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http://crypto.stackexchange.com/questions/2753/in-the-sha-hash-algorithm-why-is-the-message-always-padded/2755	# In the SHA hash algorithm, why is the message always padded? In the SHA hash algorithm the message is always padded, even if initially the correct length without padding; the padding is of the form "1" followed by the necessary number of 0s. Why is it necessary that the message always be padded? - I edited your question to make it look better, but it was unclear if you were only asking why padding is always added or if you were also asking why the particular form of padding was chosen. Feel free to roll back changes if necessary. – mikeazo♦ May 31 '12 at 18:04 ## 3 Answers Well, the entire point of a cryptographical hash function is that no one can practically devise two messages that hash to the same value. Now, the SHA family of hashes use the Merkle–Damgård construction; that is, they have an iterated hash function, and each invocation of the hash function takes as input a fixed block size (either 512 or 1024 bits in the case of the SHA family). However, we're interested in hashing messages that don't happen to be an exact multiple of 512 or 1024 bits in length. So, what we do is define a padding function that takes a message, and converts it into a form that's a multiple of the block size in length; we then run the hash compression function on this padded message. Now, there are a couple of requirements on this padding function; one of the most critical that if we had two different messages, then they must still be different after padding; if not, then the hash compression functions will work identically, and so those two messages would hash to the same value. That really is the answer to your question (a); if our padding function didn't modify (pad) messages that happened to be the correct length, then that would immediately lead to a collision. For example, if we take any message $M$ that's not a multiple of the block size, we could compute $Pad(M)$. Now, $Pad(M)$ will be a multiple of the block size, and so we'd know that $Hash(M) = Hash(Pad(M))$. As for the answer to your question (b), well, if our padding function was simply adding zero bits until the padded message was of the correct length, well, that would lead to a collision of the form $Hash(M) = Hash(M \|\| 0)$ (assuming postpending the 0 doesn't cause us to cross a block boundary). On the other hand, the SHA padding function isn't just appending 0 and 1 bits; it also involves adding a bitlength of the entire message at the very end. If the question is "why do they add a 100...000 pattern and add a message length', well, I rather suspect that's a belt-and-suspenders approach. Either would be sufficient in itself (as either would prevent two messages from padding into a common value); both are cheap, and so the designers decided to do both. - As to why two padding techniques are used: If padding was only with a 1 and enough 0 to reach the next block frontier, then an authority arbitrarily deciding the 160-bit magic value used as initialization vector could use that power to introduce a backdoor, by computing the magic value from an arbitrary 512-bit secret, that could be prefixed to messages to cause a collision. While the magic value in SHA-1 (`67452301EFCDAB8998BADCFE10325476C3D2E1F0`) is regular enough to show this is not the case, this is a nice reason to use the other padding technique (appending the length). – fgrieu Jun 4 '12 at 11:42 In many existing padding schemes, without padding always being added there is a trivial second preimage attack. For simplicity let's assume a 10 bit hash function $h_{10}$ (extending this to other size hash functions is trivial). Let $m_1=101$ and $m_2=1011000000$. I claim that $h_{10}(m_1)=h_{10}(m_2)$. Since $m_2$ is 10 bits, no padding is needed. Since $m_1$ is smaller, padding is needed. What padding will be added? The padded version of $m_1$ will be equal to $m_2$, which is why the claim is true. Thus, there are two message $m_1\neq m_2$ and yet $h_{10}(m_1)=h_{10}(m_2)$. NOTE: This attack applies to many hash functions including SHA-1 and the SHA-2 family. - This is not true. One could imagine a padding algorithm that doesn't always add padding and has no trivial second preimage attacks. For example, one could devise a padding algorithm such that if the input was a perfect fit for an even number of blocks, it isn't padded at all and in every other case, it is padded so that the padded result contains an odd number of blocks. – David Schwartz Jun 1 '12 at 15:35 @DavidSchwartz, good point. I was more referring to SHA1's padding scheme and why it specifically always needs to add padding. – mikeazo♦ Jun 1 '12 at 16:16 Strictly speaking, it's not. But you wind up with either very complex or very ugly padding algorithms. For example, say you were designing a padding algorithm such that there exists one string of length M that is "padded" by adding no padding at all. Clearly, it would be a disaster if any string of length less than M could be padded so as to produce a string that, if input, wouldn't be padded. So this leaves you two painful options: 1) You can look at the contents of the string to decide how to pad it. This makes streaming implementations difficult, and is confusing and prone to error. 2) You can reserve some lengths for unpadded strings. But this means that some lengths will have to be padded with at least two extra blocks. Basically, if you try designing an algorithm that leaves any inputs unpadded, you will find that every one you design is either broken (allowing two inputs to pad to the same output) or horrible. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 15, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9240471124649048, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/41030/why-is-int-dp-2-pi-p-rangle-langle-p-1/41246	# Why is $\int (dp/2\pi) \|p \rangle\langle p\| = 1$? In quantum mechanics, why is $\int (dp/2\pi) \|p \rangle\langle p\| = 1$ where $\|p \rangle$ represents momentum eigenstate? - Are you sure about the two pi? A family of orthonormal vectors spans the whole space (is 'complete') is the sum of all the projectors onto each single basis vector gives the identity operator. A projector onto the normalized state \|p> is \|p><p\|, the sum over all projectors $\int dp \|p><p\|$, and the identity operator is just 1 – A.O.Tell Oct 17 '12 at 16:33 It's from A.Zee's Quantum Field Theory page 11.. So it is wrong? – RRRR Oct 17 '12 at 17:51 It's not wrong, it's just a way to normalize states that is a bit unconventional. Maybe it has advantages elsewhere. I'll check Zee later – A.O.Tell Oct 17 '12 at 18:10 ## 3 Answers The factor of $2\pi$ is a convention Zee uses which normalizes the $\|p\rangle$ states so that they are not quite orthonormal: $$\langle p \| p' \rangle = 2\pi \delta(p-p')$$ which means that Zee's $p$ states are different from Ahmed's $p$ states by a factor of $\sqrt{2\pi}$. Let me call these unit normalized $p$ states $\|\|p\rangle$, then $$\sqrt{2\pi} \|\| p \rangle = \|p\rangle$$ This is something Dirac does also, and it becomes essential for relativistic states, when you want $$\int {d^3 p \over (2\pi)^3 2\omega_p} \|p\rangle\langle p\| = 1 ,$$ because then all the parts are manifestly invariant andso the integral over $p$ is an integral over the mass-shell hyperboloid: $$\int {d^4 p \over (2\pi)^4} (2\pi) \delta(p_\mu p^\mu - m ^2) = \int {d^3 p\over (2\pi)^3 (2\omega_p)} ,$$ where the delta function projects you onto the mass shell in a relativistically invariant way. But for this to be true, you need $$\|p\rangle = (\sqrt{2\pi})^3 \sqrt{2\omega_p} \|\| p \rangle$$ where $\omega_p = \sqrt{p^2 + m^2}$. This is called a relativistically normalized state. The relativistically normalized states are natural, and define relativistically normalized creation operators, $$\alpha(k) \|0\rangle = \|k\rangle ,$$ which are related to the usual nonrelativistically normalized creation operators as follows: $$\alpha(k) = (\sqrt{2\pi})^3 \sqrt{2\omega_k} a_k .$$ In terms of the $\alpha(k)$'s and their conjugates, the field Fourier expansion is manifestly invariant: $$\phi(x) = \int e^{ikx} \alpha(k) + e^{-ikx} \alpha^\dagger(x) {d^3p \over (2\pi)^3 2\omega_k} .$$ This looks trivial, but it makes everything in scattering theory and mode expansions relativistically consistent. So you can make trivial all the first chapters of any old quantum field theory books with mode expansions which look opaque, because they are noncovariant. - I don't think this question has been answered. The original poster asked: why is \begin{align} \int dp \|p\rangle \langle p\| = 1, \end{align} up to unimportant normalization issues that have been discussed extensively (but are off the point). The above statement is equivalent to the question of completeness of a set of states on an Hilbert space (which ahmed addressed but did not answer the question why the LHS of the original poster's equation is equal to the RHS). Completeness is, heuristically, the statement that a set of vectors (functions on a Hilbert space) spans the space of possible (piecewise continuous and some degree of smoothness (differentiable)) functions on that space. In other words, just about any "well-behaved" function can be expanded in a basis of states that is complete. A proof of completeness of eigenfunctions of differential equations is given in Chapter VI, Section 3, p. 424 of Courant & Hilbert's Methods of Mathematical Physics, Vol. 1, first English edition. It's too lengthy to reproduce here. But motivation (not a "proof") for the completeness relation can be given as follows. We consider the expansion of the Dirac $\delta$ function in a particular basis, say the position basis. We assume that it can be written as an expansion over momentum eigenstates in the position basis, $e^{-ipx'}$: \begin{align} \delta(x-x') &= \int_{-\infty}^\infty\!dp\,c_p(x)\frac{e^{-ipx'}}{\sqrt{2\pi}}. \end{align} Clearly the coefficient, $c_p$ in the expansion must depend on $x$ since the $\delta$ function on the LHS does. The completeness relation is determined as follows. Mulitply the above equation by $e^{ip'x'}/\sqrt{2\pi}$, interchange the order of the two integrals, and integrate over $x'$ on the interval $(-\infty,\infty)$. Use orthonormality of the plane waves, $\int_{-\infty}^\infty \frac{dx'}{2\pi} e^{i(p'-p)x'}=\delta(p'-p)$ to get \begin{align} c_{p'}(x) &= \frac{1}{\sqrt{2\pi}} e^{ip'x}. \end{align} Substitution back into the expression for $\delta(x-x')$ gives \begin{align} \delta(x-x') = \int_{-\infty}^\infty \frac{dp}{2\pi}e^{ip(x-x')}. \end{align} Rewriting this \begin{align} \delta(x-x') &= \langle x\|x' \rangle = \langle x\|1\|x' \rangle \\ &= \int_{-\infty}^\infty\! dp\, \langle x\|p\rangle\langle p\|x'\rangle \\ \langle x\|1\|x' \rangle &= \langle x\| \cdot\Big[\int_{-\infty}^\infty\! dp\, \|p\rangle\langle p\|\Big] \cdot \|x'\rangle. \end{align} And completeness follows from the arbitrariness of the values of $x,x'$. Nota Bene: This is not a proof as we have made several questionable steps (like reversing the orders of integration $dx' \leftrightarrow dp$) and essentially assumed the result we wanted (by using orthonormality). What this does, though, is to make plausible the idea that if you need to expand a function in a complete set of states, you need a relation like $\int\! dp\, \|p\rangle\langle p\| =1$ to hold. - 1 -1: OP was asking about why the $2\pi$ factors appear in some books and not others, not about why basis states are supposed to be orthonormal. – Ron Maimon Mar 28 at 3:15 You need to read the question again, Maimon: "In quantum mechanics, why is ∫(dp/2π)\|p⟩⟨p\|=1 where \|p⟩ represents momentum eigenstate?" This is asking about completeness not normalization. -1 for being off-topic. – MarkWayne Mar 28 at 4:49 1 I was here when OP asked the question, the issue was the 2pi factor, which confused OP, because it is not orthonormal, but a different convention from other books. I understand what was being asked, really, I am not offtopic. I shouldn't have downvoted you, I suppose, you didn't say wrong things, just things OP already knows. – Ron Maimon Mar 28 at 19:20 The composition of any state in terms of the momentum eigenstates is $\lvert\psi\rangle = k\int \mathrm{d}^{3}p\lvert p\rangle\langle\psi \vert p\rangle$. The set of momentum eigenstates form a complete set of orthogonal state vectors in Hilbert space (this set is uncountably infinite). The components of this decomposition are the $\langle\psi \vert p\rangle$ for each $p$ (a complex number). To simplify things, $\int \mathrm{d}^{3}p\lvert p\rangle \langle p\rvert$ is an operator. Let $\int \mathrm{d}^{3}p\lvert p\rangle\langle p\rvert = A$, then $A\lvert\psi\rangle = \lvert\psi\rangle$ for an arbitrary $\lvert\psi\rangle$ then $A$ must be the identity operator. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 36, "mathjax_display_tex": 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": 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": 0.9097495079040527, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/95/what-current-alternatives-are-there-to-the-standard-concordance-model-of-cosmolo/130	# What current alternatives are there to the standard concordance model of cosmology? The current "standard model" or concordance model of cosmology is Lambda CDM which includes late time acceleration due to a cosmological constant, cold dark matter as the missing matter component and an inflationary period at very early times. What alternative theories are also consistent with current observational data? - ## 3 Answers What alternative theories are also consistent with current observational data? There is an alternative line of thought according to which inhomogeneities in large-scale structure, known as voids, are responsible for the "perceived" or "apparent" acceleration on large time scales. This viewpoint was first advocated by Celerier and Inoue and Silk, among others. A numerical comparison of WMAP3 data with LCDM and non-Lambda model is done here. It is found that the likelihood of the non-Lambda model is comparable to that of the LCDM one. However such a fit requires a very low Hubble rate ( $\sim 0.40 - 0.45 h^{-1}$ ) compared to the widely accepted $\sim 0.73 h^{-1}$, and also requires a significant amount of negative curvature. These shortcomings aside, a very recent work by Hael Collins allows an analytic approach to the problem, albeit in a simplified setting. To quote from the abstract: This example provides an illustration of a universe where the inhomogeneities can affect its average expansion rate; and its simplicity allows a condition to be derived that tells when their presence should begin to become important. Since the averages of the non-uniform parts of the metric and the matter density grow faster than their uniform parts, the average expansion rate accelerates with the advent of the era governed by the inhomogeneities. (emphasis mine). So as they say, you free to "worship at the church of your choice" :-D [Update: Nov. 14, 2010] ... but if these (1, 2) references to papers by David Wiltshire have anything to say about the question of "apparent" vs. "actual" cosmic acceleration, it would appear that many in the cosmology community have been worshipping the false god of "dark energy" ! - Penrose's Conformal Cyclic Cosmology‎ is such an example. There is late time acceleration due to Lambda, but no inflationary period in the beginning. The late universe is equated to the beginning (of the next big bang cycle) through a conformal factor which maintains the physics. - To what extent has this been compared to observational data? – j.c. Nov 3 '10 at 15:04 It does make some predictions regarding the distribution of irregularities in the CMB. Basically black holes in the previous aeon should show up as circular patches of colder or hotter radiation. Some such irregularities have been found but more investigation is needed. – Sklivvz♦ Nov 3 '10 at 17:28 Prof. Alex Mayer has proposed a new cosmology that matches observables and which does not require dark energy. Google "Alex Mayer" (his website is the first hit) and then first review the "An Introduction to the New Cosmology" talk he gave in September. You can also download a preview of his new book. -	{"extraction_info": {"found_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": 0.18, "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": 0.9505796432495117, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/128055/how-to-show-a-sequence-is-not-uniformly-integrable	# How to show a sequence is not uniformly integrable Let $(X,\Omega,\mu)$ be a measure space. A sequence $f_n$ is said to be uniformly integrable if for every $\epsilon \gt 0$ there is a $\delta \gt 0$ such that for every measurable set $A$ with $\mu(A)\lt \delta$, $\int_A \|f_n\|~d\mu \lt \epsilon$, for every $n\in \mathbb{N}$. A sequence is said to be tight if for every $\epsilon \gt 0$ there is a measurable set $B$ of finite measure such that $\int_{X\setminus B} \|f_n\|~d\mu \lt \epsilon$, for every $n\in \mathbb{N}$. I claim the $f_n = n\cdot 1_{[0,1/n]}$ is not uniformly integrable and $g_n = 1_{[n,n+1]}$ is not tight. Proof. Fix $\epsilon \gt 0$. Pick $n$ sufficiently large so that for every $\delta \gt 0$, $n\delta \gt 1/2.$ Then there is an $n$ such that $\int_A \|f_n\| \gt 1/2.$ Let $\mu(B)\lt \infty$. Suppose to the contrary that $g_n$ were tight. Then $\mu\left((X\setminus B)\cap [n,n+1]\right) \lt \epsilon$. If I can get that $\mu(B) = \infty$, then I would have a contradiction, but I don't see how. Is what I have done above right? - – user47756 Nov 1 '12 at 5:12 @robjohn I fail to see how your comment on the deleted answer concerns anybody else than the author of the answer. – Did Nov 1 '12 at 8:42 ## 1 Answer Your arguments are not correct. It seems you are mixing up or ignoring the quantifiers. You are trying to show that $\{f_n\}$ is not uniformly integrable. The definition of uniform integrablity states that something is true for all $\epsilon>0$. So if a sequence is not uniformly integrable, that something must be false for some $\epsilon>0$. Here, as it turns out, you can take $\epsilon=1/2$; this will be the $\epsilon$ that "doesn't work" in the definition. So, set $\epsilon=1/2$. We then have to show that the following is not true: $\ \ \$There is a $\delta>0$ such that for every set $B$ with $\mu(B)<\delta$, we have $\int_B\|f_n\|<1/2$ for all $n$. So, we have to show that, given $\delta>0$, there is a set with measure less that $\delta$ but with the integral of some $f_n$ over that set greater than or equal to $1/2$. Towards this end, you can do the following: Let $\delta>0$. Choose $1/N<\delta$. Then $\mu([0,1/N])<\delta$, but $\int_{[0,1/N]} \|f_N\|=N\cdot{1\over N}=1$. Since $1>1/2$, we are done. For the second claim: I would argue directly: "We will show $\{g_n\}$ is not tight". The definition of tightness says that something is true for all $\epsilon>0$. So you have to show that that something is not true for some $\epsilon>0$. As it turns out you can take $\epsilon=1/2$; this will be the $\epsilon$ that "doesn't work" in the definition. What "doesn't work" here is the following: $\ \ \ \$There is a set $B$ of finite measure with $\int_{X\setminus B} \|g_n\|<1/2$ for all $n$. So we have to show the above statement is not true. So, we have to show that for any $B$ with finite measure, the following does not hold: $\int_{X\setminus B} \|g_n\|<1/2$ for all $n$. So, let $\mu(B)$ be finite. $B$ is fixed now, and your task is to show that for some $n$, $\int_{X\setminus B} \|g_n\|\ge1/2$. Once you've done this, you'll have your result. A hint for this: since the measure of $B$ is finite there is an $n$ with $\mu\bigl((X\setminus B)\cap[n,n+1]\bigr)\ge1/2$. - I meant to say fix $\epsilon = 1/2.$ why do you have $\int_{[1,1/n]} \|f_n\|$? – Jonjo Apr 4 '12 at 17:09 @Jonjo Sorry, that was a typo. I rewrote the answer; I hope it helps... – David Mitra Apr 4 '12 at 17:35 Thanks. I think I got it now, but why $\mu(B\cap[n,n+1])\ge1/2$ and not $\mu((X\setminus B)\cap[n,n+1])\ge1/2$? – Jonjo Apr 4 '12 at 18:18 @Jonjo Bleh, another "typo"... Thanks. – David Mitra Apr 4 '12 at 18:24 So would any $n$ such the $\mu([n,n+1]) \gt 1/2$ suffice? – Jonjo Apr 4 '12 at 18:39 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": 70, "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": 0.9689575433731079, "perplexity_flag": "head"}
http://LAJEANNA.PARKER@ssa.gov/policy/docs/statcomps/income_pop55/2004/apn.html	# Appendix Because the figures in this report are based on a sample of the older population, all reported statistics (counts, percentages, and medians) are only estimates of population parameters and may deviate somewhat from their true values—that is, from the values that would have been obtained from a complete census using the same questionnaires, instructions, and interviewers.1 The standard error is primarily a measure of sampling variability—that is, it measures the variations that occur by chance because a sample rather than the entire population is surveyed. As calculated for this report, the standard error also partly measures the effect of response and enumeration errors but does not measure systematic biases in the data. The chances are about 68 out of 100 that an estimate for the sample would differ from a complete census figure by less than the standard error. The chances are about 95 out of 100 that the difference would be less than twice the standard error. ## Standard Error of Estimated Percentages The reliability of an estimated percentage, computed by using sample data for both numerator and denominator, depends on both the size of the percentage and the size of the total on which the percentage is based. The approximate standard error Sx of an estimated percentage can be obtained using the formula $s x , p = b x p ( 100 − p )$ Here x is the total number of persons, families, or households (the base of the percentage), p is the percentage, and b is the parameter from the following table associated with the characteristic in the numerator of the percentage. Characteristic Total or white Black Hispanic Below poverty level 1,998 1,998 1,998 All income levels 1,249 1,430 1,430 Use of this formula in calculating the standard error of a single percentage is illustrated as follows: An estimated 33.8 percent of units aged 65 or older had total money income of \$30,000 or more in 2004 (Table 3.1). Because the base of this percentage is approximately 26,865,000—the number of units aged 65 or older—the standard error of the estimated 33.8 percent is approximately 0.3 percent. The chances are 68 out of 100 that the estimate would have shown a figure that differed from one resulting from a complete census by less than 0.3 percent. The chances are 95 out of 100 that the estimate would have shown a figure differing from one after a complete census by less than 0.6 percent—that is, this 95 percent confidence interval would range from 33.2 percent to 34.4 percent. For a difference between two sample estimates, the standard error is approximately equal to the square root of the sum of the squares of the standard errors of each estimate considered separately. This formula will represent the actual standard error quite accurately for the difference between separate and uncorrelated characteristics. If, however, there is a high positive correlation between the two characteristics, the formula will overestimate the true standard error. A comparison of the difference in the percentage of units aged 62 to 64 and 65 or older who had total money income of \$30,000 or more in 2004 illustrates how to calculate the standard error of a difference between two percentages: 33.8 percent of the 26,865,000 units aged 65 or older and 55.8 percent of the 4,990,000 units aged 62 to 64 had total money income of \$30,000 or more in 2004—a difference of 22 percentage points. The standard errors of those percentages are 0.3 and 0.8, respectively. The standard error of the estimated difference of 22 percentage points is about $0.9 = ( 0.3 ) 2 + ( 0.8 ) 2$ The chances are 68 out of 100 that the difference is between 21.1 and 22.9 percentage points and 95 out of 100 that it is between 20.2 and 23.8 percentage points. Because the confidence interval around the difference does not include zero, there is a statistically significant difference between the proportions of units who are aged 62 to 64 and those who are aged 65 or older with income of \$30,000 or more. ## Confidence Limits of Medians The sampling variability of an estimated median depends on the distribution as well as on the size of the base. Confidence limits of a median based on sample data may be estimated as follows: (1) using the appropriate base, the standard error of a 50 percent characteristic is determined; (2) the standard error determined in step 1 is added to and subtracted from 50 percent; and (3) the confidence interval around the median corresponding to the two points estimated in step 2 is then read from the distribution of the characteristic. A two-standard-error confidence limit may be determined by finding the values corresponding to 50 percent plus and minus twice the standard error. This procedure may be illustrated as follows: The median total money income of the estimated 26,865,000 units aged 65 or older was $20,481 in 2004 (Table 3.1). The standard error of 50 percent of those units expressed as a percentage is about 0.34 percent. As interest usually centers on the confidence interval for the median at the two-standard-error level, it is necessary to add and subtract twice the standard error obtained in step 1 from 50 percent. This procedure yields limits of approximately 49.3 percent and 50.7 percent. By interpolation, 49.3 percent of units aged 65 or older had total money income below$20,297, and 50.7 percent had total money income below $20,990. Thus, the chances are about 95 out of 100 that the census would have shown the median to be greater than$20,297 but less than \$20,990. ## Notes 1 Most of the discussion of estimation procedures has been excerpted from Current Population Reports, No. 114 (July 1978).	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 2, "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": 0.18, "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": 0.9247967004776001, "perplexity_flag": "middle"}
http://www.physicsforums.com/showthread.php?t=613040	Physics Forums ## Magnetic potential energy Dear all, I'm reading a paper on finite element magnetic field analysis. Basically there are two approaches to this. One is to use Maxwell equation and the other is to define an energy functional, discretize the problem and minimize the functional with respect to the unknowns. The paper takes the second approach. The functional is given as: $E=\int_{V}(\int_{0}^{B}H.dB)dv-\int_{V}J.Adv$ J is the source current density and A is the magnetic vector potential ( $B=\nabla \times A$) The first term is the stored energy. My question is about the second term. It seems to be the magnetic energy given to the system. I need your help to interpret and understand the second term because in electromagnetic text books, the second term (divided by 2, and without the negative sign) is proved to be equal to the first term in the linear case. I'm totally confused with this energy functional. Hassan 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 Tags energy functional Thread Tools \| \| \| \| \|------------------------------------------------\|-------------------------------\|---------\| \| Similar Threads for: Magnetic potential energy \| \| \| \| Thread \| Forum \| Replies \| \| \| Introductory Physics Homework \| 2 \| \| \| Introductory Physics Homework \| 0 \| \| \| Introductory Physics Homework \| 9 \| \| \| Classical Physics \| 1 \| \| \| Introductory Physics Homework \| 1 \|	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.883061408996582, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/211799/uniqueness-of-induced-representation/211815	# uniqueness of induced representation I am studying the book "Representation Theory" by Fulton and Harris. And I just can not understand the part where they prove the uniqueness of induced representation. If someone could explain it I'd greatly appreciate it! It's on page 33 and it goes: Choose a representative $g_{\sigma} \in G$ for each coset $\sigma \in G/H,$ with $e$ representing the trivial coset $H$. To see the uniqueness, note that each element of $V$ has a unique expression $v = \sum g_{\sigma} w_{\sigma}$ for elements $w_{\sigma}$ in $W$. Given $g \in G$ write $g \centerdot g_{\sigma} = g_{\tau} \centerdot h$ for some $\tau \in G/H$ and $h \in H.$ Then we must have $$g \centerdot (g_{\sigma} w_{\sigma})= g_{\tau}( h w_{\sigma})\;.$$ This proves The uniqueness ... I understand everything until the end, but I just don't understand how this proves the uniqueness... If someone could give me a little more explanation, I would appreciate it! Thanks! - ## 2 Answers The question of uniqueness is whether we have any freedom in defining the action of an arbitrary element $g$ on $W$. The equation shows that we don't: The action of $g$ on each summand of $W$, and thus on $W$, is entirely determined by the action of $H$ on $V$. Writing out the action for a linear combination and denoting $g_\tau$ by $g_{\sigma g}$ and $h$ by $h_g$ to mark the dependencies, we have $$g\sum g_\sigma w_\sigma=\sum g(g_\sigma w_\sigma)=\sum g_{\sigma g}(h_gw_\sigma)\;,$$ and this fully determines the action of $g$ on any element of $V$. - Thank you for your reply. I am still a little confused because you say this determines the action of g on any element of V and is entirely determined by the action of H on V. But what about the $g_{\sigma g}$ that appear in the summands? Isn't is dependent on them too? Sincerely. – J Kasahara Oct 16 '12 at 15:24 Given a group action $H \mapsto \mathrm{GL}(W)$, they want to extend to a group action $G \mapsto \mathrm{GL}\left( \sum_{\sigma \in G/H} \sigma W \right)$ acting on the direct sum over coset spaces. Is the action of $g \in G$ well-defined on any element, $v = \sum g_{\sigma} w_{\sigma}$ ? Fulton's identity says we can always find a coset representative $g \centerdot g_{\sigma} = g_{\tau} h \in g_\tau H$. The coset is unique but the coset representative is only unique up to conjugacy by elements of $H$. The action of $g$ on $g_\sigma w_\sigma$ decomposes the action of $h: W \mapsto W$ and the action of $g_\tau$ permuting the various coset spaces $W_\sigma$. -	{"extraction_info": {"found_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": 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": 0.9216176867485046, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/108624/constructing-the-lebesgue-stieltjes-integral-as-a-functional-on-c1-a-b	# Constructing the Lebesgue-Stieltjes Integral as a Functional on $C^1 ([a,b])$ This is based on an approach to a homework problem from last year that I discarded but would like to fill in the details of now. Unfortunately I can't see the necessary inequality. Suppose we are given a function $g$ of bounded variation on $[a,b]$. I would like to construct the Lebesgue-Stieltjes integral with respect to $g$ by defining a bounded linear functional on $C^1([a,b])$ by $\varphi \mapsto -\int_{\mathbb{R}} g(x) \varphi^{'}(x) dx$ then extending it to $C([a,b])$ by density and then applying the Riesz Representation Theorem. To do this, however, I need to show that $\|\int g(x) \varphi^{'}(x) dx\|\leq C\|\|\varphi\|\|_\infty$ (notice that there is no derivative here--this is the condition to make it a bounded functional on $C^0([a,b])$) which is where the trouble comes in since I am not sure how to get such a bound without integrating by parts (which we can't do since $g$ need not be absolutely continuous). I'm sure there is a nice way to handle this but I just don't see it. Thanks for the help. - – William Feb 12 '12 at 20:55 Can you elaborate a bit? I am not sure how to get a bound using the infinity norm of $\varphi$ using Holder's inequality on $g$ and $\varphi^{'}$. – Chris Janjigian Feb 12 '12 at 20:57 Holder's inequality holds for the conjugate pair $p = 1$ and $q = \infty$ (see the link I provided). Now, if $g\in L^1$, for any $\psi\in L^\infty$, you have: $\|\int g(x)\psi(x)dx\|\leq \int\|g(x)\psi(x)\|dx\leq \\|\psi\\|_\infty\int\|g(x)\|dx$. Now set $C = \int\|g(x)\|dx = \\|g\\|_{1}$. – William Feb 12 '12 at 21:00 Right but I need the bound in terms of the infinity norm of $\varphi$ and we have the derivative of $\varphi$ in the integral here. What you are saying gives me a bound of $\|\|g\|\|_1\|\|\varphi^{'}\|\|_\infty$, not $C\|\|\varphi\|\|_\infty$ which is what is necessary for this to be a bounded linear functional on $C^0$. – Chris Janjigian Feb 12 '12 at 21:01 – user21436 Feb 12 '12 at 21:01 show 3 more comments ## 1 Answer Since you are working on $C^1([a,b])$, you can make use of the usual Riemann-Stieltjes integral to see that $$\begin{align} \int_a^b \phi(x) \, dg(x) &= \left[\phi(b)g(b) - \phi(a)g(a)\right] - \int_a^b g(x) \, d\phi(x) \\ &= \left[\phi(b)g(b) - \phi(a)g(a)\right] - \int_a^b g(x)\phi'(x) \, dx \end{align}$$ And using $\left\|\int_a^b \phi(x) \, dg(x) \right\|\le \\|\phi\\|_{\infty} \; V_a^b(g)$ (where $V_a^b(g)$ denotes the total variation of $g$), we obtain $$\left\|\int_a^b g(x)\phi'(x) \, dx\right\| \le \left[\|g(a)\|+\|g(b)\|+V_a^b(g)\right] \, \\| \phi\\|_\infty = C\cdot \\|\phi\\|_\infty$$ So your functional is indeed continuous (and the Lebesgue-Stieltjes integral is an extension of the Riemann-Stieltjes integral). In case you are not familiar with the Riemann-Stieltjes integral. The first equality can be seen as follows: Given a partition $P = \{x_0,x_1, \dots, x_N\}$ of $[a,b]$; $a= x_0 < x_1 <\dots < x_N=b$ and $\xi_i \in [x_{i-1}, x_{i}]$, we can write $$\begin{align} \sum_{n=1}^N f(\xi_i) [g(x_i) - g(x_{i-1})] &= f(\xi_{N+1})g(b) - f(\xi_1)g(a) - \sum_{n=1}^N \left[f(\xi_{i+1}) - f(\xi_i)\right] g(x_{i}) \\ &= f(\xi_{N+1})g(b) - f(\xi_1)g(a) - \sum_{n=1}^N \frac{f(\xi_{i+1}) - f(\xi_i)}{\xi_{i+1} - \xi_i} g(x_{i}) \left[\xi_{i+1} - \xi_i\right] \end{align}$$ setting $\xi_{N+1} = b$. Note that $$\left\|\sum_{n=1}^N f(\xi_i) \left[g(x_i) - g(x_{i-1})\right]\right\| \le \sum_{n=1}^N \left\|f(\xi_i)\right\| \, \left\|g(x_i) - g(x_{i-1})\right\|\le \\|f\\|_\infty \, V_a^b(g)$$ for any partition $P$. So if $f \in C^1([a,b])$ and $g$ is of bounded variation, then both are Riemann-integrable and by taking the limit $\|P\|\to 0$ (finesse of the partition) we will obtain the estimate $$\left\| f(b)g(b) - f(a)g(a) - \int_a^b f'(x) g(x) \, dx \right\| \le \\|f\\|_\infty \, V_a^b(g)$$ - I don't think $f$ as defined here has to be absolutely continuous. Can you explain why the Cantor function is not a counterexample here? – Chris Janjigian Feb 13 '12 at 23:19 @Chris: You are absolutely right. I was under the impression that "bounded variation + continuity" implied "absolute continuity" (I haven't thought about Stieltjes integrals in a while, sorry). But this is not the case, as the Cantor function shows. – Sam Feb 14 '12 at 2:32 @Chris: I have corrected it now. – Sam Feb 14 '12 at 2:59 That is a very nice solution indeed, thank you! I should have added though that I was hoping to find a way to do this without going through the Riemann-Stieltjes integral. This originated as part of a question from Rudin Real and Complex which, as an intermediate step, includes the construction of the Lebesgue-Stieltjes integral. The way I solved it was to just define the Riemann-Stieltjes integral as a functional. The problem is that Rudin does not do this in R&C and so I am trying to figure out how to define a functional equivalent to Riemann-Stieltjes without using its existence. – Chris Janjigian Feb 14 '12 at 3:28 @Chris: Hmm, this seems not so easy. Have you already thought about first defining a map $g\mapsto \Lambda_g$ from some suitable dense subspace of BV (like $\{g = f+h\}$, where $f$ is absolutely continuous and $h$ is constant up to jumps, maybe) into the space of linear functionals on $C^1([a,b])$, e.g. by $\Lambda_g(\phi) = \int_a^b \phi(x) f'(x) \, dx + \sum_{x} \phi(x)[h(x+) - h(x-)]$ and proving that this (linear) mapping is continuous; so you could first extend $g\mapsto \Lambda_g$ to all of BV and then - for fixed $g$ of bounded variation - extend $\Lambda_g$ to all of $C^1([a,b])$. – Sam Feb 14 '12 at 4:51 show 4 more comments	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 48, "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": 0.945947527885437, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/223534/analyzing-a-function	# Analyzing a function I am having a problem with the following function: $f(x)=\sin^2(x)-\cos(3x)$ I need to examine the sign of $f'(x)$ I noticed that f is $2\pi$-periodic, therefore we need to analyze f(x) on $[0;2\pi]$ In addition to that $f(x)=-4\cos^3(x)-\cos^2(x)+3\cos(x)+1$ Hence $\forall x \in ]0;2\pi[, f'(x)=(-12\cos^2(x)-2\cos(x)+3)(-\sin(x))$ Let us solve $f'(x)=0$ We have $\forall x \in ]0;2\pi[ -\sin(x)=0 \Leftrightarrow x=\pi$ For the second equation, I know that there are two solution on $]0;2\pi[$ (since I visualized it on the graph), but I am unable to determine them by calculuation. Thank you in advance - ## 2 Answers You forgot to decrement the power of the $\cos^3x$ term. Once you fix that, you should have a quadratic you can solve for $\cos x$. - Ah ! Thank you!!! – user43418 Oct 29 '12 at 15:04 I have a question. I determined the 3 out of 5 zeros of $]0;2\pi[$ which are: $\pi$; $Arccos(\frac{-1-\sqrt{37}}{12})$ and $Arccos(\frac{\sqrt{37}-1 }{12})$ However there still exists 2 more which I am unable to determine. In addition to that I know that f' is $\pi$-periodic – user43418 Oct 29 '12 at 16:26 Can someone help me in order to determine the two additional values – user43418 Oct 29 '12 at 16:55 @user43418 $\cos(2\pi-x)=\cos(-x)=\cos x$. Is this what you're looking for? – Mike Oct 29 '12 at 18:05 $$\begin{align} f'(x)&=2 \cos x\sin x +3 \sin (3x) = 2 \cos x \sin x +3 (3 \sin x - 4 \sin^3 x) \\ &= \sin x \left( 2 \cos x + 9 -12 \sin^2 x \right) = \sin x \left( 2 \cos x + 9 -12 (1-\cos^2 x) \right) \end{align}$$ so that you have $\sin x=0$ and $12 \cos^2 +2 \cos x -3=0$. -	{"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": 19, "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": 0.9435010552406311, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/22648/spinor-integration	# Spinor integration I am learning on-shell methods for one loop integrals from this paper: Loop amplitudes in gauge theory: modern analytic approaches by Britto. Starting with formula (18) spinor integration is explained. One first writes the loop momentum as $$(l)_{a\dot{a}}=t\lambda_a\tilde{\lambda}_\dot{a},$$ where $t\in \mathbb{R}^+$ and $\lambda$, $\tilde{\lambda}$ are spinors with $\tilde{\lambda}=\bar{\lambda}$. Why can I write $l$ in this way? I know the decomposition $p=\lambda\tilde{\lambda}$ for momenta with $p^2=0$, but the loop momentum is not lightlike. I guess the Dirac-Delta in the integrand puts it on shell, but then where does the $t$ come from? Next the integral over the loop momentum is turned into an integral over the spinors: $$\int{d^4l\delta^{(+)}(l^2)f(l)}=-\int_0^\infty{\frac{t}{4}dt\int_{\tilde{\lambda}=\bar{\lambda}}{\langle\lambda \; d\lambda\rangle} [\tilde{\lambda} \; d\tilde{\lambda}] f(t,\lambda,\tilde{\lambda})}.$$ Can someone give the derivation of this formula or give me a reference where this is done? Britto refers to the paper MHV Vertices And Tree Amplitudes In Gauge Theory by Cachazo, Svrcek and Witten. But there I can't find an explanation that I understand. - ## 1 Answer The decomposition $p=\lambda \tilde\lambda$ is only valid for null vectors, indeed. In loop integrals, the loop momentum may generally be off-shell but because of the delta function in these integrals, only the null values of the momentum contribute, so it's enough to deal with the null momenta and they can be factorized to the spinors in this way. The variable $t$ is just a normalization factor that must be allowed to be arbitrary, real and positive, because $\lambda$ and $\tilde\lambda$ are going to play the role of homogeneous coordinates so one isn't allowed to distinguish them from their multiples. The integral is just a simple change of variables on the future light cone. The right references for these integrals are mentioned in the paper you quoted, it's [28] and [29]. http://arxiv.org/abs/hep-ph/0503132 http://arxiv.org/abs/hep-ph/0602178 - Thank you very much for your answer. This clears up my confusion about $l=t\lambda\tilde{\lambda}$. It would be very nice if you could explain how the change of variables in the integral works. That is where I am still stuck and I think it isn't explained in the references either (maybe because it is too trivial). – Johannes Mar 21 '12 at 18:54 Hi! It may be done by some straightforward algebra but it's also good to notice that both sides have the same support - they're integrals over the future light-cone - and in both cases, the delta-function imposing the support to the light cone combined with the Lorentz invariance uniquely determines the measure up to the normalization (the spinor products are Lorentz invariant much like delta of $l^2$) so they have to be equal up to the normalization one has to check (or believe). The normalization may be checked either by boring algebra or by integrating a particular example function. – Luboš Motl Mar 23 '12 at 15:37	{"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": 14, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 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": 0.9227390885353088, "perplexity_flag": "head"}
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http://mathoverflow.net/questions/61784?sort=oldest	## cohomology of BG, G compact Lie group ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) It has been stated in several papers that $H^{odd}(BG,\mathbb{R})=0$ for compact Lie group $G$. However, I've still not found a proof of this. I believe that the proof is as follows: --> $G$ compact $\Rightarrow$ it has a maximal toral subgroup, say $T$ --> the inclusion $T\hookrightarrow G$ induces inclusion $H^k(BG,\mathbb{R})\hookrightarrow H^k(BT,\mathbb{R})$ --> $H^(BT,\mathbb{R})\cong \mathbb{R}[c_1,...,c_n]$ where the $c_i$'s are Chern classes of degree $\deg(c_k)=2k$ --> Thus, any polys in $\mathbb{R}[c_1,...,c_n]$ are necessarily of even degree. Hence, $H^{odd}(BG,\mathbb{R})=0$ Is this the correct reasoning? Could someone fill in the gaps; i.e., give a formal proof of this statement? - 3 Why is this tagged mathematical-physics? – Andy Putman Apr 15 2011 at 5:48 @Andy - perhaps it is the direction the OP is coming from... which doesn't make it right however. – David Roberts Apr 15 2011 at 6:34 Yes, sorry about tagging it mp. I was thinking about where this surfaces - which happened to be some papers on Chern-Simons theory. – Kevin Wray Apr 15 2011 at 6:43 For the groups $U(n), SU(n) and Sp(n)$ the result follows from calculating the integral cohomology rings and seeing that they are polynomial rings on even degree generators. This is 4D.4 in Hatcher. But I believe the OP is thinking of the approach outlined in e.g. the first paragraph of arxiv.org/abs/0903.4865 - this is due to Borel I believe. – David Roberts Apr 15 2011 at 6:45 Retagged . – David Roberts Apr 15 2011 at 6:46 show 4 more comments ## 5 Answers For $G$ compact (and connected), $H(BG,\mathbb{R})$ is the $G$-equivariant cohomology of a point. It can be computed via infinitesimal methods and it is isomorphic to the cohomology of the Weil algebra of the Lie algebra of $G$. However there is an equivalent model computing the same cohomology, known as the Cartan model. The grading in the Cartan model is such that all cochains have even degree. Since you tagged this "mathematical physics", I should add that the Weyl algebra is the dga generated by the connection one-form (i.e., gauge field) on the universal $G$-bundle and the point behind the Cartan model is that the only appearance of the gauge field in an equivariant cocycle is via "minimal coupling" or via the curvature. There's nothing in a point for the gauge field to couple minimally to, hence all you get are curvatures. Being 2-forms all polynomial expressions in the curvatures have even degree. - ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. If you know a bit of Algebraic Topology (in particular the dreaded spectral sequences), the following is a nice way to see this. The well known Hopf Theorem states that for $G$ a compact connected Lie group the real cohomology $H^(G;\mathbb{R})\cong\wedge(y_1,\ldots , y_r)$ is an external algebra on odd dimensional generators $y_i$. This is proved using the Hopf algebra structure on $H^(G;\mathbb{R})$. A good reference is Chapter 1 of the book Algebraic Models in Geometry by Félix, Oprea and Tanré (which I believe also discusses the approach mentioned in José's answer). Since $BG$ is simply-connected, it is a nice exercise using the Serre spectral sequence of the universal $G$-bundle $G\to EG\to BG$ to see that $H^(BG;\mathbb{R})\cong\mathbb{R}[x_1,\ldots x_r]$ is a polynomial algebra on even dimensional generators $x_i$. - 1 See also McCleary's User's guide to spectral sequences, Theorem 7.30 for a result attributed to Borel: If $G$ is a topological group with $H_\ast(G;k) = \Lambda(x_1,...,x_r)$ where $deg(x_i)$ is odd for all $i$ and $k$ is a field, then $H^\ast(BG;k) = k[y_1,...,y_r]$ with $deg(y_i) = deg(x_i) + 1.$ – Ralph Apr 15 2011 at 10:33 For a reference see "Hsiang: Cohomology theory of topological transformation groups" (chapter III, §1). The results of the book that are relevant for your question can also be found in the following paper: http://www.math.uwo.ca/~rgonzal3/qfy.pdf (cf. Remark 9, Lemma 5). The idea is roughly as follows: Let $G$ be a compact Lie group and $G_0$ be the connecting component of the identity element. Then $BG_0 \to BG$ is a covering and $$H^\ast(BG;\mathbb{R}) = H^(BG_0;\mathbb{R})^{\Gamma}$$ where $\Gamma = G/G_0$ is a finite group. Let $T$ be a maximal torus of $G_0$. Then one shows that $$H^\ast(BG_0;\mathbb{R}) = H^\ast(BT;\mathbb{R})^W$$ where $W$ is the Weyl group of $G_0$. Thus $H^\ast(BG;\mathbb{R})$ can be identified with a subring of $H^\ast(BT;\mathbb{R})$ that is a polynomial ring with generators of degree two. So $H^{odd}(BG;\mathbb{R}) = 0$ follows. - 1 I added the missing asterisks: it is safer to use `\ast` than ``. – José Figueroa-O'Farrill Apr 15 2011 at 9:38 Good to know. José, thanks. – Ralph Apr 15 2011 at 9:49 The reference I know for the general fact $H^\ast(X/\Gamma;{\Bbb Q}) = H^\ast(X;{\Bbb Q})^\Gamma$ (for $\Gamma$ a finite group acting freely) is Grothendieck's Tohoku paper, which of course uses a lot of spectral sequences. Grothendieck remarks that this can be proved much more easily, presumably along the lines Allen K. indicates. Does anyone know an earlier/easier reference? (Cf. the answer here: mathoverflow.net/questions/18898/… ) – Dave Anderson Apr 17 2011 at 5:47 2 @ Dave Anderson: The isomorphism $H^\ast(X/\Gamma;{\Bbb Q}) = H^\ast(X;{\Bbb Q})^\Gamma$ is a simple application of the transfer homomorphism. For a textbook proof see my algebraic topology book, Proposition 3G.1, page 321. – Allen Hatcher Apr 17 2011 at 14:40 @Allen, thanks! – Dave Anderson Apr 17 2011 at 15:45 Recall the Chern-Weil homomorphism $Sym^{\ast} \mathfrak{g}^{\vee} \to H^{\ast}(BG; \mathbb{R})$ for each Lie group $G$. If $G$ is compact, it is an isomorphism. See Dupont, Curvature and Characteristic classes (a great book). This includes the desired statement, because the CW-homomorphism doubles the degree. Here is a sketch, taken from that book. One step is fairly easy, namely that $H^{\ast}(BG) \to H^{\ast} (BT)$ is injective: There is a fibre bundle $f:BT \to BG$ with fibre $G/T$. Let $\chi$ be the Euler class of the vertical tangent bundle of $BT$. Then the transfer $\tau_{f}: H^{\ast}(BT) \to H^{\ast}(BG)$ is defined as $x \mapsto f_{!} (x \chi)$. It is not hard to see (exercise, use $f_{!} (f^{\ast} x \cdot y) =x f_{!} (y)$ that $\tau_f \circ f^{\ast}: H^{\ast} (BG) \to H^{\ast} (BG)$ is the multiplication by Euler number of $G/T$. It is a classical result that the Euler number of $G/T$ is the order of the Weyl group, in particular positive, in particular nonzero. Hence $f^{\ast}: H^{\ast}(BG) \to H^{\ast}(BT)$ is injective. Moreover, the Chern-Weil homomorphism is an iso for a torus by a direct computation (which ultimately boils down to the computation $\frac{1}{2 \pi i} \int_{S^1} \frac{dz}{z}=1$). To prove that $f^{\ast}$ is surjective on the $W$-invariants is harder. It remains to show (write down a diagram) that the restriction $Sym^{\ast} (\mathfrak{g}^{\vee}) \to Sym^{\ast} \mathfrak{t}^{W}$ is an isomorphism. This is a theorem by Chevalley and uses quite a bit of structure theory of Lie groups. - Yes, I like your argument for injectivity and this is all that I need to show $H^{odd}(BG,\mathbb{R})=0$. Thanks. – Kevin Wray Apr 17 2011 at 6:33 Here's the argument I know that avoids spectral sequences, based on the little-known space $G/N(T)$. In between $T$ and $G$ is $N(T)$. Note that $EG$ "is an" $ET$ and $EN(T)$, since it's contractible and $T,N(T)$ act freely on it, so we can identify $BT, BN(T)$ with $EG/T, EG/N(T)$. Now consider the two maps $EG/T \to EG/N(T) \to EG/G$, with fibers $W = N(T)/T$ and $G/N(T)$ respectively. The first case divides by a free action of $W$, so we can identify $H^\ast(BN(T);{\mathbb Q}) = H^\ast(BT; {\mathbb Q})^W$ by pushing and pulling. (Actually we only need to invert $\|W\|$, and generally less; for $G=U(n)$ it's true over $\mathbb Z$.) In particular, there is only even cohomology. So let's look at the space $G/N(T) = (G/T)/W$. The space $G/T$ has a Bruhat decomposition, hence only even-degree cohomology (even over $\mathbb Z$), which you can prove via Morse theory on a generic adjoint orbit if you don't want to bring in algebraic geometry, and its Euler characteristic is $\|W\|$. Hence the space $(G/T)/W$ has (rationally) only even-degree cohomology, and Euler characteristic $1$. So it has the rational cohomology of a point! For $G=SU(2)$ this space is ${\mathbb RP}^2$. By a particularly trivial application of Leray-Hirsch (which I think is the only remainder of the spectral sequence argument Mark Grant gave), $H^\ast(EG/G; {\mathbb Q}) \cong H^\ast(EG/N(T); {\mathbb Q})$. (Oops: I guess this answer isn't so different from Ralph's.) - Allen, I can see why push-pull gives an inclusion of $H^\ast(BN(T);{\Bbb Q})$ in $H^\ast(BT;{\Bbb Q})^W$, but why is the map surjective? Am I missing something obvious? This almost proves the general fact about finite quotients from Grothendieck's Tohoku paper, and I'd like to know the "easy" proof! – Dave Anderson Apr 17 2011 at 5:54 Isn't pull-push just multiplication by $\|W\|$? – Allen Knutson Apr 17 2011 at 12:40 I like this argument. If $p:Y \to X$ is a finite Galois covering of manifolds with group $W$, then $H^{\ast} (X) \to H^{\ast}(Y)^W$ is an isomorphism (with real coeff.). Injectivity is ''pull-push''. Surjectivity: if a form $\omega$ represents and element in $H^{\ast}(Y)^G$, then $\frac{1}{\|W\|} \sum_g g^{\ast} \omega$ represents the same class and is invariant. But invariant forms on $Y$ descend to forms on $X$. $BG$ and the likes are not manifolds, but one can find Hilbert manifold models, where de Rham theory works well. – Johannes Ebert Apr 17 2011 at 14:06	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 107, "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": 0.9274453520774841, "perplexity_flag": "head"}
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http://mathoverflow.net/questions/10971?sort=newest	## Why isn’t Likelihood a Probability Density Function? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Hi everyone, first post here... I've been trying to get my head around why a likelihood isn't a probability density function. My understanding says that for an event X and a model parameter m: P(X\|m) is a probability density function P(m\|X) is not It feels like it should be, and I can't find a clear explanation of why it's not. Does it also mean that a Likelihood can take a value greater than 1? Sorry if this is a stupid question! - ## 7 Answers If $X$ is data and $m$ are the parameters, then the likelihood function $l(m) = p(X \| m)$. I.e. it's $p(X \| m)$, considered as a function of $m$. Both $p(X\|m)$ and $p(m\|X)$ are pdfs: $p(X\|m)$ is a density on $X$ and $p(m\|X)$ is a density on $m$. But the likelihood is $p(X\|m)$, not as a function of $X$ (it would indeed be a density as a function of X), but as a function of m. So it's not a pdf; in particular, it's not necessarily true that $$\sum_m p(X\|m) = 1.$$ Edit: just to clarify, $p(m\|X)$ isn't the likelihood. $p(X\|m)$ is. - Some points in this answer are wrong. Look at my answer. – Stéphane Laurent Jun 29 at 6:29 ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. The accepted answer is wrong. The likelihood is the function $\theta \mapsto L(\theta \mid x)=f(x \mid \theta)$ for a given $x$ in the observations space and $f(\cdot \mid \theta)$ is a Radon-Nikodym derivative of $P_\theta$ when the statistical model is given by a family of probabilities ${(P_\theta)}_{\theta\in\Theta}$ on the observations space. In general there's not even a $\sigma$-field in the parameter space $\Theta$, hence the question "is the likelihood a pdf ?" has not even a sense! For more information see http://stats.stackexchange.com/questions/31238/what-is-the-reason-that-a-likelihood-function-is-not-a-pdf and http://stats.stackexchange.com/questions/29682/how-to-rigorously-define-the-likelihood - Could you elaborate why the answer is wrong and what the Radon-Nikodym derivative has to do with it? It seems to me that first link you provided says the same as the accepted answer. – fabee Jun 30 at 20:31 mm difficult to elaborate more! The Radon-Nikodym derivative (=density) is nothing but the definition. See the third comment below the question of the first link, it precisely says that the present accepted answer is wrong. – Stéphane Laurent Jul 5 at 10:53 Likelihood is the chance that the reality you've hypothesized could have produced the particular data you got. Likelihood: The probability of data given a hypothesis. However Probability is the chance that the reality you're considering is true, given the data you have. Bayesian Probability: Probability of a hypothesis given data - You are essentially asking why likelihood density function cannot be used as probability density function. For a non-informative (uniform) Bayesian prior, the likelihood density function divided by its integral over entire space will give the probability density function. It is the normalization that makes the two different. - If m is a model parameter then it is not a RV. So it does not have a probability density function. m is just an unknown paramter. - A few questions were asked, so a few answers will be given. (main point: likelihood is not necessarily a product density, though this is the common interpretation.) Frequently, the likelihood is the product of densities over some provided set of examples. The examples are drawn i.i.d., and therefore this product density is the density for the corresponding product measure over the product space. What I'm saying is that yes, from this perspective, you have constructed a product density. Since you are dealing with densities, not probabilities, values are not constrained to [0,1], and your density can easily be greater than one. In fact, if you are dealing with dirac measure (which puts all mass on one point on the real line), you essentially have "infinite" density. I put that in quotes since this is not a continuous probability measure, ie it does not have a density wrt to Lebesgue measure, let alone one with infinite mass on a point. (A quick fact check: the corresponding integral wrt lebesgue measure would have value zero since it is off zero only on a set of lebesgue measure zero, which means it is not a probability distribution; but it was, which contradicts this being its density.) perhaps a more apt example: any (continuous) distribution on [0,0.5] will have to have density greater than one on a set of nonzero lebesgue measure. (you can try to construct a sequence of these which convergence to something which violates what i said, but that will be the density of something which is not continuous!) things can get a little confusing because you can write discrete probability distributions as densities wrt a measure putting 1 on each point in the support set of the probability (ie it is counting measure wrt that set). NOTE that this is a density wrt a measure which is NOT a probability measure. But anyway, the density values at each point are exactly the probability values. This allows an interchanging probability masses and densities, which can be confusing. I'll close with some further reading. A good book on machine learning is "A probabilistic Theory of Pattern Recognition" by Devroye, Gyorfi, Lugosi. Chapter 15 is on maximum likelihood and you'll notice they do NOT define likelihood as being a product probability or density, but rather as a product of functions. This is because they are careful to encompass the differing interpretations; rather, they ignore the interpretations there and work out the math. - From a Bayesian perspective, the reason the likelihood function isn't a probability density is that you haven't multiplied by a prior yet. But once you multiply by a prior distribution, the product is (proportional to) the posterior probability density for the parameters. -	{"extraction_info": {"found_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": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 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": 0.949639081954956, "perplexity_flag": "head"}
http://quant.stackexchange.com/questions/2763/one-dimensional-analog-of-cleansing-a-correlation-matrix-via-random-matrix-theor?answertab=oldest	# One dimensional analog of cleansing a correlation matrix via random matrix theory The general idea of cleansing a correlation matrix via random matrix theory is to compare its eigenvalues to that of a random one to see which parts of it are beyond normal randomness. These are then filtered out and one is left with the non-random parts. A one-dimensional analog would be to compare the return density distribution of a stock to the normal distribution and filter its normal volatility out. One should be left with the non-random distribution of that stock. I have two questions: Is my reasoning, i.e. the intuition behind using RMT and my analogy, correct and if yes, is something like the one-dimensional case done (or would it make sense to do it)? - ## 1 Answer This is correct: "The general idea of cleansing a correlation matrix via random matrix theory is to compare its eigenvalues to that of a random one to see which parts of it are beyond normal randomness." This is not correct: "These are then filtered out and one is left with the non-random parts." The term "filtering", although used extensively in the literature, is misleading because the eigenvectors from the original correlation (or covariance matrix) remain part of the matrix and the sum of the eigenvalues does not change. In all RMT filtering procedures, the matrix is decomposed and re-built via the eigenvector decomposition theorem: $$correlation \;\; matrix = eigenvectors * diag( eigenvalues ) * t(eigenvectors)$$ Notation: t() is the transpose operator, * is matrix multiplication, and diag() is an operator that creates a diagonal matrix In RMT we only perform operations on the diagonal matrix of eigenvalues resulting from the decomposed matrix. The eigenvectors are left untouched. Also, the sum of the eigenvalues is preserved pre-and post cleansing. Since the sum of the eigenvalues also equals the trace of the covariance matrix, this ensures that the variance of the system is preserved. To recap, all RMT procedures follow this four-step process: 1. The first step is, as you point out, identifying the upper noise band of eigenvalues predicted by a de-meaned random matrix of the same sigma using the Marcenko-Pastur law. Matrices that are exponentially weighted use a Power-law. 2. The matrix is decomposed via the eigenvector decomposition theorem 3. The diagonal of eigenvalues is cleansed. For example, the method of Laloux (1999, 2000) is to assign the eigenvalues for all "noisy" eigenvalues beneath the upper-noise band to the average of all such noisy eigenvalues. There are several variations in what new eigenvalues you assign to the noisy eigenvalues (power-law method, Krzanowski, etc.). Regardless - in all RMT methods the sum of all the eigenvalues is preserved post-cleansing so you cannot set these to zero. 4. Now we re-build a "cleansed" or "filtered" covariance matrix using the same eigenvectors and the revised eigenvalue via the eigenvector decomposition theorem (in reverse this time with the same eigenvectors). Digression: You could come up with your own techniques so long as the constraints above are respected. For example, since we know that the principal/top eigenvectors have significant non-normal structure or skew, you could "filter" those eigenvalues that correspond to eigenvectors that exhibit normal structure. Also, for what it's worth, in my experience filtering methods that have a successively increasing distance from one eigenvalue value to the next perform better (i.e. for example a recursive definition such as the value of the 2nd smallest eigenvalue is twice the value of the smallest eigenvalue, the value of the 3rd smallest eigenvalues is twice the value of the 2nd smallest eigenvalue, etc. up to the last noisy eigenvalue). Also, one of the most accessible introductions to Random Matrix theory are the papers by V. Plerou et al. The Capital Fund Management Team - Bouchaud, Pafka, Potters, et al has more frontier research on RMT. - 1 – Ryogi Jan 11 '12 at 23:27	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8836294412612915, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/158617/about-the-wasserstein-metric	# About the Wasserstein “metric” I've just encountered the Wasserstein metric, and it doesn't seem obvious to me why this is in fact a metric on the space of measures of a given metric space $X$. Except for non-negativity and symmetry (which are obvious), I don't know how to proceed. Do you guys have any advices or links to useful references ? Thanks in advance ! Cyril - A useful material as mentioned in my answer is the book of Cedric Villani "Topics in optimal transportation". Also one good source is Luigi Ambrosio's and Nicola Gigli's "User's guide to optimal transportation". It deals alot of nice results and detailed treatment of this topic. For instance, it considers the cases when $X$ is one of the following: Polish, geodesic metric space, Riemannian manifold, geodesic space with Alexandrov curvature (positive or nonpositive), and it defines a Metric Ricci curvature for a compact geodesic space $X$ through the study of its associated Wasserstein space. – Thomas E. Jun 15 '12 at 12:38 ## 1 Answer So I assume that what puzzles you are the triangle-inequality and $W_{p}(\mu,\mu)=0$, where $W_{p}$ denotes the $p$-Wasserstein metric. Here's some preliminary information. I will denote $\Pi(\mu,\nu)$ the collection of all transference plans from $\mu$ and $\nu$, i.e. $\pi\in\Pi(\mu,\nu)$ iff $\mu$ is the first marginal of $\pi$ and $\nu$ is the second. This can also be expressed in form $\mu=(\mathrm{pr}_{1})_{\#}\pi$ and $\nu=(\mathrm{pr}_{2})_{\#}\pi$, where $\#$ denotes the push-forward. If $(X,d)$ is Polish then for every pair of probability measures $\mu,\nu$ there exists an optimal transference plan $\pi\in\Pi(\mu,\nu)$ so that $W_{p}(\mu,\nu)=\left(\int_{X\times X}d(x,y)^{p}\,d\pi(x,y)\right)^{\frac{1}{p}}$. The proof of this can be found in the book 'Topics in optimal transportation', Cedric Villani, 2003, and the key point consists of noting that $\Pi(\mu,\nu)$ is compact in the weak-convergence of measures (which is shown by using Prokhorov's theorem). Now to the metric itself. The triangle-inequality uses a so called "Gluing lemma" (also found in Villani's book). It states that if $\mu_{1},\mu_{2},\mu_{3}$ are Borel probability measures on $X$, and $\pi_{1,2}\in\Pi(\mu_{1},\mu_{2})$ and $\pi_{2,3}\in\Pi(\mu_{2},\mu_{3})$ are optimal transference plans, then there exists a Borel probability measure $\mu$ on $X^{3}$ with marginals $\pi_{1,2}$ to the left $X\times X$ and $\pi_{2,3}$ to the right $X\times X$. This measure in a sense glues together $\pi_{1,2}$ and $\pi_{2,3}$. It follows by a simple argument using the marginal properties of each measure that the marginal of $\mu$ to $X\times X$ (the first and third $X$) denoted by $\pi_{1,3}$ is a transference plan in $\Pi(\mu_{1},\mu_{3})$ (not necessarily optimal!) $()$. Using minkovski inequality of $L^{p}(X^{3},\mu)$ $()$, marginal properties of the measures $()$, optimality of $\pi_{1,2}$ and $\pi_{2,3}$ $(*)$, we obtain \begin{align} W_{p}(\mu_{1},\mu_{3}) &\overset{()}{\leq} \bigg(\int_{X\times X}d(x,z)^{p}\,d\pi_{1,3}(x,z)\bigg)^{\frac{1}{p}}\overset{()}{=}\bigg(\int_{X\times X\times X}d(x,z)^{p}\,d\mu(x,y,z)\bigg)^{\frac{1}{p}} \\ &\leq \bigg(\int_{X\times X\times X}(d(x,y)+d(y,z))^{p}\,d\mu(x,y,z)\bigg)^{\frac{1}{p}} \\ &\overset{()}{\leq}\bigg(\int_{X\times X\times X}d(x,y)^{p}\,d\mu(x,y,z)\bigg)^{\frac{1}{p}}+\bigg(\int_{X\times X\times X}d(y,z)^{p}\,d\mu(x,y,z)\bigg)^{\frac{1}{p}} \\ &\overset{(*)}{=}\bigg(\int_{X\times X}d(x,y)^{p}\,d\pi_{1,2}(x,y)\bigg)^{\frac{1}{p}}+\bigg(\int_{X X\times X}d(y,z)^{p}\,d\pi_{2,3}(y,z)\bigg)^{\frac{1}{p}} \\ &\overset{(*)}{=}W_{p}(\mu_{1},\mu_{2})+W_{p}(\mu_{2},\mu_{3}). \end{align} So we have the triangle-inequality. About the $W_{p}(\mu,\mu)=0$, take the homeomorphism $f:X\to\Delta$ given by $x\mapsto(x,x)$, i.e. $\Delta$ is the "diagonal" of $X\times X$. Then take $\nu:=f_{\#}\mu$ (which is a Borel probability measure on the diagonal $\Delta$) and furthermore define a Borel probability measure $\pi$ on the product space $X\times X$ by setting $\pi(A)=\nu(A\cap\Delta)$ for all Borel sets $A$. Now $\pi$ is a transference plan between $\mu$ to itself (not necessarily optimal!), which is a straight-forward proof, and it vanishes outside the diagonal (i.e. $\pi(\Delta^{c})=0$). Since the diagonal is the zero set of the metric $d$, we conclude that \begin{equation} W_{p}(\mu,\mu)^{p}\leq \int_{X\times X}d(x,y)^{p}\,\pi(x,y)=\int_{\Delta}d(x,y)^{p}\,\pi(x,y)+\int_{\Delta^{c}}d(x,y)^{p}\,\pi(x,y)=0+0=0, \end{equation} whence $W_{p}(\mu,\mu)=0$. - By the way, here's some terminology explained: a $\it{transference\,\,plan}$ from $\mu$ to $\nu$ is a member of $\Pi(\mu,\nu)$. Intuitively, $d\pi(x,y)$ measures the amount of mass that $\pi$ transfers from $x$ to $y$ and $d(x,y)^{p}$ is the cost function. An optimal plan is such $\pi$ for which the infimum is reached in the definition of $W_{p}$, and for every other transference plan we have an inequality $\leq$. As mentioned above, given that $X$ is Polish guarantees the existence of optimal transference plans between any pair of Borel prob. measures. – Thomas E. Jun 15 '12 at 13:53 Thanks, that's much clearer ! I will get Villani's book to see what he says about that – Cyril Benezet Jun 15 '12 at 17:34	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 74, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 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": 0.8713982105255127, "perplexity_flag": "head"}
http://mathoverflow.net/questions/102697?sort=newest	$H$-Hopf modules equal the tensor products of their coinvariants with H Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) In a comment for this old question, it was said that > There is a theorem that Hopf modules are (up to isomorphism) the tensor products of their coinvariants with H. (This theorem is usually worded in terms of a category equivalence. I don't know a good reference.) I'm guessing that this means that given a (right) $H$-comodule $V$, for $H$ a Hopf algebra, equipped with a right $H$ action for which $\Delta_R(vh) = v_{(0)} g_{(0)} \otimes v_{(1)}g_{(1)}$, we have an isomorphism $$V \simeq V_{\text{inv}} \otimes H$$ Now it's easy to see that we have a surjective map $$V_{\text{inv}} \otimes H \to V, ~~~ v \otimes h \mapsto vh.$$ How does one show that this is an isomorphism? EDIT: Wait, I think this is actually obvious: The map $$V \to V_{\text{inv}} \otimes H, ~~~~~ v \mapsto v_{(0)} S(v_{(1)}) \otimes v_{(2)},$$ seems to have the multiplication map as its inverse. So this gives us the isomorphism. Yes? - 1 A good reference is Sweedler's book on Hopf algebras. – Mariano Suárez-Alvarez Jul 19 at 20:55 Havce you tried computing the two compositions of the maps you wrote in the question and seeing if they are identities? – Mariano Suárez-Alvarez Jul 19 at 20:57 1 Answer The result you mention is a classical result on Hopf modules, first proved by Larson and Sweedler. My favorite reference is Pareigis: When Hopf Algebras are Frobenius Algebras. J. of Alg. 18(1971), 588-596. Lemma 2. There you can also find the maps you describe in your question (so the answer is yes, they are inverse to each other). Another reference is Sweedler's book (mentioned already by Mariano), Theorem 4.1.1. However, I think in the book the base ring is always a field, while Pareigis works over a comm. ring. Somewhere in the book "Brzezinski, Wisbauer: Corings and Comodules" I read that the result in question also follows from a more general theorem on comodules over corings. But I don't know details. -	{"extraction_info": {"found_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": 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": 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": 0.9396603107452393, "perplexity_flag": "middle"}
http://www.cs.cmu.edu/~dsheehy/osu12new.html	Don Sheehy Postdoc Inria Saclay, France, don.r.sheehy@gmail.com A New Approach to Output-Sensitive Voronoi Diagrams and Delaunay Triangulations Ohio State University, Oct 31, 2012 Voronoi diagrams and their duals, Delaunay triangulations, are used in many areas of computing and the sciences. Starting in 3-dimensions, there is a substantial (i.e. polynomial) difference between the best case and the worst case complexity of these objects when starting with $n$ points. This motivates the search for algorithms that are output-senstiive rather than relying only on worst-case guarantees. In this talk, I will describe a simple, new algorithm for computing Voronoi diagrams in $d$-dimensions that runs in $O(f \log n \log \Delta)$ time, where $f$ is the output size and the spread $\Delta$ of the input points is the ratio of the diameter to the closest pair distance. For a wide range of inputs, this is the best known algorithm. The algorithm is novel in the that it turns the classic algorithm of Delaunay refinement for mesh generation on its head, working backwards from a quality mesh to the Delaunay triangulation of the input. Along the way, we will see instances of several other classic problems for which no higher-dimensional results are known, including kinetic convex hulls and splitting Delaunay triangulations.	{"extraction_info": {"found_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": 0.9062917232513428, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/73089/finite-sum-of-the-fractional-type	Finite sum of the fractional type Some special manipulations involving finite sums. How to solve this sum? $\displaystyle{\sum_{k=1}^{n}}\frac{1}{4k^2 - 1}$ - 1 Hint 1: the denominator can be factorised. Hint 2: Partial fractions. Hint 3: It telescopes. – ShreevatsaR Oct 16 '11 at 17:09 2 Answers Hint: Note that $$\frac{1}{4k^2-1}=\frac{\frac{1}{2}}{2k-1}-\frac{\frac{1}{2}}{2k+1}.$$ - The following problem below is for an infinite series, but if you can solve it you may be able to solve your problem above. 1. Let $a_n := b_{n} - b_{n+1}$, for some other sequence $b_n$. Prove that the series $\sum_{n =0}^{\infty} a_n$ converges iff the sequence $b_n$ does. 2. In the case that the sum converges, what is its sum? 3. Use (1) and (2) to show that $\sum_{k=0}^{\infty} \frac{1}{4k^2 - 1}$ converges and find its sum. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 6, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": 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": 0.9090678691864014, "perplexity_flag": "middle"}
http://www.physicsforums.com/showthread.php?p=3039031	Physics Forums Page 1 of 3 1 2 3 > Blog Entries: 4 Recognitions: Gold Member ## "Locally Lorentz" Mister Thorne Wheeler, "Gravitation" asks "What does it mean to say that the geometry of a sufficiently limited region of spacetime in the real physical world is Lorentzian?" The follow this up with two answers, neither of which appears to have much to do with the question. Instead, they give formulas to calculate proper time and proper distance. This barely scratches the surface of what it means to be Lorentz. Certainly, a body that takes an accelerated path through spacetime ages less than a body that takes an inertial path. But what MTW fails to do is to get across a deeper undestanding of the paradox involved, and more particularly the resolution to this paradox. In fact, I can't find a single example in MTW where they demonstrate any competent expanation or deeper understanding of Special Relativity. They seem to start and end with the notion that Special Relativity is completely summed up by one equation: $$s^2=-\tau^2=x^2+y^2+z^2-t^2$$ I have no problem with having one equation to sum up Special Relativity, I just think the chose the wrong one. As for me, I don't think that SR is summed up by the calculation of the space-time-interval between events. Instead, it is summed up by the Lorentz Transformation Equations. Let me offer a single example of what the Lorentz Transformations do. Then we can ask whether that operation represents a "global" or a "local" application of Lorentz, and whether it makes sense to say that physics is only "locally" Lorentz. Here is the example: I have a particle observer one foot away from a wall. Roughly 1 nanosecond ago, light came from the wall which is currently being observed by our observer. The space-time coordinates of this observer is (ct, x) = (-1, 1); that is, 1 second ago, and one foot away. Now, our observer undergoes a tremendous acceleration toward the wall of 0.99999999959c. (This corresponds to a change in rapidity of 10. tanh(10)=0.99999999959c. The event (-1,1) is transformed by Lorentz Transformation as [tex]\left( \begin{array}{cc} \cosh (\varphi ) & -\sinh (\varphi ) \\ -\sinh (\varphi ) & \cosh (\varphi ) \end{array} \right)= \left( \begin{array}{c} -22026 \\ 22026 \end{array} \right) [/tex] So this event, which happened only 1 foot away, now happened 22026 feet away; that is, over four miles away. In general with these large numbers, (velocities extremely close to c) it is a simple calculation, once you know the rapidity. If you have a rapidity of 100, then the multiple is 2cosh(100)=2.681043. With a rapidity change of 1000, the multiple is 2cosh(1000)=1.9710434 feet (6.3610414 light years.) So... Is this exampe local? The proper time between these two events was zero. The proper distance between these two events is zero. This would seem to be about as "local" as you can get. However, by applying the Lorentz Transformation to these two events, we were able to make them as far apart as we wanted in coordinate space and time. A google google google google times as big as the universe. However, the proper time, and proper distance between these events remains zero. So how can anyone justify even using the phrase "locally Lorentz" If any physics is Lorentz at all, it can be stretched over arbitrarily large swaths of spacetime--not local at all. PhysOrg.com science news on PhysOrg.com >> Front-row seats to climate change>> Attacking MRSA with metals from antibacterial clays>> New formula invented for microscope viewing, substitutes for federally controlled drug Blog Entries: 1 Recognitions: Gold Member Science Advisor Quote by JDoolin Mister Thorne Wheeler, "Gravitation" asks "What does it mean to say that the geometry of a sufficiently limited region of spacetime in the real physical world is Lorentzian?" The follow this up with two answers, neither of which appears to have much to do with the question. Instead, they give formulas to calculate proper time and proper distance. This barely scratches the surface of what it means to be Lorentz. Certainly, a body that takes an accelerated path through spacetime ages less than a body that takes an inertial path. But what MTW fails to do is to get across a deeper undestanding of the paradox involved, and more particularly the resolution to this paradox. In fact, I can't find a single example in MTW where they demonstrate any competent expanation or deeper understanding of Special Relativity. They seem to start and end with the notion that Special Relativity is completely summed up by one equation: $$s^2=-\tau^2=x^2+y^2+z^2-t^2$$ I have no problem with having one equation to sum up Special Relativity, I just think the chose the wrong one. As for me, I don't think that SR is summed up by the calculation of the space-time-interval between events. Instead, it is summed up by the Lorentz Transformation Equations. Let me offer a single example of what the Lorentz Transformations do. Then we can ask whether that operation represents a "global" or a "local" application of Lorentz, and whether it makes sense to say that physics is only "locally" Lorentz. Here is the example: I have a particle observer one foot away from a wall. Roughly 1 nanosecond ago, light came from the wall which is currently being observed by our observer. The space-time coordinates of this observer is (ct, x) = (-1, 1); that is, 1 second ago, and one foot away. Now, our observer undergoes a tremendous acceleration toward the wall of 0.99999999959c. (This corresponds to a change in rapidity of 10. tanh(10)=0.99999999959c. The event (-1,1) is transformed by Lorentz Transformation as $$\left( \begin{array}{cc} \cosh (\varphi ) & -\sinh (\varphi ) \\ -\sinh (\varphi ) & \cosh (\varphi ) \end{array} \right)= \left( \begin{array}{c} -22026 \\ 22026 \end{array} \right)$$ So this event, which happened only 1 foot away, now happened 22026 feet away; that is, over four miles away. In general with these large numbers, (velocities extremely close to c) it is a simple calculation, once you know the rapidity. If you have a rapidity of 100, then the multiple is 2cosh(100)=2.681043. With a rapidity change of 1000, the multiple is 2cosh(1000)=1.9710434 feet (6.3610414 light years.) So... Is this exampe local? The proper time between these two events was zero. The proper distance between these two events is zero. This would seem to be about as "local" as you can get. However, by applying the Lorentz Transformation to these two events, we were able to make them as far apart as we wanted in coordinate space and time. A google google google google times as big as the universe. However, the proper time, and proper distance between these events remains zero. So how can anyone justify even using the phrase "locally Lorentz" If any physics is Lorentz at all, it can be stretched over arbitrarily large swaths of spacetime--not local at all. You can say that the metric sums up everything because the complete Lorentz transform is derivable from the metric. I think the definition of locality you want to use is one phrased in terms of invariants: proper distance and proper time; also any region over which the the metric deviation from Minkowski is insignificant. In your example, despite the huge coordinate distance, the metric remains flat over this region, so it is locally Lorentz. (At least this is my understanding). Mentor Quote by PAllen You can say that the metric sums up everything because the complete Lorentz transform is derivable from the metric. I think the definition of locality you want to use is one phrased in terms of invariants: proper distance and proper time; also any region over which the the metric deviation from Minkowski is insignificant. In your example, despite the huge coordinate distance, the metric remains flat over this region, so it is locally Lorentz. (At least this is my understanding). It is my understanding too. Recognitions: Gold Member Science Advisor Staff Emeritus ## "Locally Lorentz" MTW isn't a book on SR, it's a book on GR. As PAllen points out, you can derive the Lorentz transformation from the metric or the metric from the metric from the Lorentz transformation. But in the context of GR, it's more natural to emphasize the metric, because a metric is what GR has. GR does not have frames of reference, except locally, so a transformation between frames of reference isn't a central concern. -Ben Blog Entries: 4 Recognitions: Gold Member Quote by bcrowell MTW isn't a book on SR, it's a book on GR. As PAllen points out, you can derive the Lorentz transformation from the metric or the metric from the metric from the Lorentz transformation. But in the context of GR, it's more natural to emphasize the metric, because a metric is what GR has. GR does not have frames of reference, except locally, so a transformation between frames of reference isn't a central concern. -Ben I know how to go from the Lorentz Transformation to the metric, but not the other way around. To go from the LT's to the metric, you must either select two events which (a) you could travel between; i.e. the d/c < t. or you can select two events (b) which you can't travel between, i.e. the distance/speed of light > t. In the first case, you apply the LT so that the two events are in the same position, but at different times. The time between those two events in this new reference frame is the proper time. In the second case, you apply the LT so that the two events are in different positions, but at the same time. The distance in this reference frame is the proper distance. (by the way going from lorentz transformation to the metric isn't so much derivation as definition.) There is of course, a third case, where the the distance/speed of light = 1. In which case, you cannot transform the events so they happen at the same time, or the same place. You can make them arbitrarily close in space, and arbitrarily close in time, but never zero. For this relationship, the "proper time" and "proper distance" are both zero. I'm interested to see you go the other direction, deriving the Lorentz Transformation from the definition of proper time. Blog Entries: 3 Recognitions: Gold Member It is possible to find the transformation that leaves the proper length unchanged. In 2 dimensions, $$\left[ \begin{array}{c} T' \\\ X' \end{array} \right]= \left[ \begin{array}{cc} a & b \\\ c & d\end{array} \right] \left[ \begin{array}{c} T \\\ X \end{array} \right]$$ $$\begin{align} ds^2 &= T'^2-X'^2=(aT+bX)^2-(cT+dX)^2\\ &= T^2(a^2-c^2)-X^2(b^2-d^2)+2XT(ab-cd) \end{align}$$ If $$a=d=cosh(r), \ \ c=d=sinh(r)$$ then $ds^2$ is invariant under the transformation. Quote by JDoolin Here is the example: I have a particle observer one foot away from a wall. Roughly 1 nanosecond ago, light came from the wall which is currently being observed by our observer. The space-time coordinates of this observer is (ct, x) = (-1, 1); that is, 1 second ago, and one foot away. Now, our observer undergoes a tremendous acceleration toward the wall of 0.99999999959c. (This corresponds to a change in rapidity of 10. tanh(10)=0.99999999959c. The event (-1,1) is transformed by Lorentz Transformation as $$\left( \begin{array}{cc} \cosh (\varphi ) & -\sinh (\varphi ) \\ -\sinh (\varphi ) & \cosh (\varphi ) \end{array} \right)= \left( \begin{array}{c} -22026 \\ 22026 \end{array} \right)$$ So this event, which happened only 1 foot away, now happened 22026 feet away; that is, over four miles away. In general with these large numbers, (velocities extremely close to c) it is a simple calculation, once you know the rapidity. If you have a rapidity of 100, then the multiple is 2cosh(100)=2.681043. With a rapidity change of 1000, the multiple is 2cosh(1000)=1.9710434 feet (6.3610414 light years.) So... Is this exampe local? The proper time between these two events was zero. The proper distance between these two events is zero. This would seem to be about as "local" as you can get. Do not mistake the location of the points on the manifold for the distance measure between those same points. Recognitions: Science Advisor Staff Emeritus You can think of it like this. Rotations are linear transformations that leave distances unchanged. The Lorentz transformation is a linear transformation that leaves the Lorentz interval unchanged. There are transforms other than Lorentz transforms that leave the Lorentz interval unchanged. These, however, are combinations of the Lorentz transform with standard spatial rotations. If you restrict yourself to one space and one time dimension, there is only the Lorentz transform. You might try reading "Space Time Physics" by Taylor & Wheeler for more backgorund, they explain this in more detail, taking more time to do it. But if you want to have a go at it yourself, it's not terribly hard to derive the Lorentz transforms from the invariance of the Lorentz interval. It's easiest if you choose units such that the speed of light, c, is equal to 1. You just need to look for a linear transformation x' = ax + bt t' = fx + gt such that x'^2 - t'^2 = x^2 - t^2 Expanding, you get (a^2 - f^2) x^2 + (2ab-2fg) xt + (b^2 - g^2) t^2 = x^2 - t^2 From this you conclude a^2 - f^2 = 1 , b^2 - g^2 = -1, and ab=fg. It's easy enough to confirm that the Lorentz transforms (if they look unfamiliar, it's because they're the Lorentz transforms in geometric units where c=1, so all factors of c have been omitted) satisfy this. a = gamma = 1/sqrt(1-v^2) b = vgamma = v/sqrt(1-v^2) f = vgamma = v/sqrt(1-v^2) g = gamma = 1/sqrt(1-v^2) It's a little more work (more than I care to do, and I'm not sure where to refer you to as a reference) that there aren't any other solutions that aren't equivalent to the above. (Replacing v by -v is something that falls in the category of equivalent, for instance). Recognitions: Science Advisor If the metric is everywhere diag(-1,1,1,1) then it is Minkowskian. However, in GR, the metric obeys the Einstein field equations, and is specified to have signature 2. So it is not diag(-1,1,1,1) everywhere. However, there are coordinates where the metric is diag(-1,1,1,1) at any particular point, and the first derivatives of the metric also vanish (but not second derivatives), and so the metric is said to be locally Lorentz. The deviation from local Lorentzianess as one goes away from the point is specified by terms in Taylor series, so we understand exactly how locally Lorentz it is or isn't. Quote by PAllen I think the definition of locality you want to use is one phrased in terms of invariants: proper distance and proper time Is there an invariant way of defining locality for lightlike directions? Blog Entries: 1 Recognitions: Gold Member Science Advisor Quote by Rasalhague Is there an invariant way of defining locality for lightlike directions? Not that I can think of, but even if you pick a coordinate system with lightlike basis vectors, you can discuss the size of region in terms of proper spacelike interval and proper timelike interval. Blog Entries: 4 Recognitions: Gold Member Quote by Passionflower Do not mistake the location of the points on the manifold for the distance measure between those same points. The only distances I know of are "proper distance," "proper time," and "distance." The last distance has no adjective, and it refers to the Euclidean, observer dependent distance. (It is observer dependent, because two observers with different rapidities will observe different distances.) The "distance measure" as you put it, is an ambiguous term, since relativistically traveling observers would measure different distances between the same two events. Of course, I am referring to the Euclidean, observer dependent distance, with the critera that the space is "locally lorentz," I applied a Lorentz Transform to a "local" space, and find that for all intents and purposes, local is global. Blog Entries: 4 Recognitions: Gold Member Quote by pervect You can think of it like this. Rotations are linear transformations that leave distances unchanged. The Lorentz transformation is a linear transformation that leaves the Lorentz interval unchanged. There are transforms other than Lorentz transforms that leave the Lorentz interval unchanged. These, however, are combinations of the Lorentz transform with standard spatial rotations. If you restrict yourself to one space and one time dimension, there is only the Lorentz transform. You might try reading "Space Time Physics" by Taylor & Wheeler for more backgorund, they explain this in more detail, taking more time to do it. But if you want to have a go at it yourself, it's not terribly hard to derive the Lorentz transforms from the invariance of the Lorentz interval. It's easiest if you choose units such that the speed of light, c, is equal to 1. You just need to look for a linear transformation x' = ax + bt t' = fx + gt such that x'^2 - t'^2 = x^2 - t^2 Expanding, you get (a^2 - f^2) x^2 + (2ab-2fg) xt + (b^2 - g^2) t^2 = x^2 - t^2 From this you conclude a^2 - f^2 = 1 , b^2 - g^2 = -1, and ab=fg. It's easy enough to confirm that the Lorentz transforms (if they look unfamiliar, it's because they're the Lorentz transforms in geometric units where c=1, so all factors of c have been omitted) satisfy this. a = gamma = 1/sqrt(1-v^2) b = vgamma = v/sqrt(1-v^2) f = vgamma = v/sqrt(1-v^2) g = gamma = 1/sqrt(1-v^2) It's a little more work (more than I care to do, and I'm not sure where to refer you to as a reference) that there aren't any other solutions that aren't equivalent to the above. (Replacing v by -v is something that falls in the category of equivalent, for instance). Thanks pervect. (Thank you, too, Mentz, but pervect has a few extra equal signs so I looked at his first.) So the lorentz transformation can be derived from the metric by asking "what set of transformations keep this quantity t^2 - x^2 constant? Vice versa, the metric is derived from the lorentz transformation by asking "what is the minimum distance I can make these two events?" or "What would that clock measure, that goes between those two events" or "what would that ruler measure that goes between those two events." The two things seem almost equivalent, but actually, I think there is an almost ideological difference between the two approaches. Going from the LT's to the metric, I am assuming a global geometric feature of the universe. Just as when I turn around, the universe will appear to undergo a rotational transformation, when I accelerate, the universe will undergo a Lorentz Transformation. Just as rotation applies equally to all objects at all distances, the Lorentz Transformation applies to all objects at all points in space and time. When I apply the lorentz transformation to derive the metric, I am using a universally applicable theory and finding the value of some local quantity. On the other hand, if you go from the metric to the LT's, you are only concerned with the geometric features of nearby objects, traveling fairly slowly, along geodesics within a gravitational field. The lorentz transformations, themselves, do preserve these invariants, of the clock's time, or the ruler's length, but they are merely a mathematical curiosity, of little importance, since we are only concerned with what happens on earth. bcrowell put it succinctly "GR does not have frames of reference, except locally, so a transformation between frames of reference isn't a central concern." I think bcrowell is correct, but I have often seen people claim that transformation between frames is not even a "valid" concern; they overstep the boundaries and say that Lorentz Transformations are only valid locallly. Quote by JDoolin The only distances I know of are "proper distance," "proper time," and "distance." The last distance has no adjective, and it refers to the Euclidean, observer dependent distance. (It is observer dependent, because two observers with different rapidities will observe different distances.) The metric distance between two points on the manifold is observer independent. Quote by JDoolin The "distance measure" as you put it, is an ambiguous term, since relativistically traveling observers would measure different distances between the same two events. That is not true, the metric distance is not ambiguous and observer independent. Quote by JDoolin Of course, I am referring to the Euclidean, observer dependent distance, with the critera that the space is "locally lorentz," I applied a Lorentz Transform to a "local" space, and find that for all intents and purposes, local is global. Then you clearly misapply the notion of 'locally Lorentzian'. Blog Entries: 4 Recognitions: Gold Member Quote by PAllen Not that I can think of, but even if you pick a coordinate system with lightlike basis vectors, you can discuss the size of region in terms of proper spacelike interval and proper timelike interval. I think I've seen somewhere, where they just take the space-time graph and rotateit it 45 degrees, and then the transformation becomes a contraction/expansion on the horizontal axis, and vice-vers on the vertical axis. I think this constant sized region here involves four distinct events. Two photons leave a spot, hit something to turn around, and meet again. That would enclose a region of space-time that would have constant area under lorentz transformation. But in general, no. If you have a photon going from one place to another, the best description of locality for it is zero. Effectively, there is a direct interaction between the particles at the origin, and the particles at the destination, with one caveat: The destination event definitely happens after the source event. This is totally off-topic from the idea of local lorentz, but there is another interaction that I find interesting and related; when the photon is produced in, say, a Helium atom, this occurs when the electron falls into a lower shell, or perhaps when it hits the lower shell. Does the photon arise from the interaction of two particles, or does it arise from the acceleration of one particle? Is there a direct interaction between the proton and electron (one event?) or are they separated by a finite distance when they interact (two events?), or is the photon generated in a process that takes place in a region of time and space? So (1) at the source, there are two events which occur at the same time, and (2) in between, there are events which definitely occur one after the other, and (3) at the destination, there are two events which occur at the same time. Quote by JDoolin I think I've seen somewhere, where they just take the space-time graph and rotateit it 45 degrees, and then the transformation becomes a contraction/expansion on the horizontal axis, and vice-vers on the vertical axis. Sounds like light-cone coordinates. Blog Entries: 9 Recognitions: Gold Member Science Advisor Quote by JDoolin If you have a photon going from one place to another, the best description of locality for it is zero. Effectively, there is a direct interaction between the particles at the origin, and the particles at the destination, with one caveat: The destination event definitely happens after the source event. I don't think "direct interaction" is an apt description, because it implies that there's no difference in the physics along a photon worldline regardless of which pair of events along it I pick. The fact that the spacetime interval between any two events on a photon's worldline is zero does not* imply that all events on that worldline are exactly the same in every physically relevant respect. So if I have two pairs of events, (A, B) and (A, C), that all lie on the same photon worldline, that does not imply that all the physics between A and B is exactly the same as all the physics between A and C. Simple example: a source at the origin that emits spherical wavefronts of light, and two detectors, both lying along the same radial line from the origin, one at radius R and the other at radius 2R. At time t = 0 in the frame in which all three objects (the source and both detectors) are at rest (I'm assuming flat spacetime, no gravity or other forces involved), the source emits a spherical wavefront. It arrives at detector #1 at time t = R and at detector #2 at time t = 2R. So we have three events: emission (t = 0, r = 0), detection #1 (t = R, r = R), and detection #2 (t = 2R, r = 2R). The spacetime interval between emission and detection #1 is the same as between emission and detection #2 (both are zero); however, the intensity of light measured at detection #1 is four times that measured at detection #2 (inverse square law). (In quantum terms, we would say that the amplitude for detection of a photon at detection #1 is twice the amplitude for detection of a photon at detection #2; the intensity goes as the square of the amplitude.) This difference, to me, means that saying "the locality is zero" for both pairs of events, or "direct interaction" between them, is not a good way of describing what's going on, because it gives no way of accounting for the difference in what's observed. Page 1 of 3 1 2 3 > Thread Tools Similar Threads for: "Locally Lorentz" Thread Forum Replies Introductory Physics Homework 4 Special & General Relativity 8 Special & General Relativity 13 Special & General Relativity 5 Beyond the Standard Model 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": 1, "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": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9409722089767456, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/47865/are-angles-ever-multiplied?answertab=votes	# Are angles ever multiplied? I recently explained multiplication of (non-zero) complex numbers to my Mathematics Fundamentals students, the usual bit about “multiply their lengths, and add their angles”. Of course, there is always at least one student who wants to know why the angles are added instead of multiplied – after all, this is multiplication, right? – and so I go through a non-trivial example to establish that it is at least highly plausible that addition, not multiplication, of angles is what occurs. This quiets their objection, but it always gets me to wondering: Indeed, are angles EVER multiplied anywhere in Mathematics? I don’t recall ever having seen such a thing. And since this wonderful wonderland of MSE not exists, I will pass this question on to the community. Of course, there are exercises of the form “If cos(f(x)) = sin(g(x)), solve for x.”, where this could happen, but of course what I am asking is whether this happens as part of a significant theorem, or as part of solving a problem of real physical interest. The non-multiplication of angles seems all the more counter-intuitive because they are actually dimensionless, in spite of the use of dimension-sounding phraseology such as “degrees” and “radians”. So, angles are just numbers, and surely numbers can be multiplied together, right? (Of course, not being dimensionless is no barrier to getting multiplied by objects of the same/different type. After all, we deal with the square of seconds in regard to acceleration, and with the square of grams in regard to the (statistical) variance of weights, and so on. I’m just saying that it would seem all the more plausible that they would occasionally get multiplied together if they are dimensionless.) I’m going to go out on a limb and conjecture that there is no such case. The basis of my conjecture is simply that I have never seen it happen, and also the fact that angles do not exist for vectors of zero length. I know this is quite tenuous, but what I’m guessing is that only quantities that behave well for 0 are eligible, so to speak, to be multiplied together. - 5 The reason you add angles is because that's a group homomorphism from $\mathbb{R}$ to the rotation group; you don't have to discount the possibility of multiplying angles to say that. – Qiaochu Yuan Jun 26 '11 at 23:22 2 Multiplying angles would give some special meaning to an angle of 1 radian. I can't think of any sort of geometric construction in which an angle of 1 radian arises naturally. – Corey Jun 26 '11 at 23:59 @QiaochuYuan, by the way, angles can also be subtracted: this is another way to form a field in $\mathbb{C}$, by complex conjugate multiplication. Is this operation also a group hom? – alancalvitti Sep 19 '12 at 18:00 ## 6 Answers Perhaps this is a bit too obvious to be a useful answer, but I will post it anyway: Angles are only defined up to congruence $\!\!\mod2\pi$. Addition preserves this symmetry: $$(\vartheta+2\pi k) + (\varphi+2\pi j) = \varphi + \vartheta + 2\pi(k+j) =: \xi + 2\pi l, \qquad k,j,l\in\mathbb{Z}$$ while multiplication doesn't. Therefore, the multiplication of two angles cannot be well-defined, unless you come up with some additional constraint on the range of the angles (in physics, you would call it a gauge). But this would not represent the mathematical/physical meaning of angles. - 1 Why would the product of two angles be measured in radians? See Gerry's answer; square-radians are not meaningless. – user6701 Jun 27 '11 at 7:13 Why would the product of two angles be measured in radians? I did not say it would. — As for square radians, these are not really a unit of the product of two angles, but rather the unit of the surface spanned by the exterior product of two "infinitesimal angles" (differential forms). For these, the congruence problem does not arise. – leftaroundabout Jun 27 '11 at 12:49 The question was whether angles are ever multiplied. If I understand your answer correctly, it argues that multiplication of angles wouldn't make sense if angles were closed under multiplication. Apparently angles are multiplied in some areas, and their product is not an angle. – user6701 Jun 27 '11 at 12:53 Bingo. This is the I-should-have-thought-of-that answer that I was looking for. I have up-voted this answer, and accepted it. I love MSE! – Mike Jones Jun 27 '11 at 21:44 @Tim: regardless of whether the product is again an angle or something else, there can be no equivalence relation on the space of results that takes into account any possible gauge of the factor angles. So as I said, you would need to choose a particular gauge, which would be something like $\vartheta\in[0,2\pi[$. But even then, the product of two angles would equal the solid angle of the area spanned by them only approximately, and only if they are small. – leftaroundabout Jun 27 '11 at 22:26 For what it's worth, there is such a thing as a "square degree" and a "square radian", better known as a steradian. As GEdgar notes, these come up in spherical trig. - I was wondering why no one mentioned Solid Angle and Steradians except for you. Is it not a topic in High School Physics everywhere? – Fortuon Paendrag May 9 '12 at 20:57 1 Steradians do measure solid angles, but I have only ever seen trig functions of angles actually multiplied together, even in L'Huilier's formula it is tangents of angles that are multiplied together. This preserves the modular congruence that leftaroundabout mentions. – robjohn♦ Sep 20 '12 at 11:47 Maybe look in spherical trig for more. Area of a spherical triangle, and such things. - One could define an alternate "multiplication" $$ for complex numbers $u,v$, such that $uv$ corresponds to ordinary complex multiplication but $uv = (\|u\|+\|v\|)e^{i \arg(u)\arg(v)}$, but I haven't seen anything like that in practice and am not sure what the utility of it would be. - We multiply angles in our land surveying courses at Odessa College - 3 Can you explain why you do this? – Nate Eldredge Sep 20 '12 at 4:00 http://www.math.umn.edu/~hardy/1031/handouts/March.3.pdf This is a visual explanation of why multiplication of complex numbers works the way it does. - 8 But, as the OP points out, multiplication of complex numbers involves addition of angles, not multiplication. So how does this address the question? – Pete L. Clark Jun 27 '11 at 4:22	{"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": 8, "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": 0.9526563286781311, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/15002/mathematically-oriented-treatment-of-general-relativity?answertab=active	# Mathematically-oriented Treatment of General Relativity Can someone suggest a textbook that treats general relativity from a rigorous mathematical perspective? Ideally, such a book would 1. Prove all theorems used 2. Use modern "mathematical notation" as opposed to "physics notation", especially with respect to linear algebra and differential geometry 3. Have examples that illustrate both computational and theoretical aspects 4. Have a range of exercises with varying degrees of difficulty, with answers An ideal text would read a lot more like a math book than a physics book and would demand few prerequisites in physics. Bottom line is that I would like a book that provides an axiomatic development of general relativity clearly and with mathematical precision works out the details of the theory. Thank you. Addendum (1): I did not intend to start a war over notation. As I said in one of the comments below, I think indicial notation together with the summation convention is very useful. The coordinate-free approach has its uses as well and I see no reason why the two can't peacefully coexist. What I meant by "mathematics notation" vs. "physics notation" is the following: Consider, as an example, one of the leading texts on smooth manifolds, John Lee's Introduction to Smooth Manifolds. I am very accustomed to this notation and it very similar to the notation used by Tu's Introduction to Manifolds, for instance, and other popular texts on differential geometry. On the other hand, take Frankel's Geometry of Physics. Now, this is a nice book but it is very difficult for me to follow it because 1) Lack of proofs and 2)the notation does not agree with other math texts that I'm accustomed to. Of course, there are commonalities but enough is different that I find it really annoying to try to translate between the two... Addendum (2): For the benefit of future readers, In addition to suggestions below, I have found another text that also closely-aligns with the criteria I stated above. It is, Spacetime: Foundations of General Relativity and Differential Geometry by Marcus Kriele. The author begins by discussing affine geometry, analysis on manifolds, multilinear algebra and other underpinnings and leads into general relativity at roughly the midpoint of the text. The notation is also fairly consistent with the books on differential geometry I mentioned above. - ## 5 Answers I agree with Ron Maimon that Large scale structure of space-time by Hawking and Ellis is actually fairly rigorous mathematically already. If you insist on somehow supplementing that: • For the purely differential/pseudo-Riemannian geometric aspects, I recommend Semi-Riemannian geometry by B. O'Neill. • For the analytic aspects, especially the initial value problem in general relativity, you can also consult The Cauchy problem in general relativity by Hans Ringström. • For a focus on singularities, I've heard some good things about Analysis of space-time singularities by C.J.S. Clarke, but I have not yet read that book in much detail myself. • For issues involved in the no-hair theorem, Markus Heusler's Black hole uniqueness theorems is fairly comprehensive and self-contained. • One other option is to look at Mme. Choquet-Bruhat's General relativity and Einstein's equations. The book is not really suitable as a textbook to learn from. But as a supplementary source book it is quite good. If you are interested in learning about the mathematical tools used in modern classical GR and less on the actual theorems, the first dozen or so chapters of Exact solutions of Einstein's field equations (by Stephani et al) does a pretty good job. - Thanks for the suggestions. The O'Neil book looks really good and is very close to what I'm looking for – 3Sphere Sep 24 '11 at 1:13 @3Sphere: the O'Neil book is good, but be warned that it only works through the most basic applications of GR: the Schwarzschild and Robertson-Walker solutions. It is a very rigorous and formal grounding in GR, though. – Jerry Schirmer Sep 24 '11 at 3:03 I got the impression the OP wanted something on foundations, and was willing to go to all the most modern applications later, from a second textbook, after understanding the foundations clearly. – joseph f. johnson Jan 15 '12 at 2:22 I recommend Yvonne Chocquet-Bruhat, Géométrie différentielle et systèmes extérieurs because it is so short and has exercises and is in the notation you want. I recommend it most highly (even if you can't get it with her inscribed autograph.) I also recommend very strongly her much longer (but try to get the first edition, which is still plenty long) Analysis, Manifolds and Physics by Yvonne Choquet-Bruhat, Cecile Dewitt-Morette, and Margaret Dillard-Bleick which has many exercises and much more Physics...but is too long. Goodness, it even includes Brownian Motion and path integrals.... That said, Dirac and Schroedinger have good and very short Physics books on the subject, I recommend those too even though they are not quite what you asked for. Bob Geroch's book is valuable too. - Good source, +1. – Ron Maimon Jan 15 '12 at 0:01 I don't know if it's formal enough for you, but a book I've always liked is Lilley: "Discovering Relativity for Yourself". It covers both special and general relativity. It is designed for evening-division teaching. It starts with very intuitive explanations, then gradually brings in the math, until it is, in my opinion, quite rigorous. - The Physiccs work in this field is rigorous enough. Hawking and Ellis is a standard reference, and it is perfectly fine in terms of rigor. ### Digression on notation If you have a tensor contraction of some sort of moderate complexity, for example: $$K_{rq} = F_{ij}^{kj} G_{prs}^i H^{sp}_{kq}$$ and you try to express it in an index-free notation, usually that means that you make some parenthesized expression which makes $$K = G(F,H)$$ Or maybe $$K = F(G,H)$$ Or something else. It is very easy to prove (rigorously) that there is no parentheses notation which reproduces tensor index contractions, because parentheses are parsed by a stack-language (context free grammar in Chomsky's classification) while indices cannot be parsed this way, because they include general graphs. The parentheses generate parse trees, and you always have exponentially many maximal trees inside any graph, so there is exponential redundancy in the notation. This means that any attempt at an index free notation which uses parentheses, like mathematicians do, is bound to fail miserably: it will have exponentially many different expressions for the same tensor expression. In the mathematics literature, you often see tensor spaces defined in terms of maps, with many "natural isomorphisms" between different classes of maps. This reflects the awful match between functional notation and index notation. ### Diagrammatic Formalisms fix Exponential Growth Because the parenthesized notation fails for tensors, and index contraction matches objects in pairs, there are many useful diagrammatic formalisms for tensorial objects. Diagrams represent contractions in a way that does not require a name for each index, because the diagram lines match up sockets to plugs with a line, without using a name. For the Lorentz group and general relativity, Penrose introduced a diagrammatic index notation which is very useful. For the high spin representations of SU(2), and their Clebsch-Gordon and Wigner 6-j symbols, Penrose type diagrams are absolutely essential. Much of the recent literature on quantum groups and Jones polynomial, for example, is entirely dependent on Penrose notation for SU(2) indices, and sometimes SU(3). Feynman diagrams are the most famous diagrammatic formalism, and these are also useful because the contraction structure of indices/propagators in a quantum field theory expression leads to exponential growth and non-obvious symmetries. Feynman diagrams took over from Schwinger style algebraic expressions because the algebraic expressions have the same exponential redundancy compared to the diagrams. Within the field of theoretical biology, the same problem of exponential notation blow-up occurs. Protein interaction diagrams are exponentially redundant in Petri-net notation, or in terms of algebraic expressions. The diagrammatic notations introduced there solve the problem completely, and give a good match between the diagrammatic expression and the protein function in a model. Within the field of semantics within philosophy (if there is anything left of it), the ideas of Frege also lead to an exponential growth of the same type. Frege considered a sentence as a composition of subject and predicate, and considered the predicate a function from the subject to meaning. The function is defined by attaching the predicate to the subject. So that "John is running" is thought of as the function "Is running"("John"). Then an adverb is a function from predicates to predicates, so "John is running quickly" means ("quickly"("Is running"))("John"), where the quickly acts on "is running" to make a new predicate, and this is applied to "John". But now, what about adverb modifiers, like "very", as in "John is running very quickly"? You can represent these are functions from adverbs to adverbs, or as functions from predicates to predicates, depending on how you parenthesize: (("very"("quickly"))("Is running"))("John") vs. (("very")(("quickly")("Is running"))("John") Which of these two parenthetization is correct define two schools of semantic philosophy. There is endless debate on the proper Fregian representation of different parts of speech. The resolution, as always, is to identify the proper diagrammatic form, which removes the exponential ambiguity of parenthesized functional representation. The fact that philosophers have not done this in 100 years of this type of debate on Fregian semantics shows that the field is not healthy. - 2 +1 for content, not entirely happy with tone. No need to hate on the mathmos... Be nice Ron. – genneth Sep 23 '11 at 19:11 1 Your contributions here are valuable Ron; don't debase the scientific content with random mud-slinging. As one physicist to another: please? – genneth Sep 23 '11 at 19:45 2 @3Sphere: ok, I removed the most offensive statements. But I believe that people still do not appreciate how exacly 180 degrees wrong the index-free idea is. The reason people are drawn to index free notations is because the brain for some reason prefers context free grammars, and will have a hard time following context dependent ones without training. Programming in assembly language is the only way I know to fix this human bias. I used to think that this was because human language is naturally context free, but the example of Piraha shows language is naturally regular. – Ron Maimon Sep 23 '11 at 23:15 2 @Ron: Your stuff is always so amazing. Would it be possible to file the burrs off without blunting your intelligence? You might start by being a little more forgiving of other intellectuals. They are not bad people (mostly). – Mike Dunlavey Sep 24 '11 at 0:52 2 @Mike: That's Dale Carnagie's way, and this is why American society is generally hostile to anything with mathematical precision. I don't accept Carnegie's way. I believe in Wolfgang Pauli's way, that it is your solemn duty to be an asshole. – Ron Maimon Sep 27 '11 at 0:31 show 6 more comments Sach and Wu has a yellow springer book called General Relativity for mathematician. -	{"extraction_info": {"found_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": 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": 0.9363473057746887, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/232056/expectation-inequality-with-weight-matrix	# Expectation inequality with weight matrix Given that $X$ is random matrix ($X'$ is transposed matrix) and $\Omega$ is invertible square matrix is inequality $$EX'X\cdot(EX'\Omega X)^{-1}\cdot EX'X\leq EX'\Omega^{-1}X$$ correct? All the expectations as well as inverse of $EX'\Omega X$ exist. Maybe (but it is probably wrong) $$EX'X\cdot(EX'\Omega X)^{-1}\cdot EX'X \leq EX'X\cdot(X'\Omega X)^{-1}\cdot X'X$$ Then I somehow need to show that $$X'X\cdot(X'\Omega X)^{-1}\cdot X'X = X'\Omega^{-1}X$$ Maybe expressing $\Omega$ as $\Omega^{1/2}\cdot \Omega^{-1/2}$ might somehow work. - do you mean for X to be a square matrix also ? If not, try the case where X is $n \times 1$ vector of i.i.d.s – mike Nov 7 '12 at 14:36 unfortunately $X$ is not necessarily square – jem77bfp Nov 7 '12 at 15:07	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 8, "mathjax_display_tex": 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": 0.9452087879180908, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/4763?sort=newest	## SL(2,Z/N)-decomposition of space of cusp forms for Gamma(N) ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Since $\Gamma(N)$ is normal in $\mathrm{SL}(2,\mathbb{Z})$, the quotient group $\mathrm{SL}(2,\mathbb{Z}/N)$ acts on the spaces of cusp forms $S_k(\Gamma(N))$. How do these spaces decompose into irreducible representations? I can do the case $N=2$. I'm mostly interested in the case of $N$ a prime. - I wonder if this question of mine is related: mathoverflow.net/questions/2875/… – Ilya Nikokoshev Nov 9 2009 at 20:08 ## 4 Answers See Theorem 1.0.3 of Jared Weinstein's phd thesis (it uses equivariant Riemann Roch). - ### 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 usual, once I spot a question on here I have anything useful to say about, somebody has already answered it. I can sum up that part of my thesis this way: let M be the induced representation of the character (-I) --> (-1)^k of the center up to all of SL(2,Z/NZ). Then S_k(Gamma(N)) is roughly k/12 copies of M, plus some error term which can be given precisely, with some effort. When you ask instead about Hilbert modular forms over a totally real field K, the "1/12" becomes the absolute value of zeta_K(-1). - Thank you moonface and Jared, that's perfect. I wish I could give more than one accepted answer. :) – Dan Petersen Nov 10 2009 at 8:35 You can almost do this with nothing more than Riemann-Hurwitz; in particular, by R-H you can compute the action of SL_2(Z/NZ) on H_1(X(N),C), which is just the sum of the representation you want with its dual. - The great thing about this approach is that it's completely different to the automorphic forms approach, and if the two are used together then they give information about the number of automorphic forms of a given "type". Sort of a generalisation of counting how many level 1 forms there are by computing the dimension of the space of level 1 forms by some global method. – Kevin Buzzard Nov 10 2009 at 7:38 I thought that this basically was the idea of Jared's thesis --- to count the number of forms of a given type by counting the representations in two different ways (geometric and automorphic). – Lavender Honey Nov 11 2009 at 3:34 If you think about this question in terms of automorphic representations then it sort of becomes trivial. The space $Sk(\Gamma(N))$ can be re-interpreted as the direct sum of $\pi^{U(N)}$, where $\pi$ is running through the automorphic representations of $GL_2$ which are holomorphic of weight $k$. Each factor is $SL(2,Z/NZ)$-invariant and often irreducible but sometimes has small finite length. The representation of $SL(2,Z/NZ)$ that shows up on $\pi{^U(N)}$ is the "type" of $\pi$. For explicit $\pi$s one will be able to explicitly determine this representation. - I'm not sure that they are irreducible (at least if I'm thinking about GL_2(F_p)) - e.g. can't I get things like the induction of the trivial character on the Borel showing up? – TG Nov 9 2009 at 23:48	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 16, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "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": 0.9354782700538635, "perplexity_flag": "middle"}
http://mathhelpforum.com/algebra/151651-help-needed-graphing-equations.html	# Thread: 1. ## Help Needed in Graphing Equations I honestly don't know what I would do without this forum. This time, I am stumped by a graphing problem with equations. The question is... Graph the equation 3y = 2x + 5 and calculate slope, x-intercept and y-intercept. I have been working on this one problem for a while now and so far, I have deciphered that the x and y intercept is x=-5/2 and y=5/3...is that correct? If that is correct, then the slope should be 1...again if I am doing it right. As always, any help would be greatly appreciated. 2. Originally Posted by dclary I have been working on this one problem for a while now and so far, I have deciphered that the x and y intercept is x=-5/2 and y=5/3...is that correct? They are correct. Originally Posted by dclary If that is correct, then the slope should be 1...again if I am doing it right. The slope is not 1. You need to use the points you just found to find $m = \frac{y_2-y_1}{x_2-x_1}$ or find the eqn in the form $y=mx+c$ 3. Hi Pickslides, thank you. To graph the equation for x=-5/2, I would go left 5 and up 2...right? and for the y=5/3, I would go up 5 and across to 3? I just want to make sure I am headed in the right direction. Thank you again for your help. 4. Originally Posted by dclary Hi Pickslides, thank you. To graph the equation for x=-5/2, I would go left 5 and up 2...right? and for the y=5/3, I would go up 5 and across to 3? I just want to make sure I am headed in the right direction. Thank you again for your help. No this is not correct. Draw your x and y axis, then mark the respective values you have found on each axis. Your points are and $\left(x_1,y_1\right) = \left(\frac{-5}{2},0\right)$ and $\left(x_2,y_2\right) = \left(0, \frac{5}{3}\right)$ Now find the gradient using $m = \frac{y_2-y_1}{x_2-x_1}$ 5. I finally got it figured out...thank you again Pickslides :-)	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 5, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9637966752052307, "perplexity_flag": "middle"}
http://www.physicsforums.com/showthread.php?p=2641769	Physics Forums ## Energy conservation problem 1. The problem statement, all variables and given/known data A string exerts a tension force of 55N on a 10kg block as it moves a distance of 5m up a frictionless incline. The string pulls parallel to the incline. A) Find the total external work done on the block-earth system as the block moves 5m up the incline. B)Find the change in gravitational potential energy for the block-earth system as the block moves 5m up the incline. C)Find the change in kinetic energy for the block-earth system as the block moves 5m up the incline. 2. Relevant equations $$KE=\frac{1}{2}mv^{2}$$ $$PE_{g}=mgh$$ Our teacher tells us to round gravity up to 10m/s/s. 3. The attempt at a solution A)The external force should be the tension, so: $$(55N)(5m) = 275J$$ B)I used trig (3-4-5 triangle) to find the height to go up 3m. $$mgh_{f}-mgh_{i}=\Delta KE$$ $$(10)(10)(3)-(10)(10)(0)=300J$$ C)$$KE_{f}-KE_{i}=\DeltaKE$$ I am not sure what to do here because I don't know how to solve for velocity. PhysOrg.com science news on PhysOrg.com >> Front-row seats to climate change>> Attacking MRSA with metals from antibacterial clays>> New formula invented for microscope viewing, substitutes for federally controlled drug Blog Entries: 7 Recognitions: Gold Member Homework Help What does the work-energy theorem say about change in kinetic energy? Quote by kuruman What does the work-energy theorem say about change in kinetic energy? $$W_{net}=KE_{f}-KE_{i}$$ If this is right, then I can set 275J equal to $$KE_{f}-KE_{i}$$ $$\frac{1}{2}mv_{f}^{2}-\frac{1}{2}mv_{i}^{2}=275J$$ $$5v_{f}^{2}-5v_{i}^{2}=275J$$ But then I will have two variables $$v_{f}$$ and $$v_{i}$$ I think I am on the wrong track. Blog Entries: 7 Recognitions: Gold Member Homework Help ## Energy conservation problem Quote by crybllrd $$W_{net}=KE_{f}-KE_{i}$$ This is right. If this is right, then I can set 275J equal to $$KE_{f}-KE_{i}$$ No, you cannot. Tension is not the only force that does work. What other force is there? $$\frac{1}{2}mv_{f}^{2}-\frac{1}{2}mv_{i}^{2}=275J$$ $$5v_{f}^{2}-5v_{i}^{2}=275J$$ But then I will have two variables $$v_{f}$$ and $$v_{i}$$ You don't care about the initial and final speeds. The problem is asking you to find the change in kinetic energy, not the initial and final speeds. If you find the correct Wnet, that is the change in kinetic energy. Quote by kuruman Tension is not the only force that does work. It asks what the change for the block-earth system is. On a free body diagram for the block, I have three forces: I have the Normal force (80N) by the ramp on the block, the tension force (55N) , and a weight force (80N in y-dir, 60N in x-dir). (no friction) Is the other force doing work the Normal force? If so, then I am not sure how to calculate the work because the distance is perpendicular. Blog Entries: 7 Recognitions: Gold Member Homework Help Yes there are three forces. As you say, the normal force does zero work. So what is the net work, i.e. the total work done by all three forces? Quote by kuruman So what is the net work, i.e. the total work done by all three forces? This is where I am getting confused. The problem is defining the system as block-earth, so if I am not mistaken, the weight force is the only force not doing work on the system. That would leave Normal and Tension forces. Normal is perpendicular, so equal to zero, Tension is 275J, so total work is [275+0] 275 Joules. Quote by kuruman Tension is not the only force that does work. What other force is there? You said earlier I can't set the equations equal to 275J because there are other forces. What am I missing here? Do I need to add in the weight force? I think it should be included in the block-earth system, not as external work. Blog Entries: 7 Recognitions: Gold Member Homework Help The tension increases both the kinetic energy and the gravitational potential energy of the Earth-block system. If you ignore the potential energy change, it is as if the incline is not there and the block moves across a horizontal surface. Tags energy, kinetic energy, potential energy, work Thread Tools \| \| \| \| \|--------------------------------------------------\|-------------------------------\|---------\| \| Similar Threads for: Energy conservation problem \| \| \| \| Thread \| Forum \| Replies \| \| \| Introductory Physics Homework \| 1 \| \| \| Introductory Physics Homework \| 2 \| \| \| Calculus & Beyond Homework \| 9 \| \| \| Introductory Physics Homework \| 8 \| \| \| 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": 0, "mathjax_display_tex": 18, "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": 0.9026134610176086, "perplexity_flag": "middle"}
http://mathforum.org/mathimages/index.php?title=Law_of_cosines&oldid=21125	# Law of cosines ### From Math Images Revision as of 15:43, 20 June 2011 by Rscott3 (Talk \| contribs) Law of Cosines Field: Geometry Created By: rscott3 Website: [ ] Law of Cosines The law of cosines is a trigonometric extension of the Pythagorean Theorem. # Basic Description The law of cosines is a formula that helps in solving triangles when two or three side lengths of a triangle are known. The formula combines the squares of two side lengths of a triangle and and a third term involving the cosine of a particular angle, to calculate the square of the third side. For this reason, the law of cosines is often thought of as the generalization of the Pythagorean theorem, which only applies to right triangles. The law of cosines adds an extra term to the Pythagorean theorem so that a third side length of a triangle can be determined when there is no right angle. Given a triangle with side lengths, $a,b,c$ and angle measures $A,B,C$, the law of cosines states $c^{2} = a^{2} + b^{2} - 2ab \cos C$ The law of cosines is useful in solving triangles whenever at least two side lengths are known. If the triangle that needs to be solved has less than just two known lengths, the Law of sines is more useful. ### Alternate Forms Particularly when the law of cosines is being used to find a particular side length, that is when a certain angle measure is given, the law of cosines can be written in several different ways to help set up the equation to solve for the missing element of the triangle. • $c^{2} = a^{2} + b^{2} - 2ab \cos C$ • $b^{2} = a^{2} + c^{2} - 2ac \cos B$ • $a^{2} = b^{2} + c^{2} - 2bc \cos A$ An alternate form of the law of cosines is particularly useful when solving a triangle when just the three side lengths are given. This for isolates the term with the cosine of the angle in it to make it easier to solve for the angle. • $\cos C = \frac{a^{2} + b^{2}- c^{2}}{2ab}$ • $\cos B = \frac{a^{2} + c^{2}- b^{2}}{2ac}$ • $\cos A = \frac{b^{2} + c^{2}- a^{2}}{2bc}$ # A More Mathematical Explanation [Click to view A More Mathematical Explanation] ## Proof ### By the Pythagorean Theorem An easy way to think of the law of cosines is as an extens [...] [Click to hide A More Mathematical Explanation] ## Proof ### By the Pythagorean Theorem An easy way to think of the law of cosines is as an extension of the Pythagorean theorem for a right triangle: $a^{2} + b^{2} = c^{2}$ By Pythagorean theorem, we know $c^{2} = p^{2} + h^{2}$ But $p$ is just some portion of side length $a$ which is $r$ less than the length of $a$. Substituting the difference gives us, $c^{2} = (a-r)^{2} + h^{2}$ By Pythagorean theorem, we also know that $h^{2} = b^{2} - r^{2}$ Substituting the appropriate values gives us, $c^{2} = (a-r)^{2} + ( b^{2} - r^{2})$ Expanding the squared term gives us $c^{2} = a^{2} -2ar + r^{2} + b^{2} - r^{2}$ Simplify for $c^{2} = a^{2} -2ar + b^{2}$ And by the definition of cosine, we know that $r=b \cos C$ Substituting this value in give us $c^{2} = a^{2} -2ab \cos C + b^{2}$ or $c^{2} = a^{2}+ b^{2} -2ab \cos C$ ### Using the Distance Formula Let $\vartriangle ABC$ be oriented so that $C$ is at the origin, and $B$ is at the point$(a,0)$. $\text{distance} = \sqrt {(x_{2}-x_{1})^{2} + (y_{2}-y_{1})^{2}}$ $c$ is the distance from $A$ to $B$. Substituting the appropriate points into the distance formula gives us $c = \sqrt {(a-b \cos C)^{2} + (0-b \sin C)^{2}}$ Squaring the inner terms, we have $c = \sqrt {(a^{2}-2ab \cos C+b^{2} \cos^{2} C) + (b^{2} \sin^{2} C)}$ Since $\cos^{2} C + \sin^{2} C = 1$, $c = \sqrt {a^{2}+b^{2}-2ab \cos C+b^{2}}$ Square both sides for $c^{2} = a^{2}+b^{2}-2ab \cos C+b^{2}$ ## Example Problem Solve the triangle using the law of cosines. $c^{2} = a^{2} + b^{2} - 2ab \cos C$ ### Solution To find the side length $c$, $c^{2} = 6^{2} + (6 \sqrt{2})^{2} -2 (6) (6 \sqrt{2}) \cos 45^\circ$ Simplify for $c^{2} =36 + 36 (2) - 72 \sqrt{2} \cos 45^\circ$ Since $\cos 45^\circ = \frac{1}{\sqrt{2}}$, substitution gives us $c^{2} =36 + 36 (2) - 72 \sqrt{2} (\frac{1}{\sqrt{2}})$ Simplify for $c^{2} =36 + 72 - 72$ $c^{2} =36$ Taking the square root of both sides gives us $c =6$ To solve for angle $B$, we can orient the triangle differently and use the law of cosines in the form $b^{2} = a^{2} + c^{2} - 2ab \cos B$ Substituting in the appropriate side lengths gives us $(6 \sqrt{2})^{2} = 6^{2} + 6^{2} - 2(6)(6) \cos B$ $36 (2) = 36 + 36 - 72 \cos B$ $72 = 72 - 72 \cos B$ Subtracting $72$ from both sides gives us $0 = - 72 \cos B$ Dividing both sides by $-72$ gives us $0 = \cos B$ Using inverse trigonometry, we know that $B = 90^\circ$ And we can find the last angle measure $A$ by subtracting the other two measures from $180^\circ$ $180^\circ - 90^\circ - 45^\circ = 45^\circ$ $A=45^\circ$ # Teaching Materials There are currently no teaching materials for this page. Add teaching materials. Leave a message on the discussion page by clicking the 'discussion' tab at the top of this image page.	{"extraction_info": {"found_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": 59, "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": 0.9042239785194397, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/1222/white-holes-and-time-reversed-oppenheimer-snyder-collapse	# White Holes and Time-Reversed Oppenheimer-Snyder collapse So, the canned explanation that I always hear about why the white hole solution of the extended Schwarzschild solution is non-physical is that "The matter distribution cuts off the white hole solution." If I look at Oppenheimer-Snyder collapse, however, the cosmological solution that I paste to in the dust's interior is just the closed pressureless model of the Robertson-Walker metric, which is perfectly time-symmetric about the point of maximal expansion. If I look at this naïvely, then, the fluid in the Oppenheimer-Snyder solution originated in a singularity, which then disintegrates, to form an expanded fluid, which reaches some maximal extent, and then recollapses to form a singularity in the future, after some finite time. That past singularity sure seems like a white hole to me. What am I misinterpreting here? Or is it just a pathology of the Oppenheimer-Snyder construction that it is over-simple, and more physically realistic matter models WOULD cut off the past singularity? - ## 4 Answers The point is that the configuration that you are describing as the time-reversed Oppenheimer-Snyder collapse would need very specific initial conditions to occur. A collapse process on the other hand is very generic. You can have a collapse with any number of initial configurations. In that sense a collapse process is more physical than the reverse. It is like dropping a needle and having it land on its tip. There are many ways to drop it and have it lying down on its side, but there is only one that you can drop it and have it land and stay balanced on its tip. It is a mathematical possibility but a physical improbability. - OK. Take any singularity free spacetime with a moment of time symmetry whose future development has a black hole in it. Why does that spacetime not have a white hole in its past? Alternately, why would small perturbations of the Oppenheimer-Snyder spacetime cause the past singularity to dissapear, but the future singularity to remain? – Jerry Schirmer Nov 25 '10 at 1:29 For the same reason that one can break an egg but not reconstitute a whole egg from a broken one. It's a question of entropy. – Matt Reece Nov 25 '10 at 5:21 @Jerry Schirmer: I think that you are looking at it backwards. It is not a matter of whether there should be in the past of every black hole a white hole, but whether a dust configuration you could have now has originated from the inside of a white hole. Snyder collapse is not the only way to get a black hole. Christodoulou for example has shown that you can have a convergence of gravitational waves from null infinity with general initial conditions that produces a trapped surface and ends up forming a black hole. There is no white hole in the past of that. – Vagelford Nov 25 '10 at 8:27 Sure. the whole argument collapses if there is no moment of time symmetry in the solution, but that doesn't seem to be a radical requirement to make. And one would think that most black holes came from particulate matter, and not things like geons and Brill waves. But if you can find physically realistic matter distributions that seem to have white hole solutions in their past, – Jerry Schirmer Nov 25 '10 at 11:48 The basic argument is that white holes have very low entropy, just like the big bang. – Sklivvz♦ Nov 25 '10 at 19:58 show 14 more comments So, I brought this up at my research group meeting this week. Turns out my initial guess was right--the past development of the Oppenheimer-Snyder spacetime does contain a white hole. When people say that the matter distribution cuts off the white hole, what they generally assume is that the spacetime does not contain a moment of time symmetry, and that the matter distribution is something like a star, that goes into the distant past. - Yes, Oppenheimer Snyder time reversed contains a white hole. So what. You have to set up the white hole, and with very specific conditions on the singularity that it is emitting the Oppenheimer Snyder dust shells to exit. This is considered unphysical. A better question would be for charged dust-shells collapsing to form a near-extremal Reissner Nordstrom black hole (or white hole, whichever you like, they're the same thing). – Ron Maimon Jul 28 '12 at 7:57 I can't make sense of your quoted phrase "The matter distribution cuts off the white hole solution.", but I'll try to answer the rest of the question The Oppenheimer-Synder model for black hole collapse is a solution of the equations of classical GR where a uniform speherically symmetrical dust cloud with no pressure or rotation collapses to form a black hole. The time reversal is also a valid solution of the field equations but it would represent a white hole and is therefore normally considered unphysical. Inside the sphere of dust the solution matches the well known cosmological Freidman, Robertson Walker solutions for the expanding universe. Outiside the collapsing sphere it looks like the Schwartzchild static black hole solutions. In the cosmological solution, there are different cases where the universe can expand forever or recollapse. The same thing happens in the Oppenheimer-Synder solutuon. So there are cases where it starts with a white-hole and the dust expands outwards, but the dust does not have enough energy to escap so it recollapses to form a black hole. However, there are other cases where the dust does escape from the whitehole. The time reversal of this is also the formation of a blackhole without the whitehole in the past By the way, these are all special cases of a more general class of spherically symmetric solutions called the Lamaitre-Tolman solutions. - A white hole is in a sense a time reversed black hole. If you take an Eddington-Finkelstein diagram and turn it upside down you have a white hole. A physical sense of things suggests these do not exist. The argument for “cutting off” is usually interpreted as how the Penrose conformal diagram with two square patches for time like regions and two triangular spacelike regions is reduced so the bottom spacelike region for the white hole is removed. Pretty clearly there is an asymmetry between the two halves of the pure or “eternal” solution. White holes may have played a role in the early universe though. The Hubble constant is ${\dot a}/a~=~H$, ${\dot a}~=~da/dt$, and $a$ the scale factor. The FLRW differential equation of motion for a constant vacuum for the de Sitter spacetime is $$({\dot a}/a)^2~=~8πG\Lambda/3,$$ which has an exponential solution. So we have the scale factor evolving as $$a~\simeq~\sqrt{3/8πG\Lambda}exp(\sqrt{8πG\Lambda/3}t).$$ This exponential expansion is what smoothed out anisotropies in the universe. So we consider this with some anisotropy which is rewound backwards in time. So the small anisotropies rapidly clump back together during the inflationary period, which in a time reversed setting can mean the collapse of matter into black holes. Now this is the time reversal of inflation, which means the inflationary period may have had white holes. The white holes might have been some perturbation on the inflationary cosmology and their disappearance in the exponentially expanding space some aspect of inflation. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 3, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9297085404396057, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/92272/finding-the-limit-of-a-sequence?answertab=oldest	# finding the limit of a sequence If there were a regular square root I would multiply the top by its adjacent and divide, but I've tried that with this problem and it doesn't work. Not sure what else to do have been stuck on it. $$\lim _{n\to \infty } \sqrt [3]{n^2} \left( \sqrt [3]{n+1}- \sqrt [3]{n} \right) .$$ - i don't think it matters for this problem but the limit goes to +infinity – nofe Dec 17 '11 at 16:35 use difference of cubes $a^3-b^3=(a-b)(a^2+ab+b^2)$ to simplify the expression – yoyo Dec 17 '11 at 16:41 ## 3 Answers $$\displaystyle\lim_{n\to \infty } \sqrt [3]{n^2} \left( \sqrt [3]{n+1}-\sqrt [3]{n} \right)\cdot\frac{\left(\sqrt[3] {(n+1)^2}+\sqrt[3] {n(n+1)}+\sqrt[3] {n^2}\right)}{\left(\sqrt[3] {(n+1)^2}+\sqrt[3] {n(n+1)}+\sqrt[3] {n^2}\right)}=$$ $$\displaystyle\lim_{n\to \infty }\frac{(\sqrt[3] {n^2}\cdot(n+1-n)) \div \sqrt [3] {n^2}}{\left(\sqrt[3] {(n+1)^2}+\sqrt[3] {n(n+1)}+\sqrt[3] {n^2}\right)\div \sqrt[3] {n^2}}=\frac{1}{3}$$ - thanks, this is what i'm looking for. I don't want to use l'hospital or differentiate anything. How did you know this was the adjacent? how do you find the adjacent in a polynomial higher than 2? – nofe Dec 18 '11 at 0:16 @nofeliram,because of third root terms it is obvious that you should use difference of cubes formula... – pedja Dec 18 '11 at 8:25 thanks! it's not obvious to me I'm just now returning to school after 4 years (army) of not touching any math at all since high school. – nofe Dec 18 '11 at 8:56 $$\begin{align} \lim _{n\to \infty } \sqrt [3]{n^2} \left( \sqrt [3]{n+1}- \sqrt [3]{n} \right) &= \lim _{n\to \infty } \sqrt [3]{n^2} \cdot \sqrt[3]{n} \left( \sqrt [3]{1+ \frac{1}{n}}- 1 \right) \\ &= \lim _{n\to \infty } n \left( \sqrt [3]{1+ \frac{1}{n}}- 1 \right) \\ &= \lim _{n\to \infty } \frac{\sqrt [3]{1+ \frac{1}{n}}- 1 }{\frac{1}{n}} \\ &= \lim _{h \to 0} \frac{\sqrt [3]{1+ h}- 1 }{h} \\ &= \left. \frac{d}{du} \sqrt[3]{u} \ \right\|_{u=1} \\ &= \cdots \end{align}$$ - Presumably you don't want a Taylor series expansion, since you said you "don't want to differentiate anything," but it's worth pointing out that you can apply the binomial expansion: $$\begin{eqnarray} \sqrt[3]{n+1} &=& \sqrt[3]{n}\sqrt[3]{1+n^{-1}} \\ &=& \sqrt[3]{n}\sum_{k}{{1/3}\choose{k}}n^{-k} \\ &=& \sum_{k}{{1/3}\choose{k}}n^{1/3-k} \\ &=& \sqrt[3]{n} + \frac{1}{3}n^{-2/3}+O(n^{-5/3}). \end{eqnarray}$$ So $\sqrt[3]{n^2}(\sqrt[3]{n+1}-\sqrt[3]{n}) = 1/3 + O(n^{-1}) \rightarrow 1/3$ as $n \rightarrow\infty$. -	{"extraction_info": {"found_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": 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": 0.9772817492485046, "perplexity_flag": "head"}
http://nrich.maths.org/776	### Continued Fractions I An article introducing continued fractions with some simple puzzles for the reader. ### LOGO Challenge - Circles as Bugs Here are some circle bugs to try to replicate with some elegant programming, plus some sequences generated elegantly in LOGO. ### Spirals Instead of Sunflowers Using logo to investigate spirals # Gnomon Dimensions ##### Stage: 4 Challenge Level: You may wish to try the related problem Building Gnomons first. A Gnomon is a rectangle with another rectangle cut out of one corner. The area of each Gnomon is a Fibonacci number. (The Fibonacci numbers are $1, 1, 2, 3, 5, 8$ and so on, with each new term being the sum of the previous two terms.) $G_1$ has area $3$, $G_2$ has area $5$, $G_3$ has area $8$ and so on. Draw the next three gnomons in the sequence. Look at the length and width of the large rectangle from which each gnomon is made. Is there a pattern to the lengths and widths? Can you generalise? Now look at the length and width of the rectangle cut out of each gnomon. Can you see any patterns here? Can you generalise and justify what you see? I want to group the gnomons with area 3, 8 and 21 together, and the gnomons with area 5, 13 and 34 together. Can you explain why I want to divide these into two separate groups? Can you give a convincing argument why all the gnomons fit into one of these two groups?	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 7, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 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": 0.9028740525245667, "perplexity_flag": "middle"}
http://mathhelpforum.com/calculus/43429-solved-convergence-infinite-series.html	# Thread: 1. ## [SOLVED] Convergence of an infinite series I've to show whether the following infinite series converges or not. The problem is that I tried to prove it's divergent by induction, by the root test and by trying to bound it from below. But I couldn't reach any conclusion. Here it comes : $\sum_{n=1}^{+\infty} \frac{n!}{n^n}$. I'm sure it's a famous one. It's very similar to the one that defines $e^n$ (inversed). I'd like you to tell me what test should I try. Then if I'm not able to get a conclusion, I'll ask for further help. Thanks in advance. 2. Originally Posted by arbolis I've to show whether the following infinite series converges or not. The problem is that I tried to prove it's divergent by induction, by the root test and by trying to bound it from below. But I couldn't reach any conclusion. Here it comes : $\sum_{n=1}^{+\infty} \frac{n!}{n^n}$. I'm sure it's a famous one. It's very similar to the one that defines $e^n$ (inversed). I'd like you to tell me what test should I try. Then if I'm not able to get a conclusion, I'll ask for further help. Thanks in advance. Root test...just so you know it is (high-light below) convergent Or consider that for very large n we have that (high-light) n!<<n^n by stirlings approximation 3. Does it help to know that $\left( {\sqrt[n]{{\frac{{n!}}{{n^n }}}}} \right) \to \frac{1}{e}$? 4. Or consider that for very large n we have that (high-light) n!<<n^n by stirlings approximation My intuition fails! I thought that $n!$ would have been much greater than $n^n$. But from what I did, all seemed to show the contrary and I couldn't believe it. I was trying to show that if $n^{n+1}+n \geq (n+1)^{n+1}$, then it would diverges. I recognized it was false, but I wouldn't trust my result! Does it help to know that ? Of course! Since $\frac{1}{e}<1$, the series converges... Now I will try to prove it! 5. Plato showed you, but if you wanted you could just use the ratio test: $\frac{(n+1)!}{(n+1)^{n+1}}\cdot\frac{n^{n}}{n!}=\f rac{n^{n}}{(n+1)^{n}}$ ${\rho}=\lim\frac{n^{n}}{(n+1)!}=\frac{1}{e}$ ${\rho}<1$...........convergent 6. Originally Posted by galactus Plato showed you, but if you wanted you could just use the ratio test: $\frac{(n+1)!}{(n+1)^{n+1}}\cdot\frac{n^{n}}{n!}=\f rac{n^{n}}{(n+1)^{n}}$ ${\rho}=\lim\frac{n^{n}}{(n+1)!}=\frac{1}{e}$ ${\rho}<1$...........convergent Might as well jump on the band-wagon. $\text{As}\quad{n\to\infty}\quad{n!\sim\sqrt{2\pi{n }}n^ne^{-n}}$ $\lim_{n\to\infty}\sqrt[n]{\frac{n!}{n^n}}\sim\lim_{n\to\infty}\sqrt[n]{\frac{\sqrt{2\pi{n}}n^ne^{-n}}{n^n}}=\frac{1}{e}\lim_{n\to\infty}(2\pi{n})^{\ frac{1}{2n}}$ so using the connection of the root and ratio test $\lim_{n\to\infty}(2\pi{n})^{\frac{1}{2n}}=\sqrt{\l im_{n\to\infty}(2\pi{n})^{\frac{1}{n}}}=\sqrt{\lim _{n\to\infty}\frac{2\pi(n+1)}{2\pi{n}}}=\sqrt{1}=1$ $\therefore\sqrt[n]{\lim_{n\to\infty}\frac{n!}{n^n}}=\frac{1}{e}$ 7. We certainly know it converges. Now find what it converges to.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.9761701822280884, "perplexity_flag": "head"}
http://mathoverflow.net/questions/109661/polarizable-variations-of-mixed-hodge-structures	## Polarizable variations of (mixed) Hodge structures ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I am trying to come to grips with Saito's theory of mixed Hodge modules (slightly) beyond just the basic axiomatic formalism. I will take my Hodge structures and sheaves to be rational, but I would be grateful if the experts would point out any subtleties with my questions if rational is replaced by real. Let me at the moment focus on questions related to polarizability: 1. Let $\mathcal{L}$ be a local system underlying a polarizable variation of Hodge structure on a smooth variety. Does the polarizability imply $\mathcal{L}$ is self-dual? If yes, then does every direct summand of this local system also have to be self-dual (I am guessing no to the latter)? 2. More generally, let $M$ be a polarizable mixed Hodge module on some variety (not necessarily smooth). Is polarizability equivalent to Verdier self-duality (up to Tate twist) in the derived category of mixed Hodge modules? If not equivalent, does it at least imply Verdier self-duality? - 1 Isn't your 2) (3.2.8) in "Introduction to Mixed Hodge modules"? As Donu points out, the second paragraph after Theorem 2.2 gives that the indecomposable polarisable Hodge modules are IC complexes. – Geordie Williamson Oct 23 at 8:22 @Geordie Williamson: Yes, you are right! – Reladenine Vakalwe Oct 23 at 15:24 ## 1 Answer You have my sympathies for trying come to grips with this stuff. Fortunately the answer, which is both yes and no, can be explained in the simplest case of a point. A polarization on a pure Hodge structure $H$ of weight $w$ is a pairing $$\langle, \rangle: H\otimes H\to \mathbb{Q}(-w)$$ such that $(2\pi i)^n\langle -, C -\rangle$ is positive definite and symmetric, where $C$ is the Weil operator which acts by $i^{p-q}$ on $H^{pq}$. These conditions known as the Hodge-Riemann bilinear relations imply that $H\cong H^(-w)$ as Hodge structures, so in this sense $H$ self dual. However, just having such an isomorphism would not give everything else. If $H$ is replaced by a variation of Hodge structures, then a polarization is flat pairing as above satisfying the Hodge-Riemann conditions on the fibres. For Hodge modules the story is similar but more delicate. Since the whole theory is constructed by induction on dimension of support, a polarization is also defined in this manner. Given Hodge module $H$ of weight $w$ on $X$, a polarization is a pairing $H\otimes H\to \mathbb{Q}(\dim X-w)[2\dim X]$ satisfying the inductive conditions (0.8)-(0.10) of Saito's "Modules Hodge Polarizables". Such a pairing should induce an isomorphism $H\cong DH(?)$ up to twist, but it's not equivalent. (Added I don't have a precise reference or proof for this, just a feeling that one could prove it as follows: Buried in Saito's second paper, "Mixed Hodge modules", is a proof that any simple polarizable Hodge module is $j_{!}V$, where $V$ is a VHS. This effectively reduces the duality statement to the previous case.) - This is about what I had interpreted when (trying to) wade through "Modules de Hodge...". However, there is something that is unclear to me (which is what prompted the original question): Saito requires a pairing satisfying the inductive conditions. Over a point this pairing is perfect. I don't understand why the conditions imply the pairing is "perfect" (= induces Verdier self-duality) in general. My naive picture is that the polarization should induce a polarization on each stalk, but I don't see this from the conditions (or a result to this effect in the paper). Is there a simple reason? – Reladenine Vakalwe Oct 15 at 15:38 I added some remarks about this. Unfortunately, I have to finish some other stuff. I might flesh it out later. – Donu Arapura Oct 15 at 16:04 Ah yes! This does it. – Reladenine Vakalwe Oct 15 at 16:53	{"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": 19, "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": 0.9161595106124878, "perplexity_flag": "head"}
http://mathhelpforum.com/discrete-math/17121-power-sets.html	# Thread: 1. ## Power sets I have a homework problem that I knew the answer but dont know how to do the proof. Can anyone help me? List all the subsets of the set X={a,b,c}. How many are there? Next, try with X={a,b,c,d). Now suppose that the set X has n elements: how many subsets does X have? Try to justify your answer. I do the first part as For X={a,b,c} {}, {a},{b},{c}, {a,b},{a,c},{b,c} {a,b,c} which is 1,3,3,1 For X={a,b,c,d}, in the same pattern, which is 1,4,6,4,1. I think it is the Pascal triangle's pattern and the answer is 2^n, but I have no idea how to write the proof. 2. Originally Posted by r7iris I have a homework problem that I knew the answer but dont know how to do the proof. Can anyone help me? List all the subsets of the set X={a,b,c}. How many are there? Next, try with X={a,b,c,d). Now suppose that the set X has n elements: how many subsets does X have? Try to justify your answer. I do the first part as For X={a,b,c} {}, {a},{b},{c}, {a,b},{a,c},{b,c} {a,b,c} which is 1,3,3,1 For X={a,b,c,d}, in the same pattern, which is 1,4,6,4,1. I think it is the Pascal triangle's pattern and the answer is 2^n, but I have no idea how to write the proof. It is 2^n. Consider, $\{ 1,2,...,n\}$ There is the one empty set, which is ${n\choose 0}$ There are $n$ singleton sets, which is ${n\choose 1}$ There are $n(n-1)/2$ two elements sets which is ${n\choose 2}$. And so on ... Until we use all the elements, and the full set itself which is ${n\choose n}$. In total we have, ${n\choose 0}+{n\choose 1}+...+{n\choose n}=2^n$ See if you can figure out the last line.	{"extraction_info": {"found_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": 8, "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": 0.9564427137374878, "perplexity_flag": "middle"}
http://gowers.wordpress.com/2008/08/	# Gowers's Weblog Mathematics related discussions ## Archive for August, 2008 ### Princeton Companion latest August 28, 2008 A quick post to give an update on the Princeton Companion to Mathematics. It’s been off my hands for a few weeks now. If all goes well it will be printed by the middle of September and should appear in bookshops about a month later. The illustration to the right is what the cover will be like. Click here if you want to see it in more or less full size. If you go to this page on the PUP website then you will find a podcast interview that I gave, which contains information that does not appear on this blog. UPDATE 14TH OCTOBER: here are links to a few blog posts that have discussed the Princeton Companion recently. Terence Tao had a few thoughts on receiving his copy. Brian Hayes did too. Fred Shapiro, editor of The Yale Book of Quotations, has an interest in reference books and commented about the PCM in the New York Times Freakonomics blog. Dmitry Vostokov recommends the PCM for people wanting a broad overview of mathematics. Isallaboutmath points out that the Nautilus shell on the cover does not in fact have anything to do with the golden ratio. And the book has five reviews on Amazon, some more sensible than others. FURTHER UPDATE (to which I’ll add as the occasion arises): Peter Woit has written a short review on his blog. Ian Stewart has written a review for The Times. Edmund Harriss has reviewed it on his blog. Alexander Bogomolny reviews it here. Does Antonio Cangiano like it? Perhaps you’d better judge for yourself. Having honed your interpretative skills, you’ll be in a better position to appreciate Scott Guthery’s helpful antidote. A different Scott, Scott Aaronson, reviews it on his blog. And Robin Wilson reviews it for the LMS newsletter. A nice review here, in French. 29TH OCTOBER: It’s been a long time coming, but I’ve just noticed that the PCM now appears to be available on Amazon UK, and not just from third-party sellers. According to them, it’s popular amongst geographers. If I knew how, I’d suggest they recategorized it. Posted in Princeton Companion To Mathematics \| 22 Comments » ### Just-do-it proofs August 16, 2008 This post is another sample Tricks Wiki article, which revisits a theme that I treated on my web page. Imre Leader pointed out to me that I hadn’t completely done justice to the idea, and that the notion of a “just-do-it proof” had some more specific features that I had not sufficiently emphasized. His opinion matters to me since he was the one who told me about the concept. He himself got it from Béla Bollobás: I don’t know whether it goes further back than that. This is a second attempt at explaining it. Imre, if there’s anything you don’t like about this one, then please edit it when it appears on the Tricks Wiki. Title: Just-do-it proofs. Quick description: If you are asked to prove that a sequence or a set exists with certain properties, then the best way of doing so may well be not to use any tricks but just to go ahead and do it: that is, you build the set/sequence up one element at a time, and however you do so you find that it is never difficult to continue building. (more…) Posted in Mathematical pedagogy, Mathematics on the internet \| 23 Comments » ### How to use Zorn’s lemma August 12, 2008 I am continuing my series of sample articles for the Tricks Wiki with one that is intended to represent a general class of such articles. It is common practice in lecture courses (at least if the ones I attended as an undergraduate are anything to go by) to state useful theorems, lemmas, propositions, etc., without going to much trouble to explain why they are useful. Of course, there are many ways to pick up this further understanding: taking note of where and how such results are used, doing carefully designed exercises, and so on. Nevertheless, it is often the case that more could be done to help people recognise the signs that indicate that a particular result can be applied. This article, which is principally aimed at undergraduates early on in a mathematics degree, is inspired by an experience I myself had as an undergraduate. I had a sheet of challenging problems (set by Béla Bollobás) and one of them completely stumped me. (I’m sure several of them stumped me but this is the one that sticks in my mind.) I can’t remember exactly what the question was, but it was something of similar difficulty to that of determining whether an additive function from $\mathbb{R}$ to $\mathbb{R}$ was necessarily linear. My supervision partner solved the problem using Zorn’s lemma, which we had been told about in a lecture, and I just sat there in disbelief because it hadn’t even remotely occurred to me that Zorn’s lemma might be useful. At some point in the intervening years, I “got” Zorn’s lemma and now find it straightforward to see where it is needed. This article is intended to speed up that process for other people. (more…) Posted in Mathematical pedagogy, Mathematics on the internet \| 35 Comments » ### A small countability question August 10, 2008 This is a short post to ask a simple question that arises out of the discussion in a previous post about countability. As is well-known, the familiar statement that a countable union of countable sets is countable requires the axiom of countable choice. Indeed, it comes in in the very first step of the proof, where one says something along the lines of, “For each set $A_n$ let $a_{n1},a_{n2},\dots$ be an enumeration of its elements.” This uses the axiom of choice because if we don’t know anything about the sets $A_n$ then we can’t actually define these enumerations: we just have to assert that a sequence of enumerations exists. However, if we do have explicit enumerations of the sets $A_n$ then the proof yields for us an explicit enumeration of their union. So one might take the following attitude to this particular application of the axiom of choice: the real theorem is “An explicitly enumerated union of explicitly enumerated sets is explicitly enumerated,” but because we often care only that enumerations should exist and don’t want to keep having to define artificial ones, it is convenient to appeal to the axiom of choice so that we can extend the theorem to the murky world of countable but not explicitly enumerated sets. (more…) Posted in Uncategorized \| 18 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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 6, "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": 0.9512810111045837, "perplexity_flag": "middle"}
http://jmanton.wordpress.com/tag/measure-theory/	# Jonathan Manton's Blog Thoughts and ideas on mathematics, engineering, neuroscience and academia ### Archive Posts Tagged ‘Measure Theory’ ## Measure-theoretic Probability: Still not convinced? June 22, 2010 This is a sequel to the introductory article on measure-theoretic probability and accords with my belief that learning should not be one-pass, by which I mean loosely that it is more efficient to learn the basics first at a rough level and then come back to fill in the details soon afterwards. It endeavours to address the questions: • Why a probability triple $(\Omega,\mathfrak{F},\mathbb{P})$ at all? • What if $\mathfrak F$ is not a $\sigma$-algebra? • Why is it important that $\mathbb P$ is countably additive? In addressing these questions, it also addresses the question: • Why can’t a uniform probability be defined on the natural numbers $\{0,1,2,\cdots,\infty\}$? Consider a real-life process, such as the population $X_k$ of a family of rabbits at each generation $k$. This gives us a countable family of random variables $\{X_1,X_2,\cdots\}.$ (Recall that countable means countably infinite; with only a finite number of random variables, matters would be simpler.) We can safely assume that if $X_k = 0$ for some $k$ then the population has died out, that is, $X_{k+1} = X_{k+2} = \cdots = 0.$ What is the probability that the population dies out? The key questions here are the implicit questions of how to actually define and then subsequently calculate this probability of extinction. Intuitively, we want the probability that there exists an $m$ such that $X_m = 0.$ When trying to formulate this mathematically, we may think to split this up into bits such as “does $X_1 = 0$?”, “does $X_2 = 0$?” and so forth. Because these events are not disjoint (if we know $X_1 = 0$ then we are guaranteed that $X_2 = 0$) we realise that we need some way to account for this “connection” between the random variables. Is there any better way of accounting for this “connection” other than by declaring the “full” outcome to be $\omega \in \Omega$ and interpreting each $X_k$ as a function of $\omega$? (Only by endeavouring to think of an alternative will the full merit of having an $\Omega$ become clear.) There are (at least) two paths we could take to define the probability of the population dying out. The first was hinted at already; segment $\Omega$ into disjoint sets then add up the probabilities of each of the relevant sets. Precisely, the sets $F_1 = \{\omega \in \Omega \mid X_1(\omega) = 0\}$, $F_2 = \{\omega \in \Omega \mid X_1(\omega) \neq 0, X_2(\omega) = 0\}$, $F_3 = \{\omega \in \Omega \mid X_2(\omega) \neq 0, X_3(\omega) = 0\}$ and so forth are disjoint, and we are tempted to sum the probabilities of each one occurring to arrive at the probability of extinction. This is an infinite summation though, so unless we believe that probability is countably additive (recall that this means $\mathbb{P}(\cup_{i=1}^\infty F_i) = \sum_{i=1}^\infty \mathbb{P}(F_i)$ for disjoint sets $F_k$) then this avenue is not available. Another path is to recognise that the sets $B_k = \{\omega \in \Omega \mid X_k(\omega) = 0\}$ are like Russian dolls, one inside the other, namely $B_1 \subset B_2 \subset B_3 \subset \cdots.$ This means that their probabilities, $\mathbb{P}(B_k)$, form a non-decreasing sequence, and moreover, we are tempted to believe that $\lim_{k \rightarrow \infty} \mathbb{P}(B_k)$ should equal the probability of extinction. (The limit exists because the $\mathbb{P}(B_k)$ form a bounded and monotonic sequence.) In fact, these paths are equivalent; if $\mathbb P$ is countably additive and the $B_k$ are nested as above then $\mathbb{P}(\cup_{k=1}^\infty B_k) = \lim_{k \rightarrow \infty} \mathbb{P}(B_k)$ and the converse is true too; if for any sequence of nested sets the probability and the limit operations can be interchanged (which is how the statement $\mathbb{P}(\cup_{k=1}^\infty B_k) = \lim_{k \rightarrow \infty} \mathbb{P}(B_k)$ should be interpreted) then $\mathbb P$ is countably additive. Essentially, we have arrived at the conclusion that the only sensible way we can define the probability of extinction is to agree that probability is countably additive and then carry out the calculations above. Without countable additivity, there does not seem to be any way of defining the probability of extinction in general. The above argument in itself is intended to complete the motivation for having a probability triple; the $\Omega$ is required to “link” random variables together and countable additivity is required in order to model real-world problems of interest. The following section goes further though by giving an example of when countable additivity does not hold. ### A Uniform Distribution on the Natural Numbers For argument’s sake, let’s try to define a “probability triple” $(\Omega,\mathfrak{F},\mathbb{P})$ corresponding to a uniform distribution on the natural numbers $\Omega = \{0,1,2,\cdots,\infty\}$. The probability of drawing an even number should be one half, the probability of drawing an integer multiple of 3 should be one third, and so forth. Generalising this principle, it seems entirely reasonable to define $\mathbb{P}(F)$ to be the limit, as $N \rightarrow \infty$, of the number of elements of $F$ less than $N$ divided by $N$ itself. Since this limit does not necessarily exist, we solve it by declaring $\mathfrak F$ to be the set of all $F \subset \Omega$ for which this limit exists. It can be shown directly that $\mathfrak F$ is not a $\sigma$-algebra. In fact, it is not even an algebra because it is relatively straightforward to construct two subsets of $\Omega$, call them $A$ and $B$, which belong to $\mathfrak F$ but whose intersection does not, that is, there exist $A, B \in \mathfrak F$ for which $A \cap B \not\in \mathfrak F$. Does $\mathbb{P}$ behave nicely? Let $B_k = \{0,\cdots,k\}$ and observe that $B_1 \subset B_2 \subset \cdots$ and $\Omega = \cup_{i=1}^{\infty} B_k.$ We know from the earlier discussion about extinction that it is very natural to expect that $\lim_{k \rightarrow \infty} \mathbb{P}(B_k) = \mathbb{P}(\Omega)$. However, this is not the case here; since each of the $B_k$ contain only a finite number of elements, it follows that $\mathbb{P}(B_k) = 0$. Therefore, the limit on the left hand side is zero whereas the right hand side is equal to one. In summary: • Countable additivity enables us to give meaning to probabilities of real-world events of interest to us (such as probability of extinction). • Without countable additivity, even very basic results such as $\mathbb{P}(\cup_{k=1}^\infty B_k) = \lim_{k \rightarrow \infty} \mathbb{P}(B_k)$ for nested $B_k$ need not hold. In other words, there are not enough constraints on $\mathbb P$ for a comprehensive theory to be developed if we drop the requirement of $\mathbb{P}$ being countably additive over a $\sigma$-algebra $\mathfrak F$. ## Measure-theoretic Probability: Why it should be learnt and how to get started June 4, 2010 Last Friday I gave an informal 2-hour talk at the City University of Hong Kong on measure-theoretic probability. The main points were as follows. Comments on which parts are unclear or how better to explain certain concepts are especially welcome. ### Objectives • Understand why measure-theoretic probability is useful • Learn enough to get past the initial barrier to self-learning • Motivation • Road map ### Recommended Textbooks The primary textbook I recommend is “Probability with Martingales” by David Williams. Although out of print, a secondary textbook I recommend is Wong and Hajek’s “Stochastic Processes in Engineering Systems“. ### Motivation One unattractive feature of traditional probability theory is that discrete and continuous random variables are generally treated separately and thus some care is required when studying mixtures of discrete and continuous random variables. Measure-theoretic probability provides a unified framework which is ultimately easier to work with rigorously. (In other words, and roughly speaking, fewer lines of mathematics are required and the chance of making a mistake is decreased.) A simple example of moving to a more general setting is given by the real and complex numbers. Initially, complex numbers were treated with some scepticism. Ultimately though, by generalising real numbers to complex numbers, a range of fundamental concepts became simpler and more natural. To state just one, an $n$th degree polynomial has precisely $n$ roots (counting multiplicities) over the complex field, but possibly fewer over the real field. ### Derivation Journal papers using measure-theoretic probability often start by saying, “Let $(\Omega,\mathfrak{F},\mathbb{P})$ be a probability space”. This section endeavours to derive (or re-discover) this formalism. #### The Outcome ($\omega \in \Omega$) At least for an engineer, it is benign to assume that even if a variable or process is random and not observed directly, it still has a true and actual outcome in every experiment. For the purposes of measure-theoretic probability, it is convenient (and unrestrictive) to assume that the outcomes of a series of bets placed by a gambler are known beforehand to Tyche, the Goddess of Chance. Formally, the actual outcome is denoted by a point $\omega$ drawn from the set of all possible outcomes $\Omega$. If the experiment consists of just a single coin toss, $\Omega$ might contain just two elements, say $\Omega = \{H,T\}$. (There is no reason why $\Omega$ could not contain more elements; they would merely be deemed to occur with probability zero. While this might be a silly thing to do when the set of possible outcomes is finite, there are sometimes advantages in choosing $\Omega$ larger than it needs to be in the infinite case, such as when working with stochastic processes.) Rarely though is a single toss of a coin interesting. Normally, the coin is tossed two or more times. It is important that $\Omega$ contains as many outcomes as necessary to describe the full sequence of events. So if the coin is tossed twice and only twice, it suffices to choose $\Omega = \{HH,HT,TH,TT\}$. Generally though, one would want to allow for the coin to be tossed any number of times, in which case $\Omega$ would contain all possible infinitely-long sequences of Heads and Tails. (This is denoted $\Omega = \{H,T\}^\infty$.) #### The Probability of the Outcome ($\mathbb{P}$) We must somehow characterise the fact that Tyche will choose some outcomes $\omega \in \Omega$ more frequently than others. If $\Omega$ is finite, there is an obvious way to do this. We could simply define $\mathbb P$ to be a function from $\Omega$ to the set of real numbers between $0$ and $1$ inclusively, the latter denoted by $[0,1]$. This generally does not work though if $\Omega$ consists of an infinite number of elements. To see why, assume Tyche will choose a number uniformly between $2$ and $3$. Then we may take $\Omega = [2,3]$. We would be forced to assign a probability of zero though to any particular outcome $\omega \in \Omega$. Therefore, there is not enough information to deduce that the probability of Tyche choosing a number in the set $[2.1,2.2]$ is precisely the same as the probability of choosing a number in the set $[2.8,2.9]$, for instance. Going to the other extreme, we may be tempted to solve the problem by defining the probability of occurrence of any conceivable set of outcomes. So for instance, we can define $\mathbb{P}([2.1,2.2]) = 0.1$ and $\mathbb{P}([2.8,2.9]) = 0.1$, and indeed, for any interval from $a$ to $b$ with $2 \leq a \leq b \leq 3$ we can define $\mathbb{P}([a,b]) = b-a$. Notice that now, we have made $\mathbb P$ a function which takes a subset of $\Omega$ and returns a number between $0$ and $1$. So strictly speaking, we must write $\mathbb P(\{2.5\})$ and not $\mathbb P(2.5)$ for the probability of occurrence of an individual element of $\Omega$ . Superficially, this is ok. However, it does not work for two reasons. 1. How can we define the value of $\mathbb P(A)$ for an arbitrary subset $A$ of $\Omega$ when for some sets, it is not even possible to write down a description of them? (That is, there are some subsets of the interval $[2,3]$ which we cannot even write down, so how can we even write down a definition of $\mathbb P$ which tells us what value it takes on such indescribable sets?) 2. It can be proved that there exist “bad” sets for which it is impossible to assign a probability to them in any consistent way. It is very tempting to elaborate on the second point above. However, my experience is that doing so distracts too much attention from the original aim of understanding measure-theoretic probability. It is therefore better to think that even if we could assign a probability to every possible subset, we do not want to because it would cause unnecessary trouble and complication; surely, provided we have enough interesting subsets to work with, that is enough? Therefore, ultimately we define $\mathbb P$ as a function from $\mathfrak F$ to $[0,1]$ where $\mathfrak F$ is a set of subsets of $\Omega$ which we think of as (some of) the “nice” subsets of $\Omega$, that is, subsets of $\Omega$ to which we can and want to assign probabilities of occurrence. Roughly speaking, $\mathfrak F$ should be just large enough to be useful, and no larger. #### The Set of Nice Subsets ($\mathfrak F$) Referring to what was said just before, how should we choose $\mathfrak F$? Experience suggests that if $\Omega = [2,3]$ then we would generally be interested in all open intervals $(a,b)$ and all closed intervals $[a,b]$ for starters. (Open intervals do not include their endpoints whereas closed intervals do.) We would also want to be able to take (finite) unions and intersections of such sets. This may well be enough already. However, we should also look at our requirements on $\mathbb P$ since they will have an effect on how we choose $\mathfrak F$. They are: 1. $\mathbb{P}(\{\}) = 0$. (The probability of $\omega$ being in the empty set is zero.) 2. $\mathbb{P}(\Omega) = 1$. (The probability of $\omega$ being in the set of all possible outcomes $\Omega$ is one.) 3. $\mathbb{P}( \cup_{i=1}^\infty F_i ) = \sum_{i=1}^\infty \mathbb{P}( F_i)$ whenever the $F_i$ are mutually disjoint subsets of $\Omega$. (Probability is countably additive.) In order even to be able to state these properties rigorously, we require $\mathfrak F$ to have certain properties. In particular, the first two conditions only make sense if we insist that both the empty set $\{\}$ and $\Omega$ are elements of $\mathfrak F$. (Recall that $\mathbb P$ is only defined on elements of $\mathfrak F$.) The third condition requires that if $F_i \in \mathfrak F$ then $\cup_{i=1}^\infty F_i \in \mathfrak F$. (Technically, we have only argued for this in the special case of the $F_i$ being mutually disjoint, but it ultimately turns out to be no different from requiring it to hold for non-disjoint sets too.) Note that the third condition implies (finite) additivity; just choose most of the $F_i$ to be the empty set. Therefore, if $A \in \Omega$ and if $\Omega \backslash A$ (the complement of $A$) is also in $\mathfrak F$ then properties 2 and 3 above would impy that $\mathbb{P}(\Omega \backslash A) = 1 - \mathbb{P}(A)$. It is easy to believe that this condition is fundamental enough to insist that if $A \in \mathfrak F$ then its complement $\Omega \backslash A$ is also in $\mathfrak F$. Once we have complements of sets in $\mathfrak F$ also belonging to $\mathfrak F$, then (finite and countable) intersections of sets in $\mathfrak F$ also belong to $\mathfrak F$. (Recall that $A \cap B = \Omega \backslash ( (\Omega \backslash A) \cup (\Omega \backslash B))$, for example.) To summarise the last paragraph, we have endeavoured to show that we require $\mathfrak F$ to satisfy the following conditions. 1. $\Omega \in \mathfrak F$. 2. $A \in \mathfrak F$ implies $\Omega \backslash A \in \mathfrak F$. 3. $A_i \in \mathfrak F$ implies $\cup_{i=1}^\infty A_i \in \mathfrak F$. These conditions are precisely those required for $\mathfrak F$ to be what is known as a $\sigma$-algebra. Here, $\sigma$ is used to denote the word “countable” and refers to condition 3 above. (While the alternative term $\sigma$-field is widely used, the existing definitions of “algebra” and “field” in mathematics makes the term $\sigma$-algebra the preferred term; it is not a “field” in any precise sense.) If $\Omega = [2,3]$, recall from above that we wished for $\mathfrak F$ to contain all the intervals at the very least. Therefore, we choose $\mathfrak F$ to be the smallest $\sigma$-algebra containing the intervals. (Intuitively, one could think of building $\mathfrak F$ up by starting with $\mathfrak F$ being equal to the set of all intervals, then adding all complements, then adding all countable unions, then adding all complements of these new sets, then adding all countable unions of new and old sets, and going on like this until finally $\mathfrak F$ grew no larger by repeating this process. Mathematically though, it is constructed by taking the intersection of all $\sigma$-algebras containing the intervals; it can be shown that the (uncountable) intersections of $\sigma$-algebras is still a $\sigma$-algebra.) In general, if $\Omega$ is a topological space then it is common to choose $\mathfrak F$ to be the smallest $\sigma$-algebra containing all the open sets. This is called the Borel $\sigma$-algebra generated by the open sets on $\Omega$. The elements of $\mathfrak F$ are called events. Indeed, an event $B \in \mathfrak F$ is a subset of $\Omega$ and therefore represents a set of possible outcomes or events that we might observe (we might be told that $\omega \in B$), or ask the probability of observing (we might want to know the value of $\mathbb{P}(B)$). #### How to Define $\mathbb P$ on Borel Subsets One issue remains; in general, it is not possible to write down an arbitrary Borel subset; some Borel sets are indescribable. How then can we define $\mathbb P$ on sets we cannot describe? Fortunately, we can appeal to Caratheodory’s Extension Theorem. In fact, this is a repeating theme in measure-theoretic probability; it is necessary to learn techniques for avoiding the need to work directly with indescribable sets. Caratheodory’s Extension Theorem implies that if we assign a probability to every interval in $\Omega = [2,3]$ (in a way which is consistent with the axioms for probability, e.g., respecting countable additivity) then there is one and only one way to extend the assignments of probability to arbitrary Borel subsets of $[2,3]$. In other words, by defining $\mathbb P$ just for intervals, we have implicitly defined $\mathbb P$ on all Borel subsets. (This is analogous to defining a linear function at only a handful of points; the linearity of the function means that the value of the function can be deduced at other points by using the property of linearity.) Note that $\mathbb P$ is called a probability measure. It “measures” the probability assigned to certain nice subsets of $\Omega$, or precisely, to the elements of $\mathfrak F$. (Recall that every element of $\mathfrak F$ is a subset of $\Omega$.) ### Random Variables A (real-valued) random variable is simply a function from $\Omega$ to $\mathbb R$ which satisfies a natural condition of being measurable, which will be defined presently. First though, note that a random variable gives (generally only partial) information about the outcome $\omega$. For example, if $\Omega=\{HH,HT,TH,TT\}$ and $X_1: \Omega \rightarrow \mathbb R$ is defined by $X_1(HH) = 1$, $X_1(HT) = 1$, $X_1(TH) = 0$ and $X_1(TT) = 0$ then we would describe $X_1$ as the outcome of the first coin toss (with $1$ for Heads and $0$ for Tails). We know from the previous section that when we are dealing with the set of real numbers $\mathbb R$, we would like to be able to assign a probability to any Borel subset of $\mathbb R$. Therefore, given a random variable $X: \Omega \rightarrow \mathbb R$ and a Borel subset $B$, we would like to compute the probability that the outcome $\omega$ causes $X$ to take on a value in the set $B$. Mathematically, this is written as $\mathbb{P}(\{\omega \mid X(\omega) \in B\})$, which is commonly abbreviated as $\mathbb{P}(X^{-1}(B))$. For this to make sense though, we must have $X^{-1}(B) = \{\omega \mid X(\omega) \in B\}$ being an element of $\mathfrak F$. This condition, that the inverse image of a Borel set lies in the $\sigma$-algebra $\mathfrak F$, is precisely the condition of measurability imposed on any random variable. ### Expectation is Central Although just formulating the probability triple $(\Omega, \mathfrak{F}, \mathbb{P})$ is already enough to unify discrete and continuous-valued random variables, there are other differences between measure-theoretic and “classical” probability. In particular, in measure-theoretic probability, emphasis shifts to the expectation and conditional expectation operators. One benefit of doing this is that it avoids certain unpleasantries associated with defining conditional probability; for example, Bayes rule does not apply when the denominator is zero. Note that the probability $\mathbb{P}(B)$ of an event $B \in \mathfrak F$ occurring is equal to the expected value of $I_B(\omega)$ where $I$ denotes the indicator function; $I_B(\omega)$ equals $1$ when $\omega \in B$ and $0$ otherwise. Therefore, the shift from probability being central to expectation being central is merely a change of view; it often provides a nicer view of the same underlying theory. ### Concluding Remarks • Measure-theoretic probability is initially more complicated to learn, but it is rigorous, more natural and therefore ultimately easier to work with. • Its advantages come from its different and more general viewpoint; the underlying theory is still essentially the same as classical probability. • (Rather than work with cumulative probability distributions and Riemann-Stieltjes integrals, measure-theoretic probability works with probability measures and Lebesgue integrals which are generally cleaner and easier to work with.) • When learning measure-theoretic probability: • Keep in mind that the basic ideas are straightforward; don’t let the technical detail obscure the basic ideas. • Most of the technical detail comes (at least initially) from having to work with Borel sets but not being able to describe them in general (cf., Caratheodory’s Extension Theorem mentioned earlier). • Look for and develop your own mapping between the measure-theoretic way of obtaining a result, and the classical way. (For example, Girsanov’s Change of Measure is essentially the measure-theoretic version of Bayes rule; it is stated in terms of conditional expectation rather than conditional probability and is therefore neater to work with.) 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http://unapologetic.wordpress.com/2010/11/25/induced-matrix-representations/?like=1&source=post_flair&_wpnonce=5d5a17dad5	# The Unapologetic Mathematician ## Induced Matrix Representations Sorry I missed posting this back in the morning… We want to work out the matrices of induced representations. Explicitly, if $V$ is a left $H$-module of degree $d$, where $H$ is a subgroup of $G$, then $V\!\!\uparrow_H^G$ is a left $G$-module. If we pick a basis of $V$, we get a matrix representation $X:H\to\mathrm{Mat}_d(\mathbb{C})$. We want to describe a matrix representation corresponding to $V\!\!\uparrow_H^G$. In the process, we’ll see that we were way off with our first stabs at the dimensions of tensor products over $H$. The key point is to realize that $\mathbb{C}[G]$ is a free right module over $\mathbb{C}[H]$. That is, we can find some collection of vectors in $\mathbb{C}[G]$ so that any other one can be written as a linear collection of these with coefficients (on the right) in $\mathbb{C}[H]$. Indeed, we can break $G$ up into the $\lvert G\rvert/\lvert H\rvert$ left cosets of $H$. Picking one representative $t_i$ of each coset — we call this a “transversal” for $H$ — we have essentially chopped $\mathbb{C}[G]$ up into chunks, each of which looks exactly like $\mathbb{C}[H]$. To see this, notice that the coset $t_iH$ is a subset of $G$. Thus it describes a subspace of $\mathbb{C}[G]$ — that spanned by the elements of the coset, considered as basis vectors in the group algebra. The action of $H$ on $\mathbb{C}[G]$ shuffles the basis vectors in this coset around amongst each other, and so this subspace is invariant. It should be clear that it is isomorphic to $\mathbb{C}[H]$, considered as a right $H$-module. Okay, so when we consider the tensor product $\mathbb{C}[G]\otimes_HV$, we can pull any action by $H$ across to the right and onto $V$. What remains on the left? A vector space spanned by the transversal elements $\{t_i\}$, which essentially index the left cosets of $H$ in $G$. We have one copy of $V$ for each of these cosets, and so the dimension of the induced module $V\!\!\uparrow_H^G$ is $d\lvert G\rvert/\lvert H\rvert$. How should we think about this equation, heuristically? The tensor product multiplies the dimensions of vector spaces, which gives $d\lvert G\rvert$. Then the action of $H$ on the tensor product divides by a factor of $\lvert H\rvert$ — at least in principle. In practice, this only works because in our example the action by $H$ is free. That is, no element in the bare tensor product $\mathbb{C}[G]\otimes V$ is left fixed by any non-identity element of $H$. So how does this give us a matrix representation of $G$? Well, $g$ acts on $\mathbb{C}[G]$ by shuffling around the subspaces that correspond to the cosets of $H$. In fact, this is exactly the coset representation of $G$ corresponding to $H$! If we write $g=t_ih$ for some $i$, then this uses up the transversal element $t_i$. The $h$ is left to “pass through” and act on $V$. To write this all out explicitly, we get the following block matrix: $\displaystyle X\!\!\uparrow_H^G(g)=\begin{pmatrix}X(t_i^{-1}gt_j)\end{pmatrix}=\left(\begin{array}{cccc}X(t_1^{-1}gt_1)&X(t_1^{-1}gt_2)&\cdots&X(t_1^{-1}gt_n)\\X(t_2^{-1}gt_1)&X(t_2^{-1}gt_2)&\cdots&X(t_2^{-1}gt_n)\\\vdots&\vdots&\ddots&\vdots\\X(t_n^{-1}gt_1)&X(t_n^{-1}gt_2)&\cdots&X(t_n^{-1}gt_n)\end{array}\right)$ where $n$ is the number of cosets, and we simply define $X(t_i^{-1}gt_j)$ to be a zero block if $t_i^{-1}gt_j$ does not actually fall into $H$. ## 5 Comments » 1. [...] for the induced character, we use the matrix of the induced representation that we calculated last time. If is a matrix representation of a group , which is a subgroup , [...] Pingback by \| November 29, 2010 \| Reply 2. [...] we must check that induction is additive. Here, the induced matrices will come in handy. If and are matrix representations of , then the direct sum is the matrix [...] Pingback by \| December 1, 2010 \| Reply 3. [...] we’ve defined . But if we choose a transversal for — like we did when we set up the induced matrices — then we can break down as the direct sum of a bunch of copies of [...] Pingback by \| December 3, 2010 \| Reply 4. [...] we have a matrix representation, so we look at the induced matrix representation. We have to pick a transversal for the subgroup in . Then we have the induced matrix in block [...] Pingback by \| December 6, 2010 \| Reply 5. [...] its dimension is a little easier if we recall what induction looks like for matrix representations: the direct sum of a bunch of copies of , one [...] Pingback by \| February 1, 2011 \| Reply « Previous \| Next » ## About this weblog This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”). I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 60, "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": 0.9283446669578552, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/71958/why-is-this-function-differentiable?answertab=oldest	# Why is this function differentiable? I know that the piecewise function $f(x) = \left\{ \begin{array}{lr} 2x & , x < 1\\ x+1 & , x \geq 1 \end{array} \right.$ is differentiable because $2x$ and $x + 1$ converge at different rates based on their slope but why is $f(x) = \left\{ \begin{array}{lr} \frac{x^2-1}{x-1} & , x \neq 1\\ 2 & , x=1 \end{array} \right.$ differentiable? I know that it is continuous but their slopes are different. Thanks! - 3 Your second $f$ is the same as $x\mapsto x+1$ which is clearly differentiable everywhere. – LostInMath Oct 12 '11 at 10:22 5 $\displaystyle f(x)=\left\{\begin{array}{}2x&\text{if }x<1\\x+1&\text{if }x\ge1\end{array}\right.$ is not differentiable at $x=1$. $\displaystyle\lim_{h\to0}\frac{f(1+h)-f(1)}{h}$ does not exist. – robjohn♦ Oct 12 '11 at 10:37 ## 1 Answer Don't be tricked by the seemingly complicated piecewise definition of the function $f:\mathbb{R}\to\mathbb{R}$, $$f(x) = \left\{ \begin{array}{lr} \frac{x^2-1}{x-1} & , x \neq 1\\ 2 & , x=1. \end{array} \right.$$ It is exactly the same function as $g:\mathbb{R}\to\mathbb{R}$, $g(x)=x+1$, since $$\left\{ \begin{array}{lr} \frac{x^2-1}{x-1}=\frac{(x+1)(x-1)}{x-1}=x+1=g(x) & , x \neq 1\\ 2 = 1+1 = x+1 = g(x)& , x=1. \end{array} \right.$$ Obviously the function $g$ is differentiable everywhere, and hence $f$ is too, since they are the one and same. - Oh, okay thanks! I got stuck there for a while. – Juan de la John Oct 13 '11 at 6:59	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 14, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "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": 0.9692583680152893, "perplexity_flag": "head"}
http://mathhelpforum.com/calculus/121095-finding-area.html	# Thread: 1. ## finding the area I am not sure what to do here for this problem. Sketch the region enclosed by x+ y^2= 42 and x+y = 0. Decide whether to integrate with respect to x or y. Then find the area of the region. Any help would be great 2. Originally Posted by mbrez88 I am not sure what to do here for this problem. Sketch the region enclosed by x+ y^2= 42 and x+y = 0. Decide whether to integrate with respect to x or y. Then find the area of the region. Any help would be great First thing I would do is look at the graph and notice that it will be easier to do horizontal slicing. Next find your points of intersection $42-y^2 = -y$ $\int (42 -y^2 + y)dy$	{"extraction_info": {"found_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": 0.9245606064796448, "perplexity_flag": "head"}
http://wikis.controltheorypro.com/index.php?title=Proportional_Control	• Search » • • Toolbox » • In other languages » Proportional Control From ControlTheoryPro.com \| \| \| \| \|----------------------------------------------------------------------------------------\|------\|-------------------\| \| Proportional Control \| \| \| \| \| SISO \| Controller Design \| \| \| \| \| \| In order to prevent spam, users must register before they can edit or create articles. \| \| \| 1 Introduction to Proportional Controller Design Figure 1: Standard Block Diagram A proportional controller provides fast response to system error. The draw back to the proportional controller is that it cannot drive the system to a 0 steady-state error. I've been told that calibration can significantly reduce the residual error left by a proportional controller but I haven't done this myself. 2 Proportional Control Design The equation for a proportional controller are below $LaTeX: K\left(s\right)=k_{p}$ where $LaTeX: k_{p}$ is the proportional gain. The Pole Placement design approach to controller design makes it clear that the proportional gain can change the closed loop poles. However, a simple gain can change the closed loop poles to only a very limited set of poles. My suggestion would be to use MATLAB to plot the possible closed loop poles. From that list of possible poles pick the set that produces the best performance.	{"extraction_info": {"found_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": 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": 0.8776284456253052, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/108200/where-should-the-exponent-be-written-in-numbers-with-units-of-measurement/108202	# Where should the exponent be written in numbers with units of measurement? If you are to calculate the hypotenuse of a triangle, the formula is: $h = \sqrt{x^2 + y^2}$ If you don't have any units for the numbers, replacing x and y is pretty straightforward: $h = \sqrt{4^2 + 6^2}$ But what if the numbers are in meters? $h = \sqrt{4^2m + 6^2m}$ (wrong, would become $\sqrt{52m}$) $h = \sqrt{4m^2 + 6m^2}$ (wrong, would become $\sqrt{10m^2}$) $h = \sqrt{(4m)^2 + (6m)^2}$ (correct, would become $\sqrt{52m^2}$) Or should I just ignore the unit of measurement in these cases? - @Pjennings: The Pythagorean Theorem is really about areas (the area of the square on the hypotenuse is equal to the sums of the areas of the squares on the legs), so it does make some sense to have $\text{m}^2$ along the way... – Isaac Feb 11 '12 at 18:22 – Raymond Manzoni Feb 11 '12 at 18:25 @Isaac: What I meant is that his question deals with lengths, so his final answer should be in terms a meters, a unit of length. – Holdsworth88 Feb 11 '12 at 18:27 @RaymondManzoni Thanks, I have updated sqrt() to sqrt{} – CheeseSucker Feb 11 '12 at 18:29 ## 3 Answers Suppose you have been given $x$ and $y$ in metres, and you'd like to know the quantity, $z=\sqrt{x^2+y^2}$. Then, as you have predicted this quatity will be in metres. Two things have been involved: Homogeneity of Dimension Two quantities of different dimensions cannot be added. This is one of the axioms of numerical physics. Example: It is clear that adding $5$ metres to $3$ seconds does not give a physically meaningful quantity that can be interpreted in real life. Certain functions only take values in dimensionless quantities For instance, $\sin (\sqrt{x^2+y^2})$ would not make sense even if $x$ and $y$ have same dimensions. This is a bit subtler, but this is what it is! • Coming to your question, the first quantity you tell us in dimension of $m^{1/2}$ which is against your guess! • The second quantity is dimensionally fine while numerically this is not what you want. • The third quantity is fine in all ways. My suggestion: First manipulate the numbers and then the units separately. This is a good practice in Numerical Physics. The other answers have done it all at one go. But, I don't prefer it that way! - How do you avoid unit conversion errors (and other unit errors) if you manipulate the units and numbers separately? – Rex Kerr Feb 11 '12 at 20:41 The last one: $$\begin{align} h&=\sqrt{(4\text{ m})^2+(6\text{ m})^2} \\ &=\sqrt{16\text{ m}^2+36\text{ m}^2} \\ &=\sqrt{52\text{ m}^2} \\ &=2\sqrt{13}\text{ m} \end{align}$$ - The last one is correct. The general idea is that you treat the unit as if it were another number: $$\begin{align}h &= \sqrt{(4\mathrm{m})^2+(6\mathrm{m})^2} \\ &= \sqrt{16\mathrm{m}^2 + 36\mathrm{m}^2} \\ &= \sqrt{52 \mathrm{m}^2} \\ &= \sqrt{52}\sqrt{\mathrm{m}^2} \\ &= \sqrt{52}\mathrm{m}\end{align}$$ -	{"extraction_info": {"found_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": 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": 0.9497596621513367, "perplexity_flag": "middle"}
http://johncarlosbaez.wordpress.com/2011/11/04/network-theory-part-16/	# Azimuth ## Network Theory (Part 16) We’ve been comparing two theories: stochastic mechanics and quantum mechanics. Last time we saw that any graph gives us an example of both theories! It’s a bit peculiar, but today we’ll explore the intersection of these theories a little further, and see that it has another interpretation. It’s also the theory of electrical circuits made of resistors! That’s nice, because I’m supposed to be talking about ‘network theory’, and electrical circuits are perhaps the most practical networks of all: I plan to talk a lot about electrical circuits. I’m not quite ready to dive in, but I can’t resist dipping my toe in the water today. Why don’t you join me? It’s not too cold! #### Dirichlet operators Last time we saw that any graph gives us an operator called the ‘graph Laplacian’ that’s both infinitesimal stochastic and self-adjoint. That means we get both: • a Markov process describing the random walk of a classical particle on the graph. and • a 1-parameter unitary group describing the motion of a quantum particle on the graph. That’s sort of neat, so it’s natural to wonder what are all the operators that are both infinitesimal stochastic and self-adjoint. They’re called ‘Dirichlet operators’, and at least in the finite-dimensional case we’re considering, they’re easy to completely understand. Even better, it turns out they describe electrical circuits made of resistors! Today let’s take a lowbrow attitude and think of a linear operator $H : \mathbb{C}^n \to \mathbb{C}^n$ as an $n \times n$ matrix with entries $H_{i j}$. Then: • $H$ is self-adjoint if it equals the conjugate of its transpose: $H_{i j} = \overline{H}_{j i}$ • $H$ is infinitesimal stochastic if its columns sum to zero and its off-diagonal entries are real and nonnegative: $\displaystyle{ \sum_i H_{i j} = 0 }$ $i \ne j \Rightarrow H_{i j} \ge 0$ • $H$ is a Dirichlet operator if it’s both self-adjoint and infinitesimal stochastic. What are Dirichlet operators like? Suppose $H$ is a Dirichlet operator. Then its off-diagonal entries are $\ge 0$, and since $\displaystyle{ \sum_i H_{i j} = 0}$ its diagonal entries obey $\displaystyle{ H_{i i} = - \sum_{ i \ne j} H_{i j} \le 0 }$ So all the entries of the matrix $H$ are real, which in turn implies it’s symmetric: $H_{i j} = \overline{H}_{j i} = H_{j i}$ So, we can build any Dirichlet operator $H$ as follows: • Choose the entries above the diagonal, $H_{i j}$ with $i < j$, to be arbitrary nonnegative real numbers. • The entries below the diagonal, $H_{i j}$ with $i > j$, are then forced on us by the requirement that $H$ be symmetric: $H_{i j} = H_{j i}$. • The diagonal entries are then forced on us by the requirement that the columns sum to zero: $H_{i i} = - \sum_{ i \ne j} H_{i j}$. Note that because the entries are real, we can think of a Dirichlet operator as a linear operator $H : \mathbb{R}^n \to \mathbb{R}^n$. We’ll do that for the rest of today. #### Circuits made of resistors Now for the fun part. We can easily draw any Dirichlet operator! To this we draw $n$ dots, connect each pair of distinct dots with an edge, and label the edge connecting the $i$th dot to the $j$th with any number $H_{i j} \ge 0:$ This contains all the information we need to build our Dirichlet operator. To make the picture prettier, we can leave out the edges labelled by 0: Like last time, the graphs I’m talking about are simple: undirected, with no edges from a vertex to itself, and at most one edge from one vertex to another. So: Theorem. Any finite simple graph with edges labelled by positive numbers gives a Dirichlet operator, and conversely. We already talked about a special case last time: if we label all the edges by the number 1, our operator $H$ is called the graph Laplacian. So, now we’re generalizing that idea by letting the edges have more interesting labels. What’s the meaning of this trick? Well, we can think of our graph as an electrical circuit where the edges are wires. What do the numbers labelling these wires mean? One obvious possibility is to put a resistor on each wire, and let that number be its resistance. But that doesn’t make sense, since we’re leaving out wires labelled by 0. If we leave out a wire, that’s not like having a wire of zero resistance: it’s like having a wire of infinite resistance! No current can go through when there’s no wire. So the number labelling an edge should be the conductance of the resistor on that wire. Conductance is the reciprocal of resistance. So, our Dirichlet operator above gives a circuit like this: Here Ω is the symbol for an ‘ohm’, a unit of resistance… but the upside-down version, namely ℧, is the symbol for a ‘mho’, a unit of conductance that’s the reciprocal of an ohm. Let’s see if this cute idea leads anywhere. Think of a Dirichlet operator $H : \mathbb{R}^n \to \mathbb{R}^n$ as a circuit made of resistors. What could a vector $\psi \in \mathbb{R}^n$ mean? It assigns a real number to each vertex of our graph. The only sensible option is for this number to be the electric potential at that point in our circuit. So let’s try that. Now, what’s $\langle \psi, H \psi \rangle ?$ In quantum mechanics this would be a very sensible thing to look at: it would be gives us the expected value of the Hamiltonian $H$ in a state $\psi$. But what does it mean in the land of electrical circuits? Up to a constant fudge factor, it turns out to be the power consumed by the electrical circuit! Let’s see why. First, remember that when a current flows along a wire, power gets consumed. In other words, electrostatic potential energy gets turned into heat. The power consumed is $P = V I$ where $V$ is the voltage across the wire and $I$ is the current flowing along the wire. If we assume our wire has resistance $R$ we also have Ohm’s law: $I = V / R$ so $\displaystyle{ P = \frac{V^2}{R} }$ If we write this using the conductance instead of the resistance $R$, we get $P = \textrm{conductance} \; V^2$ But our electrical circuit has lots of wires, so the power it consumes will be a sum of terms like this. We’re assuming $H_{i j}$ is the conductance of the wire from the $i$th vertex to the $j$th, or zero if there’s no wire connecting them. And by definition, the voltage across this wire is the difference in electrostatic potentials at the two ends: $\psi_i - \psi_j$. So, the total power consumed is $\displaystyle{ P = \sum_{i \ne j} H_{i j} (\psi_i - \psi_j)^2 }$ This is nice, but what does it have to do with $\langle \psi , H \psi \rangle$? The answer is here: Theorem. If $H : \mathbb{R}^n \to \mathbb{R}^n$ is any Dirichlet operator, and $\psi \in \mathbb{R}^n$ is any vector, then $\displaystyle{ \langle \psi , H \psi \rangle = -\frac{1}{2} \sum_{i \ne j} H_{i j} (\psi_i - \psi_j)^2 }$ $\displaystyle{ P = \sum_{i \ne j} H_{i j} (\psi_i - \psi_j)^2 }$ Note that this sum includes the condition $i \ne j$, since we only have wires going between distinct vertices. But the summand is zero if $i = j$, so we also have $\displaystyle{ P = \sum_{i, j} H_{i j} (\psi_i - \psi_j)^2 }$ Expanding the square, we get $\displaystyle{ P = \sum_{i, j} H_{i j} \psi_i^2 - 2 H_{i j} \psi_i \psi_j + H_{i j} \psi_j^2 }$ The middle term looks promisingly similar to $\langle \psi, H \psi \rangle$, but what about the other two terms? Because $H_{i j} = H_{j i}$, they’re equal: $\displaystyle{ P = \sum_{i, j} - 2 H_{i j} \psi_i \psi_j + 2 H_{i j} \psi_j^2 }$ And in fact they’re zero! Since $H$ is infinitesimal stochastic, we have $\displaystyle{ \sum_i H_{i j} = 0 }$ so $\displaystyle{ \sum_i H_{i j} \psi_j^2 = 0 }$ and it’s still zero when we sum over $j$. We thus have $\displaystyle{ P = - 2 \sum_{i, j} H_{i j} \psi_i \psi_j }$ But since $\psi_i$ is real, this is -2 times $\displaystyle{ \langle \psi, H \psi \rangle = \sum_{i, j} H_{i j} \overline{\psi}_i \psi_j }$ So, we’re done. █ An instant consequence of this theorem is that a Dirichlet operator has $\langle \psi , H \psi \rangle \le 0$ for all $\psi$. Actually most people use the opposite sign convention in defining infinitesimal stochastic operators. This makes $H_{i j} \le 0$, which is mildly annoying, but it gives $\langle \psi , H \psi \rangle \ge 0$ which is nice. When $H$ is a Dirichlet operator, defined with this opposite sign convention, $\langle \psi , H \psi \rangle$ is called a Dirichlet form. #### The big picture Maybe it’s a good time to step back and see where we are. So far we’ve been exploring the analogy between stochastic mechanics and quantum mechanics. Where do networks come in? Well, they’ve actually come in twice so far: 1) First we saw that Petri nets can be used to describe stochastic or quantum processes where things of different kinds randomly react and turn into other things. A Petri net is a kind of network like this: The different kinds of things are the yellow circles; we called them states, because sometimes we think of them as different states of a single kind of thing. The reactions where things turn into other things are the blue squares: we called them transitions. We label the transitions by numbers to say the rates at which they occur. 2) Then we looked at stochastic or quantum processes where in each transition a single thing turns into a single thing. We can draw these as Petri nets where each transition has just one state as input and one state as output. But we can also draw them as directed graphs with edges labelled by numbers: Now the dark blue boxes are states and the edges are transitions! Today we looked at a special case of the second kind of network: the Dirichlet operators. For these the ‘forward’ transition rate $H_{i j}$ equals the ‘reverse’ rate $H_{j i}$, so our graph can be undirected: no arrows on the edges. And for these the rates $H_{i i}$ are determined by the rest, so we can omit the edges from vertices to themselves: The result can be seen as an electrical circuit made of resistors! So we’re building up a little dictionary: • Stochastic mechanics: $\psi_i$ is a probability and $H_{i j}$ is a transition rate (probability per time). • Quantum mechanics: $\psi_i$ is an amplitude and $H_{i j}$ is a transition rate (amplitude per time). • Circuits made of resistors: $\psi_i$ is a voltage and $H_{i j}$ is a conductance. This dictionary may seem rather odd—especially the third item, which looks completely different than the first two! But that’s good: when things aren’t odd, we don’t get many new ideas. The whole point of this ‘network theory’ business is to think about networks from many different viewpoints and let the sparks fly! Actually, this particular oddity is well-known in certain circles. We’ve been looking at the discrete version, where we have a finite set of states. But in the continuum, the classic example of a Dirichlet operator is the Laplacian $H = \nabla^2$. And then we have: • The heat equation: $\frac{d}{d t} \psi = \nabla^2 \psi$ is fundamental to stochastic mechanics. • The Schrödinger equation: $\frac{d}{d t} \psi = -i \nabla^2 \psi$ is fundamental to quantum mechanics. • The Poisson equation: $\nabla^2 \psi = -\rho$ is fundamental to electrostatics. Briefly speaking, electrostatics is the study of how the electric potential $\psi$ depends on the charge density $\rho$. The theory of electrical circuits made of resistors can be seen as a special case, at least when the current isn’t changing with time. • P. G. Doyle and J. L. Snell, Random Walks and Electrical Circuits, Mathematical Association of America, Washington DC, 1984. This free online book explains, in a really fun informal way, how random walks on graphs, are related to electrical circuits made of resistors. To dig deeper into the continuum case, try: • M. Fukushima, Dirichlet Forms and Markov Processes, North-Holland, Amsterdam, 1980. This entry was posted on Friday, November 4th, 2011 at 8:11 am and is filed under mathematics, networks, physics, probability. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site. ### 64 Responses to Network Theory (Part 16) 1. Frederik De Roo says: Nice! But I believe that $G$ is the symbol for conductance, and $C$ is used for capacitance instead. In addition, though not as funny as the ‘mho’, the sievert (S) is the SI unit of conductance. 2. Arrow says: Siemens is the unit of conductance, sievert is radiation dose. One thing i keep wondering about is are there any practical applications of such approach to electrical circuits? Networks of resistors are not particularly interesting, can for example the first two circuits pictured on this page be described in this manner? Or if that is too much perhaps complex circuits with ideal capacitance and inductance in addition to resistance? • John Baez says: Electrical circuits made of resistors are incredibly interesting—there’s a whole book about them, which I referred to in this post, and it’s well worth reading. (It’s free, too.) However, electrical circuits made only of resistors don’t display any interesting time-dependent behavior; they’re just a nice discretization of electrostatics. If you want to build an oscillator, say, you need more circuit elements. Luckily, I’ll show (and it’s well-known) that the mathematical framework described here for electrical circuits for circuits made of resistors can be extended to include capacitors and resistors. As for ‘practical applications’, I’ll admit I’m theoretically driven: I’m mainly trying to organize all known approaches to complex systems made of interacting parts into a clear unified whole, not (yet) make advances in any one of these approaches. Those advances will come quite naturally when all the different theories start talking to each other in a clear framework. For example, once I saw how stochastic mechanics and quantum mechanics were related, it only took a few weeks to realize there should be a version of Noether’s theorem relating symmetries and conserved quantities (famous in quantum mechanics) for stochastic mechanics… and Brendan Fong was able to prove it in about a week. Now that I see how electrical circuits made of resistors relate to those other two subjects, other ideas will emerge. Either people know something about circuits made or resistors that can shed new light on quantum mechanics and stochastic mechanics, or vice versa, or both. And this sort of ‘cross-talk’ between subjects will grow as time goes on. I’ve learned over the years that trying to push straight to applications is not a good working style for me (though it’s great for many other people); instead, I do better when I patiently put the puzzle pieces together and figure out what’s going on. There’s a ‘problem-solving’ style of mathematics and a ‘theory-building’ style, and I follow the latter. I learned, while writing this article, that G is the official symbol for conductance, and that ‘siemens’ is more official than ‘mho’. However, ‘G’ seemed less memorable, and more likely to raise annoying questions (“why G?”) than ‘C’, and ‘siemens’ seemed less fun than ‘mho’. • Frederik De Roo says: And luckily the symbol for sievert is Sv. That was a slip of my fingers, recorded here for eternity (or until the world wide web breaks down) to remind me I should reread before I post… @John, I think that anyone who has ever worked with capacitance would be more comfortable with $G$ than $C$. • Blake Stacey says: Why $G$ instead of $C$? Well, when we get to circuits with inductors and capacitors next week, what will we use to stand for capacitance? :-) I don’t think we’ve used $K$ for anything yet. Maybe we should start defining circuits with konductance or kapacitance. Or, maybe we could use $K$ for the amplitude of an AC oscillation. If anyone asks, we can reply, “We learned this material out of German books, and $K$ is the first letter in the German word for power station.” • Frederik De Roo says: Wouldn’t it be nice to follow standard terminology and to use Y? @John: if you have another student ;) a theorem that comes to mind when talking about electrical circuits is Thévenin’s theorem • Blake Stacey says: Well, $Y$ not? • John Baez says: Now I think I’ll use the standard symbol for ‘admittance’, since eventually I’ll generalize all this stuff to circuits containing capacitors and inductors. • Eric says: Not sure if that absolves you. You’ll still want to say things like, “The admittance of a capacitor is Y = i ω C.” • John Baez says: What I’m saying is that I’ll never use C for conductance, only capacitance. I’ll use Y for admittance, and when my circuit involves only resistors, that’ll be the same as conductance. I’ve decided that for this post, since I only need a symbol for conductance in one equation, it’ll be least confusing if I call it… $\textrm{conductance}$ 3. Eugene says: P. G. Doyle and J. L. Snell, “Random Walks and Electrical Circuits,” was published as “Random Walks and Electrical Networks,” 4. kunegis says: Hi John– I’ve been enjoying your series ‘network theory’ from the beginning, and this entry really touched home for me. I can’t help but give the following quiz: If the network has edges with negative weights, how can a Dirichlet operator be defined, and what is its interpretation in terms of electrical circuits, stochastic mechanics and quantum mechanics? • John Baez says: I’m glad you’re enjoying the show! There’s no problem with negative weights for edges in quantum mechanics, since the recipe I gave still defines a self-adjoint operator. But for stochastic mechanics it makes me nervous to have negative transition probabilities. Formally, I suppose I could make a new version of probability theory where I drop the restriction that probabilities be nonnegative. But I’m not sure what use it would be. (It’s different from real quantum mechanics, which I understand reasonably well..) I can also formally work with electrical circuits built from resistors whose conductances can be arbitrary real numbers. But again I’m not sure what use this would be. There’s a causality condition in the theory of linear electrical circuits that breaks down when you build a circuit from inductors, capacitors and resistors whose conductances can be negative. I’m eager to hear your thoughts on these subjects! • westy31 says: I really should take more time to catch up with your network discussion! Negative resistances come up in various ways. One way is in discretization based on triangulations of space with obtuse triangles. Another is in network equivalents of higher order differential equations, such as the beam equation. http://westy31.home.xs4all.nl/Electric.html#Higher_order Also, in ‘space time circuits’, as shown on the same page. Imaginary capacitances show up in the electric equivalent for the Schrödinger equation. Mathematically, there is no problem with negative resistors. In real life, they would violate thermodynamics, by converting heat into electric energy. But that’s just the time reversal of dissipation. Gerard • Jérôme Kunegis says: Hi, To finally reply: in the past I’ve been studying rating networks, i.e. bipartite networks between persons and things (e.g. movies) that people can like or dislike. To predict ratings in these networks one idea is to use the resistance distance, which can in fact be extended to signed networks by slightly changing the definition of the graph Laplacian, using the sum of absolute edge weights as diagonal values. The resulting matrix is still positive semidefinite, and even positive definite when every connected component contains a cycle with an odd number of negative edges. The same signed variant of the graph Laplacian also arises when drawing signed graphs in the “obvious” way, i.e. by placing nodes at the mean coordinates of their positive neighbors. I’d like to hear your take on this! 5. Jacob Biamonte says: This is pretty exciting stuff. I just did a few calculations related to this example and wanted post them here. First, the matrix for the example is $H=\left( \begin{array}{ccccc} 0 & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & 0 \\ \frac{1}{2} & 0 & 1 & \frac{1}{2} & 0 \\ \frac{1}{2} & 1 & 0 & 0 & 1 \\ \frac{1}{2} & \frac{1}{2} & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \end{array} \right)$ From this matrix, we can define a graph and label the edges with the entries in the matrix. Of course this is not shocking news since we determined the matrix from a graph to begin with! But here is where I’m going with this. First, the graph of $H$ is not directed (e.g. all edges go both ways) since the matrix is symmetric. We call a graph strongly connected if for each pair of nodes there is a sequence of directed edges leading from any node to any other. This is the case for $H$. It turns out that $H$ is called an irreducible matrix if and only if the corresponding graph is strongly connected. This means that the so called Perron-Frobenius Theory applies (see wikipedia). In fact, since the matrix is irreducible the following (fairly strong conditions) hold. 1. There exists a real maximal eigenvalue $r > 0$. 2. The corresponding eigenspace of $r$ is simple. That is, it is one dimensional, and is hence spanned by a single eigenvector $v$. 3. The entries of $v$ are all positive real numbers. So all entries are real, positive non-zero numbers. There are more conditions too, but these already give us enough to bring up some interesting points. First, this largest eigenvalue condition, means that there is a steady state! The network would then be fault tolerant, as small perturbations $k$ form this steady state ($v + k$), would evolve back to the steady state ($v$). There is another condition that this perhaps more interesting. 4. The eigenvector $v$ is the only eigenvector that does not have at least one negative entry! Hold on! Is that even possible? We know that for stochastic mechanics, we live in an $L^1$ space. We can’t have negative probabilities! So for any such (strongly connected) stochastic operator, we are either in the unique steady state, or we are in a sum of at least two states (to make sure all entries in the state of the system are positive—a requirement for a valid probability distributions. So let’s pretend that we don’t yet trust Perron and Frobenius, but we trust computers. We would then use Mathematica to make sure this is true. Here are the eigenvalues $\{1.76601, -1.48396, 0.559161, -0.5, -0.341207\}$ and…sure enough, we have $r = 1.76601$ as expected. For the eigenvectors, first we have $v=\{1.15274,1.54241,1.76601,0.763062,1\}$ and as a reminder, we indeed confirm that all entries are strictly positive. Now what about the rest of the eigenvectors? To match the theory, we should confirm that $v$ is the only eigenvector whose components are all positive. Here are the other five eigenvectors, so we might check for negative entries directly: $\{0.298709,1.05278,-1.48396,-0.455367,1\}$ $\{-0.634202,-0.370238,0.559161,-0.898166,1\}$ $\{-2,1,0,1,0\}$ $\{0.127202,-0.947178,-0.341207,1.20158,1\}$ So we confirm that the remaining eigenvectors are exactly as expected, the do contain negative entries! and everything agrees with the Perron–Frobenius theory of non-negative matrices, as expected. So this is all very interesting! The spooky part occurs when we elevate $H$ to a quantum operator. Quantum mechanically, the state of a system could be in a unique eigenstate such as this $\psi = \{0.298709,1.05278,-1.48396,-0.455367,1\}$ since quantum mechanics lives in an $L^2$ space. In stochastic mechanics this is out of the question, since negative probabilities are not allowed. In stochastic mechanics, the best we could do is have a valid state in a mixture of $\psi$ and some other eigenvectors—to ensure that all entries are positive. For example, a valid (though not normalised) state in stochastic mechanics would be $0.84 v + \psi = \{1.26734, 2.34885, 0, 0.185826, 1.84029\}$ But $\psi$ alone is forbidden! • Note that for stochastic and quantum operators, if $H v = r v$ then $k \cdot v$ is an eigenvector of $H$ also with eigenvalue $r$. In addition, if $H v = r v$ then $k \cdot H$ has eigenvalue $k \cdot r$. This means that we could have scaled $H$ to be $4 H$ instead of $2 H$ which I did by mistake in the last comment. There is no harm in scaling operators like this — physically it will change only the time scales of the problem. • Greg Egan says: The matrix H here isn’t infinitesimal stochastic. What you’ve described is interesting, and maybe it’s related somehow to Dirichlet operators, but the Dirichlet operator for the graph associated with your matrix H would need some negative diagonal entries so that every column summed to zero. • John Baez says: When we’re computing the power used by an electrical circuit, namely $\displaystyle{ P = \sum_{i, j} H_{i j} (\psi_i - \psi_j)^2 }$ we can leave out the diagonal ($i = j$) terms without changing the answer. But when we’re using $H$ to describe a Markov process, the diagonal terms really matter—though of course they’re determined by the other terms, by the constraint that $H$ is infinitesimal stochastic. By the way, the formula I just wrote for $P$ makes it blitheringly obvious that $P$ doesn’t depend on the diagonal entries of $H$. The same fact seems almost paradoxical if we use this other formula: $\displaystyle{ P = -\frac{1}{2} \sum_{i , j} H_{i j} \psi_i \psi_j }$ But the point is that to derive this other formula, we needed to use the fact that $H$ is self-adjoint and infinitesimal stochastic! • Jacob Biamonte says: In my previous comment, I wanted to give a quick example of something I found rather perplexing. The strange oddity where you end up with forbidden eigenstates in stochastic mechanics: moreover, these forbidden states are perfectly fine in quantum mechanics! These eigenstates have negative entries in the expansion of their coefficients. Because of the $L^1$ structure of measurements in classical mechanics, eigenstates with negative coefficients are not allowed to be states of a valid physical system. Quantum mechanics enjoys the structure of an $L^2$ theory. In quantum mechanics, all eigenstates of a Hamiltonian operator are allowed. In contract to quantum mechanics, it is on the other hand impossible for a stochastic system to be in an eigenstate which has negative entries. Such states are forbidden in stochastic mechanics as they are not valid probability distributions. What’s more, is that from the Perron-Frobenius Theory of non-negative matrices we can identify certain general conditions that predict the existence of a unique eigenstate that will be the only classically allowed probability distribution. All other eigenstates would not contain only terms with positive coefficients. In other words, in stochastic mechanics, it is possible that there would be only one valid physical eigenstate of a system. I aim to give an example of these circumstances. Such an example is something rather interesting, worthy of its own complete blog post and not just a comment. Also, Greg is clearly correct, and the last example was slightly outside the class of operators that I was considering. Just to quickly review what I was after. $H$ is an intensity matrix or infinitesimally stochastic if $\sum_j H_{ij}= 0$ $\forall i\neq j, ~H_{ij}\geq 0$ In addition to this, $H$ will be a valid quantum operator if $H = H^\dagger$ Now I wanted to give an example of an operator where the Perron-Frobenius Theory applies, to find some of these perplexing eigenstates. To do this I will consider operators $H$ that are infinitesimally stochastic with the following form. $H = H' - kI$ where $I$ is the identity matrix and $k$ is real positive. Note that adding a constant $k$ to $H$ only shifts its spectrum by $k$ and in particular does not change the eigenvectors. Now we can place some conditions on $H'$. (i) $\forall i, ~H_{ii} = 0$ (ii) $H'$ is strongly connected (iii) $\sum_j H_{ij}= k$ Now all I need to do is show that such operators exist! hmmmn. It turns out to be straightforward to find some one of them in every dimension $d$. Let $Q$ be the matrix with all entries $1$, then $H' = Q - I$ $H$ is the complete graph with all edges having weight $1$, and so $H'$ is strongly connected. In such a case, $H = H' - d\cdot I$ and $H$ is readily shown to be infinitesimally stochastic. There are also other possibilities. For instance, if we let $H'=\left( \begin{array}{ccccc} 0 & 1 & 1 & 0 & 0 \\ 1 & 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 & 1 \\ 0 & 0 & 1 & 1 & 0 \end{array} \right)$ then $H' -2 I$ is infinitesimally stochastic. The corresponding graph of $H'$ is strongly connected: and so Perron-Frobenius Theory applies. We find that the Perron root or maximal eigenvalue is given as 2 and the corresponding eigenvector as $\{1, 1, 1, 1, 1\}$. All other eigenvalues are less than 2. All other eigenstates are confirmed to contain at least one negative entry. The other eigenstates individually are forbidden states of the corresponding stochastic system. This was all a bit fast, but soon I’ll post a more complete and detailed discussion of these interesting operators. 6. David Corfield says: Minor point, but when you write of a complex matrix $H$ is infinitesimal stochastic if its columns sum to zero and its off-diagonal entries are nonnegative, is it conventionally understood that if a complex number is said to be nonnegative, that it is real and nonnegative? I know the order relation makes no sense on the complex numbers as a whole, but the wording seems odd to me. • John Baez says: David wrote: Is it conventionally understood that if a complex number is said to be nonnegative, that it is real and nonnegative? I thought so. Anyway, that’s what I meant. Perhaps this usage is more common when applied to linear operators on complex Hilbert spaces: for example, everyone agrees a linear operator $T: \mathbb{C}^n \to \mathbb{C}^n$ is ‘nonnegative’ if $\langle \psi, T \psi \rangle \ge 0$ for all $\psi \in \mathbb{C}^n$. Taking $n = 1$ you see a 1×1 matrix, i.e. a complex number, is ‘nonnegative’ if it’s real and nonnegative. However, I was trying to be clear, not show off my sophistication! So, I’ll change ‘nonnegative’ to ‘real and nonnegative’. 7. Uwe Stroinski says: Thévenin’s theorem? Interesting point! In the theory of (non-)deterministic automata (http://en.wikipedia.org/wiki/Nondeterministic_finite_automaton) there is a construction that gets you an equivalent deterministic machine (the so-called power set construction). Such input-output equivalences seem to be a common theme and it indeed might be worth to check whether they have a place somewhere in your theory. • John Baez says: In a paper I’m writing on electrical circuits, I construct a category where the morphisms are electrical circuits of a certain sort. I also construct category where the morphisms are ‘input-output equivalence classes’ of such circuits. In the latter, you treat an electrical circuit as a ‘black box’: you don’t care about what’s inside, just what it does. There’s a ‘forgetful functor’ from the first category to the second, where we forget the details inside the black box. So, Thévenin’s theorem and Norton’s theorem can be seen as theorems about the functor from the first category to the second: any morphism in the image of some class of circuits is actually in the image of some smaller class. In case someone here forgot: Thévenin’s theorem says that any combination of voltage sources, current sources, and resistors with two terminals is electrically equivalent to a single voltage source and a single resistor in series. Norton’s theorem says that any collection of voltage sources, current sources, and resistors with two terminals is electrically equivalent to a single current source and a single resistor in parallel. And yes, all this should be an example of a very common theme! For any sort of gizmo made out of parts, we should have a choice as to whether we care about its inner workings or treat it as a ‘black box’. This should give two different categories $C$ and $D$ and a forgetful functor $F : C \to D$, and we can ask when morphisms in the image of some class of morphisms of $C$ are actually in the image of some smaller class. It would be nice to collect lots of known theorems of this sort, and see if lots of them are special cases of a few general results. That’s the kind of stuff I want to do. A very simple theorem of this sort, which applies to what I’m talking about today—circuits made only of resistors!—lies behind the Y-Δ transform. It says any circuit like the one at left is equivalent, ‘as a black box’, to a circuit like the one at right: This can be seen as a theorem about Dirichlet operators! 8. jamievicary says: Not much to say here, other than that I’m enjoying following along! It surprised me briefly that there was no interesting dynamics for charges moving through resistors. I suppose the reason for that is that we have no inductance in our circuits, so charge can reach its equilibrium configuration infinitely quickly. But I would expect it’s still interesting to consider how much energy would be dissipated as an un-equilibrated charge density redistributes itself. • John Baez says: Jamie wrote: It surprised me briefly that there was no interesting dynamics for charges moving through resistors. Yes, it puzzled me too—and in a way it still does, since a Dirichlet operator $H$, which describes a circuit made of resistors, can be used to formulate an equation like the heat equation $\displaystyle{ \frac{d}{d t} \psi = H \psi }$ thinking of $H$ as a discrete analogue of the Laplacian. However, as you note, we need inductance in our circuits to make current want to keep flowing in them, and get an equation like the above to hold. I posed a version of this puzzle back in ‘week294′. I said, suppose you have a loop of wire with a very small resistance, and some current flowing in it: what will happen? Some answers start here. 9. Florifulgurator says: This reference is doubly outdated: • M. Fukushima, Dirichlet Forms and Markov Processes, North-Holland, Amsterdam, 1980. • M. Fukushima, Y. Oshima, M. Takeda, Dirichlet Forms and Symmetric Markov Processes, 2nd rev. and ext. ed. (1st ed. 1994), De Gruyter, 2010. • John Baez says: I found the newer editions to be more cluttered and less clear than the old one! There’s a Zen saying “first thought best thought”, which is sometimes true for books. But thanks: our readers should be allowed to choose for themselves. 10. John Baez says: Massimo Ostilli visited the CQT yesterday and spoke about complex networks, which for him (and many people) means the study of large random graphs. He also pointed out that we can use a random walk to compute $\exp(t H)$ for any matrix $H$ with nonnegative off-diagonal entries, essentially by writing $H = A + B$ where $A$ is infinitesimal stochastic and $B$ is diagonal. To compute a matrix element $\exp(t H)_{i j}$, we can use $A$ to define a random walk and sum over paths from $j$ to $i$, weighted by their probability of occurrence, but also weighted by a factor of $\exp(s B_{k k})$ for each time the walk sits at the vertex $k$ for a time $s.$ According to Ostilli this idea is called ‘the exact probabilistic representation of lattice quantum systems’, and it’s described here: • Matteo Beccaria, Carlo Presilla, Gian Fabrizio De Angelis and Giovanni Jona-Lasinio, An exact representation of the fermion dynamics in terms of Poisson processes and its connection with Monte Carlo algorithms, Europhys. Lett. 48 (1999), 243-249. This idea was apparently be generalized to matrices with some negative off-diagonal entries, here: • Matteo Beccaria, Carlo Presilla, Gian Fabrizio De Angelis and Giovanni Jona-Lasinio, Evolution of fermionic systems as an expectation over Poisson processes, in Recent Progress in Many-Body Theories, Series on Advances in Quantum Many-Body Theory – Vol. 3, edited by R. F. Bishop, K. A. Gernoth, N. R. Walet, Y. Xian (World Scientific, Singapore, 2000), pp. 461-464. • Matteo Beccaria, Carlo Presilla, Gian Fabrizio De Angelis, Giovanni Jona-Lasinio, Probabilistic representation of fermionic lattice systems, Nucl. Phys. B – Proceed. Suppl. 83-84 (2000), 911-913. 11. [...] In Part 16 of John Baez’ series on Network Theory, he discussed electrical networks. [...] 12. We recall from Part 16 that there is a class of operators called ‘Dirichlet operators’ that are valid Hamiltonians for both stochastic and quantum mechanics [...] 13. amarashiki says: Three questions: 1st. How could you include the Dirac equation, the Maxwell equations and Einstein Field Equations in the same footing you did with Schrödinger, Poisson and heat equations? 2nd. What about a discrete nonlinear foundation of memdevices related to the whole picture? 3rd. Graph theory, network theory, matroid theory and categories. What about hypergraphs of Petri nets? • John Baez says: These questions are very big. I can’t answer them except to say that 1) I’m slowly working to unify everything that naturally fits into ‘network theory’, so stay tuned to this series of posts, and 2) I know how Maxwell theory fits into this game, but these days I’m more interested in ecology, biology and chemistry than so-called ‘fundamental’ physics, so I probably won’t try to bring the Dirac equation and Einstein’s equations into this story. 14. amarashiki says: I think I DO know how to get Maxwell equations fit into the game as well…I read long ago the discrete EM talk of one student you had…And moreover, I have read how Schrödinger-like equation for E and B can serve as a guess for Maxwell equations…Eintein’s field equations seem to be a harder task but also possible. And Dirac’s equation should be also built from the known root-of-laplacian procedure… About your answer to the other 2 questions, I am a little disturbed only, ;). But yes, I know you have moved from your fundamental issues…Not me, … I am a theoretical physicist, I don’t know how to move away from it… I AM tuned to your blog…Just subscribed long ago ;) So, you have an unbiased follower …I think I mentioned before I follow you from my undergraduate times (your TWF will be always remembered and admired). Keep doing this superb blogger job John, it is not easy at all. Finally, concerning the third question, and your mysterious comment about using network theory as a global tool/framework…Have you thought about network theory using hypergraph theory? • John Baez says: I’m glad you enjoyed This Week’s Finds in Mathematical Physics, and also my new stuff. If you haven’t read about the Feynman checkerboard discretization of the Dirac equation in 1+1 dimensions, and its extensions to higher dimensions, you might be interested in those. In 3+1 dimensions, it’s nice to use a hypercubic lattice in which each edge is lightlike. This lattice gives a coordinate system in which the Minkowski metric is $\left( \begin{array}{cccc} 0 & 1 & 1 & 1 \\ 1 & 0 & 1 & 1 \\ 1 & 1 & 0 & 1 \\ 1 & 1 & 1 & 0 \end{array} \right)$ There’s something hilarious about this, but it’s also very pretty. We get zeroes down the diagonal because as we move forward in time, 4 edges move outwards from each vertex at the speed of light. And they move outwards like the corners of an expanding regular tetrahedron, so there’s perfect symmetry: that’s why all the off-diagonal entries are equal. A recent paper is here. I don’t know anything about hypergraphs! Maybe I should fix that someday. 15. David Tanzer says: This is great stuff! For a voltage vector $V$ in $\mathbb{R}^n$, you show that $\langle H(V), V \rangle$ is the energy consumed by the circuit. We can also give a substantial interpretation of $H(V)$ itself: it is the vector $I$ of currents that are induced by the voltage vector. Here the signs are oriented so that $I_j$ is the net flow of current into the $j$th node of the circuit. Let’s illustrate with an example with three nodes $x_1, x_2, x_3$, with a 1 ohm resistor connecting each pair of points. Let the voltage vector be $(a,b,c).$ And let $H = \left( \begin{array}{ccc} -2 & 1 & 1 \\ 1 & -2 & 1 \\ 1 & 1 & -2 \end{array} \right)$ Then: $H(a,b,c) =$ $(-2a + b + c, a - 2b + c, a + b - 2c) =$ $(b-a + c-a, a-b + c-b, a-c + b-c) =$ $(\mathrm{inflow \; into \; a, \; inflow \; into \; b, \; inflow \; into c}) = \mathrm{current} I.$ Now, given that $H(V) = I$, it follows as a natural corollary that $\langle V, H(V) \rangle$ is the power consumed by the network. • David Tanzer says: .. which leads me to the following question. As an “observable,” this electrical Hamiltonian H should be described as “power,” because for a state V, $\langle H(V), V \rangle = \langle I, V \rangle =$ the power. So for the observable $H$ called power, $H(V)$ is the current. Can this perspective be carried back into quantum mechanics, so that for other observables $O$, we can give a physical interpretation of $O(x)$ as something analogous to the “current” that is induced by the “potential” $x$? Here is the root question. If $x$ is a function that assigns quantum mechanical amplitudes to conditions, and $O$ is an observable, then what can be said about the general interpretation of $O(x)$? • John Baez says: I fixed your first comment following your corrections, put it into LaTeX (in a rather ugly and lazy way), and deleted your corrections. The idea about $H(V)$ being the current associated to the voltage vector $V$ is well-known in certain circles; I can give you references if you want. But it’s definitely worth saying! This, however, seems new to me: Can this perspective be carried back into quantum mechanics, so that for other observables $O$, we can give a physical interpretation of $O(x)$ as something analogous to the “current” that is induced by the “potential” $x$? I guess my only answer is: let’s try it and see if we can extract anything interesting from this viewpoint. It sounds like a nice idea! But, we’ll need to do something with it, to see what it can do for us. Here is the root question. If $x$ is a function that assigns quantum mechanical amplitudes to conditions, and $O$ is an observable, then what can be said about the general interpretation of $O(x)$? Right! In general physicists are fairly close-mouthed about this. When as a student you hear that an observable in quantum mechanics is a self-adjoint operator $O,$ the first thing you want to know is what you get when you apply this operator to a state $x.$ That is, what’s the physical interpretation of $Ox$? But usually your teachers side-step this question. At least, all of mine did! This is especially frustrating because when you have two operators that don’t commute, like position $Q$ and momentum $P$, it would be nice to know what it means that $P Q x \ne Q P x$ As a student, you can’t help but ask: is this this telling us something about what happens if we first measure position and then momentum, versus first measuring momentum and then position? But a clear answer is rarely forthcoming. Here’s what I can say. Any observable $O$ gives a 1-parameter family of symmetries $\exp(i s O)$, that is, unitary operator depending on a real parameter $s.$ For example. if $O$ is momentum, $\exp(i s O)$ describes spatial translation by a distance $s.$ Thus the quantity $\exp(i s O) x$ has a clear meaning: it’s the state we get by applying the symmetry $\exp(i s O)$ to the state $x.$ For example, if $x$ is the state of an atom and $O$ is momentum, $\exp(i s O) x$ is the state of the atom after it’s been moved a distance $s.$ Then, we have $O x = -i \frac{d}{d s} \exp(i s O) x \vert_{s= 0}$ is basically the rate at which the state $x$ changes as we apply the symmetry $\exp(i s O)$ and change $s.$ E.g, the rate at which the atom’s state changes as we move it along. That’s the best I can do to explain the meaning of $O x.$ Maybe someone else can do better. Of course we all know what $\langle x, O x \rangle$ means: it’s the expected value of the observable $O$ in the state $x.$ But we’re wondering about $O x$ all by itself! So now it’s your turn: see if you can understand the relation between $x$ and $Ox$ as somehow analogous to the relation between voltage and current! 16. David Tanzer says: It is physically clear that any voltage vector V0 which is constant on each of the connected components will produce zero current. Hence V0 is a null vector of H. 17. David Tanzer says: To give a full physical interpretation of the Hamiltonian dynamic on this network of resistors, we should connect a capacitor between each node and a ground point. Give them all the same unit capacitance. This is what will make the current H(V)(j) into node j actually produce a rising voltage at V(j) – this condition is needed to fulfill the equation dV / dt = H(V). Intuitively, we see that any given initial state of the network, given by a voltage vector V, the network will asymptotically charge/discharge to an equilibrium state where, for each connected component of the graph, all voltages are the same. For each connected component C, the final voltage is easily calculated, by dividing the total charge in C in the network by the number of nodes in C: FinalVoltage(V) = Sum(V(j)) / n = mean(V(C)), for all j in C. The capacitor + resistor circuit gives a physical picture of eigenvectors and non-eignevectors. An eigenvector is a voltage vector V that heads in a “straight line” towards the equilibrium vector. A non-eigenvector does some “turning” as it heads towards the equilibrium vector. 18. nad says: John wrote: The idea about H(V) being the current associated to the voltage vector V is well-known in certain circles; I can give you references if you want. But it’s definitely worth saying! The $\psi$ in $\psi$ are electric potentials, so it seems that $H_{ij}(\psi_i - \psi_j)$ would be the current for the edge $ij$. Moreover I feel a little unease to call the above electrical circuit, that is mainly because I haven’t seen a discussion about Kirchhoff’s laws, but maybe I have missed that. By Kirchhoff’s second law one would have the constraints: $\sum_{loop}(\psi_i - \psi_j)=0$ and for those configurations of $\psi$‘s which satisfy the second law it seems one would need to demand for Kirchhoffs first law that: $\sum_j H_{ij}(\psi_i - \psi_j) =0,$ which gives a constrained on the entries of $H$, which may eventually not be preserved during a time evolution of the above indicated sort. ???? Warning: my physics diploma was 22 years ago and I haven’t done too much physics in the meantime. • nad says: sorry I also forgot the latex. • John Baez says: I’ll fix it. This is a good time to remind everyone that there’s a big warning when you type in a comment here, saying: You can use HTML in your comments. You can also use LaTeX, like this: \$latex E = m c^2 \$. The word ‘latex’ comes right after the first dollar sign, with a space after it. • nad says: I’ll fix it. This is a good time to remind everyone that there’s a big warning when you type in a comment here, saying: Since there are so much commercials on webpages my brain starts blanking out even big buttons. Thanks for the link to your paper on electrical circuits. You wrote: But since any real Hilbert space is equipped with a canonical isomorphism to its dual, we get isomorphisms $r:C_0(\Gamma) \to C^0(\Gamma) \qquad r:C_1(\Gamma) \to C^1(\Gamma)$ I understand what $r:C_1(\Gamma) \to C^1(\Gamma)$ means physically, but I haven’t yet understood what $r:C_0(\Gamma) \to C^0(\Gamma)$ means physically. Or in other words: Given a real engineering-like electrical circuit I thought one is usually only given $r:C_1(\Gamma)\to C^1(\Gamma)$ and the boundary conditions. ? • John Baez says: Hi! By the way, your LaTeX comments didn’t work until I fixed them because you included the UNICODE symbol → inside dollar signs; LaTeX doesn’t know how to handle that. The LaTeX symbol for a right-pointing arrow is \to so I replaced → with \to everywhere and now it’s fine. Since there are so much commercials on webpages my brain starts blanking out even big buttons. Even ones you helped create! I see now why you wanted it to be orange. I haven’t yet understood what $r:C_0(\Gamma) \to C^0(\Gamma)$ means physically. It’s hard to say what it means, because it’s very boring: it doesn’t involve any arbitrary choices. The vector space $C_0(\Gamma)$ has a basis given by the vertices of the graph $\Gamma$. In my notes, I give $C_0(\Gamma)$ an inner product where this basis is orthonormal. As usual, an inner product gives an isomorphism between a finite-dimensional vector space and its dual. The dual of $C_0(\Gamma)$ is called $C^0(\Gamma)$. So, we get an isomorphism $C_0(\Gamma) \cong C^0(\Gamma)$ which I call $r:C_0(\Gamma) \to C^0(\Gamma)$ This is very different than the case of $r:C_1(\Gamma) \to C^1(\Gamma)$ which has an interesting physical meaning. This again comes from an inner product on $C_1(\Gamma)$. But to define the inner product on $C_1(\Gamma)$, we start by arbitrarily choosing for each edge the graph $\Gamma$ a positive number called its resistance. To define the inner product on $C_0(\Gamma)$, we don’t choose anything! This may seem weird, but it works wonderfully—and by the way, it’s not something I invented. It was Hermann Weyl who first applied ideas of homology, cohomology and Hodge theory to electrical circuits made of resistors, and lots of other people have developed these ideas ever since. I give lots of references. • Nad says: Even ones you helped create! I see now why you wanted it to be orange. orange was a suggestion. In fact such a color depends also on the overall style of a blog, moreover if you want to keep a high attention factor you would probably need to change this color regularly. Thats one reason why the fashion industry keeps changing the colors which are en vogue. I meanwhile cover sometimes parts of my screen because otherwise I couldn’t read a text because of all that blinking commercials. If this commercial flooding trend continues then I could imagine that they place blinks inside texts and then I would probably cease to read those texts, like I did with TV when they started to heavily interrupt films with commercials, that is since over 20 years I don’t have a TV and thus I rarely watch it even at friends places. I understand that one likes to use ads for financing, however things can be overdone. In my notes, I give $C_0(\Gamma)$ an inner product where this basis is orthonormal. yes, but this seems to be a choice. That is – I haven’t really investigated the whole thing thoroughly, but it seems to me that you could also use non-orthonormal bases. This would of course enter the Laplacian via $d^$ and it seems on a first glance to be related to a gauge transformation? In particular I wonder in how far this could be used to include magnetism into that framework* ?. Did Weyl write something on that? I didn’t find a free copy of the paper and I currently don’t have a library account (apart from the fact that my spanish is not overly good…) I wish I had known about that work of Weyl and that I could have read a text like yours already during my studies. I actually used the discrete one-forms on graphs in one of my articles. The geometers in Berlin treated the cohomology theory of graphs as a kind of mathematical folklore, but I always wondered where it came from, so I actually cited Vladimir Arnold’s book on classical mechanics (one of the few textbooks I actually own) as a source, because that was the oldest resource I could find on that issue (at least in a decent time). However Arnold didn’t seem to have cited Weyl. Do you know where this cohomology theory on graphs was treated earliest? Did Weyl cite someone there? *I am also not sure whether you really want comments on that – it seemed to me that you wanted to develop something like this for yourself. (I usually don’t like it if I try to develop something and then someone is constantly saying you could do it in such and such way, that’s also why things like the polymath project do not look very attractive to me, but people are different.) • John Baez says: Nad wrote: yes, but this seems to be a choice. Everything we do in math is a choice, but what I meant is that this particular recipe works regardless of which circuit one is considering, without requiring any additional information like the resistance of the edges. I agree, though, that it would be fun to examine other choices, and see if there’s an interesting concept analogous to resistance that applies to the vertices of the electrical circuit! The concept obviously exists—I can see that now, thanks to you—but I don’t know what it means or how interesting it is. This would of course enter the Laplacian via $d^$ and it seems on a first glance to be related to a gauge transformation? Hmm, it seems this sort of transformation would have the effect of changing the electrostatic potential $\phi(x)$ to a new one $f(x) \phi(x)$ where $f$ is a positive function of the vertices of the graph. This would be some funny sort of ‘gauge transformation’ that’s not usually considered in electromagnetism. I am also not sure whether you really want comments on that – it seemed to me that you wanted to develop something like this for yourself. I blog because I want comments! Of course I don’t feel obliged to do what people suggest. I have this paper worked out in my mind and don’t plan to change it much—I just want to finish the darn thing and publish it. I’ve been sitting on it for over 2 years. But often comments affect what I do. Maybe this one will too, eventually. But first I’d need to understand the physical meaning of picking different inner products on $C_0(\Gamma).$ I doubt I’ll figure that out before finishing this paper—it feels like the subject of some other paper. Do you know where this cohomology theory on graphs was treated earliest? Cohomology of graphs is probably very old, but I don’t know when it was first studied. I haven’t actually found Weyl’s paper, so I don’t know who he referred to. All the interesting references I know on cohomology and electrical circuits are included in my article. • nad says: where f is a positive function of the vertices of the graph. I don’t see at the moment why f(x) needs to be positive. In particular one could perhaps include complex transformations, which I could imagine could mimick in a linearized version the term $i \cdot e \cdot \mathrm{vector \; potential}$. But as said this is just a quick guess. I blog because I want comments! I usually want comments only if I ask for them. I have usually (knock on wood) no problems to get new ideas. For me the hard part is often rather to sort out, which ideas to follow, whether I should try to get better ideas (which might be strenuous) and to sit down and to work things out. It is even often so that the pressure to let ideas come out is so big that this process hinders the working-out process. When I was a child people rather often said: “Die Fantasie geht schon wieder mit ihr durch” (“Fantasy is again bolting her”). I think this is also a reason, why I don’t always like people comment without being asked, because often I already had the same idea anyways and if people comment directly to you they somehow set a claim that they could work on the same thing and especially if this is about the development of novelties then it depends really on the person wether this ends not up in rat-race. I often rather prefer to stop working on something in this case (alone for the reason that I am slow in working out and the reason that I like to do also other things in parallel). Here by the way also the incentives play a role. If novelty is a strong incentive to work something out then a set claim may be a killer. I had also cases where people used an idea from me, sat immediately down, worked it out overnight and then presented it as their work the next day. And in fact if an idea is there it is often hard to judge how hard it was to come up with it and how much one should attribute to the person who had it. I usually try to cite people’s ideas if they do not appear too trivial (but as said that’s of course a quite debatable issue). If people made their work public than they sat a claim and depending how serious this claim is you have to take that into account. And of course often enough you may even not know about such a publication (or you may not have understood it because it is written in some obliterated language), here it is often important that the way on how you came up with your work is documented, because this may make clear that you where not just copying someones work. That’s why I think it would be good to have something as a closable archive, where the development process can be better evaluated and where you could decide for yourself whether you constantly want to set claims (and depending on the potentials eventually enter a rat race) or wether you are silent and would let others also work on the same issues in parallel. Having different treatments of a subject might be beneficial, as one can see on your paper on electrical circuits. So it seems: • H. Weyl, Repartición de corriente en una red conductora, Rev. Mat. Hisp. Amer. 5 (1923), 153-164. can only be buried in some library. I hope there are still some copies left. I am actually rather interested in this discrete treatment of physical quantities (my Ph.D. thesis had a similar background) and I had seen that you had some work on discrete Maxwell’s equations with Derek Wise, is this going to be included in your current or a follow up paper on electrical circuit work? • John Baez says: Nad wrote: Moreover I feel a little unease to call the above electrical circuit, that is mainly because I haven’t seen a discussion about Kirchhoff’s laws, but maybe I have missed that. Good point! Kirchhoff’s voltage law $\sum_{loop}(\psi_i - \psi_j)=0$ is automatically true in this formalism: it just says things like $(\psi_1 - \psi_2) + (\psi_2 - \psi_3) + (\psi_3 - \psi_1) = 0$ Kirchhoff’s current law follows automatically when the voltage minimizes the power $-2 \langle \psi, H \psi \rangle$ subject to constraints on the voltage $\psi$ at certain vertices called the ‘terminals’ of our circuit. More precisely, Kirchhoff’s current law will hold at all vertices except these terminals: current flows in and out of these terminals. To see a proof of this (known) fact, try my paper on electrical circuits. It’s just a draft, and it’s not very easy to read, but someday I’ll make it nicer and start blogging about it here. There’s a lot more to say… 19. Blake Stacey says: I think this is where the article was reprinted in Weyl’s collected works. • nad says: Thanks Blake, my google algorithm unfortunately didnt reveal that result….I should probably write a complaint letter to the head of the google books search algorithm division :) So it seems that if one doesn’t find access to a library one has to pay: 439 Euros! There go the summer holidays. The Weyl book seems to be quite a cash cow for Springer, given that this book series is almost 50 years on the market and the fact that probably every scientific library in the world had (should have?) bought it. What was wrong with Hilbert and Weyl? • Blake Stacey says: There’s an interesting anecdote about that work of Weyl: [Robert Kotiuga]: Beno, there is something I really don’t understand about Hermann Weyl. [Beno Eckmann]: What is it? RK: Well, in his collected works, there are two papers about electrical circuit theory and topology dating from 1922/3. They are written in Spanish and published in an obscure Mexican mathematics journal. They are also the only papers he ever wrote in Spanish, the only papers published in a relatively obscure place, and just about the only expository papers he ever wrote on algebraic topology. It would seem that he didn’t want his colleagues to read these papers. BE: Exactly! RK: What do you mean? BE: Because topology was not respectable! RK: Why was topology not respectable? BE: Hilbert! 20. John Baez says: Nadja wrote: I don’t see at the moment why f(x) needs to be positive. I assume the resistance of each edge is positive because then they make the space of 1-chains $C_1(\Gamma)$ into an inner product space. Similarly I would assume this function of each vertex is positive so that they make the space of 0-chains $C_0(\Gamma)$ into an inner product space. Why? Because if I have a chain complex of inner product spaces I can hit it with the usual tools of Hodge theory, as I’ve done in this paper of mine. One could try to generalize but I feel no desire to. I usually want comments only if I ask for them. I always want comments even if only 10% are useful to me. When a bunch of smart people start talking, they get to know each other and start trading ideas and attracting other smart people… and it becomes easier for me to find out answers to questions, and find good collaborators. You’re right that having a crowd of people talking about similar things causes various problems. In a way this is why I quit working on n-categories: too many very smart people were starting to work on them, and I needed more ‘open space’ and solitude. But right now I’m trying to learn lots of things and invent a slightly new branch of math that borrows ideas from dozens of existing subjects, so it’s good for me to hear lots of people’s ideas – people from different intellectual communities. I had seen that you had some work on discrete Maxwell’s equations with Derek Wise, is this going to be included in your current or a follow up paper on electrical circuit work? The electrical circuit project could easily expand to include work on discretized Maxwell equations—Derek’s work with me on chain field theory is always on my mind these days—but I don’t think I’ll go in that direction anytime soon: I’m more interested in the relation between electrical circuits and Markov processes. I’m trying to set up a theory of networks that includes lots of very practical applications. • nad says: I wrote: I usually want comments only if I ask for them. It seems I was unclear. I meant here comments to work in progress. Comments to other things are of course OK. (I thought this was clear by the context of the rest of the comment, but it seems not.) I assume the resistance of each edge is positive because then they make the space of 1-chains $C_1(\Gamma)$ into an inner product space. I don’t know enough about Hodge theory in order to be able to judge how much this Hodge theory wouldn’t be applicable when using hermitian forms as an inner product. There must be some troubles (need of symmetry?), because otherwise someone would have formulated such a generalized Hodge theory already. So I understand that you do not feel like going through the same troubles. I had actually meant some other work, which was somewhere on your homepage, about discrete Maxwell equations. You’re right that having a crowd of people talking about similar things causes various problems. In a way this is why I quit working on n-categories: too many very smart people were starting to work on them, and I needed more ‘open space’ and solitude. Most people at the n-category cafe speak in a language which is totally foreign to me. • John Baez says: Nad wrote: I meant here comments to work in progress. Comments to other things are of course OK. (I thought this was clear by the context of the rest of the comment, but it seems not.) It was clear. In my last comment I talking about why I like to blog about work in progress, like this ‘network theory’ project. One reason I blog about work in progress is because I like to build up a community of experts who tell me when I’m making mistakes or overlooking interesting facts. John wrote: I assume the resistance of each edge is positive because then they make the space of 1-chains $C_1(\Gamma)$ into an inner product space. [...] Why? Because if I have a chain complex of inner product spaces I can hit it with the usual tools of Hodge theory, as I’ve done in this paper of mine. Nad wrote: I don’t know enough about Hodge theory in order to be able to judge how much this Hodge theory wouldn’t be applicable when using hermitian forms as an inner product. There must be some troubles (need of symmetry?), because otherwise someone would have formulated such a generalized Hodge theory already. It’s not the symmetry, but the positive definiteness of the inner product that would go away if we let our resistances be negative numbers. Remember, the inner product $\langle I, I \rangle$ is the power consumed by a circuit when we run the current $I \in C_1(\Gamma)$ along the wires. The power consumed by each wire equals the square of the current along that wire times the resistance of that wire. If the resistance were negative this power could be negative! Then $\langle I , I \rangle$ could be negative. This would not only be unphysical, it would mess up the Hodge theory. If you have an ‘inner product’ that’s not positive definite on your p-chains, the inner product on p-cochains won’t be either. And then you can’t really do most of the interesting things people do with Hodge theory, like show that a harmonic p-cochain $\mu$ is both closed and coclosed: $(dd^* + d^* d) \mu = 0 \implies d \mu = 0 \; \textrm{and} \; d^* \mu = 0$ (Think of $\mu$ as a p-form and $dd^* + d^* d$ as the Laplacian.) This basic lemma of Hodge theory goes away, as far as I can tell, when the inner product isn’t positive definite! After all, the usual proof is $(dd^* + d^* d) \mu = 0 \implies$ $\langle \mu, (dd^* + d^* d) \mu \rangle = 0 \implies$ $\langle d^* \mu, d^* \mu \rangle + \langle d \mu, d \mu \rangle = 0 \implies$ $\langle d^* \mu, d^* \mu \rangle = 0 \; \textrm{and} \; \langle d \mu, d \mu \rangle = 0 \implies$ $d \mu = 0 \; \textrm{and} \; d^* \mu = 0$ but the last two steps assume the inner product is positive definite! This is why Hodge theory works so much better on Riemannian manifolds than on Lorentzian ones. What is “p-form electromagnetism” ? p-form electromagnetism is the generalization of Maxwell’s equations where the electromagnetic vector potential A is a p-form instead of a 1-form. 2-form electromagnetism shows up in string theory, and it’s the simplest example of a higher gauge theory. Just as point particles like to couple to a 1-form, strings like to couple to a 2-form. Yes, in a sense. Derek was working with me when I was interested in higher gauge theory, so he wrote this paper on the quantization of vacuum p-form electromagnetism on discretized spacetimes (for example, simplicial complexes), where you can work it all out rigorously. By ‘vacuum’ I mean that he only considers the electromagnetic field, no charged particles (or strings, or 2-branes, or….). This makes the theory linear and exactly solvable. If you only care about ordinary electromagnetism, set p = 1. • nad says: It was clear. In my last comment I talking about why I like to blog about work in progress, like this ‘network theory’ project. One reason I blog about work in progress is because I like to build up a community of experts who tell me when I’m making mistakes or overlooking interesting facts. There are of course comments which are welcome even during a work in progress, like for example if someone points out relevant literature or bad mistakes. However if a commenting situation is used for stealing ideas or if comments are of the sort: I think you should do this in that and that way (where it is obvious that one should do it in that and that way) or the whole thing you are doing here is bullshit (without specifying exactly WHAT is bullshit) then this can be rather unpleasant. I am not inventing these comments. I actually got exactly this “bullshit” comment after giving a about 2 hr presentation on some work in my Ph.D. work and this evaluation was one reason why I hesitated to publish the work. (I published it nonetheless, but only because someone else said I should publish it. I mean it is somewhat irritating if someone tells you, you are doing bullshit and you do not know what is bullshit and what to do about it) I was once in a talk with a rather big audience, where the presenter was told that what he had been doing in his work (he talked about his work) was thoroughly studied and that and that didn’t work because a.s.o. The presenter (who was if I remember correctly already in some solid position in France) broke into tears and couldn’t even talk further. The whole situation was rather disturbing. There was a similar situation during a talk, where luckily the presenter got some defense from within the audience, but nonetheless the accusations where fierce in enough that the presenter broke also into tears here. I think one should avoid such situations. I have also rather troubling recollections about Andreas Floer in this context. Of course if something is wrong then this has to be pointed out, but first there are different ways on how one can point out a mistake and secondly if a mistake would have been detected at an earlier stage the presenter would not end up in such a situation. So here commenting during a work is in progress would actually be essential. So if I say I prefer not to get comments to work in progress I actually mean these latter bullshit-types of comments , but it is usually rather complicated to explain what types of comments one wants and since it seems that the bullshit comments are more frequent than the other types I prefer to just say: please no comments. There is also the problem that if your work is (almost) in parallel on something, i.e. if you are possibly in competition for novelty then this can get very tricky. Often enough you better would not want to know, what the other is doing and in order not to bring a competitor into the bad situation of possibly needing to help you it could actually be a nice thing to say: no you don’t need to make comments. Think of $\mu$ as a p-form and $dd^* + d^* d$ as the Laplacian. The power consumed by each wire equals the square of the current along that wire times the resistance of that wire. If the resistance were negative this power could be negative! But I was talking about $C_0(\Gamma)$ and $C^0(\Gamma)$ and not $C_1(\Gamma)$ and $C^1(\Gamma)$. There could be different inner products on these spaces. Think of $\mu$ as a p-form and $dd^* + d^* d$ as the Laplacian. I don’t know why you suddenly define the Laplacian in this way, because in your circuit paper it is just $d^d$, which would be enough to ensure closedness for a “harmonic” form (I assume you mean here a form for which: $laplace \; \mu = 0$), this follows from $d \mu =0$ and positive definiteness as you had already pointed out above. This is of course rather important with respect to Kirchhoff’s laws. But why do I need coclosedness? p-form electromagnetism is the generalization of Maxwell’s equations where the electromagnetic vector potential A is a p-form instead of a 1-form. 2-form electromagnetism shows up in string theory, and it’s the simplest example of a higher gauge theory. Aha. I am still trying to get an image in my head what people may mean with higher gauge theory and string theory. I blurrily understand it as some kind of generalization of a connection (i.e. a prescription on how vectorfields are to be transformed) that allows somehow for nonlocal transformations? What is the physical interpretation of a magnetic p-form? Some multipole magnetism? Is p=0 (monopole) also allowed?!?! • John Baez says: By the way, Nad: in this particular conversation, please try to post your comments so they appear below the comment you’re replying to. You keep posting them so they show up far above it, and I keep fixing them. John wrote: The power consumed by each wire equals the square of the current along that wire times the resistance of that wire. If the resistance were negative this power could be negative! Nad wrote: But I was talking about $C_0(\Gamma)$ and $C^0(\Gamma)$ and not $C_1(\Gamma)$ and $C^1(\Gamma)$. There could be different inner products on these spaces. True. I was just pointing out the physical reason why I assume the inner product on $C_1(\Gamma)$—and thus dually on the isomorphic space $C^1(\Gamma)$—is positive definite. If you can discover or invent a physical interpretation of the inner product on $C_0(\Gamma)$ and $C^0(\Gamma)$, you will be able to decide whether you want those to be positive definite too. John wrote: Think of $\mu$ as a $p$-form and $dd^ + d^* d$ as the Laplacian. Nad replied: I don’t know why you suddenly define the Laplacian in this way, because in your circuit paper it is just $d^d$… $dd^ + d^* d$ is, up to a sign, the usual definition of the Laplacian on p-forms. But when dealing with circuits, all the spaces in our cochain complex $C^0 \to C^1 \to C^2 \to \cdots$ are zero-dimensional except $C^0$ and $C^1$: there are just vertices and edges in our graph, no higher-dimensional stuff. So in this particular case $d^$ vanishes on 0-cochains, so $(dd^ + d^* d) \mu = d^* d \mu$ for any 0-cochain $\mu.$ I took advantage of this simplification in my paper. But now we’re talking about more general things, such as electromagnetism and p-form electromagnetism, where we need 2-cochains, 3-cochains, etc. Indeed, the basic lemma of Hodge theory $(dd^* + d^* d) \mu = 0 \implies d \mu = 0 \; \textrm{and} \; d^* \mu = 0$ is completely trivial in the case of electrical circuits! In that case, either $\mu$ is a 0-cochain and $d^* \mu$ is always zero, or $\mu$ is a 1-chain and $d \mu$ is always zero. So the answer to this question of yours: But why do I need coclosedness? is simply that you don’t “need” it, you have it: 0-cochains are automatically coclosed: $d^* \mu$ is zero for any 0-cochain $\mu$, unless we’re in a weird situation where there exist nonzero (-1)-cochains! I am still trying to get an image in my head what people may mean with higher gauge theory and string theory. The basic idea is just this: you can integrate a 1-form A over the worldline of a particle, and that gives you the action for a particle coupled to the electromagnetic field. Similarly, you can integrate a 2-form B over the worldsheet of a string, and that gives you the action for a string coupled to the ‘higher electromagnetic field’, which is usually called the Kalb-Ramond field. It’s all just the usual story with the dimensions increased by one. My paper with John Huerta, An invitation to higher gauge theory, was supposed to be a friendly way to start learning about this stuff. But I’ve switched over to working on quite different subjects, so I’m not really interested in talking about it anymore. I mainly keep Derek Wise’s work in mind because it uses a lot of the same math that I’m now applying to electrical circuits. • nad says: By the way, Nad: in this particular conversation, please try to post your comments so they appear below the comment you’re replying to. You keep posting them so they show up far above it, and I keep fixing them. I am sorry. I thought I always clicked on the reply button right next to your answers, but may be not, let’s see where this here will end up. First of all thanks a lot for giving me such detailled answers. It is very rare that you find someone who are allowed to ask “holes into the stomach” (Löcher in den Bauch) (means probably you ask so much that the person you ask gets really hungry). Just as a warning: during my studies I was told by a person that one of the professors already complained to this person about my “stupid questions”, in this conversation this professor also said that I am “no big light” (“Sie ist keine grosse Leuchte”). So your answers are very likely pearls before breakfast, unless someone else reads this here too and learns by your answers or might be inspired by my stupid questions (Like an artist or so…). … is simply that you don’t “need” it, you have it: 0-cochains are automatically coclosed: $d^* \mu$ is zero for any 0-cochain $\mu$, unless we’re in a weird situation where there exist nonzero (-1)-cochains! So you say that in a theory of electricity which behaves with integrity and discretion $d^* \mu$ must be zero. But if you postulate this anyways then less then ever I understand why you would need the positive definiteness of the inner product on $C_0(\Gamma)$ or $C^0(\Gamma)$?! $dd^* + d^* d$ is, up to a sign, the usual definition of the Laplacian on p-forms. What is the problem with just taking $d^* d$ ? The possibly negative spectrum on a compact manifold? By the way in this context I may ask what I always wanted to know. If you look at the Weitzenböck tensor (difference between your usual Laplacian and the rough Laplacian) then this looks “almost” like the Einstein tensor. Can one formulate the “Einstein operator” (here I mean Einstein tensor + cosmological term – Energy momentum tensor, or may be just the Einstein tensor for simplicity) as some linear combination of the Laplacians of two different (co)derivatives ? If this would be the case then this could eventually be a useful ingredient for the discretization of the field equations (on which I believe probably tons of people have been working at, at least already when I was still a student there were a couple of people interested in this. • John Baez says: Nad wrote: So you say that in a theory of electricity which behaves with integrity and discretion $d^* \mu$ must be zero. That’s your interpretation… I prefer to repeat what I actually said, which sounds much less interesting, but has the advantage of being surely true: If $d^$ maps 0-forms to (-1)-forms, and all (-1)-forms are zero, and $\mu$ is a 0-form, then $d^ \mu = 0$. (Some people say “there are no (-1)-forms”, but a mathematically more sophisticated approach is to say all (-1)-forms are zero, and that’s the approach I usually take.) But if you postulate this anyways then less then ever I understand why you would need the positive definiteness of the inner product on $C_0(\Gamma)$ or $C^0(\Gamma)$?! I will let you worry about this; I don’t feel like it. What I know is that: 1) I have a specific formula for the inner product on $C^0(\Gamma)$ which works nicely for electrical circuits; this inner product is positive definite. 2) There’s a standard formula for the inner product on $\Omega^0(M)$ (0-forms on a compact manifold), which for Riemannian manifolds is positive definite. 3) We need the inner product on zero-forms to be positive definite for this standard calculation in Hodge theory to work: $(dd^* + d^* d) \mu = 0 \implies$ $\langle \mu, (dd^* + d^* d) \mu \rangle = 0 \implies$ $\langle d^* \mu, d^* \mu \rangle + \langle d \mu, d \mu \rangle = 0 \implies$ $\langle d^* \mu, d^* \mu \rangle = 0 \; \textrm{and} \; \langle d \mu, d \mu \rangle = 0 \implies$ $d \mu = 0 \; \textrm{and} \; d^* \mu = 0$ in the case where $\mu$ is a 1-form. What is the problem with just taking $d^* d$? The possibly negative spectrum on a compact manifold? No, that’s not the problem: $\langle \mu , d^* d \mu \rangle = \langle d \mu, d \mu \rangle \ge 0$ so in any situation where we’ve worked out the details carefully enough that $d^* d$ is self-adjoint, we can see that its spectrum is nonnegative. The problem with $d^* d$ is that it’s not the Laplacian. For example, we have a pre-existing definition of the Laplacian of a vector field on Euclidean $\mathbb{R}^n$, and thus, using the metric to turn a vector field into a 1-form, we get a definition of the Laplacian of a 1-form on $\mathbb{R}^n$. It’s just the obvious thing: take the Laplacian of each component using the standard basis of 1-forms $dx_i$. But this Laplacian does not agree with $d^* d!$ For example take $n = 1$. In this case $d^* d$ of any 1-form is zero! On the other hand, this pre-existing definition of the Laplacian of a 1-form on $\mathbb{R}^n$ does agree with $d^* d + d^* d.$ There are lots of other reasons why everyone defines $d^* d + d^* d$ to be the Laplacian. Mostly, it makes Hodge theory work (just read the section on ‘Riemannian Hodge theory’, which is nice and short and lists the key results). If you look at the Weitzenböck tensor (difference between your usual Laplacian and the rough Laplacian) then this looks “almost” like the Einstein tensor. Can one formulate the “Einstein operator” (here I mean Einstein tensor + cosmological term – Energy momentum tensor, or may be just the Einstein tensor for simplicity) as the difference of the laplacians of two different (co)derivatives ? I’ve never thought about this. I needed to look up the Weitzenböck identity to think about this. Maybe some trickery could turn the operator this reference calls $A$—I guess that’s your ‘Weitzenböck tensor’—into the Einstein tensor. But I don’t instantly see how to do it. Real experts on general relativity should know if it’s possible. It’s a nice idea. • nad says: No, that’s not the problem: $\langle \mu , d^* d \mu \rangle = \langle d \mu, d \mu \rangle \ge 0$ so in any situation where we’ve worked out the details carefully enough that d^* d is self-adjoint, we can see that its spectrum is nonnegative. If you assume positive definiteness and your other definitions yes, but as said one could try to generalize that. Eventually it would be enough to use a hermitian form then you can keep your positive definiteness. But it seems that in order to include magnetism some generalizations may eventually need to be made. The problem with $d^* d$ is that it’s not the Laplacian. It is not the Hodge Laplacian. Yes I meant that operator A. I called it Weitzenböck tensor because I don’t know how this thing is usually called in various dialects. And I called it tensor because the Wikipedia entry said: A is a linear operator of order zero involving only the curvature. So I understood this as that $A$ is at least locally linear, i.e. that it is a tensor. But may be this interpretation is wrong. I might have been lured into that interpretation by the fact that the difference of two covariant derivatives is a tensor, if I remember correctly ( : O) ). By the way, since I haven’t thought about this stuff since quite a while it feels quite awkward to talk about this in public. It feels really quite like a public Nadja-Geometry-Amnesia test. : O Maybe some trickery could turn the operator this reference calls A.. If it is a tensor then locally, lets assume that one has invertibility then it seems (again if I remember correctly) one could just compose one tensor (as a linear map) with the inverse of the other and get a linear map between them. So if this “tensor” interpretation is correct then locally this seems to work always (apart from divergencies due to noninvertibility etc.), but what about globally? I mean apart from that plus/minus sign and modulo the fact that I don’t know what is exactly meant by these alternation map and this universal derivation inverse to θ on 1-forms and what they could spoil up this Weitzenböck “tensor” looks really like the Einstein tensor. If I remember correctly ( : O ) the Einstein tensor was constructed as a generalization to the laplacian in Poisson equation for gravity, so I was hesitating to assume that it could be related to the difference of two Laplacians, thats what was meant by my “replystammer”. So I wonder wether there exists some way, with which one could express the Einstein tensor as a linear expression (meaning here a difference or a mean or something in that sense) of two different laplacians, just as the Weitzenböck “tensor” is the difference between the Hodge laplacian and the rough laplacian. • John Baez says: Nad wrote: So I understood this as that $A$ is at least locally linear, i.e. that it is a tensor. Yes, it’s a tensor, built from the Riemann tensor somehow. The Wikipedia formula for it is indeed poorly explained. Their so-called “alternation map” $\theta: T^* M \otimes \Omega^p(M) \to \Omega^{p+1}(M)$ is obviously just taking the wedge product of a 1-form and a p-form to get a (p+1)-form—what else could possibly make sense here? Some weird person just decided to call it the “alternation map” instead of its usual name, “wedge product” or “exterior multiplication”. Similarly, their so-called “universal derivation inverse to $\theta$ on 1-forms” $\#:\Omega^{p+1}(M)\rightarrow T^M\otimes\Omega^p(M)$ has got to come from interior multiplication: $i: T M \otimes \Omega^{p+1} (M) \to \Omega^p(M)$ by using duality to turn it into a map $\#:\Omega^{p+1}(M)\rightarrow T^M\otimes\Omega^p(M)$ Again, there’s nothing else it could reasonably be. I’m getting tired of this Weitzenböck stuff, so I’ll stop here. By the way, this This article claims to “demystify” the Weitzenböck tensor, but it would take a little while to read. Also by the way, I’ve never heard the term “rough Laplacian” before. • John Baez says: There’s some problem going on with the blog: comments keep appearing higher up than they should. I hope it’s just a disturbance in the Earth’s magnetic field that will go away on its own. I have moved a bunch of comments down so that the conversation makes more sense, and I’ve taken the liberty of deleting some comments by Nad and Blake that talk only about this stuff like this. I hope that if we keep hitting ‘reply’ at the very bottom of this list of comments, nothing terrible will happen. 21. John Baez says: Thanks for offering an explanation of why that paper is in Spanish, Blake! It’s very weird trying to read something by Weyl in Spanish. I can only see the first few pages on Google; maybe if I’m feeling energetic I’ll try to get ahold of his collected works, scan in this paper, and distribute it to the world. 22. nad says: I’m getting tired of this Weitzenböck stuff, so I’ll stop here. thanks anyways. It was probably the longest discussion about Laplacians I ever had in my life. This article claims to “demystify” the Weitzenböck tensor, but it would take a little while to read. This seems to be at most a draft not an article like on page 3 it starts out: 1. Lichnerowicz Laplacians (r⇤T)(X2,…,Xk) = ␣(rEiT)(Ei,X2,…,Xk) In this section …. Conventions and given the sparseness of definitions it is for me impossible to follow the paper. And by very briefly looking at it seems it wouldn’t help too much anyway, since it seems to mention only different Laplacians constructed from one connection, but I meant different Laplacians which could be constructed from different connections.	{"extraction_info": {"found_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": 397, "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": 0.9452801942825317, "perplexity_flag": "middle"}
http://planetmath.org/maximalsubgroup	# maximal subgroup Let $G$ be a group. A subgroup $H$ of $G$ is said to be a maximal subgroup of $G$ if $H\neq G$ and there is no subgroup $K$ of $G$ such that $H<K<G$. Note that a maximal subgroup of $G$ is not maximal among all subgroups of $G$, but only among all proper subgroups of $G$. For this reason, maximal subgroups are sometimes called maximal proper subgroups. Similarly, a normal subgroup $N$ of $G$ is said to be a maximal normal subgroup (or maximal proper normal subgroup) of $G$ if $N\neq G$ and there is no normal subgroup $K$ of $G$ such that $N<K<G$. We have the following theorem: ###### Theorem. A normal subgroup $N$ of a group $G$ is a maximal normal subgroup if and only if the quotient $G/N$ is simple. Type of Math Object: Definition Major Section: Reference Groups audience: ## Mathematics Subject Classification 20E28 Maximal subgroups ## Recent Activity May 17 new image: sinx_approx.png by jeremyboden new image: approximation_to_sinx by jeremyboden new image: approximation_to_sinx by jeremyboden new question: Solving the word problem for isomorphic groups by mairiwalker new image: LineDiagrams.jpg by m759 new image: ProjPoints.jpg by m759 new image: AbstrExample3.jpg by m759 new image: four-diamond_figure.jpg by m759 May 16 new problem: Curve fitting using the Exchange Algorithm. by jeremyboden new question: Undirected graphs and their Chromatic Number by Serchinnho ## Info Owner: yark Added: 2002-02-19 - 16:54 Author(s): yark ## Versions (v15) by yark 2013-03-22	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 21, "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": 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": 0.8432710766792297, "perplexity_flag": "middle"}
http://mathhelpforum.com/calculus/210056-prove-there-exists-inverse-function.html	# Thread: 1. ## prove there exists an inverse function Hello. I'm quite new to calculus and frankly speaking I'm an autodidact. Could you tell me how to solve this: $f: [\frac{\pi}{2}, \pi) \ni x \rightarrow \frac{1}{sinx} \in R$ Prove that there exists a inverese of $f$. In what set is $f^{-1}$ differentiable? Calculate $f^{-1'}$ I would really appreciate a thorough explanation. Thank you. 2. ## Re: prove there exists an inverse function Well, you could easily show that on the interval $[\frac{\pi}{2}, \pi)$, f is monotonously increasing, since $f'(x)$ exists on the interval, it is continuous on the interval. A continous function which is monotously increasing or decreasing is one to one on that interval and thus also has an inverse. Now if f is differentiable on the interval, $f^{-1}$ is also differentiable given that $f'(x) \not = 0$ at any point on the interval. You can verify that this is true. Thus $f^{-1}$ is differentiable over $[\frac{\pi}{2}, \pi)$	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 10, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 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": 0.9494888782501221, "perplexity_flag": "head"}
http://mathoverflow.net/questions/117791?sort=votes	## What axioms are stronger than the Axiom of choice? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) What other axioms in set theory are stronger than AC ? I mean what are those axioms that will imply AC ? - ## 5 Answers The axiom "every set is constructible" (denoted V = L), and the axiom "every set is definable from an ordinal parameter" (denoted often as V= HOD, and sometimes as V= OD) each implies AC, and each is provably stronger than AC. More specifically, it is well-known that: (a) (Within ZF), V = L implies V = HOD, and V = HOD implies AC. (b) Neither of the above implications is reversible. Historical Note: The axiom V = L was introduced by Gödel in his seminal work on the consistency of AC and GCH with ZF. The axiom V= HOD was first publicly introduced in a joint paper of Myhill and Scott, but their paper acknowledges that the axiom was independently considered by a number of people, including Gödel, Post, and the joint work of Vopěnka and Balcar. - Do you know a good discussion of V = HOD? I've seen it mentioned here at MO a couple of times, and I'm curious to learn more. – arsmath Jan 1 at 18:19 1 @arsmath: Kunen's chapter on ordinal definability in his textbook on set theory is a very good place to learn about V=HOD; Jech's text also has a discussion of it but with far less detail. – Ali Enayat Jan 1 at 18:44 @arsmath: You might also look at the original paper of Myhill and Scott. It's in the proceedings of the 1967 UCLA set theory conference, published by the AMS in the series "Proceedings of Symposia in Pure Mathematics" (volume 13, part 1). – Andreas Blass Jan 1 at 19:47 1 Since $V=L$ is mentioned in several answers, I should point out that more generally, the statement that $V$ is a fine-structural model (sometimes written as $V=K$, where $K$ stands for some flavor of the "core model") suffices, and is much less restrictive. For example, it is compatible with $V=L$ but also with $V=L[\mu]$, the smallest inner model for a measurable cardinal, or $V=M_1^\sharp$, etc. In all cases, if $V$ is a fine-structural model, it is self-wellorderable, so $V=HOD$ and AC holds. – Andres Caicedo Jan 1 at 23:33 @ALi: Would you explain what is "V=L" more precisely or name some references (for amateurs of course !) – Dimros Jan 7 at 9:24 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. 1. The axiom For every infinite set $X$ if $Y$ is such that $\|X\|\lt\|Y\|$ and $\|Y\|\leq\|\mathcal P(X)\|$, then $\|Y\|=\|\mathcal P(X)\|$. which is also known as the Generalized Continuum Hypothesis. 2. In turn this axiom is equivalent (as it turns out) to For every ordinal $\alpha$, $2^{\aleph_\alpha}=\aleph_{\alpha+1}$. While the equivalent may not be true in ZF, it is certainly true in ZFC. Both formulations imply the axiom of choice. (This axiom follows from $V=L$ mentioned in the other answers, but does not imply it. It does not follow from $V=HOD$.) 3. There exists an ordinal $\alpha$ such that there is no set $X$ for which there is a chain of cardinals between $X$ and $\mathcal P(X)$ of order type $\alpha$. (i.e. the "distance" between $X$ and $\mathcal P(X)$ is never longer than $\alpha$). Equally this can be required about $X$ and $X^2$. However assuming choice $\|X\|=\|X^2\|$ for infinite sets, so this turns out to be equivalent; whereas we can use class forcing to push power sets far enough so there are such sequences of any length. Generally speaking, let $\varphi$ be a proposition which is consistent with ZFC but not provable from ZFC, then AC+$\varphi$ is an axiom stronger than the axiom of choice. In the fashion above, we can think about $\varphi$ being "the axiom of choice holds, and $\lozenge_\kappa$ is true for all $\kappa<\aleph_\omega$." or something like that. - For every infinite set $X$ ... – Andres Caicedo Jan 1 at 20:04 Andres, yes, thank you. (Also $\|X\|\lt\|Y\|$ was incorrectly written as $\leq$.) – Asaf Karagila Jan 1 at 20:05 The Axiom of Constructibility implies the Axiom of Choice, as well as many other results independent of ZFC, such as the continuum hypothesis. - 1 The CH implies a well ordering of R, trivially. It says the reals can be put in 1-1 correspondence with an ordinal, namely $omega_1$. The GCH or any other global assertion or any other global assertion about cardinal exponentiation will imply full AC for the same reason. – mbsq Jan 1 at 18:15 1 I meant that the axiom of constructibility implies the continuum hypothesis, as well as the generalized continuum hypothesis. – arsmath Jan 1 at 18:17 2 CH does not imply that $\mathbb R$ can be well-ordered. That would be The Aleph Hypothesis for $\aleph_0$, whereas there are models in which $\mathbb R$ cannot be well-ordered and every set of reals is either countable or of size continuum (e.g. Solovay's model). – Asaf Karagila Jan 1 at 19:35 2 As can be seen from the comments of mbsq and Asaf Karagila, the terminology "continuum hypothesis" becomes ambiguous in the absence of the axiom of choice. It is often formulated as $2^{\aleph_0}=\aleph_1$ (which mbsq calls CH and Asaf calls the aleph hypothesis for $\aleph_0$), and it is often formulated as "every uncountable set of reals admits a bijection to $\mathbb R$ (which is what Asaf means by CH). Analogous ambiguities afflict GCH. – Andreas Blass Jan 1 at 19:42 2 As an original reference (not a textbook), see: Sierpiński, Wacław L'hypothèse généralisée du continu et l'axiome du choix. Fundamenta Mathematicae 34, (1947). 1–5 – Margaret Friedland Jan 1 at 21:04 show 14 more comments Many large cardinal hypotheses imply AC, when stated in a certain natural way. For example, the assertion over ZF that there are unboundedly many inaccessible cardinals, defined as uncountable regular strong limit cardinals, implies the axiom of choice. Indeed, even having a proper class of strong limit cardinals implies the axiom of choice. What's more, the axiom of choice is equivalent over ZF to the assertion "there are unboundedly many strong limit cardinals". The forward direction is simply the usual proof in ZFC that there is a closed unbounded class of strong limit cardinals. For the reverse implication, we define that $\kappa$ is a strong limit if and only if it is an initial ordinal for which $\beta\lt\kappa\implies P(\beta)\lt\kappa$. Note that this implies in particular that the power set $P(\beta)$ is well-orderable for $\beta\lt\kappa$, since it injects into $\kappa$, which is well-orderable. Thus, if there are a proper class of such $\kappa$, then every well-orderable set will have a well-orderable power set. It is a non-obvious fact (and in fact a quite slippery fact, which beginners often get wrong when first trying to prove it) that if every well-orderable set has a well-orderable power set, then AC holds. To see that there are several natural but inequivalent formulations of inaccessibility and of strong limit cardinals in ZF, see the treatment in A. Blass, I. Dimitriou, B. Loewe, Inaccessible cardinals without the axiom of choice. - Well a proper class of strong limit cardinals is provable in ZFC. – Asaf Karagila Jan 1 at 21:42 Asaf, yes, that is the reverse direction of the equivalence. – Joel David Hamkins Jan 1 at 21:54 Ah, so Grothendieck's axiom of universes gives AC? That's interesting. – David Roberts Jan 1 at 23:33 1 David, unfortunately, that's not quite the case, since the various forms of inaccessibility are no longer equivalent without AC, and in particular, this form of inaccessibility is not equivalent to UA in ZF alone (although it is in ZFC). – Joel David Hamkins Jan 1 at 23:42 1 Yes, I think there are analogues for UA of the issues in the Blass, Dimitriou, Loewe paper to which I link. That might make a nice topic to work out carefully... – Joel David Hamkins Jan 2 at 0:28 show 4 more comments The axiom of global choice. Technically this isn't really an axiom: global choice (GC) states that there is a formula $\phi(x, y)$ such that the relation $$A\le_\phi B:= V\models \phi(A, B)$$ is a well-ordering of the universe $V$. This can't be stated as a single formula, so in some sense it's a meta-axiom. It clearly implies choice, and is implied by $V=L$: we already have a partition of the universe into $L_0$, $L_1$, . . . , $L_\alpha$, . . ., and we can get from here to a full (class-)well-ordering of the universe by fixing at the outset some well-ordering of formulas in the language of set theory (since at each stage in the construction of $L$ we are only taking definable powersets; this is why this argument doesn't work in just $ZFC$). A couple comments on why I think global choice is interesting (even though it's not expressible in the language of set theory): • Although the relative consistency of Choice was proven rather early, it was via $L$, which satisfies $GC$ as well; the result that, assuming the consistency of $ZFC$, there is a model of $ZFC$ with no definable well-ordering of the universe came much later [NOTE: this is based on foggy memory, and I don't recall exactly when this result happened; can someone remind me?], and generally telling whether a model of $ZFC$ satisfies $GC$ is very hard. • Basically by reversing the argument that it is true in $L$, one can make an informal argument that GC implies that the universe is small (contains only "buildable" things). So there's a somewhat intuitive argument for $AC+\neg GC$: the universe should be "big enough" that for each family of sets, we have a choice function, but the universe should also be big enough that any specific definable well-order "misses something." • $GC$ is useful in other set theories. First of all, as noted above, even stating $GC$ needs a class theory like Morse-Kelley or NBG (expansions of $ZFC$ to also talk about classes). We can also ask about the status of $GC$ in really odd set theories like New Foundations (EDIT: As Ali points out below, in NF/NFU Choice and Global Choice are essentially one and the same - but the point still stands that global choice could still be interesting in set theories different than $ZFC$.) • $GC$ makes sense even outside of set theories! It doesn't make sense to ask whether a given ring $R$ satisfies the axiom of choice, since elements of $R$ aren't (at least on the face of things) sets, but it does make sense to ask whether there is a formula in the language of rings which well-orders $R$. - I've been slightly vague about my "definable" well-orderings: I think everything I said is true whether we take "definable" to mean with or without parameters. But I welcome corrections on this point! – Noah S Jan 2 at 4:55 2 Noah: (1) Despite appearances, GC as you define it (without parameters) is well-known to be equivalent to V=OD, thanks to the reflection theorem [and if you allow parameters, then this weaker version of GC is equivalent to "V is ordinal definable from a parameter], and (2) Jensen's original proof of consistency of NFU shows the consistency of NFU with global choice (in the NFU context, since there is a universal set, global choice simply stands for the statement "every set can be well-ordered"). – Ali Enayat Jan 2 at 7:10 Re: (1), that's neat, I didn't know that. Re: (2), of course - welp, that's what I get for answering while sleep-deprived. – Noah S Jan 2 at 7:37 1 I think of GC not as making an assertion about definable classes, even with parameters, but rather as one of the principle axioms of GBC, stated as the second-order assertion that some class is a well-ordering of the universe V. This can be forced over any ZFC model without adding sets, by forcing with the class of set-sized well-orderings, ordered by end-extension. This is how we know GBC is conservative over ZFC. If you insist that the order is definable, even with parameters, you wouldn't get the conservativity result. – Joel David Hamkins Jan 2 at 14:57 That is, what you've got (with or without parameters) is a strengthening of what is usually known as the global choice axiom. – Joel David Hamkins Jan 2 at 15:19	{"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": 75, "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": 0.9470309615135193, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/tagged/hyperbolic-geometry+conformal-geometry	# Tagged Questions 2answers 331 views ### How does a conformal mapping preserve angles in hyperbolic geometry? Suppose I have a sector $D = \{0 < \arg z < \alpha\}$ where $\alpha \leq 2\pi$. If I apply the function $w = \frac{\zeta - i}{\zeta + i}$ from the upper half plane to the unit disc (\$\zeta = ... 2answers 134 views ### Perpendicular in conformal disk model Firstly, please note that the related question can also be found at mathoverflow. The question is stated as following: In Euclidean Geometry, we know that from a given point there is an unique line ...	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 3, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9282543063163757, "perplexity_flag": "head"}
http://mathhelpforum.com/advanced-statistics/86351-e-x-exponential-random-variable-joint-density-elevators-cov-multinomial.html	# Thread: 1. ## E(x) of exponential random variable, joint density, and elevators +cov of multinomial So, I've been working on a bonus review for a few hours now and have hit places where I'm not sure and places where I don't know where to start. I don't need help on all of these but it's simpler if I just post the whole pdf: Beef Supreme - Now with cheese! #3b I can't figure out... it's a classic problem that I've done before, only with n people, not n-separated-by-gender. If it was just n people, I'd do $1 - \frac{(365)(364)...(364-n+1)}{365^n}$ but I don't know how to separate that population by gender. #5 I have no idea where to start. I don't know how I'll end up with a factorial. #6a I used a double integral, the outer between 0 and 1, the inner 0 and x, f(x,y)dydx... with first-time attempted LaTeX reproduction: $<br /> \int_0^1 \int_0^X f(x,y) \, dydx<br />$ and ended up with $\frac{15}{56}$ Are those the correct bounds? #6b I figured I had to find the marginal density of y first, so $\int_0^2 \frac{6}{14} (2x^2 + xy) \, dx$ gave me $\frac{16}{7} + \frac{6}{7} y$ for $0<y<2$ Now I think I have to take a (double?) integral of this to find the expected value, but I don't know what formula or bounds to start with. #7 I can't wrap my brain around at all. I was thinking along the lines of E(x) would simply be trialsprobability $10 \frac{1}{5}$ but there's no way it could be that simple, and E(X) = 2 doesn't sound right. #8 I should note there's a typo here: The cov(X1, X2) should be equal to -np1p2 and not -mp1p2. I've studied my 'Lecture 18' notes, and can't get any clarity out of it. #### Search Tags View Tag Cloud Copyright © 2005-2013 Math Help Forum. All rights reserved.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 7, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": 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": 0.9754529595375061, "perplexity_flag": "middle"}
http://mathhelpforum.com/pre-calculus/145598-solved-make-x-subject-y-2x-x-2-a.html	# Thread: 1. ## [SOLVED]Make x the subject in y = 2x - x^2. Hi it's late but I can't seem to single out x in the following equation: $y = 2x - x^2$ Can it be done? I want to evaluate the volume by using washers even though I can just use shells 2. You can't solve that equation for x can you be more specific about what you are trying to do? 3. I knew it! I was trying to find the volume enclosed by this eq. by washers instead of shells. From my book: The volume V can also be found using washers; however, the calculations would be more involved since the given equation would have to be solved for x in terms of y. The book, however, tells me that it is possible but does not elaborate It's really more a question of surmounting such a simple looking equation than actually finding the volume which I can do fine via shells. 4. Originally Posted by sponsoredwalk I knew it! I was trying to find the volume enclosed by this eq. by washers instead of shells. From my book: The book, however, tells me that it is possible but does not elaborate It's really more a question of surmounting such a simple looking equation than actually finding the volume which I can do fine via shells. $y = 2x - x^2$ $x^2 - 2x = -y$ $x^2 - 2x + 1 = 1 - y$ $(x-1)^2 = 1 - y$ $x-1 = \pm \sqrt{1-y}$ $x = 1 \pm \sqrt{1-y}$ washers ... $R(y) = 1 + \sqrt{1-y}$ $r(y) = 1 - \sqrt{1-y}$ $[R(y)]^2 = 2 + 2\sqrt{1-y} - y$ $[r(y)]^2 = 2 - 2\sqrt{1-y} - y$ $[R(y)]^2 - [r(y)]^2 = 4\sqrt{1-y}$ $V = \pi \int_0^1 4\sqrt{1-y} \, dy = \frac{8\pi}{3}$ shells ... $V = 2\pi \int_0^2 x(2x-x^2) \, dx = \frac{8\pi}{3}$ 5. Woah man you are my hero! I should be thinking like that but it's 2am Thank you	{"extraction_info": {"found_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": 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": 0.9774086475372314, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/tagged/linear-algebra?sort=active&pagesize=50	# Tagged Questions Linear algebra is concerned with vector spaces of all dimensions and linear transformations between them. This includes: systems of linear equations, Hamel basis, dimension, matrices, determinant, trace, eigenvalues and eigenvectors, diagonalization, etc. For questions specifically concerning ... 4answers 224 views ### Linear Algebra and Applications [duplicate] Possible Duplicate: Where to start learning Linear Algebra? Does anybody know of a good book in Linear Algebra for self study for a beginner? I look for a book with many exercises. 0answers 16 views ### Does convex and radially open imply open? I want to show that a convex set $A$ is radially open iff $A\cap W$ is open in W, for every finite dimensional linear subspace. 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http://mathhelpforum.com/calculus/145757-differentiation-application.html	# Thread: 1. ## Differentiation Application Thanks to those who responded yesterday to my question regarding the differentiation in this problem. As it turns out, I wasn't off as far as I thought I might be. I believed that correctly differentiating the function was the root of my problem, but as it turns out, I'm not sure how to apply the derivative correctly. Here is a restatement of the problem fleshed out: The expression for the charge entereing the upper terminal of a basic, 2-terminal circuit element, is given by, $q=\frac{1}{\alpha^2}-{(\frac{t}{\alpha}+\frac{1}{\alpha^2})}e^{-\alpha t}$ Coulombs. Find the maximum value of the current entering the terminal if $\alpha=0.03679\ s^{-1}$. I thought that finding $\lim_{t\rightarrow\infty}\frac{dq}{dt} = te^{-\alpha t}$ would give me the answer. I applied L' Hopital's Rule and ended up with zero, which obviously is incorrect. The book gives an answer of 10 Amperes. If there are any engineers out there, I could use some help. P.S. Amperes = Couloumbs/Second, which is what led me to differentiation in the first place. 2. At a given time $t$, the current is equal to $\frac{dq}{dt} = I$. So to find the maximum current, we need to take another derivative to maximize $I$. $\frac{dI}{dt} = \frac{d}{dt} \left[ te^{-\alpha t} \right] = e^{-\alpha t} - \alpha t e^{-\alpha t} = (1-\alpha t)e^{-\alpha t} = 0$ Can you finish? 3. I assume that the second derivative is taken to find relative maximum. But I still cannot figure out where to go. If I take the limit as t approaches infinity, I still end up with 0. And I might be wrong to even take that approach. $\lim_{t\rightarrow \infty}(\frac{1}{e^{\alpha t}}-\frac{\alpha t}{e^{\alpha t}})= 0 - \frac{\infty}{\infty}$ Applying L' Hopital's Rule to the indeterminate form: gives 0 - 0 = 0. Where am I going wrong??? 4. Originally Posted by drumist At a given time $t$, the current is equal to $\frac{dq}{dt} = I$. So to find the maximum current, we need to take another derivative to maximize $I$. $\frac{dI}{dt} = \frac{d}{dt} \left[ te^{-\alpha t} \right] = e^{-\alpha t} - \alpha t e^{-\alpha t} = (1-\alpha t)e^{-\alpha t} = 0$ Can you finish? I assume that the second derivative is taken to find relative maximum. But I still cannot figure out where to go. If I take the limit as t approaches infinity, I still end up with 0. And I might be wrong to even take that approach. Applying L' Hopital's Rule to the indeterminate form: gives 0 - 0 = 0. Where am I going wrong??? 5. I'm not sure why you are taking a limit. There is no limit involved in the calculation. You're correct that I was looking for relative maximums because that is usually the method used to solve for the absolute maximum. Since we are looking for a maximum for $I$, we take its derivative and set it equal to zero, and solve for $t$: $\frac{dI}{dt} = \frac{d}{dt} \left[ te^{-\alpha t} \right] = e^{-\alpha t} - \alpha t e^{-\alpha t} = (1-\alpha t)e^{-\alpha t} = 0$ Since $e^{-\alpha t}$ can never equal zero, we know that: $1-\alpha t = 0 \implies t = \frac{1}{\alpha}$ So the value of $t$ where the maximum occurs is $t=\frac{1}{\alpha}$. Therefore: $I = t e^{-\alpha t} = \frac{1}{\alpha} e^{-1} = \frac{1}{\alpha e} \approx 10$ 6. Originally Posted by drumist I'm not sure why you are taking a limit. There is no limit involved in the calculation. You're correct that I was looking for relative maximums because that is usually the method used to solve for the absolute maximum. Since we are looking for a maximum for $I$, we take its derivative and set it equal to zero, and solve for $t$: $\frac{dI}{dt} = \frac{d}{dt} \left[ te^{-\alpha t} \right] = e^{-\alpha t} - \alpha t e^{-\alpha t} = (1-\alpha t)e^{-\alpha t} = 0$ Since $e^{-\alpha t}$ can never equal zero, we know that: $1-\alpha t = 0 \implies t = \frac{1}{\alpha}$ So the value of $t$ where the maximum occurs is $t=\frac{1}{\alpha}$. Therefore: $I = t e^{-\alpha t} = \frac{1}{\alpha} e^{-1} = \frac{1}{\alpha e} \approx 10$ Thank you for your patience, and I'm sure that I'm trying it, but my reasoning was that since dq/dt represents current, I was thinking that if I differentiated "q" and took the limit as t approached infinity, that the current would be maximized. I follow the mathematics of your explanation, but I still don't understand why setting the 2nd derivative equal to zero leads to max current. I thougt that critical numbers came from the 1st derivative set equal to zero and that if the second derivative evaluated at a critical number was less than zero, that it indicated a relative max. If you would be so kind as to elaborate a little further. I really want to learn, I'm just disconnecting somewhere. Thanks 7. It's important to realize that the problem is asking you to find the maximum value of current, not the maximum value of charge. To maximize current, the first step is to find an expression for current. Since the problem told you what the charge is, you can take a derivative to find the expression for current, which is what we did in the other thread from yesterday. If we were trying to find the maximum value of charge, then we would set $\frac{dq}{dt} = 0$ because $\frac{dq}{dt}$ is the first derivative of charge. However, that's not what the problem asks for! Instead, we want the maximum value of current. The current is equal to $\frac{dq}{dt}$, so to find the maximum value, we must calculate the first derivative of the current, which is $\frac{d^2 q}{dt^2}$ and set that equal to zero. If you wanted to verify that it was a maximum and not a minimum, you would want to calculate the second derivative of current (which is the third derivative of charge), i.e., $\frac{d^3q}{dt^3}$ and determine if it is positive or negative at the value of $t$ we found. You'll notice in my work I went ahead and created a new variable $I$ to represent the current. I just did this to hopefully avoid the confusion between charge and current. You may notice that: $I = \frac{dq}{dt}$ <-- we want to maximize this $\frac{dI}{dt} = \frac{d^2q}{dt^2}$ <-- so we set this equal to zero $\frac{d^2 I}{dt^2} = \frac{d^3q}{dt^3}$ <-- and make sure this is positive 8. Thank you Drumist, I was missing that fact that what I was considering to be the 2nd derivative (of charge) was the first derivative of current! You have been a great help. I really appreciate it.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 38, "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": 0.969588041305542, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/276520/full-isomorphism-closed-reflective-subcategory-of-set	# Full isomorphism-closed reflective subcategory of Set What are the full isomorphism-closed reflective subcategories of $\textbf{Set}$? In the book "The Joy of Cats" it is mentioned (p. 58) that there are precisely three such subcategories. I can see two of them are: 1. the whole category $\textbf{Set}$ 2. the subcategory consisting of all singletons What would be the third one? - Hmm...what about $\emptyset$, i.e. the subcategory with object $\emptyset$ and the unique morphism $\emptyset \rightarrow \emptyset$? – uncookedfalcon Jan 12 at 20:30 1 @uncookedfalcon I don't think $\emptyset$ is a reflective subcategory of $\textbf{Set}$. Because given a nonempty set $A$, there is no morphism from $A$ to $\emptyset$, hence no reflection morphism. – PatrickR Jan 12 at 20:41 Ahh...fantastic point! – uncookedfalcon Jan 12 at 20:48 ## 1 Answer The subcategory containing the singletons and the empty set seems to work. - I think you are right. The category you mention works. But that makes me think, maybe I was mistaken: the subcategory consisting of only the singletons is not reflective? Because if there were a reflection arrow from $\emptyset$ to some singleton $A$, it would have to be the empty function. But the empty function $\emptyset \to \emptyset$ does not factor through that map $\emptyset \to A$. What do you think? – PatrickR Jan 12 at 21:00 sorry about the confusion: the subcategory consisting of only the singletons is reflective after all. Only the functions to a object in the subcategory need to be considered in the definition of reflection arrow. – PatrickR Jan 12 at 21:08	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.909508228302002, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/18268/consecutive-birthdays-probability?answertab=active	Consecutive birthdays probability Let $n$ be a number of people. At least two of them may be born on the same day of the year with probability: $$1-\prod_{i=0}^{n-1} \frac{365-i}{365}$$ But what is the probability that at least two of them are born on two consecutive days of the year (considering December 31st and January 1st also consecutive)? It seems a good approximation is: $$1-\prod_{i=0}^{n-1} \frac{365-2 \times i}{365}$$ However, simulating pseudo-random integers with Python, the 99%-confidence intervals may be slightly different. So do you have any closed formula? Results of the simulation with Python. Here are 99%-confidence intervals below: ````Number of people: 1 Lower bound: 0.0 Upper bound: 0.0 Number of people: 2 Lower bound: 0.00528 Upper bound: 0.00567 Number of people: 3 Lower bound: 0.01591 Upper bound: 0.01657 Number of people: 4 Lower bound: 0.03185 Upper bound: 0.03277 Number of people: 5 Lower bound: 0.0528 Upper bound: 0.05397 Number of people: 6 Lower bound: 0.07819 Upper bound: 0.07959 Number of people: 7 Lower bound: 0.10844 Upper bound: 0.11006 Number of people: 8 Lower bound: 0.14183 Upper bound: 0.14364 Number of people: 9 Lower bound: 0.17887 Upper bound: 0.18086 Number of people: 10 Lower bound: 0.21816 Upper bound: 0.2203 Number of people: 11 Lower bound: 0.25956 Upper bound: 0.26183 Number of people: 12 Lower bound: 0.30306 Upper bound: 0.30544 Number of people: 13 Lower bound: 0.34678 Upper bound: 0.34925 Number of people: 14 Lower bound: 0.39144 Upper bound: 0.39397 Number of people: 15 Lower bound: 0.43633 Upper bound: 0.4389 Number of people: 16 Lower bound: 0.48072 Upper bound: 0.48331 Number of people: 17 Lower bound: 0.52476 Upper bound: 0.52734 ```` I give here some results with a tweaked approximation formula, using Wolfram Alpha: $$\left( 1 - \frac{n-1}{2 \times 365 + n-1} \right) \times \left( 1-\prod_{i=0}^{n-1} \frac{365-2 \times i}{365} \right)$$ However, this is just a tweak, ans is clearly wrong for $n=33$ since: ````Number of people: 33 My guess: 0.91407 Number of people: 33 Lower bound: 0.94328 Upper bound: 0.94447 ```` Thanks to Jacopo Notarstefano, leonbloy, and Moron, here is the (correct) formula: $$1-\sum_{k=1}^{n}\frac{1}{365^{n-k}k}\left(\prod_{i=1}^{k-1}\frac{365-\left(k+i\right)}{365\times i}\right)\sum_{j=0}^{k-1}\left(-1\right)^{j}C_{k}^{j}\left(k-j\right)^{n}$$ And here are the results of the computations using this formula with Python: ````Number of people: 1 Probability: 0.0 Number of people: 2 Probability: 0.005479452 Number of people: 3 Probability: 0.016348283 Number of people: 4 Probability: 0.032428609 Number of people: 5 Probability: 0.053459591 Number of people: 6 Probability: 0.079104502 Number of people: 7 Probability: 0.108959718 Number of people: 8 Probability: 0.14256532 Number of people: 9 Probability: 0.179416899 Number of people: 10 Probability: 0.218978144 Number of people: 11 Probability: 0.260693782 Number of people: 12 Probability: 0.304002428 Number of people: 13 Probability: 0.34834893 Number of people: 14 Probability: 0.393195856 Number of people: 15 Probability: 0.438033789 Number of people: 16 Probability: 0.482390182 Number of people: 17 Probability: 0.525836596 Number of people: 18 Probability: 0.567994209 Number of people: 19 Probability: 0.608537602 Number of people: 20 Probability: 0.647196551 Number of people: 21 Probability: 0.683756966 Number of people: 22 Probability: 0.718059191 Number of people: 23 Probability: 0.749995532 Number of people: 24 Probability: 0.779509664 Number of people: 25 Probability: 0.806569056 Number of people: 26 Probability: 0.831211564 Number of people: 27 Probability: 0.853561895 Number of people: 28 Probability: 0.873571839 Number of people: 29 Probability: 0.892014392 Number of people: 30 Probability: 0.906106867 Number of people: 31 Probability: 0.919063161 Number of people: 32 Probability: 0.928791992 Number of people: 33 Probability: 0.944659069 ```` - 1 The reason your first formula doesn't work is this: if two people have the same birthday, then they only exclude two days between them, not four. I can't see what your second formula is trying to do. – TonyK Jan 20 '11 at 10:09 Thanks. For information, the second formula is just the result of playing with the simulation and the results. It seems okay for small n, but just an approximation of the expected result. – wok Jan 20 '11 at 10:11 7 I wish people didn't ask questions like this. I have work to do! – TonyK Jan 20 '11 at 10:16 I'll tell my friend to stop asking me intractable problems: he does not know the answer and I am not sure there is a nice formula. Sorry. – wok Jan 20 '11 at 11:18 The second formula is wrong but is a nice try: it corresponds to count all the possible ways of placing the n birthdays, taking into count that each new birthday removes two possibilities. But this is incorrect because it may happen that a new birthday just removes one possibiliy (or none!) It must be a good approximation for small n. – leonbloy Jan 20 '11 at 11:58 show 1 more comment 3 Answers I believe we can give a formula, but I would not call it "closed form". We have $\displaystyle n$ people, and $\displaystyle k$ possible birthdays to choose from (i.e. $\displaystyle k$ days in a year). Let $\displaystyle M$ be the minimum of the two. We will try and count the number of birthday assignments in which no two people have consecutive birthdays. To do this, we will try and count the assignments which use exactly $d$ distinct birthdays, for $\displaystyle d=1, 2 \dots, k$, and then add them up. Now suppose we had a set of $\displaystyle d$ distinct birthdays (don't worry about the consecutive part, just yet). How many ways can we assign these $\displaystyle d$ birthdays to $\displaystyle n$ people, so that each birthday is used at least once? This is basically the problem of finding the number of ways to partition a set of size $\displaystyle n$ in exactly $\displaystyle d$ non-empty parts and assign the $\displaystyle d$ birthdays to each part in the partition, exactly one to each. The number of ways to partition a set of size $\displaystyle n$ into $\displaystyle d$ non-empty parts is given by a Stirling Number of Second Kind, $S(n,d)$. The number of ways to assign $\displaystyle d$ birthdays is $\displaystyle d!$. Thus the number we are looking for is $\displaystyle S(n,d) \times d!$. Now suppose we managed to count the number of subsets containing $\displaystyle d$ elements (from the $\displaystyle k$ birthdays) such that no two elements of the set are consecutive, then we could multiply that number by $\displaystyle S(n,d) \times d!$ to give the number of birthday assignments which use exactly $\displaystyle d$ birthdays such that no two are consecutive. For the moment, ignore the fact that Jan 1 and Dec 31 are consecutive. We need to select $\displaystyle d$ numbers from $\displaystyle 1,2, \dots, k$ such that no two are consecutive. Now if $\displaystyle b_1 \lt b_2 \lt \dots \lt b_d$ were such numbers, then notice that $\displaystyle 1 \le b_1 \lt b_2 - 1 \lt b_3 - 2 \dots \lt b_d - (d-1) \le k-(d-1)$ gives us a way to select numbers from $\displaystyle 1, 2, \dots, k-(d-1)$ without having to bother about the consecutive issue. This can be done in $\displaystyle {k-d+1 \choose d}$ ways. Now since Jan 1 and Dec 31 are consecutive, we need to subtract from this, the number of sets which contain both $\displaystyle 1$ and $\displaystyle k$. This number is same as the number of $\displaystyle d-2$ non-consecutive subsets of $\displaystyle \{3, \dots, k-2\}$ which is $\displaystyle {k-d-1 \choose d-2}$. Thus the number of ways of selecting $\displaystyle d$ non-consecutive birthdays (assuming $\displaystyle 1$ and $\displaystyle k$ are consecutive) is $\displaystyle {k-d+1 \choose d} - {k-d-1 \choose d-2}$, with the understanding that the term being subtracted is $\displaystyle 0$ for $\displaystyle d = 1$ and that $\displaystyle {a \choose b} = 0$ if $\displaystyle a \lt b$ Thus, for $\displaystyle M = \text{min}\{n,k\}$, the total number we are looking for is, $$\sum_{d=1}^{M} S(n,d)\times d! \times \left({k-d+1 \choose d} - {k-d-1 \choose d-2}\right)$$ Note that $\displaystyle S(n,d) \ d! = \sum_{j=0}^{d} (-1)^j {d \choose j} (d-j)^n$ So we could also write the formula as $$\sum_{d=1}^{M} \left(\sum_{j=0}^{d} (-1)^j {d \choose j} (d-j)^n\right) \left({k-d+1 \choose d} - {k-d-1 \choose d-2}\right)$$ which is a bit ugly. Thus the probability you are looking for is $$1 - \frac{ \sum_{d=1}^{M} S(n,d)\times d! \times \left({k-d+1 \choose d} - {k-d-1 \choose d-2}\right)}{k^n}$$ - What I don't get is "We have n people, and k birthdays." In my "experiment", I have n (fixed) people, each with a random birthday. The set of distict birthdays (which also appeared in my deduction) is a random number, not an input. Perhaps I'm missing something. – leonbloy Jan 20 '11 at 23:37 @leon: $k = 365$ in our case. Have edited to make it clearer. – Aryabhata Jan 20 '11 at 23:40 1 @leon: I see where your confusion might be. We are using similarly named variables, but they mean different things in our answers. For instance, your $k$, is my $d$. Your $M$ is my $k$. btw, I have confirmed that your answer and my answer give the same answer. Also, in your answer, instead of $Q(n,k)$, you perhaps meant $Q(M,k)$. – Aryabhata Jan 21 '11 at 6:31 +1 for thorough explanations. I believe all three answers are great! Yet yours would be the most pedagogical one. – wok Jan 21 '11 at 15:44 1 @Moron: right, thanks for the correction, fixed. I prefered to expressed it in that way (Q individualized) to emphasize that it's a probability – leonbloy Jan 21 '11 at 18:01 show 3 more comments UPDATE - THIS IS WRONG -- SEE BELOW CORRECT ANSWER --- Let's call N1 the number of configurations that have at least one day in between birthdays (this excludes not only consecutive birthdays, but also coincident birthdays). I get (counting weak compositions) : $\displaystyle N_1 = 365 \frac{(365-n-1)!}{(365-2n)!}$ If you want to include coincident birthdays, we have $\displaystyle N_0 = \frac{365!}{(365-n)!}$ $\displaystyle P = \frac{N_0 - N_1}{365^n}$ Update: there might be is some error here, I think I'm failing to taking into account the configurations that have both coincident and consecutive birthdays, I'll revise this tonight. I suspect that N1 is correct, and that allows to compute the probability of having consecutive OR coincident birthdays. To count consecutive (exclusively) birthdays seems more difficult. UPDATE: Here's the correct (I hope) answer. The probability of having at least a pair of consecutive birthday for M (=365) days and n persons is $\displaystyle P(M,n) = 1 - \sum_{k=1}^n Q(M,k) {M \choose k} \frac{S(n,k) k! }{M^n} = 1 - \frac{1}{M^{n-1}} \sum_{k=1}^n \frac{(M-k-1)!}{(M-2k)!} S(n,k)$ where $\displaystyle Q(M,k) = \frac{{M -k - 1 \choose k -1} }{{M -1 \choose k -1} }$ and $S(n,k)$ are the Stirling numbers of the second kind Some computed values follow ````M=365 n=1 p=0.00000000 M=365 n=2 p=0.00547835 M=365 n=3 p=0.01634745 M=365 n=4 p=0.03242761 M=365 n=5 p=0.05345896 M=365 n=6 p=0.07910314 M=365 n=7 p=0.10895871 M=365 n=8 p=0.14256439 M=365 n=9 p=0.17941667 M=365 n=10 p=0.21897764 M=365 n=11 p=0.26069278 M=365 n=12 p=0.30400167 M=365 n=13 p=0.34834843 M=365 n=14 p=0.39319571 M=365 n=15 p=0.43803357 M=365 n=16 p=0.48239009 M=365 n=17 p=0.52583640 M=365 n=30 p=0.90729104 M=365 n=42 p=0.99074145 ```` Explanation follows: ------------------------------- Let $M$ (=365) number of days, and $n$ = number of persons and $k$ = number of distinct birthdays ($k$ is not fixed, it's a random variable in the range $1..n$). Let $P_{M,n}(S)$ be the probability of NOT having consecutive birthdays (S is the event: "all birthdays are separated") Then $P_{M,n}(S) = \sum_k P_{M,n}(S \; k) = \sum_k P_{M,n}( S \| k ) P_{M,n}(k )$ The following steps compute the two factors inside the summation: STEP 1: $P_{M,n}( S \| k )$ is the probability of having the events separated, given that the number of distinct birthdays is $k$. If we think of birthdays as occupied boxes in a circular list, a little reflection shows that all configurations are equiprobable ($n$ does not matter now) and we can assume without altering the result that the first box of the list is occupied. Now, the total number of possible ocurrences are the number of selection of $k-1$ boxes among $M-1$ (combination). And the number of "succesfull" arrangements are those that result in non-consecutive occupied boxes. But this is equivalent of specifying a list of $k$ numbers greater than 1 that sum up to M; and this is the same as specifying a list of $k$ numbers greater than 0 than sum up to $M-k$; and this can be counted, by a reasoning similar to this one, as the combination of $k -1$ taken from $M -k - 1$. So, $\displaystyle P_{M,n}( S \| k ) = \frac{{M -k - 1 \choose k -1} }{{M -1 \choose k -1} } = Q(M,k)$ STEP2: $P_{M,n}(k)$ : if we place $n$ balls at random in $M$ boxes, what is the probability that exactly $k$ boxes will be non-empty? The total counting is given by $M^n$. To count the "successful" cases, we multiply: • all the possible sets of occupied boxes (combinations of $k$ taken from $M$) • the total numbers of ways of putting $n$ balls in $k$ boxes leaving no empty box: that is given by the Stirling numbers of the second kind. • and we need to multiply by $k!$ (permutations of boxes) because the Stirling numbers consider nondistinct boxes. $\displaystyle P_{M,n}(k ) = {M \choose k} \frac{S(n,k) k! }{M^n}$ And from there follows the formula above. - I agree (now!) with all this. I hope you don't have anything else on tonight though :-) – TonyK Jan 20 '11 at 15:39 Thanks for trying, but it does not fit the estimated probabilities with Python. :( – wok Jan 20 '11 at 19:07 Although my symulations do not give the same probabilities for several n, I wonder whether there is a bias due to the pseudo-randomness of generated integers. – wok Jan 20 '11 at 22:08 Weird. You made the edit while I was writing my answer. They seem to be a bit different though. – Aryabhata Jan 20 '11 at 23:30 1 I have confirmed that the updated formula is same as mine. +1. Basically I have confirmed that $Q(M,k) {M \choose k} = {M-k+1 \choose k} - {M-k-1 \choose k-2}$. – Aryabhata Jan 21 '11 at 6:32 show 2 more comments NB: I worked earlier on this problem and came up with the following solution. The first answer that was posted made me think that I got something wrong, and I discarded my work. Since the new answer by Moron agrees (essentially) with my previous work here it is, a slightly different derivation of the same formula. Let $k$ be the number of days in a year. Let $m$ be the distinct number of birthdays among the $n$ friends. Let's assume that $k$ is big enough to have a non-trivial problem (say, $k > n/2$) We're interested in binary strings with these three conditions: 1. Are of length $k$, with $m$ ones and $k-m$ zeros. 2. There's at least one zero between any two ones. 3. Condition 2 holds when "wrapping around" the string. Let's count them by constructing them with the following algorithm: • Start with a string of $m$ ones: $11\dots 1$ • There are now three distinct cases: there's a birthday on the first day of the year, there is a birthday on the last day of the year, there's a birthday on neither. • In the first case we have to distribute $k-m$ zeros in $m$ non-empty buckets. Those are called compositions, and one can show that there are $\binom{k-m-1}{m-1}$ such assignments. • The second case is analogous, giving another $\binom{k-m-1}{m-1}$ possible strings. • The third case is similar, giving instead $\binom{k-m-1}{m}$ strings. Putting this together we have: $$2\cdot \binom{k-m-1}{m-1}+\binom{k-m-1}{m}$$ Since $n$ friends share $m$ birthdays we have to account for that, giving this expression: $$p(n,k) =\frac{\sum_{m=1}^n{ m! \cdot S(n,m)\cdot \Bigg (2\cdot \binom{k-m-1}{m-1}+\binom{k-m-1}{m}}\Bigg )}{k^n}$$ where $S(n,m)$ is the Stirling number of the second kind. @Moron: Why do we need to multiply by $m!$ here? Oh right! Stirling numbers of the second kind count the number of coalitions, but we care about their order, too! - Did you mean $S(n,m)$? We need to multiply by factorial as we also need to assign the birthdays. The Stirling number counts the number of unordered partitions. We could assign the birthdays to each partition in $m!$ ways. For instance for $m=2$ there are $2^{n-1} - 1$ partitions, but you can rearrange the parts of the partition to get different birthdays assigned. – Aryabhata Jan 21 '11 at 4:12 1 +1: Apart from the missing $m!$ term. We have that identity $2 \binom{k-m-1}{m-1} + \binom{k-m-1}{m} = \binom{k-m+1}{m} - \binom{k-m-1}{m-2}$, which matches my answer. So this confirms that this matches Leonbloy's answer too. – Aryabhata Jan 21 '11 at 6:39 Oh right! We do care about their order, too. Fixed the typo in $S(n,m)$. Thank you! – Jacopo Notarstefano Jan 21 '11 at 8:54 +1 for an explanation using binary strings and compositions. – wok Jan 21 '11 at 15:21 Using weak compositions makes the problem really simple to handle. Bravo! – wok Jan 21 '11 at 15:32	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 114, "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": 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": 0.9285382628440857, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/36321/why-does-the-running-of-the-gauge-couplings-show-frac1-alpha-frac1?answertab=active	# Why does the running of the gauge couplings show $\frac{1}{\alpha}$ > $\frac{1}{\alpha_w}$ at low energy? I thought the coupling constants were something like: $\alpha \approx 1/137$ $\alpha_w \approx 10^{-6}$ $\alpha_s \approx 1$ And yet if you look at any pictorial representation of the running of the couplings you see something like this (or try google): which seems to indicate that (at the energies scales we have access to) $\alpha$ is less than $\alpha_w$. Why is the ordering of the size of the couplings messed up in all of these pictures? - The plots that you are seeing and asking about are plotting the inverse of $\alpha's$. So, nothing is messed up. The larger will become smaller, and the smaller will become larger when you take the inverse of things. – stupidity Sep 13 '12 at 14:45 Yes, of course the inverse is plotted. That explains why the strong coupling is on the bottom. But why is $\alpha_w$ in the middle and not on the top? – user1247 Sep 13 '12 at 14:57 ## 1 Answer I think you're confusing the weak structure constant with the Fermi constant. The Fermi constant is $G_F=1.166\times 10^{-5}\text{ GeV}^{-2}$ and it gives us the effective strength of the four-point interaction of fermions. This four-point interaction is of course mediated by the W boson and by looking at the relevant tree-level Feynman diagrams we have $$\frac{G_F}{\sqrt{2}}=\frac{g_W^2}{8M_W^2},$$ where $g_W$ is the weak coupling and $M_W=80.4\text{ GeV}$ the mass of the W boson. Plugging in numbers, we find $$\alpha_W=\frac{g_W^2}{4\pi}\approx\frac{1}{30}.$$ - What is the weak coupling $g_W$ then? – user1247 Sep 13 '12 at 15:13 – user1247 Sep 13 '12 at 15:17 They assume that the interaction that causes the decay is the Fermi interaction, with coupling $G_F/\sqrt{2}$. In the UV completion of the Fermi theory, i.e. the weak interaction, this coupling corresponds to $g_W^2/8M_W^2$. – AndyS Sep 13 '12 at 17:17	{"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": 15, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9295579791069031, "perplexity_flag": "head"}
http://mathhelpforum.com/geometry/66831-length-side-b.html	# Thread: 1. ## Length of Side b In ΔABC, m<C = 30 and a = 24. If the area of the triangle is 42, what is the length of side b? Is this a law of sines or law of cosines problem? 2. Originally Posted by magentarita In ΔABC, m<C = 30 and a = 24. If the area of the triangle is 42, what is the length of side b? Is this a law of sines or law of cosines problem? The area of a triangle is calculated by: $A_{\Delta} = \dfrac12\cdot a\cdot b\cdot \sin(C)= \dfrac12\cdot a\cdot c\cdot \sin(B)= \dfrac12\cdot b\cdot c\cdot \sin(A)$ $\sin(30^\circ) = \dfrac12$ Plug in the values you know into the approriate equation: $A_{\Delta} =\dfrac12\cdot 24\cdot b\cdot \sin(30^\circ)=\dfrac12\cdot 24\cdot b\cdot \dfrac12= 6b$ Therefore $6b = 42~\implies~\boxed{b = 7}$ 3. ## yes........ Originally Posted by earboth The area of a triangle is calculated by: $A_{\Delta} = \dfrac12\cdot a\cdot b\cdot \sin(C)= \dfrac12\cdot a\cdot c\cdot \sin(B)= \dfrac12\cdot b\cdot c\cdot \sin(A)$ $\sin(30^\circ) = \dfrac12$ Plug in the values you know into the approriate equation: $A_{\Delta} =\dfrac12\cdot 24\cdot b\cdot \sin(30^\circ)=\dfrac12\cdot 24\cdot b\cdot \dfrac12= 6b$ Therefore $6b = 42~\implies~\boxed{b = 7}$ Yes, I forgot about those 3 formulas. #### Search Tags Copyright © 2005-2013 Math Help Forum. All rights reserved.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 8, "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": 0.8783559203147888, "perplexity_flag": "middle"}
http://mathhelpforum.com/advanced-algebra/45521-eigen-vectors.html	# Thread: 1. ## Eigen vectors I need to know how to compute T with respect to the basis b. and i need a way to know where b is a basis consisting of egeinvectors of T. Let v= p_1(R), T(a+bx)= (6a-6b)+(12a-11b)x and the basis B={3+4x,2+3x} Let V= R^3, T = $\begin{pmatrix}\ a \\ b \\ c\end{pmatrix}\$ = $\begin{pmatrix}\ 3a + 2b - 2c\\ -4a-3b+2c\\-c\end{pmatrix}\$ The basis B= { $\begin{pmatrix}\ 0 \\ 1 \\ 1\end{pmatrix}\$ $\begin{pmatrix}\ 1 \\ -1 \\ 0\end{pmatrix}\$ $\begin{pmatrix}\ 1 \\ 0 \\ 2 \end{pmatrix}\$ } general steps will defenitely help. 2. To determine if a set is a basis for a subspace, you must show that the set meets two out of the three conditions: 1. S is linearly independent 2. S is a spanning set for V 3. S contains exactly k vectors, where k is the dimension of the space you are working in. To show that the basis consists of the eigenvectors of T, you need to show that the matrix-vector product of the standard matrix of T and each vector in B is equal to the product of a scalar and the vector. Although, I think it would follow directly from proving that the assumed eigenvectors in the set that you have correspond to distinct eigenvalues, because if an n x n matrix consists of n distinct eigenvalues, and you have separate, linearly independent vectors. To determine if the vectors in the set are eigenvectors, you must show that T*v where T is the standard matrix of the linear operator and v is a vector in the basis is equal to some scalar times v, for each vector in the set. The scalars would be your eigenvalues. Sorry, this may be jumbled...making sense at all?	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 5, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.923129677772522, "perplexity_flag": "head"}
http://mathoverflow.net/questions/110273/can-one-ignore-primes-lying-over-l-in-the-fontaine-mazur-conjecture-counterexa	## Can one ignore primes lying over $l$ in the Fontaine-Mazur conjecture? Counterexamples? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) The Fontaine-Mazur conjecture predicts that an $l$-adic Galois representation of a number field is 'geometric' if it is unramified outside a finite set of primes and is De Rham for primes lying over $l$. Now, what happens if one forgets about the latter restriction; are there any counterexamples, and is there any (general?) way to understand that those are not geometric without using the De Rham restriction? - 3 Dear Mikhail, Note that any $\ell$-adic rep'n of geometric origin is de Rham locally at $\ell$, by $\ell$-adic Hodge theory (what is usually called "$p$-adic Hodge theory), and so certainly the de Rham condition in the FM conjecture is necessary. The force of the conjecture is that it is also sufficient (along with the other obvious necessary condition of being unram. outside finitely many primes). Best wishes, Matthew – Emerton Oct 22 at 2:05 ## 2 Answers To complete Kevin's good answer: the number of $\ell$-adic representations (up to isomorphism) of a number field $K$ is countable, since so are varieties over a $K$. On the other hand, we know by Mazur's theory of deformations that representations of the type you consider that is, of the Galois group of the maximal extensions of a number field $K$ unramified outside a finite set of places $S$ containing places above $\ell$ and $\infty$) form multi-dimensional $\ell$-adic family (e.g. parametrized by spaces like $\mathbb{Z}_\ell^n$ for some $n>0$, hence are uncountable. Thus not only are there counter-examples to Fontaine-Mazur's conjecture without the de Rham hypothesis, but most examples of such representations are counter-examples. Among all representations, the de Rham (or geometric) representations are expected to be dense in certain cases (e.g. representation of dimension 2 of $\mathbb Q$) but not in general (e.g. representations of dimension $2$ of a non-totally real number field) - 1 Dear Joël, Just to clarify, let me note here that "dense" means "Zariski dense" (or at least, I'm not sure whether any stronger sense of dense would be correct). Best wishes, Matthew – Emerton Oct 22 at 2:02 2 P.S. Also, in your first sentence, I guess you mean $\ell$-adic reps. of geometric origin. – Emerton Oct 22 at 2:03 ### 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 fact there are one-dimensional counterexamples: if $\chi$ is the $l$-adic cyclotomic character, and $k\in \mathbb{Z}_l \backslash \mathbb{Z}$, then $\chi^{(l-1)k}$ is unramified outside $l$, but does not arise from geometry (in particular it is not de Rham). A general way of understanding 'all' Galois representations is via Mazur's universal deformation rings. These parametrize all $l$-adic Galois representations (unramified outside a prescribed set of primes and perhaps with extra conditions) whose reduction mod $l$ is isomorphic to a given mod $l$ representation. The $R=T$ theorems you may have heard of state that certain universal deformation rings are isomorphic to rings coming from the theory of modular forms. I don't know if it's expected in general whether all Galois representations are limits of deRham representations. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 29, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9080737233161926, "perplexity_flag": "head"}
http://mathhelpforum.com/latex-help/111343-triple-lower-limits-summation-sign-urgent.html	# Thread: 1. ## Triple lower limits in summation sign (urgent) Hi! How do I get several limits below each other using the summation sign? I.e. $\sum\limits_{i \neq j} A_{ij}$, but I would also like $i, j \neq r, s$ and $i, j \neq s, r$ on two lines below the first limit... Thanks! 2. Originally Posted by Jodles Hi! How do I get several limits below each other using the summation sign? I.e. $\sum\limits_{i \neq j} A_{ij}$, but I would also like $i, j \neq r, s$ and $i, j \neq s, r$ on two lines below the first limit... Thanks! [tex]\sum_ { \tiny \begin{array}{c} x>1 \\ y<8\\ u \in A \end{array} } x+y+u[/tex] gives: $\sum_{ \tiny \begin{array}{c} x>1 \\ y<8\\ u \in A \end{array} } x+y+u$ 3. Originally Posted by Jodles How do I get several limits below each other using the summation sign? I.e. $\sum\limits_{i \neq j} A_{ij}$, but I would also like $i, j \neq r, s$ and $i, j \neq s, r$ on two lines below the first limit... [tex]\sum_{\substack{i \neq j\\ i, j \neq r, s\\ i, j \neq s, r}} A_{ij}[/tex] gives $\sum_{\substack{i \neq j\\ i, j \neq r, s\\ i, j \neq s, r}} A_{ij}$	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 11, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.838188111782074, "perplexity_flag": "middle"}
http://www.physicsforums.com/showthread.php?t=128272	Physics Forums ## Parameterization Hi, I am having a lot of trouble on the parameterization part of this problem: Suppose there is a cylindrical container of liquid with circular flow velocity given in cylindrical coordinates by v(r,$$\theta$$,z) = <-f(r,z)sin$$\theta$$, f(r,z)cos$$\theta$$,0> for some function f(r,z) which we will determine. The vector field is divergence free. Find a curve, x(t), that represents the path followed by a particle of liquid at radius r. So, x’(t) = v(x(t)) and x(0) = (r,0,z). Use this to find$$\theta$$(t), v(t) and a(t) = v’(t). Then write a in therms of r,$$\theta$$, z. This is a long problem and I understand how to do the rest of it but this section is really confusing to me and I can't think of anything that would work. Thanks for any help. PhysOrg.com science news on PhysOrg.com >> Front-row seats to climate change>> Attacking MRSA with metals from antibacterial clays>> New formula invented for microscope viewing, substitutes for federally controlled drug Recognitions: Science Advisor I'm going to assume that <-f(r,z)sin$$\theta$$, f(r,z)cos$$\theta$$,0> is in rectangular coordinates because the problem is much easier that way. To solve this problem you have to find the integral curve of a particle at radius r. Look at the velocity field, v. --What if v were <f(r,z)cos $$\theta$$, f(r,z)sin $$\theta$$,0> instead? Could you picture what that would look like? --What is the relationship between <cos $$\theta$$, sin $$\theta$$> and <-sin $$\theta$$, cos $$\theta$$>? (try the dot product) --From those two you should be able to get an idea of what v looks like, and then a good guess at what the integral curves are. Thanks for the help but I have never heard of an integral curve before, I looked around online and I couldn't get a solid definition of what one was in terms that I understood. This is for a multivariable calc class where we just finished up the divergence theorem. I don't know if you can think of another way to do it or try to explain it to me but thanks for the effort. Recognitions: Science Advisor ## Parameterization If you haven't covered integral curves yet, which would surprise me, you will soon so you might as well learn what they are. Thread Tools \| \| \| \| \|---------------------------------------\|-------------------------------\|---------\| \| Similar Threads for: Parameterization \| \| \| \| Thread \| Forum \| Replies \| \| \| Calculus & Beyond Homework \| 7 \| \| \| Calculus & Beyond Homework \| 2 \| \| \| Calculus & Beyond Homework \| 1 \| \| \| Introductory Physics Homework \| 8 \| \| \| General Math \| 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": 0, "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": 0.9425977468490601, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/tagged/integral-transforms?sort=unanswered&pagesize=15	# Tagged Questions This covers all transformations of functions by integrals, including but not limited to the Radon, X-Ray, Hilbert, Mellin transforms. Use (wavelet-transform), (laplace-transform) for those respective transforms, and use (fourier-analysis) for questions about the Fourier and Fourier-sine/cosine ... 1answer 77 views ### arguing away - complex analysis Probably a trivial question but I can't understand how to argue away the value of integrals in complex analysis. I am trying to find the inverse Laplace transform of $F(s)=\frac{1}{s(s+1)}$. 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Its Mellin transform is the function $f^\ast(z)$ given by $$f^\ast(z) = \int\limits_0^\infty x^{z-1} f(x) \, \mathrm dx.$$ What are ... 0answers 66 views ### Triple Product Integral on Real Spherical Harmonics Basis Functions Okay I know that Real Spherical Harmonics are given by If $m \lt 0$ $~$ then $\sqrt{2}$ $~$ $Im(\text{SphericalHarmonicY}[l,\|m\|])$ If $m=0$ $~$ then $~$ $\text{SphericalHarmonicY}[l,0]$ If \$m \gt ... 0answers 40 views ### nonlinear integral equation let be the integral equation for two functions $f(x)$ and $g(x)$ in the form $$g(s)= \int_{0}^{s}\sqrt{s-f(x)}dx$$ is valid to accept that in the sense of fractional calculus, the ONLY ... 0answers 68 views ### Inverse laplace transform - infinite residues I need to compute the inverse transform of the following, $f(s)= \dfrac{\sinh(k(l-x))}{\sinh(kl)}\dfrac{\omega}{\omega^2+s^2}$ where $k=\sqrt{\dfrac{s^2}{c^2}+n^2\pi^2},\ 0\leq x\leq l$. 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An ... 0answers 722 views ### Square of the sinc function integral Considering the pulse function $\text{sinc}(t)$, show that : $$\int_{-\infty}^\infty \text{sinc}^2(t) dt = 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": 78, "mathjax_display_tex": 17, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 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": 0.8607892394065857, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/30556/what-does-ads-cft-have-to-say-about-quantum-gravity-in-our-world?answertab=oldest	# What does AdS/CFT have to say about quantum gravity in our world? The Ads side of the AdS/CFT correspondence is a model of quantum gravity in 5 dimensional antidesitter space. What can it say about quantum gravity in our 4-spacetime dimensions? Or is it just a toy model to better our understanding of quantum gravity as a whole? - ## 1 Answer The holographic priciple, as I understand it, is a general feature gravity has to obey due to its non-local behavior (as seen in the black hole entropy). These features should be there in any consistent theory of quantum gravity, and in this sense are more "general" than just string theory. It can in principle be formulated on any background, and the dual QFT has to have the asymptotic symmetries of the chosen background (asymptotically AdS has conformal symmetry). The original Manldacena proposal is actually in 10-dimensions (critical superstring theory) where 5 of them are compactified (I think that the flux of higher form fields in the compactified dimensions play an important role, so its not enough to only keep the non-compact dimensions). Furthermore the duality can be generalized to other dimensions, 5 is not that special in this regard. There is for example a lot of work on $AdS_3/CFT_2$ holographic dualities these days. But the crucial thing I want to say is that, the dual boundary theory is a local field theory and not a gravitational theory. Therefore the duality doesn't say too much about gravity in one lower dimension. The dual field theory "just" provides an alternative description of the (non-local) degrees of freedom in the gravitational theory. - Could you maybe show how one can see the non-local behavior from black hole entropy? – ungerade Dec 13 '12 at 11:39	{"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": 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": 0.9325491786003113, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/217926/requirements-for-a-linear-transformation?answertab=votes	# Requirements for a Linear Transformation Let's say we have a transformation: $$T: \mathbb{R}^n \rightarrow \mathbb{R}^m.$$ This is a linear transformation iff: for all $\vec{a} , \vec{b} \in \mathbb{R}^n$ and $c \in \Bbb{R}$, 1. $T(\vec{a} + \vec{b}) = T(\vec{a}) + T(\vec{b})$ 2. $T(c \vec{a}) = cT(\vec{a})$ I've seen this kind of 'requirements' multiple times in Linear Algebra, and I wonder what the names for these requirements are. - ## 2 Answers 1. $T$ preserves addition 2. $T$ preserves multiplication by scalar Altogether: $T$ is linear. - ## Did you find this question interesting? Try our newsletter Another way to say this is that $T$ is additive and homogeneous of degree $1$. In general, a function $f:X \to Y$ is additive if $f(x+y)=f(x)+f(y)$ for all $x,y \in X$. If $X$ and $Y$ are vector spaces then we say that $f$ is homogeneous of degree $k$ if for all $a$ not equal to $0$ in the underlying scalar field of $X$ and all $x \in X$, $f(ax)=a^kf(x)$ for some integer $k$.	{"extraction_info": {"found_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": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8887951970100403, "perplexity_flag": "head"}
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http://math.stackexchange.com/questions/226371/logic-and-geometry	# Logic and geometry By delving into topos theory and sheaves one will eventually discover a "deep connection" between logic and geometry, two fields, which are superficially rather unrelated. But what if I have not the abilities or capacities of delving deeper into topos theory and sheaves? Does the deep connection between logic and geometry have to remain a mistery for me forever? At which level of abstraction and sophistication can this connection be recognized for the first time? And which seemingly superficial analogies have really to do with this "deep connection"? • What's rather easy to grasp is that there is (i) an algebra of logic and (ii) an algebra of geometry. But is this at the heart of the "deep connection"? • What comes to my mind is, that both logic (the realm of linguistic representations) and geometry (the realm of graphical representations) have to do with - representations. Is this of any relevance? - I think the Pythagoreans would very much agree with your intuition of a deep connection (as I share)...And Clearly, Euclid was an early exemplar of systematic axiomatic methods/constructs. No doubt there's more, but just wanted to note that your sentiments, I suspect, were shared by many of our ancient predecessors... – amWhy Oct 31 '12 at 22:58 1 @amWhy (= WhoAmI?): My question was not for or about intuitions and sentiments but for specific arguments - even if I should not have been able to make this clear. – Hans Stricker Oct 31 '12 at 23:27 2 That's why I didn't post my comment as an answer; I am simply sympathetic with your question and interested in viewing answers... – amWhy Oct 31 '12 at 23:45 1 The connection is nothing so superficial. On the one hand, to every logical theory of a certain type is associated a "classifying topos", which is a geometric object whose "points" correspond to models of the theory, and many properties of the classifying topos correspond to properties of the theory. This is at the heart of the "bridges" technique expounded by Olivia Caramello. On the other hand every topos is a mathematical universe unto itself and gives rise to new interpretations of intuitionistic logic, the so-called "Joyal–Kripke semantics". – Zhen Lin Nov 1 '12 at 7:49 There are other connections between logic and geometry beyond topos theory. For instance, you might take a look at Hrushovski's proof of the Mordell-Lang conjecture, which uses cutting-edge techniques from model theory. (Though I personally find this stuff even more mind boggling.) – Shawn Henry Nov 1 '12 at 15:02 show 1 more comment ## 1 Answer Both logic and geometry deal with information. Logic deals with information about the truth of statements, and geometry deals with information about location. Grothendieck toposes connect logic and geometry along this line. The simplest case it that of the topos of sheaves over a topological space: here the truth value of any proposition is an open subset of the topological space. Thus information on why a proposition is valid is connected to information on where you are in a topological space. - The point of view in your second paragraph is somewhat misleading: the internal logic of the topos of sheaves on a space does not perceive itself as having propositions which are "true" in some places and "false" in others. Rather, if we take the internal point of view, sheaves are sets "smeared out" over the base space; it is very difficult to formulate things like "if such-and-such is true at point $P$, then so-and-so is true at point $Q$" within the internal logic. – Zhen Lin Nov 12 '12 at 19:28	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.18, "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": 0.9435349106788635, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/13647/why-does-the-riemann-zeta-function-have-non-trivial-zeros/50145	## Why does the Riemann zeta function have non-trivial zeros? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) This is a very basic question of course, and exposes my serious ignorance of analytic number theory, but what I am looking for is a good intuitive explanation rather than a formal proof (though a sufficiently short formal proof could count as an intuitive explanation). So, for instance, a proof that estimated a contour integral and thereby showed that the number of zeros inside the contour was greater than zero would not count as a reason. A brief glance at Wikipedia suggests that the Hadamard product formula could give a proof: if there are no non-trivial zeros then you get a suspiciously nice formula for ζ(s) itself. But that would feel to me like formal magic. A better bet would probably be Riemann's explicit formula, but that seems to require one to know something about the distribution of primes. Perhaps a combination of the explicit formula and the functional equation would do the trick, but that again leaves me feeling as though something magic has happened. Perhaps magic is needed. A very closely related question is this. Does the existence of non-trivial zeros on the critical strip imply anything about the distribution of prime numbers? I know that it implies that the partial sums of the Möbius and Liouville functions cannot grow too slowly, and it's really this that I want to understand. - 3 If the zeta function had no zeros in the critical strip, then the usual proof of the PNT would be able to push the contour up to $\Re(s)=0$ and hence establish $\pi(x)=x/\log x+O(\log x)$. This error term can probably shown to be too small due to some trivial estimates which I can't come up with at the moment. In other words, if there weren't any non-trivial zeros, then $\pi(x)$ (and all its brothers) would behave 'suspiciously nicely'. – Thomas Bloom Feb 1 2010 at 11:11 1 It's precisely this issue -- why the error term in PNT isn't absolutely tiny -- that I want to understand. E.g. to prove that π(x) does not approximate $Li(x)$ to within $latex x^{1/3}$, the obvious method is to point to the zeros on the critical line. So I'm going round in circles. With the help of the functional equation one can say that if there are no zeros on or to the right of the critical line then there are none at all, but I don't count that as an intuitive argument. – gowers Feb 1 2010 at 13:57 2 I believe that it is still an open problem to show in an elementary way that the summatory function of $\mu(n)$ is unbounded. It seems that complex analysis is quite difficult to dispense with in these kinds of problems. Unfortunately, analytic arguments do not seem to give any arithmetical insight into the reasons for comparative largeness of the error terms. – engelbrekt Feb 1 2010 at 16:28 Part of the problem is that results on primes in short intervals are not expected to be strong enough to show this. For example, if there were infinitely many n with no primes between n and n+n^a for some a>0, this would be such a great variation in pi(x)-Li(x) as to force there to be a zero in the critical strip. But it is expected that the prime gaps are only as big as (log n)^2. – David Speyer Feb 1 2010 at 17:17 4 There are other natural arithmetical functions, such as the divisor function, where the error term in its counting function is better than square-root (for the divisor function the error is known to be better than the cube root). This example makes it difficult for me to see how to use Parseval (or some variant) to prove that large variations in the error term exist. – Matt Young Feb 2 2010 at 14:27 show 5 more comments ## 5 Answers If the Riemann zeta function had only trivial zeroes, then after multiplying by the gamma factor, it would become a zero-free entire function of finite order. Every zero-free entire function of finite order is of the form $\exp(poly(z))$. This cannot happen for the zeta function, as can be seen by considering behavior as $s\to\infty$ along the reals, for example. - ### 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's a variation on the arguments using the explicit formula. If there were no zeros in the critical strip then the explicit formula would say $$\psi_0(x) = x - \frac{\zeta'(0)}{\zeta(0)} - \frac12 \log(1-x^{-2}),$$ where $\psi_0(x)$ is the same as the usual $\psi(x)$ except when $x$ takes integer values $\psi_0(x)$ is the average of the left and right limits of $\psi(x)$. The right hand side is continuous for $x >0$ while the left hand side is not. The argument generalizes to show that there cannot be finitely many zeros in the critical strip, or even an infinitude that do not grow "too quickly." Precisely, if $$\sum_{\rho} \frac{x^{\rho}}{\rho}$$ converged fast enough to form a continuous function of $x$ then the same argument would carry over. - 2 That is a very nice argument, but it also has a magic flavour to it, since you somehow manage to bootstrap a very small error (arising from the fact that $\psi_0(x)$ is discontinuous) into a much bigger one (that the error term in PNT must be more like a square root). But perhaps the bootstrapping is done by the functional equation rather than your argument. – gowers Feb 1 2010 at 14:49 5 Yes, I think the functional equation is forcing the relatively large error term (in the sense that a zero with small but positive real part would produce another zero with real part close to $1$). However, I think of it in terms of the complexity of the two sides. The function $\psi_0(x)$ is "random" in some sense so the other side of the equation must be complicated enough to produce this erratic prime-counting function. Since $\zeta(s)$ is well-behaved to the right of $1$, and to the left of $0$ (from the functional equation), all the action has to occur in the critical strip. – Matt Young Feb 1 2010 at 15:20 (I've just typed a comment but somehow it got lost. I'll type an answer instead). Bloom is right, no zeros implies too nice a formula, except the right approximation to $\pi(x)$ is $\int_2^x dt/\log t$, not $x / \log x$. The existence of zeros implies that the error cannot be better than $O(x^{1/2})$. On the other hand, the zeta function of ${\mathbb{F}}_q[t]$ has no zeros (and thus there is a nice formula for the analogue of $\pi(x)$) so there is no philosophical reason for a zeta function to have zeros. You might interpret the existence of infinitely many zeros of the zeta function of $\mathbb{Z}$ to mean that $Spec {\mathbb{Z}}$ has "infinite genus". - Do genus-$n$ arithmetic curves have $n$ zeros for their zeta function? – Anweshi Feb 1 2010 at 20:04 3 I am not sure what you mean by a genus n arithmetic curve. In the function field case, the zeta function of a curve of genus g has 2g zeros. In the number field case, all zeta functions have infinitely many zeros. – Felipe Voloch Feb 1 2010 at 21:05 An intuitive reason is indicated by Bukh in another answer. The reason there are nontrivial zeros is because the zeta-function is known to grow in a way that it wouldn't grow if there were no nontrivial zeros. Here is some additional (but not complete) detail. Let's pass from the zeta-function to the completed zeta-function $$Z(s) = \pi^{-s/2}\Gamma\left(\frac{s}{2}\right)\zeta(s).$$ The functional equation for the zeta-function is equivalent to the equation $Z(1-s)=Z(s)$. What does this tell us? By the Euler product, $\zeta(s)$ is nonvanishing in the half-plane to the right of 1. The $\pi$ and $\Gamma$ factors in $Z(s)$ are also nonvanishing in that half-plane, so $Z(s)$ is nonzero if the real part of $s$ is greater than 1. From the equation $Z(1-s) = Z(s)$, $Z(s)$ is also nonvanishing if the real part of $s$ is less than 0. Therefore the nontrivial zeros of $\zeta(s)$ are the same thing as all zeros of $Z(s)$. Thus the question of why $\zeta(s)$ has nontrivial zeros is the same as the question of why $Z(s)$ has any zeros at all (nontrivial zeta zeros = zeros of $Z(s)$). There are (simple) poles for $Z(s)$, at $s = 0$ and $s = 1$. So consider the function $F(s) = s(s-1)Z(s)$. This is an entire function which satisfies $F(1-s) = F(s)$. Let $\|F\|_r$ be the maximum of $\|F(s)\|$ on the circle around the origin of radius $r$. Using analytic methods and the functional equation (to transfer information from right to left half-planes), $\|F(s)\|_r$ grows exponentially with $r$ and, more precisely, $\log\|F\|_r$ is asymptotic to $(1/2)r\log r$. From this growth estimate and the Hadamard product formula, if $F(s)$ had only finitely many zeros then $F(s) = e^{As}P(s)$ for some constant $A$ and some polynomial $P(s)$. The functional equation for $F(s)$ then forces $A = 0$, so $F(s)$ is a polynomial. But polynomial functions don't have their max. modulus on circles grow exponentially. So we have a contradiction, which forces $F(s)$ to have infinitely many zeros and thus $\zeta(s)$ has infinitely many nontrivial zeros. This may be looking too technical, but really the basic point is similar to the proof by Liouville's theorem of the fundamental theorem of algebra: if $p(z)$ were a nonconstant polynomial without zeros in the complex plane then we'd run into an inconsistency comparing the known behavior of $\|1/p(z)\|$ (continuous and tending to 0 as $\|z\| \rightarrow \infty$) with what would follow if there were no zeros ($1/p(z)$ is entire and nonconstant). Is that explanation of why $p(z)$ must have a zero intuitive or is it formal magic? We can't expect a formula for a zero of $p(z)$, so in some sense this growth argument with $\|1/p(z)\|$ to conjure up a zero indirectly is magical. But it has a simple elementary character to it as well: there has to be a zero because otherwise $1/p(z)$ would grow in a way inconsistent with how we know for sure it grows. In the same way, $\zeta(s)$ has to have infinitely many (not only a few) nontrivial zeros because otherwise $\zeta(s)$ would grow in a way that is inconsistent with how we know it grows. (I am glossing over the distinction between $\zeta(s)$ and $Z(s)$ and $s(s-1)Z(s)$ since we're supposed to be putting our intuitive hats on.) - May I offer an attempt at a very lowbrow answer? This answer might naturally strike anyone who knows the functional equation, who looks at a color coded graph of zeta function such as http://en.wikipedia.org/wiki/File:Complex_zeta.jpg, and who knows a little about the long-range behavior of exponential functions and $\Gamma(z)$. Nothing I will say is beyond the ken of an average serious undergraduate. $\zeta(s)\rightarrow 1$ as $\Re(s)\rightarrow +\infty$, so no zeros occur far into the right half-plane (the Euler product makes this clear too, indeed no zeros to the right of the critial strip). Then the functional equation prevents also non-trivial zeros far into the left half-plane (indeed to the left of the critical strip). Now the argument of the function oscillates as you move up vertical lines far into the left half-plane, as one learns from the factors of the functional equation, and their long-range behavior and the right-plane limiting behavior of $\zeta(s)$ itself. So think about the curves that constitute the real locus and the purely imaginary locus of $\zeta(s)$. These curves can't cross anywhere except on the negative real axis and in the critical strip. Some of these curves do cross on the negative real axis, at the trivial zeros. But there is a rightmost trivial zero, it has its two curves, and there still remain infinitely many other curves lying above those. No non-trivial zeros would mean no crossing so no doubling back for all these infinitely many curves. What else could they do? They could only head up into the critical strip, packing tightly together. That would imply faster than exponential decay along some locus as you moved up the strip. And why is that impossible? [I see now that my attempt at a punchline was based on an unjustified assumption. I'm still thinking about a valid and simple replacement. Suggestions welcome.] - "One can calculate $\zeta(s)$ in the strip by series acceleration..." - interesting, I haven't heard of this approach before (the method I'm used to uses Riemann-Siegel); would you happen to have a reference for this on hand? – J. M. Dec 22 2010 at 10:32 @J.M. Gregg Zuckerman taught me this 35 years ago while he had me reading out of Serre's Course in Arithmetic. But I just checked there and it's not something in the book. But look here en.wikipedia.org/wiki/Series_acceleration under Euler's transform. That and $(1-2\cdot 2^{-s})\zeta(s)= \sum_n (-1)^{n+1} n^{-s}.$ – David Feldman Dec 22 2010 at 10:58 Ah, I see, just accelerating the convergence of the series for Dirichlet $\eta$. Nowadays, though, the Euler transform has been supplanted by more modern algorithms, e.g. the Levin transformation or Cohen-Rodriguez Villegas-Zagier. I talked about them in this m.SE thread: math.stackexchange.com/questions/3271 – J. M. Dec 22 2010 at 11:04 3 I have a mental picture. As you move left, first ζ(s) converges, then it starts to spiral, the spirals getting ever wider. Roughly speaking the spirals have involutes which have involutes, etc. to convergence. So nature picks out a value you can see even without convergence, hence analytic continuation. Each iteration gives you convergence in one more unit strip to the left. The "smoothness" of the terms $n^{-s}$ makes the method work, so that's the arithmetic "meaning" of analytic continuation and helps you see why most similar looking Dirichlet series have natural boundaries at $Re(s)=1$. – David Feldman Dec 22 2010 at 11:05	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 90, "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": 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": 0.9509458541870117, "perplexity_flag": "head"}
http://mathematica.stackexchange.com/questions/982/how-to-easily-find-location-of-hidden-characters-that-causes-hard-to-find-syntax	# how to easily find location of hidden characters that causes hard to find syntax error in notebook? For last 30 minutes (and I was lucky today, sometimes I spend 5 times as much time) I have been removing code little by little from my large one cell, (have to be one cell, it is Manipulate demo stylesheet) in order to find where the syntax error is (since Mathematica, with all its might and power, can't tell the user which line number the syntax error is located at). Each time I see the red line show up on the right side of the notebook, indicating a syntax error, I know that I will now waste another hour at least looking for the syntax error. Why is it that hard for Mathematica to tell the user where the syntax error is? I zoomed down to this little 5 character code, and I copy it to new notebook to examine it: So, there is a hidden bad character there. Sometimes, with my big fingers and my small keyboard (can't find big keyboards any more, everything is so tiny), I must have hit wrong key combination, which must have injected some bad character to the notebook, and this happens. I will keep this cell on its own in order to find what the problem is. I tried to select the cell, and do convert to input form, but nothing happened. no output came of it. Then I did cell->ShowExpression, and this came up ````Cell[BoxData[ RowBox[{"(", RowBox[{"k", "h"}], SuperscriptBox[")", "2"]}]], "Input", CellChangeTimes->{3.536893041212217^9}, EmphasizeSyntaxErrors->True] ```` So, what does all this mean? Do you see from the above what is the problem? I think I entered the superscript as I always do, using Ctrl ^ then 2. I think the notebook interface is powerful and allows many things not possible otherwise, but because of problems such the above, I wish sometimes I am using plain text editor like emacs, where I can see what I have and not worry about hidden things I do not see, but I tried that once, and it is not practical overall setup to work with. btw, originally, I had lots of code like the above inside strings, i.e. inside `" "` for formatting, and now I am removing all the strings to do some other way of formatting. I do not know if this has anything to do with it. But as you can see, now there is no string around the expression. Here is a screen shot of the earlier version of the notebook, from the same location of the code, showing how the above was inside a string, and I simply was removing the outside string quotes, that is all: - 2 – Simon Jan 30 '12 at 7:04 Btw - there might be a problem in the `ToExpression` code that your example reveals. `ToExpression["(h k)^2"]` works fine, but the version with a superscript, $\text{ToExpression["}(h\ k)^2\text{"]}$, produces an error. For reasons very similar to the example that you gave: The string with a superscript is interpretted using the string representation of boxes `"(k h\!$\SuperscriptBox[\()$, $2$]\)"` – Simon Jan 30 '12 at 7:11 Out of curiosity, why do you think all your code has to be in a single cell for demonstrations? (I've split out utility functions into separate cells for demonstrations I've written in the past.) – Brett Champion Jan 31 '12 at 5:07 @BrettChampion, I am not sure how you do that? How do you break a Manipulate into many cells? In the Demonstration style sheet, there is the initialization section, which is one cell, and the Manipulate itself, which is one cell. Which one of your demos you did that? thanks – Nasser Jan 31 '12 at 10:42 btw, I do not use the `Initialization` cell (at the top of the notebook). Instead, I use the Manipulate cell and then type `Initialization:>{....}` to add the initialization functions in there. When using the separate initialization cell, one must also add SaveDefinitions->True to Manipulate (it says so in the notebook). This has caused me strange problems when I do that. So that is why just use the Manipulate cell. That works fine and never had problems with that setup. I use the Demonstration stylesheet ofcourse. – Nasser Jan 31 '12 at 10:51 show 2 more comments ## New answer There was actually an old post on mathgroup, Strange Syntax problem, that looks like it was exactly the same problem you're experiencing. A power/superscript being associated with the closing parenthesis instead of the whole parenthesized expression (see the box forms presented in my original answer below for more details). Unfortunately, this problem was not resolved in the the mathgroup post. The best I can suggest at the moment is to simple press the `+` button to get the expanded form This should help narrow down the problem. Alternatively, if it is this particular problem, you could Show Expression, then search for the problematic code snippet `SuperscriptBox[")"`. Then you can modify the box structure directly. This is normally easiest if you past it into a new input cell so that the Extend Selection mechanism works. ## Original answer There is no "hidden bad character" in the cell that you posted. Rather, the box structure is crazy and can't be interpreted by Mathematica - I have trouble imagining how you managed to create it... ````Cell[BoxData[ RowBox[{"(", RowBox[{"k", "h"}], SuperscriptBox[")", "2"]}]], "Input"] ```` and here is how it should look ````Cell[BoxData[ SuperscriptBox[ RowBox[{"(", RowBox[{"k", " ", "h"}], ")"}], "2"]], "Input"] ```` So, your question does not really a work with the example you provided. However, a simple way to maybe clean your code (at least in this case) is to copy it as plain text or input text, both of which produce `(k h)^2` when pasted. - `However, a simple way to maybe clean your code (at least in this case) is to copy it as plain text or input text`. Thanks for the suggestion. But I can't really do this, I have 10,000 line of code cell. If I copy it to notepad and copy it back to the Mathematica notebook, it will break all sort of formatting I had? I tired such things before in the past, they do not really work for me. The problem is how to find location of bad characters in the code in the first place. Once the location is found, then the solution is easy. – Nasser Jan 30 '12 at 7:38 @Nasser: As I said in my answer, there are no "bad characters" in your code - just bad box structures. – Simon Jan 30 '12 at 7:48 – Simon Jan 30 '12 at 7:50 Ok, then how to find location of `bad box structure`? what it is called is not really important for me, what is important, is how to find such location. Regarding the initialization code, it is about 6,000 lines of code. The UI control code is about 4,000 or so. I have both in one cell now. But even if make the initialization code a separate cell, it is still too large. The demo stylesheet does not allow one to break things so that each function is in one cell on its own. I wish that was possible, it will make life much easier. thanks – Nasser Jan 30 '12 at 7:56 Regadring the slowness part when cell gets large. I know about that. But I found if I un-select the `Enable automatic syntax coloring` from preferences, then the notebook is no longer as slow as before, and this really helped allot, otherwise, I would not be able to do this. When I need to check something, then I enable this back – Nasser Jan 30 '12 at 7:59 show 4 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": 0.18, "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": 0.9387160539627075, "perplexity_flag": "middle"}
http://nrich.maths.org/962/note	nrich enriching mathematicsSkip over navigation ### Bracelets Investigate the different shaped bracelets you could make from 18 different spherical beads. How do they compare if you use 24 beads? ### Is a Square a Rectangle? How many rectangles can you find in this shape? Which ones are differently sized and which are 'similar'? ### Part the Polygons Draw three straight lines to separate these shapes into four groups - each group must contain one of each shape. # Quadrilaterals ## Quadrilaterals Quadrilaterals are shapes that have four straight sides. How many DIFFERENT quadrilaterals can be made by joining the dots on the circle? (There are eight evenly-spaced dots.) Can you work out the angles of all your quadrilaterals? For a printable set of dotted circles to use with this resource click here. ### Why do this problem? This problem will help learners extend their knowledge of properties of quadrilaterals. It requires visualisation, a systematic approach and is a good context for generalisation and symbolic representation of findings. ### Possible approach To start with, you could pose the problem orally, asking children to imagine a circle with eight equally-spaced dots placed on its circumference. How many quadrilaterals do they think it might be possible to draw by joining four of the dots? Take a few suggestions and then ask how they think they could go about finding out. You could use this interactivity or draw an eight-point circle on the board. Invite them each to imagine a quadrilateral on this circle. How would they describe their quadrilateral to someone else? Let the class offer some suggestions e.g. by numbering the dots and describing a quadrilateral by the numbers at its vertices. Then return to the problem of the number of different quadrilaterals. Discuss ways in which they will be able to keep track of the quadrilaterals and how they will know they have them all. Some children may wish to draw quadrilaterals in a particular order, for example those with a side of $1$ first (i.e. adjacent pegs joined), then $2$ etc. Others may feel happy just to list the quadrilaterals as numbers. This sheet of blank eight-point circles may be useful. Encourage children to work in small groups to find the total number. After a short time, you could stop the group and focus their attention on one of the quadrilaterals, using the interactivity. How could they work out the angles of this quadrilateral? At first it looks very difficult, but marking the centre of the circle with a dot (this can be done on the interactivity) makes it more accessible. Knowing the angles of each quadrilateral may help when it comes to checking there are no duplicates. After giving the class more time to work, bring them together to share findings and systems, using the interactivity to aid visualisation. You may also want to discuss the angles of each quadrilateral and, in particular, how pairs went about calculating these angles. You can read about one teacher's experience of using this task in the classroom on the NCETM website. ### Key questions How do you know your quadrilaterals are all different? How do you know you have all the different quadrilaterals? ### Possible extension You could challenge pupils to think about whether they could predict the number of different quadrilaterals which are possible for different point circles. How would they go about finding out? (Here is a sheet of nine-point circles which may be useful.) ### Possible support Children could start by investigating the quadrilaterals on circles with smaller numbers of dots. These sheets might be useful: four peg, five peg, six peg, seven peg. The NRICH Project aims to enrich the mathematical experiences of all learners. To support this aim, members of the NRICH team work in a wide range of capacities, including providing professional development for teachers wishing to embed rich mathematical tasks into everyday classroom practice. More information on many of our other activities can be found here.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": false, "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": 0.9332082867622375, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/15217?sort=newest	## Proof of the Reidemeister theorem ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) While preparing for my introduction to topology course, I've realized that I don't know where to find a detailed proof of the Reidemeister theorem (two link diagrams give isotopic links, iff they can be connected by a sequence of Reidemeister moves). The students in my class are not really very advanced and I'm not sure all of them would be able to reconstruct all the details of the proof given in Burde and Zieschang's Knot theory or in other sources I know of. And, being a lazy man, I would like to avoid having to type a detailed proof with pictures when I can refer the students to a book. So does anybody know of a source which would have a proof with all details spelled out? - ## 3 Answers Kunio Murasugi's Knot Theory and its Applications contains what you are looking for, I believe. I taught the proof of Reidemeister's theorem as part of a short course on knot theory, from that book. Definitely suitable for undergraduates. - Thanks, Daniel! Will try to borrow this tomorrow from the library. – algori Feb 14 2010 at 16:57 ### 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. Reidemeister's proof involves a single move: replacing 2 (or 1) edges of a triangle with the other edge (edges). It is in the English translation of his book. I don't know if you can ferret out the details of our first movie move theorem in JKTR, or even the CRS version, but when we did those, I felt that we had addressed the general position issues that Dan mentions. - Scott, thanks! Your second paragraph is a bit cryptic. For the record: I know how to prove Reidemeister's theorem, I'm looking for a detailed proof accessible to undergraduate students with a limited experience of mathematics, let alone topology. – algori Feb 14 2010 at 5:30 1 I did not mean to be cryptic at all, but it was late when I wrote it. Masahico and I proved a "movie move theorem" that was published in Journ. of Knot Thy. and its Ram. There were problems with the interpretations that people could make of it. So we proved a more precise result with Joachim Rieger (CRS). A lot of the transversality arguments are worked out in both places. In particular, when your knot diagram has a height function, you have to deal with critical interchange, zig-zag, and $\psi$ moves. I just don't remember if we gave a proof of the R thm, or not. Probably not. – Scott Carter Feb 14 2010 at 13:48 I taught knot theory last semester and ran into the same problem. I looked in every book I could get my hands on, and could not find an undergraduate level proof. In the end, I wrote up my own notes (which I would be happy to scan when I get back into the office). The key ideas for the case-by-case analysis are in the book "Knots, links, braids, and 3-manifolds" by Prasolov and Sosinsky. I also found Louis Kauffman's book "On knots" to be helpful. There are two lemmas I could not find anywhere: (1) the general position argument, which says that there is a nice projection and (2) the argument which says that you can find a general projection so that the associated diagram is equivalent to the original diagram (most books skip this issue). The point of the second lemma is that it is not enough to show that there exist two projections that differ by Reidemeister moves, rather, you want to show that the two given diagrams differ by Reidemeister moves. - Dan, thanks! I've seen the proof Prasolov and Sossinsky give (my impression is that they give more details than most sources but not all of them). It would be very interesting to see your notes, if you decide to scan them. I can in turn scan mine and post them somewhere. – algori Feb 14 2010 at 5:22 Hi Dan, the general position argument (1) has a nice proof using the transversality technique in the proof of the weak Whitney embedding theorem given in Guillemin and Pollack. I'm not quite understanding your lemma (2) statement. – Ryan Budney Feb 14 2010 at 9:31 Hi Ryan. As for lemma (2): You start with two diagrams of the equivalent PL knots. Equivalent means they differ by a finite sequence of triangle moves. So now you need to either rotate the projection plane (or modify the triangle moves) so that each move gives a knot with a regular projection. By regular position (lemma 1), you can do this. However, if you rotate too much, the new diagrams of your original two knots are not equivalent to the original diagrams. Therefore, you are just showing that there exist two diagrams of the original knots that differ by R moves. Does that make sense? – Dan Margalit Feb 14 2010 at 12:37 Ah, okay. I take it everyone here is thinking about proofs in the PL category. I prefer the proof in the smooth category as there's less details to fuss over. Of course you have to use transversality so you have a bigger machine at your disposal. – Ryan Budney Feb 14 2010 at 23:07	{"extraction_info": {"found_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": 0.9559940099716187, "perplexity_flag": "head"}
http://mathhelpforum.com/differential-geometry/163240-homeomorphism-circle.html	# Thread: 1. ## Homeomorphism of the circle Let $f: \mathbb{S}^1 \rightarrow \mathbb{S}^1$ be an orientation preserving homeomorphism with at least one fixed point. Show that every periodic point of $f$ has period 1, i.e. is a fixed point. Anybody got an idea? 2. Let p0 be a fixed point and p be a periodic point with period n, that is, f^n(p) = f(f(...f(p))) = p while f^k(p) not equals p if k < n. Let A be the shorter arc between p0 and p, since f is a homeomorphism, f(A) is also an arc with f(p) being the second end point. ( p0 is the first one). Suppose the length of f(A) is greater than A. Denote this fact simply as f(A) > A. Since f(A) contains A, f(f(A)) contains f(A). And since f(f(p)) does not equal to f(p), we get f(f(A)) > f(A). With this process we get a series of arcs, f(A) < f(f(A)) < ... f^n(A). Now f^n(p) = p, so f^n(A) must cover the circle S^1 one or more times, ending at p. So A is covered at least twice. While f^n is also a homemorphism, a contradiction. 3. ok thanks. Can we somehow expand this proof to prove the following fact? If $f: \mathbb{S}^1 \rightarrow \mathbb{S}^1$ is an orientation preserving homeomorphism with at least one of minimal period MATH]n > 1[/tex], dhow that then every periodic point of $f$ has minimal period n. 4. g=f^n is an orientation preserving homemorphism having at least one fixed point. According to your first question, every periodic point of g is a fixed point, that is, a periodic point of period n for f. DONE.	{"extraction_info": {"found_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": 0.9223456382751465, "perplexity_flag": "middle"}
http://unapologetic.wordpress.com/2010/07/23/product-measures/?like=1&source=post_flair&_wpnonce=4ca8fc8c32	# The Unapologetic Mathematician ## Product Measures We continue as yesterday, considering the two $\sigma$-finite measure spaces $(X,\mathcal{S},\mu)$ and $(Y,\mathcal{T},\nu)$, and the product measure space $(X\times Y,\mathcal{S}\times\mathcal{T})$. Last time we too a measurable set $E\subseteq X\times Y$ and defined the functions $f(x)=\nu(E_x)$ and $g(y)=\mu(E^y)$. We also showed that $\displaystyle\int f\,d\mu=\int g\,d\nu$ That is, for every measurable $E$ we can define the real number $\displaystyle\lambda(E)=\int\nu(E_x)\,d\mu=\int\mu(E^y)\,d\nu$ I say that this function $\lambda$ is itself a $\sigma$-finite measure, and that for any measurable rectangle $A\times B$ we have $\lambda(A\times B)=\mu(A)\nu(B)$. Since measurable rectangles generate the $\sigma$-ring $\mathcal{S}\times\mathcal{T}$, this latter condition specifies $\lambda$ uniquely. To see that $\lambda$ is a measure, we must show that it is countably additive. If $\{E_n\}$ is a sequence of disjoint sets then we calculate $\displaystyle\begin{aligned}\lambda\left(\biguplus\limits_{n=1}^\infty E_n\right)&=\int\nu\left(\left(\biguplus\limits_{n=1}^\infty E_n\right)_x\right)\,d\mu\\&=\int\nu\left(\biguplus\limits_{n=1}^\infty(E_n)_x\right)\,d\mu\\&=\int\sum\limits_{n=1}^\infty\nu\left((E_n)_x\right)\,d\mu\\&=\sum\limits_{n=1}^\infty\int\nu\left((E_n)_x\right)\,d\mu\\&=\sum\limits_{n=1}^\infty\lambda(E_n)\end{aligned}$ where we have used the monotone convergence theorem to exchange the sum and the integral. We verify the $\sigma$-finiteness of $\lambda$ by covering each measurable set $E$ by countably many measurable rectangles with finite-measure sides. Since the sides’ measures are finite, the measure of the rectangle itself is the product of two finite numbers, and is thus finite. We call the measure $\lambda$ the “product” of the measures $\mu$ and $\nu$, and we write $\lambda=\mu\times\nu$. We thus have a $\sigma$-finite measure space $(X\times Y,\mathcal{S}\times\mathcal{T},\mu\times\nu)$ that we call the “cartesian product” of the spaces $(X,\mathcal{S},\mu)$ and $(Y,\mathcal{T},\nu)$. ### Like this: Posted by John Armstrong \| Analysis, Measure Theory ## 6 Comments » 1. Professor Armstrong, I think that Feynman’s path integrals of quantum field theory might provide the kind of product measures that we were discussing although I am not so sure how rigorous they are considered to be mathemnatically. I wished to email this note to you, but I could not figure out where to find your email. It is probably not posted. Comment by \| July 25, 2010 \| Reply 2. They’re completely unrigorous, mathematically; that’s the whole issue with them! The path-integral is a wonderful heuristic, that can be interpreted in many cases to give a mathematically-sensible statement. But in general there is no known coherent analogue of something like Lebesgue measure on function spaces — no way to “integrate over all paths” as the plain reading of the path-integral formulation suggests. Finding a universally-applicable interpretation of the path-integral heuristic instead of problem-specific ad hoc methods is the single greatest puzzle of mathematical physics. Comment by \| July 25, 2010 \| Reply 3. [...] of . Now if we have a measure on and Lebesgue measure on the Borel sets, we can define the product measure on . Since we know and are both measurable, we can investigate their measures. I assert [...] Pingback by \| July 26, 2010 \| Reply 4. [...] and Iterated Integrals Let and be two -finite measure spaces, and let be the product measure on [...] Pingback by \| July 27, 2010 \| Reply 5. [...] We continue our assumptions that and are both -finite measure spaces, and we consider the product space [...] Pingback by \| July 28, 2010 \| Reply 6. I can suggest to see the following papers where can be found constructions of partial analogs of Lebesgue measure on some function spaces. There are 1) Baker, R., “Lebesgue measure” on \$R\sp \infty\$. Proc. Amer. Math. Soc., 113(4) (1991), 1023–1029. 2) Baker, R., “Lebesgue measure” on \$\Bbb R\sp \infty\$. II.Proc. Amer. Math. Soc., 132(9), (2004), 2577–2591 3) Pantsulaia,G., On ordinary and Standard Lebesgue Measures on \$R^{\infty}\$, Bull. Polish Acad. Sci.73(3) (2009), 209-222. 4) Pantsulaia,G., On a standard product of an arbitrary family of sigma-finite Borel measures with domain in Polish spaces, Theory Stoch. Process, vol. 16(32), 2010, no 1, p.84-93. Comment by Gogi Pantsulaia \| December 17, 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": 31, "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": 0.912553071975708, "perplexity_flag": "middle"}
http://mathematica.stackexchange.com/questions/tagged/linear-algebra+options	# Tagged Questions 1answer 319 views ### Linear Solve with Modular Arithmetic I am interested in using LinearSolve[m,b] which will find a solution to the equation $m.x=b$, where I am in mod 2 arithmetic. Is there any way to perform 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": 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": 0.8919593095779419, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/8770/if-you-put-a-latex-balloon-in-a-vacuum-how-much-would-it-expand?answertab=active	# If you put a latex balloon in a vacuum, how much would it expand? If you put a latex balloon in a vacuum, how much would it expand? And would it pop? Assume it doesn't leak. EDIT: Some numbers: Ambient pressure is 100 KPa, balloon is perfectly spherical with a diameter of 300 mm, deflated it has a diameter of 25mm, temperature is always at equilibrium. Question components stated more formally: • What pressure does the balloon need to have been inflated to to reach the 300 mm radius? • What is the relationship between the (gauge) pressure and volume? I think this simply comes down to PV=nRT • If there is some sort of spring constant involved, what is this value for a typical latex balloon? • What is the relationship between the pressure (or volume) and the tension on the material the balloon is made from? • What is the limit the tension can reach before popping? Semi-related: how much would a balloon expand if you sealed it deflated and put it in a vacuum? - 3 Is there anything in the balloon? – Greg P Apr 17 '11 at 2:02 A normal balloon filled with air or helium. – Random832 Apr 17 '11 at 2:10 How much air or helium? If the amount is small enough, I don't think it would pop. – Greg P Apr 17 '11 at 2:23 and is it closed or not? Even if closed, I'd expect diffusion of gasses through the latex material to equalise pressure during the evacuation process, at least to an extent. – jwenting Apr 17 '11 at 14:39 ## 2 Answers It would depend on two things - how much air is in the balloon and the tensile strength of latex. To see why, I hope you'll find the following useful. In a balloon on earth near the surface, the pressure inside depends on how much air you blow into it. But in order for the balloon to be in static equilibirium, that internal pressure has to be matched by the (constant) atmospheric pressure that opposes outward expansion PLUS the elastic surface tension of the surface, which also tends to oppose outward expansion (wanting to minimize the surface area). $$P_i = P_0 + S$$ If you blow more air into it, $P_i$ increases, and since $P_0$ is constant at the same height, the surface tension has to increase for LHS and RHS to match. This will continue until the surface tension exceeds the tensile strength of the balloon, at which point the balloon pops. Now let's say you have blown only enough air into the balloon to make it taut -- if you let this balloon go so that it floats up, the atmospheric pressure $P_0$ starts dropping with height, and so again, the surface tension has to keep rising so that the RHS matches the constant internal pressure $P_i$. Two things could happen now. 1. Even before $P_0$ has dropped to zero (in the vacuum of outer space), the surface tension has had to become so high that it exceeds the tensile strength and makes the balloon pop. This is what happens to most helium balloons let go - they eventually pop even before reaching vacuum - because whatever material the surface is made of has a tensile strength too low to be able to sustain the minimum amount of internal pressure required to make the balloon taut once the external pressure has dropped enough. Note that this minimum amount of internal pressure could be made smaller by reducing the size of the balloon but you can't do this beyond the limit at which the balloon's mass overcompensates and makes it impossible for it to float in the first place! 2. The other possibility is that the tensile strength is high enough that even after $P_0$ is zero, the surface tension required to match the internal pressure doesn't exceed the tensile strength and you'll have a nice happy balloon floating in the vacuum. This is what happens with space stations, for instance :) - ""If you blow more air into it, Pi increases,"" for the first "blow" into a balloon it takes more pressure than for later blows. I think that this is due to the small radius at first. Pressure is inversly related to balloon radius, but at the same time "surface tension" of the balloon increases. The pressure/radius relation for soap bubbles is well known, the same thing for rubber balloons I do not know. – Georg Apr 17 '11 at 9:42 "if you let this balloon go so that it floats up... the surface tension has to keep rising so that the RHS matches the constant internal pressure Pi." I think you have a small error here. As the atmospheric pressure drops, the balloon will expand, the tension term will increase, but Pi will also drop. I'm not sure of the exact relationship describing this, which probably has a lot to do with the spring constant of the balloon, but I'm pretty sure that the internal pressure will not be exactly constant. – Colin K Apr 17 '11 at 20:11 If I can manage to turn this into a concrete answer with the numbers I've arbitrarily selected in the edit, I'll accept this answer once I've done the math. – Random832 Apr 20 '11 at 21:44 it would expand untill it popped because the pressure from inside the balloon would be greater than the vacum since there is no air (pressure) the only reason a balloon doesnt pop in a setting like earth is because the atmospheric pressure is pushing on the balloon with the same force as the helium is pushing outward on the balloon - 4 Balloons are elastic. – Mark Eichenlaub Apr 17 '11 at 2:28 There's still a finite amount of pressure, and as the volume increased the pressure would decrease - I'm not going to accept this answer... and i'd still like to know what pressure it would pop at even so. – Random832 Apr 17 '11 at 4:21 balloons are also not gas sealed. The pressure differential causes gas to leak through the material, just as happens with an inflated balloon in air. This will delay and may even prevent the balloon from popping. – jwenting Apr 17 '11 at 14:41	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 6, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 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": 0.9607601761817932, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/29469/spontaneous-time-reversal-symmetry-breaking?answertab=votes	# Spontaneous Time Reversal Symmetry Breaking? It is known that you can break P spontaneously--- look at any chiral molecule for an example. Spontaneous T breaking is harder for me to visualize. Is there a well known condensed matter system which is uncontroversial example where T is broken spontaneously? I remember vaguely articles of Wen, Wilczek, and Zee from 1989 or so on standard High Tc hopping models, electrons which singly-occupy lattice sites, double-occupation repulsion, small amount of p-doping (holes running around), where they made the claim that T is spontaneously broken. Unfortunately I didn't understand how this happened or if it actually happened. If somebody understands the Zee example, that's good, but I would be happy with any example. I am not looking for explicit T breaking, only spontaneous T breaking. I would also like an example where the breaking is thermodynamically significant in the large system limit, so mesoscopic rings with permanent currents caused by electron discreteness is not a good example. - I guess you are thinking about the papers by Wen, Wilczek and Zee. I remember this is explained in Xiao-Gang Wens book (chapter 9), under chiral spin liquids. Sadly I don't remember the details or have the book to look in right now. – Heidar Jun 4 '12 at 9:04 1 Better fix it quickly before Wen reads it! (I apologize, I was citing from memory) – Ron Maimon Jun 4 '12 at 9:09 1 And I suppose that you're not talking about spontaneous CP breaking that then implies spontaneous T breaking via CPT, right? – Jerry Schirmer Jun 4 '12 at 11:35 3 The simplest example in condensed matter physics that spontaneously breaks time reversal symmetry is a ferromagnet. Because spins (angular momentum) change sign under time reversal, the spontaneous magnetization in the ferromagnet breaks the symmetry. This is a macroscopic example. – Everett You Jun 4 '12 at 14:28 I was about to mention the same example as Everett You. So let me just add that the same applies roughly speaking to any many-body system T-invariant on the microscopic (Hamiltonian) level where some continuous symmetry is spontaneously broken and the associated Nambu-Goldstone boson has a dispersion relation quadratic in momentum. The reason is that the effective Lagrangian for the NG boson then contains a term with a single time derivative. – Tomáš Brauner Jun 4 '12 at 14:54 show 5 more comments ## 1 Answer The simplest example in condensed matter physics that spontaneously breaks time reversal symmetry is a ferromagnet. Because spins (angular momentum) change sign under time reversal, the spontaneous magnetization in the ferromagnet breaks the symmetry. This is a macroscopic example. The chiral spin liquid (Wen-Wilczek-Zee) mentioned in the question is a non-trivial example that breaks time reversal but with out any spontaneous magnetization. Its order parameter is the spin chirality $E_{123}=\mathbf{S}_1\cdot(\mathbf{S}_2\times\mathbf{S}_3)$, which measures the Berry curvature (effective magnetic field) in the spin texture. Because $E_{123}$ also changes sign under time reversal, so the T symmetry is broken by spontaneous development of the spin chirality. Chiral spin liquid can be consider as a condensation of the skyrmion which carries the quantum of spin chirality but is spin neutral as a whole. In fact, within the spin system, one can cook up any order parameter consisting of odd number of spin operators ($\mathbf{S}_1$ for ferromagnets and $E_{123}$ for chiral spin liquid are both examples of such constructions). Then by ordering such order parameter, the time reversal symmetry can be broken spontaneously. Beyond the spin system, it is still possible to break time reversal symmetry by the development of orbital angular momentum (loop current) ordering. Just think of spins and loop currents are both angular momenta, what can be done with spins can also be done with loop currents. Indeed, the spinless fermion system can break the time reversal symmetry using the loop current (Note the word "spinless", so there is no spin SU(2) nor spin-orbit coupling involved in the following discussion). Simply consider the spinless fermion $c_i$ on a square lattice coupling to a U(1) gauge field $a_{ij}$, the Hamiltonian reads $$H=-t\sum_{\langle ij\rangle}e^{ia_{ij}}c_i^\dagger c_j+g\sum_\square \prod_{\langle ij\rangle\in\partial\square}e^{ia_{ij}}+h.c.$$ With zero flux per plaquette and with the filling of 1/2 fermion per site, the system has a fermi surface and the fermi level rest on a Van Hove singularity, which is very unstable energetically. The fermions wish to develop any kind of order as long as a it helps to open a gap at the fermi level, such that the fermi energy can be reduced. It is found that the stagger flux is a solution, in which the U(1) flux $\pm\phi$ goes through the plaquette alternately following the checkboard pattern. The corresponding gauge connection is $a_{i,i+x}=0, a_{i,i+y}=(\phi/2)(-)^{i_x+i_y}$. One can show that the energy dispersion for the fermion is given by $$E=\pm\sqrt{\cos^2k_x+\cos^2k_y+2\cos\frac{\phi}{2}\cos k_x\cos k_y},$$ which removes the Van Hove singularity and opens up a pseudo gap (like Dirac cones) as long as $\phi\neq 0$. Therefore driven by the fermi energy, $\phi$ wishes to grow toward the maximum flux $\pi$. However due to the $g$ term in the Hamiltonian, the development of stagger flux consumes magnetic energy (the energy of orbital angular momentum), which grows as $\phi^2$ for small $\phi$. The competition between the fermi energy $t$ and the magnetic energy $g$ will eventually agree on a saddle point value for $\phi$ which is between 0 and $\pi$ and its specific value can be tuned by the $t/g$ ratio. In terms of fermions, the stagger flux $\phi$ is interpreted as loop currents alternating between clockwise and counterclockwise around each plaquette following the check board pattern. Such a state is also call the orbital antiferromagnet (an antiferromagnetic arrangement of orbital angular momentum) or d-wave density wave (DDW) in high-Tc context. Here $\phi$ serves as the order parameter of the stagger flux state. Because $\phi$ changes sign under time reversal symmetry (like any other magnetic flux), the spontaneous development of the stagger flux pattern in the spinless fermion system will break the time reversal symmetry. In solid-state materials, such phenomenon has not been observed due to the too small $t/g$ ratio which is unable to drive $\phi$ away from 0. However considering the fast development of cold atom physics, the spontaneous time reversal symmetry broken in spinless fermion system may be realized in the future in the optical lattice. - +1, accepted, and thanks. – Ron Maimon Jun 16 '12 at 16:52 @RonMaimon Welcome. I have learnt much from you in your other posts. – Everett You Jun 16 '12 at 19:47	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 24, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9137944579124451, "perplexity_flag": "middle"}
http://www.physicsforums.com/showthread.php?p=3945182	Physics Forums Significance of kinematic and dynamic viscosity It is stated that kinematic viscosity is the ratio of dynamic viscosity to density. can anyone elaborate it further that what are the uses of both types of viscosities and why we differentiate them.does dynamic viscosity reates with static fluid? PhysOrg.com engineering news on PhysOrg.com >> Researchers use light projector and single-pixel detectors to create 3-D images>> GPS solution provides 3-minute tsunami alerts>> Single-pixel power: Scientists make 3-D images without a camera It is really just a term of convenience. Oftentimes, the equations of fluid mechanics are manipulated such that you end up with $\mu/\rho$ terms, so it is simply much easier to carry around a $\nu$ instead of a fraction. This is commonly seen with the Reynolds number, which has a $\mu/\rho$ term in it. then how kinematic viscosity is termed as momentum diffusivity, i m very confused with these terms. Significance of kinematic and dynamic viscosity Well, as with any sort of viscosity, it is really a measure of how momentum is diffused through a fluid. If you are familiar with the heat equation, you should notice some analogs between it and the Navier-Stokes equations. The heat equation: $$\frac{\partial \phi}{\partial t} = c^2\nabla^2 \phi$$ The incompressible Navier-Stokes equation: $$\frac{\partial \mathbf{v}}{\partial t} + \mathbf{v}\cdot\nabla\mathbf{v}=-\frac{1}{\rho}\nabla p + \nu\nabla^2\mathbf{v} + \mathbf{f}$$ The heat equation is a simplified version of the diffusion equation that describes the diffusion of basically any quantity through a material. In heat transfer, $c^2=\alpha=\frac{k}{\rho c_p}$ is the thermal diffusivity. In the Navier-Stokes equations, notice that the $\nu\nabla^2\mathbf{v}$ term takes the same form, only the N-S equations are a momentum balance, so the kinematic viscosity, $\nu$, is essentially a diffusivity constant that describes how momentum diffuses through the medium. In other words, it describes one particle's ability to affect the momentum of the adjacent particles. i have understand. thanxx for your help boneh3ad. :) Thread Tools \| \| \| \| \|----------------------------------------------------------------------\|------------------------\|---------\| \| Similar Threads for: Significance of kinematic and dynamic viscosity \| \| \| \| Thread \| Forum \| Replies \| \| \| General Physics \| 5 \| \| \| Classical Physics \| 3 \| \| \| Mechanical Engineering \| 0 \| \| \| Mechanical Engineering \| 3 \| \| \| General Physics \| 5 \|	{"extraction_info": {"found_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": 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": 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": 0.8974674940109253, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/52974/what-is-the-last-digit-of-pi?answertab=oldest	# What is the last digit of $\pi$? I want to know: what is the last digit of $\pi$? Some people say there are no such thing, but they fail to mention why. - 2 Sometimes even rational numbers don't have a "last digit", think of 1/3=0.33333... – Vhailor Jul 21 '11 at 22:12 37 Nice Poem. ${}$ – jspecter Jul 21 '11 at 22:13 1 If $\pi$ has a last digit, then $0.999\ldots\neq 1$. – Asaf Karagila Jul 21 '11 at 22:13 5 This question should specify "base 10". The "no last digit" phenomenon depends on how $\pi$ is represented. To take a contrived setting, base-$\pi$ numbers, then $\pi$ is written as $1$. I'm not trying to be pedantic here: representation is a fundamental part of this question. – Fixee Jul 22 '11 at 6:09 1 @Vhailor, Surely the last digit of 1/3 is 3? – Jonathan. May 31 '12 at 17:06 show 3 more comments ## 6 Answers There is no "last" digit of $\pi$. If there was a last digit, then there could only be finitely many digits in front so that $\pi$ would be a rational number. However $\pi$ was shown to be irrational in the 18th century by Lambert. (This Meta.StackExchange post is a joke based on the impossibility of finding such a last digit) - 3 Thanks for the link to the Meta.StackExchange post: hilarious. I've got my laughs in for the day. We all need a good dose of humor now and then! – amWhy Jul 21 '11 at 22:16 Proving that $\pi$ is irrational is more difficult than proving that $e$ or $\sqrt{2}$ or $\log_2 3$ is irrational. See http://en.wikipedia.org/wiki/Proof_that_pi_is_irrational . Proving that an irrational number has no last digit is easier than that: http://en.wikipedia.org/wiki/Irrational_number - Since you may have never seen the topics in my colleagues' answers, I'll try to explain them in some detail. Suppose for the sake of argument that when $\pi$ is written as a decimal expansion ($3.1415 \dots$) it does have a final digit. This would clearly imply that there is a finite number of terms in the expansion. All real numbers with finite decimal expansions can be written in the form $\frac{a}{b}$ where $a$ and $b$ are integers (whole numbers). By this reasoning we conclude that $\pi = \frac{a}{b}$ for some positive integers $a$ and $b$, i.e., that $\pi$ is rational. This is the starting point for this short proof given by I. Niven in 1946, which is especially easy to follow if you've had a little trigonometry and even less differential calculus. The proof concludes with an absurdity like the existence of an integer between $0$ and $1$, which implies that $a$ and $b$ do not exist and $\pi$ is irrational (and has an infinite decimal expansion). It should be noted that the irrationality of $\pi$ was first established by Lambert in 1761 by studying the continued fraction expansion of the tangent function and using the identity $\tan \frac{\pi}{4} = 1$. More generally, he proved that if $x$ is rational, then $\tan x$ is irrational. In short, there is no final digit in the decimal expansion of $\pi$ because it is irrational. - Sadly, your link to Niven's proof doesn't work. – jprete Jul 22 '11 at 4:53 Ok, try it now. – user02138 Jul 22 '11 at 5:25 HINT $\rm\ \pi = 3.1415\ \Rightarrow\ 10^4\: \pi = 31415\ \Rightarrow\ \pi = 31415/10^4\$ is rational, contra Lambert's proof. - Even rational numbers usually have no "last digit": what is the last digit of $$0.1313131\dots = 0.\overline{13} = \frac{13}{99} ?$$ So what sort of numbers have a last digit? One, numbers with a terminating decimal expansion: numbers like $\displaystyle 2.23627 = \frac{223627}{100000}$. As you can see, all such numbers can be written as a fraction with denominator being a power of $10$. Two, depending on your definition of "last digit", numbers like $0.4677777\dots = \frac1{100}46.77777$ = $\displaystyle \frac1{100}\left(46 + \frac79\right) = \frac1{100}\frac{421}{9}$. These numbers can be written as $\displaystyle \frac1{10^k} \frac{n}9$ for some integers $k$ and $n$. So a number $x$ has a "last digit" if and only if $(9\cdot 10^k)x$ is an integer for some $k$. Only very special numbers are of this form, and it should be no surprise that $\pi$ is not. (Admittedly, I don't actually see how to prove this without invoking $\pi$'s irrationality, but it's a much weaker property.) - The last known digit of $\pi$ is 1, which is the ten trillion and fiftieth digit. - – Jonas Meyer Jun 20 '12 at 14:48 3 It is oddly arbitrary to talk about the last "known" digit of $\pi$ since a subsequent and presumably unrelated digit can be found with little effort using the results of the previous computation, i.e. BBP. In that sense it is sort of like saying the largest known number is a googleplex. – Dan Brumleve Nov 2 '12 at 10:02 ## protected by Qiaochu YuanJul 21 '11 at 22:17 This question is protected to prevent "thanks!", "me too!", or spam answers by new users. To answer it, you must have earned at least 10 reputation on this site.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 50, "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": 0.947620689868927, "perplexity_flag": "head"}
http://mathoverflow.net/questions/tagged/fans	## Tagged Questions 0answers 379 views ### Reference for this polyhedral lemma Recall the definition of a fan: Let $U$ be a finite dimensional real vector space. Then a fan is a collection $\mathcal{F}$ of cones in $U$ such that (1) If \$\sigma \in \mathcal{F … 0answers 224 views ### A question about a blue fan and a red fan and their common refinement Is the following conjecture true? Conjecture: Let $M_1$ be a red map and let $M_2$ be a blue map drawn in general position on $S^n$, and let $M$ be their common refinement. There …	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 7, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.7927571535110474, "perplexity_flag": "head"}
http://mathoverflow.net/questions/118479/orientations-for-pseudoholomorphic-curves-with-totally-real-boundary-condition/118483	## Orientations for pseudoholomorphic curves with totally real boundary condition ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I am trying to understand what the obstructions are to orienting moduli spaces of pseudoholomorphic curves with totally real boundary condition. I believe that Fukaya-Oh-Ohta-Ono have shown that if a Lagrangian is relatively spin, the moduli spaces of disks with boundary in it can be oriented. My question has 3 related parts: 1. is there any sense in which the FOOO relative spin condition is also necessary? 2. if I consider curves of higher genus and/or more boundary components, do I need to impose additional conditions on the Lagrangian to guarantee orientability of the moduli spaces? 3. There has been a fair bit of recent work in the case in which the Lagrangian is the fixed point set of an anti-symplectic involution (Crétois, Georgieva-Zinger). Are the orientation difficulties in this case the same as in the general case, or do some special features appear here? EDIT: Penka Georgieva pointed out that I was mistaken. Her paper (arxiv/1207.5471) deals with the general case of curves with boundary on a Lagrangian. - ## 2 Answers 1) The problem of orienting moduli spaces of pseudo-holomorphic discs with totally real boundary conditions is really a problem in index theory. It was solved Vin de Silva in his (unpublished) D. Phil. thesis, using Atiyah's Real K-theory, and independently by FOOO. There's an excellent account in Seidel's book (section 11, especially Lemma 11.7). A totally real Cauchy-Riemann problem is by definition a loop in the totally real Grassmannian $\mathrm{Gr}(V)$ of a complex vector space $V$ with a given real structure. The resulting space $L\mathrm{Gr}(V)$ parametrizes a family of Fredholm operators (the Cauchy-Riemann operator for functions on the closed disc, valued in $V$, with boundary conditions specified by the loop). Hence there is a determinant index bundle $$\underline{det} \to L \mathrm{Gr}(V),$$ and the basic orientation problem is to describe $w_1(\underline{det}) \in H^1(L \mathrm{Gr}(V);\mathbb{Z}/2)$. For the component $L_k\mathrm{Gr}(V)$ of Maslov index $k$ loops, $H^1(L_k \mathrm{Gr}(V);\mathbb{Z}/2)$ is 2 dimensional, as one calculates using a homotopy equivalence $\mathrm{Gr}(\mathbb{C}^n) \simeq U(n)/O(n)$, so there are just four possible answers to the orientation question for each $k$. The answer is simpler to state assuming $k$ is even. It is then as follows (I learned this from de Silva's thesis): Take a loop $\gamma\colon S^1\to L_k\mathrm{Gr}(V)$. Then $\langle w_1(\underline{det}), \gamma\rangle = \langle w_2, T_\gamma \rangle$, where $w_2$ is the second SW class of the universal totally real bundle on $\mathrm{Gr}(V)$, and $T_\gamma\colon S^1\times S^1\to \mathrm{Gr}(V)$ is the torus of boundary values swept out by $\gamma$. So, in a space of pseudo-holomorphic discs in a symplectic manifold $X$ attached to an orientable Lagrangian $\Lambda$, $w_1$ of the determinant bundle evaluates on a loop $\gamma$ by evaluating the torus of boundary values $T_\gamma$ on $w_2(T\Lambda)$. Essentially for this reason, it's natural to trivialize the determinant bundle by trivializing $w_2(\Lambda)$, i.e. specifying a Pin structure. More generally, one can trivialize $w_2(\Lambda)$ relative to a fixed background class $b\in H^2(\Lambda;\mathbb{Z}/2)$ which restricts to $w_2(\Lambda)$, i.e. specify a relative (or "twisted") Pin structure. That suffices essentially because the torus of boundary values is the boundary of a 3-chain in $X$, and so vanishes if $w_2$ is the restriction of $b$. Choosing Pin structures relative to a fixed background class $b$ gives a uniform way to orient moduli spaces of pseudo-holomorphic discs for relatively Pin Lagrangians. 2) For Cauchy-Riemann operators on other curves, one can degenerate to a nodal union of discs and closed Riemann surfaces, combining the orientations for the space of discs with the complex orientation of the determinant line bundle over the moduli of closed curves. Thus no additional obstructions to orientation appear. To be precise, what goes into this is a gluing theorem for the index bundle, which is part of the linear analysis that underpins Floer theory and Gromov-Witten theory. It implies that the determinant index line over a connected sum is canonically isomorphic to the tensor product of the determinant index lines on the summands. Again, see Seidel's book, section 11 for the argument. For the underlying analysis, I'd recommend Donaldson's Floer homology book, chapter 3. 3) I don't know, but there are further concrete calculations for real loci in the work of Welschinger and also Solomon. - @Perutz: Where can I find more information on your point (2) ? This seems useful. – Chris Gerig Jan 10 at 0:35 Chris, I think there should be several sources for this, among them FOOO's book, but right now I can't remember which do this in detail. I'll try to get back to you on that. – Tim Perutz Jan 10 at 4:45 1 Chris, I've now added some details. – Tim Perutz Jan 11 at 23:25 (Perhaps I should add in light of Aleksey's answer that, as I think is usual for an MO answer, this is an informal exposition of the situation as I understand it, and is certainly not comprehensive, mathematically or bibliographically. I mentioned what I take to be the two earliest sources, but among the possible later references I am biased towards the source that I myself learned from.) – Tim Perutz Jan 21 at 16:33 Also, for the statement at the end of 1) the relatively Pin Lagrangians should still be orientable (thanks to Penka Georgieva for noting this). – Tim Perutz Jan 21 at 16:34 ### 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. (0) I do not know what is contained in a thesis which is not published and is not even available online (12 years after the defense). (1) By Proposition 8.1.4 in the 1000-page FOOO book, a relatively spin structure on a (necessarily orientable) Lagrangian submanifold $L$ of a symplectic manifold $M$ determines an orientation on the moduli spaces of $J$-holomorphic disks $(D^2,S^1)\longrightarrow (M,L)$. The same reasoning applies to a family of real Cauchy-Riemann operators induced (as in Remark 1.3 of 1207.5471) by a bundle pair $(E,F)\longrightarrow(M,L)$, where $L$ is any submanifold of any manifold $M$. Proposition 8.1.7 gives an example of a non-orientable family of real Cauchy-Riemann operators, crediting it to Vin de Silva, and including a proof. I do not believe this book contains other, substantially different, statements on orientability in open GW-theory. Thus, this book discusses orienting moduli spaces of disks in some cases, but says fairly little about their orientability in general. (2) By Theorem 1.1 in 0606429, a relatively Pin structure on a non-orientable Langrangian induces an isomorphism between the orientation line bundle of the moduli space of open $J$-holomorphic maps from Riemann surfaces with a fixed complex structure and a product of pull-backs of the orientation line bundle of the Lagrangian by evaluation maps. I do not believe this paper contains other, substantially different, statements on orientability in open GW-theory. Thus, this paper contains a number of results on both orienting and orientability of moduli spaces. (3) Lemma 11.7 in Seidel's book does what Tim says in (1). Unfortunately, it requires more than a quick look to understand and see that it implies Proposition 8.1.4 in FOOO and the disk case of Theorem 1.1 in 0606429. (4) Theorem 1.1 in 1207.5471 describes the holonomy of the orientation bundle of a family of real Cauchy-Riemann operators over bordered Riemann with varying complex structures. Its statement is absolutely clear from looking at the preceding half a page, at the beginning of the introduction. In particular, it is almost immediately clear that this theorem implies Proposition 8.1.4 in FOOO and the full statement of Theorem 1.1 in 0606429. The proof, contained in Section 3, is beautifully simple and uses no K-theory or even homotopy exact sequences. (5) In the case of anti-symplectic involutions, one often wants to orient moduli spaces of real maps, not their halves, even if they are halvable. In the case of maps from $S^2$ with the standard conjugation, if the corresponding moduli space of disk maps is orientable, the space of real maps is orientable if the flip map on the disk space is orientation-preserving. So, this becomes a problem about computing the sign of this flip map. Results on this are Proposition 5.1 in 0606429 and Theorems 1.1 and 1.3 in 0912.2646. If the involution on $S^2$ has no fixed points, the orientation problem has nothing to do with any Lagrangians. Results on orientability in this case are Theorem 1.3 and Example 2.5 in 1205.1809 and Theorem 1.1 and Corollary 1.8 in 1301.1074. (6) 1301.1074 says less about the Lagrangian case than 1207.5471. The point of 1301.1074 is to study the orientability problem for $J$-holomorphic maps that commute with involutions on the domain and the target. These maps need not be halvable to a map from a bordered Riemann surface with Lagrangian boundary conditions. An application is Corollary 1.8. - Thank you for clearing up my confusion about real maps vs their halves. – Sam Lisi Jan 22 at 9:06	{"extraction_info": {"found_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": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 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": 0.9135379195213318, "perplexity_flag": "head"}
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http://wiki.xiph.org/index.php?title=Intra&oldid=13862&diff=prev	# Intra ### From XiphWiki (Difference between revisions) \| \| \| \| \| \|--------------------------------------------------------------------------------------\|---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------\|-------------------------------------------------\|---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------\| \| Unlord (Talk \| contribs) (Adding results from 4x4 and 8x8 intra predictor training) \| \| Gmaxwell (Talk \| contribs) m ( vp8 not theora) \| \| \| Line 24: \| \| Line 24: \| \| \| \| Because the decoder must know the coefficients for each of the block modes, there is a tradeoff between how well we can predict the values in <math>y</math> (number of block modes) with decoder complexity. Our hypothesis is that the optimal number of block modes will correspond to how closely we can fit different geometries inside the block. For 4x4 blocks, this might correspond to a mode for each of the 8 cardinal directions (ask Jason to clarify) with larger blocks potentially supporting a larger number of directions. In addition, we would like to support the DC and True Motion modes from WebM/VP8 as they are often the best fit. For common video sequences, anywhere from 20% to 45% of the intra frames use the True Motion mode [1]. \| \| Because the decoder must know the coefficients for each of the block modes, there is a tradeoff between how well we can predict the values in <math>y</math> (number of block modes) with decoder complexity. Our hypothesis is that the optimal number of block modes will correspond to how closely we can fit different geometries inside the block. For 4x4 blocks, this might correspond to a mode for each of the 8 cardinal directions (ask Jason to clarify) with larger blocks potentially supporting a larger number of directions. In addition, we would like to support the DC and True Motion modes from WebM/VP8 as they are often the best fit. For common video sequences, anywhere from 20% to 45% of the intra frames use the True Motion mode [1]. \| \| \| \| \| \| \| - \| This is slightly complicated by the fact that our coefficients are not in the time-domain, but are the result of a lapped transform, see [[TDLT]]. In practice all of these modes have an equivalent in the frequency domain. As a starting point, we are using the 10 modes from Theora to classify the blocks from a set of sample images. Each category of blocks will be used to construct a set of predictors <math>\beta</math> which is then used to reclassify the blocks based on Sum of Absolute Transform Differences (SATD). <i>Tim suggested weighting each blocks contribution based on SATD_bestfit-SATD_nearest</i>. \| + \| This is slightly complicated by the fact that our coefficients are not in the time-domain, but are the result of a lapped transform, see [[TDLT]]. In practice all of these modes have an equivalent in the frequency domain. As a starting point, we are using the 10 modes from VP8 to classify the blocks from a set of sample images. Each category of blocks will be used to construct a set of predictors <math>\beta</math> which is then used to reclassify the blocks based on Sum of Absolute Transform Differences (SATD). <i>Tim suggested weighting each blocks contribution based on SATD_bestfit-SATD_nearest</i>. \| \| \| \| \| \| \| \| == <math>L^2</math>-norm == \| \| == <math>L^2</math>-norm == \| ## Latest revision as of 03:48, 9 February 2013 In Daala, intra-frame predictors are coefficients that are used to predict the contents of a block based on its three neighboring blocks. Consider a 4x4 block structure: $\left( \begin{array}{cccc\|cccc} x_{1}&x_{2}&x_{3}&x_{4}&x_{5}&x_{6}&x_{7}&x_{8}\\ x_{9}&x_{10}&x_{11}&x_{12}&x_{13}&x_{14}&x_{15}&x_{16}\\ x_{17}&x_{18}&x_{19}&x_{20}&x_{21}&x_{22}&x_{23}&x_{24}\\ x_{25}&x_{26}&x_{27}&x_{28}&x_{29}&x_{30}&x_{31}&x_{32}\\ \hline x_{33}&x_{34}&x_{35}&x_{36}&y_{1}&y_{2}&y_{3}&y_{4}\\ x_{37}&x_{38}&x_{39}&x_{40}&y_{5}&y_{6}&y_{7}&y_{8}\\ x_{41}&x_{42}&x_{43}&x_{44}&y_{9}&y_{10}&y_{11}&y_{12}\\ x_{45}&x_{46}&x_{47}&x_{48}&y_{13}&y_{14}&y_{15}&y_{16}\\ \end{array} \right)$ Assuming a linear predictor, we would like to find the 768 coefficients that best predict yi from the neighboring xi. That is, for a given block with coefficients $\hat x$ and $\hat y$, we would like a 48x16 matrix β such that the residual $\hat e=\hat y-\hat x \beta$ is minimized (for some p-norm). Because it is this residual $\hat e$ that is quantized and coded in the bitstream, we would prefer that most of its coefficients are zero. This will be discussed further when comparing the different p-norms. ## Block Modes These predictors often follow certain geometry in the time-domain image space. Imagine a vertical edge that runs through y. A natural predictor might be to simply copy the values $\begin{array}{cccc}x_{29}&x_{30}&x_{31}&x_{32}\end{array}$ down into each row of the 4x4 block y. In order to account for the different possible geometries, each block is assigned a mode which indicates what set of prediction coefficients to use (the β). When encoding a block y, the encoder chooses a mode, computes $\hat e$ and writes both the block mode and the residual $\hat e$. Because the decoder must know the coefficients for each of the block modes, there is a tradeoff between how well we can predict the values in y (number of block modes) with decoder complexity. Our hypothesis is that the optimal number of block modes will correspond to how closely we can fit different geometries inside the block. For 4x4 blocks, this might correspond to a mode for each of the 8 cardinal directions (ask Jason to clarify) with larger blocks potentially supporting a larger number of directions. In addition, we would like to support the DC and True Motion modes from WebM/VP8 as they are often the best fit. For common video sequences, anywhere from 20% to 45% of the intra frames use the True Motion mode [1]. This is slightly complicated by the fact that our coefficients are not in the time-domain, but are the result of a lapped transform, see TDLT. In practice all of these modes have an equivalent in the frequency domain. As a starting point, we are using the 10 modes from VP8 to classify the blocks from a set of sample images. Each category of blocks will be used to construct a set of predictors β which is then used to reclassify the blocks based on Sum of Absolute Transform Differences (SATD). Tim suggested weighting each blocks contribution based on SATD_bestfit-SATD_nearest. ## L2-norm Suppose for some mode we have n blocks. We can then construct the matrices X and Y where the ith row of each contains the zero-mean values from the ith block. $X=\begin{pmatrix} x_{1,1}-\bar{x_1} & x_{1,2}-\bar{x_2} & \cdots & x_{1,m}-\bar{x_m}\\ x_{2,1}-\bar{x_1} & x_{2,2}-\bar{x_2} & \cdots & x_{2,m}-\bar{x_m}\\ \vdots & \vdots & \ddots & \vdots \\ x_{n,1}-\bar{x_1} & x_{n,2}-\bar{x_2} & \cdots & x_{n,m}-\bar{x_m} \end{pmatrix}$ $Y=\begin{pmatrix} y_{1,1}-\bar{y_1} & y_{1,2}-\bar{y_2} & \cdots & y_{1,l}-\bar{y_l}\\ y_{2,1}-\bar{y_1} & y_{2,2}-\bar{y_2} & \cdots & y_{2,l}-\bar{y_l}\\ \vdots & \vdots & \ddots & \vdots \\ y_{n,1}-\bar{y_1} & y_{n,2}-\bar{y_2} & \cdots & y_{n,l}-\bar{y_l} \end{pmatrix}$ In the case of 4x4 blocks, m = 48 and l = 16. Let C = XTX and D = XTY. Then β = C − 1D. ## 4x4 Intra Predictors Using the set of 50 images in subset1-y4m to train sparse (444 = 64 multiplies per block) intra predictors: ```VP8 Intra Predictors Mode 0 Blocks 379965 SATD 1030.8 Bits 136.827 Mean 1725.16 Var 591085 CgRef 12.9338 CgPred 14.4075 Pg 1.47375 Mode 1 Blocks 328826 SATD 1410.93 Bits 140.398 Mean 1755.67 Var 615743 CgRef 10.7723 CgPred 14.4396 Pg 3.66727 Mode 2 Blocks 364320 SATD 1318.99 Bits 140.038 Mean 1771.33 Var 613037 CgRef 12.2597 CgPred 14.321 Pg 2.0613 Mode 3 Blocks 461738 SATD 1266.24 Bits 138.956 Mean 1799.99 Var 581067 CgRef 12.4228 CgPred 14.4985 Pg 2.07569 Mode 4 Blocks 317707 SATD 1320.78 Bits 140.709 Mean 1738.33 Var 615735 CgRef 11.4334 CgPred 13.989 Pg 2.55553 Mode 5 Blocks 276650 SATD 1222.78 Bits 139.462 Mean 1691.97 Var 612193 CgRef 11.5441 CgPred 14.2428 Pg 2.69867 Mode 6 Blocks 264186 SATD 1238.89 Bits 139.503 Mean 1664.18 Var 603871 CgRef 11.7692 CgPred 14.2162 Pg 2.44696 Mode 7 Blocks 354682 SATD 1293.8 Bits 139.885 Mean 1654 Var 603002 CgRef 12.1736 CgPred 14.2243 Pg 2.05075 Mode 8 Blocks 278080 SATD 1322.45 Bits 140.806 Mean 1706.37 Var 634358 CgRef 11.6018 CgPred 14.0524 Pg 2.4506 Mode 9 Blocks 253209 SATD 1321.22 Bits 140.924 Mean 1722.95 Var 625202 CgRef 11.4638 CgPred 13.7727 Pg 2.30894 Pooled Blocks 3279363 SATD 1270.74 Bits 139.871 Mean 1727.99 Var 619488 CgRef 11.4416 CgPred 14.2047 Pg 2.76305 ``` ```Daala Intra Predictors Mode 0 Blocks 472906 SATD 1017.46 Bits 138.832 Mean 1728.18 Var 553398 CgRef 12.9443 CgPred 14.6488 Pg 1.7045 Mode 1 Blocks 261910 SATD 1519.87 Bits 146.67 Mean 1779.45 Var 471328 CgRef 9.70601 CgPred 12.4896 Pg 2.78355 Mode 2 Blocks 198050 SATD 1818.89 Bits 149.906 Mean 1778.28 Var 440574 CgRef 10.0855 CgPred 11.5142 Pg 1.42869 Mode 3 Blocks 400658 SATD 1091.01 Bits 138.022 Mean 1765.68 Var 699810 CgRef 14.9146 CgPred 17.0632 Pg 2.14857 Mode 4 Blocks 440888 SATD 1056.23 Bits 138.926 Mean 1768.35 Var 669611 CgRef 13.8652 CgPred 15.8732 Pg 2.008 Mode 5 Blocks 123873 SATD 2646.11 Bits 158.928 Mean 1761.66 Var 348360 CgRef 6.56612 CgPred 7.681 Pg 1.11488 Mode 6 Blocks 389669 SATD 1128.95 Bits 140.725 Mean 1607.85 Var 604298 CgRef 12.8995 CgPred 14.7494 Pg 1.84991 Mode 7 Blocks 364446 SATD 1132.17 Bits 139.413 Mean 1649.46 Var 674713 CgRef 14.0093 CgPred 16.1988 Pg 2.18954 Mode 8 Blocks 373419 SATD 1116.84 Bits 140.019 Mean 1720.85 Var 707029 CgRef 13.9352 CgPred 15.8526 Pg 1.91734 Mode 9 Blocks 253544 SATD 1571.98 Bits 147.453 Mean 1797.04 Var 502237 CgRef 10.7593 CgPred 12.3135 Pg 1.55419 Pooled Blocks 3279363 SATD 1261.89 Bits 142.641 Mean 1727.99 Var 619488 CgRef 12.3321 CgPred 14.4918 Pg 2.15973 ``` Using the set of 1000 images in subset3-y4m to test the sparse (444 = 64 multiplies per block) intra predictors: ```VP8 Intra Predictors Mode 0 Blocks 8440618 SATD 773.069 Bits 129.891 Mean 1987.82 Var 714490 CgRef 16.1033 CgPred 17.7858 Pg 1.68249 Mode 1 Blocks 7177017 SATD 1097.06 Bits 133.975 Mean 1835.81 Var 682211 CgRef 13.2692 CgPred 17.2354 Pg 3.96624 Mode 2 Blocks 7486531 SATD 1129.14 Bits 136.141 Mean 1849.6 Var 699452 CgRef 14.1344 CgPred 16.3106 Pg 2.17612 Mode 3 Blocks 9420648 SATD 1030.04 Bits 133.978 Mean 1856.55 Var 667016 CgRef 14.7014 CgPred 16.9182 Pg 2.21677 Mode 4 Blocks 6537593 SATD 1074.28 Bits 135.633 Mean 1877.16 Var 703018 CgRef 13.7951 CgPred 16.4451 Pg 2.65005 Mode 5 Blocks 5764375 SATD 1037.61 Bits 135.162 Mean 1862.93 Var 714729 CgRef 13.7467 CgPred 16.5772 Pg 2.83053 Mode 6 Blocks 5526915 SATD 1047.34 Bits 135.235 Mean 1853.98 Var 725463 CgRef 13.9811 CgPred 16.5951 Pg 2.61399 Mode 7 Blocks 7493442 SATD 1061.93 Bits 134.918 Mean 1797.8 Var 701592 CgRef 14.467 CgPred 16.6788 Pg 2.21183 Mode 8 Blocks 5635636 SATD 1077.08 Bits 135.666 Mean 1850.12 Var 724307 CgRef 13.9346 CgPred 16.4812 Pg 2.54658 Mode 9 Blocks 5148581 SATD 1100.52 Bits 136.283 Mean 1895.9 Var 726050 CgRef 13.7549 CgPred 16.1284 Pg 2.37353 Pooled Blocks 68631356 SATD 1035.13 Bits 134.722 Mean 1868.07 Var 721805 CgRef 13.8553 CgPred 16.77 Pg 2.91471 ``` ```Daala Intra Predictors Mode 0 Blocks 9675144 SATD 841.841 Bits 133.86 Mean 1877.93 Var 662343 CgRef 15.701 CgPred 17.4209 Pg 1.71986 Mode 1 Blocks 4809122 SATD 1246.09 Bits 141.563 Mean 1764.73 Var 466747 CgRef 12.3822 CgPred 14.2618 Pg 1.8796 Mode 2 Blocks 3895161 SATD 1620.32 Bits 146.753 Mean 1763.23 Var 408176 CgRef 10.9493 CgPred 12.402 Pg 1.45271 Mode 3 Blocks 8512887 SATD 895.992 Bits 133.096 Mean 1838.33 Var 781431 CgRef 17.0033 CgPred 19.2762 Pg 2.27292 Mode 4 Blocks 10224556 SATD 788.516 Bits 131.961 Mean 2073.27 Var 777684 CgRef 16.8937 CgPred 18.964 Pg 2.07024 Mode 5 Blocks 2492854 SATD 2306.98 Bits 155.1 Mean 1781.56 Var 306540 CgRef 7.42632 CgPred 8.54354 Pg 1.11722 Mode 6 Blocks 7839240 SATD 960.763 Bits 136.271 Mean 1723.23 Var 684064 CgRef 15.2766 CgPred 17.1139 Pg 1.83731 Mode 7 Blocks 8127384 SATD 913.37 Bits 133.886 Mean 1865.64 Var 797013 CgRef 16.528 CgPred 18.8681 Pg 2.34004 Mode 8 Blocks 8289385 SATD 856.357 Bits 133.506 Mean 1963.06 Var 814308 CgRef 16.8012 CgPred 18.6876 Pg 1.88641 Mode 9 Blocks 4765623 SATD 1420.7 Bits 144.479 Mean 1773.33 Var 478656 CgRef 11.868 CgPred 13.3845 Pg 1.51648 Pooled Blocks 68631356 SATD 1030.34 Bits 137.446 Mean 1868.07 Var 721805 CgRef 14.964 CgPred 17.0421 Pg 2.07809 ``` ## 8x8 Intra Predictors Using the set of 50 images in subset1-y4m to train sparse (488 = 256 multiplies per block) intra predictors: ```VP8 Intra Predictors Mode 0 Blocks 153189 SATD 6082.74 Bits 633.298 Mean 1720.67 Var 559647 CgRef 12.0272 CgPred 12.3392 Pg 0.311931 Mode 1 Blocks 49244 SATD 6482.51 Bits 631.517 Mean 1752.33 Var 634379 CgRef 12.983 CgPred 14.5119 Pg 1.52891 Mode 2 Blocks 81386 SATD 6317.27 Bits 633.888 Mean 1772.67 Var 617045 CgRef 13.2505 CgPred 13.8038 Pg 0.55328 Mode 3 Blocks 149817 SATD 6142.2 Bits 629.138 Mean 1821.6 Var 577836 CgRef 13.4642 CgPred 14.0662 Pg 0.601936 Mode 4 Blocks 59095 SATD 6488.09 Bits 639.403 Mean 1738.64 Var 635175 CgRef 12.8858 CgPred 13.546 Pg 0.660174 Mode 5 Blocks 51138 SATD 6321.64 Bits 636.812 Mean 1680.52 Var 614762 CgRef 12.5191 CgPred 13.2898 Pg 0.770625 Mode 6 Blocks 46860 SATD 6214.38 Bits 635.479 Mean 1636.45 Var 606974 CgRef 13.0011 CgPred 13.5919 Pg 0.590785 Mode 7 Blocks 101899 SATD 6579.86 Bits 637.096 Mean 1621 Var 597451 CgRef 12.9896 CgPred 13.5883 Pg 0.598636 Mode 8 Blocks 55425 SATD 6946.74 Bits 644.887 Mean 1714.81 Var 646845 CgRef 12.5421 CgPred 13.1933 Pg 0.651179 Mode 9 Blocks 43833 SATD 6717.76 Bits 640.445 Mean 1738.18 Var 634206 CgRef 12.5208 CgPred 13.1502 Pg 0.629351 Pooled Blocks 791886 SATD 6356.01 Bits 635.537 Mean 1728.58 Var 619153 CgRef 12.4832 CgPred 13.326 Pg 0.842891 ``` ```Daala Intra Predictors Mode 0 Blocks 62290 SATD 8842.88 Bits 695.702 Mean 1762.94 Var 546310 CgRef 10.527 CgPred 11.2558 Pg 0.728816 Mode 1 Blocks 51719 SATD 8525.04 Bits 687.553 Mean 1814.55 Var 507208 CgRef 10.9023 CgPred 11.7365 Pg 0.834129 Mode 2 Blocks 36010 SATD 12451.8 Bits 728.279 Mean 1791.12 Var 384211 CgRef 7.39062 CgPred 7.865 Pg 0.474379 Mode 3 Blocks 112574 SATD 5202.09 Bits 631.108 Mean 1783.42 Var 690506 CgRef 15.7576 CgPred 16.3784 Pg 0.620821 Mode 4 Blocks 115633 SATD 5215.31 Bits 633.79 Mean 1769.83 Var 629411 CgRef 14.1115 CgPred 14.8214 Pg 0.709915 Mode 5 Blocks 54821 SATD 9203.44 Bits 696.669 Mean 1744.5 Var 463757 CgRef 9.45701 CgPred 10.0708 Pg 0.613816 Mode 6 Blocks 87676 SATD 5900.82 Bits 646.94 Mean 1578.28 Var 591176 CgRef 13.1847 CgPred 13.8014 Pg 0.616685 Mode 7 Blocks 92133 SATD 5419.48 Bits 636.372 Mean 1634.71 Var 692905 CgRef 15.1441 CgPred 15.8044 Pg 0.660312 Mode 8 Blocks 76551 SATD 6501.37 Bits 658.66 Mean 1736.97 Var 612091 CgRef 12.6443 CgPred 13.4359 Pg 0.791668 Mode 9 Blocks 102479 SATD 5847.21 Bits 646.934 Mean 1733.7 Var 570790 CgRef 12.9057 CgPred 13.583 Pg 0.677333 Pooled Blocks 791886 SATD 6625.85 Bits 659.058 Mean 1728.58 Var 619153 CgRef 12.6445 CgPred 13.4208 Pg 0.776311 ``` Using the set of 1000 images in subset3-y4m to test the sparse (488 = 256 multiplies per block) intra predictors. ```VP8 Intra Predictors Mode 0 Blocks 3285321 SATD 4831.25 Bits 609.544 Mean 1942.13 Var 681922 CgRef 15.1468 CgPred 15.5036 Pg 0.356785 Mode 1 Blocks 1097934 SATD 4966.25 Bits 600.994 Mean 1847.45 Var 682361 CgRef 16.1138 CgPred 17.8094 Pg 1.69564 Mode 2 Blocks 1708487 SATD 5380.91 Bits 616.147 Mean 1869.28 Var 707993 CgRef 15.3892 CgPred 15.978 Pg 0.58875 Mode 3 Blocks 3106373 SATD 5029.6 Bits 608.275 Mean 1879.34 Var 663156 CgRef 15.8358 CgPred 16.4784 Pg 0.642574 Mode 4 Blocks 1208481 SATD 5328.85 Bits 618.814 Mean 1882.27 Var 709893 CgRef 15.2482 CgPred 15.8695 Pg 0.621321 Mode 5 Blocks 1049455 SATD 5420.14 Bits 619.549 Mean 1855.61 Var 719825 CgRef 14.9456 CgPred 15.7082 Pg 0.762665 Mode 6 Blocks 959802 SATD 5356.04 Bits 618.585 Mean 1832.64 Var 732730 CgRef 15.2935 CgPred 15.878 Pg 0.584477 Mode 7 Blocks 2216886 SATD 5374.46 Bits 613.848 Mean 1782.23 Var 703311 CgRef 15.4919 CgPred 16.1339 Pg 0.64197 Mode 8 Blocks 1089024 SATD 5614.08 Bits 620.635 Mean 1830.07 Var 723888 CgRef 15.031 CgPred 15.6673 Pg 0.636324 Mode 9 Blocks 870162 SATD 5664.15 Bits 621.395 Mean 1877.86 Var 725235 CgRef 14.8465 CgPred 15.4287 Pg 0.582189 Pooled Blocks 16591925 SATD 5205.41 Bits 613.921 Mean 1868.35 Var 720473 CgRef 15.1122 CgPred 15.9821 Pg 0.869956 ``` ```Daala Intra Predictors Mode 0 Blocks 1179905 SATD 7866.76 Bits 682.077 Mean 1726.99 Var 519986 CgRef 12.0984 CgPred 12.4692 Pg 0.370783 Mode 1 Blocks 1008452 SATD 7410.96 Bits 669.185 Mean 1815.4 Var 486451 CgRef 12.7369 CgPred 13.1166 Pg 0.379717 Mode 2 Blocks 676403 SATD 11047.9 Bits 713.149 Mean 1806.12 Var 353773 CgRef 8.53864 CgPred 9.00691 Pg 0.46827 Mode 3 Blocks 2540750 SATD 4183.79 Bits 607.011 Mean 1934.57 Var 794608 CgRef 18.4004 CgPred 19.0367 Pg 0.636302 Mode 4 Blocks 2656446 SATD 3991.2 Bits 605.507 Mean 2030 Var 728302 CgRef 17.1992 CgPred 17.8889 Pg 0.68974 Mode 5 Blocks 1011671 SATD 8495.58 Bits 687.746 Mean 1744.3 Var 449801 CgRef 10.8417 CgPred 11.4205 Pg 0.578879 Mode 6 Blocks 1710619 SATD 5227.87 Bits 632.837 Mean 1648.81 Var 645320 CgRef 15.2445 CgPred 15.8622 Pg 0.617689 Mode 7 Blocks 2352651 SATD 3996.86 Bits 601.981 Mean 2013.93 Var 842396 CgRef 18.4642 CgPred 19.1573 Pg 0.693167 Mode 8 Blocks 1482503 SATD 5447.2 Bits 637.809 Mean 1810.08 Var 641515 CgRef 14.9565 CgPred 15.5448 Pg 0.588273 Mode 9 Blocks 1972525 SATD 5184.71 Bits 633.59 Mean 1822.5 Var 633855 CgRef 14.9489 CgPred 15.5997 Pg 0.650804 Pooled Blocks 16591925 SATD 5466.77 Bits 636.877 Mean 1868.35 Var 720473 CgRef 15.3899 CgPred 16.0271 Pg 0.63723 ``` [1] http://blog.webmproject.org/2010/07/inside-webm-technology-vp8-intra-and.html Main Page Icecast	{"extraction_info": {"found_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}, 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http://mathhelpforum.com/calculus/137436-showing-tangent-curve-also-tangent-curve-another-point.html	# Thread: 1. ## Showing that the tangent to the curve is also tangent to curve at another point. Show that the tangent to the curve y=(x^2 +x -2)^2 +3 at the point where x=1 is also tangent to the curve at another point. So far this is what I did: Subbed in 1 to the oringinal curve to find out that the point is (1,3) Than I derived it to 4x^3 +6x^2 -6x -4 and when I subbed in 1 into the derivative I got 0. Than I found out the equation of the line which is y=3, but i'm completely lost on what to do from there 2. Originally Posted by kmjt Show that the tangent to the curve y=(x^2 +x -2)^2 +3 at the point where x=1 is also tangent to the curve at another point. So far this is what I did: Subbed in 1 to the oringinal curve to find out that the point is (1,3) Than I derived it to 4x^2 +6x^2 -6x -4 and when I subbed in 1 into the derivative I got 0. Than I found out the equation of the line which is y=3, but i'm completely lost on what to do from there Yes, that is correct except for a small typo. $f'(x)=2\left(x^2+x-2\right)(2x+1)=4x^3+6x^2-6x-4$ At x=1, f'(x)=0. x=1, y=3. The tangent at x=1 is parallel to the x-axis (it's a local minimum). If this tangent is tangent at another point, then there is a 2nd turning point at y=3. Hence, we find a second solution for x, if y=3 corresponds to f'(x)=0 for more than one x. This also means we can examine f(x) to see how many x causes f(x)=3. $f(x)=\left(x^2+x-2\right)^2+3=3\ \Rightarrow\ \left(x^2+x-2\right)=0$ $\Rightarrow\ (x+2)(x-1)=0$ f(x)=3 for x=1 and x=-2 only. To check the tangent, if x=-2, f'(x)=-32+16+16+8-4-4=0. Hence, The tangent is also tangent to the curve at (-2,3). 3. Hmm i'm kind of confused at this point. How are you going from (x^2 +x -2)^2 +3 to (x^2 +x -2)? 4. $f(x)=(x^2 +x-2)^2 + 3 = 3 \Rightarrow (x^2+x-2)^2 = 3-3 \Rightarrow (x^2 + x-2)^2 = 0$ $(x^2 + x-2)^2 = (x+2)^2 (x-1)^2$ Also $(x^2 + x-2) = (x+2) (x-1)$ They have the same roots, therefore it will yield the same answer, so this is why he is jumping from: $(x^2 + x-2)^2 + 3 = 3 \Rightarrow (x^2 + x-2) = 0$ Hope this helps. 5. Originally Posted by kmjt Hmm i'm kind of confused at this point. How are you going from (x^2 +x -2)^2 +3 to (x^2 +x -2)? Yes, $f(x)=\left(x^2+x-2\right)^2+3=3$ means $\left(x^2+x-2\right)^2=0\ \Rightarrow\ \left(x^2+x-2\right)\left(x^2+x-2\right)=0$ $\Rightarrow\ \left(x^2+x-2\right)=0\ \Rightarrow\ (x+2)(x-1)=0$ $x=1$ $x=-2$ 6. Why wouldn't you expand (x^2 +x -2)(x^2 +x -2)? 7. Originally Posted by kmjt Why wouldn't you expand (x^2 +x -2)(x^2 +x -2)? Hi kmjt, we could, but there is no need to! it's just extra work... $f(x)=\left(x^2+x-2\right)\left(x^2+x-2\right)+3$ If f(x)=3, then f(x)=0+3, the product of the brackets must be zero, hence even if we multiply out the brackets, all that will still be zero at the second point on the graph that that particular tangent touches. Since it is the same factor multiplied by itself, we are left to solve $x^2+x-2=0.$ since 0(0)=0 As this is a quadratic, it gives us at most 2 values of x causing f(x)=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": 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": 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": 0.9421060681343079, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/250194/is-the-optimal-solution-to-this-problem-to-row-straight-to-the-store	# Is the optimal solution to this problem to row straight to the store? For my homework, I was given this brainteaser: You’re sunbathing on the island shown on the map below. The island is six miles from shore at the closest point, and the nearest store is a convenience store seven miles down the beach. If you can row at 4 miles an hour, and you can walk at 5 miles an hour, where should you land to get a bag of tortilla chips in the least possible time? (Ignore tides, currents, and sharks. No fair renting a helicopter.) I found the equation for the time it takes to get to the store to be: $\frac{x}{4} + \frac{7-\sqrt{x^2 - 36}}{5}$ I then found the derivative: $\frac{1}{4} - \frac{2x}{10 \sqrt{x^2-36}}$ I didn't find any zeroes for the derivative inside the domain of the function [6, $\sqrt{85}]$, so I answered that the shortest time would be achieved by rowing straight to the store. The teacher said the answer was eight. Where did I make a mistake? edit: Given image: edit: Some closure: My email to him: On the brainteaser, the store is only 7 miles away, how can the answer be 8 :P For my formula, 10 gave the same answer as 8 in the formula the answer uses. i.e., rowing 10 miles lands you 8 miles from the point across from the island. Of course, sqrt(85) is less than 10, so it's only faster if you can then walk 7 - 8 = -1 miles to the store and turn back time a little. His response: It assumes you can go back in time. - Perhaps you could provide the map or at least describe the geometry of the problem? – espen180 Dec 3 '12 at 19:46 @espen180, Sure, there you go – Walkerneo Dec 3 '12 at 19:48 8 is the answer to a related question: if the store is very far down the beach (or 10 miles down the beach, for instance), how far down the beach should you aim for? If you have the question right, your teacher is simply wrong. – TonyK Dec 3 '12 at 21:00 1 There is an error in your derivative calculation. Also if it were me I would not set x to be the distance rowed, but rather the horizontal distance from the landing site to the store. That said your method should work (once you correct the derivative), it is just tougher to interpret – Morgan Sherman Dec 3 '12 at 22:56 1 @Wolkerneo You are missing an $x$ in the numerator. – Morgan Sherman Dec 3 '12 at 23:35 show 4 more comments ## 3 Answers Putting the origin on the left, your travel time is $d(x)=\cfrac{1}{4} \sqrt{6^2+x^2} +\cfrac{1}{5}(7 - x)$ The derivative is $d'(x) = \cfrac{1}{4} \cfrac{x}{\sqrt{6^2+x^2}} -\cfrac{1}{5}$ $d'$ is negative on $(-\infty,8)$ so $d$ is strictly decreasing on $[0,7]\subset(-\infty,8)$. Since we want the minimal value of $d$ on $[0,7]$, we'll get it at $x=7$, $f(7)=\cfrac{\sqrt{85}}{4}$. I'm not sure about your second term but your $x/4$ means your $x$ is on the vertical axe... So it'd be like doing it this way: - Hmm, I see how that works, but I don't understand why mine doesn't work – Walkerneo Dec 3 '12 at 19:56 @Walkerno: I updated. – xavierm02 Dec 3 '12 at 20:06 2 Wait, 8 isn't even in your domain. If x is the distance between the point straight across the island and the store, then your domain is [0,7]. – Walkerneo Dec 3 '12 at 20:10 True. I'll doublecheck. I just asked maple to solve for $0$. Stupid lazy me. – xavierm02 Dec 3 '12 at 20:19 Walkerneo: I edited. You were right, the fastest way is to go straight to the store. – xavierm02 Dec 3 '12 at 20:51 show 2 more comments For fun, we do the problem the hard way. Forget about the location of the store for now, say it is far away. We can row to the nearest point on the shore, then walk. Imagine setting out instead at an angle $\theta$ to the shortest line to the shore. Then we row an extra distance of $6\sec\theta -6$, but we walk less by $6\tan\theta$, so the amount of time "saved" over the direct to shore way is $$\frac{6\tan\theta}{5} -\frac{6\sec\theta-6}{4}.$$ Now we could use calculus, the differentiation is in some ways more pleasant than if we use $x$'s. But let's see what can be done without calculus. Change to sines and cosines. After a little manipulation we arrive at $$\frac{6}{\cos\theta} \left(\frac{\sin \theta}{5} -\frac{1-\cos\theta}{4}\right).$$ Using half-angle identities, and letting $\phi=\theta/2$, we find that the time saved is $$\frac{12\sin\phi}{\cos\theta}\left(\frac{\cos\phi}{5}-\frac{\sin\phi}{4}\right).$$ We save time if the above expression is positive, that is if $\tan \phi\lt \frac{4}{5}$. Thus increasing $\phi$ is always a good idea until $\tan\phi=\frac{4}{5}$. This corresponds to $\tan\theta=\frac{40}{9}$. But $\tan\theta=\frac{40}{9}$ is $\frac{240}{9}$ miles down the beach, a lot more than $7$ miles. So the best strategy is to row directly to the store. (The whole calculation is probably unnecessary. The speeds $4$ and $5$ are close to each other, as are the distances $6$ and $7$. So the diagonal is the best way.) - This is at odds with xaviermo2's answer, so one of you must be wrong! I checked xaviermo2's answer, and it seemed right to me... – TonyK Dec 3 '12 at 23:05 Where did those numbers come from? 6 sec theta and 6 tan theta? – Walkerneo Dec 3 '12 at 23:14 1 @Walkerneo: Draw a right triangle $DNX$, with $D$ point of departure, $N$ nearest point on shore, and $X$ a variable point on the shore somewhere in the direction of the store. The right angle is at $N$. The angle $\theta$ is $\angle NDX$. Then $DN=6$. So by trigonometry $NX=6\tan\theta$ and $DX=6\sec\theta$. – André Nicolas Dec 3 '12 at 23:24 @TonyK: The same go directly to the store conclusion is in xaviermo2's answer, as the answer currently stands. – André Nicolas Dec 3 '12 at 23:29 Yes, but you differ in how far away the store can be for this solution to remain optimal. – TonyK Dec 4 '12 at 0:19 show 2 more comments Let $y$ be the landing position relative to the island (so the store is at $y=7$). Then the distance from the island to the landing position is $\sqrt{36+y^2}$, and the walking distance to the store from there is $\|y-7\|$. The total time is $$T(y)=\frac{1}{4}\sqrt{36+y^2}+\frac{1}{5}\|y-7\|,$$ and the derivative with respect to $y$ is $$T'(y)=\frac{y}{4\sqrt{36+y^2}}+\frac{1}{5}\text{sgn}({y-7})$$ for $y\neq 7$; the function $T(y)$ is not differentiable at $y=7$, but has the value $\frac{1}{4}\sqrt{85}$ there. It remains to check for any other local extrema. There are two cases: $y>7$ and $y<7$. Clearly $T'(y)$ is positive for all $y>7$, since both terms are positive in that case. For $y<7$, we need to check for solutions to $$\frac{y}{4\sqrt{36+y^2}}-\frac{1}{5}=0.$$ The only solution, $y=8$, is not in the range where this form of $T(y)$ applies. We conclude that there is no local minimum of $T(y)$ except the one at $y=7$, and so you are correct that the fastest way to reach the store is by rowing straight to it. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 61, "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": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9418702721595764, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/86874?sort=votes	## Is every finite group a quotient of the Grothendieck-Teichmuller group? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) The Grothendieck-Teichmuller conjecture asserts that the absolute Galois group $Gal(\mathbb{Q})$ is isomorphic to the Grothendieck-Teichmuller group. I was wondering, would this conjecture imply the Inverse Galois Problem? I.e. is every finite group a quotient of the Grothendieck-Teichmuller group? - ## 1 Answer It's only a partial answer since this survey is already 5 years old, but it suggest that almost nothing is known about (non-abelian) finite quotients of $\widehat{GT}$ (question 1.7). Edit: I should maybe recall what happen in the abelian case, and why it's encouraging: elements of $\widehat{GT}$ are pairs $(f,\lambda)$ where $f$ is in the derived subgroup of $\hat{F}_2$, and $\lambda \in \hat{\mathbb{Z}}^{\times}$, satisfying some complicated equations. It turns out that the set theoretic map $(f,\lambda) \mapsto \lambda$ induces a surjective group morphism $\widehat{GT}\rightarrow \hat{\mathbb{Z}}^{\times}$. And the good news is that the composite $$G_{\mathbb{Q}} \hookrightarrow \widehat{GT} \rightarrow \hat{\mathbb{Z}}^{\times}$$ is nothing but the cyclotomic character. - 2 I have the feeling that their is very little known about $\widehat{GT}$ period! – Spice the Bird Jan 28 2012 at 3:38 1 I too think this problem sounds extremely hard. To give some indication -- as far as I know (and I could certainly be wrong) -- the only reason we know the map GThat -> Zhat^* is surjective is that G_Q embeds in GThat! Without that, I don't think I even know how to show that the image is infinite! – JSE Jan 28 2012 at 22:19	{"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": 10, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": 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": 0.9456269145011902, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/7795/what-is-5746-divided-by-17	# What is $57^{46}$ divided by 17? $$\frac{57^{46}}{17}$$ What is best and quickest way to evaluate an approximate answer to this by hand? - 1 Do you want the result $57^{46} / 17 \approx 3.5 \times 10^{79}$, the quotient, or the remainder? – KennyTM Oct 25 '10 at 13:00 3 Some of the answers below are interpreting the question as being $57^{46}$ modulo $17$, or in other words the remainder upon division by $17$. The OP should of course clarify, but given that s/he asks for an "approximate answer", it sounds like s/he really wants the quotient. – Pete L. Clark Oct 25 '10 at 14:15 7 @Pete, if the OP has the remainder in mind, an easy approximate answer would be $0\pm 9$ :) – Mariano Suárez-Alvarez♦ Oct 25 '10 at 14:29 3 @Mariano: I can get a sharper bound with $8\pm 8$. =) – Willie Wong♦ Oct 25 '10 at 15:48 @Mariano: agreed. – Pete L. Clark Oct 26 '10 at 5:35 show 1 more comment ## 9 Answers It really depends what you mean by "approximate answer". Since you tagged arithmetic... Notice 57 = 3 . 19, you can say that $57^{46} / 17 = 3 \cdot 57^{45} \cdot \frac{19}{17}$. So if you know what $57^{45}$ is, you can approximate to within 2% if you multiply that by 3.3. Now, compute by hand $57^5 = 601692057$, which is less than 0.3% off from $6\cdot 10^8$. Observe that $57^{45} = (57^5)^9$. Using the binomial theorem you see that you can approximate that by $(6\cdot 10^8)^9 = 10 077 696 \cdot 10^{72}$ to within 3%. Rounding off the lower digits won't matter much in the error, so you have $57^{45} \sim 10^{79}$ to 3%. So putting it all together you have that to within 5%, $$57^{46} \sim 3.3 \cdot 10^{79}$$ - Actually, I was the one who tagged it `arithmetic`, since the original tags didn't exactly fit... in any event, this looks something like what Fermi would have done if asked this question. +1! – J. M. Oct 25 '10 at 12:23 1 Hum, I must have made a small error propagating the error. Wolfram tells me that the actual answer is roughtly 3.466 . 10^79, so the error is 5.03%, so not within 5%. But I hope that's close enough to be forgiven. – Willie Wong♦ Oct 25 '10 at 12:41 ## Did you find this question interesting? Try our newsletter email address Why not simply do the computation? The most time-consuming part is to compute the $46$th power of $57$, but that can be done quite fast using repeated squaring in 6 multiplications. That results in a number with 81 digits. Dividing by 17 using long division is faster, and it takes about 80 easy steps. - Does Google compute that for you? For me it doesn't. – Rasmus Oct 25 '10 at 14:46 @Rasmus: I've fixed the link (I don't know what was wrong...) – Mariano Suárez-Alvarez♦ Oct 25 '10 at 15:01 Ah, nice! $\mbox{}$ – Rasmus Oct 25 '10 at 15:46 With mental arithmetic alone I would try the following. Undoubtedly you have needed the value of $$\tan\frac\pi6=\frac1{\sqrt3} \approx0.57\ldots$$ enough many times to have it memorized. Therefore $57\approx 100/\sqrt3$. This means that $57^{46}\approx 10^{92}/3^{23}$. Because $\sqrt{10}\approx3$ we also have $\log_{10}3\approx 0.5$ Thereforer $3^{23}\approx10^{11.5}$. So $57^{46}$ is probably somewhere between $10^{80}$ and $10^{81}$. Dividing this by 17 gives a result most likely between $10^{79}$ and $10^{80}$. If such a ballpark figure is enough, then we're done. If you want a significant digit, I need to work a bit harder, and use more accurate estimates. - Euler's theorem gives you the remainder. - You can do it by Fermats little theorem as well. - HINT $\rm \mod\ 17:\ \ 57\equiv 6,\ \ 6^{46}\equiv 1/6^2,\ \ 1/6\equiv 3\ \$ which immediately yields the result - The below concerns methods for obtaining the most significant digits, the roughest approximation to the magnitude or mantissa of the quantity you want. Other posts can help you use the some results of number theory to find the least significant digits (in any base). Arbitrary precision methods for hand calculation are also available, or perhaps you can find someone like Gauss to make the exact calculation in their head. With a slide rule (if you literally need it by hand and can accept only a few decimal places of accuracy) or hand calculator (if 10 or so places will suffice), you can explicitly approximate such a number using the principle of logarithms. If $x=\frac{57^{46}}{17}$, then $$\eqalign{ \log x &= 46\log 57-\log 17 \\ &\approx 46(4.04305126783455)-(2.83321334405622) \approx 183.147144976333 \qquad\text{base }e \\ &\approx 46(1.75587485567249)-(1.23044892137827) \approx 79.5397944395563 \qquad\text{base }10 }$$ so that $$\eqalign{ x&=10^{\log_{10}x}=e^{\log_{e}x} \\ &\approx e^{183.147144976333} \\ &\approx 10^{79.5397944395563} \\ &\approx 3.46572771657154 \times 10^{79} }$$ The values of the logarithms vary depending on which base you are using. Calculators usually have the natural logarithm written as $\ln=\log_e$ (logarithm naturale: in French, the adjective follows the noun), whereas computers often use the "natural" base $e\approx 2:71828182845905$. The unspecified base is unfortunately ambiguous, and is a frequent source of confusion much like radians and degrees for $\sin$ and $\cos$. Whether $\log$ means $\log_{10}$ or $\ln$ must be made clear by a statement as there is no universally accepted convention. Computers might return the final answer as $3.46572771657154$ E $79$, where the E is an abbreviation for scientific notation "$\times10$^". In this notation, the quantity $x$ is represented as a mantissa $M$ and an exponent $E$: $$x \approx M \times 10^E$$ where $M$ is typically either in the interval $[0,1)$ or $[1,10)$ and $E$ is any integer (up to the storage size used by the data type on the computer or built-in size on the calculator). More info if you search for IEEE float/double. The accuracy of the above calculation is practically as good as the size (number of digits of accuracy) of the mantissa. Depending on the exponent storage size of your calculator or computer system, you could also just directly calculate the number, and the mantissa will be automatically truncated to the available precision. There are also of course methods of calculating this to more than ten places by pen(cil) and paper; is this what you want? - By Euler's theorem $57^{16}\equiv1 \pmod{17}$. It could be done using this. - I think the question is for the quotient, not the remainder, and this answer has been given before. – Ross Millikan May 9 at 19:34 The result is exactly: $34657277165715299429841134098673896005946880584360167911317357675111364398875720+\frac{9}{17}$ - 6 The question asks for the quickest way to calculate an approximate answer by hand. Did you do this by hand, and quickly? – Graphth Apr 6 '12 at 16:01 (-1) for the answer and (+1) for @Grapth's comment... – Gottfried Helms Apr 6 '12 at 19:38 1 This answer is hilarious! – Jase Nov 10 '12 at 4:45 3 @Jase: I am a mental arithmetic amateur. – dot dot Nov 10 '12 at 9:16	{"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": 48, "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": 0.9331737756729126, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/85329?sort=votes	## Hovey’s unit axiom in monoidal model categories ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Let $\mathcal{C}$ be a monoidal model category in the sense of Hovey's book. He assumes the following unit axiom not considered in other references (e.g. Schwede-Shipley): given a cofibrant replacement of the monoidal unit $q\colon QI\stackrel{\sim}\rightarrow I$ and a cofibrant object $X$, the morphisms $q\otimes X$ and $X\otimes q$ are weak equivalences. This axiom obviously holds if $I$ is cofibrant. Moreover, if I'm not mistaken, a stronger version holds in all examples I know: $q\otimes X$ and $X\otimes q$ are weak equivalences for any $X$. Does any of you know any example where this stronger version is false? Is it always true under some 'standard' assumptions on $\mathcal{C}$? - Hi Fernando. If you're thinking about the same Schwede-Shipley paper as I am (Algebras and Modules in Monoidal Model Categories), then they do mention the Unit Axiom. In my version it's Remark 3.2, right before the definition of the Monoid Axiom. Still, this doesn't answer your question about the stronger statement – David White Jan 10 2012 at 16:05 One more thing: my gut says there must be examples where the stronger version fails, but I haven't come up with any so far. I keep feeling this stronger version would imply some other property on $\mathcal{C}$. The comments to this old question of mine might help: mathoverflow.net/questions/73704 – David White Jan 10 2012 at 16:23 I have a question. This map $q\otimes X$ is the pushout product of $f:0\hookrightarrow QI$ and $id_X$, right? So shouldn't the pushout product axiom say that for $X$ cofibrant this map must be a trivial cofibration (since $id_X$ is a trivial cofibration)? I guess what I'm asking is why we need the Unit Axiom at all. According to Schwede-Shipley's remark, it's to be sure $I$ represents the unit on the homotopy level, but it seems to me that it should be implied by the Pushout Product axiom. – David White Jan 10 2012 at 18:48 The pushout product you say is just the identity in $QI\otimes X$ – Fernando Muro Jan 10 2012 at 19:58 Oops, I meant to write $f:QI \rightarrow I$ and $0\rightarrow X$. And with this it's clear why the pushout product axiom doesn't imply the unit axiom, since the map $f$ is not a cofibration. Thanks for finding my error – David White Jan 10 2012 at 20:03 ## 1 Answer I can only answer half of your question: namely, a standard condition under which the more general unit axiom holds. I don't know of any examples where Hovey's unit axiom holds but this more general one does not. The hypothesis is that cofibrant objects are flat, i.e. smashing with cofibrant objects preserves weak equivalences. This hypothesis comes up all the time in Hovey's work, and also appears in the paper of Schwede-Shipley where the Monoid Axiom is first introduced. There you need cofibrant objects to be flat in order to conclude that if $R\rightarrow S$ is a weak equivalence of ring objects, then $Ho(R-mod)\cong Ho(S-mod)$. Suppose that cofibrant objects are flat. Then we know Hovey's unit axiom automatically, since $QI\rightarrow I$ is a weak equivalence and so for any cofibrant $Y$, $Y$ smashed with this map is still a weak equivalence. To see that the more general unit axiom holds, let $X$ be any object (not necessarily cofibrant). Then we have the following commutative diagram: $$\begin{array}{rrrr} QI\otimes QX & \rightarrow & QX & \rightarrow & X\\ \downarrow & & & & \downarrow \\ QI \otimes X & & \rightarrow & & X \end{array}$$ Here the top maps are weak equivalences by Hovey's unit axiom and by the definition of $QX$. The left vertical map is a weak equivalence because cofibrant objects are flat and $QI$ is cofibrant. The right vertical is clearly a weak equivalence because it's the identity. Thus, by the 2-out-of-3 property, the bottom horizontal is a weak equivalence. EDIT: It should go without saying, but in a non-symmetric setting "cofibrant objects are flat" means both $K\otimes f$ and $f\otimes K$ are weak equivalences. So the argument above works if you apply the twist natural transformation to everything, and we also have $X\otimes QI \rightarrow X$ is a weak equivalence. - 1 At this link you can find an example where cofibrant objects are not flat. My gut says that if you tweak the example to be $T$-alg, then it'll be a monoidal model category and Hovey's unit axiom will hold. Since cofibrant objects are not flat, maybe this is a natural place to look for a counter-example to your claim. books.google.com/… – David White Jan 18 2012 at 14:41	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 33, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9152365326881409, "perplexity_flag": "head"}
http://stats.stackexchange.com/questions/16742/how-to-explain-certain-patterns-appearing-after-kernel-averaging	# How to explain certain patterns appearing after kernel averaging? Having a `2D map` filled `uniformly by random values` (Figure:top-left), the next maps are kernel averaged with a kernel of sizes 3x3, 5x5, ..., 11x11. The questions are: What are the patterns appeared in the kernel averaged maps? What is the statistical basis of creation of such clusters/pattern? How is the pattern related to the original dispersion model i.e., uniform? Note: To generate the maps a kernel based averaging system kind C was applied on the original data. An answer from a different aspect has been given at this link. If you can develop based on more statistical/spatial statistics please do it. - Smoothing is in general an implementation of diffusion, so what you see are more less the frames of this noise evolving into constant value (plus an evidence of fixed boundary condition). – mbq♦ Oct 9 '11 at 18:07 @mbq Could you please provide me some links to the diffusion concept with emphasizing on the points you mentioned? – Developer Oct 10 '11 at 9:12 1 You can look around with keywords like heat equation or diffusion equation -- it is not the same, but can lead you to some clues. About patterns, you could look into CG papers about making noise textures (Perlin noise and following stuff). Or also ask this on Physics.SE, people from statistical mechanics may know something about it. You are asking very hard questions though. – mbq♦ Oct 10 '11 at 11:23 1 @mbq Actually, this is a simple GIS/time-series question with a straightforward answer. Because the kernel is square and not circular, it's really a pair of independent "diffusions" in 1D, where its analysis (as a moving average) is a fundamental part of time series methodology. – whuber♦ Oct 10 '11 at 15:02 1 @mbq I don't know what you mean by "characteristics" or by "density patterns"! The illustration shows moving averages of uniform noise using a sequence of $k\times k$ square kernels; those are straightforward to characterize as spatial random processes. The question hints at a generalization whereby some generic spatial process is smoothed in this way; the relationships among the process and its smooths are again straightforward to describe, especially in terms of the multivariate distributions to which the process gives rise. – whuber♦ Oct 10 '11 at 18:28 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": 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": 0.9363681674003601, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/10083/computing-lengths-in-the-a-2-affine-weyl-group	computing lengths in the A_2 affine weyl group Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) The A_2 affine Weyl group is the symmetry group of the triangulation of the plane by equilateral triangles. As Sean points out, it may be generated by reflections $r_1, r_2, r_3$ about the edges of a single equilateral triangle. Since A_n is a Coxeter group, every element $\alpha \in A_2$ may be assigned a length $l(\alpha)$ with respect to the generators $r_i$. How might one compute this length with respect to this presentation? Any generating set has to contain a reflection about a line in each of the three directions ($0, 2\pi/3, 4\pi/3$) of the lattice. Is that condition sufficient? - Do you mean something other than the Coxeter group presentation? Also, in what sense do you mean "compute"? Finally, the affine Weyl group of type A2 has 3 generators: you pick one of the triangles and then take the group generated by reflections across each of the 3 lines bounding the triangle. – Sean Rostami Dec 30 2009 at 1:13 why don't you post this as an answer? – John Mangual Dec 30 2009 at 2:19 I don't know, it was 2/3 request for clarification, 1/3 answer so I thought it should be a comment. Is that not right? – Sean Rostami Dec 30 2009 at 2:25 Your answer would be more helpful if you wrote out the Coxeter presentation in the case of A_2 (and perhaps in the case of A_n). Given an element of this group - which we now know is a symmetry of the triangular lattice in the plane - we would like to compute the length of the shortest word in the Coxeter generators, representing this symmetry. Other people might find such a clear and thoughtful response to be helpful. – John Mangual Dec 30 2009 at 3:29 2 Please do not edit the text of the question as much as that. You have more or less completely changed it! – Mariano Suárez-Alvarez Dec 30 2009 at 4:09 1 Answer Ok, here's a reply in "answer" format. I asked first whether you knew about the Coxeter group presentation because I didn't want to write a bunch of stuff that you already knew (and therefore would not answer your question). I also didn't (and still don't) understand how strong of a concept you mean by "compute". It would have been helpful if you said something to clear up these two issues in your most recent reply. Anyway here's some basic stuff I can hammer out quickly... The presentation of a Coxeter group (by definition) is that has a finite list of order 2 generators (called "simple reflections" in the Weyl group case) and otherwise the only relations are to specify the orders of products of pairs of generators, and some pairs can be omitted (=infinite order). Affine Weyl groups are Coxeter groups so all this applies. An affine Weyl group $W$ is an infinite group and there are arbitrarily long reduced expressions (=shortest words). Even though there are only a finite number of generating reflections, there are an infinite number of reflections and the hyperplanes they fix (in your $A_2$ case, they are lines) divide the space up into connected pieces usually called "alcoves" (in your $A_2$ case, they are the triangles). A fundamental fact is that $W$ acts simply-transitively on the set of these alcoves, which means if you pick a "basepoint" alcove $\mathcal{A}$, you can associate (bijectively) to each element $w \in W$ the alcove $w(\mathcal{A})$. Again, I'm not sure what you mean by "compute", but here's an attempt at an answer anyway: A "geometric" way to compute the length $\ell(w)$ of $w \in W$ is to count the number of hyperplanes separating $\mathcal{A}$ from $w(\mathcal{A})$. In a $2$-dimensional (or perhaps even $3$-dimensional) case, e.g. $A_2$, I guess you could quickly compute shortest words for a particular "geometrically defined" element $w$ as follows: • print out a drawing of the triangulated space and label $\mathcal{A}$ (this is probably the longest step) • figure out which alcove $w(\mathcal{A})$ is (you should already know this by definition of "geometrically defined") • count hyperplanes to determine $\ell(w)$ (this is quick and easy) • label all the alcoves with their lengths, up to and including $\ell(w)-1$ (quick and easy) • check all 3 simple reflections until you find one sending $w(\mathcal{A})$ to an alcove whose labeled length is $< \ell(w)$ (quick and easy) • write down that simple reflection on a piece of paper • now repeat the previous 2 steps for the new alcove • when you get to the base alcove, you're finished and the list of simple reflections you wrote down is a reduced word It's a long list of steps because I like to be explicit but it really doesn't take that long. - Step 1 can be made faster, thanks to John Stembridge's Coxeter graph paper: math.lsa.umich.edu/~jrs/archive.html#cox – David Speyer Dec 30 2009 at 4:22 Haha =) Thanks! – Sean Rostami Dec 30 2009 at 4:30 You shouldn't worry about duplicating information. I don't know much about Coxeter groups and neither may some readers. – John Mangual Dec 30 2009 at 4:58 eh, whatever, please just clarify your questions if people ask for them to be clarified – Sean Rostami Dec 30 2009 at 7:20	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": false, "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": 0.9376446008682251, "perplexity_flag": "middle"}
http://mathhelpforum.com/number-theory/133283-prime-decomposition-number-rings.html	# Thread: 1. ## Prime decomposition in number rings Assume that the number field K contains a primitive n'th root of unity $u$. Let $a$ in the integral ring of K and let I be an prime ideal of the integral ring of K such that $na$ is not in I. Let L=K( $a^{1/n}$). Show that $x^{n}-a$ is separable modulo I. Thank you!!!	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 5, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8495466709136963, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/34891/derivative-of-a-position-eigenket?answertab=oldest	# Derivative of a Position Eigenket I was flicking through Zettili's book on quantum mechanics and came across a 'derivation' of the momentum operator in the position representation on page 126. The author derived that $\langle\vec{r}\|\hat{\vec{P}}\|\psi\rangle = -i\hbar\vec{\nabla}\langle\vec{r}\|\psi\rangle$ (I've omitted the full derivation). However, from this relationship he concluded that $\hat{\vec{P}} = -i\hbar\vec{\nabla}$. I'm sure this is very basic but why can you immediately conclude this? Surely this assumes that $\langle\vec{r}\|(-i\hbar\vec{\nabla})\|\psi\rangle = -i\hbar\vec{\nabla}\langle\vec{r}\|\psi\rangle$. I'm not sure why this is necessarily true. - ## 2 Answers $-i\hbar\vec{\nabla}\|\psi\rangle$ is not a valid notation. The nabla operator is defined in three-dimensional Euclidean space, not the in the Hilbert space of quantum states. When the author says $\hat{\boldsymbol{P}}=-i\hbar\vec{\nabla}$ he does not mean the momentum operator defined in the state space, but the space of wavefunctions. Then $\hat{\boldsymbol{P}}\psi(\boldsymbol{r})=-i\hbar\vec{\nabla}\psi(\boldsymbol{r})$. - +1! If it's a precise formula you're after, note you can "cancel" the $\|\psi\rangle$ to get $\langle\vec{r}\|\hat{\boldsymbol{P}}=-i\hbar\vec{\nabla}\langle\vec{r}\|$ (which should always be understood as $\langle\vec{r}\|$ acting before $\nabla$). – Emilio Pisanty Aug 25 '12 at 13:46 The $\hat{P}$ operator is a matrix, and you can write it out as a continuous analog of a matrix between two continuous indices x and x' $$\hat{P}_{xx'} = i\delta'(x-x')$$ Then the action of $\hat{P}$ on a ket $\psi(x)$ is by the continuous analog of matrix multiplication: $$\hat{P}\psi(x) = \int \hat{P}_{xx'} \psi_{x'} dx' = -i \psi'(x)$$ after formally integrating by parts and popping the delta function. For learning these things, Dirac's book is best. To understand this, always think of a lattice, in which case $$\hat{P}_{xx'} = \Delta'(x-x')$$ Where $\Delta(0)=-{1\over \epsilon}$ and $\Delta(\epsilon) ={1\over \epsilon}$ on a lattice of step $\epsilon$. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 16, "mathjax_display_tex": 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": 0.9543529748916626, "perplexity_flag": "head"}
http://mathoverflow.net/questions/110721/invertibility-of-a-matrix-with-a-gaussian-perturbation	## invertibility of a matrix with a Gaussian perturbation ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Suppose that $A$ is an arbitrary fixed $n\times n$ matrix and $G$ a random $n\times n$ matrix with i.i.d. $N(0,1)$ entries. Is there a simple proof that $A+G$ is invertible with probability 1? What if $G$ is a random Wigner matrix (symmetric, upper diagonal entries are i.i.d. $N(0,1)$)? Is $A+G$ still invertible with probability 1? Is there a simple proof? - 7 You can write $det(A+G)$ has a polynomial in $G_{ij}$ and the set of zeroes of $det(A+X)$ has null Lebesgue measure on $\mathbb{R}^{n^2}$. Since $(G_{ij})_{ij}$ can be seen as a standard Gaussian in $\mathbb{R}^{n^2}$, the result follows. – Alekk Oct 26 at 9:06 3 What you describe is a nondegenerate Gaussian measure on the vector space of $n\times n$ matrices. As indicated somewhere else on this site, such a measure is absolutely continuous w.r.t. the Lebesgue measure. Since the Lebesgue measure of the hypersurface $\det A=0$ is zero, so will be its Gaussian measure. Thus with probability $0$ a a random matrix (of the first type you indicated) is not invertible. – Liviu Nicolaescu Oct 26 at 10:02 ## 1 Answer We use the idea suggested by Alekk. Let $A_{i,j}$ the entries of $A$. Then in the first case, the entries of $M:=A+G$ are Gaussian independent random variables, that is, $M_{i,j}\sim N(A_{i,j},1)$. Denote $N$ the set of elements of $\Bbb R^{n^2}$ such that the matrix of generic term $x_{i,j}$ is not invertible. This set has null Lebesgue measure in $\Bbb R^{n^2}$ as it's the zeros of a polynomial. By independence, the family $(M_{i,j},i,j\in [n])$ is Gaussian, so $$p:=P(A+G\mbox{ is not invertible})=P((M_{i,j})_{i,j=1}^n\in S).$$ As the law of $(M_{i,j},i,j\in [n])$ is absolutely continuous with respect to Lebesgue measure in $\Bbb R^{n^2}$, we conclude that $p=0$. In the second case, write $\det(A+G)$ as a polynomial of the $G_{i,j},i\leqslant j$, and use the fact that $(G_{i,j},1\leqslant i\leqslant j\leqslant n)$ is Gaussian to conclude in the same way as in the first case. • The result doesn't depend on the choice of the deterministic matrix $A$. • We don't need i.i.d.ness, just the fact that $(G_{i,j},i,j\in [n])$ is Gaussian in the first case, $(G_{i,j},i\leqslant j,i,j\in [n])$ in the second. • We can have a more general result when the law of $(A_{i,j}+G_{i,j})_{i,j\in [n]}$ is absolutely continuous with respect to Lebesgue measure on $\Bbb R^{n^2}$. -	{"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": 38, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9389673471450806, "perplexity_flag": "head"}
http://mathhelpforum.com/pre-calculus/205325-getting-root-equation.html	# Thread: 1. ## Getting to the root of the equation The equation has two roots. Find the smaller and larger one? so I came up with 2x2e-5x(3-3x)=0 Just not sure where to go from here? 2. ## Re: Getting to the root of the equation Originally Posted by M670 The equation has two roots. Find the smaller and larger one? so I came up with 2x2e-5x(3-3x)=0 Just not sure where to go from here? You need to use the zero product principle. This would give $2x^2=0 \quad e^{-5x}=0 \quad 3-3x=0$ Only two of these equations have solutions.	{"extraction_info": {"found_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": 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": 0.9513422846794128, "perplexity_flag": "middle"}
http://en.wikipedia.org/wiki/Basis_(topology)	# Base (topology) (Redirected from Basis (topology)) In mathematics, a base (or basis) B for a topological space X with topology T is a collection of open sets in T such that every open set in T can be written as a union of elements of B. We say that the base generates the topology T. Bases are useful because many properties of topologies can be reduced to statements about a base generating that topology, and because many topologies are most easily defined in terms of a base which generates them. ## Simple properties of bases Two important properties of bases are: 1. The base elements cover X. 2. Let B1, B2 be base elements and let I be their intersection. Then for each x in I, there is a base element B3 containing x and contained in I. If a collection B of subsets of X fails to satisfy either of these, then it is not a base for any topology on X. (It is a subbase, however, as is any collection of subsets of X.) Conversely, if B satisfies both of the conditions 1 and 2, then there is a unique topology on X for which B is a base; it is called the topology generated by B. (This topology is the intersection of all topologies on X containing B.) This is a very common way of defining topologies. A sufficient but not necessary condition for B to generate a topology on X is that B is closed under intersections; then we can always take B3 = I above. For example, the collection of all open intervals in the real line forms a base for a topology on the real line because the intersection of any two open intervals is itself an open interval or empty. In fact they are a base for the standard topology on the real numbers. However, a base is not unique. Many bases, even of different sizes, may generate the same topology. For example, the open intervals with rational endpoints are also a base for the standard real topology, as are the open intervals with irrational endpoints, but these two sets are completely disjoint and both properly contained in the base of all open intervals. In contrast to a basis of a vector space in linear algebra, a base need not be maximal; indeed, the only maximal base is the topology itself. In fact, any open sets in the space generated by a base may be safely added to the base without changing the topology. The smallest possible cardinality of a base is called the weight of the topological space. An example of a collection of open sets which is not a base is the set S of all semi-infinite intervals of the forms (−∞, a) and (a, ∞), where a is a real number. Then S is not a base for any topology on R. To show this, suppose it were. Then, for example, (−∞, 1) and (0, ∞) would be in the topology generated by S, being unions of a single base element, and so their intersection (0,1) would be as well. But (0, 1) clearly cannot be written as a union of the elements of S. Using the alternate definition, the second property fails, since no base element can "fit" inside this intersection. Given a base for a topology, in order to prove convergence of a net or sequence it is sufficient to prove that it is eventually in every set in the base which contains the putative limit. ## Objects defined in terms of bases • The order topology is usually defined as the topology generated by a collection of open-interval-like sets. • The metric topology is usually defined as the topology generated by a collection of open balls. • A second-countable space is one that has a countable base. • The discrete topology has the singletons as a base. ## Theorems • For each point x in an open set U, there is a base element containing x and contained in U. • A topology T2 is finer than a topology T1 if and only if for each x and each base element B of T1 containing x, there is a base element of T2 containing x and contained in B. • If B1,B2,...,Bn are bases for the topologies T1,T2,...,Tn, then the set product B1 × B2 × ... × Bn is a base for the product topology T1 × T2 × ... × Tn. In the case of an infinite product, this still applies, except that all but finitely many of the base elements must be the entire space. • Let B be a base for X and let Y be a subspace of X. Then if we intersect each element of B with Y, the resulting collection of sets is a base for the subspace Y. • If a function f:X → Y maps every base element of X into an open set of Y, it is an open map. Similarly, if every preimage of a base element of Y is open in X, then f is continuous. • A collection of subsets of X is a topology on X if and only if it generates itself. • B is a basis for a topological space X if and only if the subcollection of elements of B which contain x form a local base at x, for any point x of X. ## Base for the closed sets Closed sets are equally adept at describing the topology of a space. There is, therefore, a dual notion of a base for the closed sets of a topological space. Given a topological space X, a base for the closed sets of X is a family of closed sets F such that any closed set A is an intersection of members of F. Equivalently, a family of closed sets forms a base for the closed sets if for each closed set A and each point x not in A there exists an element of F containing A but not containing x. It is easy to check that F is a base for the closed sets of X if and only if the family of complements of members of F is a base for the open sets of X. Let F be a base for the closed sets of X. Then 1. ∩F = ∅ 2. For each F1 and F2 in F the union F1 ∪ F2 is the intersection of some subfamily of F (i.e. for any x not in F1 or F2 there is an F3 in F containing F1 ∪ F2 and not containing x). Any collection of subsets of a set X satisfying these properties forms a base for the closed sets of a topology on X. The closed sets of this topology are precisely the intersections of members of F. In some cases it is more convenient to use a base for the closed sets rather than the open ones. For example, a space is completely regular if and only if the zero sets form a base for the closed sets. Given any topological space X, the zero sets form the base for the closed sets of some topology on X. This topology will be finest completely regular topology on X coarser than the original one. In a similar vein, the Zariski topology on An is defined by taking the zero sets of polynomial functions as a base for the closed sets. ## Weight and character We shall work with notions established in (Engelking 1977, pp. 12, 127--128). Fix $X\,$ a topological space. We define the weight as $w(X)\triangleq\,$ the minimum cardinality of a basis; we define the network weight as $nw(X)\triangleq\,$ the minimum cardinality of a network; the character of a point $\chi(x,X)\triangleq\,$ the minimum cardinality of a neighbourhood basis for $x\,$ in $X\,$; and the character of $X\,$ to be $\chi(X)\triangleq\sup\{\chi(x,X):x\in X\}\,$. Here, a network is a family $\mathcal{N}\,$ of sets, for which, for all points $x\,$ and open neighbourhoods $U\ni x\,$, there is a $B\in\mathcal{N}\,$ for which $x\in B\subseteq U\,$. The point of computing the character and weight is useful to be able to tell what sort of bases and local bases can exist. We have following facts: • $nw(X)\leq w(X)\,$. • if $X\,$ is discrete, then $w(X)=nw(X)=\|X\|\,$. • if $X\,$ is Hausdorff, then $nw(X)\,$ is finite iff $X\,$ is finite discrete. • if $B\,$ a basis of $X\,$ then there is a basis $B'\subseteq B\,$ of size $\|B'\|\leq w(X)\,$. • if $N\,$ a neighbourhood basis for $x\in X\,$ then there is a neighbourhood basis $N'\subseteq N\,$ of size $\|N'\|\leq \chi(x,X)\,$. • if $f:X\to Y\,$ is a continuous surjection, then $nw(Y)\leq w(X)\,$. (Simply consider the $Y\,$-network $f'''B\triangleq\{f''U:U\in B\}\,$ for each basis $B\,$ of $X\,$.) • if $(X,\tau)\,$ is Hausdorff, then there exists a weaker Hausdorff topology $(X,\tau')\,$ so that $w(X,\tau')\leq nw(X,\tau)\,$. So a fortiori, if $X\,$ is also compact, then such topologies coincide and hence we have, combined with the first fact, $nw(X)=w(X)\,$. • if $f:X\to Y\,$ a continuous surjective map from a compact metrisable space to an Hausdorff space, then $Y\,$ is compact metrisable. The last fact comes from the fact that $f(X)\,$ is compact Hausdorff, and hence $nw(f(X))=w(f(X))\leq w(X)\leq\aleph_{0}\,$ (since compact metrisable spaces are necessarily second countable); as well as the fact that compact Hausdorff spaces are metrisable exactly in case they are second countable. (An application of this, for instance, is that every path in an Hausdorff space is compact metrisable.) ### Increasing chains of open sets Using the above notation, suppose that $w(X)\leq\kappa\,$ some infinite cardinal. Then there does not exist a strictly increasing sequence of open sets (equivalently strictly decreasing sequence of closed sets) of length $\geq\kappa^{+}\,$. To see this (without the axiom of choice), fix $(U_{\xi})_{\xi\in\kappa}\,$ a basis of open sets. And suppose per contra, that $(V_{\xi})_{\xi\in\kappa^{+}}\,$ were a strictly increasing sequence of open sets. This means $(\forall{\alpha<\kappa^{+}})V_{\alpha}\setminus\bigcup_{\xi<\alpha}V_{\xi}\,$ is non-empty. If $x\in V_{\alpha}\setminus\bigcup_{\xi<\alpha}V_{\xi}\,$, we may utilise the basis to find some $U_{\gamma}\,$ with $x\in U_{\gamma}\subseteq V_{\alpha}\,$. In this way we may well-define a map, $f:\kappa^{+}\to\kappa\,$ mapping each $\alpha\,$ to the least $\gamma\,$ for which $U_{\gamma}\subseteq V_{\alpha}\,$ and meets $V_{\alpha}\setminus\bigcup_{\xi<\alpha}V_{\xi}\,$. This map can be seen to be injective. (For otherwise there would be $\alpha<\beta\,$ with $f(\alpha)=f(\beta)=\gamma\,$, say, which would further imply $U_{\gamma}\subseteq V_{\alpha}\,$ but also meets $V_{\beta}\setminus\bigcup_{\xi<\alpha}\subseteq V_{\beta}\setminus V_{\alpha}\,$ which is a contradiction.) But this would go to show that $\kappa^{+}\leq\kappa\,$, a contradiction. ## References • Engelking, Ryszard (1977). General Topology. PWN, Warsaw. • James Munkres (1975) Topology: a First Course. Prentice-Hall. • Willard, Stephen (1970) General Topology. Addison-Wesley. Reprinted 2004, Dover Publications.	{"extraction_info": {"found_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": 60, "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": 0.9349278211593628, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/132967/history-of-f-circ-g	# History of $f \circ g$ $f \circ g$ is usually interpreted as $f(g(x))$ although, as Google shows, $g(f(x))$ is used frequently too. My question: Does anybody know who was the first mathematician to use this symbol and what was his interpretation? - 2 $f\circ g$ is always interpreted as $f(g(x))$; more precisely, the rule for the function $f\circ g$ is $(f\circ g)(x)=f(g(x))$. $g(f(x))$ would correspond to $g\circ f$: $(g\circ f)(x)=g(f(x))$. $f\circ g$ and $g\circ f$ are generally different functions. – David Mitra Apr 17 '12 at 15:12 7 $f\circ g$ is not interpreted as $f(g(x))$. Rather $(f\circ g)(x)$ is interpreted as $f(g(x))$, and $(f\circ g)(w)$ is interpreted as $f(g(w))$, and $(f\circ g)(5.2)$ is interpreted as $f(g(5.2))$, and so on. – Michael Hardy Apr 17 '12 at 15:30 6 it depends on whether you write $f$ as $f(x)$ or $(x)f$. although the latter notation is rare, it does occur (often in the context of permutation mappings, or other algebraic contexts in which "we do as we parse"). – David Wheeler Apr 17 '12 at 16:25 4 If you put the symbol for a function after arguments, then f∘g means (x)(f∘g) or equivalently ((x)f)g which means the same thing as g(f(x)) when you put the symbol for a function before arguments. – Doug Spoonwood Apr 17 '12 at 16:28 3 Back in graduate school, in our Galois Theory text, the author wrote $x^\sigma$ for the result of applying automorphism $\sigma$ to the element $x$. So naturally composition was $x^{\sigma\circ\tau} = (x^\sigma)^\tau$. That is: $\sigma\circ\tau$ means: first do $\sigma$, then do $\tau$. – GEdgar Apr 17 '12 at 18:17 show 7 more comments ## 4 Answers N. Bourbaki used $f\circ g$ with the interpretation $(f\circ g)(x)=f(g(x))$ in 1949 (Fonctions d'une variable réelle). Looking at the Bourbaki papers, I found this example from 1944 (middle of page 5), with the same interpretation. I haven't found any older examples, although I haven't tried very hard. (Van der Waerden does not use this notation in his Moderne Algebra from 1930.) It is certainly conceivable that the notation $f\circ g$ was invented by someone from the Bourbaki group. They were certainly very occupied with good notation, and André Weil introduced the modern symbol for the empty set in 1939 to be able to distinguish between $\emptyset$ and $0$. This notation for composition could have appeared from a similar discussion about $f(g(x))$ and $f(x)g(x)$. - I have never seen "$(f\circ g)(x)=g(f(x))$" in a math paper or book. Looking at a few of the results you quote from Google, the only academic papers I find there are either in computer science or engineering. Maybe the confusion arises from the following fact: in Spanish (and maybe in French?) which is the language I took my undergrad classes in, one reads "$f\circ g$" as "$g$ composed with $f$". I remember this was a source of confusion for many students. - I know of a professor in category theory/computer science who uses $(f\circ g)(x)$ to mean $g(f(x))$. He does this because when drawing a commutative diagram, the composition $f\circ g$ would be rendered $X\overset g \to Y\overset f \to Z$ and so he writes the composition as $gf$. – SL2 Apr 17 '12 at 18:03 Doesn't Herstein, in his book Topics in Algebra, use $(f\circ g)(x)=g(f(x))?$ I don't have a copy of his book with me now, but I seem to recall that in his book $fg$ means apply $f$ first, then apply $g$ to the result. (I had a 2-semester undergraduate algebra sequence from Herstein's text, but this was during the 1976-77 academic year.) Perhaps Herstein didn't use the $\circ$ notion, however ... I don't remember. – Dave L. Renfro Apr 17 '12 at 18:30 2 @AMPerrine: that makes sense, when one writes everything on the other side. – Martin Argerami Apr 17 '12 at 22:46 2 @ Martin Argerami: AMPerrine's quote can be found on p. 13 of the 2nd edition (1975) of Herstein's text. I recall that we used to say something like you have to drive on the other side of the road when reading Herstein. – Dave L. Renfro Apr 18 '12 at 19:17 1 FWIW, I don't think it is standard even for Spanish-speaking countries to read "$f\circ g$" as "$g$ composed with $f$"... In Uruguay, at least, we say "$f$ compuesto con $g$". Also, my secondary education was in French, and I remember reading "$f\circ g$" as "$f$ rond $g$" (which in retrospect is quite ugly!) – Bruno Stonek Jun 15 '12 at 1:30 show 4 more comments Typing $$\rm history\ mathematical\ notation$$ into Google turned up some sites that might be worth checking out, and it also gave Florian Cajori's book, A History of Mathematical Notations, which ought to be worth a look. - Unfortunately Cajori's two volumes do not contain the solution (unless I am too blind). – Wolfgang Mueckenheim Apr 24 '12 at 8:04 In dealing with categories and groupoids it is natural to write the composition of $$A \xrightarrow{f} B \xrightarrow{g} C$$ as $$A \xrightarrow{fg} C.$$ This convention is used in the book Higgins, P.J., Categories and Groupoids, Van Nostrand Reinhold Mathematical studies 32, Van Nostrand Reinhold, London, (1971); Reprints in Theory and Applications of Categories, No. 7 (2005) pp 1--195. (downloadable) and might be called the "algebraist's convention". I have found it very useful in dealing with double and higher categories and groupoids. It involves writing functions on the right as $(x)f$ as mentioned in other answers. This goes against the grain of course in dealing with the functions sin and log ! These ideas and notations, for example $x \mapsto x^2 +1$, evolved through trying to clarify the notion of function, and eventually finding both the domain and codomain were important, leading to a function being $f: A \to B$, with domain $A$ and codomain $B$. This arrow notation is one of the impetuses behind the notion of category. A further complication is that ordinary real number analysis and calculus is largely about partial functions $\mathbb R \to \mathbb R$, i.e. functions whose domain of definition is a subset of $\mathbb R$. - 4 How does that relate to the question who invented the $\circ$ notation is beyond me... – Asaf Karagila Jun 14 '12 at 21:41	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 68, "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": 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": 0.9612712860107422, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/74897/small-cycles-in-an-undirected-graph/75069	# Small cycles in an undirected graph Let $G=(V,E)$ be an undirected graph. I am using the usual convention that $n=\|V\|, m=\|E\|$. For $v \in V$, let $deg(v)$ be the degree of the vertex $v$. I am trying to show that if we have $m > \frac{n^{3/2}}{2}$, it follows that $G$ has either a triangle $C_3$ or a square $C_4$. I found a neat way that almost works, but it is not tight enough... Here is what I have done so far: First of all (not concerning the solution but a related problem) it is a neat example of the Cauchy-Riemann inequality to show that every graph $G$ with more than $n^2/4$ edges contains a triangle $C_3$ (this is Mantel's theorem). Now I did this (basically I reproduce a proof given by Erdős): I try to show if $G$ has no square $C_4$, then $m \leq \frac{n^{3/2}}{2}$. We can consider the amount $k$ of subgraphs of $G$ that look like $K_{1,2}$ (a "cherry"). Certainly if we (double) count those, considering a vertice in the center of the cherry, we get: $$k=\sum_{v\in V}\binom{\text{deg}(v)}{2}$$ Now if we don't allow squares a pair of vertices $\{v,w\}$ can be endpoints of only one cherry, otherwise we would have a cycle, therefore $$\binom{n}{2} \geq k=\sum_{v\in V}\binom{\text{deg}(v)}{2}$$ Then due to convexity of $\binom{x}{2}$ and the Jensen Inequality one might conclude now that $m \leq \frac{1}{4} \cdot \left(\sqrt{n^2 (4 n-3)}+n\right)$. Sadly after all the work I see that asymtotically this is enough but the bound I want to show is still a big tighter: I would be very glad if someone could give me an alternate proof that gives me the tight bounds or maybe refine what I did (I did not use the fact yet that $G$ is also triangle-free). - ## 1 Answer A proof can be found in: Garnick, David K., Kwong, Y. H. Harris, Lazebnik, Felix, Extremal graphs without three-cycles or four-cycles. J. Graph Theory 17 (1993), no. 5, 633–645. - You are right, thank you! I will try to understand the proof when I am back home, how did you find this :-)? I basically searched for hours and came up with nothing. – Listing Oct 23 '11 at 11:23	{"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": 22, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9508063793182373, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/14363/hydrostatic-friction-why-do-water-droplets-stay-at-rest-on-an-inclined-glass-su/14369	# Hydrostatic friction: why do water droplets stay at rest on an inclined glass surface? Tjis is a non-expert question on a (seemingly simple) text-book topic. The question is about "hydrostatic friction", defined as follows. Consider a drop of water resting on a flat surface. If the surface is slightly inclined, then the drop will not run off but just stay in place. Does this phenonemon have a simple description? "Simple" as in "surface tension is simply described by a constant $\gamma$ which gives energy per unit area, $dE = \gamma \ dA$" or "Coulombic friction force is equal to normal reaction force times the coefficient of static friction $\mu_s$." EDIT-1: First answer revived my hope for a simple gravity + surface-tension solution. If the glass plate were horizontal, the droplet is known "chooses" energetically optimal contact area. Now the same with tilt (gravity): 1. impose no-slip condition, 2. minimize total energy with fixed contact area, 3. compare two optimal shapes with slightly different contact areas. I hope there will be a critical angle beyond which the gain of gravitational energy overcomes losses to surface tension. Need more effort to write down an solve the variational problem (in cylindrical geometry for simplicity). EDIT-2: Found a recent review article the relevant subject: Rev.Mod.Phys. 81, 739 (2009); full text available on author's website. If this helps, will post an answer. - Now that's a good question! Compare to a meniscus (energy issues, which you are already thinking about) and to boundary layer (viscus friction). // too lazy to really think about this right now – dmckee♦ Sep 5 '11 at 17:41 – Slaviks Sep 5 '11 at 18:11 I was thinking that the drops shape would be driven in part by the tendency of the outward parts to get ahead of the inner parts if there was any motion (thus not quite minimum energy), but you could be right. – dmckee♦ Sep 5 '11 at 18:19 ## 2 Answers I know this is an old question, but for the benefit of people visiting here wondering what the answer was, here it goes: A droplet can stay at rest on an inclined plate because of small heterogeneities on the surface. This can either be a small roughness (of the order of nano/micrometers) or `dirty' spots where the surface chemistry is locally different. The existence of these heterogeneities allows the droplet to have a different, bigger, contact angle at the downhill side then at the uphill side (this means that a perfectly smooth, clean surface will not be able to hold any droplets, this can most easily be shown by numerical simulation, because a perfectly smooth and clean surface is very hard to find/make experimentally). This difference in contact angle, thus surface energy, translates into a force pointing upwards, which is therefore able to counteract the gravitational force. Because there is a certain maximum to the front and rear contact angles (which depends on the surface roughness/dirtyness, higher roughness/dirtyness will give a larger difference between the two) the droplet will at some point start to move, at which point you get into the paradoxical discussion that Ron Maimon was talking about. The balance that you get looks like this: $$\rho V g \sin \alpha = k \gamma w (\cos\theta_u - \cos \theta_d)$$ where, in order of appearance, you have the density of the liquid, the volume of the droplet, the gravitational acceleration, the sin of the tilt angle, a `fudge' factor of O(1) depending on the detailed shape of the droplet, the width of the base of the droplet and the cosines of the uphill contact angle and the downhill angle. The phenomenon of the difference in downhill and uphill contact angle is called contact angle hysteresis. If you want to know more about it, you can google for it: there are plenty of good sources out there. To summarize: local heterogeneities on the surface allow a difference in surface energy at the front and rear of the droplet, thus introducing a force that can counteract gravity. - 1 Fantastic! Thanks for a really useful answer. "contact angle hysteresis" is really the key. – Slaviks Feb 11 at 19:34 This is a well known paradox: a drop in continuum mechanics can never slide. The reason is that there are no-slip boundary conditions between the fluid and the solid, and this no-slip condition plus the contact-angle condition requires the drop cannot move. The surprise is not that it doesn't move for a while, but that it starts to move at all. This means that there is a violation of no slip in the boundary line, and it is a manifestation of the finite atomic scale, the nonzero Boltzmann's constant. - Aren't these two related but different paradoxes? In the completely static case, I wonder if gravity has any sizebale effect on the contact angle? This probablyis related to the violation of non-slip conditon as the thing starts moving, and thus has something to do with the breakdown of macroscopic continuous dynamics approximations at the contact line. But why do you say "nonzero Boltzmann's constant"? Does it imply that thermal fluctuation are vital here? – Slaviks Sep 5 '11 at 19:10 I am saying this from memory--- I saw a seminar on this a long time ago, I haven't worked it out myself. It is clear that the no-slip has to be violated to get the droplet to move, since no slip holds on the contact region and bending the surface to extend it requires overcoming the surface tension. But the continuum analysis I saw did not take into account Van-der-Waals attraction of the bulk water to the glass, and this might account for the rolling without requiring a violation of the continuum limit (like a no-slip violation) which reveals the mean-free-path. I will think about it. – Ron Maimon Sep 6 '11 at 1:41 Maybe gravity alone is able to overcome surface tension and bend the liquid over the critical angle if the drop is large enough? Will try to construct a variational energy balance staying with surface (surface tension) and bulk (gravity) terms. – Slaviks Sep 6 '11 at 5:56 2 "It can never slide" does not imply that it will stay at rest. It is intuitive that the drop will roll rather than slide. You seem to argue this is prevented by the contact angle, but how? – leftaroundabout Sep 7 '11 at 1:45 The argument I saw said that the contact angle never can become flat, because of the balance of surface tension and gravity. I don't know if it is true, because Van-der-Waals might allow it to roll. I am not certain of the answer yet, I am still thinking about it--- sorry. – Ron Maimon Sep 7 '11 at 3:06 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": 3, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9252856969833374, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/74188/finding-centre-and-radius-of-circle/74205	# Finding centre and radius of circle Let $a,c \in \mathbb R$ with $a \neq 0$, and let $b \in \mathbb C$. Define $$S=\{z\in \mathbb C: az\bar{z}+b\bar{z}+\bar{b}z+c=0\}.$$ a. Show that $S$ is a circle, if $\|b\|^2 > ac$. Determine its centre and radius. b. What is $S$ if $a=0$ and $b \neq 0$? How would I get started in this? I'm completely stuck. Most appreciated. - 2 Start by writing $z=x+yi$, $b=p+qi$ and expand. – Henning Makholm Oct 20 '11 at 2:29 ## 2 Answers Where $z = x + iy$, your equation can be rewritten as $$a(x^2 + y^2) + 2Re(b\bar{z}) + c = 0$$ Writing $b = b_1 + b_2i$ this is the same as $$a(x^2 + y^2) + 2b_1x + 2b_2y + c = 0$$ Complete the square now.... - Thanks got it, appreciated! – John Southall Oct 21 '11 at 9:02 One way is to work your way backward to the problem from what you know about circles. A circle has a center and a radius. Let's call the center $w$, the radius, $r$. The circle is all the points $z$ whose distance from $w$ is $r$. The square of the distance between two points $u$ and $v$ in the complex plane is $\|u-v\|^2$. If $s$ is a complex number, then $\|s\|^2=s\overline s$. If you put all that together, you should come up with a formula looking very similar to the one in the problem, mostly just with different letters. -	{"extraction_info": {"found_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": 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": 0.9461515545845032, "perplexity_flag": "head"}
http://mathhelpforum.com/trigonometry/153358-solutions-triangle.html	# Thread: 1. ## solutions of a triangle prove that: $\:\dfrac{c}{a+b}\:\ = \:\dfrac{1-tan\:\dfrac{A}{2}\:\tan\:\dfrac{B}{2}\ }{1 +tan\:\dfrac{A}{2}\:\tan\:\dfrac{B}{2} }$ where $\:\boxed{a,b,c}$ are the sides opposite to the angles $\:\boxed{A,B,C}$ respectively in a triangle 2. Let $P , Q , R$ be the points of tangency of the incircle to the sides $BC , CA , AB$ respectively . Then $CP = CQ = (a+b-c)/2 = r \cot(\frac{C}{2})$ where $r$ is the inradius . Also , $c = AR + BR = r [ \cot(\frac{A}{2}) + \cot(\frac{B}{2}) ]$ But $\cot(\frac{C}{2}) = \tan( \frac{A+B}{2} ) = \tan(\frac{A}{2}) \tan(\frac{B}{2}) ~ \frac{ \cot(\frac{A}{2}) + \cot(\frac{B}{2}) }{ 1 - \tan(\frac{A}{2}) \tan(\frac{B}{2}) }$ Therefore , $\frac{ a+b-c}{c} = 2r \tan(\frac{A}{2}) \tan(\frac{B}{2}) ~ \frac{ \cot(\frac{A}{2}) + \cot(\frac{B}{2}) }{ 1 - \tan(\frac{A}{2}) \tan(\frac{B}{2}) } ~ \frac{1}{ r[ \cot(\frac{A}{2}) + \cot(\frac{B}{2}) ]}$ $= \frac{2\tan(\frac{A}{2}) \tan(\frac{B}{2}) }{ 1 - \tan(\frac{A}{2}) \tan(\frac{B}{2}) }$ $\frac{ a+b}{c} = \frac{1 + \tan(\frac{A}{2}) \tan(\frac{B}{2}) }{1-\tan(\frac{A}{2}) \tan(\frac{B}{2}) }$ or $\boxed{ \displaystyle{ \frac{c}{a+b} = \frac{ 1 -\tan(\frac{A}{2}) \tan(\frac{B}{2}) }{1+ \tan(\frac{A}{2}) \tan(\frac{B}{2}) } }}$ 3. Originally Posted by simplependulam to grgrsanjay; Let $\:\boxed{P , Q , R }$ be the points of tangency of the incircle to the sides $BC , CA , AB$ respectively . what does tangency to a circle mean actually??? 4. Originally Posted by grgrsanjay what does tangency to a circle mean actually???	{"extraction_info": {"found_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": 15, "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": 0.851590633392334, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/64420/is-there-a-faster-way-to-calculate-a-few-diagonal-elements-of-the-inverse-of-a-h	# Is there a faster way to calculate a few diagonal elements of the inverse of a huge symmetric positive definite matrix? I asked this on SO first, but decided to move the math part of my question here. Consider a $p \times p$ symmetric and positive definite matrix $\bf A$ (p=70000, i.e. $\bf A$ is roughly 40 GB using 8-byte doubles). We want to calculate the first three diagonal elements of the inverse matrix $({\bf A^{-1}})_{11}$, $({\bf A^{-1}})_{22}$ and $({\bf A^{-1}})_{33}$. I have found this paper by James R. Bunch who seems to solve this exact problem without calculating the full inverse $\bf A^{-1}$. If I understand it correctly he first calculates the Cholesky decomposition, i.e. the upper triangular matrix $\bf R$ which satisfies $\bf A=R^T R$, which needs $\frac16p^2+\frac12p^2-\frac23p$ floating point operations (multiplications/divisions) using the LINPACK function SPOFA. He then proceeds to calculate individual diagonal elements of the inverse $({\bf A^{-1}})_{ii}$ using an expression which exploits the sparsity of ${\bf R}^T{\bf y}={\bf e}_j$ and which requires $\frac12(p-i)^2+\frac52(p-i)+2$ floating point operations. (I don't understand the full details of this, so I can't currently sum it up correctly). The paper is based on LINPACK; it isn't cited by anyone, so it seems nobody cared for the last 23 years? After reading this, I'm wondering whether this is still the best way of doing things, or whether a modern LAPACK-based approach could avoid the Cholesky decomposition? In short, is there a quicker way to calculate those diagonal elements of the inverse? - 1 Maybe you can give some details on the actual problem you're trying to solve. There's a chance it might have some special structure that you could take advantage of. – cardinal Sep 14 '11 at 9:44 @Jonas Heidelberg: I am terribly sorry, but I think I may have given you an incorrect solution to your problem. I hope it didn't waster too much of your time. – talonmies Sep 14 '11 at 15:01 2 You matrix is sparse? – leonbloy Sep 14 '11 at 17:14 @leonbloy: I suspect that the matrix is approximately sparse, i.e. that $\bf A_{10}\approx 0$ (using notation from talonmies' answer). In fact in my current implementation I just invert $\bf A_{00}$, but this might not be precise enough. – Jonas Heidelberg Sep 14 '11 at 17:30 – leonbloy Sep 14 '11 at 17:58 show 1 more comment ## 3 Answers I suggested that a partial Cholesky decomposition should be possible here, but based on a comment I received, I started thinking about my answer again and realised that my suggestion was incorrect. Apologies if I lead anyone astray. You will need to use a full Cholesky decomposition, but the problem to deduce the inverse can be reduced in scale to save redundant computation. If your SPD matrix $\mathbf{A}$ is written in block form as $\mathbf{A}=\begin{bmatrix} \mathbf{A}_{00} & \mathbf{A}_{10}^T \\ \mathbf{A}_{10} & \mathbf{A}_{11} \end{bmatrix}$ with $\mathbf{A}_{00}$ being a $3\times3$ SPD block, its inverse will be given by $\mathbf{A}^{-1}=\begin{bmatrix} \mathbf{B}_{00} & \mathbf{B}_{10}^T \\ \mathbf{B}_{10} & \mathbf{B}_{11} \end{bmatrix}$ The equivalent Cholesky decomposition of $\mathbf{A}$ is then given by $\mathbf{R}=\begin{bmatrix} \mathbf{R}_{00} & \\ \mathbf{R}_{10} & \mathbf{R}_{11} \end{bmatrix}$ The resulting matrix equation to solve for the inverse block matrix $\mathbf{B}_{00}$ can then be reduced to $\begin{bmatrix} \mathbf{R}_{00} & \\ \mathbf{R}_{10} & \mathbf{R}_{11} \end{bmatrix} \begin{bmatrix} \mathbf{R}_{00}^T & \mathbf{R}_{10}^T\\ & \mathbf{R}_{11}^T \end{bmatrix} \begin{bmatrix} \mathbf{B}_{00} \\ \mathbf{B}_{10} \end{bmatrix}=\begin{bmatrix}\mathbf{I}_{00} \\ \mathbf{0} \end{bmatrix}$ and solved via $\begin{bmatrix} \mathbf{R}_{00} & \\ \mathbf{R}_{10} & \mathbf{R}_{11} \end{bmatrix} \begin{bmatrix} \mathbf{X}_{0} \\ \mathbf{X}_{1} \end{bmatrix} =\begin{bmatrix}\mathbf{I}_{00} \\ \mathbf{0} \end{bmatrix}$ and $\begin{bmatrix} \mathbf{R}_{00}^T & \mathbf{R}_{10}^T\\ & \mathbf{R}_{11}^T \end{bmatrix} \begin{bmatrix} \mathbf{B}_{00} \\ \mathbf{B}_{10} \end{bmatrix}=\begin{bmatrix}\mathbf{X}_{0} \\ \mathbf{X}_{1} \end{bmatrix}$ The solution you are looking for will be the diagonal entries of $\mathbf{B}_{00}$. The right hand side of this expression is only $n\times 3$, and this combined with lower triangular structure of $\mathbf{R}$ should be about the most efficient way to get the solution you require. - I find it funny that Google gives your answer as the second hit for `"partial Cholesky decomposition"`, 36 minutes after you posted... – Jonas Heidelberg Sep 14 '11 at 9:49 Using Wikipedia and numerical recipes it should be trivial to interrupt the decomposition after 3 columns, so I'm marking yours as the accepted answer :-). – Jonas Heidelberg Sep 14 '11 at 10:15 Okay, I think I understand you new explanation (revision 6), that makes sense. So your last equation shows that $R_{00}$ is not enough to calculate $B_{00}$, since the equations for $B_{00}$ also involve $R_{10}$ :-/. – Jonas Heidelberg Sep 14 '11 at 17:49 Not quite. The incomplete factorization also gives $R_{10}$, but $R_{11}$ is also required to solve for $B_{00}$. That implies the full factorization is necessary. – talonmies Sep 14 '11 at 18:34 @talonmies: I've followed your model and have now a little follow-up-question. I'll put it in a formal "Answer" because it seems, this comment-field is too small – Gottfried Helms Sep 15 '11 at 12:08 show 2 more comments I suggest you look at the book Matrices, Moments, and Quadrature with Applications. The basic idea is that you have a $n \times n$ positive semi-definite matrix $A$, and you wish to compute the quadratic form $u^T f(A)u$, where $u \in \mathbb{R}^n$. In your case you wish to compute $$A^{-1}_{ii} = e_i^T f(A)e_i,$$ where $e_i \in \mathbb{R}^n$ is a vector of all zeros, with a 1 in the $i$th position, and the function $f$ is defined as $f(\lambda) = \lambda^{-1}$. You can turn the quadratic form into an Stieltjes integral, since $$u^T f(A)u = u^T Q^T f(\Lambda) Qu = \alpha^T f(\Lambda) \alpha = \sum_{i=1}^n f(\lambda_i) \alpha_i^2 = \int_a^b f(\lambda) d\alpha(\lambda),$$ where the eigenvalue decomposition of $A$ is given by $A = Q\Lambda Q^T$ and $\Lambda$ is a diagonal matrix containing the eigenvalues of $A$, and the vector $\alpha = Qu$. The integral can be estimated using Gaussian quadrature via $$I[f] = \int_a^b f(\lambda) d \alpha(\lambda) = \sum_{i=1}^N \omega_i f(t_i) + R[f],$$ where $\{\omega_i\}_{i=1}^N$ are a set of weights, and $\{t_i\}_{i=1}^N$ are a set of nodes at which to evaluate the function $f$, and $R[f]$ is a remainder or error term. The values of the $\omega$'s and $t$'s are unknown and must be solved for. The values for the weights may be computed via an iterative algorithm similar to the Lanczos algorithm for computing eigenvalues. The values of the nodes may be obtained from components of an eigenvector of a matrix derived from $A$. This computation may be done efficiently. For more details see the book, as well as this lecture on the topic by James Lambers. The underlying mathematics and linear algebra may seem a little scary at first, but I assure you this leads to a fairly simple and efficient algorithm. I wrote Matlab code to calculate the weights and nodes for a class in graduate school. It wasn't very difficult. Take a look at the book. Good luck. - This looks interesting as well. Too bad I won't have the time to go into this method right now :-(. Just to make sure I get the basics right - to implement this you repeatedly need to evaluate $x^TAx$ for different x, right? And this is faster than the Bunch algorithm at the expense of calculating an approximation? – Jonas Heidelberg Sep 14 '11 at 23:08 – J. M. Sep 15 '11 at 0:31 – codehippo Sep 15 '11 at 6:19 The first part here below is not an "answer" but an extended comment/followup-question to talonmies' previous answer/comment I've extended your notation a bit and got a clearer exposition: First let us extend the X-matrix to be the inverse of the full R-matrix, such that $\small \mathbf A^{-1}= \mathbf B =\mathbf X^\tau \mathbf X$ and let us also extend the indexes for X , then $\qquad \small \mathbf X_{00} =\mathbf R_{00}^{-1}$, $\small \qquad \mathbf X_{11} =\mathbf R_{11}^{-1} \quad$ but unfortunately $\small \mathbf X_{10} \ne \mathbf R_{10}^{-1}$ . By the inverse-relation $\small \mathbf X =\mathbf R^{-1}$ we have, that $\small \mathbf X_{10} \mathbf R_{00}+ \mathbf X_{11} \mathbf R_{10} =\mathbf 0$ and thus $\small \mathbf X_{10} = - \mathbf X_{11} \mathbf R_{10} \mathbf R_{00}^{-1}$ (from which we want to receive the top-left diagonal elements from B by $\small \mathbf X^\tau_{00} \mathbf X_{00} + \mathbf X^\tau_{10} \mathbf X_{10} = \mathbf B_{00}$ ) Here $\small \mathbf X_{11}$ represents the part of the problem for which we want to reduce the computational effort, because this is the inversion of the (huge) remaining part of the cholesky-factor.I thought to introduce the concept of the pseudoinverse pinv(X) by which we could proceed: $\small \begin{array} {rcl} \mathbf Y &=& - ( \mathbf X_{11} \mathbf R_{10} \mathbf R_{00}^{-1} )^{-1} \\ &=& - \mathbf R_{00} \operatorname{pinv} (\mathbf R_{10}) \mathbf R_{11} \\ \mathbf X_{10} &=& \operatorname{pinv} ( \mathbf Y) \end{array}$ This would allow to invert small matrices only, but although the pseudoinverse-operations works well in many cases I could not make it work correctly for the pinv(R10)-part; for all variants of my computations I always needed the full version R11 in this or that version. Do you see any general reason, why the pinv(R10) is not working sufficiently here? A practical computation Hmm, as far as we accept that we need the full cholesky-decomposition anyway, the process might even be described shorter: a) consider the symmetric SPD-matrix A , we look for the diagonal-elements of the 3x3-submatrix $\small \mathbf B_{00}$ of $\small \mathbf B = \mathbf A^{-1}$ b) perform the cholesky-decomposition from bottom up into the matrix R; denote $\small \mathbf R_{00}$ the top left 3x3 upper triangular submatrix (thinking in terms of a correlation-matrix A this represents the partial or "unexplained" variance/covariance) c) invert $\small \mathbf R_{00}$ to get $\small \mathbf X_{00}$ (this is also cheap because it's already triangular) d) in $\small \mathbf B_{00} = \mathbf X_{00}^\tau \mathbf X_{00}$ we find the required diagonal elements. (This last operation can even be replaced by a simple summing of squares along the columns in X ) Here is a complete example usable for Pari/GP. To have it as clear as possible no optimizations, errorchecks etc are done: ```Crop(M,dim) = matrix(dim,dim,r,c,M[r,c]) \\ reduces size of a matrix \\ procedure to show the first dim diagonal-entries of A^-1 {invdiag(A,dim=3)=local(rs=rows(A),lV,R,X); \\ reduce A by a cholesky-process to a dim x dim-residual matrix A_00 \\ the cholesky-process goes bottom-up forstep( d = rs, dim+1, -1, lV = A[,d]/sqrt(A[d,d]); A = Crop( A - lVlV~ , d-1 ); ); \\after this A is the dim x dim residual-matrix \\ compute the cholesky-factor of that A-residual into matrix R \\ we just proceed working bottom up, but A is no more cropped now R = matrix(dim,dim); forstep( d = dim, 1, -1, R[,d] = A[,d]/sqrt(A[d,d]); A = A - R[,d] R[,d]~; ); X = R^-1; \\ inverse of R of size dim x dim \\ and to extract values of diag(A^-1) =diag( R^-1~ * R^-1 ) \\ it suffices to sum the squares along the columns lv = vector(dim,c,sum(r=1,c, X[r,c]^2) ); return(lv); } \\ create a sample matrix A A = matrix(8,8,r,c,binomial(r-1,c-1)/2^(c-1)) A = A* A~ \\ make it a symmetric positive definite one print(invdiag(A,3)) \\ show the first 3 results by the partial cholesky-method print(diag(A^-1)) \\ show the true results by inversion of the full matrix \\ output: [21845.0000000, 980612.000000, 8259152.00000] [21845, 980612, 8259152, 21815360, 21017856, 7373824, 806912, 16384]~ ``` -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 73, "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": 0.9314477443695068, "perplexity_flag": "head"}
http://en.wikipedia.org/wiki/Logistic_map	# Logistic map The logistic map for 100 generations of x (plotted left to right) as r moves from 0 to 4 The logistic map is a polynomial mapping (equivalently, recurrence relation) of degree 2, often cited as an archetypal example of how complex, chaotic behaviour can arise from very simple non-linear dynamical equations. The map was popularized in a seminal 1976 paper by the biologist Robert May, in part as a discrete-time demographic model analogous to the logistic equation first created by Pierre François Verhulst.[1] Mathematically, the logistic map is written $(1)\qquad x_{n+1} = r x_n (1-x_n)$ where: $x_n$ is a number between zero and one, and represents the ratio of existing population to the maximum possible population at year n, and hence x0 represents the initial ratio of population to max. population (at year 0) r is a positive number, and represents a combined rate for reproduction and starvation. This nonlinear difference equation is intended to capture two effects. • reproduction where the population will increase at a rate proportional to the current population when the population size is small. • starvation (density-dependent mortality) where the growth rate will decrease at a rate proportional to the value obtained by taking the theoretical "carrying capacity" of the environment less the current population. However, as a demographic model the logistic map has the pathological problem that some initial conditions and parameter values lead to negative population sizes. This problem does not appear in the older Ricker model, which also exhibits chaotic dynamics. The r=4 case of the logistic map is a nonlinear transformation of both the bit shift map and the $\mu =2$ case of the tent map. ## Behavior dependent on r By varying the parameter r, the following behavior is observed: • With r between 0 and 1, the population will eventually die, independent of the initial population. • With r between 1 and 2, the population will quickly approach the value $\frac{r-1}{r}$, independent of the initial population. • With r between 2 and 3, the population will also eventually approach the same value $\frac{r-1}{r}$, but first will fluctuate around that value for some time. The rate of convergence is linear, except for r=3, when it is dramatically slow, less than linear. • With r between 3 and $1+\sqrt{6}$ (approximately 3.44949), from almost all initial conditions the population will approach permanent oscillations between two values. These two values are dependent on r. • With r between 3.44949 and 3.54409 (approximately), from almost all initial conditions the population will approach permanent oscillations among four values. The latter number is a root of a 12th degree polynomial (sequence in OEIS). • With r increasing beyond 3.54409, from almost all initial conditions the population will approach oscillations among 8 values, then 16, 32, etc. The lengths of the parameter intervals which yield oscillations of a given length decrease rapidly; the ratio between the lengths of two successive such bifurcation intervals approaches the Feigenbaum constant δ = 4.66920$\dots$. This behavior is an example of a period-doubling cascade. • At r approximately 3.56995 (sequence in OEIS) is the onset of chaos, at the end of the period-doubling cascade. From almost all initial conditions we can no longer see any oscillations of finite period. Slight variations in the initial population yield dramatically different results over time, a prime characteristic of chaos. • Most values beyond 3.56995 exhibit chaotic behaviour, but there are still certain isolated ranges of r that show non-chaotic behavior; these are sometimes called islands of stability. For instance, beginning at $1+\sqrt{8}$ (approximately 3.82843) there is a range of parameters r which show oscillation among three values, and for slightly higher values of r oscillation among 6 values, then 12 etc. • The development of the chaotic behavior of the logistic sequence as the parameter r varies from approximately 3.56995 to approximately 3.82843 is sometimes called the Pomeau–Manneville scenario, which is characterized by a periodic (laminar) phase interrupted by bursts of aperiodic behavior. Such a scenario has an application in semiconductor devices.[2] There are other ranges which yield oscillation among 5 values etc.; all oscillation periods occur for some values of r. A period-doubling window with parameter c is a range of r-values consisting of a succession of sub-ranges. The kth sub-range contains the values of r for which there is a stable cycle (a cycle which attracts a set of initial points of unit measure) of period $c2^{k}.$ This sequence of sub-ranges is called a cascade of harmonics.[3] In a sub-range with a stable cycle of period $c2^{k^{}},$ there are unstable cycles of period $c2^{k}$ for all $k<k^{}.$ The r value at the end of the infinite sequence of sub-ranges is called the point of accumulation of the cascade of harmonics. As r rises there is a succession of new windows with different c values. The first one is for c = 1; all subsequent windows involving odd c occur in decreasing order of c starting with arbitrarily large c.[3][4] • Beyond r = 4, the values eventually leave the interval [0,1] and diverge for almost all initial values. For any value of r there is at most one stable cycle. A stable cycle attracts almost all points.[5]:13 For an r with a stable cycle of some period, there can be infinitely many unstable cycles of various periods. A bifurcation diagram summarizes this. The horizontal axis shows the values of the parameter r while the vertical axis shows the possible long-term values of x. Bifurcation diagram for the logistic map The bifurcation diagram is a self-similar: if you zoom in on the above mentioned value r = 3.82843 and focus on one arm of the three, the situation nearby looks like a shrunk and slightly distorted version of the whole diagram. The same is true for all other non-chaotic points. This is an example of the deep and ubiquitous connection between chaos and fractals. ## Chaos and the logistic map Two- and three-dimensional phase diagrams show the stretching-and-folding structure of the logistic map. A cobweb diagram of the logistic map, showing chaotic behaviour for most values of r > 3.57. Logistic function for r=3.5 after first 3 iterations The relative simplicity of the logistic map makes it an excellent point of entry into a consideration of the concept of chaos. A rough description of chaos is that chaotic systems exhibit a great sensitivity to initial conditions—a property of the logistic map for most values of r between about 3.57 and 4 (as noted above). A common source of such sensitivity to initial conditions is that the map represents a repeated folding and stretching of the space on which it is defined. In the case of the logistic map, the quadratic difference equation (1) describing it may be thought of as a stretching-and-folding operation on the interval (0,1). The following figure illustrates the stretching and folding over a sequence of iterates of the map. Figure (a), left, gives a two-dimensional phase diagram of the logistic map for r=4, and clearly shows the quadratic curve of the difference equation (1). However, we can embed the same sequence in a three-dimensional phase space, in order to investigate the deeper structure of the map. Figure (b), right, demonstrates this, showing how initially nearby points begin to diverge, particularly in those regions of Xt corresponding to the steeper sections of the plot. This stretching-and-folding does not just produce a gradual divergence of the sequences of iterates, but an exponential divergence (see Lyapunov exponents), evidenced also by the complexity and unpredictability of the chaotic logistic map. In fact, exponential divergence of sequences of iterates explains the connection between chaos and unpredictability: a small error in the supposed initial state of the system will tend to correspond to a large error later in its evolution. Hence, predictions about future states become progressively (indeed, exponentially) worse when there are even very small errors in our knowledge of the initial state. Since the map is confined to an interval on the real number line, its dimension is less than or equal to unity. Numerical estimates yield a correlation dimension of 0.500 ± 0.005 (Grassberger, 1983), a Hausdorff dimension of about 0.538 (Grassberger 1981), and an information dimension of 0.5170976... (Grassberger 1983) for r=3.5699456... (onset of chaos). Note: It can be shown that the correlation dimension is certainly between 0.4926 and 0.5024. It is often possible, however, to make precise and accurate statements about the likelihood of a future state in a chaotic system. If a (possibly chaotic) dynamical system has an attractor, then there exists a probability measure that gives the long-run proportion of time spent by the system in the various regions of the attractor. In the case of the logistic map with parameter r = 4 and an initial state in (0,1), the attractor is also the interval (0,1) and the probability measure corresponds to the beta distribution with parameters a = 0.5 and b = 0.5. Specifically,[6] the invariant measure is $\pi ^{-1}x^{-1/2}(1-x)^{-1/2}$. Unpredictability is not randomness, but in some circumstances looks very much like it. Hence, and fortunately, even if we know very little about the initial state of the logistic map (or some other chaotic system), we can still say something about the distribution of states a long time into the future, and use this knowledge to inform decisions based on the state of the system. ## Solution in some cases The special case of r = 4 can in fact be solved exactly, as can the case with r = 2;[7] however the general case can only be predicted statistically.[8] The solution when r = 4 is,[7][9] $x_{n}=\sin^{2}(2^{n} \theta \pi)$ where the initial condition parameter $\theta$ is given by $\theta = \tfrac{1}{\pi}\sin^{-1}(x_0^{1/2})$. For rational $\theta$, after a finite number of iterations $x_n$ maps into a periodic sequence. But almost all $\theta$ are irrational, and for irrational $\theta$ $x_n$ never repeats itself – it is non-periodic. This solution equation clearly demonstrates the two key features of chaos – stretching and folding: the factor 2n shows the exponential growth of stretching, which results in sensitive dependence on initial conditions, while the squared sine function keeps $x_n$ folded within the range [0, 1]. By contrast, the solution when r=2 is $x_n = \frac{1}{2} - \frac{1}{2}(1-2x_0)^{2^{n}}$ for $x_0 \in [0,1)$. Since $(1-2x_0)\in (-1,1)$ for any value of $x_0$ other than the unstable fixed point 0, the term $(1-2x_0)^{2^{n}}$ goes to 0 as n goes to infinity, so $x_n$ goes to the stable fixed point $\tfrac{1}{2}.$ ## Finding cycles of any length when r = 4 For the r = 4 case, from almost all initial conditions the iterate sequence is chaotic. Nevertheless, there exist an infinite number of initial conditions which lead to cycles, and indeed there exist cycles of length k for all integers k ≥ 1. We can exploit the relationship of the logistic map to the dyadic transformation (also known as the bit-shift map) to find cycles of any length. If x follows the logistic map $x_{n+1} = 4 x_n(1-x_n) \,$ and y follows the dyadic transformation $y_{n+1}=\begin{cases}2y_n & 0 \le y_n < 0.5 \\2y_n -1 & 0.5 \le y_n < 1, \end{cases}$ then the two are related by $x_{n}=\sin^{2}(2 \pi y_{n})$. The reason that the dyadic transformation is also called the bit-shift map is that when y is written in binary notation, the map moves the binary point one place to the right (and if the bit to the left of the binary point has become a "1", this "1" is changed to a "0"). A cycle of length 3, for example, occurs if an iterate has a 3-bit repeating sequence in its binary expansion (which is not also a one-bit repeating sequence): 001, 010, 100, 110, 101, or 011. The iterate 001001001... maps into 010010010..., which maps into 100100100..., which in turn maps into the original 001001001...; so this is a 3-cycle of the bit shift map. And the other three binary-expansion repeating sequences give the 3-cycle 110110110... → 101101101... → 011011011... → 110110110.... Either of these 3-cycles can be converted to fraction form: for example, the first-given 3-cycle can be written as 1/7 → 2/7 → 4/7 → 1/7. Using the above translation from the bit-shift map to the r = 4 logistic map gives the corresponding logistic cycle .611260467... → .950484434... → .188255099... → .611260467... . We could similarly translate the other bit-shift 3-cycle into its corresponding logistic cycle. Likewise, cycles of any length k can be found in the bit-shift map and then translated into the corresponding logistic cycles. However, since almost all numbers in [0, 1) are irrational, almost all initial conditions of the bit-shift map lead to the non-periodicity of chaos. This is one way to see that the logistic r = 4 map is chaotic for almost all initial conditions. Amount of cycles of (minimal) length k for logistic map with r = 4 (tent map with $\mu =2$) is a known integer sequence (sequence in OEIS): 2, 1, 2, 3, 6, 9, 18, 30, 56, 99, 186, 335, 630, 1161 ... It tells us that logistic map with r = 4 has 2 fixed points, 1 cycle of length 2, 2 cycles of length 3 and so on. This sequence takes a particularly simple form for prime k: $2 (2^{k-1}-1)/k$. For example: $2 (2^{13-1}-1)/13 = 630$ is the number of cycles of length 13. ## See also • Malthusian growth model • Chaos theory • List of chaotic maps • Logistic function the continuous version • Stiff equation • Lyapunov stability for iterated systems • Radial basis function network This article illustrates the inverse problem for the logistic map. • Complex quadratic map • Schröder's equation ## References 1. Carson Jeffries; Jose Perez (1982). "Observation of a Pomeau–Manneville intermittent route to chaos in a nonlinear oscillator". 26 (4): 2117–2122. Bibcode:1982PhRvA..26.2117J. doi:10.1103/PhysRevA.26.2117. 2. ^ a b R.M. May (1976). "Simple mathematical models with very complicated dynamics". Nature 261 (5560): 459–67. Bibcode:1976Natur.261..459M. doi:10.1038/261459a0. PMID 934280. 3. Baumol, William, and Benhabib, Jess, "Chaos: Significance, mechanism, and economic applications," Journal of Economic Perspectives 3, Winter 1989, 77-106. 4. Collet, Pierre, and Jean-Pierre Eckmann, Iterated Maps on the Interval as Dynamical Systems, Birkhauser, 1980. 5. Jakobson, M.,"Absolutely continuous invariant measures for one-parameter families of one-dimensional maps," Communications in Mathematical Physics 81, 1981, 39-88. 6. ^ a b Schröder, Ernst (1870). "Über iterierte Funktionen". Math. Ann. 3 (2): 296–322. doi:10.1007/BF01443992. 7. Little, M., Heesch, D. (2004). "Chaotic root-finding for a small class of polynomials" (PDF). Journal of Difference Equations and Applications 10 (11): 949–953. doi:10.1080/10236190412331285351. 8. Lorenz, Edward (1964), "The problem of deducing the climate from the governing equations," Tellus 16 (February): 1-11. ### Textbooks • Sprott, Julien Clinton (2003). Chaos and Time-Series Analysis. Oxford University Press. ISBN 0-19-850840-9. • Strogatz, Steven (2000). Nonlinear Dynamics and Chaos. Perseus Publishing. ISBN 0-7382-0453-6. • Tufillaro, Nicholas; Tyler Abbott, Jeremiah Reilly (1992). An experimental approach to nonlinear dynamics and chaos. Addison-Wesley New York. ISBN 0-201-55441-0. ### Journal articles • P. Grassberger and I. Procaccia (1983). "Measuring the strangeness of strange attractors". Physica D 9 (1–2): 189–208. Bibcode:1983PhyD....9..189G. doi:10.1016/0167-2789(83)90298-1. • P. Grassberger (1981). "On the Hausdorff dimension of fractal attractors". Journal of Statistical Physics 26 (1): 173–179. Bibcode:1981JSP....26..173G. doi:10.1007/BF01106792.	{"extraction_info": {"found_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": 35, "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": 0.8743958473205566, "perplexity_flag": "middle"}
http://physics.aps.org/articles/large_image/f1/10.1103/Physics.3.107	Credit: Alan Stonebraker Figure 1: (Top) Experimental configuration used by Minardi et al. to produce light bullets. $1550$-$nm$ laser pulses with $170$-$fs$ duration go into a hexagonal array of glass waveguides. The nonlinear optical interaction in the glass causes formation of solitons that are stable in time and space. Part of the original laser beam is split off and is used to map out the pulse shape. (Bottom) Upper panel shows how a pulse undergoes spreading in time during propagation in the waveguide array. Lower panel shows how nonlinear effects cause the light packet to remain stable and compact as it moves.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 4, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "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": 0.9062619805335999, "perplexity_flag": "middle"}
http://gilkalai.wordpress.com/2008/09/16/a-diameter-problem-5/?like=1&source=post_flair&_wpnonce=81d32a40f7	Gil Kalai’s blog ## A Diameter Problem (5) Posted on September 16, 2008 by ### 6. First subexponential bounds. Proposition 1: $F(d,n) \le F_k(d,n) \times F(d-k,n-k).$ How to prove it: This is easy to prove: Given two sets $S$ and $T$ in our family $\cal F$, we first find a path of the form $S=S_0, S_1, S_2, \dots, S_t = T$ where, $\|S_{i-1} \cap S_i\| \ge k$ and $t \le F_k(d,n)$. We let $B_i \subset (S_{i-1} \cap S_i )$ with $\|B_i\|=k$ and consider the family ${\cal F}'[B_i]$. This is a family of $(d-k)$-subsets of an $(n-k)$ set ($N \backslash B_i$) . It follows that we can have a path from $S_{i-1}$ to $S_i$ in $G({\cal F}[B_i])$ of length at most $F(d-k,n-k)$. Putting all these paths together gives us the required result. (We remind the notations at the end of this post.) How to use it: It is not obvious how to use Proposition 1. Barnette’s argument from part 3 was about $k=1$, and it used something a bit more sophisticated. Applying Proposition 1 directly for $k=1$ does not give anything non trivial. However, when $n$ is small compared to $d$ and $k$ is a small fraction of $d$ we can say something. Let us start with an example: $n = 3d$. let us choose $k= d/4$. Consider a path $S=S_0,S_1,\dots S_t=T$ in $G({\cal F})$ from two sets $S$ and $T$. Suppose also that in this path () $S_i \cap T \subset S_{i+1}\cap T$, for every $i$. Let $U_1$ be the last set in the path whose intersection with $S$ has at least $d/4$ elements. Let $U_2$ be the last set in the path whose intersection with $U_1$ has at least $d/4$ elements. I claim that $S, U_1, U_2, T$ is a path in $G_{d/4}({\cal F})$. To see this note that $U_1$ must contain $3d/4$ new elements not already in $S$. Next $U_2$ must contain at least d/2 elements not already in $S$ and $U_1$. Together the three sets $S, U_1, U_2$ must therefore contain at least $d+3d/4+d/2$ elements. This means that their union has at least $9/4d$ elements, hence their union contains at least $d/4$ elements from $T$ and by () $U_3$ and $T$ share at least $d/4$ elements. Sababa. This argument extends to the following proposition: Proposition 2: $F_{d/(2r-2)} F(d,rd) \le r$. So what? How to use these propositions: Remember that the bound to beat was $3^d n$ (actually, Larman improved it to $2^d n$, but in any case, it is exponential in $d$.) Applying the two propositions and the trivial bound $F(d,n) \le {{n} \choose {d}}$ we can get Proposition 3: $F(d,n) \le n^{2 \sqrt n}$. Can we automatize it? Let me return to the question of whether such arguments can be automatized. The above argument for Proposition 2 was simple but somewhat ad hoc. You can get a slightly worse upper bound by noting that if $\cal F$ is a family of $d$-subsets of an n-set and $n \le rd$, then $G_{d/2r-1}({\cal F})$ does not contain an independent set of size $2r-1$. Just use the fact that $S_1 \cup S_2 \cup \dots \cup S_t \ge td-{{t} \choose {2}} d/(2r-1)$. This implies that its diameter is not larger than $4r-4$. The bound on the independence set based on a standard estimetes for union of sets and the connection between the diameter and the independence number both look rather automatable. So is “thinking about” and proving Proposition 1 and deducing Proposition 3. (But I must admit that overall I am less optimistic about the ability to make these very elementary attacks on the problem automatic.) What else? There is a little more to be said. The problem we face using Propositions 1 and 2 is that the ratio between $d$ and $n$ may deteriorate. Once $n$ is large compared with $d$ the situation is hopeless. But if we force the ratio between $n$ and $d$ to be bounded also for families ${\cal F}'[A]$ we can get better (polynomial!!) bounds. I will state these bounds for polytopes keeping in mind the simple connection between the abstract problem and the diameter problem for graphs of polytopes. Proposition: Let $P$ be a simple $d$-polytope and suppose that for every face $F$ of $P$ the number of facets of $F$ is at most $r \dim F$. Then the diameter of the graph of $P$ is at most $d^{c(r)}$. Here $c(r) = K r \log r$. For $r=2$ it is not hard to see that $G_{d/2}$ has a diameter at most 2 and to then deduce that the graph of the polytope has diameter at most $d$. ### Reminder: Our Diameter problem for families of sets and some notations and basic observations. Consider a family $\cal F$ of subsets of size d of the set N={1,2,…,n}. Associate to $\cal F$ a graph $G({\cal F})$ as follows: The vertices of $G({\cal F})$ are simply the sets in $\cal F$. Two vertices $S$ and $T$ are adjacent if $\|S \cap T\|=d-1$. For a subset $A \subset N$ let ${\cal F}[A]$ denote the subfamily of all subsets of $\cal F$ which contain $A$. MAIN ASSUMPTION: Suppose that for every $A$ for which ${\cal F}[A]$ is not empty $G({\cal F}[A])$ is connected. MAIN QUESTION: How large can the diameter of $G({\cal F})$ be in terms of $d$ and $n$. Let us denote the answer by $F(d,n)$. ### More notations and a basic observation from previous parts: Let ${\cal F'}[A]$ be the family obtained from ${\cal F}[A]$ by removing the elements of A from every set. Note that $G({\cal F}[A]) = G({\cal F}' [A])$. Therefore, the diameter of $G({\cal F}[A])$ is at most $F(d',n')$, where $d'= d-\|A\|$ and $n'$ is the number of elements in the union of all the sets in $G({\cal F}'[A])$. We associated more general graphs to $\cal F$ as follows: For an integer $k$ $1 \le k \le d$ define $G_k({\cal F})$ as follows: The vertices of $G_k({\cal F})$ are simply the sets in $\cal F$. Two vertices $S$ and $T$ are adjacent if $\|S \cap T\| \ge k$. Our original problem dealt with the case $k=d-1$. Thus, $G({\cal F}) = G_{d-1}({\cal F})$. Let $F_k(d,n)$ be the maximum diameter of $G_k({\cal F})$ in terms of $k,d$ and $n$, for all families $\cal F$ of $d$-subsets of $N$ satisfying our connectivity relations.	{"extraction_info": {"found_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": 130, "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": 0.9273973703384399, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/49960/do-spheres-exist-in-nature/50017	# Do spheres exist in nature? Often in physics, Objects are approximated as spherical. However do any perfectly spherical objects actually exist in nature? - 1 I believe black holes are considered to be perfectly spherical, though I don't have the appropriate knowledge to elaborate on this. – Wouter Jan 11 at 21:18 2 @Wouter Isn't even that a model? – gerrit Jan 11 at 21:27 @gerrit I'm not entirely sure, which is why I put it in a comment, hoping someone else with more knowledge on the subject would elaborate. – Wouter Jan 11 at 21:29 1 @Wouter In some sense, in certain models, with careful definitions of coordinates, the event horizon of an ideal, nonrotating black hole damps out asphericities on very short timescales. However, astrophysicists would be very surprised to find a nonrotating black hole in nature. – Chris White Jan 11 at 22:58 @ChrisWhite Thanks for the explanation. I heard about it at a lecture and thought it was something like that but couldn't remember exactly. What order of timescales are we talking about here? – Wouter Jan 12 at 0:16 show 3 more comments ## 5 Answers No, but it doesn't matter. The theories that approximate things using spheres are ones in which the final result (the number you measure, the reading on your meter, whatever) depends continuously in some sense on the deviations from sphericity. More symbolically, for any $\varepsilon$ tolerance you allow in your measurement (none of our measurements are infinitely precise), there exists a $\delta$ such that any real object "within $\delta$" of being a sphere will give the same measurement to within $\varepsilon$. It is not that theories are invalid because they assume something "wrong" about nature. Instead, you have to understand that there is always an implicit statement about how "real" behavior approaches the model as deviations from the model's assumptions get smaller. - If we are considering the states of matter we are familiar with (fermions), since these structures are inherently discretized (solids, liquids, gases), they will not exhibit perfect spherical symmetry. We can loosen the definition of perfect and construct a cutoff where variations in the radius are negligible below some length scale and call that perfect. Replying to wouter, the black hole will have an event horizon that is spherical if and only if the black hole has zero angular momentum. It ultimately depends on the interpretation of Cactus' question about what an "object" constitutes and if an event horizon is considered a valid answer. - I suppose it depends on how perfect your perfect sphere is. A drop of water in space with no other gravity effects would look spherical, but if you zoomed in enough to see edges of molecules, then it wouldn't be. From there you can just keep getting smaller (atoms, protons and neutrons, quarks, etc.) until you get into things like string theorem and quantum foam. I would say no, like straight lines, there will not be a perfect spherical physical object in the natural world. - The word `perfect` is based on your choice of scales. On super precise scale, nothing is perfect spherical (not even in artificial setups like freely-falling water droplet in vacuum chamber). The Reason: Every system in nature is dynamic and the problem originates from microscopic level. All denizens of quantum world are followers of probability (we don't know why). The answer can be YES if you also make your measurement time interval super precise. At some instant, you can find water bubbles perfect spherical, for example. - The other answers highlight the importance of a model vs observations, but we do have plenty of very, very spherical objects, as far as experimental measurements go. I only know of this through the excellent sixty symbols video on the topic, but the electron's distribution of charge has a dipole moment of less than $10^{-28}$ C m, which is pretty dang spherical. See also: http://www.nature.com/news/2011/110525/full/news.2011.321.html -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 5, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9392368793487549, "perplexity_flag": "middle"}
http://en.m.wikipedia.org/wiki/Young's_inequality	# Young's inequality In mathematics, Young's inequality is either of two inequalities: one about the product of two numbers,[1] and one about the convolution of two functions.[2] They are named after William Henry Young. Young's inequality for products can be used to prove Hölder's inequality. It is also used widely to estimate the norm of nonlinear terms in PDE theory, since it allows one to estimate a product of two terms by a sum of the same terms raised to a power and scaled. ## Young's inequality for products ### Standard version for conjugate Hölder exponents In its standard form, the inequality states that if a and b are nonnegative real numbers and p and q are positive real numbers such that 1/p + 1/q = 1, then $ab \le \frac{a^p}{p} + \frac{b^q}{q}.$ Equality holds if and only if ap = bq. This form of Young's inequality is a special case of the inequality of weighted arithmetic and geometric means and can be used to prove Hölder's inequality. Proof The claim is certainly true if a = 0 or b = 0. Therefore, assume a > 0 and b > 0 in the following. Put t = 1/p, and (1 − t) = 1/q. Then since the logarithm function is strictly concave $\log(t a^p + (1-t) b^q) \ge t \log(a^p) + (1-t) \log(b^q) = \log(a) +\log(b) = \log(ab)$ with equality if and only if ap = bq. Young's inequality follows by exponentiating. ### Elementary case An elementary case of Young's inequality is the inequality with exponent 2, $ab \le \frac{a^2}{2} + \frac{b^2}{2},$ which also gives rise to the so-called Young's inequality with ε (valid for every ε > 0), sometimes called the Peter–Paul inequality. This name refers to the fact that tighter control of the second term is achieved at the cost of losing some control of the first term – one must "rob Peter to pay Paul" $ab \le \frac{a^2}{2\varepsilon} + \frac{\varepsilon b^2}{2}.$ Proof Young's inequality with exponent 2 is the special case p = q = 2. However, it has a more elementary proof, just observe that $0\le (a-b)^2=a^2+b^2-2ab,$ add 2ab to every side and divide by 2. Young's inequality with ε follows by applying Young's inequality with exponent 2 to $a'=a/\sqrt{\varepsilon},\text{ }b'=\sqrt{\varepsilon}b.$ ### Standard version for increasing functions The area of the rectangle a,b can't be larger than sum of the areas under the functions $f$ (red) and $f^{-1}$ (yellow) For the standard version[3][4] of the inequality, let f denote a real-valued, continuous and strictly increasing function on [0, c] with c > 0 and f(0) = 0. Let f−1 denote the inverse function of f. Then, for all a ∈ [0, c] and b ∈ [0, f(c)], $ab \le \int_0^a f(x)\,dx + \int_0^b f^{-1}(x)\,dx$ with equality if and only if b = f(a). ### Generalization using Legendre transforms If f is a convex function and its Legendre transform (convex conjugate) is denoted by g, then $ab \le f(a) + g(b). \,$ This follows immediately from the definition of the Legendre transform. This inequality also holds — in the form a ·b ≤ f(a) + g(b) — if f is a convex function taking a vector argument (Arnold 1989, §14). #### Examples • The Legendre transform of f(a) = ap/p is g(b) = bq/q with q such that 1/p + 1/q = 1, and thus Young's inequality for conjugate Hölder exponents mentioned above is a special case. • The Legendre transform of f(a) = ea – 1 is g(b) = 1 − b + b ln b, hence ab ≤ ea − b + b ln b for all non-negative a and b. This estimate is useful in large deviations theory under exponential moment conditions, because b ln b appears in the definition of relative entropy, which is the rate function in Sanov's theorem. ↑Jump back a section ## Young's inequality for convolutions In real analysis, the following result is also called Young's inequality:[5] Suppose f is in Lp(Rd) and g is in Lq(Rd) and $\frac{1}{p} + \frac{1}{q} = \frac{1}{r} +1$ with 1 ≤ p, q, r ≤ ∞. Then $\\|fg\\| _r\le\\|f\\|_p\\|g\\|_q.$ Here the star denotes convolution, Lp is Lebesgue space, and $\\|f\\|_p = \Bigl(\int \|f(x)\|^p\,dx \Bigr)^{1/p}$ denotes the usual Lp norm. An example application is that Young's inequality can be used to show that the heat semigroup is a contracting semigroup using the L2 norm (i.e. the Weierstrass transform does not enlarge the L2 norm). In case p, q > 1 Young's inequality can be strengthened to a sharp form, viz $\\|fg\\| _r\le c_{p,q} \\|f\\|_p\\|g\\|_q.$ where the constant cp,q < 1.[6] ↑Jump back a section ## See also ↑Jump back a section ## Notes 1. Young, W. H. (1912), "On classes of summable functions and their Fourier series", Proc. Roy. Soc. Lond. Series A 87 (594): 225–229, doi:10.1098/rspa.1912.0076, JSTOR 93236, Zbl 43.1114.12 2. Young, W. H. (1912), "On the multiplication of successions of Fourier constants", Proc. Roy. Soc. Lond. Series A 87 (596): 331–339, doi:10.1098/rspa.1912.0086, JSTOR 93120, Zbl 44.0298.02 3. Hardy, G. H.; Littlewood, J. E.; Pólya, G. (1952) [1934], Inequalities, Cambridge Mathematical Library (2nd ed.), Cambridge: Cambridge University Press, ISBN 0-521-05206-8, MR 0046395, Zbl 0047.05302 , Chapter 4.8 4. Henstock, Ralph (1988), Lectures on the Theory of Integration, Series in Real Analysis Volume I, Singapore, New Jersey: World Scientific, ISBN 9971-5-0450-2, MR 0963249, Zbl 0668.28001 , Theorem 2.9 5. Bogachev, Vladimir I. (2007), Measure Theory I, Berlin, Heidelberg, New York: Springer-Verlag, ISBN 978-3-540-34513-8, MR 2267655, Zbl 1120.28001 , Theorem 3.9.4 6. Fournier, John J. F. (1977), "Sharpness in Young's inequality for convolution", Pacific J. Math. 72 (2): 383–397, MR 0461034, Zbl 0357.43002 ↑Jump back a section ## References • Arnold, V. I. (1989), Mathematical Methods of Classical Mechanics (2. ed.), Springer, ISBN 978-0-387-96890-2 . • Witkowski, Alfred (2006), "On Young's inequality", Journal of Inequalities in Pure and Applied Mathematics 7 (5): 3 p., Zbl pre05224679 ↑Jump back a section	{"extraction_info": {"found_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": 14, "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": 0.8153436779975891, "perplexity_flag": "middle"}

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