous quasi-AP discrepancy. This result is weaker than the Vijay result, but uses a different method. + + *Vijay, Sujith. [http://www.combinatorics.org/Volume_15/PDF/v15i1r104.pdf On the discrepancy of quasi-progressions]. Electronic Journal of Combinatorics, R104 of Volume 15(1), 2008. + + A quasi-progression is a sequence of the form <math>\lfloor k \alpha \rfloor </math>, with <math>1\leq k \leq K</math> for some K. The main theorem of interest is: If the integers from 0 to n are 2-coloured, there exists a quasi-progression that has discrepancy at least <math> (1/50)n^{1/6}</math>. + + * Doerr, Benjamin; Srivastav, Anand; Wehr, Petra. [http://www.combinatorics.org/Volume_11/Abstracts/v11i1r5.html Discrepancy of Cartesian products of arithmetic progressions]. Electron. J. Combin. 11 (2004), no. 1, Research Paper 5, 16 pp. (electronic). + + Abstract: We determine the combinatorial discrepancy of the hypergraph H of cartesian + products of d arithmetic progressions in the <math> [N]^d</math> –lattice. + The study of such higher dimensional arithmetic progressions is motivated by a + multi-dimensional version of van derWaerden’s theorem, namely the Gallai-theorem + (1933). We solve the discrepancy problem for d–dimensional arithmetic progressions + by proving <math>disc(H) = \Theta�(N^{d/4})</math> for every fixed positive integer d. This extends the famous + lower bound of <math> \Omega(N^{1/4})</math> of Roth (1964) and the matching upper bound <math> O(N^{1/4})</math> + of Matoušek and Spencer (1996) from d = 1 to arbitrary, fixed d. To establish + the lower bound we use harmonic analysis on locally compact abelian groups. For + the upper bound a product coloring arising from the theorem of Matoušek and + Spencer is sufficient. We also regard some special cases, e.g., symmetric arithmetic + progressions and infinite arithmetic progressions. + + *Cilleruelo, Javier; Hebbinghaus, Nils [http://www.ams.org/mathscinet-getitem?mr=2547932 Discrepancy in generalized arithmetic progressions]. European J. Combin. 30 (2009), no. 7, 1607--1611. MR 2547932 + + *Valkó, Benedek. [Discrepancy of arithmetic progressions in higher dimensions]. (English summary) + J. Number Theory 92 (2002), no. 1, 117--130. MR1880588 (2003b:11071) + + Abstract: + K. F. Roth (1964, Acta. Arith.9, 257–260) proved that the discrepancy of arithmetic progressions contained in [N] is at least <math> cN^{1/4} </math>, and later it was proved that this result is sharp. We consider the d-dimensional version of this problem. We give a lower estimate for the discrepancy of arithmetic progressions on <math> [N]^d </math> and prove that this result is nearly sharp. We use our results to give an upper estimate for the discrepancy of lines on an N×N lattice, and we also give an estimate for the discrepancy of a related random hypergraph. + + *Roth, K. F., Remark concerning integer sequences. Acta Arith. 9 1964 257--260. MR0168545 (29 #5806) + + Shows that the discrepancy on APs is at least <math>cN^{1/4}</math>. ## Current revision If you want to you can jump straight to the main experimental results page. ## Introduction and statement of problem Let (xn) be a $\scriptstyle \pm 1$ sequence and let C be a constant. Must there exist positive integers d,k such that $\left| \sum_{i=1}^k x_{id} \right| > C$? Known colloquially as "The Erdős discrepancy problem", this question has remained unanswered since the 1930s (Erdős, 1957) and Erdős offered \$500 for an answer. It was also asked by Chudakov (1956). It is extremely easy to state, and a very appealing problem, so perhaps Polymath can succeed where individual mathematicians have failed. If so, then, given the notoriety of the problem, it would be a significant coup for the multiply collaborative approach to mathematics. But even if a full solution is not reached, preliminary investigations have already thrown up some interesting ideas which could perhaps be developed into publishable results. It seems likely that the answer to Erdős's question is yes. If it is, then an easy compactness argument tells us that for every C there exists n such that for every $\scriptstyle \pm 1$ sequence of length n there exist d,k with $dk\leq n$ such that $\left| \sum_{i=1}^k x_{id} \right| > C$. In view of this, we make some definitions that allow one to talk about the dependence between C and n. For any finite set A of integers, we define the error on A by $E(A) := \sum_{a\in A} x_a$, and for a set $\mathcal{A}$ of finite sets of integers, we define the discrepancy $\delta(\mathcal{A},x) := \sup_{A\in \mathcal{A}} |E(A)|.$ We can think of the values taken by the sequence (xn) as a red/blue colouring of the integers that tries to make the number of reds and blues in each $\scriptstyle A\in\mathcal{A}$ as equal as possible. The discrepancy measures the extent to which the sequence fails in this attempt. Taking $\scriptstyle \mathcal{HAP}(N)$ to be the set of homogeneous arithmetic progressions {d,2d,3d,...,nd} contained in {1,2,...,N}, we can restate the question as whether $\scriptstyle \delta(\mathcal{HAP}(N),x) \to \infty$ for every sequence x. Two related questions have already been solved in the literature. Letting $\scriptstyle \mathcal{AP}(N)$ be the collection of all (not necessarily homogeneous) arithmetic progressions in {1,2,...,N}, Roth (1964) proved that $\scriptstyle \delta(\mathcal{AP}(N),x) \geq c n^{1/4}$, independent of the sequence x. Letting $\scriptstyle \mathcal{HQAP}(N)$ be the collection of all homogeneous quasi-arithmetic progressions $\scriptstyle\{\lfloor \alpha \rfloor,\lfloor 2\alpha \rfloor,\dots,\lfloor k\alpha \rfloor\}$ contained in {1,2,...,N}, Vijay (2008) proved that $\scriptstyle \delta(\mathcal{HQAP}(N),x) \geq 0.02 n^{1/6}$, independent of the sequence x. ## Some definitions and notational conventions A homogeneous arithmetic progression, or HAP, is an arithmetic progression of the form {d,2d,3d,...,nd}. When x is clear from context, we write δ(N) in place of $\delta(\mathcal{HAP}(N),x)$. The discrepancy of a sequence (xn) is the supremum of $|\sum_{n\in P}x_n|$ over all homogeneous arithmetic progressions P. (Strictly speaking, we should call this something like the HAP-discrepancy, but since we will almost always be talking about HAPs, we adopt the convention that "discrepancy" always means "HAP-discrepancy" unless it is stated otherwise.) Insertformulahere A sequence (xn) has discrepancy at most φ(n) if for every natural number N the maximum value of $|\sum_{n\in P}x_n|$ over all homogeneous arithmetic progressions P that are subsets of {1,2,...,N} is at most φ(N). The EDP is the Erdős discrepancy problem. (This may conceivably be changed if enough people don't like it.) A sequence (xn) is completely multiplicative if xmn = xmxn for any two positive integers m and n. We shall sometimes abbreviate this to "multiplicative", but the reader should be aware that the word "multiplicative" normally refers to the more general class of sequences such that xmn = xmxn whenever m and n are coprime. A completely multiplicative function is determined by the values it takes at primes. The Liouville function λ is the unique completely multiplicative function that takes the value -1 at every prime: if the prime factorization of n is $\prod p_i^{a_i}$ then λ(n) equals $(-1)^{\sum a_i}$. Another important multiplicative function is μ3, the multiplicative function that’s -1 at 3, 1 at numbers that are 1 mod 3 and -1 at numbers that are 2 mod 3. This function has mean-square partial sums which grow logarithmically, see this calculation. A HAP-subsequence of a sequence (xn) is a subsequence of the form xd,x2d,x3d,... for some d. If (xn) is a multiplicative $\pm 1$ sequence, then every HAP-subsequence is equal to either (xn) or ( − xn). We shall call a sequence weakly multiplicative if it has only finitely many distinct HAP-subsequences. Let us also call a $\pm 1$ sequence quasi-multiplicative if it is a composition of a completely multiplicative function from $\mathbb{N}$ to an Abelian group G with a function from G to {-1,1} (This definition is too general. See this and the next comment). It can be shown that if (xn) is a weakly multiplicative sequence, then it has an HAP-subsequence that is quasi-multiplicative. It is convenient to define the maps Tk for $k \geq 1$ by Tk(x)n = xkn, where x = (x1,x2,...) is a sequence. ## Simple observations • To answer the problem negatively, it is sufficient to find a completely multiplicative sequence (that is, a sequence (xn) such that xmn = xmxn for every m,n) such that its partial sums $x_1+\dots+x_n$ are bounded. First mentioned in the proof of the theorem in this post. • The function that takes n to 1 if the last non-zero digit of n in its ternary representation is 1 and -1 if the last non-zero digit is 2 is completely multiplicative and the partial sum up to n is easily shown to be at most log3n + 1. Therefore, the rate at which the worst discrepancy grows, as a function of the length of the homogeneous progression, can be as slow as logarithmic. More examples of this sort are here. [1] It turns out that the base can be made significantly higher than 3, so this example is not best possible. Correction • To answer the problem negatively, it is also sufficient to find a completely multiplicative function f from $\mathbb{N}$ to a finite Abelian group G (meaning that f(mn)=f(m)+f(n) for every m,n) and a function h from G to {-1,1} such that for each g in G there exists d with f(d)=g and with the partial sums hf(d)+hf(2d)+...+hf(nd) bounded. In that case, one can set xn = hf(n). (Though this observation is simple with the benefit of hindsight, it might not have been made had it not been for the fact that this sort of structure was identified in a long sequence that had been produced experimentally). First mentioned here. Se also this post and this comment. • It can be shown that if a $\pm 1$ sequence (xn) starts with 1 and has only finitely many distinct subsequences of the form $(x_d,x_{2d},x_{3d},\dots)$, then it must have a subsequence that arises in the above way. (This was mentioned just above in the terminology section.) • The problem for the positive integers is equivalent to the problem for the positive rationals. [2](Sketch proof: if you have a function f from the positive integers to {-1,1} that works, then define fq(x) to be f(q!x) whenever q!x is an integer. Then take a pointwise limit of a subsequence of the functions fq. Click here for a more detailed discussion of this construction and what it is good for. • Define the drift of a sequence to be the maximal value of |f(md)+...+f(nd)|. The discrepancy problem is equivalent to showing that the drift is always infinite. It is obvious that it is at least 2 (because |f(2)+f(4)|, |f(2)+f(3)|, |f(3)+f(4)| cannot all be at most 1); one can show that it is at least 3 (which implies as a corollary that the discrepancy is at least 2). One can also show that the upper and lower discrepancy are each at least 2. • Using Fourier analysis, one can reduce the problem to one about completely multiplicative functions. • Here is a page whose aim is to record a human proof that completely multiplicative sequences have discrepancy at least 2. • Here is an argument that shows that bounded discrepancy multiplicative functions do not correlate with characters. • The answer to a function field version of the problem seems to be negative. • The problem can be generalized to pseudointegers. Here we have found problems similar to EDP with a negative answer. ## Experimental evidence Main page: Experimental results A nice feature of the problem is that it lends itself well to investigation on a computer. Although the number of $\pm 1$ sequences grows exponentially, the constraints on these sequences are such that depth-first searches can lead quite quickly to interesting examples. For instance, at the time of writing they have yielded several examples of sequences of length 1124 with discrepancy 2. (This is the current record.) See this page for more details and further links. Another sort of evidence is to bound the discrepancy for specific sequences. For example, setting x1 = − x2 = 1 and xn = − xn / 3 or xn = xn − 3 according to whether n is a multiple of 3 or not, yields a sequence with $\delta(N) \leq \log_9(N)+1$ for sufficiently large N. Currently, this is the record-holder for slow growing discrepancy. The Thue-Morse sequence has discrepancy $\delta(N) \gg N^{\log_4(3)}$. A random sequence (each xi chosen independently) has discrepancy at least (asymptotically) $\sqrt{2N \log\log N}$ by the law of the iterated logarithm. Determining if the discrepancy of Liouville's lambda function is O(n1 / 2 + ε) is equivalent to solving the Riemann hypothesis. However, this growth cannot be bounded above by n1 / 2 − ε for any positive ε. ## Interesting subquestions Given the length of time that the Erdős discrepancy problem has been open, the chances that it will be solved by Polymath5 are necessarily small. However, there are a number of interesting questions that we do not know the answers to, several of which have arisen naturally from the experimental evidence. Good answers to some of these would certainly constitute publishable results. Here is a partial list -- further additions are very welcome. (When the more theoretical part of the project starts, this section will probably grow substantially and become a separate page.) • Is there an infinite sequence of discrepancy 2? (Given how long a finite sequence can be, it seems unlikely that we could answer this question just by a clever search of all possibilities on a computer.) • The long sequences of low discrepancy discovered by computer all have some kind of approximate weak multiplicativity. If we take a hypothetical counterexample (xn) (which we could even assume has discrepancy 2), can we prove that it, or some other counterexample derived from it by passing to HAP-subsequences and taking pointwise limits, has some kind of interesting multiplicative structure? • Closely related to the previous question. If (xn) is a hypothetical counterexample, must it satisfy a compactness property of the following kind: for every positive c there exists M such that if you take any M HAP-subsequences, then there must be two of them, (yn) and (zn), such that $\lim\inf N^{-1}\sum_{n=1}^Ny_nz_n\geq 1-c$? • An even weaker question. If (xn) is a hypothetical counterexample, must it have two HAP-subsequences (yn) and (zn) such that the lim inf of $N^{-1}\sum_{n=1}^N y_nz_n$ is greater than 0? • A similar question, perhaps equivalent to the previous one (this should be fairly easy to check). Given a sequence (xn) define f(N) to be the average of xaxbxcxd over all quadruples (a,b,c,d) such that ab=cd and $a,b,c,d\leq N$. If (xn) is a counterexample, does that imply that $\lim\inf f(N)$ is greater than 0? See here for a computation of this average for the first 1124 sequence. • Is there any completely multiplicative counterexample? (This may turn out to be a very hard question. If so, then answering the previous questions could turn out to be the best we can hope for without making a substantial breakthrough in analytic number theory.) • Does there exist a $\pm 1$ sequence such that the corresponding discrepancy function grows at a rate that is slower than clogn for any positive c? • Does the number of sequences of length n and discrepancy at most C grow exponentially with n or more slowly than exponentially? (Obviously if the conjecture is true then it must be zero for large enough n, but the hope is that this question is a more realistic initial target.) ## General proof strategies This section contains links to other pages in which potential approaches to solving the problem are described. First obtain multiplicative structure and then obtain a contradiction Find a different parameter, show that it tends to infinity, and show that that implies that the discrepancy tends to infinity Prove the result for shifted HAPs instead of HAPs Find a good configuration of HAPs Generalize to a graph-theoretic formulation Representation of the diagonal Bounding the discrepancy in terms of the common difference ## Annotated Bibliography • Blog Discussion on Gowers's Weblog • Zeroth Post (Dec 17 - Jan 6). • First Post (Jan 6 - Jan 12). • Second Post (Jan 9 - Jan 11). • Third Post (Jan 11 - Jan 14). • Fourth Post (Jan 14 - 16). • Fifth Post (Jan 16-19). • First Theoretical Post (Jan 19-21) • Second Theoretical Post (Jan 21-26) • Third Theoretical Post (Jan 26 -?) • Fourth Theoretical Post (Jan 30- Feb 2) • Fifth Theoretical Post (Feb 2 - Feb 5) • Sixth Theoretical Post (Feb 5- Feb 8) • Seventh Theoretical Post (Feb 8 - 19) • Eighth Theoretical Post (Feb 19-Feb 24) • Ninth Theoretical Most (Feb 24-Mar 2) • Tenth Theoretical Post (Mar 2-Mar 7) • Eleventh Theoretical Post (Mar 7-Mar 12) • Twelfth Theoretical Post (Mar 13-Mar 22) • Thirteenth Theoretical Post (Mar 23-Apr 24) • Fourteenth Theoretical Post (Apr 25-?) • Mathias, A. R. D. On a conjecture of Erdős and Čudakov. Combinatorics, geometry and probability (Cambridge, 1993). This one page paper establishes that the maximal length of sequence for the case where C=1 is 11, and is the starting point for our experimental studies. • Tchudakoff, N. G. Theory of the characters of number semigroups. J. Indian Math. Soc. (N.S.) 20 (1956), 11--15. MR0083515 (18,719e) The Mathias paper states that one of Erdős's problem lists states that this paper "studies related questions". The Math Review for this paper states that it summarizes results from seven other papers that are in Russian. The Math Review leaves the impression that the topic concerns characters of modulus 1. • Borwein, Peter, and Choi, Stephen. A variant of Liouville's lambda function: some surprizing formulae. Explicit formulas for the discrepancy of some completely multiplicative functions whose discrepancy is logarithmic. A typical example is: Let $\lambda_3(n) = (-1)^{\omega_3(n)}$ where ω3(n) is the number of distinct prime factors congruent to − 1mod 3 in n (with multiple factors counted multiply). Then the discrepancy of the λ3 sequence up to N is exactly the number of 1's in the base three expansion of N. (This turns out not to be so surprising after all, since λ3(n) is precisely the same as the ternary function defined in the second item of the Simple Observations section.) • Borwein, Peter, Choi, Stephen and Coons, Michael. Completely multiplicative functions taking values in {-1,1}. The published version of the above paper, which explains the results and proofs more clearly than the preprint, and is more explicit about the relationship with Erdős's question. • Granville and Soundararajan. Multiplicative functions in arithmetic progressions In this blog post Terry Tao wrote: "Granville and Soundararajan have made some study of the discrepancy of bounded multiplicative functions. The situation is remarkably delicate and number-theoretical (and is closely tied with the Granville-Soundararajan theory of pretentious characters)." • Wirsing, E. Das asymptotische Verhalten von Summen über multiplikative Funktionen. II. (German) [The asymptotic behavior of sums of multiplicative functions. II.] Acta Math. Acad. Sci. Hungar. 18 1967 411--467. MR0223318 (36 #6366) (partial) English translation • Newman, D. J. On the number of binary digits in a multiple of three. Proc Amer Math Soc. 21(1969): 719--721. Proves that the Thue-Morse sequence has discrepancy $\gg N^{\log_4(3)}$ • Halász, G. On random multiplicative functions. Hubert Delange colloquium (Orsay, 1982), 74–96, Publ. Math. Orsay, 83-4, Univ. Paris XI, Orsay, 1983. The discrepancy of a random multiplicative function is close to \$\sqrt{N}\$. ### Problem lists on which this problem appears • Erdős, P. and Graham, R. Old and New Problems and Results in Combinatorial Number Theory: van der Waerden's Theorem and Related Topics, L'Enseignement Math. 25 (1979), 325-344. Our problem is mentioned at the bottom of page 331, where they indicate knowledge of a coloring with logarithmic discrepancy. On pages 330-1, they review work on the non-homogeneous problem. • Erdős, Paul, Some of my favourite unsolved problems. A tribute to Paul Erdős, 467--478, Cambridge Univ. Press, Cambridge, 1990. MR1117038 (92f:11003) "Let $f(n)=\pm1$. Is it true that for every c there is a d and an m so that $\left| \sum_{k=1}^m f(kd) \right| > c$? I will pay \$500 for an answer. Let $f(n)=\pm1$ and f(ab) = f(a)f(b). Is it then true that $\left| \sum_{k=1}^m f(kd) \right|$ cannot be bounded?" • P. Erdős. Some unsolved problems, Michigan Math. J. 4 (1957), 291--300 MR20 #5157; Zentralblatt 81,1. Problem 9 of this paper is ours. Erdős dates the problem to around 1932, and notes what we know about Liouville's function (lower and upper bound). • Finch, S. Two-colorings of positive integers. Dated May 27, 2008. Summarizes knowledge of discrepancy of two colorings when the discrepancy is restricted to homogeneous progressions, nonhomogeneous progressions, and homogeneous quasi-progressions. Contains bibliography with 17 entries, including most of those above. ### non-homogeneous AP, quasi-AP, and other related discrepancy papers • Hochberg, Robert. Large Discrepancy In Homogenous Quasi-Arithmetic Progressions. Combinatorica, Volume 26 (2006), Number 1. Roth's method for the n1 / 4 lower bound on nonhomogeneous AP discrepancy is adapted to give a lower bound for homogeneous quasi-AP discrepancy. This result is weaker than the Vijay result, but uses a different method. • Vijay, Sujith. On the discrepancy of quasi-progressions. Electronic Journal of Combinatorics, R104 of Volume 15(1), 2008. A quasi-progression is a sequence of the form $\lfloor k \alpha \rfloor$, with $1\leq k \leq K$ for some K. The main theorem of interest is: If the integers from 0 to n are 2-coloured, there exists a quasi-progression that has discrepancy at least (1 / 50)n1 / 6. • Doerr, Benjamin; Srivastav, Anand; Wehr, Petra. Discrepancy of Cartesian products of arithmetic progressions. Electron. J. Combin. 11 (2004), no. 1, Research Paper 5, 16 pp. (electronic). Abstract: We determine the combinatorial discrepancy of the hypergraph H of cartesian products of d arithmetic progressions in the [N]d –lattice. The study of such higher dimensional arithmetic progressions is motivated by a multi-dimensional version of van derWaerden’s theorem, namely the Gallai-theorem (1933). We solve the discrepancy problem for d–dimensional arithmetic progressions by proving disc(H) = Θ(Nd / 4) for every fixed positive integer d. This extends the famous lower bound of Ω(N1 / 4) of Roth (1964) and the matching upper bound O(N1 / 4) of Matoušek and Spencer (1996) from d = 1 to arbitrary, fixed d. To establish the lower bound we use harmonic analysis on locally compact abelian groups. For the upper bound a product coloring arising from the theorem of Matoušek and Spencer is sufficient. We also regard some special cases, e.g., symmetric arithmetic progressions and infinite arithmetic progressions. • Cilleruelo, Javier; Hebbinghaus, Nils Discrepancy in generalized arithmetic progressions. European J. Combin. 30 (2009), no. 7, 1607--1611. MR 2547932 • Valkó, Benedek. [Discrepancy of arithmetic progressions in higher dimensions]. (English summary) J. Number Theory 92 (2002), no. 1, 117--130. MR1880588 (2003b:11071) Abstract: K. F. Roth (1964, Acta. Arith.9, 257–260) proved that the discrepancy of arithmetic progressions contained in [N] is at least cN1 / 4, and later it was proved that this result is sharp. We consider the d-dimensional version of this problem. We give a lower estimate for the discrepancy of arithmetic progressions on [N]d and prove that this result is nearly sharp. We use our results to give an upper estimate for the discrepancy of lines on an N×N lattice, and we also give an estimate for the discrepancy of a related random hypergraph. • Roth, K. F., Remark concerning integer sequences. Acta Arith. 9 1964 257--260. MR0168545 (29 #5806) Shows that the discrepancy on APs is at least cN1 / 4.

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http://mathoverflow.net/questions/26079?sort=votes

## Algorithm for determining if a path exists in a graph or if not, the closest edit distance. ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Given a directed acyclic graph `G` and a path made up from its set of nodes `N`, what is the closest approximate match to N, equipped with an intuitive notion of distance? A path in a directed acyclic graph is essentially a string (that uses the set of nodes as the alphabet), so when comparing paths one can make use of edit distances developed for approximate string matching: There are many algorithms for approximate string matching: http://en.wikipedia.org/wiki/Approximate_string_matching This string matching answers the question when the `G` itself is also a path.. then we're merely asking to compare two strings. Asking if an arbitrary graph `A` built from the same set of nodes is a sub-graph of `G` is the general case of the problem, but I'm only interested in the case where `A` is a path and `G` is directed & acyclic. Any general pointers are also welcome. -- Edit: I ended up using a dynamic programming algorithm, independently also suggested in the accepted answer. Good call! It is probably the most accurate option as well, and barely more "complex" than the string-to-string case when the average # of edges per node is low. - 3 for q1, I'm not seeing the problem. given the sequence of nodes in the path, why can't you just verify that directed edges exist between each adjacent pair of nodes ? – Suresh Venkat May 27 2010 at 1:09 good point, I guess q2 is what I'd like know more about. – Deniz May 27 2010 at 3:53 ## 1 Answer If graph is acyclic you can use some sort of dynamic programming. Let $a_{u,k}$ be the best distance you can get if you start from vertex $u \in G$ and consider only $k$ last vertices of your given path. It's quite straightforward how to calculate all $a_{u,k}$ based on all values of $a_{u',k'}$ with $u'$ "after" $u$ and $k' < k$: you just iterate over all edges going from vertex $u$. This approach works in $O(|G| \times strlen)$ time. -

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http://mathoverflow.net/revisions/68849/list

## Return to Answer 2 added 29 characters in body Oh sure, this is quite well-known. The closure of an element is the smallest closed element which is greater than or equal to the given element. Dually for the interior operator. In general, a closure operator $\phi: P \to P$ on a poset $P$ is an order-preserving, inflationary ($x \leq \phi(x)$), idempotent ($\phi(\phi(x)) = \phi(x)$) operation, and if you want a topological closure operator, then you demand $\phi(x \vee y) = \phi(x) \vee \phi(y)$ as well. Alternatively, a closure operator can be specified by an inclusion $i: C \hookrightarrow P$ such that every $p \in P$ has a least upper bound $c \in C$. The assignment $p \mapsto c$ gives an order-preserving mapping $j: P \to C$ with the property $$j(q) \leq d \Leftrightarrow q \leq i(d)$$ (where the left side is to be read in $C$ and the right in $P$), and the closure operator $\phi$ is the composite $i \circ j$; it is easy to check the order-preserving inflationary idempotent properties. In the context of Boolean algebras $P$, as in Sikorski, you demand that $C$ be closed under joins as well (and that $i$ preserve them), to make $\phi$ a topological closure. I am writing this answer to bear out a far wider connection with category theory. The posets here are special cases of categories, where there is at most one arrow in any hom-set $\hom(x, y)$, which we write as $x \leq y$. Functors between poset categories amount to order-preserving maps. A closure operator amounts to a monad on a poset. The subcollection $i: C \subseteq P$ of closed elements for that operator is the category of algebras for that monad. The corresponding mapping $j: P \to C$ is the left adjoint to $i$. Any adjoint pair (known under another name as a Galois connection) between posets always induces a closure operator. And so on. There is an embarrassment of riches of references (i.e., so many that it's hard to think of one that stands out). But a trade secret among category theorists is to understand what general concepts mean in the simplified case of poset categories, and this point of view is explicitly declared in Paul Taylor's Practical Foundations of Mathematics. I think just about any introductory book on category theory will refer to this, though, and you will also find it used heavily in more specialized treatises like Johnstone's Stone Spaces that involve looking at lattices from a categorical point of view. 1 Oh sure, this is quite well-known. The closure of an element is the smallest closed element which is greater than or equal to the given element. Dually for the interior operator. In general, a closure operator $\phi: P \to P$ on a poset $P$ is an order-preserving, inflationary ($x \leq \phi(x)$), idempotent ($\phi(\phi(x)) = \phi(x)$) operation, and if you want a topological closure operator, then you demand $\phi(x \vee y) = \phi(x) \vee \phi(y)$ as well. Alternatively, a closure operator can be specified by an inclusion $i: C \hookrightarrow P$ such that every $p \in P$ has a least upper bound $c \in C$. The assignment $p \mapsto c$ gives an order-preserving mapping $j: P \to C$ with the property $$j(q) \leq d \Leftrightarrow q \leq i(d)$$ (where the left side is to be read in $C$ and the right in $P$), and the closure operator $\phi$ is the composite $i \circ j$; it is easy to check the order-preserving inflationary idempotent properties. In the context of Boolean algebras $P$, as in Sikorski, you demand that $C$ be closed under joins as well, to make $\phi$ a topological closure. I am writing this answer to bear out a far wider connection with category theory. The posets here are special cases of categories, where there is at most one arrow in any hom-set $\hom(x, y)$, which we write as $x \leq y$. Functors between poset categories amount to order-preserving maps. A closure operator amounts to a monad on a poset. The subcollection $i: C \subseteq P$ of closed elements for that operator is the category of algebras for that monad. The corresponding mapping $j: P \to C$ is the left adjoint to $i$. Any adjoint pair (known under another name as a Galois connection) between posets always induces a closure operator. And so on. There is an embarrassment of riches of references (i.e., so many that it's hard to think of one that stands out). But a trade secret among category theorists is to understand what general concepts mean in the simplified case of poset categories, and this point of view is explicitly declared in Paul Taylor's Practical Foundations of Mathematics. I think just about any introductory book on category theory will refer to this, though, and you will also find it used heavily in more specialized treatises like Johnstone's Stone Spaces that involve looking at lattices from a categorical point of view.

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http://pediaview.com/openpedia/Quadrilateral

# Quadrilateral Quadrilateral Six different types of quadrilaterals Edges and vertices 4 Schläfli symbol {4} (for square) Area various methods; see below Internal angle (degrees) 90° (for square) In Euclidean plane geometry, a quadrilateral is a polygon with four sides (or edges) and four vertices or corners. Sometimes, the term quadrangle is used, by analogy with triangle, and sometimes tetragon for consistency with pentagon (5-sided), hexagon (6-sided) and so on. The origin of the word "quadrilateral" is the two Latin words quadri, a variant of four, and latus, meaning "side." Quadrilaterals are simple (not self-intersecting) or complex (self-intersecting), also called crossed. Simple quadrilaterals are either convex or concave. The interior angles of a simple (and planar) quadrilateral ABCD add up to 360 degrees of arc, that is $\angle A+\angle B+\angle C+\angle D=360^{\circ}.$ This is a special case of the n-gon interior angle sum formula (n − 2) × 180°. In a crossed quadrilateral, the four interior angles on either side of the crossing add up to 720°.[1] All convex quadrilaterals tile the plane by repeated rotation around the midpoints of their edges. ## Convex quadrilaterals – parallelograms[] Euler diagram of some types of quadrilaterals. (UK) denotes British English and (US) denotes American English. A parallelogram is a quadrilateral with two pairs of parallel sides. Equivalent conditions are that opposite sides are of equal length; that opposite angles are equal; or that the diagonals bisect each other. Parallelograms also include the square, rectangle, rhombus and rhomboid. • Rhombus or rhomb: all four sides are of equal length. An equivalent condition is that the diagonals perpendicularly bisect each other. An informal description is "a pushed-over square" (including a square). • Rhomboid: a parallelogram in which adjacent sides are of unequal lengths and angles are oblique (not right angles). Informally: "a pushed-over rectangle with no right angles." • Rectangle: all four angles are right angles. An equivalent condition is that the diagonals bisect each other and are equal in length. Informally: "a box or oblong" (including a square). • Square (regular quadrilateral): all four sides are of equal length (equilateral), and all four angles are right angles. An equivalent condition is that opposite sides are parallel (a square is a parallelogram), that the diagonals perpendicularly bisect each other, and are of equal length. A quadrilateral is a square if and only if it is both a rhombus and a rectangle (four equal sides and four equal angles). • Oblong: a term sometimes used to denote a rectangle which has unequal adjacent sides (i.e. a rectangle that is not a square). ## Convex quadrilaterals – other[] • Kite: two pairs of adjacent sides are of equal length. This implies that one diagonal divides the kite into congruent triangles, and so the angles between the two pairs of equal sides are equal in measure. It also implies that the diagonals are perpendicular. • Right kite: a kite with two opposite right angles. • Trapezoid (North American English) or Trapezium (British English): at least one pair of opposite sides are parallel. • Trapezium (NAm.): no sides are parallel. (In British English this would be called an irregular quadrilateral, and was once called a trapezoid.) • Isosceles trapezoid (NAm.) or isosceles trapezium (Brit.): one pair of opposite sides are parallel and the base angles are equal in measure. Alternative definitions are a quadrilateral with an axis of symmetry bisecting one pair of opposite sides, or a trapezoid with diagonals of equal length. • Tangential trapezoid: a trapezoid where the four sides are tangents to an inscribed circle. • Tangential quadrilateral: the four sides are tangents to an inscribed circle. A convex quadrilateral is tangential if and only if opposite sides have equal sums. • Cyclic quadrilateral: the four vertices lie on a circumscribed circle. A convex quadrilateral is cyclic if and only if opposite angles sum to 180°. • Bicentric quadrilateral: it is both tangential and cyclic. • Orthodiagonal quadrilateral: the diagonals cross at right angles. • Equidiagonal quadrilateral: the diagonals are of equal length. • Ex-tangential quadrilateral: the four extensions of the sides are tangent to an excircle. ## More quadrilaterals[] • An equilic quadrilateral has two opposite equal sides that, when extended, meet at 60°. • A Watt quadrilateral is a quadrilateral with a pair of opposite sides of equal length.[2] • A quadric quadrilateral is a convex quadrilateral whose four vertices all lie on the perimeter of a square.[3] • A geometric chevron (dart or arrowhead) is a concave quadrilateral with bilateral symmetry like a kite, but one interior angle is reflex. • A self-intersecting quadrilateral is called variously a cross-quadrilateral, crossed quadrilateral, butterfly quadrilateral or bow-tie quadrilateral. A special case of crossed quadrilaterals are the antiparallelograms, crossed quadrilaterals in which (like a parallelogram) each pair of nonadjacent sides has equal length. The diagonals of a crossed or concave quadrilateral do not intersect inside the shape. • A non-planar quadrilateral is called a skew quadrilateral. Formulas to compute its dihedral angles from the edge lengths and the angle between two adjacent edges were derived for work on the properties of molecules such as cyclobutane that contain a "puckered" ring of four atoms.[4] See skew polygon for more. Historically the term gauche quadrilateral was also used to mean a skew quadrilateral.[5] A skew quadrilateral together with its diagonals form a (possibly non-regular) tetrahedron, and conversely every skew quadrilateral comes from a tetrahedron where a pair of opposite edges is removed. ## Special line segments[] The two diagonals of a convex quadrilateral are the line segments that connect opposite vertices. The two bimedians of a convex quadrilateral are the line segments that connect the midpoints of opposite sides.[6] They intersect at the "vertex centroid" of the quadrilateral (see Remarkable points below). The four maltitudes of a convex quadrilateral are the perpendiculars to a side through the midpoint of the opposite side.[7] ## Notations in metric formulas[] In the metric formulas below, the following notations are used. A convex quadrilateral ABCD has the sides a = AB, b = BC, c = CD, and d = DA. The diagonals are p = AC and q = BD, and the angle between them is θ. The semiperimeter s is defined as $s=\tfrac{1}{2}(a+b+c+d)$. ## Area of a convex quadrilateral[] There are various general formulas for the area K of a convex quadrilateral. ### Trigonometric formulas[] The area can be expressed in trigonometric terms as $K = \tfrac{1}{2} pq \cdot \sin \theta,$ where the lengths of the diagonals are p and q and the angle between them is θ.[8] In the case of an orthodiagonal quadrilateral (e.g. rhombus, square, and kite), this formula reduces to $K=\tfrac{1}{2}pq$ since θ is 90°. Bretschneider's formula[9] expresses the area in terms of the sides and two opposite angles: $\begin{align} K &= \sqrt{(s-a)(s-b)(s-c)(s-d) - \tfrac{1}{2} abcd \; [ 1 + \cos (A + C) ]} \\ &= \sqrt{(s-a)(s-b)(s-c)(s-d) - abcd \left[ \cos^2 \left( \tfrac{A + C}{2} \right) \right]} \\ \end{align}$ where the sides in sequence are a, b, c, d, where s is the semiperimeter, and A and C are two (in fact, any two) opposite angles. This reduces to Brahmagupta's formula for the area of a cyclic quadrilateral when A+C = 180°. Another area formula in terms of the sides and angles, with angle C being between sides b and c, and A being between sides a and d, is $K = \tfrac{1}{2}ad \cdot \sin{A} + \tfrac{1}{2}bc \cdot \sin{C}.$ In the case of a cyclic quadrilateral, the latter formula becomes $K = \tfrac{1}{2}(ad+bc)\sin{A}.$ In a parallelogram, where both pairs of opposite sides and angles are equal, this formula reduces to $K=ab \cdot \sin{A}.$ Alternatively, we can write the area in terms of the sides and the intersection angle θ of the diagonals, so long as this angle is not 90°:[10] $K = \frac{|\tan \theta|}{4} \cdot \left| a^2 + c^2 - b^2 - d^2 \right|.$ In the case of a parallelogram, the latter formula becomes $K = \tfrac{1}{2}|\tan \theta|\cdot \left| a^2 - b^2 \right|.$ Another area formula including the sides a, b, c, d is[11] $K=\tfrac{1}{4}\sqrt{(2(a^2+c^2)-4x^2)(2(b^2+d^2)-4x^2)}\sin{\varphi}$ where x is the distance between the midpoints of the diagonals and φ is the angle between the bimedians. ### Non-trigonometric formulas[] The following two formulas expresses the area in terms of the sides a, b, c, d, the semiperimeter s, and the diagonals p, q: $K = \sqrt{(s-a)(s-b)(s-c)(s-d) - \tfrac{1}{4}(ac+bd+pq)(ac+bd-pq)},$ [12] $K = \frac{1}{4} \sqrt{4p^{2}q^{2}- \left( a^{2}+c^{2}-b^{2}-d^{2} \right) ^{2}}.$ [13] The first reduces to Brahmagupta's formula in the cyclic quadrilateral case, since then pq = ac + bd. The area can also be expressed in terms of the bimedians m, n and the diagonals p, q: $K=\tfrac{1}{2}\sqrt{(m+n+p)(m+n-p)(m+n+q)(m+n-q)},$ [14] $K=\tfrac{1}{2}\sqrt{p^2q^2-(m^2-n^2)^2}.$ [15] ### Vector formulas[] The area of a quadrilateral ABCD can be calculated using vectors. Let vectors AC and BD form the diagonals from A to C and from B to D. The area of the quadrilateral is then $K = \tfrac{1}{2} |\mathbf{AC}\times\mathbf{BD}|,$ which is half the magnitude of the cross product of vectors AC and BD. In two-dimensional Euclidean space, expressing vector AC as a free vector in Cartesian space equal to (x1,y1) and BD as (x2,y2), this can be rewritten as: $K = \tfrac{1}{2} |x_1 y_2 - x_2 y_1|.$ ### Area inequalities[] If a convex quadrilateral has the consecutive sides a, b, c, d and the diagonals p, q, then its area K satisfies[16] $K\le \tfrac{1}{4}(a+c)(b+d)$ with equality only for a rectangle. $K\le \tfrac{1}{4}(a^2+b^2+c^2+d^2)$ with equality only for a square. $K\le \tfrac{1}{4}(p^2+q^2)$ with equality only if the diagonals are perpendicular and equal. $K\le \tfrac{1}{2}\sqrt{(a^2+c^2)(b^2+d^2)}$ with equality only for a rectangle.[11] From Bretschneider's formula it directly follows that the area of a quadrilateral satisfies $K \le \sqrt{(s-a)(s-b)(s-c)(s-d)}$ with equality if and only if the quadrilateral is cyclic or degenerate such that one side is equal to the sum of the other three (it has collapsed into a line segment, so the area is zero). The area of any quadrilateral also satisfies the inequality[17] $\displaystyle K\le \tfrac{1}{2}\sqrt[3]{(ab+cd)(ac+bd)(ad+bc)}.$ ## Diagonals[] ### Properties of the diagonals in some quadrilaterals[] In the following table it is listed if the diagonals in some of the most basic quadrilaterals bisect each other, if their diagonals are perpendicular, and if their diagonals have equal length.[18] The list applies to the most general cases, and excludes named subsets. Quadrilateral Bisecting diagonals Perpendicular diagonals Equal diagonals Trapezoid No See note 1 No Isosceles trapezoid No See note 1 Yes Parallelogram Yes No No Kite See note 2 Yes See note 2 Rectangle Yes No Yes Rhombus Yes Yes No Square Yes Yes Yes Note 1: The most general trapezoids and isosceles trapezoids do not have perpendicular diagonals, but there are infinite numbers of (non-similar) trapezoids and isosceles trapezoids that do have perpendicular diagonals and are not any other named quadrilateral. Note 2: In a kite, one diagonal bisects the other. The most general kite has unequal diagonals, but there is an infinite number of (non-similar) kites in which the diagonals are equal in length (and the kites are not any other named quadrilateral). ### Length of the diagonals[] The length of the diagonals in a convex quadrilateral ABCD can be calculated using the law of cosines. Thus $p=\sqrt{a^2+b^2-2ab\cos{B}}=\sqrt{c^2+d^2-2cd\cos{D}}$ and $q=\sqrt{a^2+d^2-2ad\cos{A}}=\sqrt{b^2+c^2-2bc\cos{C}}.$ Other, more symmetric formulas for the length of the diagonals, are[19] $p=\sqrt{\frac{(ac+bd)(ad+bc)-2abcd(\cos{B}+\cos{D})}{ab+cd}}$ and $q=\sqrt{\frac{(ab+cd)(ac+bd)-2abcd(\cos{A}+\cos{C})}{ad+bc}}.$ ### Generalizations of the parallelogram law and Ptolemy's theorem[] In any convex quadrilateral ABCD, the sum of the squares of the four sides is equal to the sum of the squares of the two diagonals plus four times the square of the line segment connecting the midpoints of the diagonals. Thus $a^2 + b^2 + c^2 + d^2 = p^2 + q^2 + 4x^2$ where x is the distance between the midpoints of the diagonals.[20]:p.126 This is sometimes known as Euler's quadrilateral theorem and is a generalization of the parallelogram law. A corollary is the inequality $a^2 + b^2 + c^2 + d^2 \ge p^2 + q^2$ where equality holds if and only if the quadrilateral is a parallelogram. Euler also generalized Ptolemy's theorem, which is an equality in a cyclic quadrilateral, into an inequality for a convex quadrilateral. It states that $pq \le ac + bd$ where there is equality if and only if the quadrilateral is cyclic.[20]:p.128–129 This is often called Ptolemy's inequality. The German mathematician Carl Anton Bretschneider derived in 1842 the following generalization of Ptolemy's theorem, regarding the product of the diagonals in a convex quadrilateral[21] $p^2q^2=a^2c^2+b^2d^2-2abcd\cos{(A+C)}.$ This relation can be considered to be a law of cosines for a quadrilateral. In a cyclic quadrilateral, where A + C = 180°, it reduces to pq = ac + bd. Since cos (A + C) ≥ -1, it also gives a proof of Ptolemy's inequality. ### Other metric relations[] If X and Y are the feet of the normals from B and D to the diagonal AC = p in a convex quadrilateral ABCD with sides a = AB, b = BC, c = CD, d = DA, then[22]:p.14 $XY=\frac{|a^2+c^2-b^2-d^2|}{2p}.$ In a convex quadrilateral ABCD with sides a = AB, b = BC, c = CD, d = DA, and where the diagonals intersect at E, $efgh(a+c+b+d)(a+c-b-d) = (agh+cef+beh+dfg)(agh+cef-beh-dfg)$ where e = AE, f = BE, g = CE, and h = DE.[23] The shape of a convex quadrilateral is fully determined by the lengths of its sides in sequence and of one diagonal between two specified vertices. The two diagonals p, q and the four side lengths a, b, c, d of a quadrilateral are related[24] by the Cayley-Menger determinant, as follows: $\det \begin{bmatrix} 0 & a^2 & p^2 & d^2 & 1 \\ a^2 & 0 & b^2 & q^2 & 1 \\ p^2 & b^2 & 0 & c^2 & 1 \\ d^2 & q^2 & c^2 & 0 & 1 \\ 1 & 1 & 1 & 1 & 0 \end{bmatrix} = 0.$ ## Bimedians[] The Varignon parallelogram EFGH The midpoints of the sides of any quadrilateral (convex, concave or crossed) are the vertices of a parallelogram called the Varignon parallelogram. It has the following properties: • Each pair of opposite sides of the Varignon parallelogram are parallel to a diagonal in the original quadrilateral. • The length of a side in the Varignon parallelogram is half as long as the diagonal in the original quadrilateral it is parallel to. • The area of the Varignon parallelogram equals half the area of the original quadrilateral. This is true in convex, concave and crossed quadrilaterals provided the area of the latter is defined to be the difference of the areas of the two triangles it is composed of.[25] • The perimeter of the Varignon parallelogram equals the sum of the diagonals of the original quadrilateral. The diagonals of the Varignon parallelogram are the bimedians of the original quadrilateral. The two bimedians in a quadrilateral and the line segment joining the midpoints of the diagonals in that quadrilateral are concurrent and are all bisected by their point of intersection.[20]:p.125 In a convex quadrilateral with sides a, b, c and d, the length of the bimedian that connects the midpoints of the sides a and c is $m=\tfrac{1}{2}\sqrt{-a^2+b^2-c^2+d^2+p^2+q^2}$ where p and q are the length of the diagonals.[26] The length of the bimedian that connects the midpoints of the sides b and d is $n=\tfrac{1}{2}\sqrt{a^2-b^2+c^2-d^2+p^2+q^2}.$ Hence[20]:p.126 $\displaystyle p^2+q^2=2(m^2+n^2).$ This is also a corollary to the parallelogram law applied in the Varignon parallelogram. The length of the bimedians can also be expressed in terms of two opposite sides and the distance x between the midpoints of the diagonals. This is possible when using Euler's quadrilateral theorem in the above formulas. Whence[15] $m=\tfrac{1}{2}\sqrt{2(b^2+d^2)-4x^2}$ and $n=\tfrac{1}{2}\sqrt{2(a^2+c^2)-4x^2}.$ Note that the two opposite sides in these formulas are not the two that the bimedian connects. In a convex quadrilateral, there are the following dual connection between the bimedians and the diagonals:[22] • The two bimedians have equal length if and only if the two diagonals are perpendicular • The two bimedians are perpendicular if and only if the two diagonals have equal length ## Trigonometric identities[] The four angles of a simple quadrilateral ABCD satisfy the following identities:[27] $\sin{A}+\sin{B}+\sin{C}+\sin{D}=4\sin{\frac{A+B}{2}}\sin{\frac{A+C}{2}}\sin{\frac{A+D}{2}}$ and $\frac{\tan{A}\tan{B}-\tan{C}\tan{D}}{\tan{A}\tan{C}-\tan{B}\tan{D}}=\frac{\tan{(A+C)}}{\tan{(A+B)}}.$ Also,[28] $\frac{\tan{A}+\tan{B}+\tan{C}+\tan{D}}{\cot{A}+\cot{B}+\cot{C}+\cot{D}}=\tan{A}\tan{B}\tan{C}\tan{D}.$ In the last two formulas, no angle is allowed to be a right angle, since then the tangent functions are not defined. ## Maximum and minimum properties[] Among all quadrilaterals with a given perimeter, the one with the largest area is the square. This is called the isoperimetric theorem for quadrilaterals. It is a direct consequence of the area inequality[17]:p.114 $K\le \tfrac{1}{16}L^2$ where K is the area of a convex quadrilateral with perimeter L. Equality holds if and only if the quadrilateral is a square. The dual theorem states that of all quadrilaterals with a given area, the square has the shortest perimeter. The quadrilateral with given side lengths that has the maximum area is the cyclic quadrilateral.[29] Of all convex quadrilaterals with given diagonals, the orthodiagonal quadrilateral has the largest area.[17]:p.119 This is a direct consequence of the fact that the area of a convex quadrilateral satisfies $K=\tfrac{1}{2}pq\sin{\theta}\le \tfrac{1}{2}pq,$ where θ is the angle between the diagonals p and q. Equality holds if and only if θ = 90°. If P is an interior point in a convex quadrilateral ABCD, then $AP+BP+CP+DP\ge AC+BD.$ From this inequality it follows that the point inside a quadrilateral that minimizes the sum of distances to the vertices is the intersection of the diagonals. Hence that point is the Fermat point of a convex quadrilateral.[30]:p.120 ## Remarkable points and lines in a convex quadrilateral[] The centre of a quadrilateral can be defined in several different ways. The "vertex centroid" comes from considering the quadrilateral as being empty but having equal masses at its vertices. The "side centroid" comes from considering the sides to have constant mass per unit length. The usual centre, called just centroid (centre of area) comes from considering the surface of the quadrilateral as having constant density. These three points are in general not all the same point.[31] The "vertex centroid" is the intersection of the two bimedians.[32] As with any polygon, the x and y coordinates of the vertex centroid are the arithmetic means of the x and y coordinates of the vertices. The "area centroid" of quadrilateral ABCD can be constructed in the following way. Let Ga, Gb, Gc, Gd be the centroids of triangles BCD, ACD, ABD, ABC respectively. Then the "area centroid" is the intersection of the lines GaGc and GbGd.[33] In a general convex quadrilateral ABCD, there are no natural analogies to the circumcenter and orthocenter of a triangle. But two such points can be constructed in the following way. Let Oa, Ob, Oc, Od be the circumcenters of triangles BCD, ACD, ABD, ABC respectively; and denote by Ha, Hb, Hc, Hd the orthocenters in the same triangles. Then the intersection of the lines OaOc and ObOd is called the quasicircumcenter; and the intersection of the lines HaHc and HbHd is called the quasiorthocenter of the convex quadrilateral.[33] These points can be used to define an Euler line of a quadrilateral. In a convex quadrilateral, the quasiorthocenter H, the "area centroid" G, and the quasicircumcenter O are collinear in this order, and HG = 2GO.[33] There can also be defined a quasinine-point center E as the intersection of the lines EaEc and EbEd, where Ea, Eb, Ec, Ed are the nine-point centers of triangles BCD, ACD, ABD, ABC respectively. Then E is the midpoint of OH.[33] Another remarkable line in a convex quadrilateral is the Newton line. ## Other properties of convex quadrilaterals[] • Let exterior squares be drawn on all sides of a quadrilateral. The segments connecting the centers of opposite squares are (a) equal in length, and (b) perpendicular. Thus these centers are the vertices of an orthodiagonal quadrilateral. This is called Van Aubel's theorem. • The internal angle bisectors of a convex quadrilateral either form a cyclic quadrilateral[20]:p.127 or they are concurrent. In the latter case the quadrilateral is a tangential quadrilateral. • For any simple quadrilateral with given edge lengths, there is a cyclic quadrilateral with the same edge lengths.[29] • The four smaller triangles formed by the diagonals and sides of a convex quadrilateral have the property that the product of the areas of two opposite triangles equals the product of the areas of the other two triangles.[34] ## Taxonomy[] A taxonomy of quadrilaterals is illustrated by the following graph. Lower forms are special cases of higher forms. Note that "trapezium" here is referring to the British definition (the North American equivalent is a trapezoid), and "kite" excludes the concave kite (arrowhead or dart). Inclusive definitions are used throughout. ## References[] 1. G. Keady, P. Scales and S. Z. Németh, "Watt Linkages and Quadrilaterals", The Mathematical Gazette Vol. 88, No. 513 (Nov., 2004), pp. 475–492. 2. A. K. Jobbings, "Quadric Quadrilaterals", The Mathematical Gazette Vol. 81, No. 491 (Jul., 1997), pp. 220–224. 3. M.P. Barnett and J.F. Capitani, Modular chemical geometry and symbolic calculation, International Journal of Quantum Chemistry, 106 (1) 215–227, 2006. 4. William Rowan Hamilton, , Proceedings of the Royal Irish Academy, 4 (1850), pp. 380–387. 5. E.W. Weisstein. "Bimedian". MathWorld - A Wolfram Web Resource. 6. E.W. Weisstein. "Maltitude". MathWorld - A Wolfram Web Resource. 7. Harries, J. "Area of a quadrilateral," Mathematical Gazette 86, July 2002, 310–311. 8. R. A. Johnson, Advanced Euclidean Geometry, 2007, Dover Publ., p. 82. 9. Mitchell, Douglas W., "The area of a quadrilateral," Mathematical Gazette 93, July 2009, 306–309. 10. ^ a b Josefsson, Martin (2013), "Five Proofs of an Area Characterization of Rectangles", Forum Geometricorum 13: 17–21. 11. J. L. Coolidge, "A historically interesting formula for the area of a quadrilateral", American Mathematical Monthly, 46 (1939) 345–347. 12. E.W. Weisstein. "Bretschneider's formula". MathWorld - A Wolfram Web Resource. 13. Archibald, R. C., "The Area of a Quadrilateral", American Mathematical Monthly, 29 (1922) pp. 29–36. 14. ^ a b Josefsson, Martin (2011), "The Area of a Bicentric Quadrilateral", Forum Geometricorum 11: 155–164. 15. O. Bottema, Geometric Inequalities, Wolters-Noordhoff Publishing, The Netherlands, 1969, pp. 129, 132. 16. ^ a b c Alsina, Claudi; Nelsen, Roger (2009), When Less is More: Visualizing Basic Inequalities, Mathematical Association of America, p. 68. 17. Rashid, M. A. & Ajibade, A. O., "Two conditions for a quadrilateral to be cyclic expressed in terms of the lengths of its sides", Int. J. Math. Educ. Sci. Technol., vol. 34 (2003) no. 5, pp. 739–799. 18. Altshiller-Court, Nathan, College Geometry, Dover Publ., 2007. 19. Andreescu, Titu & Andrica, Dorian, Complex Numbers from A to...Z, Birkhäuser, 2006, pp. 207–209. 20. ^ a b Josefsson, Martin (2012), "Characterizations of Orthodiagonal Quadrilaterals", Forum Geometricorum 12: 13–25. 21. Hoehn, Larry (2011), "A New Formula Concerning the Diagonals and Sides of a Quadrilateral", Forum Geometricorum 11: 211–212. 22. E.W. Weisstein. "Quadrilateral". MathWorld - A Wolfram Web Resource. 23. H. S. M. Coxeter and S. L. Greitzer, Geometry Revisited, MAA, 1967, pp. 52-53. 24. C. V. Durell & A. Robson, Advanced Trigonometry, Dover, 2003, p. 267. 25. MathPro Press, "Original Problems Proposed by Stanley Rabinowitz 1963–2005", p. 23, [2] 26. ^ a b Thomas Peter, "Maximizing the Area of a Quadrilateral", The College Mathematics Journal, Vol. 34, No. 4 (September 2003), pp. 315–316. 27. Alsina, Claudi and Nelsen, Roger, Charming Proofs. A Journey Into Elegant Mathematics, Mathematical Association of America, 2010, pp. 114, 119, 120, 261. 28. King, James, Two Centers of Mass of a Quadrilateral, [3], Accessed 2012-04-15. 29. Honsberger, Ross, Episodes in Nineteenth and Twentieth Century Euclidean Geometry, Math. Assoc. Amer., 1995, pp. 35–41. 30. ^ a b c d Myakishev, Alexei (2006), "On Two Remarkable Lines Related to a Quadrilateral", Forum Geometricorum 6: 289–295. 31. Martin Josefsson, "Characterizations of Trapezoids", Forum Geometricorum 13 (2013) 23–35. ## Source Content is authored by an open community of volunteers and is not produced by or in any way affiliated with ore reviewed by PediaView.com. Licensed under the Creative Commons Attribution-ShareAlike 3.0 Unported License, using material from the Wikipedia article on "Quadrilateral", which is available in its original form here: http://en.wikipedia.org/w/index.php?title=Quadrilateral • ## Finding More You are currently browsing the the PediaView.com open source encyclopedia. Please select from the menu above or use our search box at the top of the page. • ## Questions or Comments? If you have a question or comment about material in the open source encyclopedia supplement, we encourage you to read and follow the original source URL given near the bottom of each article. You may also get in touch directly with the original material provider. This open source encyclopedia supplement is brought to you by PediaView.com, the web's easiest resource for using Wikipedia content. All Wikipedia text is available under the terms of the Creative Commons Attribution-ShareAlike 3.0 Unported License. Wikipedia® itself is a registered trademark of the Wikimedia Foundation, Inc.

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http://www.nag.com/numeric/CL/nagdoc_cl23/html/S/s21bcc.html

NAG Library Function Documentnag_elliptic_integral_rd (s21bcc) 1 Purpose nag_elliptic_integral_rd (s21bcc) returns a value of the symmetrised elliptic integral of the second kind. 2 Specification #include <nag.h> #include <nags.h> double nag_elliptic_integral_rd (double x, double y, double z, NagError *fail) 3 Description nag_elliptic_integral_rd (s21bcc) calculates an approximate value for the integral $R D x,y,z = 3 2 ∫ 0 ∞ dt t+x t+y t+z 3$ where $x$, $y\ge 0$, at most one of $x$ and $y$ is zero, and $z>0$. The basic algorithm, which is due to Carlson (1979) and Carlson (1988), is to reduce the arguments recursively towards their mean by the rule: $x 0 = x , y 0 = y , z 0 = z μ n = x n + y n + 3 z n / 5 X n = 1 - x n / μ n Y n = 1 - y n / μ n Z n = 1 - z n / μ n λ n = x n y n + y n z n + z n x n x n+1 = x n + λ n / 4 y n+1 = y n + λ n / 4 z n+1 = z n + λ n / 4$ For $n$ sufficiently large, $ε n = max X n , Y n , Z n ∼ 1 / 4 n$ and the function may be approximated adequately by a 5th-order power series $R D x,y,z = 3 ∑ m=0 n-1 4 -m z m + λ n z m + 4 -n μ n 3 1 + 3 7 S n 2 + 1 3 S n 3 + 3 22 S n 2 2 + 3 11 S n 4 + 3 13 S n 2 S n 3 + 3 13 S n 5 ,$ where ${S}_{n}^{\left(m\right)}=\left({X}_{n}^{m}+{Y}_{n}^{m}+3{Z}_{n}^{m}\right)/2m$. The truncation error in this expansion is bounded by $3{\epsilon }_{n}^{6}/\sqrt{{\left(1-{\epsilon }_{n}\right)}^{3}}$ and the recursive process is terminated when this quantity is negligible compared with the machine precision. The function may fail either because it has been called with arguments outside the domain of definition, or with arguments so extreme that there is an unavoidable danger of setting underflow or overflow. Note: ${R}_{D}\left(x,x,x\right)={x}^{-3/2}\text{,}$ so there exists a region of extreme arguments for which the function value is not representable. . 4 References Abramowitz M and Stegun I A (1972) Handbook of Mathematical Functions (3rd Edition) Dover Publications Carlson B C (1979) Computing elliptic integrals by duplication Numerische Mathematik 33 1–16 Carlson B C (1988) A table of elliptic integrals of the third kind Math. Comput. 51 267–280 5 Arguments 1: x – doubleInput 2: y – doubleInput 3: z – doubleInput On entry: the arguments $x$, $y$ and $z$ of the function. Constraint: x, ${\mathbf{y}}\ge 0.0$, ${\mathbf{z}}>0.0$ and only one of x and y may be zero. 4: fail – NagError *Input/Output The NAG error argument (see Section 3.6 in the Essential Introduction). 6 Error Indicators and Warnings NE_REAL_ARG_EQ On entry, ${\mathbf{x}}+{\mathbf{y}}$ must not be equal to 0.0: ${\mathbf{x}}+{\mathbf{y}}=〈\mathit{\text{value}}〉$. Both x and y are zero and the function is undefined. NE_REAL_ARG_GE On entry, x must not be greater than or equal to $〈\mathit{\text{value}}〉$: ${\mathbf{x}}=〈\mathit{\text{value}}〉$. On entry, y must not be greater than or equal to $〈\mathit{\text{value}}〉$: ${\mathbf{y}}=〈\mathit{\text{value}}〉$. On entry, z must not be greater than or equal to $〈\mathit{\text{value}}〉$: ${\mathbf{z}}=〈\mathit{\text{value}}〉$. There is a danger of setting underflow and the function returns zero. NE_REAL_ARG_LE On entry, z must not be less than or equal to 0.0: ${\mathbf{z}}=〈\mathit{\text{value}}〉$. The function is undefined. NE_REAL_ARG_LT On entry, either z is too close to zero or both x and y are too close to zero: there is a danger of setting overflow. On entry, $〈\mathit{parameters}〉$ must not be less than $〈\mathit{\text{value}}〉$: $〈\mathit{parameters}〉=〈\mathit{\text{value}}〉$. On entry, x must not be less than 0.0: ${\mathbf{x}}=〈\mathit{\text{value}}〉$. On entry, y must not be less than 0.0: ${\mathbf{y}}=〈\mathit{\text{value}}〉$. The function is undefined. 7 Accuracy In principle the function is capable of producing full machine precision. However, round-off errors in internal arithmetic will result in slight loss of accuracy. This loss should never be excessive as the algorithm does not involve any significant amplification of round-off error. It is reasonable to assume that the result is accurate to within a small multiple of the machine precision. 8 Further Comments Symmetrised elliptic integrals returned by functions nag_elliptic_integral_rd (s21bcc), nag_elliptic_integral_rc (s21bac), nag_elliptic_integral_rf (s21bbc) and nag_elliptic_integral_rj (s21bdc) can be related to the more traditional canonical forms (see Abramowitz and Stegun (1972)), as described in the s Chapter Introduction. 9 Example This example program simply generates a small set of nonextreme arguments which are used with the function to produce the table of low accuracy results. 9.1 Program Text Program Text (s21bcce.c) None. 9.3 Program Results Program Results (s21bcce.r)

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http://physics.stackexchange.com/questions/44209/ground-state-energy-e-0-and-evaluation-of-physical-energies/44215

# Ground state energy $E_{0}$ and evaluation of physical energies Given the lowest eigenvalue $E_0$ of an Schrödinguer operator, do the other energies $E_{n}$ for $n >0$ depend strongly on the lowest eigenvalue of the system? I mean, if we somehow fixed the eigenvalue $E_{0}$, could we get more or at least better approximations to the other eigenenergies of the system? thanks - ## 1 Answer No. The differences between the eigenvalues (giving the spectral frequencies) are highly specific for each chemical substance, so knowledge of $E_0$ tells very little about the remainder of the spectrum. -

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http://www.sciforums.com/showthread.php?113574-Has-James-Gates-Discovered-Computer-Code-in-String-Theory-Equations&p=2936937

• Forum • New Posts • FAQ • Calendar • Ban List • Community • Forum Actions • Encyclopedia • What's New? 1. If this is your first visit, be sure to check out the FAQ by clicking the link above. You need to register and post an introductory thread before you can post to all subforums: click the register link above to proceed. To start viewing messages, select the forum that you want to visit from the selection below. # Thread: 1. ## Has James Gates Discovered Computer Code in String Theory Equations? Is physical reality ultimately constructed of computer code? http://www.transcend.ws/?p=3020 Dr. S. James Gates, Jr., a theoretical physicist, the John S. Toll Professor of Physics at the University of Maryland, and the Director of The Center for String & Particle Theory, is reporting that certain string theory, super-symmetrical equations, which describe the fundamental nature of the Universe and reality, contain embedded computer codes. These codes are digital data in the form of 1′s and 0′s. Not only that, these codes are the same as what make web browsers work and are error-correction codes! Gates says, “We have no idea what these ‘things’ are doing there”. http://being.publicradio.org/program...sofpower.shtml Physicists have long sought to describe the universe in terms of equations. Now, James Gates explains how research on a class of geometric symbols known as adinkras could lead to fresh insights into the theory of supersymmetry — and perhaps even the very nature of reality. [...] However, with the observation that structures from information theory — codes — control the structure of equations with the SUSY property, we may be crossing a barrier. I know of no other example of this particular intermingling occurring at such a deep level. Could it be that codes, in some deep and fundamental way, control the structure of our reality? In asking this question, we may be ending our "treasure hunt" in a place that was anticipated previously by at least one pioneering physicist: John Archibald Wheeler. [...] As for my own collaboration on adinkras, the path my colleagues and I have trod since the early 2000s has led me to conclude that codes play a previously unsuspected role in equations that possess the property of supersymmetry. This unsuspected connection suggests that these codes may be ubiquitous in nature, and could even be embedded in the essence of reality. If this is the case, we might have something in common with the Matrix science-fiction films, which depict a world where everything human beings experience is the product of a virtual-reality-generating computer network. http://arxiv.org/abs/0806.0051 Which leads to the spooky idea that our universe is a computer simulation... http://www.simulation-argument.com/simulation.html http://rationalwiki.org/wiki/Simulation_argument 2. Originally Posted by khan Which leads to the spooky idea that our universe is a computer simulation... Most Cosmologists acknowledge the plausible (yet not probable) possibility that we exist in a computer-generated universe. 3. Originally Posted by Xotica Most Cosmologists acknowledge the plausible (yet not probable) possibility that we exist in a computer-generated universe. I don't know why you're singling out cosmologists because all scientists would admit we cannot disprove that notion but that's a long way from saying it's plausible. Atheists acknowledge they (we) cannot prove there is no deity or deities but that doesn't mean we consider the notion to be plausible. As for the work in question he's dealing with spinor structures, specifically spin 1/2 structures so you get binary representations naturally. This $\pm \frac{1}{2}$ structure can be written as 0,1 structure instead, so it's not like binary representations are unknown things in physics. In this case it happens the particular spinor constructs he's considering also can be put in a graph structure. Again, this is not a new concept, using graphs to represent decompositions of algebras is a centuries old procedure, known to anyone studying theoretical physics, particularly gauge theory. In this case the graph structure happens to be equivalent to error correcting code. It's important not to fall into the trap of confirmation bias. As I said, there's many many years of physics involving graph structures, spinors, decompositions and symmetries. It's not terribly shocking that at some point someone comes across a structure which has been seen elsewhere. Saying "Wow, this is clear justification for the Matrix view of the universe!" is a bit like me asking someone to think of a number between 1 and 1000, trying to guess it and when I finally guess right after many failed tries I declare it evidence I'm telepathic. Part of being a good mathematician is noticing structures which link seemingly different concepts. Realising "Oh, if I just write this like that then it becomes a well known expression whose solutions are known" is something every mathematician will do at some point in their research. In this case it's connected error correcting code with string theory and so people have started reading all sorts of things from it. If this is reason to consider the universe as just a computer simulation can we therefore use the same reasoning to say the thousands of things not related to computer programming in string theory are pieces of evidence against that notion? Because if we can't then you're making a logical fallacy. 4. Originally Posted by AlphaNumeric I don't know why you're singling out cosmologists because all scientists would admit we cannot disprove that notion but that's a long way from saying it's plausible. plau·si·ble adj \ˈplȯ-zə-bəl\ Definition of PLAUSIBLE 1 :superficially fair, reasonable, or valuable but often specious The notion is indeed plausible, albeit fantastical. 5. Math and reality ...what is the connection? http://arxiv.org/abs/0704.0646 I explore physics implications of the External Reality Hypothesis (ERH) that there exists an external physical reality completely independent of us humans. I argue that with a sufficiently broad definition of mathematics, it implies the Mathematical Universe Hypothesis (MUH) that our physical world is an abstract mathematical structure. I discuss various implications of the ERH and MUH, ranging from standard physics topics like symmetries, irreducible representations, units, free parameters, randomness and initial conditions to broader issues like consciousness, parallel universes and Godel incompleteness. I hypothesize that only computable and decidable (in Godel's sense) structures exist, which alleviates the cosmological measure problem and help explain why our physical laws appear so simple. I also comment on the intimate relation between mathematical structures, computations, simulations and physical systems. 6. How many scientists have claimed the universe computes? What does it mean? Does it just underline that any information we record (or observe) is the result of some physical process which can be argued is like an algorithm? Why do scientists balk at the idea of the universe (and all the information in it) being algorithmic? Or why do they mostly take the view that such a claim is somehow "lowering" the status of physical theories, to "just a computer simulation", isn't that like saying "the universe is just a universe"? Well, yeah, I guess that's what it is, just a universe. Maybe only just . . . It seems a bit contrary: every physical equation I've ever seen is arguably algorithmic, it describes what happens to a given input, or what some physical system does to an input--it either gives an output or perhaps goes into a nonterminating loop of some kind. 7. It depends on how far you take it. Someone like Wolfram (of Mathematica fame) believes the universe is, at it's most basic, a cellular automaton which is computing the passage of time. This is a somewhat extreme view lacking justification. On the other hand the rules of the universe do seem to be consistent and allow for systems to be set up and then just allowed to evolve under said rules to then spit out some 'answer' to a question encoded within the initial conditions. This is precisely what a computer does, we set up all the electrons and silicon to compute a new state from which we extract an answer. Using this we can even compute the answer to some set of equations which we model the universe (or some part of it) by, so you could argue that the universe itself computes the exact answer to such questions as it evolves in time. The reason a lot of scientists don't like phrasing it that way is that it seems to imply there's some kind of intent behind the universe doing what it does, ie what or who is asking the questions or that the universe isn't just doing whatever it is it does but actually 'computing' something. For example, when I jump in the air I don't have to compute the acceleration imparted on me by my leg muscles so I can compute how high to rise, it just happens. I can indeed make a machine or some algorithm for me to do in my head which will compute those answers but the universe doesn't compute "You will rise 0.442 metres in the air", things just bounce off one another, interact, exchange energy and momentum etc and that's it. So while anyone doing physics is obviously implementing an algorithm to allow them to determine the outcome of some set of dynamics based on initial conditions saying the universe computes is saying something else and it's that scientists don't like the connotations of. 8. Originally Posted by AlphaNumeric For example, when I jump in the air I don't have to compute the acceleration imparted on me by my leg muscles so I can compute how high to rise, it just happens. I can indeed make a machine or some algorithm for me to do in my head which will compute those answers but the universe doesn't compute "You will rise 0.442 metres in the air", things just bounce off one another, interact, exchange energy and momentum etc and that's it. You're missing something here, I think. You're implying that "compute" how high you rise when you jump in the air means "return an answer in metres" say, but despite that your height above the ground still changes. The change (whatever you want to call it, or however you measure it) is the output. That is, the height your body rises above the ground is the output "computed" by the energy you impart to it, via leg muscles, a bungy cord or whatever. You know your height changes because you experience the change, so you therefore "measure" the output, you don't need to specify it in metres or any other standard. So the universe does compute that you will rise in the air; that it doesn't "tell" you what that is in some arbitrary measurement standard is clearly irrelevant. But: Criticism The critics of digital physics—including physicists[citation needed] who work in quantum mechanics—object to it on several grounds. Physical symmetries are continuous One objection is that extant models of digital physics are incompatible[citation needed] with the existence of several continuous characters of physical symmetries, e.g., rotational symmetry, translational symmetry, Lorentz symmetry, and electroweak symmetry, all central to current physical theory. Proponents of digital physics claim that such continuous symmetries are only convenient (and very good) approximations of a discrete reality. For example, the reasoning leading to systems of natural units and the conclusion that the Planck length is a minimum meaningful unit of distance suggests that at some level space itself is quantized.[29] Locality Some argue[citation needed] that extant models of digital physics violate various postulates of quantum physics. For example, if these models are not grounded in Hilbert spaces and probabilities, they belong to the class of theories with local hidden variables that some deem ruled out experimentally using Bell's theorem. This criticism has two possible answers. First, any notion of locality in the digital model does not necessarily have to correspond to locality formulated in the usual way in the emergent spacetime. A concrete example of this case was recently given by Lee Smolin.[30] Another possibility is a well-known loophole in Bell's theorem known as superdeterminism (sometimes referred to as predeterminism).[31] In a completely deterministic model, the experimenter's decision to measure certain components of the spins is predetermined. Thus, the assumption that the experimenter could have decided to measure different components of the spins than he actually did is, strictly speaking, not true. Physical theory requires the continuum It has been argued[weasel words] that digital physics, grounded in the theory of finite state machines and hence discrete mathematics, cannot do justice to a physical theory whose mathematics requires the real numbers, which is the case for all physical theories having any credibility. But computers can manipulate and solve formulas describing real numbers using symbolic computation, thus avoiding the need to approximate real numbers by using an infinite number of digits. Before symbolic computation, a number—in particular a real number, one with an infinite number of digits—was said to be computable if a Turing machine will continue to spit out digits endlessly. In other words, there is no "last digit". But this sits uncomfortably with any proposal that the universe is the output of a virtual-reality exercise carried out in real time (or any plausible kind of time). Known physical laws (including quantum mechanics and its continuous spectra) are very much infused with real numbers and the mathematics of the continuum. "So ordinary computational descriptions do not have a cardinality of states and state space trajectories that is sufficient for them to map onto ordinary mathematical descriptions of natural systems. Thus, from the point of view of strict mathematical description, the thesis that everything is a computing system in this second sense cannot be supported".[32] For his part, David Deutsch generally takes a "multiverse" view to the question of continuous vs. discrete. In short, he thinks that “within each universe all observable quantities are discrete, but the multiverse as a whole is a continuum. When the equations of quantum theory describe a continuous but not-directly-observable transition between two values of a discrete quantity, what they are telling us is that the transition does not take place entirely within one universe. So perhaps the price of continuous motion is not an infinity of consecutive actions, but an infinity of concurrent actions taking place across the multiverse.” January, 2001 The Discrete and the Continuous, an abridged version of which appeared in The Times Higher Education Supplement. --http://en.wikipedia.org/wiki/Digital_physics 9. I agree with AlphaNumeric, I think that it is a coincidence. Sort of like the Bible codes found in the Bible, that it correlates to known codes doesn't mean that it was intentionally put there or that they even function the way they function in computers. If they do function as error-correcting methods in reality then you might be on to something though but I don't see from the quote that this is the case. Another explanation could be that the theory is, in fact, man-made and structures could have been put there that are borrowed from other fields (such as the technology of computers), that a theory works to describe reality doesn't mean that all the structures within the theory has to apply to structures in reality. In fact, if the theory auto-corrects itself through those structures then it could be a indication that it is false. 10. mathematics it can prove anything it wants so is string theory the of everything ? 11. Originally Posted by AlphaNumeric I don't know why you're singling out cosmologists because all scientists would admit we cannot disprove that notion but that's a long way from saying it's plausible. Atheists acknowledge they (we) cannot prove there is no deity or deities but that doesn't mean we consider the notion to be plausible. As for the work in question he's dealing with spinor structures, specifically spin 1/2 structures so you get binary representations naturally. This $\pm \frac{1}{2}$ structure can be written as 0,1 structure instead, so it's not like binary representations are unknown things in physics. In this case it happens the particular spinor constructs he's considering also can be put in a graph structure. Again, this is not a new concept, using graphs to represent decompositions of algebras is a centuries old procedure, known to anyone studying theoretical physics, particularly gauge theory. In this case the graph structure happens to be equivalent to error correcting code. It's important not to fall into the trap of confirmation bias. As I said, there's many many years of physics involving graph structures, spinors, decompositions and symmetries. It's not terribly shocking that at some point someone comes across a structure which has been seen elsewhere. Saying "Wow, this is clear justification for the Matrix view of the universe!" is a bit like me asking someone to think of a number between 1 and 1000, trying to guess it and when I finally guess right after many failed tries I declare it evidence I'm telepathic. Part of being a good mathematician is noticing structures which link seemingly different concepts. Realising "Oh, if I just write this like that then it becomes a well known expression whose solutions are known" is something every mathematician will do at some point in their research. In this case it's connected error correcting code with string theory and so people have started reading all sorts of things from it. If this is reason to consider the universe as just a computer simulation can we therefore use the same reasoning to say the thousands of things not related to computer programming in string theory are pieces of evidence against that notion? Because if we can't then you're making a logical fallacy. Well said, Alphanumeric. It still is fundamentally remarkable that our mathematical observation mimic the microcosms we create. The question which is more genuinely remarkable (in my view). Is this observer bias? Or is our fundamental human experience merely a replica of our physical reality? Both solution's are remarkable, but for greatly different reasons. #### Posting Permissions • You may not post new threads • You may not post replies • You may not post attachments • You may not edit your posts • • BB code is On • Smilies are On • [IMG] code is On • [VIDEO] code is On • HTML code is Off Forum Rules All times are GMT -5. The time now is 06:54 PM. sciforums.com

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http://physics.stackexchange.com/questions/29285/rotational-speed-of-a-discus/29287

# Rotational speed of a discus I was wondering whether the rotational speed of a discus has any influence on the flight of the discus. Would slowing the rotation or speeding it up change the trajectory in any way or would the flight simply become unstable when slowing down? - ## 3 Answers Allegedly the rotation has two effects. I say "allegedly" because although I was told this in physics lectures at university I'm not sure if anyone has ever rigorously proved it. Anyhow, with the disclaimer behind me, the first effect is that the angular momentum stabilises the angle of the discus as it travels through the air. That allows the angle of attack to be maintained at the optimum value and hence increases the lift and therefore the range. The second effect is that a high rotational speed makes the boundary layer turbulent, and this decreases aerodynamic drag and once again increases the range. Given this I guess higher rotational speed is better, though presumably an athlete is limited to how high a rotational speed they can generate without compromising the speed they can throw at. - + This reminds me of a related issue - how does a frisbee work - but in that case there's more of an airfoil effect, I would think. – Mike Dunlavey May 31 '12 at 20:21 I take it then that accelerating the angular moment influences only the distance and stability of the flight, but I cannot "steer" the discus to the left or the right by accelerating or slowing down the angular moment, right? – cdecker May 31 '12 at 20:46 @cdecker: don't forget, it's a gyroscope, and gyroscopes precess. So if there is a force to cause it to "pitch up", that will make it roll right or left, and vice versa. Which way it goes depends on the direction and rate of spin. – Mike Dunlavey Jun 1 '12 at 0:36 The faster it spins, the greater the aerodynamic side force on it; see Magnus effect. Also, higher rotation increases the $\mu$ (ratio of edge speed relative to body to airspeed of the body) of the disc; the higher airspeed of the advancing edge relative to the retreating edge creates asymmetric lift & drag. The former would impart a rolling moment, while the latter would impart a moment opposing the in-plane rotation of the discus. All that said, I doubt either are particularly significant effects. - 1 Simply on one side the airspeed is higher than the other side, affecting the lift and drag on each side. This causes the discus to drift or slice. – ja72 Jun 1 '12 at 1:55 Given the physical conditions, this seems like an appropriate explanation: The faster the discus rotates, the more violently and quickly it displaces the air around it. Now the absence or scarcity of air causes a reduction in air friction or viscosity around the discus and this allows it to move onward in the direction of propulsion; now that depends on what angle the athlete projects it. After a certain distance there begins a constant deceleration of rotational speed because at some point, the air friction starts overpowering the rotation and this results in the discus entering the second half of its trajectory, i.e., moves downward along a curved path. I hope that answers your question. - Maybe I'm not used to the lingo, but "violently and quickly it displaces the air" and "absence or scarcity of air" doesn't sound like the kind of aerodynamics I'm used to. Want to give it another shot? – Mike Dunlavey May 31 '12 at 20:25 What I was trying to convey by the use of the quoted phrases was that Bernoulli's Principle comes into play in this situation. – Graviton Jun 1 '12 at 6:53 Let me suggest alternate wording, then you do what you want: The disk is an airfoil whose orientation is gyroscopically stabilized. As it follows its (nominally parabolic) arc, as it goes into the descending portion of the arc, its angle of attack increases. The rotational speed could have an effect of delaying the onset of aerodynamic stall, thus lengthening the trajectory. It also could have increased lift on the forward-moving side, causing precession that could also affect the angle of attack. – Mike Dunlavey Jun 1 '12 at 13:22

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http://physics.stackexchange.com/questions/45644/can-bosons-that-are-composed-of-several-fermions-occupy-the-same-state/45645

Can bosons that are composed of several fermions occupy the same state? It is generally assumed that there is no limit on how many bosons are allowed to occupy the same quantum mechanical state. However, almost every boson encountered in every-day physics is not a fundamental particle (with the photon being the most prominent exception). They are instead composed of a number of fermions, which can not occupy the same state. Is it possible for more than one of these composite bosons to be in the same state even though their constituents are not allowed to be in the same state? If the answer is "yes", how does this not contradict the more fundamental viewpoint considering fermions? - 2 Answers This is a nice puzzle--- but the answer is simple: the composite bosons can occupy the same state when the state is spatially delocalized on a scale larger than the scale of the wavefunction of the fermions inside, but they feel a repulsive force which prevents them from being at the same spatial point, so that they cannot sit at the same point at the same time. The potential energy of this force is always greater than the excitation energy of the composite system, so if you force the bosons to sit at the same point, you will excite one of them, so that the composing fermions are no longer in the same state, and the two particles become distinguishable. The scale for this effective repulsion is the decay-length of the wavefunction of the composing fermions, and this repulsion is what leads matter to feel hard. The reason you haven't heard this is somewhat political--- there are people who say that the exclusion principle is not the cause of the repulsive contact forces in ordinary matter, that this force is electrostatic, and despite this being ridiculously false, nobody wants to get into the mud and argue with them. So people don't explain the fermionic exclusion principle forces properly. If you have a two-fermion composite which is net bosonic, like a H atom with a proton nucleus and spin-polarized electron, when you bring the H-atoms close, the energy of the electronic ground state is the effective Hamiltonian potential energy for the nuclei. When the nuclei are close enough so that the electronic wavefunctions have appreciable overlap, you get a strong repulsion. You can see that this repulsion is pure Pauli, because if the electrons have opposite spins, you don't get repulsion at short distances, you get attraction, and the result is that you form an H2 molecule of the two H atoms. You can see this exclusion force emerge in an exactly solvable toy model. Consider a 1d line with two attractive unit delta function pontetials at positions a and -a, each with a fermion attached in the ground state. Each one has an independent ground state wavefunction that has the shape $exp(-|x|)$, but when the two are together at separation 2a, the two states are deformed, and the ground state energy for the fermions goes up. The effect is quadratic in the separation, because the ground state (one fermion) goes down in energy, and the first excited state goes up in energy, and to leading order in perturbations, the two are cancelling when both states are occupied. To next leading order, the effect is positive potential energy, a repulsion. This potential is the effective potnetial of the two delta functions when you make them dynamical instead of fixed. The maximum value of the repulsive potential in this model is exactly where the model breaks down, which is at a=1. At this point, the ground state is exp(-2x) to the left of -1, constat between the two delta functions, then exp(2x) to the right, with energy -2, and the first excited state is constant to the left of -1, a straight line from -1 to 1, and constant past 1, with energy 0. The result is a net energy of -1 unit. This is half the binding energy of the two separated delta functions, which is -2. This effect is the exclusion repulsion, and it reconciles the fermionic substructure with the net bosonic behavior of the particle. You can only see the substructure when the wavefunction of the boson is concentrated enough to have appreciable overlap on the scale of the composing fermion wavefunctions, and this is why you need high energies to probe the compositeness of the Higgs (or for that matter, the alpha particle). To get the wavefunctions to sit at the same point to this accuracy, you need to localize them at high energy. - You state that "if you force the bosons to sit at the same point, you will excite one of them". There is the problem I am having. When the two bosons are in the same state, they are on the same point, in the sense that their wavefunctions are the same and therefore overlap entirely. It all comes down to this: If the wavefunctions of two bosons are the same, the wavefunctions of their constituents should pairwise be the same, which is impossible. What am I missing? – Friedrich Dec 3 '12 at 0:10 @Friedrich: Not exactly--- the wavefunction is only exactly the same for two bosons if they are non-interacting. If they have a repulsion at ultra-short distances, the wavefunction is entangled so that it is very close to the product of the wavefunction with itself at most points, but the entanglement zeros out the diagonal (so they aren't ever at the same point). This is a property of higher dimensional multi-particle wavefunctions. The best definition for the condensate wavefunction by giving the effective field for the boson a VEV, and this only produces a product state for a free field. – Ron Maimon Dec 3 '12 at 3:26 Ron, how can the volume of the wavefunction of a composite boson be larger than the volume of the wavefunction of its fermions? – lurscher Dec 4 '12 at 19:44 @lurscher: The volume of the wavefunction of the fermions is in the dimension of the relative coordinate, while the volume of the composite is in the center of mass coordinate. – Ron Maimon Dec 4 '12 at 23:37 @Ron: I hope you will return safely in four days. If two interacting bosons can not be in a product state, is it even meaningful to speak about them being in "the same state"? – Friedrich Dec 31 '12 at 1:39 show 2 more comments Yes, they can, an experimental example of that is Bose-Einstein Condensate of fermions. And that is possible because actually they will have the same wave function, in sense that nature no more capable of making any distinguish between them. Regarding everyday life, actually saying that it is bosonic is just a formal statement, in sense that Pauli exclusion principle not working here, but that not because the composite things are bosons, but because in everyday life there is almost nothing in the same state, because accomplishing that is very hard, and if done, it will reproduce Bose-Einstein condensate. -

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http://mathhelpforum.com/differential-geometry/86622-integrable-functions.html

Thread: 1. Integrable Functions Let f be a function on the interval [a, b] such that for every epsilon > 0 there exist integrable functions g, h on [a, b] with h(x) <= f(x) <= g(x) and ((integral from a to b g(x)dx) - (integral from a to b h(x)dx)) < epsilon. Show that f is integrable on [a, b]. Any ideas on where to start and/or how I would go about showing this? Thanks. 2. Riemann integrable? Assuming that, take some partition of [a, b] and look at the Riemann sums for g, f, and h. You should be able to show that [itex]R(g)\le R(f)\le r(h)[/tex] (R(f) is the Riemann sum of f for the particular partition). "((integral from a to b g(x)dx) - (integral from a to b h(x)dx)) < epsilon" is a bit peculiar since the two integrals are specific numbers. That really just says the integrals of g and h are the same. That is that the lim, as the number of intervals in the partition goes to infinity, of the Riemann sums, R(g) and R(h), are the same so the limit of R(f) exists and is that joint value. 3. Originally Posted by kjwill1776 Let f be a function on the interval [a, b] such that for every epsilon > 0 there exist integrable functions g, h on [a, b] with h(x) <= f(x) <= g(x) and ((integral from a to b g(x)dx) - (integral from a to b h(x)dx)) < epsilon. Show that f is integrable on [a, b]. Any ideas on where to start and/or how I would go about showing this? Thanks. $\int h \leq \int_{-} f \leq \int g$ and $\int h \leq \int^{-} f \leq \int g.$ thus $0 \leq \int^{-}f - \int_{-}f \leq \int g - \int h < \epsilon.$ since $\epsilon > 0$ is arbitrary, we must have $\int^{-}f - \int_{-}f=0.$

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http://crypto.stackexchange.com/questions/1767/proving-knowledge-of-a-preimage-of-a-hash-without-disclosing-it/1771

Proving knowledge of a preimage of a hash without disclosing it? We consider a public hash function $H$, assumed collision-resistant and preimage-resistant (for both first and second preimage), similar in construction to SHA-1 or SHA-256. Alice discloses a value $h$, claiming that she (or/and parties she can communicate with or/and devices they have access to) knows a message $m$ such that $H(m)=h$. Can some protocol convince Bob of this claim without help of a third party/device that Bob trusts, nor allowing Bob to find $m$? At Crypto 98 rump session, Hal Finney made a 7-minutes presentation A zero-knowledge proof of possession of a pre-image of a SHA-1 hash which seems to be intended for that. This remarkable result is occasionally stated as fact, including recently here and next door. But I do not get how it is supposed to work. Update: This talk mentions using the protocol in the Crypto'98 paper of Ronald Cramer and Ivan B. Damgård: Zero-Knowledge Proofs for Finite Field Arithmetic or: Can Zero-Knowledge be for Free? (this freely downloadable version is very similar, or there is this earlier, longer version). - I could not find any article that supported his talk in the rump session. Do you have an access to any such article. I would be very much interested in that because it sounds like a challenging problem especially when one sees hash function in the random oracle model. Maybe, then I could be of some help. – Jalaj Jan 28 '12 at 17:58 1 I went beyond what was in the talk. I think that considering a RO-model and solving this problem will be a theoretically great result. – Jalaj Jan 29 '12 at 2:51 My current intuition is that no practical protocol can rightly convince Bob of Alice's claim, for that very protocol would be able to distinguish $H$ from a random oracle, something that was never done for SHA-1 (restricted to messages of constant length). I reason that in the random oracle model, the only way for Bob to be convinced that the oracle outputs $h$ for message $m$ is to submit it himself, which implies knowledge of $m$. – fgrieu Jan 29 '12 at 9:38 1 The protocol can allow Bob to distinguish $H$ from a random oracle, but it requires Alice's input. I don't think that contradicts indistinguishability. For example, Elgamal is IND-CPA-secure but also admits proofs of plaintext knowledge. Alice could allow Bob to distinguish ciphertexts he couldn't on his own. In any case, no implemented hash function can act as a random oracle in the real world. – PulpSpy Jan 30 '12 at 2:02 2 The hash design needs to be known so both parties can follow the progress through the circuit. I don't think we can always have something real-world practical (it isn't claimed in the talk that we can; only that we can for SHA-1). It depends on how well the function lends itself to certain tricks for speeding up the proof. What he did was as much art as science in converting SHA-1 into a blend of boolean/arithmetic circuitry. – PulpSpy Jan 30 '12 at 15:12 show 2 more comments 2 Answers I'm not sure what I can add that wasn't covered in the talk. The approach is that Alice commits to the preimage and sends the commitment to Bob. The commitment has homomorphic properties, meaning it is possible to do computation on the value. For example, if Alice commits to $x$ and $y$, Bob may be able to compute a commitment to $z$ where $z=f(x,y)$ for some function $f$. For example, if the commitments are additively homomorphic, Bob can add two commitments or multiply by a constant for free. Alternatively, $f$ may not be directly computable. In this case, Alice who knows the actual values computes $z$, sends a commitment to $z$ to Bob asserting that it is correct. Bob holds commitments to $x$, $y$, and $z$; as well as knowing $f$. Alice can then interactively prove that for $f$, the commitment containing $z$ does contain the correct output for the inputs contained in the commitments to $x$ and $y$. The Cramer-Damgaard paper shows how to do these proofs for a simple Turning-complete set of both boolean and arithmetic gates (for example, NAND and modular addition/multiplication). The size of the circuit implementing SHA-1, expressed with only NAND gates for example, will be very large and infeasible. The art to doing the proof in practice is breaking it into subprotocols that are best represented by either boolean or arithmetic circuits and switching between the proof systems as appropriate. For certain SHA operations, he works at the bit level with boolean operations and for other operations, he works with integers in a finite field. - Do you feel confident that the answer to the question is a clear "yes, using the techniques in the Cramer-Damgård paper"? Or something weaker? I get that there is no evidence of impossibility. – fgrieu Jan 30 '12 at 12:54 2 An unbounded Alice can always prove. A bounded Alice can only prove when the circuit complexity of the hash has the same bound. A real-life Alice can only prove if she can generate a relatively simple description of the circuit with operations that have efficient zero-knowledge proofs. The latter is the case with SHA-1, but necessarily other hash functions. – PulpSpy Jan 30 '12 at 15:08 I guess the above is intended to read "but not necessarily other hash functions", and otherwise is supportive that a practical protocol can be devised in the case of SHA-1 as claimed in the talk. I'm still a tad skeptical, on these (admittedly weak) arguments: 1) such a protocol is something we can't devise in general for a perfect hash, yet we are computationally unable to distinguish SHA-1 (with fixed message size) from a perfect hash, other than by the fact that it is SHA-1; 2) details on the talk's method are scarce; 3) this remarkable result does not appear to have been reproduced. – fgrieu Jan 30 '12 at 20:10 @fgrieu: 3) Do you have use cases where this construction would be useful (and more desirable than a ZKPoK of a plaintext)? – bob Oct 13 '12 at 19:49 @bob: to be honest, no I do not have a use case. My interest at that point is purely theoretical. – fgrieu Oct 15 '12 at 11:14 I wanted to comment on PulpSpy's answer, but my comment turned out to be too large! I have an intuitive understand why they does not stick to the idea of using Boolean circuits alone for all the proofs which I am writing it down here. I might be wrong and I would like to be cross examined on this. There will be an issue with the boolean and arithmetic circuits. I view this from a linear algebra point of view. We can write the boolean circuit in form of a corresponding matrix that performs the transformation. Since every matrix will be linear transformation, there are known lower bounds which are of form $\Omega(n^2/r^c)$, where $n$ is the input size, $c$ is an arbitrary constant, and $r$ is the size of the biggest sub-matrix that is linear dependent. Since SHA-1 does mixing very well, I am sure we won't be able to reject even a constant fraction of input for any small (in matrix form, a submatrix of large size, comparable to $n$). This further implies that computing the circuit will require a lot of computation one that we are not willing to take. I feel this argument works well for most of the candidate hash functions that uses linear arithmetic circuit to do the mixing. -

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http://mathhelpforum.com/advanced-algebra/192784-irreducible-polynomials-over-ring-integers.html

Thread: 1. Irreducible polynomials over ring of integers ? Is it true that polynomials of the form : $f_n= x^n+x^{n-1}+\cdots+x^{k+1}+ax^k+ax^{k-1}+\cdots+a$ where $\gcd(n+1,k+1)=1$ , $a\in \mathbb{Z^{+}}$ , $a$ is odd number , $a>1$, and $a_1\neq 1$ are irreducible over the ring of integers $\mathbb{Z}$? Note that general form of $f_n$ is : $f_n=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$ , so condition $a_1 \neq 1$ is equivalent to the condition $k \geq 1$ . Also polynomial can be rewritten into form : $f_n=\frac{x^{n+1}+(a-1)x^{k+1}-a}{x-1}$ Eisenstein's criterion , Cohn's criterion , and Perron's criterion cannot be applied to the polynomials of this form. Example : The polynomial $x^4+x^3+x^2+3x+3$ is irreducible over the integers but none of the criteria above can be applied on this polynomial. 2. Re: Irreducible polynomials over ring of integers ? Originally Posted by princeps Is it true that polynomials of the form : $f_n= x^n+x^{n-1}+\cdots+x^{k+1}+ax^k+ax^{k-1}+\cdots+a$ where $\gcd(n+1,k+1)=1$ , $a\in \mathbb{Z^{+}}$ , $a$ is odd number , $a>1$, and $a_1\neq 1$ are irreducible over the ring of integers $\mathbb{Z}$? Note that general form of $f_n$ is : $f_n=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$ , so condition $a_1 \neq 1$ is equivalent to the condition $k \geq 1$ . Also polynomial can be rewritten into form : $f_n=\frac{x^{n+1}+(a-1)x^{k+1}-a}{x-1}$ Eisenstein's criterion , Cohn's criterion , and Perron's criterion cannot be applied to the polynomials of this form. Example : The polynomial $x^4+x^3+x^2+3x+3$ is irreducible over the integers but none of the criteria above can be applied on this polynomial. i don't know the answer to the general case but it's easy to prove that if $n = 2^m - 1, \ m \geq 2, \ 2 \mid k$ and $a \equiv 3 \mod 4,$ then $f$ is irreducible. to see this, note that the numbers $\binom{2^m}{i}, \ 0 < i < 2^m,$ and $a - 1$ are even and hence $f(x+1) = \frac{(x+1)^{n+1}+(a-1)(x+1)^{k+1}-a}{x}=x^{n} + a_{n-1}x^{n-1}+ \ldots +$ $a_1x + a_0,$ where all $a_j$ are even and $a_0 = 2^m + (k+1)(a-1) \not \equiv 0 \mod 4.$ thus $f(x+1),$ and so $f(x),$ is irreducible by the Eisenstein's criterion. 3. Re: Irreducible polynomials over ring of integers ? Originally Posted by NonCommAlg $f(x+1) = \frac{(x+1)^{n+1}+(a-1)(x+1)^{k+1}-a}{x}=x^{n} + a_{n-1}x^{n-1}+ \ldots +$ $a_1x + a_0,$ where all $a_j$ are even and $a_0 = 2^m + (k+1)(a-1) \not \equiv 0 \mod 4.$ There is condition in the text of the question $a_k=a_{k-1}=\ldots =a_1=a_0=a$ , where $a$ is an odd number so $a_j$ cannot be even number. 4. Re: Irreducible polynomials over ring of integers ? Originally Posted by princeps There is condition in the text of the question $a_k=a_{k-1}=\ldots =a_1=a_0=a$ , where $a$ is an odd number so $a_j$ cannot be even number. in my solution, $a_j$ are the coefficients of $f(x+1)$ not $f(x)$. 5. Re: Irreducible polynomials over ring of integers ? Can you give me an example of f(x+1) ?

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http://math.stackexchange.com/questions/216049/lim-x-to-infty-fracxex-1x-1ex

# $\lim_{x \to \infty} \frac{xe^{x-1}}{(x-1)e^x}$ $$\lim_{x \to \infty} \frac{xe^{x-1}}{(x-1)e^x}$$ I don't know what to do. At all. I've read the explanations in my book at least a thousand times, but they're over my head. Oh, and I'm not allowed to use L'Hospital's rule. (I'm guessing it isn't needed for limits of this kind anyway. This one is supposedly simple - a beginners problem.) Most of the answers I've seen on the Internet simply says "use L'Hospital's rule". Any help really appreciated. I'm so frustrated right now... - What is $e^{x-1}/e^x$? – wj32 Oct 18 '12 at 2:50 Do you know any of the limit laws? Such as $\lim cf = c\lim f$? They are quite helpful in this case. – Raymond Cheng Oct 18 '12 at 2:53 ## 4 Answers Multiply the fraction by $1$ in the carefully chosen disguise $\dfrac{e^{-x}}{e^{-x}}$ and do a bit of algebra: $$\begin{align*} \lim_{x \to \infty} \frac{xe^{x-1}}{(x-1)e^x}&=\lim_{x\to\infty}\left(\frac{xe^{x-1}}{(x-1)e^x}\cdot\frac{e^{-x}}{e^{-x}}\right)\\ &=\lim_{x\to\infty}\frac{xe^{-1}}{x-1}\\ &=\lim_{x\to\infty}\frac{x}{e(x-1)}\\ &=\frac1e\lim_{x\to\infty}\frac{x}{x-1}\\ &=\frac1e\lim_{x\to\infty}\frac{x-1+1}{x-1}\\ &=\frac1e\lim_{x\to\infty}\left(1+\frac1{x-1}\right)\;. \end{align*}$$ That last limit really is easy. You may wonder how I came up with some of the steps. The very first one was simply recognizing that if I divided numerator and denominator by $e^x$, the resulting fraction would be a lot simpler, in that all of the exponentials would be gone. Pulling constant factors outside the limit is usually useful and almost never hurts. The simplification of $\frac{x}{x-1}$ could also have been achieved by doing a straightforward polynomial long division of $x-1$ into $x$, but the trick of subtracting and adding the same amount (here $1$) in order to get an expression that can be split in some nice way is a pretty common one that’s worth remembering. - Clearly, $$\frac{{x{e^{x - 1}}}}{{(x - 1){e^x}}} \sim \frac{{x{e^{x - 1}}}}{{x{e^x}}} = \frac{{{e^{x - 1}}}}{{{e^x}}}.$$ So ... - $$\lim_{x\to\infty}\frac{xe^{x-1}}{(x-1)e^x}=\lim_{x\to\infty}\frac{x}{x-1}\cdot\lim_{x\to\infty}\frac{e^{x-1}}{e^x}=1\cdot\frac{1}{e}=\frac{1}{e}$$ the first equality being justified by the fact that each of the right hand side limits exists finitely. - first downvote didn't come from me. I usually don't downvote neither questions, nor answers. – Chris's wise sister Oct 20 '12 at 5:42 THere seems to be a serial downvoter around here. I could almost have sworn it was you but, of course, I didn't since I'm not sure. I don't care either, to be honest, as in any case the serial downvotes are reversed back by the system and also because 2,4, or 40 points less or more make no real difference. – DonAntonio Oct 20 '12 at 17:20 honestly, if I wanted to downvote you then I'd have no problem telling you this direclty. I didn't downvote you this time. I downvote you ONLY when you downvote me. I do not like to downvote people (neither do I, nor my sister). – Chris's wise sister Oct 20 '12 at 17:25 then after I saw that you downvoted me I downvoted you. I mean that the first downvote didn't come from me. – Chris's wise sister Oct 20 '12 at 17:27 As before, I won't get into long discussions with you and I'll end this one after this message: I've no idea what you're talking about. I didn't downvote you, and I really don't care whether you downvoted me because you thought I did with you or because any other reason. I'm not interested in this kids game and please do feel as free as you want to downvote me: I don't care. Have a good day. – DonAntonio Oct 20 '12 at 18:25 show 1 more comment This is straightforward $$\lim_{x \to \infty} \frac{xe^{x-1}}{(x-1)e^x}=\lim_{x \to \infty} \frac{x}{(x-1)e}=\frac{1}{e}$$ (Chris) -

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http://sumidiot.wordpress.com/tag/approximation/

∑idiot's Blog The math fork of sumidiot.blogspot.com Posts Tagged ‘approximation’ Hardy and Wright, Chapter 11 (part 2) June 16, 2009 Today we finished off Chapter 11. We worked through some of the proofs, and discussing the meaning of some of theorems, and a good time was had by all. In 11.10 it is mentioned that there are some notable constant multiples besides $\sqrt{5}$ and $2\sqrt{2}$ in the bounding inequalities on approximating irrationals by rationals. However, the text doesn’t mention what they are, which I thought was unfortunate. I also wondered what sort of numbers are the troublesome examples for the constant $2\sqrt{2}$. That is, the troublesome number for $\sqrt{5}$ is the golden ratio (or anybody whose continued fraction ends in a string of 1s), so what numbers do it for $2\sqrt{2}$. I think we decided that probably it was not a single number, but more like… any number whose continued fraction expansion is just lots of 1s and 2s. The more ones, the worse the number, in some sense. But as long as there are infinitely many twos, maybe you start running into, or getting close to, this $2\sqrt{2}$ bound. We talked a little bit of our way through the proof that almost all numbers have arbitrarily large “quotients” (the $a_n$ in the continued fraction). I tried to dig up some memories from my reading of Khinchin’s book about how to picture some of the intervals and things in the proof. I have this pictures in my head of rectangles over the interval $[1/(n+1),1/n]$ of height the length of the interval (I guess that makes them squares, huh?). So the biggest rectangle is the one between 1/2 and 1, and they get smaller as you move left. Then each rectangle is split up again, this time with the rectangles getting smaller as you move to the right (within one of the first-stage rectangles). The first set of rectangles correspond, somehow, to the first term of continued fractions, and the second (smaller) rectangles correspond to the second term. Probably I should dig out that book and try to figure out what this picture actually says, but for now… that’s the picture I have in my head. We were all a little bit slow in understanding some of the later proofs about things like the discussion in 11.11: “Further theorems concerning approximations”. But we also didn’t seem interested enough to really dive in to it. In the section on simultaneous approximations, Eric mentioned that similar things are done in other contexts (like, perhaps, $p$-adics). When you have valuations, you prove a weak (single) and strong (simultaneous) theorem about approximations. While we were talking about it, I wondered if there was some analogy to the distinction between continuous (at each point in an interval) and uniformly continuous (on that interval). It seems like there maybe should be. Finally, we spent a while digging through the proof that $e$ is transcendental. Mostly because I was stubbornly refusing to believe I wasn’t being lied to throughout the proof. Setting $h^r=r!$ and then “plugging $h$ into” polynomials really made me uncomfortable. As we went, I joked about things not having any actual meaning. Eventually Chris and Eric pointed out that they do, actually, have meaning. This “plugging $h$ in” thing is actually giving you an integer (if your polynomial has integer coefficients). That calmed me down a bit. I still feel like I don’t understand the proof at all, and certainly couldn’t explain even an outline of it. Eric said similar things, but asked if we should have expected that somehow. Eric also mentioned that these sorts of formal manipulations with things that look wrong can sometimes be ok, and that it was something related to umbral calculus. He showed us an identity (Vandermonde’s) associated with binomial coefficients that does similar sorts of symbolic trickery. Which apparently I should now go read some more about. I had printed out a paper about the continued fraction expansion of $e$ (which maybe was pointed out to me in this comment), which talked about Pade approximations. Some of the things looked somewhat similar to what was going on in the proof that $e$ is transcendental (which the paper said was where they came from), but I couldn’t explain the paper well during out meeting (since I don’t understand it well enough), and we ran out of time. Tags:approximation, continued fraction, hardy and wright, number theory, transcendental, umbral calculus

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http://physics.stackexchange.com/questions/5815/does-string-theory-disagree-with-general-relativity

# Does String Theory disagree with General Relativity? I would like to expand on what I mean by the title of this question to focus the answers. Normally whenever a theory (e.g. General Relativity) replaces another (e.g. Newtonian Gravity) there is a correspondence requirement in some limit. However there is also normally some experimental area where the new larger theory makes predictions which are different from the older theory which made predictions of the same phenomena. This is ultimately because the newer theory has a deeper view of physics with its own structures which come into play in certain situations that the old theory didnt cover well. Additionally the newer theory will make predictions based on its novel aspects which the older theory did not consider. I know that String Theory is quite rich in this regard, but am not interested in that here. Nor am I concerned as to whether experiment has caught up, as I know that ST (and Quantum Gravity in general) is not easy to test. So for the GR to Newtonian example an answer to this question would be: bending of light rays; Mercury perhelion movement - GR had a different results to Newton. What would not count as an answer would be new structures which GR introduces like Black Holes or even gravitational curvature per se. So does ST have anything like Mercury perhelion movement waiting to be experimentally verified, and thus "improving" on GR within GR's own back yard? - ## 2 Answers String theory implies new physics in - and only in - the quantum regime. In particular, at distances that are very short - comparable to the string scale or Planck scale - there are new effects. The black holes decay (while they preserve the information), effective actions have higher-derivative terms, e.g. $R^2$, there are strings, branes, fluxes, extra dimensions, and so on. The new physics at the Planck scale - which is very far - is almost certainly not testable by the naive 19th century observations such as the Mercury perihelion's precession. This is not a problem of string theory in any sense: it is a tautological consequence of the questions that string theory addresses - namely the behavior of the Universe at the most fundamental scale - and any other theory that addresses the same questions inevitably shares the inaccessibility of the phenomena by direct tests. When you only look at the classical limit or classical physics, string theory exactly agrees with general relativity. In some sense, this is true even in the quantum regime: string theory is the only consistent quantum completion of general relativity. - 1 I think the higher derivative terms are a good answer to the question - if you had sensitive enough probes, even without using any new "stringy" physics, you'd be able to detect them. Also - they are a result of classical string theory ($\alpha'$ corrections). – user566 Feb 24 '11 at 19:36 1 Motl: So this is saying that String Theory doesnt replace General Relativity - in its predictive and testable aspects - even in principle, at the larger-than-Planck scale? – Roy Simpson Feb 24 '11 at 19:45 @Moshe: I dont understand what the higher derivative terms are doing in this answer, but in other theories higher derivative terms correspond to non-local effects. If "local" means Planck scale, then "non-local" could mean "greater than Planck scale" - perhaps much greater. Thus these effects would be measurable - was that your point? – Roy Simpson Feb 24 '11 at 20:26 1 @Roy: Higher derivative terms are not "non-local", they correspond to additional terms beyond the Einstein-Hilbert action, which modify in a small and specific way any solution of GR. One example is $R^2$ terms, which by dimensional analysis are suppressed by a power of high mass scale (string scale in this case). Those occur with calculable coefficients in string theory. Generally, GR occurs automatically at distances larger than Planck length in string theory, with corrections that are small in that regime or become more dramatic and qualitative at short distances. – user566 Feb 24 '11 at 20:32 1 @Raindrop yes, is the short answer. – anna v Feb 28 at 7:28 show 1 more comment The higher derivative terms are there just like in semiclassical gravity simply because when you introduce quantum mechanics, any term that is not forbidden by some symmetry principle or other physical criteria must be present, albeit Planck suppressed. So it's not s that urprising that string theory predicts them. The interesting thing is that there are additional objects beyond the usual spectrum of GR. Things like the dilaton, supersymmetric objects (gravitino et al) and so on. -

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http://physics.stackexchange.com/questions/31882/why-does-string-theory-require-9-dimensions-of-space-and-one-dimension-of-time?answertab=votes

# Why does string theory require 9 dimensions of space and one dimension of time? String theorists say that there are many more dimensions out there, but they are too small to be detected. 1. However, I do not understand why there are ten dimensions and not just any other number? 2. Also, if all the other dimensions are so coiled up in such a tiny space, how do we distinguish one dimension from the other? 3. If so, how do we define dimension? - 1 (1) It's all in the mathematics. (2) Can you distinguish your everyday 3 dimensions? Nope. So then there's no problem if the curled up ones are indistinguishable (not saying they are, though). [Atleast, I think it's this.] – Manishearth♦ Jul 12 '12 at 12:27 – Dilaton Jul 12 '12 at 13:53 – Dilaton Jul 12 '12 at 13:53 1 – Dilaton Jul 12 '12 at 13:54 – Rody Oldenhuis Jul 12 '12 at 18:58 ## 2 Answers One can posit mathematical string theories in any dimensions of any kind. However, I do not understand why there are ten dimensions and not just any other number? The specific dimensions arise from the requirements of the known physics encapsulated in the Standard Model and other data coming from particle physics, plus the requirement of General Relativity and its quantization. The Special Unitary groups whose representations accommodate the SM need at least these dimensions. There are models with more dimensions than this. Also, if all the other dimensions are so coiled up in such a tiny space, how do we distinguish one dimension from the other? We cannot move into the coiled ones, only in x.y,z. We do not need to distinguish them, as we do not distinguish the molecules in the air. The predictions from this type of theory on the behavior of particles is the only way of checking for their existence: consistency of theory with data. If so, how do we define dimension? A space variable ( centimeters) or time one ( seconds) that is continuous and maps the real numbers, each dimension at 90degrees to the rest, an extension of how we define normal x,y,z.That some are curled should not bother one. The coordinates over the earth are curled over the sphere's surface, for example, though the 90degree does not hold there. It would hold on the surface of a cylinder , z from -infinity to infinity, x from 0 to 2*r*pi. - 2 This answer is not accurate, you cannot formulate any sort of string theory in 60 dimensions, if you have too many spatial dimensions, there are too many degrees of freedom on the horizon. Regarding spherical coordinates, these are orthogonal. – Ron Maimon Jul 13 '12 at 1:45 – anna v Jul 13 '12 at 4:35 1 I see what you mean. But if you have ghosts in the theory, and divergent loop integrals, what does the mathematics mean? I agree that in all dimensions less than 10 and in all dimensions less than 26 you can formulate a fermionic/bosonic string theory (if you use a Polyakov noncritical string or linear dilaton), but in general I don't like to say this for dimensions higher than 26, because ghosts are not the same kind of problem qualitatively. BTW, only the poles are bad in spherical coordinates. – Ron Maimon Jul 13 '12 at 4:52 (1) String Theory is a very mathematical theory based on some natural assumptions, and this ends up relating Quantum Mechanics and General Relativity, as we want. Some of the equations in String Theory, however, have a proportionality constant $c$ in it, called the central charge. And when we manipulate these equations and set them equal to each other, we see that they ONLY make sense if $c=26$. This $c$ is the dimension of space that String Theory is a priori defined over, so now we see that we need 26 dimensions to not have absurdities... BUT that only made use of the bosonic particles in the world -- we forgot about fermions!! This is where Supersymmetry comes into play, and it throws in the fermions, and the equations are perturbed and leads to a new dimension of 10 for everything to make sense. (2) Just because we can't see it, doesn't mean it's not there... we can't see atoms with the eye, but we can use tools to see them... same thing happens here, our current technology can't see them, but we hope to change this in the future. EVEN BETTER though, is that the formula for gravitational force should actually be different because of these extra small dimensions -- thus we plan to figure these extra dimensions out by testing the gravitational force at small distances and seeing a perturbation to the standard inverse-square law of Newton. These extra dimensions are what is supposed to make gravity so weak compared to the other forces of nature. (3) a dimension is just a coordinate axis... so time is a dimension too. And just like your clock, this axis can repeat itself and not stretch to infinity. - I don't know how to make sense of compactified time, especially in string theory. Even if you compactify euclidean time, thermal string theory is hard to make sense of too, because gravity doesn't allow thermal ensembles of infinite extent. Regarding the central charge argument, it's fine, but it doesn't require 10 dimensions per-se, just an equivalent central charge, so you can have a non-geometric compactification. – Ron Maimon Jul 13 '12 at 1:42

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http://openwetware.org/index.php?title=User:Pranav_Rathi/Notebook/OT/2011/03/01/Laser_Shutter_.2&diff=next&oldid=657478

# User:Pranav Rathi/Notebook/OT/2011/03/01/Laser Shutter .2 ### From OpenWetWare (Difference between revisions) | | | | | |----------|-----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------|----------|-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------| | () | | () | | | Line 42: | | Line 42: | | | | | | | | | === Construction=== | | === Construction=== | | - | The buildup is divided is the following categories. In the start make sure that you have a good set of screw driver, players, wire cutter & stripper and solder. | + | The buildup is divided is the following categories. In the start make sure that you have a good set of screw driver, pliers, wire cutter, stripper and solder. | | | ==== Shutter==== | | ==== Shutter==== | | - | I will start with the choice of the motor. A 12V DC motor is a good choice because it can provide a wide rang of torques. The motor and the spring works against each other so it important to have the right set. So choose a 12V DC motor with a shaft length of at least 15mm. Mount the motor on the cage plate as shown in the picture, you will need 4/40 screws to tight it. once this is done, unscrew and take the motor out. | + | I will start with the choice of the motor. A 12V DC motor is a good choice because it can provide a wide rang of torques. The motor and the spring works against each other so it important to have the right set. So choose a 12V DC motor with a shaft length of at least 15mm. Mount the motor on the cage plate or platform as shown in the picture, you will need 4/40 screws to tight it (in present setup). once this is done, unscrew and take the motor out. | | | | | | | - | Now next task is to choose the right spring. The spring should be half the length of the shaft and the shaft should easily fit through the spring. I want to keep all this really simple, so there is a very easy way to choose the right spring. To choose the spring first need to know the torque of the motor. This is really simple, it can be mathematically calculate if the voltage and the current is known. But I like the experimental way; the setup is shown in the slide2. In the slide I have a DC motor with a leverage pivoted at the shaft and a free weight hanging with a string on the other side. In this, all I am doing is balancing the weigh at that point by giving the motor just enough power. | + | Now next task is to choose the right spring. The spring should be half the length of the shaft and the shaft should easily fit through the spring. I want to keep the spring-choice procedure really simple and easy. To choose the spring first need to know the torque of the motor. This is really simple, it can be mathematically calculate if the voltage and the current is known. But I like the experimental way; the setup is shown in the slide2. In the slide I have a DC motor with a leverage pivoted at the shaft and a free weight hanging with a string on the other side. In this, all I am doing is balancing the weigh at that point by giving the motor just enough power. | | | so the torque is equal to the weight at the distance from the shaft: | | so the torque is equal to the weight at the distance from the shaft: | | | | + | | | | *:: <math> \mathbf{\tau}=r \times F = r \times m .g </math> | | *:: <math> \mathbf{\tau}=r \times F = r \times m .g </math> | | | | | | | - | Once this is know we can choose the spring with less torque than this. The restoring torque applied by the spring can also be measure in the same fashion (as motor) as shown in slide1, with hanging known weight. The spring stiffness and the torque can be related through: | + | Once this is know we can choose the spring with less torque. The restoring torque applied by the spring can also be measure in the same fashion (as motor) as shown in slide1, with hanging known weight. The spring stiffness and the torque can be related through: | | | | + | | | | *:: <math> \mathbf{\tau}=2{\pi}n.K.r^2 </math> | | *:: <math> \mathbf{\tau}=2{\pi}n.K.r^2 </math> | | | | | | ## Motivation Motivation behind designing this shutter is speed, accuracy and variability (activation-time). In our optical tweezers we need to center the trap over the tethered bead, to get the geometry right (this affects the force measurement) and make our feedback program run. To center the tether we need to turn the laser intensity (trap) on/off quickly for various time intervals. To do this we used to use AOM (acoustooptic module), because it’s extremely quick (nano second on/off time). But AOM has some inherent oscillation problem which makes it use as a switch, unsuccessful. So I needed something else, (a shutter) which can replace this AOM function. This shutter does it exact and it is fast (activation-time) with an opening time of 4 and closing time of 2 μsec. The shutter runs on a +5 volts voltage. It moves to an on-position with an active voltage (controlled by the toggle foot switch) working against a restore-spring and remains on when the voltage is applied and turns to off when the voltage is removed. This gives a freedom of choice for the duration shutter is active with very simple electronics used to control the speed (activation time.) In the design the laser passes through an aperture on the only moving part a cylinder. No gears and no electronics in the design make it very simple, stable and accurate, even under the heat produced by a laser beam. The shutter needs no special power supply and can be run on a cell phone charge with an output of roughly 300mA/5V. Cost and construction time is also important. With this design, a shutter can be prepared under \$40 with 10 hours of construction time (10 x 25 (hourly wage of a technician) =\$250+40=\$290), still better than many commercially available shutter systems with same performance. ## Design & Construction ### Components There are three major parts of the system. • Shutter. • Control box. • Power supply. The components used: #### Shutter • 12V DC motor • Wood rotation stage • Spring with torque of .1 N m • Rubber padding • Pillar Post Extension, Length=1" from Thorlabs (shutter cylinder) • 30mm Cage Plate Optic Mount from Thorlabs • Post-holder, base-plate ext... #### Control Box • 1 power jack M&F • 1 1/4" mono Panel-Mount Audio Jack M&F • 1 Foot Paddle • 1 100Ω pot with, 1 220Ω resistor • 1 on/off toggle switch • 1 LED • 1 box enclosure • some connection wires, solder gun and solder wire #### Power Supply Any power-supply which can provide 300mA at 5V and above. The motor torque is power dependent and the shutter speed is resorting spring's stiffness (torque) dependent. So choose the spring carefully before decide on the power supply. I would recommend a variable power-supply which can be bought easily from any where. ### Construction The buildup is divided is the following categories. In the start make sure that you have a good set of screw driver, pliers, wire cutter, stripper and solder. #### Shutter I will start with the choice of the motor. A 12V DC motor is a good choice because it can provide a wide rang of torques. The motor and the spring works against each other so it important to have the right set. So choose a 12V DC motor with a shaft length of at least 15mm. Mount the motor on the cage plate or platform as shown in the picture, you will need 4/40 screws to tight it (in present setup). once this is done, unscrew and take the motor out. Now next task is to choose the right spring. The spring should be half the length of the shaft and the shaft should easily fit through the spring. I want to keep the spring-choice procedure really simple and easy. To choose the spring first need to know the torque of the motor. This is really simple, it can be mathematically calculate if the voltage and the current is known. But I like the experimental way; the setup is shown in the slide2. In the slide I have a DC motor with a leverage pivoted at the shaft and a free weight hanging with a string on the other side. In this, all I am doing is balancing the weigh at that point by giving the motor just enough power. so the torque is equal to the weight at the distance from the shaft: • $\mathbf{\tau}=r \times F = r \times m .g$ Once this is know we can choose the spring with less torque. The restoring torque applied by the spring can also be measure in the same fashion (as motor) as shown in slide1, with hanging known weight. The spring stiffness and the torque can be related through: • $\mathbf{\tau}=2{\pi}n.K.r^2$ Where K is the stiffness, r distance from the shaft center and n number of times spring twisted one complete 360 degrees circle before it was hacked. Since the torque is n dependent, it is a matter of great convenience, because now it is really easy to pick the right spring. All you have to do, is twist the required number of times before hacked to get the right restoring torque. Now we are ready to put the the shutter together. Start with the motor, most of the motors have little holes for the screws to hold slide3. Choose a screw which can fit into that but still sticking out few millimeters slide4. Next is the rotation stage which joins the cylinder to the motor and holds the spring assembly. I choose the wood for this function, because its easy to machine and a good thermal-insulator slide5. I drilled two hole on both sides; to fit the shaft and 1/4 screw on either sides. To drill for 1/4 I used 15/64 size drill-bit slide6,7,8 & 9. Now the motor can be attached to one side of the stage and cylinder to other slide10 & 11. Now we can machine the spring assembly. Slide 12 shows the parts of spring assembly. We need two metal strips of .75 and 1 cm long with a hole in the end for the screws. This strips with the screw on the motor will define the boundaries of the movement. Cylinder will rotate end to end between these strips. To screw the strips directly on the stage we need to drill two holes for the screws. The holes are 135 degree apart over the face of the stage as shown in the slide13. We also drill a small hole for the spring attachment near the edge. Now all parts are ready to get together slide 14. This shutter is little different of what is shown in the video but follows the same concept. Put the shorter end of the spring in the hole on the stage like slide 15. Now the motor shaft will go through the spring into the stage. The other side of the spring is hold against the screw. You can twist the required number of turns before it holds to get the right restoring torque. Now the spring wants to move the stage clockwise but the strip against the screw does not let it (this is position 1). As the voltage is applied the motor turns counterclockwise working against the spring until the second strip stops it. When the voltage is applied the stage remains in the second position. As soon as the voltage is turned down the spring restores the stage into the previous position. These two positions can be chosen for the shutter to be on/off slide 16,17 & 18. A fully assembled shutter is ready. Now the next task is to prepare a control box. #### Control Box I used a 2X3 inch aluminium box to house the electronics slide20. The electronics is really simple. The circuit is shown in the slide21. The foot switch is used to connect the motor ground to the power-supply hence activates the shutter. Since the shutter works on active voltage, the voltage is always on, when the foot paddle is active. So it is very important to give the right voltage to the motor to avoid damage to the motor over the long duration of active voltage. To do this i use a 100 ohm POT. POT let me choose the right voltage and current. This can be done once the shutter is ready. Now connect the shutter to the shutter input and foot-paddle to its input. #### Power Supply The good thing about the system is that it can use any power supply which can give enough power to operate the motor. The POT inside the control box let choose the appropriate voltage and current and hence makes easy to choose a power-supply. ## Comments Motor shutter buildup View more presentations from pranavrathi. Full Shutter System Shutter Control box Shutter different view Control box different view

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http://mathoverflow.net/revisions/78891/list

## Return to Answer 5 Added property P_d A simple example is obtained by taking $P$ to mean "has positive dimension". Every local domain of positive dimension $(A,\mathfrak m)$ has $P$ at all maximal ideals (i.e. just at $\mathfrak m$ !) since $A_{\mathfrak m}=A$ , but $P$ fails at the generic point $\eta=(0)$ since $A_\eta=Frac(A)$ has dimension zero, being a field. Edit In order to address Chuck's comment, let me emphasize that the answer above is very easily adapted to non local non-local rings. For example any finitely generated domain $A$ of positive dimension $d$ over a field has property $P$ when localized at a maximal ideal $\frak m$ but not at the zero ideal. More precisely, $dim A_{(0)}=0$ and $dimA_{\frak m}=d$ for any maximal ideal ${\frak m}$ : this equidimensionality result follows from Noether's normalization theorem. This shows that if property $P_d$ is " has dimension $d$ ", then $P_d$ holds for $A_{\frak m}$ if ${\frak m}$ is maximal and does not hold for $A_{\frak p}$ if the prime $\frak p$ is not maximal. 4 added non local example. A simple example is obtained by taking $P$ to mean "has positive dimension". Every local domain of positive dimension $(A,\mathfrak m)$ has $P$ at all maximal ideals (i.e. just at $\mathfrak m$ !) since $A_{\mathfrak m}=A$ , but $P$ fails at the generic point $\eta=(0)$ since $A_\eta=Frac(A)$ has dimension zero, being a field. Edit In order to address Chuck's comment, let me emphasize that the answer above is very easily adapted to non local rings. For example any finitely generated domain $A$ of positive dimension $d$ over a field has property $P$ when localized at a maximal ideal $\frak m$ but not at the zero ideal. More precisely, $dim A_{(0)}=0$ and $dimA_{\frak m}=d$ for any maximal ideal ${\frak m}$ : this equidimensionality result follows from Noether's normalization theorem. 3 added 8 characters in body A simple example is obtained by taking $P$ to mean "has positive dimension". Every local domain of positive dimension $(A,\mathfrak m)$ has $P$ at all maximal ideals (i.e. just at $\mathfrak m$ !) since $A_{\mathfrak m}=A$ , but $P$ fails at the generic point $\eta=(0)$ since $A_\eta$ A_\eta=Frac(A)\$ has dimension zero, being a field. 2 Replaced "integral ring" by "domain" A simple example is obtained by taking $P$ to mean "has positive dimension". Every integral local ring domain of positive dimension $(A,\mathfrak m)$ has $P$ at all maximal ideals (i.e. just at $\mathfrak m$ !) since $A_{\mathfrak m}=A$ , but $P$ fails at the generic point $\eta=(0)$ since $A_\eta$ has dimension zero, being a field. 1 A simple example is obtained by taking $P$ to mean "has positive dimension". Every integral local ring of positive dimension $(A,\mathfrak m)$ has $P$ at all maximal ideals (i.e. just at $\mathfrak m$ !) since $A_{\mathfrak m}=A$ , but $P$ fails at the generic point $\eta=(0)$ since $A_\eta$ has dimension zero, being a field.

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http://www.physicsforums.com/showthread.php?p=3814788

Physics Forums ## Conservation of momentum and energy contradiction? Suppose we have two objects and we're only talking about rectilinear motion. Initially, one object has mass m and is moving at velocity V. The other has mass M and is standing still. Then they hit each other and suppose that all kinetic energy is conserved and they stick together and move at velocity v. Then this follows from the conservation of energy: 1/2mV$^{2}$ = 1/2(m+M)v$^{2}$ And this follows from the conservation of momentum: mV = (m+M)v But when we divide the first equation by the second one we get: 1/2V = 1/2v V = v Which implies that M is zero but I haven't stated that anywhere, at least not directly. What did I do wrong? PhysOrg.com physics news on PhysOrg.com >> Promising doped zirconia>> New X-ray method shows how frog embryos could help thwart disease>> Bringing life into focus Mentor If they stick together, then the collision must be inelastic ie kinetic energy cannot be conserved. Remember that momentum is always conserved in a collision (provided there are no external forces). So, to see that what I'm saying is true, first apply conservation of momentum to figure out the final velocity in terms of the initial one. Then use these velocities to compare the final kinetic energy to the initial. Quote by cepheid If they stick together, then the collision must be inelastic ie kinetic energy cannot be conserved. Can you elaborate on this? This is kind of too abstract for me. Does this mean that for these masses and velocity always the same amount of kinetic energy is lost? I'd imagine it would also depend on the materials or contact area etc. Mentor Blog Entries: 1 ## Conservation of momentum and energy contradiction? Quote by Fyreth Can you elaborate on this? This is kind of too abstract for me. Just do the calculation. Momentum is always conserved. And since they stick together, it's easy to calculate the final velocity. Then you can directly compare the final KE with the initial. Does this mean that for these masses and velocity always the same amount of kinetic energy is lost? Right. Since they stick together, the maximum amount of KE is lost. I'd imagine it would also depend on the materials or contact area etc. All you need to know is that they stick together. Quote by Doc Al Just do the calculation. Momentum is always conserved. And since they stick together, it's easy to calculate the final velocity. Then you can directly compare the final KE with the initial. Right. Since they stick together, the maximum amount of KE is lost. All you need to know is that they stick together. I understand it mathematically but not physically. It's my first time looking into these collisions. How does this work? Which process turns the uniform motion into other forms of energy? Why is it not dependent on other properties? Why does sticking together leads to KE loss? Quote by Fyreth I understand it mathematically but not physically. It's my first time looking into these collisions. How does this work? Which process turns the uniform motion into other forms of energy? Why is it not dependent on other properties? Why does sticking together leads to KE loss? Hey, Try this. Do NOT assume that the balls stick together afterwards. Assume the initial conditions you gave. Conserve BOTH momentum and energy, and see what you expect the system to do. Now, ask yourself: Why do the balls not naturally stick together. The answer is, you need extra energy to force the balls to stick together. This energy comes from the kinetic energy of the ball. Thus, some of the energy is lost there and we cannot account for it. Of course, total energy = KE of both balls afterwards + Energy lost in forcing balls together is still conserved. But since we cannot calculate the latter (atleast directly), we cannot use energy conservation. You may ask of course, Why is MOMENTUM not lost in forcing the balls to stick together. The answer is that momentum is ALWAYS conserved if there are no net force. (Energy conservation is a more subtle issue). In this case, the balls experience forces that make it stick together, but all these forces are internal and effectively cancel out leaving zero net force. Thus, you may use momentum conservation, but not energy conservation here. Quote by Fyreth I understand it mathematically but not physically. It's my first time looking into these collisions. How does this work? Which process turns the uniform motion into other forms of energy? Why is it not dependent on other properties? Why does sticking together leads to KE loss? When the balls stick together, some energy is converted to heat (assuming no other losses, like sound waves, etc.) Suppose you had two balls one coming from the left with a speed of v, one coming from the right at the same speed. Suppose they have the same mass, and they stick together. Then when they stick together, they will stop dead. The total momentum is zero before and after, because their velocities were equal and opposite, no problem, but their kinetic energy has gone to zero. It all turned into heat, the balls are warmer after the collision. This usually happens because the two balls deform when they collide, and the different parts of the ball moving past each other creates friction, which creates the heat. You could also have the collision create sound waves inside the balls, and the sound waves would bounce around inside the balls, eventually getting absorbed and turned into heat. If there is a sound when they collide, and that sound gets radiated away, and carries some of the kinetic energy with it. What I'm saying below is parasitic on Rap's post. As Doc Al suggested, use momentum conservation to find v. As you wrote: mV = (m+M)v. So v = mV/(m+M). Now imagine that you view the collision from a car which is moving with this just this velocity. In your new frame of reference, m and M collide and each becomes stationary. That's what makes sticking together so special: there is a frame of reference in which both bodies lose all their KE. This won't happen for all colliding bodies, but there are quite a few materials (the obvious one is putty) for which this does happen, owing to the nature of the forces between the particles making up the material. But losing all KE in this special frame equates to sticking together in other frames of reference, and the loss of some of the original KE. The fact that only a certain fraction of the KE is lost in these frames, and that this fraction can be calculated with the aid of momentum conservation, makes it seem mysterious – as if the properties of the material aren't involved. But they are – as we saw by looking at the collision from the frame of reference (called the 'centre of mass frame') in which both bodies finish up stationary. Mentor Quote by Fyreth I understand it mathematically but not physically. It's my first time looking into these collisions. How does this work? Which process turns the uniform motion into other forms of energy? Why is it not dependent on other properties? Why does sticking together leads to KE loss? Think of different kinds of collisions with the ground. If something collides with the ground and maintains its KE then it bounces back up. If something collides with the ground and loses its KE then it splats on the ground. So, just by experience you know that "bouncy" collisions maintain KE and "sticky" collisions do not. Now, think about the result of a bouncy or a sticky collision on the bouncy or sticky object. A rubber ball is largely unaffected by a collision with the ground. It temporarily deforms, and then bounces right back to its original shape. A ball of mud or clay on the other hand deforms permanently. After the sticky collision it is no longer a ball. So, what happened to the KE of the clay? "1/2mV2 = 1/2(m+M)v2" This equation is wrong,because when they hit each other,some energy chenge into heat Q,which is about the movement of molecule in the body.So he have: So, conservation of energy is also true,but conservation of kinetic energy is not satisfied. With the equation which contain the heat ,you won't get the contradiction which make m=0.So conservation of momentum is also not broken. Tags conservation laws, contradiction Thread Tools | | | | |-------------------------------------------------------------------------|-------------------------------|---------| | Similar Threads for: Conservation of momentum and energy contradiction? | | | | Thread | Forum | Replies | | | Introductory Physics Homework | 18 | | | Classical Physics | 12 | | | General Physics | 1 | | | Engineering Systems & Design | 9 | | | Introductory Physics Homework | 2 |

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http://math.stackexchange.com/questions/68220/what-makes-simple-groups-so-special

# What makes simple groups so special? The classification of finite simple groups was one of the most important problem in group theory. But what makes simple groups so interesting and special? - 5 – user1729 Sep 28 '11 at 13:31 1 What makes prime numbers so interesting and special? Since I have class starting in a minute, I'll just post a quote from the Wikipedia page for simple groups: "The finite simple groups are important because in a certain sense they are the "basic building blocks" of all finite groups, somewhat similar to the way prime numbers are the basic building blocks of the integers. This is expressed by the Jordan–Hölder theorem which states that any two composition series of a given group have the same length and the same factors, up to permutation and isomorphism." – Alex Sep 28 '11 at 13:31 2 – Bruno Stonek Sep 28 '11 at 13:32 Simple groups, and their relatives the quasi-simple and almost simple groups, tend to come up a lot. Volume and form preserving matrix groups tend to be quasi-simple, and automorphisms of highly symmetric geometries tend to be almost simple. While simple groups are building blocks in general, the non-abelian ones tend to build very interesting things with only one (non-abelian) block. – Jack Schmidt Sep 28 '11 at 14:26 ## 1 Answer Simple groups are like prime numbers........ One must not let the account of the matter end with a full stop after the last word above. If one could say that all finite groups are products of finite simple groups, then the analogy would be simpler. But one can say that simple groups are as far as you can take the process of taking quotient groups without going to the very smallest quotient group: the group with only one element. The complication is that just taking Cartesian products of simple groups doesn't give all finite groups, and in fact, I think the understanding the ways in which other finite groups are built out of finite simple groups is quite a substantial problem in itself. However. Say you have a composite number like 299. You can form quotients: $$299/13 = 23.$$ $$299/23 = 13.$$ But from the prime number 23, you can't form any quotients except 23 and 1. Similarly there are no quotient groups of a simple group except itself and the group with only one element. -

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http://math.stackexchange.com/questions/76566/what-is-transcendental-algebraic-geometry?answertab=active

# What is 'Transcendental algebraic geometry'? What is 'Transcendental algebraic geometry'? Could you give me some good references in this field? Thanks. - ## 3 Answers Transcendental Algebraic Geometry is the study of the algebraic geometry of a variety defined over the complex numbers $\mathbb C$ by concentrating on its undelying structure as a holomorphic manifold or variety. This allows one to study the variety through the powerful tools of topology, analysis and differential geometry like : characteristic classes, elliptic partial differential equations, Kähler structure, ... Serre wrote a remarkable article (always quoted as GAGA!) giving a very precise functorial correspondence between a projective algebraic variety $X$ over $\mathbb C$ and the corresponding complex analytic variety $X^{an}$: here it is. Voisin's book mentioned by sweetjazz is indeed masterful but rather advanced. Two good introductory, fairly reader friendly, books to that area are Taylor's textbook and Wells's classic Differential Analysis on Complex Manifolds . The latter book culminates in the profound solution by Kodaira of Hodge's problem of characterizing the Kähler complex manifolds underlying a projective algebraic variety. - Griffith & Harris - "Principles of algebraic geometry" is also very good reference. - Transcendental Algebraic Geometry generally refers to algebraic geometry studied using techniques from the theory of complex variables, so that the results generally only apply to varieties defined over $\mathbb{C}$. One of the basic references for this area is Claire Voisin's two volume series Hodge Theory and Complex Algebraic Geometry. -

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http://mathhelpforum.com/calculus/59466-real-valued-function.html

# Thread: 1. ## Real-valued function Hi ! Do you know how to solve this problem: Find a real-valued function on R possessing derivatives of all orders whose Taylor series at a certain point converges to the function only at that point. 2. Originally Posted by Milus Hi ! Do you know how to solve this problem: Find a real-valued function on R possessing derivatives of all orders whose Taylor series at a certain point converges to the function only at that point. Consider $x^{x^x}$ at $x=0$ 3. Thank,but still I do not know what means that up side down L inside the log parentasice. Also, do you have any idea how to prove it. 4. Originally Posted by Milus Thank,but still I do not know what means that up side down L inside the log parentasice. Also, do you have any idea how to prove it. That is my signature, that is not part of the post. And for the proof try finding the general form for the n-th derivative of x. 5. Can you help me a little more? I am still strugling with general form of the derivative 6. Originally Posted by Milus Hi ! Do you know how to solve this problem: Find a real-valued function on R possessing derivatives of all orders whose Taylor series at a certain point converges to the function only at that point. I didn't know MathStud28's example. The example that I would have called the "classical" one is: $f(x)=e^{-\frac{1}{|x|}}$ for $x\neq 0$ and $f(0)=0$. I can't remember which famous mathematician introduced it, perhaps Cauchy, anyway it has a long history. And it seems simpler to deal with. For $x>0$, $f(x)=e^{-\frac{1}{x}}$. First notice that $f(x)\to_{x\to 0^+} 0$. Notice as well $\frac{e^{-\frac{1}{x}}}{x^n}\to_{x\to 0^+}0$ for any integer $n$. To convince yourself of this, substitute $u=\frac{1}{x}$, so that the limit is $\lim_{u\to\infty} u^n e^{-u}$. Now, the $n$-th derivative of $f$ is (easily) seen to be a rational function of $x$ (a quotient of two polynomials) times $e^{-\frac{1}{x}}$. This is proved by induction (suppose there is a rational function $R$ such that the $n$-th derivative is, for $x>0$, $R(x)e^{-\frac{1}{n}}$, and prove that the next derivative is again of the same kind). Because of the previously mentioned limit, you obtain that the n-th derivative converges to 0 as $x$ tends to 0 from the right. By symmetry, the same holds from the left. Finally, there is a theorem (that you probably know if you're asked this problem) telling you that, as a consequence, $f$ is indefinitely differentiable at 0 and the derivatives at 0 are the limits of the derivatives at $x$ when $x$ tends to 0. That is to say, all derivatives at 0 exist and are 0. Yet, the function is only zero at zero...

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http://blogs.ubc.ca/math101sec210portal/2013/02/05/lecture-10/

Home › Essential › News › Lecture › Lecture #10 # Lecture #10 Posted on February 5, 2013 by Today we covered some tough examples of trigonometric substitution, for easier examples, read examples 1-7 in p.479 – p.482 of the textbook. The examples I gave are unusual not in the techniques of trig sub itself, because the first step of identifying of the correct substitution is not really so hard. However, they are unusual second step, where we need to evaluate the weird trigonometric integrals that come out, and by now I have shown pretty much all special cases of $$\int \tan^m x \sec^n x dx$$. Note that the examples I gave doesn’t fall into the following cases • m is not odd (otherwise, substitute $$u = \sec \theta$$ will work) • n is not even (otherwise, substitute $$u = \tan \theta$$ will work) Situations when m and n are super large numbers AND does not conform to the rules will MOST LIKELY NOT APPEAR again in MATH101, because it requires reduction formulas to be solved with a reasonable amount of work. Similarly, for integrals of sines and cosines, when the powers of sine and cosine are odd, we can substitute cosine and sine respectively. But the case when both powers are even and large will most likely not appear again in this course. (for small even powers, though, we will use half angle formulas) So before it confuses you finally, we isolate these really evil special cases: • $$\int \sec x \ dx$$ • $$\int \sec^3 x dx$$ • $$\int \tan^2 x \sec x dx$$ The lecture note is here: wk5day10, and the promised extra discussion on \int\sec x dx[\latex] is here: wk5day10_spec_int. Posted in Lecture Search the Weblog

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http://mathoverflow.net/questions/27987/are-all-probabilities-conditional-probabilities/27993

## Are all probabilities conditional probabilities? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) We know that $P(A|B) = \frac{P(A \cap B)}{P(B)}$. So $P(B) = P(A|B)P(A \cap B)$. Thus are all probabilities conditional probabilities? Can one make a probability more accurate by introducing a conditional component? For example, the probability of rolling a six on a fair die is $1/6$. But can we make this more accurate by assuming prior events? - The definition of the words "fair die" is that the probability that each face comes up is $\frac{1}{6}$. What would it mean to make this more accurate? (Of course, in the physical world it is likely that there is no such thing as a fair die.) – JBL Jun 13 2010 at 2:53 @JBL, that's the kind of accuracy the OP is referring to: introducing a bias based on prior knowledge about the die etc. – Suresh Venkat Jun 13 2010 at 3:27 ## 5 Answers There's an obvious algebra error in Tony's question. In one trivial sense all probabilities are conditional: just condition on the whole "sample space". In (another?) sense, probabilities are necessarily conditional on what is known. - ### 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. In a sense this is precisely the Bayesian persective on what probabilities mean. Rather than being the limit fraction of a number of times an event occurs, it is a strength of belief conditioned on a prior. Bayesian approaches to statistical reasoning always start with a prior belief and condition based on that. - It is possible to develop probability theory taking conditional probability as one of the basic definitions; see section 3.2 in this book and the references mention there. Renyi was one of the first mathematicians to favor this approach, which is described in his book Probability Theory. Part of the motivation for this approach is to directly build in the ability to condition on measure-zero events without having to make a limiting argument. Another key word related to this idea is disintegration. So, as Kjetil mentions, it all depends on what one takes as axioms. But certainly it is possible to develop theories that take conditional probability as the centerpiece. - I think that's the main difference between the Bayesian approach vs. the frequentist approach, having prior knowledge is supposed to make your predictions more accurate. See [LessWrong][1] Bailey was trained in statistics, and when he joined an insurance company he was horrified to see them using Bayesian techniques developed in 1918. They asked not "What should the new rates be?" but instead "How much should the present rates be changed?" But after a year of trying different things, he realized that the Bayesian actuarial methods worked better than frequentist methods. Bailey "realized that the hard-shelled underwriters were recognizing certain facts of life neglected by the statistical theorists." For example, Fisher's method of maximum likelihood assigned a zero probability to nonevents. But since many businesses don't file insurance claims, Fisher's method produced premiums that were too low to cover future costs.cover future costs. [1]: http://lesswrong.com/lw/774/a_history_of_bayes_theorem/cover future costs. - I think this is a tricky question! and the answer depends on the exact meaning of the question. In the usual standard probability theory, based on Kolmogoro'vs axioms, in the axioms itself there are no conditional probabilities, and only later conditional probability are separately axiomatized. This seems to me a most strange procedure, and I am not sure why it is done that way. But in the lighjt of this theory, there are in fact nn-conditional probabilities. From an applied point of view, that of course do not make sense. Others can maybe comment on alternative axiomatizations. -

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http://mathoverflow.net/questions/23741?sort=votes

## How well does a truncated fourier expansion of a stepfunction perform near the expansionpoint ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Hey, i'am currently trying to make a stability analysis of a binary fluid at its phase border. As the governing equations in this double-diffusive problem are rather complicated i have to do a series expansion of my density profile. My problem is, that my PDEs contain a hevyside step function right at the border and i want to do a fourier expansion (i can then use symmetries of my problem and do some horrible algebra to actually get a "solution" which can be computed for up to 5 or 6 modes). Now i know that fourier expansion performs quite horribly at a discontinous jump if only the first few modes are kept. Is it nervertheless possible to obtain a "margin of error"? Numerical precision is not of utmost concern, but the physics behind my problem should not all be thrown out by such an expansion. Do you know of any criteria or "better" expansion where its possible to make heavy use of symmetry/antisymmetry? all the best, jan - SFAIK the standard way of handling the "ringing" that comes up in Fourier approximants of discontinuous functions requires multiplication by a sine cardinal (sinc(x)=sin(x)/x), as recommended by Lanczos. Have you tried this already? – J. M. Aug 12 2010 at 8:02 ## 3 Answers Too lazy to think about the precise answer, but using something like a Cesaro sum should improve matters. There are some useful references in the Wikipedia pages for Gibbs phenomenon, Fejer's theorem, etc. - ### 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. As Nate pointed out, Fejer's summation should be better than the standard Fourier sum. The standard Fourier sum will converge pointwise like 1/n near the discontinuity. - Isn't it just the Gibbs phenomenom you are asking for? If there is a discontinuity (as for the Heaviside function) the discontinuity in the partial Fourier series does not die out for $n \to \infty$ , but will be about 18% larger than the discontinuity in the original function. This does not give you the precise value for 5 or 6 modes, but tells you that for Heaviside function the overshot will be 0,18 for $n \to \infty$, which will give you the order of magnitude also for 5 or 6 modes. (Please see also http://en.wikipedia.org/wiki/Gibbs_phenomenon.) - Is this an expansion of Nate Eldridge's answer? – S. Carnahan♦ Jun 18 2010 at 19:16

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http://mathhelpforum.com/advanced-applied-math/28887-heat-problems.html

# Thread: 1. ## Heat problems 1) Determine the amount of heat needed to turn 500g of ice in a 100g aluminum pan into steam. 2) What will be the phase of 1 liter of water in a 200g copper pot if it is placed in a coolant that absorbs 100,000J of heat? 3) How much heat is required to turn 50g of ice with temperature of -50 C to steam at 150 C? Specific Heat: Water = 4180 J/kg*K or 1.000 cal/g*C Copper = 392 J/kg*K or 0.093 cal/g*C Aluminum = 895 J/kg*K or 0.214 cal/g*C Latent Heat: = Heat of fusion Water = $3.34 * 10^5 J/kg$ or $80 cal/g$ Copper = $2.05 * 10^5 J/kg$ or $48.9 cal/g$ Aluminum = $3.76 * 10^5 J/kg$ or $89.9 cal/g$ = Heat of vaporization Water = $22.6 * 10^5 J/kg$ or $540 cal/g$ Copper = $48 * 10^5 J/kg$ or $1150 cal/g$ Aluminum = $113.71 * 10^5 J/kg$ or $2717.7 cal/g$ 2. Originally Posted by ihmth 1) Determine the amount of heat needed to turn 500g of ice in a 100g aluminum pan into steam. My answer is 360,000 cal If you are going to specify that we have an aluminum pan then we need to know what temperature the pan is starting at. Additionally we need to know what temperature the ice starts at as well. There is not enough information present to do this problem. -Dan 3. Originally Posted by ihmth 2) What will be the phase of 1 liter of water in a 200g copper pot if it is placed in a coolant that absorbs 100,000J of heat? I dont have an answer Specific Heat: Water = 4180 J/kg*K or 1.000 cal/g*C Copper = 392 J/kg*K or 0.093 cal/g*C Latent Heat: = Heat of fusion Water = $3.34 * 10^5 J/kg$ or $80 cal/g$ The maximum amount of heat energy the water can have is when its initial temperature is at the boiling point, 373 K. (Otherwise it is steam, not water.) Assume the copper pot also has this temperature at the beginning. Let's find out how much heat needs to be absorbed from the system to turn the water at the boiling point into ice: $c_wm_w(373 - 273) + c_{Cu}m_{Cu}(373 - 273) + L_wm_w = 7.5984 \times 10^5~J$ (The copper pot must be a maximum of 273 K for the water to freeze, so I'm assuming that it drops at the same temperature rate as the water. This is not quite physically true, but will do as an approximation.) The first two terms take care of the heat needed to be absorbed to take the water and copper from 100 C to 0 C, the third term is the heat to be absorbed from the water at 0 C to turn it into ice. The coolant absorbs $1 \times 10^5~J$ of heat energy. This is less than the heat we can absorb from the water and copper at 100 C to freeze the ice, so we would need a starting temperature for the water-ice system in order to give a definitive answer to this question. (Now if the coolant absorbed, say, $1 \times 10^6~J$ then we would know for sure that the water would freeze.) -Dan 4. Originally Posted by ihmth 3) How much heat is required to turn 50g of ice with temperature of -50 C to steam at 150 C? My answer is 41,000 cal Specific Heat: Water = 4180 J/kg*K or 1.000 cal/g*C Latent Heat of fusion: Water = $3.34 * 10^5 J/kg$ or $80 cal/g$ = Heat of vaporization Water = $22.6 * 10^5 J/kg$ or $540 cal/g$ This one we can do: Heat needed to turn the 50 g of ice at -50 C to ice at 0 C: $c_im_i(0 - -50) = 5125~J$ ( $c_i = 2.050~J/g ^o C$. You didn't give this, I had to look it up. It's Wikipedia so I'm only assuming the numbers are correct.) Heat needed to turn 50 g of ice at 0 C to water at 0 C: $L_fm_i = 16700~J$ Heat needed to turn 50 g water at 0 C to water at 100 C: $c_wm_w(100 - 0) = 20900~J$ Heat needed to turn 50 g water at 100 C to steam at 100 C: $L_vm_w = 113000~J$ Finally, heat need to turn 50 g of steam at 100 C to steam at 150 C: $c_sm_s(150 - 100) = 5200~J$ ( $c_s = 2.080~J/g ^oC$. Again, you didn't give us this information.) Adding these up gives a total of 160925 J of heat required. In the future, please provide us with all the information you are given. -Dan 5. Originally Posted by topsquark If you are going to specify that we have an aluminum pan then we need to know what temperature the pan is starting at. Additionally we need to know what temperature the ice starts at as well. There is not enough information present to do this problem. -Dan ice and the aluminum pan is at 0 C 6. Originally Posted by ihmth ice and the aluminum pan is at 0 C Originally Posted by ihmth 1) Determine the amount of heat needed to turn 500g of ice in a 100g aluminum pan into steam. My answer is 360,000 cal Specific Heat: Water = 4180 J/kg*K or 1.000 cal/g*C Aluminum = 895 J/kg*K or 0.214 cal/g*C Latent Heat: = Heat of fusion Water = $3.34 * 10^5 J/kg$ or $80 cal/g$ = Heat of vaporization Water = $22.6 * 10^5 J/kg$ or $540 cal/g$ So $L_f m_i + c_wm_i(100 - 0) + c_{Al}m_{Al}(100 - 0) + L_vm_i$ -Dan

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http://physics.stackexchange.com/questions/tagged/schrodinger-equation?page=1&sort=votes&pagesize=15

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http://mathoverflow.net/questions/114245/i-know-that-you-know/114265

## I know that you know… ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) A bit unsure if the following vague question has enough mathematical content to be suitable upon here. In the case, please feel free to close it. In several circumstances of competition, a particular situation of partial information occurs, usually described as "I know that you know that I know... something". We may distinguish a whole hierarchy of more and more complicate situations closer and closer to a complete information. E.g. : • $I_0$: I know $X$, but you don't know that I know. • $I_1$: I know $X$, you know that I know, but I don't know that you know that I know. • $I_1$: I know $X$, you know that I know, I know that you know that I know, but you don't know that I know that you know that I know. • .... &c. For small values of $k$, I can imagine simple situations where passing from $I_k$ to $I_ {k+1}$ really makes a difference (for instance: you are Grandma Duck, and $X$ is : "you left a cherry pie to cool on the window ledge". Clearly, $I_0$ is quite agreeable position; $I_1$ may lead to an unpleasant end (for me); $I_2$ leaves me some hope, if I behave well, and so on). But, I can't imagine how passing from $I_6$ to $I_7$ may affect my strategy, or Grandma's. Are there situations, real or factitious, concrete or abstract, where $I_k$ implies a different strategy than $I_{k+1}$ for the competitors? What about $I_{\omega}$ and, more generally, $I_\alpha$ for an ordinal $\alpha$ (suitably defined by induction)? How these situations are modeled mathematically? - 9 I assume you have heard of the blue eyed islander problem? This is a classic example of where arbitrarily high orders of knowledge of this form become relevant. – Steven Gubkin Nov 23 at 15:01 2 There is a (not serious) mention of this in "A canticle for Leibowitz", but sadly I don't remember where exactly. Can somebody find the exact quotation? – Laurent Berger Nov 23 at 15:35 3 A whole load of fictitious examples can be found at tvtropes.org/pmwiki/pmwiki.php/Main/… – Terry Tao Nov 23 at 18:03 There are some papers on "reflexive game theory" by Alexander Chkhartishvili, see e.g. link.springer.com/article/…, where this hierarchy of reflexive levels is formalized and he shows how this affects the Nash equilibrium etc. Chkhartishvili apparently did not put any of his work on arXiv, so unfortunately one has to resort to English translations of some Russian journals where this is published. – ansobol Nov 23 at 19:49 ## 6 Answers My wife and I have a standing agreement where I pick up our son Horatio from school and she picks up our daughter Hypatia. One day, because I knew I would be near Hypatia's school, it was convenient to swap duties. I emailed her a message, "I'll pick up Hypatia today, and you get Horatio. Please confirm; otherwise it is as usual." She texted me back, "Let's do it. Let me know if you get this message, so that I know we're really on." I left her a voicemail, "OK, we're definitely on for the swap! ....as long as I know you get this message." She emailed me back, "Got the message. We're on! But let me know that you get this message so I can count on you." You see, without confirmation she couldn't be sure that I knew she had gotten my earlier confirmation of her acknowledgement of my first message, and she may have worried that the plan to swap was consequently off. And so on ad infinitum...... How truly frustrating it was for us that at no stage of our conversation could we seem to know for certain that the other person had all the necessary information to ensure that the plan would be implemented! The result, of course, since we had time to exchange only at most a finite number of messages, was that the only rational course of action was for us each to abandon the plan to swap: we both independently decided just to pick up the usual child. To see that this was rational, observe that clearly the first message needed to have been confirmed in order for the plan to be implemented properly. Furthermore, if the $n$-th message need not have been confirmed, then it wasn't important to know that it had been received and the algorithm should have worked whether or not it was received, meaning that it didn't actually need to have been sent. So by induction, no number of confirmations suffices to implement the common knowledge that we both needed, namely, that we had each agreed to make the swap. See also the two generals problem. - 1 Incidentally, it seems to me that the same problem still arises even if we were to speak in person, because one would have to make sure that the previous comment was received and understood, before its acknowledgement would rise to the level of common knowledge. I am left to wonder how it is possible ever for us to attain common knowledge... – Joel David Hamkins Nov 23 at 21:25 Tweetie-bird, of course my intention here was to provide a comparatively concrete example exhibiting the principle issues of common knowledge. I am actually fascinated by the logic and mathematics of this kind of situation, which I view as a genunine, sophisticated mathematical topic. – Joel David Hamkins Nov 23 at 22:50 I deleted my previous comment so as not to confuse others. This logic puzzle is already confusing enough. – tweetie-bird Nov 23 at 23:18 .....principal. – Joel David Hamkins Nov 24 at 1:55 5 As someone who didn't have the easiest time in school, I am glad that I was not named horatio or hypatia... – Bern Oay Nov 24 at 5:15 show 4 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. Here is another puzzle I like which is in the spirit of the question. 100 people play a game as follows. Each person secretly writes a number between 1 and 1000000. The numbers are then all revealed and the person who is closest to 2/3 of the average wins a prize. If there is a tie, the prize is shared between the winners. What number should one write down? Now, it is clearly foolish to write down any number greater than 666667, since 2/3 of the average cannot be more than 666667. But now we can view the game as being played on the interval from 1 to 666667 instead of from 1 to 1000000. Now we can iterate again and conclude that it is foolish to choose any number greater than 444444. Ultimately (but this requires many iterations of knowledge), the only rational choice is for all players to choose 1 and to split the prize. - 1 A agree that choosing $1$ is the only rational choice, but still I would like to actually do this experiment with a bunch of non-mathematicians and see what $2/3$ of the average turns out to be. I suspect it changes after the first go, or if you give them much time to think, but such empirical information could have been useful in a pub quiz I used to go to. – Paul Reynolds Nov 23 at 19:43 1 IIRC, I’ve read that in human experiments the 2/3 of the average tends to be around 20% of the maximum. – Emil Jeřábek Nov 23 at 19:58 Interesting! I wonder how this varies with the fraction $r = 2/3$... – Paul Reynolds Nov 23 at 20:02 3 Paul & Emil: such "beauty contest" experiments reveal baffling play: for r = 0, not everyone plays 0. Some do not play increasing functions of r. Many people seems to play roughly linearly in r relative to their play at r=1. Here is my analysis of some data I recently collected from web surveys: faculty.chicagobooth.edu/richard.hahn/…. Plots of observed strategies: faculty.chicagobooth.edu/richard.hahn/… – R Hahn Nov 23 at 21:09 1 ---Now, it is clearly foolish to write down any number greater than 666667, since 2/3 of the average cannot be more than 666667.--- Not if you playing with a partner. Of course, you won't take the prize yourself, but you can raise your partner's chances quite a bit by writing a large number if you agree upon a decent strategy. Now, how do I know that there are no coalitions out there? And if I don't, the second step of the induction fails... – fedja Nov 26 at 0:38 show 6 more comments Check out the wikipedia page on common knowlege, which includes some mathematical formalizations of the concept as well as the famous blue-eyed islander problem mentioned by Steven Gubkin. This problem was also discussed by Terrence Tao on his blog. - The classic model of these things is due to Robert Aumann and was introduced in his 1976 Agreeing to Disagree. There is a famous example due to Ariel Rubinstein, the electronic mail game, in which the behavior of $I_\omega$ radically differs from $I_n$ for any $n$. Here is a way to show that one might have to apply this reasoning up to an arbitrarily large ordinal. Let $\alpha$ be a successor ordinal. Ann and Bob play the game of picking ordinals in $\alpha$ simultaneously. The one who chooses the highest number wins. One can never win by choosing $0$, so rationality rules out choosing $0$. It is however possible to win by choosing $1$ if the other player plays $0$. But if both Ann and Bob know that they are both rational, they have to choose at least $2$. It is clear that one has to iterate this reasoning up to the predecessor of $\alpha$. - I don’t get the example. Whatever the other player knows, thinks, or plays, I’m always better off playing the predecessor of $\alpha$ than any other ordinal, hence rationality dictates I do just that, without any iteration. – Emil Jeřábek Nov 23 at 16:03 @Emil It is correct that playing the predecessor is always compatible with common knowledge of rationality up to some ordinal $\beta<\alpha$. But there are many more choices that are compatible with common rationality up to a lower level. If you want to role them out, you have to go all the way. The notion of rationality being used is not using a choice you know is not a best response. – Michael Greinecker Nov 23 at 16:15 3 Playing an ordinal $\beta$ smaller than the predecessor of $\alpha$ is never rational, because there is the possibility that the opponent may play something larger than $\beta$, which would make me lose, whereas if I play the predecessor, it is either a win for me or a draw. I know this a priori, without considering what the opponent knows. – Emil Jeřábek Nov 23 at 16:43 @Emil: No using weakly dominated strategies is stronger requirement than rationality. And having common knowledge of not playing weakly dominated strategies might be impossible. There is a well known paper by L. Samuelson, in which he show that there are games in which common knowledge of not-playing weakly dominated strategies is inconsistent (alturl.com/d2s5m). The notion I used coincides with point-rationalizability in the sense of Bernheim (alturl.com/fc9oe). – Michael Greinecker Nov 23 at 17:55 But I am not assuming any common knowledge, I am in fact not assuming anything whatsoever about the opponent. You defined rationality as “not using a choice you know is not a best response”, and I know choosing an ordinal smaller than the predecessor is not the best response, because choosing the predecessor is a better response. I have no idea whether this conforms to one formal definition or another, but if not, then IMHO the example merely illustrates that the definition is contrived rather than the phenomena mentioned by the OP. – Emil Jeřábek Nov 23 at 18:39 show 3 more comments A good source for this sort of reasoning is: J.-J. Meyer and W. van der Hoek, Epistemic Logic for AI and Computer Science, Number 41 in Cambridge Tracts in Theoretical Computer Science, Cambridge University Press, 1995. The question of common knowledge, and of the algorithmics of epistemic modal logics, provides a lot of good material on this sort of problem. (It is good fun to try the mathematical side of this in Popularisation activities. I used Muddy Children with various groups with a lot of laughter and appreciation of the mathematical models.) - The article http://plato.stanford.edu/entries/common-knowledge/ is more extensive than wikipedia's. However, it doesn't deal with the issue of infinite information. -

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http://mathhelpforum.com/advanced-statistics/79415-joint-probability-question.html

# Thread: 1. ## Joint Probability Question Hi, I am having trouble with the following question: Five balls are drawn without replacement from an urn containing 7 blue balls, 4 green balls, and 8 yellow balls. Let B, G, and Y be the total number of blue, green and yellow balls drawn, respectively. Find a) The joint probability mass function of B, G, and Y. b) The joint probability mass function of Y and B c) THe marginal probability mass function for B, G, and Y separately. Do these marginal probabilities look familiar? ID them. For a) I tried modelling the joint probability mass function using a combinatorial approach. c(7 choose B)c(4 choose G)c(8 choose Y)/c(19 choose 5). And B+G+Y=5 (because you can only choose 5 balls). However, after this I'm stuck. 2. these are all hypergeometrics joint is $P(B=b,G=g,Y=y)={ {7 \choose b} {4 \choose g} {8 \choose y}\over {19 \choose 5}}$. b is really the same... $P(B=b,G=g)={ {7 \choose b} {4 \choose g} {8 \choose 5-b-g}\over {19 \choose 5}}$. I'll only do B's distribution. It's B vs the world so... $P(B=b)={ {7 \choose b} {12 \choose 5-b} \over {19 \choose 5}}$. you do the other two. 3. I understand how you got part c, but I don't really understand how you got part b. Is it because you are focusing on B and G, so Y is just what's left over? 4. I was in a rush. I wrote the joint distribution of B and G. Here's the joint of Y and B... $P(B=b,Y=y)={ {7 \choose b} {8 \choose y} {4 \choose 5-b-y}\over {19 \choose 5}}$. This is because b+y+g=5. So, if you know two of them, you know the third one.

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http://math.stackexchange.com/questions/44694/understanding-algebraic-manipulation

Understanding algebraic manipulation If $a+b+c \neq 0$ where $a,b$ and $c$ are three non-zero distinct integers, then find the value of: $$\frac{ab+ca}{a^2+ab+ca} + \frac{ab+cb}{b^2+ab+bc} + \frac{ac+cb}{c^2+ac+bc}$$ What confusing me here, is the not so obvious hint which is given with the problem,which says that that form could be written as: $$3- \frac{a^2}{a^2+ab+ca} - \frac{b^2}{b^2+ab+bc} - \frac{c^2}{c^2+ac+bc}$$ But how is this possible? - What have you tried? Where are you stuck? Have you tried seeing what you get when you plug in numbers (say $a = b = c = 1$)? – JavaMan Jun 11 '11 at 3:19 @DJC:Please read my question carefully. – Quixotic Jun 11 '11 at 3:20 I have read your question carefully. So where are you stuck? What have you tried? – JavaMan Jun 11 '11 at 3:21 $\quad$@DJC:Edited. – Quixotic Jun 11 '11 at 3:25 3 Answers All you need to do is "simplify" the fractions, for example by dividing "top" and "bottom" of the first one by $a$, of the second by $b$, of the third by $c$. We get $$\frac{b+c}{a+b+c}+\frac{a+c}{a+b+c}+\frac{a+b}{a+b+c}$$ We have a common denominator $a+b+c$. The numerators add up to $2(a+b+c)$. Cancel. We get $2$. Comment: If you find this not obvious, let's look at the first term, that is, at $$\frac{ab+ca}{a^2+ab+ca}$$ The "top" is $a(b+c)$. The "bottom" is $a(a+b+c)$. So the fraction is $$\frac{a(b+c)}{a(a+b+c)}$$ Divide top and bottom by $a$, or equivalently, "cancel" the $a$'s. We are using the "algebra" version of the familiar fact that $$\frac{2\cdot 3}{2 \cdot 5}=\frac{3}{5}$$ Additional comment The hint is strange. True, we can rewrite the expression as $$\left(1-\frac{a^2}{a^2+ab+ca}\right) +\left(1-\frac{b^2}{b^2+ab+bc}\right)+\left(1-\frac{c^2}{c^2+bc+ca}\right)$$ which is equivalent to what was given. And then we could divide top and bottom by $a$ in the first fraction, by $b$ in the second, by $c$ in the third, and end up with $$3-\frac{a+b+c}{a+b+c}$$ But it sure seems like a lot of work when we can cancel immediately! You are sure that you quoted the problem correctly? - Thanks this is a really easy approach the hint given confused me. – Quixotic Jun 11 '11 at 3:27 I can understand your answer,but I don't understand the hint given in my module (my edited question),Of-course the same thing could be done after that step given $3-1=2$. – Quixotic Jun 11 '11 at 3:32 Yes,I am sure I have quoted the problem correctly. – Quixotic Jun 12 '11 at 23:55 @Debanjan: The reason I wondered is that the hint seemed unreasonable in view of the obvious cancellations. – André Nicolas Jun 13 '11 at 0:06 André's approach is pretty simple. In case you really must follow the hint, here's how to get it: note, for example, that in the first fraction, $$\frac{ab+ca}{a^2+ab+ca}$$ the only thing that is missing from having the numerator exactly equal to the denominator is an $a^2$; so adding it and taking it away again we get $$\frac{ab+ca}{a^2+ba+ca} = \frac{ab+ca+a^2-a^2}{a^2+ba+ca} = \frac{a^2+ab+ca}{a^2+ab+ca} - \frac{a^2}{a^2+ba+ca} = 1-\frac{a^2}{a^2+ba+ca}.$$ The same thing can be done with $$\frac{ab+cb}{b^2+ab+bc}$$ by adding and subtracting $b^2$ to the numerator. And also with $$\frac{ac+cb}{c^2+ac+bc}$$ by adding and subtracting $c^2$ to the numerator. Doing all three, we have: $$\begin{align*} \frac{ab+ca}{a^2+ba+ca} &+ \frac{ab+cb}{b^2+ab+bc} + \frac{ac+cb}{c^2+ac+bc} \\&= \left(1 - \frac{a^2}{a^2+ba+ca}a\right)\\ &\qquad\quad\mathop{+}\left(1 - \frac{b^2}{b^2+ab+bc}\right)\\ &\qquad\quad\mathop{+}\left(1 - \frac{c^2}{c^2+ac+bc}\right)\\ &= 3 - \frac{a^2}{a^2+ba+ca}a - \frac{b^2}{b^2+ab+bc} - \frac{c^2}{c^2+ac+bc}, \end{align*}$$ which is exactly the formula your hint gives. As to how the hint gives the answer any better than the straightforward method suggested by André (which would be my own approach, to simplify first), I couldn't tell you, but that is how the equation can be rewritten in that form. - It is obvious that we have to use the cancellation after the hint step,thanks for the explanation and I agree this hint is not necessary and lengthy to some extent. – Quixotic Jun 11 '11 at 3:38 Or just add and subtract $a^2$, $b^2$, $c^2$ to the numerators of the three fractions respectively. -

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http://mathhelpforum.com/calculus/152044-determining-if-series-convergent-divergent.html

# Thread: 1. ## Determining if a series is convergent or divergent Hi I was wondering if anyone could explain this to me, I have Series (-1)^n n/ln(n) sum : infinite n:2 I took the derivative and got : 1/ln(x) - 1/(ln(x)^2) How do I determine if this is a decreasing or increasing function. From what I could tell 2 is the only value to make it go <0 , so is this neither increasing or decreasing? 2. Originally Posted by illidari Hi I was wondering if anyone could explain this to me, I have Series (-1)^n n/ln(n) sum : infinite n:2 I took the derivative and got : 1/ln(x) - 1/(ln(x)^2) How do I determine if this is a decreasing or increasing function. From what I could tell 2 is the only value to make it go <0 , so is this neither increasing or decreasing? You wrote $\sum\limits^\infty_{n=2}(-1)^n\frac{n}{\ln n}$ . This series diverges since the general term' sequence doesn't converge to zero. I can't understand what you're trying to derivate and wrt what... Tonio 3. Yeah that is the series I wrote, I was trying to apply the alternating series test. The 2nd part of the equation, excluding (-1)^n, needs to be decreasing I took the derivative of it and need to determine if that derivative is <0 . That is the part I got stuck :/ 4. Originally Posted by illidari Yeah that is the series I wrote, I was trying to apply the alternating series test. The 2nd part of the equation, excluding (-1)^n, needs to be decreasing I took the derivative of it and need to determine if that derivative is <0 . That is the part I got stuck :/ Are you trying to derivate wrt n, which is a discrete variable?! It can't be done, of course. Besides this the series does not converge at all, Leibnitz or not. Tonio

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http://cogsci.stackexchange.com/questions/tagged/experimental-psychology+reading

# Tagged Questions 5answers 296 views ### How long does it take to read X number of characters? How does the time needed to read a sentence scale with the number of characters? Or does this time scaling depend on something more than just character count? For example, let $X$ be the number of ...

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http://www.mathplanet.com/education/algebra-1/quadratic-equations/completing-the-square

# Completing the square If you've got a quadratic equation on the form of $\\ ax^{2}+c=0 \\$ Then you can solve the equation by using the square root of $\\ x=\pm \sqrt{\frac{-c}{a}} \\$ Example: $\\ 3x^{2}-243=0 \\\\3x^{2}=243 \\\\x^{2}=\frac{243}{3} \\\\x^{2}=81 \\x=\pm \sqrt{81} \\x=9\: \: or\: \: x=-9 \\$ This method can only be used if b = 0. If we instead have an equation on the form of $\\ x^{2}+bx=0 \\$ we can't use the square root initially since we do not have c-value. But we can add a constant d to both sides of the equation to get a new equivalent equation that is a perfect square trinomial. Remember that a perfect square trinomial can be written as $\\ x^{2}+bx + d=\left ( x+d \right )^{2}=0 \\$ This process is called completing the square and the constant d we're adding is $\\ d=\left (\frac{b}{2} \right )^{2} \\$ Example: $\\ x^{2}+12x=0 \\$ We begin by finding the constant d that can be used to complete the square. $\\ d=\left (\frac{b}{2} \right )^{2}=\left ( \frac{12}{2} \right )^{2}=6^{2}=36 \\\\\\ x^{2}+12x+d=0+d\Rightarrow \\x^{2}+12x+36=0+36\Rightarrow \\\\\begin{pmatrix}x+6 \end{pmatrix}^{2}=36 \\\\\\\sqrt{\begin{pmatrix} x+6 \end{pmatrix}^{2}}=\pm \sqrt{36} \\\\\\\begin{matrix} x+6=6\: \: &or\: \: & x+6=-6\\ x=0 & & x=-12 \end{matrix} \\$ The completing the square method could of course be used to solve quadratic equations on the form of $\\ ax^{2}+bx+c=0 \\$ In this case you will add a constant d that satisfy the formula $\\ d=\left ( \frac{b}{2} \right )^{2}-c \\$ Video lesson: Solve the equation by completing the squares $\\x^{2} - 3x - 10=0 \\$ Next Class: Quadratic equations, The quadratic formula • Pre-Algebra • Algebra 1 • Algebra 2 • Geometry • Sat • Act

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http://www.encyclopediaofmath.org/index.php/Analytic_function

# Analytic function From Encyclopedia of Mathematics A function that can be locally represented by power series. The exceptional importance of the class of analytic functions is due to the following reasons. First, the class is sufficiently large; it includes the majority of functions which are encountered in the principal problems of mathematics and its applications to science and technology. Secondly, the class of analytic functions is closed with respect to the fundamental operations of arithmetic, algebra and analysis. Finally, an important property of an analytic function is its uniqueness: Each analytic function is an "organically connected whole" , which represents a "unique" function throughout its natural domain of existence. This property, which in the 18th century was considered as inseparable from the very notion of a function, became of fundamental significance after a function had come to be regarded, in the first half of the 19th century, as an arbitrary correspondence. The theory of analytic functions originated in the 19th century, mainly due to the work of A.L. Cauchy, B. Riemann and K. Weierstrass. The "transition to the complex domain" had a decisive effect on this theory. The theory of analytic functions was constructed as the theory of functions of a complex variable; at present (the 1970's) the theory of analytic functions forms the main subject of the general theory of functions of a complex variable. There are different approaches to the concept of analyticity. One definition, which was originally proposed by Cauchy, and was considerably advanced by Riemann, is based on a structural property of the function — the existence of a derivative with respect to the complex variable, i.e. its complex differentiability. This approach is closely connected with geometric ideas. Another approach, which was systematically developed by Weierstrass, is based on the possibility of representing functions by power series; it is thus connected with the analytic apparatus by means of which a function can be expressed. A basic fact of the theory of analytic functions is the identity of the corresponding classes of functions in an arbitrary domain of the complex plane. Exact definitions are given below. Let $D$ be a domain in the complex plane $\mathbb C$. If to each point $z\in D$ there has been assigned some complex number $w$, then one says that on $D$ a (single-valued) function $f$ of the complex variable $z$ has been defined and one writes: $w=f(z), z\in D$ (or $f:D\to\mathbb C$). The function $w=f(z)=f(x+iy)$ may be regarded as a complex function of two real variables $x$ and $y$, defined in the domain $D\subset\mathbb R^2$ (where $\mathbb R^2$ is the Euclidean plane). To define such a function is tantamount to defining two real functions \begin{equation*} u=\phi(x,y),\quad v=\psi(x,y),\quad (x,y)\in D\quad (w = u+iv). \end{equation*} Having fixed a point $z\in D$, one gives $z$ the increment $\Delta z = \Delta x+ i\Delta y$ (such that $z+\Delta z \in D$) and considers the corresponding increment of the function $f$: \begin{equation} \Delta f(z) = f(z+\Delta z) - f(z). \end{equation} If \begin{equation} \Delta f(z) = A\Delta z + o(\Delta z) \end{equation} as $\Delta z\to 0$, or in other words, if \begin{equation} \lim_{\Delta z\to 0}\frac{\Delta f(z)}{\Delta z} = A \end{equation} exists, the function $f$ is said to be differentiable (in the sense of complex analysis or in the sense of $\mathbb C$) at $z$; $A = f'(z)$ is the derivative of $f$ at $z$, and \begin{equation} A\Delta z = f'(z)dz = df(z) \end{equation} is its differential at that point. A function $f$ which is differentiable at every point of $D$ is called differentiable in the domain $D$. One may compare the concepts of differentiability of $f$ considered as a function of two variables (in the sense of $\mathbb R^n$) and in the sense of $\mathbb C$. In the former case the differential $df$ has the form \begin{equation} \frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y}dy, \end{equation} where \begin{equation} \frac{\partial f}{\partial x} = \frac{\partial \phi}{\partial x} + i\frac{\partial \psi}{\partial x},\quad \frac{\partial f}{\partial y} = \frac{\partial \phi}{\partial y} + i\frac{\partial \psi}{\partial y}, \end{equation} are the partial derivatives of $f$. Passing from the independent variables $x, y$ to the variables $z, \overline{z}$, which may formally be considered as new independent variables, related to the old ones by the equations $z = x+iy$, $\overline{z}=x-iy$ (from this point of view, the function $f$ may also be written as $f(z,\overline{z})$) and expressing $dx$ and $dy$ in terms of $dz$ and d\overline{z} according to the usual rules of differential calculus, one can write $df$ in its complex form: \begin{equation} df = \frac{\partial f}{\partial z}dz + \frac{\partial f}{\partial \overline{z}}z\overline{z} \end{equation} where \begin{equation} \frac{\partial f}{\partial z} = \frac{1}{2}\left(\frac{\partial f}{\partial x} - i\frac{\partial f}{\partial y}\right), \quad \frac{\partial f}{\partial \overline{z}} = \frac{1}{2}\left(\frac{\partial f}{\partial x} + i\frac{\partial f}{\partial y}\right), \end{equation} are the (formal) derivatives of $f$ with respect to $z$ and $\overline{z}$, respectively. It is seen, accordingly, that $f$ is differentiable in the sense of $\mathbb C$ if and only if it is differentiable in the sense of $\mathbb R^2$ and if the equation $\partial f /\partial\overline{z}=0$ is satisfied, which in expanded form may be written as If is differentiable in the sense of in , the latter relations are satisfied at all point of the domain; they are called the Cauchy–Riemann equations. These equations occurred already in the 18th century in J.L. d'Alembert's and L. Euler's studies on functions of a complex variable. The initial definition may now be rendered more precise as follows. A function , defined in a domain , is said to be holomorphic (analytic) at a point if there exists a neighbourhood of this point in which may be represented by a power series: If this requirement is satisfied at every point of , the function is said to be holomorphic (analytic) in the domain . A function which is holomorphic at a point is differentiable at that point. In addition, the sum of a convergent power series has derivatives of all orders (is infinitely differentiable) with respect to the complex variable ; the coefficients of the series may be expressed in terms of the derivatives of at by the formulas . The power series, written in the form is known as the Taylor series of at . Thus, holomorphy of a function in a domain means that it is infinitely differentiable at any point in and that its Taylor series converges to it in some neighbourhood of this point. On the other hand, the following noteworthy fact is established in the theory of analytic functions: If a function is differentiable in a domain , it is holomorphic in this domain (for a single point this statement is not true: is differentiable at , but is nowhere holomorphic). Thus, the concepts of complex differentiability and holomorphy of a function in a domain are identical; each one of the following properties of a function in a domain — differentiability in the sense of , differentiability in the sense of together with satisfaction of the Cauchy–Riemann equations, holomorphy — may serve as definition of analyticity of in this domain. One other characteristic of an analytic function is connected with the notion of an integral. The integral of a function along an (oriented rectifiable) curve : , , may be defined by the formula: or by means of a curvilinear integral: A key result in the theory of analytic functions is Cauchy's integral theorem: If is an analytic function in a domain , then for any closed curve bounding a domain inside . The converse result (Morera's theorem) is also true: If is continuous in a domain and if for any such curve , then is an analytic function in . In particular, in a simply-connected domain, those and only those continuous functions are analytic, whose integral along any closed curve is zero (or, which is the same thing, the integral along any curve connecting two arbitrary points does depend only on the points and themselves and not on the shape of the curve). This characterization of analytic functions forms the basis of many of their applications. Cauchy's integral theorem yields Cauchy's integral formula, which expresses the values of an analytic function inside a domain in terms of its values on the boundary: Here, is a domain whose boundary consists of a finite number of non-intersecting rectifiable curves (the orientation of is assumed to be positive with respect to ), and is a function which is analytic in some domain . This formula makes it possible, in particular, to reduce the study of many problems connected with analytic functions to the corresponding problems for a very simple function — the Cauchy kernel , , . For more details see Integral representation of an analytic function. A very important property of analytic functions is expressed by the following uniqueness theorem: Two functions which are analytic in a domain and which coincide on some set with a limit point in , coincide throughout (are identical). In particular, an analytic function , , which is not identically zero can only have isolated zeros in . If, in addition, is a zero of , then one has, in some neighbourhood of , , where is a natural number (called the multiplicity of the zero of at ), while is a analytic function in . An important role in the theory of analytic functions is played by the points at which the function is not analytic — the so-called singular points of the analytic function. Here, only isolated singular points of (single-valued) analytic functions are considered; for more details cf. Singular point. If is an analytic function in an annulus of the form , it may be expanded there in a Laurent series which contains, as a rule, not only positive but also negative powers of . If there are no terms with negative powers in the series ( for ), is called a regular point of (a removable singular point). At a regular point there also exists a finite limit Putting , one obtains an analytic function in the whole disc . If the Laurent series of the function contains only a finite number of terms with negative powers of : the point is called a pole of (of multiplicity ); a pole is characterized by The function has a pole of multiplicity at the point if and only if the function has a zero of multiplicity at this point. If the Laurent series contains an infinite number of negative powers of ( for an infinite set of negative indices ), then is called an essential singular point; at such points there is no finite and no infinite limit for . The coefficient in the Laurent series for with centre at the isolated singular point is called the residue of at : It can be defined by the formula where and is sufficiently small (so that the disc does not contain singular points of other than ). The important role of residues is made clear by the following theorem: If is an analytic function in a domain , except for some set of isolated singular points, if is a contour bounding a domain and not passing through any singular points of , and if are all the singular points of inside , then This theorem is an effective tool in calculating integrals. See also Residue of an analytic function. The sum of the terms of the Laurent series for at corresponding to the negative indices , is known as the principal part of the Laurent series (or of the function ) at the point . This principal part determines the nature of the singularity of at . Functions which are representable as a quotient of two functions that are holomorphic in a domain are called meromorphic in the domain . A function which is meromorphic in a domain is holomorphic in that domain, except possibly at a finite or countable set of poles; at the poles the values of a meromorphic function are considered to be infinite. If such values are allowed, then meromorphic functions in a domain may be defined as functions that in a neighbourhood of each point can be represented by a Laurent series in with a finite number of terms involving negative powers of (depending on ) in a neighbourhood of each point . Both holomorphic and meromorphic functions in a domain are often designated as analytic in the domain . In this a case holomorphic functions are also said to be regular analytic or simply regular functions. The simplest class of analytic functions consists of the functions which are holomorphic in the whole plane; such functions are called entire functions. Entire functions are represented by series which are convergent in the whole plane. This class includes the polynomials in , the functions etc. Weierstrass' theorem states that, for any sequence of complex numbers , without limit points in , there exists an entire function that vanishes at the points and only at these points (among the there may be coincident points, to which correspond zeros of of corresponding multiplicity). Here, the function may be represented as a (generally infinite) product of entire functions each one of which has only one zero. Functions that are meromorphic in the plane (i.e. that may be represented as quotients of entire functions) are called meromorphic functions. These include rational functions, , , elliptic functions, etc. According to Mittag-Leffler's theorem, for any sequence , without limit points in , there exists a meromorphic function with poles at the points and only at those points, such that its principal parts at the points coincide with pre-assigned polynomials in . The function may be represented as a (usually infinite) sum of meromorphic functions, each one with a pole at a single point only. Theorems on the existence of a holomorphic function with pre-assigned zeros and of meromorphic functions with pre-assigned poles and principal parts are also valid for an arbitrary domain . In the study of analytic functions the related geometric notions are also of importance. If is an analytic function, the image of the domain is also a domain (principle of preservation of domains). , If the mapping preserves angles at both in value and in sign, i.e. it is conformal. Thus, there exists a close connection between analyticity and the important geometric notion of conformal mapping. If is an analytic function in and for (such functions are called univalent), then in and defines a one-to-one and conformal mapping of the domain onto the domain . Riemann's theorem, which is the fundamental theorem in the theory of conformal mappings, says that on any simply-connected domain whose boundary contains more than one point there exist univalent analytic functions which conformally map this domain onto a disc or a half-plane (cf. Conformal mapping; Univalent function). The real and imaginary parts of a function which is holomorphic in a domain satisfy the Laplace equation in that domain: i.e. they are harmonic functions (cf. Harmonic function). Two harmonic functions which are connected by the Cauchy–Riemann equations are called conjugate. In a simply-connected domain any harmonic function has a conjugate function and is thus the real part of some holomorphic function in . The connections with conformal mappings and harmonic functions form the basis of many applications of the theory of analytic functions. A function ( being an arbitrary set), is called analytic at a point if there exists a neighbourhood of this point such that may be represented by a convergent power series on the intersection of this neighbourhood with . The function is called analytic on the set if it is analytic on some open set which contains (or, more exactly, if there exist both an open set containing and an analytic function on this set which coincides with on ). For open sets the notion to analyticity coincides with the notion of differentiability with respect to the set. However, this is not the case in general; in particular, on the real line there exist functions which not only have a derivative, but which are infinitely differentiable at every point and are not analytic even at a single point of this line. The property of connectedness of the set is necessary in order that the uniqueness theorem for analytic functions holds. This is why analytic functions are usually considered in domains, i.e. on connected open sets. All the preceding refers to single-valued analytic functions , considered in a given domain (or on a given set ) of the complex plane. In considering the possible extension of a function , as an analytic function, to a larger domain, one arrives at the concept of the analytic function considered as a whole, i.e. throughout its whole natural domain of existence. If the function is thus extended, its domain of analyticity becomes larger, and may overlap itself, supplying new values of the function at points in the plane where it already was defined. Accordingly, an analytic function considered as a whole is generally multi-valued. Many problems in analysis (inversion of a function, the determination of a primitive and the construction of an analytic function with a given real part in multiply-connected domains (cf. Multiply-connected domain), the solution of algebraic equations with analytic coefficients, etc.) require the study of multi-valued functions; such functions include , , , , algebraic functions, etc. (cf. Algebraic function). A regular process which yields the complete analytic function, considered throughout its natural domain of existence, was proposed by Weierstrass; it is known as Weierstrass' method of analytic continuation. The initial concept is that of an element of an analytic function, viz. a power series with a non-zero radius of convergence. Such an element : defines a certain analytic function on its disc of convergence . Let be a point of different from . Expanding in a series with centre at , one obtains a new element : whose disc of convergence will be denoted by . In the intersection of and the series converges to the same function as the series . If extends beyond the boundary of , the series defines the function determined by on some set outside (where the series is divergent). In such a case the element is called a direct analytic continuation of the element . Let be a chain of elements in which is a direct analytic continuation of (); the element is then said to be an analytic continuation of the element (by means of the given chain of elements). When the centre of the disc belongs to it may happen that the element is not a direct analytic continuation of the element . In such a case the sums of the series and will have different values in the intersection of and ; thus analytic continuation may lead to new values of the function inside . The totality of all elements which may be obtained by analytic continuation of an element forms the complete analytic function (in the sense of Weierstrass) generated by ; the union of their discs of convergence represents the (Weierstrass) domain of existence of this function. It follows from the uniqueness theorem for analytic functions that an analytic function in the sense of Weierstrass is completely determined by the given element . The initial element may be any other element belonging to this function; the complete analytic function will not be affected. A complete analytic function , considered as a function of the points in the plane belonging to its domain of existence , is generally multi-valued. In order to eliminate this feature, is considered not as a function of the points in the plane domain , but rather as a function of the points on some multi-sheeted surface (lying above ) such that to each point of there correspond as many points of the surface (projecting onto the given point of ) as there are different elements with centre at this point in the complete analytic function ; on the surface the function becomes single-valued. The idea of passing to such surfaces is due to Riemann, and the surfaces themselves are known as Riemann surfaces. The abstract definition of the notion of a Riemann surface has made it possible to replace the theory of multi-valued analytic functions by the theory of single-valued analytic functions on Riemann surfaces (cf. Riemann surface). Now fix a domain belonging to the domain of existence of the complete analytic function , and fix some element of with centre at a point in . The totality of all elements which may be obtained by analytic continuation of by means of chains with centres belonging to is called a branch of the analytic function . A branch of a multi-valued analytic function may turn out to be a single-valued analytic function in the domain . Thus, arbitrary branches of the functions and which correspond to an arbitrary simply-connected domain not containing the point 0, are single-valued functions. The function has exactly different branches in such a domain, while has an infinite set of such branches. The selection of single-valued branches (using some cuts in the domain of existence) and their study by the theory of single-valued analytic functions constitute one of the principal methods of studying specific multi-valued analytic functions. A.A. Gonchar ## Analytic functions of several complex variables. The complex space , consisting of the points , , is a vector space over the field of complex numbers with the Euclidean metric It differs from the -dimensional Euclidean space by a certain asymmetry: on passing from to (i.e. on introducing a complex structure in ), the coordinates are subdivided into pairs which appear in the complex combinations . If a complex function is defined in a domain and is differentiable at all points in the sense of (i.e. as a function of the real variables and ), its differential may be represented in the form Here while the symbols and are defined as in the case of the plane. If is of the form i.e. is a complex linear function in , the function is said to differentiable in the sense of or holomorphic or analytic in the domain . Thus, the condition of holomorphy of in a domain consists of the condition that it be differentiable in the sense of and that it satisfy the system of complex equations (), which is equivalent to the system of first-order partial differential equations (Cauchy–Riemann system). In the case of space () as distinct from that of the plane () this system is overdetermined, since the number of equations is larger than that of the unknown functions. It remains overdetermined on passing to the (geometrically more natural) spatial analogue of a holomorphic function of one complex variable — a holomorphic mapping , which is realized by a system of functions which are holomorphic in a domain . The mapping is called biholomorphic if it is one-to-one and if it is holomorphic together with its inverse . The conditions for holomorphy of a mapping are expressed by a system of real equations involving real functions. The overdeterminacy of the conditions of holomorphy for is the cause of a number of effects typical of the spatial case — such as the absence of a spatial analogue of the Riemann theorem on the existence of conformal mappings. According to Riemann's theorem, if , any two simply-connected domains whose boundaries do not reduce to a single point are isomorphic. However, if , even such simple simply-connected domains as the ball and the product of discs (polydisc) are non-isomorphic. The non-isomorphism is brought to light on comparing the groups of automorphisms of these domains (i.e. their biholomorphic mappings onto themselves, cf. Biholomorphic mapping) — the groups prove to be algebraically non-isomorphic, whereas a biholomorphic mapping of one domain onto another, if it existed, would establish an isomorphism of these groups. Owing to this, the theory of biholomorphic mappings of domains in complex space is substantially different from the theory of conformal mappings in the plane. A function is called holomorphic at a point if it is holomorphic in some neighbourhood of this point. According to the Cauchy–Riemann criterion, a function of several variables which is holomorphic at a point is holomorphic with respect to each variable (if the values of the other variables are fixed). The converse proposition is also true: If, in a neighbourhood of some point, a function is holomorphic with respect to each variable separately, then it is holomorphic at this point (Hartogs' fundamental theorem). In analogy with the case of the plane the holomorphy of a function at a point is equivalent to its expandability in a multiple power series in a neighbourhood of this point or, in abbreviated notation, where is a multi-index of integers , and A holomorphic function is infinitely differentiable, and the above series is its Taylor series, i.e. the derivatives being taken at the point . The fundamental facts of the theory of holomorphic functions of one variable extend to holomorphic functions of several variables, sometimes in an altered form. An instance of this is the Weierstrass preparation theorem (cf. Weierstrass theorem), which extends the property of holomorphic functions of one variable to become zero as an integral power of to the spatial case. The theorem is formulated as follows: If a function which is holomorphic at a point is equal to zero at that point, then it may be represented, in a certain neighbourhood (possibly after a non-degenerate linear transformation of the independent variables) in the form where is an integer, are functions of which are holomorphic in a neighbourhood of the point (a "prime" preceding a letter denotes the projection on the space of the first coordinates) and which are equal to zero at , while is holomorphic and zero-free in . This theorem is of fundamental importance in the study of analytic sets (cf. Analytic set), which are described locally, in a neighbourhood of each one of their points, as sets of common zeros of a certain number of functions which are holomorphic at this point. By Weierstrass' preparation theorem such sets may be locally described as sets of common zeros of polynomials in the variable , with coefficients from the ring of holomorphic functions in the other variables . This circumstance permits extensive use of algebraic methods in the local study of analytic sets. Cauchy's integral theorem must also be slightly modified in the spatial case, and is then known as the Cauchy–Poincaré theorem: Let a function be holomorphic in a domain ; then, for any -dimensional surface compactly imbedded in , with piecewise-smooth boundary , As in the planar case, this integral is defined by a parametric representation of the given set: If has the equation , where the parameter varies over an -dimensional cell , then, by definition, The difference between the spatial and the planar cases consists in the fact that in the former case the dimension of the surface is less than that of the domain , while in the planar case the dimensions are the same . The spatial analogue of the Cauchy integral formula can be written in a particularly simple form for polycylinder domains, i.e. for products of plane domains. Let be a domain in which is a domain in the complex plane with a piecewise-smooth boundary (), while the function is holomorphic in a domain which compactly contains . Successive application of Cauchy's formula for one variable then yields, for any point , where is an -dimensional surface in the boundary , , and However, polycylinder domains are only a very special class, and in general domains such a separation of variables is not possible. The role of Cauchy's integral for arbitrary domains with a piecewise-smooth boundary is played by the Martinelli–Bochner integral formula: For any function which is holomorphic in a domain containing , and for any point , where , and This is Green's formula for a pair of functions, one of which is holomorphic in , while the other is a fundamental solution of the Laplace equation in the space with singular point . If , this is the ordinary Cauchy integral. If , the formula differs from Cauchy's multiple integral for a product of plane domains in that, first, the integration is not over an -dimensional part of the boundary, but over the whole -dimensional boundary of the domain, and, secondly, its kernel (the factor multiplying under the integral sign) does not depend analytically on the parameter . An analytic kernel, however, is essential in a number of problems, and it is therefore desirable to construct an integral formula with such a kernel for as large a class of domains as possible. An ample supply of integral formulas, including formulas with an analytic kernel for many domains, is contained in the general Leray formula. This formula is where is a smooth vector function depending also on , and are defined above, and ; it is assumed that for any fixed and running over . The value of the integral in this formula does not depend on the choice of the vector function (provided that for all , ), and if , this integral coincides with the Martinelli–Bochner integral. By varying the choice of for different classes of domains, the Leray formula will yield various integral formulas. In the theory of analytic functions of several variables other integral representations, which are valid only for certain classes of domains, are also considered. An important class of this kind consists of the so-called Weil domains, which are a generalization of the product of plane domains. For such domains one has the Bergman–Weil representation with a kernel which also depends analytically on the parameter. As in the planar case, the study of the singularities of analytic functions is of fundamental interest; the main difference between the planar and the spatial cases is expressed by the Osgood–Brown theorem on the removability of compact singularities, according to which any function which is holomorphic in , where is a domain in () and is compact subset of which does not subdivide , extends holomorphically to the whole domain . By this theorem, holomorphic functions of several variables cannot have isolated singular points. These are replaced in () by singular sets which are analytic if their dimension is lower than . This fact is essential in the theory of multi-dimensional residues. This theory deals with the problem of computing the integral of a function , which is holomorphic everywhere in a domain except for an analytic set , over a closed -dimensional surface not intersecting . Since the dimension of the singular set is lower than the dimension of by at least two, does not subdivide . If the surface is not linked with , i.e. bounds an -dimensional surface compactly belonging to , then by the Cauchy–Poincaré theorem . In order to calculate this integral in the general case, it is necessary to clarify how is linked with the singular set , and to calculate the integrals over special -dimensional surfaces associated with separate portions of the set (residues). The solution of this problem involves considerable topological and analytic difficulties. These may often be overcome by the methods proposed by E. Martinelli and J. Leray. The Martinelli method is based on the use of the topological Aleksander–Pontryagin duality principle, and reduces the study of the -dimensional homologies of the set to the study of the -dimensional homologies of the singular set . The Leray method is more general: it is based on the examination of special homology classes and on the calculation of certain differential forms (residue forms). The multi-dimensional theory of residues has also found applications in theoretical physics (cf. Feynman integral). The Osgood–Brown theorem reveals an important fundamental difference between the spatial and the planar theories. In the plane one can, for any domain , construct a function which is holomorphic in but which cannot be extended analytically beyond its boundary, i.e. is the natural domain of existence. In space the situation is different: thus, the spherical shell cannot be the domain of existence of any holomorphic function, since by the Osgood–Brown theorem, any function which is holomorphic in it will certainly extend analytically to the entire ball . Thus arises the problem of the characterization of the natural domains of existence of holomorphic functions — the so-called domains of holomorphy. A simple sufficient condition may be formulated with the aid of the concept of a barrier at a boundary point of the domain, i.e. a function which is holomorphic in this domain and which increases without limit as tends to . will be a domain of holomorphy if it is possible to construct a barrier for an everywhere-dense set of points in its boundary. This condition is satisfied, in particular, by any convex domain; for any point it is sufficient to select, in the -dimensional supporting plane to at the point , a -dimensional plane of the form the function will then be a barrier. Consequently, every convex domain in is a domain of holomorphy. However, convexity is not a necessary condition for holomorphy: A product of plane domains is always a domain of holomorphy, and such a product need not be convex. Nevertheless, if the notion of convexity is suitably generalized, it is possible to arrive at a necessary and sufficient condition. One such generalization is based on the observation that the convex hull of a set may be described as the set of points at which the value of any linear function does not exceed the supremum of the values of this function on . By analogy, the holomorphically convex hull of a set is defined by where denotes the set of all functions which are holomorphic in . A domain is called holomorphically convex if, for every compact subset of , the hull also is a compact subset of . Holomorphic convexity is a necessary and sufficient condition for a domain of holomorphy. However, this criterion is not very effective, since holomorphic convexity is difficult to verify. Another generalization is connected with the notion of a plurisubharmonic function, which is the complex analogue of a convex function. A convex function in a domain of may be defined as a function for which the restrictions to the segments in of the straight lines (where and is a real parameter) are convex functions of . A real function , defined and upper semi-continuous in a domain , is said to be plurisubharmonic in if for each complex line (, is a complex parameter), its restriction to the parts of this line in is a subharmonic function of . If is twice continuously differentiable, then the condition of plurisubharmonicity, in accordance with the rules of differentiation of composite functions, is that the Hermitian form — the so-called Levi form — be non-negative. A domain is called pseudo-convex if the function , where denotes the Euclidean distance from the point to the boundary , is plurisubharmonic in this domain. Pseudo-convexity is also a necessary and sufficient condition for a domain to be a domain of holomorphy. In some cases it is possible to verify effectively the pseudo-convexity of a domain. As regards domains which are not domains of holomorphy, there arises the problem of describing their envelope of holomorphy, i.e. the smallest domain of holomorphy to which any function holomorphic in extends analytically. For domains of the simplest types envelopes of holomorphy can be effectively constructed, but in the general case the problem is unsolvable within the class of single-sheeted domains. Under analytic continuation of functions beyond the boundary of a given domain , multi-valuedness may result, which can be avoided by introducing multi-sheeted covering domains over , analogous to Riemann surfaces (cf. Riemann surface). In the class of covering domains, the problem of constructing envelopes of holomorphy is always solvable. This problem also has applications in theoretical physics, to wit in quantum field theory. The transition from the plane to a complex space substantially increases the variety of geometrical problems related to holomorphic functions. In particular, such functions are naturally considered not only in domains, but also on complex manifolds — smooth manifolds of even real dimension, the neighbourhood relations of which are biholomorphic. Among these, Stein manifolds (cf. Stein manifold) — natural generalizations of domains of holomorphy — play a special role Several problems in analysis may be reduced to the problem of constructing, in a given domain, a holomorphic function with given zeros or a meromorphic function with given poles and principal parts of the Laurent series. In the plane case, these problems have been solved for arbitrary domains by the theorems of Weierstrass and Mittag-Leffler and their generalizations. The spatial case is different — the solvability of the corresponding problems, the so-called Cousin problems, depends on certain topological and analytic properties of the complex manifolds considered. The key step in the solution of the Cousin problems is to construct — starting from locally-defined functions with given properties — a global function, defined on the whole manifold under consideration and having the same local properties. Such kinds of constructions are very conveniently effected using the theory of sheaves, which arose from the algebraic-topological treatment of the concept of an analytic function, and which has found important applications in various branches of mathematics. The solution of the Cousin problems by methods of the theory of sheaves was realized by H. Cartan and J.-P. Serre. B.V. Shabat ## The contemporary theory of analytic functions and their generalizations. This is one of the most important branches of analysis, it is closely connected with quite diverse branches of mathematics and it has numerous applications in theoretical physics, mechanics and technology. Fundamental investigations on the theory of analytic functions have been carried out by Soviet mathematicians. Extensive interest in the theory of functions of a complex variable emerged in the Soviet Union at the beginning of the 20th century. This was in connection with noteworthy investigations by Soviet scientists on applications of the theory of analytic functions to various problems in the mechanics of continuous media. N.E. Zhukovskii and S.A. Chaplygin solved very important problems in hydrodynamics and aerodynamics by using methods of the theory of analytic functions. In the works of G.V. Kolosov and N.I. Muskhelishvili these methods were applied to fundamental problems in the theory of elasticity. In subsequent years the theory of functions of a complex variable underwent extensive development. The development of various aspects of the theory of analytic functions was determined by the fundamental research of, among others, V.V. Golubev, N.N. Luzin, I.I. Privalov and V.I. Smirnov (boundary properties), M.A. Lavrent'ev (geometric theory, quasi-conformal mappings and their applications to gas dynamics), M.V. Keldysh, M.A. Lavrent'ev and L.I. Sedov (applications to problems in the mechanics of continuous media), D.E. Men'shov (theory of monogeneity), M.V. Keldysh, M.A. Lavrent'ev and S.N. Mergelyan (approximation theory), I.N. Vekua (theory of generalized analytic functions and their applications), A.O. Gel'fond (theory of interpolation), N.N. Bogolyubov and V.S. Vladimirov (theory of analytic functions of several variables and its application to quantum field theory). The development of the theory of analytic functions of one and several complex variables and their generalizations is continuing. See Boundary properties of analytic functions; Quasi-conformal mapping; Boundary value problems of analytic function theory; Approximation of functions of a complex variable. #### References [1] I.I. [I.I. Privalov] Priwalow, "Einführung in die Funktionentheorie" , 1–3 , Teubner (1958–1959) (Translated from Russian) MR0342680 MR0264037 MR0264036 MR0264038 MR0123686 MR0123685 MR0098843 Zbl 0177.33401 Zbl 0141.26003 Zbl 0141.26002 Zbl 0082.28802 [2] V.I. Smirnov, "A course of higher mathematics" , III-2 , Addison-Wesley (1964) (Translated from Russian) MR0182690 MR0182688 MR0182687 MR0177069 MR0168707 Zbl 0122.29703 Zbl 0121.25904 Zbl 0118.28402 Zbl 0117.03404 [3] A.I. Markushevich, "Theory of functions of a complex variable" , 1–3 , Chelsea (1977) (Translated from Russian) MR0444912 Zbl 0357.30002 [4] M.A. Lavrent'ev, B.V. Shabat, "Methoden der komplexen Funktionentheorie" , Deutsch. Verlag Wissenschaft. (1967) (Translated from Russian) [5] G.M. Goluzin, "Geometric theory of functions of a complex variable" , Transl. Math. Monogr. , 26 , Amer. Math. Soc. (1969) (Translated from Russian) MR0247039 Zbl 0183.07502 [6] M.A. Evgrafov, "Analytic functions" , Saunders , Philadelphia (1966) (Translated from Russian) MR0197686 Zbl 0147.32605 [7] A.G. Sveshnikov, A.N. Tikhonov, "The theory of functions of a complex variables" , Moscow (1967) (In Russian) MR0722295 MR0349962 MR0349961 [8] B.A. Fuks, "Theory of analytic functions of several complex variables" , 1–2 , Amer. Math. Soc. (1963–1965) (Translated from Russian) MR0188477 MR0174786 MR0168793 MR0155003 MR0037915 MR0027069 Zbl 0146.30802 Zbl 0138.30902 Zbl 0040.19002 [9] V.S. Vladimirov, "Methods of the theory of functions of several complex variables" , M.I.T. (1966) (Translated from Russian) [10] A.I. Markushevich, "Notes on the history of the theory of analytic functions" , Moscow-Leningrad (1951) (In Russian) [11] , Mathematics in the USSR during thirty years: 1917–1947 , Moscow-Leningrad (1948) pp. 319–414 (In Russian) [12] , Mathematics in the USSR during 40 years: 1917–1957 , 1 , Moscow (1959) pp. 381–510 (In Russian) [13] B.V. Shabat, "Introduction of complex analysis" , 1–2 , Moscow (1976) (In Russian) Zbl 0799.32001 Zbl 0732.32001 Zbl 0732.30001 Zbl 0578.32001 Zbl 0574.30001 [14] A.V. Bitsadze, "Fundamentals of the theory of analytic functions of a complex variable" , Moscow (1972) (In Russian) A.A. GoncharB.V. Shabat #### Comments Complex analysis, the theory of analytic functions, has been an active field of research up to the present time also in the West. Some of the history may be found in the extensive article in the Encyclopaedia Britannica entitled Analysis, complex. There is a large number of textbooks on complex analysis besides those listed above. Some good current books are listed below: [a1]–[a10] for the case of one variable, [a11]–[a18] for the case of several variables. Some final comments on complex analysis of several variables are in order. The "Osgood–Brown theorem on the removability of compact singularities" is usually called the Hartogs extension theorem in the West. The modern proof, due to L. Ehrenpreis, makes use of the inhomogeneous Cauchy–Riemann or -equations. The so-called -method, extensively developed by J.J. Kohn and L. Hörmander, has become one of the three most powerful techniques available in complex analysis today. Among other things, it can be used to obtain solutions of the Levi problem (are domains of holomorphy the same as pseudo-convex domains?) and the Cousin problems, cf. [a15]. These fundamental problems had been settled previously by methods now belonging to sheaf theory, and which go back to K. Oka (1936–1954), and have culminated in the sheaf-cohomology theory of H. Cartan, J.-P. Serre, H. Grauert and others, cf. the elegant treatment in [a12]. The third and most recent technique consists in the use of suitable integral representations as developed by G.M. Henkin [G.M. Khenkin] and E. Ramirez de Arellano, cf. [a14]. #### References [a1] L.V. Ahlfors, "Complex analysis" , McGraw-Hill (1966) MR0188405 Zbl 0154.31904 [a2] J.B. Conway, "Functions of one complex variable" , Springer (1973) MR0447532 Zbl 0277.30001 [a3] A. Dinghas, "Vorlesungen über Funktionentheorie" , Springer (1961) MR0179329 Zbl 0102.29301 [a4] P. Henrici, "Applied and computational complex analysis" , 1–3 , Wiley (1974–1986) MR1164865 MR1541308 MR1008928 MR0822470 MR0453984 MR0372162 Zbl 1107.30300 Zbl 0925.30003 Zbl 0635.30001 Zbl 0578.30001 Zbl 0363.30001 Zbl 0313.30001 [a5] E. Hille, "Analytic function theory" , 1–2 , Chelsea, reprint (1974) MR1532177 MR0201608 MR1530816 MR0107692 Zbl 0273.30002 Zbl 0102.29401 Zbl 0088.05204 [a6] Z. Nehari, "Conformal mapping" , Dover, reprint (1975) pp. 2 MR0377031 Zbl 0071.07301 Zbl 0052.08201 Zbl 0048.31503 Zbl 0041.41201 [a7] R. Nevanilinna, "Analytic functions" , Springer (1970) (Translated from German) [a8] R. Nevanlinna, V. Paatero, "Introduction to complex analysis" , Addison-Wesley (1969) (Translated from German) MR0239056 Zbl 0169.09001 [a9] W. Rudin, "Real and complex analysis" , McGraw-Hill (1974) pp. 24 MR0344043 Zbl 0278.26001 [a10] E.C. Titchmarsh, "The theory of functions" , Oxford Univ. Press (1939) MR0593142 MR0197687 MR1523319 Zbl 65.0302.01 [a11] H. Grauert, K. Fritzsche, "Several complex variables" , Springer (1976) (Translated from German) MR0414912 Zbl 0381.32001 [a12] H. Grauert, R. Remmert, "Theory of Stein spaces" , Springer (1979) (Translated from German) MR0580152 Zbl 0433.32007 [a13] R.C. Gunning, H. Rossi, "Analytic functions of several complex variables" , Prentice-Hall (1965) MR0180696 Zbl 0141.08601 [a14] G.M. [G.M. Khenkin] Henkin, J. Leiterer, "Theory of functions on complex manifolds" , Birkhäuser (1984) MR0795028 MR0774049 [a15] L. Hörmander, "An introduction to complex analysis in several variables" , North-Holland (1973) MR0344507 Zbl 0271.32001 [a16] S.G. Krantz, "Function theory of several complex variables" , Wiley (Interscience) (1982) MR0635928 Zbl 0471.32008 [a17] R. Narasimhan, "Several complex variables" , Univ. Chicago Press (1971) MR0342725 Zbl 0223.32001 [a18] W. Rudin, "Function theory in the unit ball in " , Springer (1980) MR601594 Zbl 0495.32001 [a19] R.M. Range, "Holomorphic functions and integral representation in several complex variables" , Springer (1986) MR0847923 How to Cite This Entry: Analytic function. Encyclopedia of Mathematics. URL: http://www.encyclopediaofmath.org/index.php?title=Analytic_function&oldid=29387 This article was adapted from an original article by A.A. Gonchar, B.V. Shabat (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article

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http://mathhelpforum.com/calculus/64839-solved-triple-integral-issue.html

# Thread: 1. ## [SOLVED] triple integral issue I need some help with a triple integral problem that I keep getting wrong in the last step somehow. the problem is: Evaluate $\int\int\int$ on E xyz dV where E lies between the spheres $\rho$=2 and $\rho$=4 and above the cone $\phi$= $\pi$/3. There work I've done so far is set up the integral integrated it and got: 1/4 sin^4 $\phi$ from 0 to $\pi$ * 1/2sin^2 $\theta$ from 0 to 2 $\pi$ * 1/6 $\rho$^6 from 2 to 4. I get that the middle term should be zero, but for some reason my prof gives the answer to be (1562 $\pi$)/15 2. That is a large value your professor got. I am not so sure he/she is correct. 3. That's what I was thinking as I am pretty sure the answer should be zero, but I thought that it wouldn't hurt to check. 4. Consider symmetry. The region of integration is rotationally symmetric about the z axis, and so if we rotate the function being integrated about the z axis, the integral should be unchanged. Now, let us rotate the integrand by 90 degrees about the z axis: we transform x and y to x' and y' via x'=-y and y'=x. Then the integrand is f(x',y',z)=xyz=(y')(-x')z=-x'y'z. Thus, we have $\iiint_{E}xyz\,dx\,dy\,dz=\iiint_{E'}-x'y'z\,dx'\,dy'\,dz$. But, if we rename the variables x' and y' as x and y, (and as E'=E), we have $\iiint_{E}xyz\,dx\,dy\,dz=-\iiint_{E}xyz\,dx\,dy\,dz$ which means $\iiint_{E}xyz\,dx\,dy\,dz=0$. Thus, due to the symmetry, the integral is zero. [We can also look at this as the fact that xyz's dependence on the azimuthal angle $\theta$ is of the form $\sin(2\theta)$, and the region has no dependence on the azimuthal angle (the rotational symmetry), so the integral over $\theta$ will be 0] --Kevin C. 5. Thanks everyone. I got an e-mail back from my professor and he said that it was a typo. Thanks again!

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http://crypto.stackexchange.com/questions/6417/why-do-we-assume-un-security-of-communication-channel-on-every-cryptography-syst

# Why do we assume un-security of communication channel on every cryptography system While reading about a few cryptographic systems, I noticed that we always assume the communication channel is not secured. Why is this assumption made? And, why the effort is being put into designing cryptographic systems rather than working on making these communication channels more secure? I know the question is basic, but really cannot find a complete answer. - 3 I suppose because you need cryptography to make those channels secure. And you can't just create a mega-channel and say "this is secure, use this for encrypted communication", because a "secure channel" requires authentication of both parties, which is inherently an individual process. – Thomas Feb 21 at 2:56 Additionally, if we can find ways of solving problems without having to rely on a secure channel, that's one fewer point of failure. What happens when our "secure" channel turns out not to be? – Stephen Touset Feb 21 at 3:54 Mathematics in general and cryptology in particular is about getting strong conclusions from weak assumptions. Public key cryptography for example can convert an insecure channel from A to B into a secret one assuming there was an authentic channel from B to A before. – jug Feb 21 at 8:43 ## 3 Answers The goal of cryptography is to create these secure communication channels. However, keep in mind that a secure channel is more than just an encrypted channel. A secure channel should be able to provide confidentiality (using encryption), data integrity (using something like signatures), and data authentication (using something like certificates, MACs). The goal of cryptography is to create these properties to make a "secure channel" out of an unsecured channel. Note that some cryptosystems build on top of other types of channels. For instance, in public-key cryptography, you generally need to assume that you have access to a authenticated channel for key exchanges (which is a stronger assumption than a completely unsecured channel). - For public key cryptography you need at least a trusted certificate, and this has to be secured through a secure channel. In this case the secured channel could however be e.g. a browser installation. After that you don't need secure channels anymore. – owlstead Feb 21 at 22:56 The point of the trusted certificate is to ensure that you are talking to who you think you're talking to. In other words, the point of this CA is to create an authentication channel to exchange keys. After the keys are exchange, the rest of the communication can occur over the fully unsecured channel. – Oleksi Feb 22 at 0:31 +1 although the second paragraph is still a bit vague to me - which is understandable given the question – owlstead Feb 22 at 13:09 If you already had a secure channel, you wouldn't need any cryptography. So if you're reading about cryptographic systems, you're reading about how to make things secure. Now, if you want to make a secure communications channel, the first thing you need is a communications channel. - 1 There would still be digital signatures, multi-party computation, and string commitment. $\hspace{0.95 in}$ – Ricky Demer Feb 21 at 5:02 1 Don't forget fair coin tosses. – Stephen Touset Feb 21 at 18:12 Any time you transmit information you have have to send it through some kind of medium ( air, wire, optical fiber, ect.). The medium can be tapped into at any point by an adversary and is thus considered insecure. -

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http://unapologetic.wordpress.com/2009/08/10/the-complex-spectral-theorem/

# The Unapologetic Mathematician ## The Complex Spectral Theorem We’re now ready to characterize those transformations on complex vector spaces which have a diagonal matrix with respect to some orthonormal basis. First of all, such a transformation must be normal. If we have a diagonal matrix we can find the matrix of the adjoint by taking its conjugate transpose, and this will again be diagonal. Since any two diagonal matrices commute, the transformation must commute with its adjoint, and is therefore normal. On the other hand, let’s start with a normal transformation $N$ and see what happens as we try to diagonalize it. First, since we’re working over $\mathbb{C}$ here, we can pick an orthonormal basis that gives us an upper-triangular matrix and call the basis $\left\{e_i\right\}_{i=1}^n$. Now, I assert that this matrix already is diagonal when $N$ is normal. Let’s write out the matrices for $N$ $\displaystyle\begin{pmatrix}a_{1,1}&\cdots&a_{1,n}\\&\ddots&\vdots\\{0}&&a_{n,n}\end{pmatrix}$ and $N^*$ $\displaystyle\begin{pmatrix}\overline{a_{1,1}}&&0\\\vdots&\ddots&\\\overline{a_{1,n}}&\cdots&\overline{a_{n,n}}\end{pmatrix}$ Now we can see that $N(e_1)=a_{1,1}e_1$, while $N^*(e_1)=\overline{a_{1,1}}e_1+\dots+\overline{a_{1,n}}e_n$. Since these bases are orthonormal, it’s easy to calculate the squared-lengths of these two: $\displaystyle\begin{aligned}\lVert N(e_1)\rVert^2&=\lvert a_{1,1}\rvert^2\\\lVert N^*(e_1)\rVert^2&=\lvert a_{1,1}\rvert^2+\dots+\lvert a_{1,n}\rvert^2\end{aligned}$ But since $N$ is normal, these two must be the same. And so all the entries other than maybe $a_{1,1}$ in the first row of our matrix must be zero. We can then repeat this reasoning with the basis vector $e_2$, and reach a similar conclusion about the second row, and so on until we see that all the entries above the diagonal must be zero. That is, not only is it necessary that a transformation be normal in order to diagonalize it, it’s also sufficient. Any normal transformation on a complex vector space has an orthonormal basis of eigenvectors. Now if we have an arbitrary orthonormal basis — say $N$ is a transformation on $\mathbb{C}^n$ with the standard basis already floating around — we may want to work with the matrix of $N$ with respect to this basis. If this were our basis of eigenvectors, $N$ would have the diagonal matrix $\Lambda=\Bigl(\lambda_i\delta_{ij}\Bigr)$. But we may not be so lucky. Still, we can perform a change of basis using the basis of eigenvectors to fill in the columns of the change-of-basis matrix. And since we’re going from one orthonormal basis to another, this will be unitary! Thus a normal transformation is not only equivalent to a diagonal transformation, it is unitarily equivalent. That is, the matrix of any normal transformation can be written as $U\Lambda U^{-1}$ for a diagonal matrix $\Lambda$ and a unitary matrix $U$. And any matrix which is unitarily equivalent to a diagonal matrix is normal. That is, if you take the subspace of diagonal matrices within the space of all matrices, then use the unitary group to act by conjugation on this subspace, the result is the subspace of all normal matrices, which represent normal transformations. Often, you’ll see this written as $U\Lambda U^*$, which is really the same thing of course, but there’s an interesting semantic difference. Writing it using the inverse is a similarity, which is our notion of equivalence for transformations. So if we’re thinking of our matrix as acting on a vector space, this is the “right way” to think of the spectral theorem. On the other hand, using the conjugate transpose is a congruence, which is our notion of equivalence for bilinear forms. So if we’re thinking of our matrix as representing a bilinear form, this is the “right way” to think of the spectral theorem. But of course since we’re using unitary transformations here, it doesn’t matter! Unitary equivalence of endomorphisms and of bilinear forms is exactly the same thing. ### Like this: Posted by John Armstrong | Algebra, Linear Algebra ## 9 Comments » 1. Should your first sentence end “with respect to some orthonormal basis”? Comment by Å | August 11, 2009 | Reply 2. Yeah, probably… Comment by | August 11, 2009 | Reply 3. I’m really REALLY enjoying these. I’m going nuts over it being 2 weeks until I theoretically am teaching full-time High School Math someplace in Los Angeles County. But no interviews, as school districts are paralyzed by the legislature having rejected several Billion dollars (yes, \$10^9) of Federal Funds because they don’t like the strings attached. Really. The legislators are lobbied by the teachers unions, who would rather their members be laid off than accept money tainted with the willingness to allow ANY use of student test score in evaluating teacher merit. Tactically, it’s excruciating for me to do on-line applications with district-by-district tweaks of a State-based “EdJoin” online system, that requires endless attachments to everything, but won’t take .doc files or TFF files, so that I physically had to scan 50 pages of hardcopy to blurry GIF files and attach them. It’s easy of newly graduated teachers, but I have 1/3 century of teaching and professorship and research and letters of recommendation to drag around, and they don’t imagine that this makes length-limit problems for online applications. Great. Chase away the over-qualified such as me and thee, and then complain that it’s hard to find good Math profs and secondary school teachers… Comment by | August 11, 2009 | Reply 4. It’s intruiging how all the anglo societies are setting themselves up ASAP for their own Century of Humiliation – I guess the only bright spot is that with the speed at which things move these days, it will only last a few decades Comment by Avery Andrews | August 12, 2009 | Reply 5. [...] the last couple lemmas we’ve proven and throw them together to prove the real analogue of the complex spectral theorem. We start with a self-adjoint transformation on a finite-dimensional real inner-product space [...] Pingback by | August 14, 2009 | Reply 6. [...] Singular Value Decomposition Now the real and complex spectral theorems give nice decompositions of self-adjoint and normal transformations, [...] Pingback by | August 17, 2009 | Reply 7. [...] Here’s a neat thing we can do with the spectral theorems: we can take square roots of positive-semidefinite transformations. And this makes sense, [...] Pingback by | August 20, 2009 | Reply 8. [...] we work with a normal transformation on a complex inner product space we can pick orthonormal basis of eigenvectors. That is, we can find a unitary transformation and a diagonal transformation so that [...] Pingback by | August 24, 2009 | Reply 9. [...] We're now ready to characterize those transformations on complex vector spaces which have a diagonal matrix with respect to some orthonormal basis. First of all, such a transformation must be normal. If we have a diagonal matrix we can find the matrix of the adjoint by taking its conjugate transpose, and this will again be diagonal. Since any two diagonal matrices commute, the transformation must commute with its adjoint, and is therefore normal. … Read More [...] Pingback by | November 18, 2010 | Reply « Previous | Next » ## About this weblog This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”). I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.

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http://mathoverflow.net/questions/120121?sort=newest

Mean value theorems for the Haar integral? Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Let $G$ be a compact topological group (feel free to add hypotheses if necessary). Is there any mean value theorem for its (normalized to 1) Haar integral? In general, are there mean value theorems for abstract spaces with measures? (Or at least for Borel measures?) Later edit: After reading the first two comments, let me be more precise; I'm looking for a theorem giving something like: for any continuous $f$ on $G$, there exist $x \in G$ such that $\int_G f(g) \mathrm{d}g = f(x)$. Does such an $x$ really exist? Can anything else be said about it (the integral being so special, maybe this $x$ can be made more precise)? - What is your substitute for the notion of an interval when you move outside the setting of R? Already for domains in R^2 one has to be a bit careful in finding a correct generalization of R^2? What notions have you tried thus far? – Yemon Choi Jan 28 at 16:20 Maybe it also helps if you indicate what do you need such a result for. – András Bátkai Jan 28 at 16:45 1 Answer Say $\mu$ is a Borel probability measure on a connected set $A$ in a topological space. Let $f : A \to \mathbb R$ be continuous. Then the mean value $\int_A f\;d\mu$ is equal to $f(a)$ for some $a \in A$. Proof: the mean value is between the sup of all values and the inf of all values, so (by connectedness) it is a value of the function. Of course the desired result fails for non-connected sets. Even in the two-point group we get a counterexample. - Excellent answer, thank you. But please note that your proof works in general, while I require more: if there is an underlying group structure and the integral is translation-invariant, do I get any extra information about the points where the mean value is attained? One minor remark: one must have $\mu (A)<\infty$ in order for the above proof to work. – Alex M Jan 28 at 20:40 One seems to need $\mu(A)=1$ in fact! Either that or Gerald forgot to divide by $\mu(A)$; both are trivial ways to fix things. – wccanard Jan 28 at 22:10 3 $\mu(A)=1$ is implied: $\mu$ was a probability measure on $A$. – Vince Jan 29 at 1:32 How would one extend the proof presented by Gerald Edgar above to the case of complex-valued functions? Separating into the real and imaginary part is a wrong approach, since this would give two different points. – Alex M Feb 7 at 17:16

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http://unapologetic.wordpress.com/2009/01/28/the-characteristic-polynomial/?like=1&_wpnonce=76a2d9cd23

# The Unapologetic Mathematician ## The Characteristic Polynomial Given a linear endomorphism $T:V\rightarrow V$ on a vector space $V$ of dimension $d$, we’ve defined a function — $\det(T-\lambda1_V)$ — whose zeroes give exactly those field elements $\lambda$ so that the $\lambda$-eigenspace of $T$ is nontrivial. Actually, we’ll switch it up a bit and use the function $\det(\lambda1_V-T)$, which has the same useful property. Now let’s consider this function a little more deeply. First off, if we choose a basis for $V$ we have matrix representations of endomorphisms, and thus a formula for their determinants. For instance, if $T$ is represented by the matrix with entries $t_i^j$, then its determinant is given by $\displaystyle\sum\limits_{\pi\in S_d}\mathrm{sgn}(\pi)\prod\limits_{k=1}^dt_k^{\pi(k)}$ which is a sum of products of matrix entries. Now, the matrix entries for the transformation $\lambda1_V-T$ are given by $t_i^j-\lambda\delta_i^j$. Each of these new entries is a polynomial (either constant or linear) in the variable $\lambda$. Any sum of products of polynomials is again a polynomial, and so our function is actually a polynomial in $\lambda$. We call it the “characteristic polynomial” of the transformation $T$. In terms of the matrix entries of $T$ itself, we get $\displaystyle\sum\limits_{\pi\in S_d}\mathrm{sgn}(\pi)\prod\limits_{k=1}^d(\lambda\delta_k^{\pi(k)}-t_k^{\pi(k)})$ What’s the degree of this polynomial? Well, first let’s consider the degree of each term in the sum. Given a permutation $\pi\in S_d$ the term is the product of $d$ factors. The $k$th of these factors will be a field element if $k\neq\pi(k)$, and will be a linear polynomial if $k=\pi(k)$. Since multiplying polynomials adds their degrees, the degree of the $\pi$ term will be the number of $k$ such that $k=\pi(k)$. Thus the highest possible degree happens if $k=\pi(k)$ for all index values $k$. This only happens for one permutation — the identity — so there can’t be another term of the same degree to cancel the highest-degree monomial when we add them up. And so the characteristic polynomial has degree $d$, equal to the dimension of the vector space $V$. What’s the leading coefficient? Again, the degree-$d$ monomial can only show up once, in the term corresponding to the identity permutation. Specifically, this term is $\prod\limits_{k=1}^d(\lambda-t_k^k)$ Each factor gives $\lambda$ a coefficient of ${1}$, and so the coefficient of the $\lambda^d$ term is also ${1}$. Thus the leading coefficient of the characteristic polynomial is ${1}$ — a fact which turns out to be useful. ### Like this: Posted by John Armstrong | Algebra, Linear Algebra ## 13 Comments » 1. [...] from the Characteristic Polynomial This one’s a pretty easy entry. If we know the characteristic polynomial of an endomorphism on a vector space of finite dimension , then we can get its determinant from [...] Pingback by | January 29, 2009 | Reply 2. [...] let’s take a linear endomorphism on a vector space of finite dimension . We know that its characteristic polynomial can be defined without reference to a basis of , and so each of the coefficients of is independent [...] Pingback by | January 30, 2009 | Reply 3. [...] does this assumption buy us? It says that the characteristic polynomial of a linear transformation is — like any polynomial over an algebraically closed field [...] Pingback by | February 2, 2009 | Reply 4. [...] remember that we can always calculate the characteristic polynomial of . This will be a polynomial of degree — the dimension of . Further, we know that a field [...] Pingback by | February 10, 2009 | Reply 5. [...] So we’ve got a linear transformation on a vector space of finite dimension . We take its characteristic polynomial to find the eigenvalues. If all of its roots are distinct (and there are of them, as there must be [...] Pingback by | February 11, 2009 | Reply 6. please send me the coefficient in the characteristic polynomial of signed graph of order n. Comment by | February 13, 2009 | Reply 7. A little demanding, aren’t we? Comment by | February 13, 2009 | Reply 8. [...] we saw that if the entries along the diagonal of an upper-triangular matrix are , then the characteristic polynomial [...] Pingback by | February 19, 2009 | Reply 9. [...] we can calculate the characteristic polynomial of , whose roots are the eigenvalues of . For each eigenvalue , we can define the generalized [...] Pingback by | March 4, 2009 | Reply 10. [...] of an Eigenpair An eigenvalue of a linear transformation is the same thing as a root of the characteristic polynomial of . That is, the characteristic polynomial has a factor . We can evaluate this polynomial at to [...] Pingback by | April 3, 2009 | Reply 11. [...] we found this in the complex case we saw that the characteristic polynomial had to have a root, since is algebraically closed. It’s the fact that isn’t [...] Pingback by | August 12, 2009 | Reply 12. [...] on the remaining -dimensional space. This tells us that all of our eigenvalues are , and the characteristic polynomial is , where . We can evaluate this on the transformation to find [...] Pingback by | January 19, 2010 | Reply 13. [...] the Jordan-Chevalley decomposition. We let have distinct eigenvalues with multiplicities , so the characteristic polynomial of [...] Pingback by | August 28, 2012 | Reply « Previous | Next » ## About this weblog This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”). I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.

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http://math.stackexchange.com/questions/7185/simple-solution-to-diffusion-equation

# Simple Solution to Diffusion equation A very simple solution to the diffusion equation is $u(x,t)=x^2+2 t$ My question: How can this be a solution to the diffusion equation when nothing really diffuses, but just stays the same - see plot: Here I can see that it is a solution but can't believe my eyes... Is there some kind of intuition other than: "it satisfies the PDE"? Are there other examples of solutions to well-known equations that behave completely un-intuitive in some situations? - ## 3 Answers Here are a few pointers, from the more obtuse to the more direct. 1. "Meeting a friend in a corridor, Wittgenstein said: 'Tell me, why do people always say it was natural for men to assume that the sun went round the earth, rather than that the earth was rotating?' His friend said, 'Well, obviously, because it looks as if the sun is going round the earth.' To which the philosopher replied, 'Well, what would it have looked like if it had looked as if the earth was rotating?'" Apparently from Tom Stoppard's play Jumpers. More to the point: what does your intuition think the solution to the diffusion equation with initial conditions $u(x,0) = x^2$ should have looked like? 2. $u_1(x,t) = \text{const}$, and $u_2(x,t) = x$ are also solutions of the diffusion equation. Do you find them more or less surprising than the $x^2 + t$ solution? 3. The standard intuitive interpretation of the diffusion equation is that it "smooths out" variations in the solution. The way it finds these variations is by looking at the second derivative. But the second derivative of $x^2$ is constant everywhere, so what can the diffusion equation do? 4. Locally, every point "wants to be" equal to the average of the value in its neighbourhood. If the second derivative at a point is positive, the function looks like a valley, so the point increases its value over time; if the second derivative is negative, the point is on a peak, and tries to decrease. With $x^2$, all of $\mathbb{R}$ is a valley. - Well, concerning 1. I would have thought that some flattening out takes place ending in a flat line, but for 2.: This really dropped my jaw to the floor! Esp. with $u_2$ there is such an disequilibrium (and it could be multiplied by any const. to make it even worse) and the whole situation doesn't change over time - now, where is the intuition here?!? – vonjd Oct 20 '10 at 9:54 @vonjd: Well, diffusion is certainly taking place, in that heat is flowing at a constant rate from the right to the left. But the temperature $u$ cannot change on any region, because exactly as much heat flows out from the left as flows in from the right. A more intuitive variant may be to think of diffusion on an interval $[-a,a]$ with the boundaries fixed at temperatures $-a$ and $a$. Then $u(x,t) = x$ on $x \in [-a,a]$ is a solution: heat comes in from the right, goes out the left. My example is what you get when you let $a \rightarrow \infty$. – Rahul Narain Oct 20 '10 at 18:58 But there is diffusion! Heat – or whatever quantity $u$ represents – flows from the far left and right ($x\to\pm\infty$, where there's lots of it!) towards the middle ($x=0$). For every fixed point $x$ there is always more heat flowing towards it from the outside than what it emits inwards, and the result is that the heat keeps increasing at each point. Now this might seem unphysical, and that's because it is. For this particular solution, the total amount of heat, $\int_{-\infty}^{\infty} u(x,t) dx$, is infinite for each $t$. - This is a very intuitive explanation - the fascinating thing is that it keeps it form which is lifted upwards over time. – vonjd Oct 20 '10 at 9:56 It is difficult to interpret a PDE without knowing its boundary conditions and/or initial values. But, at first guess, I'd say you should look at individual time slices of the solution on the same graph and see if that helps your interpretation. The 3D image is nice for showing everything at once, but that can obscure an interpretation. -

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

# Tagged Questions The collision tag has no wiki summary. 6answers 1k views ### What would happen if Large Hadron Collider would collide electrons? After some reading about the Large Hadron Collider and it's very impressive instruments to detect and investigate the collision results, there is a remaining question. What would happen if the ... 6answers 5k views ### Newton's cradle Why, when one releases 2 balls in Newton's cradle, two balls on the opposite side bounce out at approximately the same speed as the 1st pair, rather than one ball at higher speed, or 3 balls at lower ... 6answers 3k views ### Is two cars colliding at 50mph the same as one car colliding into a wall at 100 mph? I was watching a youtube video the other day where an economist said that he challenged his physics professor on this question back when he was in school. His professor said each scenario is the same, ... 7answers 2k views ### Would it help if you jump inside a free falling elevator? 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When the board breaks in two due to my force, the halves have a ... 1answer 191 views ### Anti-Matter for Neutrons The anti-particle corresponding to a proton or an electron is a particle with an equal mass, but an opposite charge. So what is the anti-particle corresponding to a neutron (which does not possess a ... 2answers 190 views ### Spinning spheres colliding In an ideal environment with no friction, in a vacuum, what happens to the velocity of the spin of two spheres spinning in perfect parity at two different velocities when they come into contact? 2answers 795 views ### What causes destruction in car crash? Suppose a car crashes at a speed $v$ against a wall and comes to a stop. Now if the car crashes at $2v$, does that mean it suffers twice as much destruction, if that can be objectively measured? If ... 2answers 903 views ### Conservation of Momentum/Energy collision Problem I'm working on a physics problem in preparation for the MCAT and there's this particular problem that's troubling me. I don't know if it's a bad question or if I'm not understanding some sort of ... 1answer 160 views ### What is the function of the top point of a bouncing ball? A ball is thrown away as parallel to x axis from M(0,h) point with speed V . After each jumping on x axis , it can reach half of previous height as shown in the figure.(Assume that no any air ... 1answer 75 views ### Angular momentum after elastic collision If two balls collide (elastically) and there is no friction between them, will their angular momentum change after the collision? 1answer 86 views ### Why do we hear a higher pitched sound out of water when we hit two stones against each other in the water? The observer is outside the water; the stones are underwater (say, 1 m below surface, if that matters). This produces a higher pitched sound for the observer than when both the observer and the stones ... 1answer 368 views ### Simple 2D Vehicle collision physics I'm trying to create a simplified GTA 2 clone to learn. I'm onto vehicle collisions physics. The basic idea I would say is, To apply force F determined by vehicle A's position and velocity onto point ... 1answer 502 views ### What phrases describe collisions with coefficients of restitution less than zero or greater than one? The coefficient of restitution describes the elasticity of a collision: 1 = perfectly elastic, kinetic energy is conserved 0 = perfectly inelastic, the objects move at the same speed post impact ... 2answers 671 views ### Is there a 2D generalization of the coefficient of restitution? The coefficient of restitution characterizes a collision in one dimension by relating the initial and final speeds of the particles involved, $$C_R = -\frac{v_{2f} - v_{1f}}{v_{2i} - v_{1i}}$$ In a ... 0answers 57 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 ... 3answers 173 views ### Simple elastic collision If a particle with mass $m$ collides with a wall at right angles, and the collision is perfectly elastic. The particle hits the wall at $v\ ms^{-1}$. There is no friction or gravity. So the particle ... 3answers 251 views ### Train crash: are these situations alike? I was just wondering... I believe that if a car travelling 50 miles per hour crashes into a wall, the result should be the same as crashing to another car also travelling 50 miles per hour (but in the ... 3answers 204 views ### What is the strange event in this simulation of a galactic collision? I was watching this video on YouTube: 2 Spiral Galaxies w/Supermassive Black Holes Collide Around half way, and again almost at the end, the black holes seem to suddenly give off some sort of force ... 2answers 214 views ### Small car colliding with large truck A small car collides with a large truck. Why do both vehicles experience the same magnitude of force? Wouldn't the large vehicle experience less force than the small one? 1answer 437 views ### Physics needed to build a top down billiards game [duplicate] Possible Duplicate: How are these balls reflected after they hit each other? I was wondering what sort of physics equations would I need in order to build a top down billiards game? I tried ... 1answer 64 views ### Is Joule heating only between charged particles? The Wikipedia page for Joule heating explains "It is now known that Joule heating is caused by interactions between the moving particles that form the current (usually, but not always, electrons) and ... 2answers 109 views ### Elastic collision and spring Bodies $A$ and $B$ are moving in the same direction in a straight line with a constant velocities on a frictionless surface. The mass and the velocity of $A$ are $2 \text{kg}$ and $10 \text{m/s}$. ... 0answers 88 views ### Elastic collision of rotating bodies How would you explain in detail elastic collision of two rotating bodies to someone with basic understanding of classical mechanics? I'm writing simple physics engine, but now only simulating ... 1answer 94 views ### Swords, impacts and elasticity for a noob So I'm a game developer and I'm trying to understand some (extremely) basic facts of impact mechanics. I had read something entitled Dynamics of Hand-Held Impact Weapons, but it was a bit too ... 0answers 251 views ### kinetic energy in collisions We were hoping you could help us understand collision energy. Vehicle A is driving West at 35mph and weighs 1437kg. Vehicle B is driving North at 35 mph and weighs 1882kg. Vehicle B crashes into the ... 1answer 118 views +50 ### Firing machine question Suppose we have a firing machine on a frictionless surface at point $x=0$. It fires a bullet of mass $m$ every $T$ seconds. Each bullet has the same constant velocity $v_0$. There's a body of mass ... 9answers 2k views ### Should you really lean into a punch? There's a conventional wisdom that the best way to minimize the force impact of a punch to the head is to lean into it, rather than away from it. Is it true? If so, why? EDIT: Hard to search for ... 3answers 159 views ### The theory of moon creation when a Mars size planet hit Earth As we know the predominant theory where does the moon come from is that a Mars size planet hit the earth and took a chunk out of it which eventually materialized into moon. My question is that if a ... 1answer 68 views ### How to get the new direction of 2 disks colliding? I'm developing a 2D game including collisions between many disks. I would like to know how I can get the angle corresponding to the new direction of each disk. For every disk I have this information ... 1answer 98 views ### Could ions emitted by an ion thruster represent any realistic danger? As ion thurster designs improve, the ions emitted could have a velocity (relative to the spacecraft) of well above 10^5 or even 10^6 m/s. It the likelihood of any such ions ever hitting a human ... 1answer 150 views ### Momentum And A Car Collision I am studying an example problem, concerning the very topic mentioned in the title. In this example problem, a car has a head-on collision with the wall; the initial and final velocity are known, as ... 2answers 957 views ### Inelastic collision and conservation of linear and angular momentum Is it possible for two spheres (a & b) to have an inelastic collision with BOTH the total linear and angular momentum preserved? I'm doing some physics simulation of some spheres attracting each ... 2answers 939 views ### What are all the equations we use to calculate how bounces work? I mean, what is the object's final displacement, or the function that describes the object's height over time (see [1]) of an object thrown by a height $h$ with a speed of $\vec{v_0}$, a mass of $m$, ... 1answer 35 views ### Finding the coffecient of restitution A ball moving with velocity $1 \hat i \ ms^{-1}$ and collides with a friction less wall, afetr collision the velocity of ball becomes $1/2 \hat j \ ms^{-1}$. Find the coefficient of restitution ... 1answer 79 views ### Simulation of broken object I was thinking and the following question came out: how an object that is falling is simulated once it hits the ground? Specifically, I would like to understand how one would be able to simulate the ... 1answer 414 views ### Scattering problem: Converting the two-body lab frame problem into a one-body center-of-mass frame problem I'm reading the section on scattering in Goldstein's Classical Mechanics, and I have a rather basic question about this. It says that scattering in the laboratory is a two-body problem because of ... 1answer 188 views ### Inelastic Collision and incident angle This problem is puzzling me. Scenario 1: A certain sphere is having an innelastic collision with a plane of a very large mass. Let the angle between the sphere's velocity and the surface be 90 ...

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

# Divergent series In mathematics, a divergent series is an infinite series that is not convergent, meaning that the infinite sequence of the partial sums of the series does not have a limit. If a series converges, the individual terms of the series must approach zero. Thus any series in which the individual terms do not approach zero diverges. However, convergence is a stronger condition: not all series whose terms approach zero converge. The simplest counterexample is the harmonic series $1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \cdots =\sum_{n=1}^\infty\frac{1}{n}.$ The divergence of the harmonic series was proven by the medieval mathematician Nicole Oresme. In specialized mathematical contexts, values can be usefully assigned to certain series whose sequence of partial sums diverges. A summability method or summation method is a partial function from the set of sequences of partial sums of series to values. For example, Cesàro summation assigns Grandi's divergent series $1 - 1 + 1 - 1 + \cdots$ the value 1/2. Cesàro summation is an averaging method, in that it relies on the arithmetic mean of the sequence of partial sums. Other methods involve analytic continuations of related series. In physics, there are a wide variety of summability methods; these are discussed in greater detail in the article on regularization. ## Theorems on methods for summing divergent series A summability method M is regular if it agrees with the actual limit on all convergent series. Such a result is called an abelian theorem for M, from the prototypical Abel's theorem. More interesting and in general more subtle are partial converse results, called tauberian theorems, from a prototype proved by Alfred Tauber. Here partial converse means that if M sums the series Σ, and some side-condition holds, then Σ was convergent in the first place; without any side condition such a result would say that M only summed convergent series (making it useless as a summation method for divergent series). The operator giving the sum of a convergent series is linear, and it follows from the Hahn–Banach theorem that it may be extended to a summation method summing any series with bounded partial sums. This fact is not very useful in practice since there are many such extensions, inconsistent with each other, and also since proving such operators exist requires invoking the axiom of choice or its equivalents, such as Zorn's lemma. They are therefore nonconstructive. The subject of divergent series, as a domain of mathematical analysis, is primarily concerned with explicit and natural techniques such as Abel summation, Cesàro summation and Borel summation, and their relationships. The advent of Wiener's tauberian theorem marked an epoch in the subject, introducing unexpected connections to Banach algebra methods in Fourier analysis. Summation of divergent series is also related to extrapolation methods and sequence transformations as numerical techniques. Examples for such techniques are Padé approximants, Levin-type sequence transformations, and order-dependent mappings related to renormalization techniques for large-order perturbation theory in quantum mechanics. ## Properties of summation methods Summation methods usually concentrate on the sequence of partial sums of the series. While this sequence does not converge, we may often find that when we take an average of larger and larger initial terms of the sequence, the average converges, and we can use this average instead of a limit to evaluate the sum of the series. So in evaluating a = a0 + a1 + a2 + ..., we work with the sequence s, where s0 = a0 and sn+1 = sn + an+1. In the convergent case, the sequence s approaches the limit a. A summation method can be seen as a function from a set of sequences of partial sums to values. If A is any summation method assigning values to a set of sequences, we may mechanically translate this to a series-summation method AΣ that assigns the same values to the corresponding series. There are certain properties it is desirable for these methods to possess if they are to arrive at values corresponding to limits and sums, respectively. 1. Regularity. A summation method is regular if, whenever the sequence s converges to x, A(s) = x. Equivalently, the corresponding series-summation method evaluates AΣ(a) = x. 2. Linearity. A is linear if it is a linear functional on the sequences where it is defined, so that A(k r + s) = k A(r) + A(s) for sequences r, s and a real or complex scalar k. Since the terms an = sn+1 − sn of the series a are linear functionals on the sequence s and vice versa, this is equivalent to AΣ being a linear functional on the terms of the series. 3. Stability. If s is a sequence starting from s0 and s′ is the sequence obtained by omitting the first value and subtracting it from the rest, so that s′n = sn+1 − s0, then A(s) is defined if and only if A(s′) is defined, and A(s) = s0 + A(s′). Equivalently, whenever a′n = an+1 for all n, then AΣ(a) = a0 + AΣ(a′). The third condition is less important, and some significant methods, such as Borel summation, do not possess it. One can also give a weaker alternative to the last condition. 1. Finite Re-indexability. If s and s′ are two sequences such that there exists a bijection $f: \mathbb{N} \rightarrow \mathbb{N}$ such that si = s′f(i) for all i, and if there exists some $N \in \mathbb{N}$ such that si = s′i for all i > N, then A(s) = A(s′). (In other words, s′ is the same sequence as s, with only finitely many terms re-indexed.) Note that this is a weaker condition than Stability, because any summation method that exhibits Stability also exhibits Finite Re-indexability, but the converse is not true. A desirable property for two distinct summation methods A and B to share is consistency: A and B are consistent if for every sequence s to which both assign a value, A(s) = B(s). If two methods are consistent, and one sums more series than the other, the one summing more series is stronger. There are powerful numerical summation methods that are neither regular nor linear, for instance nonlinear sequence transformations like Levin-type sequence transformations and Padé approximants, as well as the order-dependent mappings of perturbative series based on renormalization techniques. ## Axiomatic methods Taking regularity, linearity and stability as axioms, it is possible to sum many divergent series by elementary algebraic manipulations. For instance, whenever r ≠ 1, the geometric series $\begin{align} G(r,c) & = \sum_{k=0}^\infty cr^k & & \\ & = c + \sum_{k=0}^\infty cr^{k+1} & & \mbox{ (stability) } \\ & = c + r \sum_{k=0}^\infty cr^k & & \mbox{ (linearity) } \\ & = c + r \, G(r,c), & & \mbox{ whence } \\ G(r,c) & = \frac{c}{1-r} , & & \\ \end{align}$ can be evaluated regardless of convergence. More rigorously, any summation method that possesses these properties and which assigns a finite value to the geometric series must assign this value. However, when r is a real number larger than 1, the partial sums increase without bound, and averaging methods assign a limit of ∞. ## Nörlund means Suppose pn is a sequence of positive terms, starting from p0. Suppose also that $\frac{p_n}{p_0+p_1 + \cdots + p_n} \rightarrow 0.$ If now we transform a sequence s by using p to give weighted means, setting $t_m = \frac{p_m s_0 + p_{m-1}s_1 + \cdots + p_0 s_m}{p_0+p_1+\cdots+p_m}$ then the limit of tn as n goes to infinity is an average called the Nörlund mean Np(s). The Nörlund mean is regular, linear, and stable. Moreover, any two Nörlund means are consistent. The most significant of the Nörlund means are the Cesàro sums. Here, if we define the sequence pk by $p_n^k = {n+k-1 \choose k-1}$ then the Cesàro sum Ck is defined by Ck(s) = N(pk)(s). Cesàro sums are Nörlund means if k ≥ 0, and hence are regular, linear, stable, and consistent. C0 is ordinary summation, and C1 is ordinary Cesàro summation. Cesàro sums have the property that if h > k, then Ch is stronger than Ck. ## Abelian means Suppose λ = {λ0, λ1, λ2, ...} is a strictly increasing sequence tending towards infinity, and that λ0 ≥ 0. Recall that an = λn+1 − λn is the associated series whose partial sums form the sequence λ. Suppose $f(x) = \sum_{n=0}^\infty a_n \exp(-\lambda_n x)$ converges for all real numbers x>0. Then the Abelian mean Aλ is defined as $A_\lambda(s) = \lim_{x \rightarrow 0^{+}} f(x).$ A series of this type is known as a generalized Dirichlet series; in applications to physics, this is known as the method of heat-kernel regularization. Abelian means are regular, linear, and stable, but not always consistent between different choices of λ. However, some special cases are very important summation methods. ### Abel summation See also: Abel's theorem If λn = n, then we obtain the method of Abel summation. Here $f(x) = \sum_{n=0}^\infty a_n \exp(-nx) = \sum_{n=0}^\infty a_n z^n,$ where z = exp(−x). Then the limit of ƒ(x) as x approaches 0 through positive reals is the limit of the power series for ƒ(z) as z approaches 1 from below through positive reals, and the Abel sum A(s) is defined as $A(s) = \lim_{z \rightarrow 1^{-}} \sum_{n=0}^\infty a_n z^n$ Abel summation is interesting in part because it is consistent with but more powerful than Cesàro summation: A(s) = Ck(s) whenever the latter is defined. The Abel sum is therefore regular, linear, stable, and consistent with Cesàro summation. ### Lindelöf summation If λn = n ln(n), then (indexing from one) we have $f(x) = a_1 + a_2 2^{-2x} + a_3 3^{-3x} + \cdots .$ Then L(s), the Lindelöf sum (Volkov 2001), is the limit of ƒ(x) as x goes to zero. The Lindelöf sum is a powerful method when applied to power series among other applications, summing power series in the Mittag-Leffler star. If g(z) is analytic in a disk around zero, and hence has a Maclaurin series G(z) with a positive radius of convergence, then L(G(z)) = g(z) in the Mittag-Leffler star. Moreover, convergence to g(z) is uniform on compact subsets of the star. ## References • Arteca, G.A.; Fernández, F.M.; Castro, E.A. (1990), Large-Order Perturbation Theory and Summation Methods in Quantum Mechanics, Berlin: Springer-Verlag . • Baker, Jr., G. A.; Graves-Morris, P. (1996), Padé Approximants, Cambridge University Press . • Brezinski, C.; Zaglia, M. Redivo (1991), Extrapolation Methods. Theory and Practice, North-Holland . • Hardy, G. H. (1949), Divergent Series, Oxford: Clarendon Press . • LeGuillou, J.-C.; Zinn-Justin, J. (1990), Large-Order Behaviour of Perturbation Theory, Amsterdam: North-Holland . • Volkov, I.I. (2001), "Lindelöf summation method", in Hazewinkel, Michiel, , Springer, ISBN 978-1-55608-010-4 . • Zakharov, A.A. (2001), "Abel summation method", in Hazewinkel, Michiel, , Springer, ISBN 978-1-55608-010-4 .

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http://physics.stackexchange.com/questions/36128/laplacian-of-a-delta-function-as-an-interaction-potential-for-laughlin-state/36130

# Laplacian of a delta function as an interaction potential for Laughlin state I am reading Xiao-Gang Wen's paper "Pattern-of-zeros approach to Fractional quantum Hall states and a classification of symmetric polynomial of infinite variables", on page 8, he gives three interaction potentials for various Laughlin state and claim that these Laughlin state are exactly the zero-energy ground state of the given interaction potential. I do have some questions about these interaction potentials. 1. I do believe that the second term in the interaction potential for $\nu=1/4$ is from Kivelson-Trugman's delta function expansion ("Exact results for the fractional quantum Hall effect with general interactions"), but I think in their paper, the Laplacian is with respect to the relative coordinate which is $z_{rel}=z_1-z_2$, but why in Xiao-Gang's paper that Laplacian is only with respect to $z_1$? 2. Why the delta function only involves $z$, but no conjugate $z^*$? Isn't a two dimensional delta function usually defined as $\delta(z,z^*)$? - – Qmechanic♦ Sep 11 '12 at 7:45 ## 1 Answer I haven't read your paper but is Laplacian not invariant by translation ? Z=z1-z2 For the notation for δ(z,z∗), you are right to say that to recover x, y you need z and z*, but it might just be a shorthening of the writing. - The situation I am concerned is that when you treat the delta function as a limiting case of a Gaussian function(say if we want to investigate the interaction range for this Laplacian of the delta function), then if z is the only variable wouldn't make sense. – huyichen Sep 11 '12 at 6:16 – Shaktyai Sep 11 '12 at 8:21 I am aware of that definition, but I don't think that is quite equivalent to a two dimensional delta function, maybe a delta function for analytic complex functions. – huyichen Sep 11 '12 at 18:36

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http://www.physicsforums.com/showthread.php?t=53270

Physics Forums ## Fluid Mechanics - Pressure measurements I'm having difficulty with one problem and was hoping someone could help me out. Here's the problem: Mercury is poured into a U-tube as in Figure P14.18a. The left arm of the tube has a cross-sectional area A1 of 107.0 cm2, and the right arm has a cross-sectional area A2 of 4.10 cm2. One hundred grams of water are then poured into the right arm, as in Figure P14.18b. (a) Determine the length of the water column in the right arm of the U-tube (b) Given that the density of mercury is 13.6 g/cm3, what distance h does the mercury rise in the left arm? I've attached the picture of the figure to this post For part (a), given the radius of A2 and the volume of water, I easily got the length of the water column: 24.5cm Part (b) is giving me a problem. How do I equate the change in height of the mercury due to the pressure of the water on it? I can determine the pressure of the water (P=F/A) on the mercury and thereby determine the upward pressure on the left arm (A2/A1=F2/F1). Once I get the change in height (dy) of the right arm I can use it to figure out the subsequent change on the left arm (A1/dy1=A2/dy2) but getting to that point is difficult for me. I would greatly appreciate any suggestions or a point in the right direction. Thanks! 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: Gold Member Science Advisor Open the archive attached, I have modified the yours one a bit. Pay attention to the new references z, P_1, and L. P_1 is the pressure just at the interface water-mercury, and L is the lenght obtained in a). Pressure at the interface (Hydrostatics): $$P_1=P_{at}+\rho_w g L$$ that pressure is the same at the same height inside the mercury. So that, we can relate the jump of pressures at the left side: $$P_1=P_{at}+\rho_{Hg} g (h+z)$$ So that: $$\rho_w g L=\rho_{Hg} g (h+z)$$ (1) In addition and due to the mass conservation: $$\rho_{Hg} z A_2=\rho_{Hg} h A_1$$ (2) With (1) and (2) you can solve for h. Attached Images untitled.bmp (47.2 KB, 167 views) Thank you very much! The solution you provided worked perfectly. I don't think I would have ever come to the same conclusion on my own but with your explanation I understand the physics involved and how you arrived at the answer. Thanks again! Thread Tools Similar Threads for: Fluid Mechanics - Pressure measurements Thread Forum Replies Introductory Physics Homework 1 Classical Physics 8 Introductory Physics Homework 1 Introductory Physics Homework 12 Introductory Physics Homework 14

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http://cs.stackexchange.com/questions/439/which-combinations-of-pre-post-and-in-order-sequentialisation-are-unique

# Which combinations of pre-, post- and in-order sequentialisation are unique? We know post-order, ````post L(x) => [x] post N(x,l,r) => (post l) ++ (post r) ++ [x] ```` and pre-order ````pre L(x) => [x] pre N(x,l,r) => [x] ++ (pre l) ++ (pre r) ```` and in-order traversal resp. sequentialisation. ````in L(x) => [x] in N(x,l,r) => (in l) ++ [x] ++ (in r) ```` One can easily see that neither describes a given tree uniquely, even if we assume pairwise distinct keys/labels. Which combinations of the three can be used to that end and which can not? Positive answers should include an (efficient) algorithm to reconstruct the tree and a proof (idea) why it is correct. Negative answers should provide counter examples, i.e. different trees that have the same representation. - ## 1 Answer First, I'll assume that all elements are distinct. No amount of sequentialisations is going to tell you the shape of a tree with elements `[3,3,3,3,3]`. It is possible to reconstruct some trees with duplicate elements, of course; I don't know what nice sufficient conditions exist. Continuing on the negative results, you can't fully rebuild a binary tree from its pre-order and post-order sequentializations alone. `[1,2]` preorder, `[2,1]` post-order has to have `1` at the root, but `2` can be either the left child or the right child. If you don't care about this ambiguity, you can reconstruct the tree with the following algorithm: • Let $[x_1,\dots,x_n]$ be the pre-order traversal and $[y_n,\ldots,y_1]$ be the post-order traversal. We must have $x_1=y_1$, and this is the root of the tree. • $x_2$ is the leftmost child of the root, and $y_2$ is the rightmost child. If $x_2 = y_2$, the root node is unary; recurse over $[x_2,\ldots,x_n]$ and $[y_n,\ldots,y_2]$ to build the single subtree. • Otherwise, let $i$ and $j$ be the indices such that $x_2 = y_i$ and $y_2 = x_j$. $[x_2,\ldots,x_{j-1}]$ is the pre-order traversal of the left subtree, $[x_j,\ldots,x_n]$ that of the right subtree, and similarly for the post-order traversals. The left subtree has $j-2=n-i+1$ elements, and the right subtree has $i-2=n-j+1$ elements. Recurse once for each subtree. By the way, this method generalizes to trees with arbitrary branching. With arbitrary branching, find out the extent of the left subtree and cut off its $j-2$ elements from both lists, then repeat to cut off the second subtree from the left, and so on. As stated, the running time is $O(n^2)$ with $\Theta(n^2)$ worst case (in the case with two children, we search each list lineraly). You can turn that into $O(n\,\mathrm{lg}(n))$ if you preprocess the lists to build an $n\,\mathrm{lg}(n)$ finite map structure from element values to positions in the input lists. Also use an array or finite map to go from indices to values; stick to global indices, so that recursive calls will receive the whole maps and take a range as argument to know what to act on. With the pre-order traversal $[x_1,\ldots,x_n]$ and the in-order traversal $[z_1,\ldots,z_n]$, you can rebuild the tree as follows: • The root is the head of the pre-order traversal $x_1$. • Let $k$ be the index such that $z_k = x_1$. Then $[z_1,\ldots,z_{k-1}]$ is the in-order traversal of the left child and $[z_{k+1},\ldots,z_n]$ is the in-order traversal of the right child. Going by the number of elements, $[x_2,\ldots,x_k]$ is the pre-order traversal of the left child and $[x_{k+1},\ldots,x_n]$ that of the right child. Recurse to build the left and right subtrees. Again, this algorithm is $O(n^2)$ as stated, and can be performed in $O(n\,\mathrm{lg}(n))$ if the list is preprocessed into a finite map from values to positions. Post-order plus in-order is of course symmetric. - Is there a typo here: "[1,2] preorder, [1,2] post-order has to have 1 at the root, but 2 can be either the left child or the right child. " The post order of such a tree would be [2,1] not [1,2] whether 2 is a left or right child. Also, do you mean if both preorder and postorder are given, we cannot reconstruct the tree, or do you mean if we are given only one of them then we cannot reconstruct the tree? – CEGRD Jun 20 '12 at 14:55 @CEGRD Indeed, the postorder was a typo. The example shows that you cannot fully reconstruct the tree in this case: you can't know whether `2` is a left child or a right child. This corresponds to the “single subtree” case of the reconstruction algorithm. – Gilles♦ Jun 20 '12 at 19:55

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http://math.stackexchange.com/questions/159700/complex-analysis-exercises?answertab=oldest

# Complex analysis exercises These two questions are driving me mad as I need to help my daughter but I can't remember all this stuff. $\,(1)\,\,$ Let $\,p(z)\,,\,q(z)\,$ be two non-constant complex polynomials of the same degree s.t. $$\text{whenever |z|=1}\,\,,\,|p(z)|=|q(z)|$$ If all the zeros of both $\,p(z)\,,\,q(z)\,$ are within the open unit disk $\,|z|<1\,$ , prove that $$\forall z\in\mathbb{C}\,\,,\,q(z)=\lambda\, p(z)\,,\,\lambda\in\mathbb{C}\,\,\text{a constant}$$ What've I thought: since the polynomials are of the same degree, I know that $$\lim_{|z|\to\infty}\frac{q(z)}{p(z)}$$ exists finitely, so we can bound $\,\displaystyle{\frac{q(z)}{p(z)}}\,$ say in $\,|z|>1\,$ . Unfortunately, I can't use Liouville's Theorem to get an overall bound as the rational function is not entire within the unit disk... $\,(2)\,\,$ Let $\,f(z)\,$ be analytic in the punctured disk $\,\{z\in\mathbb{C}\;|\;0<|z-a|<r\,,\,\text{for some}\,\,0<r\in\mathbb R\}\,$. Prove that if $\,\displaystyle{\lim_{z\to a}f'(z)}\,$ exists and finite, then $\,a\,$ is a removable singularity of $\,f(z)\,$ . My thoughts: We have a Laurent expansion in the above disk$$f(z)=\ldots +\frac{a_{-n}}{(x-a)^n}+\ldots +\frac{a_{-1}}{z-a}+a_0+a_1(z-a)+\ldots$$ so taking the derivative term-term (is there any special condition that must be fulfilled in this particular case to do so?) we get$$f'(z)=\ldots -\frac{na_{-n}}{(z-a)^{n+1}}-\ldots -\frac{a_{-1}}{(z-a)^2}+a_1+2a_2(z-a)+\ldots$$Now, as the limit of the above when $\,z\to a\,$ exists finitely, it must be that all the terms with negative power of $\,z-a\,$ vanish, thus $$\ldots =\,a_{-n}=a_{-n+1}=\ldots =a_{-1}=0$$ and we get that the above Laurent series for $\,f(z)\,$ is, in fact, a Taylor one and, thus, the function's limit (not the derivative's!) exists and finite when $\,z\to a\,$ and $\,a\,$ is then a removable singularity. Any help in (1) if I got right (2), or in both if there's some problem with the latter will be much appreciated. - 1 +1 for your carefully formulated question, your solution of (2) and above all for being such a great daddy! – Georges Elencwajg Jun 18 '12 at 11:16 Hehe...thanks, @George! But this girl's exercises in complex functions are making me realize I should have taught that course or, at least, follow it more closely. Most of the time it was linear, abstract algebra, calculus...I need now to get into shape in order to be a good dad... – DonAntonio Jun 18 '12 at 12:33 ## 1 Answer The function $g(z)= \frac{p(\frac{1}{z})}{q(\frac{1}{z})}$ has a removable singularity at $0$ since the $\lim_{z \rightarrow \infty }\frac{p(z)}{q(z)}$ exists. Furthermore since $p(z),q(z)$ have all their zeros inside $|z|=1\,$ , the functions $g(z)$ and $\frac{1}{g(z)}$ are analytic in $|z|<1+\epsilon\,$ , for some $\,\epsilon >0\,$ , so by the maximum principle both functions $\left(|g(z)|, \left|\frac{1}{g(z)}\right|\right)$ are bounded by $1$ for $|z| \leq 1$ since the condition $|p(z)|=|q(z)|$ for $|z|=1$ that means $|g(z)|=1$ for $|z| \leq 1$ and you are done. - you needed that $\epsilon >0$ in order to apply the Maxixmum principle, also you can find it because the zeros inside the circe are finite so they have a positive distance from the circe. – clark Jun 18 '12 at 0:13 thanks for the answer but I still don't follow. At the end of your answer, did you mean $\,\displaystyle{|f(z)|:=\left|\frac{p(z)}{q(z)}\right|=1\,\,\,for\,\,\,|z|\leq 1}\,$? Or what is that function $\,f(z)\,$? And how exactly does any of the above end problem (1)? – DonAntonio Jun 18 '12 at 2:05 I fixed it, it was a typo mistake I meant $g(z)$ instead of $f(z)$ and you conclude $|g(z)|=1$ for $|z| \leq 1$ because it is analytic you can coclude that $g(z)$ is constant you can go there using the cauchy riemann equations for instance. And then $\frac{p(z)}{q(z)} = e^{i \theta} = g(\frac{1}{z}) \forall z \geq 1$ and since they are polynomials and they take the same value for infinite $z$ they must coincide – clark Jun 18 '12 at 10:42

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

TCP tuning From Wikipedia, the free encyclopedia (Redirected from TCP Tuning) Jump to: navigation, search Internet protocol suite Application layer Transport layer Internet layer Link layer TCP tuning techniques adjust the network congestion avoidance parameters of TCP connections over high-bandwidth, high-latency networks. Well-tuned networks can perform up to 10 times faster in some cases.[1] However, blindly following instructions without understanding their real consequences can hurt the performance as well. Network and system characteristics Bandwidth-delay product (BDP) Bandwidth-delay product (BDP) is a term primarily used in conjunction with the TCP to refer to the number of bytes necessary to fill a TCP "path", i.e. it is equal to the maximum number of simultaneous bits in transit between the transmitter and the receiver. High performance networks have very large BDPs. To give a practical example, two nodes communicating over a geostationary satellite link with a round trip delay of 0.5 seconds and a bandwidth of 10 Gbit/s can have up to 0.5×1010 bits, i.e., 5 Gbit = 625 MB of unacknowledged data in flight. Despite having much lower latencies than satellite links, even terrestrial fiber links can have very high BDPs because their link capacity is so large. Operating systems and protocols designed as recently as a few years ago when networks were slower were tuned for BDPs of orders of magnitude smaller, with implications for limited achievable performance. Buffers The original TCP configurations supported buffers of up to 65,535 Bytes (64 KiB-1) TCP receive window size, which was adequate for slow links or links with small round trip times (RTTs). Larger buffers are required by the high performance options described below. Buffering is used throughout high performance network systems to handle delays in the system. In general, buffer size will need to be scaled proportionally to the amount of data "in flight" at any time. For very high performance applications that are not sensitive to network delays, it is possible to interpose large end to end buffering delays by putting in intermediate data storage points in an end to end system, and then to use automated and scheduled non-real-time data transfers to get the data to their final endpoints. TCP speed limits Maximum achievable throughput for a single TCP connection is determined by different factors. One trivial limitation is the maximum bandwidth of the slowest link in the path. But there are also other, less obvious limits for TCP throughput. Bit errors can create a limitation for the connection as well as round-trip time. Window size See also: TCP window scale option In computer networking, RWIN (TCP Receive Window) is the amount of data that a computer can accept without acknowledging the sender. If the sender has not received acknowledgement for the first packet it sent, it will stop and wait and if this wait exceeds a certain limit, it may even retransmit. This is how TCP achieves reliable data transmission. Even if there is no packet loss in the network, windowing can limit throughput. Because TCP transmits data up to the window size before waiting for the acknowledgements, the full bandwidth of the network may not always get used. The limitation caused by window size can be calculated as follows: $\mathrm{Throughput} \le \frac {\mathrm{RWIN}} {\mathrm{RTT}} \,\!$ where RWIN is the TCP Receive Window and RTT is the round-trip time for the path. At any given time, the window advertised by the receive side of TCP corresponds to the amount of free receive memory it has allocated for this connection. Otherwise it would risk dropping received packets due to lack of space. The sending side should also allocate the same amount of memory as the receive side for good performance. That is because, even after data has been sent on the network, the sending side must hold it in memory until it has been acknowledged as successfully received, just in case it would have to be retransmitted. If the receiver is far away, acknowledgments will take a long time to arrive. If the send memory is small, it can saturate and block emission. A simple computation gives the same optimal send memory size as for the receive memory size given above. Packet loss When packet loss occurs in the network, an additional limit is imposed on the connection.[2] In the case of light to moderate packet loss when the TCP rate is limited by the congestion avoidance algorithm, the limit can be calculated according to the formula (Mathis et al.): $\mathrm{Throughput} \le \frac {\mathrm{MSS}} {\mathrm{RTT} \sqrt{ P_{\mathrm{loss} }}}$ where MSS is the maximum segment size and Ploss is the probability of packet loss.[3] If packet loss is so rare that the TCP window becomes regularly fully extended, this formula doesn't apply. TCP Options for High Performance A number of extensions have been made to TCP over the years to increase its performance over fast high-RTT links ("long fat networks", or LFNs for short). TCP timestamps (RFC 1323) play a double role: they avoid ambiguities due to the 32-bit sequence number field wrapping around, and they allow more precise RTT estimation in the presence of multiple losses per RTT. With those improvements, it becomes reasonable to increase the TCP window beyond 64 kB, which can be done using the window scaling option (RFC 1323). The TCP selective acknowledgment options (SACK, RFC 2018) allows a TCP receiver to precisely inform the TCP server about which segments have been lost. This increases performance on high-RTT links, when multiple losses per window are possible. Path MTU discovery avoids the need for in-network fragmentation, which increases performance in the presence of losses. References Retrieved from "http://en.wikipedia.org/w/index.php?title=TCP_tuning&oldid=531135675"

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http://math.stackexchange.com/questions/151081/gcd-of-rationals/151090

# GCD of rationals Disclaimer: I'm an engineer, not a mathematician Somebody claimed that $\gcd$ only is applicable for integers, but it seems I'm perfectly able to apply it to rationals also: $$\gcd\left(\frac{13}{6}, \frac{3}{4} \right) = \frac{1}{12}$$ I can do this for a number of cases on sight, but I need a method. I tried Euclid's algorithm, but I'm not sure about the end condition: a remainder of 0 doesn't seem to work here. So I tried the following: $$\gcd\left(\frac{a}{b}, \frac{c}{d} \right) = \frac{\gcd(a\cdot d, c \cdot b)}{b \cdot d}$$ This seems to work, but I'm just following my intuition, and I would like to know if this is a valid equation, maybe the correct method. (It is in any case consistent with $\gcd$ over the natural numbers.) I'm not a mathematician, so please type slowly :-) - @WillieWong - Why did you remove the [GCD] tag? Because it's a new one? But it is about GCD, and the [GCD] tag may help people who're looking for information. – stevenvh May 29 '12 at 11:25 the gcd tag is too specific and too broad at the same time (it is about a particular operation which has interpretation in many different levels/specialties of mathematics). And any question about gcd will have the phrase "gcd" or "greatest common denominator" in the question body, which can be easily found using the search function. – Willie Wong♦ May 29 '12 at 11:44 To illustrate: math.stackexchange.com/search?q=gcd returns over 1000 questions about greatest common denominators in different contexts. It is much more useful to use the tags to help specify in which context your question about gcd is asked. – Willie Wong♦ May 29 '12 at 11:46 Since you are an engineer: Think about the numbers as representing some quantities. Divide your unit in smaller units so that both quantities become integers. Apply the gcd, and then switch back to the original unit....This is exactly what Greek's understanding mentioned in Andre's answer means.... – N. S. May 31 '12 at 8:27 ## 3 Answers In the ancient Greek sense, if we have two quantities $x$ and $y$, then the quantity $z$ is a common measure of $x$ and $y$ if each of $x$ and $y$ is an integer multiple of $z$. The quantities involved were not thought of as numbers, but of course they were what we think of as positive. So for example the diagonal of a square, and the diagonal of a square whose sides are $50\%$ bigger, have a common measure. But some pairs of quantities are incommensurable, like the side and diagonal of a square. Any two rationals, unless they are both $0$, have a greatest common measure. You are therefore perfectly correct. The two rationals also have a least number that they both measure. So if you had further argued that two rationals, not both $0$, have a least positive common multiple, you would also be right. Your calculation method is also correct. One brings the two rationals to a common denominator $q$, say as $\frac{x}{q}$ and $\frac{y}{q}$. Then the $\gcd$ is $\frac{\gcd(x,y)}{q}$. You chose the common denominator $q=bd$. Any common denominator will do, and will produce the same $\gcd$. Most of the standard theorems about $\gcd$ and lcm for integers extend in a straightforward way to results about the $\gcd$ and lcm for rationals. For example, a mild variant of what we nowadays call the Euclidean algorithm will compute the $\gcd$ of two positive rationals. Actually, this variant is the original Euclidean algorithm! Remark: A number of programs, including Mathematica, accept rationals as inputs to their $\gcd$ function. The same appears to be true of WolframAlpha, which has the great advantage of being freely accessible. One way, perhaps, to (sort of) settle the argument in your favour is to type $\gcd(3/7, 12/22)$ into Alpha. It will, probably, return $\frac{3}{77}$ (it did when I used it). With other rationals, test first, Alpha is not bug-free. - Thanks for the Wolfram Alpha reference too. Interesting site, though I guess you guys would use Mathematica. Also gives additional information like series representation and prime factorization. I took the liberty of adding a link to it. – stevenvh May 31 '12 at 7:56 Euclid's algorithm (originally formulated using repeated subtractions rather than divisions) can be applied to any pair of comparable quantities (two lengths, two masses, two frequencies): just keep subtracting the smaller of a pair from the larger and then replace the larger one by the difference found; stop when the difference becomes $0$, returning the smaller (which is now also the larger) retained quantity as GCD. Although I haven't actually read Euclid, I would be surprised if the original formulation wasn't actually in terms of geometric lengths, given that Greek mathematicians were more at ease with geometry than with numbers. The only caveat is that with lengths it is not an algorithm in the sense that it is guaranteed to terminate; two lengths are by definition "commensurable" if the procedure does terminate, and it then returns the largest unit of length for which both intitial lengths are integer multiples. One can easily show (without using any arithmetic) that the procedure does indeed never terminate (because one gets back to the original ratio of lengths, but at a smaller scale) when starting with the lengths of a side and a chord of a regular pentagon (giving the golden ratio) or for a side and a diagonal of a square (ratio $1:\sqrt2$); this is a nice exercise in geometry. $\qquad$ Now if you apply this to rational numbers, you see there is no problem, since they can be compared, subtracted, and any pair of them is necessarily commensurable (one over the product of the denominators gives common measure, though not necessarily the largest one). Bringing both rational numbers to this common denominator, one sees that the formula you guessed is indeed the correct one. If you have prime factorizations of $a,b,c,d$ available, you can also find the $\gcd(\frac ab,\frac cd)$ by taking every prime number $p$ to a power that is the minimum of the valuations of $p$ in $\frac ab$ and in $\frac cd$, where the valuation in $\frac ab$ is its multiplicity in $a$ minus its multiplicity in $b$ (a possibly negative integer). However your formula combined with Euclid's algorithm for integers (or directly Euclid's algorithm for rational numbers) is a more efficient method of computing. Added in reply to comment by @Michael Hardy: When I say that the Greek mathematicians were more at ease with geometry than with numbers, I did not mean to say the did not know about numbers, but that they would prefer stating and treating a statement about numbers in geometric terms, quite opposite to our modern tendency to translate geometric problems into numerical terms. A case in point is Euclid's proof of what we would formulate as the infiniteness of the set of prime numbers (he never uses terms "finite" or "infinite", and I don't think these notions were even used at the time, but that is not my point here). I think most people would agree this is a purely number theoretic statement with absolutely no geometric content, yet Euclid states and proves it geometrically (essentially by constructing a least common multiple of a given (finite) set of prime multiples of a given unit length). And for this reason of geometric preference I think it is probably historically inaccurate to use the term "Euclidean algorithm" for a procedure restricted to the case of a pair of positive integers (and therefore assured to terminate) rather than a pair of lengths (which could fail to terminate); however I'll repeat my ignorance of the precise sources here. Added later Thank you @Robert Isreal for the reference to the translation of Elements. So Euclid operates in his description of his algorithm, in Book VII Propositions 2,3, on numbers, which are integer multiples some unit veiled in mystery ("that by virtue of which each of the things that exist is called one"). Yet the language suggests that numbers are thought of as lengths; if not, why denote individual numbers by $AB$, $CD$, talk about one number measuring another, etc.? Moreover, in book X, Proposition 2 one finds the same procedure, but with numbers replaced by "magnitudes", treated exactly like numbers are (also denoted $AB$ etc.) but with the difference that they need not have a common measure. So rather than the relatively prime/relatively composite distinction, one has a (rather different) commensurable/incommensurable distinction for magnitudes. And in the incommensurable case, during the procedure "that which is left never measures the one before it", implying that the procedure never terminates. Maybe I haven't looked well enough, but it would seem that Euclid isn't very strong in proving the existence of incommensurable magnitudes. The example given (the same one I gave above) doesn't seem to be Euclid's, and Book X, Proposition 10, apart from being non genuine, fails to prove the existence of (integer) ratios that are not the ratio of two square numbers. For instance he does not show (at least not at that proposition) why $1:2$ cannot also be the ratio of two square numbers, even though the proof would be easy. - Right: The Greeks didn't think of A mod B; they thought of "the line segment left over when you start with a line segment of length A and subtract as many line segments of length B as possible". – Dan May 29 '12 at 14:09 "Greek mathematicians were more at ease with geometry than with numbers" isn't a very precise statement of the matter to say the least. Numbers, i.e. 2,3,4,5,6,... were something Euclid was quite comfortable with; he used the word "arithmoi" frequently. The Greeks had no concept at all of what we now call "real numbers". That's not just "discomfort" with them. They had a concept of congruence of line segments; they had a concept of laying a specified number of segments end to end; the had a concept of the ratio of two lengths being the same as.... – Michael Hardy May 29 '12 at 17:48 ....another ratio of two lengths even when the ratios were of incommensuable pairs of segments, etc. They could multiply two lengths and get an area. But you couldn't say the area was greater than or less than or equal to any particular length: these were not real numbers. – Michael Hardy May 29 '12 at 17:49 – Robert Israel May 30 '12 at 7:51 He is talking about numbers here, not lengths. He needs two propositions, one for the case of relatively prime numbers and one for the greatest common measure of two numbers that are not relatively prime, because for the Greeks $1$ was not a "number". – Robert Israel May 30 '12 at 7:54 Yes, your formula yields the unique extension of $\rm\:gcd\:$ from integers to rationals (fractions), presuming the natural extension of the divisibility relation from integers to rationals, i.e. for rationals $\rm\:r,s,\:$ we define $\rm\:r\:$ divides $\rm\:s,\:$ if $\rm\ s/r\:$ is an integer, in symbols $\rm\:r\:|\:s\:$ $\!\iff\!$ $\rm\:s/r\in\mathbb Z.\:$ [Such divisibility relations induced by subrings are discussed further here] Essentially your formula for the gcd of rationals works by scaling the gcd arguments by a factor that yields a known gcd (of integers), then performing the inverse scaling back to rationals. Even in more general number systems (integral domains), where gcds need not always exist, this scaling method still works to compute gcds from the value of a known scaled gcd, namely $\rm{\bf Lemma}\ \ \ gcd(a,b)\ =\ gcd(ac,bc)/c\ \ \ if \ \ \ gcd(ac,bc)\$ exists $\rm\quad$ Therefore $\rm\ \ gcd(a,b)\, c = gcd(ac,bc) \ \ \ \ \ if\ \ \ \ gcd(ac,bc)\$ exists $\quad$ (GCD distributive law) The reverse direction fails, i.e. $\rm\:gcd(a,b)\:$ exists does not generally imply that $\rm\:gcd(ac,bc)\:$ exists. $\$ For a counterexample see my post here, which includes further discussion and references. More generally, as proved here, we have these dual formulas for reduced fractions $$\rm\ gcd\left(\frac{a}b,\frac{c}d\right) = \frac{gcd(a,c)}{lcm(b,d)}\ \ \ if\ \ \ \gcd(a,b) = 1 = \gcd(c,d)$$ $$\rm\ lcm\left(\frac{a}b,\frac{c}d\right) = \frac{lcm(a,c)}{gcd(b,d)}\ \ \ if\ \ \ \gcd(a,b) = 1 = \gcd(c,d)$$ Some of these ideas date to Euclid, who computed the greatest common measure of line segments, by anthyphairesis (continually subtract the smaller from the larger), i.e. the subtractive form of the Euclidean algorithm. The above methods work much more generally since they do not require the existence of a Euclidean (division) algorithm but, rather, only the existence of (certain) gcds. - I don't follow- $\mathbb Q$ is a field. What is the meaning of $gcd(x,y)$ where $x,y$ are members of a field ? – users31526 May 30 '12 at 4:10 @Kuashik As mentioned in the first paragraph, divisibility is w.r.t. to a specified subdomain Z of "integers". In an integral domain, $\rm\:d = gcd(a,b)\:$ is a greatest common divisor of $\rm\:a,b\:$ if it is a common divisor $\rm\:d\:|\:a,b\:$ that is (divisibly) greatest, i.e. divisible by every common divisor, i.e. $\rm\: c\:|\:a,b\Rightarrow c\:|\:d.\:$ Said more universally $\rm\:c\:|\:a,b\iff c\:|\:gcd(a,b),\:$ analogous to $\rm\:c\le a,b\iff c\le min(a,b).\:$ See here for more on such universal definitions of GCD and LCM. – Gone May 30 '12 at 4:31 @ Bill Dubuque: There should be a sense of ordering in a Integral Domain otherwise no one can say which one is greater among $x,y$ where $x,y$ are members of an Integral Domain. In a field except additive identity (usually we say $0$) "everyone divisible by everyone". So in that sense, I can say $gcd(\frac{13}{2},\frac{3}{4})=100$ or $1000$.... I have doubt that's why I am asking don't get angry – users31526 May 30 '12 at 4:47 @Kuashik The (partial) order is that given by divisibility, as I said. This is the standard definition of gcd in a general integral domain. It specializes to the well-known definition in $\mathbb Z$ (or any Euclidean domain). Again, the divisibility relation used in $\mathbb Q$ is the natural extension of integer divisibility - see the first paragraph of my answer. – Gone May 30 '12 at 4:53 @ Bill Dubuque: Yes, any Integral domain $R$ has natural partial order (antisymmetric, antireflexive, transitive) say $x\leq y\iff \exists$ non unit $c\in R$ such that $y=cx$. Then one can talk about chain and one can talk about $gcd(x,y)$ where $x,y\in R$ – users31526 May 30 '12 at 5:19 show 2 more comments

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http://math.stackexchange.com/questions/tagged/taylor-expansion+trigonometry

# Tagged Questions 2answers 40 views ### Taylor Series of $\sin 2x$ finding $f^{(n)} (a)$ where $a = 0$ ok so i get; f (x) = sin 2x f ' = 2cos 2x f '' = -4sin 2x f ''' = -8cos 2x f '''' = 16sin 2x f ''''' = 32cos 2x f (0) = 0 f '(0) = 2 f ''(0) = 0 f '''(0) = -8 f ''''(0) = 0 f '''''(0) = 32 ... 2answers 165 views ### Finding the 9th derivative of $\frac{\cos(5 x^2)-1}{x^3}$ How do you find the 9th derivative of $(\cos(5 x^2)-1)/x^3$ and evaluate at $x=0$ without differentiating it straightforwardly with the quotient rule? The teacher's hint is to use Maclaurin Series, ... 2answers 82 views ### Infinite (Taylor) Series What is the general term for a Taylor series of $\sin(x)$ centered at $\pi/4$? It should be $(-1)^{[??]} \times \sqrt{2}/2 \times \frac{(x-\pi/4)^n}{n!}$ What power is $(-1)$ supposed to be raised ... 1answer 253 views ### Approximating $\arctan x$ for large $|x|$ I would like to know if there is reasonably fast converging method for computing large arguments of arctan. Until now I've came across Taylor series that converges only on interval $(-1,1)$ and for ... 2answers 304 views ### How do we know Taylor's Series works with complex numbers? Euler famously used the Taylor's Series of $\exp$: $$\exp (x)=\sum_{n=0}^\infty \frac{x^n}{n!}$$ and made the substitution $x=i\theta$ to find $$\exp(i\theta) = \cos (\theta)+i\sin (\theta)$$ How ... 2answers 416 views ### Is the Maclaurin series expansion of $\sin x$ related to the inclusion-exclusion principle? When I see the alternating signs in the infinite series expansion of $\sin x$, I'm reminded of the inclusion-exclusion principle. Could there be any way to visualize it in such a way? Also, is there ... 1answer 192 views ### Taylor series expansion of $\sec(x +y^2)$ We have $f(x,y) = \sec(x+y^2)$ I want to find the first two non-zero terms of $f$ at $(0,0)$ starting by Taking the first few terms of $\cos x$ centered at zero, $1 - \frac{x^2}{2!}$ Using this ...

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http://mathoverflow.net/questions/49849/decomposition-of-an-abelian-von-neumann-algebra

## Decomposition of an abelian von Neumann algebra ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Hi, I came across the statement below and I couldn't figure out why it is true. I was hoping someone could explain it or give me a good reference. Thank you in advance. "Let $\pi$ be a non-degenerate representation of a separable C*-algebra $A$ on a separable Hilbert space $H$. Suppose $M$ is an abelian von Neumann algebra contained in the commutant of $\pi(A)$. Then there exists a unique sequence of mutually orthogonal projections $(e_{n})$ in $M$ such that $\sum e_{n}=I$ and $M'e_{n}$ is spatially isomorphic to $L^{\infty}(E_{n})\otimes M_{n}(C)$. Moreover, each $e_{n}$ is invariant under spatial automorphisms of $M$. - ## 2 Answers Apparently, the data of π and A is irrelevant here. We might as well start with an arbitrary representation of an abelian von Neumann algebra M on a Hilbert space H. Geometrically, commutative von Neumann algebras are just measurable spaces (the opposite category of the category of commutative von Neumann algebras is equivalent to the category of (localizable) measurable spaces). Likewise, representations of von Neumann algebras on Hilbert spaces can be thought of geometrically as bundles of Hilbert spaces over measurable spaces, i.e., there is a similar equivalence of categories here. Spatial automorphisms of representations of von Neumann algebras are precisely automorphisms f of the underlying measurable space together with an isomorphism f*V→V, where V is the bundle of Hilbert spaces corresponding to the representation. More generally, a morphism of von Neumann algebras M→N, where M is abelian, can be thought of as a bundle of von Neumann algebras over the spectrum (i.e., the corresponding measurable space) of M. Again, this can be stated as an equivalence of categories. In our case we have a morphism M→M' and M is abelian. The corresponding bundle of von Neumann algebras is precisely the bundle of (fiberwise) endomorphisms of the bundle V of Hilbert spaces corresponding to the representation of M on H. Denote by E_n the measurable subset of Spec M consisting of points with the dimension of the fiber of V being equal to n. Denote by e_n the corresponding projection in M. Observe that E_n are disjoint and ⋃_n E_n=Spec M, i.e., e_n are mutually orthogonal and ∑_n e_n=1. Moreover, M'e_n is the algebra of endomorphisms of the restriction of V to E_n. The latter bundle is trivial and has dimension n, hence its algebra of endormophisms is just L^∞(E_n)⊗M_n(C) as required. From the geometric interpretation of spatial isomorphisms it follows that every e_n is invariant under any spatial isomorphism. Finally, the sequence e is unique because the property M'f_n=L^∞(E_n)⊗M_n(C) forces every F_n to be a subset of E_n, and because ⋃_n F_n=Spec M we have F_n=E_n. - ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. It's a standard result in von Neumann algebras, called the homogeneous decomposition of a type I von Neumann algebra. To quote results from the literature, I will use Takesaki's book (volume I). Since your $M$ is an abelian von Neumann algebra, it is type I. Then it's commutant $M'$ is also type I (V.1.30). Then Theorem V.1.27 expresses exactly the result you are asking about. -

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http://mathhelpforum.com/calculus/204600-problems-spivak-calculus-chapter-6-problem-6-a.html

1Thanks # Thread: 1. ## Problems with Spivak Calculus Chapter 6 Problem 6 I have the solutions book and see the answers, but they make no sense to me. Any/All help is appreciated. a) Find a function that is discontinuous at 1, 1/2, 1/4, 1/4... but continuous at all other points. b) Find a function that is discontinuous at 1, 1/2, 1/4, 1/4... and at 0 but continuous at all other points. Problem 7 doesn't seem to have enough explanation in the solution either. How is f being continuous 0 used? 7) Suppose f satisfies f(x+y) = f(x) + f(y) and f is continuous at 0. Prove it is continuous at A for all values of A. 2. ## Re: Problems with Spivak Calculus Chapter 6 Problem 6 Attached Thumbnails 3. ## Re: Problems with Spivak Calculus Chapter 6 Problem 6 Well that certainly didn't help at all. Thanks, anyway. 4. ## Re: Problems with Spivak Calculus Chapter 6 Problem 6 For 6a, you can also consider $f(x)=\begin{cases}\frac{1}{k}& \frac{1}{2^k}<x\le \frac{1}{2^{k-1}}, k=1,2,\dots\\0&\text{otherwise}\end{cases}$ 5. ## Re: Problems with Spivak Calculus Chapter 6 Problem 6 Problem #7: Ignoring issues of the domain, $f$ continuous at $a$ means: $\lim_{x\to a} f(x) = f(a)$. Now observe that, always, $\lim_{x\to a} f(x) = \lim_{h\to 0} f(a + h)$. 6. ## Re: Problems with Spivak Calculus Chapter 6 Problem 6 Originally Posted by johnsomeone Problem #7: Ignoring issues of the domain, $f$ continuous at $a$ means: $\lim_{x\to a} f(x) = f(a)$. Now observe that, always, $\lim_{x\to a} f(x) = \lim_{h\to 0} f(a + h)$. I think that one also has to prove that f(0) = 0. 7. ## Re: Problems with Spivak Calculus Chapter 6 Problem 6 Thanks, I managed something with 7, but I still don't understand 6. Is there any way to show how you construct an answer? 8. ## Re: Problems with Spivak Calculus Chapter 6 Problem 6 Originally Posted by patrickmanning I still don't understand 6. Is there any way to show how you construct an answer? Can you construct a function that is discontinuous at x = 0? For example, f(x) = 0 for x <= 0 and f(x) = 1 for x > 0. If this is clear, then what is difficult about problem 6? I suggested a function whose graph looks like a staircase. 9. ## Re: Problems with Spivak Calculus Chapter 6 Problem 6 Originally Posted by emakarov Can you construct a function that is discontinuous at x = 0? For example, f(x) = 0 for x <= 0 and f(x) = 1 for x > 0. If this is clear, then what is difficult about problem 6? I suggested a function whose graph looks like a staircase. In your example in this post, f(x) is really discontinuous at x = 0? It seems I don't even understand continuity properly. At the point where there is a jump, that point is discontinuous? Because it is different when approached from left and right sides, right? Wouldn't that mean in your first example, f(x) was also discontinuous at x 0? 10. ## Re: Problems with Spivak Calculus Chapter 6 Problem 6 Originally Posted by patrickmanning At the point where there is a jump, that point is discontinuous? Because it is different when approached from left and right sides, right? Yes. Would you agree that a person's spacial position as a function of time is a continuous function? If f(t) = 0 for t <= 0 and f(1) = 1 for t > 0 described my spacial position, then I would be at position 0 before time 0 and then suddenly move to position 1 (say, 1 mile from position 0). If you know how to construct such transportation, I would certainly like to hear! Formally, f(t) is not continuous at 0 because any interval containing t = 0 is mapped by f to a set that contains some $y_1$ and $y_2$ such that $|y_1 - y_2| = 1$. In other words, however small you make an interval around 0, the image of that interval contains numbers that are sufficiently far apart, i.e., not arbitrarily close. It does not matter how precisely you approximate t = 0: if you change t just a little bit, f(t) can change by 1. 11. ## Re: Problems with Spivak Calculus Chapter 6 Problem 6 Thank you. I hate to be a bother, but can you continue helping by answering my question about your first example? Isn't f(x) discontinuous at x = 0 in your example with f(x) = 1/k? 12. ## Re: Problems with Spivak Calculus Chapter 6 Problem 6 Originally Posted by patrickmanning sn't f(x) discontinuous at x = 0 in your example with f(x) = 1/k? As defined in post #4, f(x) = 0 for $x\le 0$. As far as x > 0, f(x) decreases as x decreases. In fact, f(x) tends to 0 as x tends to 0 from the right. Since $\lim_{x\to0}f(x)=f(0)$, f is continuous at 0. 13. ## Re: Problems with Spivak Calculus Chapter 6 Problem 6 Thank you. I understand well now. Your example actually is not discontinuous at 1/3 then. Changing it a bit to: f(x) = 1/b , 1/a < x <= 1/b where a = b+1 and b = 1, 2, 3... will fix that. 14. ## Re: Problems with Spivak Calculus Chapter 6 Problem 6 Originally Posted by patrickmanning Your example actually is not discontinuous at 1/3 then. Changing it a bit to: f(x) = 1/b , 1/a < x <= 1/b where a = b+1 and b = 1, 2, 3... will fix that. I though that "1, 1/2, 1/4, 1/4..." from post #1 was supposed to mean "1, 1/2, 1/4, 1/8...," but you are right if it should be "1, 1/2, 1/3, 1/4..." 15. ## Re: Problems with Spivak Calculus Chapter 6 Problem 6 Haha, you are right. I didn't realise I typed 1/4 twice. Thank you so much for everything. I understand this all now. For 6b, you can just change it to "otherwise, x = -1" and now x = 0 will be discontinuous as from the left, f(x) approaches -1 but from the right, f(x) still approaches 0.

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http://mathoverflow.net/questions/46541/how-to-introduce-notions-of-flat-projective-and-free-modules/46627

## How to introduce notions of flat, projective and free modules? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) In the coming spring semester I will be teaching for the first time an introductory (graduate) course in Commutative Algebra. As many people know, I have been plugging away for a while at this subject and keeping my own lecture notes. So I feel relatively prepared to teach the course in the sense that I know more than enough theorems and proofs to cover. However, there is more to teaching a graduate course than just theorems and proofs. Commutative algebra has a reputation for being somewhat dry and unmotivated. After something like 10 years of hard work (over a period of about 15 years!), I am at a point where I find the subject both interesting in and of itself and useful. But how to communicate this to students? Looking through my notes, the technicalities begin with the introduction of various fundamental classes of modules, especially free, projective and flat modules (but also including other things like finite generation and finite presentation). I remember well that when I first learned this material, projective and (especially) flat modules were a tough sell: for instance, the first time I picked up Bourbaki's Commutative Algebra out of curiosity, I saw that the very first chapter was on flat modules, and I put it down in horror. How would you introduce these concepts to an early career graduate student? What are the fundamental differences between these classes of modules, and why do we care? Here are some of my preliminary thoughts: 1) Free modules are of course an easy sell: for such things the usual notions of linear algebra work well, including that of dimension (called "rank" in this case: note that my rings are commutative!). One can easily prove that for a commutative ring R, all R-modules are free iff R is a field, so the need to move beyond free modules is easily linked to the ideal theory of R. 2) Projective modules are important for at least the following reasons. a) Geometric: A finitely generated module over a ring R is projective iff it is locally free (in the stronger sense of an open cover of $\operatorname{Spec} R$). In other words, projective modules are the way to express vector bundles in algebraic language. I plan to drive this point home by discussing Swan's theorem on modules over $C^{\infty}(M)$. b) Homological: projective modules play a distinguished role in homological algebra, i.e., in the construction of left-derived functors. c) K-theoretic: It is of interest to know to what extent finitely generated projective modules must be free. This leads to the construction of $K_0(R)$ -- i.e., stable isomorphism classes of projective modules -- and in the rank one case, to the Picard group $\operatorname{Pic}(R)$. (After writing the above, I feel that in some sense it is a good enough answer -- certainly these are three important reasons for studying projective modules. But I'm not sure how to explain them to beginning students. Let's be clear that this is part of the question.) 3) Flat modules: this is harder to explain! a) Geometric: Flatness is the "right" condition for things to vary nicely in families, but this is more of a mantra than an explanation. I think that most people hear this at one point and come to believe and understand it slowly over time. b) Homological: Flat modules are those which are acyclic for the Tor functors. But this is not a homological algebra course: I would be happiest not to mention Tor at all. c) Near equivalence with projective modules in the finitely generated case: the difference between finitely generated flat and finitely generated projective modules is very subtle. Recall for instance: Theorem: For a finitely generated flat module $M$ over a commutative ring $R$, TFAE: (i) $M$ is projective. (ii) $M$ is finitely presented. (iii) The associated rank function is locally constant. Thus in almost all of the cases of importance to a beginning algebra student -- e.g. if $R$ is Noetherian or is an integral domain -- finitely generated flat modules are projective. d) However infinitely generated flat modules are a much bigger class, since flatness is preserved under direct limits. In particular, there are large classes of domains $R$ for which a module is flat iff it is torsionfree (namely this holds iff $R$ is a Prufer domain; even the case $R = \mathbb{Z}$ is useful). Perhaps this is significant? Your insight will be much appreciated. Update: the prevailing sentiment of the answers thus far seems to be that a very little bit of homological algebra will go a long way in presenting the basic definitions in a unified and useful way. Duly noted. - 3 In my opinion, the very basic projective non-free modules are (fractional) ideals in number rings (and Dedekind domains); this particular case of Pic is very intriguing for those who are interested in algebraic number theory. By the way, do you want to introduce the defintion of a Dedekind domain (early in your course)? – Mikhail Bondarko Nov 18 2010 at 21:55 2 For projective modules, you might start with the observation that many of the nice properties of free modules follow only from the lifting property. – Kevin Ventullo Nov 18 2010 at 22:51 1 This may or may not be helpful to you, but there are many different characterizations of flatness in Lam's book Lectures on Modules and Rings, Ch. 4. Flatness can be viewed purely in terms of linear equations as in Thm. 4.24, the "equational criteria for flatness." This may not add much intuition, but it at least avoids any mention of homological algebra. – Manny Reyes Nov 18 2010 at 23:10 1 Why would you avoid mentioning "homological algebra"(whereby one means, in this context "preservation of exactness properties") when introducing a property which is essentially of homological nature? – Mariano Suárez-Alvarez Nov 19 2010 at 0:13 2 IMHO, the motivation of flatness based on the single case in which one considers the problem of when extension of scalars along a map of rings $A\to B$ preserves exactness is quite good already. In fact, I think that extension of scalars is a great way to motivate tensor products, because it is very very concrete and natural. – Mariano Suárez-Alvarez Nov 19 2010 at 0:39 show 3 more comments ## 11 Answers Hi Pete, this sounds like a lot of fun! I wish I could be there (-: Here is a concrete and useful property of flatness, you can explain it without using Tor. Suppose $R\to S$ is a flat extension. Then if $I$ is an ideal of $R$, tensoring the exact sequence: $$0 \to I \to R \to R/I \to 0$$ with $S$ gives that $I\otimes_RS = IS$. The left hand side is somewhat abstract object, but the right hand side is very concrete. There are very natural extensions which are flat but not projective. For example, if $R$ is Noetherian and $\dim R>0$, then $S=R[[X]]$ is flat but never projective over $R$. - 3 @Hailong: you can be there as far as I'm concerned. Please feel free to drop by, especially if you don't mind being handed the chalk. – Pete L. Clark Nov 19 2010 at 3:23 Dear Long, Thanks for the interesting reference! – Emerton Dec 3 2010 at 8:15 This is indeed a very nice motivation of flatness. If you wanted even less homological algebra and your students already believed free and projective modules were interesting then I suppose you could introduce flat modules as direct limits of f.g. free modules. After talking about direct summands of free modules, I can't imagine people would be shocked when you asked about direct limits. – David White Jun 22 2011 at 14:35 ### 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. All of this is coming from the point of view of being a student. I took a commutative algebra course last winter and am currently taking a homological algebra course, so maybe I can touch on your differentiation between those two perspectives. It sounds like you want the motivating ideas, and I think those make the most sense homologically. I completely understand your not wanting to introduce Tor and/or Ext, but you don't have to. $Hom_R (P,-)$ is exact iff $P$ is projective, $Hom_R (-,I)$ is exact iff $I$ is injective, and $F \otimes_R -$ is exact iff $F$ is flat. These can even be taken as definitions (these are equivalent to $Tor_1$ and $Ext^1$ vanishing which is each equivalent to all the higher derived functors vanishing). Edit: What if you don't know why you should care about exact sequences as Pete asked in a comment? Well, you should still care about when functors preserve injections (tensor-ing with flat modules) and surjections (Hom-ing out of projective modules), and this is precisely what flat and projective do for you. This is the only place where exactness might fail, so just skip it and talk about injections and surjections and when they are preserved. This is how I feel most comfortable thinking about these concepts. It is pretty concrete and not far away from Commutative Algebra (I think, meaning that this is how we talked about it in my commutative algebra class). I should also mention I appreciate you posting all of those course notes, it is a great resource for us students out there. - 1 @Sean: thanks. I agree that these are nice, clean definitions. But I think more motivation would be helpful: why do we care whether some functor is exact? – Pete L. Clark Nov 18 2010 at 22:41 4 @Pete: In my mind, the exactness of the tensor product is relatively easy to motivate through change of rings. Say you've got your sequence over $R$ and you've got an ideal $I$. Is it still exact when we think of everything as $R/I$ modules? Well, that's what the exactness of the tensor product will tell you. Of course this becomes very useful once you have localization and completions to play with. For $Hom$, it seems easiest to motivate $Hom(-,R)$ by analogy with vector space duals, and you can build up from there to $Hom(-,M)$ and $Hom(M,-)$. – Justin DeVries Nov 18 2010 at 23:42 Please forgive these very naïve remarks. I am enjoying the chance to learn something about flatness in trying to contribute to this question. First of all, since as pointed out here, the definition of flatness is that the object behaves as simply as possible under tensor product, it follows that the primary use of the concept is in applications of the tensor product. Now there are three of these that come to mind, (after some review of the literature), namely 1) forming fibers, 2) localizing, (and changing rings) 3) completions. These are all local operations, looking at the inverse image of one point, restricting to a Zariski neighborhood of a point, and restriction to an analytic or formal neighborhood of one point. Thus one wants to compare the geometry at a point with the geometry near that point. E.g. given an algebraic subvariety through a point, one can take its algebraic germ, formal germ, or analytic germ, and then ask whether one can recover the original germ from these. This is equivalent to asking whether one recovers the original ideal after extending to the localization or completion, and then restricting back to the original ring. Flatness is the property that tells us yes to all these questions. I.e. both the localization and the completion are flat over the original ring. Moreover, the algebraic and analytic local rings form a “flat pair”, slightly stronger than one being flat over the other, and this apparently implies they have the same completions. This lets us compare analytic and algebraic local rings, by comparing both with their completions. It follows e.g. that the algebraic dimension of an algebraic variety equals the analytic dimension of the associated analytic variety. Reasoning of this sort allows Serre to prove geometric results such as those mentioned above as well as homological ones. Homological results desired are of the sort that compare the analytic cohomology to the algebraic cohomology. The simplest way to do this is to show that the analytic sheaves and their analytic cohomology is obtained from the algebraic ones by tensoring with flat objects, i.e. changing rings in the simplest way. Then the desired results say that homomorphisms of sheaves, and cohomology of sheaves, commutes with this process of tensoring, i.e. of applying the functor of making these objects analytic. Flatness is the key to all these results. Thus flatness of one object over another seems to imply a relation between their local geometric structures. This is my take on Serre’s lovely paper GAGA, available free online at NUMDAM, after a brief perusal. It certainly looks worth a careful read. We have mentioned before the result that a surjective morphism of smooth varieties is flat if and only if the fibers have constant dimension. From perusing Matsumura, other geometric properties of flat maps seem to be: the dimension of (non empty) fibers is always the expected one, i.e. dim(source) - dim(target); the going down theorem holds, hence every subvariety through the point f(p) is the image of a subvariety through p, in particular a flat map cannot separate branches at a point; indeed flat maps cannot remove singularities in any fashion, i.e. f is flat and if f(p) is a regular point, so is p; and a birational flat map is an open immersion. Another remark building on those above is that flatness is a natural weakening of the property of projectiveness, and hence for local rings, of freeness. If finite map behaves well locally when it defines a locally free module, what about a map that lowers dimension? What is the closest thing to locally free that still holds when the fibers vary nicely but not smoothly? I.e. a flat map is a slight weakening of a smooth map. I am struggling here from ignorance. Thanks for the question! - 6 Dear Roy: Last spring I asked Serre about the history. He viewed it as a purely algebraic notion, inspired by usefulness for the back-and-forth passage between analytic and algebraic local rings through their common completion (noetherian property of the former being pretty serious, of course) and clarifying localization. It came as a total surprise to him later that flatness is an important concept in algebraic geometry, and he said that all credit for its geometric significance belongs to Grothendieck; he (Serre) did not anticipate it would be of interest beyond pure algebra. – BCnrd Nov 19 2010 at 4:43 Thanks. Do you have a favorite reference for where Grothendieck exploits it, analogous to this one I just perused by Serre? It seems to me that even in GAGA Serre shows how it relates local geometric concepts like dimension, but maybe I am stretching, if he says otherwise! – roy smith Nov 19 2010 at 4:53 BC, I agree there is no visible occurrence in GAGA of the geometry of a general flat map. I was just to imagine how the idea extends. So that was Grothendieck's contribution, that's nice to know. – roy smith Nov 19 2010 at 5:02 3 Dear Roy: My impression is that Serre viewed local analytic dimension as part of algebra. Usefulness of flatness for geometric problems (Hilbert schemes, deformation theory, etc.) is what I think surprised him. There are too many amazing geometric uses of flatness by Grothendieck, so no "favorite": fpqc descent (to work intuitively with $G/H$ and etale topology, etc.), fibral flatness criteria, openness results for loci defined by fibral properties relative to proper fppf maps, flatness of formally etale lfp maps,... Some of your examples are really about excellence (another miracle!), btw. – BCnrd Nov 19 2010 at 5:22 Gee, this is nice. In line with the idea that flat maps are a slight weakening of smooth ones, one can just read the titles of the first few sections of SGA 1: i.e. etale maps, smooth maps, flat maps,..... Thanks for prompting me to look at Grothendieck. – roy smith Nov 19 2010 at 5:24 show 1 more comment For me, the most intuitive characterization of flatness of an $R$-module $M$ is the "principle of sufficient reason": Whenever elements `$x_i$` of $M$ satisfy a linear equation with coefficients in $R$, say `$\sum_{i=1}^n r_ix_i=0$`, then this is "because" of linear equations holding in $R$; that is, the $x_i$ can be expressed as linear combinations of other elements `$y_j\in M$`, say `$x_i=\sum_j s_{ij}y_j$` (where `$s_{ij}\in R$`), such that `$\sum_i r_is_{ij}=0$` for all $j$. Thus the original linear relation among the $x$'s follows from the expressions in terms of the $y$'s plus properties in $R$ of the coefficients `$r_i$` and `$s_{ij}$`. Admittedly, this description of flatness doesn't help much with geometry, but it clarifies the algebraic situation. The same intuition also applies, for example, to the notion of "flat functor" that occurs in the study of classifying topoi. - 1 @Andreas: right, this is the "equational criterion of flatness" that Manny Reyes referred to above. I do plan on covering it (it's in my private copy of my notes, although not yet the one posted on my webpage) because it's used to prove that a finitely generated flat module over a local ring is projective. I admit though that I don't have a lot of intuition for what this criterion "really means". Any further hints would be appreciated. – Pete L. Clark Nov 19 2010 at 3:19 2 The equational criterion more or less amounts to the statement: "$M$ is flat IFF every homomorphism from a finitely presented module factors through a free module". (With the instant consequence that flat & finitely presented imply projective.) – Charles Rezk Nov 19 2010 at 3:39 (every homomorphism from a finitely presented module to $M$, of course) – Charles Rezk Nov 19 2010 at 3:41 Pete I agree that locally free is a good geometric intuition for projective. In my algebra class I gave an exercise for the students to prove that the module of tangent vector fields on a 2-sphere is locally free over the ring of smooth functions, and challenged them to show it is not free. If you introduce direct limits, maybe the fact you mentioned about flatness being preserved is a way to motivate it. I.e. you want to take direct limits of finite generated projectives, but they lose the property of projectivity. So what property do they keep? But that may be artificial. it is hard to get away from the points made by Sean Tilson. I guess the difficulty there is that Hom is for many of us more intuitive than tensor. So lifting and extension problems are more intuitive than the distinction between IM and ItensorM. You might review your reasons why flat modules are in the course. If you have an application for them in mind, maybe the application can motivate the concept. In the spirit of "flat families", I used to cling to the geometric fact that a surjective morphism of smooth algebraic varieties is flat if and only if the fiber dimension is constant, but I don't know if you can use that. Those "local criteria for flatness" are usually among the more advanced flatness topics. E.g. that is treated in the books of Matsumura near the end. But this suggests that in some contexts flatness is the algebraic version of locally constant geometry. I have heard that Serre focused on the concept in some of his work relating algebraic and analytic varieties. Maybe the properties he used are illuminating. Great question. - 6 @Roy: +1 for the audacious suggestion that I track down why I am discussing flat modules in the first place. (If I found out that I had no good application, my natural reaction would be to find one, rather than cut this important concept.) – Pete L. Clark Nov 18 2010 at 22:50 This was one of the aspects of algebra that I enjoyed the most while first learning it. This is the way I would develop the subject: 1) Introduction to exact sequences with an emphasis on short exact sequences. We can use these to illustrate how the properties of any module in such a sequence are related to those of others. Relevant here would be discussions of theorems such as, length of a module is additive, a module has ACC/DCC if and only if a submodule and quotient modulo the submodule have ACC/DCC. We can also develop direct sums via short exact sequences, the short five lemma, the snake lemma here. 2) Projective modules: Given a short exact sequence, $0\to L\to M\to N\to 0$, of $R$-modules, and some other $R$-module $T$ and a homomorphism $T\to L$ there exists a homomorphism $T\to M$ via composition. We can ask the opposite question. When does giving a homomorphism $T\to M$ give a homomorphism $T\to L$. This can be used to motivate $Hom_R(T,-)$. Then we can go on to show this functor is left exact. Then we define projective modules as those which make this functor exact. 3) At this point I would introduce Free modules and motivating them via vector spaces as you suggested. I would also discuss free resolutions since this is a very elegant machinery. Then, I would show that projective modules are just direct summands of free modules. 4) Injective modules: A similar development as for projective modules, this time via the functor $Hom_R(-,T)$. I would also develop the characterizations of injective modules via Baer's theorem and divisible modules. 5) Flat modules: I shall then introduce the functors $T\otimes -$ and $- \otimes T$ and show these are right exact. Then define flat modules as those which make the above functors exact. I shall also discuss that projective modules are flat. To tie all of this together, I shall then discuss the adjointness of Hom and $\otimes$ followed by Homological algebra if that is part of the course. - @Timothy: thanks for your response. Interestingly (to me!), in my notes I currently do not discuss injective modules at all. Obviously in some sense I should, but I think there is something to ponder here in the notion that in commutative algebra itself, projective modules are much more important than injective modules. (Or do people disagree with this? Let me know!) – Pete L. Clark Nov 18 2010 at 22:52 @Timothy: thinking out loud...whether a projective module is free is one of the great questions of commutative algebra. Asking whether an injective module is cofree does not have the same allure. In my experience, although it is very useful to know that an abelian category has enough injectives, this does not necessarily make injective objects worthy of study. I think that I have never met an injective sheaf in person, for instance. – Pete L. Clark Nov 18 2010 at 22:57 2 @Pere: As someone who studies algebra, I find injective modules are quite ubiquitous. They show up as injective hulls which is a central notion in local cohomology. – Timothy Wagner Nov 18 2010 at 22:59 @Timothy: fair enough. I don't know anything about local cohomology. I do remember injective envelopes coming up in Part III of Serre's book on representations of finite groups, albeit for modules over non-commutative rings. I would be happy to hear of a reference, so as to broaden my horizons... – Pete L. Clark Nov 19 2010 at 3:59 1 @Timothy: thanks very much for the link, and please don't worry about calling me "Pere": I knew you were talking to me and not to your father. – Pete L. Clark Nov 19 2010 at 4:56 show 1 more comment Another student answer: Considering 2b, it seems a bit strange that you want to mention derived functors and not to go into $\mathbf{Tor}$ and $\mathbf{Ext}$. The problem with derived functors is that first of all, categories usually have enough injective objects. Secondly, projectives are just a suitable collection of objects to use, but not always. However, the universal property is really important here. Lifting morphisms is something extremely natural to ask (which is the exactness of $\mathbf{Hom}$). This is also a way to mention that projective modules are direct summands of free (which fits perfectly into you sequence of presentation). Speaking of flat modules, I'd avoid saying about "varying nicely". It's not very convincing the first time you see it (like telling a child that to have a baby one should kiss his partner). I'd also be nice to somehow sneak in the adjointness of $\otimes$ and $\mathbf{Hom}$. - @Anton: sorry, maybe I wasn't clear. When I say that I would ideally like to avoid derived functors, that includes Tor and Ext. I am however willing to reconsider that -- a lot of people are saying that it's very important. – Pete L. Clark Nov 19 2010 at 3:22 +1 for "like telling a child that to have a baby one should kiss his partner." – Qiaochu Yuan Nov 20 2010 at 0:15 You might perhaps mention Lazard's theorem. - @Greg: Thanks. The Govorov-Lazard Theorem (a module is flat iff it can be expressed as the direct limit of finitely generated free modules) is in the latest copy of my notes. – Pete L. Clark Nov 19 2010 at 20:58 In the question there have been already mentioned many deep applications of these notions for modules (flat, projective, free). However, I think that an introductory course should not focus on these aspects. It would be very nice (at least for the ones who can follow...) if some applications are at least mentioned and are partially worked out as exercises. The students will learn the "big story" later, perhaps even in the same course of commutative algebra. But as for the introduction of these notions, I would just use the standard definitions (see Sean Tilson's answer). They later also apply in abelian categories (whereas, "free" generalizes rather to left adjoint functors). I think an introductory course should show the students how to work with the new objects and what can be done with them. For example, it is pretty useful that a short exact sequence $0 \to I \to M \to P \to 0$ splits if $P$ is projective, and that the converse is also true: If $P$ has the property that every short exact sequence of the above type splits, then $P$ is projective. By duality, there is a dual statement characterizing the injectivity of $I$. A propos, it should be partially pointed out that the notions of projective and injective are dual to each other, and the reason that they differ so horribly in $R$-Mod comes from the fact that this category has no autoequivalence. Concerning flatness, it should be pointed out that they are stable under directed colimits and have various other nice local properties which fails for the notions of projective and free. Here is a beautiful theorem which fascinated me when I started to learn commutative algebra and algebraic geometry. It basically says that freeness of a module "extends from points to open neighborhoods". Let $M$ be a finitely presented $A$-module and $\mathfrak{p} \subseteq A$ a prime ideal such that $M_\mathfrak{p}$ is free. Then there is some $f \in A \backslash \mathfrak{p}$, such that $M_f$ is free. I would certainly include this in the course. The proof is also pretty illuminating, using some of the devissage arguments which are important in commutative algebra and algebraic geometry. - The question reminds me of the time when I was studying mathematics. I had attended a course on algebra (some basic theory of group, rings, categories, loads of Galois theory and some valuation theory, which was the main area of research of the lecturer). Commutative algebra was the first subject I studied on my own, because I wanted to attend a course on algebraic curves, and because I needed some commutative algebra for my diploma thesis. This is not a direct answer to your question, but maybe some thoughts from that time are of use... Literature: As a student I found Bourbaki, Nagata (local rings) and Matsumura (Comm. rings) too difficult as a starting point. I liked the book of Atiyah-MacDonald and I loved the book "Kommutative Algebra" of Brüske, Ischebeck, Vogel. The Brüske-Ischebeck-Vogel book is out of print and available online http://wwwmath.uni-muenster.de/u/ischebeck/SkriptBrskeIschebeckVogel.pdf. If I had to teach a course on commutative algebra, then I would surely have this book on my desk again. Unfortunately it is written in German, but nevertheless it might be worth to have a look... Free, projective and flat modules: I remember that I needed them for my thesis project and my thesis contained a section summarizing these things. I think at that time I was fine with the account in the BIV book. They define M projective iff $Hom(M, -)$ is exact, flat iff $M\otimes -$ is exact and injective if $Hom(-, M)$ is exact. I found this natural at a first reading, and later on the step to $Ext$ and $Tor$ was quite natural as well. (Maybe I was influenced a bit by the fact that categories and functors were always in the air in Munich at that time and were mentioned in my algebra course.) Having these definitions at hand, one can directly go into the proof of theorems comparing these classes of modules... - Coincidentally, I was on the verge of posing a flatness question. But it concerned flat homomorphisms of commutative rings, rather than flat modules. I finally slogged through Durov's proof of the "affine base change theorem" for his generalized rings. As far as I can tell, the only use he makes of flatness is the fact that it forces base change to preserve finite products. No one has mentioned that motivation so far. Is that a typical use? - Yes flatness is a typical ingredient for base change theorems in algebraic geometry. – Martin Brandenburg Jan 31 2012 at 10:22

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http://en.wikipedia.org/wiki/Repeat-accumulate_code

# Repeat-accumulate code In computer science, repeat-accumulate codes (RA codes) are a low complexity class of error-correcting codes. They were devised so that their ensemble weight distributions are easy to derive. RA codes were introduced by Divsalar et al. In an RA code, an information block of length ${N}$ is repeated ${q}$ times, scrambled by an interleaver of size ${qN}$, and then encoded by a rate 1 accumulator. The accumulator can be viewed as a truncated rate 1 recursive convolutional encoder with transfer function ${1/(1 + D)}$, but Divsalar et al. prefer to think of it as a block code whose input block ${(z_1, \ldots , z_n)}$ and output block ${(x_1, \ldots , x_n)}$ are related by the formula ${x_1 = z_1}$ and $x_i = x_{i-1}+z_i$ for $i > 1$. The encoding time for RA codes is linear and their rate is $1/q$. They are nonsystematic. ## References • D. Divsalar, H. Jin, and R. J. McEliece. "Coding theorems for ‘turbo-like’ codes." Proc. 36th Allerton Conf. on Communication, Control and Computing, Allerton, Illinois, Sept. 1998, pp. 201–210.

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http://mathoverflow.net/revisions/96102/list

## Return to Question 4 added 7 characters in body Suppose you are given a single unit square, and you are permitted to cut it into $k$ (connected) pieces (where $k=1$ means just the square). Your task is to construct the largest volume convex body in $\mathbb{R}^3$ by pasting the $k$ pieces together as its complete surface; so its surface area is $\le 1$. Denote its volume by $V_\max(k)$. (This is a more general version of my earlier question, Covering a Cube with a Square.) The best one could hope for is to create a sphere with unit surface area, when the radius satisfies $4 \pi r^2 =1$ and so $r=\frac{1}{2\sqrt{\pi}} \approx 0.28$, and the volume is $V_\max(\infty)= V_\max = \frac{4}{3} \pi r^3 = \frac{1}{6 \pi} sqrt{\pi}} \approx 0.094$. In response to a question of Joe Malkevitch, with students I computed the optimum for $k=1$, with no cuts and no overlap, and found a maximum volume of $V_\max(1)\approx 0.056$, which is about 60% of the sphere volume. [p.418 in Geometric Folding Algorithms: Linkages, Origami, Polyhedra, Demaine, O'Rourke, Cambridge, 2007.]: The seeming randomness of this shape discouraged my further pursuit. The case $k=2$ resembles the famous tea-bag problem, but here I am restricting attention to convex bodies. As $k \to \infty$, one should be able to approach $V_\max$ by, for example, cutting the square into many nearly equilateral triangles and constructing a geodesic dome. I am wondering if anyone has heard of this problem in any guise before? And in any case, can anyone see any clear hypothesis, for any particular $k$? Presumably one could establish that $V_\max(k+1) > V_\max(k)$, but beyond that, I do not see how to make inroads. Is there an obvious candidate for $k=2$? Update. Here is my interpretation of Gerhard's suggestion for $k=2$: 3 added 233 characters in body Suppose you are given a single unit square, and you are permitted to cut it into $k$ (connected) pieces (where $k=1$ means just the square). Your task is to construct the largest volume convex body in $\mathbb{R}^3$ by pasting the $k$ pieces together as its complete surface; so its surface area is $\le 1$. Denote its volume by $V_\max(k)$. (This is a more general version of my earlier question, Covering a Cube with a Square.) The best one could hope for is to create a sphere with unit surface area, when the radius satisfies $4 \pi r^2 =1$ and so $r=\frac{1}{2\sqrt{\pi}} \approx 0.28$, and the volume is $V_\max(\infty)= V_\max = \frac{4}{3} \pi r^3 = \frac{1}{6 \pi} \approx 0.094$. In response to a question of Joe Malkevitch, with students I computed the optimum for $k=1$, with no cuts and no overlap, and found a maximum volume of $V_\max(1)\approx 0.056$, which is about 60% of the sphere volume. [p.418 in Geometric Folding Algorithms: Linkages, Origami, Polyhedra, Demaine, O'Rourke, Cambridge, 2007.]: The seeming randomness of this shape discouraged my further pursuit. The case $k=2$ resembles the famous tea-bag problem, but here I am restricting attention to convex bodies. As $k \to \infty$, one should be able to approach $V_\max$ by, for example, cutting the square into many nearly equilateral triangles and constructing a geodesic dome. I am wondering if anyone has heard of this problem in any guise before? And in any case, can anyone see any clear hypothesis, for any particular $k$? Presumably one could establish that $V_\max(k+1) > V_\max(k)$, but beyond that, I do not see how to make inroads. Is there an obvious candidate for $k=2$? Update. Here is my interpretation of Gerhard's suggestion for $k=2$: 2 added 80 characters in body Suppose you are given a single unit square, and you are permitted to cut it into $k$ (connected) pieces (where $k=1$ means no cuts)just the square). Your task is to construct the largest volume convex body in $\mathbb{R}^3$ by pasting the $k$ pieces together as its complete surface; so its surface area is $\le 1$. Denote its volume by $V_\max(k)$. (This is a more general version of my earlier question, Covering a Cube with a Square.) The best one could hope for is to create a sphere with unit surface area, when the radius satisfies $4 \pi r^2 =1$ and so $r=\frac{1}{2\sqrt{\pi}} \approx 0.28$, and the volume is $V_\max(\infty)= V_\max = \frac{4}{3} \pi r^3 = \frac{1}{6 \pi} \approx 0.094$. In response to a question of Joe Malkevitch, with students I computed the optimum for $k=1$, with no cuts and no overlap, and found a maximum volume of $V_\max(1)\approx 0.056$, which is about 60% of the sphere volume. [p.418 in Geometric Folding Algorithms: Linkages, Origami, Polyhedra, Demaine, O'Rourke, Cambridge, 2007.]: The seeming randomness of this shape discouraged my further pursuit. The case $k=2$ resembles the famous tea-bag problem, but here I am restricting attention to convex bodies. As $k \to \infty$, one should be able to approach $V_\max$ by, for example, cutting the square into many nearly equilateral triangles and constructing a geodesic dome. I am wondering if anyone has heard of this problem in any guise before? And in any case, can anyone see any clear hypothesis, for any particular $k$? Presumably one could establish that $V_\max(k+1) > V_\max(k)$, but beyond that, I do not see how to make inroads. Is there an obvious candidate for $k=2$? 1 # Maximum volume convex body coverable by a unit square Suppose you are given a single unit square, and you are permitted to cut it into $k$ (connected) pieces (where $k=1$ means no cuts). Your task is to construct the largest volume convex body in $\mathbb{R}^3$ by pasting the $k$ pieces together as its surface. Denote its volume by $V_\max(k)$. (This is a more general version of my earlier question, Covering a Cube with a Square.) The best one could hope for is to create a sphere with unit surface area, when the radius satisfies $4 \pi r^2 =1$ and so $r=\frac{1}{2\sqrt{\pi}} \approx 0.28$, and the volume is $V_\max(\infty)= V_\max = \frac{4}{3} \pi r^3 = \frac{1}{6 \pi} \approx 0.094$. In response to a question of Joe Malkevitch, with students I computed the optimum for $k=1$, and found a maximum volume of $V_\max(1)\approx 0.056$, which is about 60% of the sphere volume. [p.418 in Geometric Folding Algorithms: Linkages, Origami, Polyhedra, Demaine, O'Rourke, Cambridge, 2007.]: The seeming randomness of this shape discouraged my further pursuit. The case $k=2$ resembles the famous tea-bag problem, but here I am restricting attention to convex bodies. As $k \to \infty$, one should be able to approach $V_\max$ by, for example, cutting the square into many nearly equilateral triangles and constructing a geodesic dome. I am wondering if anyone has heard of this problem in any guise before? And in any case, can anyone see any clear hypothesis, for any particular $k$? Presumably one could establish that $V_\max(k+1) > V_\max(k)$, but beyond that, I do not see how to make inroads. Is there an obvious candidate for $k=2$?

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http://unapologetic.wordpress.com/2008/03/05/functions-of-bounded-variation-i/?like=1&source=post_flair&_wpnonce=bbf4ba6a50

# The Unapologetic Mathematician ## Functions of Bounded Variation I In our coverage of the Riemann-Stieltjes integral, we have to talk about Riemann-Stieltjes sums, which are of the form $\displaystyle\sum\limits_{i=1}^nf(t_i)(\alpha(x_i)-\alpha(x_{i-1}))$ How can we choose a function $f$ to make sums like this get really big? Well, we could make the function $f$ get big, but that’s sort of a cheap trick. Let’s put a cap that $|f(x)|\leq1$. Now how can we make a Riemann-Stieltjes sum big? Well, sometimes $\alpha(x_i)>\alpha(x_{i-1})$ and sometimes $\alpha(x_i)<\alpha(x_{i-1})$. In the first case, we can try to make $f(t_i)=1$, and in the second we can try to make $f(t_i)=-1$. In either case, we’re adding as much as we possibly can, subject to our restriction. We’re saying a lot of words here, but what it boils down to is this: given a partition we can get up to $\sum\limits_{i=1}^n|\alpha(x_i)-\alpha(x_{i-1})|$ in our sum. As we choose finer and finer partitions, it’s not hard to see that this number can only go up. So, if there’s an upper bound $M$ for our $\alpha$, then we’re never going to get a Riemann-Stieltjes sum bigger than $\max\limits_{x\in\left[a,b\right]}(f(x))M$. That will certainly make it easier to get nets of sums to converge. So let’s make this definition: a function $\alpha$ is said to be of “bounded variation” on the interval $\left[a,b\right]$ if there is some $M$ so that for any partition $a=x_0<...<x_n=b$ we have $\displaystyle\sum\limits_{i=1}^n|\alpha(x_i)-\alpha(x_{i-1})|\leq M$ That is, $M$ is an upper bound for the set of variations as we look at different partitions of $\left[a,b\right]$. Then, by the Dedekind completeness of the real numbers, we will have a least upper bound $V_\alpha(a,b)$, which we call the “total variation” of $\alpha$ on $\left[a,b\right]$ Let’s make sure that we’ve got some interesting examples of these things. If $\alpha$ is monotonic increasing — if $c<d$ implies that $\alpha(c)\leq\alpha(d)$ — then we can just drop the absolute values here. The sum collapses, and we’re just left with $\alpha(b)-\alpha(a)$ for every partition. Similarly, if $\alpha$ is monotonic decreasing then we always get $\alpha(a)-\alpha(b)$. Either way, $\alpha$ is clearly of bounded variation. An easy consequence of this condition is that functions of bounded variation are bounded! In fact, if $\alpha$ has variation $V_\alpha(a,b)$ on $\left[a,b\right]$, then $|\alpha(x)|\leq|\alpha(a)|+V_\alpha(a,b)$ for all $x\in\left[a,b\right]$. Indeed, we can just use the partition $a<x<b$ and find that $\displaystyle|\alpha(x)-\alpha(a)|\leq|\alpha(x)-\alpha(a)|+|\alpha(b)-\alpha(x)|\leq V_\alpha(a,b)$ from which the bound on $|\alpha(x)|$ easily follows. ### Like this: Posted by John Armstrong | Analysis ## 12 Comments » 1. The strict inequality |f|<1 later turns out to be not so strict in your post. Comment by Johan Richter | March 5, 2008 | Reply 2. Thanks for catching that. Comment by | March 5, 2008 | Reply 3. [...] of Bounded Variation II Let’s consider the collection of functions of bounded variation on a little more deeply. It turns out that they form a subring of the ring of all real-valued [...] Pingback by | March 6, 2008 | Reply 4. [...] continue our study of functions of bounded variation by showing that total variation is “additive” over its interval. That is, if is of [...] Pingback by | March 9, 2008 | Reply 5. [...] over subintervals As I noted when I first motivated bounded variation, we’re often trying to hold down Riemann-Stieltjes sums to help them converge. In a sense, [...] Pingback by | March 21, 2008 | Reply 6. [...] Oscillation in a function is sort of a local and non-directional version of variation. If is a bounded function on some region , and if is a nonempty subset of , then we define the [...] Pingback by | December 7, 2009 | Reply 7. [...] explicitly, let be a continuous function on the rectangle , and let be of bounded variation on . Define the function on [...] Pingback by | January 12, 2010 | Reply 8. Why is the upper bound M on alpha, and not on the set of all possible “distances” (i.e. absolute values of differences) between successive alpha values? Comment by Kizi | November 14, 2012 | Reply 9. It may not be completely clear; if there’s an upper bound on the $\alpha$ values then there’s an upper bound on their differences. That’s in the motivation, but the actual definition that follows is written in terms of an upper bound on the sums of absolute differences. Comment by | November 14, 2012 | Reply • Thank you for the prompt response and clarification. In the motivation, do the alpha values need to be bounded below as well (or is there some property of the alpha values that I am missing, like non-negativity)? I do apologise if I am missing something obvious. Thanks. Comment by Kizi | November 14, 2012 | Reply 10. Yeah, they’d probably need to be bounded above and below; the motivation is meant to give the idea, but the definition is precise. Comment by | November 14, 2012 | Reply • Thank you Comment by Kizi | November 14, 2012 | Reply « Previous | Next » ## About this weblog This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”). I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.

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http://mathoverflow.net/questions/28470/infinite-system-of-stochastic-ordinary-differential-equations-coupled-by-infinite

## Infinite System of Stochastic Ordinary Differential Equations Coupled by Infinite Graphs ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) $\ \ \$ In Time Evolution of Infinite Anharmonic Systems, Lanford,Lebowitz and Lieb, roughly speaking, proved that for some families of functions $F_v$ $(v\in\mathbb Z^d)$ and a large set of initial condition, there exist a unique solution defined on a non-degenerated interval $[0,T]$ for the infinite system ```$$ \left\{ \begin{array}{rcl} \frac{d}{dt}q_v(t)&=&p_v(t)\\ \frac{d}{dt}p_v(t)&=&F_v(X(t))\\ X(0)&=&X_0, \end{array} \right. $$``` where $X_0$ and $X(t)$ belongs to some Banach space contained in $(\mathbb R^n\times\mathbb R^n)^{\mathbb Z^d}$ and as mentioned before $v\in\mathbb Z^d$. $\ \ \$ Although the authors do not mention, their results can be generalized for any infinite uniformly bounded degree graph. But a more interesting generalization, would be made by considering the stochastic case. Because it seems to be physically relevant to consider that the forces in each vertex of the graph are not completely known, due some random perturbations. $\ \ \$ Could you help me to find a reference where this kind of model is considered ? -

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http://mathhelpforum.com/calculus/198661-related-rates-cylinder.html

# Thread: 1. ## Related Rates of a Cylinder Water is being pumped into a vertical cylinder of radius 4 meters and height 15 meters at a rate of 2 meters3/min. How fast is the water level rising when the cylinder is half full? 2. ## Re: Related Rates of a Cylinder Originally Posted by titansfreak93 Water is being pumped into a vertical cylinder of radius 4 meters and height 15 meters at a rate of 2 meters3/min. How fast is the water level rising when the cylinder is half full? $V = \pi r^2 h$ since radius is constant ... $V = 16\pi h$ you are given $\frac{dV}{dt}$ , take the time derivative of the volume equation and determine $\frac{dh}{dt}$ ... you'll find that the amount in the cylinder doesn't matter.

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http://math.stackexchange.com/questions/59588/where-can-i-find-information-on-the-calculations-of-limits

Where can I find information on the calculations of limits? I'm doing some research for an algorithm, and I have a lot of curiosity, too. I'd like to learn how limits are calculated by mathematicians and mathematical software. I understand calculus somewhat, but I'm a litty rusty on it and I sometimes get perplexed by calculations involving limits. What's my best bet for learning about how limits are calculated? Should I review my calculus, or is there more science to limits (like numerical methods?!?)? - 1 – anon Aug 25 '11 at 2:17 – JavaMan Aug 25 '11 at 2:25 On the numerical front, things get a little hazy... you'll want to look into the use of sequence transformation methods. Briefly, you construct an appropriate sequence for your limit; e.g. for something like $\lim\limits_{x\to 0} f(x)$, you might consider the sequence $f(x_k)$ where $x_k$ is a sequence that tends to 0 like $k^-n,\quad n>0$ or $c^k,\quad 0 < c < 1$, and then apply a transformation that accelerates the convergence. If that's what you're interested in, I'll write up something. – J. M. Aug 25 '11 at 2:51 @J.M. Yes-I'm very interested in sequence transformation methods. Anything relating to limits is good, and this seems prime territory to explore! I'd like to thank you in advance for bringing this up. :-) – Matt Groff Aug 25 '11 at 3:27 – J. M. Aug 25 '11 at 3:29 show 1 more comment

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http://mathoverflow.net/questions/42065?sort=oldest

## Posets of finite sequences are highly connected ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I need the following result for an example in a paper I'm writing. It's easy enough to prove, but I'd prefer to just give a reference. Does anyone know one? Fix $1 \leq k \leq n$. Define $X_{n,k}$ to be the following poset. The elements of $X_{n,k}$ are ordered sequences $\omega = (x_1,\ldots,x_m)$, where the $x_i$ are distinct elements of the $n$-element set `$\{1,\ldots,n\}$` and $m \geq k$. The order relation is that $\omega_1 \leq \omega_2$ if $\omega_1$ is a subsequence of $\omega_2$. The theorem then is that the geometric realization of $X_{n,k}$ is $(n-1-k)$-connected. - what's the geometric realization of a poset ? – Suresh Venkat Oct 13 2010 at 20:47 1 The set of chains is an abstract simplicial complex, so the geometric realization is just the realization of this complex. – Richard Kent Oct 13 2010 at 20:54 ah ok. i was wondering if that is what it was. – Suresh Venkat Oct 13 2010 at 21:44 ## 1 Answer If I understand your question correctly, an answer should appear in the paper "On lexicographically shellable posets" of Anders Bj\"orner and Michelle Wachs, in Transactions of the AMS 277, pp. 323-341. - Yes, this paper gives me exactly what I want. Thanks! – Andy Putman Oct 14 2010 at 1:39

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http://mathoverflow.net/questions/8874/what-are-some-slogans-that-express-mathematical-tricks/40743

## What are some slogans that express mathematical tricks? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Many "tricks" that we use to solve mathematical problems don't correspond nicely to theorems or lemmas, or anything close to that rigorous. Instead they take the form of analogies, or general methods of proof more specialized than "induction" or "reductio ad absurdum" but applicable to a number of problems. These can often be summed up in a "slogan" of a couple of sentences or less, that's not fully precise but still manages to convey information. What are some of your favorite such tricks, expressed as slogans? (Note: By "slogan" I don't necessarily mean that it has to be a well-known statement, like Hadamard's "the shortest path..." quote. Just that it's fairly short and reasonably catchy.) Justifying blather: Yes, I'm aware of the Tricki, but I still think this is a useful question for the following reasons: 1. Right now, MO is considerably more active than the Tricki, which still posts new articles occasionally but not at anything like the rate at which people contribute to MO. 2. Perhaps causally related to (1), writing a Tricki article requires a fairly solid investment of time and effort. The point of slogans is that they can be communicated without much of either. If you want, you can think of this question as "Possible titles for Tricki articles," although that's by no means its only or even main purpose. - 2 I realize this question's borderline (and I wasn't sure whether I should ask it), but I don't think it's any less MO-appropriate than "Fundamental examples" or the mathematical joke thread. People who downvoted, care to explain why you disagree? – Harrison Brown Dec 14 2009 at 15:31 5 I wasn't the downvoter, but I think some people are getting a little annoyed at the number of these "produce a ginormous list of answers" soft questions. You're entirely right that there are worse offenders on the site, but I think the issue is in part the density of them, not any particular ones. – Ben Webster♦ Dec 14 2009 at 16:07 1 @Ben: That makes sense, although a better solution might be to ignore the "soft-question" tag. Or make a new tag for "ginormous list" questions. But this would be more appropriate on meta, so I'll start a topic there. – Harrison Brown Dec 14 2009 at 16:11 5 Ben: I really like the ginormous list of answer questions. It's nice to have some big picture responses from a variety of mathematicians. Peter: If you don't like soft answer questions, you don't have to read them. There's even a box on the front page for ignoring tags you don't like. – Tom LaGatta Jan 13 2010 at 9:51 2 @Tom: soft-answer is not the same as big-list; the meta discussion that came from this thread actually inspired the "big-list" tag, for exactly the reason that it can be ignored. But if you have more to say, feel free to contribute on meta.MO. – Harrison Brown Jan 13 2010 at 11:01 show 2 more comments ## 21 Answers If something does not hold, make it true! Examples: - Sobolev spaces (not necessarily differentiable functions satisfy differential equations) - distribution theory (think of identities involving the delta "function") - no converging? take the closure of your vector space (analysis) or compactify your space (geometry) - 8 Related to that, "Look in a bigger bag." For example, to find an integer solution, first find a rational or even complex solution then try to show it's an integer. Or find a solution in a Sobolev space then prove the solution is actually a classical solution. – John D. Cook Dec 14 2009 at 21:15 3 Another famous example of this is the notion of stacks: quotients of schemes by group actions do not necessarily exist? We make them exist by sheer brute force! (And it's funny that the definition of a stack also mimics the definition of a distribution somehow -- in both situations, the idea is to forget the object itself and only remember how it acts on some class of "test objects".) – Dan Petersen Dec 15 2009 at 8:19 ### 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 analyst's toolbox consists of three things: 1. The Cauchy-Schwarz inequality 2. Changing the order of integration/summation 3. Integration by parts (I'm not saying I believe that; it's just a very common saying.) - 17 You forgot 4. Adding and substracting something. – Mariano Suárez-Alvarez Jan 13 2010 at 7:37 8 Would you allow me the principle that if the average of some numbers is at least C then one of the numbers must be at least C? – gowers Sep 23 2010 at 21:42 3 An analyst once told me: "When in doubt, integrate by parts." – Micah Milinovich Sep 23 2010 at 22:42 5 I've also heard the following cited as a tool: if $a \leq b + \epsilon$ for all $\epsilon > 0$, then $a \leq b$. – S. Carnahan♦ Sep 24 2010 at 3:02 2 When Peter Lax went to receive the national medal of science, he was asked by the other recipients about his merits. His answer was (apocryph) I integrated by parts. – Denis Serre Apr 7 2011 at 9:33 show 3 more comments Try to replace a structure on an object with a map to a clasifying object. E.g., replace a cohomology class of a space with a map to an Eilenberg-MacLane space. Replace a vector/general bundle on a manifold with a map to the Grassmanian/other classifying space. There must also be plenty of examples outside algebraic topology, though this technique seems to be most popular there... - Along these lines, replace a subset of a finite set by an element of a vector space over GF(2). – Harrison Brown Dec 15 2009 at 2:50 Devissage is a useful tool when proving something holds for a general class of objects, at least in algebraic geometry, like all schemes/stacks/morphisms. - 3 What is devissage? – Ilya Grigoriev Dec 15 2009 at 18:13 8 I don't know if there is a formal definition. For me it roughly means that, when proving something in general, one reduces step by step to special cases. Some people say that it's an important feature in Grothendieck's style proofs. Here's an "example". Suppose we want to prove something for any morphism $f:X\to Y$ of schemes of finite type over a field k. One can check if this property is local in the sense that, if it's true for all fibers of $f,$ then it's true for $f.$ If it is local then we reduce to the case where Y is a point Spec k. (too long to fit in one comment; to be continued) – shenghao Dec 15 2009 at 20:37 3 Let U be any open subset of X with complement Z. Suppose that if the property holds for both U and Z, then it holds for X. Then we may shrink X to any open subset and use noetherian induction, so we can assume X is affine of equidimension d. There exists a map $X\to X'$ with fibers of dimension <=1, and dim X'=d-1, so by induction on d we reduce to two cases: X is a curve, or X is a point. Finally we prove these special cases. – shenghao Dec 15 2009 at 20:39 I forget who this is attributed to, but someone said something like "A technique is a trick used twice." - Weil's "three columns": Number fields over $\mathbb{Q}$ behave like function fields of curves over finite fields which are related to the field of algebraic functions over $\mathbb{C}$. (This is far removed from my comfort zone, so please fix it if I'm off the mark.) - "Think homologically, prove cohomologically!" definitely sounds like a slogan. One argument for this is that homology has a nice explanation in terms of geometry, think singular simplices or cells, so you can think about a space in terms of its cellular homology. When proving things you might want to have more structure around, like a product, and this is where cohomology comes in. - If you have to chose some auxiliary object and that object is not unique, it's better to make all choices simultaneously. I think there are many examples of this, but for me it first hit home when I learned about crystalline cohomology. There you want to lift varieties in positive characteistic to characteristic zero. Locally there are many nonisomorphic lifts, and rather than picking one, it's better to work with the category of all of them. I've absorbed this lesson pretty fully, to the point where I don't need to remind myself of it, but at first it seemed revolutionary. - You must exchange the order of summation in order to prove any identity involving multiple sums. - The best way to solve a problem is to define it out of existence. Typical example: Weil constructed abelian varieties over finite fields, and at first he did not know if these were varieties because it was not clear that they were projective. Weil defined this problem out of existence by changing the definition of variety and inventing abstract varieties. - There are two interesting tricks in K-theory / operator algebras / homotopy theory - one attached to an amusing slogan and the other with an amusing name - that I think foot the bill. The first is "uniqueness is a relative form of existence", due apparently to Shmuel Weinberger. This slogan seems to appear frequently in operator theory. Take, for example the problem of proving that K-theory commutes with direct limits (say, of C* algebras $A_1 \subseteq A_2 \subseteq \ldots \subseteq A$). There are two components to the proof: surjectivity (the "existence" part) which amounts to showing that every element of $K_0(A)$ lies in the image of some $K_0(A_j) \to K_0(A)$, and injectivity (the "uniqueness" part) which involves proving that if two elements of $K_0(A_j)$ are equivalent in $K_0(A)$ then they are equivalent in $K_0(A_j)$. Once you have proven existence you can verify uniqueness by joining representatives of your chosen $K_0(A_j)$ classes by a homotopy in the space of generators for $K_0(A)$ and then use your existence argument to lift to a homotopy in $A_j$. In other words, prove uniqueness by applying your existence argument to a pair. The second is the (in)famous "Eilenberg Swindle" which seems to come up everywhere. I first encountered it in K-theory, but I think the canonical example is the argument which proves that the $n$-sphere is prime with respect to connected sum (which I will denote +). Suppose that $M$ and $N$ are manifolds such that $M + N = S^n$. We have that $(M + N) + (M + N) + (M + N) + \ldots$ is homeomorphic to $\mathbb{R}^n$ (it is a cylinder with the left opening glued shut), and similarly so is $(N + M) + (N + M) + \ldots$. Since $M + (N + M) + \ldots = (M + N) + (M + N) + \ldots$, we have shown that $M + \mathbb{R}^n = \mathbb{R}^n$ which forces $M$ to be homeomorphic to $S^n$. - 1 This is definitely older than Shmuel, but i guess you are referring specifically to the quote. The main point is that uniqueness in homotopy theory is just saying everything you want to construct is connected by a homotopy. So you need that homotopy to exist, but you have an existence result that you can usually apply to construct your homotopy. Also think of the example provided by whitney's embedding theorem and how it shows any two embeddings are homotopic. – Sean Tilson Sep 26 2010 at 6:02 Look at flabbier objects. This seems to be especially useful in complex algebraic geometry. Hard to prove something for varieties? See if there's a version that's true for schemes. Or maybe Kahler manifolds. Or worse: stacks. Vector bundles giving you trouble? Try coherent sheaves. Try quasi-coherent sheaves. In fact, try complexes of them. This is really just a special case of "Generalize the question as far as you can" but in this specific case, it's rather clarifying, here are some examples in algebraic geometry: 1. It's hard to say anything about fundamental groups of complex projective varieties that isn't also true about compact Kahler manifolds. Perhaps the proof should focus on using the Kahler structure, when you're working on these. 2. Want to parameterize subvarieties of a projective variety? Tough, it doesn't work. SubSCHEMES, however, gives the Hilbert Scheme. 3. Proving things about ideals is often easier to do with modules in general - can you elaborate on 3? – Ho Chung Siu Jan 13 2010 at 17:04 Well, that's pretty much one of the main themes of commutative algebra books: that looking at ideals is the wrong point of view, really they're just submodules of the free module on one generator, and many (most, perhaps) results about ideals are actually true about modules, and how to prove them is easier to see when you realize just what the right way to look at things is. – Charles Siegel Jan 13 2010 at 17:10 "If you count something two different ways, you get the same result." This is related to the trick of changing the order of integration (or summation) discussed above, but discrete and more general. This method is used all the time in combinatorics. I think it has also been phrased differently, but I don't remember the exact phrasing. - If you want to show that a graph has few edges, prove that not too many vertices can have large degree. (The complementary statement is the main trick in the solution to this MO question, by way of example. It's also used in the proof of the Stanley-Wilf conjecture.) - is there away of enlarging this to other fields? or rather, how should it be phrased? – Sean Tilson Sep 26 2010 at 6:04 Something like "If the average is small, and you have controlled the large outliers, then everything must be small" but that's not very snappy. – Matthew Daws Oct 1 2010 at 14:04 I'm not sure if this is a bit too general, but it is a slogan/heuristic that I find very useful and that I think most people will be able to come up with plenty of examples of: "Extremalities always arise from symmetry." - 1 I would personally change 'always' to 'often'. Life for geometric analysts will be somewhat easier if this were actually a theorem. In the theoreical physics literature this is related to the so-called Coleman's Principle. However, it was shown by Kapitanski and Ladyzhenskaya that unless one makes additional assumptions, this principle is generally incorrect. ams.org/mathscinet-getitem?mr=711846 – Willie Wong Oct 1 2010 at 15:14 "When in doubt, differentiate." I've heard this attributed to Chern. - A perfect example are the twelve heuristics listed on page 1 of L. Larson, "Problem solving through problems": http://books.google.com/books?id=qFNZIUQ_MYUC&lpg=PP1&dq=larson%20problem%20solving&pg=PA1#v=onepage&q&f=false - One of the slogans in T. W. Körner's book Fourier Analysis that is definitely in the harmonic analyst's toolbox: The function $f*g$ has the good properties both of f and g. An example of its use is in approximating functions by trigonometric polynomials: convolving the function with any trigonometric polynomial gives you a trigonometric polynomial, and if you pick the polynomial carefully the resulting function will have similar properties to the original one. - 2 With one exception: support in physical space. If $f$ has compact support and $g$ does not, the convolution does not have compact support. Of course, another way of looking at it is that having compact support is such an unstable property that it is actually not good in harmonic analysis. – Willie Wong Oct 1 2010 at 15:06 ### "It is easy to prove existence when there is only one, or when there are many" explanation: If there is only one object with a certain property, you can sometimes use it to define it. For example, in geometric situations, you can sometimes define it locally and glue the patches since uniqueness guarantees compatibility on overlaps. It suggests that you should try proving uniqueness before proving existence and if uniqueness fails, maybe you should add constraints (thus, paradoxically, adding constrains can help in proving existence). On the other hand, sometimes it is easier to prove that there are many than to point out one specific example (transcendental numbers, continues nowhere differentiable functions,...). Therefore, you may want to seek for the right notion of "many" in your universe (cardinality, measure, "topological bigness" like the baire property,...) and try to prove that actually there are "few" objects that don't have the required property. comment: This relates to the answer saying that when you can't avoid making a choice, make all of them simultaneously. This happens when there are more than one, but not many... - 1. Pick a random example. 2. If you add lots of small and reasonably independent things together then the result will be highly concentrated about its mean. - Jacobi's famous quote that "one must always invert." He had elliptic integrals in mind. -

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http://mathoverflow.net/revisions/25973/list

## Return to Answer 2 added 1 characters in body; deleted 47 characters in body I would also go for $\Pi(t)$ or $t!$, but a possible reason to prefer the shifted version, $\Gamma(t)$, is the following. The gamma densities $\gamma_t$, $t\in \mathbb{R}$ defined as $\gamma_t(x):=\frac{x_+^{t-1}e^{-x}}{\Gamma(t)},$ are a convolution semigroup, so that $\Gamma(t)$ appears naturally as the normalization factor of $\gamma_t$. (And, of course, the semigroup relation $\gamma_t*\gamma_s=\gamma_{t+s}$ would be destroyed shifting from t-1 to t in the definition of $\gamma_t$) Also note that the expression of the Beta function $B(t,s):=\int_0^1 x^{t-1}(1-x)^{s-1}dx$ in terms of the $\Gamma$ function, if shifted, would also loose the useful form $B(t,s)=\frac{\Gamma(t)\Gamma(s)}{\Gamma(t+s)}.$ (incidentally note that this relation follows plainly from the semigroup property since as a general fact, the integral of a convolution of two functions is the product of their integrals, it you feel like tryin the short computation)integrals). 1 I would also go for $\Pi(t)$ or $t!$, but a possible reason to prefer the shifted version, $\Gamma(t)$, is the following. The gamma densities $\gamma_t$, $t\in \mathbb{R}$ defined as $\gamma_t(x):=\frac{x_+^{t-1}e^{-x}}{\Gamma(t)},$ are a convolution semigroup, so that $\Gamma(t)$ appears naturally as the normalization factor of $\gamma_t$. (And, of course, the semigroup relation $\gamma_t*\gamma_s=\gamma_{t+s}$ would be destroyed shifting from t-1 to t in the definition of $\gamma_t$) Also note that the expression of the Beta function $B(t,s):=\int_0^1 x^{t-1}(1-x)^{s-1}dx$ in terms of the $\Gamma$ function, if shifted, would also loose the useful form $B(t,s)=\frac{\Gamma(t)\Gamma(s)}{\Gamma(t+s)}.$ (incidentally note that this relation follows plainly from the semigroup property since as a general fact, the integral of a convolution of two functions is the product of their integrals, it you feel like tryin the short computation).

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http://mathhelpforum.com/advanced-algebra/178902-normalizing-4x4-matrix-unknown-functions-elements.html

Thread: 1. Normalizing a 4x4 matrix with unknown functions as elements Hi All! This is my first time putting up my own thread on MF. I can usually find what I'm looking for, but this time: no go. As the title says, I'm trying to find the normalization constant of this 4x4 matrix (g and f are functions): (1-g^2) 0 0 0 0 (1+f^2) (-g^2-f^2) 0 0 (-g^2-f^2) (1+f^2) 0 0 0 0 (1-g^2) It's a matrix that's in a research paper which gives the normalization constant (they use the word 'factor'. I'm not sure if that means something diff.) as: N=4-2g^2+2f^2. 1]I've been looking up online and found that N can be found with: N=\sqrt{\sum{X^2}} where X represents the elements of the matrix. 2]I also found somewhere which said that I need to find the determinant. I'm not sure who's right, but I'm not getting what's in the paper. For method [1] I'm getting as far as: N^2 = 4(1+f^4+f^2g^2+f^2) and got stuck trying to find the square root (it's been a while since I've done multinomial theorem). So I backtracked to see if their N^2 matches my N^2. But their N^2=16+4g^4+4f^4+16g^2-8g^2f^2+16f^2. and method [2] is giving me something so long, with so many variables of (g^2), (f^2), (g^4), (f^4),(g^2f^4) (and it keeps going for about 3tysomething variables) that I've given up. So I'm wrong all over the place. Can someone help me out? booklist05 confused 2. I don't know what is meant by the normalisation constant of a matrix. But it looks from your suggested definitions 1] and 2] as though it is meant to be some function of the eigenvalues of a (positive?) matrix. The definition 1] gives the root mean square sum of the eigenvalues. The suggestion in 2] is to find the determinant (product of the eigenvalues). It looks to me as though the formula for N given here ( $N=4-2g^2+2f^2$) is simply that N is the sum of the eigenvalues. This is equal to the trace of the matrix, which is the sum of its diagonal elements. That is easily seen to be $4-2g^2+2f^2$. Or is that too simple-minded a suggestion? 3. Omg! can it really be that simple? Later on in the paper, they take each of the elements and divide them each by N. This is why I think they call it "normalization constant". I'm still in shock. so simple? Well since it works, that's what I'm going with. No one else I've asked has been able to give me an answer. Thanks!

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http://physics.stackexchange.com/questions/25802/why-does-a-black-hole-have-a-finite-mass/25803

# Why does a black hole have a finite mass? I mean besides the obvious "it has to have finite mass or it would suck up the universe." A singularity is a dimensionless point in space with infinite density, if I'm not mistaken. If something is infinitely dense, must it not also be infinitely massive? How does a black hole grow if everything that falls into it merges into the same singularity, which is already infinitely dense? - 3 Note that not all theories explain a black hole as a singularity. There are some exotic ideas out there, but in some cases the black hole is simply extremely dense, but not necessarily infinitely dense (but still dense enough to have an event horizon). The singularity arises from relativity. – voithos Jun 14 '11 at 6:33 ## 4 Answers If something is infinitely dense, must it not also be infinitely massive? Nope. The singularity is a point where volume goes to zero, not where mass goes to infinity. It is a point with zero volume, but which still holds mass, due to the extreme stretching of space by gravity. The density is $\frac{mass}{volume}$, so we say that in the limit $volume\rightarrow 0$, the density goes to infinity, but that doesn't mean mass goes to infinity. The reason that the volume is zero rather than the mass is infinite is easy to see in an intuitive sense from the creation of a black hole. You might think of a volume of space with some mass which is compressed due to gravity. Normal matter is no longer compressible at a certain point due to Coulomb repulsion between atoms, but if the gravity is strong enough, you might get past that. You can continue compressing it infinitely (though you'll probably have to overcome some other force barriers along the way) - until it has zero volume. But it still contains mass! The mass can't just disappear through this process. The density is infinite, but the mass is still finite. - 2 Right. I guess the division by zero means there are infinite solutions, not necessarily that the parameters are infinite. I blurred that distinction. – Carson Myers Jun 14 '11 at 2:34 1 Also, black holes are constructed from a finite amount of mass-energy. – Andrew Jun 14 '11 at 17:20 A black hole has an infinite density; since its volume is zero, it is compressed to the very limit. So it also has infinite gravity, and sucks anything which is near it! Not everything there is!! Now above all when it sucks things it adds up to its mass, which remains finite and it always will, even if it did suck in the whole universe!! It's all for the formula: density in a black hole is mass divided by volume (0) so the density is infinite, not the mass!!! So thus a black hole has mass which is finite and will always be finite. - if it has infinite density, that would mean it has infinite gravity (or infinite space curvature in special relativity, which i believe in). so, it obviously doesn't have infinite density because that would suck the whole entire universe into itself. that would mean it would have to have an unmeasurably small amount of volume. and even if it did have infinite density, space curvature should be more powerful nearer to the singularity, but how can we add more to an infinite amount of gravity? if it had infinite density, its curvature should be equal everywhere, which is also impossible. What do you think of this theory? - I'm not going to claim any real knowledge on the topic, or most of what I bring up, so feel free to correct me for any extremely rash or impossible aspects, but here it goes. A black hole to me, seems like it's just the epitomy of nothing. There's something, then general space, and then a black hole. It appears to me that something to nothing is relatively possible, as at one point nothing came to be something, however I feel that the universe itself wouldn't allow such a flaw, as the universe seems quite perfect in its design. One theory I've stumbled upon is a theory that all atoms have a negative and positive form of which they can swap between. Kind of a matter and anti-matter deal, but all atoms experience this, and that we live in the positive. This kind of interacts with the idea of a black hole, a massive explosion of something that created energy to powers beyond numbers we have words for (googlplex's), for a length of years we barely use the number for (billions). And it eventually explodes, and in the wake of such energy, a black hole is formed (If that's still what people believe). So a rediculous mass of positive energy goes out from a point in the universe, and in this point, it sucks in mass, everything, a perfect unescapable gravity pit. Could it possibly be that after so much positive energy radiates from one point, it explodes, then begins sucking in all the positive energy that comes into it, to rebalance the rediculous amount of negative energy that would then be in the wake of such a thing? I can kind of imagine it, more like an explosion under water (an implosion really). First it explodes, and then everything gets sucked in until it's at its natural point. Except this is on a massive (much bigger than any bomb we could ever make or imagine), yet scale so small (fills itself in to a sub-atomic level, if not to an electron level) it takes an unbelievably long time to go back to normal. And even then that doesn't factor in to say that all this energy its taking in, isn't being vortex'd to another space or even time. So, as a summary, I suppose what I'm concluding black holes to be, is an enormous sphere of negative. And as all things want to be neutral, it sucks in as much positive (the reality we live in) as possible, practically, indefinitely, all positive matter that goes near it to be sucked in without flaw, no possible escape. To anwser whether it has mass, I really don't know for sure, but from what I can tell, it has a Finite mass. But the time required for it to reach a neutral point is absolutely rediculous, seeing as how it would have to fill all that space flawlessly, to the absolute with the little bit of energy that it gets coming in from light, or small fragments of atoms. But again, I have no real knowledge on the topic, its just my educated guess on what I've read and learned over a little while. - This isn't really coherent. There's also no science behind it. – Daniel Blay Dec 11 '12 at 22:55

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http://mathoverflow.net/questions/68746/riemannian-length-of-an-element-of-the-fundamental-group-of-a-manifold/68750

## riemannian length of an element of the fundamental group of a manifold ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) It is a stupid question i guess but like they say if you ask you are stupid for 5 minutes and if you don't ask you are stupid forever . here is the question given a closed manifold $(M,g)$ and $\alpha$ in $\pi_1(M,p)$ we can define the length of $\alpha$ as the minimum riemannian length of a representative now obviousle length $\alpha^2$ is less or equal then $2\mathrm{length}(\alpha)$ but it seems it is always equal $2\mathrm{length}(\alpha)$ i want to know why ? - 4 The statement is false. Think projective plane. – Richard Kent Jun 24 2011 at 16:01 under what conditions the statement is true ? and can the length of \alpha^2 be less the length of \alpha ? – unkown Jun 24 2011 at 16:34 3 When $\pi_1$ has an element $\alpha$ of order 2, $\alpha^2$ can have length 0 (that's what R.Kent was suggesting). There are intermediate possibilities too, such as $\alpha$ of order 3 implies $\alpha$ and $\alpha^2$ have the same length; this doesn't happen in two dimensions but does in three (e.g. lens space). – Noam D. Elkies Jun 24 2011 at 17:02 8 Generically, a geodesic loop realizing the minimum length does not close up smoothly at $p$, and then the double loop is not a geodesic and hence not a minimizer. So the statement is almost never true unless you minimize in a free homotopy class. – Sergei Ivanov Jun 24 2011 at 17:40 (sorry, I see I have unwillingly down-voted this question) – Pietro Majer Jun 27 2011 at 16:16 ## 1 Answer There are geometric hypotheses that ensure the property you want. For example, suppose that $(M,g)$ has negative curvature. Then every $\alpha \in \pi_1(M)$ is freely homotopic to a unique geodesic representative $\alpha^*$. Usually people write $\ell_g(\alpha)$ for the length of $\alpha^*$. Finally, uniqueness of geodesic representatives implies that $\ell_g(\alpha^k) = k \cdot \ell_g(\alpha)$. This is just the beginning of an important area in Riemannian geometry. (When are geodesic representatives unique? What is the interaction between the metric and the variational properties of geodesics? And in a different direction: How does the fundamental group act on the universal cover? What does the metric tell us about the algebraic topology of $M$ and the universal cover? Etc.) It is amusing to contemplate all the ways in which the real projective plane, or more generally any closed manifold with finite fundamental group, differs from a negatively curved manifold. - is there is some algebraic properties of the fundamental group that can garuentee the property without talking about curvature say for example if the fundamental group is torsion free – unkown Jun 24 2011 at 17:22 From another hand thanks for your answer but it doesn't answer the question because the length i am talking about is for \alpha=[c] in pi_1(M,p) length(\alpha) is the infimum length of loops that are homotpic to c and not freely homotopic – unkown Jun 24 2011 at 17:26 I personally seriously doubt that there exists such an algebraic condition, as there are many manifolds with a given fundamental group and you can perturb the metric on a Riemannian manifold in several ways. I would be very surprised if the property you require was stable under those perturbations (except in the case of `$S^1$`), see the comment by Sergei Ivanov. Btw, the only examples of manifolds with that property I can think of are tori (with flat metric). And simply connected manifolds. – Alessandro Sisto Jun 24 2011 at 18:47 @unknown - I assumed that free homotopy classes were meant and I explicitly wrote free homotopy in my answer. In the setting of based homotopy classes (as others have pointed out) length is basically never multiplicative with respect to powers... Somethings can still be shown even in the based case. For example, if you look at $\ell_g(\alpha^k)/k$ using based length (where $g$ is negatively curved) then this approaches the free length. – Sam Nead Jun 24 2011 at 22:24 @unknown - Regarding your first comment. The Klein bottle has torsion free fundamental group. At the same time there is a based loop whose (based) square has length less than twice the original. @Alessandro - and products of spheres and tori, I guess. Are there other examples? – Sam Nead Jun 24 2011 at 22:34 show 3 more comments

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http://nrich.maths.org/4935

### Area L Draw the graph of a continuous increasing function in the first quadrant and horizontal and vertical lines through two points. The areas in your sketch lead to a useful formula for finding integrals. ### Integral Equation Solve this integral equation. ### Integral Sandwich Generalise this inequality involving integrals. # Integral Inequality ##### Stage: 5 Challenge Level: (i)Suppose that $a$, $b$ and $t$ are positive. Which of the following two expressions is the larger $$P=\left(\int_0^t x^{a+b}dx\right)^2, \qquad Q=\left(\int_0^t x^{2a}dx \right) \left(\int_0^t x^{2b}dx\right)\ ?$$ (ii)By considering the inequality $$\int_0^t [f(x)+\lambda g(x)]^2 dx \geq 0,$$ prove that, for all functions $f(x)$ and $g(x)$, $$\left(\int_0^t f(x)g(x)dx\right)^2 \leq \left(\int_0^t f(x)^2 dx\right) \left(\int_0^t g(x)^2 dx\right).$$ The NRICH Project aims to enrich the mathematical experiences of all learners. To support this aim, members of the NRICH team work in a wide range of capacities, including providing professional development for teachers wishing to embed rich mathematical tasks into everyday classroom practice. More information on many of our other activities can be found here.

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http://math.stackexchange.com/questions/106604/min-number-of-vertices-in-graph-as-function-of-kappag-and-diamg

# Min. number of vertices in graph as function of $\kappa(G)$ and $diam(G)$ Question I found in "Introduction to Graph Theory" by Douglas B. West: Let $G$ graph on $n$ vertices with connectivity $\kappa(G)=k \geq 1$. Prove that $n \geq k(diam(G)-1)+2$, when $diam(G)$ is the the maximal (edge) distance between a pair of vertices in $G$ Seems easy and I've seen the answer, too, which goes the same as thought: by taking the vertices the end vertices of the path of length $diam(G)$, and applying Menger's Theorem. I'm having a hard time understanding one part - it seems like all the paths need to be of the same length for the $n \geq k(diam(G)-1)+2$ to apply. Am I missing some part here? Thanks in advance. - ## 1 Answer Let's write $d$ for the diameter of $G$. Then there is some pair of vertices such that one path between them has exactly $d$ edges, and each path between them has at least $d$ edges. Menger says there are $k$ independent paths. So the number of vertices is at least $k(d-1)+2$. I'm not sure why you think the paths all have to be the same length. They just have to have length at least $d$ - and, they do. - I don't think that all the paths have the same length, but that the maximal path length is $d$ (or am I misunderstanding the notion of $diam(G)$?). – Pavel Feb 8 '12 at 19:25 My understanding is that the diameter of a graph is the maximum over all pairs of the minimum distance between the pair. Once you have that maximizing pair, all paths joining those two vertices have length at least the maximum (and at least one path has length exactly the maximum). Maybe I'm wrong - find a definition of diameter in some textbook or on the web, and show me. – Gerry Myerson Feb 8 '12 at 23:10 Oh, now I see that I was wrong about the definition. Thanks for the answer! – Pavel Feb 9 '12 at 11:19

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http://torus.math.uiuc.edu/cal/math/cal?year=2010&month=10&day=28&interval=day

Seminar Calendar for events the day of Thursday, October 28, 2010. . events for the events containing More information on this calendar program is available. Questions regarding events or the calendar should be directed to Tori Corkery. ``` September 2010 October 2010 November 2010 Su Mo Tu We Th Fr Sa Su Mo Tu We Th Fr Sa Su Mo Tu We Th Fr Sa 1 2 3 4 1 2 1 2 3 4 5 6 5 6 7 8 9 10 11 3 4 5 6 7 8 9 7 8 9 10 11 12 13 12 13 14 15 16 17 18 10 11 12 13 14 15 16 14 15 16 17 18 19 20 19 20 21 22 23 24 25 17 18 19 20 21 22 23 21 22 23 24 25 26 27 26 27 28 29 30 24 25 26 27 28 29 30 28 29 30 31 ``` Thursday, October 28, 2010 Group Theory Seminar 1:00 pm in 347 Altgeld Hall, Thursday, October 28, 2010 Del Edit Copy Submitted by clein. Enric Ventura (Universitat Politecnica de Catalunya and CRM-Montreal)The conjugacy problem for some extensions of F_n, Z^m, B_n, and FAbstract: We will review the main idea in the solution of the conjugacy problem (CP) for free-by-cyclic groups given by Bogopolski-Martino-Maslavova-Ventura in 2006. A close analysis of this argument having the classical Miller's groups in mind (which are free-by-free and have unsolvable conjugacy problem) gave rise to a subsequent and stronger result by the same authors, giving an explicit characterization of the solvability of the conjugacy problem within the family of free-by-free groups. It turns out that the freeness of the base group is irrelevant in the whole proof, and the only crucial property is the solvability of the so-called twisted conjugacy problem (TCP). This way, we shall give characterizations of the solvability of the conjugacy problem for certain families of extensions of groups with solvable TCP. We will discuss the particular cases of extensions of free groups $F_n$, free abelian groups $Z^m$, Braid groups $B_n$, and Thomson's group $F$. Number Theory 1:00 pm in 241 Altgeld Hall, Thursday, October 28, 2010 Del Edit Copy Submitted by berndt. Kevin Ford (Illinois)R. I. P.Abstract: Recently, Candes and Tao showed that a k-sparse vector x in R^N (think of x as a signal) can be effectively recovered from a set of n-dimensional linear measurements with n much smaller than N, if the matrix of the vectors v_i satisfies the Restricted Isometry Property (RIP for short) of order k. All known explicit constructions of RIP matrices make use of number theory, and all have order k=O(sqrt(n)) (we will review these). We give a new explicit construction of RIP matrices of order n^{1/2+c} for some positive c, based on additive combinatorics. This is joint work with J. Bourgain, S. J. Dilworth, S. Konyagin and D. Kutzarova. Mathematics Colloquium 4:00 pm in 245 Altgeld Hall, Thursday, October 28, 2010 Del Edit Copy Submitted by clein. David Dumas (University of Illinois at Chicago)Complex projective structures and character varietiesAbstract: A complex projective structure is a way to build a surface from pieces of the Riemann sphere glued together using Mobius transformations. After giving a more precise definition and some examples, we will describe the moduli space of such structures and discuss how this space is related to the character variety of maps from a surface group in PSL(2,C).

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http://www.physicsforums.com/showthread.php?t=517665

Physics Forums ## Finding Moment of Interia of a 'Loop' when given Density & Cross sectional area... 1. The problem statement, all variables and given/known data I eventually have to solve for maximum angular acceleration of the loop in a magnetic field, and I have gotten everything with the exception of the moment of inertia, so I won't include the emf and B known variables. Known: a copper wire with a density of $\rho$ = 8960 kg/m3 is formed into a circular loop of radius 0.50 m. Cross sectional area of the wire is 1.00 x 10-5m2. 2. Relevant equations I=MR2 (and eventually) $\tau$ = $\alpha$I 3. The attempt at a solution I know since mass isn't given, I need to integrate something so I can use the density. However, it's been a really long time since I've integrated, so I'm not very familar with it. I've been unable to find an equation to find the volume of the 'loop,' so I know integration is the only way. 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 No integration required. A loop is just a cylinder bent into a circle. Imagine bending the loop back into a "normal" cylinder, and find the volume of that object. (If you don't remember the formula for the volume if a cylinder, it is easy to find). You can approximate the loop as a circle (line) because the cross section is much smaller than the radius. ## Finding Moment of Interia of a 'Loop' when given Density & Cross sectional area... Yes, I forgot to add that part. But first, OP needs to find the mass, which requires finding the volume. Once that is done you would forget about the finite width and simply use I = MR2 Tags angular acceleration, density, magnetic field, moment of interia Thread Tools | | | | |---------------------------------------------------------------------------------------------------------|----------------------------------|---------| | Similar Threads for: Finding Moment of Interia of a 'Loop' when given Density & Cross sectional area... | | | | Thread | Forum | Replies | | | Precalculus Mathematics Homework | 1 | | | General Math | 11 | | | Classical Physics | 0 | | | Introductory Physics Homework | 2 | | | Introductory Physics Homework | 3 |

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http://math.stackexchange.com/questions/127964/about-probability/127967

# About probability Is a probability function in $\Omega{(a1, a2, a3)}$, find $P(a1)$ whether $P({a2, a3}) = 2P(a1)$. I know that $1 = a1 + a2 + a3$. from where I have to start ? - Do you mean to say that the sample space is a three element set? That is $\Omega=\{a_1,a_2,a_3\}$? – user21436 Apr 4 '12 at 12:48 This is exactly, the sample space is Ω={a1,a2,a3}. – mastergoo Apr 4 '12 at 12:50 You have to start with a comprehensible question. What does $P(a_2,a_3)$ mean? – Gerry Myerson Apr 4 '12 at 12:52 $a2 \Cup a3$, this is ? – mastergoo Apr 4 '12 at 12:53 ## 1 Answer Start by using additivity: $P(\{a_2, a_3\} \cup \{a_1\}) = P(\{a_2,a_3\})+P(\{a_1\})$. Since $\{a_2, a_3\} \cup \{a_1\} = \Omega$, and $P(\Omega) = 1$, we have found that $$P(\{a_2,a_3\})+P(\{a_1\}) = 1.$$ Now substitute $P(\{a_2,a_3\}) = 2P(\{a_1\})$ into this equation, and you'll find $P(\{a_1\}) = \frac{1}{3}$. - Oh yes, I've find the following result: $1 = a1 + a2 + a3$, so $1 = a1 + (2a1)$, so $a1 = \frac{1}{3}$ – mastergoo Apr 4 '12 at 12:59 THANK YOU SO MUCH. Very good explanation. – mastergoo Apr 4 '12 at 13:02 1 You're welcome, but you really need to be more precise with your notation: $a_1$ is the event (or rather $\{a_1\}$ is), which is very different from its probability $P(\{a_1\})$. – lazyhaze Apr 4 '12 at 13:10 By the way, don't put "Solved" in the title, but instead accept my answer (if you find it acceptable, of course). – lazyhaze Apr 4 '12 at 13:11 Ok ok, I undertood.. – mastergoo Apr 4 '12 at 13:23

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http://chemistry.stackexchange.com/questions/151/why-do-elements-in-columns-6-and-11-assume-abnormal-electron-configurations

# Why do elements in columns 6 and 11 assume 'abnormal' electron configurations? When I look around for why copper and chromium only have one electron in their outermost s orbital and 5/10 in their outermost d orbital, I'm bombarded with the fact that they are more stable with a half or completely filled d orbital, so the final electron enters that orbital instead of the 'usual' s orbital. What I'm really looking for is why the d orbital is more stable this way. I assume it has to do with distributing the negative charge of the electrons as evenly as possible around the nucleus since each subshell of the d orbital is in a slightly different location, leading to a more positive charge in the last empty or half-filled d orbital. Putting the final electron in the s orbital would create a more negative charge around the atom as a whole, but still leave that positive spot empty. Why does this not happen with the other columns as well? Does this extra stability work with all half or completely filled orbitals, except columns 6 and 11 are the only cases where the difference is strong enough to 'pull' an electron from the s orbital? It seems like flourine would have a tendency to do do this as well, so I suppose the positive gap left in the unfilled p orbital isn't strong enough to remove an electron from the lower 2s orbital. - ## 5 Answers As I understand this, there are basically two effects at work here. When you populate an s orbital, you add a significant amount of electron density close to the nucleus. This screens the attractive charge of the nucleus from the d orbitals, making them higher in energy (and more radially diffuse). The difference in energy between putting all the electrons in d orbitals and putting one in an s orbital increases as you fill the d orbitals. Additionally, pairing electrons in one orbital (so, adding the second s electron) carries a significant energy cost in terms of Coulombic repulsion because you're adding an electron essentially in exactly the same space as there's already an electron. I'm assuming that the effect isn't strong enough to avert fluorine having a $2s^2$ occupation, and if you look at gadolinium, the effect there isn't strong enough to stop the s from filling (large nuclear charge and orbital extent at the nucleus is a good combination energy-wise), it does manage to make it more favourable to add the electron into the 5d instead of the 4f orbitals. Also, if you take a look at tungsten vs gold, there the effect isn't strong enough for tungsten to avoid a $6s^2$ occupation, but is for gold - more d electrons making the screening effect overcome the strong nuclear charge and enhanced nuclear penetration of an s orbital. - ## Did you find this question interesting? Try our newsletter email address This is just a confirmation to Aesin's answer... Say, we take copper. The expected electronic configuration (as we blindly fill the d-orbitals along the period) is $[Ar]\ 3d^9\ 4s^2$, whereas the real configuration is $[Ar]\ 3d^{10}\ 4s^1$. There is a famous interpretation for this, that d-orbitals are more stable when half-filled and completely-filled. That's a complete myth. There are very few pages explaining this myth. As we fill the electrons starting from $3d^1$, we'd be stuck at Chromium and also at Copper. While filling Chromium and Copper, it has been observed that the energies of $4s$ and $3d$ orbitals are fairly close to each other. The increasing nuclear charge (as we go along the period) and the size & shape of d-orbital should be a reason. This similarity makes the energy for pairing up electrons in d-orbital very less than that of pairing up in s-orbital (i.e.) the energy difference between these orbitals is much less than the pairing energy required to fill the electrons in 4s orbital. Moreover, the energy for the configuration $3d^5\ 4s^1$ is much less than that of $3d^4\ 4s^2$. Since we usually fill electrons in the order of increasing energy, the next electron (in case of Mn) goes into the 4s-orbital. The same reason for effective nuclear charge makes the 3d-orbitals somewhat lower in energy than 4s-orbitals and hence, the unusual configuration of Cr and Cu. Here's a paper which supports that this is quite untrue... In the case of chromium, this means that $4s^1\ 3d^5$ will be lower in energy than $4s^2\ 3d^4$, because in the second case you have to "pay" the electron pairing energy. Since this pairing energy is larger than any difference in the energies of the 4s and 3d orbitals, the lowest energy electron configuration will be the one which has one electron in each of the six orbitals that are available. Effectively this is Hund's rule applying not just to strictly degenerate orbitals (orbitals with the same energy), but to all orbitals that are (significantly) closer in energy than the electron pairing energy. In the case of copper, the 3d orbital has dropped in energy below that of the 4s, so that it is better to have the paired electrons in the d and the unpaired one in the s. The reason why the 3d is lower than 4s is tied to the high effective nuclear charge. The high effective nuclear charge gives rise to the small size of Cu compared with the earlier transition metals, and also means that orbitals in inner shells are more stabilised with respect to those further out for copper than for earlier elements. - We know that the electronic configuration of chromium is $\ce{[Ar] 3d^5 4s^1}$. It is because promotion occurs in case of $3d$ and $4s$ orbitals -- in other words, the electron is shifted from a lower energy level to a higher one (also known as excitation). Promotional energy and pairing energy both are endergonic so the process which requires less energy would preferably take place. - 1 Could you explain a bit more? For example, why does that process have less energy? Elaborate a bit and your post will be improved greatly :) – ManishEarth♦ Nov 19 '12 at 18:51 The experimental data has 6-unpaired e's for Cr and 1 for Cu, therefore, the spdf model must "adapt". It is easy to see Cu (s1d10), but less for Cr (s1d5). The MCAS orbital model presents a different explanation as to why these are different. The MCAS atomic orbital model provides a more logical sstem for the periodic table, too. See the Cu, Cr issue and the periodic table according to this orbital model at this site You can see more about Modeling the MCAS Way at this other site which is the sharper version of the original that is in the arxiv.org database as html/physics/9902046v2. Of course, the physics and chemistry communities stand entrenched behind the spdf model that is heavily steeped in math. - 1 Welcome to Chemistry.SE. Your answer would be improved if you could summarize the content of the links you present rather than just putting the url into the text. – Ben Norris Apr 17 at 1:32 Thanks for the advise. I wrote much as it was. The two links have abstracts. It would be difficult to "summarize" something when a reader would undoubtedly have little knowledge of the model that is different that the spdf one. The URL links let the reader know where to look for what the model is and, as I pointed out, addresses the Cu, Cr electron issue explicitedly, which is why I chose to answer the question. – Joel M Williams Apr 17 at 2:21 Perhaps this shouldn't be counted as an answer, but since this topic has been resurrected, I'd like to point to Cann (2000). He explains the apparent stability of half-filled and filled subshells by invoking exchange energy (actually more of a decrease in destabilization due to smaller-than-expected electron-electron repulsions). According to him, there is a purely quantum mechanical energy term proportional to $$\frac{n_{↑}(n_{↑}-1)}{2} + \frac{n_{↓}(n_{↓}-1)}{2}$$, where $n_{↑}$ and $n_{↓}$ represent the number of spin-up and spin-down electrons in a subshell. This term decreases the potential energy of the atom, and it can be shown that the difference in exchange energy between two consecutive subshell populations (such as $p^{3}/p^{4}$, $f^{9}/f^{10}$, etc) has local maxima at half-filled and filled subshell configurations. This can be used to defend the favourability of $s^{1}d^{5}$ and $s^{1}d^{10}$ configurations relative to $s^{2}d^{4}$ and $s^{2}d^{9}$, even though there would be a slight increase of electron density in the more compact $d$ subshells. However, this clashes with Crazy Buddy's reference, which seems to deny any stabilization effect. So, which is (more) true? Or is neither? -

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http://mathhelpforum.com/calculus/114574-intersecting-points-polar-coordinates.html

# Thread: 1. ## intersecting points - polar coordinates Hi C1 r= 3cos x C2 r= 1 + cos x First question is find the polar coordinates of the intersection points of the two curves. I have found (3/2, pi/3) and (3/2, -pi/3) now what I am wondering is the point that we would normally consider (0,0) also an intersecting point and how would I find the coordinates for this point? Thanks Calculus beginner 2. Originally Posted by calcbeg Hi C1 r= 3cos x C2 r= 1 + cos x First question is find the polar coordinates of the intersection points of the two curves. I have found (3/2, pi/3) and (3/2, -pi/3) now what I am wondering is the point that we would normally consider (0,0) also an intersecting point and how would I find the coordinates for this point? Thanks Calculus beginner Your question seems to be a bit unclear. (0, 0) is not in the solution set. In fact neither of your given curves passes through this point. As to the rest we have $r = 3~cos( \theta)$ and $r = 1 + cos( \theta)$ Since r = r this means $3~cos( \theta) = 1 + cos( \theta)$ so $2~cos( \theta) = 1$ $cos( \theta ) = \frac{1}{2}$ You have mentioned the only two correct solutions. ( pi / 3 is the reference angle so there is another solution 4 pi / 3, or -pi / 3 which you already gave. -Dan

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http://mathematica.stackexchange.com/questions/11542/numerical-comparisons-of-matrices

# Numerical comparisons of matrices I have a matrix which should be equal to a null matrix. However due to the numerical precision, a brutal equality test with a matrix initialized with zeros does not work. How should I perform the numerical equality test (with a given threshold for the precision) ? - ## 1 Answer A simpler way than to adjust the threshold for `Equal` is to use `Chop` which replaces approximate real numbers in expr that are close to zero by the exact integer 0. Adding the suggestions from the comments you have the following possibilities: • Use `Chop` as it is. Here, you may only chop the `Norm` at the end by `Chop[Norm[mat, 1]] == 0`. • Look at the second argument to `Chop` when you want to adjust the default tolerance. Ideally, it should correspond to the "sizes" of the matrices from which the putative null was constructed. For instance, if those matrices involve numbers in the millions, then any matrix whose coordinates are all less than about 0.0001 would likely need to be considered null. Typically, the second argument will be somewhere between the smallest and largest absolute values of the eigenvalues of those matrices. (These eigenvalues can reliably be found with `SingularValueDecomposition`; in many applications, they are already available from previous computations.) • Look at the `Internal`$EqualTolerance` (which is probably not the best idea in your case). - 1 Note, also, that one can control the tolerance used by `Chop[]` through its second argument. See the docs for details. – J. M.♦ Oct 4 '12 at 10:03 1 Another possible route to check if the matrix `mat` is a null matrix: `Chop[Norm[mat, 1]] == 0`; only a null matrix has zero norm. – J. M.♦ Oct 4 '12 at 10:14 1 It might be good to not compare against zero but test if the norm is smaller than an epsilon. – ruebenko Oct 4 '12 at 11:44 1 Be careful. There does not exist any absolute universal test like this. The comparison of a matrix to zero needs to account for the matrices used to create it. As an example, emulate the `SingularValueDecomposition` help by creating a random matrix `m` with entries on the order of, say, $10^{12}$, reconstruct `m` via its SVD, and subtract the reconstruction from `m` to see whether the two are equal. They're not--due to imprecision--but `chop` won't help. How much imprecision should one expect? Use the sizes of the eigenvalues of `m` (the diagonal of the SVD) to estimate the tolerance. – whuber Oct 4 '12 at 14:44 1 @whuber I gave you comment +1 and I would be happy if you insert your concerns into the answer. – halirutan Oct 4 '12 at 15:01 show 3 more comments lang-mma

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http://stats.stackexchange.com/questions/16349/how-to-compute-the-confidence-interval-of-the-ratio-of-two-normal-means

# How to compute the confidence interval of the ratio of two normal means I want to derive the limits for the $100(1-\alpha)\%$ confidence interval for the ratio of two means. Suppose, $X_1 \sim N(\theta_1, \sigma^2)$ and $X_2 \sim N(\theta_2, \sigma^2)$ being independent, the mean ratio $\Gamma = \theta_1/\theta_2$. I tried to solve: $$\text{Pr}(-z(\alpha/2)) \leq X_1 - \Gamma X_2 / \sigma \sqrt {1 + \gamma^2} \leq z(\alpha/2)) = 1 - \alpha$$ but that equation couldn't be solved for many cases (no roots). Am I doing something wrong? Is there a better approach? Thanks - – mbq♦ Oct 2 '11 at 8:58 1 @mbq - the Cauchy distribution presents no problems for confidence intervals, as the CDF is the inverse tangent function. Variance need not be defined for CIs to work. And the ratio of two normal RVs with zero mean is Cauchy, but not necessarily two normal RVs with non-zero mean. – probabilityislogic Oct 2 '11 at 10:58 @probabilityislogic Sure, I must stop trying to think on Sunday mornings. – mbq♦ Oct 2 '11 at 12:13 ## 2 Answers Fieller's method does what you want -- compute a confidence interval for the quotient of two means, both assumed to be sampled from Gaussian distributions. The original citation is: Fieller EC: The biological standardization of Insulin. Suppl to J R Statist Soc 1940, 7:1-64. The Wikipedia article does a good job of summarizing, and an appendix to the first edition of my Intutitive Biostatistics is even more concise. I've created an online calculator that does the computation. - It's very good references, I also like that you actually made a calculator for it (+1). As expected though, in your calculator you clearly state that when the confidence interval of the denominator includes zero, it is not possible to compute the CI of the quotient. I think it's the same that happens when I try solving the quadratic equation. suppose variance is 1, mu1 = 0 and mu2=1, N=10000. It's unsolvable. – francogrex Oct 2 '11 at 16:03 1 thanks for the online calculator Harvey, I'm a typical biologist with insufficient background in statistics and your calculator was exactly what I needed. – Timtico Jul 5 '12 at 8:29 R has the package `mratios` with the function `t.test.ratio`. Gemechis Dilba Djira, Mario Hasler, Daniel Gerhard and Frank Schaarschmidt (2011). mratios: Inferences for ratios of coefficients in the general linear model. R package version 1.3.15. http://CRAN.R-project.org/package=mratios See also http://www.r-project.org/user-2006/Slides/DilbaEtAl.pdf -

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http://mathhelpforum.com/algebra/86465-compounded-intrest-continuously.html

# Thread: 1. ## Compounded intrest continuously Suppose an investor deposits \$4,000 in an account bearing 2.5% intrest compounded continuously. Find the total amount in the account for the following time period. 29 years The total amount of money is=? I am not asking for the answer, I just want to know if I need to know how to get the answer....do I multiply the 2.5 by 29 then divide? I can't figure out how to set it up???? Please help. Thanks 2. You should be studying the formulas for compound interest if you are given this sort of problem. It isn't something students usually can derive on their own. When the interest in compounded continuously, than the number "e" comes into play and the formula for calculating the total value after t-time is: $P=P_{0}e^{rt}$ Look familiar? 3. ## compounded It does, and i understand the formula. What I did not know was that I was able to plug that into my calculator. Unfortunatly I do not have the e on the calculator I am using. So that brings me to my next question. Am I screwed because I do not have that capability??? 4. Originally Posted by tpoma It does, and i understand the formula. What I did not know was that I was able to plug that into my calculator. Unfortunatly I do not have the e on the calculator I am using. So that brings me to my next question. Am I screwed because I do not have that capability??? Really? Have you tried looking near the ln key - it might be shift+ln for example. On my Casio fx-85ES is it Code: ``` img.top {vertical-align:15%;} $Alpha \rightarrow \text { x }10^x$``` Alternatively google has a calculator function: below is my search for e.

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http://mathhelpforum.com/advanced-algebra/108926-change-basis.html

# Thread: 1. ## change of basis find the change of basis matrix from the basis B = (1, t-3, (t-3)^2) to the standard basis U = (1, t, t^2) 2. Originally Posted by noles2188 find the change of basis matrix from the basis B = (1, t-3, (t-3)^2) to the standard basis U = (1, t, t^2) The "change of basis matrix" maps the coefficients of vectors written in terms of the first basis to the coefficients of the same vectors written in the second basis. Further, a simple way of finding the matrix corresponding to a linear transformation in terms of bases for the "domain" and "range" spaces is: Apply the linear transformation to each of the basis vectors for the domain in turn. Write the result in terms of the range basis. The coefficients for that linear combination form the column of the matrix. For example, 1 is just mapped into 1 since the first "basis vector" in each basis is the same: 1= 1(1)+ 0(t)+ 0(t^2) so the first column of the matrix is $\begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix}$. t- 3 is mapped into -3(1)+ 1(t)+ 0(t^2) so the second column of the matrix is $\begin{bmatrix}-2 \\ 1 \\ 0\end{bmatrix}$. Can you do the third column? 3. slight typo: it's $-3.$ 4. I got 9, -6, and 1 for the third column

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http://math.stackexchange.com/questions/276522/how-to-solve-this-nonlinear-equation-system-by-changing-this-system-to-linear-sy

# how to solve this nonlinear equation system by changing this system to linear system consider following nonlinear equation system how solve it? $$x'=|y|$$ $$y'=x$$ and whats the matrix that associated to this system - I haven't worked out the details, but here's a thought: at first glance, your system appears to be nonlinear (in fact, the vector field isn't even C^1, which makes me nervous) but I suspect (and this will be easy to check) that an initial condition in a given quadrant (e.g. x>0, y>0) will remain in that quadrant for all times. So my gut now tells me that the fundamental solution matrix is going to have a piecewise definition based on the quadrant your initial conditions lie in (this makes sense, because your solutions should not be C^1). – A Blumenthal Jan 12 at 20:35 all none linear system change to this form AX+G(t)=0 by Jacobian matrix i mean whats this matrix and this question is exercise of book and doesn't define initial condition – Maisam Hedyelloo Jan 12 at 20:56 What's the book? On what pages does it discuss your $AX+G(t)=0$? On what page is the exercise? – Gerry Myerson Jan 25 at 5:49 ## 1 Answer It seems like you would have to be given some ICs that keep you in a particular quadrant or that you would require that with your solution. In order to solve, we would have to break this up into two systems as: $$x'=y$$ $$y'=x$$ This would give us eigenvalues $\lambda_{1,2} = \pm 1$, which yields a solution: $x(t) = c_1 e^{-t} (e^{2 t}+1)+c_2 e^{-t} (e^{2 t}-1)$, and $y(t) = c_1 e^{-t} (e^{2 t}-1)+c_2 e^{-t} (e^{2 t}+1)$ $$x'=-y$$ $$y'=x$$ This would give us eigenvalues $\lambda_{1,2} = \pm i$, which yields a solution: $x(t) = c_2 \sin(t)+c_1 \cos(t)$, and $y(t) = c_1 \sin(t)+c_2 \cos(t).$ Of course, there would have to be restrictions placed on the ICs and I suppose you can just plot them with different values of x and y and see if they are consistent for all time. It is possible that you would have to glue together the solutions for both systems based on the the values of $x$ and $y$. Regards - $+1\;$You look good in green $\;\ddot\smile\;$ – amWhy Apr 22 at 0:11 @amWhy: Thank you, I never feel like I get enough of those and some of you seem to be in the stratosphere with the number of answers and accepts (green)! :-) Yourself included! regards – Amzoti Apr 22 at 0:16

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http://nrich.maths.org/5876

### Circles Ad Infinitum A circle is inscribed in an equilateral triangle. Smaller circles touch it and the sides of the triangle, the process continuing indefinitely. What is the sum of the areas of all the circles? ### Climbing Powers $2\wedge 3\wedge 4$ could be $(2^3)^4$ or $2^{(3^4)}$. Does it make any difference? For both definitions, which is bigger: $r\wedge r\wedge r\wedge r\dots$ where the powers of $r$ go on for ever, or $(r^r)^r$, where $r$ is $\sqrt{2}$? ### Flexi Quads A quadrilateral changes shape with the edge lengths constant. Show the scalar product of the diagonals is constant. If the diagonals are perpendicular in one position are they always perpendicular? # Root Hunter ##### Stage: 5 Challenge Level: . This graph is positive for $x = 5$ and negative for $x = 3$. This means that the graph must cut the $x$ axis somewhere between $3$ and $5$. Although in this case the result is obvious (because we have the whole graph to look at!), we can also use this idea to show that more tricky functions also have roots. Use this idea to show that these functions possess at least one solution $f(x) = 0$: $$f(x)=\frac{1}{x-2}+\frac{1}{x-3}$$ $$f(x)= x^x - 1.5 x$$ $$f(x)= x^{1000000}+{1000000}^x - 17$$ $$f(x)=\cos(\sin(\cos x)) - \sin(\cos(\sin x))$$ Optional extension activity: Can you make a spreadsheet that helps you find the numerical values of the roots to, say, four decimal places? The NRICH Project aims to enrich the mathematical experiences of all learners. To support this aim, members of the NRICH team work in a wide range of capacities, including providing professional development for teachers wishing to embed rich mathematical tasks into everyday classroom practice. More information on many of our other activities can be found here.

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http://nrich.maths.org/6276

### I'm Eight Find a great variety of ways of asking questions which make 8. ### More Mods What is the units digit for the number 123^(456) ? ### Expenses What is the largest number which, when divided into 1905, 2587, 3951, 7020 and 8725 in turn, leaves the same remainder each time? # Galley Division ##### Stage: 4 Challenge Level: The 'Galley' method of division was probably the most widely used method of division prior to the 1600s. It is called 'Galley' because, when completed, a division looks a bit like a galley (boat). The Galley method is thought to be Hindu in origin and an early version of this method was known to be used in the ninth century by a Persian called Al-Khowarizmi. It is still taught in parts of North Africa and the Middle East today. This is how you could work out $65284$ divided by $594$ using the Galley method: Can you see the two numbers I started with? The calculation tells me that $625284 \div 594 = 109$ remainder $538$. Can you see the answer and the remainder? Here's another division done using the Galley method: What division do you think is being done here? What is the answer to this division? Here are two video clips of divisions using Galley arithmetic: This text is usually replaced by the Flash movie. This text is usually replaced by the Flash movie. If you are struggling to see what is happening, try the division using your own method and see if you can recognise numbers that are appearing in the galley method that are also in your own recording. If you need further guidance, there is a division, showing each of the main steps and some explanation, in the hints section of this problem. Send us answers to the following divisions showing that you have used the same method. $$2986 \div 47$$ $$76254 \div 235$$ Why does the method work? How does this method compare with our modern methods of division? The NRICH Project aims to enrich the mathematical experiences of all learners. To support this aim, members of the NRICH team work in a wide range of capacities, including providing professional development for teachers wishing to embed rich mathematical tasks into everyday classroom practice. More information on many of our other activities can be found here.

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

# Menger sponge An illustration of M4, the fourth iteration of the construction process. In mathematics, the Menger sponge is a fractal curve. It is a universal curve, in that it has topological dimension one, and any other curve (more precisely: any compact metric space of topological dimension 1) is homeomorphic to some subset of it. It is sometimes called the Menger-Sierpinski sponge or the Sierpinski sponge. It is a three-dimensional extension of the Cantor set and Sierpinski carpet. It was first described by Karl Menger (1926) while exploring the concept of topological dimension. The Menger sponge simultaneously exhibits an infinite surface area and encloses zero volume. ## Construction The construction of a Menger sponge can be described as follows: 1. Begin with a cube (first image). 2. Divide every face of the cube into 9 squares, like a Rubik's Cube. This will sub-divide the cube into 27 smaller cubes. 3. Remove the smaller cube in the middle of each face, and remove the smaller cube in the very center of the larger cube, leaving 20 smaller cubes (second image). This is a level-1 Menger sponge (resembling a Void Cube). 4. Repeat steps 2 and 3 for each of the remaining smaller cubes, and continue to iterate ad infinitum. The second iteration will give you a level-2 sponge (third image), the third iteration gives a level-3 sponge (fourth image), and so on. The Menger sponge itself is the limit of this process after an infinite number of iterations. An illustration of the iterative construction of a Menger sponge up to M3, the third iteration. The number of cubes is 20n, with n being the number of iterations performed on the first cube. A sculptural representation of the previous illustration. ## Properties Each face of the Menger sponge is a Sierpinski carpet; furthermore, any intersection of the Menger sponge with a diagonal or medium of the initial cube M0 is a Cantor set. The Menger sponge is a closed set; since it is also bounded, the Heine–Borel theorem implies that it is compact. It has Lebesgue measure 0. It is an uncountable set. The topological dimension of the Menger sponge is one, the same as any curve. Menger showed, in the 1926 construction, that the sponge is a universal curve, in that any possible one-dimensional curve is homeomorphic to a subset of the Menger sponge, where here a curve means any compact metric space of Lebesgue covering dimension one; this includes trees and graphs with an arbitrary countable number of edges, vertices and closed loops, connected in arbitrary ways. In a similar way, the Sierpinski carpet is a universal curve for all curves that can be drawn on the two-dimensional plane. The Menger sponge constructed in three dimensions extends this idea to graphs that are not planar, and might be embedded in any number of dimensions. The Menger sponge simultaneously exhibits an infinite surface area and encloses zero volume. In spite of this, there exists a homeomorphism of the cube having finite distortion that "squeezes the sponge" in the sense that the holes in the sponge go to a Cantor set of zero measure (Iwaniec & Martin 2001, §6.5.6). The sponge has a Hausdorff dimension of (log 20) / (log 3) (approximately 2.726833). ## Formal definition Formally, a Menger sponge can be defined as follows: $M := \bigcap_{n\in\mathbb{N}} M_n$ where M0 is the unit cube and $M_{n+1} := \left\{\begin{matrix} (x,y,z)\in\mathbb{R}^3: & \begin{matrix}\exists i,j,k\in\{0,1,2\}: (3x-i,3y-j,3z-k)\in M_n \\ \mbox{and at most one of }i,j,k\mbox{ is equal to 1}\end{matrix} \end{matrix}\right\}.$ ## References • Iwaniec, Tadeusz; Martin, Gaven (2001), Geometric function theory and non-linear analysis, Oxford Mathematical Monographs, The Clarendon Press Oxford University Press, ISBN 978-0-19-850929-5, MR1859913 . • Menger, Karl (1926), "Allgemeine Räume und Cartesische Räume. I.", Communications to the Amsterdam Academy of Sciences English translation reprinted in Edgar, Gerald A., ed. (2004), Classics on fractals, Studies in Nonlinearity, Westview Press. Advanced Book Program, Boulder, CO, ISBN 978-0-8133-4153-8, MR2049443 • Karl Menger, Dimensionstheorie, (1928) B.G Teubner Publishers, Leipzig. • Zhou, Li (2007), "Problem 11208: Chromatic numbers of the Menger sponges", 114 (9): 842

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http://math.stackexchange.com/questions/203640/ratio-between-two-sets-of-numbers

# Ratio between two sets of numbers? I have two sets of numbers. The first set is 1 to 4. The second set is 0 to 190. What is the calculation to get the proportionate number in the second set if a number in the first set is, for example, 1.2? Math is not my forte! - ## 1 Answer It is $\displaystyle\frac{1.2-1}{4-1}\cdot (190-0) +0$. - Can you express that equation without the fraction? For a programming equation? – user1342133 Sep 27 '12 at 23:19 If your input is $x$, then to go from $[1,4]$ to $[0,190]$, compute $(x-1)*190/3$. – Théophile Sep 27 '12 at 23:32 That's it, thanks! – user1342133 Sep 27 '12 at 23:37

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http://mathhelpforum.com/advanced-algebra/95583-algebra-problems-fun-35-a.html

# Thread: 1. ## Algebra, Problems For Fun (35) Characterize all subgroups of $\mathbb{Q}^{\times}.$ 2. here's an idea: let $G=\{1,-1 \},$ considered as a multiplicative group. show that $\mathbb{Q}^{\times}$ and $G \oplus \mathbb{Z} \oplus \mathbb{Z} \oplus \cdots$ are isomorphic. 3. My guess would be that $~ \mathbb{Q}^{\times}_{>0} = \displaystyle\bigoplus_{p ~ prime} \langle p \rangle ~$. Clearly each $\langle p \rangle \simeq \mathbb{Z}$ Then $\mathbb{Q}^{\times} \simeq G \oplus \mathbb{Q}^{\times}_{>0}$ as we can define the trivial isomorphism $\theta$ : $(g,q) \rightarrow gq$ Just the first step needs some justification... 4. Originally Posted by pomp My guess would be that $~ \mathbb{Q}^{\times}_{>0} = \oplus_p \langle p \rangle ~$ for p prime. Clearly each $\langle p \rangle \simeq \mathbb{Z}$ Then $\mathbb{Q}^{\times} \simeq G \oplus \mathbb{Q}^{\times}_{>0}$ as we can define the trivial isomorphism $\theta$ : $(g,q) \rightarrow gq$ Just the first step needs some justification... well, every $r \in \mathbb{Q}^{\times}$ can be written uniquely as $r=\pm p_1^{n_1}p_2^{n_2} \cdots p_k^{n_k},$ where $p_j$ is the j-th prime number and $n_j \in \mathbb{Z}.$ now define $f(r)=(\text{sgn}(r),n_1,n_2, \cdots, n_k, 0, 0, \cdots ).$ 5. EDIT: Aw man, ninja'd >< Original post: As a consequence of uniqueness of prime factorization, every nonzero rational number can be expressed uniquely as $q=(-1)^{r_0}p_1^{r_1}p_2^{r_2}\ldots p_s^{r_s}$ where $r_i \in \mathbb{Z}^\times$. Then given a particular enumeration of the primes $P=\{p_1, p_2, \ldots\}$, there are only a finite number of prime divisors, so with respect to our enumeration there is a unique such representation with $q=(-1)^{r_0}p_1^{r_1}p_2^{r_2}\ldots =(-1)^{r_0}\prod_{i=1}^{\infty} p_i^{r_i}$ where $r_i \in \mathbb{Z}$ such that only a finite number of $r_i$ are nonzero This would suggest an isomorphism $Q^\times \rightarrow G\times\displaystyle\bigoplus_{i=1}^\infty \mathbb{Z}$ given by $(-1)^{r_0}\prod_{i=1}^{\infty} p_i^{r_i} \mapsto ((-1)^{r_0}, r_1, r_2, \ldots)$. Uniqueness of factorization assures this is a well defined and 1-1 function. The finite number of prime factors assures this indeed maps into and onto the direct sum. Finally the morphism property follows from the properties of exponents. Hand wavy but conveys the ideas I think. 6. so, to answer the question you only need to find the subgroups of $G \oplus \mathbb{Z} \oplus \mathbb{Z} \oplus ...$. it shouldn't be hard! 7. Either the subgroups are finitely generated or not. If they are finitely generated, then the fundamental theorem of finitely generated abelian groups tells us that up to isomorphisms the subgroups are the direct product of copies of the integers with the 2 element group. If a subgroup is not finitely generated, then intuition says we should expect an isomorphic copy of the group, or an isomorphic copy of the infinite direct sum of integers. 8. let $N$ be a subgroup of $\mathbb{Q}^{\times}.$ then either $\{1,-1 \}=G \subseteq N$ or $N \cap G = \{1 \}.$ if $G \subseteq N,$ then $N/G$ would be a subgroup of the free abelian group $\mathbb{Q}^{\times}/G \cong \mathbb{Z} \oplus \mathbb{Z} \oplus \cdots.$ we know that any (non-trivial} subgroup of a free abelian group is free. so $N$ is isomorphic to $G \oplus K,$ where $K$ is the direct sum of finitely or infinitely (countable) many copies of $\mathbb{Z}.$ if $N \cap G = \{1 \},$ then $NG/G \cong N/(N \cap G) \cong N.$ thus $NG/G \cong N$ is a subgroup of the free abelian group $\mathbb{Q}^{\times}/G \cong \mathbb{Z} \oplus \mathbb{Z} \oplus \cdots.$ so $N$ is just a direct sum of finitely or infinitely (countable) many of copies of $\mathbb{Z}.$ briefly subgroups of $\mathbb{Q}^{\times}$ are in one of these forms: $\{1 \}, \ \{1,-1 \}, \ K, \ \{1,-1 \} \oplus K,$ where $K$ is any free abelian group of at most countable rank.

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http://mathoverflow.net/questions/114881/a-consequence-of-convexity/114885

## A consequence of convexity ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Let $f:\mathbb{R}\to\mathbb{R}$ a convex decreasing function. Let $x_0 < x_1 < x_2$. Studying the behaviour of the difference quotient, it is clear that $$f(x_0)-f(x_2) \leq M (f(x_0)-f(x_1))$$ with $M=\frac{x_2-x_0}{x_1-x_0}>0$. Now take $F:\mathbb{R}^2\to\mathbb{R}$ convex and decreasing with respect to each variable. Let $x_0 < x_1 < x_2$ and $y_0 < y_1 < y_2$. I ask if a similar condition holds, say for example $$F(x_0,y_0)-F(x_2,y_2) \leq M (F(x_0,y_0)-F(x_1,y_1))$$ with $M=\max ( \frac{x_2-x_0}{x_1-x_0}, \frac{y_2-y_0}{y_1-y_0} )$ or $M=\frac{x_2-x_0}{x_1-x_0}+\frac{y_2-y_0}{y_1-y_0}$. Edit after Brian's answer: you may add the hypothesis that $F$ is $C^2$ and also the mixed derivative $\frac{\partial^2F}{\partial x\partial y}$ is non-negative. Edit: under the additional assumption the answer is yes with $M=\max(\frac{x_2-x_0}{x_1-x_0},\ \frac{x_2-x_0}{x_1-x_0})$. But is it possible to reach the same conclusion without the additional assumption on the mixed derivatives? - Do I understand correctly that you want your inequality for $f$ to hold with SOME $M$ that depends only on $x_j,y_j$ but not of $f$ ? Or you only want one of the two expressions you proposed for $M$ ? – Alexandre Eremenko Nov 29 at 13:34 Yes I want an inequality with some $M$ that depends only on the points $x_,y_j$, not necessarely one of those I proposed. – xn--qwertyuiop-86a Nov 29 at 17:01 ## 3 Answers You can use the triangle inequality to solve this by looking at each coordinate separately. $F(x_0,y_0)-F(x_2,y_2)=F(x_0,y_0)-F(x_2,y_0)+F(x_2,y_0)-F(x_2,y_2)$ $\leq M_1(F(x_0,y_0)-F(x_1,y_0))+M_2(F(x_2,y_0)-F(x_2,y_1))$. Replacing $F(x_1,y_0)$ with $F(x_1,y_1)$ only increases the right side, and replacing both $x_2$'s on the far right side with $x_0$ only makes the right hand side larger by convexity. Finally, replace the last $x_0$ that we just added with $x_1$ without ncreasing the right hand side. This gives us $F(x_0,y_0)-F(x_2,y_2)\leq M(F(x_0,y_0)-F(x_1,y_1))$, where $M$ is $2\max(\frac{x_2-x_0}{x_1-x_0},\frac{y_2-y_0}{y_1-y_0})$. $M$ can also be the sum instead of the max, in which case we can drop the 2. Edit: This argument doesn't work without additional assumptions; see the comments. - Thank you Brian. I think the statement after the triangular inequality should be "Repalcing $F(x_1,y_0)$ with $F(x_1,y_1)$". Please could you explain how do you use convexity to prove that $F(x_2,y_0)-F(x_2,y_1)\leq F(x_0,y_0)-F(x_0,y_1)$ ? – xn--qwertyuiop-86a Nov 29 at 16:59 1 @xn I can't explain it because it's not true! I had a false picture in my head. I was essentially claiming (in the smooth case) that having positive `unmixed' second partial derivatives implies that the mixed partial derivative is positive, which is not true. I apologize! – Brian Rushton Nov 29 at 17:25 Thank you as well, maybe I can extend my assumptions. If I assume all the second partial derivatives (mixed and pures) of F are $\geq0$, all should work right? – xn--qwertyuiop-86a Nov 29 at 17:49 1 As long as the derivatives exist, it should work. If the derivatives do not exist, then you would have to assume that $F(x,y)-F(x,y+\Delta Y)$ is a decreasing function of $x$ (you could also interchange $x$ and $y$ and have this work). This is analagous to regular convexity where $F(x)-F(x+\Delta x)$ is a decreasing function of $x$, and it probably has a name. But yes, it should work if all second derivatives are positive and $F$ is decreasing in each variable separately. – Brian Rushton Nov 29 at 18:43 1 To improve the bound one may use: 1) triangular inequality; 2) convexity in $x$ and $y$; 3) $\frac{\partial F}{\partial x\partial y}\geq0$. And to sum up one obtains the bound $M=\max(\frac{x_2-x_0}{x_1-x_0},\ \frac{y_2-y_0}{y_1-y_0})$. In this way there's no $2$, so that $M$ does not depend on the dimension of the space. – xn--qwertyuiop-86a Nov 30 at 9:48 show 1 more comment ### 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. Without loss of generality, we can assume that $$\frac{y_2-y_0}{y_1-y_0}\ge\frac{x_2-x_0}{x_1-x_0}.$$ Let $x_2'$ be the real number number such that $$\frac{y_2-y_0}{y_1-y_0}=\frac{x'_2-x_0}{x_1-x_0}.$$ Clearly $x_2'\ge x_2$; therefore $$F(x_2',y_2)\le F(x_2,y_2).$$ From the one-dimensional case, you get $$F(x_0,y_0)-F(x_2',y_2) \leq M\cdot (F(x_0,y_0)-F(x_1,y_1))$$ with $$M=\frac{y_2-y_0}{y_1-y_0}=\max\left\{\frac{y_2-y_0}{y_1-y_0},\frac{x_2-x_0}{x_1-x_0}\right\}.$$ Hence the result follows. It seems to be the optimal bound. P.S. It turned out that the term "convex" was used in a nonstandard way; namely, xm wants to consider $F$ such that the restriction to any line parallel to the $x$-axis or to the $y$-axis is convex. In particular $F(x,y)= -x\cdot y$ is "xm"-convex in positive quadrant. Note that in this example $F(t,t)=- t^{2}$, so $M$ has to be much bigger; maybe $$M=\left(\max\left\{\frac{y_2-y_0}{y_1-y_0},\frac{x_2-x_0}{x_1-x_0}\right\}\right)^2$$ will do. - Thank you Anton. Please could you explain how you use the one dimensional case to get the inequality? I found the same bound but adding the hypothesis $\frac{\partial F}{\partial x\partial y}\geq0$ (see the last comment to Brian's answer). Instead you don't use this hypothesis, right? – xn--qwertyuiop-86a Nov 30 at 10:01 @xn--qwertyuiop-86a, consider function $$f(t)=F(x_0+(x_1-x_0)\cdot t, y_0+(y_1-y_0)\cdot t).$$ – Anton Petrunin Nov 30 at 23:46 @Anton Petrunin, let me see if I've understood. The inequality $F(x_0,y_0)-F(x_2',y_2)\leq M\cdot(F(x_0,y_0)-F(x_1,y_1))$ is equivalent to $f(0)-f(\frac{y_2-y_0}{y_1-y_0}\leq M\cdot(f(0)-f(1)))$ and this is true if $f$ is convex. But to have $f$ convex one need also $\frac{\partial^2F}{\partial x\partial y}\geq0$: it isn't enough to assume $F$ convex w.r.t. each variable separately. Correct? – xn--qwertyuiop-86a Dec 1 at 10:26 I understood the question as "convex" and "decreasing with respect to each variable". Convex usually means convex on any line. It seems that you wanted to say "restriction to any line t↦(x,t) and t↦(t,y) is convex". Let me know if this is the case. – Anton Petrunin Dec 1 at 19:16 Yes, I wanted to say $F$ restricted to any line parallel to the $x$-axis or to the $y$-axis is convex. In other words (assuming $F$ regular enough) my hypotesis is that $\frac{\partial^2 F}{\partial x^2}\geq0$ and $\frac{\partial^2 F}{\partial y^2}\geq0$ . – xn--qwertyuiop-86a Dec 2 at 12:09 show 1 more comment In general, whenever you have a (separately) convex function $f :\mathbb{R}^N \to \mathbb{R}$, which means it is convex in each variable, this implies the function is locally Lipschitz. The fact that it is decreasing allows you to make some explicit calculations of the constant, as Brian Rushton does, but in general the inequality looks like this: $|f(x)-f(y)| \leq \frac{N}{r} (\sup_{B_{2r}} f - \inf_{B_{2r}} f) |x-y|$ for all $x,y \in B_{r}$. -

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http://stats.stackexchange.com/questions/16998/accuracy-of-a-random-classifier

# Accuracy of a random classifier I was wondering how to compare accuracy of my classifier to a random one. I'm going to elaborate further. Let's say we have a binary classification problem. We have $n^+$ positive examples and $n^-$ negative examples in the test set. I say that and record is positive with a probability of $p$. I can estimate that, on average, I get: $$TP = pn^+ \ TN=(1-p)n^- \ FN = pn^- \ FP = (1-p)n^+$$ thus $$\mbox{acc} = \frac{TP+TN}{TP+TN+FP+FN} = \frac{pn^+ + (1-p)n^-}{pn^+(1-p)n^-pn^- + (1-p)n^+}$$ that is $$= \frac{pn^+ + (1-p)n^-}{n^+ + n^-}$$ For example if we have $n^+=n^-$ accuracy is always $1/2$ for any $p$. This can be extended in multiclass classification: $$\mbox{acc} = \frac{\sum_{i=1}^c p_i n_i}{\sum n_i}$$ where $p_i$ is the probability to say "it is in the $i$th class", and $n_i$ is the count of records of class $i$. Also in this case, if $n_i = n/c \ \forall i$ then $$\mbox{acc} = 1/c$$ But, how can I compare the accuracy of my classifier without citing a test set? For example if I said: my classifier accuracy is 70% (estimated somehow, e.g. Cross-Validation), is it good or bad compared to a random classifier? - ## 1 Answer I am not sure I understand the last part of your question But, how can I compare the accuracy of my classifier without citing a test set? but I think I understand your concern. A given binary classifier's accuracy of 90% may be misleading if the natural frequency of one case vs the other is 90/100. If the classifier simply always chooses the most common case then it will, on average, be correct 90% of the time. A useful score to account for this issue is the Information score. A paper describing the score and its rationale can be found here. I learned about this score because it is part of the cross-validation suite in the excellent Orange data mining tools (you can use no-coding-needed visual programming or call libraries from Python). - Thank you. You made me realize that you have always to take into account prior probabilities. Even if the accuracy is estimated by Cross-Validation, the classes do have prior probabilities. And, for example, you can exstimate them with $n_i$ that I wrote in my question. – Simone Oct 16 '11 at 10:01 2 – Simone Oct 16 '11 at 10:19 Last comment. Classifying the positive class with probability $p$ means that your classifier lies along the main diagonal of the ROC curve (if you are able to compute it, because not all the classifier gives probabilities for output, as state in Kononenko and Bratko paper). Because $TPR = \frac{pn^+}{pn^+ + (1-p)n^+} = p$ and $FPR = \frac{pn^-}{pn^- + (1-p)n^-} = p$. With a diagonal ROC curve you can understand your performance is similar to a random one. Instead, changing $p$ changes $acc$, and it is somehow misleading. Thus, there should be better performance measures. – Simone Oct 16 '11 at 10:51 @Simone: +1 for finishing your thoughts and the useful link, but mostly for exposing me to a new word (maieutics). – Josh Hemann Oct 17 '11 at 4:53

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http://math.stackexchange.com/questions/182704/a-question-on-iterated-sums?answertab=active

# A question on iterated sums We are to prove that the set $$\Bigg\{ \sum^m_{i=1}\sum^n_{j=1}|a_ib_j| :m,n\in \mathbb N \Bigg\}$$ is bounded. Also, We are to use this to show that the iterated sum $\sum^\infty_{i=1} \sum^\infty_{j=1}|a_ib_j|$ converges. I could prove that the set is bounded, however, I can't solve the second part. I become very confused when I have to deal with iterated sums. Could anyone please help me with this problem. - 2 There are two hints: (1) The infinite summation of a non-negative sequence is equal to the supremum of the set of partial sums: $$\sum_{i=1}^{\infty} |a_i| = \sup_{n\in\mathbb{N}} \sum_{i=1}^{n} |a_i|,$$ and (2) supremum can be distributed: $$\sup_{(x,y)\in A\times B}f(x, y) = \sup_{x\in A}\sup_{y\in B} f(x,y).$$ – sos440 Aug 15 '12 at 5:16 ## 2 Answers Although by first proving that the sum is bounded, it can be proved that the sum converges. But, there is a way to directly prove that it converges. For that, let $$\sum^\infty_{i=1}|a_i|=L~~~~~\text{and}~~~~~\sum^\infty_{j=1}|b_j|=M$$ For each fixed $i \in \mathbb N$, the Algebraic Limit Theorem will allow us to write $\sum^\infty_{j=1}|a_ib_j|$=$|a_i|\sum^\infty_{j=1}|b_j|$. If one continues the process, we will see that: $$\sum^\infty_{i=1}\sum^\infty_{j=1}|a_ib_j|=\sum^\infty_{i=1}|a_i|\sum^\infty_{j=1}|b_j|=ML$$ Hence,$\sum^\infty_{i=1}\sum^\infty_{j=1}|a_ib_j|$ converges (to $ML$). - In order to show that $\sum_{i=1}^\infty\sum_{j=1}^\infty|a_ib_j|$ converges, you must show that there is some real number $a$ such that for every $\epsilon>0$ there is some $n_0\in\Bbb Z^+$ such that $$\left|a-\sum_{i=1}^m\sum_{j=1}^n|a_ib_j|\right|<\epsilon$$ whenever $m,n\ge n_0$. For $m,n\in\Bbb Z^+$ let $$s_{mn}=\sum_{i=1}^m\sum_{j=1}^n|a_ib_j|\;.$$ You know that the set $S=\{s_{mn}:m,n\in\Bbb Z^+\}$ is bounded. You also know that $s_{k\ell}\le s_{mn}$ whenever $k\le m$ and $\ell\le n$, since the terms $|a_ib_j|$ are non-negative. For $n\in\Bbb Z^+$ let $d_n=s_{nn}$; then $s_{k\ell}\le d_n$ whenever $k,\ell\le n$. In other words, the sums $d_n$ are cofinal in $S$: for each $s\in S$ there is an $n\in\Bbb Z^+$ such that $s\le d_n$. Consider the sequence $\langle d_n:n\in\Bbb Z^+\rangle$: it’s bounded and non-decreasing, so it converges to some limit $a$; I’ll leave to you the easy verification that $s\le a$ for all $s\in S$. Fix $\epsilon>0$. There is some $n_0\in\Bbb Z^+$ such that $|a-d_n|<\epsilon$ whenever $n\ge n_0$. Suppose that $m,n\ge n_0$; then $a-\epsilon<d_{n_0}\le s_{mn}\le a$, so $$\left|a-\sum_{i=1}^m\sum_{j=1}^n|a_ib_j|\right|<\epsilon$$ whenever $m,n\ge n_0$, exactly as we wanted. -

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http://math.stackexchange.com/questions/22928/limit-of-algebraic-function-avoiding-lhopitals-rule

# Limit of algebraic function avoiding l'Hôpital's rule Find $$\lim_{x \to \infty} \sqrt[3]{1 + x^{2} + x^{3}} -x$$ - 1 @white: I have edited the question, please if this is what you wanted to post or not. – anonymous Feb 20 '11 at 13:39 Hi ? Please ? Homework ? Any thoughts ? – Sam Feb 20 '11 at 13:39 1 @white: As Arturo, always keeps mentioning, please pose the question in a more polite form. This is like asking a homework question. We would like to know what you have tried and where you are finding difficulty. – anonymous Feb 20 '11 at 13:40 lol..ok i am sorry if i am rude.. – user7284 Feb 20 '11 at 13:43 3 In general, $\lim_{x\rightarrow\infty} \sqrt[n]{x^n+ax^{n-1}+O(x^{n-2})}-x=\frac{a}{n}$. – Eric♦ Feb 20 '11 at 23:59 show 6 more comments ## 5 Answers HINT $\$ It's simply a first derivative: $\:$ changing variables $\rm\ x\to 1/x\$ transforms it to $$\rm\displaystyle\ \lim_{x\to\ 0^{+}}\ \frac{f(x)-f(0)}x\ =\ f\:'(0) \ \ \ for\ \ \ f(x) = (1+x+x^3)^{1/3}$$ Now it is easy to calculate $\rm\ f\:'(0)\ =\ 1/3\$ by direct evaluation (it's not indeterminate). Namely $$\rm f\:'(x)\ =\ \frac{d}{dx}\ (1+x+x^3)^{1/3}\ =\ \frac{1+3\ x^2}{3\ (1+x+x^3)^{2/3}}\ \ \Rightarrow\ \ f\:'(0)\ =\ \frac{1}3$$ Note that this method employs only knowledge of the definition of the derivative and basic rules for calculating derivatives of polynomial and powers. It does not employ more advanced techniques such as L'Hospital's rule, or (binomial) power series expansions, etc. - is this lopital?? – user7284 Feb 20 '11 at 19:29 @white: The above easy calculation of $\rm\ f\:'(0)\$ doesn't use L'Hopital's rule (it's not indeterminate). While there is indeed a close relationship between L'Hospital's rule, derivatives, Taylor's formula, the mean-value-theorem, etc., that doesn't imply that use of the latter implies use of L'Hospital's rule. The above answer uses only the definition of the derivative and the basic rules for calculating derivatives. – Gone Feb 20 '11 at 19:53 I like this answer, +1. To explain something white, this actually is L'hopitals rule, but not quite. I say not quite because Bill's argument actually shows that L'Hopitals Rule follows immediately from the definition of the derivative when the denominator is $x$. – Eric♦ Feb 20 '11 at 22:41 Here is a more elementary way using the difference of cubes identity. (It is not as elegant as the Taylor series presented by Willie Wong, but requires less background) Since $x=\sqrt[3]{x^{3}}$, we are looking at $\lim_{x\rightarrow\infty}\sqrt[3]{1+x^{2}+x^{3}}-\sqrt[3]{x^{3}}$.Recall that the cubic identity $a^{3}-b^{3}=\left(a-b\right)\left(a^{2}+ab+b^{2}\right)$, which tells us that $$\left(\sqrt[3]{1+x^{2}+x^{3}}\right)^{3}-\left(\sqrt[3]{x^{3}}\right)^{3}$$ $$=\left(\sqrt[3]{1+x^{2}+x^{3}}-\sqrt[3]{x^{3}}\right)\cdot \left(\left(1+x^{2}+x^{3}\right)^{\frac{2}{3}}+\sqrt[3]{x^{3}}\cdot\left(1+x^{2}+x^{3}\right)^{\frac{1}{3}}+\left(x^{3}\right)^{\frac{2}{3}}\right)$$ and hence $$\frac{1+x^{2}}{\left(1+x^{2}+x^{3}\right)^{\frac{2}{3}}+x\cdot\left(1+x^{2}+x^{3}\right)^{\frac{1}{3}}+x^{2}}=\sqrt[3]{1+x^{2}+x^{3}}-\sqrt[3]{x^{3}}.$$ Divide the top and bottom of the left hand side by $x^{2}$ to find $$\frac{1+\frac{1}{x^{2}}}{\left(1+\frac{1}{x^{2}}+\frac{1}{x^{3}}\right)^{\frac{2}{3}}+\left(1+\frac{1}{x^{2}}+\frac{1}{x^{3}}\right)^{\frac{1}{3}}+1}=\sqrt[3]{1+x^{2}+x^{3}}-\sqrt[3]{x^{3}}.$$ Since $$\lim_{x\rightarrow\infty}1+\frac{1}{x^{2}}=1$$ and $$\lim_{x\rightarrow\infty}\left(1+\frac{1}{x^{2}}+\frac{1}{x^{3}}\right)^{\frac{2}{3}}+\left(1+\frac{1}{x^{2}}+\frac{1}{x^{3}}\right)^{\frac{1}{3}}+1=3$$ we see by the quotient rule for limits $$\lim_{x\rightarrow\infty}\sqrt[3]{1+x^{2}+x^{3}}-\sqrt[3]{x^{3}}=\frac{1}{3}.$$ Hope that helps, Edit: This faq question: Limits: How to evaluate $\lim\limits_{x\rightarrow \infty}\sqrt[n]{x^{n}+a_{n-1}x^{n-1}+\cdots+a_{0}}-x$ was made shortly after this post to help answer it more generally. - hey..thx for the info..but unfortunately i'm not so understand about this method..abit confusing as from the 3rd part of ur solution,the x has been forfited n i duno where to cancel it off..sry bro i not so good in maths..XD – user7284 Feb 20 '11 at 17:19 Which part exactly? I might be able to explain better. – Eric♦ Feb 20 '11 at 22:43 hint: you can rewrite the terms inside the limit as $$x \cdot \left( \sqrt[3]{\frac{1}{x^3} + \frac{1}{x} + 1} - 1\right)$$ for the term underneath the cube-root, use the Taylor expansion of the cube-root function near the value 1: $$\sqrt[3]{1 + y} = 1 + \frac{1}{3}y - \frac{1}{9} y^2 + \ldots$$ - bionomial???ok..i try it nw..thx for ur help.. – user7284 Feb 20 '11 at 14:32 ermm,@Willie Wong♦ ,i am not so sure about the taylor expansion as u mention..can u explain more on hw to apply it on the cube roots?? – user7284 Feb 20 '11 at 14:41 1 The Taylor expansion of (1+x)^k starts with 1+kx, even if k is not an integer. For this problem you only need the $y/3$ term. – Ross Millikan Feb 20 '11 at 16:39 – Willie Wong♦ Feb 20 '11 at 17:45 A very similar question was asked recently here... The limit follows immediately upon showing, using the mean value theorem, that $$\sqrt[3]{{\bigg(x + \frac{1}{3}\bigg)^3 }} - \sqrt[3]{{\bigg(x + \frac{1}{3}\bigg)^3 - \bigg(\frac{x}{3} - \frac{{26}}{{27}}\bigg)}} \to 0$$ as $x \to \infty$. - I hope you understand where the term $(x + \frac{1}{3})^3 - (\frac{x}{3} - \frac{{26}}{{27}})$ comes from. – Shai Covo Feb 20 '11 at 19:02 nope..i am noob maths..lol – user7284 Feb 20 '11 at 19:28 @white: It is equal to $x^3 + x^2 +1$. – Shai Covo Feb 20 '11 at 19:37 For such problems, quite often it pays off to use the identity $a^b = \exp[ b * \log(a)]$ because then one can apply all his/her knowledge about exp and log. - Can you explain? I don't see how you use this for this example unless you expand $e^{b\log a}=1+b\log a+\frac{1}{2}(b\log a)^2+\cdots$ and then expand the $\log$'s into their power series as well. Am I missing something? – Eric♦ Feb 20 '11 at 16:38 The comment was for an earlier version of the question which looked quite a bit different [$\lim_{x\to \infty} (1 + x^2 + x^3)^{1/3 -x}$]. – Fabian Feb 20 '11 at 20:27

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http://new.wikipedia.org/wiki/%E0%A4%97%E0%A5%8D%E0%A4%B0%E0%A5%81%E0%A4%AA_%E0%A4%B8%E0%A4%BF%E0%A4%A6%E0%A5%8D%E0%A4%A7%E0%A4%BE%E0%A4%A8%E0%A5%8D%E0%A4%A4

# ग्रुप सिद्धान्त विकिपिडिया नं ग्रुप सिद्धान्त गणितयागु छगु ख्यः ख। थुकिगु छ्येलेज्या भौतिकशास्त्र व रसायनशास्त्रय् जुइ। पुचः यागु छ्येलेज्या गणितय् यक्व थासे अप्वयाना इन्टर्नल सिमेट्रि क्याप्चर यायेत अटोमर्फिज ग्रुप यागु रुपे जुइ। An internal symmetry of a structure is usually associated with an invariant property; the set of transformations that preserve this invariant property, together with the operation of composition of transformations, form a group called a symmetry group. In Galois theory, which is the historical origin of the group concept, one uses groups to describe the symmetries of the equations satisfied by the solutions to a polynomial equation. The solvable groups are so-named because of their prominent role in this theory. Abelian groups underlie several other structures that are studied in abstract algebra, such as rings, fields, and modules. In algebraic topology, groups are used to describe invariants of topological spaces (the name of the torsion subgroup of an infinite group shows the legacy of this field of endeavor). They are called "invariants" because they are defined in such a way that they do not change if the space is subjected to some deformation. Examples include the fundamental group, homology groups and cohomology groups. The concept of the Lie group (named after mathematician Sophus Lie) is important in the study of differential equations and manifolds; they combine analysis and group theory and are therefore the proper objects for describing symmetries of analytical structures. Analysis on these and other groups is called harmonic analysis. In combinatorics, the notion of permutation group and the concept of group action are often used to simplify the counting of a set of objects; see in particular Burnside's lemma. An understanding of group theory is also important in physics and chemistry and material science. In chemistry, groups are used to classify crystal structures, regular polyhedra, and the symmetries of molecules. In physics, groups are important because they describe the symmetries which the laws of physics seem to obey. Physicists are very interested in group representations, especially of Lie groups, since these representations often point the way to the "possible" physical theories. Physics examples: Standard Model, Gauge theory ## इतिहास There are three historical roots of group theory: the theory of algebraic equations, number theory and geometry. Euler, Gauss, Lagrange, Abel and French mathematician Galois were early researchers in the field of group theory. Galois is honored as the first mathematician linking group theory and field theory, with the theory that is now called Galois theory.[१] An early source occurs in the problem of forming an $m$th-degree equation having as its roots m of the roots of a given $n$th-degree equation ($m < n$). For simple cases the problem goes back to Hudde (1659). Saunderson (1740) noted that the determination of the quadratic factors of a biquadratic expression necessarily leads to a sextic equation, and Le Sœur (1748) and Waring (1762 to 1782) still further elaborated the idea.[१] A common foundation for the theory of equations on the basis of the group of permutations was found by mathematician Lagrange (1770, 1771), and on this was built the theory of substitutions. He discovered that the roots of all resolvents (résolvantes, réduites) which he examined are rational functions of the roots of the respective equations. To study the properties of these functions he invented a Calcul des Combinaisons. The contemporary work of Vandermonde (1770) also foreshadowed the coming theory.[१] Ruffini (1799) attempted a proof of the impossibility of solving the quintic and higher equations. Ruffini distinguished what are now called intransitive and transitive, and imprimitive and primitive groups, and (1801) uses the group of an equation under the name l'assieme delle permutazioni. He also published a letter from Abbati to himself, in which the group idea is prominent.[१] Galois found that if $r_1, r_2, \ldots, r_n$ are the $n$ roots of an equation, there is always a group of permutations of the $r$'s such that (1) every function of the roots invariable by the substitutions of the group is rationally known, and (2), conversely, every rationally determinable function of the roots is invariant under the substitutions of the group. Galois also contributed to the theory of modular equations and to that of elliptic functions. His first publication on group theory was made at the age of eighteen (1829), but his contributions attracted little attention until the publication of his collected papers in 1846 (Liouville, Vol. XI).[१] Arthur Cayley and Augustin Louis Cauchy were among the first to appreciate the importance of the theory, and to the latter especially are due a number of important theorems. The subject was popularised by Serret, who devoted section IV of his algebra to the theory; by Camille Jordan, whose Traité des Substitutions is a classic; and to Eugen Netto (1882), whose Theory of Substitutions and its Applications to Algebra was translated into English by Cole (1892). Other group theorists of the nineteenth century were Bertrand, Charles Hermite, Frobenius, Leopold Kronecker, and Emile Mathieu.[१] It was Walther von Dyck who, in 1882, gave the modern definition of a group. The study of what are now called Lie groups, and their discrete subgroups, as transformation groups, started systematically in 1884 with Sophus Lie; followed by work of Killing, Study, Schur, Maurer, and Cartan. The discontinuous (discrete group) theory was built up by Felix Klein, Lie, Poincaré, and Charles Émile Picard, in connection in particular with modular forms and monodromy. The classification of finite simple groups is a vast body of work from the mid 20th century, which is thought to classify all the finite simple groups. Other important mathematicians in this subject area include Emil Artin, Emmy Noether, Sylow, and many others. ## ग्रुप सिद्धान्त विचातः ### ग्रुपयागु अर्थ मू पौ: पुचः (गणित) A group (G, *) is a set G with a binary operation * : G × G → G (one that assigns each ordered pair (a,b) in G an element in G denoted by a*b) that satisfies the following 3 axioms: • Associativity: For all a, b and c in G, (a * b) * c = a * (b * c). • Identity element: There is an element e in G such that for all a in G, e * a = a * e = a. • Inverse element: For each a in G, there is an element b in G such that a * b = b * a = e, where e is an identity element. ### Order of groups and elements The order of a group G is the number of elements in the set G. If the order is not finite, then the group is an infinite group. The order of an element a in a group G is the least positive integer n such that an=e, where an is multiplication of a by itself n times. For finite groups, it can be shown that the order of every element in the group must divide the order of the group. ### Subgroups A set H is a subgroup of a group G if it is a subset of G and is a group using the operation defined on G. In other words, H is a subgroup of (G, *) if the restriction of * to H is a group operation on H. If G is a finite group, the order of H divides the order of G. A subgroup H is a normal subgroup of G if for all h in H and g in G, ghg-1 is also in H. An alternate definition is that a subgroup is normal if its left and right cosets coincide. Normal subgroups are useful because they can be used to create quotient groups. ### Special classes of groups A group is abelian (or commutative) if the operation is commutative (that is, for all a, b in G, a * b = b * a). A non-abelian group is a group that is not abelian. The term "abelian" is named after the mathematician Niels Abel. A cyclic group is a group that is generated by a single element. A simple group is a group that has no nontrivial normal subgroups. A solvable group , or soluble group, is a group that has a normal series whose quotient groups are all abelian. The fact that S5, the symmetric group in 5 elements, is not solvable proves that some quintic polynomials cannot be solved by radicals. ### Operations on groups A homomorphism is a map between two groups that preserves the structure imposed by the operator. If the map is bijective, then it is an isomorphism. An isomorphism from a group to itself is an automorphism. The set of all automorphisms of a group is a group called the automorphism group. The kernel of a homomorphism is a normal subgroup of the group. A group action is a map involving a group and a set, where each element in the group defines a bijective map on a set. Group actions are used to prove the Sylow theorems and to prove that the center of a p-group is nontrivial. ## Some useful theorems • Some basic results in elementary group theory • Lagrange's theorem: if G is a finite group and H is a subgroup of G, then the order (that is, the number of elements) of H divides the order of G. • Cayley's Theorem: every group G is isomorphic to a subgroup of the symmetric group on G. • Sylow theorems: perhaps the most useful of the group theorems. Among them, that if pn (and p prime) divides the order of a finite group G, then there exists a subgroup of order pn. • The Butterfly lemma is a technical result on the lattice of subgroups of a group. • The Fundamental theorem on homomorphisms relates the structure of two objects between which a homomorphism is given, and of the kernel and image of the homomorphism. • Jordan-Hölder theorem: any two composition series of a given group are equivalent. • Krull-Schmidt theorem: a group G, subjected to certain finiteness conditions of chains of subgroups, can be uniquely written as a finite product of indecomposable subgroups. • Burnside's lemma: the number of orbits of a group action on a set equals the average number of points fixed by each element of the group ## Miscellany James Newman summarized group theory as follows: The theory of groups is a branch of mathematics in which one does something to something and then compares the results with the result of doing the same thing to something else, or something else to the same thing. One application of group theory is in musical set theory. Group theory is also very important to the field of chemistry, where it is used to assign symmetries to molecules. The assigned point groups can then be used to determine physical properties (such as polarity and chirality), spectroscopic properties (particularly useful for Raman spectroscopy and Infrared spectroscopy), and to construct molecular orbitals. In Philosophy, Ernst Cassirer related the theory of group to the theory of perception as described by Gestalt Psychology; Perceptual Constancy is taken to be analogous to the invariants of group theory. ## लिधंसा 1. ↑ Smith, D. E., History of Modern Mathematics, Project Gutenberg, 1906. • Rotman, Joseph (1994). An introduction to the theory of groups. New York: Springer-Verlag. ISBN 0-387-94285-8. A standard modern reference. • Scott, W. R. [1964] (1987). Group Theory. New York: Dover. ISBN 0-486-65377-3. An inexpensive and fairly readable textbook (somewhat outdated in emphasis, style, and notation). • Livio, M. (2005). The Equation That Couldn't Be Solved: How Mathematical Genius Discovered the Language of Symmetry. Simon & Schuster. ISBN 0-7432-5820-7. Pleasant to read book that explains the importance of group theory and how its symmetries lead to parallel symmetries in the world of physics and other sciences. Really helps congeal the importance of group theory as a practical tool). पुच:

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http://mathematica.stackexchange.com/questions/19699/badly-conditioned-matrix-generalluc

# badly conditioned matrix (General::luc) With some matrices I am receiving the following message Inverse::luc Result for Inverse of badly conditioned matrix (M) may contain significant numerical errors. How can I tell to Mathematica to isolate(or detect) that result (or badly conditioned matrix)? For instance, using the "If" statement, which condition should I put into the argument of If statement when my matrix is badly conditioned? p.s. Some advice to tackle with these kind of matrices would be good. - 3 You can use `LinearAlgebra`MatrixConditionNumber` to find the condition number of your matrix. What value you choose as the threshold depends on your particular application... In general, a condition number that is $\mathcal{O}(10^n)$ can make you lose up to $n$ digits of accuracy (in addition to FP errors). – rm -rf♦ Feb 16 at 19:28 3 Numerical inversion of a matrix is both slow and poorly conditioned, a fact mentioned in most books on numerical analysis. Typically, a solution to an applied problem can be formulated in such a way that matrix inversion is expressed in terms of the solution of a system of equations. Naturally, it would be much easier to provide details if you provided more details of your problem. – Mark McClure Feb 16 at 19:36 1 – whuber Feb 16 at 19:44 ## 2 Answers This might be as good a time as any to distill the collective wisdom of Messrs. Huber, McClure, and Toad R. M. As already mentioned, there is this quantity of great interest to people in the business of solving simultaneous linear equations, called the condition number, and conventionally denoted by the symbol $\kappa$. This is usually associated with a matrix $\mathbf A$ that figures as the matrix of coefficients in the linear system, and thus, we have the symbol $\kappa(\mathbf A)$. Further, there is not just one condition number, but a number of them, all dependent on the underlying matrix norm used in their definition. One then speaks of the "2-norm condition number", $\kappa_2(\mathbf A)$, the "$\infty$-norm condition number", $\kappa_\infty(\mathbf A)$... and so on. Now, why should we be interested in this condition number? One for instance cannot depend on the determinant, as it is not a reliable measure of how badly a coefficient matrix will behave in a linear system (see also this answer). In inexact arithmetic, the condition number is pretty much the only nice diagnostic you have. For applications, usually the choice of condition number does not matter much, since if some matrix $\mathbf A$ has a large/small $p$-norm condition number, the $q$-norm condition number of the same matrix will be of comparable magnitude. In any event, one usually wants matrices whose condition numbers are as near to unity as can be, as this is the case of well-conditioning. Conversely, if your matrix's condition number is "huge" (for some application-dependent definition of "huge"), then your matrix might as well be singular. (A matrix that is truly singular has a condition number of $\infty$.) The two-norm condition number is easily computed in Mathematica: ````cond2[mat_?MatrixQ] := Divide @@ Through[{Max, Min} SingularValueList[mat, Tolerance -> 0]] ```` As you might ascertain from this routine, this condition number requires the computation of the singular value decomposition (SVD). This is quite expensive, and sometimes one instead uses condition number estimators, which require much less computational effort. (One thing it can do that the next method I am about to describe can't is that it can be applied to non-square matrices as well.) Mathematica has a condition number estimator built-in, in the form of the function `LinearAlgebra`MatrixConditionNumber[]`. This function can be set to estimate either the $1$-norm or $\infty$-norm condition number, depending on the setting of its `Norm` option. The Hager-Higham condition estimator, which is the underlying algorithm, almost always gives a result that is equal to or near the exact condition number, though there are matrices that can defeat it. (Luckily, these counterexamples are rather contrived and do not seem to crop up in practice.) So, what to do in Mathematica? As R. M. notes, this is application dependent, but you will want to use the heuristic that when solving a linear system with coefficient matrix $\mathbf A$, you stand to lose $\approx\log_b(\kappa(\mathbf A))$ base-$b$ digits in the solution of your linear system. You will then want to do something like `If[LinearAlgebra`MatrixConditionNumber[A] < 1/$MachineEpsilon, (* code for well-conditioned case *), (* code for ill-conditioned case *)]`, to use a typical example. Mark's advice is only slightly related to the matter at hand, but he is of course right: one often does not need to compute inverses, and what an inverse can do, an appropriate matrix decomposition can do as well or better. (There are instances where inverses are genuinely needed, like variance-covariance matrices, but they are few and far between.) For instance, if you are doing something like ````x = Inverse[A].b ```` this is better done using the functionality of `LinearSolve[]`, which internally stores a matrix decomposition: ````lf = LinearSolve[A]; x = lf[b] ```` There are still a number of details to say (or I have forgotten), but this post is getting too long already, and I think it best to stop here. - ## Did you find this question interesting? Try our newsletter Use `Check` with 3 arguments, possibly in conjunction with `Quiet`. For example: ````Quiet[Check[Inverse[N[HilbertMatrix[20]]], "Badly conditioned", Inverse::luc], Inverse::luc] ````

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http://mathoverflow.net/questions/99549/how-does-my-radio-work/100249

## How Does My Radio Work? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Bear with me for a moment while I invoke the real world; the main question at the end is purely mathematical. I live in an area with $n$ AM radio stations and $m$ FM radio stations. AM station number $j$ wants to send me the signal $\phi_j(t)$. FM station number $k$ wants to send me the signal $\psi_k(t)$. Of course if they just sent those signals, my radio would recieve their sum and have no idea how to disentangle them. Therefore, the signals are first encoded. My (possibly ill-informed) understanding is that (modulo a gazillion bells and whistles), AM station $j$ sends the signal $\phi_j(t)\sin(\omega_j t)$. where $\omega_j$ is some constant, and FM station $k$ sends the signal $A_k \sin(\psi_k(t))$ where $A_k$ is some constant. My radio then receives the signal $$\sum_{j=1}^n\phi_j(t)\sin(\omega_j t)+\sum_{k=1}^mA_k\sin(\psi_k(t))\quad\quad(1)$$ Having received this signal, and knowing the values of the $\omega_j$ and the $A_k$, my radio is then somehow able to compute any one of the signals $\phi_j(t)$ or $\psi_k(t)$ and play it for me on request. (In fact, I'm pretty sure it can recover $\phi_j$ on the basis of $\omega_j$ alone, without knowing the values of the other $\omega$'s.) It's not obvious to me that this is mathematically possible, though my radio seems to have no problem doing it. Question 1 (Pure Mathematics). For what values of $\omega_1,\ldots,\omega_n,A_1,\ldots,A_m$ is it possible to recover the functions $\phi_1,\ldots,\phi_n,\psi_1,\ldots\psi_m$ from expression (1) alone? And what assumptions are being made on the class of allowable functions from which the $\phi_j$ and $\psi_k$ are drawn? Question 2 (Part Engineering, part Pure Mathematics). If (as is not impossible), AM and/or FM works entirely differently than I think it does, thus rendering Question 1 entirely unmotivated, then how do AM and FM work, what is the correct analogue of expression (1), and what is the right answer to the corresponding new version of Question 1? Edited to add: I'm aware that there are all sorts of issues with distorted transmissions, error-correcting, etc. I want to abstract away from all of these and understand the basics. - 9 Follow up question: "... and can you guys fix it?" – Vidit Nanda Jun 14 at 4:42 23 Have you tried turning it off and on again? – François G. Dorais♦ Jun 14 at 4:58 5 ...or just take a very simple circuit diagram for a crystal diode radio and write down the ODE for the response in the earphones from the earphones with the given signal. This should extract out frequencies close to the one to which you are tuned and perform the (amplitude) demodulation. – George Lowther Jun 14 at 8:57 6 @David Roberts' points out that you did not quite express FM correctly – you did not incorporate the base frequency, whereas I care whether I'm listening to 88.5 or 89.1. According to Wikipedia, FM radio transmits the data $\psi_k(t)$ via an encoding of the form $A_k\sin(\lambda_k t + \Psi_k(t))$, where $\lambda_k$ is a frequency much higher (three orders of magnitude, or so) than any of the Fourier modes in $\psi_k(t)$, and $\Psi_k(t)$ is (a multiple of – the units would be off otherwise) an antiderivative of $\psi_k(t)$. – Theo Johnson-Freyd Jun 15 at 2:35 4 Also, again according to Wikipedia, AM is more accurately described via a signal of the form $\Phi_j(t) \sin(\omega_j t)$, where $\Phi_j(t) = A_j(1 + M_j\phi_j(t))$. Here $A$ and $M$ are unit-full quantities, usually set so that $1 + M\phi_j(t)$ is always positive (otherwise the signal is distorted), but has minimum value $1/c$ for some single-digit number $c$. – Theo Johnson-Freyd Jun 15 at 2:49 show 10 more comments ## 4 Answers Modulation is used to reduce antenna height, noise distortion, to avoid interference... The low frequency signal (e.g., human voice) is superimposed to a high frequency signal (the carrier) and transmited. Every radio station have its own high frequency (carrier frequency) of transmission. When you tune your radio (choosing the carrier frequency of the radio station), you're also indicating the electronic circuit to be used to demodulate (extract information from the carrier) AM or FM signals to recover the (low frequency) human audible signal. See any book from B.P. Lathi (e.g. Modern Digital and Analog Communication System) or from Oppenheim (e.g Signals and Systems). These are classical books used in Electric Enginnering courses. According to Lathi, the AM signal can be demodulated coherently (demodulation synchronous) or noncoherently (demodulation assynchronous). In practice, two demodulation noncoherent methods are used: (1) rectifier detection and (2) envelope detection. (1) Rectifier Detector: If an AM signal is applied to a diode and resistor circuit, the negative part of the AM wave will be supressed. The rectified output, $v_r(t)$, is: $$v_r(t) = [A+m(t)]\cos \omega_ct [1/2 + 2/\pi(\cos\omega_ct- 1/3 \cos3\omega_ct + \ldots)] = 1/\pi[A + m(t)] + hft$$ where $hft$ are high frequency terms, $m(t) = B \cos\omega_mt$ is the information (low frequency signal), $\omega_m$, information frequency, $\omega_c$, carrier frequency, $A$ and $B$, amplitude of carrier and low frequency signals, respectively. When $v_r(t)$ is applied to a low-pass filter of cutoff $B$ Hz, the output is $[A +m(t)]/\pi$, and all the other terms in $v_r$ of frequencies higher than $B$ Hz are supressed. The dc term $a/\pi$ may be blocked by a capacitor to give the desired output $m(t)/\pi$. (2) Envelope Detector: the output of the detector follows the envelop of the modulated signal. The circuit is a diode followed by a RC-filter. Mathematical details in Lathi. The information (low frequency signal) in FM resides in the instantaneous frequency $\omega_i = \omega_c + k_f m(t)$, $k_f$ is a modulation index. A frequency-selective network with a transfer function $$|H(\omega)| = a\omega + b$$ over the FM band would yield an output proportional to the instantaneous frequency, $\omega_i$. There are several possible networks with such characteristics, the simplest is an ideal differentiator with transfer function $j\omega$. The mathematical details can be seen in Lathi. There is a tutorial here or here. BTW, this is a very beautiful (real) application of Fourier Theory. ADDED: In relation to your question, if we have $$\phi_{AM}(t) = A \cos\omega_ct + f(t)\cos\omega_ct$$ representing the AM signal and $$\phi_{FM}(t) = A \cos\omega_ct - A k_f g(t)\sin\omega_ct$$ representing the FM signal, then $$\phi_{AM}(t) \leftrightarrow \frac{1}{2}[F(\omega + \omega_c) + F(\omega - \omega_c)] + \pi A [\delta(\omega + \omega_c) + \delta(\omega - \omega_c)]$$ and $$\phi_{FM}(t) \leftrightarrow \pi A [\delta(\omega + \omega_c) + \delta(\omega - \omega_c)] + j\frac{Ak_f}{2}[G(\omega - \omega_c) - G(\omega + \omega_c)]$$ Considering $$v(t) = \phi_{FM}(t) + \phi_{AM}(t)$$ we have the Fourier transform: $$\frac{1}{2}[F(\omega + \omega_c) + F(\omega - \omega_c)] + 2 \pi A [\delta(\omega + \omega_c) + \delta(\omega - \omega_c)] + j\frac{Ak_f}{2}[G(\omega - \omega_c) - G(\omega + \omega_c)]$$ The next step is to adapt your equation to these equations. Then, procedures (1) or (2) should be used to recover the low frequency signal in the receiver side. - Contrary to your second sentence, the low-frequency signal is not superimposed on the high-frequency signal. Superimposing means addition. Modulation maintains the signal within a small neighborhood of the carrier in frequency space. – S. Carnahan♦ Jun 20 at 8:47 Sorry I disagree with you. The "superimposing" word was used in the sentence with the meaning of this definition: Modulation is the addition of information (or the signal) to an electronic or optical signal carrier as you can see at searchnetworking.techtarget.com/definition/…. Furthermore, in any part of my answer can be found the word "modulation" meaning "addition". This also applies to the equations I wrote. Maybe I used the wrong word. My apologies, English is not my first language. Thank you for your comments. – PaPiro Jun 20 at 14:38 ### 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. Entirely engineering: I believe "different frequency bands" means that the Fourier transform of $\sin (\phi_k(t))$ is small in some interval containing the $\omega_j$. Note that I am completely unqualified to speak about the engineering here. Both engineering and mathematics: I believe that the radio recovers, approximately, the functions $\phi_j$ by convolving the signal with $e^{(-\beta+\omega_j i)t}$. Mathematics: One cannot recover the signal exactly without strong assumptions on the $\phi_j$. Take the signal $\sin(\omega_1 t )\sin(\omega_2 t)$. This could represent $\phi_1=\sin(\omega_2 t)$, $\phi_2=0$, or $\phi_1=0$, $\phi_2=\sin(\omega_1 t)$ or any linear combination. Assume that the $\phi_i$ have Fourier transforms supported on the interval $[-\delta,\delta]$ and the $\omega_i$ are separated by gaps of size at least $2\delta$. Assume no FM stations. Then we can recover each AM station by taking the Fourier transform and setting everything outside $[\omega_i -\delta,\omega_i+\delta]$ to $0$. I don't know the right solution for FM. - I've been trained as an engineer, and I can tell you that engineers have a somewhat simplified view of the matter. (But, not only a simplified view, of course.) The other answers fill in some detail, but I think a higher-level view is useful. There is no such thing as perfect recovery of the transmitted signal. The best you can hope is to bound the error. For most modulation techniques the basic idea is that the spectrum $X$ of a signal $x$ is nearly 0 outside a narrow band: $X(f)\approx0$ when $|f \pm f_0| < B$. Both AM and FM are essentially means of transforming a spectrum centered around $0$ into one centered around $f_0$. So, in order to recover a signal, the main concern is to make sure that the spectrums $X_1$, $X_2$, …, $X_n$ do not overlap. This is achieved in a rather uninteresting way: regulation. Then you can extract one signal by shifting $f_0$ to $0$ (convolution with a Dirac impulse in frequency domain, meaning multiplication with a harmonic signal in the time domain), and then applying a low pass filter (multiplication with a rectangular function in the frequency domain, meaning convolution with a sinc in the time domain). See also this related question. There are broad-spectrum modulation techniques, which are used for example in fourth generation mobile-phone networks, that do not rely on the assumption that the signal covers a narrow band. The two main ones are frequency hoping (use some narrow-band modulation technique but change $f_0$ often in some pseudorandom sequence) and spread spectrum (multiply the signal with a pseudorandom sequence before using a narrow-band modulation technique). The signals obtained thru such methods have a wide band, but are bounded $|X(f)| < c$ for some $c$ for all $f$. This way they behave as background noise as far as demodulating any narrow-band signal is concerned. - Your expression for FM transmissions is not quite right - it's missing the radio frequency! The simple model that captures the essentials of what FM station $k$ is sending you is the function $$B_k\sin\left((\omega_k+\gamma_k\psi_k(t))t\right),$$ where $\gamma_k\psi_k$ never gets close to $\omega_k$ (so you're only modulating the frequency and not completely disrupting it). If the interesting signal $\psi_k$ is a pure note at frequency $\omega$, then the spectrum of the actual radio signal can be found in terms of Bessel functions and consists of sidebands separated from the carrier by spacing $\omega$. (The number of sidebands is controlled by how large $\gamma_k$ is.) The real radio signal your device is getting, then, is $$F(t)=\sum_{j=1}^n\phi_j(t)\sin(\omega_j t)+\sum_{k=1}^m B_k\sin\left((\omega_k+\gamma_k\psi_k(t))t\right).$$ Because of the modulation, none of the stations' radio signals are single peaks; instead they are spread over a bandwidth roughly given by the frequency content of the audio signals they encode. (For comparison, human hearing can detect 16 Hz to roughly 20,000 Hz, AM frequencies are medium-wave radio at 520 kHz to 1,610 kHz, and FM stations run at 87.5 to 108 MHz. Thus in reality the peaks are quite narrow!) To detect a signal, your device uses a combination of antennas, loops of wire, parallel plates, and the like, which contrive to give to the decoding device (the one that takes a radio signal and gives you an audio one) a voltage $f$ that's controlled by a damped harmonic oscillator equation of the form $$\frac{d^2}{dt^2}f-2\gamma\frac{d}{dt}f+\omega_0^2f=F,$$ where the resonance frequency $\omega_0$ is controlled by a knob on the device. The spectral response of this dynamical system is routinely evaluated in college ODE courses, and comes out as a Lorentzian bell-shaped curve centred at $\omega_0$ and of width $\gamma$. Choose $\gamma$ to match the spectral width of the typical radio station, and you've got a fantastic filter! EDIT: After doing some looking up, I find that the $\psi_k$ here is not exactly the audio signal the station is trying to encode, but rather something like its average over the interval $[0,t]$, so it is equivalent to it up to simple mathematical operations performed at the decoder. -

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http://math.stackexchange.com/questions/44983/how-do-i-combine-multiple-sets-of-ratios-in-order-to-meet-an-overall-target-ra?answertab=votes

# How do I combine multiple sets of ratios in order to meet an overall “target” ratio? (word problem supplied to illustrate) (I had no idea how to tag this, any help would be great) :) I am a programmer with little math experience beyond high school - this will be demonstrated by my explanation. I am developing an application that needs a formula. The formula involves calculating what amounts of food to eat based on factors like calories and concentrations of macronutrients (carbs, protein, fats). Since there are other calculations need that aren't relevant to main problem, I have abstracted it into the following word problem. Here it is: Your job is to keep your store, Colorful Widgets, Inc., supplied with widgets. There are four suppliers that you can order from every month. There is a set ratio of colors that must be met as closely as possible to match sales expectations. The ratio is: Target Color Ratios: 40% Red widgets 40% Blue widgets 20% Yellow widgets The catch is that each supplier offers shipments of mixed colors in different ratios, with each company specializing in one color. Here are the ratios available from the suppliers: Mostly Red Widgets, Inc. 86% Red 11% Blue 03% Yellow Acme Widgets, Inc (Limit 3 boxes) 79% Red 09% Blue 20% Yellow Mostly Blue Widgets, Inc. 15% Red 77% Blue 08% Yellow Mostly Yellow Widgets, Inc. 10% Red 00% Blue 90% Yellow All four companies sell their widgets in boxes containing 100 widgets each. Boxes cannot be split into smaller amounts. You need to order 24,000 widgets, with the total ratio of colors matching the targets laid out above. Restrictions: You have a contract with the first three companies - they must be ordered from every month, no exceptions. The fourth (Mostly Yellow Widgets, Inc.) is a stop gap - if your total ratio is low on yellow widgets, you can order from them. Additionally, Acme Widget, Inc. makes the best widgets, but only in limited quantity. You must order 3 boxes from them each ordering cycle. I understand that this problem probably will not have a neat and tidy catch-all formula. Even if there is no formula for this problem with it's restrictions, there should be a formula or something that will help with the problem of combining three different ratios in such a way to come as close as possible to the target ratio. I also understand that if the target ratios or ratios of suppliers change too much, a solution may not be possible. - I don't really understand the restrictions. Are you saying that you must order at least one box from each of the first three companies? What is meant by "best"? – Qiaochu Yuan Jun 12 '11 at 18:09 The numbers for Acme don't add up to 100%. – Robert Israel Jun 12 '11 at 18:53 @Qiaochu: This word problem was my poor attempt to abstract what is actually a food calculation. The "Acme Widget" actually represents green veggies in my real problem - something that cannot be skipped. :) – Matt Miller Jun 13 '11 at 18:00 @Robert: That's what I get for making a last minute change to my question. – Matt Miller Jun 13 '11 at 18:00 ## 2 Answers For Acme widgets, you may order either $0$ or $1$ or $2$ or $3$ boxes; this integer will be called $M$. It's just four possibilities and all of them are small numbers of boxes and it is easy to manually check which of these four values will you the best results. Most of the adjustment is done by and most of the boxes will come from the mostly red, mostly blue, and mostly yellow companies. You must order $R, B, Y$ boxes from each, respectively. The conditions, neglecting $M$ for a while, are $$R + B + Y = 240,$$ $$0.86 R + 0.15 B + 0.10 Y = 96,$$ $$0.11 R + 0.77 B + 0 Y = 96,$$ $$0.03 R + 0.08 B + 0.90 = 48.$$ The four equations are not independent: the sum of the last three gives the first. So eliminate the first. We have a set of three linear equations for three unknowns $R,B,Y$. Define matrix $A$ as the $3\times 3$ matrix of the entries on the left hand side of the three equations. Then the solution is $$(R,B,Y) = A^{-1} (96,96,48)^T \approx (87.36,112.20, 40.45)$$ That's the number of boxes you would ideally want to order if $M=0$: you would have to round it to $(87,112,41)$. Now, solve the same equations with the extra terms for $M=1,2,3$ and decide which of them is the best according to your best criteria (which, I guess, is either the sum of the deviation of the three percentages from $.4,.4,.2$ or the sum of the squares of these numbers). At any rate, the number of boxes $R,B,Y$ won't deviate by more than 3 from the numbers written above. - – Matt Miller Jun 13 '11 at 18:05 It's basically a linear programming problem: if $x_1$, $x_2$, $x_3$, $x_4$ are the numbers of boxes ordered from the four companies, you would like $$\eqalign{ x_1 + x_2 + x_3 + x4 &= 240 \cr 86 x_1 + 79 x_2 + 15 x_3 + 10 x_4 &= 40 \times 240\cr 11 x_1 + 9 x_2 + 77 x_3 &= 40 \times 240\cr 3 x_1 + 20 x_2 + 8 x_3 + 90 x_4 &= 20 \times 240\cr x_2 \le 3,\ {\rm all}\ x_i \ge 0}$$ However, this may not be possible to attain precisely, so you allow some slack. One way is to minimize $r$ subject to $$\eqalign{ x_1 + x_2 + x_3 + x_4 &= 240 \cr 86 x_1 + 79 x_2 + 15 x_3 + 10 x_4 &\le 40 \times 240 + r \cr 86 x_1 + 79 x_2 + 15 x_3 + 10 x_4 &\ge 40 \times 240 - r \cr 11 x_1 + 9 x_2 + 77 x_3 &\le 40 \times 240 + r\cr 11 x_1 + 9 x_2 + 77 x_3 &\ge 40 \times 240 - r\cr 3 x_1 + 20 x_2 + 8 x_3 + 90 x_4 &\le 20 \times 240 + r\cr 3 x_1 + 20 x_2 + 8 x_3 + 90 x_4 &\ge 20 \times 240 - r\cr x_2 \le 3,\ {\rm all}\ x_i \ge 0}$$ Now usually the solution values will not be integers. If you want integer values, this becomes an integer linear programming problem. Many software packages will solve these. In this case the optimal solution, as given by Maple's Optimization package, is $x_1 = 85$, $x_2 = 3$, $x_3 = 112$, $x_4 = 40$, $r = 27$. -

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http://mathhelpforum.com/advanced-math-topics/28696-solved-can-anyone-help.html

# Thread: 1. ## [SOLVED] Can anyone help? I was wondering if there is any kind of formula that will allow me to do this: I have about 20 numbers. I want a combination of some of the numbers to add up to be (let's say) \$25,768.45. I don't have to use all the numbers to make them add up to that number. I can use 2 or I can use all 20. Is there any kind of formula I can use to help me do this? The numbers are all big and the final number I need them to add up to is big too so I've just been plugging in numbers and guessing but it is taking forever AND I cannot make it work. I realize it may not work, but I thought maybe there is a trick or a shortcut to this. Thanks. Sorry if this makes no sense! 2. Given a set of twenty numbers, there are a maximum of $2^{20} - 1 = 1048575$ possible sums. (There may be several equal sums, but that is the number of different sums.) Given the power of today’s computers, it should be possible to actually check out each possible sum. But there is no mathematical formula to do this for an arbitrary set.

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http://mathoverflow.net/questions/17295/results-about-the-order-of-a-group-forcing-a-particular-property/17304

## Results about the order of a group forcing a particular property. ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Given a group of order $n$ where $n$ is either a specific number, or a number of a particular form, e.g. square-free, when does $n$ completely determine a particular group property among all groups of that order? Vipul's group theory wiki has several stubs on this topic, and in the language of his wiki, I will call this a $P$-forcing number, where $P$ is a particular group theoretic property. We already have quite a few easy examples, for instance, orders $pq$, $pqr$, and $p^2q$ force solvability, and $p^2$ forces abelian. Then there are more specific results like 99 is an abelian-forcing number. I am interested in general, in any results of this flavor beyond what would be considered a common result in a standard graduate-level group theory book. - 1 See Pete Clark's answer at: mathoverflow.net/questions/11001/… – S. Carnahan♦ Mar 8 2010 at 5:20 ## 4 Answers The numbers n such that every group of order n is cyclic, abelian, nilpotent, supersolvable, or solvable are known. Most are described in an easy to read survey: Pakianathan, Jonathan; Shankar, Krishnan. "Nilpotent numbers." Amer. Math. Monthly 107 (2000), no. 7, 631-634. MR 1786236 DOI: 10.2307/2589118 If you want to go beyond results like this, you may have better luck looking at a slightly more refined version of the order: the isomorphism type of the Sylow subgroups. Sometimes a p-group P has the property that every group G containing it as a Sylow p-subgroup has a normal subgroup Q of order coprime to p such that G is the semi-direct product of P and Q. An easy version of this that does appear in many group theory texts is that if n=4k+2, then in each group G of order n there is a normal subgroup Q of order 2k+1 so that G is the semi-direct product of any of its Sylow 2-subgroups and Q. Groups all of whose Sylow p-subgroups are cyclic have very nice properties, subsuming those of groups of square-free order. Groups all of whose Sylows are abelian have more flexibility, but are still basically under control. - ### 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 haven't had time to read it yet, but The influence of conjugacy-class sizes on the structure of finite groups seems tailor made to answer these questions. It's a survey article, too. - Very famous one is the Feit-Thompson theorem: if n is odd, then G is solvable. Though I suppose this is stated (but not proved) in most modern algebra texts. - $(n,\phi(n)) = 1$ forces $G$ cyclic. $p^nq^m$ forces $G$ solvable. -

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