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http://ams.org/bookstore?fn=20&arg1=mmonoseries&ikey=MMONO-88	New Titles \| FAQ \| Keep Informed \| Review Cart \| Contact Us Quick Search (Advanced Search ) Browse by Subject General Interest Logic & Foundations Number Theory Algebra & Algebraic Geometry Discrete Math & Combinatorics Analysis Differential Equations Geometry & Topology Probability & Statistics Applications Mathematical Physics Math Education Return to List Fewnomials SEARCH THIS BOOK: Translations of Mathematical Monographs 1991; 139 pp; hardcover Volume: 88 ISBN-10: 0-8218-4547-0 ISBN-13: 978-0-8218-4547-9 List Price: US\$69 Member Price: US\$55.20 Order Code: MMONO/88 The ideology of the theory of fewnomials is the following: real varieties defined by "simple," not cumbersome, systems of equations should have a "simple" topology. One of the results of the theory is a real transcendental analogue of the Bezout theorem: for a large class of systems of $$k$$ transcendental equations in $$k$$ real variables, the number of roots is finite and can be explicitly estimated from above via the "complexity" of the system. A more general result is the construction of a category of real transcendental manifolds that resemble algebraic varieties in their properties. These results give new information on level sets of elementary functions and even on algebraic equations. The topology of geometric objects given via algebraic equations (real-algebraic curves, surfaces, singularities, etc.) quickly becomes more complicated as the degree of the equations increases. It turns out that the complexity of the topology depends not on the degree of the equations but only on the number of monomials appearing in them. This book provides a number of theorems estimating the complexity of the topology of geometric objects via the cumbersomeness of the defining equations. In addition, the author presents a version of the theory of fewnomials based on the model of a dynamical system in the plane. Pfaff equations and Pfaff manifolds are also studied. • An Analogue of the Bezout Theorem for a System of Real Elementary Equations • Two Simple Versions of the Theory of Fewnomials • Analogues of the Theorems of Rolle and Bezout for Separating Solutions of Pfaff Equations • Pfaff Manifolds • Real-Analytic Varieties with Finiteness Properties and Complex Abelian Integrals AMS Home \| Comments: webmaster@ams.org © Copyright 2012, American Mathematical Society Privacy Statement	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.860558032989502, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/3330/how-can-i-generate-random-permutations-of-n-with-k-cycles-where-k-is-much-larg/3518	## How can I generate random permutations of [n] with k cycles, where k is much larger than log n? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I've been thinking a lot lately about random permutations. It's well-known that the mean and variance of the number of cycles of a permutation chosen uniformly at random from Sn are both asymptotically log n, and the distribution is asymptotically normal. I want to know what a typical permutation of [n] with k(n) cycles "looks like" (in terms of cycle structure), where k(n)/(log n) → ∞ as n → ∞. The special case I have in mind is permutations of [n] with n1/2 cycles, since I've come across such permutations in another context, but I'm also curious about the more general problem. In order to do this I would like an algorithm that generates permutations of n with k cycles uniformly at random -- that is, it generates each one with probability 1/S(n,k) where S(n,k) is a Stirling number of the first kind -- so that I can experiment on them. (I'd be willing to settle for a Markov chain that converges to this distribution if it does so reasonably quickly.) Unfortunately the only way I know to do this is to take a permutation of [n] uniformly at random (this is easy) and then throw it out if it doesn't have k cycles. If k is far from log(n) this is very inefficient, since those permutations are rare. A few references I've come across that are related: This paper of Granville looks at permutations with o(n1/2-ε) cycles or Ω(n1/2+ε) cycles and shows that their cycle lengths are "Poisson distributed", but right around n1/2 is a transitional zone. And this paper of Kazimirov studies "the asymptotic behavior of various statistics" under the distribution I've claimed, but I haven't read it yet because I can't read Russian and I'm waiting for the English translation. Finally, the algorithm I'm looking for might be in one of the fascicles of volume 4 of Knuth, but our library doesn't have them. - ## 9 Answers Youll find the answer on page 38 of my lecture notes East Side, West Side, which are a free download from my web site. It's a complete, short Maple program. The problem was originally solved in 1978 in Combinatorial Algorithms, by Albert Nijenhuis and myself. Herb Wilf - Thanks for the pointer. The permutation version of the problem is on page 33, not 38. – Michael Lugo Apr 19 2010 at 21:29 ### 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. Various algorithms for the unbiased generation of random permutations of certain types (such as permutations into k cycles) are given in this thesis submitted just yesterday: http://www.jjj.de/pub/ Hope this helps, jj - The algorithm you give in Section 6.5 looks like exactly what I need. – Michael Lugo Oct 31 2009 at 12:15 One Markov chain that works: Pick uniformly two cycles A,B, and repartition their union uniformly into two new cycles. This is reversible, hence the uniform measure is stationary. It would be interesting to know how fast this mixes. - Thanks! At the very least this will give me something to play around with and generate conjectures. And this chain has the nice property that if all I care about is the cycle type of the result (which is true for the original purpose I want this for), all I need to do is keep track of the cycle type. Of course the mixing time is different for the Markov chain on cycle types (i. e. partitions of n) than the Markov chain on permutations themselves. – Michael Lugo Oct 30 2009 at 0:48 If it mixes much faster for cycle types than for cycles, one idea would be to use this to generate a random cycle type with the correct probability and then fill in the elements in a random order (much easier) to get a random permutation with that cycle type. – David Eppstein Dec 8 2009 at 21:53 Another Markov chain which converges to the uniform distribution is a variation on the "random to random" shuffling: pick a random element, take it out of the permutation and then stick it back in a random place, unless the random element you picked was a unicycle in which case you have to put it back as a unicycle to preserve the number of cycles. This probably converges much more slowly than Omer's chain (at least for k << n), but it does have the advantage that, unless I'm mistaken, one can use path coupling to bound the mixing time by something like n^2 log n. If I had to guess I'd say the real mixing time is n log n. Luckily, I don't. Perhaps path coupling could work for Omer's chain as well. - One can sample from the uniform distribution on permutations on $n$ elements with $k$ cycles in expected time $O(n\sqrt{k})$. For large $n$ and $k$ this may be more feasible than the method Herb Wilf refers to in his answer, which, if I understand right, requires the generation of the Stirling cycle numbers $S(m,r)$ for $m\leq n$ and $r\leq k$. The idea is as follows: consider a Poisson-Dirichlet($\theta$) partition of $[n]$. The blocks of such a partition have the distribution of the cycles of a random permutation of $[n]$ from the distribution which gives weight proportional to $\theta^r$ to any permutation with $r$ cycles. In particular, conditioned on the number of cycles, the permutation is uniformly distributed. (Once one has a partition into blocks corresponding to the cycles, one can just fill in the elements of $[n]$ into the positions in the cycles uniformly at random). Choose $\theta$ in such a way that the mean number of cycles is around $k$. One can sample a PD($\theta$) partition of $[n]$ in time $O(n)$ (see below). Keep generating independent samples of the partition until you get one with exactly $k$ blocks. The variance of the number of cycles will be $O(k)$ (see below) and the probability that the number of cycles is precisely $k$ will be on the order of $1/\sqrt{k}$, so one will need to generate $O(\sqrt{k})$ such samples before happening upon one with precisely $m$ cycles. So this is really not so far from what you suggested in the question (generate random permutations until you find one that fits) with the twist that instead of generating from the uniform distribution (which corresponds to PD(1)) you choose a better value of $\theta$ and generate from PD($\theta$). Here are two nice ways to sample a PD($\theta$) partition of $[n]$: (1) "Chinese restaurant process" of Dubins and Pitman. We add elements to the partition one by one. Element 1 starts in a block on its own. Thereafter, when we add element $r+1$, suppose there are currently $m$ blocks whose sizes are $n_1, n_2, ... n_m$. Add element $r+1$ to block $i$ with probability $n_i/(r+\theta)$, for $1\leq i\leq m$, and put element $r+1$ into a new block on its own with probability $\theta/(r+\theta)$. (2) "Feller representation". Generate independent Bernoulli random variables $B_1, \dots, B_n$ with $P(B_i=1)=\theta/(i-1+\theta)$. Write a string of length $n$ divided up into blocks, with the rule that we start a new block before position $i$ whenever $B_i=1$. So for example if $n=10$ with $B_1=B_5=B_6=B_9=1$ and the other $B_i$ equal to 0, then the pattern is (a b c d)(e)(f g h)(i j). (Note that always $B_1=1$). Then assign the elements of $[n]$ to the positions in the blocks uniformly at random. The expected number of blocks is $\sum_{i=1}^n \mathbb{E}B_i$, which is $\sum_{i=1}^n \theta/(i+\theta)$, which is approximately $\theta(\log n-\log \theta)$. If this is equal to $k$ with $1 << k << n$, then the number of blocks will be approximately normal with mean $k$ and variance $O(k)$. For details of some of the things mentioned here to do with Poisson-Dirichlet partitions, random permutations etc, see e.g. Pitman's lecture notes from Saint-Flour: http://bibserver.berkeley.edu/csp/april05/bookcsp.pdf - Knowing next to nothing about randomness, this is how I'd go about picking a random permutation of [k] with exactly n cycles: 1) Pick n elements of k, removing each element from the pool you pick out of. 2) For the remaining elements, pick out an element at random, then pick an index i from [n] at random. Append the element to the cycle at i. Alas, I don't really know how to analyze probabilities for this setup, nor to check what kind of distribution it gives. - 2 Consider the case where I want permutations of [4] with 2 cycles. There are eleven of these: three that are conjugate to (12)(34) and eight that are conjugate to (123)(4). I want to generate each one of these with the same probability, 1/11. The fact that 11 is prime shows that this won't work -- the probabilities that come from this model are going to be sums of products of fractions with 2, 3, or 4 in the denominator. (Actually finding those probabilities is not that hard but won't fit in this box.) – Michael Lugo Oct 30 2009 at 12:05 I'm not sure this will help you, but you can compute the relative probabilities of the cycle types via Ewens's sampling formula (with θ=1). - Sure, but the cases I'm interested in are precisely those where there are too many cycle types for that to be practical. – Michael Lugo Oct 30 2009 at 14:22 Look at the following paper: Moni Naor, Omer Reingold: Constructing Pseudo-Random Permutations with a Prescribed Structure. J. Cryptology 15(2): 97-102 (2002) You can find it on Moni's site. - Naor and Reingold's paper, mentioned in this answer, only explains how to construct a random permutation with a fixed prescribed structure from a given random permuation P. This is done by taking the simplest possible permutation with the structure you wish to get, and conjugating it by P. This reduces the problem to choosing a k-cycles cycle structure with the correct distribution, which does not seem to be easier to me. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 55, "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": 0.9421172142028809, "perplexity_flag": "head"}
http://en.wikipedia.org/wiki/Problem_of_points	# Problem of points The problem of points, also called the problem of division of the stakes, is a classical problem in probability theory. One of the famous problems that motivated the beginnings of modern probability theory in the 17th century, it led Blaise Pascal to the first explicit reasoning about what today is known as an expectation value. The problem concerns a game of chance with two players who have equal chances of winning each round. The players contribute equally to a prize pot, and agree in advance that the first player to have won a certain number of rounds will collect the entire prize. Now suppose that the game is interrupted by external circumstances before either player has achieved victory. How does one then divide the pot fairly? It is tacitly understood that the division should depend somehow on the number of rounds won by each player, such that a player who is close to winning will get a larger part of the pot. But the problem is not merely one of calculation; it also includes deciding what a "fair" division should mean in the first place. ## Early solutions Luca Pacioli considered such a problem in his 1494 textbook Summa de arithmetica, geometrica, proportioni et proportionalità. His method was to divide the stakes in proportion to the number of rounds won by each player, and the number of rounds needed to win did not enter his calculations at all.[1] In the mid-16th century Niccolò Tartaglia noticed that Pacioli's method leads to counterintuitive results if the game is interrupted when only one round has been played. In that case, Pacioli's rule would award the entire pot to the winner of that single round, though a one-round lead early in a long game is far from decisive. Tartaglia constructed a method that avoids that particular problem by basing the division on the ratio between the size of the lead and the length of the game.[1] This solution is still not without problems, however; in a game to 100 it divides the stakes in the same way for a 65–55 lead as for a 99–89 lead, even though the former is still a relatively open game whereas in the latter situation victory for the leading player is almost certain. Tartaglia himself was unsure whether the problem was solvable at all in a way that would convince both players of its fairness: "in whatever way the division is made there will be cause for litigation".[2] ## Pascal and Fermat The problem arose again around 1654 when Chevalier de Méré posed it to Blaise Pascal. Pascal discussed the problem in his ongoing correspondence with Pierre de Fermat. Through this discussion Pascal and Fermat not only came up with a convincing, self-consistent solution to the division of the stakes, but also developed concepts that continue to be fundamental in probability to this day. The starting insight for Pascal and Fermat was that the division should not depend so much on the history of the part of the interrupted game that actually took place, as on the possible ways the game might have continued, were it not interrupted. It is intuitively clear that a player with a 7–5 lead in a game to 10 has the same chance of eventually winning as a player with a 17–15 lead in a game to 20, and Pascal and Fermat therefore thought that interruption in either of the two situations ought to lead to the same division of the stakes. In other words, what is important is not the number of rounds each player has won yet, but the number of rounds each player still needs to win in order to achieve overall victory. Fermat now reasoned thus:[3] if one player needs r more rounds to win and the other needs s, the game will surely have been won by someone after $r+s-1$ additional rounds. Therefore, imagine that the players were to play $r+s-1$ more rounds; in total these rounds have $2^{r+s-1}$ different possible outcomes. In some of these possible futures the game will actually have been decided in fewer than $r+s-1$ rounds, but it does no harm to imagine the players continuing to play with no purpose. Considering only equally long futures has the advantage that one easily convinces oneself that each of the $2^{r+s-1}$ possibilities is equally likely. Fermat was thus able to compute the odds for each player to win, simply by writing down a table of all $2^{r+s-1}$ possible continuations and counting how many of them would lead to each player winning. Fermat now considered it obviously fair to divide the stakes in proportion to those odds. Fermat's solution, certainly "correct" by today's standards, was improved by Pascal in two ways. First, Pascal produced a more elaborate argument why the resulting division should be considered fair. Second, he showed how to calculate the correct division more efficiently than Fermat's tabular method, which becomes completely impractical (without modern computers) if $r+s-1$ is more than about 10. Instead of just considering the probability of winning the entire remaining game, Pascal devised a principle of smaller steps: Suppose that the players had been able to play just one more round before being interrupted, and that we already had decided how to fairly divide the stakes after that one more round (possibly because that round lets one of the players win). The imagined extra round may lead to one of two possible futures with different fair divisions of the stakes, but since the two players have even chances of winning the next round, they should split the difference between the two future divisions evenly. In this way knowledge of the fair solutions in games with fewer rounds remaining can be used to calculate fair solutions for games with more rounds remaining.[4] It is easier to convince oneself that this principle is fair than it is for Fermat's table of possible futures, which are doubly hypothetical because one must imagine that the game sometimes continues after having been won. Pascal's analysis here is one of the earliest examples of using expectation values instead of odds when reasoning about probability. Shortly after, this idea would become a basis for the first systematic treatise on probability by Christiaan Huygens. Later the modern concept of probability grew out of the use of expectation values by Pascal and Huygens. The direct application of Pascal's step-by-step rule is significantly quicker than Fermat's method when many rounds remain. However, Pascal was able to use it as a starting point for a developing even slicker computation methods. Through clever manipulation of identities involving what is today known as Pascal's triangle (including several of the first explicit proofs by induction) Pascal finally showed that in a game where one player needs r points to win and the other needs s points to win, the correct division of the stakes is in the ratio of (using modern notation) $\sum_{k=0}^{s-1} \binom{r+s-1}{k} \mbox{ to } \sum_{k=s}^{r+s-1} \binom{r+s-1}{k}.$ The problem of dividing the stakes became a major motivating example for Pascal in his Treatise on the arithmetic triangle.[4] [5] Though Pascal's derivation of this result was independent of Fermat's tabular method, it is clear that it also describes exactly the counting of different outcomes of $r+s-1$ additional rounds that Fermat suggested. ## References • Anders Hald: A history of Probability and Statistics and their Applications before 1750. Wiley 2003, ISBN 978-0-471-47129-5, p. 35, 54	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 9, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9759826064109802, "perplexity_flag": "middle"}
http://mathhelpforum.com/calculus/168911-limits.html	# Thread: 1. ## Limits 1. Show that lim (n^n)/(n!e^n) exists without finding the limit. n->oo 2. Show that lim (1/n){(24...(2n))/(13...(2n-1))}^2 exists without finding the limit. n->oo I guess I have to prove this by least upper bounds, however hard to start! 2. Originally Posted by mathsohard 1. Show that lim (n^n)/(n!e^n) exists without finding the limit. n->oo 2. Show that lim (1/n){(24...(2n))/(13...(2n-1))}^2 exists without finding the limit. n->oo I guess I have to prove this by least upper bounds, however hard to start! For 1, the denominator grows faster than the numerator. Hence, the limit is 0. 3. Originally Posted by dwsmith For 1, the denominator grows faster than the numerator. Hence, the limit is 0. The same could be said about $\displaystyle{\frac{n}{n+1}}$ , yet this seq. does not converge to zero... Tonio 4. Originally Posted by tonio The same could be said about $\displaystyle{\frac{n}{n+1}}$ , yet this seq. does not converge to zero... Tonio Limits of a simple polynomial we just evaluate the fraction of the highest power variable. I would have changed my answer to take n/n then but for this case and since it doesn't say solve the limit, I gave the poster a correct answer for his limit. 5. Originally Posted by mathsohard 1. Show that lim (n^n)/(n!e^n) exists without finding the limit. n->oo Show that your sequence is monotone descending, which is surprisingly easy if you already know that the sequence $\displaystyle{\left(1+\frac{1}{n}\right)^n$ converges monotonically ascending to $e$ 2. Show that lim (1/n){(24...(2n))/(13...(2n-1))}^2 exists without finding the limit. n->oo I guess I have to prove this by least upper bounds, however hard to start! You can try here the above, but it'll be much more involved. First, be sure you can show that $\displaystyle{2\cdot 4\cdot\ldots\cdot (2n)=2^n\,n!\,,\,\,1\cdot 3\cdot\ldots\cdot (2n-1)=\frac{(2n)!}{2\cdot 4\cdot\ldots\cdot (2n)}}$ , so that now you can write your sequence as $\displaystye{a_n:=\frac{1}{n}\left(\frac{(2^n\,n!) ^2}{(2n)!}\right)^2$ , and now show that $a_{n+1}\leq a_n$ ...very carefully and slowly! Tonio . 6. Originally Posted by dwsmith Limits of a simple polynomial we just evaluate the fraction of the highest power variable. I would have changed my answer to take n/n then but for this case and since it doesn't say solve the limit, I gave the poster a correct answer for his limit. I'm not trying to start an argument here but your answer isn't correct since my example contradicts it. You didn't explain anything about polynomials, and even if you did that doesn't prove that a sequence which is given in form of a fraction will necessarily converge to zero if simply, as you wrote, "the denominator grows faster than the numerator". Tonio 7. Originally Posted by mathsohard 1. Show that lim (n^n)/(n!e^n) exists without finding the limit. n->oo 2. Show that lim (1/n){(24...(2n))/(13...(2n-1))}^2 exists without finding the limit. n->oo I guess I have to prove this by least upper bounds, however hard to start! Try using the ratio test. 8. Originally Posted by tonio . To show it is a monotone descending sequence, I just have to show that n < n+1 right?? and then how can I prove it converges??? need little more help on both of them 9. Originally Posted by mathsohard To show it is a monotone descending sequence, I just have to show that n < n+1 right?? and then how can I prove it converges??? need little more help on both of them Well, as both sequences are obviously positive, showing they are monotonee descending automatically makes them monotone and bounded and thus converging. Tonio 10. Originally Posted by mathsohard To show it is a monotone descending sequence, I just have to show that n < n+1 right?? and then how can I prove it converges??? need little more help on both of them To show that a sequence is monotone ascending you would need to show that $a_n< a_{n+1}$. To show a sequence is monotone descending, you would need to show that $a_n> a_{n+1}$. If a monotone ascending sequence has an upper bound, then it converges. If a monotone descending sequence has a lower bound, then it converges. Sequences of positive numbers have, by definition of "positive", 0 as a lower bound. 11. I have another question, I understand that if a sequence is monotone ascending and have an upper bound, there is a limit, and if monotone descending and have a lower bound there is a limit. And to show a sequence is ascending I have to show An < An+1 and for descending An > An+1. But for this problem n^n/(n!e^e), I know it is descending and has lower bound of 0 since it is positive, however what is the proper way of showing that An > An+1 ??? I can't think of anything else but plugging in... 12. Originally Posted by mathsohard I have another question, I understand that if a sequence is monotone ascending and have an upper bound, there is a limit, and if monotone descending and have a lower bound there is a limit. And to show a sequence is ascending I have to show An < An+1 and for descending An > An+1. But for this problem n^n/(n!e^e), I know it is descending and has lower bound of 0 since it is positive, however what is the proper way of showing that An > An+1 ??? I can't think of anything else but plugging in... As in my last post, if you put $\displaystyle{a_n:=\frac{n^n}{n!\,e^n}}$ , then we get $\displaystyle{a_{n+1}\leq a_n\Longleftrightarrow \frac{(n+1)^{n+1}}{(n+1)!\,e^{n+1}}\leq \frac{n^n}{n!\,e^n}}\Longleftrightarrow \frac{(n+1)^n}{n^n}\leq e}$ , and we get that this last inequality is true since, as was observed in my past message, the left side in it converges monotonically ascending to the right side. Tonio 13. ## Cauchy Convergence Theorem It seems to me if you have to show something converges without finding a limit then Cauchy's convergence theorem is worth a try. 14. Use ratio test proposed by Drexel28 above: An/An+1= (n+1)!/n! X n^n/(n+1)^(n+1) X e^(n+1)/e^n = (n+1/n)! X (n/n+1)^n X 1/(n+1) X e^(n+1)/e^n = (n+1/n)^n X e > 1 An+1/An < 1 => monotonically decreaing, and > 0. (Sorry about Cauchy. Didn't read everything and assumed ratio test had been tried.) 15. Originally Posted by tonio The same could be said about $\displaystyle{\frac{n}{n+1}}$ , yet this seq. does not converge to zero... Tonio I wouldn't say that $n+1$ grows faster than $n$. I would say that $n+1$ is greater than $n$ for all $n$, but that they grow at the same rate. I'm not saying that this is wrong, but maybe a better counterexample would be something like $\frac{n^2}{n^2+n}$. I also think that dwsmith's statement is correct if we add the word "much." That is, the denominator grows "much faster" than the numerator (in the original question). When I say that a function $f$ grows much faster than a function $g$, by that I mean $\lim_{x\rightarrow \infty\frac{g(x)}{f(x)}=0$. Although I suppose this is just rephrasing the original question.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 21, "wp_latex": 0, "mimetex.cgi": 0, "/images/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": 0.9523406624794006, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/55638/condition-for-closed-orbit/55646	# Condition for closed orbit [closed] I'd like to know when an orbit is closed. I know that, to have a closed orbit, there is a ratio that must be a rational number, but I don't know other things.. - 1 what closed orbit are you taking about. Give us the detailed information about what do you want to know – Akash Mar 2 at 15:18 @Akash There isn't a general rule to apply for any case? For example.... we can consider a mass point on a cone: what is the condition to have closed orbit? thanks! – sunrise Mar 2 at 15:48 @sunrise Isn't it just a case of having the eccentricity less than 1? If $e=0$, the orbit is a circle, if $0 < e < 1$, the orbit is an ellipse, if $e=1$, the orbit is a parabola, is $e>1$, the orbit is a hyperbola. Only for the circle and ellipse are the orbits closed. – user12345 Mar 2 at 19:11 ## closed as not a real question by Manishearth♦Mar 2 at 17:50 It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, see the FAQ. ## 3 Answers There's a branch of study called dynamical systems theory that deals with questions like this. The answer is not straightforward, except in a handful of cases. Trajectories can appear to be highly chaotic and yet, when observed for a sufficiently long time, will turn out to be perfectly periodic. The notion of Poincairé recurrence time arises, which suggests that over a long enough time, any orbit must be considered periodic. This, in turn, leads to a debate over whether there is such a thing as an irrational number in a physical system with lower limits to measurable lengths and times. The straightforward answer to this very broad question is that there isn't a straightforward answer. - Your answer is very very interesting! :) – sunrise Mar 2 at 16:50 To add to KDN's nice answer, there is a theorem called Bertrand's theorem which states that in the case of a particle moving in a central potential, the only potentials that produce stable, closed orbits are the inverse square and radial harmonic oscillator potentials. - Thank you! mmmmh.. any idea to theorems about ratios? – sunrise Mar 2 at 16:52 @sunrise No, not offhand. – joshphysics Mar 2 at 16:55 I believe that there are also circular orbits in a $r^{-5}$ potential, but they go through the center of force where things are singular. It does not contradict Bertrand's theorem because it is not a general result for that power law. See Goldstein (Ed. 2) problem 3.6. – dmckee♦ Mar 2 at 17:10 Your reference to rational ratios suggests that you might be thinking of the simple dynamical system on the torus---a favourite toy example of mathematicians. See, for example, the wikipedia article "Linear flow on the torus". - I'm thinking to a quiet simple system, but I don't know the torus.. other analogous cases that can be answer to my question? – sunrise Mar 2 at 16:49	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.95182204246521, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/205765/fibonacci-matrix?answertab=active	# Fibonacci Matrix I'm looking for a good way to find the matrix form of fibonacci equation, and also a more general implementation. I've looked all other the web and haven't found it. I'll be really thankfull to any one that can point me in the right direction. - 2 – Qiaochu Yuan Oct 2 '12 at 2:24 ## 2 Answers The Fibonacci's sequence satisfies $F_{n+2} = F_{n+1} + F_n$. Write this as $$\begin{array}{cc} F_{n+2} &=& F_{n+1} + F_{n} \cr F_{n+1} &=& F_{n+1} \phantom{+F_n} \end{array} \implies \begin{pmatrix} F_{n+2} \cr F_{n+1} \end{pmatrix} = \underbrace{\begin{pmatrix} 1 & 1 \\ 1 & 0\end{pmatrix}}_M \cdot \begin{pmatrix} F_{n+1} \cr F_{n} \end{pmatrix}$$ Denoting vector $v_{n} = (F_{n+1}, F_n)$ the above recurrence equation reads $v_{n+1} = M v_n$, having a general solution $v_n = M^n v_0$. - You know that $$f_{n+1}=f_n+f_{n-1} \,,$$ to discover the matrix, all you need is to figure that $f_n=f_n$. Thus $$f_{n+1}=f_n+f_{n-1} \,,$$ $$f_n=f_n$$ This means that $(f_{n+1}, f_n)^T=F (f_{n}, f_{n-1})^T$, where $F$ is the coefficient matrix... -	{"extraction_info": {"found_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": 4, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9298072457313538, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/tagged/big-numbers	Tagged Questions The big-numbers tag has no wiki summary. 1answer 69 views Examples of Diophantine equations with a large finite number of solutions I wonder, if there are examples of Diophantine equations (or systems of such equations) with integer coefficients fitting on a few lines that have been proven to have a finite, but really huge number ... 1answer 368 views How do I calculate the 2nd term of continued fraction for the power tower ${^5}e=e^{e^{e^{e^{e}}}}$ I need to find the 2nd term of continued fraction for the power tower ${^5}e=e^{e^{e^{e^{e}}}}$ ( i.e. $\lfloor\{e^{e^{e^{e^{e}}}}\}^{-1}\rfloor$), or even higher towers. 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Probably a proof (if any exist) that calls upon Knuth's up-arrow notation or Busy Beaver.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 23, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9266719222068787, "perplexity_flag": "middle"}
http://mathhelpforum.com/calculus/98067-point-intersection.html	Thread: 1. Point of intersection What is the point of intersection between the curves y=sec²x and y=4? I need it in order to find the area of the region bounded by the two curves. Thanks. 2. Originally Posted by sun1 What is the point of intersection between the curves y=sec²x and y=4? I need it in order to find the area of the region bounded by the two curves. Thanks. The y-values must be equal: $\sec^2(x)=4$ $\|\sec(x)\|=2$ $\left\|\dfrac1{\cos(x)}\right\|=2$ $\cos(x) = -\dfrac12~\vee~\cos(x)=\dfrac12$ Can you take it from 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": 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": 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": 0.9479519128799438, "perplexity_flag": "head"}
http://mathematica.stackexchange.com/questions/6822/tiling-a-square?answertab=votes	# Tiling a square I wondered if there was a way to automate the process of finding a way to tile a tile into a square. The idea is to represent the tile with a matrix of 0s for blank space and 1s for filled spaces like in {{0,1,1},{1,0,0}}. The square (or a general rectangular area) is represented by an array as well. The function should then have as arguments two lists(representing the tile and the area) and should find the ways(if they exist) to do the tiling. This is quite an hard problem for me,i have tried playing with Plus, so that if the result matrix has 2 it means that the tile overlaps while if it is 0 there is a gap but i can't find a way to tell Mathematica that they should fit the area. - 6 – belisarius Jun 14 '12 at 18:30 Man thanks so much!I'll read it now but at a first glance seems very useful. Thanks again! – Gianpiero Cea Jun 14 '12 at 18:31 1 @belisarius I think this is nice and should be posted as an asnwer – Vitaliy Kaurov Jun 14 '12 at 19:48 @VitaliyKaurov Ok, done. I usually don't post as an answer something that came up from a Google search ... but that also depends on how long it took me to find the answer. – belisarius Jun 14 '12 at 21:50 ## 2 Answers In this article the author solves the problem of tiling a rectangle by using pieces taken from a set of polyominoes, which are plane geometric figures formed by joining one or more equal squares edge to edge. For example, these are the pentaminoes, polyominoes formed by joining 5 squares: Of course this problem is more difficult than the one you asked for, but it is also more interesting and ... there is Mathematica code in the article! - 5 +1 Thank you for posting. I think that Mathematica Journal is great and people can find many gems there. I wish more people would read it. – Vitaliy Kaurov Jun 14 '12 at 22:03 Some time ago i had found that great resource,the Mathematica Journal,but my skills in Mathematica were really poor(and still they are :D)and so i eventually forgot it.Anyway i'm getting better and better and i think i'll start giving a deeper look to it.Thank you all for remembering me this great source! – Gianpiero Cea Jun 15 '12 at 8:20 +1 For letting me know about the existance of the Mathematica Journal. – Gustavo Bandeira Oct 20 '12 at 3:58 With some diffidence (because there appears to be a Mathematica bug: see below), I would like to offer an answer in the spirit of the OP's original attempt to solve the problem algebraically. ## Solution This problem can be formulated as a binary integer linear program. The reformulation represents the square (or more generally, a rectangle as implemented below) as a vector. Its coefficients represent occupancy: $0$ for empty, $1$ for full. In this representation, every possible tile placement corresponds to a unique vector indicating the cells occupied by the tile. Let's begin by generating the vectors corresponding to all possible placements of tiles in all possible orientations. It will be convenient to specify as little as possible, beginning with a set of unique tile shapes. As a running example I use the L, C, T, and Z tiles: ````tileSet = {{{1, 1, 1}, {1, 0, 0}}, {{1, 1, 1}, {1, 0, 1}}, {{1, 1, 1}, {0, 1, 0}}, {{0, 1, 1}, {1, 1, 0}}}; ArrayPlot[#, Frame -> False, ImageSize -> 10 Max[Length[First[#]]]] & /@ tileSet ```` (There is no restriction that tiles have a common bounding box, but--to make sure they can be properly positioned--it is important that their matrix descriptions not have any zero padding along outside rows or columns.) We need to rotate (and, if we wish, flip) all these tiles. An elegant way (perhaps not the simplest) is to generate the group action abstractly in terms of generators (`alpha` and `beta`). ````apply[s_List, alpha] := Reverse /@ s; (* Horizontal reflection ) apply[s_List, beta] := Transpose[s]; ( Diagonal reflection ) apply[s_List, g_List] := Fold[apply, s, g]; ( Group composition ) group = FoldList[Append, {}, Riffle[ConstantArray[alpha, 4], beta]]; ```` (`Riffle` and `ConstantArray` are just a trick, specific to Coxeter groups, to generate a unique set of group elements. If you don't want to allow reflections, then eliminate the four orientation-reversing group elements, which are the ones using odd numbers of generators.) Apply the group action to the tiles and eliminate any duplicates: ````tiles = Union[Flatten[Outer[apply[#1, #2] &, tileSet, group, 1], 1]]; ArrayPlot[#, Frame -> False, ImageSize -> 10 Max[Length[First[#]]]] & /@ tiles ```` These tiles need to be converted from arrays (suitable for graphical display) into vectors (for input to the linear program). Because these vectors represent all possible positions (translations) of the oriented tiles into the square, here is where we specify the square's dimensions, $m$ rows by $n$ columns. I use $6$ by $9$ for this example (which is a significant size). `makeAllTiles` creates all possible positions of a single tile within a rectangle; mapping it over the set of all oriented tiles and flattening the result gives what we want. (No need any more to remove duplicates, because there is no possibility of an accidental match.) ````makeAllTiles[tile_, m_Integer, n_Integer] := With[{m0 = Length[tile], n0 = Length[First[tile]]}, Flatten[Table[ArrayPad[tile, {{i, m - m0 - i}, {j, n - n0 - j}}], {i, 0, m - m0}, {j, 0, n - n0}], 1]]; allTiles = Flatten[makeAllTiles[#, 6, 9] & /@ tiles, 1]; ArrayPlot[options = Transpose[Flatten /@ allTiles], ImageSize -> 800] ```` Each possible position of each possible orientation of each tile shape is represented by a column vector in this array. The rows correspond to cells in the rectangle. It's now surprisingly straightforward to complete the setup: we seek a subset of columns summing to the vector representing a filled square: all the coefficients in this sum must be $1$'s. The subset can be picked out by an indicator vector $x$. We therefore require its entries to be integral, bounded below by $0$, and bounded above by $1$. A feasible vector, let's say, will be one where the corresponding sum fills the square, but sometimes fills its cells multiple times (due to overlap of tiles): that is, a lower bound $b$ for the result should be a $1$ in each cell of the square. Finally, we minimize the sum of the cell values: if a solution exists, this sum will equal the number of squares. The sum is a linear combination with unit coefficients, represented as a vector $c$. ````b = ConstantArray[1, Length[options]]; ( Lower bounds on cell values ) c = ConstantArray[1, Length[First[options]]]; ( Sum of cell values ) lu = ConstantArray[{0, 1}, Length[First[options]]]; ( 0-1 constraints *) x = Quiet[LinearProgramming[c, options, b, lu, Integers, Tolerance -> 0.0005], {LinearProgramming::lpip, LinearProgramming::lpsnf}]; If[!ListQ[x] \|\| Max[options.x] > 1, x = {}]; solution = allTiles[[Select[x Range[Length[x]], # > 0 &]]]; ArrayPlot[z = Sum[i solution[[i]], {i, 1, Length[solution]}], PlotRange -> {0, Max[z]}, Mesh -> True, ColorFunction -> "Rainbow"] ```` `Solution` is a list of tiles (in their orientations and positions). ## Limitations `LinearProgramming` (apparently) can return only one solution even when many are optimal. To explore all solutions, or a range of solutions, may require different methods. (There are still some tricks we can play: e.g., by permuting the order of columns in `Options`, we might wind up with a different solution.) As with the systematic searching methods, when this problem gets big, the computing time grows rapidly. You can see this is so, just by considering the quadratic growth in the list of possible positions of the tiles as the square's dimensions increase. Computing time for the example is around one second (increasing as `Tolerance` decreases). Finally, I had to fudge something to get this solution: `LinearProgramming`, with its default setting for `Tolerance`, returned a non-optimal solution! (`options.x` had some 2's in it, indicating some double overlaps.) By starting with `Tolerance` set to $0.5$ and gradually decreasing it, within a few steps a valid solution was obtained. (With the value set to `10^-1`, though, I raised an Assert in the underlying C code :-(. That is a bug, at least in version 8.0.0.0.) - 1 Wow! I love answers like this. It's humbling and enlightening all at once. – Mr.Wizard♦ Jun 16 '12 at 15:29 Great answer! Perhaps your dihedral groups mastering may help me in this question on math.se related to this question in this site – belisarius Jun 16 '12 at 15:35 1 @belisarius Actually, I had been using these methods to study your peg solitaire game (and I have even simplified it using exactly the same `LinearProgramming` treatment described here). – whuber Jun 16 '12 at 18:43 I think the problem is that you are trying to minimize the number of pieces used to cover the square when you use `c={1,1,1,1...}`. If instead you tried to minimize the number of total squares by using `c=Total@options` it seems to work fine without the `Tolerance` specification. What do you think? – Rojo Aug 8 '12 at 1:51 @Rojo That's a good idea. What I was really seeking was any feasible solution. I could not find an option in `LinearProgramming` to do that and so, as a work-around, was forced to find a feasible solution by optimizing something. – whuber Aug 8 '12 at 13:01 show 6 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": 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": 0.9068300127983093, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/31789/having-trouble-verifying-absolute-conditional-convergence-for-these-series/31803	# Having trouble verifying absolute/conditional convergence for these series Greetings, I'm having trouble applying the tests for convergence on these series; I can never seem to wrap my head around how to determine if they're absolutely convergent, conditionally convergent or divergent. a) $\displaystyle \sum_{k=1}^{\infty}\frac{\sqrt{k}}{e^{k^3}}$. b) $\displaystyle \sum_{k=2}^{\infty}\frac{(-1)^k}{k(\ln k)(\ln\ln k)}$. - ## 3 Answers Hint: I imagine that the Ratio Test is part of your toolbox. Problem (a) should yield fairly easily to that. For (b), note that the given series is (almost) an alternating series. (The first two terms have the same sign.) To prove that the series does not converge absolutely, a tool that works is the Integral Test. Try differentiating $\ln(\ln(\ln x))$. - 2 I should have put in some heuristic comments. The Ratio Test is good if, informally, the terms decrease "real fast." As for (b), presumably it screams alternating series. And probably, in course notes or the book, $\sum 1/(k\ln k)$ was discussed. Imitate. – André Nicolas Apr 8 '11 at 20:06 Thanks for the hints, the community here is very helpful. I'm making progress, albeit slow progress. – sephy7324 Apr 8 '11 at 20:22 The first series is positive, so it either converges absolute, or diverges. My gut reaction was to try some back-of-the-envelope comparison. Let's see: $\sum \frac{1}{e^{k}}$ converges, and $\frac{1}{e^{k^3}}\leq \frac{1}{e^k}$, so $\sum\frac{1}{e^{k^3}}$ converges. The exponential dominates $\sqrt{k}$, so personally, I would expect the series to converge. I would try an integral test, but it looks like an annoying function to integrate. So perhaps we can find a series $\sum\frac{p(k)}{e^{k^3}}$ for some function $p(k)$, such that $\int_1^{\infty}p(x)e^{-x^3}\,dx$ is easy to integrate. This is doable, but as user6312 pointed out, the Ratio Test will do the job as well. The second series is (eventually) alternating. Try the Alternating Series Test. To check the absolute convergence, try an Integral test. You'll need a change of variable. Hint. If you set $u=\ln x$, what happens? - a) Since $e^x >x$ then $$\frac{\sqrt{k}}{e^{k^3}}<\frac{\sqrt{k}}{k^3}=\frac{1}{k^{5/2}}$$ But $\sum_{k=1}^\infty \frac{1}{k^{5/2}}$ is convergent ($p$-series test). Hence the original series is convergent (hence absolutely convergent since it is a positive series). -	{"extraction_info": {"found_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": 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": 0.9280316829681396, "perplexity_flag": "middle"}
http://all-science-fair-projects.com/science_fair_projects_encyclopedia/Quantum	# All Science Fair Projects ## Science Fair Project Encyclopedia for Schools! Search Browse Forum Coach Links Editor Help Tell-a-Friend Encyclopedia Dictionary # Science Fair Project Encyclopedia For information on any area of science that interests you, enter a keyword (eg. scientific method, molecule, cloud, carbohydrate etc.). Or else, you can start by choosing any of the categories below. # Quantum The word quantum, pl. "quanta", comes from the Latin "quantus", for "how much". In general, it refers to an "amount of something". But, the term is often used in the more specific sense it has in physics, where a quantum refers to an indivisible, and perhaps elementary entity. For instance, a "light quantum", being a unit of light (that is, a photon). In combinations like "quantum mechanics", "quantum optics", etc., it distinguishes a more specialized field of study. Behind this one finds the fundamental notion that a physical property may be "quantized", referred to as "quantization". This means that the magnitude can take on only certain numerical values, rather than any value, at least within a range. For example, the energy of an electron bound to an atom (at rest) is quantized. This accounts for the stability of atoms, and matter in general. Contents ## How quantization was discovered Quantum physics, being the branch of physics based on quantization, was founded in the year 1900, with the theory, due to M. Planck, explaining the so-called black body radiation. His work incorporated quantization in essentially the same way that it is used today. But, the break with classical mechanics was so significant that it took another 30 years of investigation, before quantum theory became correctly formulated and understood. Some would maintain that it remains not fully understood until the present day. There is much to be learned about the nature of science from seeing how this came about. Not least Planck himself felt disturbed by the new idea of quantization, and with good reason. But, finding no alternative, he nevertheless ended up using it, and the work was well received. After "a few weeks of the most strenuous work of his life", as he recalled it in the lecture on the occasion of receiving the Nobel Prize for physics, 18 years later. In the course of those weeks he even had to discard much of his own theoretical work from the preceding years. Quantization turned out to be the only way to describe the new, and detailed experiments which were just then being performed. He did this practically overnight, openly reporting his change of mind to his scientific colleages, in the October, November, and December meetings of the German Physical Society, in Berlin, where the black body work was being intensely discussed. In this way, careful experimentalists (including F. Paschen, O.R. Lummer, E. Pringsheim, H.L. Rubens, and F. Kurlbaum), and a reluctant theorist, ushered in the greatest revolution science has ever seen. ### The quantum black-body radiation formula When a body is heated, it emits heat radiation, which is infrared light, being a form of electromagnetic waves. All of this was well understood at the time, and of considerable practical importance. When the body becomes red-hot, the red wavelength parts start to become visible. This had been studied over the previous years, as the instruments were being developed. However, most of the heat radiation remains infrared, until the body becomes as hot as the surface of the Sun (about 6000 °C, where most of the light is green in color). This was not achievable in the laboratory at that time. What is more, to measure specific infrared wavelengths was only then becoming feasible, due to newly developed experimental techniques. Until then, most of the black body spectrum was not measurable, and therefore not mapped out in detail. The quantum black-body radiation formula, being the very first piece of quantum mechanics, appeared sunday evening October 7, 1900, in a so-called back-of-the-envelope calculation by Planck. It was based on a report by Rubens (visiting with his wife) of the very latest experimental findings in the infrared. Later that evening, Planck sent the formula on a postcard, which Rubens had the following morning. A couple of days later, he could tell Planck that it worked perfectly. As it does to this day. At first, it was just a fit to the data. Only weeks later did it turn out to enforce quantization. That the latter became possible involved a certain amount of luck (or skill, even though Planck himself called it "a fortuitous guess at an interpolation formula"). It only had that drastic "side effect" because the formula happened to become fundamentally correct, in regard to the as yet non-existent quantum theory. And normally, that much is not at all expected. The skill lay in simplifying the mathematics, so that this could happen. And here Planck used hard won experience from the previous years. Briefly stated, he had two mathematical expressions: • (i) from the previous work on the red parts of the spectrum, he had x; • (ii) now, from the new infrared data, he got x². Combining these as x(a+x), he still has x, approximately, when x is much smaller that a ( the red end of the spectrum). But now also x², again approximately, when x is much larger than a (in the infrared). The luck part is that, this procedure turned out to actually give something completely right, far beyond what could reasonably be expected. The formula for the energy E, in a single mode of radiation at frequency f, and temperature T, can be written $E = \frac{h f}{e^{\frac{h f}{k T}} - 1}$ This is (essentially) what is being compared with the experimental measurements. There are two parameters to determine from the data, written in the present form by the symbols used today: h is the new Planck's constant, and k is Boltzmann's constant. Both have now become fundamental in physics, but that was by no means the case at the time. The "elementary quantum of energy" is hf. But such a unit does not normally exist, and is not required for quantization. ### The birthday of quantum mechanics From the experiments, Planck deduced the numerical values of h and k. Thus he could report, in the German Physical Society meeting on December 14, 1900, where quantization (of energy) was revealed for the first time, values of the Avogadro-Loschmidt number, the number of real molecules in a mole, and the unit of electrical charge, which were more accurate than those known until then. This event has been referred to as "the birthday of quantum mechanics". ## Quantization in antiquity In a sense, it can be said that the quantization idea is very old. A string under tension, and fixed at both ends, will vibrate at certain quantized frequencies, corresponding to various standing waves. This, of course, is the basis of music. The basic idea was regarded as essential by the Pythagoreans, who are reported to have held numbers in high esteem. It is a curious fact that, the famous formula, named after Pythagoras, for the side lengths of a right triangle, today serves as a cornerstone of quantum mechanics as well. The very existence of atoms, molecules, solids, and so on, can be ascribed to various forms of quantization. Contrary to notions of matter as some form of continuous medium. This was also understood already in antiquity, particularly by Leucippos and Democritos, although not generally appreciated, even by physicists, really, until the invention of quantum mechanics. In should mentioned, though, that later works within the Epicurean school of thought played a significant role in forming the physics and chemistry of the Renaissance period in Europe. ## References • M. Planck, A Survey of Physical Theory, transl. by R. Jones and D.H. Williams, Methuen & Co., Ltd., London 1925 (Dover editions 1960 and 1993) including the Nobel lecture. • J. Mehra and H. Rechenberg, The Historical Development of Quantum Theory, Vol.1, Part 1, Springer-Verlag New York Inc., New York 1982. ## See also 03-10-2013 05:06:04	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 1, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9660210609436035, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/52394/is-there-a-non-trivial-example-for-a-1-homogeneous-function-satisfying-a-specific/90997	## Is there a non-trivial example for a 1-homogeneous function satisfying a specific inequality of second order? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Hello everybody, Actually, I'm thinking about the following problem: Let $\mathbb{R}^n$ be the $n$-dimensional real vector space with Cartesian coordinates $x=(x^1,\ldots, x^n)\in \mathbb{R}^n$. I'm searching for a non-trivial example of a function $A:\mathbb{R}^n \rightarrow \mathbb{R}$, which is continuously differentiable on $\mathbb{R}^n\backslash \lbrace 0\rbrace$ (i.e. $A\in C^\infty(\mathbb{R}^n\backslash \lbrace 0\rbrace)$), such that it satisfies \begin{align} A(tx)&=tA(x),\newline \sum^{n}_{i,j=1}\xi^i A(x)\frac{\partial^2 A(x)}{\partial x^i \partial x^j} \xi^j &\le 0, \end{align} for all $t\in \mathbb{R}$, $x, \xi \in \mathbb{R}^n$. Obviously, if $A$ is chosen to be a linear function, it is a solution to the problem above. But are there any non-linear solutions for dimension $n\ge 3$? Is there some topological argument against the existence of such solutions? - Can you phrase what you want in terms of real projective space of dimension one less? The Hessian would need to be "reduced" but there is one degree of freedom that can be removed, e.g. by means of Euler's theorem on homogeneous functions. The topology of the projective space is of course well known, and given that it has only torsion in homology in the middle dimensions, it isn't immediately apparent that there is an obstruction. – Charles Matthews Jan 18 2011 at 11:52 In the definition, do you want $t\in\mathbb{R}$ or $t\in\mathbb{R}_+$? – Willie Wong Jan 18 2011 at 15:20 @Charles: I have already tried to rewrite it as a problem on the unit-sphere (for $m=3$) to get rid of the homogenity condition, so I think it would also work for the real projective space. In the unit sphere case, I used a polarcoordinate representation. But the resulting polar coordinate version of the differential inequality condition does not have a nice form. @Willie: I do actually want $t\in \mathbb{R}$. So $A$ is an function for which holds \begin{align} A(-x)&=-A(x)\\ A(\lambda x)&= \lambda A(x) \end{align} for all $x\in\mathbb{R}^n$ and $\lambda\in\mathbb{R}_{+}$. – Patrick Jan 19 2011 at 15:23 ## 2 Answers You can define $A(x_1,x_2,\dots,x_n)=B(x_1,x_2)$ where $B$ is a solution for $n=2$. As for $B$, one can e.g. define it on the unit circle by $B(\sin t,\cos t)=\cos 3t$ and extend by homogeneity. - ### 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. Take $A(x)=\vert x\vert$ the Euclidean norm in $\mathbb R^n$, which is obviously homogeneous of degree 1. We have $$A'(x)=\frac{x}{A(x)},\quad A''(x)=\frac{Id}{A(x)}-\frac{x\otimes x}{A(x)^3}=A(x)^{-1} \underbrace{\bigl(Id-\frac{x\otimes x}{\vert x\vert^2}\bigr)}_{L(x)}.$$ The matrix $L(x)$ is the matrix of the projection onto the orthogonal of $x$ and is thus a non-negative matrix. That example does not answer your question, but nevertheless gives an example of an homogeneous $A$ with degree 1 such that $$A(x) A''(x)\ge 0,$$ which has also the linear forms as solutions. Bazin. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 27, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.918494701385498, "perplexity_flag": "head"}
http://mathoverflow.net/revisions/33822/list	## Return to Answer 3 added 1205 characters in body LATER EDIT: The very nice survey article by Steuding that David Speyer mentions in his comment actually refers for greater detail to the book by Mark Kac, an M.A.A. Carus Monograph, called "Statistical Independence in Probability, Analysis and Number Theory." Chapter 4 is called "Primes play a game of chance" and section 2 is called "The statistics of the Euler $\phi$-function." That begins on page 54. In the section Problems, pages 62-64, we learn that $$\frac{\sigma(n)}{n}$$ does in fact have a limiting distribution (proved by Davenport, methods improved by Erdos), and this density$$D \left\{ \frac{\sigma(n)}{n} < \omega \right\} = \tau(\omega)$$ is a continuous function of $\omega$. There is not much more to hope for in details, as Erdos showed that the analogous density for $$\log \frac{\phi(n)}{n}$$ is continuous but "singular," that is has derivative 0 almost everywhere. However, Davenport's result does show that the abundant and deficient numbers both have densities, while the perfect numbers have density 0. While no "variance" is mentioned, a mean for the distribution is given, $$M \left\{ \frac{\sigma(n)}{n} \right\} = \frac{\pi^2}{6}$$ORIGINAL: There is a nice survey on fairly elementary methods here by J. L. Nicolas, in a 1988 book called "Ramanujan Revisited." Meanwhile, there is an unconditional result which has not been mentioned, for $N \geq 3$ we have 2 added 254 characters in body There is a nice survey on fairly elementary methods here by J. L. Nicolas, in a 1988 book called "Ramanujan Revisited." Meanwhile, there is an unconditional result which has not been mentioned, for $N \geq 3$ we have $$\sigma(N) < e^\gamma \; N \log \log N + \frac{ 0.6482 N}{\log \log N}$$ I hope I am reporting this correctly, it is from a secondary source, attribution is to G.Robin, Grandes valeurs de la fonction somme des diviseurs et hypothese de Riemann, J. Math. Pures Appl. (9) 63 (1984) 187-213. The overall methodology is to consider the colossally abundant numbers of Alaoglu and Erdos (1944), http://en.wikipedia.org/wiki/Colossally_abundant_number which were eventually discovered to have also been present in the original version of Ramanujan's paper Highly Composite Numbers (1915). There is some history about why that section was initially omitted, I forgetevidently a paper shortage. Here is a link to the first page of a related recent article, also apparently a survey, by Nicolas: http://www.springerlink.com/content/p8311481mh32145v/ 1 There is a nice survey on fairly elementary methods here by J. L. Nicolas, in a book called "Ramanujan Revisited." Meanwhile, there is an unconditional result which has not been mentioned, for $N \geq 3$ we have $$\sigma(N) < e^\gamma \; N \log \log N + \frac{ 0.6482 N}{\log \log N}$$ I hope I am reporting this correctly, it is from a secondary source, attribution is to G.Robin, Grandes valeurs de la fonction somme des diviseurs et hypothese de Riemann, J. Math. Pures Appl. (9) 63 (1984) 187-213. The overall methodology is to consider the colossally abundant numbers of Alaoglu and Erdos (1944), which were eventually discovered to have also been present in the original version of Ramanujan's paper Highly Composite Numbers (1915). There is some history about why that section was omitted, I forget.	{"extraction_info": {"found_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": 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": 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": 0.9446113705635071, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/289128/geometric-interpretation-of-a-vector-space-and-subspace?answertab=votes	# Geometric interpretation of a vector space and subspace? I understand how to manipulate vector spaces and subspaces and how to prove various statements about them, but I still don't fully understand what they represent geometrically. I just need an intuitive grasp as to what these are. Is a vector space a geometric area that contains all possible vectors of the field $\Bbb F$? For example, if $V$ is a vector space over $\Bbb R^3$ then does that mean $V$ contains all vectors in three-dimensions that are part of $\Bbb R$? But then what is a subspace of $V$? Would that perhaps be a plane? Would it be a vector space in $\Bbb R^2$? - The way I think about it is that, vector space is a collection of objects. We have defined (our own) way of "linearly Combining" given any $N$ objects and any $N$ numbers from field $F$ for every $N$. Then, if every such possible linear combination gives a object inside the set, then its a vector space. Now as for a subspace. It is a strict subset of the original set, which has the same properties as the orginal set. – dineshdileep Jan 28 at 17:56 – Fly by Night Jan 28 at 17:56 ## 1 Answer The classic example of a $3$-dimensional real vector space is the Euclidean space $\mathbb R^3$. Its $2$-dimensional subspaces are precisely all planes passing through the origin, and its $1$-dimensional subspaces are precisely all lines passing through the origin. The origin itself forms the singleton $\{0\}$ which is the only $0$-dimensional subspace. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 17, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9312070608139038, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/89796/prove-the-following-function-is-riemann-integrable-on-0-1	# Prove the following function is Riemann Integrable on $[0,1]$. Suppose that $f:[0,1] \to \mathbb{R}$ is defined by $f(x) = 1$ when $x = \frac{1}{n}$ for some positive integer $n$ and $f(x) = 0$ otherwise. How can I prove that $f$ is Riemann integrable on $[0,1]$? - 3 Cover each $x = \frac1n$ by interval of size $\frac{\epsilon}{2^n}$ i.e. consider intervals around $\frac1n$ as $\left(\frac1n - \frac{\epsilon}{2^{n+1}}, \frac1n + \frac{\epsilon}{2^{n+1}} \right)$ – user17762 Dec 9 '11 at 2:51 @Sivaram Perhaps I am missing something, but I am unsure how to proceed from your hint. Doesn't Riemann integration consider only finite partitions of $[0,1]$? – Srivatsan Dec 9 '11 at 2:55 2 @Srivatsan: This should go with the previous hint. Choose $N$ such that $\frac1N < \frac{\epsilon}{2}$. For all $n < N$, cover each $x = \frac1n$ by interval of size $\frac{\epsilon}{2^{n+1}}$ i.e. consider intervals around $\frac1n$ as $\left(\frac1n - \frac{\epsilon}{2^{n+2}}, \frac1n + \frac{\epsilon}{2^{n+2}} \right)$ and for the rest use an interval of the form $[0,\epsilon/2)$ – user17762 Dec 9 '11 at 2:59 – Henry Shearman Dec 9 '11 at 3:00 ## 2 Answers Here's a more elementary argument. We show that for every tagged partition $P$ of $[0,1]$ of norm at most $\varepsilon \gt 0$, the lower and upper Riemann sums $L(f; P)$ and $U(f;P)$ of $f$ satisfy $$0 \leqslant L(f, P) \leqslant U(f;P) \leqslant 2\sqrt{\varepsilon} . \tag{1}$$ The Riemann integrability of $f$ follows as a consequence. The lower bound is obvious; we show only the upper bound of $O(\sqrt{\varepsilon})$ here. Fix a parameter $t \in (0,1)$ (to be determined later). Without loss of generality With a modest loss in generality, assume that $t$ is the endpoints of two intervals in $P$. [I leave it to the OP to remove this assumption.] The strategy is to bound the contributions of the two types of intervals individually: • Those that fall to the left of $t$ (i.e., those that are a subset of $[0, t]$). The contribution from these intervals to the Riemann sum is at most $t$, the total length of the left part (since the function is always bounded above by $1$ in magnitude). • Those that fall to the right of $t$ (i.e., those that are a subset of $[t, 1]$). Of all such intervals, at most $1/t$ of them contain a point of the form $1/n$. Therefore, all but $1/t$ of them contribute zero to the Riemann sum. Since each of these intervals is of length at most $\varepsilon$, the total contribution of these to the Riemann sum is at most $\varepsilon \cdot \frac{1}{t} = \frac{\varepsilon}{t}$. Adding the two contributions above, we get that the total Riemann sum is at most $t + \frac{\varepsilon}{t}$. Picking $t = \sqrt{\varepsilon}$ to optimise this upper bound, we get $(1)$. - The discontinuities of the function are in the form $x=\frac{1}{n}$. Using the given hint by Sivaram Ambikasaran in the comments to construct covering intervals, you can show that the set $$D=\left\{x\in[0,1]:x=\frac{1}{n}\right\}$$ has measure zero which satisfies the Lebesgue criterion for integrability. - 4 Applying the Lebesgue criterion for this problem seems like an overkill. :-) – Srivatsan Dec 9 '11 at 3:01 2 This is a nice and probably the efficient way to go provided the OP knows Lebesgue integral in the first place. – user17762 Dec 9 '11 at 3:07 Probably is overkill, but the interval construction works very nicely in showing that the sets have measure zero. – Henry Shearman Dec 9 '11 at 3:08 2 The extra work to get a finite partition is so small, why not go ahead and do it? – GEdgar Dec 9 '11 at 20:07	{"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": 47, "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": 0.9154782891273499, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/tagged/quantum-optics%20quantum-computation	# Tagged Questions 0answers 37 views ### What is three-photon interference? Whilst reading this paper on a quantum processor that performs a type of matrix computation, I came across the concept of 'three-photon interference'. A quick Google search shows that this process is ... 1answer 87 views ### Using wavepackets instead of photons in quantum computer Why does a photonic quantum computer require photons? Why wouldn't wave packets work just as well, better in fact since it would get away from the use of fragile single photons? (Article) 0answers 74 views ### Is Photonics going to be the next phase in Electronics? Considering the recent Nobel Prize 2012 award for Physics [closed] Photonics seems to be a rather closed field involving only specialists and experts working on it. Will there be a period which is going to define the power of Photonics in which Electronics is ... 1answer 154 views ### Quantum Cryptography First question was a little bit ambiguous. Photons are passed through a linear polarizer that is oriented $\theta$ degrees again the photon passes through another linear polarizer that also have a ... 2answers 813 views ### Quantum memories: What are they? Searching the literature for the term "quantum memory" seems to bring up results from two different communities. On the one hand there are quantum opticians, who see a quantum memory as something ... 1answer 255 views ### Which qubit states are accessible with linear optics operations? Given a quantum state of $n$ qubits, and being restricted to linear optics (that is, the output annihilation operators are linear combinations of the input annihilation operators): Which states are ...	{"extraction_info": {"found_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": 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": 0.947799026966095, "perplexity_flag": "middle"}
http://en.wikibooks.org/wiki/Trigonometry/Introduction	# Trigonometry/Introduction Trigonometry is the study of the properties of triangles. • "Tri" is Ancient Greek word for three, • "gon" means side, • "metry" measurement Together they make "measuring three sides". If you know some facts about a triangle, such as the lengths of its sides, then using trigonometry you can find out other facts about it. If you know the lengths of sides then you can find what the angles are. If you know the length of one side and of two of the angles, then you can work out what the remaining angle is, and also what the lengths of the other two sides are. As a consequence the Ancient Greeks were able to use trigonometry to calculate the distance from the Earth to the Moon. ## Starting Learning Trigonometry One of the first things we learn in trigonometry is how to calculate what the third angle in a triangle is when given the other two angles. For example, in a triangle if two of the angles are $60^{\circ}\,$ then the third angle must be $60^{\circ}\,$ too. However, if we're instead told two lengths, say 7cm and 10cm, and if we're not told anything about the angles, then the third length has a range of possibilities. It could be 5cm. It could be 15cm. To calculate lengths we start out by examining a right-angled triangle, i.e. one where one of the angles is $90^{\circ}\,$. A right-angle triangle We'll be using right-angled triangles a lot. In a right-angled triangle we can work out the length of the longest side if given the lengths of the two shorter sides. In a right-angled triangle the lengths of the sides have a particularly nice relationship to each other. The formula for this relationship is written: $a^2 + b^2 = c^2\!\,$ How you use this formula to calculate the third side when you know two of the sides in a right-angled triangle is explained in the page on the 'Pythagoras' Theorem' What if the triangle isn't a right-angled triangle, though? What if it doesn't have a right angle? More complex triangles can be built by joining right-angled triangles together. What we learn about right-angled triangles enables us to work with other kinds of triangles too. Trigonometry and trigonometric functions can also be used with more complex shapes such as squares, hexagons, circles and ellipses. Ultimately the most important mathematical tools we have for measuring the universe are based on the study of the mathematics of triangles. Trigonometry is a fundamental step in your mathematical education. From the seemingly simple shape, the right triangle, we gain tools and insight that help us in practical and theoretical endeavors. The subtle mathematical relationships between the right triangle, the circle, the sine wave, and the exponential curve can only be fully understood with a firm foundation in trigonometry. ## Applications Triangles used in architecture. This is part of a roof. Triangles and trigonometry are important in engineering. Trigonometry is a branch of mathematics concerned with the lengths of the sides and size of the angles of triangles. Because many physical questions can be framed in terms of triangles, trigonometry has found wide use in the physical sciences and engineering. Trigonometry is needed in surveying and architecture since relationships between lengths and angles in triangles are used directly. 3 phase AC waveform. Trigonometry functions are used in Electrical Engineering Trigonometry is also used in electrical engineering. The functions that relate angles and side lengths in right triangles turn out to also be useful in expressing how AC electric current varies with time.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 4, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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": 0.9224475026130676, "perplexity_flag": "head"}
http://stats.stackexchange.com/questions/26431/chi2-test-correcting-for-not-fully-independent-sample	# $\chi^2$ test - correcting for not fully independent sample I have carried out a study in which I have gathered 3 sentences per participant. These sentences were then classified in 2 ways. I want to test if there are significant interactions between the classifications and find out the effect size. I have done a chi square test for independence and calculated Cramer's $V$ (on the counts of sentences for the cross of the 2 classifications). However, I am concerned that this is not correct, since the sentences are not fully independent (because each participant has entered 3 sentences). Is there a good way to correct for this (or are they other statistical tests that I should carry out instead of $\chi^2$ and/or Cramer's $V$)? A colleague of mine suggested removing the sentences that were entered by participants who had sentences in more than 1 category of a classification. - 2 I'm not quite clear on what your data look like. How does a "query" (in your last paragraph) relate to a "sentence" in the rest of the question? But I've answered it on the basis of what I think you are asking. – Peter Ellis Apr 14 '12 at 2:52 Sorry I meant to write 'sentences' - edited. Thanks! – Lars Grammel Apr 14 '12 at 5:45 ## 1 Answer I read this to mean that each sentence is classified in both ways, so you are interested in whether there is any relationship between the two classification methods. If so I don't think it matters where the sentences came from - even if they are random or not. The randomness that makes this a statistical question comes from the idea that under the null hypothesis the two classifications of each sentence are random and unrelated, and you are looking for evidence that this is not the case. You are making inferences about the population of classifications, not of sentences. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 4, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.970932126045227, "perplexity_flag": "head"}
http://mathhelpforum.com/advanced-applied-math/84175-converting-sinosoids-complex-exponentials-print.html	# Converting Sinosoids to complex exponentials Printable View • April 17th 2009, 05:15 AM briwel Converting Sinosoids to complex exponentials I am doing some revision exercises and have come across this question: Write the following sinusoid as a complex exponential and plot both in time and frequency domains. x(t) = 2sin(200 pi t)+ 3sin(400 pi t) I'm not sure how to do this. Is it something to do with the complex exponential version of Fourier transform? • April 17th 2009, 06:28 AM Mush Quote: Originally Posted by briwel I am doing some revision exercises and have come across this question: Write the following sinusoid as a complex exponential and plot both in time and frequency domains. x(t) = 2sin(200 pi t)+ 3sin(400 pi t) I'm not sure how to do this. Is it something to do with the complex exponential version of Fourier transform? No it's more to do with the fact that $\sin(x) = \frac{1}{2i}(e^{ix} - e^{-ix})$ • April 17th 2009, 07:04 AM briwel Yeah I thought Fourier was a bit complicated. In the example I have given is there anything else that can be done to further simplify the formula after substituting x for 2sin(200pi t) ? • April 17th 2009, 07:08 AM Mush Quote: Originally Posted by briwel Yeah I thought Fourier was a bit complicated. In the example I have given is there anything else that can be done to further simplify the formula after substituting x for 2sin(200pi t) ? $x(t) = 2\sin(200 \pi t)+ 3\sin(400 \pi t) = \frac{2}{2i}(e^{i 200 \pi t} - e^{-i 200 \pi t}) + \frac{3}{2i}(e^{i 400\pi t} - e^{-i 400 \pi t})$ Now... $e^{i 200 \pi t} = \bigg(e^{i \pi}\bigg)^{200t}$ And $e^{i \pi} = -1$ so $e^{i 200 \pi t} = \bigg(-1\bigg)^{200t}$ The same logic can be applied to the other 3 terms. • April 17th 2009, 07:14 AM Mush Actually. The LaPlace transform of this would be a good idea. • April 17th 2009, 07:45 AM briwel Thanks Mush. Any idea about the plotting of the graphs? I'm assuming the time domain is pretty simple, just plotting the formula against increasing values of t. I'm not sure about frequency though... • April 17th 2009, 07:59 AM Mush Quote: Originally Posted by briwel Thanks Mush. Any idea about the plotting of the graphs? I'm assuming the time domain is pretty simple, just plotting the formula against increasing values of t. I'm not sure about frequency though... When you perform the LaPlace transform you will be given the function in terms of the complex variable s. • April 17th 2009, 08:40 AM briwel I didn't actually use laplace transform as its not something that I have been taught so I assume I'm not meant to do it that way. Is there another way to plot the frequency graph? • April 17th 2009, 12:59 PM briwel also, the equaion always seems to equat to 0. Is that correct? All times are GMT -8. The time now is 02:15 AM.	{"extraction_info": {"found_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": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9331254363059998, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/122807?sort=oldest	## Class of functions that the Fourier inversion holds ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) The following is from Stein and Shakarchi's Complex Analysis: For each $a>0$ we denote by ${\mathcal F}_a$ the class of all functions $f$ that satisfy the following two conditions: 1. The function $f$ is holomorphic in the horizontal strip `$$S_a=\{z\in{\Bbb C}:\|Im(z)\|<a\}$$` 2. There exists a constant $A>0$ such that ```$$ \|f(x+iy)\|\leq\frac{A}{1+x^2}\quad\text{for all}\quad x\in{\Bbb R}, \|y\|<a. $$``` Denote by ${\mathcal F}$ the class of all functions that belong to ${\mathcal F}_a$ for some $a$. Then the Fourier inversion holds in this class. My questions are: is there a name for this class? Does it have anything to do with the Schwartz space on which the Fourier transform is a linear isomorphism? - 1 The Fourier transform is an isomorphism also on $\mathscr S'(\mathbb R^d)$, the space of tempered distributions. – Jochen Wengenroth Feb 24 at 17:14 ## 3 Answers This space has no name (no that I know), and it is probably invented by Stein and Shakarchi for some pedagogical purpose, or for convenience of exposition. As Serge says it is not contained in the Schwartz space $S$ (of test functions). Neither it contains the Schwartz space $S$. It is contained in the space of Schwartz distributions $S^\prime$ but probably the authors did not want to introduce distributions in the elementary Complex Analysis textbook. As it contains all examples they want to consider in this book, they introduced it. There are several other spaces where Fourier transform can be defined, and even the image can be explicitly described, for example, $L^2$, or more general distributions than Schwartz, but I agree with Bazin that Schwartz distributions is perhaps the most useful class. In an elementary textbook, one has to begin with something, so they invented this artificial class. In the later volumes of the course they address $L^2$ and Schwartz distributions. - ### 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 function $f(x)=1/(1+x^2)$ belongs to this class but not to the Schwartz space of test functions. Any $C^\infty$ function with bounded support belongs to the Schwartz space but not to this class. - The optimal space for the Fourier transform is the space of tempered distributions $\mathscr S'(\mathbb R^n)$, i.e. the dual space of the Schwartz functions $\mathscr S(\mathbb R^n)$. The latter is the Fréchet space of $C^\infty$ functions on $\mathbb R^n$ decreasing faster than any polynomial as well as all their derivatives; the semi-norms are $$p_{\alpha\beta}(\phi)=\sup_{\mathbb R^n}\vert x^\beta\partial^\alpha\phi\vert.$$ Defining $\hat \phi(\xi)=\int e^{-2i\pi x\cdot\xi} \phi(x) dx$ for $\phi\in\mathscr S(\mathbb R^n)$, it is easy to prove that the Fourier transform is an isomorphism of $\mathscr S(\mathbb R^n)$ and $$\phi(x)=\int e^{2i\pi x\cdot\xi} \hat\phi(\xi) d\xi.\tag{1}$$ The dual space of $\mathscr S(\mathbb R^n)$ is huge, contains for instance all the spaces $L^p(\mathbb R^n)$ and all the distribution derivatives of any function in these spaces. The space $\mathscr S'(\mathbb R^n)$ contains also the homogeneous distributions and the distributions with compact support. It is easy to define by duality the Fourier transform on $\mathscr S'(\mathbb R^n)$. With brackets of duality, we define $$\langle\hat T,\phi\rangle=\langle T,\hat \phi\rangle\quad\text{and we have $\hat{\hat T}= \tilde T$,}$$ where $\tilde T(x)=T(-x)$, that is the very same formula as in (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": 28, "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": 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": 0.9256459474563599, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/219775/how-to-approach-probability-questions	# How to approach probability questions? I have four questions: The time that it takes to assemble a piece of machinery is well modeled by the normal distribution with mean of 72.9 minutes and standard deviation of 8.55 minutes. What is the probability that it will take less than an hour to assemble the next piece of machinery? What is the probability that it will take between 65 minutes and 75 minutes to assemble the next piece of machinery? What is the probability that it will take more than 80 minutes to assemble the next piece of machinery Find the 16th percentile of the random variable, time taken to assemble. This is the time that is exceeded 84% of the time I'm not sure how to approach these questions, can someone guide me through them possibly? - ## 2 Answers You want to know the following : $P(X\leq60)=P(\frac {X-\mu}{\sigma}\leq \frac {60-72.9}{8.55})$.Then you will find (after calculations...) :$P(Z\leq -1.5)$.Using the symmetry of the normal curve, you must find: $0.5- P(0\leq Z\leq 1,5)$.For a better understanding, take a look at the curve. Then you just have to look at the Normal table (look at the interval you found ), which will give you the probability. For the other questions you will do just the same.Paying attention to the area.Try to draw the curve and the desired probability.It always helps. - oh of course, the normal-curve area tables, I forgot I even had them. – Unknown Oct 24 '12 at 4:47 although, can you explain why it's $0.5-P$ – Unknown Oct 24 '12 at 4:50 Yes,I can.if you look the graph, as i said , the whole area is one, you know it is symmetrical,then the half of it has area 0.5.you want the area under the graph correspondind to less than x, for example.your table only gives you probability to values greater than zero and less than the given interval.by the symmetry you have positive values .then look in the table with $P(0\leq Z\leq x)$, how. the half of the graph is 0.5 and you have the previous area the implies 0.5- $P(0\leq Z\leq x)$.if you did not understood you can email me. – Charlie Oct 24 '12 at 16:26 Write $X=$time to assemble a piece of machinery. We model $X$~$N(72.9, 8.55^{2})$. For Q1. $P(X < 60) = P(Z < (60 - 72.9)/8.55)$. Then, obtain this probability either with tables or software. (Here $Z$ is a random variable with the standard normal distribution). The other two are approached similarly. The last one is slightly different. Here, we want to find $t$ such that $P(X > t) = 0.84$. Then, we know that $P(Z > (t - 72.9)/8.55) = 0.84$. Again, consult tables or software to find the value of (t - 72.9)/8.55 which satisfies 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": 14, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9409635066986084, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/283893/finding-total-amount-of-permutations-between-edges-and-nodes-in-a-cycle/283899	# Finding total amount of permutations between edges and nodes in a cycle I am interested in the graph theory/combinatorics problem: if you have a cycle, how many ways can you rearrange the edges and nodes such that you still only have one cycle? any discussion is welcome To state it another way If the graph itself was simply one large cycle, how many ways could I rearrange the edges while maintaining a single cycle? Only one incoming and outgoing edge per node and therefore the length of the cycle must be the same with each permutation. nodes are marked edges are not. Does that may things clearer? - Are you asking about rearranging a graph without touching that specific cycle? (Otherwise the fact it had a cycle originally means nothing more than a min length requirement) – Guest 86 Jan 21 at 23:59 If the graph itself was simply one large cycle, how many ways could I rearrange the edges while maintaining a single cycle? Only one incoming and outgoing edge per node and therefore the length of the cycle must be the same with each permutation. Does that may things clearer? – user1340048 Jan 22 at 0:08 Yes, are both the edges and nodes marked? – Guest 86 Jan 22 at 0:14 nodes are marked edges are not – user1340048 Jan 22 at 0:14 ## 1 Answer If you have a cycle graph with $N$ nodes and edges, you can permute the nodes among themselves in $N!$ ways, without altering the shape (Is the word here topology?). But as this results in similar graphs up to rotation, you should either treat it as holding a node constant and performing the permutation on the remaining $N-1$ nodes for a $(N-1)!$ result or simply divide the previous result by $N$ for there are this many equivalence classes. - This overcounts the possibilities because a cycle can be rotated without giving a different graph. Hanging our hats on the OP's commment (not entirely consistent with how the question is asked) that "nodes are marked[;] edges are not", one wants a count of permuting "the nodes among themselves" that gives distinct cycles. – hardmath Jan 22 at 0:20 hardmath, that is exact what I want. How many distinct cycles can I obtain. – user1340048 Jan 22 at 0:22 1 Ah my bad, should've held one node as a constant permuting the remaining $N-1$ or simply divide the result by $N$ as there are N equivalent classes. I'll fix it, sorry. – Guest 86 Jan 22 at 0:29	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.9415320754051208, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/7505/are-jacobians-principally-polarized-over-non-algebraically-closed-fields	## Are Jacobians principally polarized over non-algebraically closed fields? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) How does one define the Torelli map $M_g \to A_g$ of moduli stacks? On geometric points a curve maps to its principally polarized Jacobian. So what I am asking is: if I have a curve $C$ over a non-algebraically closed field $k$ such that $C(k)$ is empty, is the Jacobian of C still principally polarized? After base change to $\bar{k}$ one has a theta divisor; does it descend? Also, is the relative Jacobian of a family of curves principally polarized? The thing I am confused about is that the theta divisor naturally lives on $Pic^{g-1}$ as the image of the map from the symmetric power $C^{g-1}$; this is a torsor under $Pic_0$, but not itself an abelian variety. Also, the classical Torelli theorem says that this map is an injection on field valued points. Is this actually a locally closed immersion of stacks? - ## 2 Answers There's a more down to earth way to deal with this, which is already explained in Mumford's GIT: make an fppf (or even etale) surjective base change to acquire a section, use that to define the principal polarization, and then show it is independent of the choice. (Short reason: varying the choice amounts to a morphism from the smooth proper curve to a Hom or Isom scheme that is unramified over the base, hence constant.) Thus, by descent theory one gets the polarization over the original base. This is related to the same issue which comes up in explaining why a polarization of an abelian scheme need not arise as the "Mumford construction" $\phi_{\mathcal{L}}$ even though it automatically does so on geometric fibers (due to the special nature of $k$-simple finite commutative $k$-groups when $k = \overline{k}$). That is, a definition of "polarization" which is better-suited to the relative case is not to mimic what one traditionally does over an algebraically closed field (the Mumford construction) but rather something which makes more effective use of the Poincar\'e bundle. The possible lack of $\mathcal{L}$ over the base is analogous to the possible lack of a section of the curve to define the principal polarization. See the Wikipedia page on abelian varieties'' for more on this. :) - ### 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. Incidentally, as I posted this question someone who knew the answer wandered into my office. The map $M_g \to A_g$ factors through the moduli space $\tau_g$ of pairs (A,P,L) where A is an abelian variety, P is an A torsor, and L is an ample line bundle on P which is geometrically a principal polarization. The map $M_g \to \tau_g$ is given by $C \mapsto (Pic_0, Pic_{g-1}, L(\theta))$, where the theta divisor on $Pic_{g-1}$ is given by the image of $C^{g-1}$. To construct the map $\tau_g \to A_g$, note that $Pic_0(A) \cong Pic_0(P)$, so that L indeed gives a map $A \to A^{\vee}$ given by $a \mapsto t^*_aL \otimes L^{-1}$. The point is one doesn't need to descend the theta divisor. The reference to this is 5.1 of Martin Olsson's book Compactifying moduli spaces of abelian varieties. - This is really cool! Thank you for sharing. – Yuhao Huang Dec 1 2009 at 23:12 3 "The reference to this is 5.1 of Martin Olsson's book Compactifying moduli spaces of abelian varieties": which in turn refers to Alexeev "Compact moduli..." for this construction, which in turn is based on Raynaud's "Faisceaux ample..." – VA Feb 13 2010 at 15:56	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 23, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9362395405769348, "perplexity_flag": "head"}
http://anhngq.wordpress.com/2011/12/	# Ngô Quốc Anh ## December 31, 2011 ### A Hardy-Moser-Trudinger inequality: A conjecture by Wang and Ye Filed under: PDEs — Tags: Hardy's inequality, Moser-Trudinger's inequality — Ngô Quốc Anh @ 21:48 Let $B$ denote the standard unit disk in $\mathbb R^2$. The famous Moser–Trudinger inequality says that $\displaystyle\int_B {\exp \left( {\frac{{4\pi {u^2}}}{{\left\\| {\nabla u} \right\\|_2^2}}} \right)dx} \leqslant C < \infty ,\quad\forall u \in H_0^1(B)\backslash \{ 0\}$ holds. There is another important inequality in analysis, the Hardy inequality which claims that $\displaystyle H(u) = \int_B {\|\nabla u{\|^2}dx} - \int_B {\frac{{{u^2}}}{{{{(1 - \|x{\|^2})}^2}}}dx} \geqslant 0,\quad\forall u \in H_0^1(B)$ holds. The one $H$ is usuall called the Hardy functional. One can immediately see that $\displaystyle\frac{{4\pi {u^2}}}{{\left\\| {\nabla u} \right\\|_2^2}} \leqslant \dfrac{{4\pi {u^2}}}{{\displaystyle\int_B {\|\nabla u{\|^2}dx} - \int_B {\frac{{{u^2}}}{{{{(1 - \|x{\|^2})}^2}}}dx} }}$ for any $u \in H_0^1(B)\backslash \{ 0\}$. Recently, in a paper accepted in Advances in Mathematics journal, Wang and Ye proved that there exists a constant $C_0 >0$ such that the following $\displaystyle\int_B {\frac{{4\pi {u^2}}}{{H(u)}}dx} \leqslant C_0 < \infty ,\quad\forall u \in \mathcal H(B^n)\backslash \{ 0\}$ where $B^n$ is the unit ball in $\mathbb R^n$, $n \geqslant 2$ and $\mathcal H=\mathcal H(B^n)$ is the complement of $C_0^\infty(B^n)$ with respect to the following norm $\\|u\\|_{\mathcal H}=\sqrt{H(u)}$. Let us go back to the case $n=2$. They then defined $\displaystyle {H_d}(u) = \int_\Omega {\|\nabla u{\|^2}dx} - \frac{1}{4}\int_\Omega {\frac{{{u^2}}}{{d{{(x,\partial \Omega )}^2}}}dx} > 0,\quad \forall u \in H_0^1(\Omega )\backslash \{ 0\}$ where $\Omega$ is a regular, bounded and convex domain sitting in $\mathbb R^2$. They then conjectured that the following $\displaystyle\int_\Omega {\frac{{4\pi {u^2}}}{{{H_d}(u)}}dx} \leqslant C(\Omega ) < \infty ,\quad\forall u \in {\mathcal H_d}(\Omega )\backslash \{ 0\}$ still holds for some constant $C(\Omega)>0$ where ${\mathcal H_d}(\Omega )$ denotes the completion of $C_0^\infty (\Omega)$ with the corresponding norm associated with $H_d$. Apparently, the conjecture holds true for $\Omega = B$. Theme: Shocking Blue Green. Blog at WordPress.com.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 25, "mimetex.cgi": 0, "/images/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": 0.9270058274269104, "perplexity_flag": "head"}
http://mathhelpforum.com/algebra/200290-mathematical-induction-problem-2-ib-math-hl-print.html	# Mathematical Induction Problem 2 (IB Math HL) Printable View • June 23rd 2012, 12:27 AM alireza1992 1 Attachment(s) Mathematical Induction Problem 2 (IB Math HL) Hello. I have another question about mathematical induction. I have problem dealing with factorial notation in mathematical induction problems. Here are two question which I couldn't solve. I'd be grateful if you could help me to figure out how to solve them. Attachment 24139 • June 23rd 2012, 01:41 AM Prove It Re: Mathematical Induction Problem 2 (IB Math HL) Quote: Originally Posted by alireza1992 Hello. I have another question about mathematical induction. I have problem dealing with factorial notation in mathematical induction problems. Here are two question which I couldn't solve. I'd be grateful if you could help me to figure out how to solve them. Attachment 24139 You wish to prove $\displaystyle \begin{align} 1 \cdot 1! + 2\cdot 2! + 3\cdot 3! + \dots + n \cdot n! = (n + 1)! - 1 \end{align}$. Base step, where $\displaystyle \begin{align} n = 1 \end{align}$: $\displaystyle \begin{align} LHS &= 1 \cdot 1! \\ &= 1 \\ \\ RHS &= (1 + 1)! - 1 \\ &= 1 \\ &= LHS \end{align}$ Inductive step, assume the statement $\displaystyle \begin{align} 1 \cdot 1! + 2\cdot 2! + 3 \cdot 3! + \dots + n \cdot n! = (n + 1)! - 1 \end{align}$ holds true for $\displaystyle \begin{align} n = k \end{align}$, and use this to show it holds true for $\displaystyle \begin{align} n = k + 1 \end{align}$, i.e. show that $\displaystyle \begin{align} 1 \cdot 1! + 2\cdot 2! + 3\cdot 3! + \dots + k\cdot k! + (k + 1)\cdot (k + 1)! = (k + 2)! - 1 \end{align}$. When we let $\displaystyle \begin{align} n = k + 1 \end{align}$ we have $\displaystyle \begin{align} LHS &= 1 \cdot 1! + 2 \cdot 2! + 3\cdot 3! + \dots + k \cdot k! + (k + 1)\cdot (k + 1)! \\ &= (k + 1)! - 1 + (k + 1)\cdot (k + 1)! \\ &= (k + 1)! + (k + 1)(k + 1)! - 1 \\ &= (k + 1)!( 1 + k + 1 ) - 1 \\ &= (k + 1)!( k + 2) - 1 \\ &= (k + 2)! - 1 \\ &= RHS \end{align}$ Q.E.D. • June 23rd 2012, 01:49 AM Reckoner Re: Mathematical Induction Problem 2 (IB Math HL) Quote: Originally Posted by alireza1992 Here are two question which I couldn't solve. I'd be grateful if you could help me to figure out how to solve them. For the first one, Base case: $1\times1! = 1 = (1+1)! - 1.$ Now, assume that the statement is true for $n = k.$ So, we have $1\times1! + 2\times2! + 3\times3! +\cdots+ k\times k! = (k+1)! - 1$ $\Rightarrow1\times1! + 2\times2! + 3\times3! +\cdots+ k\times k! + (k+1)(k+1)!$ . . . . $= (k+1)! - 1 + (k+1)(k+1)!$ $\Rightarrow1\times1! + 2\times2! + 3\times3! +\cdots+ k\times k! + (k+1)(k+1)!$ . . . . $= (k+1)! + (k+1)(k+1)! - 1$ . . . . $= (k+1)!\left[1 + (k+1)\right] - 1$ . . . . $= (k+1)!(k+2) - 1$ . . . . $= (k+2)! - 1$ • June 23rd 2012, 08:04 AM alireza1992 Re: Mathematical Induction Problem 2 (IB Math HL) Thanks a lot! :) • July 7th 2012, 10:22 AM GGBcn Re: Mathematical Induction Problem 2 (IB Math HL) Proof to part b: Base: Where n=1: LHS=1/2! RHS=((1+1)!-1)/(1+1)!=(2!-1)/2!=1/2!=LHS Now we assume the statement holds true for n=k, and show that it holds true for n=k+1 So we have 1/2!+2/3!+3/4!+4/5!+⋯+k/(k+1)!=((k+1)!-1)/(k+1)! And want to show that 1/2!+2/3!+3/4!+4/5!+⋯+k/(k+1)!+((k+1))/(k+2)!=((k+2)!-1)/(k+2)! LHS=1/2!+2/3!+3/4!+4/5!+⋯+k/(k+1)!+((k+1))/(k+2)! =((k+1)!-1)/(k+1)!+((k+1))/(k+2)! =((k+2)!(k+1)!-(k+2)!+(k+1)(k+1)!)/(k+1)!(k+2)! =((k+2)!(k+1)!-(k+2)(k+1)!+(k+1)(k+1)!)/(k+1)!(k+2)! =((k+2)!-(k+2)+(k+1))/(k+2)! =((k+2)!-k+k-2+1)/(k+2)! =((k+2)!-1)/(k+2)! Q.E.D. All times are GMT -8. The time now is 04:00 AM.	{"extraction_info": {"found_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": 0.8700075745582581, "perplexity_flag": "middle"}
http://mathhelpforum.com/calculus/102721-how-would-i-work-problem-out.html	Thread: 1. How would i work this problem out we have the function f(x)=2x^3-3x^2-12x+20 first we have to find the zeroes. we cant use a calculator on this problem so i was thinking about using the anti derivative or something but you would need a point. any ideas? also we need to write an equation for the line normal to the graph of f at x=0. I know the first step is to take the derivative of the equation and plug in the 0 but i forget if that solves for the tangent or normal slope. lastly we need to find the x and y coordinates of all the points on the graph of f where the line tangent to the graph is parallel to the x-axis. Im not sure how to aproach that at all. 2. Originally Posted by calc_help123 we have the function f(x)=2x^3-3x^2-12x+20 first we have to find the zeroes. we cant use a calculator on this problem so i was thinking about using the anti derivative or something but you would need a point. any ideas? also we need to write an equation for the line normal to the graph of f at x=0. I know the first step is to take the derivative of the equation and plug in the 0 but i forget if that solves for the tangent or normal slope. lastly we need to find the x and y coordinates of all the points on the graph of f where the line tangent to the graph is parallel to the x-axis. Im not sure how to aproach that at all. Solving $f(x)=2x^3-3x^2-12x+20 = 0$ has nothing to do with the anti-derivative. Start by noting that f(2) = 0.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 1, "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": 0.941077709197998, "perplexity_flag": "head"}
http://mathematica.stackexchange.com/questions/tagged/graphics3d+group-theory	# Tagged Questions 2answers 278 views ### Plotting All Possible Points Belonging to a Group Orbit Given that $$X = \{(x,y,z) \in \mathbb{R}^3 \|\, x^2 + y^2 + z^2 - 2(xy + xz + yz) = k\}\,,$$ where $k$ is a constant. Also given that a group $G$ is represented by \langle g_1,g_2,g_3\|\, g_1^2 = ...	{"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": 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": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8713749647140503, "perplexity_flag": "middle"}
http://en.wikipedia.org/wiki/Modular_group	Modular group For a group whose lattice of subgroups is modular, see Iwasawa group. Group theory • Integers (Z) • Lattice Modular groups • PSL(2,Z) • SL(2,Z) • General linear GL(n) • Special linear SL(n) • Orthogonal O(n) • Euclidean E(n) • Special orthogonal SO(n) • Unitary U(n) • Special unitary SU(n) • Symplectic Sp(n) Infinite dimensional Lie group • O(∞) • SU(∞) • Sp(∞) In mathematics, the modular group Γ is a fundamental object of study in number theory, geometry, algebra, and many other areas of advanced mathematics. The modular group can be represented as a group of geometric transformations or as a group of matrices. Definition The modular group Γ is the group of linear fractional transformations of the upper half of the complex plane which have the form $z\mapsto\frac{az+b}{cz+d}$ where a, b, c, and d are integers, and ad − bc = 1. The group operation is function composition. This group of transformations is isomorphic to the projective special linear group PSL(2, Z), which is the quotient of the 2-dimensional special linear group SL(2, Z) over the integers by its center {I, −I}. In other words, PSL(2, Z) consists of all matrices $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ where a, b, c, and d are integers, ad − bc = 1, and pairs of matrices A and −A are considered to be identical. The group operation is the usual multiplication of matrices. Some authors define the modular group to be PSL(2, Z), and still others define the modular group to be the larger group SL(2, Z). However, even those who define the modular group to be PSL(2, Z) use the notation of SL(2, Z), with the understanding that matrices are only determined up to sign. Some mathematical relations require the consideration of the group SL(2, Z) of matrices with determinant plus or minus one. (SL(2, Z) is a subgroup of this group.) Similarly, PSL(2, Z) is the quotient group SL(2,Z)/{I, −I}. A 2 × 2 matrix with unit determinant is a symplectic matrix, and thus SL(2, Z) = Sp(2, Z), the symplectic group of 2x2 matrices. One can also use the notation GL(2, Z) for SL(2, Z), because an integer matrix is invertible if and only if it has determinant equal to ±1. Alternatively, one may use the explicit notation SL±(2, Z). Number-theoretic properties The unit determinant of $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ implies that the fractions a/b, a/c, c/d and b/d are all irreducible, that is have no common factors (provided the denominators are non-zero, of course). More generally, if p/q is an irreducible fraction, then $\frac{ap+bq}{cp+dq}$ is also irreducible (again, provided the denominator be non-zero). Any pair of irreducible fractions can be connected in this way, i.e.: for any pair p/q and r/s of irreducible fractions, there exist elements $\begin{pmatrix} a & b \\ c & d \end{pmatrix}\in\operatorname{SL}(2,\mathbf{Z})$ such that $r = ap+bq \quad \mbox{ and } \quad s=cp+dq.$ Elements of the modular group provide a symmetry on the two-dimensional lattice. Let $\omega_1$ and $\omega_2$ be two complex numbers whose ratio is not real. Then the set of points $\Lambda (\omega_1, \omega_2)=\{ m\omega_1 +n\omega_2 : m,n\in \mathbf{Z} \}$ is a lattice of parallelograms on the plane. A different pair of vectors $\alpha_1$ and $\alpha_2$ will generate exactly the same lattice if and only if $\begin{pmatrix}\alpha_1 \\ \alpha_2 \end{pmatrix} = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} \omega_1 \\ \omega_2 \end{pmatrix}$ for some matrix in SL(2, Z). It is for this reason that doubly periodic functions, such as elliptic functions, possess a modular group symmetry. The action of the modular group on the rational numbers can most easily be understood by envisioning a square grid, with grid point (p, q) corresponding to the fraction p/q (see Euclid's orchard). An irreducible fraction is one that is visible from the origin; the action of the modular group on a fraction never takes a visible (irreducible) to a hidden (reducible) one, and vice versa. If $p_{n-1}/q_{n-1}$ and $p_{n}/q_{n}$ are two successive convergents of a continued fraction, then the matrix $\begin{pmatrix} p_{n-1} & p_{n} \\ q_{n-1} & q_{n} \end{pmatrix}$ belongs to SL(2, Z). In particular, if bc − ad = 1 for positive integers a,b,c and d with a < b and c < d then a/b and c/d will be neighbours in the Farey sequence of order min(b, d). Important special cases of continued fraction convergents include the Fibonacci numbers and solutions to Pell's equation. In both cases, the numbers can be arranged to form a semigroup subset of the modular group. Group-theoretic properties Presentation The modular group can be shown to be generated by the two transformations $S: z\mapsto -1/z$ $T: z\mapsto z+1$ so that every element in the modular group can be represented (in a non-unique way) by the composition of powers of S and T. Geometrically, S represents inversion in the unit circle followed by reflection about the line Re(z)=0, while T represents a unit translation to the right. The generators S and T obey the relations S2 = 1 and (ST)3 = 1. It can be shown [1] that these are a complete set of relations, so the modular group has the presentation: $\Gamma \cong \langle S, T \mid S^2=I, (ST)^3=I \rangle$ This presentation describes the modular group as the rotational triangle group (2,3,∞) (∞ as there is no relation on T), and it thus maps onto all triangle groups (2,3,n) by adding the relation Tn = 1, which occurs for instance in the congruence subgroup Γ(n). Using the generators S and ST instead of S and T, this shows that the modular group is isomorphic to the free product of the cyclic groups C2 and C3: $\Gamma \cong C_2 * C_3$ Braid group The braid group B3 is the universal central extension of the modular group. The braid group B3 is the universal central extension of the modular group, with these sitting as lattices inside the (topological) universal covering group $\overline{\mathrm{SL}_2(\mathbf{R})} \to \mathrm{PSL}_2(\mathbf{R})$. Further, the modular group has trivial center, and thus the modular group is isomorphic to the quotient group of B3 modulo its center; equivalently, to the group of inner automorphisms of B3. The braid group B3 in turn is isomorphic to the knot group of the trefoil knot. Quotients The quotients by congruence subgroups are of significant interest. Other important quotients are the (2,3,n) triangle groups, which correspond geometrically to descending to a cylinder, quotienting the x coordinate mod n, as Tn = (z ↦ z+n). (2,3,5) is the group of icosahedral symmetry, and the (2,3,7) triangle group (and associated tiling) is the cover for all Hurwitz surfaces. Relationship to hyperbolic geometry See also: PSL2(R) The modular group is important because it forms a subgroup of the group of isometries of the hyperbolic plane. If we consider the upper half-plane model H of hyperbolic plane geometry, then the group of all orientation-preserving isometries of H consists of all Möbius transformations of the form $z\mapsto \frac{az + b}{cz + d}$ where a, b, c, and d are real numbers and ad − bc = 1. Put differently, the group PSL(2, R) acts on the upper half-plane H according to the following formula: $\begin{pmatrix} a & b \\ c & d \end{pmatrix}\cdot z \,= \,\frac{az + b}{cz + d}$ This (left-)action is faithful. Since PSL(2, Z) is a subgroup of PSL(2, R), the modular group is a subgroup of the group of orientation-preserving isometries of H. Tessellation of the hyperbolic plane A typical fundamental domain for the action of Γ on the upper half-plane. The modular group Γ acts on H as a discrete subgroup of PSL(2, R), i.e. for each z in H we can find a neighbourhood of z which does not contain any other element of the orbit of z. This also means that we can construct fundamental domains, which (roughly) contain exactly one representative from the orbit of every z in H. (Care is needed on the boundary of the domain.) There are many ways of constructing a fundamental domain, but a common choice is the region $R = \left\{ z \in \mathbf{H}: \left\| z \right\| > 1,\, \left\| \,\mbox{Re}(z) \,\right\| < \frac{1}{2} \right\}$ bounded by the vertical lines Re(z) = 1/2 and Re(z) = −1/2, and the circle \|z\| = 1. This region is a hyperbolic triangle. It has vertices at 1/2 + i√3/2 and −1/2 + i√3/2, where the angle between its sides is π/3, and a third vertex at infinity, where the angle between its sides is 0. By transforming this region in turn by each of the elements of the modular group, a regular tessellation of the hyperbolic plane by congruent hyperbolic triangles is created. Note that each such triangle has one vertex either at infinity or on the real axis Im(z)=0. This tiling can be extended to the Poincaré disk, where every hyperbolic triangle has one vertex on the boundary of the disk. The tiling of the Poincaré disk is given in a natural way by the J-invariant, which is invariant under the modular group, and attains every complex number once in each triangle of these regions. This tessellation can be refined slightly, dividing each region into two halves (conventionally colored black and white), by adding an orientation-reversing map; the colors then correspond to orientation of the domain. Adding in (x, y) ↦ (−x, y) and taking the right half of the region R (Re(z) ≥ 0) yields the usual tessellation. This tessellation first appears in print in (Klein 1878/79a),[2] where it is credited to Richard Dedekind, in reference to (Dedekind 1877).[2][3] Visualization of the map (2,3,∞) → (2,3,7) by morphing the associated tilings.[4] The map of groups (2,3,∞) → (2,3,n) (from modular group to triangle group) can be visualized in terms of this tiling (yielding a tiling on the modular curve), as depicted in the video at right. Noncompact hyperbolic uniform tilings in [∞,3] family Symmetry: [∞,3], (∞32) [∞,3]+, (∞32) [1+,∞,3], (∞33) [∞,3+], (3∞) {∞,3} t01{∞,3} t1{∞,3} t12{∞,3} t2{∞,3} t02{∞,3} t012{∞,3} s{∞,3} h{∞,3} h1,2{∞,3} Uniform duals V∞3 V3.∞.∞ V(3.∞)2 V6.6.∞ V3∞ V4.3.4.∞ V4.6.∞ V3.3.3.3.∞ V(3.∞)3 V3.3.3.3.3.∞ Congruence subgroups Main article: Congruence subgroup Important subgroups of the modular group Γ, called congruence subgroups, are given by imposing congruence relations on the associated matrices. There is a natural homomorphism SL(2, Z) → SL(2, Z/nZ) given by reducing the entries modulo N. This induces a homomorphism on the modular group PSL(2, Z) → PSL(2, Z/nZ). The kernel of this homomorphism is called the principal congruence subgroup of level N, denoted Γ(N). We have the following short exact sequence: $1\to\Gamma(N)\to\Gamma\to\mbox{PSL}(2,\mathbf{Z}/n\mathbf{Z})\to 1$. Being the kernel of a homomorphism Γ(N) is a normal subgroup of the modular group Γ. The group Γ(N) is given as the set of all modular transformations $z\mapsto\frac{az+b}{cz+d}$ for which a ≡ d ≡ ±1 (mod N) and b ≡ c ≡ 0 (mod N). The principal congruence subgroup of level 2, Γ(2), is also called the modular group Λ. Since PSL(2, Z/2Z) is isomorphic to S3, Λ is a subgroup of index 6. The group Λ consists of all modular transformations for which a and d are odd and b and c are even. Another important family of congruence subgroups are the modular group Γ0(N) defined as the set of all modular transformations for which c ≡ 0 (mod N), or equivalently, as the subgroup whose matrices become upper triangular upon reduction modulo N. Note that Γ(N) is a subgroup of Γ0(N). The modular curves associated with these groups are an aspect of monstrous moonshine – for a prime p, the modular curve of the normalizer is genus zero if and only if p divides the order of the monster group, or equivalently, if p is a supersingular prime; see details at congruence subgroup. Dyadic monoid One important subset of the modular group is the dyadic monoid, which is the monoid of all strings of the form STkSTmSTn ... for positive integers k, m, n, ... . This monoid occurs naturally in the study of fractal curves, and describes the self-similarity symmetries of the Cantor function, Minkowski's question mark function, and the Koch curve, each being a special case of the general de Rham curve. The monoid also has higher-dimensional linear representations; for example, the N = 3 representation can be understood to describe the self-symmetry of the blancmange curve. Maps of the torus The group GL(2, Z) is the linear maps preserving the standard lattice Z2, and SL(2, Z) is the orientation-preserving maps preserving this lattice; they thus descend to self-homeomorphisms of the torus (SL mapping to orientation-preserving maps), and in fact map isomorphically to the (extended) mapping class group of the torus, meaning that every self-homeomorphism of the torus is isotopic to a map of this form. The algebraic properties of a matrix as an element of GL(2, Z) correspond to the dynamics of the induced map of the torus. Hecke groups The modular group can be generalized to the Hecke groups, named for Erich Hecke, and defined as follows.[5] The Hecke group Hq is the discrete group generated by $z \mapsto -1/z$ $z \mapsto z + \lambda_q,$ where $\lambda_q=2\cos(\pi/q). \,$ The modular group Γ is isomorphic to H3 and they share properties and applications – for example, just as one has the free product of cyclic groups $\Gamma \cong C_2 C_3,$ more generally one has $H_q \cong C_2 * C_q,$ which corresponds to the triangle group (2,q,∞). There is similarly a notion of principal congruence subgroups associated to principal ideals in Z[λ]. For small values of q, one has: $\lambda_3 = 1,$ $\lambda_4 = \sqrt{2},$ $\lambda_5 = \tfrac{1}{2}(1+\sqrt{5}),$ $\lambda_6 = \sqrt{3}.$ History The modular group and its subgroups were first studied in detail by Richard Dedekind and by Felix Klein as part of his Erlangen programme in the 1870s. However, the closely related elliptic functions were studied by Joseph Louis Lagrange in 1785, and further results on elliptic functions were published by Carl Gustav Jakob Jacobi and Niels Henrik Abel in 1827. References 1. Gunning, Robert C. (1962). Lectures on Modular Forms. Princeton, NJ: Princeton University Press. pp. 5–7. ISBN 978-0-691-07995-0. 2. ^ a b 3. Stillwell, John (2001-01). "Modular Miracles". The American Mathematical Monthly 108 (1): 70–76. ISSN 0002-9890. JSTOR 2695682. 4. Combinatorial group theory, discrete groups, and number theory, by Gerhard Rosenberger, Benjamin Fine, Anthony M. Gaglione, Dennis Spellman p. 65 • Tom M. Apostol, Modular Functions and Dirichlet Series in Number Theory, Second Edition (1990), Springer, New York ISBN 0-387-97127-0 See chapter 2. • Klein, Felix (1878/79), "Über die Transformation der elliptischen Funktionen und die Auflösung der Gleichungen fünften Grades (On the transformation of elliptic functions and ...)", Math. Annalen 14: 13–75 (in Oeuvres, Tome 3) • Dedekind, Richard (September 1877), "Schreiben an Herrn Borchard uber die Theorie der elliptische Modulfunktionen", Crelle's Journal 83: 265–292 .	{"extraction_info": {"found_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": 34, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8571993112564087, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/83591/does-this-map-extend-to-a-unique-automorphism-of-the-polynomial-ring-of-an-integ?answertab=active	# Does this map extend to a unique automorphism of the polynomial ring of an integral domain? Suppose that $A$ is an integral domain. If $a,b$ are element of $A$ and $a$ is a unit, then how come the change of variable $X\mapsto aX+b$ extends to a unique automorphism of the polynomial ring $A[X]$, which equals the identity map when restricted to $A$? Thanks for any ideas or ways to get started. - ## 1 Answer Well, to start with the polynomial Ring $A[X]$ has the universal property, that for every morphism $f : A \rightarrow B$ of Rings and every $c \in B$ there is a unique homomorphism $\overline{f}: A[X] \rightarrow B$ which extends $f$ and maps $X$ to $c$. That is, if $p = \sum_{i = 1}^n a_i X^i$ is given, we have $$\overline{f}(p) = \sum_{i = 1}^n f(a_i) c^i.$$ In your case, you can take $B = A[X]$, $f: A \rightarrow A[X]$ the canonical inclusion and $c = a X + b$. Now the only thing left to be proved is that this is indeed a bijection. This can be done by explicitly stating the inverse and here you will need that $a$ is a unit. - Thank you. ${}{}$ – Clara Nov 22 '11 at 20:24	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 20, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9130785465240479, "perplexity_flag": "head"}
http://mathoverflow.net/revisions/58771/list	## Return to Answer 2 added 1118 characters in body Now, it is not quite true that $P_n$ has more divisors than any $N < P_n$. It is a good strategy for making a number with many divisors, but soon enough it is better to add more factors of $p_1$, then eventually more factors of $p_2$, etc., than to keep adding new prime factors. At To understand this situation better, we can make many numbers (but not all numbers) that have more divisors than their predecessors with the "threshold method". The idea is to optimize the ratio $\log(d(N))/\log(N)$ globally by optimizing it locally (with respect to prime factorization). Pick a constant $t > 0$, the threshold, and say that $N$ should have at least qualitatively$k > 0$ factors of a prime $p$ if and only if$$\frac{\log(k+1) - \log(k)}{\log(p)} \ge t.$$Then I think that $d(M) < d(N)$ when $M < N$. In fact, finding large values of $cd(G)$ (which I will use to mean the same thing eventually happens for cardinality of the group character degrees rather than the set) is a very similar problem when $H_n$. G$is nilpotent. A finite group is nilpotent if and only if it is a the product of$p$-groups. You can make a nilpotent group that looks similar to a number with a record number its Sylow subgroups. The main idea of divisors, by slowly increasing the$p$-group factor for small values of construction above is that in this case$p$. In factcd(G)$ is multiplicative, a particularly good choice i.e., the product of its values for $p$-groups. Following the comment by Frieder Ladisch, $p$ cd(G)$is the maximized for$k p$-groups by$C_{p^m} \times k$upper-triangular group over ltimes C_{p^{m+1}}$. (In the first version of the answer I used other $\mathbb{Z}/p$. This p$-groups that aren't as good.) I.e., this group has dimension character degrees$p^{O(k^2)}$, 1,p,\ldots,p^m$, and it was shown by Isaacs that it has no $O(k^2)$ distinct dimensions of irreducible representations (p$group with a different constant factor)$p^{2m}$or fewer elements can have an irrep with$p^m$elements. So , if you regularize the divisor function can find many record values of$d(n)$by taking cd(G)$ for nilpotent groups using instead the smallest monotonic function threshold formula$d'(n)$\frac{\log(k+1) - \log(k)}{\min(4-k,2)\log(p)} \ge d(n)$, then it is conceptually similar to t.$\$ Let's incorporate the same regularization concept of a "record value" by defining $d'(N)$ to be the lower boundmaximum of $d(M)$ with $M \le N$. I'd have Likewise define $cd'(N)$ to be the maximum of $cd(G)$ with $\|G\| \le N$. Then I think about how far apart that the bounds above constructions show that $d'(N)$ and $cd'(N)$ are at least similar functions, thoughand that$$d'(N) > cd'(N) > \sqrt[3]{d'(N)}$$when $N$ is large enough. In fact I might add think that the exponent of the second inequality climbs from $1/3$ to some higher value, although for nilpotent groups one also has$$\sqrt{d'(N)} > cd'_{\text{nil}}(N).$$ Let me also mention that the bound $O(\sqrt[3]{\|G\|})$ follows immediately from the fact that $\|G\|$ is the sum of the squares of the dimensions of the irreducible representations --- maybe that's what you have in mind with your bound. 1 The number of divisors at least roughly resembles the best achievable lower bound. For each prime $p$, there is a group $G_p$ with $p^3$ elements which has an irreducible representation of dimension 1 (the trivial representation) and an irreducible representation of dimension $p$ (because it's non-abelian). Now let $p_n$ be the $n$th prime. The number $$P_n = p_1p_2\cdots p_n$$ is a type of number with a lot of divisors. If you likewise let $$H_n = G_{p_1} \times G_{p_2} \times \cdots \times G_{p_n},$$ then $H_n$ has an irreducible representation for every $d$ that divides $P_n$, and it has $P^3_n$ elements. Now, it is not quite true that $P_n$ has more divisors than any $N < P_n$. It is a good strategy for making a number with many divisors, but soon enough it is better to add more factors of $p_1$, then eventually more factors of $p_2$, etc., than to keep adding new prime factors. At least qualitatively, the same thing eventually happens for the group $H_n$. A finite group is nilpotent if and only if it is a product of $p$-groups. You can make a nilpotent group that looks similar to a number with a record number of divisors, by slowly increasing the $p$-group factor for small values of $p$. In fact, a particularly good choice for the $p$ is the $k \times k$ upper-triangular group over $\mathbb{Z}/p$. This group has dimension $p^{O(k^2)}$, and it was shown by Isaacs that it has $O(k^2)$ distinct dimensions of irreducible representations (with a different constant factor). So, if you regularize the divisor function $d(n)$ by taking the smallest monotonic function $d'(n) \ge d(n)$, then it is conceptually similar to the same regularization of the lower bound. I'd have to think about how far apart the bounds are, though. I might add that the bound $O(\sqrt[3]{\|G\|})$ follows immediately from the fact that $\|G\|$ is the sum of the squares of the dimensions of the irreducible representations --- maybe that's what you have in mind with your bound.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 71, "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": 0.95442134141922, "perplexity_flag": "head"}
http://www.jirka.org/ra/errata.html	# Basic Analysis: Introduction to Real Analysis: Errata This page lists errors in the various editions. Nonmathematical typos, misspellings, and grammar or style problems are not listed here. Also not listed are things that were correct but simply deserved improvement. Of course, older editions may suffer from recently discovered errata as well. December 16th 2012 edition: 1. In exercise 5.3.5, page 137, it should be OK to assume "continuously differentiable". The exercise is possible, but it is harder than intended. (Thanks to Sonmez Sahutoglu) 2. On top of page 148, the second equality is an inequality. It is not difficult to see, but it requires a proof. I will be adding a relevant exercise. (Thanks to Sonmez Sahutoglu) 3. Page 150, exercises 6.2.10 and 6.2.11. The assumption of "continuous" on $f_n$ is missing in both of these exercises. They're not true without it. (Thanks to Sonmez Sahutoglu) 4. Page 174, exercise 7.3.10: part b) is not correct. The metric given is not the right metric for this problem. The problem will be replaced with a different and more comprehensive one. (Thanks to Sonmez Sahutoglu) October 1st 2012 edition: 1. In exercise 2.5.9 on page 78, the two series should be absolutely convergent, or part a) is not true. (Thanks to Sonmez Sahutoglu) 2. On page 73, second equation from the top we say "triangle inequality," but we write an equality. Of course that should be an inequality. (Thanks to Yi Zhang) 3. On page 76, 6th line from the bottom, an $r^n$ should be $r^k$, we get it right in the line just below. (Thanks to Yi Zhang) 4. In Theorems 3.3.7 and 3.3.8, c is really in the open interval (a,b), though that's obvious. It's correct as stated, but stupid. So change to $c \in (a,b)$. Similarly in the application on page 95. (Thanks to Yi Zhang) 5. In remark 5.1.5, we meant integrals from a to b, not 0 to 1. April 8th 2012 edition: No known errata. December 25th 2011 edition: No known errata. December 15th 2011 edition: 1. Yaikes! The triangle inequality on page 157 has a typo. Of course it is an inequality, otherwise it would be called the triangle equality. So it should say $d(x,z) \leq d(x,y) + d(y,z)$. Thanks to students from my class. 2. When renaming t to x in section 7.6, in the definition of T(f) on page 185, I forgot to change t to x. It says T(f)(t) when it should say T(f)(x). November 18th 2011 edition: 1. On page 164 in the definition of an open set. It should say "... if for every $x \in V$, there exists ..." Instead it says for every $x \in X$ which is of course wrong. (Thanks to Steve Hoerning) 2. Page 183, in the contraction mapping principle, of course X should be nonempty. I don't feel too bad about making this mistake as Rudin does it also, even in the third edition. 3. Page 181, in Theorem 7.5.4, f is bounded not compact, it is f(X) that is compact. 4. Exercise 7.1.8 works only for nonempty bounded sets. The definition works if you use extended reals, but you won't get a metric space. 5. Page 166, in the proof of proposition 7.2.8 in the middle of the estimate there is $-\delta$ when there should be $+\delta$. 6. In the proof of proposition 7.2.13 we actually need the opposite direction of Proposition 7.2.11. So 7.2.11 needs to be made into an if and only if. The other part of the if and only if simply refers to exercise 7.2.12. To do this we also need to change the assumption on the intersection to $U_1 \cap U_2 \cap S \not= \emptyset$. Consequently in the proof we show that $U_1 \cap S$ and $U_2 \cap S$ have a point in common (which we actually do). 7. In the proof of proposition 7.2.13 we are working with [x,y] in the end not with S, that's the whole point of defining x and y. 8. On page 177 in the middle when defining $n_{j+1}$ it should be in the ball of radius 1/(j+1) not 1/j. Or alternatively we should start with $n_{j-1}$ defined and define $n_j$ to be in the ball of radius 1/j. 9. In proof of proposition 7.3.7, on the last line we say $n \geq N$ when of course that is $n \geq M$ (there is no N in the whole proof). 10. In exercise 7.4.1, A is of course finite subset of X. (Thanks to Jeremy Sutter) 11. In the proof of Theorem 7.4.6 (sequentially compact is equivalent to compact) we define a sequence starting at $x_0$ and $\lambda_0$, but then several times we forget it later and write down the elements starting at $x_1$ and $\lambda_1$. Of course we should consistently start with 0 (or even better with 1). 12. In exercise 7.6.5, the "best" k is of course the smallest one, not the largest one (there is no largest one of course). 13. Definition 1.1.1 on page 14 was not well stated. An ordered set is the set together with the relation. (Thanks to Paul Vojta) 14. On page 106, in the statement of the chain rule, $c \in I_1$ not $c \in I_2$. (Thanks to Paul Vojta) 15. On page 107, in exercises 4.1.3 and 4.1.8 it was not specified what is n. So specify that it is an integer, in which case also specify that x and f(x) should not be 0. (Thanks to Paul Vojta) 16. On page 112, in the proof of Proposition 4.2.6, we assume that $x < y$ but then we use that $x - y > 0$ when we should use $x - y < 0$. (Thanks to Paul Vojta) 17. On page 134, the second displayed equation is only true when $c < x$, otherwise the opposite inequalities hold. (Thanks to Paul Vojta) October 16th 2011 edition: 1. On page 96, in the proof of Proposition 3.3.10, near the end there is a missing absolute value sign, or at least a negative sign. That is, after the "or in other words", there should be $-(b_{d-1}M^{d-1}+\cdots+b_1M+b_0) < M^d$. Just above, there should be absolute value signs, as in $\| ... \| < 1$. Similarly in the following paragraph the typo was propagated. So for the $K$ it should say $b_{d-1} (-K)^{d-1} + \cdots + b_1 (-K) + b_0 < K^d$ 2. On page 93, in proof of Theorem 3.3.2. When we say "there exist convergent subsequences $\{ x_{n_i} \}$ and $\{ y_{n_i} \}$". The indices should be different. Now it is possible to pick one $n_i$ for both, so it is not really wrong, but there is no need to do this. Simply change $y_{n_i}$ to $y_{m_i}$. 3. Exercise 3.3.7, says image of a closed and bounded interval is a closed and bounded interval. I suppose it is OK, but some would not consider $[c,c] = \{c\}$ an interval. So best to suppose that f is nonconstant. 4. On page 94, about 3/4 down. d is the limit of $\{ b_n \}$ which is decreasing, so it's the infimum, not the supremum (Thanks to Daniel Alarcon). 5. Page 106, the quotient rule is of course missing a minus sign. (Thanks to Jeremy Sutter) 6. On page 59, line 5, that should be $\{ x_n : n \geq n_k+1 \}$. (Thanks to Eliya Gwetta) 7. On page 120, in the proof of 5.1.7, the $\tilde{m}_j$ definition is not quite right, a tilde is missing. It should be $\tilde{m}_j := \inf \{ f(x) : \tilde{x}_{j-1} \leq x \leq \tilde{x}_j \}$. 8. On page 128, in the proof of Lemma 5.2.6, we wish to take an $n$ such that $\frac{b-a}{n} < \delta$, the (b-a) was missing. 9. On page 122, when computing the difference of upper and lower integral it is $(1+\epsilon)-(1-\epsilon)$ of course. 10. On page 128, in the definition of "finitely many discontinuities". Of course we mean that f is continuous at all points of A, and not that the restriction of f to A is continuous. 11. On page 143, there were absolute value signs missing in the estimate towards the end of proof of 6.1.13. That is there should be absolute value signs around $f(x)-f_k(x)$. 12. On page 145 and 147, it says that $n > \frac{1}{x}$ implies $x < \frac{1}{n}$, while that should of course be $x > \frac{1}{n}$, which is why $f_n(x)=0$. 13. On page 149, Exercise 6.2.3, of course the function f should be Riemann integrable to be able to take the integral. Similarly for 6.2.8 and 6.2.9 on page 150, the functions $f_n$ should be Riemann integrable. 14. On page 136, the definition of the erf function is wrong. A minus sign and a square root is missing. It is simply an example, and not used elsewhere in the book. 15. On page 75, we mention that $\sum \frac{1}{n^2}$ converges to $\frac{\pi^2}{2}$ when in fact it converges to $\frac{\pi^2}{6}$ (Thanks to Daniel Alarcon for spotting this). 16. In the examples on pages 154 and 155, the h guaranteed by the proof of the theorem was computed for an older version of the proof. For exponential, $h=\frac{1}{2}$ will work and for $y'=y^2$, $h=1-\frac{\sqrt{3}}{2}$ will work. 17. On page 143, the strict inequality in the displayed equation in the middle of the page should of course be a nonstrict inequality as we took a limit. 18. In example 5.3.2, the computation has a typo, while the final answer is correct. 19. In example 5.1.12, at the end of the example, there is $\int_0^1 f$ when there should of course be $\int_0^2 f$ (Thanks to Jeremy Sutter). 20. On page 135, the difference quotient is less than or equal to epsilon, not strictly less than. 21. In exercise 6.1.3, A is of course a subset of S, not just a subset of the real numbers, otherwise the restrictions don't make sense. 22. In exercise 5.2.10, f should be bounded as well, otherwise it obviously does not make any sense (Thanks to students in my class) April 26th 2011 edition: 1. Proposition 1.2.7, page 28: The set A should be bounded and nonempty, it is true without this hypothesis, but we just made a big deal about not doing arithmetic with ${\mathbb{R}}^*$. 2. Proposition 1.2.8, page 29: A, B must be nonempty sets! 3. In the proof on top of page 29, all inequalities should be nonstrict. 4. Exercise 2.3.6, page 63: The sequences should be bounded. The exercise works with unbounded, but we have not defined limsup and liminf for unbounded sequences. 5. Example 2.5.18, page 77 is misleading. It is not true that the series converges because the terms go to zero, but by the way we proved that the terms go to zero (using the ratio test). Better replace the justification by the actual ratio $\lim_{n\to\infty} \frac{2^{n+1}/(n+1)!}{2^n/n!} = \lim_{n\to\infty} \frac{2}{n+1} = 0 .$ 6. Exercises 1.2.9 and 1.2.10 should say that A and B are nonempty (Found by Paul Vojta) 7. Exercise 2.4.3 has a typo. The rational numbers need to be dense in F. Here is a rewritten version that is also a lot more explicit. Exercise 2.4.3: Suppose that F is an ordered field that contains the rational numbers $\mathbb{Q}$, such that $\mathbb{Q}$ is dense, that is: whenever $x,y \in F$ are such that $x < y$, then there exists a $q \in \mathbb{Q}$ such that $x < q < y$. Say a sequence $\{ x_n \}_{n=1}^\infty$ of rational numbers is Cauchy if given any $\epsilon \in \mathbb{Q}$ with $\epsilon > 0$, there exists an M such that for all $n,k \geq M$ we have $\|x_n-x_k\| < \epsilon$. Suppose that any Cauchy sequence of rational numbers has a limit in F. Prove that F has the least-upper-bound property. 8. In exercise 2.5.6 it should say "converges absolutely" in a) it just says "converges". Of course that's not wrong (it's weaker), but it could be misleading, and the way to prove convergence here is to prove absolute convergence. 9. On page 85, second line it says "thus c is a cluster point of A", of course that is supposed to be "cluster point of S" as that's what we are trying to prove. 10. On page 110, end of proof of Theorem 4.2.3. Is says "Hence the relative minimum is 0 and the relative maximum is 0" Those are of course the "absolute" min and max. 11. On page 112: In proposition 4.2.8, the second interval is (c,b) not (c,d). There is no d in the proposition. February 28th 2011 edition: 1. On top of page 102, the $=\frac{1}{2}\|x-y\|$ should of course be $\leq\frac{1}{2}\|x-y\|$. December 26th 2010 edition: 1. Exercise 1.1.7 has a extra subset sign. $A \subset \subset \mathbb{N}$ should be $A \subset \mathbb{N}$. 2. On page 12, the sentence about induction hypothesis was missing the word "in", that is "assumption in" rather than just "assumption". It is "P(n) is true" that is the hypothesis. November 1st 2010 edition: 1. On page 11, "So let us assume that $x \in A \cap (B \cup C)$" should be "So let us assume that $x \in A \setminus (B \cup C)$". Thanks to Dan Stoneham. 2. On page 80, first line of subsection 3.1.2 c is of course a cluster point of S, not A. October 3rd 2010 edition: 1. In exercise 1.3.4, the functions f and g are of course bounded as in proposition 1.3.7. (Glen Pugh) 2. Proof of proposition 2.2.5 part iii was off if $y=0$. We should use $\frac{\epsilon}{2(\|y\| + 1)}$ instead of $\frac{\epsilon}{2\|y\|}$. (Glen Pugh) 3. The function defined in the beginning of Example 6.2.3, was not the one on Figure 6.3 nor the rest of the example (though it could also be used to show the same idea). 4. In example 4.2.10, the argument showing that f' is not continuous at zero was insufficient. That the limit does not exist is not as obvious as it seemed. September 6th 2010 edition: The following errata were found by Glen Pugh. 1. In example 0.3.14, $f^{-1}(\{0\}) = \pi \mathbb{Z}$ not $\mathbb{Z}$. 2. In proposition 0.3.16, C and D are of course subsets of A, not B. 3. In exercise 0.3.4 part b), the explanatory sentence has a typo, it should be that "$f( C \cap D)$ is a proper subset of $f(C) \cap f(D)$" August 12th 2010 edition: 1. On bottom of page 11, the counter-example for swapping intersection and union doesn't work due to a typo. The set should be defined by $\{ k \in \mathbb{N} \mid mk < n \}$ not by $\{ k \in \mathbb{N} \mid k < nm \}$. Thanks to Glen Pugh for spotting this. July 15th 2010 edition: 1. Exercise 5.2.11 mentioned as a side note that the Thomae function is "everywhere discontinuous," which is a typo of course. It is only discontinuous on a dense set. (Thanks to an anonymous reader) June 23rd 2010 edition: No known errata. April 8th 2010 edition: All the below errata were found thanks to Jana Maříková. • p.13, line 6: the denominator should be $1-c^{n+1} + (1-c)c^{n+1}$ • p.15, Def. 0.3.18: f and g were switched in the definition. • p.21, Def. 1.1.2: the definition of lower bound is missing • p.22, Def. 1.1.3: need to assume that E is nonempty • p.61, $y=\frac{b_k - a_k}{2}$ should be $y=\frac{a_k + b_k}{2}$. • p.66, "b:=liminf xn" should be "b:=limsup xn" • p.87, line -5: "Fix $c \in (0,\infty)$." should be "Fix $c \in (-\infty , \infty)$." • p.87, line -4: "x" is missing the lower index "n" • p.102, Exercise 3.4.5: what is to be proved is false as stated - need additional assumptions • p.105, line -6: numerator of second fraction should be "(f(x)+g(x))-(f(c)+g(c))" • p.107: line 5: equality missing between "...(g(x)-g(c))" and "u(g(x))..." December 23rd 2009 edition: No known errata. December 11th 2009 edition: Proposition 5.2.4 had a typo in it. There was one too many alphas in 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": 91, "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": 0.9327394962310791, "perplexity_flag": "head"}
http://sbseminar.wordpress.com/2009/08/21/a-question-for-the-combinatorialists/	jump to navigation ## A question for the combinatorialists August 21, 2009 Posted by Ben Webster in blegs, combinatorics. trackback One of the points of combinatorics I never really learned is how to play correctly with Mobius functions. I mean, I can state Mobius inversion for an arbitrary poset if you give me a moment, but it all ends up a bit hard to manipulate. This is particularly frustrating since I know that there are a certain number of people out there in the world who know all these tricks by heart. One of them should make a cheat sheet for all these identities. So, here’s a question that might be easy to answer, but that I can’t quite muddle my way through. Let $P$ be a ranked poset (in my example, it’s flats of a matroid) with unique minimal element 0 and maximal 1 and Mobius function $\mu$. Is there any better expression for the sum $\sum_{p\in P} (-1)^{\mathrm{rk}(p)}\mu(0,p)\mu(p,1)$? About these ads Like Loading... ## Comments» 1. David Speyer - August 21, 2009 I thought about this for a bit this morning, but didn’t reach any strong conclusions — I couldn’t do anything with it, but it looks like the sort of thing that should simplify. Maybe this note will inspire someone else to come out of the woodwork. I find the easiest way to think about mobius function identities is to think about the matrix M whose (i,j) entry is mu(i,j). So M^{-1}, as you no doubt know, is the (0,1) matrix encoding the poset relation. You are interested in MDM, where D is the diagonal matrix keeping track of the sign twist. That didn’t get me anywhere, though… 2. Gil - August 21, 2009 It does not look to me as something that should simplify for general posets. (But I will be happy to learn otherwise.) 3. Harrison - August 21, 2009 Have you worked through some nontrivial proportion of the exercises in the appropriate chapter of Stanley? (Yeah, I haven’t either, but there’s no royal road to combinatorics, and that’s probably the best way — maybe the only way — to really learn it. Certainly it’s the only way to learn algebraic geometry.) Alternatively, I guess, you could try asking Stanley himself… 4. Hugh Thomas - August 28, 2009 A couple of thoughts: * the Mobius function of the lattice of flats of a matroid (a “geometric lattice”) is described by Rota’s NBC basis theorem. See, for example, Blass and Sagan’s 1997 paper in Advances, which generalizes it. (Though Rota’s result could be in EC1 too; I don’t have my copy around.) * geometric lattices are EL-shellable. This means that there is a labelling of the edges of the Hasse diagram of the poset such that the Mobius function of an interval is the number of unrefinable chains from the bottom to the top of the interval such that the labels strictly decrease as you read up the chain. Your sum is capturing chains which increase at most once (though with what seems an odd choice of sign) and with the chains which are decreasing everywhere counting either once or zero times, depending on the parity of the rank of the whole poset. (I guess there is a way to state your formula in terms of the “flag f-vector” of the poset, but so far as I know, that doesn’t help.) Neither of these ideas seems to obviously yield something useful, but maybe it’ll make more sense in context. If you want more details/references, let me know. %d bloggers like this:	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 3, "mimetex.cgi": 0, "/images/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": 0.9355035424232483, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/175406/structures-on-torus?answertab=votes	Structures on torus Quotienting $\mathbb R^2$ by different lattices isomorphic to $\mathbb Z^2$, we get different tori. Somehow I think of the tori as having different "structures", but thinking more about it, I am not quite sure what different structures I am really thinking of. Two structures I am guessing at are complex structures and metrics. Could someone explain how these differ? Also, am I thinking of the right kind of structure? Are there other structures which vary with the lattice chosen? - 1 – Willie Wong♦ Jul 26 '12 at 10:46 Thanks. If I understand it correctly, points in the Teichmüller space for the torus correspond to complex structures, but the various metrics on the Teichmüller space don't correspond to metrics on the torus... How about the Riemannian metric inherited from $\mathbb R^2$. This one should vary with the lattice, I would assume. I just don't know how... – user1205935 Jul 26 '12 at 11:41 In so far as the metric is concerned, what you are looking for is actually the existence of isometries between the various tori. One quick way to see that you have different metrics is by consider the geodesic loops in various homotopy classes. In particular, if they have different lengths the two tori cannot be isometric. – Willie Wong♦ Jul 26 '12 at 12:12 A torus gives a complex elliptic curve, which is a complex variety with a (compatible) structure of an abelian group. Changing the lattice changes the group structure, and hence the elliptic curve under consideration. – M Turgeon Jul 26 '12 at 12:54 2 @MTurgeon: yes, but the surprising thing is that the complex variety also changes. (Actually, the group structure doesn't add anything to the picture, since an elliptic curve is just a complex torus with a marked point). – PseudoNeo Jul 26 '12 at 13:11 show 1 more comment 1 Answer You're mostly right. The relation between complex structures and metrics comes from their common passion about angles. Basically, a complex structure on a Riemann surface is just a procedure for turning tangent vectors 90° counterclockwise. (Well, actually, that is an almost complex structure but they are the same thing in (complex) dimension 1). But that is one of the things a metric (with an orientation) allows you to do! So a metric on a surface defines canonically a complex structure. Of course, many metrics give the same structure (example 1: you can rescale the metric; example 2: the sphere $S^2$ has a lot of metrics but only one complex structure.) The good notion is the notion of conformally equivalent metrics. Roughly speaking, two metrics are conformally equivalent if they define the same notions of angles between two tangent vectors. That means that they are proportional to each other (the ratio being a positive smooth function on the manifold.) So you get an important fact about surfaces: complex structures and conformal classes of metrics are basically the same thing. So, most of the Riemann surface theory can be stated equivalently in the holomorphic world or in the conformal world. (Example: the uniformisation theorem says either “Any Riemann surface is a quotient of $S^2$, $\mathbb C$ or $\mathbb H$” or “Any Riemannian metric on a surface is conformally equivalent to a metric of constant curvature.”) This polyvalence clearly is one of the riches of Riemann surface theory and explains partially the huge number of dedicated textbooks, as there is room for lots of different approaches. So, here are two possible answers to your question: the relevant structures are “complex structures” or “conformal Riemannian structures”. This gives the same notion of “equivalent” lattices: two of them are equivalent if there is an affine map sending one to the other. (One can imagine variants: for example, one could choose to call two lattices equivalent if there is a volume-preserving map sending one to the other. In that case, one has to enrich the relevant structures on the quotient. Of course, the finer the equivalence relation on the lattices, the richer the structure, but affine equivalence is a popular choice, because the two structures I mentioned are very important and very natural.) - Maybe you should mention that every Riemannian metric on a surface is locally conformal to a flat metric, so we can say that these two structures on surfaces are equivalent. – jerrysciencemath Jul 26 '12 at 12:35 That is much more than I expected from any answer. Thank you so much! – user1205935 Jul 26 '12 at 12:42 I should said something about flat structures, I agree. (That's one of the reasons I mentioned the uniformisation theorem, but I should have mentioned that in the torus case, the curvature is 0). But it is not true that these two structures are equivalent: there are pairs of nonisometric flat tori that are conformally equivalent (you can rescale them, but I don't think that this the only possible construction). – PseudoNeo Jul 26 '12 at 12:42 1 @PseudoNeo But if you consider the conformal class of Riemannian metrics, then two metrics are in the same conformal class up to a orient-preserving (or orient-reversing) diffeomorphism if and only if they define the same (or conjugate) complex structure up to the same diffeomorphism. – jerrysciencemath Jul 26 '12 at 12:48 Ah, OK, you meant that “conformal classes of flat metrics” was still another equivalent answer! You're right, of course. Sorry for the confusion. – PseudoNeo Jul 26 '12 at 12:50	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 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": 0.9264879822731018, "perplexity_flag": "head"}
http://unapologetic.wordpress.com/2011/03/19/mappings-between-presheaves/?like=1&source=post_flair&_wpnonce=ae4d6bb289	# The Unapologetic Mathematician ## Mappings Between Presheaves As ever, we want our objects of study to be objects in some category, and presheaves (and sheaves) are no exception. But, luckily, this much is straightforward. Remember that we ended up defining a presheaf as a functor. Given our topological space $X$ we set up the partial order category $\mathrm{Subset}(X)$, flipped it around to $\mathrm{Subset}(X)^\mathrm{op}$ so the arrows pointed the opposite way, and then said a presheaf of sets is a functor $\mathcal{F}:\mathrm{Subset}(X)^\mathrm{op}\to\mathbf{Set}$. So the natural home for them is the functor category $\mathbf{Set}^{\mathrm{Subset}(X)^\mathrm{op}}$, where the morphisms are natural transformations. So what does this mean for our usual case where we consider presheaves of sets, or of sets equipped with some algebraic structure? Well, it means that we map from one presheaf $\mathcal{F}$ to another one $\mathcal{G}$ by picking a map for each and every open set: $\phi_U:\mathcal{F}(U)\to\mathcal{G}(U)$. But these maps must be compatible with the restrictions: if $V\subseteq U$ then we must have $\phi_V\circ\cdot\vert_V=\cdot\vert_V\circ\phi_U$. That is, given an element in $\mathcal{F}(U)$, we can either first restrict it to $V$ and then map it by $\phi_V$ to $\mathcal{G}(V)$, or we can first map it by $\phi_U$ to $\mathcal{G}_U$ and then restrict the result to $V$. In either case, we should get the same answer. ### Like this: Posted by John Armstrong \| Topology ## 1 Comment » 1. [...] Direct Image Functor So far our morphisms only let us compare presheaves and sheaves on a single topological space . In fact, we have a [...] Pingback by \| March 21, 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": 17, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9133879542350769, "perplexity_flag": "head"}
http://www.physicsforums.com/showthread.php?p=3355885	Physics Forums ODE now made me think about derivatives and partial derivatives 1. The problem statement, all variables and given/known data Let's say I have a function for a circle $$x^2 + y^2 = C$$ where C is a constant. Then this is a cylinder with the z-axis. Now in my ODE book, we would normally define it as $$F(x,y) = C = x^2 + y^2$$ as a level surface. Now my question is about what the partial derivative with respect to x mean as opposed to (single-variable calculus) derivative with respect to x mean. Am I losing anything if I take one derivative over the other? I should mention that many of these problems assume that F(x,y(x)). $$\frac{\partial F}{\partial x} = 2x$$ $$\frac{\partial F}{\partial y} = 2y$$ $$\frac{\mathrm{d} F}{\mathrm{d} x} = 2x + 2y\frac{\mathrm{d} y}{\mathrm{d} x} = 0$$ So now my question is, what exactly is this $$2x + 2y\frac{\mathrm{d} y}{\mathrm{d} x} = 0$$ as opposed to 2x PhysOrg.com science news on PhysOrg.com >> Front-row seats to climate change>> Attacking MRSA with metals from antibacterial clays>> New formula invented for microscope viewing, substitutes for federally controlled drug Recognitions: Gold Member Science Advisor Staff Emeritus $$2x+ 2y\frac{dy}{dx}$$ is the rate of change of the function $f(x, y(x))= x^2+ y(x)^2$ with respect to x- it measures how fast f(x,y(x)) changes as x changes. Of course that will depend upon exactly how y(x) changes as x changes- and that is what dy/dx tells you. Suppose y(x)= 3x. Then $d(x^2+ y^2)/dx= 2x+ 2y dy/dx= 2x+ 2y(3)= 2x+ 2(3x)(3)= 20x$.That is exactly the same as if you had replaced y with 3x from the start: $x^2+ (3x)^2= x^2+ 9x^2= 10x^2$ so [itex]df/dx= 20x[/tex] Quote by HallsofIvy $$2x+ 2y\frac{dy}{dx}$$ is the rate of change of the function $f(x, y(x))= x^2+ y(x)^2$ with respect to x- it measures how fast f(x,y(x)) changes as x changes. I thought that was what the partial derivative with respect to x is Recognitions: Gold Member Science Advisor Staff Emeritus ODE now made me think about derivatives and partial derivatives No, the partial derivative of f with respect to x is the rate of change as x change assuming y does not change. Recognitions: Science Advisor Quote by flyingpig I should mention that many of these problems assume that F(x,y(x)). Right. given a small change in x, dx, the change is x^2 is close to 2xdx and the change in y^2 is close to 2ydy. But dy = (dy/dx)dx. If y were independent of x then dy/dx would be zero. This would just mean that x can change without a change in y. But the constraint C = x^2 + y^2 tells you that y is a function of x, at least locally. Quote by lavinia Right. given a small change in x, dx, the change is x^2 is close to 2xdx and the change in y^2 is close to 2ydy. But dy = (dy/dx)dx. If y were independent of x then dy/dx would be zero. This would just mean that x can change without a change in y. But the constraint C = x^2 + y^2 tells you that y is a function of x, at least locally. If I were to graph all three of those "derivatives" what would they look like? How do yuo even graph Fx alone? Actualyl could I get a unit representation? Say F(x,y) was a position function of time (perhaps x(t) = x and y = y(t)) what would the different derivatives tell me? Thread Tools \| \| \| \| \|--------------------------------------------------------------------------------------\|----------------------------\|---------\| \| Similar Threads for: ODE now made me think about derivatives and partial derivatives \| \| \| \| Thread \| Forum \| Replies \| \| \| Differential Equations \| 4 \| \| \| Calculus \| 10 \| \| \| Calculus & Beyond Homework \| 7 \| \| \| Calculus & Beyond Homework \| 1 \| \| \| Calculus & Beyond 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": 4, "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": 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": 0.9500477910041809, "perplexity_flag": "middle"}
http://www.exampleproblems.com/wiki/index.php/Vector_(spatial)	# Vector (spatial) ### From Exampleproblems In physics and in vector calculus, a spatial vector is a concept characterized by a magnitude, which is a scalar, and a direction (which can be defined in a 3-dimensional space by the Euler angles). Although it is often described by a number of "components", each of which is dependent upon the particular coordinate system being used, a vector is an object with properties which do not depend on the coordinate system used to describe it. A common example of a vector is force — it has a magnitude and an orientation in three dimensions (or however many spatial dimensions one has), and multiple forces sum according to the parallelogram law. A spatial vector can be formally defined by its relationship to the spatial coordinate system under rotations. Alternatively, it can be defined in a coordinate-free fashion via a tangent space of a three-dimensional manifold in the language of differential geometry. These definitions are discussed in more detail below. A spatial vector is a special case of a tensor and is also analogous to a four-vector in relativity (and is sometimes therefore called a three-vector in reference to the three spatial dimensions, although this term also has another meaning for p-vectors of differential geometry). Vectors are the building blocks of vector fields and vector calculus. The word vector is also now used for more general concepts (see also vector and generalizations below), but this article describes the original spatial meaning except where otherwise noted. ## Definitions Informally, a vector is a quantity characterized by a magnitude (in mathematics a number, in physics a number times a unit) and a direction, often represented graphically by an arrow. Examples are "moving north at 90 km/h" or "pulling towards the center of Earth with a force of 70 newtons". The notion of having a "magnitude" and "direction" is formalized by saying that the vector has components that transform like the coordinates under rotations. That is, if the coordinate system undergoes a rotation described by a rotation matrix R, so that a coordinate vector x is transformed to x' = Rx, then any other vector v is similarly transformed via v' = Rv. This ensures the invariance of the operations dot product, Euclidean norm, cross product, gradient, divergence, curl, and scalar triple product, and trivially for vector addition and subtraction, and scalar multiplication. The terms scalar and vector as used here include pseudoscalars and pseudovectors or axial vectors (see also below). Accordingly, let, for example, each of two vectors be expressed as three space coordinates, and apply the formula for the cross product, resulting in three coordinates, which represent a third vector. If we rewrite the two vectors in rotated coordinates, and apply the formula for the cross product again, then the result is the original cross product in terms of rotated coordinates. Also, let, for example, a vector field be expressed as three space coordinate functions of three variables, and apply the formula for the curl based on these functions, resulting in three additional functions, which represent a second vector field. If we rewrite the original vector field in terms of rotated position coordinates and correspondingly rotated coordinates for the vector function values, and apply the formula for the curl based on these functions, then the result is the rewritten version of the original curl: also in terms of rotated position coordinates and correspondingly rotated coordinates for the vector function values. The same applies for dot product, gradient, divergence, vector addition and scalar multiplication. For these, also reflection in a plane can be applied. The scalars involved should not be transformed (e.g. in the case of a rotation by 180°, the scalar should not be multiplied by -1). Thus even in 1D we have to distinguish scalars and vectors: 2 × 3 = 6 can be interpreted as a scalar multiplication or a dot product, but not as a product of two vectors. Similarly differentiation in 1D can be interpreted as a gradient or a divergence: one of the two functions is scalar and one a vector, and the argument is a vector, ensuring invariance under inversion of the vectors without changing the scalars. Since rotation of the three Cartesian coordinate axes changes the formulas the same as an inverse rotation of the field itself, we can also conclude: • if the same rotation is applied to two vectors, then the cross product is correspondingly rotated, but the dot product remains the same • rotation of a scalar field results in a correspondingly rotated vector field for the gradient • rotation of a vector field results in a correspondingly rotated scalar field for the divergence and a correspondingly rotated vector field for the curl where rotation of a scalar field involves only rotation of the position vectors, while rotation of a vector field involves also a corresponding rotation of the vector field values. Note that the concept of corresponding rotations applies even if different coordinate systems are used for field values and position vectors, so that e.g. for one we multiply by an orthogonal matrix and for the other we add an angle to an angle coordinate. In order to use the usual formulas, e.g. to compute mechanical work, the x-axis of forces should be in the same direction as the x-axis of position, etc. When, as described above, coordinate rotations of position are accompanied by corresponding coordinate rotations of forces, this property is preserved. On the other hand, the origin of forces is simply at the zero force (no force), while the origin of position can be chosen as desired. For example, work depends on displacement, which is the difference of positions and therefore does not depend on the origin. Position and function value of a vector field are often, but not necessarily, expressed in similar coordinate systems. For example gravitational field strength due to a particular point mass may be gr = − 9.8(r / r0) − 2m / s2, with both the function value and the position vector in spherical coordinates. For the position vector the origin is chosen here at the center of the point mass; for the field strength the origin is simply at "zero field strength" anyway. How the other two coordinates are chosen does not matter in this case, because the field does not depend on them, and the field has no components in their directions. More generally, a vector is a tensor of contravariant rank one. In differential geometry, the term vector usually refers to quantities that are closely related to tangent spaces of a differentiable manifold (assumed to be three-dimensional and equipped with a positive definite Riemannian metric). (A four-vector is a related concept when dealing with a 4 dimensional spacetime manifold in relativity.) Examples of vectors include displacement, velocity, electric field, momentum, force, and acceleration. Vectors can be contrasted with scalar quantities such as distance, speed, energy, time, temperature, charge, power, work, and mass, which have magnitude, but no direction (they are invariant under coordinate rotations). The magnitude of any vector is a scalar. A related concept is that of a pseudovector (or axial vector). This is a quantity that transforms like a vector under proper rotations, but gains an additional sign flip under improper rotations. Examples of pseudovectors include magnetic field, torque, and angular momentum. (This distinction between vectors and pseudovectors is often ignored, but it becomes important in studying symmetry properties.) To distinguish from pseudo/axial vectors, an ordinary vector is sometimes called a polar vector. See also parity (physics). Sometimes, one speaks informally of bound or fixed vectors, which are vectors additionally characterized by a "base point". Most often, this term is used for position vectors (relative to an origin point). More generally, however, the physical interpretation of a particular vector can be parameterized by any number of quantities. ### Examples in one dimension A force may be "15N to the right", with coordinate 15N if the basis vector is to the right, and −15N if the basis vector is to the left. The magnitude of the vector is 15N in both cases. A displacement may be "4m to the right", with coordinate 4m if the basis vector is to the right, and −4m if the basis vector is to the left. The magnitude of the vector is 4m in both cases. The work done by the force in the case of this displacement is 60J in both cases. The force and displacement are vectors, the magnitudes are scalars, and the coordinates are neither. ### Generalizations In mathematics, a vector is any element of a vector space over some field. The spatial vectors of this article are a very special case of this general definition (they are not simply any element of Rd in d dimensions), which includes a variety of mathematical objects (algebras, the set of all functions from a given domain to a given linear range, and linear transformations). Note that under this definition, a tensor is a special vector! ## Representation of a vector Symbols standing for vectors are usually printed in boldface as a; this is also the convention adopted in this encyclopedia. Other conventions include $\vec{a}$ or a, especially in handwriting. Alternately, some use a tilde (~) placed under the vector. The length or magnitude or norm of the vector a is denoted by \|a\|. Vectors are usually shown in graphs or other diagrams as arrows, as illustrated below: File:Vecab.png Here the point A is called the tail, base, start, or origin; point B is called the head, tip, endpoint, or destination. The length of the arrow represents the vector's magnitude, while the direction in which the arrow points represents the vector's direction. If a vector is itself spatial, the length of the arrow depends on a dimensionless scale. If it represents e.g. a force, the "scale" is of physical dimension length/force. Thus there is typically consistency in scale among quantities of the same dimension, but otherwise scale ratios may vary; for example, if "1 newton" and "5 m" are both represented with an arrow of 2cm, the scales are 1:250 and 1m:50N respectively. Equal length of vectors of different dimension has no particular significance unless there is some proportionality constant inherent in the system that the diagram represents. Also length of a unit vector (of dimension length, not length/force, etc.) has no coordinate-system-invariant significance. In the figure above, the arrow can also be written as $\overrightarrow{AB}$ or AB In order to calculate with vectors, the graphical representation is too cumbersome. Vectors in a n-dimensional Euclidean space can be represented as a linear combination of n mutually perpendicular unit vectors. In this article, we will consider R3 as an example. In R3, we usually denote the unit vectors parallel to the x-, y- and z-axes by i, j and k respectively. Any vector a in R3 can be written as a = a1i + a2j + a3k with real numbers a1, a2 and a3 which are uniquely determined by a. Sometimes a is then also written as a 3-by-1 or 1-by-3 matrix: ${a} = \begin{bmatrix} a_1\\ a_2\\ a_3\\ \end{bmatrix}$ ${a} = \begin{pmatrix} a_1 & a_2 & a_3 \\ \end{pmatrix}$ even though this notation suppresses the dependence of the coordinates a1, a2 and a3 on the specific choice of coordinate system i, j and k. ## Length of a vector The length of the vector a = a1i + a2j + a3k can be computed with the Euclidian norm $\left\\|\mathbf{a}\right\\|=\sqrt{a_1^2+a_2^2+a_3^2}$ which is a consequence of the Pythagorean theorem. ## Vector equality Two vectors are said to be equal if they have the same magnitude and direction. However if we are talking about bound vector, then two bound vectors are equal if they have the same base point and end point. For example, the vector i + 2j + 3k with base point (1,0,0) and the vector i+2j+3k with base point (0,1,0) are different bound vectors, but the same (unbounded) vector. ## Vector addition and subtraction Let a=a1i + a2j + a3k and b=b1i + b2j + b3k. The sum of a and b is: $\mathbf{a}+\mathbf{b} =(a_1+b_1)\mathbf{i} +(a_2+b_2)\mathbf{j} +(a_3+b_3)\mathbf{k}$ The addition may be represented graphically by placing the start of the arrow b at the tip of the arrow a, and then drawing an arrow from the start of a to the tip of b. The new arrow drawn represents the vector a + b, as illustrated below: This addition method is sometimes called the parallelogram rule because a and b form the sides of a parallelogram and a + b is one of the diagonals. If a and b are bound vectors, then the addition is only defined if a and b have the same base point, which will then also be the base point of a + b. One can check geometrically that a + b = b + a and (a + b) + c = a + (b + c). The difference of a and b is: $\mathbf{a}-\mathbf{b} =(a_1-b_1)\mathbf{i} +(a_2-b_2)\mathbf{j} +(a_3-b_3)\mathbf{k}$ Subtraction of two vectors can be geometrically defined as follows: to subtract b from a, place the ends of a and b at the same point, and then draw an arrow from the tip of b to the tip of a. That arrow represents the vector a − b, as illustrated below: If a and b are bound vectors, then the subtraction is only defined if they share the same base point which will then also become the base point of their difference. This operation deserves the name "subtraction" because (a − b) + b = a. In physics, vectors of different physical dimension may occur in the same diagram. However, adding or subtracting them (graphically or otherwise) is meaningless. ## Scalar multiplication A vector may also be multiplied by a real number r. In mathematics numbers are often called scalars to distinguish them from vectors, and this operation is therefore called scalar multiplication. The resulting vector is: $r\mathbf{a}=(ra_1)\mathbf{i} +(ra_2)\mathbf{j} +(ra_3)\mathbf{k}$ The length of ra is \|r\|\|a\|. If the scalar is negative, it also changes the direction of the vector by 180o. Two examples (r = -1 and r = 2) are given below: Here it is important to check that the scalar multiplication is compatible with vector addition in the following sense: r(a + b) = ra + rb for all vectors a and b and all scalars r. One can also show that a - b = a + (-1)b. The set of all geometrical vectors, together with the operations of vector addition and scalar multiplication, satisfies all the axioms of a vector space. Similarly, the set of all bound vectors with a common base point forms a vector space. This is where the term "vector space" originated. In physics, scalars also have a unit. The scale of acceleration in the diagram is e.g. 2 m/s² : cm, and that of force 5 N : cm. Thus a scale ratio of 2.5 kg : 1 is used for mass. Similarly, if displacement has a scale of 1:1000 and velocity of 0.2 cm : 1 m/s, or equivalently, 2 ms : 1, a scale ratio of 0.5 : s is used for time. ## Unit vector Main article: Unit vector A unit vector is any vector with a length of one. If you have a vector of arbitrary length, you can use it to create a unit vector. This is known as normalizing a vector. To normalize a vector a = [a1, a2, a3], scale the vector by the inverse of its length \|\|a\|\|. That is: $\mathbf{\hat{a}}=\frac{\mathbf{a}}{\left\\|\mathbf{a}\right\\|}=\frac{a_1}{\left\\|\mathbf{a}\right\\|}\mathbf{\hat{i}}+\frac{a_2}{\left\\|\mathbf{a}\right\\|}\mathbf{\hat{j}}+\frac{a_3}{\left\\|\mathbf{a}\right\\|}\mathbf{\hat{k}}$ ## Dot product Main article: Dot product The dot product of two vectors a and b (sometimes called inner product, or, since its result is a scalar, the scalar product) is denoted by a·b and is defined as: $\mathbf{a}\cdot\mathbf{b} =\left\\|\mathbf{a}\right\\|\left\\|\mathbf{b}\right\\|\cos(\theta)$ where \|\|a\|\| and \|\|b\|\| denote the norm (or length) of a and b, and θ is the measure of the angle between a and b (see trigonometric function for an explanation of cosine). Geometrically, this means that a and b are drawn with a common start point and then the length of a is multiplied with the length of that component of b that points in the same direction as a. This operation is often useful in physics; for instance, work is the dot product of force and displacement. ## Cross product The cross product (also vector product or outer product) differs from the dot product primarily in that the result of a cross product of two vectors is a vector. While everything that was said above can be generalized in a straightforward manner to more than three dimensions, the cross product is only meaningful in three dimensions (although a related product exists in seven dimensions - see below). The cross product, denoted a×b, is a vector perpendicular to both a and b and is defined as: $\mathbf{a}\times\mathbf{b} =\left\\|\mathbf{a}\right\\|\left\\|\mathbf{b}\right\\|\sin(\theta)\,\mathbf{n}$ where θ is the measure of the angle between a and b, and n is a unit vector perpendicular to both a and b. The problem with this definition is that there are two unit vectors perpendicular to both b and a. Which vector is the correct one depends upon the orientation of the vector space, i.e. on the handedness of the coordinate system. The coordinate system i, j, k is called right handed, if the three vectors are situated like the thumb, index finger and middle finger (pointing straight up from your palm) of your right hand. Graphically the cross product can be represented by this figure File:Crossproduct.png In such a system, a×b is defined so that a, b and a×b also becomes a right handed system. If i, j, k is left-handed, then a, b and a×b is defined to be left-handed. Because the cross product depends on the choice of coordinate systems, its result is referred to as a pseudovector. Fortunately, in nature cross products tend to come in pairs, so that the "handedness" of the coordinate system is undone by a second cross product. The length of a×b can be interpreted as the area of the parallelogram having a and b as sides. ## Scalar triple product The scalar triple product (also called the box product or mixed triple product) isn't really a new operator, but a way of applying the other two multiplication operators to three vectors. The scalar triple product is denoted by (a b c) and defined as: $(\mathbf{a}\ \mathbf{b}\ \mathbf{c}) =\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c})$ It has three primary uses. First, the absolute value of the box product is the volume of the parallelepiped which has edges that are defined by the three vectors. Second, the scalar triple product is zero if and only if the three vectors are linearly dependent, which can be easily proved by considering that in order for the three vectors to not make a volume, they must all lie in the same plane. Third, the box product is positive if and only if the three vectors a, b and c are oriented like the coordinate system i, j and k. In coordinates, if the three vectors are thought of as rows, the scalar triple product is simply the determinant of the 3-by-3 matrix having the three vectors as rows. The scalar triple product is linear in all three entries and anti-symmetric in the following sense: $(\mathbf{a}\ \mathbf{b}\ \mathbf{c})$ $=(\mathbf{c}\ \mathbf{a}\ \mathbf{b})$ $=(\mathbf{b}\ \mathbf{c}\ \mathbf{a})$ $=-(\mathbf{a}\ \mathbf{c}\ \mathbf{b})$ $=-(\mathbf{b}\ \mathbf{a}\ \mathbf{c})$ $=-(\mathbf{c}\ \mathbf{b}\ \mathbf{a})$ Technically, the scalar triple product is not a scalar, it is a pseudoscalar: under a coordinate inversion (x goes to −x), it flips sign. ## Vectors as directional derivatives A vector may also be defined as a directional derivative: consider a function f(xα) and a curve xα(σ). Then the directional derivative of f is a scalar defined as $\frac{df}{d\sigma} = \frac{dx^\alpha}{d\sigma}\frac{\partial f}{\partial x^\alpha}.$ where the index α is summed over the appropriate number of dimensions (e.g. from 1 to 3 in 3-dimensional Euclidian space, from 0 to 3 in 4-dimensional spacetime, etc.). Then consider a vector tangent to xα(σ): $t^\alpha = \frac{dx^\alpha}{d\sigma}.$ We can rewrite the directional derivative in differential form (without a given function f) as $\frac{d}{d\sigma} = t^\alpha\frac{\partial}{\partial x^\alpha}.$ Therefore any directional derivative can be identified with a corresponding vector, and any vector can be identified with a corresponding directional derivative. We can therefore define a vector precisely: $\mathbf{a} \equiv a^\alpha \frac{\partial}{\partial x^\alpha}.$ ## See also • Four-vector, the specialization to space-time in relativity • Pseudovector • Vector calculus • Vector bundle • Unit vector • Null vector • Normal vector ##### Toolbox Get A Wifi Network Switcher Widget for Android	{"extraction_info": {"found_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": 22, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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": 0.924567461013794, "perplexity_flag": "head"}
http://mathoverflow.net/questions/tagged/taylor-series	## Tagged Questions 0answers 37 views ### Series expansion with remaining $log n$ Hi, I'm studying the asymptotic behavior $(n \rightarrow \infty)$ of the following formula, where $k$ is a given constant. $$\frac{1}{n^{k(k+1)/(2n)}(2kn−k(1+k) \ln n)^2}$$ I'm … 0answers 68 views ### Two Series Questions [closed] 1) iF $f(x) = x^2 +x$, find the taylor series for f centered at a = 2. 2)what is the sum from 1 to infinity of $(.95)^n$ I got these questions wrong on my last test, and I'm not … 1answer 121 views ### Inequality of Partial Taylor Series Hi, For a given $\theta < 1$, and $N$ a positive integer, I am trying to find an $x > 0$ (preferably the smallest such $x$) such that the following inequality holds: \sum_{k … 0answers 55 views ### Taylor expansion of a one-parameter transformation If we have a $\phi: \mathbb{R} \times \mathbb{R}^n \times \mathbb{R} \to \mathbb{R}$, $\phi = \phi(t, \mathbf{q},\alpha)$ one-parameter group of infinitesimal transformation which … 0answers 50 views ### Find a bounded function with a supporting point Given, $g(Z)=\operatorname{tr}\phi(Z)$, where $\phi(Z)= Z^T\left( \operatorname{diag}(ZZ^T\mathbf{1}) - ZZ^T\right) Z$ where $Z$ is a real rectangular matrix with more rows than co … 1answer 168 views ### A question about approximation of Real analytic functions Define $B$ to be the set of functions $f:[0,1]\rightarrow \mathbb{R}$ for which there exists a dense set $C\subset [0,1]$ of computables numbers and an algorithm $F$ such that for … 0answers 75 views ### Calculating entropy of adjacency matrix using eigenvalue decomposition? How to calculate entropy using the eigenvalues when the eigenvalues are negative? http://mathoverflow.net/questions/102569/is-there-a-simple-relation-between-the-entropy-of-a-mat … 2answers 112 views ### Taylor expansion convergence relation to power-spectrum Is there some connection between the power-spectrum of a real function $f:\mathbb{R}\to\mathbb{R}$ (that is, its Fourier transform) and the convergence radius of its Taylor expansi … 0answers 128 views ### The bound on derivatives of a function [closed] Let $f(x)$ be a smooth function from $\mathbb{R}$ to $\mathbb{R}$. Is there any way to find an upper bound (any bound, not necessary supremum) on The $n^{th}$ derivative of $f$ … 2answers 754 views ### Numerical Computation of arcsin and arctan for real numbers [closed] I'm coding some numerical methods and I do not know what the correct analysis would be for choosing the implementation for $arcsin$ and $arctan$ for real numbers. Here's what I kno … 1answer 127 views ### nonlinear delay differential equation Consider the delay differential equation: $y_x(x) = \sqrt{y(x-\bar{x})}$ where $y$ is the unknown function of $x$, and where $\bar{x}$ is a fixed parameter. This equation doe … 2answers 914 views ### Integral representation of higher order derivatives I'm quite curious about the following phenomena, that still puzzle me although I have a proof, and I'd be really glad if someone may shred some light, showing an interpretation or … 3answers 2k views ### Is there an “elegant” non-recursive formula for these coefficients? Also, how can one get proofs of these patterns? Hi. Not sure if this is a "good" question for this forum or if it'll get panned, but here goes anyway... Consider this problem. I've been trying to find a formula to expand the " … 5answers 2k views ### Roots of truncations of e^x - 1 During a talk I was at today, the speaker mentioned that if you truncate the Taylor series for $e^x - 1$, you'll get lots of roots with nonzero real part, even though the full Tayl … 1answer 325 views ### Approximation:- Algorithmic considerations Hello I want to approximate a function $f$ on $(a,b)$. The function is singular at the points $a$ and $b$, however I have asymptotic expansions at these points. I can also constr … 15 30 50 per 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": 36, "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": 0.8832272291183472, "perplexity_flag": "middle"}
http://stats.stackexchange.com/questions/30159/is-it-possible-to-have-a-pair-of-gaussian-random-variables-for-which-the-joint-d	# Is it possible to have a pair of Gaussian random variables for which the joint distribution is not Gaussian? Somebody asked me this question in a job interview and I replied that their joint distribution is always Gaussian. I thought that I can always write a bivariate Normal with their means and variance and covariances. I am wondering if there can be a case for which the join probability of two Gaussians is not Gaussian? - 1 – user10525 Jun 9 '12 at 23:30 – Stéphane Laurent Jun 10 '12 at 9:01 ## 2 Answers The bivariate normal distribution is the exception, not the rule! It is important to recognize that "almost all" joint distributions with normal marginals are not the bivariate normal distribution. That is, the common viewpoint that joint distributions with normal marginals that are not the bivariate normal are somehow "pathological", is a bit misguided. Certainly, the multivariate normal is extremely important due to its stability under linear transformations, and so receives the bulk of attention in applications. Examples It is useful to start with some examples. The figure below contains heatmaps of six bivariate distributions, all of which have standard normal marginals. The left and middle ones in the top row are bivariate normals, the remaining ones are not (as should be apparent). They're described further below. The bare bones of copulas Properties of dependence are often efficiently analyzed using copulas. A bivariate copula is just a fancy name for a probability distribution on the unit square $[0,1]^2$ with uniform marginals. Suppose $C(u,v)$ is a bivariate copula. Then, immediately from the above, we know that $C(u,v) \geq 0$, $C(u,1) = u$ and $C(1,v) = v$, for example. We can construct bivariate random variables on the Euclidean plane with prespecified marginals by a simple transformation of a bivariate copula. Let $F_1$ and $F_2$ be prescribed marginal distributions for a pair of random variables $(X,Y)$. Then, if $C(u,v)$ is a bivariate copula, $$F(x,y) = C(F_1(x), F_2(y))$$ is a bivariate distribution function with marginals $F_1$ and $F_2$. To see this last fact, just note that $$\renewcommand{\Pr}{\mathbb P} \Pr(X \leq x) = \Pr(X \leq x, Y < \infty) = C(F_1(x), F_2(\infty)) = C(F_1(x),1) = F_1(x) \>.$$ The same argument works for $F_2$. For continuous $F_1$ and $F_2$, Sklar's theorem asserts a converse implying uniqueness. That is, given a bivariate distribution $F(x,y)$ with continuous marginals $F_1$, $F_2$, the corresponding copula is unique (on the appropriate range space). The bivariate normal is exceptional Sklar's theorem tells us (essentially) that there is only one copula that produces the bivariate normal distribution. This is, aptly named, the Gaussian copula which has density on $[0,1]^2$ $$c_\rho(u,v) := \frac{\partial^2}{\partial u \partial v} C_\rho(u,v) = \frac{\varphi_{2,\rho}(\Phi^{-1}(u),\Phi^{-1}(v))}{\varphi(\Phi^{-1}(u)) \varphi(\Phi^{-1}(v))} \>,$$ where the numerator is the bivariate normal distribution with correlation $\rho$ evaluated at $\Phi^{-1}(u)$ and $\Phi^{-1}(v)$. But, there are lots of other copulas and all of them will give a bivariate distribution with normal marginals which is not the bivariate normal by using the transformation described in the previous section. Some details on the examples Note that if $C(u,v)$ is am arbitrary copula with density $c(u,v)$, the corresponding bivariate density with standard normal marginals under the transformation $F(x,y) = C(\Phi(x),\Phi(y))$ is $$f(x,y) = \varphi(x) \varphi(y) c(\Phi(x), \Phi(y)) \> .$$ Note that by applying the Gaussian copula in the above equation, we recover the bivariate normal density. But, for any other choice of $c(u,v)$, we will not. The examples in the figure were constructed as follows (going across each row, one column at a time): 1. Bivariate normal with independent components. 2. Bivariate normal with $\rho = -0.4$. 3. The example given in this answer of Dilip Sarwate. It can easily be seen to be induced by the copula $C(u,v)$ with density $c(u,v) = 2 (\mathbf 1_{(0 \leq u \leq 1/2, 0 \leq v \leq 1/2)} + \mathbf 1_{(1/2 < u \leq 1, 1/2 < v \leq 1)})$. 4. Generated from the Frank copula with parameter $\theta = 2$. 5. Generated from the Clayton copula with parameter $\theta = 1$. 6. Generated from an asymmetric modification of the Clayton copula with parameter $\theta = 3$. - +1 for the remark that the bivariate normal density is the exceptional case! – Dilip Sarwate Jun 10 '12 at 23:40 +1, great answer. – NRH Jun 11 '12 at 14:47 It is true that each element of a multivariate normal vector is itself normally distributed, and you can deduce their means and variances. However, it is not true that any two Guassian random variables are jointly normally distributed. Here is an example: Edit: In response to the consensus that a random variable that is a point mass can be thought of as a normally distributed variable with $\sigma^2=0$, I'm changing my example. Let $X \sim N(0,1)$ and let $Y = X \cdot (2B-1)$ where $B$ is a ${\rm Bernoulli}(1/2)$ random variable. That is, $Y = \pm X$ each with probability $1/2$. We first show that $Y$ has a standard normal distribution. By the law of total probability, $$P(Y \leq y) = \frac{1}{2} \Big( P(Y \leq y \| B = 1) + P(Y \leq y \| B = 0) \Big)$$ Next, $$P(Y \leq y \| B = 0) = P(-X \leq y) = 1-P(X \leq -y) = 1-\Phi(-y) = \Phi(y)$$ where $\Phi$ is the standard normal CDF. Similarly, $$P(Y \leq y \| B = 1) = P(X \leq y) = \Phi(y)$$ Therefore, $$P(Y \leq y) = \frac{1}{2} \Big( \Phi(y) + \Phi(y) \Big) = \Phi(y)$$ so, the CDF of $Y$ is $\Phi(\cdot)$, thus $Y \sim N(0,1)$. Now we show that $X,Y$ are not jointly normally distributed. As @cardinal points out, one characterization of the multivariate normal is that every linear combination of its elements is normally distributed. $X,Y$ do not have this property, since $$Y+X = \begin{cases} 2X &\mbox{if } B = 1 \\ 0 & \mbox{if } B = 0. \end{cases}$$ Therefore $Y+X$ is a $50/50$ mixture of a $N(0,4)$ random variable and a point mass at 0, therefore it cannot be normally distributed. - 2 I don't agree with this answer. A degenerate point mass of $1$ at $\mu$ is usually considered to be a degenerate Gaussian random variable with zero variance. Also,$(X, -X)$ are not jointly continuous though they are marginally continuous. For an example of two jointly continuous random variables that are marginally Gaussian but not jointly Gaussian, see, for example, the latter half of this answer. – Dilip Sarwate Jun 10 '12 at 0:02 2 – Macro Jun 10 '12 at 0:04 2 A standard characterization of the multivariate Gaussian is that $X \in \mathbb R^{n}$ is multivariate Gaussian if and only if $a^T X$ is Gaussian for all $a \in \mathbb R^n$. As @Dilip hints at, it is worth considering if this true for your example. – cardinal Jun 10 '12 at 0:12 3 Since you apparently don't like appeals to rationality ;-), how about appeals to authority? (That's a joke, if it's not apparent.) I just happened upon this purely by accident as I was looking something else up: Example 2.4, page 22 of G. A. F. Seber and A. J. Lee, Linear Regression Analysis, 2nd. ed., Wiley. It quoteth: "Let $Y \sim \mathcal N(\mu,\sigma^2)$ and put $\mathbf Y' = (Y, -Y)$...Thus $\mathbf Y$ has a multivariate normal distribution." – cardinal Jun 10 '12 at 0:51 4 The discussion is about definitions. Clearly, if the covariance matrix by definition is required to be non-singular Macro provides an example, but this is not a example according to the more liberal definition that @cardinal refers too. One good reason to prefer the more liberal definition is that then all linear transformations of normal variables are normal. In particular, in linear regression with normal errors the residuals have a joint normal distribution but the covariance matrix is singular. – NRH Jun 10 '12 at 20:10 show 30 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": 57, "mathjax_display_tex": 9, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "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": 0.9196414947509766, "perplexity_flag": "head"}
http://mathforum.org/mathimages/index.php?title=Henon_Attractor&diff=7676&oldid=7674	# Henon Attractor ### From Math Images (Difference between revisions) \| \| \| \| \| \|-----------\|------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------\|-----------\|------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------\| \| \| \| \| \| \| Line 41: \| \| Line 41: \| \| \| \| \| \| \| \| \| ==Chaotic System== \| \| ==Chaotic System== \| \| - \| \|1=[[Image:Henon2.jpg\|150px]] \| + \| \| \| - \| \|2=[[Image:Henon2.jpg\|thumb\|200px\|left\|Original Henon Attractor , a = 1.4, b = 0.3]] \| + \| \| \| \| The Henon system can be described as [[Chaos\|chaotic]] and random. However, the system does have structure in that its points settle very close to an underlying pattern called a '''chaotic attractor''' or '''basin of attraction'''. The basic Henon Attractor can be described by the equations: \| \| The Henon system can be described as [[Chaos\|chaotic]] and random. However, the system does have structure in that its points settle very close to an underlying pattern called a '''chaotic attractor''' or '''basin of attraction'''. The basic Henon Attractor can be described by the equations: \| \| \| \| \| \| \| Line 120: \| \| Line 118: \| \| \| \| Here are some more examples of Henon Attractors with different ''a'' and ''b'' values. \| \| Here are some more examples of Henon Attractors with different ''a'' and ''b'' values. \| \| \| \| \| \| \| - \| <nowiki><gallery> \| + \| <gallery widths="200px" heights="200px" perrow="3"> \| \| - \| Image: \| + \| Image:Henon2.jpg\|Original Henon Attractor, a = 1.4, b = 0.3 \| \| - \| Image: \| + \| </gallery> \| \| - \| Image: \| + \| \| \| - \| </gallery></nowiki> \| + \| \| \| \| \| \| \| \| \| \| \| \| ## Revision as of 11:46, 8 July 2009 Henon Attractor This image is a Henon Attractor (named after astronomer Michel Hénonn), which is a fractal in the division of the chaotic strange attractor. The Henon Attractor emerged from Hénon's attempt to model the orbits of celestial objects. Henon Attractor Fields: Dynamic Systems and Fractals Created By: SiMet # Basic Description The Henon Attractor is a special kind of fractal that belongs in a group called Strange Attractors. A characteristic of this strange fractal is that it is drawn irregularly. The Henon Attractor is described by two equations. Let us say that we take a starting value (x,y) and apply the equations to the starting values and then the resulting outcome over and over (a process called iteration). If we plot every outcome from this iteration one at a time, we would observe that the points jump from one random location to another within the image. If you take a look at the animation, you can see the irregularity of a number of plotted points. Eventually, the individual points become so numerous that they appear to form lines and an image emerges. This image results from an iterated function, meaning that the equations that describe it can be applied to itself an infinite amount of times. In fact, if you magnify this image, you would find that the lines (really many, many points) that appear to be single lines on the larger image are actually sets or bundles of lines, who, if magnified closer, are bundles of lines and so on. This property is called self-similarity, which means that even as you look closer and closer into the image, it continues to look the same. In other words, the larger view of the image is similar to a magnified part of the image. 1X 8X 64X 512X # A More Mathematical Explanation Note: understanding of this explanation requires: *Algebra ## Fractal Properties [Show] [Hide] Zooming in on the Henon Attractor The Henon Attractor is often described as being similar to the Cantor Set. Let us zoom into the Henon Attractor near the doubled-tip of the fractal (as seen in the animation). We can see that as we continue to magnify the lines that form the structure of the Henon Attractor, these lines become layers of increasingly deteriorating lines that appear to resemble the Canter Set. The Fractal Dimension of the Henon Attractor is not calculable using a single equation$D = \frac{log(n)}{log(e)}$, but it is estimated to be about 1.261. ## Chaotic System The Henon system can be described as chaotic and random. However, the system does have structure in that its points settle very close to an underlying pattern called a chaotic attractor or basin of attraction. The basic Henon Attractor can be described by the equations: $x_{n+1} = y_n + 1 - ax^2_n$ $y_{n+1} = bx_n\,$ The original Henon Attractor (the featured image at the top of the page) created by astronomer Michel Hénon uses the values a = 1.4 and b = 0.3 and begins with a starting point (1,1). However, by changing the values of a and b, we can obtain Henon Attractors that look slightly different. ## Fixed Points [Show] [Hide] Original Henon Attractor with fixed points Looking at the system of equations that describe the fractal, the Henon Attractor uses only two variables (x and y) that are evaluated into themselves. This results in two equilibrium or fixed points for the attractor. Fixed points are such that if the system of equations are applied to the fixed points, the resulting output would be the same fixed points. Therefore, if the system ever plotted onto the fixed points, the fractal would become stagnant. The two fixed points of the Henon Attractor must satisfy $x_{n+1} = x\,$ and $y_{n+1} = y\,$. Using the Henon Attractor's system of equations, the fixed points are (0.6314 , 0.1894) and (-1.1314 , -0.3394). [Show Solving the System of Equations][Hide Solving the System of Equations] To solve the system of equations: $x_{n+1} = y_n + 1 - ax^2_n$ $y_{n+1} = bx_n\,$ If $x_{n+1} = x\,$ and $y_{n+1} = y\,$ then $x = y + 1 - ax^2$ $y = bx\,$ $x = bx + 1 - ax^2$ Using the quadratic equation $x_{1,2} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $x_{1,2} = \frac{-(b-1) \pm \sqrt{(b-1)^2 - 4(-a)(1)}}{2(-a)}$ $x_{1,2} = \frac{-(b-1) \pm \sqrt{(b-1)^2 + 4a}}{-2a}$ Using a = 1.4, b = 0.3: $x_{1,2} = 0.6314,-1.1314 \,$ Using y = bx: $y_{1,2} = 0.1894,-0.3394 \,$ There are two types of fixed points, stable and unstable. The first fixed point (0.6314, 0.1894), labeled "1" on the image, is located within the bounds of the attractor and is unstable. This means that if the system gets close to the point, it will exponentially move away from the fixed point to continue chaotically. The second fixed point, labeled "2", is considered stable, and it is located outside of the bounds of the attractor. ## Changing "a" and "b" Although the original Henon Attractor uses the values a = 1.4 and b = 0.3, we can alter those values within a range to produce various-looking Henon Attractors. The image to the right is a Henon Attractor with the values a = 1 and b = 0.542. Range... Here are some more examples of Henon Attractors with different a and b values. Original Henon Attractor, a = 1.4, b = 0.3 # Teaching Materials There are currently no teaching materials for this page. Add teaching materials. # About the Creator of this Image The images created by this author were found on the author's (username SiMet) Picasa Web Album under the category "Computer Art". # References Glenn Elert, The Chaos Hypertextbook Heinz-Otto Peitgen, Hartmut Jürgens, Dietmar Saupe, Chaos and fractals Bill Casselman, Simple Chaos-The Hénon Map www.ibiblio.org Henon Strange Attractors # Future Directions for this Page A better, less vague description of how sections of the Henon Attractor resembles the Cantor Set 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": 17, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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": 0.9117051362991333, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/36460/force-to-use-in-harmonic-oscillation-through-the-inside-of-a-planet?answertab=oldest	# Force to use in harmonic oscillation through the inside of a planet I am to find an equation for the time it takes when one falls through a planet to the other side and returns to the starting point. I have seven different sets of values - mass of object falling, mass of planet, radius of the planet, and time. I'm not including friction in the calculations. I think this qualifies as a harmonic oscillator, and thus I work with the formula $$T = 2\pi \sqrt{\frac{m}{k}}$$ To find the spring constant $k$ I need force $F$, and this is where I get uncertain. Should I work with the gravitational force between the object and the planet when the fall begins? In other words $$F = G\times\frac{m \times M}{R^2}$$ When I try this I find that $$F = kx \iff k = \frac{F}{x}$$ $$\iff k = \frac{G\times\frac{m \times M}{R^2}}{2R} = \frac{G \times m \times M}{2R^3}$$ $$\Rightarrow T = 2\pi \sqrt{\frac{m}{\frac{G \times m \times M}{2R^3}}} \iff T = 2\pi \sqrt{\frac{2R^3}{G \times M}}$$ Using this equation for the values I have, however, I get the wrong results - $T = 7148$ instead of $T = 5055$. What am I doing wrong? - Just as an aside, you could have edited the earlier question to make it clear that it was not a duplicate then flagged for moderator attention. We try to be responsive to such things. – dmckee♦ Sep 15 '12 at 2:13 The first one really was a duplicate. Finding that original question helped me start to visualise the solution I later needed help finishing (in this question). Thank you though. – Quispiam Sep 15 '12 at 14:34 ## 2 Answers The key to this problem is the fact that the planet's mass $M$ as it appears in Newton's law of gravitation, $$F=\frac{GMm}{r^2},$$ is not actually constant. This is because the layers of the planet that are above you cause zero net force: if you are inside of a hollow spherical shell of mass then diametrically opposite elements of solid angle exert equal forces in opposite directions. Thus, the effective mass of the planet in this problem is only that of a sphere of radius $r$ and density $3M_0/4\pi R^3$, i.e. $M(r)=\frac{r^3}{R^3}M_0$. The force is then $$F=\frac{GM_0m}{R^3}r$$ and it of course causes harmonic motion, with "spring constant" $k=GM_0m/R^3$. - Wow, this solution gives me exactly the right answer. Thank you. I'm still not sure I grasp what exactly the solution does, but I'm getting there. It's just about mentally visualising it now... :) – Quispiam Sep 15 '12 at 14:29 @Quispiam If you still are having trouble with concepts, then we have failed. The purpose of this site is not to do your homework for you, but to help you understand the physics behind it (which is not a bad philosophy for all your studies). If you have any further, specific conceptual issues, you should bring them up, since walking away with the right answer but no deeper understanding is the same as walking away with nothing. – Chris White Sep 15 '12 at 16:24 I understand mathematically how you get $M_r = \frac{r^3}{R^3}M_0$, but I can't grasp why this is done. Why is the density of the sphere multiplied by the volume of the planet, and not the volume of the sphere itself? – Quispiam Sep 15 '12 at 16:27 @ChrisWhite I know. I want to understand the concepts. If I was happy with the solution I would have written down what Emilio wrote and been happy with it. I'm not happy until I understand what is happening. The mathematical solution is just a step closer to understanding the concept. – Quispiam Sep 15 '12 at 16:29 The important concept is that the mass above you does not add to the gravitational attraction, so that you are effectively on the surface of a smaller planet of radius $r$ which has constant density $\rho=M_0/\frac{4\pi}{3}R^3$ for $R$ the planet's radius. Thus the mass of the "effective planet" is $\frac{4\pi}{3}r^3\rho$. In your terms, the density of the planet is multiplied by the volume of the sphere of radius $r$. – Emilio Pisanty Sep 15 '12 at 16:56 show 1 more comment The period is indeed $T = 2\pi \sqrt{m/k}$, when $k$ is the proportionality constant between displacement and force, as in your third equation. So far so good. Now, why did you replace $x$ with $2R$? $x$ is the displacement from equilibrium at which you evaluated $F$. There are two ways to go. Either say leave $x$ unknown and evaluate $F$ in terms of it, or choose a value for $x$ and find the force in that particular case. Both methods should agree. In the former, you'll know you have a simple harmonic oscillator if the $x$-dependence drops out when you find $k = F/x$. If you choose the latter, remember what $x$ is: displacement from equilibrium. Where is the equilibrium position of your intra-planet traveler, and how far away from that point are you when you evaluate the force on said traveler? - To the downvoter: it would be great if you could leave a comment suggesting improvements or saying how this answer is bad. – Chris White Sep 15 '12 at 4:14 The answer misses the point of the problem, which is the breakdown of the simple inverse square formula when inside the planet. The harmonic motion is around the centre of the planet and therefore the displacement from equilibrium $x$ coincides with the radial coordinate $r$ up to a sign, but taking $k=F/x=F/r$ while keeping $F\propto1/r^2$ will always fail - motion outside the planet is not harmonic! – Emilio Pisanty Sep 15 '12 at 11:58 Just to be clear, neither I nor the OP ever said $F \propto 1/r^2$, nor was motion outside the planet brought up. – Chris White Sep 15 '12 at 16:19	{"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": 32, "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": 0.9411444664001465, "perplexity_flag": "head"}
http://mathoverflow.net/questions/88938?sort=oldest	Doubt in the proof of $\lim_{s \to 1^{+}} L(s,\chi) = L(1,\chi)$ Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Theorem 12 of the following link asserts the following: $\textbf{Theorem.}$ Let $\chi \in X_{N}$ with $\chi \neq \epsilon$. There exists $C > 0$ such that $$L(s,\chi) = L(1,\chi) + O(s-1)$$ as $s \to 1^{+}$. In particular, $$\lim_{s \to 1^{+}} L(s,\chi) = L(1,\chi).$$ The proof is as follows: Let $1< s < 2$. From the proof of $\textbf{Theorem 9}$ we have $$L(s,\chi) - L(1,\chi) = \sum\limits_{n=1}^{\infty} a_{n} \Biggl[\biggl(\frac{1}{n^s} - \frac{1}{(n+1)^{s}}\biggr) - \biggl(\frac{1}{n} - \frac{1}{n+1}\biggr)\Biggr]$$ where the sequence $\{a_{n}\}$ is bounded. Applying the mean value theorem to the function $s \mapsto n^{-s} - (n+1)^{-s}$ gives a sequence $\{s_{n}\}$ with $1 < s_{n} < s$ and $$L(s,\chi) - L(1,\chi) = (s-1) \sum\limits_{n=1}^{\infty} a_{n} \Biggl[\frac{\log\:(n+1)}{(n+1)^{s_n}} - \frac{\log\:(n)}{n^{s_n}} \Biggr] \qquad \qquad \cdots\cdots (1)$$ I don't understand how $(1)$ is derived. When I applied the Mean-Value-Theorem to the function $f(s)=x^{-s} - (x+1)^{-s}$ on $[n,n+1]$ i get $$f'(s) = -x^{-s}\log\:(x) + (x+1)^{-s}\log\:(x+1).\hspace{40pt}(\ast)$$ So by the Mean-Value-Theorem i get an $s_{n} \in (n,n+1)$ such that $$f'(s_{n}) = -n^{-s_n}\log\:(n) + (n+1)^{-s}\log\:(n+1) - (n+1)^{-s_n}\log\:(n+1) + (n+2)^{-s_n}\log\:(n+2)$$ which gives \begin{align} f'(s_{n}) &= \frac{\log(n+2)}{(n+2)^{s_n}} - \frac{\log\:(n)}{n^{s_n}} \\ &= \frac{f(b)-f(a)}{b-a} = (n+1)^{-s} - (n+2)^{-s} - (n+1)^{-s} + n^{-s} \\ &= \frac{1}{n^s} - \frac{1}{(n+2)^s} \end{align} Am I making a mistake. I am not able to see how the author get's to that step. • Are there any other nice proofs of the above theorem which you people would like to recommend? - 2 If I'm not missing something, it looks fine. For, let $f(s) = n^{-s} - (n+1)^{-s}$. Then $f(s) - f(1) = (s-1)f'(s_n)$. Setting in your computation of $f'$ in $(\ast)$ gives just the formula over the phrase "I don't understand ...". – Ralph Feb 19 2012 at 14:52 2 Also note that the paper wants $s_n \in (1,s)$ and not $s_n \in (n,n+1)$ (see the line after the $(\ast)$). – Ralph Feb 19 2012 at 15:24 @Ralph: Dear Ralph, thanks for your response. I shall work through this and get back to you. – Chandrasekhar Feb 19 2012 at 16:12 Why the downvote without a reason. – Chandrasekhar Feb 20 2012 at 10:38 The downvote was probably because your question was not of research level. You asked for clarifying a certain step in a certain proof of a well-known fact. E.g. see my response below for two more proofs available in introductory textbooks. – GH Feb 20 2012 at 19:44 show 5 more comments 2 Answers Since you edited the question but did not say that it is clear now, I assume you are hoping for some details in addition to what Ralph said. So: Let `$f_n(s) = \frac{1}{n^s}- \frac{1}{(n+1)^{s}}$` the derivative of this with respect to $s$ is as you computed `$\frac{-\log n}{n^s}+ \frac{\log(n+1)}{(n+1)^{s}}$`. This is a function of $s$, and you do not consider it on $[n,n+1]$, but rather $[1,s]$ for each $n$. The MVT tells you that `$\frac{f_{n}(s) - f_n(1)}{s -1} = f'(s_n) $` for some `$s_n\in [1,s]$`. Multiplying by $(s-1)$ and plugging in the explicit expressions for the functions this means `$$(\frac{1}{n^s}- \frac{1}{(n+1)^{s}}) - (\frac{1}{n} -\frac{1}{n+1}) = (s-1)(\frac{-\log n}{n^{s_n}}+ \frac{\log (n+1)}{(n+1)^{s_n}}).$$` The left hand side appears in the original sum, and the result is obtained by instead plugging in the right hand side. - 1 @quid: Thanks a lot for the help. – Chandrasekhar Feb 20 2012 at 10:39 You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. It is a standard fact that $L(s,\chi)$, initially defined as a locally uniformly convergent Dirichlet series in $\Re s>1$, extends to a holomorphic function on $\mathbb{C}$. See for example Chapter 9 in Davenport: Multiplicative Number Theory, especially the first half of Page 69. Now basic complex analysis tells us that the Taylor series of $L(s,\chi)$ around $s=1$ converges absolutely on $\mathbb{C}$, of which $$L(s,\chi)=L(1,\chi)+O(\|s-1\|)$$ is a consequence. In fact for the last equation we only need to know that $L'(s,\chi)$ is continuous on $\mathbb{C}$, which follows directly from Cauchy's integral formula. A quick proof of the holomorphicity of $L(s,\chi)$ in $\Re s>0$ follows from the fact that the partial sums $\sum_{n\leq x}\chi(n)$ are bounded. See Proposition 9 in Section VI.2 of Serre: A course in arithmetic, or Theorem 1.3 in Montgomery-Vaughan: Multiplicative number theory I. - @GH: Thanks for your answer. – Chandrasekhar Feb 20 2012 at 20:21	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 40, "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": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9222499132156372, "perplexity_flag": "head"}
http://mathoverflow.net/questions/68778/complexity-of-evd	## Complexity of EVD ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) What is the computational complexity of Eigen Value decomposition of a correlation matrix? - maybe google helps; also, on what model of computation? maybe the fact that your matrix is symmetric matters more than that it is a correlation matrix. – S. Sra Jun 25 2011 at 5:30 And what is your motivation? – András Bátkai Jun 26 2011 at 20:13 ## 2 Answers In practice using algorithms in EISPACK or LAPACK on floating point single or double precision symmetric matrices, computing the EVD takes $O(n^3)$ time. The constant hidden within the big $O$ is considerably larger than for Cholesky factorization. In theory (but such algorithms are not practically useful for typical double precision floating point computations on matrices with dimensions in the thousands to 10's of thousands), you can do as well as matrix multiplication, for which the best current complexity is if I recall correctly $O(n^{2.376})$. - Hi Brian, I don't see it immediately, but is it easy to prove that complexity of eigenvalue decomposition is the same as matrix multiplication? Matrix inversion is certainly there, but since EVD is inherently iterative, and inexact, I find it hard to believe your second paragraph; am I missing something? – S. Sra Jun 26 2011 at 17:49 Like many other similar computations such as solving a linear programming problem, we're talking about the complexity of obtaining an epsilon approximate solution for some accuracy level. Igor's answer includes a link to a cstheory.stackexchange question whose answers include a link to a STOC paper that appears to discuss how to reduce the eigenvalue decomposition to matrix multiplication. I'm certainly not an expert on this stuff- I'm much more interested in what can be done practically. – Brian Borchers Jun 26 2011 at 19:33 If you'd like an explanation of why Strassen-Winograd and other more sophisticated matrix multiplication algorithms don't work well in practice for matrices of size in the thousands to tens of thousands, I'd be happy to explain that- the basic reason is that Strassen's algorithm trades off matrix multiplications for matrix additions, but matrix additions are relatively more expensive than they should be because matrix addition speed is limited by the poor memory bandwidth of contemporary computers. – Brian Borchers Jun 26 2011 at 19:33 Thanks Brian; actually I had totally forgotten---the cs.SE answer links to a STOC paper that I had once skimmed. But indeed, it is not immediate to see that EVD complexity also boils down to matrix multiplication (in the $\epsilon$-accurate sense) – S. Sra Jun 27 2011 at 3:03 -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 4, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9389939904212952, "perplexity_flag": "middle"}
http://en.wikipedia.org/wiki/Grzegorczyk_hierarchy	Grzegorczyk hierarchy The Grzegorczyk hierarchy (pronounced: ), named after the Polish logician Andrzej Grzegorczyk, is a hierarchy of functions used in computability theory (Wagner and Wechsung 1986:43). Every function in the Grzegorczyk hierarchy is a primitive recursive function, and every primitive recursive function appears in the hierarchy at some level. The hierarchy deals with the rate at which the values of the functions grow; intuitively, functions in lower level of the hierarchy grow slower than functions in the higher levels. Definition First we introduce an infinite set of functions, denoted Ei for some natural number i. We define $E_0(x,y)=x+y$ and $E_1(x)=x^2+2$. I.e., E0 is the addition function, and E1 is a unary function which squares its argument and adds two. Then, for each n greater than 1, we define $E_n(x)=E^{x}_{n-1}(2)$. From these functions we define the Grzegorczyk hierarchy. $\mathcal{E}^n$, the n-th set in the hierarchy, contains the following functions: 1. Ek for k < n 2. the zero function (Z(x) = 0); 3. the successor function (S(x) = x + 1); 4. the projection functions ($p_i^m(t_1, t_2, \dots, t_m) = t_i$); 5. the (generalized) compositions of functions in the set (if h, g1, g2, ... and gm are in $\mathcal{E}^n$, then $f(\bar{u}) = h(g_1(\bar{u}), g_2(\bar{u}), \dots, g_m(\bar{u}))$ is as well)[note 1]; and 6. the results of limited (primitive) recursion applied to functions in the set, (if g, h and j are in $\mathcal{E}^n$ and $f(t, \bar{u}) \leq j(t, \bar{u})$ for all t and $\bar{u}$, and further $f(0, \bar{u}) = g(\bar{u})$ and $f(t+1, \bar{u}) = h(t,\bar{u},f(t,\bar{u}))$, then f is in $\mathcal{E}^n$ as well)[note 1] In other words, $\mathcal{E}^n$ is the closure of set $B_n = \{Z, S, (p_i^m)_{i \le m}, E_k : k < n\}$ with respect to function composition and limited recursion (as defined above). Properties These sets clearly form the hierarchy $\mathcal{E}^0 \subseteq \mathcal{E}^1 \subseteq \mathcal{E}^2 \subseteq \cdots$ because they are closures over the $B_n$'s and $B_0 \subseteq B_1 \subseteq B_2 \subseteq \cdots$. They are strict subsets (Rose 1984; Gakwaya 1997). In other words $\mathcal{E}^0 \subsetneq \mathcal{E}^1 \subsetneq \mathcal{E}^2 \subsetneq \cdots$ because the hyper operation $H_n$ is in $\mathcal{E}^n$ but not in $\mathcal{E}^{n-1}$. • $\mathcal{E}^0$ includes functions such as x+1, x+2, ... • $\mathcal{E}^1$ provides all addition functions, such as x+y, 4x, ... • $\mathcal{E}^2$ provides all multiplication functions, such as xy, x4 • $\mathcal{E}^3$ provides all exponentiation functions, such as xy, 222x, and is exactly the elementary recursive functions. • $\mathcal{E}^4$ provides all tetration functions, and so on. Relation to primitive recursive functions The definition of $\mathcal{E}^n$ is the same as that of the primitive recursive functions, RP, except that recursion is limited ($f(t, \bar{u}) \leq j(t, \bar{u})$ for some j in $\mathcal{E}^n$) and the functions $(E_k)_{k<n}$ are explicitly included in $\mathcal{E}^n$. Thus the Grzegorczyk hierarchy can be seen as a way to limit the power of primitive recursion to different levels. It is clear from this fact that all functions in any level of the Grzegorczyk hierarchy are primitive recursive functions (i.e. $\mathcal{E}^n \subseteq RP$) and thus: $\bigcup_n{\mathcal{E}^n} \subseteq RP$ It can also be shown that all primitive recursive functions are in some level of the hierarchy (Rose 1984; Gakwaya 1997), thus $\bigcup_n{\mathcal{E}^n} = RP$ and the sets $\mathcal{E}^0, \mathcal{E}^1 - \mathcal{E}^0, \mathcal{E}^2 - \mathcal{E}^1, \dots, \mathcal{E}^n - \mathcal{E}^{n-1}, \dots$ partition the set of primitive recursive functions, RP. Extensions Main article: Fast-growing hierarchy The Grzegorczyk hierarchy can be extended to transfinite ordinals. Such extensions define a fast-growing hierarchy. To do this, the generating functions $E_\alpha$ must be recursively defined for limit ordinals (note they have already been recursively defined for successor ordinals by the relation $E_{\alpha+1}(n) = E_\alpha^n(2)$). If there is a standard way of defining a fundamental sequence $\lambda_m$, whose limit ordinal is $\lambda$, then the generating functions can be defined $E_\lambda(n) = E_{\lambda_n}(n)$. However, this definition depends upon a standard way of defining the fundamental sequence. Rose (1984) suggests a standard way for all ordinals α < ε0. The original extension was due to Martin Löb and Stan S. Wainer (1970) and is sometimes called the Löb–Wainer hierarchy. Notes 1. ^ a b Note: here $\bar{u}$ represents a tuple of inputs to f. The notation $f(\bar{u})$ means that f takes some arbitrary number of arguments and if $\bar{u} = (x, y, z)$, then $f(\bar{u}) = f(x, y, z)$. In the notation $f(t, \bar{u})$, the first argument, t, is specified explicitly and the rest as the arbitrary tuple $\bar{u}$. Thus, if $\bar{u} = (x, y, z)$, then $f(t, \bar{u}) = f(t, x, y, z)$. This notation allows composition and limited recursion to be defined for functions, f, of any number of arguments. References • Brainerd, W.S., Landweber, L.H. (1974), Theory of Computation, Wiley, ISBN 0-471-09585-0 • Cichon, E. A.; Wainer, S. S. (1983), "The slow-growing and the Grzegorczyk hierarchies", The Journal of Symbolic Logic 48 (2): 399–408, doi:10.2307/2273557, ISSN 0022-4812, MR704094 • Gakwaya, J.–S. (1997), A survey on the Grzegorczyk Hierarchy and its Extension through the BSS Model of Computability • Grzegorczyk, A. (1953), Some classes of recursive functions, Rozprawy matematyczne, Vol 4, pp. 1–45. • Löb, M.H. and Wainer, S.S., "Hierarchies of Number Theoretic Functions I, II: A Correction," Arch. Math. Logik Grundlagenforschung 14, 1970 pp. 198–199. • Rose, H.E., "Subrecursion: Functions and hierarchies", Oxford University Press, New York, USA, 1984. ISBN 0-19-853189-3 • Wagner, K. and Wechsung, G. (1986), Computational Complexity, Mathematics and its Applications v. 21. ISBN 978-90-277-2146-4	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 49, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.81095951795578, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/53556/definition-of-ring	# Definition of Ring I'm studying Abstract Algebra right now, currently covering rings. In the introduction of the subject, I am curious as to why there is no need for there to be a multiplicative identity. I understand that in order to be a ring, we require the set to be an abelian group under addition operation and a monoid under multiplication. But what is the reason for the monoid, rather than group under multiplication--or lack of multiplication? - 6 – Qiaochu Yuan Jul 24 '11 at 23:18 3 Related questions here and here. – Henry Jul 24 '11 at 23:20 It's the former. Thanks both for the links. – hdtv1104 Jul 24 '11 at 23:35 – GEdgar Jul 24 '11 at 23:37 – Gone Jul 25 '11 at 0:17 show 1 more comment ## 5 Answers Basically, you want to encompass a large enough class of objects that the study is useful, while at the same time having "enough" structure to be able to say interesting things. This leads to a difficult balance: the more requirements you put on the structure, the fewer structures that the definition will cover. For example, if you require that the structure be an abelian group under both addition and multiplication, then $\mathbb{Z}$, the integers, don't qualify. So, the fewer conditions (requirements), the more structures you expect to satisfy them. On the other hand, in order to be able to say interesting things you usually need assumptions. That means that the more conditions you put on the structure, the more things you have to "play with", and the more likely you are to be able to say interesting (or far reaching) things. Consider for example "finite groups". We are very far from having a satisfactory answer to the question "What are all the finite groups?"; but throw in the simple condition that multiplication commute, and the question "What are all the finite abelian groups?" already has a very good answer (the structure theorem for finitely generated abelian groups). So, the more conditions you place on the objects you want to study, the more you expect to be able to say about them. And so, we usually study semigroups and groups rather than magmas; why? Because groups and semigroups are prevalent enough that a lot of objects satisfy the conditions, and at the same time the requirements are strong enough to let us say lots of interesting things about them (we have a harder with semigroups than with groups, and an even harder with loops and magmas...) For rings, a good balance turns out to be when we require the multiplicative structure to be either a semigroup or a monoid, rather than full-fledged group. (We do study the cases where you get a group: you get fields and division rings). A monoid structure (with a multiplicative identity) used to be prefered, but it turns out to exclude a lot of interesting cases (many of which arise in places like functional analysis). So an expansion of the class of structures to those in which you only have a semigroup structure has been prefered, though many authors still assume all rings have an identity and that homomorphisms between rings map identities to identities (e.g., Lam's books all make this assumption). So... it's a balancing act between trying to "cover" a lot and at the same time being able to "say" a lot. Asking for the multiplicative structure to be a semigroup or a monoid is a good balance. There are other, weaker structures (such as near-rings and semi-rings) as well, just like with groups you have semigroups, quasigroups, loops, and magmas. - The point of algebraic structures is not just to have algebraic structures, but rather to have algebraic structures that reflect the things that are already out there. One of the simplest algebraic objects out there is the collection of integers, $\mathbb{Z}$. We have two operations, multiplication and addition, and they satisfy all sorts of properties. However, if we demanded that we have multiplicative inverses, then $\mathbb{Z}$ would not fit into our mold, and we want some sort of definition that describes the structure present in $\mathbb{Z}$. What if we want to consider $2\mathbb{Z}\subset \mathbb{Z}$, the even integers? We still have multiplication and addition, but we no longer have a multiplicative identity. Should this be considered a ring? Or something else? This depends on who you ask, as a lot of people require rings to have multiplicative identities, and they will explicitly say "ring without unit" otherwise. Still, regardless what we call it, it is clear that we should have some sort of algebraic structure that models the properties of $2\mathbb{Z}$. If, when we ignore $0$, we have a group under multiplication, we get the notion of a division ring, and if the group is abelian we get this notion of a field, which is very important. The rational numbers, the real numbers, the complex numbers are all fields. Some of the most beautiful mathematics comes from working over fields, as there are things you can do over fields that you cannot do over arbitrary rings. However, as we have seen, it is important that we consider rings, because otherwise there would be basic algebraic objects out there begging for a name. - "If, when we ignore 0, we have a group under multiplication, we get the notion of a field" I think you mean an abelian group. Otherwise you get a division ring. – JSchlather Jul 25 '11 at 0:41 @Jacob Yes, for some reason I was thinking about commutative rings. Although, division rings are also called "skew fields", and share many similar properties. If history had been slightly different, there would be fields and commutative fields. Still, making the change. – Aaron Jul 25 '11 at 0:56 @Aaron: This is something that I've thought about with regard to requiring or not requiring a multiplicative identity in rings: requiring a multiplicative identity doesn't allow $n\mathbb{Z} \subset \mathbb{Z}$, $n \ge 2$ to be an ideal (since an ideal is a subring), which doesn't seem very convenient or natural. Do you agree that this is a good motivator for not requiring a multiplicative identity? – Alex Petzke Jan 17 at 18:06 @AlexPetzke: No, I think language usage trumps this consideration. Most of the non-unital rings I see are $k$-algebras for some (unital, often a field) ring $k$. Moreover, this additional $k$-algebra structure ends up being important, so "$k$-algebra" couldn't really be omitted. For people in algebraic geometry or rep theory, explicitly calling them unital rings every single time would be cumbersome. Papers would have blanket disclaimers that "every ring is assumed unital unless noted" and most algebaists would start assuming that rings are unital. – Aaron Jan 17 at 21:42 Taking your "rather than group under multiplication" phrase: It is interesting to consider things which have some properties similar to the integers under addition and multiplication. So you want a group under addition but not under multiplication. - If lots of interesting stuff can be proved without assuming there's a unit, then assuming there's a unit would merely complicate the proofs with an irrelevant assumption. So if you're getting this stuff from a textbook, see which results the book proves without assuming a unit. - Many mathematicians only assume that the set is a semigroup not monoid under multiplication for example Nathan Jacobson,I. N. Herstein,Seth Warner,N. McCoy, and Van der Werden. There are some theorems on rings which require a multiplicative (right or left or an) identity however. For an interesting nonunital ring see "Planetmath.org" - Klein 4 ring. See Hopkins theorem (an Artinian ring with identity is a Noetherian ring) for a need for identity element for a ring. Also rings with an identity element have a maximal ideal,a maximal right ideal, and a maximal left ideal using Zorn's lemma.Many early results in Ring Theory did not need the asumption that a ring was unital. Many fascinating non unital rings and subrings are found in matrix rings over a non unital ring. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 13, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9467251300811768, "perplexity_flag": "head"}
http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.em/1048515870	previous :: next ### Chebyshev's bias Michael Rubinstein and Peter Sarnak Source: Experiment. Math. Volume 3, Issue 3 (1994), 173-197. #### Abstract The title refers to the fact, noted by Chebyshev in 1853, that primes congruent to 3 modulo 4 seem to predominate over those congruent to 1. We study this phenomenon and its generalizations. Assuming the Generalized Riemann Hypothesis and the Grand Simplicity Hypothesis (about the zeros of the \hbox{Dirichlet} $L$-function), we can characterize exactly those moduli and residue classes for which the bias is present. We also give results of numerical investigations on the prevalence of the bias for several moduli. Finally, we briefly discuss generalizations of the bias to the distribution to primes in ideal classes in number fields, and to prime geodesics in homology classes on hyperbolic surfaces. First Page: Primary Subjects: 11N13 Secondary Subjects: 11N69, 11Y35 Full-text: Open access	{"extraction_info": {"found_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": 0.8819947838783264, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/54038?sort=oldest	## Graphs with many triangles but few complete graphs on 4 vertices ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Let $G$ be a graph on $n$ vertices with $an^2$ edges containing at most $an^2/2$ copies of $K_4$. If there are cubically many triangles, say $cn^3$, then there is at least one edge that is not contained in any $K_4$. Note that it is necessary to require that the number of $K_4$'s is significantly smaller than the number of edges. Otherwise we could take a complete tripartite graph and add an edge in each of the three colour classes. So far my proof attempts using averaging arguments failed. Maybe there are counterexamples? If so, does it help to assume in addition that $a$ is bigger than $1/4$? ## Background/Motivation If the statement is true it implies that asymptotically $5n^2/16$ is the smallest size of an antichain in $2^{[n]}$ that is maximal among the antichains containing only 2-sets and 4-sets: basically, the $K_4$'s and the nonedges in the graph are the 4-sets and the 2-sets in the antichain. - ## 1 Answer Your statement may be true for large enough values of $c$, but it is not true for all constants $c>0$. Specifically, for small enough values of $c$, form a counterexample $G$ consisting of the disjoint union of two subgraphs: • $K_{n\sqrt 2,n\sqrt 2,n\sqrt 2}$ together with one extra edge in each component of the tripartition as in your example • $K_{n\sqrt{12},n\sqrt{12}}$ together with a perfect matching in each component of its bipartition. (Obviously these numbers of vertices are not all going to be integers, so round them.) Then there are approximately $18n^2$ edges ($6n^2$ in the first subgraph and $12n^2$ in the second), approximately $9n^2$ $K_4$'s ($6n^2$ in the first subgraph and $3n^2$ in the second), approximately $(3\sqrt 2 + 4\sqrt 3)n$ vertices, and approximately $(2\sqrt 2)n^3$ triangles (mostly in the first subgraph). So this example has only half as many $K_4$'s as edges, as you ask, and every edge belongs to at least one $K_4$, with $c\approx\frac{2\sqrt 2}{(3\sqrt 2 + 4\sqrt 3)^3}\approx 0.00203$. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 31, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": 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": 0.939268171787262, "perplexity_flag": "head"}
http://www.scholarpedia.org/article/Delay_differential_equations	# Delay-Differential Equations From Scholarpedia Skip Thompson (2007), Scholarpedia, 2(3):2367. (Redirected from Delay differential equations) Curator and Contributors 1.00 - Skip Thompson Delay differential equations differ from ordinary differential equations in that the derivative at any time depends on the solution (and in the case of neutral equations on the derivative) at prior times. The simplest constant delay equations have the form $\tag{1} y'(t) = f(t, y(t), y(t-\tau_1), y(t-\tau_2),\ldots, y(t-\tau_k))$ where the time delays (lags) $$\tau_j$$ are positive constants. More generally, state dependent delays may depend on the solution, that is $$\tau_i = \tau_i (t,y(t)) \ .$$ ## Introduction Systems of delay differential equations now occupy a place of central importance in all areas of science and particularly in the biological sciences (e.g., population dynamics and epidemiology). Baker, Paul, & Willé (1995) contains references for several application areas. Interest in such systems often arises when traditional pointwise modeling assumptions are replaced by more realistic distributed assumptions, for example, when the birth rate of predators is affected by prior levels of predators or prey rather than by only the current levels in a predator-prey model. The manner in which the properties of systems of delay differential equations differ from those of systems of ordinary differential equations has been and remains an active area of research; see Martin & Ruan (2001) and Raghothama & Narayanan (2002) for typical examples of such studies. See also Shampine, Gladwell, and Thompson (2003) for a description of several common models. ## Initial History Function Additional information is required to specify a system of delay differential equations. Because the derivative in (1) depends on the solution at the previous time $$t - \tau_j \ ,$$ it is necessary to provide an initial history function to specify the value of the solution before time $$t = 0 \ .$$ In many common models the history is a constant vector; but non constant history functions are encountered routinely. For most problems there is a jump derivative discontinuity at the inital time. In addition, any solution or derivative discontinuity in the history function at points prior to the initial time need to be handled appropriately since such discontinuities are propagated to future times. ## Derivative Discontinuities In most models, the delay differential equation and the initial history are incompatible: for some derivative order, usually the first, the left and right derivatives are not equal. For example, the simple model $$y'(t) = y(t-1)$$ with constant history $$y(t) = 1$$ has the property that $$y'(0^{+}) = 1 \ne y'(0^{-}) = 0 \ .$$ One of the most fascinating properties of delay differential equations is the manner in which such derivative discontinuities are propagated in time. For the equation and history just described, for example, the initial first discontinuity is propagated as a second degree discontinuity at time $$t = 1 \ ,$$ as a third degree discontinuity at time $$t = 2 \ ,$$ and, more generally, as a discontinuity in the $${(n+1)}^{st}$$ derivative at time $$t = n \ .$$ This behavior is typical of that for a wide class of delay differential equations: generalized smoothing occurs as the initial derivative discontinuity is propagated successively to higher order derivatives. Smoothing need not occur for neutral equations or for non-neutral equations with vanishing delays. Neves & Feldstein (1976) characterized the tree of derivative discontinuity times for state dependent delay differential equations as the zeroes with odd multiplicity of equations $\tag{2} \tau_i (t,y(t)) - T = 0$ where $$T$$ is the initial time or any later discontinuity time. ## Continuous Extensions Several of the solvers discussed in the next section use explicit Runge-Kutta methods to integrate systems of delay differential equations. An important question in this case is that of interpolation. Unlike ordinary differential equation solvers that are based on linear multistep methods possessing natural extensions, early Runge-Kutta solvers did not incorporate interpolation; rather they stepped exactly to the next output point instead of stepping beyond it and obtaining interpolated solutions. Interest in the issues of obtaining dense output without limiting the step size in this fashion and by the desire to incorporate root finding led to the development of Runge-Kutta methods endowed with suitable interpolants. Interpolation is handled in one of two ways in modern Runge-Kutta solvers, Hermite interpolation and continuously imbedded methods. For example, the solver dde23 which is based on a third order Runge-Kutta method uses Hermite interpolation of the old and new solution and derivative to obtain an accurate interpolant. By way of contrast, the solver dde_solver uses a sixth order Runge-Kutta method based on a continuously embedded $$C^1$$ interpolant derived from the same derivative approximations used by the basic method. In addition to providing accurate and efficient solutions, either type of interpolant can be used in conjunction with a root finder to locate derivative discontinuity times. ## Available Delay Differential Equation Software A number of issues must be taken into account by software for delay differential equations. Baker, Paul, & Willé (1995), Shampine & Thompson (2001), and Thompson & Shampine (2006) discuss the various issues. The well known dmrode solver (Neves (1975)) was the first effective software for delay differential equations. Many of the central ideas on which this solver was based were used in later f77 solvers dklag5 (Neves & Thompson (1992)) and dklag6 (Corwin, Sarafyan, and Thompson (1997)), and the Fortran 90/95 dde_solver (Thompson & Shampine (2006)). Although the state of the art for numerical software for delay differential equations is not as advanced as that for ordinary differential equation software, several high quality solvers have recently been developed. The effectiveness of the software is determined in large part by the manner in which propagated derivative discontinuities are handled. Some delay differential equation solvers such as those in Paul (1995), and Thompson & Shampine (2006) explicitly track and locate the zeroes of (2) and include them as integration mesh points. Different approaches are used in other software. For example, the ddverk solver (Enright & Hayashi (1997)) uses derivative defect error control to implicitly locate discontinuity times. It then uses special interpolants to step cross the discontinuities. The ddesd solver (Shampine (2005)) uses residual error control to avoid the use of embedded local error estimates near discontinuity times. Effective delay differential equation software must deal with other difficulties peculiar to systems of delay differential equations. Early software, for example, limited the step sizes used to be no larger than the smallest delay. But small delays are encountered in many problems; and this artificial restriction on the step size can have a drastic effect on the efficiency of a solver. Most of the solvers mentioned above are based on pairs of explicit continuously embedded Runge-Kutta methods (Shampine (1994)). When the step size exceeds a delay, the underlying interpolation polynomials are iterated in a manner somewhat akin to a predictor-corrector iteration for linear multistep methods. Refer to Baker & Paul (1996), Baker, Paul, & Willé (1995), Enright & Hayashi (1998), and Shampine & Thompson (2001) for details of various aspects of this issue. The solvers dde23, ddesd, and dde_solver contain a very useful provision for finding zeroes of event functions (Shampine (1994)) that depend on the solution. In addition to solving a system of delay differential equations, they simultaneously locate zeroes of state dependent functions $$g(t,y(t)) = 0 \ .$$ Such special events may signal problem changes requiring integration restarts. The use of event functions is illustrated in the next section. Although much recent delay differential equation software utilizes explicit continuously embedded Runge-Kutta methods, software based on other methods has been developed. For example, Jackiewicz & Lo (2006) and Willé & Baker (1992) utilize generalized Adams linear multistep methods; and the radar5 solver (http://www.unige.ch/~hairer/software.html) is based on collocation methods. Another well known and widely used program with the ability to solve delay differential equations is the xppaut program (Ermentrout (2002)). The use of software based on a class of general linear methods (diagonally implicit multistage integration methods) is discussed in Hoppensteadt & Jackiewicz (2006) in conjunction with the problem considered in the next section. Bellen & Zennaro (2003) discuss the commonly used methods for delay differential equations in considerable detail. ## An Example Hoppensteadt & Jackiewicz (2006) investigated a model which generalizes previously studied models for infectious diseases. Solving this model requires the determination of a threshold time at which the accumulated dosage of infection reaches a prescribed level. Once this time is determined, the relevant equations may be integrated to obtain the desired solution. The minimum threshold time $$t_0$$ is the unique value for which $\int_{0}^{t_0} \rho(t) I_0(t) dt = m .$ The defining delay differential equations are $\tau'(t) = \frac{\rho(t) I(t)}{\rho(\tau(t)) I(\tau(t))}, \quad \tau(t_0) = 0$ $S'(t) = -r(t) I(t) S(t), \quad S(0) = 0$ Here the function $$I(t)$$ is $I(t) = \begin{cases} I_0(t), & -\sigma \le t \le t_0, \\ I_0(t) + S_0 - S(\tau(t)), & t_0 \le t \le t_0 + \sigma \\ S(\tau(t-\sigma)) - S(\tau(t)), & t_0 + \sigma \le t \end{cases}$ For Example 1 of the reference, the relevant variables and functions are given by $$m = 0.1, \sigma = 1, S_0 = 10, \rho(t) = 1, r(t) = r_0, S_0 = 10,$$ and $I_0 = \begin{cases} 0.4(1+t), & t \le 0 \\ 0.4(1-t), & 0 \le t \le 1 \\ 0, & 1 < t \end{cases}$ Following is a brief description of the manner in which this problem can be solved. As the description suggests, considerable dexterity may be required to solve a realistic system of delay differential equations. The solution of this problem involves three solution phases. Three delay differential equations are solved in each phase, one for $$\tau'(t) \ ,$$ one for $$S'(t) \ ,$$ and one for the accumulated dosage. The accumulated dosage $$y_3(t)$$ is obtained by solving the equation $y_3'(t) = \rho(t) I_0(t) .$ Three additional delay functions, $$t-1 \ ,$$ $$\tau(t) \ ,$$ and $$\tau(t) - \sigma \ ,$$ can be used to facilitate interpolations that must be performed during the different phases of the solution. During the first phase of the integration, $$\tau'(t) = 0 \ .$$ A solution dependent event function $$g(t) = y_3(t) - m$$ may be used to locate the threshold time $$t_0 \ ,$$ i.e., the time $$t_0$$ for which $$y_3(t_0) = m \ .$$ Once $$t_0$$ is located, the relevant equations for the second phase of the solution can be solved. The event function is changed to $$g(t) = t - (t_0 + \sigma) \ .$$ This allows the program to locate $$t_0 + \sigma$$ as an event. Once the second event is located, the full set of equations can be integrated to the desired final time of $$t = 8 \ .$$ Two examples are considered in the paper. For the first example, $$\rho(t) = 1$$ and $$r(t) = r_0 \ .$$ For the second example, $$\rho(t) = e^{-t}$$ and $$r(t) = r_0 (1 + \sin(5t)) \ .$$ Plots of $$S(t)$$ and $$I(t)$$ are given for several values of the parameter $$r_0 \ .$$ To obtain these graphs, the problem is solved for each value of $$r_0 \ .$$ Plots of the solutions thus obtained are displayed in Figures 1 and 2; they replicate those given in the paper. Figure 1: Infected Individuals Figure 2: Susceptible Individuals ## References • Baker, C.T.H. and Paul, C.A.H., A global convergence theorem for a class of parallel continuous explicit Runge-Kutta methods and vanishing lag delay differential equations, SIAM J. Numer. Anal., 33:1559-1576, 1996. • Baker, C.T.H., Paul, C.A.H., and Willé, D.R., A bibliography on the numerical solution of delay differential equations, Numerical Analysis Report 269, Mathematics Department, University of Manchester, U.K., 1995. • Baker, C.T.H., Paul, C.A.H., and Willé, D.R., Issues in the numerical solution of evolutionary delay differential equations, Adv. Comput. Math., 3:171-196, 1995. • Bellen, A. and Zennaro, M., Numerical Methods for Delay Differential Equations, Oxford Science Publications, Clarendon Press, Oxford, 2003. • Corwin, S.P., Sarafyan, D. and Thompson S., DKLAG6: a code based on continuously imbedded sixth order Runge-Kutta methods for the solution of state dependent functional differential equations, Appl. Numer. Math., 24:319-333, 1997. • Enright, W.H. and Hayashi, H., A delay differential equation solver based on a continuous Runge-Kutta method with defect control, Numer. Alg., 16:349-364, 1997. • Enright, W.H. and Hayashi, H., Convergence analysis of the solution of retarded and neutral differential equations by continuous methods, SIAM J. Numer. Anal., 35:572-585, 1998. • Ermentrout, B., Simulating, Analyzing, and Animating Dynamical Systems: A Guide to XPPAUT for Researchers and Students, SIAM, Philadelphia, USA, 2002. • Hoppensteadt, F.C. and Jackiewicz, Z. , Numerical solution of a problem in the theory of epidemics, Appl. Numer. Math., 56:533-543, 2006. • Jackiewicz, Z. and Lo E., Numerical solution of neutral functional differential equations by Adams methods in divided difference form, J. Comput. Appl. Math., 18:592-605, 2006. • Martin, A. and Ruan, S., Predator-prey models with delay and prey harvesting, J. Math. Biol., 43:247-267, 2001. • Neves, K.W., Automatic integration of functional differential equations: an approach, ACM Trans. Math. Softw., 1:357-368, 1975. • Neves, K.W. and Feldstein A., Characterization of jump discontinuities for state dependent delay differential equations, J. Math. Anal. and Appl., 5:689-707, 1976. • Neves, K.W. and Thompson S., Software for the numerical solution of systems of functional differential equations with state dependent delays, Appl. Numer. Math., 9:385-401, 1992. • Paul, C.A.H., A user-guide to ARCHI, Numerical Analysis Report 283, Mathematics Department, University of Manchester, U.K., 1995. • Raghothama, A. and Narayanan, S., Periodic response and chaos in nonlinear systems with parametric excitation and time delay, Nonlin. Dyn., 27:341-365, 2002. • Shampine, L.F., Numerical Solution of Ordinary Differential Equations, Chapman & Hall, New York, NY, 1994. • Shampine, L.F., Solving ODEs and DDEs with residual control, Appl. Numer. Math., 52:113-127, 2005. • Shampine, L.F., Gladwell, I., and Thompson, S., Solving ODEs with MATLAB, Cambridge Univ. Press, Cambridge, 2003. • Shampine, L.F. and Thompson, S., Solving DDEs in Matlab, Appl. Numer. Math., 37:441-458, 2001. • Thompson, S. and Shampine, L.F., A friendly fortran 90 DDE solver, Appl. Numer. Math., 56:503-516, 2006. • Willé, D.R. and Baker, C.T.H., DELSOL - a numerical code for the solution of systems of delay-differential equations, Appl. Numer. Math., 9:223-234, 1992. Internal references • Zdzislaw Jackiewicz (2007) General linear methods. Scholarpedia, 2(4):2852. • Frank Hoppensteadt (2006) Predator-prey model. Scholarpedia, 1(10):1563. • John Butcher (2007) Runge-Kutta methods. Scholarpedia, 2(9):3147. • Bard Ermentrout (2007) XPPAUT. Scholarpedia, 2(1):1399.	{"extraction_info": {"found_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": 40, "mathjax_asciimath": 0, "img_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": 0.8872387409210205, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/109637/uniform-convergence-of-measures	# Uniform convergence of measures Let $A\subset[0,1]^n\subset\mathbb{R}^n$ be a closed semi-algebraic set. Let $f_k: [0,1]\rightarrow\mathbb{R}$, $f_k(x_i)=\mu^{n-1}(A_{\|_{x_i}}+B^{n-1}_{1/k})$ where $A_{\|_{x_i}}=A\cap H^i_{x_i}$ with $H^i_{x_i}$ the hyperplane orthogonal to the $i$-axis through $x_i$, $\mu^{n-1}$ the $n-1$-dimensional Lebesgue measure and $B^{n-1}_\varepsilon$ the $n-1$-dimensional $\varepsilon$-ball around zero. I.e. for any $x_i \in [0,1]$ it measures the cross section of $A$ that is widened by an $\varepsilon$-ball. Show that $f_k$ converges uniformly to $f(x_i)=\mu^{n-1}(A_{\|_{x_i}})$ for $x_i\in [0,1]$. $A_{\|_{x_i}}$ is closed, so $A_{\|_{x_i}}$ is the limit of $A_{\|_{x_i}}+B^{n-1}_{1/k}$. From the properties of the Lebesgue measure we get that $f_k$ converges pointwise. But I have no idea how to prove uniform convergence. EDIT: Some more thoughts of mine: The series above converges uniformly, if $f_k'(x_i)=\mu^{n-1}(A_{\|_{x_i}}+B_{1/k}) -\mu^{n-1}(A_{\|_{x_i}})$ converges uniformly to zero. Since $f_k''(x_i)=\mu^{n-1}(\delta(A_{\|_{x_i}})+B_{1/k})\ge f_k'(x_i)$ for all $x_i$ it would suffice that $f_k''$ converges uniformly to zero. Here, discontinuities of $f_k''$ are a problem. It seems that $f_k''$ is discontiuous in $x\in[0,1]$ iff the set-valued map $M^A:[0,1]\subset\mathbb{R}\rightrightarrows\mathbb{R}^{n-1}; M^A(x_i)=A_{\|_{x_i}}$ is Hausdorff discontiuous in $x_i$. Uniform convergence should follow if $M^A$ is Hausdorff discontiuous in finitely many points only. $A$ is semi-algebraic. For a set defined by a single polynomial (in-) equality the number of discontinuities is finite. Does it stay finite under unions / intersections (complement seems obvious)? Any help is greatly appreciated! I have the feeling that I'm missing the easy path and taking a detour through the djungle... Thank you J.P. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 33, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": 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": 0.9242895841598511, "perplexity_flag": "head"}
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Is ...	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 28, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 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": 0.8893547654151917, "perplexity_flag": "middle"}
http://en.wikipedia.org/wiki/Trigonometric_functions	# Trigonometric functions "Cosine" redirects here. For the similarity measure, see Cosine similarity. Base of trigonometry: if two right triangles have equal acute angles, they are similar, so their side lengths are proportional. Proportionality constants are written within the image: sin θ, cos θ, tan θ, where θ is the common measure of five acute angles. In mathematics, the trigonometric functions (also called the circular functions) are functions of an angle. They are used to relate the angles of a triangle to the lengths of the sides of a triangle. Trigonometric functions are important in the study of triangles and modeling periodic phenomena, among many other applications. The most familiar trigonometric functions are the sine, cosine, and tangent. In the context of the standard unit circle with radius 1, where a triangle is formed by a ray originating at the origin and making some angle with the x-axis, the sine of the angle gives the length of the y-component (rise) of the triangle, the cosine gives the length of the x-component (run), and the tangent function gives the slope (y-component divided by the x-component). More precise definitions are detailed below. Trigonometric functions are commonly defined as ratios of two sides of a right triangle containing the angle, and can equivalently be defined as the lengths of various line segments from a unit circle. More modern definitions express them as infinite series or as solutions of certain differential equations, allowing their extension to arbitrary positive and negative values and even to complex numbers. Trigonometric functions have a wide range of uses including computing unknown lengths and angles in triangles (often right triangles). In this use, trigonometric functions are used, for instance, in navigation, engineering, and physics. A common use in elementary physics is resolving a vector into Cartesian coordinates. The sine and cosine functions are also commonly used to model periodic function phenomena such as sound and light waves, the position and velocity of harmonic oscillators, sunlight intensity and day length, and average temperature variations through the year. In modern usage, there are six basic trigonometric functions, tabulated here with equations that relate them to one another. Especially with the last four, these relations are often taken as the definitions of those functions, but one can define them equally well geometrically, or by other means, and then derive these relations. ## Right-angled triangle definitions (Top): Trigonometric function sinθ for selected angles θ, π − θ, π + θ, and 2π − θ in the four quadrants. (Bottom) Graph of sine function versus angle. Angles from the top panel are identified. The notion that there should be some standard correspondence between the lengths of the sides of a triangle and the angles of the triangle comes as soon as one recognizes that similar triangles maintain the same ratios between their sides. That is, for any similar triangle the ratio of the hypotenuse (for example) and another of the sides remains the same. If the hypotenuse is twice as long, so are the sides. It is these ratios that the trigonometric functions express. To define the trigonometric functions for the angle A, start with any right triangle that contains the angle A. The three sides of the triangle are named as follows: • The hypotenuse is the side opposite the right angle, in this case side h. The hypotenuse is always the longest side of a right-angled triangle. • The opposite side is the side opposite to the angle we are interested in (angle A), in this case side a. • The adjacent side is the side having both the angles of interest (angle A and right-angle C), in this case side b. In ordinary Euclidean geometry, according to the triangle postulate, the inside angles of every triangle total 180° (π radians). Therefore, in a right-angled triangle, the two non-right angles total 90° (π/2 radians), so each of these angles must be in the range of (0°,90°) as expressed in interval notation. The following definitions apply to angles in this 0° – 90° range. They can be extended to the full set of real arguments by using the unit circle, or by requiring certain symmetries and that they be periodic functions. For example, the figure shows sin θ for angles θ, π − θ, π + θ, and 2π − θ depicted on the unit circle (top) and as a graph (bottom). The value of the sine repeats itself apart from sign in all four quadrants, and if the range of θ is extended to additional rotations, this behavior repeats periodically with a period 2π. Rigorously, in metric space, one should express angle, defined as scaled arc length, as a function of triangle sides. It leads to inverse trigonometric functions first and usual trigonometric functions can be defined by inverting them back. The trigonometric functions are summarized in the following table and described in more detail below. The angle θ is the angle between the hypotenuse and the adjacent line – the angle at A in the accompanying diagram. Function Abbreviation Description Identities (using radians) Sine sin opposite / hypotenuse $\sin \theta = \cos \left(\frac{\pi}{2} - \theta \right) = \frac{1}{\csc \theta}$ Cosine cos adjacent / hypotenuse $\cos \theta = \sin \left(\frac{\pi}{2} - \theta \right) = \frac{1}{\sec \theta}\,$ Tangent tan (or tg) opposite / adjacent $\tan \theta = \frac{\sin \theta}{\cos \theta} = \cot \left(\frac{\pi}{2} - \theta \right) = \frac{1}{\cot \theta}$ Cotangent cot (or cotan or cotg or ctg or ctn) adjacent / opposite $\cot \theta = \frac{\cos \theta}{\sin \theta} = \tan \left(\frac{\pi}{2} - \theta \right) = \frac{1}{\tan \theta}$ Secant sec hypotenuse / adjacent $\sec \theta = \csc \left(\frac{\pi}{2} - \theta \right) = \frac{1}{\cos \theta}$ Cosecant csc (or cosec) hypotenuse / opposite $\csc \theta = \sec \left(\frac{\pi}{2} - \theta \right) = \frac{1}{\sin \theta}$ The sine, tangent, and secant functions of an angle constructed geometrically in terms of a unit circle. The number θ is the length of the curve; thus angles are being measured in radians. The secant and tangent functions rely on a fixed vertical line and the sine function on a moving vertical line. ("Fixed" in this context means not moving as θ changes; "moving" means depending on θ.) Thus, as θ goes from 0 up to a right angle, sin θ goes from 0 to 1, tan θ goes from 0 to ∞, and sec θ goes from 1 to ∞. The cosine, cotangent, and cosecant functions of an angle θ constructed geometrically in terms of a unit circle. The functions whose names have the prefix co- use horizontal lines where the others use vertical lines. ### Sine, cosine and tangent The sine of an angle is the ratio of the length of the opposite side to the length of the hypotenuse. (The word comes from the Latin sinus for gulf or bay,[1] since, given a unit circle, it is the side of the triangle on which the angle opens.) In our case $\sin A = \frac {\textrm{opposite}} {\textrm{hypotenuse}} = \frac {a} {h}.$ Note that this ratio does not depend on size of the particular right triangle chosen, as long as it contains the angle A, since all such triangles are similar. The cosine of an angle is the ratio of the length of the adjacent side to the length of the hypotenuse: so called because it is the sine of the complementary or co-angle.[2] In our case $\cos A = \frac {\textrm{adjacent}} {\textrm{hypotenuse}} = \frac {b} {h}.$ The tangent of an angle is the ratio of the length of the opposite side to the length of the adjacent side: so called because it can be represented as a line segment tangent to the circle, that is the line that touches the circle, from Latin linea tangens or touching line (cf. tangere, to touch).[3] In our case $\tan A = \frac {\textrm{opposite}} {\textrm{adjacent}} = \frac {a} {b}.$ The acronyms "SOHCAHTOA" ("Soak-a-toe", "Sock-a-toa", "So-kah-toa") and "OHSAHCOAT" are commonly used mnemonics for these ratios. ### Reciprocal functions The remaining three functions are best defined using the above three functions. The cosecant csc(A), or cosec(A), is the reciprocal of sin(A), i.e. the ratio of the length of the hypotenuse to the length of the opposite side: $\csc A = \frac {1}{\sin A} = \frac {\textrm{hypotenuse}} {\textrm{opposite}} = \frac {h} {a}.$ The secant sec(A) is the reciprocal of cos(A), i.e. the ratio of the length of the hypotenuse to the length of the adjacent side: $\sec A = \frac {1}{\cos A} = \frac {\textrm{hypotenuse}} {\textrm{adjacent}} = \frac {h} {b}.$ It is so called because it represents the line that cuts the circle (from Latin: secare, to cut).[4] The cotangent cot(A) is the reciprocal of tan(A), i.e. the ratio of the length of the adjacent side to the length of the opposite side: $\cot A = \frac {1}{\tan A} = \frac {\textrm{adjacent}} {\textrm{opposite}} = \frac {b} {a}.$ ### Slope definitions Equivalent to the right-triangle definitions, the trigonometric functions can also be defined in terms of the rise, run, and slope of a line segment relative to horizontal. The slope is commonly taught as "rise over run" or rise⁄run. The three main trigonometric functions are commonly taught in the order sine, cosine, tangent. With a line segment length of 1 (as in a unit circle), the following mnemonic devices show the correspondence of definitions: 1. "Sine is first, rise is first" meaning that Sine takes the angle of the line segment and tells its vertical rise when the length of the line is 1. 2. "Cosine is second, run is second" meaning that Cosine takes the angle of the line segment and tells its horizontal run when the length of the line is 1. 3. "Tangent combines the rise and run" meaning that Tangent takes the angle of the line segment and tells its slope; or alternatively, tells the vertical rise when the line segment's horizontal run is 1. This shows the main use of tangent and arctangent: converting between the two ways of telling the slant of a line, i.e., angles and slopes. (Note that the arctangent or "inverse tangent" is not to be confused with the cotangent, which is cosine divided by sine.) While the length of the line segment makes no difference for the slope (the slope does not depend on the length of the slanted line), it does affect rise and run. To adjust and find the actual rise and run when the line does not have a length of 1, just multiply the sine and cosine by the line length. For instance, if the line segment has length 5, the run at an angle of 7° is 5 cos(7°) ## Unit-circle definitions The unit circle The six trigonometric functions can also be defined in terms of the unit circle, the circle of radius one centered at the origin. The unit circle definition provides little in the way of practical calculation; indeed it relies on right triangles for most angles. The unit circle definition does, however, permit the definition of the trigonometric functions for all positive and negative arguments, not just for angles between 0 and π/2 radians. It also provides a single visual picture that encapsulates at once all the important triangles. From the Pythagorean theorem the equation for the unit circle is: $x^2 + y^2 = 1. \,$ In the picture, some common angles, measured in radians, are given. Measurements in the counterclockwise direction are positive angles and measurements in the clockwise direction are negative angles. Let a line through the origin, making an angle of θ with the positive half of the x-axis, intersect the unit circle. The x- and y-coordinates of this point of intersection are equal to cos θ and sin θ, respectively. The triangle in the graphic enforces the formula; the radius is equal to the hypotenuse and has length 1, so we have sin θ = y/1 and cos θ = x/1. The unit circle can be thought of as a way of looking at an infinite number of triangles by varying the lengths of their legs but keeping the lengths of their hypotenuses equal to 1. Note that these values (sin 0°, sin 30°, sin 45°, sin 60° and sin 90°) can easily be memorized in the form $\frac{1}{2}\sqrt{0},\quad \frac{1}{2}\sqrt{1},\quad \frac{1}{2}\sqrt{2},\quad \frac{1}{2}\sqrt{3},\quad \frac{1}{2}\sqrt{4},$ but the angles are not equally spaced. The values for 15°, 18º, 36º, 54°, 72º, and 75° are derived as follows: $\sin 15^\circ = \cos 75^\circ = \dfrac{\sqrt6 - \sqrt2}{4}\,\!$ $\sin 18^\circ = \cos 72^\circ = \frac{\sqrt5 - 1}{4}$ $\sin 36^\circ = \cos 54^\circ = \frac{\sqrt{10 - 2\sqrt5}}{4}$ $\sin 54^\circ = \cos 36^\circ = \dfrac{\sqrt5 + 1 } {4}\,\!$ $\sin 72^\circ = \cos 18^\circ = \frac{\sqrt{10 + 2\sqrt5}}{4}$ $\sin 75^\circ = \cos 15^\circ = \dfrac{\sqrt6 + \sqrt2}{4}.\,\!$ From these, the values for all multiples of 3º can be analytically computed. For example: $\sin 3^\circ = \cos 87^\circ = \dfrac{\sqrt{30} + \sqrt{10} + \sqrt{20 + 4 \sqrt5} - \sqrt6 - \sqrt2 - \sqrt{60 + 12 \sqrt5}}{16}\,\!$ $\sin 6^\circ = \cos 84^\circ = \dfrac{\sqrt{30 - 6 \sqrt5} - \sqrt5 - 1 }{8}\,\!$ $\sin 9^\circ = \cos 81^\circ = \dfrac{\sqrt{90} + \sqrt{18} + \sqrt{10} + \sqrt2 - \sqrt{20 - 4 \sqrt5} - \sqrt{180 - 36 \sqrt5}}{32}\,\!$ $\sin 84^\circ = \cos 6^\circ =\frac{\sqrt{10 - 2\sqrt5} + \sqrt{15} + \sqrt3}{8}$ $\sin 87^\circ = \cos 3^\circ =\frac{\sqrt{60 + 12\sqrt5}+\sqrt{20 + 4\sqrt5}+\sqrt{30}+\sqrt2-\sqrt6-\sqrt{10}}{16}$ The sine and cosine functions graphed on the Cartesian plane. For angles greater than 2π or less than −2π, simply continue to rotate around the circle; sine and cosine are periodic functions with period 2π: $\sin\theta = \sin\left(\theta + 2\pi k \right),\,$ $\cos\theta = \cos\left(\theta + 2\pi k \right),\,$ for any angle θ and any integer k. The smallest positive period of a periodic function is called the primitive period of the function. The primitive period of the sine or cosine is a full circle, i.e. 2π radians or 360 degrees. Above, only sine and cosine were defined directly by the unit circle, but other trigonometric functions can be defined by: $\begin{align} \tan\theta & = \frac{\sin\theta}{\cos\theta},\ \cot\theta = \frac{\cos\theta}{\sin\theta} = \frac{1}{\tan\theta} \\ \sec\theta & = \frac{1}{\cos\theta},\ \csc\theta = \frac{1}{\sin\theta} \end{align}$ So : • The primitive period of the secant, or cosecant is also a full circle, i.e. 2π radians or 360 degrees. • The primitive period of the tangent or cotangent is only a half-circle, i.e. π radians or 180 degrees. Trigonometric functions: Sine, Cosine, Tangent, Cosecant (dotted), Secant (dotted), Cotangent (dotted) The image at right includes a graph of the tangent function. • Its θ-intercepts correspond to those of sin(θ) while its undefined values correspond to the θ-intercepts of cos(θ). • The function changes slowly around angles of kπ, but changes rapidly at angles close to (k + 1/2)π. • The graph of the tangent function also has a vertical asymptote at θ = (k + 1/2)π, the θ-intercepts of the cosine function, because the function approaches infinity as θ approaches (k + 1/2)π from the left and minus infinity as it approaches (k + 1/2)π from the right. All of the trigonometric functions of the angle θ can be constructed geometrically in terms of a unit circle centered at O. Alternatively, all of the basic trigonometric functions can be defined in terms of a unit circle centered at O (as shown in the picture to the right), and similar such geometric definitions were used historically. • In particular, for a chord AB of the circle, where θ is half of the subtended angle, sin(θ) is AC (half of the chord), a definition introduced in India[5] (see history). • cos(θ) is the horizontal distance OC, and versin(θ) = 1 − cos(θ) is CD. • tan(θ) is the length of the segment AE of the tangent line through A, hence the word tangent for this function. cot(θ) is another tangent segment, AF. • sec(θ) = OE and csc(θ) = OF are segments of secant lines (intersecting the circle at two points), and can also be viewed as projections of OA along the tangent at A to the horizontal and vertical axes, respectively. • DE is exsec(θ) = sec(θ) − 1 (the portion of the secant outside, or ex, the circle). • From these constructions, it is easy to see that the secant and tangent functions diverge as θ approaches π/2 (90 degrees) and that the cosecant and cotangent diverge as θ approaches zero. (Many similar constructions are possible, and the basic trigonometric identities can also be proven graphically.[6]) ## Series definitions The sine function (blue) is closely approximated by its Taylor polynomial of degree 7 (pink) for a full cycle centered on the origin. Trigonometric functions are analytic functions. Using only geometry and properties of limits, it can be shown that the derivative of sine is cosine and the derivative of cosine is the negative of sine. (Here, and generally in calculus, all angles are measured in radians; see also the significance of radians below.) One can then use the theory of Taylor series to show that the following identities hold for all real numbers x:[7] $\begin{align} \sin x & = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots \\ & = \sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)!}, \\ \cos x & = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots \\ & = \sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{(2n)!}. \end{align}$ These identities are sometimes taken as the definitions of the sine and cosine function. They are often used as the starting point in a rigorous treatment of trigonometric functions and their applications (e.g., in Fourier series), since the theory of infinite series can be developed, independent of any geometric considerations, from the foundations of the real number system. The differentiability and continuity of these functions are then established from the series definitions alone. The value of π can be defined as the smallest positive number for which sin = 0. Other series can be found.[8] For the following trigonometric functions: Un is the nth up/down number, Bn is the nth Bernoulli number, and En (below) is the nth Euler number. Tangent $\begin{align} \tan x & {} = \sum_{n=0}^\infty \frac{U_{2n+1} x^{2n+1}}{(2n+1)!} \\ & {} = \sum_{n=1}^\infty \frac{(-1)^{n-1} 2^{2n} (2^{2n}-1) B_{2n} x^{2n-1}}{(2n)!} \\ & {} = x + \frac{1}{3}x^3 + \frac{2}{15}x^5 + \frac{17}{315}x^7 + \cdots, \qquad \text{for } \|x\| < \frac{\pi}{2}. \end{align}$ When this series for the tangent function is expressed in a form in which the denominators are the corresponding factorials, the numerators, called the "tangent numbers", have a combinatorial interpretation: they enumerate alternating permutations of finite sets of odd cardinality.[9] Cosecant $\begin{align} \csc x & {} = \sum_{n=0}^\infty \frac{(-1)^{n+1} 2 (2^{2n-1}-1) B_{2n} x^{2n-1}}{(2n)!} \\ & {} = x^{-1} + \frac{1}{6}x + \frac{7}{360}x^3 + \frac{31}{15120}x^5 + \cdots, \qquad \text{for } 0 < \|x\| < \pi. \end{align}$ Secant $\begin{align} \sec x & {} = \sum_{n=0}^\infty \frac{U_{2n} x^{2n}}{(2n)!} = \sum_{n=0}^\infty \frac{(-1)^n E_{2n} x^{2n}}{(2n)!} \\ & {} = 1 + \frac{1}{2}x^2 + \frac{5}{24}x^4 + \frac{61}{720}x^6 + \cdots, \qquad \text{for } \|x\| < \frac{\pi}{2}. \end{align}$ When this series for the secant function is expressed in a form in which the denominators are the corresponding factorials, the numerators, called the "secant numbers", have a combinatorial interpretation: they enumerate alternating permutations of finite sets of even cardinality.[10] Cotangent $\begin{align} \cot x & {} = \sum_{n=0}^\infty \frac{(-1)^n 2^{2n} B_{2n} x^{2n-1}}{(2n)!} \\ & {} = x^{-1} - \frac{1}{3}x - \frac{1}{45}x^3 - \frac{2}{945}x^5 - \cdots, \qquad \text{for } 0 < \|x\| < \pi. \end{align}$ From a theorem in complex analysis, there is a unique analytic continuation of this real function to the domain of complex numbers. They have the same Taylor series, and so the trigonometric functions are defined on the complex numbers using the Taylor series above. There is a series representation as partial fraction expansion where just translated reciprocal functions are summed up, such that the poles of the cotangent function and the reciprocal functions match:[11] $\pi \cdot \cot (\pi x) = \lim_{N\to\infty}\sum_{n=-N}^N \frac{1}{x+n}.$ This identity can be proven with the Herglotz trick.[12] By combining the $-n$-th with the $n$-th term, it can be expressed as an absolutely convergent series: $\pi \cdot \cot (\pi x) = \frac{1}{x} + \sum_{n=1}^\infty \frac{2x}{x^2-n^2}.$ ### Relationship to exponential function and complex numbers Euler's formula illustrated with the three dimensional helix, starting with the 2-D orthogonal components of the unit circle, sine and cosine (using θ = t ). It can be shown from the series definitions[13] that the sine and cosine functions are the imaginary and real parts, respectively, of the complex exponential function when its argument is purely imaginary: $e^{i\theta} = \cos\theta + i\sin\theta. \,$ This identity is called Euler's formula. In this way, trigonometric functions become essential in the geometric interpretation of complex analysis. For example, with the above identity, if one considers the unit circle in the complex plane, parametrized by e ix, and as above, we can parametrize this circle in terms of cosines and sines, the relationship between the complex exponential and the trigonometric functions becomes more apparent. Euler's formula can also be used to derive some trigonometric identities, by writing sine and cosine as: $\sin\theta = \frac{e^{i \theta} - e^{-i \theta}}{2i} \;$ $\cos\theta = \frac{e^{i \theta} + e^{-i \theta}}{2} \;$ Furthermore, this allows for the definition of the trigonometric functions for complex arguments z: $\sin z = \sum_{n=0}^\infty \frac{(-1)^{n}}{(2n+1)!}z^{2n+1} = \frac{e^{i z} - e^{-i z}}{2i}\, = \frac{\sinh \left( i z\right) }{i}$ $\cos z = \sum_{n=0}^\infty \frac{(-1)^{n}}{(2n)!}z^{2n} = \frac{e^{i z} + e^{-i z}}{2}\, = \cosh \left(i z\right)$ where i 2 = −1. The sine and cosine defined by this are entire functions. Also, for purely real x, $\cos x = \operatorname{Re}(e^{i x}) \,$ $\sin x = \operatorname{Im}(e^{i x}) \,$ It is also sometimes useful to express the complex sine and cosine functions in terms of the real and imaginary parts of their arguments. $\sin (x + iy) = \sin x \cosh y + i \cos x \sinh y,\,$ $\cos (x + iy) = \cos x \cosh y - i \sin x \sinh y.\,$ This exhibits a deep relationship between the complex sine and cosine functions and their real (sin, cos) and hyperbolic real (sinh, cosh) counterparts. #### Complex graphs In the following graphs, the domain is the complex plane pictured, and the range values are indicated at each point by color. Brightness indicates the size (absolute value) of the range value, with black being zero. Hue varies with argument, or angle, measured from the positive real axis. (more) \| \| \| \| \| \| \| \|-----------------------------------------------------------------------------\|-----------------------------------------------------------------------------\|-----------------------------------------------------------------------------\|-----------------------------------------------------------------------------\|-----------------------------------------------------------------------------\|-----------------------------------------------------------------------------\| \| \| \| \| \| \| \| \| $\sin z\,$ \| $\cos z\,$ \| $\tan z\,$ \| $\cot z\,$ \| $\sec z\,$ \| $\csc z\,$ \| ## Definitions via differential equations Both the sine and cosine functions satisfy the differential equation: $y'' = -y.\,$ That is to say, each is the additive inverse of its own second derivative. Within the 2-dimensional function space V consisting of all solutions of this equation, • the sine function is the unique solution satisfying the initial condition $\scriptstyle \left( y'(0), y(0) \right) = (1, 0)\,$ and • the cosine function is the unique solution satisfying the initial condition $\scriptstyle \left( y'(0), y(0) \right) = (0, 1)\,$. Since the sine and cosine functions are linearly independent, together they form a basis of V. This method of defining the sine and cosine functions is essentially equivalent to using Euler's formula. (See linear differential equation.) It turns out that this differential equation can be used not only to define the sine and cosine functions but also to prove the trigonometric identities for the sine and cosine functions. Further, the observation that sine and cosine satisfies y′′ = −y means that they are eigenfunctions of the second-derivative operator. The tangent function is the unique solution of the nonlinear differential equation $y' = 1 + y^2\,$ satisfying the initial condition y(0) = 0. There is a very interesting visual proof that the tangent function satisfies this differential equation.[14] ### The significance of radians Radians specify an angle by measuring the length around the path of the unit circle and constitute a special argument to the sine and cosine functions. In particular, only sines and cosines that map radians to ratios satisfy the differential equations that classically describe them. If an argument to sine or cosine in radians is scaled by frequency, $f(x) = \sin kx, \,$ then the derivatives will scale by amplitude. $f'(x) = k\cos kx. \,$ Here, k is a constant that represents a mapping between units. If x is in degrees, then $k = \frac{\pi}{\textstyle 180^\circ}.$ This means that the second derivative of a sine in degrees does not satisfy the differential equation $y'' = -y\,$ but rather $y'' = -k^2 y.\,$ The cosine's second derivative behaves similarly. This means that these sines and cosines are different functions, and that the fourth derivative of sine will be sine again only if the argument is in radians. ## Identities Main article: List of trigonometric identities Many identities interrelate the trigonometric functions. Among the most frequently used is the Pythagorean identity, which states that for any angle, the square of the sine plus the square of the cosine is 1. This is easy to see by studying a right triangle of hypotenuse 1 and applying the Pythagorean theorem. In symbolic form, the Pythagorean identity is written $\sin^2 x + \cos^2 x = 1$ where $\textstyle \sin^2 x + \cos^2 x = 1$ is standard notation for $\textstyle (\sin x)^2 + (\cos x)^2 = 1$ Other key relationships are the sum and difference formulas, which give the sine and cosine of the sum and difference of two angles in terms of sines and cosines of the angles themselves. These can be derived geometrically, using arguments that date to Ptolemy. One can also produce them algebraically using Euler's formula. $\sin \left(x+y\right)=\sin x \cos y + \cos x \sin y, \,$ $\cos \left(x+y\right)=\cos x \cos y - \sin x \sin y, \,$ $\sin \left(x-y\right)=\sin x \cos y - \cos x \sin y, \,$ $\cos \left(x-y\right)=\cos x \cos y + \sin x \sin y. \,$ These in turn lead to the following three-angle formulae: $\sin \left(x+y+z\right)=\sin x \cos y \cos z + \sin y \cos z \cos x + \sin z \cos y \cos x - \sin x \sin y \sin z, \,$ $\cos \left(x+y+z\right)=\cos x \cos y \cos z - \cos x \sin y \sin z - \cos y \sin x \sin z - \cos z \sin x \sin y, \,$ When the two angles are equal, the sum formulas reduce to simpler equations known as the double-angle formulae. $\sin 2x = 2 \sin x \cos x, \,$ $\cos 2x = \cos^2 x - \sin^2 x = 2 \cos^2 x - 1. \,$ When three angles are equal, the three-angle formulae simplify to $\sin 3x = 3 \sin x - 4 \sin^3 x . \,$ $\cos 3x = 4 \cos^3 x - 3 \cos x . \,$ These identities can also be used to derive the product-to-sum identities that were used in antiquity to transform the product of two numbers into a sum of numbers and greatly speed operations, much like the logarithm function. ### Calculus For integrals and derivatives of trigonometric functions, see the relevant sections of Differentiation of trigonometric functions, Lists of integrals and List of integrals of trigonometric functions. Below is the list of the derivatives and integrals of the six basic trigonometric functions. The number C is a constant of integration. \| \| \| \| \|-----------------------------------------------------------------------------\|-----------------------------------------------------------------------------\|-----------------------------------------------------------------------------\| \| $\ \ \ \ f(x)$ \| $\ \ \ \ f'(x)$ \| $\int f(x)\,dx$ \| \| $\,\ \sin x$ \| $\,\ \cos x$ \| $\,\ -\cos x + C$ \| \| $\,\ \cos x$ \| $\,\ -\sin x$ \| $\,\ \sin x + C$ \| \| $\,\ \tan x$ \| $\,\ \sec^2 x = 1+\tan^2 x$ \| $-\ln \left \|\cos x\right \| + C$ \| \| $\,\ \cot x$ \| $\,\ -\csc^2 x = -(1+\cot^2 x)$ \| $\ln \left \|\sin x\right \| + C$ \| \| $\,\ \sec x$ \| $\,\ \sec x\tan x$ \| $\ln \left \|\sec x + \tan x\right \| + C$ \| \| $\,\ \csc x$ \| $\,\ -\csc x \cot x$ \| $\ -\ln \left \|\csc x + \cot x\right \| + C$ \| ### Definitions using functional equations In mathematical analysis, one can define the trigonometric functions using functional equations based on properties like the difference formula. Taking as given these formulas, one can prove that only two real functions satisfy those conditions. Symbolically, we say that there exists exactly one pair of real functions — $\scriptstyle \sin\,$ and $\scriptstyle \cos\,$ — such that for all real numbers $\scriptstyle x\,$ and $\scriptstyle y\,$, the following equation hold:[15] $\cos(x- y) = \cos x\cos y + \sin x\sin y\,$ with the added condition that $0 < x\cos x < \sin x < x\hbox{ for }0 < x < 1.\,$ Other derivations, starting from other functional equations, are also possible, and such derivations can be extended to the complex numbers. As an example, this derivation can be used to define trigonometry in Galois fields. ## Computation The computation of trigonometric functions is a complicated subject, which can today be avoided by most people because of the widespread availability of computers and scientific calculators that provide built-in trigonometric functions for any angle. This section, however, describes details of their computation in three important contexts: the historical use of trigonometric tables, the modern techniques used by computers, and a few "important" angles where simple exact values are easily found. The first step in computing any trigonometric function is range reduction—reducing the given angle to a "reduced angle" inside a small range of angles, say 0 to π/2, using the periodicity and symmetries of the trigonometric functions. Main article: Generating trigonometric tables Prior to computers, people typically evaluated trigonometric functions by interpolating from a detailed table of their values, calculated to many significant figures. Such tables have been available for as long as trigonometric functions have been described (see History below), and were typically generated by repeated application of the half-angle and angle-addition identities starting from a known value (such as sin(π/2) = 1). Modern computers use a variety of techniques.[16] One common method, especially on higher-end processors with floating point units, is to combine a polynomial or rational approximation (such as Chebyshev approximation, best uniform approximation, and Padé approximation, and typically for higher or variable precisions, Taylor and Laurent series) with range reduction and a table lookup—they first look up the closest angle in a small table, and then use the polynomial to compute the correction.[17] Devices that lack hardware multipliers often use an algorithm called CORDIC (as well as related techniques), which uses only addition, subtraction, bitshift, and table lookup. These methods are commonly implemented in hardware floating-point units for performance reasons. For very high precision calculations, when series expansion convergence becomes too slow, trigonometric functions can be approximated by the arithmetic-geometric mean, which itself approximates the trigonometric function by the (complex) elliptic integral.[18] Main article: Exact trigonometric constants Finally, for some simple angles, the values can be easily computed by hand using the Pythagorean theorem, as in the following examples. For example, the sine, cosine and tangent of any integer multiple of $\pi / 60$ radians (3°) can be found exactly by hand. Consider a right triangle where the two other angles are equal, and therefore are both $\pi / 4$ radians (45°). Then the length of side b and the length of side a are equal; we can choose $a = b = 1$. The values of sine, cosine and tangent of an angle of $\pi / 4$ radians (45°) can then be found using the Pythagorean theorem: $c = \sqrt { a^2+b^2 } = \sqrt2\,.$ Therefore: $\sin \frac{\pi}{4} = \sin 45^\circ = \cos \frac{\pi}{4}= \cos 45^\circ = {1 \over \sqrt2},\,$ $\tan \frac{\pi}{4} = \tan 45^\circ = { {\sin \frac{\pi}{4} } \over {\cos \frac{\pi}{4} } }= {1 \over \sqrt2} \cdot {\sqrt2 \over 1} = {\sqrt2 \over \sqrt2} = 1. \,$ To determine the trigonometric functions for angles of π/3 radians (60 degrees) and π/6 radians (30 degrees), we start with an equilateral triangle of side length 1. All its angles are π/3 radians (60 degrees). By dividing it into two, we obtain a right triangle with π/6 radians (30 degrees) and π/3 radians (60 degrees) angles. For this triangle, the shortest side = 1/2, the next largest side =(√3)/2 and the hypotenuse = 1. This yields: $\sin \frac{\pi}{6} = \sin 30^\circ = \cos \frac{\pi}{3} = \cos 60^\circ = {1 \over 2}\,,$ $\cos \frac{\pi}{6} = \cos 30^\circ = \sin \frac{\pi}{3} = \sin 60^\circ = {\sqrt3 \over 2}\,,$ $\tan \frac{\pi}{6} = \tan 30^\circ = \cot \frac{\pi}{3} = \cot 60^\circ = {1 \over \sqrt3}\,.$ ### Special values in trigonometric functions There are some commonly used special values in trigonometric functions, as shown in the following table. Function $0 \ (0^\circ)$ $\frac{\pi}{12} \ (15^\circ)$ $\frac{\pi}{8} \ (22.5^\circ)$ $\frac{\pi}{6} \ (30^\circ)$ $\frac{\pi}{4} \ (45^\circ)$ $\frac{\pi}{3} \ (60^\circ)$ $\frac{5\pi}{12} \ (75^\circ)$ $\frac{\pi}{2} \ (90^\circ)$ sin $0$ $\frac{ \sqrt{6} - \sqrt{2} } {4}$ $\frac{ \sqrt{2 - \sqrt{2}} } {2}$ $\frac{1}{2}$ $\frac{\sqrt{2}}{2}$ $\frac{\sqrt{3}}{2}$ $\frac{ \sqrt{6} + \sqrt{2} } {4}$ $1$ cos $1$ $\frac{\sqrt{6}+\sqrt{2}}{4}$ $\frac{ \sqrt{2 + \sqrt{2}} } {2}$ $\frac{\sqrt{3}}{2}$ $\frac{\sqrt{2}}{2}$ $\frac{1}{2}$ $\frac{ \sqrt{6} - \sqrt{2}} {4}$ $0$ tan $0$ $2-\sqrt{3}$ $\sqrt{2} - 1$ $\frac{\sqrt{3}}{3}$ $1$ $\sqrt{3}$ $2+\sqrt{3}$ $\infty$[19] cot $\infty$[19] $2+\sqrt{3}$ $\sqrt{2} + 1$ $\sqrt{3}$ $1$ $\frac{\sqrt{3}}{3}$ $2-\sqrt{3}$ $0$ sec $1$ $\sqrt{6} - \sqrt{2}$ $\sqrt{2} \sqrt{ 2 - \sqrt{2} }$ $\frac{2\sqrt{3}}{3}$ $\sqrt{2}$ $2$ $\sqrt{6}+\sqrt{2}$ $\infty$[19] csc $\infty$[19] $\sqrt{6}+\sqrt{2}$ $\sqrt{2} \sqrt{ 2 + \sqrt{2} }$ $2$ $\sqrt{2}$ $\frac{2\sqrt{3}}{3}$ $\sqrt{6} - \sqrt{2}$ $1$ The symbol $\infty$ here represents the point at infinity on the real projective line, the limit on the extended real line is $+\infty$ on one side and $-\infty$ on the other. ## Inverse functions Main article: Inverse trigonometric functions The trigonometric functions are periodic, and hence not injective, so strictly they do not have an inverse function. Therefore to define an inverse function we must restrict their domains so that the trigonometric function is bijective. In the following, the functions on the left are defined by the equation on the right; these are not proved identities. The principal inverses are usually defined as: Function Definition Value Field $\arcsin x = y \,$ $\sin y = x \,$ $-\frac{\pi}{2} \le y \le \frac{\pi}{2} \,$ $\arccos x = y \,$ $\cos y = x \,$ $0 \le y \le \pi \,$ $\arctan x = y \,$ $\tan y = x \,$ $-\frac{\pi}{2} < y < \frac{\pi}{2} \,$ $\arccsc x = y \,$ $\csc y = x \,$ $-\frac{\pi}{2} \le y \le \frac{\pi}{2}, y \ne 0 \,$ $\arcsec x = y \,$ $\sec y = x \,$ $0 \le y \le \pi, y \ne \frac{\pi}{2} \,$ $\arccot x = y \,$ $\cot y = x \,$ $0 < y < \pi \,$ The notations sin−1 and cos−1 are often used for arcsin and arccos, etc. When this notation is used, the inverse functions could be confused with the multiplicative inverses of the functions. The notation using the "arc-" prefix avoids such confusion, though "arcsec" can be confused with "arcsecond". Just like the sine and cosine, the inverse trigonometric functions can also be defined in terms of infinite series. For example, $\arcsin z = z + \left( \frac {1} {2} \right) \frac {z^3} {3} + \left( \frac {1 \cdot 3} {2 \cdot 4} \right) \frac {z^5} {5} + \left( \frac{1 \cdot 3 \cdot 5} {2 \cdot 4 \cdot 6 } \right) \frac{z^7} {7} + \cdots\,.$ These functions may also be defined by proving that they are antiderivatives of other functions. The arcsine, for example, can be written as the following integral: $\arcsin z = \int_0^z (1 - x^2)^{-1/2}\,dx, \quad \|z\| < 1.$ Analogous formulas for the other functions can be found at Inverse trigonometric functions. Using the complex logarithm, one can generalize all these functions to complex arguments: $\arcsin z = -i \log \left( i z + \sqrt{1 - z^2} \right), \,$ $\arccos z = -i \log \left( z + \sqrt{z^2 - 1}\right), \,$ $\arctan z = \frac12i \log\left(\frac{1-iz}{1+iz}\right).$ ### Connection to the inner product In an inner product space, the angle between two non-zero vectors is defined to be $\operatorname{angle}(x,y) = \arccos \frac{\langle x, y \rangle}{\\|x\\| \cdot \\|y\\|}.$ ## Properties and applications Main article: Uses of trigonometry The trigonometric functions, as the name suggests, are of crucial importance in trigonometry, mainly because of the following two results. ### Law of sines The law of sines states that for an arbitrary triangle with sides a, b, and c and angles opposite those sides A, B and C: $\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c},$ or, equivalently, $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R,$ where R is the triangle's circumradius. A Lissajous curve, a figure formed with a trigonometry-based function. It can be proven by dividing the triangle into two right ones and using the above definition of sine. The law of sines is useful for computing the lengths of the unknown sides in a triangle if two angles and one side are known. This is a common situation occurring in triangulation, a technique to determine unknown distances by measuring two angles and an accessible enclosed distance. ### Law of cosines The law of cosines (also known as the cosine formula or cosine rule) is an extension of the Pythagorean theorem: $c^2=a^2+b^2-2ab\cos C, \,$ or equivalently, $\cos C=\frac{a^2+b^2-c^2}{2ab}.$ In this formula the angle at C is opposite to the side c. This theorem can be proven by dividing the triangle into two right ones and using the Pythagorean theorem. The law of cosines can be used to determine a side of a triangle if two sides and the angle between them are known. It can also be used to find the cosines of an angle (and consequently the angles themselves) if the lengths of all the sides are known. ### Law of tangents Main article: Law of tangents The following all form the law of tangents[20] $\begin{align} \frac{\tan \frac{A-B}{2 }}{\tan \frac{A+B}{2 } } & = \frac{a-b}{a+b} \\[6pt] \frac{\tan \frac{A-C}{2 }}{\tan \frac{A+C}{2 } } & = \frac{a-c}{a+c} \\[6pt] \frac{\tan \frac{B-C}{2 }}{\tan \frac{B+C}{2 } } & = \frac{b-c}{b+c} \end{align}$ The explanation of the formulae in words would be cumbersome, but the patterns of sums and differences; for the lengths and corresponding opposite angles, are apparent in the theorem. ### Law of cotangents Main article: Law of cotangents If $\zeta = \sqrt{\frac{1}{s} (s-a)(s-b)(s-c)}$ (the radius of the inscribed circle for the triangle) and $s = \frac{a+b+c}{2 }$ (the semi-perimeter for the triangle), then the following all form the law of cotangents[21] $\cot{ \frac{A}{2 }} = \frac{s-a}{\zeta }$ $\cot{ \frac{B}{2 }} = \frac{s-b}{\zeta }$ $\cot{ \frac{C}{2 }} = \frac{s-c}{\zeta }$ It follows that $\frac{\cot(A/2)}{s-a} = \frac{\cot(B/2)}{s-b} = \frac{\cot(C/2)}{s-c}.$ In words the theorem is: the cotangent of a half-angle equals the ratio of the semi-perimeter minus the opposite side to the said angle, to the inradius for the triangle. ### Other useful properties #### Sine and cosine of sums of angles See also: Angle addition formula ### Periodic functions An animation of the additive synthesis of a square wave with an increasing number of harmonics Sinusoidal basis functions (bottom) can form a sawtooth wave (top) when added. All the basis functions have nodes at the nodes of the sawtooth, and all but the fundamental (k = 1) have additional nodes. The oscillation seen about the sawtooth when k is large is called the Gibbs phenomenon The trigonometric functions are also important in physics. The sine and the cosine functions, for example, are used to describe simple harmonic motion, which models many natural phenomena, such as the movement of a mass attached to a spring and, for small angles, the pendular motion of a mass hanging by a string. The sine and cosine functions are one-dimensional projections of uniform circular motion. Trigonometric functions also prove to be useful in the study of general periodic functions. The characteristic wave patterns of periodic functions are useful for modeling recurring phenomena such as sound or light waves.[22] Under rather general conditions, a periodic function ƒ(x) can be expressed as a sum of sine waves or cosine waves in a Fourier series.[23] Denoting the sine or cosine basis functions by φk, the expansion of the periodic function ƒ(t) takes the form: $f(t) = \sum _{k=1}^\infty c_k \varphi_k(t).$ For example, the square wave can be written as the Fourier series $f_\text{square}(t) = \frac{4}{\pi} \sum_{k=1}^\infty {\sin \left ( (2k-1)t \right ) \over 2k-1}.$ In the animation of a square wave at top right it can be seen that just a few terms already produce a fairly good approximation. The superposition of several terms in the expansion of a sawtooth wave are shown underneath. ## History Main article: History of trigonometric functions While the early study of trigonometry can be traced to antiquity, the trigonometric functions as they are in use today were developed in the medieval period. The chord function was discovered by Hipparchus of Nicaea (180–125 BC) and Ptolemy of Roman Egypt (90–165 AD). The functions sine and cosine can be traced to the jyā and koti-jyā functions used in Gupta period Indian astronomy (Aryabhatiya, Surya Siddhanta), via translation from Sanskrit to Arabic and then from Arabic to Latin.[24] All six trigonometric functions in current use were known in Islamic mathematics by the 9th century, as was the law of sines, used in solving triangles.[25] al-Khwārizmī produced tables of sines, cosines and tangents. They were studied by authors including Omar Khayyám, Bhāskara II, Nasir al-Din al-Tusi, Jamshīd al-Kāshī (14th century), Ulugh Beg (14th century), Regiomontanus (1464), Rheticus, and Rheticus' student Valentinus Otho[citation needed] Madhava of Sangamagrama (c. 1400) made early strides in the analysis of trigonometric functions in terms of infinite series.[26] The first published use of the abbreviations 'sin', 'cos', and 'tan' is by the 16th century French mathematician Albert Girard. In a paper published in 1682, Leibniz proved that sin x is not an algebraic function of x.[27] Leonhard Euler's Introductio in analysin infinitorum (1748) was mostly responsible for establishing the analytic treatment of trigonometric functions in Europe, also defining them as infinite series and presenting "Euler's formula", as well as the near-modern abbreviations sin., cos., tang., cot., sec., and cosec.[5] A few functions were common historically, but are now seldom used, such as the chord (crd(θ) = 2 sin(θ/2)), the versine (versin(θ) = 1 − cos(θ) = 2 sin2(θ/2)) (which appeared in the earliest tables[5]), the haversine (haversin(θ) = versin(θ) / 2 = sin2(θ/2)), the exsecant (exsec(θ) = sec(θ) − 1) and the excosecant (excsc(θ) = exsec(π/2 − θ) = csc(θ) − 1). Many more relations between these functions are listed in the article about trigonometric identities. Etymologically, the word sine derives from the Sanskrit word for half the chord, jya-ardha, abbreviated to jiva. This was transliterated in Arabic as jiba, written jb, vowels not being written in Arabic. Next, this transliteration was mis-translated in the 12th century into Latin as sinus, under the mistaken impression that jb stood for the word jaib, which means "bosom" or "bay" or "fold" in Arabic, as does sinus in Latin.[28] Finally, English usage converted the Latin word sinus to sine.[29] The word tangent comes from Latin tangens meaning "touching", since the line touches the circle of unit radius, whereas secant stems from Latin secans — "cutting" — since the line cuts the circle. ## See also • List of periodic functions • Generating trigonometric tables • Aryabhata's sine table • Madhava's sine table • Madhava series • Bhaskara I's sine approximation formula • Hyperbolic function • Unit vector (explains direction cosines) • Table of Newtonian series • List of trigonometric identities • Proofs of trigonometric identities • Euler's formula • Polar sine — a generalization to vertex angles • All Students Take Calculus — a mnemonic for recalling the signs of trigonometric functions in a particular quadrant of a Cartesian plane • Gauss's continued fraction — a continued fraction definition for the tangent function • Generalized trigonometry ## Notes 1. Oxford English Dictionary, sine, n.2 2. Oxford English Dictionary, cosine, n. 3. Oxford English Dictionary, tangent, adj. and n. 4. Oxford English Dictionary, secant, adj. and n. 5. ^ a b c See Boyer (1991). 6. See Maor (1998) 7. See Ahlfors, pages 43–44. 8. Abramowitz; Weisstein. 9. Stanley, Enumerative Combinatorics, Vol I., page 149 10. Stanley, Enumerative Combinatorics, Vol I 11. Aigner, Martin; Ziegler, Günter M. (2000). Proofs from THE BOOK (Second ed.). Springer-Verlag. p. 149. ISBN 978-3-642-00855-9. 12. Remmert, Reinhold (1991). Theory of complex functions. Springer. p. 327. ISBN 0-387-97195-5. , Extract of page 327 13. Needham, p. Visual Complex Analysis. ISBN 0-19-853446-9. 14. Kannappan, Palaniappan (2009). Functional Equations and Inequalities with Applications. Springer. ISBN 978-0387894911. 15. Kantabutra. 16. 17. ^ a b c d Abramowitz, Milton and Irene A. Stegun, p.74 18. The Universal Encyclopaedia of Mathematics, Pan Reference Books, 1976, page 529. English version George Allen and Unwin, 1964. Translated from the German version Meyers Rechenduden, 1960. 19. The Universal Encyclopaedia of Mathematics, Pan Reference Books, 1976, page 530. English version George Allen and Unwin, 1964. Translated from the German version Meyers Rechenduden, 1960. 20. Stanley J Farlow (1993). Partial differential equations for scientists and engineers (Reprint of Wiley 1982 ed.). Courier Dover Publications. p. 82. ISBN 0-486-67620-X. 21. See for example, Gerald B Folland (2009). "Convergence and completeness". Fourier Analysis and its Applications (Reprint of Wadsworth & Brooks/Cole 1992 ed.). American Mathematical Society. pp. 77 ff. ISBN 0-8218-4790-2. 22. Owen Gingerich (1986). Islamic Astronomy 254. Scientific American. p. 74. Retrieved 2010-37-13. 23. J J O'Connor and E F Robertson. "Madhava of Sangamagrama". School of Mathematics and Statistics University of St Andrews, Scotland. Retrieved 2007-09-08. 24. Nicolás Bourbaki (1994). Elements of the History of Mathematics. Springer. 25. See Maor (1998), chapter 3, regarding the etymology. 26. ## References • Abramowitz, Milton and Irene A. Stegun, Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, Dover, New York. (1964). ISBN 0-486-61272-4. • Lars Ahlfors, Complex Analysis: an introduction to the theory of analytic functions of one complex variable, second edition, McGraw-Hill Book Company, New York, 1966. • Boyer, Carl B., A History of Mathematics, John Wiley & Sons, Inc., 2nd edition. (1991). ISBN 0-471-54397-7. • Gal, Shmuel and Bachelis, Boris. An accurate elementary mathematical library for the IEEE floating point standard, ACM Transaction on Mathematical Software (1991). • Joseph, George G., The Crest of the Peacock: Non-European Roots of Mathematics, 2nd ed. Penguin Books, London. (2000). ISBN 0-691-00659-8. • Kantabutra, Vitit, "On hardware for computing exponential and trigonometric functions," IEEE Trans. Computers 45 (3), 328–339 (1996). • Maor, Eli, Trigonometric Delights, Princeton Univ. Press. (1998). Reprint edition (February 25, 2002): ISBN 0-691-09541-8. • Needham, Tristan, "Preface"" to Visual Complex Analysis. Oxford University Press, (1999). ISBN 0-19-853446-9. • O'Connor, J.J., and E.F. Robertson, "Trigonometric functions", MacTutor History of Mathematics archive. (1996). • O'Connor, J.J., and E.F. Robertson, "Madhava of Sangamagramma", MacTutor History of Mathematics archive. (2000). • Pearce, Ian G., "Madhava of Sangamagramma", MacTutor History of Mathematics archive. (2002). • Weisstein, Eric W., "Tangent" from MathWorld, accessed 21 January 2006.	{"extraction_info": {"found_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": 207, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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": 0.8632164001464844, "perplexity_flag": "middle"}
http://en.wikibooks.org/wiki/Communication_Systems/Wired_Transmission	# Communication Systems/Wired Transmission This page will discuss the topic of signal propagation through physical mediums, such as wires. ## Transmission Line Equation Wikipedia has related information at Transmission_line Many kinds of communication systems require signals at some point to be conveyed over copper wires. The following analysis requires two assumptions: • A transmission line can be decomposed into small, distributed passive electrical elements • These elements are independent of frequency (i.e. although reactance is a function of frequency, resistance, capacitance and inductance are not) These two assumptions limit the following analysis to frequencies up to the low MHz region. The second assumption is particularly difficult to defend since it is well known that the resistance of a wire increases with frequency because the conduction cross-section decreases. This phenomenon is known as the skin effect and is not easy to evaluate. The purpose behind the following mathematical manipulation is to obtain an expression that defines the voltage (or current) at any time (t) along any portion (x) of the transmission line. Later, this analysis will be extended to include the frequency domain. Recall the characteristic equations for inductors and capacitors: $v = L\frac{{\partial i}}{{\partial t}}$ and $i = C\frac{{\partial v}}{{\partial t}}$ ### Kirchoff's Voltage Law Kirchoff's voltage law (KVL) simply states that the sum of all voltage potentials around a closed loop equal zero. Or in other words, if you walked up a hill and back down, the net altitude change would be zero. Applying KVL in the above circuit, we obtain: $v\left( {x,t} \right) = R\Delta xi\left( {x,t} \right) + L\Delta x\frac{{\partial i}}{{\partial t}}\left( {x,t} \right) + v\left( {x + \Delta x,t} \right)$ Rearranging: $v\left( {x,t} \right) - v\left( {x + \Delta x,t} \right) = R\Delta xi\left( {x,t} \right) + L\Delta x\frac{{\partial i}}{{\partial t}}\left( {x,t} \right)$ But the LHS (left hand side) of the above equation, represents the voltage drop across the cable element $\Delta v$, therefor: $\Delta v = R\Delta xi\left( {x,t} \right) + L\Delta x\frac{{\partial i}}{{\partial t}}\left( {x,t} \right)$ Dividing through by $\Delta x$, we obtain: $\frac{{\Delta v}}{{\Delta x}} = Ri\left( {x,t} \right) + L\frac{{\partial i}}{{\partial t}}\left( {x,t} \right)$ The LHS is easily recognized as a derivative. Simplifying the notation: $\frac{{\partial v}}{{\partial x}} = Ri + L\frac{{\partial i}}{{\partial t}}$ This expression has both current and voltage in it. It would be convenient to write the equation in terms of current or voltage as a function of distance or time. ### Simplifying the Equation (trust me) The first step in separating voltage and current is to take the derivative with respect to the position x (Equation 1): $\frac{{\partial ^2 v}}{{\partial x^2 }} = R\frac{{\partial i}}{{\partial x}} + L\frac{{\partial ^2 i}}{{\partial x\partial t}}$ The next step is to eliminate the current terms, leaving an expression with voltage only. The change in current along the line is equal to the current being shunted across the line through the capacitance C and conductance G. By applying KCL in the circuit, we obtain the necessary information (Equation 2): $\frac{{\partial i}}{{\partial x}} = Gv + C\frac{{\partial v}}{{\partial t}}$ Taking the derivative with respect to time, we obtain (Equation 3): $\frac{{\partial ^2 i}}{{\partial x\partial t}} = G\frac{{\partial v}}{{\partial t}} + C\frac{{\partial ^2 v}}{{\partial t^2 }}$ Substituting (Equation 2) and (Equation 3) into (Equation 1), we obtain the desired simplification: $\frac{{\partial ^2 v}}{{\partial x^2 }} = R\left[ {Gv + C\frac{{\partial v}}{{\partial t}}} \right] + L\left[ {G\frac{{\partial v}}{{\partial t}} + C\frac{{\partial ^2 v}}{{\partial t^2 }}} \right]$ Collecting the terms, we obtain: The Transmission Line Equation for Voltage $\frac{{\partial ^2 v}}{{\partial x^2 }} = RGv + \left( {RC + LG} \right)\frac{{\partial v}}{{\partial t}} + LC\frac{{\partial ^2 v}}{{\partial t^2 }}$ This equation is known as the transmission line equation. Note that it has voltage at any particular location x as a function of time t. Similarly for current, we obtain: The Transmission Line Equation for Current $\frac{{\partial ^2 i}}{{\partial x^2 }} = RGi + \left( {RC + LG} \right)\frac{{\partial i}}{{\partial t}} + LC\frac{{\partial ^2 i}}{{\partial t^2 }}$ But we're not quite done yet. ### Solving the Transmission Line Equation Historically, a mathematician would solve the transmission line equation for v by assuming a solution for v, substituting it into the equation, and observing whether the result made any sense. An engineer would follow a similar procedure by making an “educated guess” based on some laboratory experiments, as to what the solution might be. Today there are more sophisticated techniques used to find solutions. In this respect, the engineer may lag behind the mathematician by several centuries in finding applications for mathematical tools. To solve the transmission line equation, we shall guess that the solution for the voltage function is of the form: $v\left( t \right) = e^{j\omega t} e^{ - \gamma x}$ The first term represents a unity vector rotating at an angular velocity of $\omega$ radians per second, in other words, a sine wave of some frequency. The second term denotes the sinusoid being modified by the transmission line, namely its amplitude decaying exponentially with distance. If we let $\gamma$ be a complex quantity, we can also include any phase changes which occur as the signal travels down the line. The sine wave is used as a signal source because it is easy to generate, and manipulate mathematically. Euler’s Identity shows the relationship between exponential notation and trigonometric functions: Euler's Identity $e^{j\omega t} = \cos \left( {\omega t} \right) + j\sin \left( {\omega t} \right)$ Going back to our educated guess, we will let $\gamma = \alpha + j\beta$, therefore: $v\left( t \right) = e^{j\omega t} e^{ - \left( {\alpha + j\beta } \right)x} = e^{ - \alpha x} e^{\left( {\omega t - \beta x} \right)j}$ The term $e^{ - \alpha x}$ represents the exponential amplitude decay as this signal travels down the line. $\alpha$ is known as the attenuation coefficient and is expressed in Nepers per meter. The term $e^{\left( {\omega t - \beta x} \right)j}$ represents the frequency of the signal at any point along the line. $\beta$ component is known as the phase shift coefficient, and is expressed in radians per meter. Substituting our educated guess $v\left( t \right) = e^{j\omega t} e^{ - \left( {\alpha + j\beta } \right)x}$ into the transmission line equation for voltage, we obtain: $\frac{{\partial ^2 }}{{\partial x^2 }}\left[ {e^{j\omega t} e^{ - \left( {\alpha + j\beta } \right)x} } \right] = RG\left[ {e^{j\omega t} e^{ - \left( {\alpha + j\beta } \right)x} } \right] + \left( {RC + LG} \right)\frac{\partial }{{\partial t}}\left[ {e^{j\omega t} e^{ - \left( {\alpha + j\beta } \right)x} } \right] + LC\frac{{\partial ^2 }}{{\partial t^2 }}\left[ {e^{j\omega t} e^{ - \left( {\alpha + j\beta } \right)x} } \right]$ This looks pretty intimidating, but if you can do basic differentials and algebra, you can do this! #### Simplifying the Equation (trust me) The idea now is to work through the math to see if we come up with a reasonable solution. If we arrive at a contradiction or an unreasonable result, it means that our educated guess was wrong and we have to do more experimenting and come up with a better guess as to how voltage and current travel down a transmission line. Let's look at this equation one term at a time: LHS = RHS Term 1 + RHS Term 2 + RHS Term 3 Starting with the left hand side (LHS) we get the following simplification: $\frac{{\partial ^2 }}{{\partial x^2 }}\left[ {e^{j\omega t} e^{ - \left( {\alpha + j\beta } \right)x} } \right] = \frac{\partial }{{\partial x}}\left[ { - \left( {\alpha + j\beta } \right)e^{j\omega t} e^{ - \left( {\alpha + j\beta } \right)x} } \right] = \left( {\alpha + j\beta } \right)^2 e^{j\omega t} e^{ - \left( {\alpha + j\beta } \right)x}$ Believe it or not, the RHS Term 1 does not need simplifying. Simplifying the RHS Term 2, we obtain: $\left( {RC + LG} \right)\frac{\partial }{{\partial t}}\left[ {e^{j\omega t} e^{ - \left( {\alpha + j\beta } \right)x} } \right] = \left( {RC + LG} \right)j\omega \left( {e^{j\omega t} e^{ - \left( {\alpha + j\beta } \right)x} } \right)$ Simplifying the RHS Term 3, we obtain: $LC\frac{{\partial ^2 }}{{\partial t^2 }}\left[ {e^{j\omega t} e^{ - \left( {\alpha + j\beta } \right)x} } \right] = LC\frac{\partial }{{\partial t}}\left[ {j\omega e^{j\omega t} e^{ - \left( {\alpha + j\beta } \right)x} } \right] = - LC\omega ^2 e^{j\omega t} e^{ - \left( {\alpha + j\beta } \right)x}$ Let's put it all back together: $\left( {\alpha + j\beta } \right)^2 e^{j\omega t} e^{ - \left( {\alpha + j\beta } \right)x} = RG\left( {e^{j\omega t} e^{ - \left( {\alpha + j\beta } \right)x} } \right) + \left( {RC + LG} \right)j\omega \left( {e^{j\omega t} e^{ - \left( {\alpha + j\beta } \right)x} } \right) - LC\omega ^2 \left( {e^{j\omega t} e^{ - \left( {\alpha + j\beta } \right)x} } \right)$ Note that each of the four terms contain the expression $e^{j\omega t} e^{ - \left( {\alpha + j\beta } \right)x}$. Therefore we end up with: $\left( {\alpha + j\beta } \right)^2 = RG + \left( {RC + LG} \right)j\omega - LC\omega ^2$ And this can be further simplified to: Attenuation and Phase Shift Coefficients $\alpha + j\beta = \gamma = \sqrt {\left( {R + j\omega L} \right)\left( {G + j\omega C} \right)}$ This result is not self contradictory or unreasonable. Therefore we conclude that our educated guess was right and we have successfully found an expression for attenuation and phase shift on a transmission line as a function of its distributed electrical components and frequency. ### Lossless Transmission Line Signal loss occurs by two basic mechanisms: signal power can be dissipated in a resistor [or conductance] or signal currents may be shunted to an AC ground via a reactance. In transmission line theory, a lossless transmission line does not dissipate power. Signals, will still gradually diminish however, as shunt reactances return the current to the source via the ground path. For the power loss to equal zero, R = G = 0. This condition occurs when the transmission line is very short. An oscilloscope probe is an example of a very short transmission line. The transmission line equation reduces to the voltage equation: $\frac{{\partial ^2 v}}{{\partial x^2 }} = LC\frac{{\partial ^2 v}}{{\partial t^2 }}$ and the current equation: $\frac{{\partial ^2 i}}{{\partial x^2 }} = LC\frac{{\partial ^2 i}}{{\partial t^2 }}$ To determine how sinusoidal signals are affected by this type of line, we simply substitute a sinusoidal voltage or current into the above expressions and solve as before, or we could take a much simpler approach. We could start with the solution for the general case: $\alpha + j\beta = \gamma = \sqrt {\left( {R + j\omega L} \right)\left( {G + j\omega C} \right)}$ Let R = G = 0, and simplify: $\alpha + j\beta = \sqrt {\left( {j\omega L} \right)\left( {j\omega C} \right)} = \omega \left( {LC} \right)^2 j$ Equating the real and imaginary parts: $\alpha = 0$ $\beta = \omega \sqrt {LC}$ This expression tells us that a signal travelling down a lossless transmission line, experiences a phase shift directly proportional to its frequency. #### Phase Velocity A new parameter, known as phase velocity, can be extracted from these variables: $V_p = \frac{1}{{\sqrt {LC} }} = \frac{\omega }{\beta }$ meters per second Phase velocity is the speed at which a fixed point on a wavefront, appears to move. In the case of wire transmission lines, it is also the velocity of propagation., typically: 0.24c < Vp < 0.9c . The distance between two identical points on a wavefront is its wavelength ($\lambda$) and since one cycle is defined as 2$\pi$ radians: $\lambda = \frac{{2\pi }}{\beta }$ and $\omega = 2\pi f$ therefore: $V_p = \lambda f$ In free space, the phase velocity is 3 x 108 meters/sec, the speed of light. In a cable, the phase velocity is somewhat lower because the signal is carried by electrons. In a waveguide transmission line, the phase velocity exceeds the speed of light. ### Distortionless Transmission Line A distortionless line does not distort the signal phase, but does introduce a signal loss. Since common transmission lines are not super conductors, the signal will decrease in amplitude but retain the same shape as the input. This characteristic is essential for long cable systems. Phase distortion does not occur if the phase velocity Vp is constant at all frequencies. By definition, a phase shift of 2$\pi$ radians occurs over one wavelength $\lambda$. Since $V_p = \lambda f\quad \quad \lambda = \frac{{2\pi }}{\beta }\quad \quad f = \frac{\omega }{{2\pi }}$ Then: $V_p = \frac{{2\pi }}{\beta } \times \frac{\omega }{{2\pi }} = \frac{\omega }{\beta }$ This tells us that in order for phase velocity Vp to be constant, the phase shift coefficient $\beta$, must vary directly with frequency $\omega$. Recall $\gamma = \sqrt {\left( {R + j\omega L} \right)\left( {G + j\omega C} \right)} = \alpha + j\beta$ The problem now is to find $\beta$. This can be done as follows: $\gamma = \sqrt {\left( {\frac{{R + j\omega L}}{{j\omega L}}} \right)\left( {j\omega L} \right)\left( {\frac{{G + j\omega C}}{{j\omega C}}} \right)\left( {j\omega C} \right)} = j\omega \sqrt {LC} \sqrt {1 + \frac{R}{{j\omega L}}} \sqrt {1 + \frac{G}{{j\omega C}}}$ It may seem that we have lost $\beta$, but do not give up. The 2nd and 3rd roots can be expanded by means of the Binomial Expansion. Recall: $\left( {1 + x} \right)^n = 1 + nx + \frac{{n\left( {n - 1} \right)}}{{2!}}x^2 + \frac{{n\left( {n - 1} \right)\left( {n - 2} \right)}}{{3!}}x^3 + \cdots$ In this instance n = 1/2. Since the contribution of successive terms diminishes rapidly, $\gamma$ is expanded to only 3 terms: $\gamma \approx j\omega \sqrt {LC} \left( {1 + \frac{1}{2}\frac{R}{{j\omega L}} - \frac{1}{8}\left( {\frac{R}{{j\omega L}}} \right)^2 } \right)\left( {1 + \frac{1}{2}\frac{G}{{j\omega C}} - \frac{1}{8}\left( {\frac{G}{{j\omega C}}} \right)^2 } \right)$ This may seem complex, but remember it is only algebra and it will reduce down to simple elegance. Expanding the terms we obtain: $\gamma \approx j\omega \sqrt {LC} \left\{ \begin{array}{l} 1 + \frac{1}{2}\frac{G}{{j\omega C}} - \frac{1}{8}\left( {\frac{G}{{j\omega C}}} \right)^2 + \frac{1}{2}\frac{R}{{j\omega L}} - \frac{1}{4}\frac{{RG}}{{\omega ^2 LC}} \\ - \frac{1}{{16}}\frac{R}{{j\omega L}}\left( {\frac{G}{{j\omega C}}} \right)^2 - \frac{1}{8}\left( {\frac{R}{{j\omega L}}} \right)^2 \\ - \frac{1}{{16}}\left( {\frac{R}{{j\omega L}}} \right)^2 \frac{G}{{j\omega C}} + \frac{1}{{64}}\left( {\frac{R}{{j\omega L}}} \right)^2 \left( {\frac{G}{{j\omega C}}} \right)^2 \\ \end{array} \right\}$ Since $\gamma = \alpha + j\beta$, we merely have to equate the real and imaginary terms to find $\beta$. $\beta \approx \omega \sqrt {LC} \left\{ {1 + \underbrace {\frac{1}{8}\left( {\frac{G}{{\omega C}}} \right)^2 - \frac{1}{4}\frac{{RG}}{{\omega ^2 LC}} + \frac{1}{8}\left( {\frac{R}{{\omega L}}} \right)^2 }_{{\rm{Difference}}\;{\rm{of}}\;{\rm{squares}}} + \underbrace {\frac{1}{{64}}\left( {\frac{R}{{\omega L}}} \right)^2 \left( {\frac{G}{{\omega C}}} \right)^2 }_{{\rm{Very}}\;{\rm{small}}}} \right\}$ Or $\beta \approx \omega \sqrt {LC} \left\{ {1 + \frac{1}{8}\left( {\frac{R}{{\omega L}} - \frac{G}{{\omega C}}} \right)^2 } \right\}$ Note that if $\frac{R}{{\omega L}} = \frac{G}{{\omega C}}$ then $\beta \approx \omega \sqrt {LC}$ From this we observe that $\beta$ is directly proportional to $\omega$. Therefore the requirement for distortionless transmission is: RC = LG Wikipedia has related information at Heaviside_condition This is one of the essential design characteristics for a broadband coax cable network. If we equate the real terms, we obtain: $\alpha \approx \sqrt {RG}$ So there is a reason to study algebra after all! ## The Frequency Domain Signal analysis is often performed in the frequency domain. This tells us how the transmission line affects the spectral content of the signals they are carrying. To determine this, it is necessary to find the Fourier Transform of the transmission line equation. Recall: $\frac{{\partial ^2 v}}{{\partial x^2 }} = RGv + \left( {RC + LG} \right)\frac{{\partial v}}{{\partial t}} + LC\frac{{\partial ^2 v}}{{\partial t^2 }}$ and recall (hopefully) the Fourier Transform (which converts the time domain to the frequency domain): $\mathbb{F}\left\{ {f\left( t \right)} \right\} = F\left( \omega \right) = \int\limits_{ - \infty }^\infty {e^{ - j\omega t} f\left( t \right)dt}$ To prevent this analysis from ‘blowing up’, we must put a stipulation on the voltage function namely, that it vanishes to zero at an infinite distance down the line. This comprises a basic boundary condition. ${\text{let}}\quad v \to 0\quad {\text{as}}\quad x \to \infty$ This stipulation is in agreement with actual laboratory experiments. It is well known that the signal magnitude diminishes as the path lengthens. Likewise, a time boundary condition, that the signal was zero at some time in the distant past and will be zero at some time in the distant future, must be imposed. ${\text{let}}\quad v \to 0\quad {\text{as}}\quad t \to \infty$ Although engineers have no difficulty imposing these restrictions, mathematical purists, are somewhat offended. For this and other reasons, other less restrictive transforms have been developed. The most notable in this context, is the Laplace transform, which does not have the same boundary conditions. Having made the necessary concessions in order to continue our analysis, we must find the Fourier Transform corresponding to the following terms: $\mathbb{F}\left\{ v \right\}\quad \quad \quad \mathbb{F}\left\{ {\frac{{\partial v}} {{\partial t}}} \right\}\quad \quad \quad \mathbb{F}\left\{ {\frac{{\partial ^2 v}} {{\partial t^2 }}} \right\}$ ${\text{Let:}}\quad \quad \quad \mathbb{F}\left\{ v \right\} = V$ Then applying the transform on the derivative, we obtain: $\mathbb{F}\left\{ {\frac{{\partial v}} {{\partial t}}} \right\} = \int\limits_{ - \infty }^\infty {e^{ - j\omega t} \frac{{\partial v}} {{\partial t}}dt}$ This equation can be solved by using integration by parts: $\int {u\,dv} = uv - \int {v\,du}$ ${\text{let}}\quad u = e^{ - j\omega t} \quad \quad \quad \therefore du = - j\omega e^{ - j\omega t}$ ${\text{and}}\quad dv = \frac{{\partial v}}{{\partial t}}\quad \quad \quad \therefore v = v$ $\therefore \mathbb{F}\left\{ {\frac{{\partial v}} {{\partial t}}} \right\} = e^{ - j\omega t} \left. v \right\|_{ - \infty }^\infty - \int\limits_{ - \infty }^\infty {v\left( { - j\omega e^{ - j\omega t} } \right)dt}$ Applying the boundary conditions when t goes to infinity makes the 1st term disappear. $\therefore \mathbb{F}\left\{ {\frac{{\partial v}} {{\partial t}}} \right\} = j\omega \int\limits_{ - \infty }^\infty {e^{ - j\omega t} v\,dt}$ Note that the resulting integral is simply the Fourier Transform. In other words: $\mathbb{F}\left\{ {\frac{{\partial v}} {{\partial t}}} \right\} = j\omega \mathbb{F}\left\{ v \right\} = j\omega V$ similarly: $\mathbb{F}\left\{ {\frac{{\partial ^2 v}} {{\partial t^2 }}} \right\} = \left( {j\omega } \right)^2 \mathbb{F}\left\{ v \right\} = \left( {j\omega } \right)^2 V$ We can now write the transmission line equation in the frequency domain: $\frac{{\partial ^2 V}} {{\partial x^2 }} = RGV + \left( {RC + LG} \right)j\omega V + LC\left( {j\omega } \right)^2 V$ where: $V = V\left( \omega \right) = \mathbb{F}\left\{ {v\left( t \right)} \right\}$ Rearranging the terms, we obtain: $\frac{{\partial ^2 V}} {{\partial x^2 }} = \left[ {RG + \left( {RC + LG} \right)j\omega + \left( {j\omega L} \right)\left( {j\omega C} \right)} \right]V$ or $\frac{{\partial ^2 V}} {{\partial x^2 }} = \left[ {\left( {R + j\omega L} \right)\left( {G + j\omega C} \right)} \right]V$ since: $\sqrt {\left( {R + j\omega L} \right)\left( {G + j\omega C} \right)} = \alpha + j\beta = \gamma$ then $\frac{{\partial ^2 V}}{{\partial x^2 }} = \gamma ^2 V$ or $\frac{{\partial ^2 V}}{{\partial x^2 }} - \gamma ^2 V = 0$ This represents the most general form of the transmission line equation in the frequency domain. This equation must now be solved for V to observe how voltage (or current) varies with distance and frequency. This can be done by assuming a solution of the form: $V = \underbrace {Ae^{ - \gamma x} }_{{\text{forward}}\;{\text{wave}}} + \underbrace {Be^{\gamma x} }_{{\text{reverse}}\;{\text{wave}}}$ These terms represent an exponential decay as the signal travels down the transmission line. If we ignore any reflections, assuming that the cable is infinitely long or properly terminated, this simplifies to: $V = V_0 e^{ - \gamma x}$ To verify whether this assumption is correct, substitute it into the equation, and see if a contradiction occurs. If there is no contradiction, then our assumption constitutes a valid solution. $\frac{{\partial ^2 }}{{\partial x^2 }}V_0 e^{ - \gamma x} - \gamma ^2 V_0 e^{ - \gamma x} = 0$ $\frac{\partial }{{\partial x}}\left( { - \gamma ^2 V_0 e^{ - \gamma x} } \right) - \gamma ^2 V_0 e^{ - \gamma x} = 0$ $\gamma ^2 V_0 e^{ - \gamma x} - \gamma ^2 V_0 e^{ - \gamma x} = 0$ $0 = 0$ Thus we validate the assumed solution. This tells us that in the frequency domain, the voltage or current on a transmission line decays exponentially: $V = V_0 e^{ - \gamma x}$ where: $\gamma = \sqrt {\left( {R + j\omega } \right)\left( {G + j\omega } \right)} = \left\| \gamma \right\|\angle \varphi = \alpha + j\beta$ $\gamma = {\text{ propagation}}\;{\text{constant}}$ $\alpha = {\text{ attenuation}}\;{\text{coeficient}}$ $\beta = {\text{phase}}\;{\text{coefficient}}$ In exponential notation, a sinusoid may be represented by a rotating unity vector, of some frequency: $e^{j\omega t} = \cos \omega t + j\sin \omega t$ Note that the magnitude of this function is 1, but the phase angle is changing as a function of t. If we let: $V_0 = e^{j\omega t}$ Then: $V_0 = e^{j\omega t} e^{ - \gamma x} = e^{j\omega t} e^{ - \left( {\alpha + j\beta } \right)x} = \underbrace {e^{ - \alpha x} }_{{\text{attenuation}}\;vs.\;x}\overbrace {e^{j\left( {\omega t - \beta x} \right)} }^{{\text{phase}}\;vs{\text{.}}\;t\;{\text{and}}\;x}$ This result is quite interesting because it is the same solution for the transmission line equation in the time domain. The term ${e^{ - \alpha x} }$ represents an exponential decay. The signal is attenuated as length x increases. The amount of attenuation is defined as: Attenuation in Nepers: $N = \left\| {\ln e^{ - \alpha x} } \right\| = \alpha x$ Attenuation in dB: $= 20\log e^{ - \alpha x} \approx 8.68589\alpha x$ This allows us to determine the attenuation at any frequency at any point in a transmission line, if we are given the basic line parameters of R, L, G, & C. The term ${e^{j\left( {\omega t - \beta x} \right)} }$ represents a rotating unity vector since: $e^{j\left( {\omega t - \beta x} \right)} = \cos \left( {\omega t - \beta x} \right) + j\sin \left( {\omega t - \beta x} \right)$ The phase angle of this vector is ${\beta x}$ radians. ## Characteristic Impedance The characteristic impedance of a transmission line is also known as its surge impedance, and should not be confused with its resistance. If a line is infinitely long, electrical signals will still propagate down it, even though the resistance approaches infinity. The characteristic impedance is determined from its AC attributes, not its DC ones. Recall from our earlier analysis:	{"extraction_info": {"found_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": 110, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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": 0.9257345199584961, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/44086/the-synthesis-of-254-textno/44088	# The synthesis of $^{254}\text{No}$ How is $^{254}\text{No}$ synthesised? Could you explain the reaction where it is preceded by $^{208}\text{Pb}(^{48}\text{Ca}, 2\text{n})$? References to articles are well enough—I was somehow unable to find anything sufficiently detailed and informative. - ## 1 Answer The notation `X(Y,Z)W` is a compact way of describing nuclear and particle experiments. • Particles that appear to the left of the comma (`,`) are in the initial state and those that appear to the right are in the final state. • The energy and/or momenta of particles that appear inside the parenthesis are measured. Particles that appear outside have unmeasured energy and/or momentum. One caveat here: some (or all) unmeasured initial energy and momentum may be deduced on the basis that they represent a fixed target material. Unobserved final-state particles are often omitted, and sometimes a notation like $X$ is used to imply many possible final state (i.e. an inclusive measurement). So, when I say that my dissertation looked at $A(e,e'p)$, I mean that I fired an electron beam at fixed nuclear targets (we used $^1\mathrm{H}$ for calibration and acceptance; $^2\mathrm{H}$; $^{12}\mathrm{C}$; and $^{56}\mathrm{Fe}$) and measured the coincident electrons and protons emerging from quasi-elastic scattering events. The recoiling nucleus was unobserved, and other events were cut during analysis. Similarly the notation above suggests that calcium nuclei were accelerated into a lead target, and the fast ejecta was observed. Those events with exactly two ejected neutrons were selected, leaving a unobserved heavy nucleus assumed to be $^{254}\mathrm{No}$ (the assumption is good if you really understand the measured ejecta). - Excellent, thank you very much for your answer. – Harold Cavendish Nov 12 '12 at 20:44	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 10, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 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": 0.9591802954673767, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/303906/identify-a-formula-for-each-entry-of-the-matrix-beginpmatrix-1-3-2-6/303917	# Identify a formula for each entry of the matrix $\begin{pmatrix} 1 & 3 \\ 2 & 6 \end{pmatrix}^n$ Identify a formula for each entry of the matrix $\begin{pmatrix} 1 & 3 \\ 2 & 6 \end{pmatrix}^n$. It's easy to find a solution by just looking at the first few results: \begin{pmatrix} 7^{n-1} & 3\cdot 7^{n-1} \\ 2\cdot 7^{n-1} & 6\cdot 7^{n-1} \end{pmatrix} But how would I do it if it wasn't so obvious? - 1 Diagonalize.${}$ – Chris Eagle Feb 14 at 11:30 1 with Cayley-Hamilton theorem – Mercy Feb 14 at 11:30 6 If $M = A^{-1}DA$ then $M^n = A^{-1}D^nA$. If $D$ is diagonal, $D^n$ can be easily calculated. – Alfonso Fernandez Feb 14 at 11:32 1 Diagonalization is the "standard" way to solve this kind of problems. In your particular case, however, your matrix is equal to $uv^T$, where $u^T=(1,2)$ and $v^T=(1,3)$. Therefore $(uv^T)^n=u(v^Tu)(v^Tu)v^T\cdots u(v^Tu)v^T=(v^Tu)^{n-1}uv^T$. – user1551 Feb 14 at 11:41 Or you can use inducion. For $n=1$ it is clearly true. Now suppose it is true for some natural $n$. And prove that it is true for $n+1$. – Tomas Feb 14 at 11:43 show 1 more comment ## 4 Answers Fleshing out a little what the comments hint you: $$\det(tI-A)=\begin{vmatrix}t-1&-3\\-2&t-6\end{vmatrix}=t(t-7)\Longrightarrow$$ the matrix's eigenvalues are $\,0\,,\,7\,$ , and eigenvectors for these values can be found as follows: $$(1)\,\,\lambda=0:\;\;\;\;\begin{cases}\;\,-x-3y=0\\{}\\-2x-6y=0\end{cases}\;\;\Longrightarrow x=-3y\Longrightarrow\,\,\text{for example}\,\,\binom{3}{\!\!-1}$$ $$(1)\,\,\lambda=t:\;\;\;\;\begin{cases}\;6x-3y=0\\{}\\-2x+y=0\end{cases}\;\;\Longrightarrow y=2y\Longrightarrow\,\,\text{for example}\,\,\binom{1}{2}$$ Form now the matrix $$P:=\begin{pmatrix}3&1\\\!\!\!-1&2\end{pmatrix}\Longrightarrow P^{-1}=\begin{pmatrix}\frac{2}{7}&\!\!\!-\frac{1}{7}\\{}\\\frac{1}{7}&\;\frac{3}{7}\end{pmatrix}$$ so that $$P^{-1}AP=\begin{pmatrix}0&0\\0&7\end{pmatrix}=:D$$ And now it is a piece of cake: $$A^n=\left(PDP^{-1}\right)^n=PD^nP^{-1}\;\;,\;\;\text{and}\;\;\;D^n=\begin{pmatrix}0&0\\0&7^n\end{pmatrix}\ldots$$ - Nice formula for $D^n$ + – Babak S. Feb 15 at 9:07 Diagonalization works generally, but it's overkill here. Simpler, note that the 2nd row is twice the 1st, which must stay true for $\rm\,A^n\,$ too since $\rm\, vA = (2,-1)A = 0\,$ $\Rightarrow$ $\rm\,vA^n = (vA)A^{n-1}\! = 0.\:$ So $$\rm \left[\begin{array}{rr}\rm a_{n+1}\!\!\!\! &\rm b_{n+1}\\ \rm 2a_{n+1}\!\!\!\! &\rm 2b_{n+1}\end{array}\!\right] = \left[\begin{array}{cc}1 \!\!&\rm 3\\ 2\!\! &\rm 6\end{array}\!\right] \left[\begin{array}{rr}\rm a_{n}\!\!\!\! &\rm b_{n}\\ \rm 2a_{n}\!\!\!\! &\rm 2b_{n}\end{array}\right]\Rightarrow \begin{array}{ll} \rm a_{n+1}\! = 7 a_n\!= 7^n a_0\! = 7^n\\ \rm b_{n+1}\! = 7 b_n\! = 7^n b_0\! = 3\!\cdot\! 7^n\end{array} \!\Rightarrow A^{\!n+1} \! = \left[\begin{array}{rr}\rm 7^n\!\!\! &\rm 3\!\cdot\! 7^n\\ \rm 2\!\cdot\! 7^n\!\!\!&\rm 6\!\cdot\! 7^n\end{array}\right]$$ - $$A^2-Tr(A)\cdot A +\det{A}\cdot I_{2}=O_{2}.$$ $$\det{A}=6-6=0.$$ $$A^{2}=Tr(A) \cdot A=7 \cdot A.$$ $$A^{2}=7A.$$ $$A^{3}=7A^{2}=7^{2}A.$$ $$\vdots$$ $$A^{n}=7^{n-1}\cdot A$$ for me, this proof is more simple. - I expand the Mercy's idea with Cayley-Hamilton theorem because I like it. Let $p(\lambda) = \text{det}(A-\lambda I)$ characteristic polynomial of A. Than you can write $\lambda^n = p(\lambda)g(\lambda) + q(\lambda)$, where $\deg(q)<\deg(p)$. By Cayley-Hamilton theorem you get $A^n = q(A)$. So you need only calculate $A^1,..,A^{m-1}$ where $m$ is dimension of matrix. -	{"extraction_info": {"found_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": 14, "mathjax_asciimath": 0, "img_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": 0.8875111937522888, "perplexity_flag": "middle"}
http://www.haskell.org/haskellwiki/Maximal_free_expression	# Maximal free expression ### From HaskellWiki This is within the context of a given expression, and subexpressions are partially ordered with respect to containment, and have finite length, so there will always be maximal (but possibly not unique) free (sub-)expressions. Note that there is a subtle but important difference between the words maximal and maximum. An element x of a partially ordered set $(S, \le)$ is called maximal if there is no $y \in S$ such that $x \le y$, and it is called a maximum if $\forall y \in S, x \le y$. If a maximum exists, it is unique, but there can be many maximal (but not maximum) elements.	{"extraction_info": {"found_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": 4, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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": 0.9115240573883057, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/tagged/reference-request?page=7&sort=newest&pagesize=50	# Tagged Questions Use this tag for questions seeking a single specific paper or a short, non-open-ended list of references, like "What paper first discovered X?", "Where can I find the original derivation of X?", or "What is the canonical source for X?" etc. 0answers 140 views ### Readings of the detectors at Japan and Izu-Bonin-Mariana Trenches Given the relevance of this subduction system, I would expect that a wide range of detectors (temperature, vibration, seismometers, whatever) are deployed in the depth of these trenches. What would be ... 3answers 184 views ### References on the physics of anyons Anyone know some good introductory references on the physics of anyons? 2answers 352 views ### What's a good reference for the electrodynamics of moving media? 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It may be the case that I learn it before but forget the name, or, there is no name for the equation ... 7answers 760 views ### Evolution in the interpretation of the Dirac equation As I understand, Dirac equation was first interpreted as a wave equation following the ideas of non relativistic quantum mechanics, but this lead to different problems. The equation was then ... 3answers 467 views ### Books for linear operator and spectral theory I need some books to learn the basis of linear operator theory and the spectral theory with, if it's possible, physics application to quantum mechanics. Can somebody help me? 2answers 463 views ### Quantum Field Theory cross sections integrals Where can I find some examples of cross sections calculations in QFT done step-by-step? Those integrals are a little horror. For example - a simple scalar+scalar -> scalar+scalar at the tree level in ...	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 3, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9199712872505188, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/18102?sort=oldest	## lebesgue measure and countable sets [closed] ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) the lebesgue integral $\int_{[0,1]} 1_{\mathbb{Q}} dm = 0$ . and if we integrate the complement $\int_{[0,1]} 1_{\mathbb{Q}^C} dm =1$ which is the same as $\int_{[0,1]} dm$ to me this is still a bit of a mystery to there are $\infty$ rationals in the interval $[0,1]$ yet if we exclude all these rational the measure is still the same as if we included the set $\mathbb{Q} \cap [0,1]$ . why is that, i know the measure of a countable set is zero, but why i cant find an explanation for this, but how can a set with cardinality of $\infty$ still be zero - 1 MathOverflow is a site for questions and answers of interest to research mathematicians. Your question would be appropriate for a more student-oriented site. See e.g. mathoverflow.net/faq#whatnot for a list of some such sites. – Pete L. Clark Mar 13 2010 at 23:31 ## 2 Answers A single point has measure zero. A countable set is the countable union of points, and since the measure is countably additive, you have that the measure is the sum of the measure of the single points. - so if the measure of a single point was not zero then measure of a countable set would be $\infty$ which would be the same as the measure of set \$[0, \infty) which wouldnt make sense. – curious Mar 13 2010 at 21:44 Positive isn't enough, perhaps the measures of the points in some countable set are the terms of a convergent series. – Gerald Edgar Mar 14 2010 at 0:06 @curious: If all points had the same positive measure, then you would be correct: a countably infinite set would have infinite measure. However, there are measures in which (some) single points have positive measure, but countable sets may still have finite measure. For example, take a bijection between $f\colon\mathbb{Q}\to\mathbb{N}$, and define a measure $\mu$ on $\mathcal{P}(\mathbb{R})$ by letting $\mu(X) = \sum_{q\in \mathbb{Q}\cap A}2^{-f(q)}$. Then every subset of $\mathbb{R}$ is measurable, and has finite measure; some points (the rationals) have positive measure, some have measure 0. – Arturo Magidin Mar 14 2010 at 20:24 ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. It is perhaps instructive to note that there is an open subset of `$\mathbb{R}$` which has finite measure and contains all rational numbers. Just number the rational numbers as `$\mathbb{Q}=\{q_1,q_2,\ldots,\}$` and let `$$U=\bigcup_{i=1}^\infty (q_i-2^{-i-1},q_i+2^{-i-1}).$$` The measure of $V$ is less than 1 (you can make it less than some prescribed `$\epsilon$` easily enough). And the measure of open sets seems easier to understand, since any open set is the disjoint union of open intervals and the measure is the sum of the length of these intervals. (But note that the intervals defining $U$ above are not disjoint; there will be a great deal of overlap, so overlapping intervals join together to make bigger intervals …) This is a bit counterintuitive of course, since there doesn't seem to be a whole lot of room outside of $U$, yet the complement of $U$ has infinite measure. If you think about it, it must mean that even though $U$ is the countable union of open intervals, the complement of $U$ has an uncountable number of components, each consisting of just one point. (Any connected subset of `$\mathbb{R}$` is an interval, but an interval containing no rational point must be a singleton.) -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 19, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9701860547065735, "perplexity_flag": "head"}
http://mathoverflow.net/questions/76474?sort=oldest	## Applications of Hilbert’s metric ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Among the fascinating constructions in mathematics is the Hilbert metric on a bounded convex subset of ${\mathbb R}^n$. Where, within mathematics, is it used ? I know at least a proof of the Perron--Frobenius Theorem for non-negative matrices. What are its applications in other sciences ? - Excellent question Denis! I'd wanted to ask a question along these lines for quite some time now. – S. Sra Sep 27 2011 at 9:08 How about a link or definition of the Hilbert distance? – Gerald Edgar Sep 27 2011 at 14:48 1 Dear Denis, I would like to ask you, if you don't mind, to explain a bit more why you find this construction fascinating. Also, could you explain a bit how one can prove Perron-Frobenius theorem using Hilbert metric? Would this be some proof that is more conceptual, than a usual one? – Dmitri Sep 27 2011 at 21:35 ## 4 Answers Update I found the following nice survey paper that lists lots of applications of Hilbert's metric (Birkhoff's version): Birkhoff's version of Hilbert's metric and its applications in analysis by Bas Lemmens, Roger Nussbaum, April, 2013. Not Hilbert's metric directly, but immediately related is the Thompson metric which is frequently used. I list below two interesting examples. 1. Solving nonlinear matrix equations: Invariant metrics, contractions and nonlinear matrix equations by H. Lee and Y. Lim 2. Dynamic Programming: Thompson metric, contraction property and differentiability of policy functions by L. Montrucchio - ### 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. Garrett Birkhoff used the Hilbert metric (he called it the projective metric) to prove that every $n$-by-$n$ matrix $A$ with positive entries is a Hilbert metric contraction on the cone of nonnegative vectors. He gave a formula for the contraction constant which is $\frac{1}{4} \arctan \Delta$ where $\Delta$ is the maximum Hilbert's (projective) metric distance between $Ae_i$ and $Ae_j$ where $e_i$ and $e_j$ are distinct elementary basis vectors. This immediately implies one aspect of the Perron-Frobenius theorem: that matrices with positive entries have a unique Perron eigenvector. G. Birkhoff, Extensions of Jentzsch’s theorem. Trans. Amer. Math. Soc. 85 (1957), 219–227. The Hilbert metric proof of the Perron-Frobenius theorem also extends to nonlinear maps which are monotone and homogeneous of degree one (i.e., $f(\lambda x) = \lambda f(x)$, $\forall \lambda > 0$). This is the primary advantage of using the Hilbert metric to prove the Perron-Frobenius theorem. - This is precisely what I mentionned in my question. – Denis Serre Sep 29 2011 at 5:43 See e.g. pages 167-169 of "Topics in nonlinear analysis & applications" By Donald H. Hyers, George Isac, Themistocles M. Rassias. You can read it on Google books. - This might not be considered an application, but Hilbert metrics have been studied geometrically and dynamically. Here are several examples of questions that have been partially or totally answered: • when is a convex domain endowed with its Hilbert metric $\delta$-hyperbolic? • what is the volume entropy of Hilbert metrics? • does there exist convex sets that admit a cocompact group of isometries (relative to their Hilbert metric)? (the answer is yes!) I guess that amoung these, the last part can be considered an application: Hilbert metrics yields interesting subgroups of $\mathrm{PSL}(n,\mathbb{R})$. For more details you can look at the works of Yves Benoist (in particular the "convexes divisibles" series and "Convexes hyperboliques et fonctions quasisymétriques", in french), Constantin Vernicos, Ludovic marquis and Mickaël Crampon. -	{"extraction_info": {"found_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": 0.900318443775177, "perplexity_flag": "middle"}
http://planetmath.org/anosovdiffeomorphism	# Anosov diffeomorphism If $M$ is a compact smooth manifold, a diffeomorphism $f\colon M\to M$ (or a flow $\phi\colon\mathbb{R}\times M\to M$) such that the whole space $M$ is an hyperbolic set for $f$ (or $\phi$) is called an Anosov diffeomorphism (or flow). Anosov diffeomorphisms were introduced by D.V. Anosov, who proved that they are $\mathcal{C}^{1}$-structurally stable. Not every manifold admits an Anosov diffeomorphism; for example, there are no such diffeomorphisms on the sphere $S^{n}$. The simplest examples of compact manifolds admiting them are the tori $\mathbb{T}^{n}$: they admit the so called linear Anosov diffeomorphisms, which are isomorphisms of $\mathbb{T}^{n}$ having no eigenvalue of modulus $1$. It was proved that any other Anosov diffeomorphism in $\mathbb{T}^{n}$ is topologically conjugate to one of this kind. It is not known which manifolds support Anosov diffeomorphisms. The only known examples of are nilmanifolds and infranilmanifolds, and it is conjectured that these are the only ones. Anosov flows are more abundant; for example, if $M$ is a Riemannian manifold of negative sectional curvature, then its geodesic flow is an Anosov flow. Another famous conjecture is that the nonwandering set of any Anosov diffeomorphism is the whole manifold $M$. This is known to be true for linear Anosov diffeomorphisms (and hence for any Anosov diffeomorphism in a torus). For Anosov flows, there are examples where the nonwandering set is a proper subset of $M$. Type of Math Object: Definition Major Section: Reference ## Mathematics Subject Classification 37D20 Uniformly hyperbolic systems (expanding, Anosov, Axiom A, etc.) ## Comments ### Infranil manifolds & Riemann surfaces I have in front of me a textbook example of an Anosov flow; its the spliting of the tangent manifold of the upper half plane (or of a Riemann surface of negative curvature) into three parts: the geodesic flow plus two horocycle flows, one expanding, the other contracting. I'm having some trouble identifying how this is an infranil manifold. I guess the group structure is rotations O(2) semi-direct translations; and te finite group must be the covering group. So I guess that means every tangent bundle to a Riemann surface is an infranil manifold ?? It'd be nice for this article to clarify. --linas ### Re: Infranil manifolds & Riemann surfaces The problem is that you're looking at an Anosov flow. The conjecture is that the only manifolds supporting Anosov diffeomorphisms are nilmanifolds or infranilmanifolds. Anosov flows are much more common; in fact the example you mention is a particular case of the more general fact that in any riemannian manifold with negative sectional curvature, the geodesic flow is in fact an Anosov flow. I will add a comment on that. ## 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: Koro Added: 2003-07-26 - 20:13 Author(s): Koro ## Versions (v9) by Koro 2013-03-22	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 15, "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": 0.9115152359008789, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/54784/uses-of-chevalley-warning	# Uses of Chevalley-Warning In the recent IMC 2011, the last problem of the 1st day (no. 5, the hardest of that day) was as follows: We have $4n-1$ vectors in $F_2^{2n-1}$: $\{v_i\}_{i=1}^{4n-1}$. The problem asks : Prove the existence of a subset $A \subseteq [4n-1], \|A\| = 2n$ such that $\sum_{i \in A} v_i = \vec{0}$. Is there a solution that uses the Chevalley-Warning theorem about divisibility of number of solutions by the field's characteristic? The statement of the problem seems as a "xor" analog of Erdos-Ginzburg-Ziv, that is usually proved by Chevally-Warning. My idea is to formulate homogeneous polynomial equations in $a_1,\cdots,a_{4n-1} \in F_2$: $P_j(\vec{a}) = (\sum a_i v_i)_j$ for $1 \le j \le 2n-1$. Let $A(\vec{a}) = \{ i \| a_i = 1\}$ and $g(\vec{a}) = \|A(\vec{a})\|$ (we want to choose $A = A(\vec{a})$ for a solution $a$ of the system). Ideas for completing the solution: • We can also create $Q_{k}(\vec{a}) =$ Symmetric polynomial of degree $2^k$ = $a_1 \cdots a_{2^k} + \cdots$. We notice that $Q_k(a) = \binom{g(\vec{a})}{2^k}$. By Lucas's Theorem, mod 2 we have $\binom{m}{2^k} =$ k'th digit of $m$ in base-$2$. So we choose the equations $Q_k(a) = \binom{2n}{2^k}$ mod 2. The problem is that we get $\sum deg(Q_k) \ge 2n$ (sum over number of digits of $4n-1$). Also, $0$ is not a trivial solution. • Another idea was to define the equations $R_i(a) = a_{2i-1} + a_{2i} - a_{2i+1} - a_{2i+2} = 0, 1 \le i \le 2n$. A solution implies that $a_{2i-1} + a_{2i}$ is constant. If it is $1$, we get that from each pair of even and odd vector, one is chosen - $2n$ vectors in total. But it has some holes. • Another idea is to create equations that force $g(\vec{a}) = 0$ mod n, so $\vec{0}$ is a trivial solution, and another solution would imply that a $A$ exists of size $n, 2n$ or $3n$. A nice observation is that one of the vectors is (wlog) $\vec{0}$ (can be assumed since $2n$ is even, so replacing $v_i$ by $v_i - v_{1}$ is fine). If you have idea of how to use the theorem, and examples in general of its non-trivial uses - it would be welcome. - I guess $F_2^{2-1}$ should be $F_2^{2n-1}$. (I don't want to edit it yet since it's <6 characters.) – Srivatsan Jul 31 '11 at 17:58 Ofir: could you please reread your question, there are some obvious typos, but maybe some less obvious ones. – Peter Patzt Jul 31 '11 at 19:12 I've fixed the typos and added an idea, thanks. – Ofir Aug 1 '11 at 15:41 (Re: analog of Erdos-Ginzburg-Ziv) a variant of Kemnitz's problem, I'd say... – Grigory M Aug 19 '11 at 11:01	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 38, "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": 0.9528852105140686, "perplexity_flag": "head"}
http://mathoverflow.net/questions/4695?sort=oldest	## K3 surfaces with good reduction away from finitely many places ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Let S be a finite set of primes in Q. What, if anything, do we know about K3 surfaces over Q with good reduction away from S? (To be more precise, I suppose I mean schemes over Spec Z[1/S] whose geometric fibers are (smooth) K3 surfaces, endowed with polarization of some fixed degree.) Are there only finitely many isomorphism classes, as would be the case for curves of fixed genus? If one doesn't know (or expect) finiteness, does one have an upper bound for the number of such K3 surfaces X/Q of bounded height? - I will remark that the paper "Families of K3 surfaces" by Borchers, Katzarkov, Pantev, and Shepherd-Barron shows that under certain circumstances you don't have any non-isotrivial K3 surfaces over proper curves. – JSE Nov 9 2009 at 0:29 Added "number-theory" tag. – Lavender Honey Nov 9 2009 at 4:15 @JSE: "proper" is crucial here, right? (and hence the application of the result to the question at hand is limited). There are no non-isotrivial families of elliptic curves over a proper curve, but in my mind this tells us little about elliptic curves with bad reduction outside, say, {2}. – Kevin Buzzard Nov 9 2009 at 11:16 ## 4 Answers Some thoughts. There are no such varieties when S = 1. This is a consequence of a theorem of Fontaine, MR1274493 (Schémas propres et lisses sur Z). I think that one should only expect finitely many such varieties for any fixed S. Let me give an argument that uses every possible conjecture I know. There may be an unconditional proof, but that would probably require knowing something about K3-surfaces. I first want to claim that the ramification at primes q\|S is "bounded" independently of X. The corresponding fact for elliptic curves will be that the power of the conductor for each q\|N is bounded by 2 (if p > 3) or (if p = 2 or 3) by some fixed number I can't remember. The most obvious argument along these lines is to consider the representation on inertia I`_`q acting on the p-adic etale cohomology groups H^2(X). These correspond to Galois representations with image in GL`_`22(Z`_`p). The argument I have in mind for elliptic curves works directly in this case, providing that one has "independence of p" statement for the Weil-Deligne representations at q (quick hint: the image of wild inertia divides the gcd of the orders of GL`_`22(F_p) over all primes p). This may require the existence of semi-stable models, which one certainly has for elliptic curves, but I don't know for K3-surfaces. The next step is to use a Langlands-type conjecture. The p-adic representation V on H^2(X) may be reducible, but at least we know that each irreducible chunk will correspond to an irreducible Galois representation of Q into GL`_`n(Z`_`p) for some n (at most 22). Each of these, conjecturally, will correspond to a cuspidal automorphic form of fixed weight and level divisible only by q\|S. Moreover, from the previous paragraph, the level will be bounded at q\|S. Thus there will only be finitely many representations which can occur as H^2(X) for any K3-surface X/Z[1/S]. (Maybe I am assuming here that the Galois representation acting on H^2(X) is semi-simple --- let us do so, since this is a conjecture of Grothendieck and Serre.) Finally, I want to deduce from any equality H^2(X) = H^2(X') that X is (essentially) X'. From the Tate conjecture we deduce the existence of correspondences X~~>X' and X'~~>X over Q whose composition induces an isomorphism on H^2(X) --- and now hopefully some knowledge of the geometry of K3 surfaces is enough to show that these sets of "isogenous" K3 surfaces form a finite set. EDIT: As Buzzard points out, I obscured the fact in the last paragraph that some more arithmetic may be necessary. What I meant to say is that understanding isogeny classes of K3's over Q will first require understanding isogeny classes over C, and hopefully this second task will be the hard part. As David points out, the Torelli theorem for K3 will surely be relevant here. I think there can be non-isomorphic isogenous K3s, however. If one takes an isogeny of abelian surfaces A->B then one can presumably promote this to an isogeny of the associated Kummer surfaces. EDIT: Here is another thought. Deligne proves the Weil conjecture for K3 surfaces: http://www.its.caltech.edu/~clyons/DeligneWeilK3trans.pdf The philosophy is that there should be an inclusion of motives H^2(X) --> H^1(A) tensor H^1(A) for some abelian variety A (possibly of some huge but uniformly bounded dimension, like 2^19). It may be possible (conjecturally or otherwise) to reduce your question to the analogous statement for A, for which it is known. (Prop 6.5 is relevant here). It may well be possible to show that the variety A is defined over Z[1/2S], for example. I could make this edit more coherent but I'm off to lunch, so treat this as a thought fragment. - If you have an isomorphism H^2(X,Z) = H^2(X',Z) of lattices in H^2(X,C) = H^2(X',C), then your last statement is simply Torelli for K3 surfaces (which is classical); is this what you tried to do in the last paragraph ? – David Lehavi Nov 9 2009 at 11:17 @FC: a few almost content-free comments. "S=1" at the beginning is of course "S is empty". Second, you're implicitly asserting what seems to me to be a rather natural conjecture, which I've never seen before: let's say a K3 surface over Q is modular if its H^2 can be naturally associated with an algebraic automorphic representation of GL_{22}(adeles). We presumably want to conjecture that all K3 surfaces are modular---did anyone explicitly write this down anywhere? – Kevin Buzzard Nov 9 2009 at 11:23 ...The conjecture would not (arguably) follow from Fontaine-Mazur, who "only" conjecture that the H^2 would come from geometry, which it clearly does. Finally, last para you implicitly seem (to me) to say that the missing part of the argument is complex-geometric. But I don't think it is: given an ell curve over Q there will be infinitely many isogenous curves over C. Hence the remaining problem in this para (finiteness of "isogeny classes") is still arithmetic. – Kevin Buzzard Nov 9 2009 at 11:25 @buzzard: Isn't is basically a conjecture of Langlands that everything motivic is automorphic? – Lavender Honey Nov 9 2009 at 13:59 @buzzard: If you are going to complain to me about "S=1" when S is a set of primes, why didn't you complain to JSE about "Spec Z[1/S]" :P – Lavender Honey Nov 9 2009 at 14:03 show 10 more comments ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. One classical trick that may be useful for reduction to the abelian scheme case is the Kuga-Satake construction which takes a weight 2 Hodge structure of K3 type to a weight 1 Hodge structure built out of its Clifford algebra. Although I've never read it, there is a paper of Rizov (http://arxiv.org/abs/math/0608497) which is supposed to make this a priori transcendental construction work over a general base (when you start with an honest family of polarized K3 surfaces). There may still be an issue about working up to isogeny, but at least this may let you shortcut the arithmetic discussion. - I was alluding to exactly this construction in my last edit. – Lavender Honey Nov 9 2009 at 22:22 Here is a result about Kummer surfaces (one of the K's in K3 stands for Kummer) proved by Tetsushi Ito in his unpublished Master's thesis (Tokyo). I have a copy dated January 2001. Theorem 5.2 . Let $K$ be a number field, $S$ a finite set of places of $K$ containing all archimedean places, and $d$ an integer. There are only finitely many Kummer surfaces over $K$, polarised of degree $d$, which have a rational point and which have good reduction outside $S$. He uses Faltings' proof of the Shafarevich Conjecture for abelian varieties. - Yves Andre has proved the finiteness of the number of K3 surfaces over a number field $K$ with a polarisation of fixed degree $d$ and having good reduction (as a polarised variety) outside a fixed finite set of primes. The reference is: "On the Shafarevich and Tate conjectures for hyper-Kähler varieties".Math. Ann. 305 (1996), no. 2, 205–248. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 9, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9297782778739929, "perplexity_flag": "middle"}
http://en.wikipedia.org/wiki/Graham's_number	Graham's number Graham's number, named after Ronald Graham, is a large number that is an upper bound on the solution to a certain problem in Ramsey theory. The number gained a degree of popular attention when Martin Gardner described it in the "Mathematical Games" section of Scientific American in November 1977, writing that, "In an unpublished proof, Graham has recently established ... a bound so vast that it holds the record for the largest number ever used in a serious mathematical proof." The 1980 Guinness Book of World Records repeated Gardner's claim, adding to the popular interest in this number. According to physicist John Baez, Graham invented the quantity now known as Graham's number in conversation with Gardner himself. While Graham was trying to explain a result in Ramsey theory which he had derived with his collaborator B. L. Rothschild, Graham found that the quantity now known as Graham's number was easier to explain than the actual number appearing in the proof. Because the number which Graham described to Gardner is larger than the number in the paper itself, both are valid upper bounds for the solution to the Ramsey-theory problem studied by Graham and Rothschild.[1] Graham's number is unimaginably larger than other well-known large numbers such as a googol, googolplex, and even larger than Skewes' number and Moser's number. Indeed, like the last three of those numbers, the observable universe is far too small to contain an ordinary digital representation of Graham's number, assuming that each digit occupies at least one Planck volume. Even power towers of the form $\scriptstyle a ^{ b ^{ c ^{ \cdot ^{ \cdot ^{ \cdot}}}}}$ are useless for this purpose, although it can be easily described by recursive formulas using Knuth's up-arrow notation or the equivalent, as was done by Graham. The last ten digits of Graham's number are ...2464195387. Specific integers known to be far larger than Graham's number have since appeared in many serious mathematical proofs (e.g., in connection with Friedman's various finite forms of Kruskal's theorem). Context Example of a 2-colored 3-dimensional cube containing one single-coloured 4-vertex coplanar complete subgraph. The subgraph is shown below the cube. Note that this cube would contain no such subgraph if, for example, the bottom edge in the present subgraph were replaced by a blue edge – thus proving by counterexample that N* > 3. Graham's number is connected to the following problem in Ramsey theory: Consider an n-dimensional hypercube, and connect each pair of geometric vertices, to obtain a complete graph on 2n graph vertices. Then colour each of the edges of this graph either red or blue. What is the smallest value of n for which every such colouring contains at least one single-coloured complete subgraph on 4 coplanar vertices? In 1971, Graham and Rothschild proved that this problem has a solution N, giving as a bound 6 ≤ N ≤ N, with N being a large but explicitly defined number $\scriptstyle F^7(12) \;=\; F(F(F(F(F(F(F(12)))))))$, where $\scriptstyle F(n) \;=\; 2\uparrow^n 3$ in Knuth's up-arrow notation; the number is between 4 → 2 → 8 → 2 and 2 → 3 → 9 → 2 in Conway chained arrow notation.[2] This was reduced in 2013 via upper bounds on the Hales–Jewett number to $\scriptstyle N' \;=\; 2\;\uparrow\uparrow\;2\;\uparrow\uparrow\;2\;\uparrow\uparrow\;9$.[3] The lower bound of 6 was later improved to 11 by Geoff Exoo in 2003, and to 13 by Jerome Barkley in 2008. Thus, the best known bounds for N* are 13 ≤ N* ≤ N'. Graham's number, G, is much larger than N: $\scriptstyle f^{64}(4)$, where $\scriptstyle f(n) \;=\; 3 \uparrow^n 3$. This weaker upper bound for the problem, attributed to an unpublished work of Graham, was eventually published and named by Martin Gardner in Scientific American in November 1977.[4] Definition Using Knuth's up-arrow notation, Graham's number G (as defined in Gardner's Scientific American article) is $\left. \begin{matrix} G &=&3\underbrace{\uparrow \uparrow \cdots\cdots\cdots\cdots\cdots \uparrow}3 \\ & &3\underbrace{\uparrow \uparrow \cdots\cdots\cdots\cdots \uparrow}3 \\ & &\underbrace{\qquad\;\; \vdots \qquad\;\;} \\ & &3\underbrace{\uparrow \uparrow \cdots\cdot\cdot \uparrow}3 \\ & &3\uparrow \uparrow \uparrow \uparrow3 \end{matrix} \right \} \text{64 layers}$ where the number of arrows in each layer, starting at the top layer, is specified by the value of the next layer below it; that is, $G = g_{64},\text{ where }g_1=3\uparrow\uparrow\uparrow\uparrow 3,\ g_n = 3\uparrow^{g_{n-1}}3,$ and where a superscript on an up-arrow indicates how many arrows are there. In other words, G is calculated in 64 steps: the first step is to calculate g1 with four up-arrows between 3s; the second step is to calculate g2 with g1 up-arrows between 3s; the third step is to calculate g3 with g2 up-arrows between 3s; and so on, until finally calculating G = g64 with g63 up-arrows between 3s. Equivalently, $G = f^{64}(4),\text{ where }f(n) = 3 \uparrow^n 3,$ and the superscript on f indicates an iteration of the function, e.g., f 4(n) = f(f(f(f(n)))). Expressed in terms of the family of hyperoperations $\scriptstyle \text{H}_0, \text{H}_1, \text{H}_2, \cdots$, the function f is the particular sequence $\scriptstyle f(n) \;=\; \text{H}_{n+2}(3,3)$, which is a version of the rapidly growing Ackermann function A(n,n). (In fact, $\scriptstyle f(n) \;>\; A(n,\, n)$ for all n.) The function f can also be expressed in Conway chained arrow notation as $\scriptstyle f(n) \;=\; 3 \rightarrow 3 \rightarrow n$, and this notation also provides the following bounds on G: $3\rightarrow 3\rightarrow 64\rightarrow 2 < G < 3\rightarrow 3\rightarrow 65\rightarrow 2.\,$ Magnitude To convey the difficulty of appreciating the enormous size of Graham's number, it may be helpful to express—in terms of exponentiation alone—just the first term (g1) of the rapidly growing 64-term sequence. First, in terms of tetration ($\scriptstyle \uparrow\uparrow$) alone: $g_1 = 3 \uparrow \uparrow \uparrow \uparrow 3 = 3 \uparrow \uparrow \uparrow (3 \uparrow \uparrow \uparrow 3) = 3 \uparrow\uparrow (3 \uparrow\uparrow (3 \uparrow\uparrow \ \dots \ (3 \uparrow\uparrow 3) \dots ))$ where the number of 3s in the expression on the right is $3 \uparrow \uparrow \uparrow 3 \ = \ 3 \uparrow \uparrow (3 \uparrow \uparrow 3).$ Now each tetration ($\scriptstyle\uparrow\uparrow$) operation reduces to a "tower" of exponentiations ($\scriptstyle \uparrow$) according to the definition $3 \uparrow\uparrow X \ = \ 3 \uparrow (3 \uparrow (3 \uparrow \dots (3 \uparrow 3) \dots )) \ = \ 3^{3^{\cdot^{\cdot^{\cdot^{3}}}}} \quad \text{where there are X 3s}.$ Thus, $g_1 = 3 \uparrow\uparrow (3 \uparrow\uparrow (3 \uparrow\uparrow \ \dots \ (3 \uparrow\uparrow 3) \dots )) \quad \text{where the number of 3s is} \quad 3 \uparrow \uparrow (3 \uparrow \uparrow 3)$ becomes, solely in terms of repeated "exponentiation towers", $g_1 = \left. \begin{matrix}3^{3^{\cdot^{\cdot^{\cdot^{\cdot^{3}}}}}}\end{matrix} \right \} \left. \begin{matrix}3^{3^{\cdot^{\cdot^{\cdot^{3}}}}}\end{matrix} \right \} \dots \left. \begin{matrix}3^{3^3}\end{matrix} \right \} 3 \quad \text{where the number of towers is} \quad \left. \begin{matrix}3^{3^{\cdot^{\cdot^{\cdot^{3}}}}}\end{matrix} \right \} \left. \begin{matrix}3^{3^3}\end{matrix} \right \} 3$ and where the number of 3s in each tower, starting from the leftmost tower, is specified by the value of the next tower to the right. In other words, g1 is computed by first calculating the number of towers, n = 3↑3↑3↑...↑3 (where the number of 3s is 3↑3↑3 = 7625597484987), and then computing the nth tower in the following sequence: ``` 1st tower: 3 2nd tower: 3↑3↑3 (number of 3s is 3) = 7625597484987 3rd tower: 3↑3↑3↑3↑...↑3 (number of 3s is 7625597484987) = … ⋮ g1 = nth tower: 3↑3↑3↑3↑3↑3↑3↑...↑3 (number of 3s is given by the n-1th tower) ``` where the number of 3s in each successive tower is given by the tower just before it. Note that the result of calculating the third tower is the value of n, the number of towers for g1. The magnitude of this first term, g1, is so large that it is practically incomprehensible, even though the above display is relatively easy to comprehend. Even n, the mere number of towers in this formula for g1, is far greater than the number of Planck volumes (roughly 10185 of them) into which one can imagine subdividing the observable universe. And after this first term, still another 63 terms remain in the rapidly growing g sequence before Graham's number G = g64 is reached. Rightmost decimal digits Graham's number is a "power tower" of the form 3↑↑n (with a very large value of n), so its rightmost decimal digits must satisfy certain properties common to all such towers. One of these properties is that all such towers of height greater than d (say), have the same sequence of d rightmost decimal digits. This is a special case of a more general property: The d rightmost decimal digits of all such towers of height greater than d+2, are independent of the topmost "3" in the tower; i.e., the topmost "3" can be changed to any other nonnegative integer without affecting the d rightmost digits. The following table illustrates, for a few values of d, how this happens. For a given height of tower and number of digits d, the full range of d-digit numbers (10d of them) does not occur; instead, a certain smaller subset of values repeats itself in a cycle. The length of the cycle and some of the values (in parentheses) are shown in each cell of this table: Number of different possible values of 3↑3↑…3↑x when all but the rightmost d decimal digits are ignored Number of digits (d) 3↑x 3↑3↑x 3↑3↑3↑x 3↑3↑3↑3↑x 3↑3↑3↑3↑3↑x 1 4 (1,3,9,7) 2 (3,7) 1 (7) 1 (7) 1 (7) 2 20 (01,03,…,87,…,67) 4 (03,27,83,87) 2 (27,87) 1 (87) 1 (87) 3 100 (001,003,…,387,…,667) 20 (003,027,…387,…,587) 4 (027,987,227,387) 2 (987,387) 1 (387) The particular rightmost d digits that are ultimately shared by all sufficiently tall towers of 3s are in bold text, and can be seen developing as the tower height increases. For any fixed number of digits d (row in the table), the number of values possible for 3$\scriptstyle\uparrow$3↑…3↑x mod 10d, as x ranges over all nonnegative integers, is seen to decrease steadily as the height increases, until eventually reducing the "possibility set" to a single number (colored cells) when the height exceeds d+2. A simple algorithm[5] for computing these digits may be described as follows: let x = 3, then iterate, d times, the assignment x = 3x mod 10d. Except for omitting any leading 0s, the final value assigned to x (as a base-ten numeral) is then composed of the d rightmost decimal digits of 3↑↑n, for all n > d. (If the final value of x has fewer than d digits, then the required number of leading 0s must be added.) Let k be the numerousness of these stable digits, which satisfy the congruence relation G(mod 10k)≡[GG](mod 10k). k=t-1, where G(t):=3↑↑t.[6] It follows that, g63 ≪ k ≪ g64. The algorithm above produces the following 500 rightmost decimal digits of Graham's number (or of any tower of more than 500 3s): ``` …02425950695064738395657479136519351798334535362521 43003540126026771622672160419810652263169355188780 38814483140652526168785095552646051071172000997092 91249544378887496062882911725063001303622934916080 25459461494578871427832350829242102091825896753560 43086993801689249889268099510169055919951195027887 17830837018340236474548882222161573228010132974509 27344594504343300901096928025352751833289884461508 94042482650181938515625357963996189939679054966380 03222348723967018485186439059104575627262464195387 ``` Notes 1. Lavrov, Mikhail; Lee, Mitchell; Mackey, John (2013). "Graham's Number is Less Than 2 ↑↑↑ 6" (PDF). arXiv:1304.6910. Retrieved 2013-04-29. 2. Martin Gardner (1977) "In which joining sets of points leads into diverse (and diverting) paths". Scientific American, November 1977 References • Graham, R. L.; Rothschild, B. L. (1971). "Ramsey's Theorem for n-Parameter Sets". Transactions of the American Mathematical Society 159: 257–292. doi:10.2307/1996010. JSTOR 1996010. The explicit formula for N appears on p. 290. This is not the "Graham's number" G published by Martin Gardner. • Graham, R. L.; Rothschild, B.L. (1978). "Ramsey Theory", Studies in Combinatorics, Rota, G.-G., ed., Mathematical Association of America, 17:80–99. On p. 90, in stating "the best available estimate" for the solution, the explicit formula for N is repeated from the 1971 paper. • Gardner, Martin (November 1977). "Mathematical Games". Scientific American 237: 18–28. ; reprinted (revised) in Gardner (2001), cited below. • Gardner, Martin (1989). Penrose Tiles to Trapdoor Ciphers. Washington, D.C.: Mathematical Association of America. ISBN 0-88385-521-6. • Gardner, Martin (2001). The Colossal Book of Mathematics: Classic Puzzles, Paradoxes, and Problems. New York, NY: Norton. ISBN 0-393-02023-1. • Exoo, Geoffrey (2003). "A Euclidean Ramsey Problem". Discrete and Computational Geometry 29 (2): 223–227. doi:10.1007/s00454-002-0780-5. Exoo refers to the Graham & Rothschild upper bound N by the term "Graham's number". This is not the "Graham's number" G published by Martin Gardner. • Barkley, Jerome (2008). "Improved lower bound on an Euclidean Ramsey problem". arXiv:0811.1055v1 [math.CO].	{"extraction_info": {"found_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": 23, "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": 0.8965920209884644, "perplexity_flag": "middle"}
http://mathhelpforum.com/discrete-math/64363-recurrence-relation-problem.html	# Thread: 1. ## Recurrence relation problem Find a recurrence relation for the number of ternary strings (0 s, 1 s and 2 s) of length n that contain none of the strings 101, 202, 102, or 201. What are the initial conditions? Use the recurrence to compute the number of such sequences of length 5. I am trying this for the past 7 days. Please help me in solving this. help would be greatly appreciated. 2. Originally Posted by kkkkkk Find a recurrence relation for the number of ternary strings (0 s, 1 s and 2 s) of length n that contain none of the strings 101, 202, 102, or 201. What are the initial conditions? Use the recurrence to compute the number of such sequences of length 5. Let $a_n$ be the number of such strings of length n. Now think about how to construct a string of length n+1 by taking such a string, call it s, and adding an additional number at the end of s. If s ends in a 1 or a 2 then we can add any of 0,1,2 and still have an allowable string. But if s ends in a 0 we need to look at the last two elements of s. If they are 00, we can add anything and still have an allowable string. But if s ends in 10 or 20 then we have to add a 0 as the (n+1)th element. Another way to say that is that we can add a 0 to any string and still have an allowable string. But we have to be more careful about adding a 1 or a 2. So the number of strings of length n ending in a 0 is $a_{n-1}$ and therefore the number ending in a 1 or a 2 is $a_n-a_{n-1}$. Now, how many strings of length n+1 are there? For each of the $a_{n-1}$ n-strings ending in a 0 we can get three (n+1)-strings. For each of the $a_{n-2}$ n-strings ending in 00 we can also get three (n+1)-strings. But for each of the $a_{n-1} - a_{n-2}$ n-strings ending in x0 (where x = 1 or 2) we can only get one (n+1)-string. Take that information and make it into a recurrence relation. 3. thankyou for your great help I would appreciate it. But Finding the recurrence relation was a great difficulty. So, can you suggest me the sequence so that I can make a recurrence relation because I couldn't follow the sequence. 4. Just to add : The initial conditions are : a1 = 3 a2 = 9 a3 = 23 5. thanks for u r help. But I couldn't establish the recurrence relation. I tried some. They were working for 1 and 2 but not from 3. So, suggest a recurrence relation that would work for all 6. If i am not wrong the recurrence relation is $a_n = a_{n-1}+4a_{n-2}+2a_{n-3}$. 7. Originally Posted by tester85 If i am not wrong the recurrence relation is $a_n = a_{n-1}+4a_{n-2}+2a_{n-3}$. The relation I arrived at (following my previous hints) is $a_n = 3a_{n-1}-2a_{n-2}+2a_{n-3}$. With the initial conditions $a_1=3$, $a_2=9$, $a_3=23$, that gives the prediction $a_4=57$, which agrees with what I get by a direct count. 8. thank u very much for u r help. I will appreciate it greatly.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 13, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.943683385848999, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/16596/about-the-ether-theory-acceptance/16598	# About the Ether Theory acceptance Why was the Ether Theory refused by Modern Physics? If you please explain me, I just wanted to understand it more. - It was a conspiracy. :=( – Georg Nov 5 '11 at 14:56 The ether was rejected for philosophical reasons. The philosophy is called "logical positivism" and it is the only thing in the whole of the field of philosophy that isn't worthless drivel. See this question: physics.stackexchange.com/questions/3798/… . The philosophy of logical positivism was too clear and resolved too many questions of philosophy definitively, so it is rejected by philosophers. This makes the whole field bankrupt, and all the practitioners incompetent. – Ron Maimon Feb 29 '12 at 16:36 4 I have to disagree. "Logical positivism" is worthless drivel, and set back the theory of quantum information by decades. – Peter Shor Mar 4 '12 at 18:05 Not to nit-pick but if someone has found any "empty space" to transmit em radiation ie light through to prove its speed is consistent please let me know. Unfortunately in all my searches I have still to find any evidence of empty space either naturally occurring or man made. – Argus Nov 20 '12 at 17:54 ## 5 Answers It's because the luminiferous aether was, by definition, composed out of some particles or elementary building blocks with a well-defined location in space. Consequently, it picks a privileged reference frame, the rest frame of the aether. In this rest frame, the speed of light – vibrations of the aether – could be constant, $c$. However, things moving relatively to this aether by the speed $v$ should detect a different speed of the light relatively to them – the speed would go from $c-v$ to $c+v$, depending on the direction. However, this modification of the light speed, the so-called aether wind, was shown to be non-existent by the Morley-Michelson experiment which measured the speed to be $c$ regardless of the source and the observer. This falsifies the existence of the aether. The equivalent but even more robust refutation of the aether came from the theory. A physicist named Albert Einstein built a whole new theory of spacetime, the so-called special theory of relativity (a picture of this physicist is often being shown by the ordinary people as well), that also assumes/guarantees that the speed of light is always constant and there can't be any privileged reference frame. Relativity has been backed by the Morley-Michelson experiment as well as hundreds of much more specific experiments. One of the things it guarantees is that light (electromagnetic radiation) has to be made out of disturbances of the empty space, the vacuum itself, and not a localized material carrier. - I'm going to argue with the wording "picks a privileged reference frame". I'm fairly sure you're implying that it is a well-founded expectation that physics does not pick a preferred reference frame. Physics, yes, but the universe, no. The reference frame defined by the co-moving coordinates (or stationary relative to the CMB) is preferred in some sense. I don't understand how anyone can argue otherwise. So I ask: what is the justification behind the requirement that no preferred reference frame exist, aside from experimental evidence relevant to E&M (a la Michelson and Morley). – AlanSE Nov 5 '11 at 22:52 Dear @Zassounotsukushi, the breaking of the local Lorentz symmetry by the cosmic microwave background is a spontaneous breaking. It's very tiny and what's more important, it can be completely undone. If you screen a vessel against the microwaves, the interior will be completely empty and no preferred inertial system may be identified by experiments inside the vessel. As long as you don't measure some very weak radiation that has nothing to do with typical other processes, it's true even in the Universe where the CMB is present. – Luboš Motl Nov 6 '11 at 7:58 "So I ask: what is the justification behind the requirement that no preferred reference frame exist" - It's the validity of a principle known as the "principle of relativity". It's one of the postulates of special relativity (the other's being the constancy of the light speed) but it's been a part of physics since Galileo. In non-relativistic mechanics, it implied the existence of the Galileo symmetry (one may add $\vec v$ to all velocities and nothing changes) which was deformed to Lorentz symmetry in the age of relativity. – Luboš Motl Nov 6 '11 at 7:59 1 Ultimately, the principle of relativity boils down to observations, and I don't mean Morley-Michelson's experiment. The relevant observations are that on the Earth's surface, we don't even detect that the Earth is moving 30 km/s in the solar system, by other speeds around its axis. In the moving train, we can't detect that we're in a moving train as long as the motion is uniform, etc. This is known from informal as well as very accurate observations and is summarized as the principle of relativity, a cornerstone of physics: physical laws have the same form if we add the velocity $\vec v$... – Luboš Motl Nov 6 '11 at 8:00 Ha-ha! Disturbances of the empty space! How convincing! – Vladimir Kalitvianski Nov 7 '11 at 10:52 show 2 more comments It's not entirely true that ether theory was refused by modern physics, but seen as superfluous and over-complicated compared to special relativity's notion of time and space. There's a version called Lorentz ether theory which postulates that it's not possible to detect the absolute ether, and gives the same results as predicted by special relativity. Hence it's not possible to experimentally determine the difference between the two theories. But whereas LET is constructed in an hoc way by adding time dilation and Lorenz contraction into the theory to make the ether undetectable, these follow from the two postulates of SR. So the sheer elegance and simplicity in favour of SR is why it's taken more seriously than LET. - There are a number of experiments that disproved that there exists an inertial frame against which everything is moving, and electromagnetic waves propagate, the luminiferous aether. There exists a wikipedia article that covers these questions. - Anna, that's a really good reference, well worth reading. I liked in particular its matter-of-fact explanation of Einstein's later introduction of an updated ether concept, which often has not been discussed or even mentioned in courses on relativity. Open discussion is better, I think. – Terry Bollinger Jul 7 '12 at 16:31 There have been many attempts over the past century to answer this question, all of them incorrect as far as I can see. The most common refutation of ether (or aether) theory is based on the null result of the Michelson-Morley experiment (about which there is no doubt, the experiment having been refined and repeated many times with the same null result). M-M could be excused for it, but there is no excuse for modern physicists making the same mistake. What everyone seems to forget is that solid matter is mostly "empty" space - the separation between atomic nuclei is about 10^5 times their diameter, as discovered by Rutherford in 1910. Also, we now know that the atoms are held in place by electromagetic fields, so that the lengths of the arms of M-M's interferometer would be subject to the same effects of different light (e-m field) speeds as the interfering light beams themselves. This ("Lorentz contraction") is not an ad hoc effect, but follows from the finite speed of field changes within solids, limited to the same speed as the light beams bouncing to and fro among the interferometer mirrors. This physical explanation for Lorentz contraction is set out well by Feynman in his Lectures on Physics, Vol II, Sec. 21-6 on retarded potential effects. The only physical constraint needed to explain relativistic length and clock/time effects in Lorentz Ether Theory (LET) is that light is the fastest signal in empty space. In practice, the ether as a unique reference frame for e-m field (e.g. light) propagation can be ignored because the experimental predictions of LET are always identical with those of Special Relativity (SR). But whereas SR offers no physical explanation for relativity beyond asserting the principle, LET does, which should surely be more satisfactory for physicists! Any inertial observer can assume that the speed of light is the same in all directions around him/her, but so can any other inertial observer, and they can not all be right, as is shown by the fact that they will disagree about estimates of the timing of simultaneous events, for example! LET shows exactly how this works in terms of the speed of light as an upper limit, as was known by Poincare in his 1904 St Louis World's Fair lecture "Science et Hypothese", in which he speculated about the collapse of relativity as a universal principle if anything can go faster than light. By about 1920, Einstein had realised his error in declaring the ether superfluous as a concept, because he had realised that a unique, universal reference frame is implied by General Relativity in which the speed of light becomes locally dependent on gravitational potential around masses, instead of being absolutely fixed in value everywhere as in SR and LET. But noone seems to have been ready for that backflip. Instead, conceptual nonsenses like "curved space" are now preferred to a straighforward dependence of light speed on position in a gravitational field! In summary, LET = SR in all predictions, but in my view LET is superior because it contains a physical explanation (light speed in a unique inertial reference frame as an upper limit), rather than relying on a statement of a principle - how do atoms know about principles? The next interesting question is how does "empty space" (or "ether", to give a name for the concept of a space with physical properties) control the speed of light, between and inside atoms, and as slowed down near large masses? Does Higgs Boson theory shed any light on this (sorry!)? (See trevm42 on Youtube) - – Terry Bollinger Jul 7 '12 at 15:55 There never was really any experimental or observation evidence that the luminous aether ever existed in the first place. It was merely an invention to pave over a gap in the Newtonian model of light. Up until the mid-19th century everyone thought that Newton's experiments with prisms and slits had conclusively demonstrated that light was a wave phenomena i.e. it behaved just like waves in water. That, however, raised the question of "waves in what?" The answer was the "aether", something that couldn't be seen, touched or measured but whose motion transmitted light just like the motion of water transmitted the force of a wave. Basically, the luminous aether was just a concept that the era's scientist pulled out of their collective… er, hats. By contrast, Leibniz argued, purely on philosophical grounds, that light must be made of particles. Leibniz lost the argument because his hypothesis didn't have the experimental predictive power that Newton's did. However, if Leibniz's idea had been the dominate idea, the concept of luminous aether would have never been invented in the first place. Prior to the late-19th century the idea of "subtle fluids" was evoked to explain many phenomena such as heat and electricity. That is why even today we talk of electricity "flowing" in "currents." The history of science is filled with such inventions that paper over gaps in existing knowledge. - ## protected by Qmechanic♦Mar 3 at 18:54 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": 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": 0.3, "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": 0.9609200358390808, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/21729/charge-of-a-field-under-the-action-of-a-group	# Charge of a field under the action of a group What does it mean for a field (say, $\phi$) to have a charge (say, $Q$) under the action of a group (say, $U(1)$)? - ## 1 Answer It means that 1. the charge operator $Q$ is a Lie algebra generator for some Lie group $G$. 2. the field $\phi\in V$ takes values/transform in a representation $V$ of the Lie group. (Note that any Lie group representation $V$ is also Lie algebra representation of the corresponding Lie algebra.) 3. the charge operator $\rho(Q)$ in the representation $\rho: G \to GL(V)$ is proportional to the identity. The proportionality factor/eigenvalue is the actual charge of the field. To see an example of this, say the (strong) $u(1)$ hypercharge $Y$, see this answer. - Thank you, it helped a lot! – John Smith Mar 2 '12 at 16:05	{"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": 12, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8808901309967041, "perplexity_flag": "head"}
http://mathhelpforum.com/calculus/154462-book-explaining-calculus.html	# Thread: 1. ## Book Explaining Calculus Hi all, I am searching for a book on Calculus which explains in detail and with diagrams how the equation/formula/method was derived. All books I've read in the past merely explain a systematic approach towards solving a mathematical problem, but I have not learned how it works; a simple example, what is sine and how does one go about re-discovering sine? Does such a book exist? I don't want a book on the history of mathematics, I would like a book which teaches it to me and explains how and why it works. I want to UNDERSTAND mathematics, not just do it! 2. I don't really understand what you're asking. You want to "discover" sine, but that's not calculus (at least, not to begin with). Calculus is the study of limits, rates, integrals etc. What are you asking for exactly? 3. A book on mathematics then I suppose; I saw sine as a component of calculus. Have you never questioned how sine was derived and why it works? That is the stage in life I am at and I want answers . I watched a good lecture about linear algebra and it showed me that there exists people who explain mathematics and why the methods work the way they do. 4. The definition of sine comes from the unit circle. Draw a circle of radius 1 on a set of Cartesian axes, centred at the origin. Draw a segment from the origin to the circle (i.e. a radius). This creates an angle with the positive $x$ axis in the anticlockwise direction, call it $\theta$. Draw a segment from where your radius touches the circle to the $x$ axis (perpendicular to the $x$ axis). This length is $\sin{\theta}$. Similarly, the length that goes from where sine touches the $x$ axis to the origin along the $x$ axis is $\cos{\theta}$. It is called cosine, because as you change your angle and the sine changes, this length changes along with it. Since it works with sine, it's the cosine. Does that make sense so far? 5. ...how sine was derived... But it's not exactly derived from more basic principles. It's a definition: in a right triangle that also has an angle $0<\theta<\pi/2$, the sine of $\theta$ is defined to be the length of the side opposite to $\theta$ divided by the length of the hypotenuse. That is the definition of sine. One book you might try is Mathematics for the Nonmathematician, by Morris Kline. Kline (1908-1992) was an excellent expositor. 6. Why must the radius be 1? 7. Originally Posted by Ackbeet But it's not exactly derived from more basic principles. It's a definition: in a right triangle that also has an angle $0<\theta<\pi/2$, the sine of $\theta$ is defined to be the length of the side opposite to $\theta$ divided by the length of the hypotenuse. That is the definition of sine. One book you might try is Mathematics for the Nonmathematician, by Morris Kline. Kline (1908-1992) was an excellent expositor. I am definitely going to look into that book. It sounds like it achieves what I want to know. Any others like this you can recommend? 8. Originally Posted by webguy Why must the radius be 1? Because $\sin{\theta}$ is defined to be the length of the perpendicular on the UNIT circle - a circle of radius 1.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 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": 0.9523962736129761, "perplexity_flag": "head"}
http://mathhelpforum.com/number-theory/159292-algebraic-numbers.html	# Thread: 1. ## Algebraic numbers Here goes: I have proven that $\sqrt[3]{2}+\sqrt{3}$ is algebraic number by definition (constructed a polynomial of sixth degree having that number as a root). Could someone please give me a sketch of proof (or the proof itself) for the following claim. I'd appreciate it. The claim is: If $x$ is algebraic number prove that $\frac{1}{x}$ also is algebraic number. Now I know a little bit of higher math/algebra and I know that algebraic numbers form a field and thus a quotient of two algebraic numbers is algebraic number. Is there a proof that does not require the axioms of field? Can someone give me an advice (or two)? Thx in advance. 2. Originally Posted by MathoMan Here goes: I have proven that $\sqrt[3]{2}+\sqrt{3}$ is algebraic number by definition (constructed a polynomial of sixth degree having that number as a root). I would start with something like $(x- \sqrt[3]{2}- \sqrt{3})(x- \sqrt[3]{2}+ \sqrt{3})= (x- \sqrt[3]{2})^2- 3= x^2- 2\sqrt[3]{2}x+ 3/4- 2= x^2- 5/4- 2\sqrt[3]{2}$ where I have used " $(a- b)(a+ b)= a^2- b^2$". Now use the fact that $(a- b)(a^2+ ab+ b^2)= a^3- b^3$ to get rid of that cube root. Could someone please give me a sketch of proof (or the proof itself) for the following claim. I'd appreciate it. The claim is: If $x$ is algebraic number prove that $\frac{1}{x}$ also is algebraic number. Now I know a little bit of higher math/algebra and I know that algebraic numbers form a field and thus a quotient of two algebraic numbers is algebraic number. Is there a proof that does not require the axioms of field? Can someone give me an advice (or two)? Thx in advance. You say you know that the "algebraic number form a field" and the field axioms but do you know the definition of "algebraic number". If x is an algebraic number, then there exist in integer, n, and integers, $a_n$, $a_{n-1}$, ..., $a_1$ and $a_0$ such that $a_nx^n+ a_{n-1}a^{n-1}+ \cdot\cdot\cdot+ a_1x+ a_0= 0$. Okay, what do you get if you divide each term in that equation by $x^h$? Or, since you know "a quotient of two algebraic numbers is algebraic number", 1/x is the quotient of 1 and x. Do you know that 1 is an algebraic number. By the way, "a quotient of two algebraic numbers is algebraic number" is NOT strictly true. The denominator has to be non-zero. In fact, the original claim "If "x" is algebraic number then 1/x also is algebraic number" is NOT true. A counter example is x= 0. What is true is "If "x" is a non-zero algebraic number, then 1/x is also an algebraic number". 3. Thanks for the post. I am aware of all the things you've mentioned. Maybe I wasn't precise enough. As you said: What is true is "If "x" is a non-zero algebraic number, then 1/x is also an algebraic number". I was looking for a way to prove that statement without using the obvious: that 1 is algebraic and that x is assumed to be algebraic and since algebraic numbers form a field hence 1/x must be algebraic number. I hope we understand each other now. Thanks anyway.	{"extraction_info": {"found_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": 0.9584085941314697, "perplexity_flag": "head"}
http://mathoverflow.net/revisions/51263/list	## Return to Question 8 edited body If $P(x,y,...,z)$ is a polynomial with integer coefficients then every integer solution of $P=0$ corresponds to a homomorphism from $\mathbb{Z}[x,y,...,z]/(P)$ to $\mathbb{Z}$. So there are infinitely many solutions iff there are infinitely many homomorphisms. If $P$ is homogeneous, we consider solutions up to a scalar factor. Now if $G$ is a finitely generated group and $\Gamma=\langle X\mid r_1,...,r_n\rangle$ is another group, then any solution of the system of equations $r_1=1,...,r_n=1$ in $G$ corresponds to a homomorphism $\Gamma\to G$. We usually consider homomorphisms up to conjugation in $G$. Now it is well known that if there are infinitely many homomorphisms from $\Gamma$ to $G$, then $\Gamma$ acts non-trivially (without a globally fixed point) by isometries on the asymptotic cone of $G$ (i.e. the limit of metric spaces $G,G/d_1, G/d_2,...$ where $d_i\to \infty$, $G/d_i$ is the set $G$ with the word metric rescaled by $d_i$; here $d_i$ is easily computed from the $i$-th homomorphism $\Gamma\to G$: given a homomorphism $\phi$, we can define an action of $\Gamma$ on $G$: $\gamma\circ g=\phi(\gamma)g$; the number $d_i$ is the largest number such that every $g\in G$ is moved by one of the generators by at least $d_i$ in the word metric). Question. Is there a similar object for Diophantine equations, that is if $P$ has infinitely many integral solutions, then $\mathbb{Z}[x,y,...,z]/(P)$ "acts" on something, not necessarily a metric space. The reason for my question is that several statements in group theory and the theory of Diophantine equations assert finiteness of the number of solutions. For example, Faltings' theorem states that if the genus of $P$ is high enough, then the equation $P=0$ has only finite number of solutions. Similarly, if $\Gamma$ is a lattice in a higher rank semi-simple Lie group, then the number of homomorphisms from $\Gamma$ to a hyperbolic group is finite (the latter fact follows because the asymptotic cone of a hyperbolic group is an $\mathbb{R}$-tree, and a group with Kazhdan property (T) cannot act non-trivially on an $\mathbb{R}$-tree). Update: To make the question more concrete, consider one of the easiest (comparing to the other statements) finiteness results about Diophantine equations. Let $P(x,y)$ be a homogeneous polynomial. If the degree of $P$ is at least 3 and $P$ is not a product of two polynomials with integer coefficients, then for every integer $p\ne m\ne 0$ the equation $P(x,y)=p$ P(x,y)=m$has only finitely many integer solutions. It is the Thue's theorem. Note that for degree 2 the statement is false because of the Pell equation$x^2-2y^2=1\$. The standard proof of Thue's theorem is this. Let the degree of $P$ be 3 (the general case is similar). Represent $P$ as $d(x-ay)(x-by)(x-cy)$ where $a,b,c$ are the roots none of which is rational by assumption. Then we should have $\|(x/y-a)\|\cdot \|(x/y-b)\|\cdot \|(x/y-c)\|=O(1)/\|y^3\|$ for infinitely many integers $x,y$. Then the right hand side can be made arbitrarily small. Note that if one of the factors in the left hand side is small, the other factors are $O(1)$ (all roots are different). Hence we have that $\|x/y-a\|=O(1)/y^3$ (or the same with $b$ or $c$). But for all but finitely many $x,y$ we have $\|x/y-a\|\ge C/y^{5/2+\epsilon}$ for any $\epsilon>0$ by another theorem of Thue (a "bad" approximation property of algebraic numbers), a contradiction. The question is then: is there an asymptotic geometry proof of the Thue theorem. 7 added 4 characters in body If $P(x,y,...,z)$ is a polynomial with integer coefficients then every integer solution of $P=0$ corresponds to a homomorphism from $\mathbb{Z}[x,y,...,z]/(P)$ to $\mathbb{Z}$. So there are infinitely many solutions iff there are infinitely many homomorphisms. If $P$ is homogeneous, we consider solutions up to a scalar factor. Now if $G$ is a finitely generated group and $\Gamma=\langle X\mid r_1,...,r_n\rangle$ is another group, then any solution of the system of equations $r_1=1,...,r_n=1$ in $G$ corresponds to a homomorphism $\Gamma\to G$. We usually consider homomorphisms up to conjugation in $G$. Now it is well known that if there are infinitely many homomorphisms from $\Gamma$ to $G$, then $\Gamma$ acts non-trivially (without a globally fixed point) by isometries on the asymptotic cone of $G$ (i.e. the limit of metric spaces $G,G/d_1, G/d_2,...$ where $d_i\to \infty$, $G/d_i$ is the set $G$ with the word metric rescaled by $d_i$; here $d_i$ is easily computed from the $i$-th homomorphism $\Gamma\to G$: given a homomorphism $\phi$, we can define an action of $\Gamma$ on $G$: $\gamma\circ g=\phi(\gamma)g$; the number $d_i$ is the largest number such that every $g\in G$ is moved by one of the generators by at least $d_i$ in the word metric). Question. Is there a similar object for Diophantine equations, that is if $P$ has infinitely many integral solutions, then $\mathbb{Z}[x,y,...,z]/(P)$ "acts" on something, not necessarily a metric space. The reason for my question is that several statements in group theory and the theory of Diophantine equations assert finiteness of the number of solutions. For example, Faltings' theorem states that if the genus of $P$ is high enough, then the equation $P=0$ has only finite number of solutions. Similarly, if $\Gamma$ is a lattice in a higher rank semi-simple Lie group, then the number of homomorphisms from $\Gamma$ to a hyperbolic group is finite (the latter fact follows because the asymptotic cone of a hyperbolic group is an $\mathbb{R}$-tree, and a group with Kazhdan property (T) cannot act non-trivially on an $\mathbb{R}$-tree). Update: To make the question more concrete, consider one of the easiest (comparing to other statements) finiteness results about Diophantine equations. Let $P(x,y)$ be a homogeneous polynomial. If the degree of $P$ is at least 3 and $P$ is not a product of two polynomials with integer coefficients, then for every integer $p\ne 0$ the equation $P(x,y)=p$ has only finitely many integer solutions. It is the Thue's theorem. Note that for degree 2 the statement is false because of the Pell equation $x^2-2y^2=1$. The standard proof of Thue's theorem is this. Let the degree of $P$ be 3 (the general case is similar). Represent $P$ as $d(x-ay)(x-by)(x-cy)$ where $a,b,c$ are the roots none of which is rational by assumption. Then we should have $\|(x/y-a)\|\cdot \|(x/y-b)\|\cdot \|(x/y-c)\|=O(1)/\|y^3\|$ for infinitely many integers $x,y$. Then the right hand side can be made arbitrarily small. Note that if one of the factors in the left hand side is small, the other factors are $O(1)$ (all roots are different). Hence we have that $\|x/y-a\|=O(1)/y^3$ (or the same with $b$ or $c$). But for all but finitely many $x,y$ we have $\|x/y-a\|\ge C/y^{5/2+\epsilon}$ for any $\epsilon>0$ by another theorem of Thue (a "bad" approximation property of algebraic numbers), a contradiction. The question is then: is there an asymptotic geometry proof of the Thue theorem. 6 added 1334 characters in body; added 49 characters in body Update: To make the question more concrete, consider one of the easiest (comparing to other statements) finiteness results about Diophantine equations. Let $P(x,y)$ be a homogeneous polynomial. If the degree of $P$ is at least 3 and is not a product of two polynomials with integer coefficients, then for every integer $p\ne 0$ the equation $P(x,y)=p$ has only finitely many integer solutions. It is the Thue's theorem. Note that for degree 2 the statement is false because of the Pell equation $x^2-2y^2=1$. The standard proof of Thue's theorem is this. Let the degree of $P$ be 3 (the general case is similar). Represent $P$ as $d(x-ay)(x-by)(x-cy)$ where $a,b,c$ are the roots none of which is rational by assumption. Then we should have $\|(x/y-a)\|\cdot \|(x/y-b)\|\cdot \|(x/y-c)\|=O(1)/\|y^3\|$ for infinitely many integers $x,y$. Then the right hand side can be made arbitrarily small. Note that if one of the factors in the left hand side is small, the other factors are $O(1)$ (all roots are different). Hence we have that $\|x/y-a\|=O(1)/y^3$ (or the same with $b$ or $c$). But for all but finitely many $x,y$ we have $\|x/y-a\|\ge C/y^{5/2+\epsilon}$ for any $\epsilon>0$ by another theorem of Thue (a "bad" approximation property of algebraic numbers), a contradiction. The question is then: is there an asymptotic geometry proof of the Thue theorem. 5 edited tags 4 deleted 1 characters in body If $P(x,y,...,z)$ is a polynomial with integer coefficients then every integer solution of $P=0$ corresponds to a homomorphism from $\mathbb{Z}[x,y,...,z]/(P)$ to $\mathbb{Z}$. So there are infinitely many solutions iff there are infinitely many homomorphisms. If $P$ is homogeneous, we consider solutions up to a scalar factor. Now if $G$ is a finitely generated group and $\Gamma=\langle X\mid r_1,...,r_n\rangle$ is another group, then any solution of the system of equations $r_1=1,...,r_n=1$ in $G$ corresponds to a homomorphism $\Gamma\to G$. We usually consider homomorphisms up to conjugation in $G$. Now it is well known that if there are infinitely many homomorphisms from $\Gamma$ to $G$, then $\Gamma$ acts non-trivially (without a globally fixed point) by isometries on the asymptotic cone of $G$ (i.e. the limit of metric spaces $G,G/d_1, G/d_2,...$ where $d_i\to \infty$, $G/d_i$ is the set $G$ with the word metric rescaled by $d_i$; here $d_i$ is easily computed from the $i$-th homomorphism $\Gamma\to G$: given a homomorphism $\phi$, we can define an action of $\Gamma$ on $G$: $\gamma\circ g=\phi(\gamma)g$; the number $d_i$ is the smallest largest number such that every $g\in G$ is moved by one of the generators by at least $d_i$ in the word metric). Question. Is there a similar object for Diophantine equations, that is if $P$ has infinitely many integral solutions, then $\mathbb{Z}[x,y,...,z]/(P)$ "acts" on something, not necessarily a metric space. The reason for my question is that several statements in group theory and the theory of Diophantine equations assert finiteness of the number of solutions. For example, Faltings' theorem states that if the genus of $P$ is high enough, then the equation $P=0$ has only finite number of solutions. Similarly, if $\Gamma$ is a lattice in a higher rank semi-simple Lie group, then the number of homomorphisms from $\Gamma$ to a hyperbolic group is finite (the latter fact follows because the asymptotic cone of a hyperbolic group is an $\mathbb{R}$-tree, and a group with Kazhdan property (T) cannot act non-trivially on an $\mathbb{R}$-tree). 3 added 207 characters in body If $P(x,y,...,z)$ is a polynomial with integer coefficients then every integer solution of $P=0$ corresponds to a homomorphism from $\mathbb{Z}[x,y,...,z]/(P)$ to $\mathbb{Z}$. So there are infinitely many solutions iff there are infinitely many homomorphisms. If $P$ is homogeneous, we consider solutions up to a scalar factor. Now if $G$ is a finitely generated group and $\Gamma=\langle X\mid r_1,...,r_n\rangle$ is another group, then any solution of the system of equations $r_1=1,...,r_n=1$ in $G$ corresponds to a homomorphism $\Gamma\to G$. We usually consider homomorphisms up to conjugation in $G$. Now it is well known that if there are infinitely many homomorphisms from $\Gamma$ to $G$, then $\Gamma$ acts non-trivially (without a globally fixed point) by isometries on the asymptotic cone of $G$ (i.e. the limit of metric spaces $G,G/d_1, G/d_2,...$ where $d_i\to \infty$, $G/d_i$ is the set $G$ with the word metric rescaled by $d_i$; here $d_i$ is easily computed from the $i$-th homomorphism $\Gamma\to G$(it : given a homomorphism $\phi$, we can define an action of $\Gamma$ on $G$: $\gamma\circ g=\phi(\gamma)g$; the number $d_i$ is the "displacement" number)smallest number such that every $g\in G$ is moved by one of the generators by at least $d_i$ in the word metric). Question. Is there a similar object for Diophantine equations, that is if $P$ has infinitely many integral solutions, then $\mathbb{Z}[x,y,...,z]/(P)$ "acts" on something, not necessarily a metric space. The reason for my question is that several statements in group theory and the theory of Diophantine equations assert finiteness of the number of solutions. For example, Faltings' theorem states that if the genus of $P$ is high enough, then the equation $P=0$ has only finite number of solutions. Similarly, if $\Gamma$ is a lattice in a higher rank semi-simple Lie group, then the number of homomorphisms from $\Gamma$ to a hyperbolic group is finite (the latter fact follows because the asymptotic cone of a hyperbolic group is an $\mathbb{R}$-tree, and a group with Kazhdan property (T) cannot act non-trivially on an $\mathbb{R}$-tree). 2 added 8 characters in body If $P(x,y,...,z)$ is a polynomial with integer coefficients then every integer solution of $P=0$ corresponds to a homomorphism from $\mathbb{Z}[x,y,...,z]/(P)$ to $\mathbb{Z}$. So there are infinitely many solutions iff there are infinitely many homomorphisms. If $P$ is homogeneous, we consider solutions up to a scalar factor. Now if $G$ is a finitely generated group and $\Gamma=\langle X\mid r_1,...,r_n\rangle$ is another group, then any solution of the system of equations $r_1=1,...,r_n=1$ in $G$ corresponds to a homomorphism $\Gamma\to G$. We usually consider homomorphisms up to conjugation in $G$. Now it is well known that if there are infinitely many homomorphisms from $\Gamma$ to $G$, then $\Gamma$ acts non-trivially (without a globally fixed point) by isometries on the asymptotic cone of $G$ (i.e. the limit of metric spaces $G,G/d_1, G/d_2,...$ where $d_i\to \infty$, $G/d_i$ is the set $G$ with the word metric rescaled by $d_i$; here $d_i$ is easily computed from the $i$-th homomorphism $\Gamma\to G$ (it is the "displacement" number). Question. Is there a similar object for Diophantine equations, that is if $P$ has infinitely many integral solutions, then $\mathbb{Z}[x,y,...,z]/(P)$ "acts" on something, not necessarily a metric space. The reason for my question is that several statements in group theory and the theory of Diophantine equations assert finiteness of the number of solutions. For example, Faltings' theorem states that if the genus of $P$ is high enough, then the equation $P=0$ has only finite number of solutions. Similarly, if $\Gamma$ is a lattice in a higher rank semi-simple Lie group, then the number of homomorphisms from $\Gamma$ to the free a hyperbolic group is finite (the latter fact follows because the asymptotic cone of the free a hyperbolic group is an $\mathbb{R}$-tree, and a group with Kazhdan property (T) cannot act non-trivially on an $\mathbb{R}$-tree). 1 # Infinitely many solutions of a diophantine equation If $P(x,y,...,z)$ is a polynomial with integer coefficients then every integer solution of $P=0$ corresponds to a homomorphism from $\mathbb{Z}[x,y,...,z]/(P)$ to $\mathbb{Z}$. So there are infinitely many solutions iff there are infinitely many homomorphisms. If $P$ is homogeneous, we consider solutions up to a scalar factor. Now if $G$ is a finitely generated group and $\Gamma=\langle X\mid r_1,...,r_n\rangle$ is another group, then any solution of the system of equations $r_1=1,...,r_n=1$ in $G$ corresponds to a homomorphism $\Gamma\to G$. We usually consider homomorphisms up to conjugation in $G$. Now it is well known that if there are infinitely many homomorphisms from $\Gamma$ to $G$, then $\Gamma$ acts non-trivially (without a globally fixed point) by isometries on the asymptotic cone of $G$ (i.e. the limit of metric spaces $G,G/d_1, G/d_2,...$ where $d_i\to \infty$, $G/d_i$ is the set $G$ with the word metric rescaled by $d_i$; here $d_i$ is easily computed from the $i$-th homomorphism $\Gamma\to G$ (it is the "displacement" number). Question. Is there a similar object for Diophantine equations, that is if $P$ has infinitely many integral solutions, then $\mathbb{Z}[x,y,...,z]/(P)$ "acts" on something, not necessarily a metric space. The reason for my question is that several statements in group theory and the theory of Diophantine equations assert finiteness of the number of solutions. For example, Faltings' theorem states that if the genus of $P$ is high enough, then the equation $P=0$ has only finite number of solutions. Similarly, if $\Gamma$ is a lattice in a higher rank semi-simple Lie group, then the number of homomorphisms from $\Gamma$ to the free group is finite (the latter fact follows because the asymptotic cone of the free group is an $\mathbb{R}$-tree, and a group with Kazhdan property (T) cannot act non-trivially on an $\mathbb{R}$-tree).	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 270, "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": 0.9414748549461365, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/51491/whats-the-role-of-classically-forbidden-paths-in-path-integral	# What's the role of classically forbidden paths in path integral? I'm interested in how and how much classically-forbidden paths contribute to a path integral? Is there any good reference on the issue? Any discussion in QM or QFT context would be appreciated. EDIT: I have a more specific question than the above but initially it was a reference request so I decided to make it more general. Now it doesn't seem appropriate any more so let me reformulate the question: First of all let me apologize for the bad terminology. By "classically forbidden" I actually meant "differentiable"(i.e. for QM differentiable in time direction, for QFT in both space and time direction) instead of "forbidden by classical dynamics". My motivation comes from path integral of QED, if we only integrate the fermionic degrees of freedom under some smooth gauge field, we will get a quantized theory of many electrons with a classical gauge background, and the fully quantized theory will emerge after we also integrate over gauge fields. This seems to be a reasonable way of thinking, but some of my subsequent derivations seem to suggest some quantum effects will disappear such as photon-photon scattering, but something is still preserved like the many-body feature of QFT(I'm not very sure about calculation yet so I'd rather not show it here). It occured to me it might be because I'm only including smooth backgrounds. This motivates me to ask, what exactly is the role of smooth and non-smooth paths in path integral? Do they result in different and isolated features of QFT so that it's ok to consider them separately, or do their effects just mix with each other so that we always have to consider them as a whole? Last but not least, the comments and answers below remind me of another question, if the classical path(this time I mean path predicted by classical dynamics) always contributes 0 to the path integral for any value of $\hbar$, then what do we mean by saying the classical path will dominate in $\hbar\to0$ limit? After all a simple fact of math is that a sequence of 0's cannot give you a limit of 1. - What do you mean by classically forbidden? In quantum mechanics, for example, the set of classical paths is a set of measure zero for the path integral. In fact, even the set of differentiable paths has measure zero. – user1504 Jan 17 at 21:59 @Gugg: I think your answer looks correct, but the question seems a little vague. (Not sure why someone downvoted it. Have an upvote.) My comment isn't technically an answer: I asked Jia to clarify the question because I wasn't completely sure if we're discussing support of path integral measures or something like tunnelling. – user1504 Jan 17 at 22:57 Dear @Jia Yiyang. I suggest to roll back the question formulation to version 2. The notion of classically forbidden traditionally refers to quantum tunneling, but here I would (if I had no prior information) assume it refers to situations where there are no classical paths. In any case, the 4 answers so far have paid no attention to your later revisions $\geq 3$, and only few readers would associate classically forbidden with non-smoothness. After a roll-back, you could post your remaining question about non-smoothness in a new separate post using correct terminology. – Qmechanic♦ Feb 8 at 22:46 ## 4 Answers Other people have already addressed quantum mechanics, so let me comment on the field theory case. In all of the QFTs which have been rigorously constructed, in spacetime dimension 2 & 3, the Euclidean path integral is supported on a space of distributions. The set of continuous classical fields sits inside this space of distributions, but it has measure zero with respect to the path integral. I see no reason to expect the QFTs that describe real world physics to be any better behaved. The path integral measure has to be supported on distributions to give an OPE with short distance singularities. So yes, just summing over classical fields will probably not give you a good approximation to the path integral. The only reference I know on this stuff is Glimm & Jaffe. (There may be more accessible references somewhere in the literature. I just don't know them.) - So if we ignore all the differentiable paths, the physics remains the same? These just all seem too fuzzy to me, I can't help wondering if there's some subtlety. – Jia Yiyang Jan 21 at 16:43 It's really a physical thing. If we didn't ignore the differentiable paths, we'd find that the commutation relations for position and momentum were altered. – user1504 Jan 21 at 17:22 I'm more confused, are you saying not just that differentiable paths can be ignored, but they have to be ignored? Or else the commutation relation will be wrong? – Jia Yiyang Jan 22 at 6:00 Yes. The commutation relations can be derived from the path integral. If the path integral gives non-zero weight to the differentiable paths, the commutation relations will be altered. (The portion of the measure supported on differentiable paths predicts that position and momentum commute. Non-differentiability of paths is essential for deriving the commutation relations.) – user1504 Jan 22 at 14:36 I see what you mean, and I agree with the commutation relation part. However, does the fact that differentiable paths are of measure 0 necessarily implies these paths contribute nothing to the physics? One thing for example, how does the classical path (which is differentiable) emerge from the theory in $\hbar\to0$ limit? – Jia Yiyang Jan 22 at 19:06 Unfortunately, after the edit to the question, this answer only refers to the final "last but not least" part of it. In certain experiments there is no classical path that leads to the outcome. In such cases classically-forbidden paths contribute 100% to the path integral. See Feynman's QED, fig. 27. The picture is on the internet. The reverse "might" be said of (most) macroscopic objects (in particular not QM-experimental set-ups). There the classical path would in itself describe the total path integral. (But they would't contribute. See below.) But I take it that whenever there are non-classical paths, you could always take out the exact classical paths and still end up with the same path integral, because the classical paths would have measure $0$. So that would mean that, if there are (a measurable amount of) non-classical paths, they contribute 100%. But there are always (a measurable amount of) non-classical paths, therefore: Classically forbidden paths contribute 100% to a path integral. (Plenty of handwaving here, e.g., you might want to show that all non-classical paths together don't destructively interfere. But they don't, because then there wouldn't be any amplitude left.) Added after the edit to the question: I gather this picture, also taken from Feynman's QED, should clear up the "last but not least" part. (And in the limit, amplitudes are amplitude densities.) To me, "dominate" appears to be the wrong word. I think it is the other way around: The nearby paths determine the classical path (in situations where it is OK to use CM). - I don't know if I understand the picture correctly, it seems to suggest all the paths sum up to the classical path, and it tells us nothing about the how the transition from quantum to classical happens as $\hbar\to 0$ – Jia Yiyang Jan 21 at 16:49 In Fig. 32, the path sum (of amplitudes) consists of 7 paths represented by the 7 small arrow sticked together in the picture on the right. The little arrow $D$ represents the classical path. As you start to move towards the path integral the number of paths and therefore little arrows will increase, however combined they will keep roughly forming the same "S" shape. Therefore, more arrows means that they each become smaller, that is, their individual contribution to the path sum will decrease. [nxt cmt] – Gugg Jan 21 at 22:27 In particular, the arrow that represents the classical path will decrease in amplitude (and it will stay roughly where it is). In the limit, the amplitude of that arrow will have zero contribution, but we still have the "S" shape. So, we still have full amplitude for moving from $S$ to $P$, which means that the "illusion" of CM seems retrieved, but it remains an illusion since the contribution of the classical path $D$ has measure zero... I hope that helps. – Gugg Jan 21 at 22:33 The point being: Once your in the limit, you can remove the arrow associated with the classical path from the "S" shape, and it wouldn't change the "S" shape. (And keep in mind that sometimes no classical path even exists.) – Gugg Jan 21 at 22:39 You might want to check Quantum mechanics and classical physics, Correspondence principle, Classical limit, Quantum decoherence. – Gugg Jan 22 at 8:16 show 3 more comments I think I know what you are thinking. Here's how it goes: in classical mechanics, we have a Lagrangian describing the system. Our principle of least action says that the system will follow a path that extremizes $S = \int L dt$. This amounts to taking the E-L equations on it. The resulting path is called the classical path and we say that it is the only path that the system follows. In QM, the Feynman path integral says, let us talk about the amplitude for a particle to go from $a$ to $b$. Let us call this $K(b,a)$. Heuristically this is given by $K(b,a) = \sum_{all paths} \phi[x(t)]$, where the contribution of each path that goes from $a$ to $b$ is $\phi[x(t)] = \text{const} e^{i/\hbar S[x(t)]}$, where $S[x(t)]$ is the classical action of that path. So to answer your question, how much do classically forbidden paths contribute (and I'll add, to the transition amplitude)? It's just $e^{i/\hbar S[x(t)]}$. Now how do we recover classical mechanics? Send $\hbar \to 0$, and we see that the most contribution to the sum comes from the smallest value of $S$, as $e^{i/\hbar S[x(t)]}$ oscillates the least. (This is, or is like, the saddle point approximation). The path that gives the most contribution is the classical path, and it is that path that the particle takes classically. Of course that wasn't very rigorous, but I would recommend 'Quantum mechanics and path integrals' by the guy who came up with this himself, Richard P. Feynman. Cheers. - Doesn't this assume that there is a classical path to the outcome? – Gugg Jan 17 at 22:38 No? since when? If you're asking about whether there will be solutions to the EL eqn, I think that's given by some regularity conditions of the Lagrangian. there are systems where we start with some Lagrangian which doesn't have a classical analogue, but the path integral formalism still holds. – nervxxx Jan 17 at 22:57 Since "Now how do we recover classical mechanics?" seems to assume that CM always holds. And, more importantly, even when CM can be recovered, it doesn't mean that the classical path contributes to the path integral measurably. (And I agree with "[…] the path integral formalism still holds.") – Gugg Jan 17 at 23:21 It doesn't have to be the "smallest value of S", classical paths must only be extrema of the action functional – Raphael R. Jan 18 at 4:08 lol you guys are reading too much into this. I did say it's not very rigorous, but I was trying to convey the idea of the path integral in the spirit that Feynman might have. And yes, it's true that it should be more appropriately named principle of stationary action, but I think for many cases the action functional also happens to be the least. With regard to recovering CM, the path integral should, or rather, must reduce to CM in the cases when the Lagrangian is 'derived' from a classical system. i was just showing heuristically how it does so. – nervxxx Jan 18 at 4:23 QM deals with waves, and for waves every point of space takes part in creating the resulting wave. The same is valid for a path-integral formulation. It is difficult to present a general weight of these or those paths. It depends. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 23, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9469743967056274, "perplexity_flag": "head"}
http://mathoverflow.net/questions/21616/provability-of-termination-whats-wrong-with-my-reasoning/21626	Provability of termination. Whats wrong with my reasoning? Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Let A be a pair (i, x) $H(A)$ <=> program i halts on input x $P(A)$ <=> (there exists a proof for $H(A)$) $\vee$ (there exists a proof for $\neg H(A)$) Assume $\forall A: P(A)$ then we can solve the halting problem (write an algorithm that enumerates all proofs and checks whether they proof $H(A)$ or $\neg H(A)$) Therefore $\exists A: \neg P(A)$. For such an A: Assume $H(A)$: Then we have a proof for $H(A)$ (the sequence of steps that lead to the halting state). Therefore $\neg H(A)$. But now we have a proof for $\neg H(A)$ and therefore $P(A)$ which contradicts the definition of A??? I assume I have a basic misconception about something. Please tell me what went wrong? Thanks - 4 Answers I believe that the error sneaks in with the phrase "for such an A." It is indeed true that for any $A$ such that $\neg P(A)$ we have that $\neg H(A)$; it follows that to be able to prove $\neg P(A)$ actually requires us to be able to prove $\neg H(A)$. You've substituted the a priori knowledge "this is such an $A$" for the proof that this $A$ has the property $\neg P(A)$, but it's the latter of these that you actually need. This avoids any paradox. - +1 . – Joel David Hamkins Apr 17 2010 at 21:31 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. What you have (re)discovered is a proof of Goedel's first incompleteness theorem via the halting problem. Let us suppose that we have already established that the halting problem is undecidable, which is not difficult to prove. Theorem.(Goedel) There is no computational algorithm to determine whether a given statement is true or not in the natural numbers. Proof. If there were such an algorithm, then for any program e and input n, we could form the statement "program e halts on input n" and check by the supposed computational procedure whether this statement is true or not. This would enable us to solve the halting problem, which is impossible. QED The argument can be refined to the following: Corollary. For any computable collection of axioms of arithmetic T, there is a statement φ that is true, but not provable in T. Furthermore, φ can be taken to be of the form: "a certain Turing machine program e never halts on a particular input n". Proof. Let us suppose that all true statements of that form were provable in T. Then we would be able to solve the halting problem, as follows: given program e and input n, then during the day, we simultate the progress of e on n. If it ever halts, we say, Yes, it halted. Meanwhile, at night, we search through all possible proofs from T, which is a computably enumerable procedure since T has a computable set of axioms, and look for a proof that e does not halt on n. If we ever find such a proof, then we say, No, it won't halt. Our assumptions ensure that exactly one of these situations will eventually occur, and so we will solve the halting problem, which is impossible. QED In particular, the error in your reasoning was your assumption that if a particular program did not halt, then this was a provable fact. You gave an argument for ¬H(A) for the A you found with ¬P(A), and this much is right, for when a program does halt, then this is witnessed by a particular halting computation, and even very weak axiomatic systems can prove that the program does indeed halt. But this proof does not take place within the same formal system as your proof notion used in P. I view this part of your argument as an argument by cases: for the A such that ¬P(A), you have ruled out the case that H(A). The remaining case is that A does not halt, and you have no reason to assume that this fact is provable in your system. What the argument shows is that it is difficult (or impossible in principle) for a system to prove all instances of non-halting behavior. I like this version of the proof of the Incompleteness theorem very much, because one can explain it to anyone who is familiar with the undecidability of the halting problem. In my view, it makes the otherwise mysterious claims of the Incompleteness theorem relatively accessible. One drawback of this method, however, is that it does not so easily extend to a proof of the second incompletness theormem, the statement that no computably axiomatizable theory can prove its own consistency. - (In particular, the error...) I thought of deleting my answer since yours makes the same point but is much more elaborate but then I saw this sentence. No, epthh hasn't made this particular error. His actual error was exactly in the opposite direction. – fedja Apr 16 2010 at 22:55 "One drawback of this method, however, is that it does not so easily extend to a proof of the second incompletness theormem, the statement that no computably axiomatizable theory can prove its own consistency. " Is there any way to do this with the recursion theorem? I know Sipser uses it to construct an explicit statement in arithmetic which is not provable (from any recursively enumerable proof system, I suppose). Is there any connection between the self-reference in the recursion theorem and in the second incompleteness theorem? – Akhil Mathew Apr 16 2010 at 23:31 2 The error in the argument begins with the phrase "But now we have a proof that not H(A)...", which is exactly the error that I mentioned, of assuming that since A does not halt that there is a proof that A does not halt. The argument up to that point is that if H(A) is true, then this is provable (which is fine), but this doesn't mean that if H(A) is not true, then THAT fact is provable. – Joel David Hamkins Apr 16 2010 at 23:38 1 You are right: once you have an erratic argument, there is more than one way to find the closest correct one and decide what exactly went wrong. Anyway, I guess we've covered all possible interpretations by now and there is no real point in trying to discuss which meaning was originally intended :-). – fedja Apr 17 2010 at 1:20 1 Epthh, in fact one can always find a particular A, which does not halt, but such that the original theory T does not prove this. For example, let A be the program that searches for a proof of a contradiction in T, halting when one is found. If T is consistent, then this program does not halt, but T cannot prove that it does not halt, since then it would prove its own consistency. The proof I just gave that A does not halt--analogous to your proof of not-H(A)--is not a proof in T, however, but a proof in T+Con(T), a stronger theory. – Joel David Hamkins Apr 17 2010 at 11:13 show 7 more comments The subtlety is that the halting problem asks whether i actually stops on x, not whether we can give a formal proof that i stops or doesn't stop on x. The whole point is that if $\forall A P(A)$ is true you must have proofs of false statements but not the solution of the halting problem. Your first "then" is thus flawed and what you actually proved is just the classical weak form of Godel's theorem. - It is true that $H(A)$ implies that we have a proof of $H(A)$, and we can find it by enumeration: as you noted, a proof of $H(A)$ is just a finite sequence of configurations or states of program $i$ on input $x$, each one of them reachable from the previous one according to the rules described by program $i$ itself (hence, such a proof is algorithmically verifiable). The problem with your argument is that $\neg H(A)$ does not necessarily imply that we have a proof of it, as you assume when saying “But now we have a proof of $\neg H(A)$”. Some programs can indeed have a non-termination proof; for instance, a finite sequence of configurations of the program that repeat cyclically. But other programs can go through an infinite number of non-recurring configurations: thus, the notion of non-termination proof as a finite sequence of configurations is flawed. Even by giving a lot of thought to the question, you won’t probably find any satisfying notion of “non-termination proof” applicable to all programs. The reason is that such a notion is provably nonexistent (formally, the set of non-halting programs is not recursively enumerable). -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 31, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9490441083908081, "perplexity_flag": "head"}
http://nrich.maths.org/359/clue	### Baby Circle A small circle fits between two touching circles so that all three circles touch each other and have a common tangent? What is the exact radius of the smallest circle? ### Pericut Two semicircle sit on the diameter of a semicircle centre O of twice their radius. Lines through O divide the perimeter into two parts. What can you say about the lengths of these two parts? ### Kissing Two perpendicular lines are tangential to two identical circles that touch. What is the largest circle that can be placed in between the two lines and the two circles and how would you construct it? # Orthogonal Circle ##### Stage: 5 Challenge Level: Call the centre of the orthogonal circle $(h, k)$ and the radius $r$ (equal to $CA$ in triangle $ABC$). Use Pythagoras' Theorem for triangle $ABC$ to write down an equation. In a similar way find equations using the other two circles. The NRICH Project aims to enrich the mathematical experiences of all learners. To support this aim, members of the NRICH team work in a wide range of capacities, including providing professional development for teachers wishing to embed rich mathematical tasks into everyday classroom practice. More information on many of our other activities can be found here.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 5, "mathjax_display_tex": 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": 0.9201143383979797, "perplexity_flag": "middle"}
http://mathhelpforum.com/calculus/136644-finding-intersection-parametric-equation-axis.html	# Thread: 1. ## Finding Intersection of Parametric Equation and Axis Hi, I've been trying to learn the C4 paper ahead of schedule and have gotten down to one last question I can't quite get. Q5) A curve C has parametric equations $x = at^2$ , $y = 2at$. Show that the equation of the normal to C at the point P, whose parameter is $p$, is: $px + y - 2ap - ap^3 = 0$ The normal to C at P meets the x-axis at Q. The perpendicular from P to the x-axis meets the x-axis at R. Find the length of QR. This is my working so far: $dx/dt = 2at$ $dy/dt = 2a$ $dy/dx = dy/dt * dt/dx = 1/t$ (Chain Rule) Normal Gradient = $-1/m$ = $-p$ (Use p as parameter for normal equation) $y - y1 = m(x - x1)$ $y - 2ap = -p(x - ap^2)$ $y - 2ap = -px + ap^3$ $px + y - 2ap - ap^3 = 0$ (As required) And thats about as far as I got, I've been trying to let y = 0 and let q be the parameter, so: $qx - 2aq - aq^3 = 0$ But do you then make them equal to each other and figure it out from there? I don't really know, its a WJEC paper too so I can't get any marking schemes without paying, so I thought this way would be better to get an understanding. Thanks 2. Originally Posted by steve989 Hi, I've been trying to learn the C4 paper ahead of schedule and have gotten down to one last question I can't quite get. Q5) A curve C has parametric equations $x = at^2$ , $y = 2at$. Show that the equation of the normal to C at the point P, whose parameter is $p$, is: $px + y - 2ap - ap^3 = 0$ The normal to C at P meets the x-axis at Q. The perpendicular from P to the x-axis meets the x-axis at R. Find the length of QR. This is my working so far: $dx/dt = 2at$ $dy/dt = 2a$ $dy/dx = dy/dt * dt/dx = 1/t$ (Chain Rule) Normal Gradient = $-1/m$ = $-p$ (Use p as parameter for normal equation) $y - y1 = m(x - x1)$ $y - 2ap = -p(x - ap^2)$ $y - 2ap = -px + ap^3$ $px + y - 2ap - ap^3 = 0$ (As required) And thats about as far as I got, I've been trying to let y = 0 and let q be the parameter, so: $qx - 2aq - aq^3 = 0$ But do you then make them equal to each other and figure it out from there? I don't really know, its a WJEC paper too so I can't get any marking schemes without paying, so I thought this way would be better to get an understanding. Thanks So you do the hard part and get stuck in an almost trivial part? Putting $y=0$ in the line you get $x=2a+ap^2\Longrightarrow Q=(2a+ap^2,0)$ , and the perpendicular from the point P to the x-axis meets the axis at $(ap^2,0)$ , so... Tonio 3. Oh yeah, haha that put me in my place, thanks a lot, was a long day, spent 4-5 hours researching unknown topics Thanks again	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 31, "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": 0.9495443105697632, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/289715/counting-a-bipartite-graph-in-two-ways	# Counting a bipartite graph in two ways I've got a bipartite graph where the left side corresponds to points $X = \{x_1,\ldots,x_n\}$ and the right side corresponds to subsets $\mathcal F = \{A_1,\ldots,A_m\}$ of $X$. There's an edge between the left and right side if a point is a member of a subset. $d(x)$ is the out-degree of a vertex on the left side of the graph, and indicates the number of members of $\mathcal F$ that contain $x$. While it's easy to count the bipartite graph in two ways and show: $$\sum_{A\in \mathcal F}^m\big\| A \big\| = \sum_{x\in X}d(x)$$ I'm trying to show that: $$\sum_{i,j=1}^m\big\| A_i\cap A_j \big\| = \sum_{x\in X}d(x)^2$$ This seems intuitive, but where do I start to get a more rigorous proof? - ## 2 Answers For the first equation (and title), you're not double counting a graph. You are counting, in two different ways, the number of pairs $(x,A)$ with $x\in X$ and $A\in\mathcal P$ satisfying $x\in A$. For the second equation, count in two different ways the number of triples $(x,A,B)$ with $x\in X$ and $A,B\in\mathcal P$ satisfying $x\in A\cap B$. - i thought "double counting" and "counting in two ways" referred to the same technique -- would it be better to update the title and question with "count in two ways"? – aaronstacy Jan 29 at 15:02 1 "Double counting" means "counting in two ways". But "double counting graphs" does not mean "double counting pairs $(x,A)$ with $x\in A$". – Colin McQuillan Jan 29 at 15:43 ah, i see. i'll update the question's wording. – aaronstacy Jan 29 at 17:27 We can view double counting as counting pairs $(x_i,A_j)$ where $x_i\in A_j$. For your proposed extension, I'd count triples $(x_i,A_j,A_k)$ where $x_i\in A_k$ and $x_i\in A_k$. -	{"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": 21, "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": 0.9262731671333313, "perplexity_flag": "head"}
http://www.physicsforums.com/showthread.php?p=4194584	Physics Forums ## Can dy/dx=x^2+Y^2 be solved analytically? I found this initial value problem and was supposed to comment on the accuracy of Runge Kutta method. Please enlighten me on the analytic solution. Find y(2) given the differential equation $\frac{dy}{dx}$=y$^{2}$+x$^{2}$ and the initial value y(1)=0. Thank you. PhysOrg.com science news on PhysOrg.com >> Front-row seats to climate change>> Attacking MRSA with metals from antibacterial clays>> New formula invented for microscope viewing, substitutes for federally controlled drug Recognitions: Homework Help Your equation has the form of a Riccati Equation. As per that wikipedia article, you can transform your particular equation into a linear second order ODE using the change of variables y = -u'(x)/u(x). Have a look at : http://www.physicsforums.com/showthread.php?t=637384 ## Can dy/dx=x^2+Y^2 be solved analytically? Thanks, Mute and Jacquelin for your support. The question was initially meant to be an easy exercise. However I have been on it since without much success. With your suggestions I was able to do the following. However my result, on comparison with the Runge Kutta estimate, does not look good. I have limited knowledge on the operations on Bessel functions and could be making a mistake. I really want to know how to get around this. Thank you, again. Attached Files Riccati-Bessel equation.pdf (175.3 KB, 9 views) Thanks, JJacquelin, for the advice. I followed your method and have since posted a pdf of what I did. I do not know whether I messed it up since my exact solution is not even close to the Runge Kutta estimate. Please, check it out and advise me. Thank you again. Hi ! As far as I can see with a quick look only, there are two main mistakes: First : BesselJ[-n,z] = (-1)^n BesselJ[n,z] is true only if n is an integer, which is not the case. So, you have the choice : Continue with BesselJ[1/4,z] AND BesselJ[-1/4,z] or, if you want positive orders, use the relationship between BesselJ[-1/4,z] and BesselJ[3/4,z]and BesselJ[7/4,z], but it will be more complicated. Anyways, in both cases, the two constants A and B must remain. You cannot eliminate one at this stage of the calculus. Second : At the end, it is not correct to add a new constant D, because -(df/dx)/f is not an integral. Finally, you will obtain an expression y(x) with, into it, one parameter on the form C=(A/B), or C=(B/A), which can be determined according to y(1)=0. You are right, Jacquelin. Let me try that. I tried the solution suggested in the book "Handbook for Exact Solutions for Ordinary Differential Equations" in the combinations of J and Y functions I have worked with these suggested solutions and they dont look good either. Once I'm done I shall try to post my solutions and hope for guidance. Thank you very much for your patience. Heres what I have done since (attached pdf). I know I shouldn't persue it this long, but it is a good challenge. There are two possible solutions: a combination of Bessel J and Y, and a combination of Bessel J+ and J-. I worked them both out and if I didn't bungle it again, they neither match each other nor match the numerical solution. Thank you Physics Forums for the guidance and patience. Attached Files Riccati Differential Equation.pdf (215.5 KB, 7 views) Hi ! It seems at first sight that you have done a good job. But, I am awfully sorry, it would take too much time and it is a too boring task to verify all these equations just after the New Year's Day ! I suggest that you first compare your result to the WolframAlpha results : http://www.wolframalpha.com/input/?i...C+y%281%29%3D0 Best wishes ! The solution given by WolframAlpha leads to y(2)=6.703786022295645 which is in rather good agreement with the numerical result (Runge-Kutta process). This make think that probably there are some mistakes in your analytical calculus. Note that, if all was correct, the slope of the curves drawn on your graph should be equal to 1 at (x=1, y=0), which clearly is not the case for the blue curve. Attached Thumbnails I did take a look at the WolframAlpha solution. It seems that they took the solution into the more negative direction. As I'm using the basic Excel worksheet to solve the Bessel functions, I have been avoiding the differentiation that takes me to the more negative Bessel functions. This being that I have to perform reconversion to positive Bessels for the Excel worksheet and i have no calculator that can perform calculations involving Bessel functions. I shall try to solve it and try to reproduce the Wolfram solution. It however shall be more tedious. Thanks, JJacquelin, for the time. I am truly grateful. I did take a look at the WolframAlpha solution. It seems that they took the solution into the more negative direction. As I'm using the basic Excel worksheet to solve the Bessel functions, I have been avoiding the differentiation that takes me to the more negative Bessel functions. This being that I have to perform reconversion to positive Bessels for the Excel worksheet and i have no calculator that can perform calculations involving Bessel functions. I shall try to solve it and try to reproduce the Wolfram solution. It however shall be more tedious. Thanks, JJacquelin, for the time. I am truly grateful. It is not essential to reproduce the Wolfram solution. This was only to compare to the numerical results. There is only one solution, but many manner to express it, with or without negative order for the Bessel functions. The method that you used in your last pdf seems good. If you find where is the mistake, I am sure that all the curves on the drawing will be perfectly the same. I have just checked the Excel worksheet. It truncates all non-integral Bessel functions so that $J_{\frac{1}{4}}=J_{\frac{3}{4}}=J_0$. My bad. Sorry. Let me seek a different Besselcalculator. Thanks all. Well, guys, I did get a good calculator and was able to use my two formulae for exact solution of the Ricatti equation to reach the same result as wolfram did i.e $y\left(2\right)=6.70378602229564586367$. Great thanks to all of you. I was even able to come up with a graph that fits sufficiently well with the numerical solution. The site of the online calculator is:http://keisan.casio.com/has10/Free.cgi Thread Tools \| \| \| \| \|----------------------------------------------------------------\|------------------------\|---------\| \| Similar Threads for: Can dy/dx=x^2+Y^2 be solved analytically? \| \| \| \| Thread \| Forum \| Replies \| \| \| Calculus \| 4 \| \| \| Calculus \| 10 \| \| \| General Math \| 2 \| \| \| Electrical Engineering \| 0 \| \| \| General Math \| 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": 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": 0.9430065751075745, "perplexity_flag": "middle"}
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Suppose that we are given data in ... 0answers 27 views ### What to do when the predictors do not accord with common sense/literature, but the model is fine/best according to log likelihood and LRT? down vote favorite I would try to clarify the problem and then ask the questions. The problem (variable names are masked due to confidentiality): I ran a binary logistic regression, in which there ... 0answers 22 views ### Multivariate linear regression/aproximation I'm solving a problem which features a function $f: \mathbb{R}^4 \rightarrow \mathbb{R}^4.$ I don't know the function, but I assume it's linear and it can be expressed as \$ \mathbf{y} = \mathbf{A} ... 0answers 30 views ### differences between linear regression and generalized linear regression I have found the R program of the linear regression for the consumer price index of a country, expressed in terms of years and quarters division. The formula obtained is: ...	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 66, "mathjax_display_tex": 12, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "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": 0.9203819632530212, "perplexity_flag": "middle"}
http://physics.aps.org/articles/v2/91	# Viewpoint: Flipping a photonic shock wave and , 3112 Etchverry Hall, NSF Nanoscale Science and Engineering Center (SINAM), University of California, Berkeley, CA 94720-1740, USA Published November 2, 2009 \| Physics 2, 91 (2009) \| DOI: 10.1103/Physics.2.91 Čerenkov radiation with the emission cone reversed has been observed in a metamaterial with negative refractive index. #### Experimental Verification of Reversed Cherenkov Radiation in Left-Handed Metamaterial Sheng Xi, Hongsheng Chen, Tao Jiang, Lixin Ran, Jiangtao Huangfu, Bae-Ian Wu, Jin Au Kong, and Min Chen Published November 2, 2009 \| PDF (free) When charged particles such as electrons travel through a dielectric medium with a speed greater than the phase velocity of the light in the medium, electromagnetic radiation is emitted that falls into a cone fanning out in the forward direction [Fig. 1, top left]. This phenomenon is called Čerenkov radiation, named after the Russian scientist who first characterized it rigorously and was awarded the Nobel Prize in Physics in 1958. Most people are familiar with Čerenkov radiation from the blue glow of an underwater nuclear reactor as it emits energetic charged particles. The angle of the Čerenkov emission cone is related in a simple way to the particle velocity. This unique feature enables a wide range of applications, from the measurement of fast particles in high-energy physics, the characterization of fission rate in nuclear reactors, to the detection of labeled biomolecules. Now in a paper appearing in Physical Review Letters, Sheng Xi and colleagues at Zhejiang University, China, and the Massachusetts Institute of Technology, US, experimentally demonstrate that the direction of the cone of Čerenkov radiation can be reversed in artificially engineered composite media, namely, metamaterials [1]. Backward Čerenkov radiation is one among a number of other unusual phenomena associated with so called left-handed materials, first proposed by Victor Veselago in 1968 [2]. A left-handed medium has simultaneous negative permittivity $ε$ and negative permeability $μ$. In such a medium, the electric field, magnetic field, and wave vector form a left-handed triad, in contrast to the “right-hand rule” in natural materials known so far, such as water and glass. As a result, the direction of the advancing phase (the wave vector of the electromagnetic wave) is opposite to its energy flow (or Poynting vector) and the phase index of refraction is negative. The exotic phenomena in a left-handed medium include negative refraction, inverse Doppler effect, backward Čerenkov radiation, and negative radiation pressure, as predicted by Veselago more than forty years ago. The deduction for backward Čerenkov radiation in a left-handed medium is straightforward. The half angle $α$ of the Čerenkov radiation cone is determined by the ratio between the phase velocity of the light and the particle speed as $cosα=c′/ν$, where $c′$ and $ν$ are the phase velocity of the electromagnetic wave and the particle speed in the medium, respectively. Since the phase velocity in a left-handed medium is negative, i.e., opposite to the direction of the energy flow, the cone angle is obtuse and hence the radiation cone is facing toward the backward direction [Fig.1, bottom left]. While exhibiting very interesting features, media with negative refractive index are not found in nature. Thanks to the discoveries by Pendry and others about a decade ago, composite materials with effective negative refractive index have been realized [3, 4], which are called metamaterials. Negative index metamaterials normally consist of two types of functional elements: “split-ring resonators” with a negative magnetic response and thin metal wires with a negative electric response. This opens the door to the testing of some of those unconventional phenomena proposed by Veselago. Many groups around the world have routinely observed a negative refractive angle at the interface between a positive index medium and a negative index metamaterial [3, 5]. There have also been a few experiments exhibiting the inverse Doppler effect [6]. There have been a few studies in the past to investigate the Čerenkov radiation in left-handed metamaterials, but a convincing experimental demonstration of reversed Čerenkov radiation turns out to be challenging. This is largely due to the fact that the Čerenkov radiation power peaks in the ultraviolet region of the spectrum, where negative index metamaterials are not currently available. The Čerenkov radiation power falls off rapidly at longer wavelength and becomes extremely weak at microwave frequencies (where the technologies for the fabrication of metamaterials are very mature). In a recent study, Antipov et al. at the Illinois Institute of Technology observed microwave radiation when a charged particle beam was directed through a waveguide loaded with negative index metamaterials operating in the microwave frequency range [7]. The spectrum of the radiation exhibited a peak at the same frequency as the left-handed band of the metamaterial, providing some evidence of Čerenkov radiation in the negative index medium. However, the propagation direction of the radiation could not be determined conclusively, thus backward Čerenkov radiation was not observed in their study. The successful demonstration of backward Čerenkov radiation by Xi et al. largely relied on two key experimental innovations. The first is a novel negative index metamaterial design, which operates for electromagnetic waves with the magnetic field component perpendicular to the plane of wave propagation in the two-dimensional experimental setting [Fig. 1, right]. Until now, most of the negative index metamaterials were designed for waves with out-of-plane electric field, so the electromagnetic wave can be confined simply by two parallel metal plates. However, this configuration is not suitable for the investigation of Čerenkov radiation since the electric field of Čerenkov radiation lies in the plane formed by the particle trajectory and the radiation wave vector. In this work, Xi et al. develop a new design in which both the split-ring resonators and metal wires are in-plane. This configuration works for the electromagnetic waves that have in-plane electric field (along the wires) and out-of-plane magnetic field (perpendicular to the split-ring resonators). It also allows an isotropic propagation of the electromagnetic waves in the metamaterials and thus makes it possible to determine the direction of Čerenkov radiation. The second innovation is to emulate a charged-particle beam by means of a waveguide with a periodic array of slots, instead of using real charged particles [Fig. 1, right]. By doing so, Xi et al. solved the problem of extremely weak Čerenkov radiation in the microwave frequencies associated with moving charged particles. As an electromagnetic wave travels inside the waveguide, it emerges at each slot with a fixed phase retardation relative to the neighboring slots. This leaking radiation from the waveguide is equivalent to the radiation from a phased antenna array. The Fourier transform of the electric current carried by a moving charged particle results in a broad spectrum in the frequency domain. But as long as a single frequency is concerned, the current of the charged particle is equivalent to that of a phased dipole array, as mathematically proved by the authors. In this analog, the phase velocity of the electromagnetic wave propagating inside the waveguide corresponds to the moving speed of the charged particles in a regular Čerenkov radiation configuration. In the experiment, the authors designed a waveguide with comparatively low refractive index of $n$ $∼0.5$, which is equivalent to a charged particle moving with a speed twice as great as that of light in a vacuum. With this new experimental configuration, the radiation signal can be many orders of magnitude stronger than the traditional Čerenkov radiation induced by a fast charged-particle beam, and thus the Čerenkov radiation was directly observed along the backward direction within the left-handed frequency range of the metamaterials. It is worth noting that Grbic and Eleftheriades carried out an experimental attempt earlier at the University of Toronto, in which they observed the backward radiation from a low-index left-handed microwave transmission line into free space [8]. However, considering the fact that it is the phase velocity rather than group velocity of the electromagnetic wave propagating in the transmission line that corresponds to the speed of an equivalent moving charge, the Čerenkov radiation observed by Grbic et al. was indeed in the forward direction with respect to the direction of the equivalent moving charge. From one point of view, the experiment by Xi et al. could be considered as a special case of negative refraction, in which the light is negatively refracted from a low positive index medium (or waveguide) to a high negative index medium at a grazing incident angle. This alternative view highlights the intrinsic connection between the phenomena of backward Čerenkov radiation and negative refraction. This should be of no surprise, however, since both phenomena are coherent processes involving a phase matching condition along a certain direction. On the other hand, a more direct demonstration of backward Čerenkov radiation with charged particles passing through a left handed metamaterial may be highly desirable for the purpose of both fundamental investigations and practical applications, such as detectors. In the last several years, the negative index metamaterials have been extended from microwave frequencies to terahertz [9], infrared, and even optical frequencies [10, 11, 12], where the Čerenkov radiation is expected to be orders of magnitude more intense than that at the microwave frequencies. In particular, the recent realization of bulk negative index optical metamaterials may offer us the opportunity to observe backward Čerenkov radiation of actual moving charged particles in the foreseeable future [13]. ## Acknowledgments Support from US Army Research Office ARO MURI program 50432-PH-MUR is acknowledged. ### References 1. S. Xi, H. Chen, T. Jiang, L. Ran, J. Huangfu, B-I. Wu, J. A. Kong, and M. Chen, Phys. Rev. Lett. 103, 194801 (2009). 2. V. G. Veselago, Soviet Physics Uspekhi 10, 509 (1968). 3. J. B. Pendry et al., IEEE Trans. Microwave Theory Tech. 47, 2075 (1999). 4. R. A. Shelby, D. R. Smith, and S. Schultz, Science 292, 77 (2002). 5. A. A. Houck, J. B. Brock, and I. L. Chuang, Phys. Rev. Lett. 90, 137401 (2003). 6. N. Seddon and T. Bearpark, Science, 302, 5650 (2003)10.1126/science.1089342. 7. S. Antipov et al., J. Appl. Phys. 104, 014901 (2008). 8. A. Grbic and G. V. Eleftheriades, J. Appl. Phys. 92, 5930 (2002). 9. T. J. Yen, W. J. Padilla, N. Fang, D. C. Vier, D. R. Smith, J. B. Pendry, D. N. Basov, and X. Zhang, Science 303, 1494 (2004). 10. S. Zhang, W. Fan, N. C. Panoiu, K. J. Malloy, R. M. Osgood, and S. R. J. Brueck, Phys. Rev. Lett. 95, 137404 (2005). 11. V. M. Shalaev, W. Cai, U. K. Chettiar, H-K. Yuan, A. K. Sarychev, V. P. Drachev, and A. V. Kildishev, Opt. Lett. 30, 3356 (2005). 12. G. Dolling, G. M. Wegener, C. M. Soukoulis, and S. Linden, Opt. Lett. 32, 53 (2007). 13. J. Valentine, S. Zhang, T. Zentgraf, E. Ulin-Avila, D. A. Genov, G. Bartal, and X. Zhang, Nature 455, 376 (2008). ### About the Author: Shuang Zhang Shuang Zhang received his Ph.D. (2005) in electrical engineering at the University of New Mexico. From 2005 to 2006 he worked as a Postdoctoral Associate at the University of llinois, Urbana-Champaign. He is currently a Postdoctoral Associate at the University of California, Berkeley. His main research interests include metamaterials, plasmonics, nanofabrication, and optoelectronics. ### About the Author: Xiang Zhang Xiang Zhang is the Ernest S. Kuh Endowed Chair Professor at the University of California, Berkeley, the Director of NSF Nano-scale Science and Engineering Center (SINAM) and a faculty scientist in the Lawrence Berkeley National Laboratory. He received his Ph.D. from Berkeley in 1996 and was on faculty at The Pennsylvania State University and the University of California, Los Angeles, before joining Berkeley in 2004. Professor Zhang’s current research is focused on nanoscale physics, metamaterials, nanophotonics, and biotechnologies. ## Related Articles ### More Optics Nanostructures Put a Spin on Light Synopsis \| May 16, 2013 Wave-Shaping Surfaces Viewpoint \| May 6, 2013 ### More Particles and Fields Positrons Galore Viewpoint \| Apr 3, 2013 A Year-Long Search for Dark Matter Synopsis \| Mar 28, 2013 ### More Metamaterials Invisibility Cloak for Heat Focus \| May 10, 2013 Wave-Shaping Surfaces Viewpoint \| May 6, 2013 ## New in Physics Wireless Power for Tiny Medical Devices Focus \| May 17, 2013 Pool of Candidate Spin Liquids Grows Synopsis \| May 16, 2013 Condensate in a Can Synopsis \| May 16, 2013 Nanostructures Put a Spin on Light Synopsis \| May 16, 2013 Fire in a Quantum Mechanical Forest Viewpoint \| May 13, 2013 Insulating Magnets Control Neighbor’s Conduction Viewpoint \| May 13, 2013 Invisibility Cloak for Heat Focus \| May 10, 2013 Desirable Defects Synopsis \| May 10, 2013	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 8, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8943958878517151, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/38663/a-chain-64-meters-long-whose-mass-is-20-kilograms-is-hanging/38668	# A chain 64 meters long whose mass is 20 kilograms is hanging [closed] A chain 64 meters long whose mass is 20 kilograms is hanging over the edge of a tall building and does not touch the ground. How much work is required to lift the top 3 meters of the chain to the top of the building? Use that the acceleration due to gravity is 9.8 meters per second squared. Your answer must include the correct units. - 1 Hi Ryan, and welcome to Physics Stack Exchange! We don't allow homework-like questions like this where you just ask the question directly; you should narrow it down to focus on a specific physics concept first. See our FAQ and homework policy for more information. – David Zaslavsky♦ Sep 29 '12 at 23:35 ## closed as too localized by David Zaslavsky♦Sep 29 '12 at 23:34 This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, see the FAQ. ## 3 Answers $$W =\int^b_a \! F \, \mathrm{d}x$$ This is the relationship between work and force, you'll need to work out what the force might be. Where a and b are the positions where you are moving from and to against the force. If you've never done integration before $$\int^b_a \! F \, \mathrm{d}x = [bF - aF]$$ - the density of the chain is $\rho = \frac{20}{64}$ the amount of work is made with respect to the force $F=mg$ but teh mass is chaning according we are lifting the chain i think the answer is $\rho g \int_{0}^{3}(64-x)dx$ since the work made (absolute value) is $W= \int_{a}^{b}F.dx$ - Since the gravity is uniform and the chain is freely suspended, we just need to consider the change in height and the change in potential energy. $$W = \Delta P.E. = mg(h_2 - h_1)$$ -	{"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": 4, "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": 0.9550023674964905, "perplexity_flag": "middle"}
http://nrich.maths.org/6206&part=	### Strange Numbers All strange numbers are prime. Every one digit prime number is strange and a number of two or more digits is strange if and only if so are the two numbers obtained from it by omitting either its first or its last digit. Find all strange numbers. ### Stars Can you find a relationship between the number of dots on the circle and the number of steps that will ensure that all points are hit? ### Weekly Problem 45 - 2013 Weekly Problem 45 - 2013 # Weekly Problem 41 - 2008 ##### Stage: 3 Challenge Level: How many pairs of numbers of the form $x$, $2x+1$ are there in which both numbers are prime numbers less than $100$ If you liked this problem, here is an NRICH task which challenges you to use similar mathematical ideas. This problem is taken from the UKMT Mathematical Challenges. View the previous week's solution View the current weekly problem 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": 3, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.90399169921875, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/65387?sort=votes	## products in the category of banach spaces ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Let `$\{X_{\alpha} \}_{\alpha \in A}$` be a collection of Banach spaces. It is easy to show that `$ P = \{(x_{\alpha}) : {\rm sup}_{\alpha} \\|x_{\alpha} \\| < \infty \} $` with `$\\| (x_{\alpha} ) \\| = {\rm sup}_{\alpha} \\| x_{\alpha} \\|$` is a banach space. If the indexing set $A$ is finite, then it is easy to show that $P$ (with the natural projection maps) is the product of `$\{X_{\alpha} \}_{\alpha \in A}$` in the category of banach spaces. Assume $Y$ is a banach space and $f_{\alpha} : Y \to X_{\alpha}$ is a linear continuous map for each $\alpha$. If `${\rm sup}_{\alpha} \\|f_{\alpha} \\| < \infty$` then it is easy to see that the induced linear map $Y \to \Pi X_{\alpha}$ has image a subset of $P$ and is continuous. If this condition is not satisfied then it is not clear to me that there exists a continuous linear map $g:Y \to P$ such that $\pi_{\alpha} \circ g = f_{\alpha}$ for all $\alpha$. Is there any way to prove that $P$ is the categorical product of the collection `$\{X_{\alpha} \}_{\alpha \in A}$` when $A$ is infinite? - 2 In the category of Banach spaces and bounded linear maps, $P$ is NOT the product of $(X_\alpha)$ - that category does not have infinite products. – Yemon Choi May 19 2011 at 1:16 Just take each $X_\alpha$ and $Y$ to be a copy of the ground field and let $y_\alpha$ be multiplication by $\lambda(\alpha)$ where $\lambda: A \to {\bf R}$ is your favourite unbounded function. – Yemon Choi May 19 2011 at 1:47 (Sorry, in previous comment I meant "let $f_\alpha$ be multiplication...") – Yemon Choi May 19 2011 at 1:47 thanks Yemon. If you post this as an answer I will accept it – Daniel Barter May 19 2011 at 2:04 2 @Zen Harper: the categorical product would be a Banach space $B$ with the property that a collection of maps $\{f_\alpha\colon Y\to X_\alpha\}$ was the same as a single map $f\colon Y\to B$. This is the "universal mapping property" of the product, hence the term "categorical". In the case of Banach spaces such a space $B$ cannot exist, and you've seen why: the condition on the $f_\alpha$ is that they be individually bounded, but their norms can be unbounded in $\alpha$, and it's not possible to bundle this up into a single space $B$ and bounded maps to it. – Tom Church May 19 2011 at 3:30 show 5 more comments ## 1 Answer [Promoted from comments, as requested] The category of Banach spaces and bounded linear maps does not have arbitrary infinite products. Here is a simple example, using the notation of the question: take each $X_\alpha$ and $Y$ to be a copy of the ground field, fix an unbounded function $\lambda:A\to{\mathbb R}$, and let `$f_\alpha: Y \to X_\alpha$` be multiplication by `$\lambda(\alpha)$`. It is then easy to see that the cone `$(f_\alpha: Y\to X_\alpha)_{\alpha\in A}$` cannot factor through the cone `$(\pi_\alpha: P \to X_\alpha)_{\alpha\in A}$`. This shows that $P$ is not the product of the $X_\alpha$, and a little more work shows that no such product can exist (assume it does and then show it would have to be isomorphic to $P$). -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 33, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": 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": 0.9444721341133118, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/86777/fubini-tonelli-proof-purely-using-complex-analysis?answertab=active	# Fubini-Tonelli proof purely using complex analysis? Let $F:[a,b]\times[c,d]\rightarrow \mathbb{R}$ be continuous. Show that:$$\int_{a}^{b}\left(\int_{c}^{d}F(x,y)dy\right)dx = \int_{c}^{d}\left(\int_{a}^{b}F(x,y)dx\right)dy$$ Lebesgue integral hasn't been introduced, and this is from a book concerning complex analysis. Is there a way to show this without the usage of the Lebesgue integral? Using complex analysis? Merci for all hints. - ## 1 Answer Yes, there is a version for Riemann-Integrals, too. A quite straight-forward proof can be found in "The College Mathematics Journal" Vol. 33, No. 2 p.126-130, it is available online too. The main part is a clever application of the mean value theorem. Note also that the term Fubini-Tonelli is mostly used for the Lebesgue-Integral "version" of the theorem. - Thanks!……………………………. – VVV Dec 1 '11 at 14:08 No problem :).................... – Listing Dec 1 '11 at 14:11	{"extraction_info": {"found_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": 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": 0.8578821420669556, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/10818/why-should-the-observed-probability-distributions-in-quantum-mechanics-always-al	# Why should the observed probability distributions in quantum mechanics always align with the pointer basis of decoherence? It has always been claimed decoherence solves the problem of the preferred-basis for observed probability distributions, but why should this be the case? If there is only one world, and there are probabilities for certain outcomes, why should the basis in which the probabilities are observed coincide with the pointer basis determined dynamically by decoherence? - Or why don't we see superposed cats? – Raskolnikov Jun 6 '11 at 10:52 If we consider the heisenberg uncertainty principle a cat is a superposition of many cats, as we all are superpositions of many us. It is the size of hbar that makes this superposition irrelevant because we live in a world many orders of magnitude larger than this fuzziness. – anna v Jun 6 '11 at 14:15 @Raskolnikov the comment above should have been addressed to you. – anna v Jun 6 '11 at 15:44 @anna I do not see how it addresses the issue, but why don't you make it a reply to the OP? – Raskolnikov Jun 7 '11 at 7:47 ## 4 Answers It's because the probability that a given mixed state $\rho$ is found in a particular normalized state $\psi$ is only "sharp" and well-defined if $\psi$ is actually an eigenstate of $\rho$. So whatever basis in which $\rho$ is diagonal the decoherence produces, is also the basis of the states that have well-defined probabilities. For all their general linear superpositions, one may calculate the expectation values of the probabilities but they're not probabilities that can be measured or "perceived". The prescription above was described e.g. at http://motls.blogspot.com/2011/06/density-matrix-and-its-classical.html It treats the density matrix in a similar way as the observables - even though the density matrix is not an observable. The rule that only the eigenstates of the density matrix may be "perceived" is pretty much equivalent to the "consistent histories" approach to quantum mechanics that tells you which questions may be asked in quantum mechanics and which can't. - If I understand you correctly, the preferred basis is the eigenbasis of the density matrix and has thus a priori nothing to do with decoherence. Except that a posteriori, decoherence has produced a certain density matrix, and therefore its eigenbasis is the basis of pointer states. Correct? – Raskolnikov Jun 7 '11 at 8:07 If I understand you correctly, then yes! ;-) Decoherence is just a particular process, and/or a calculation imitating this process, that just allows one to trace what the eigenstates of the density matrix will be. But the fact that the basis in which $\rho$ is diagonal is preferred for defining "alternative histories or outcomes" whose probabilities may be nicely checked and "perceived" is an independent fact. – Luboš Motl Jun 8 '11 at 6:47 Let's hear it from the great quantum philosopher Niels Bohr. During his time, no one used the term preferred basis, but the term complementary observables covers pretty much the same issue. If you wish to object, please point out why. So let's say we have an electron floating around. Do we use the position basis, or the momentum basis? Bohr's answer is very ingenious and cuts right to the heart of the matter. Search and search into the dynamics of the electron in itself, and you will never find the answer. All experiments have to include the description of the experimental apparatus as part of the complete description. It is the choice of apparatus and its settings which determines the preferred basis. Is an electron passing through a double slit a particle or a wave? If you don't measure which slit it went through, the wave basis is preferred. If you do measure which slit it went through, the particle basis is preferred. Is the spin of an electron aligned along the z-axis, or the x-axis? The answer is not found in the electron itself. The answer is found in the orientation of the Stern-Gerlach magnetic field of the experiment. The so-called pointer basis is none other than the basis that the apparatus picks out. The pointer basis is very sensitive to the nature of the apparatus and its settings. - Most people misunderstand what decoherence is all about. Decoherence is not really about how the environment affects the system. Decoherence is about how the system imprints itself irreversibly in the environment. Obviously, you can't observe a system unless information about the system has already been imprinted within the environment for you to pick up. The pointer information of decoherence are precisely those information about the system which becomes entangled with the environment for you to pick up. When stated in this manner, the OP's question becomes almost tautological. - It is an utter misunderstanding that the preferred basis is given by the eigenstates of the reduced density matrix. If this false definition is correct, decoherence is tautological. Decoherence is defined to be a suppression of off-diagonal components in the basis of choice, but the basis is chosen to be the eigenstates? tautological. Besides, taking this definition seriously, as Motl does, the basis depends upon where the boundary of the system is drawn. Take Schrodinger's cat. Place the boundary at Pluto when the signals haven't had time to reach it yet and there is no entanglement assuming stosszahlansatz and the eigenstates form some funny bizarre unphysical states. When the signal reaches Pluto, it turns into a live or dead cat. Place the boundary around the cat so that the rest of the box is in the environment and the basis is more well behaved. When two eigenvalues nearly match, the eigenstates can mix into unphysical vectors. Also, the nondominant eigenvalue eigenstates are always funny, but stick to the dominant probable ones for the moment. It is also sensitive to how thick or thin the boundary wall is. Too thin, not enough smearing, and the eigenstates also become funny and bad. Too much focus upon quantum fluctuations across the boundary. There is an art to choosing boundaries. No, the true tenet of decoherence is maximal predictivity. What basis, such that if we know which basis vector allows us to predict the future with greatest accuracy, i.e. least increase in entropy? Informational patterns which survive over time, maybe in a transmuted encoding form. The basis comes from the interaction Hamiltonian term factors, and are the most stable over time. Many seemingly good bases suck because they mix up rapidly in time with no stability. The better the survival of information over time and the longer it survives, the higher the predictability and the better the basis. Then check if in this maximally predictable basis, the off diagonal terms are suppressed. This is decoherence. Of course, some people would just claim the preferred basis is the basis of consciousness, but this is highly controversial. - 1 Without collapse all information survives. Unitarity guarantees this. – wnoise Nov 23 '11 at 19:42	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 6, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9340441226959229, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/9049/why-doesnt-the-moon-fall-upon-earth/9115	# Why doesn't the Moon fall upon Earth? Why doesn't the Moon, or for that matter anything rotating another larger body, ever fall into the larger body? - 15 It is falling. That is why it moves. – belisarius Apr 24 '11 at 5:40 5 – dmckee♦ Apr 24 '11 at 14:42 2 @belisarius - the fall is not the reason it moves. But it is the reason for it constantly making the turn... – ysap May 17 '11 at 23:11 ## 8 Answers The moon does not fall to Earth because it is in an orbit. One of the most difficult things to learn about physics is the concept of force. Just because there is a force on something does not mean it will be moving in the direction of the force. Instead, the force influences the motion to be a bit more in the direction of the force than it was before. For example, if you roll a bowling ball straight down a lane, then run up beside it and kick it towards the gutter, you apply a force towards the gutter, but the ball doesn't go straight into the gutter. Instead it keeps going down the lane, but picks up a little bit of diagonal motion as well. Imagine you're standing at the edge of a cliff 100m tall. If you drop a rock off, it will fall straight down because it had no velocity to begin with, so the only velocity it picks up is downward from the downward force. If you throw the rock out horizontally, it will still fall, but it will keep moving out horizontally as it does so, and falls at an angle. (The angle isn't constant - the shape is a curve called a parabola, but that's relatively unimportant here.) The the force is straight down, but that force doesn't stop the rock from moving horizontally. If you throw the rock harder, it goes further, and falls at a shallower angle. The force on it from gravity is the same, but the original velocity was much bigger and so the deflection is less. Now imagine throwing the rock so hard it travels one kilometer horizontally before it hits the ground. If you do that, something slightly new happens. The rock still falls, but it has to fall more than just 100m before it hits the ground. The reason is that the Earth is curved, and so as the rock traveled out that kilometer, the Earth was actually curving away underneath of it. In one kilometer, it turns out the Earth curves away by about 10 centimeters - a small difference, but a real one. As you throw the rock even harder than that, the curving away of the Earth underneath becomes more significant. If you could throw the rock 10 kilometers, the Earth would now curve away by 10 meters, and for a 100 km throw the Earth curves away by an entire kilometer. Now the stone has to fall a very long way down compared to the 100m cliff it was dropped from. Check out the following drawing. It was made by Isaac Newton, the first person to understand orbits. IMHO it is one of the greatest diagrams ever made. What it shows is that if you could throw the rock hard enough, the Earth would curve away from underneath the rock so much that the rock actually never gets any closer to the ground. It goes all the way around in the circle and might hit you in the back of the head! This is an orbit. It's what satellites and the moon are doing. We can't actually do it here close to the surface of the Earth due to wind resistance, but on the surface of the moon, where there's no atmosphere, you could indeed have a very low orbit. This is the mechanism by which things "stay up" in space. Gravity gets weaker as you go further out. The Earth's gravity is much weaker at the moon than at a low-earth orbit satellite. Because gravity is so much weaker at the moon, the moon orbits much more slowly than the International Space Station, for example. The moon takes one month to go around. The ISS takes a few hours. An interesting consequence is that if you go out just the right amount in between, about six Earth radii, you reach a point where gravity is weakened enough that an orbit around the Earth takes 24 hours. There, you could have a "geosynchronous orbit", a satellite that orbits so that it stays above the same spot on Earth's equator as Earth spins. Although gravity gets weaker as you go further out, there is no cut-off distance. In theory, gravity extends forever. However, if you went towards the sun, eventually the sun's gravity would be stronger than the Earth's, and then you wouldn't fall back to Earth any more, even lacking the speed to orbit. That would happen if you went about .1% of the distance to the sun, or about 250,000 km, or 40 Earth radii. (This is actually less than the distance to the moon, but the moon doesn't fall into the Sun because it's orbiting the sun, just like the Earth itself is.) So the moon "falls" toward Earth due to gravity, but doesn't get any closer to Earth because its motion is an orbit, and the dynamics of the orbit are determined by the strength of gravity at that distance and by Newton's laws of motion. note: adapted from an answer I wrote to a similar question on quora - The first sentence contradicts itself in that orbital motion is a state of free fall toward the center of attraction. Then, in the final paragraph, the first sentence contradicts the very first sentence. A beginner would have a difficult time with this logic. – user11266 Jan 14 at 17:05 No, it doesn't contradict itself. "Fall to Earth" means, to any cognizant reader, the same as "crash into Earth" in that sentence. Further, this phrasing mimics the language of the question. Words like "fall" can mean different things in contexts. Most people are able to understand this. – Mark Eichenlaub Jan 14 at 20:45 It is indeed contradictory. "Falling to Earth" and "crashing into Earth" are two entirely different things. A ball can fall without crashing into Earth, but it cannot crash into Earth without first falling. Words, especially in physics, must be as unambiguous as possible. And by the way, I'm quite cognizant thank you. – user11266 Jan 15 at 14:09 Moon is continuously falling towards earth but missing all the time! Same with other planets too. In general, in an inverse square central force field one can calculate the trajectory of a particle and verify that the trajectory is either a parabola or ellipse or hyperbola (conic sections) depending upon the initial position and initial momentum of the particle. For a two body system with certain initial conditions, it is a stable elliptical orbit. In case of the sun and the earth it is an ellipse (ignoring gravitation of other objects and also ignoring the relativistic precision of orbit). - Another way to look at it: In the frame of reference of earth, the moon has angular momentum. Angular momentum is preserved if no torque is applied ($\tau=dL/dt$). The gravity forces between earth and the moon are in the direction of the center of mass, so they produce no torque ($\tau=mv\times R$), so the angular momentum ($L$) cannot change. Gravity is perpendicular to the moon's speed, so it changes the direction and not the magnitude of the speed itself. $L=mv\times R$ and if $L$, $m$ and $v$ are constant, $R$ must remain constant as well, so the radius doesn't change. - 2 I agree, but there is some sleight of hand going on in the phrase "gravity is perpendicular to the moon's speed", which itself essentially assumes a circular orbit. – Mark Eichenlaub Apr 24 '11 at 19:39 The truth is that the moon IS constantly trying to fall upon the earth, due to the force of gravity; but it is constantly missing, due to its tangential velocity. To understand this, think of whirling a rock, tied to the end of a string, around and around, with your hand just above your head. As the rock travels in circles it is constantly being pulled toward you by the force on the string (which is like Earth's pull of gravity on the moon). Why doesn't the rock come bonk you on the head, if you are constantly pulling it toward your head? The answer is that the rock is always trying to change its velocity vector to come do just that; but the change is only enough to just keep it in a circular path, like the pull on the Moon is just enough to keep it in a circular orbit around Earth. - You should change "trying to fall" to "falling" (which is what it does in Newtonian mechanics). Also, the path is elliptic, not circular (but it eccentricity is indeed very close to 1). – ysap May 17 '11 at 23:10 The Moon does not fall towards Earth right now because Earth rotates itself. The energy from the Earth's own rotation around its axis is gradually tranferred into energy of the Moon's orbital motion. That's why the Earth's rotating speed decreases but the distance to the Moon increases. This process will continue until Earth's proper rotation will slow to the point where it has the same angular velocity as the Moon's orbital motion. From that moment on the Moon will start to gradually approach Earth. - 4 This is not an answer to the question. The Moon's orbit would be virtually unchanged, even if the Earth weren't rotating about its axis. – Ted Bunn Apr 24 '11 at 13:10 If the Earth weren't rotating about its axis, the Moon would start to fall on Earth. – Anixx Apr 24 '11 at 13:21 1 But it'd never get there ... the Earth would start rotating faster until it was in lunasyncronous rotation (always keeping the same face to the moon), at which point the Moon would stop falling. – Peter Shor Apr 24 '11 at 15:37 To become lunasynchrous the Earth should rotate slower, not faster. After the rotation becomes lunasynchrous, the Moon will start approach Earth. – Anixx Apr 24 '11 at 18:21 @Anixx: Would you be so kind as to explain what mechanism you believe would be responsible for that? – dmckee♦ Apr 26 '11 at 17:24 show 3 more comments The best simple answer I can think of are this: An orbit of one body of an another is essentially a degree of balance among forces, real and fictitious. These would include centripetal force (gravity) attracting the orbiting body ("the fall") and centrifugal force which arises out of the orbiting body's inertia (the orbiting body's tendency to remain in a constant linear motion away from the body it is orbiting). In General Relativistic terms, the orbit is the result of a body moving in a straight line through the curved space that exists around the more massive body. If the lesser body moves with the sufficient combination of momentum and distance it will continue to pass the more massive body onto other regions of space. If that combination is not sufficient to overcome the curvature of space in the region around the more massive body then the lesser body will continue it's tendency to travel in a straight line but it must it do so in a curved space that it cannot "escape". If it has a sufficient minimum momentum, it's tendency to move in a straight line away from the more massive body will overcome the downward curvature. These two conditions will cause the lesser body, per Newton, to become the perpetual satellite of the more massive body because the lesser body must stay in motion unless an equal and opposite force is applied to it's motion. The lesser body doesn't experience resistance from friction or air in space and the gravitational force is perpendicular, not opposite to the motion of the lesser body so absent an equal and opposite force, the lesser body continues its trek about the more massive body indefinitely while it's momentum is in balance with the massive body's gravity. - as we know that the moon is revolving around the earth in circular path where centripetal force is developed by the gravitation and then outward force the outcome of the circular motion "centrifugal force is balancing the centripetal force. - 3 Maybe you could edit some more details into your answer? – Manishearth♦ Jan 14 at 14:27 The original question is: Why doesn't the Moon, or for that matter anything rotating another larger body, ever fall into the larger body? Others have answered that the centrifugal forces equal the centripetal forces, so the moon stays in an orbit of the earth. Satellites orbit the earth for the same reason. However, satellite orbits sometimes decay, so the satellite "orbit" changes to a collapsing spiral, and eventually the satellites do come back down to earth (normally burning up from atmospheric friction). Orbits can end in the other direction too, where the satellite moves away from earth in an enlarging spiral, eventually escaping earth's gravity entirely. - ## protected by Qmechanic♦Jan 15 at 7:54 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": 8, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9461731910705566, "perplexity_flag": "middle"}
http://www.all-science-fair-projects.com/science_fair_projects_encyclopedia/Potential_energy	# All Science Fair Projects ## Science Fair Project Encyclopedia for Schools! Search Browse Forum Coach Links Editor Help Tell-a-Friend Encyclopedia Dictionary # Science Fair Project Encyclopedia For information on any area of science that interests you, enter a keyword (eg. scientific method, molecule, cloud, carbohydrate etc.). Or else, you can start by choosing any of the categories below. # Potential energy Potential energy (U, or Ep), a kind of scalar potential, is energy by virtue of matter being able to move to a lower-energy state, releasing energy in some form. For example a mass released above the Earth has energy resulting from the gravitational attraction of the Earth which is transferred in to kinetic energy. Contents ## Types ### Gravitational potential energy This energy is stored as a result of the elevated position of an object such as a rock on top of a hill or water behind a dam. It is written as $U_g = m g h \,$ where m is the mass of the object, g the acceleration due to gravity and h the height above a chosen reference level (typical units would be kilograms for m, metres/second2 for g, and metres for h). In relation to spacecraft and astronomy g is not constant and the formula becomes an integral. In the case of a sphere of uniform mass (such as a planet), with h measured above the surface, the integral takes the form: $U_g = \int_{h_0}^{h + h_0} {GmM \over r^2} dr$ Where h0 is the radius of the sphere, M is the mass of the sphere, and G is the gravitational constant. If h is instead taken to be the distance from the center of the sphere, then outside the sphere the potential energy relative to that at the center has two terms: $U_g = \int_{h_0}^h {GmM \over r^2} dr + \int_0^{h_0} {GmM \over h_0^2} {r \over h_0} dr$ which evaluates to: $U_g = GmM \left[{1 \over h_0} - {1 \over h}\right] + {1 \over 2} {GmM \over h_0} = GmM \left[{3 \over 2h_0} - {1 \over h}\right]$ [We may also want to link to an explanation of that second term (the gravitational forces created by hollow spherical shells)] A frequently adopted convention is that an object infinitely far away from an attracting source has zero potential energy. Relative to this, an object at a finite distance r from a source of gravitation has negative potential energy. If the source is approximated as a point mass, the potential energy simplifies to: $U_g = - {GmM \over r}$ See also Gravitational binding energy. ### Elastic potential energy This energy is stored as the result of a deformed solid such as a stretched spring. As a result of Hooke's law, it is given by: $U_e = {1\over2}kx^2$ where k is the spring constant (a measure of the stiffness of the spring), expressed in N/m, and x is the displacement from the equilibrium position, expressed in metres (see Main Article: Elastic potential energy). ### Chemical energy Chemical energy is a form of potential energy related to the breaking and forming of chemical bonds. ### Rest mass energy Albert Einstein's famous equation, derived in his special theory of relativity, can be written: $E_0 = m c^2 \,$ where E0 is the rest mass energy, m is mass of the body, and c is the speed of light in a vacuum. (The subscript zero is used here to distinguish this form of energy from the others that follow. In most other contexts, the equation is written with no subscript.) The rest mass energy is the amount of energy inherent in the mass when it is at rest. This equation quantifies the equivalence of mass and energy: A small amount of mass is equivalent to a very large amount of energy. (i.e., 90 petajoule/kg ≈ 21 megatons/kg) ### Electrical potential energy The electrical potential energy per unit charge is called electrical potential. It is expressed in volts. The fact that a potential is always relative to a reference point is often made explicit by using the term potential difference. The term voltage is also common. The electrical potential energy between two charges q1 and q2 is: $U = - \frac{q_1 q_2} {4 \pi \epsilon_o r}$ The electric potential generated by charges q1 (denoted V1) and q2 (denoted V2) is: $V_1 = - \frac{q_1} {4 \pi \epsilon_o r}$ $V_2 = - \frac{q_2} {4 \pi \epsilon_o r}$ ## Relation between potential energy and force Potential energy is closely linked with forces. If the work done going around a loop is zero, then the force is said to be conservative and it is possible to define a numerical value of potential associated with every point in space. A force field can be re-obtained by taking the vector gradient of the potential field. For example, gravity is a conservative force. The work done by a unit mass going from point A with U = a to point B with U = b by gravity is (b - a) and the work done going back the other way is (a - b) so that the total work done from $U_{A \to B \to A} = (b - a) + (a - b) = 0 \,$ The nice thing about potential energy is that you can add any number to all points in space and it doesn't affect the physics. If we redefine the potential at A to be a + c and the potential at B to be b + c [where c can be any number, positive or negative, but it must be the same number for all points] then the work done going from $U_{A \to B} = (b + c) - (a + c) = b - a \,$ as before. In practical terms, this means that you can set the zero of U anywhere you like. You might set it to be zero at the surface of the Earth or you might find it more convenient to set it zero at infinity. A thing to note about conservative forces is that the work done going from A to B does not depend on the route taken. If it did then it would be pointless to define a potential at each point in space. An example of a non-conservative force is friction. With friction, the route you take does affect the amount of work done, and it makes no sense at all to define a potential associated with friction. All the examples above are actually force field stored energy (sometimes in disguise). For example in elastic potential energy, stretching an elastic material forces the atoms very slightly further apart. Powerful electromagnetic forces try to keep the atoms at their optimal distance and so elastic potential is actually electromagnetic potential. Having said that, scientists rarely talk about forces on an atomic scale. Everything is phrased in terms of energy rather than force. You can think of potential energy as being derived from force or you can think of force as being derived from potential energy. ## Graphical representation A graph of a 1D or 2D potential function with the function value scale increasing upward is useful to visualize the potential field: a ball rolling to the lowest part corresponds to a mass or charge, etc. being attracted. E.g. a mass, being an area of attraction, may be called a gravitational well. See also potential well. ## See also 03-10-2013 05:06:04	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 12, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9291680455207825, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/27479/calculating-correlation-functions-of-exponentials-of-fields?answertab=active	# Calculating correlation functions of exponentials of fields In their book Condensed Matter Field Theory, Altland and Simons often use the following formula for calculating thermal expectation values of exponentials of a real field $\theta$: $$\langle e^{i(\theta(x,\tau)-\theta(0,0))} \rangle = e^{-\frac12 \langle (\theta(x,\tau)-\theta(0,0))^2 \rangle}$$ An example can be found in chapter 4.5, problem "Boson-fermion duality", part c). (This refers to the second edition of the book, page 185.) In other words, expectation values of exponentials can be cast as exponentials of expectation values under certain conditions. Unfortunately, I seem to be unable to find an explanation of why this can be done and what the conditions on the Lagrangian of $\theta$ are. Hence, my question is: How to derive the above formula? What do we need to know about $\theta$ for it to be valid in the first place? Ideally, I am looking for a derivation using the path integral formalism. (I managed to rederive a very special case in terms of operators and the Baker-Campbell-Hausdorff formula, but I'd like to gain a more thorough understanding.) - ## 2 Answers This is just a property of Gaussian averaging analogous to the finite dimensional case: $\langle e^{ix} \rangle=\frac{1}{\sqrt{2\pi}\sigma}\int_{-\infty}^{\infty} e^{ix}e^{-\frac{x^2}{2\sigma^2}}=e^{-\frac{\sigma^2}{2}}= e^{-\frac{\langle x^2 \rangle}{2}}$ The field can be decomposed into its independent Gaussian modes and integrated for each mode separately. - The simplest answers are simply the best. – Greg Graviton Oct 7 '11 at 7:50 So this formula only holds in the noninteracting case. – Arnold Neumaier May 2 '12 at 18:48 David Bar Moshe's derivation is of course right. Let me offer you a Taylor-expansion-based alternative proof: $$\left\langle e^{ix} \right \rangle = \left\langle \sum_{n=1}^\infty \frac{(ix)^n}{n!} \right \rangle = \left\langle \sum_{k=1}^\infty \frac{(ix)^{2k}}{(2k)!} \right \rangle$$ Here, I just used that by some odd-ness, the odd powers have a vanishing expectation value. In the formula above, $x$ is whatever linear function of the elementary fields you want, including your coefficient. But the expectation value of $x^{2k}$ is $$\left\langle x^{2k} \right \rangle = \frac{ \int_{-\infty}^\infty {\rm d}x \,x^{2k}\exp(-x^2/2x_0^2)}{ \int_{-\infty}^\infty {\rm d}x \,\exp(-x^2/2x_0^2)} = 1\times 3\times \cdots \times (2k-1) \times x_0^{2k}$$ which may be computed by integrating by parts or by converting it to the Euler integral (which is also evaluated by parts) so when you combine it with the $1/(2k)!$ factor, the odd integers cancel, only the product of the even integers which is equal to $2^k k!$ is left, and the original sum from the first line is $$\sum_{k=1}^\infty \frac{(-x_0^2)^k}{2^k k!} = \exp(-x_0^2/2)$$ where $x_0^2$ is the expectation value of $\langle x^2\rangle$ because I used it in the probabilistic distribution. Again, you may set $x=\theta(x_1,t_1)-\theta(x_2,t_2)$ or whatever linear function of variables and my derivation still holds. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 11, "mathjax_display_tex": 4, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9158428907394409, "perplexity_flag": "head"}
http://quant.stackexchange.com/questions/373/what-is-a-cubature-scheme	# What is a cubature scheme? Ideally an intuitive explanation with an example, please. - ## 1 Answer Cubature (of a given order) is a general method that allows you to do some approximate integration by being exact on a subset of integrand. If you are given a measure $M$ over for example $\mathbb R^n$ then will approach $M$ by (typically) a discrete measure $M^d=\sum_{i=1}^m \lambda_i\delta(x_i)$ such that polynomials $P$ of degree less or equal to $\gamma$ you have : $$\int_{\mathbb R^n}P(x)dM(x)=\sum_{i=1}^m \lambda_i.P(x_i)$$ In the context of a Stochastic Diffusion processes $X_t$ defined by an SDE (ideally in a Stratonovitch form), if you have the to calculate the expectation of a functional of the diffusion path, then you can think of this as an integration over the Wiener measure. Formally this looks like : $$E_{\mathbb{W}}[F(X_.)]=\int_{p\in Path}F(p)d\mathbb{W}(p)$$ Of course here the problem is infinite dimensional so quite hard to address in its full generallity and in a numerically tractable form. Anyway by using Cubature over Wiener Space you can "in a way" approximate the problem by switching to another (and simpler to use) measured space over finite variation paths (recall that Wiener measure doesn't weigh finite variation path !!!) and this approximate measure is such that it matches the values Wiener Measure moments of Iterated Wiener Integral. This transforms then the SDE into a classical ODE (that can be eventually solved analyticaly or numercaly), and finaly your expectation of your functional becomes hopefully numerically tractable. Regards -	{"extraction_info": {"found_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": 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": 0.9231355786323547, "perplexity_flag": "middle"}
http://en.wikipedia.org/wiki/Degree_of_a_polynomial	# Degree of a polynomial The degree of a polynomial is the highest degree of its terms, when the polynomial is expressed in canonical form (i.e. as a linear combination of monomials). The degree of a term is the sum of the exponents of the variables that appear in it. The word degree is now standard, but in some older books, the word order may be used instead. For example, the polynomial $7x^2y^3 + 4x - 9$ has three terms. (Notice, this polynomial can also be expressed as $7x^2y^3 + 4x^1y^0 - 9x^0y^0$.) The first term has a degree of 5 (the sum of the powers 2 and 3), the second term has a degree of 1, and the last term has a degree of 0. Therefore, the polynomial has a degree of 5 which is the highest degree of any term. To determine the degree of a polynomial that is not in standard form (for example $(x+1)^2-(x-1)^2$), one has to put it first in standard form by expanding the products (by distributivity) and combining the like terms; for example $(x+1)^2-(x-1)^2= 4x$, and its degree is 1, although each summand has degree 2. However, this is not needed when the polynomial is expressed as a product of polynomials in standard form, because the degree of a product is the sum of the degrees of the factors. ## Names of polynomials by degree The following names are assigned to polynomials according to their degree:[1] • Degree 0 – constant • Degree 1 – linear • Degree 2 – quadratic • Degree 3 – cubic • Degree 4 – quartic (or, less commonly, biquadratic) • Degree 5 – quintic • Degree 6 – sextic (or, less commonly, hexic) • Degree 7 – septic (or, less commonly, heptic) • Degree 8 – octic • Degree 9 – nonic • Degree 10 – decic The degree of the zero polynomial is either left undefined, or is defined to be negative (usually −1 or −∞). ## Other examples • The polynomial $3 - 5 x + 2 x^5 - 7 x^9$ is a nonic polynomial • The polynomial $(y - 3)(2y + 6)(-4y - 21)$ is a cubic polynomial • The polynomial $(3 z^8 + z^5 - 4 z^2 + 6) + (-3 z^8 + 8 z^4 + 2 z^3 + 14 z)$ is a quintic polynomial (as the $z^8$ are cancelled out) The canonical forms of the three examples above are: • for $3 - 5 x + 2 x^5 - 7 x^9$, after reordering, $- 7 x^9 + 2 x^5 - 5 x + 3$; • for $(y - 3)(2y + 6)(-4y - 21)$, after multiplying out and collecting terms of the same degree, $- 8 y^3 - 42 y^2 + 72 y + 378$; • for $(3 z^8 + z^5 - 4 z^2 + 6) + (-3 z^8 + 8 z^4 + 2 z^3 + 14 z)$, in which the two terms of degree 8 cancel, $z^5 + 8 z^4 + 2 z^3 - 4 z^2 + 14 z + 6$. ## Behavior under addition, subtraction, multiplication and function composition The degree of the sum (or difference) of two polynomials is equal to or less than the greater of their degrees, i.e. $\deg(P + Q) \leq \max(\deg(P),\deg(Q))$. $\deg(P - Q) \leq \max(\deg(P),\deg(Q))$. E.g. • The degree of $(x^3+x)+(x^2+1)=x^3+x^2+x+1$ is 3. Note that 3 ≤ max(3, 2) • The degree of $(x^3+x)-(x^3+x^2)=-x^2+x$ is 2. Note that 2 ≤ max(3, 3) The degree of the product of a polynomial by a non-zero scalar is equal to the degree of the polynomial, i.e. $\deg(cP)=\deg(P)$. E.g. • The degree of $2(x^2+3x-2)=2x^2+6x-4$ is 2, just as the degree of $x^2+3x-2$. Note that for polynomials over a ring containing divisors of zero, this is not necessarily true. For example, in $\mathbf{Z}/4\mathbf{Z}$, $\deg(1+2x) = 1$, but $\deg(2(1+2x)) = \deg(2+4x) =\deg(2) = 0$. The collection of polynomials with coefficients from a given field F and degree smaller than or equal to a given number n thus forms a vector space. (Note, however, that this collection is not a ring, as it is not closed under multiplication, as is seen below.) The degree of the product of two polynomials over a field is the sum of their degrees $\deg(PQ) = \deg(P) + \deg(Q)$. E.g. • The degree of $(x^3+x)(x^2+1)=x^5+2x^3+x$ is 3 + 2 = 5. Note that for polynomials over an arbitrary ring, this is not necessarily true. For example, in $\mathbf{Z}/4\mathbf{Z}$, $\deg(2x) + \deg(1+2x) = 1 + 1 = 2$, but $\deg(2x(1+2x)) = \deg(2x) = 1$. The degree of the composition of two polynomials over a field or integral domain is the product of their degrees $\deg(P \circ Q) = \deg(P)\deg(Q)$. E.g. • If $P = (x^3+x)$, $Q = (x^2+1)$, then $P \circ Q = P \circ (x^2+1) = (x^2+1)^3+(x^2+1) = x^6+3x^4+4x^2+2$, which has degree 6. Note that for polynomials over an arbitrary ring, this is not necessarily true. For example, in $\mathbf{Z}/4\mathbf{Z}$, $\deg(2x) \deg(1+2x) = 1\cdot 1 = 1$, but $\deg(2x\circ(1+2x)) = \deg(2+4x)=\deg(2) = 0$. ## The degree of the zero polynomial This section does not cite any references or sources. Please help improve this section by adding citations to reliable sources. Unsourced material may be challenged and removed. (October 2012) Like any constant value, the value 0 can be considered as a (constant) polynomial, called the zero polynomial. It has no nonzero terms, and so, strictly speaking, it has no degree either. The above rules for the degree of sums and products of polynomials do not apply if any of the polynomials involved is the zero polynomial. It is convenient, however, to define the degree of the zero polynomial to be minus infinity, −∞, and introduce the rules $\max(a,-\infty) = a, \,$ and $a + -\infty = -\infty. \,$ For example: • The degree of the sum $\ (x^3+x)+(0)=x^3+x$ is 3. Note that $3 \le \max(3, -\infty)$. • The degree of the difference $\ x-x = 0$ is $-\infty$. Note that $\ -\infty \le \max(1,1)$. • The degree of the product $\ (0)(x^2+1)=0$ is $\ (-\infty)+2 = -\infty$. The price to be paid for saving the rules for computing the degree of sums and products of polynomials is that the general rule $\ a+b=a \quad \Rightarrow \quad b=0, \,$ breaks down when $\ a = -\infty$. ## The degree computed from the function values The degree of a polynomial f can be computed by the formula $\deg f = \lim_{x\rarr\infty}\frac{\log \|f(x)\|}{\log x}.$ This formula generalizes the concept of degree to some functions that are not polynomials. For example: • The degree of the multiplicative inverse, $\ 1/x$, is −1. • The degree of the square root, $\sqrt x$, is 1/2. • The degree of the logarithm, $\ \log x$, is 0. • The degree of the exponential function, $\ \exp x$, is ∞. Another formula to compute the degree of f from its values is $\deg f = \lim_{x\to\infty}\frac{x f'(x)}{f(x)}.$ ## Extension to polynomials with two or more variables For polynomials in two or more variables, the degree of a term is the sum of the exponents of the variables in the term; the degree of the polynomial is again the maximum of the degrees of all terms in the polynomial. For example, the polynomial x2y2 + 3x3 + 4y has degree 4, the same degree as the term x2y2. However, a polynomial in variables x and y, is a polynomial in x with coefficients which are polynomials in y, and also a polynomial in y with coefficients which are polynomials in x. x2y2 + 3x3 + 4y = (3)x3 + (y2)x2 + (4y) = (x2)y2 + (4)y + (3x3) This polynomial has degree 3 in x and degree 2 in y. ## Degree function in abstract algebra Given a ring R, the polynomial ring R[x] is the set of all polynomials in x that have coefficients chosen from R. In the special case that R is also a field, then the polynomial ring R[x] is a principal ideal domain and, more importantly to our discussion here, a Euclidean domain. It can be shown that the degree of a polynomial over a field satisfies all of the requirements of the norm function in the euclidean domain. That is, given two polynomials f(x) and g(x), the degree of the product f(x)•g(x) must be larger than both the degrees of f and g individually. In fact, something stronger holds: deg( f(x) • g(x) ) = deg(f(x)) + deg(g(x)) For an example of why the degree function may fail over a ring that is not a field, take the following example. Let R = $\mathbb{Z}/4\mathbb{Z}$, the ring of integers modulo 4. This ring is not a field (and is not even an integral domain) because 2•2 = 4 (mod 4) = 0. Therefore, let f(x) = g(x) = 2x + 1. Then, f(x)•g(x) = 4x2 + 4x + 1 = 1. Thus deg(f•g) = 0 which is not greater than the degrees of f and g (which each had degree 1). Since the norm function is not defined for the zero element of the ring, we consider the degree of the polynomial f(x) = 0 to also be undefined so that it follows the rules of a norm in a euclidean domain. ## See also • Degree — for other meanings of degree in mathematics	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 54, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9296350479125977, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/228910/is-the-product-of-the-slopes-of-2-perpendicular-lines-1?answertab=oldest	# Is the product of the slopes of $2$ perpendicular lines $-1$? If I have $2$ lines$AB$ and $PQ$ perpendicular to each other drawn on a graph- then will the slope of $AB$ times the slope of $PQ$ be equal to $-1$? i.e $$[(y2-y1)/(x2-x1)] \times [(b2-b1)/(a2-a1)] = -1 ?$$ - ## 1 Answer Yes, that is so. A small trigonometrical proof will be like this: $$\tan( \theta + \frac{\pi}{2} ) = - \cot {\theta} = - \frac{1}{\tan{\theta}}$$ - So long as neither line is vertical, of course. – Cameron Buie Nov 4 '12 at 18:10 @CameronBuie I see what you did there. – hjpotter92 Nov 4 '12 at 18:27	{"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": 8, "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": 0.8905017375946045, "perplexity_flag": "middle"}
http://cs.stackexchange.com/questions/7340/why-does-every-member-fn-in-thetagn-and-gn-have-to-be-asymptotica	Why does every member $f(n) \in \Theta(g(n))$, and $g(n)$ have to be asymptotically non-negative? The following is an excerpt from CLRS: The definition of $g(n)$ requires that every member $f(n) \in \Theta(g(n))$ be asymptotically nonnegative, that is, that $f(n)$ be nonnegative whenever n is sufficiently large. (An asymptotically positive function is one that is positive for all sufficiently large $n$). Consequently, the function $g(n)$ itself must be asymptotically nonnegative, or else the set $g(n)$ is empty. Intuition suggests that having one function with a positive domain while the other with a negative one perverts the purpose of asymptotic analysis (a measure of the order of growth) as the positive function can be an upper asymptotic bound of the negative function simply on merit of it being positive, even in cases where the negative function grows faster. In cases where both functions have negative domains, asymptotic analysis would still be a valid measure of the order of growth, making the restriction of both functions having to be positive appear useless. - 1 Answer Mostly for convenience's sake. Since big-theta notation specifically tends to be used mostly for algorithmics (as opposed to $O()$ and especially $o()$, which make occasional appearances in more 'traditional' mathematics for talking about goodness of estimates) and is generally used to measure 'usage' quantities of some sort (e.g. CPU cycles, memory, comparisons, etc) there's generally little lost by requiring the quantities involved to be non-negative, and it simplifies some of the definitions and results. Note that unlike with $\Theta()$, the definitions of $O()$ and $o()$ are often phrased in terms of absolute value, making this point moot - can you see why that idea doesn't make sense for $\Theta()$? - Sorry, I don't. Could you please explain why $\Theta()$ shouldn't be phrased in terms of absolute value? – Farhad Yusufali Dec 11 '12 at 22:56 1 @farhadYusufali Consider the function $f(n) = (-1)^nn^2$. Then $f(n)\in O(n^2)$ and $f(n)\in o(n^3)$ both 'make sense' at some level, but $f(n)\in\Theta(n^2)$ doesn't match with what we intuitively 'want' $\Theta()$ to represent. – Steven Stadnicki Dec 11 '12 at 23:02 Thank you for the help. – Farhad Yusufali Dec 11 '12 at 23:05	{"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": 20, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9333606362342834, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/250781/distribution-of-the-sum-of-iid-beta-negative-binomial-random-variables?answertab=active	# Distribution of the sum of iid Beta-Negative-Binomial random variables I am facing a problem when trying to calculate the distribution of the sum of iid Beta-Negative-Binomial random variables or for that matter if only parameter $r$ is different. To get a hint to how they might be distributed I calculated the characteristic function of the BNB distribution: $$\varphi_X(t)=\int_0^1 \! \sum_{x=0}^{\infty} \binom{x+r-1}{x}p^r(1-p)^x\frac{p^{\alpha-1}(1-p)^{\beta-1}}{B(\alpha,\beta)}e^{itx} \, dp$$ where $B$ is the Beta-function. One can now factor out the terms that do not contain $x$ and use the generating function $\sum_{n=0}^\infty \binom{n+k}{k}x^n=\frac{1}{(1-x)^{k+1}}$ to get the following expression for the characteristic function: $$\varphi_X(t)=\int_0^1 \! \frac{p^{\alpha-1}(1-p)^{\beta-1}}{B(\alpha,\beta)}\frac{p^r}{(1-(1-p)e^{it})^r} \, dp.$$ The characteristic function of the sum of two independent variables $X$ and $Y$ is in this case the product of their corresponding characteristic functions. $$\varphi_{X+Y}(t)=\int_0^1 \! \frac{p^{\alpha-1}(1-p)^{\beta-1}}{B(\alpha,\beta)}\frac{p^{r_1}}{(1-(1-p)e^{it})^{r_1}} \, dp \,\,\,\,\int_0^1 \! \frac{p^{\alpha-1}(1-p)^{\beta-1}}{B(\alpha,\beta)}\frac{p^{r_2}}{(1-(1-p)e^{it})^{r_2}} \, dp$$ How can one now deduce the distribution of the sum? Thank you in advance. EDIT: Probability mass function: $\int_0^1 \! \binom{x+r-1}{x}p^r(1-p)^x\frac{p^{\alpha-1}(1-p)^{\beta-1}}{B(\alpha,\beta)}=\binom{x+r-1}{x}\frac{B(\alpha+r,\beta+x)}{B(\alpha,\beta)}=\frac{\Gamma(x+r)\Gamma(\alpha+r)\Gamma(\beta+x)\Gamma(\alpha+\beta)}{x!\Gamma(r)\Gamma(\alpha+\beta+r+x)\Gamma(\alpha)\Gamma(\beta)}$ Thus, the characteristic function can also be written as: $\varphi_X(t) = \sum_{x=0}^{\infty}e^{itx} \binom{x+r-1}{x}\frac{\Gamma(\alpha+r)\Gamma(\beta+x)\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)\Gamma(\alpha+\beta+r+x)}$ EDIT2: I made the non-trivial error to forget the integral in the pmf, which is also the reason why $\varphi_X(0)\ne0$. $\varphi_X(0)=\frac{1}{B(\alpha,\beta)} \int_0^1 p^{\alpha-1}(1-p)^{\beta-1}\frac{p^r}{(1-(1-p))^r} \, dp = 1$ with $B(x,y)=\int_0^1 p^{x-1}(1-p)^{y-1} \, dp$ - 1 QUOTE: The characteristic function of the sum of two variables $X$ and $Y$ is defined as the product of their corresponding characteristic functions. END OF QUOTE. That's not true. If they're independent, then the charateristic function of their sum can be shown to be the product of their characteristic functions, but it is not defined to be that. Rather, that is a demonstrable fact, not a definition. The definition is still the same as for characteristic functions in general: it is $t\mapsto \mathbb E(e^{t(X+Y)})$. – Michael Hardy Dec 4 '12 at 17:34 1 Some of your characteristic functions $\varphi$ are such that $\varphi(0)\ne1$. This is odd. – Did Dec 4 '12 at 18:52 @MichaelHardy : That is obviously correct. I will correct that in my question. – Mark Dec 4 '12 at 19:26 @did: I will add the pmf of the BNB distribution too so it will be more obvious why i tried it this way. Trying to find the characteristic function when using the textbook definition of the pmf has proven to be too hard for me. – Mark Dec 4 '12 at 19:29 1 The trouble is not with what you tried or did not try but with the fact that characteristic functions are defined as $\varphi_X(t)=\mathbb E(\mathrm e^{itX})$. Hence $\varphi_X(0)=1$ for every random variable $X$. If $\psi(0)\ne1$, then $\psi\ne\varphi_X$ for every $X$. Thus, there is a problem with your formulas, which I suggest to get rid of before going any further. – Did Dec 4 '12 at 23:23 show 2 more comments	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.9230960011482239, "perplexity_flag": "head"}
http://www.stanford.edu/class/cs221/homeworks/homework2.4.html	1. [2 points] Let $h_1$ and $h_2$ be consistent heuristics. Define a new heuristic $h(s) = \max\{h_1(s), h_2(s)\}$. Prove that $h$ is consistent. 2. [2 points] Suppose a heuristic $h$ is consistent for a particular state space model. If new edges are added to the search space graph, does $h$ remain consistent for the new model? What if edges are removed? In both cases, give a proof or a counterexample. 3. [2 points] Consider a search problem with start state $s_\text{start}$ and goal state $s_\text{goal}$, where each action is reversible with the same cost (i.e., if an action takes you from $s$ to $t$ with cost $c$, then some action can take you from $t$ to $s$ with cost $c$). Suppose you've run Uniform Cost Search (UCS) and found the correct answer. Now consider the problem of finding the minimum cost path from a different start state $s_\text{new}$ to a the goal state $s_\text{start}$. Define a consistent heuristic for A* for the new model. 4. [2 points] Assume all edge costs are strictly positive. Define a heuristic $h(s)$ based on breadth-first search and show that it is consistent. Now suppose you are given a new model with the same graph structure but different costs. How would you adapt the old heuristic so that it is consistent for the new model? Your adaptation should take constant time.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 17, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9166388511657715, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/9998/is-there-any-interacting-quantum-field-theory-of-massless-spin-1-fields-expresse	# Is there any interacting quantum field theory of massless spin-1 fields expressed locally entirely in terms of F, with no vector potential? Is there any interacting quantum field theory of massless fields with helicity $\pm 1$ which can be expressed entirely locally in terms of the field strength Fμν with no reference to vector potentials at all? Clearly, quantum electrodynamics doesn't fall into this category. The Aharonov-Bohm effect is the reason. Classically, it's easy to come up with many such theories, but unfortunately, they happen to be nonrenormalizable. But do asymptotically safe models exist? - i think the word on the street is that the fact that the gauge field constrained to $U(1)$ covariance would somehow be connected to renormalizability, of course you are concerned with the converse affirmation; does the absence of a interesting gauge imply nonrenormalizable? – lurscher May 18 '11 at 20:43 1 Before even attempting to discuss QFT and renormalizability, I would recommend you to be more specific about, e.g., the classical field content ($\vec{E}$ and $\vec{B}$ as opposed to $A_{\mu}$?), and the classical equations of motion (Maxwell equations?), and perhaps mention an action for one of the many classical theories, that it's easy to come up with, cf. v2 of the question. Then we might get a better idea of what sort of Aharonov-Bohm-free theory, that you are looking for, at the quantum level. – Qmechanic♦ May 19 '11 at 14:46 ## 2 Answers There are simple non- renormalizable examples: the easiest is QED with elementary neutrons only. The neutrons have a magnetic dipole moment, and interact with the photon locally through the F tensor only contracted with the antisymmetric product of gamma matrices. There are no renormalizable local examples in 4d because we have a list of all of these, and the vectors always interact with charged particles, not magnetically polarized ones. (I missed that the question asks specifically for asymptotically safe examples, and the OP was probably aware of this system. But the answer might be useful to others) - The answer is yes. There are many examples of emergent gauge theory from quantum spin models on lattice. For example, see Phys. Rev. D68, 024501 (2003), and references there. The quantum spin models have no gauge symmetry and all their degrees of freedom are physical and, in some sense, correspond to $F_{\mu\nu}$. - But these models are not Lorentz invariant. If you make Lorentz invariant helicity 1 theory, can you get a pure F coupling in any renormalizable model? Perturbative considerations for local actions say no. What is the low-energy limit of the model you are considering? – Ron Maimon May 28 '12 at 7:00 @Ron: The low-energy limit of the model has emergent Lorentz symmetry, and photons with only helicity $\pm 1$ modes. At lattice scale, the model is just some quantum spin model with no Lorentz symmetry and no gauge symmetry. – Xiao-Gang Wen May 28 '12 at 7:43	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.9232187271118164, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/220985/how-many-additive-abelian-groups-g-of-order-16-have-the-property-such-xxxx?answertab=active	# How many additive abelian groups $G$ of order 16 have the property such $x+x+x+x=0$ Up to isomorphic, how many additive abelian groups $G$ of order 16 have the property such that $x+x+x+x=0$, for each $x$ in G? My question is that which theorem can be used? My answer is 3. Is that right? - – blitzer Oct 25 '12 at 17:44 ## 3 Answers It follows from the hypothesis (and the Structure Theorem for Abelian Groups) that such groups will be products of $\Bbb Z_2$ and/or $\Bbb Z_4$. In particular, it will be one of $\Bbb Z_4\times\Bbb Z_4$, $\Bbb Z_4\times\Bbb Z_2\times\Bbb Z_2$, or $\Bbb Z_2\times\Bbb Z_2\times\Bbb Z_2\times\Bbb Z_2$. Your answer is correct. - Do you know the classification theorem for finitely generated abelian groups? It states that $$G \simeq \mathbb{Z}^j \oplus \mathbb{Z}_{p_1^{r_1}}\oplus \mathbb{Z}_{p_2^{r_2}}\oplus ... \oplus \mathbb{Z}_{p_k^{r_k}}$$ where $k$ and the $r_i$'s are positive and the $p_i$'s are non necessarily distinct primes. Thus, there are only a finite number of abelian groups with order 16, and only a subset of these satisfy your condition, which is that every element has order less than or equal to 4. - Using the big gun "classification of finite abelian groups", we see that there are $(\mathbb Z/4\mathbb Z)^2$, $\mathbb Z/4\mathbb Z\times (\mathbb Z/2\mathbb Z)^2$ and $(\mathbb Z/2\mathbb Z)^4$. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 16, "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": 0.9552257061004639, "perplexity_flag": "head"}
http://mathoverflow.net/questions/28040/de-rham-decomposition-theorem-generalisations-and-good-references/63046	## De Rham decomposition theorem, generalisations and good references ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) De Rham decomposition theorem states that every simply-connected Riemannian manifold $M$ that admits complementary sub-bundles $T'(M)$ and $T''(M)$ of its tangent bundle parallel with respect to the levi-Chevita connection is isometric to the direct product of two Riemanninan manifolds $M'\times M''$. Question 1. In the first place I would like to have a good reference for a clear "modern" and complete proof of this theorem, if it exists (more recent than Kobayshi-Nomizu pp. 187-193) (Note, that Besse 10.44 claimed that no simple proof exists yet). Eddited. Question 2. Secondly it seems to me that there should be some statement much more general than de Rham theorem. Namely, suppose we have a metric space $X$ that is locally decomposable as an isometric product of two in such a way that this decomposition is "coherent" in a appropriate sense, i.e. forms something like a presheaf. When will we be able to say that $X=Y\times Z$? (I am interested only in the cases when this will work, not when this will fail). As corollary of such a general statement one should be able to deduce de Rham theorem for example, for Finsler of polyhedral manifold, ect. - intlpress.com/JDG/archive/1972/7-1&2-161.pdf – Steve Huntsman Jun 13 2010 at 16:44 ams.org/journals/proc/1998-126-10/… – Steve Huntsman Jun 13 2010 at 16:46 emis.ams.org/journals/GM/vol5/reck.ps – Steve Huntsman Jun 13 2010 at 16:47 springerlink.com/content/k41141205w063k47 – Steve Huntsman Jun 13 2010 at 16:50 10 Steve, it appears to me that you're just doing a search on Riemannian decomposition on mathscinet, google, or somewhere and listing some of the more promising results. I think your response would be more helpful, if you could actually reveal how you're finding these papers and saying more about what you know about them. Otherwise, you're not providing any more help to Dmitri than what he could easily do himself. – Deane Yang Jun 13 2010 at 18:40 show 3 more comments ## 4 Answers For the first question, I can provide some literature:de Rham's original proof, Wu Hongxi's Ph.D thesis On the de Rham decomposition theorem. I don't know which reference would be earliest one. http://dspace.mit.edu/handle/1721.1/11601 - ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. I suppose we have to ask the Riemannian manifold to be complete. Otherwise $\mathbb R^3 - \lbrace 0 \rbrace$ would be a counterexample. I do not have an answer to question 2, but you might be interested in variations of De Rham decomposition Theorem to the realm of compact complex manifolds. There, it is natural to ask, as Beauville did, if a holomorphic decomposition of the holomorphic tangent bundle implies that the universal covering is isomorphic to a product(no metric assumption). Without further assumptions there is no hope since Hopf surfaces provide examples with decomposable tangent bundle but with universal covering isomorphic to $\mathbb C^2 - \lbrace 0 \rbrace$. If one assume that the manifold is projective or Kahler then there are some positive results, the first of which can be found in Beaviulle's paper linked to above. In the projective case you can also look at this, this, and this paper. In the Kahler case you can look at here. The general problem seems to be wide open, in the above results either one assumes that one of the factors is one-dimensional or imposes strong conditions on the ambient variety itself (dimension $\le 3$ or uniruledness). - what does it means that a sub-bundle of the tangent bundle is parallel with respect to the levi-Civita connection? I found that the for each couple of vector fields $X,Y$ of the subbundle the covariant derivative is in the subbundle too. Is it correct? - This means, that if $X'$ ($X''$) is in the sub-bundle $TM'$ ($TM''$) then its covariant derivative along any other field is also in this sub bundle. – Dmitri Apr 26 2011 at 15:09 what does it means that a sub-bundle of the tangent bundle is parallel with respect to the levi-Civita connection? I found that the for each couple of vector fields $X,Y$ of the subbundle the covariant derivative is in the subbundle too. Is it correct? It is stupid, it is a question! Sorry -	{"extraction_info": {"found_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": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.925518810749054, "perplexity_flag": "head"}
http://mathhelpforum.com/calculus/74745-finding-area-region.html	Thread: 1. Finding the area of this region I've attached the problem I'm trying to solve. I know I'd want to set the two equal to find the points of intersection, but would I leave them in terms of X or Y? And then I would just take the integral of the top minus the bottom curve but I can't figure out WHAT is the top and bottom curve in this problem since they both kind of cross at the same point. Attached Thumbnails 2. w/r to y ... (right curve) - (left curve) $\int_0^5 (3y-y^2) - (y^2-7y) \, dy$ 3. Ok my final answer was 112/3, and this may be a stupid follow up question, but would my unit just be units squared? It just seems weird to use units squared when dealing with odd shapes like this, but it's still correct right? 4. Originally Posted by fattydq Ok my final answer was 112/3, and this may be a stupid follow up question, but would my unit just be units squared? It just seems weird to use units squared when dealing with odd shapes like this, but it's still correct right? Yes, area is always in (linear units) squared.	{"extraction_info": {"found_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": 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": 0.9802923202514648, "perplexity_flag": "middle"}
http://www.physicsforums.com/showthread.php?t=479218	Physics Forums cardinality of sets of functions Hello, let's consider the set $\Omega$ of all the continuous and integrable functions $f:R \to R$. Suppose we now take two subsets A and B, where: - A is the subset of all the gaussian functions centered at the origin: $\exp(-ax^2)$, where a>0 - B is the set of all the even functions: $f(x)=f(-x)$ It is trivial to prove that $A\subset B$. However, my question is: is it possible to "quantify" how bigger the subset B is compared to A ? How should I count the cardinality of A and B ? My goal is computing the ratio of cardinalities: \|B\|/\|A\|. Is it possible? Thanks. 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: 27 Recognitions: Gold Member Homework Help Science Advisor hello mnb96! the set A can be indexed by one parameter, a … how many parameters are needed to index the set B ? uhm...how many parameters do I need to form a continuous and integrable even-function? I don't know! intuitively I would guess an infinite amount of parameters: the cardinality of $\mathbb{R}^+$, but I might be wrong, as I don't know how to (dis)prove it, and I have no idea on how to constrain the parameters so that the function they represent is in $\Omega$ (= is continuous and integrable). EDIT: oh wait...maybe I could represent them with Taylor series: in that case all the coefficients of the even-powers are potentially non-zero (still they are infinite, and who knows if they generate an integrable function?). Btw...how would you use the number of parameters to identify the cardinality of a set? For example A is identified with one parameter. Fine, so what is its cardinality? Recognitions: Homework Help Science Advisor cardinality of sets of functions Let $\mathfrak{c}$ denote the cardinality of the continuum. Then it's easy to see that: $$\|\mathbb{R}\| = \|\mathbb{R} ^+\| = \|A\| = \|B\| = \|\Omega \| = \mathfrak{c}$$ The following equalities/inequalities are obvious: $$\mathfrak{c} = \|\mathbb{R}\| \geq \|\mathbb{R}^+\| = \|A\| \leq \|B\| \leq \|\Omega \|$$ So it'll suffice to show: $$\|\mathbb{R}\| = \|\mathbb{R} ^+\|,\ \|\mathbb{R}\| = \|\Omega \|$$ The first is perhaps a good exercise for you to do. For the second, first observe that a continuous function is completely determined by how it acts on the rationals. Moreover, since every real number is the limit of a (countable) sequence of rationals, every function $\mathbb{Q} \to \mathbb{R}$ is the pointwise limit of some (countable) sequence of functions $\mathbb{Q} \to \mathbb{Q}$. So we get an upper bound on $\|\Omega \|$ if we can count the number of countable sequences of functions from the rationals to the rationals. Since the rationals and naturals have the same cardinality, we want to count the number of countable sequences of functions $\mathbb{N} \to \mathbb{N}$. Such a sequence $f_0, f_1, \dots$ can be regarded as a function $f: \mathbb{N} \times \mathbb{N} \to \mathbb{N}$ (where $f(m,n) = f_m(n)$). If you haven't seen it already, you should do the following: Exercise: $\|\mathbb{N} \times \mathbb{N}\| = \|\mathbb{N}\|$ So we've reduced this to counting the number of fucntions $\mathbb{N} \to \mathbb{N}$. Since such a function can be regarded as a subset of $\mathbb{N} \times \mathbb{N}$, we can reduce the problem to counting the number of subsets of $\mathbb{N}$ by the above exercise. The following exercise finishes it: Exercise: $\|\mathbb{R}\| = \|\mathcal{P}(\mathbb{N})\|$ where $\mathcal{P}$ denotes the power set. -------------------- So $\|A\| = \|B\| = \mathfrak{c}$, but it's also easy to see that $\|B \setminus A\| = \mathfrak{c}$, so that's one way in which you might want to quantify the intuitive feeling that "$B$ is much bigger than $A$." It doesn't make sense to divide or subtract infinite cardinals, so the ratio $\|B\|/\|A\|$ doesn't make sense. In fact for infinite cardinals, addition and multiplication aren't that interesting: $$\kappa \times \lambda = \kappa + \lambda = \mbox{max}\{\kappa , \lambda \}$$ Exponentiation however is of great interest, and studying the possibilities of $\kappa ^{\lambda}$ is indeed its own sub-discipline of set theory. -------------------- Cardinality is the most basic way to measure size, but there are other ways in which you may wish to measure size which let you say more precisely how much bigger $\|B\|$ is than $A$. Another way is topologically: $\Omega$ has a natural structure as a metric space, using the bounded sup metric. You may want to ask whether $A$ is meager with respect to this topology, and whether $B$ is not. Yet another way is algebraically. $\Omega$ has a natural vector space structure, where $B$ is a subspace and $A$ spans a proper subspace of $B$. You may want to ask about the codimension of $\mbox{Span}(A)$ in $B$. Hi AKG, thanks for your precise and interesting explanation. After having "meditated" a bit on your answer, I came up to the conclusion that for my purposes it would be probably more interesting to consider $\Omega$ as a vector space and try to find the codimension of Span(A) in B. However, I have two questions: 1) B are continuous integrable even function: how would you find a finite amount of basis functions to represent this subspace? 2) Doesn't the concept of codimension have meaning only for finite-dimensional vector spaces? Thanks again! Recognitions: Homework Help Science Advisor Quote by mnb96 Hi AKG, thanks for your precise and interesting explanation. You're welcome! After having "meditated" a bit on your answer, I came up to the conclusion that for my purposes it would be probably more interesting to consider $\Omega$ as a vector space and try to find the codimension of Span(A) in B. However, I have two questions: 1) B are continuous integrable even function: how would you find a finite amount of basis functions to represent this subspace? 2) Doesn't the concept of codimension have meaning only for finite-dimensional vector spaces? Thanks again! 1) First observe quickly that $B$ is a subspace of $\Omega$, i.e. it satisfies the required closure properties, so it can be regarded as a vector space by itself. Thus it makes sense to ask whether it has a basis. You're correct, it won't have a finite basis, thus $B$ is an example of an infinite dimensional vector space. Since $\|B\| = \mathfrak{c}$, the largest its dimension could possibly be is $\mathfrak{c}$. I can find a set of $\mathfrak{c}$ linearly independent vectors in $B$, hence its dimension is also at most $\mathfrak{c}$. Thus its dimension is precisely $\mathfrak{c}$ (as a vector space over $\mathbb{R}$). 2) The definition extends to the infinite-dimensional case: If $W$ is a subspace of $V$, then the codimension of $W$ in $V$ is defined to be the dimension of the quotient space $V/W$. It's not hard to show that if you can find a basis $\alpha$ for $W$ and a basis $\beta$ for $V$ such that $\alpha \subset \beta$, then: Exercise (if you're comfortable with the notion of quotient spaces): $$\mbox{dim}(V/W) = \|\beta \setminus \alpha\|$$ So in our case, we have $W = \mbox{Span}(A)$, and $V = B$. Exercise: $A$ is in fact linearly independent, hence it forms a basis for $\mbox{Span}(A)$. So we may set $\alpha = A$. Note also that this justifies the claim I made earlier about being able to find a $\mathfrak{c}$-sized linearly independent subset of $B$. So in theory, what we want to do now is extend $\alpha$ to a basis $\beta \supset \alpha$ of $B$ and then count $\|\beta \setminus \alpha \|$. In practice, since we already have the guess that $B$ is much bigger than $A$, our guess is that this count will be as large as it can be, namely $\mathfrak{c}$. To prove this guess, it'll be easier to find an easy-to-describe class of functions $\gamma$ such that $\alpha \cup \gamma$ is linearly independent and contained in $B$, and $\|\gamma \| = \mathfrak{c}$. The best rule of thumb I have for undergrads is, when coming up with examples or counterexamples, DON'T THINK TOO HARD! A classic example: if you're asked to find an invertible matrix that's not diagonalizable, try to think of a 2x2 matrix with only 0-1 entries, rather than a 3x3 or 4x4 matrix with all sorts of various numerical entries. In this case, we already know that the family of functions $x \mapsto \exp (-ax^2)$ is a linearly independent family of size $\mathfrak{c}$ belonging to $B$. Exercise: The family $\gamma$ of functions $x \mapsto \exp(-ax^4)$ is contained in $B$, has size $\mathfrak{c}$, and $\alpha \cup \gamma$ is linearly independent. Recognitions: Homework Help Science Advisor By the way, by "integrable" do you mean that the integral $\int _{-\infty} ^{\infty} f(x)dx$ exists and equals a finite number, or simply that the function $f$ has an antiderivative? If it's the latter, then saying "integrable continuous" is redundant since continuity implies that kind of integrability. Hi AKB, thanks a lot! your answers are very dense and complete. I am still going through it, so let's try to clarify some points. Quote by AKG ...First observe quickly that $B$ is a subspace of $\Omega$, ... ... ... ..., the largest its dimension could possibly be is $\mathfrak{c}$. I can find a set of $\mathfrak{c}$ linearly independent vectors in $B$, hence its dimension is also at most $\mathfrak{c}$. Ok, are you referring, for example, to the Fourier bases (only real part), or perhaps to the Taylor polynomials (only even grade), and so on...? Thus its dimension is precisely $\mathfrak{c}$ (as a vector space over $\mathbb{R}$). How did you deduce, that if the dimension is at most $\mathfrak{c}$, then it must be exactly $\mathfrak{c}$? Intuitively it makes sense, but still, it would be nice to understand it with a bit of rigor. The definition extends to the infinite-dimensional case: If $W$ is a subspace of $V$, then the codimension of $W$ in $V$ is defined to be the dimension of the quotient space $V/W$ I have to review the notion of quotient space. When I have refreshed the topic, I will come back to the proof & exercises. By the way, by "integrable" do you mean that the integral $\int _{-\infty} ^{\infty} f(x)dx$ exists and equals a finite number Yes, I meant this one. --- By the way, what is the main difference between what you proved now (using the notion of vector space), and the statement that a set is meager with respect to another set? Thanks again! Recognitions: Homework Help Science Advisor We agree that $B$ is a subspace right? The sum of two even functions is even, a scalar multiple of an even function is even, and the zero function is even. Therefore, it makes sense to ask for a basis for $B$. Like with $\mathbb{R}^2$, if we consider the line through the origin $y = x$, it's a subspace, so we can ask to find a basis of it. On the other hand, the line $y = x + 1$ is not a subspace, so it doesn't make sense to ask for a basis for it. So at this point, all I'm saying is that $B$ has a basis, I'm not saying what it is. Let's quickly review what a basis is: For $\beta \subset B$, we say $\beta$ is [b]linearly independent[/itex] iff for every finite linear combination of vectors in $\beta$ the linear combination is $\vec{0}$ iff all the coefficients appearing in the linear combination are 0. $\beta$ spans $B$ iff every vector in $B$ can be written as a finite linear combination of vectors in $\beta$. $\beta$ is a basis for $B$ iff it's linearly independent and spans $B$ or, equivalently $\beta$ is a basis for $B$ iff every vector in $B$ can be written in a unique way as a finite linear combination of vectors in $\beta$ I emphasized finite throughout my definitions. If you take the second equivalent formulation of "basis" and replace "finite" with "countable," you get a slightly different notion of basis known as a Schauder basis. Sometimes, to make the distinction more clear, the notion of basis where "finite" is used is called a Hamel basis. But when you see the word "basis" used somewhere and it's not specified whether they mean Hamel basis or Schauder basis, i.e. they don't specify whether only finite linear combinations are allowed or whether countably infinite combinations are being allowed, the standard meaning is Hamel basis, i.e. only finite combinations. So the $\beta$ we're (in theory) looking for is a basis (= Hamel basis) but the examples you gave, such as the real parts of the Fourier basis, are Schauder bases. So, in theory, we're looking for a set of functions $\beta$ such that every even continuous integrable function can be written as a finite linear combination of those functions, in a unique way. I can't give you a nice, explicit list of vectors that'll do the trick. The reason is that the very existence of a basis for $B$ depends on the axiom of choice, and typically anything whose existence require AC cannot be described in an explicit manner (otherwise it wouldn't have required AC!). However, AC is used ubiquitously in mathematics and this is no reason to doubt the existence of a basis. In fact: Fact: The statement "every vector space has a basis" is equivalent to the axiom of choice. Also, without AC, the notion of cardinality itself changes drastically. Anyways, using AC, we have the following fact: Fact: Every vector space has a basis, and any two bases of the same vector space have the same cardinality, hence "dimension" is a well-defined concept. Recall the dimension of a vector space is the cardinality one of/all of its bases. How did you deduce, that if the dimension is at most $\mathfrak{c}$, then it must be exactly $\mathfrak{c}$? Intuitively it makes sense, but still, it would be nice to understand it with a bit of rigor. First I said that its dimension is at most $\mathfrak{c}$ since the cardinality of the whole space $B$ itself is $\mathfrak{c}$. Next I said its dimension is at least $\mathfrak{c}$ since it contains a linearly independent subset of size $\mathfrak{c}$. Hence, its dimension equals $\mathfrak{c}$. To get the at least part, I used the following: Fact: If $V$ is a vector space and it has a linearly independent subset of size $\kappa$, then $\mbox{dim}(V) \geq \kappa$. This fact should be familiar to you in the case $\kappa$ finite. It happens to hold true in general. By the way, what is the main difference between what you proved now (using the notion of vector space), and the statement that a set is meager with respect to another set? Thanks again! Again, you're welcome! There's a very big difference between what I proved here and the statement that one set is meager with respect to another. Firstly, not every topological space has a natural vector space structure, so there may be two sets where it makes sense to compare them topologically (e.g. decide if one is meager as a topological subspace of the other) but not be able to compare them algebraically (e.g. decide if one has large codimension as a vector subspace of the other). Likewise, not every vector space has a natural topology. Furthermore, there are examples of spaces with a natural topology and a natural vector space structure, such that a given subset is large in one sense but small in the other. For example, if we take the space $\mathbb{R}^{\infty}$ of countably infinite sequences of real numbers that are eventually 0, then there's a natural vector space structure: $$(x_0, x_1, \dots ) + (y_0, y_1, \dots ) = (x_0 + y_0, x_1 + y_1, \dots )$$ and you can guess how scalar multiplication works. The subset: $$A = \{ (x_0, x_1, \dots ) \in \mathbb{R}^{\infty} : x_0 = 0 \}$$ has codimension 1, hence it's a relatively large algebraically speaking. However $\mathbb{R}^{\infty}$ has some natural topological structures it inherits as a subspace of $\mathbb{R}^{\omega}$, where the latter set can be given the product topology or the box topology as the $\omega$-fold product of the topological space $\mathbb{R}$. Never mind if you don't know what some of those words mean. The bottom line is that, in either of those topologies, A forms a meager subspace of $\mathbb{R}^{\infty}$, in fact it forms a nowhere dense subset. Finally, it's worth noting that there are different ways to measure relative size, it all depends on the context. If there's a vector space structure, there are algebraic ways of measuring relative size, like codimension, and if there's a topology, the notion of meagerness plays the role of "relatively small" or "negligible." If there's a natural measure, we can ask whether a certain set has measure 0. If we have an ultrafilter on the larger set in question, we can ask whether the subset in question belongs to said ultrafilter. If the larger set in question has a natural well-order, we can ask whether the subset is closed-unbounded, or whether it's stationary. Etc, etc, etc. Recognitions: Homework Help Science Advisor We agree that $B$ is a subspace right? The sum of two even functions is even, a scalar multiple of an even function is even, and the zero function is even. Therefore, it makes sense to ask for a basis for $B$. Like with $\mathbb{R}^2$, if we consider the line through the origin $y = x$, it's a subspace, so we can ask to find a basis of it. On the other hand, the line $y = x + 1$ is not a subspace, so it doesn't make sense to ask for a basis for it. So at this point, all I'm saying is that $B$ has a basis, I'm not saying what it is. Let's quickly review what a basis is: For $\beta \subset B$, we say $\beta$ is linearly independent iff for every finite linear combination of vectors in $\beta$ the linear combination is $\vec{0}$ iff all the coefficients appearing in the linear combination are 0. $\beta$ spans $B$ iff every vector in $B$ can be written as a finite linear combination of vectors in $\beta$. $\beta$ is a basis for $B$ iff it's linearly independent and spans $B$ or, equivalently $\beta$ is a basis for $B$ iff every vector in $B$ can be written in a unique way as a finite linear combination of vectors in $\beta$ I emphasized finite throughout my definitions. If you take the second equivalent formulation of "basis" and replace "finite" with "countable," you get a slightly different notion of basis known as a Schauder basis. Sometimes, to make the distinction more clear, the notion of basis where "finite" is used is called a Hamel basis. But when you see the word "basis" used somewhere and it's not specified whether they mean Hamel basis or Schauder basis, i.e. they don't specify whether only finite linear combinations are allowed or whether countably infinite combinations are being allowed, the standard meaning is Hamel basis, i.e. only finite combinations. So the $\beta$ we're (in theory) looking for is a basis (= Hamel basis) but the examples you gave, such as the real parts of the Fourier basis, are Schauder bases. So, in theory, we're looking for a set of functions $\beta$ such that every even continuous integrable function can be written as a finite linear combination of those functions, in a unique way. I can't give you a nice, explicit list of vectors that'll do the trick. The reason is that the very existence of a basis for $B$ depends on the axiom of choice, and typically anything whose existence require AC cannot be described in an explicit manner (otherwise it wouldn't have required AC!). However, AC is used ubiquitously in mathematics and this is no reason to doubt the existence of a basis. In fact: Fact: The statement "every vector space has a basis" is equivalent to the axiom of choice. Also, without AC, the notion of cardinality itself changes drastically. Anyways, using AC, we have the following fact: Fact: Every vector space has a basis, and any two bases of the same vector space have the same cardinality, hence "dimension" is a well-defined concept. Recall the dimension of a vector space is the cardinality one of/all of its bases. How did you deduce, that if the dimension is at most $\mathfrak{c}$, then it must be exactly $\mathfrak{c}$? Intuitively it makes sense, but still, it would be nice to understand it with a bit of rigor. First I said that its dimension is at most $\mathfrak{c}$ since the cardinality of the whole space $B$ itself is $\mathfrak{c}$. Next I said its dimension is at least $\mathfrak{c}$ since it contains a linearly independent subset of size $\mathfrak{c}$. Hence, its dimension equals $\mathfrak{c}$. To get the at least part, I used the following: Fact: If $V$ is a vector space and it has a linearly independent subset of size $\kappa$, then $\mbox{dim}(V) \geq \kappa$. This fact should be familiar to you in the case $\kappa$ finite. It happens to hold true in general. By the way, what is the main difference between what you proved now (using the notion of vector space), and the statement that a set is meager with respect to another set? Thanks again! Again, you're welcome! There's a very big difference between what I proved here and the statement that one set is meager with respect to another. Firstly, not every topological space has a natural vector space structure, so there may be two sets where it makes sense to compare them topologically (e.g. decide if one is meager as a topological subspace of the other) but not be able to compare them algebraically (e.g. decide if one has large codimension as a vector subspace of the other). Likewise, not every vector space has a natural topology. Furthermore, there are examples of spaces with a natural topology and a natural vector space structure, such that a given subset is large in one sense but small in the other. For example, if we take the space $\mathbb{R}^{\infty}$ of countably infinite sequences of real numbers that are eventually 0, then there's a natural vector space structure: $$(x_0, x_1, \dots ) + (y_0, y_1, \dots ) = (x_0 + y_0, x_1 + y_1, \dots )$$ and you can guess how scalar multiplication works. The subset: $$A = \{ (x_0, x_1, \dots ) \in \mathbb{R}^{\infty} : x_0 = 0 \}$$ has codimension 1, hence it's a relatively large algebraically speaking. However $\mathbb{R}^{\infty}$ has some natural topological structures it inherits as a subspace of $\mathbb{R}^{\omega}$, where the latter set can be given the product topology or the box topology as the $\omega$-fold product of the topological space $\mathbb{R}$. Never mind if you don't know what some of those words mean. The bottom line is that, in either of those topologies, A forms a meager subspace of $\mathbb{R}^{\infty}$, in fact it forms a nowhere dense subset. Finally, it's worth noting that there are different ways to measure relative size, it all depends on the context. If there's a vector space structure, there are algebraic ways of measuring relative size, like codimension, and if there's a topology, the notion of meagerness plays the role of "relatively small" or "negligible." If there's a natural measure, we can ask whether a certain set has measure 0. If we have an ultrafilter on the larger set in question, we can ask whether the subset in question belongs to said ultrafilter. If the larger set in question has a natural well-order, we can ask whether the subset is closed-unbounded, or whether it's stationary. Etc, etc, etc. Thread Tools \| \| \| \| \|-------------------------------------------------------\|--------------------------------------------\|---------\| \| Similar Threads for: cardinality of sets of functions \| \| \| \| Thread \| Forum \| Replies \| \| \| Calculus & Beyond Homework \| 3 \| \| \| Set Theory, Logic, Probability, Statistics \| 5 \| \| \| Set Theory, Logic, Probability, Statistics \| 1 \| \| \| Set Theory, Logic, Probability, Statistics \| 6 \| \| \| Set Theory, Logic, Probability, Statistics \| 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": 188, "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": 0.9481604695320129, "perplexity_flag": "head"}
http://nrich.maths.org/1954/note	### Just Touching Three semi-circles have a common diameter, each touches the other two and two lie inside the biggest one. What is the radius of the circle that touches all three semi-circles? ### Bendy Quad Four rods are hinged at their ends to form a convex quadrilateral. Investigate the different shapes that the quadrilateral can take. Be patient this problem may be slow to load. ### Biggest Bendy Four rods are hinged at their ends to form a quadrilateral with fixed side lengths. Show that the quadrilateral has a maximum area when it is cyclic. # Hexy-metry ### Why do this problem? This problem requires the solver to reason geometrically and make use of symmetry. By re-presenting the information in a different way, for example by adding additional lines (a useful technique in geometrical problems) more structure can be revealed. It is an interesting idea that adding something, and therefore apparently making it more complex, can sometimes make a problem more accessible. Then of course there is an opportunity to use the cosine rule in a non-standard context. ### Possible approach Use the images in this document to make cards. Display the problem and ask learners to work in groups rearranging the cards in ways which help to make connections for them. Share ideas and relationships that groups notice before going on to solve the problem. Encourage careful reasoning and convincing arguments. For example: • "Why can you say those two angels are equal? There are several ways of solving the problem so share different approaches and discuss what helped to move thinking forward. Two observations may be worth drawing attention to : • the reflection symmetry between $a$ followed by $b$ and $b$ followed by $a$, • the angle at the circumference as half the angle subtended at the centre onto the same arc (in this case an arc greater than a semicircle) . ### Key questions • How can you use the symmetry of the figure to determine some of the angles ? • What things in the images helped to make connections for you? • What do we need to find, and can that be found directly ? ### Possible extension For a similar level challenge try Bendy Quad And for something much harder try Biggest Bendy ### Possible support For less able students concentrate on the interesting property that all angles in the hexagon are equal. This might be approached practically : drawing a circle, split into thirds to emphasise the symmetry, then splitting each third into the same two unequal proportions before measuring all six angles. Ask the students if they can explain why these should all be equal . The NRICH Project aims to enrich the mathematical experiences of all learners. To support this aim, members of the NRICH team work in a wide range of capacities, including providing professional development for teachers wishing to embed rich mathematical tasks into everyday classroom practice. More information on many of our other activities can be found here.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 4, "mathjax_display_tex": 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": 0.9247798919677734, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/158357/calculate-alpha-from-alpha-sin-alpha-k/158375	# Calculate alpha from $\alpha + \sin(\alpha)$ = K Sorry for the dumb question, but I'm not involved in math. I need to reverse the following formula, to calculate $\alpha$: $$a = b(\alpha + \sin \alpha)/c$$ So I have: $$(\alpha + \sin \alpha)=ac/b = K$$ Since $a$, $b$, $c$ are constant, I put equal to $K$. $\alpha$ is measured in radians. I need to find the value of $\alpha$ (in radians or degree). Thanks to all!! - ## 4 Answers There is no "closed-form" solution to this equation. You can use numerical methods to solve it for any given value of $K$. If $K$ is small, you can use a series: $$\alpha = {\frac {1}{2}}K+{\frac {1}{96}}{K}^{3}+{\frac {1}{1920}}{K}^{5}+{\frac {43}{1290240}}{K}^{7}+{\frac {223}{92897280}}{K}^{9} + \ldots$$ The error in the approximation using the terms above is less than about $2 \times 10^{-7}$ for $-1 \le K \le 1$. - Unfortunately K is at list $300 \cdot 600$. But thanks, you remembered me the use of series! – Tommaso Jun 14 '12 at 17:53 4 @Tommaso: A similar series can be used in other ranges of $K$. If you write $\alpha = 2 n \pi + t$, $\alpha + \sin(\alpha) = 2 n \pi + t + \sin(t) = K$ if $t + \sin(t) = K - 2 n \pi$. So if $K$ is near $2 n \pi$, use my series with $K$ replaced by $K - 2 n \pi$ and then add $2 n \pi$ to the result. – Robert Israel Jun 14 '12 at 18:00 3 @Tommaso: If $K$ is that big, the approximation $\alpha \approx K$ shouldn't be too bad... – Hans Lundmark Jun 14 '12 at 18:01 @RobertIsrael Thanks for the suggestion! I will try to create a code in order to get at least an approximation of the solution – Tommaso Jun 14 '12 at 18:18 These are what we call Transcendental Equations. Solving these involve graphical or numerical analysis, both of which yield approximate results. In graphical analysis, you first rearrange the equation as: $\sin\alpha = K - \alpha$ Now, on a graph, plot the curve $y=\sin\alpha$. On the same graph, plot the straight line $y=K-\alpha$ The point at which the two intersect is basically the solution of the equation. From the $y$ coordinate of the intersection point, you can then easily calculate the value of $\alpha$ using $y=K-\alpha$ Numeric solutions involve methods like the Newton Raphson Method, Bisection Method, etc. Here is a nice wikipedia article enlisting all such methods. - A useful link with a useful terminology as well http://en.wikipedia.org/wiki/Lagrange_inversion_theorem This equation is often present in the mathematical problems of celestial mechanics - You are trying to find the root of the equation $f(\alpha) = \sin(\alpha) + \alpha - K$. Just use Newton-Raphson to get to the solution. If your $K$ is fairly small, then initializing Newton-Raphson with $\frac{K}{2}$ should be good and if its quite large, then initializing $\alpha$ with $K$ should do. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 35, "mathjax_display_tex": 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": 0.8964861035346985, "perplexity_flag": "head"}
http://crypto.stackexchange.com/questions/2214/can-you-make-a-hash-out-of-a-stream-cipher	# Can you make a hash out of a stream cipher? A comment on another question made me wonder about something: Assume you're on a rather constrained platform — say, a low-end embedded device — with no built-in crypto capabilities, but you do have access to a simple stream cipher; say, RC4 or one of the eSTREAM ciphers. What other crypto primitives can you build out of that stream cipher? In particular, are there any practical ways to build a cryptographic hash function and/or a MAC out of just a stream cipher? We already have questions about turning a hash into a stream cipher and about turning a stream cipher into a block cipher, but this particular transformation doesn't seem to have been covered yet. Obviously, if the platform constraints permit it, one could ignore the stream cipher and just implement a standard hash function from the ground up. What I'm wondering is whether having the stream cipher available might let one do better than that in terms of code size, memory usage and/or speed. While a construction that treats the stream cipher as a black box would be nice, schemes that only use parts of the stream cipher (like RC4-Hash, which, alas, has practical collision attacks) would be interesting too, at least if they're simple enough. Edit: What about Carter–Wegman MACs? I'm not too familiar with those, but it seems like it should be possible to instantiate them with a stream cipher. Of course, implementing the universal hash family will require some non-trivial amount of code, but it might still be less than implementing a conventional MAC from scratch. - If you will look at the security requirement in an universal hash function, you will notice the difference from the well-known security requirements of cryptographic hash functions, viz collision resistance. The reason why UOWHF were introduced was to construct much efficient signature scheme without relying on stronger properties like collision resistance. – Jalaj Mar 31 '12 at 15:34 Carter-Wegman MACs are okay for making authenticated encryption, but they're not successful for fast, secure hash functions. As Jalaj points out, there's a different set of security requirements. – Jon Callas Apr 3 '12 at 20:21 ## 4 Answers A PRG and a hash function are both PRFs, but they have different security considerations. (At least, historically they do. Moving forward some of them may be more closely aligned.) Due to this, building a hash from an existing PRG designed under older security constraints probably isn't a good idea for the intuitive construction of using the plaintext as the key to the PRG and the PRG output as the hash. See this nice summary: Why stream ciphers shouldn’t be used for hashing . - This is a really bad idea. Most stream ciphers, certainly RC4 and anything in a Counter Mode construction are ultimately a PRNG XORed onto the plaintext. This is fine for encryption -- the information theory is pretty easy to understand -- but is no basis for a hash function. The goals of a hash function and a cipher are different. As it turns out, you can make a block cipher into a hash function, but even then not any block cipher makes for a good hash function. Think of it this way -- a cipher is like a game of cards. It relies upon the fact that you can't see what's in your opponent's hand, and what makes it hard or intractable is the uncertainty. But a hash function is like a game of chess. Everyone knows everything, and the goal of the hash function is to make it so complex that you can't go backwards and that there's no easy way to predict forwards. - One could see the PRNG part of the stream cipher as a random function of the key, and use the message to be hashed as a key, not in the plaintext position. – Paŭlo Ebermann♦ Apr 13 '12 at 23:06 I am pretty sure that it is not advisable to use a stream model to construct a cryptographic hash. I do not have a full theoretic proof in hand, but can think about it from the perspective of a famous stream cipher, RC4. It is susceptible to related-key attack. Now to use RC4 for constructing a hash function, you need to argue that it should be resistant to a stronger adversarial model, viz chosen-key attack, whereupon the adversary can target the weakness in key scheduling. I am sure full chosen-key is not under consideration for RC4 (correct me if I am wrong). On the other hand, you allow the adversary to pick the key in the very first part of the definition of any form of resistance (apart from chosen-target-forced-prefix resistance which is not yet formally defined) for cryptographic hash function! So, I am pretty sure that it is not advisable. However, if we look at few results from the theoretical point of view, Coron et al. showed the equivalence between ideal cipher model and the random oracle model. This give some hope: convert the stream cipher to block cipher secure in the ideal cipher model and then by equivalence it is secure in random oracle model; but we get stuck here (how to move from RO model to an instantiation of a hash function. After all, the million dollar question is whether we trust random oracle model after Canetti et al worked showed weakness in the random oracle model! - The negative vote down, can you state the reason why you are not satisfied with the answer?! – Jalaj Mar 31 '12 at 15:34 I didn't downvote you, but I do feel that both your and B-Con's answers have only touched on part of the question. As both of you correctly point out, there's no a priori reason to assume that $H(m) = S(m)$ is a secure hash, even if $S(m)$ is a secure stream cipher, due to the different security requirements; indeed, Indesteege and Preneel's analysis of RC4-Hash seems to prove that this naïve approach does not yield a secure hash when applied to RC4. However, this doesn't mean there couldn't be some less naïve construction that would yield a secure hash (or MAC) when applied to... – Ilmari Karonen Apr 1 '12 at 14:53 ...a secure stream cipher, such that e.g. finding a collision for the hash or forging the MAC with better than brute force probability would imply a better than brute force distinguisher for the stream cipher. Of course, the answer might well turn out to be that no such construction is currently known (although, if I'm not mistaken, at least for MACs the Carter–Wegman construction would seem to provide an example), but I'd rather keep waiting a bit longer yet before accepting such a negative answer. – Ilmari Karonen Apr 1 '12 at 15:01 @Illmari: You are correct, I only sought to address one straight-forward construction. I edited my answer to emphasis that. @ Jalaj: I was also downvoted, so it seems like someone disliked our answers. (Maybe because they were more informative and less conclusive.) – B-Con Apr 3 '12 at 19:33 Edit: See the comments, this construction is not provably secure After some more thought, the answer is "yes", you can construct a hash from a stream cipher. (My other answer from a few days ago was only negative, eg, how not to construct it.) Here is a construction. First build a block cipher $B$ from the stream cipher $S$: • Take the PRG $P$ from $S$. • You can build a PRF family $F$ from a PRG using the CGM construction. Use $P$ to build $F$. • You can build a PRP (block cipher) $B$ from a PRF using a Fiestel network. (The Luby-Rackoff theorem states that a secure PRF can be used to construct a secure PRP using 3 (or 4 depending on security desires) rounds of a Fiestel network.) Use $F$ to build $B$. Now we have a PRP from a PRG. (And we didn't use a hash to do so, as some stream cipher to block cipher constructions do; using a hash in this construction seems uselessly circular). Now build a hash from a block cipher: • You can build a collision resistant compression function $h$ from a PRP using the Davies-Meyer compression function. Use $B$ to build $h$. (Edit: Not necessarily - DM requires an ideal cipher, which is stronger than what $B$ is. The security proof doesn't follow in this step. This may be fixable by finding a different $h$ construction or $B$ construction, but I have not found one.) • You can build a collision resistant hash function $H$ from a collision resistant compression function using the Merkle–Damgård construction. Use $h$ to build $H$. As to your practical considerations: Is it efficient? I can't speak in general, so let's analyze this construction. Popular hash functions like MD5 and the SHA family use the same Merkle–Damgård construction, so to compare our $M$ against them we'll just focus on analyzing our $h$. • Speed: Every $h$ iteration, we have 3+ rounds of a Fiestel network, each of which consists of keying $F$ and generating a small output from it. Keying our $F$ and generating $n$ bits of output itself requires $n$ re-keys of $P$ and $n$ number of $2n$ bit outputs from $P$ due to how the CGM construction works, and none of it is parrallelizable. So our $n$ bit $h$ requires at least $3n$ re-keys of $P$ and $6n^2$ bits of output from $P$. Considering the speed of modern stream ciphers this is very slow compared to modern hash algorithms, but might be practical in a situation where: hashing is rare, hashed input is small, or hashing is not very time critical. • Implementation overhead: We need to implement the CGM construction, Feistel network, and DM function. Building $F$ from $P$ re-uses $P$'s implementation and adds a feedback loop pushing $n$ bits of output back into the key for $P$ with a branch condition on 1 bit, and no more than $n$ additional bits beyond the input will need to be stored in memory at a time. The Fiestel network construction is minimal overhead, just a few XORs and array swapping. The DM construction is just a couple inputs into the Fiestel network and another XOR. That looks like it is less overhead than a SHA-2 compression function. We should be able improve on this scheme. We might be able to build a compression function from a PRF more directly than going through the PRP. And the biggest loss of efficiency here is CGM, maybe another construction might turn a PRG into a PRF more efficiently than CGM does. Edit: This may work if we can build a compression function differently. Until then, it's just an interesting construction with no security proof. I'm leaving it largely as-is as a starting point for a future self or other to fix. - 1 Well, that is certainly a construction; however, you haven't shown that it is a secure construction. For example, Luby-Rackoff generates a PRF that is secure using the standard block cipher assumptions (that the key is secret and related key attacks do not apply); the Davies-Meyer construction assumes that we have an ideal block cipher (that the output is essentially a random function of the cipher and the plaintext). Does the Luby-Rackoff construction also provide the additional properties that Davies-Meyer requires? I wouldn't assume that. – poncho Apr 4 '12 at 22:11 You're right. I unwisely glossed over this step (it slipped my mind that DM required IC) and it appears fatal: PRPs are insufficient for compression functions(pg 4). The best I can find is 10 rounds of Luby-Rackoff provides security under full CPCA. On the other side, I can't find a compression function that requires less than Ideal Cipher. Since security from $S$ to $B$ and $h$ to $H$ seem clear, I'll leave this in case a future self or other can bridge the gap. – B-Con Apr 13 '12 at 18:24	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 48, "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": 0.9399541020393372, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/286982/solve-8-logx-x-0	# Solve $8 \log(x) - x = 0$ Someone came to me recently with this seemingly simple equation to solve: $$8 \log(x) - x = 0$$ So far, everything I have tried has been a dead end. Is there a symbolic solution to this kind of equation? If so, how do I get there? - 3 See Lambert W Function. Using WA. Regards – Amzoti Jan 25 at 22:08 This really isn't a precalculus thing anymore, although it seems to be. – DoctorBatmanGod Jan 26 at 0:04 @DoctorBatmanGod: Logarithms and inverse functions are typical pre-calculus topics. It seems plausible that solving this equation might be investigated. – robjohn♦ Jan 26 at 0:07 @robjohn I absolutely agree, but its solution involves topics not covered in precalculus. You would certainly know more than me about proper classification though, Mr. Moderator. – DoctorBatmanGod Jan 26 at 0:11 – Mhenni Benghorbal Jan 26 at 0:25 ## 1 Answer $$8\log(x)=x\Rightarrow -x/8\ e^{-x/8}=-1/8$$ Therefore, $$-x/8=\mathrm{W}(-1/8)\Rightarrow x=-8\mathrm{W}(-1/8)$$ where $\mathrm{W}$ is the Lambert W function, the inverse to $x=we^w$. Mathematica yields $1.1553708251000778334$, and $26.093485476611910215$ as the two real solutions: `N[-8 LambertW[0, -1/8], 20]` and `N[-8 LambertW[-1, -1/8], 20]`. For those lacking a CAS like Mathematica, I have outlined an algorithm to compute Lambert W. - Or you could just use wolfram alpha, producing this and this. – DoctorBatmanGod Jan 26 at 0:03 Wolfram Alpha is essentially a remote CAS. – robjohn♦ Jan 26 at 0:11	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 5, "mathjax_display_tex": 3, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9097535014152527, "perplexity_flag": "middle"}
http://www.physicsforums.com/showthread.php?s=a1af0661208251867a81098959f2502a&p=4270035	Physics Forums ## The birth of statistical mechanics This topic is about history of physics so I decided to post it in general physics section but it would be nice to have a history of physics(or maybe science)section. Anyway,during my statistical mechanics course,I realized QM is being used from the beginning,in contrast to other parts of physics where a classical theory is developed first and then there are quantum corrections.So I wondered whether there was a time that there was classical statistical mechanics.I know,you now tell "of course there was" but by classical statistical mechanics I mean not considering energy levels and degeneracies and considering energy as a continuous parameter. I found Boltzmann's 1877 paper at http://www.trivialanomaly.com/ and took a look at it.In it,boltzmann assumes that particles can take velocities of the form $\frac{p}{q}$ and also he assumes that the energy(he uses the term "alive force" which I think he means energy)of any particle is an integer multiple of a constant factor. Also in http://arxiv.org/pdf/physics/9710007.pdf , it is said that Max Planck was inspired by Boltzmann's ideas in his theory about black body radiation. We know that maxwell independently had discovered maxwell-boltzmann distribution.I wanna know had maxwell also have the idea of energy quantization or he just derived the distribution experimentally? Also I will appreciate any ideas about classical statistical mechanics and whether there is a distribution which considers energy as a continuous parameter. Thanks PhysOrg.com science news on PhysOrg.com >> US scientist not involved in classified research: witnesses>> Corruption influences migration of skilled workers>> Researchers suggest Victorian-era people more intelligent modern-day counterparts Recognitions: Homework Help Yes, there is a "classical statistical mechanics" in which energy is considered as a continuous parameter - or, more precisely, energy is a function of the continuous state variables of position and momentum. However, it turns out that in order to write down sensible densities of states, etc., you need to bin the positions and momentum. The bin widths ##\Delta x## and ##\Delta p## end up entering into the density of states as the product ##\Delta x \Delta p##, so modern treatments tend to use our knowledge of quantum mechanics to identify this bin area with ##\hbar## (raised to the appropriate power if in 2 or 3 dimensions). See, for example, sections 3.6 and 4.3.2 of Tobochnik and Gould, available online here: http://stp.clarku.edu/notes/ Yeah,my thoughts also led me to the result that classically,a particle has infinite number of choices for its energy content.So I concluded that for finding a classical energy distribution,a different approach should be taken. Maybe we can tell that every energy between 0 and E is equally probable and probability distribution is 1/E. Thread Tools \| \| \| \| \|---------------------------------------------------------\|---------------------------\|---------\| \| Similar Threads for: The birth of statistical mechanics \| \| \| \| Thread \| Forum \| Replies \| \| \| Advanced Physics Homework \| 1 \| \| \| Academic Guidance \| 5 \| \| \| Advanced Physics Homework \| 7 \| \| \| Advanced Physics Homework \| 14 \| \| \| Quantum Physics \| 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": 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": 0.9222038984298706, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/20493/what-is-torsion-in-differential-geometry-intuitively	## What is torsion in differential geometry intuitively? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Hi, given a connection on the tangent space of a manifold, one can define its torsion: $$T(X,Y):=\triangledown_X Y - \triangledown_Y X - [X,Y]$$ What is the geometric picture behind this definition—what does torsion measure intuitively? - 2 The wikipedia page en.wikipedia.org/wiki/… has some geometric picture of torsion in terms of the "twisting" of reference frames along geodesics. – José Figueroa-O'Farrill Apr 6 2010 at 14:03 See also: mathoverflow.net/questions/122729/… – Ryan Budney Feb 24 at 20:03 ## 13 Answers The torsion is a notoriously slippery concept. Personally I think the best way to understand it is to generalize past the place people first learn about torsion, which is usually in the context of Riemannian manifolds. Then you can see that the torsion can be understood as a sort of obstruction to integrability. Let me explain a little bit first. The torsion really makes sense in the context of general G-structures. Here $G \subseteq GL_n(\mathbb{R}) = GL(V)$ is some fixed Lie group. Typical examples are $G = O(n)$ and $G = GL_n(\mathbb{C})$. We'll see that these will correspond to Riemannian metrics and complex structures respectively. Now given this data, we have an exact sequence of vector spaces, $$0 \to K \to \mathfrak{g} \otimes V^\ast \stackrel{\sigma}{\to} V \otimes\wedge^2 V^\ast \to C \to 0$$ Here $\sigma$ is the inclusion $\mathfrak{g} \subseteq V \otimes V^\ast$ together with anti-symmetrization. K and C are the kernel and cokernel of $\sigma$. If we are given a manifold with $G$-structure, we then get four associated bundles, which fit into an exact sequence: $$0 \to \rho_1P \to ad(P) \otimes T^M \to \rho_3P \to \rho_4P \to 0$$ Now the difference of two connections which are both compatible with the G-structure is a tensor which is a section of the second space $\rho_2P = ad(P) \otimes T^M$. This means that we can write any connection as $$\nabla + A$$ where $A$ is a section of $\rho_2(P)$. Now the torsion of any G-compatible connection is a section of this third space. Suppose that we have two compatible connections. Then their torsions are sections of this third space. However since we can write the connections as $\nabla$ and $\nabla + A$, the torsion differ by $\sigma(A)$. Thus they have the same image in the fourth space $\rho_4(P)$. The section of this fourth space is the intrinsic torsion of the G-structure. It measures the failure of our ability to find a torsion free connection. If this obstruction vanishes, then the torsion free connections are a torsor over sections of the smaller bundle $\rho_1P$. Now some examples: 1. $G = O(n)$. This is the case of a Riemannian structure. In this case $\sigma$ is an isomorphism so that the there is always a unique torsion free connection. The Levi-Civita connection. 2. $G = GL_m(\mathbb{C})$. This is the case of a complex structure. More precisely a $GL_m(\mathbb{C})$-structure is the same as an almost complex structure. In this case the intrinsic torsion can be identified with the Nijenhuis tensor. So it vanishes precisely when the almost-complex structure is integrable (i.e. a ordinary complex structure). 3. $G = Sp(n)$. Having an $Sp(n)$-structure on a manifold for which the intrinsic torsion vanishes is equivalent to having a symplectic manifold. From these examples you can see that the vanishing of torsion can be viewed as a sort of integrality condition. In these latter two cases the space of torsion free connections consists of more then a single point. There are many such connections. That's one reason why we don't see them popping up more often. - 2 This answer is just great, thank you very much! – Jan Weidner Apr 7 2010 at 7:29 1 Chris -- your answer is very interesting but a bit difficult to follow, in particular since you use some notation that you do not define ($P,\rho_1 P,\ldots$), so I'd like to ask: is this notion of torsion the same as in Sebastian's answer below (i.e. the same as the one given e.g. in Milnor-Stasheff, Characteristic classes, appendix C)? More precisely, we take a connection on the (co)tangent bundle compatible with a given $G$-structure on a manifold. In example 1 the answer is yes, so I was wondering about the other two. – algori Aug 2 2010 at 5:43 2 @ algori: P is the G-principal bundle coming from the G-structure (see the link I provided) and $\rho_i(P)$ are the associated bundles induced by the representations $\rho_i$, which come from the short exact sequence I mention. Sorry if that was confusing. It seemed clear enough from context, but perhaps it wasn't. Anyway, I don't have my copy of Milnor and Stasheff handy, but I'm fairly certain this torsion is the same. Most of what I say is explained in more detail in D. Joyce's book "Compact Manifolds with Special Holonomy". I suggest looking there for more details. – Chris Schommer-Pries Aug 2 2010 at 11:58 Thanks Chris! This has helped clarify things. – algori Aug 2 2010 at 18:59 1 Don't you mean GL_n(C) in GL_{2n}(R)? – David Corwin Dec 16 at 1:36 ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. Here is an example which I found useful when learning about torsion. Consider $\mathbb{R}^3$. Let $X$, $Y$ and $Z$ be the coordinate vector fields, and take the connection for which ```$$\begin{matrix} \nabla_X(Y)=Z & \nabla_Y(X)=-Z \\ \nabla_X(Z)=-Y & \nabla_Z(X)=Y \\ \nabla_Y(Z)=X & \nabla_Z(Y)=-X \end{matrix}$$``` A body undergoing parallel translation for this connection spins like an American football: around the axis of motion with speed proportional to its velocity. So the geodesics are straight lines, and this connection preserves the standard metric, but it has torsion and is thus not the Levi-Cevita connection. - This example is indeed very helpful, thanks! – Jan Weidner Apr 6 2010 at 19:57 Almost too basic an approach to give, but I think the only way to intuitively get under the hood of torsion (at least in the Levi-Civita sense) is to really understand the ideas of Lie bracket and connection: We're used to the fact, working on $\mathbb{R}^n$, that partial derivatives commute: $\frac{\partial}{\partial x_i}\circ\frac{\partial}{\partial x_j}=\frac{\partial^2}{\partial x_ix_j}=\frac{\partial}{\partial x_j}\circ\frac{\partial}{\partial x_i}$. But not only is this untrue in the setting of general $C^2$ manifolds, it also makes no sense- with no global coordinates to turn to, we need some other way of defining a 'direction of differentiation' globally. Fortunately that's exactly what vector fields do, so now our updated equation $\frac{\partial}{\partial X}\circ\frac{\partial}{\partial Y}=\frac{\partial}{\partial Y}\circ\frac{\partial}{\partial X}$ makes sense (modulo some issues of notation)- our only problem being its falsehood in general, which we measure with the Lie bracket. Now it might be tempting to blame our vector fields for the Lie bracket's general non-zero nature- perhaps we get non-zero Lie brackets just when we pick a really weird vector field... but close examination (of, say, the image of the coordinate vector fields under the differential of your faourite chart map) reveals this is not the case. In fact the $C^2$ness of the vector field ensures that on an infinitessimal level our vector fields are never really very pathological: what the Lie bracket is measuring is something much more intrinsic about our manifold- about how vector fields must locally twist as they move along each other to keep time with the metric. But telling us how vector fields do move along one another is the job of a connection- which, by giving us $\nabla_X Y$, prescribes $\frac{\partial}{\partial X}Y$, but $Y$ is really $\frac{\partial}{\partial Y}$ so this 'prescribes a value' for the Lie bracket as $\nabla_X Y-\nabla_Y X$. Subrtracting the former from the latter gives the actual infinitessimal twist minus the neccessary infinitessimal twist to give the 'unneccessary twist' of the connection. - 2 This is a really nice explanation. If someone asks about torsion then the answer should be in terms of things of similar or less complexity than torsion itself, and answers in terms of sections of bundles and Lie algebras add to confusion. Like using the example of a car to explain to a caveman what a wheel is. (Not that Jan is a caveman!) I just wanted to pick up on your comment about the Lie bracket measuring the necessary twist of vector fields to keep time with the metric. What about the general case, where M doesn't necessarily have a metric, and ∇ isn't necessarily Levi-Civita? – – Fly by Night Jun 28 2011 at 17:56 There was no metric in Tom Boardman's answer. – Al-burcas Jul 31 2011 at 10:19 Great answer, of the kind I'd like to see more often on MO. – Filippo Alberto Edoardo Nov 2 at 1:10 Here's another reinterpretation of the torsion tensor which seems perhaps more natural. Consider the identity endomorphism $\mathrm{id}:TM \to TM$, but thought of as a 1-form with values in $TM$; that is, $$\mathrm{id} \in \Omega^1(M;TM).$$ The connection $\nabla$ defines an exterior covariant derivative: $$d^\nabla : \Omega^1(M;TM) \to \Omega^2(M;TM)$$ and the torsion of $\nabla$ is precisely $$T^\nabla = d^\nabla(\mathrm{id}).$$ - Perhaps, the following two facts help to understand torsion: 1. Two connections are equal if and only if they have the same geodesics and equal torsions. 2. For any connection there is a unique torsion-free connection with the same geodesics. This is proved in Spivak, volume II, page 249. - Similar to José's answer, one can consider the following: for each connection $\nabla$ on the tangent bundle (or its dual), one can consider the induced connection $\nabla\colon\Gamma(M;\Lambda^k T^* M)\to\Gamma(M; T^* M\otimes \Lambda^k T^* M).$ Denote by $\Lambda\colon T^* M \otimes \Lambda^k T^* M\to \Lambda^{k+1} T^* M$ the antisymmetrising map, and by $d_\nabla=\Lambda\circ\nabla$ some kind of exterior derivative. Then $d_\nabla$ is the exterior derivative if and only if $\nabla$ is torsion free. Moreover $d_\nabla^2=0$ if and only if $\nabla$ is torsion free. This is very similar to the equation of the curvature of a connection $\tilde\nabla$ of an arbitary bundle in terms of its absolute exterior derivative $d^{\tilde\nabla}.$ The torsion of a connection is the obstruction to the induced calculus of the connection to be the usual/natural calculus on a manifold. - ere is a review article by Hehl and Obukhov about the role of torsion in geometry and physics. The article contains the intutive geometric explanation of the torsion tensor as stated by Deane Yang as a measure (figure-1) of the failure to close an infinitesimal parallel transport parallelogram. The article also contains an interpretation of the torsion tensor in three dimensions as the dislocation density of a dislocated crystal. Here are a few additional properties of the torsion tensor. In dynamically generated gravity theories and fluid dynamics, the generated torsion tensor is proportional to the anti-symmeterized spin density and vorticity respectively. In harmonic analysis on (vector bundles over) homogeneous spaces G/H, the Levi-Civita Lagrangian, based on the torsionless connection is not diagonal in the spaces of sections beonging to irreducible G representations (except for the trivial representation). On the other hand there exists an H-connection which is not torsion free whose Laplacian is diagonal. The explanation of this result is that the information about the inducing H-representation defining the vector bundle is contained in the torsion tensor. - I'm afraid that the torsion is not motivated by any picture. It's just the skew-symmetric part of $\nabla$. Let $M$ be your manifold and $p\in M$. Consider two tangent vectors $v,w\in T_pM$. You can extend them to commuting vector fields $V$ and $W$ in a neighborhood of $p$. Then $$T(v,w) = \nabla_vW-\nabla_wV ,$$ so in this case $T$ measures non-symmetry of $\nabla$. In general (for non-commuting vector fields), the formula $\nabla_XY-\nabla_YX$ does not define a tensor and the term $[X,Y]$ fixes this problem. - I'm convinced that there is geometric explanation analogous to curvature measuring infinitesimal holonomy, but I haven't been able to work it out yet. In any case, at least in the context of Riemannian geometry, what's geometrically natural is zero torsion, so it's not surprising that a geometric interpretation of nonvanishing torsion is a little elusive. Here are some things that are implied by (and are essentially equivalent to) zero torsion: 1) The ability to define the Hessian of a function as a symmetric tensor 2) A parameterized curve is a constant speed geodesic if and only if its velocity curve is parallel along the curve This extends some useful properties of Euclidean space to a Riemannian manifold. These properties (and probably some others) along with its uniqueness make the Levi-Civita connection very powerful and useful. - Let me expand a little the answer of José Figueroa-O'Farrill. Suppose that $\nabla$ is a linear connection on a vector bundle $E\to M$, and that there is $\sigma\in \Omega^1(M;E)$, a 1-form on $M$ with values in $E$ such that $\sigma_x:T_xM\to E_x$ is a linear isomorphism. This is called a soldering form. It identifies $E$ with $TM$. The torsion is then $d^{\nabla}\sigma\in\Omega^2(M;E)$. It is an obstruction against the soldering form being parallel for $\nabla$. Maybe this explains, that space is twisting along geodesics if the torsion is non-zero. So torsion can be viewed either as a property of the soldering form (choose it better if you want to get rid of torsion), or as a property of $\nabla$ (if you identify $TM$ with $E$ with the given soldering form). This works also with $G$-structures on $M$. Consider a principal $G$-bundle $P\to M$ and a representation $\rho:G\to GL(V)$ where $\dim(V)=\dim(M)$. A soldering form is now a $G$-equivariant and horizontal 1-form $\sigma\in\Omega^1(P,V)^G_{hor}$ which is fiberwise surjective. This induces a form $\bar\sigma\in\Omega^1(M,P\times_G V)$ which is a soldering form in the sense above. You can compute torsion either on $P$ or on $M$ and they correspond to each other. This ties in with the answer of Chris Schommer-Pries. - Well, one should think in term of Euclidean motions, i.e. rotations AND translations (see Cartan connections) - hence the name affine connection. The (Cartan) curvature of this (Cartan) connection splits into two parts: one measuring infinitesimal rotations (i.e. the ordinary Riemannian curvature) and one measuring infinitesimal translations ("slipping") (i.e. the torsion). Maybe one should elaborate on this in more detail. (This explanation is related to Jose's) - Torsion is easy to understand but this knowledge seems to be lost. I had to go back to Elie Cartan's articles to find an intuitive explanation (for example, chapter 2 of http://www.numdam.org/numdam-bin/fitem?id=ASENS_1923_3_40__325_0). Let $M$ be a manifold with a connection on its tangent bundle. The basic idea is that any path $\gamma$ in $M$ starting at $x\in M$ can be lifted as a path $\tilde\gamma$ in $T_xM$, but is the $\gamma$ is a loop $\tilde \gamma$ need not be a loop. The resulting translation of the end point is the torsion (or its macroscopic version). The situation is easy in a Lie group $G$ (which I imagine Cartan had in mind). $G$ has a canonical flat connection for which the parallel vectors fields are left invariant vectors fields. For this connection the parallel transport is simply the left translation. The Maurer-Cartan form $\alpha$ is then the parallel transport to the tangent space $T_1G$ at the identity $1\in G$. If $\gamma:[0,1]\to G$ is a path in $G$ starting at $1$. $\gamma'$ is a path in $TG$ and $\alpha(\gamma')$ is a path in $T_1M$. $\alpha(\gamma')$ can be integrated to another path $\tilde \gamma$ in $T_1M$. Let $\gamma_{\leq x}$ be the path $\gamma:[0,x]\to G$, then we define $$\tilde \gamma(x) = \int_0^x\alpha(\gamma'(t))dt = \int_{\gamma_{\leq x}}\alpha.$$ In the sense given by the connection, $\gamma$ and $\tilde\gamma$ have the same speed and the same starting point, so they are the same path (but in different spaces). If $\gamma$ is a loop and $D$ a disk bounding $\gamma$, $\tilde\gamma$ is a loop iff $\tilde\gamma(1)=0\in T_1G$. We have $$\tilde\gamma(1) = \int_\gamma\alpha = \int_Dd\alpha.$$ $\tilde\gamma$ is a loop iff this integral is zero. Now, $\alpha$ can be viewed as the solder form for $TG$, so the torsion is the covariant differential $T=d^\nabla\alpha$. As the connection is flat $T$ reduces to $T=d\alpha$. The Maurer-Cartan equation gives an explicit formula: $T=d\alpha = -\frac{1}{2}[\alpha,\alpha]$. The previous integral is then the integral of the torsion $$\tilde\gamma(1) = \int_Dd\alpha = -\frac{1}{2}\int_D[\alpha,\alpha]$$ and may not be zero. The situation is the same for a general manifold, but the parallel transport is not explicit and formulas are harder. The notion behing this is that of affine connection. As I understand it, an affine connection is a data that authorize to picture the geometry of $M$ inside the tangent space $T_xM$ of some point $x$. If I move away from $x$ in $M$, there will be a corresponding movement away from the origin in $T_xM$ (this is the above lifting of path). If I transport in parallel a frame with me, the frame will move in $T_xM$. Globally the movement of my point and frame is encoded by a family of affine transformations in $T_xM$. Of course this picture of the geometry of $M$ in $T_xM$ is not faithful. Because of the torsion, if I have two paths in $G$ starting at $x$ and ending at the same point, they may not end at the same point in $T_xM$. Because of curvature, even if my two lifts end at the same point, my two frames may not be parallel. The picture is faithful if $M$ is an affine space iff both torsion and curvature vanish (Cartan's structural equations for affine space). I think torsion is beautiful :) - One more interpretation: The torsion is the curvature of the smooth functions (as a vector bundle over your manifold). - 1 What does this mean? Can you please elaborate? – Spiro Karigiannis Aug 28 at 2:33 1 I believe the point is that if the connection has torsion, then the Hessian of a function is no longer symmetric. So you can call the anti-symmetric part of the Hessian the "curvature" of the function. – Deane Yang Aug 28 at 12:29 Ah, I see. Thanks, Deane. – Spiro Karigiannis Aug 28 at 13:23	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 141, "mathjax_display_tex": 11, "mathjax_asciimath": 0, "img_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": 0.9328736662864685, "perplexity_flag": "head"}

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