Datasets:

keirp
/

tiny_math

Dataset card Data Studio Files Files and versions Community

Split (1)

train · 17.1k rows

url stringlengths 17 172	text stringlengths 44 1.14M	metadata stringlengths 820 832
http://mathoverflow.net/questions/19587/diophantine-equation-of-first-degree/19593	## Diophantine equation of first degree ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) a1x1 + a2x2 + ... + anxn = S, where: 1. a1 through an are positive bounded integers 2. x1 through xn are positive bounded integers 3. 'S' is the sum of the expression for n=2 say a1x1 + a2x2=S we know when S>a1a2-a1-a2 the equation has solution. Do any of you know such kind of condition when n is in general? - What does 'positive bounded' mean exactly? Your condition for $n=2$ doesn't always work if $x_1$ and $x_2$ are bounded. It also doesn't work if when $S = 17$ and $a_1 = a_2 = 2$. – François G. Dorais♦ Mar 28 2010 at 5:34 ## 2 Answers It sounds to me like the OP is asking about the Diophantine Problem of Frobenius. This is as follows: let $(a_1,\ldots,a_n)$ be positive integers which generate the unit ideal (i.e., their setwise gcd is $1$). The Frobenius number $f(a_1,\ldots,a_n)$ is the largest positive integer $N$ such that there do not exist non-negative integers $x_1,\ldots,x_n$ such that $a_1 x_1 + \ldots + a_n x_n = N$. In the case of $n = 2$, the Frobenius number was explicitly computed by J.J. Sylvester (before Frobenius!): it is $a_1 a_2 - a_1 - a_2$, as the OP mentioned. Using this fact, it is a nice exercise to show by induction on $n$ that every sufficiently large integer $N$ can indeed be represented as a non-negative integer linear combination of the $a_i$'s. Perhaps the two most famous results on the Frobenius problem are as follows: I. Schur's Theorem: if we define `$r(a_1,\ldots,a_n;N) = \# \ \{(x_1,\ldots,x_n) \in \mathbb{N}^n \ \| \ a_1 x_1 + \ldots + a_n x_n = N\}$` to be the number of representations of $N$, then as $N \rightarrow \infty$ we have $r(a_1,\ldots,a_n;N) = \frac{N^{n-1}}{(a_1 \cdots a_n) (n-1)!} + O(N^{n-2})$. II. (Alfred) Brauer's theorem: for $1 \leq i \leq n$, put $e_i = \operatorname{gcd}(a_1,\ldots,a_i)$. Then $f(a_1,\ldots,a_n) \leq \sum_{i=2}^n a_i \frac{e_{i-1}}{e_i} - \sum_{i=1}^n a_i + 1$, with equality iff for all $i \geq 2$, $\frac{e_{i-1}}{e_i} a_i$ can be represented as a non-negative integer combination of the integers $(a_1,\ldots,a_i)$. There have been on the order of a thousand papers written about various aspects of this problem and as well as a rather authoritative recent book: Ramírez Alfonsín, J. L. The Diophantine Frobenius problem. Oxford Lecture Series in Mathematics and its Applications, 30. Oxford University Press, Oxford, 2005. - ### 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. Your question is a little imprecise. In general, when the integers $x_i$'s are bounded above and below, such problems are very difficult to decide whether there is a solution see the Knapsack Problem, the Subset Sum Problem, and Integer Linear Programming. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 26, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.930301308631897, "perplexity_flag": "head"}
http://mathoverflow.net/questions/16745?sort=newest	## When is Br(X) = H^2(X,G_m)? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) In Milne, Étale cohomology, it is proved that $\mathrm{Br}(X) = H^2(X,\mathbf{G}_m)$ for $X$ regular of dimension $\leq 2$. Are there in the meantime further results for $X$ regular? - ## 2 Answers When $X$ is quasi-projective over an affine scheme (or more generally if $X$ has an ample [EDIT: invertible] sheaf), then its Brauer group is isomorphic to the torsion part of $H^2(X, {\mathbb G}_m)$. This is an unpublished result of Gabber, and J. de Jong wrote down a different proof. - Are there criteria when $H^2(X,\mathbf{G}_m)$ is torsion free? – norondion Mar 1 2010 at 14:37 In Lichtenbaum, Zeta-Functions of Varieties Over Finite Fields at s = 1, he proves that it is torsion for varieties over finite fields. – norondion Mar 1 2010 at 14:39 You mean X has an ample line bundle? – Hailong Dao Mar 1 2010 at 14:49 Yes, thanks, I corrected it! – Qing Liu Mar 2 2010 at 17:09 It is torsion for $X$ regular, Noetherian: Consider $0 \to \mathbf{G}_m \to j_\mathbf{G}_m \to Div_X \to 0$ and the long exact sequence; since $X$ is regular, we have $Div_X = \oplus_{x \in X^{(1)})i_{x,}\mathbf{Z}$. – norondion Mar 2 2010 at 17:35 ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. Dear norondion: It appears to me from your comments to Qing Liu's answer that you are interested in when this cohomological Brauer group is actually $0$. If that is true, then these two related MO questions (and Emerton's answer to one of them) may be of interest: Two conjectures by Gabber. Flat cohomology and Picard groups. (Of course, the punctured spectrum of a regular local ring is a regular scheme. Also, you can probably get some statements for projective $X$ by looking at the local ring of the cone over $X$). My apology if this is not relevant. - 1 Thanks. Actually, I'm interested in cases where it is finite. – norondion Mar 6 2010 at 18:14	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 15, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9040522575378418, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/23469/is-fire-plasma?answertab=oldest	# Is fire plasma? Is Fire a Plasma? If not, what is it then? If yes why, don't we teach kids this basic example? UPDATE: I probably meant a regular commonplace fire of the usual temperature. That should simplify the answer. - 5 Home experiment: put a strong (i.e. rare earth) magnet on the end of a pole and see if it affects the behavior of a flame more than a pole without the magnet does. Why? Because the magnet will affect ions and free electrons in motion much more strongly than neutral components. – dmckee♦ Apr 9 '12 at 20:56 3 As far as I know ordinary flames are not significantly ionized: they're just hot gas with chemistry going on in them. But I'm not certain enough to post an answer, so don't quote me. – dmckee♦ Apr 9 '12 at 21:18 2 – Greg P Apr 10 '12 at 2:20 1 – Qmechanic♦ Apr 10 '12 at 4:36 2 @dmckee What about a simpler experiment of conductivity: can a circuit be closed with a flame? If there is plasma there will be free electrons and ions so two conducting leads to a flame ccould be closing a circuit and current would appear. – anna v Apr 10 '12 at 4:49 show 2 more comments ## 6 Answers Broadly speaking, fire is a fast exothermic oxidation reaction. The flame is composed of hot, glowing gases, much like a metal that is heated sufficiently that it begins to glow. The atoms in the flame are a vapor, which is why it has the characteristic wispy quality we associate with fire, as opposed to the more rigid structure we associate with hot metal. Now, to be fair, it is possible for a fire to burn sufficiently hot that it can ionize atoms. However, when we talk about common examples of fire, such as a candle flame, a campfire, or something of that kind, we are not dealing with anything sufficiently energetic to ionize atoms. So, when it comes to using something as an example of a plasma for kids, I'm afraid fire wouldn't be an accurate choice. - 8 I don't think so--- the fire itself has a significant proportion of ionized atoms, it is not just hot gas, because the glow is due to the recombination in particular lines which are dependent on the chemical emission lines (you can see this in burning salt). – Ron Maimon Apr 10 '12 at 4:01 @Ron, are you sure it's recombination or maybe it's just transitions? I don't have a strong opinion on that, the reason I ask is that I think this might make the difference between fire and plasma (if any). – Lev Levitsky Apr 11 '12 at 14:54 @LevLevitsky: I am not sure--- it might be just outer shell transitions--- but once you ionize once, it is so much easier to eject the electron out teh atom. I am confused now on the issue--- I have found sources going both ways. I though the best thing would be to test conductivity, but somebody did, and said he didn't see conductivity (but this might be a low conductivity), the sharp boundary of the flame, the lack of gradual cooling, suggests that it is some sort of steady-state phase of combustion. It is hard to say if steady state non-equilibrium stuff is this or that. I don't know. – Ron Maimon Apr 11 '12 at 15:35 @Ron, even cold gas is glowing, just in infrared spectrum. – Anixx Sep 4 '12 at 16:53 @Anixx: Of course, but why the sharp change in color? Does the gas abruptly cool down? Why there? I figured it was ionization transition of some kind, but I am not sure anymore. – Ron Maimon Sep 4 '12 at 20:49 Nope. Fire is a thermal phenomenon, plasma is more of electrical. ## What's plasma? Plasma is the state when you strip off electrons/add electrons to a gas--so plasma consists of charged gas ions. It usually glows due to electron transitions and whatnot. ## What's fire? In a flame, you basically have hot soot/&c molecules flying up. Any hot material emits photons, which are usually in the infrared range for normal temperatures. At higher temperatures, they can go into the visible range. One way to explain this is by blackbody radiation-- the soot must emit photons since it has a nonzero temperature. What's actually going on is that the electrons are "thermally excited"--they have extra energy and are prone to making transitions. Transitions lead to absorption/emission of light, and this is what causes the color. You can see that there aren't any ions involved in fire, so it's not plasma. But ionization will occur if you heat it to even higher temperatures, and it can become plasma. - Would an oxy-acetylene (welding) torch be a plasma – Martin Beckett Apr 10 '12 at 1:59 5 "Fire is a thermal phenomenon, plasma is more of electrical." I can't say I like this formulation much...heat alone is enough to ionize atoms if things get hot enough. – dmckee♦ Apr 10 '12 at 2:23 @dmckee: But ionization will occur if you heat it to even higher temperatures, and it can become a plasma. I know, but it's not hot enough in fire.. – Manishearth♦ Apr 10 '12 at 3:19 – anna v Apr 10 '12 at 4:43 @annav, yes a plasma cutter is a plasma! Oxy-acetylene is about 3500C which is the bottom end of a K type star surface – Martin Beckett Apr 10 '12 at 14:46 Back of the envelope calculation: The Saha equation for a Hydrogen plasma says $$\frac{N_i^2}{N_H} = V \left(\frac{2 \pi m_e k_b T}{h^2}\right)^{3/2} \exp\left(\frac{-R}{k_b T}\right)$$ where $N_i$ is the number of ions, $N_H$ is the number of Hydrogen atoms, $V$ is the volume of the plasma, and $R$ is the Hydrogen ionization energy (13.6eV). Defining the degree of ionization $\xi = N_i / N_0$, where $N_0 = N_i + N_H$ is the total number of atoms in the system, this can be written $$\frac{\xi^2}{1-\xi} = \frac{V}{N_0} \left(\frac{2 \pi m_e k_b T}{h^2}\right)^{3/2} \exp\left(\frac{-R}{k_b T}\right)$$ A candle burns at 1000 Celsius, and the flame has a volume of around 1cm^3, with probably 10^20 atoms in the flame. For simplicity, let's assume it's mainly Hydrogen in the flame (the ionization energy of other elements is of the same order of magnitude anyway, so we won't be far off). Then I make the right hand side of the equation (we'll call it $f$) to be around 10^-54. Then we can solve $\frac{\xi^2}{1-\xi} = f$ using the quadratic formula: $$\xi = \frac{\sqrt{f^2 + 4f} - f}{2}$$ This gives us $\xi = 10^{-27}$: none of the particles in a candle flame are ionized (remember, we guessed there were only 10^20 particles). This makes perfect sense, because 1000C is only around 0.1eV, a good two orders of magnitude less than the ionization potential. The particle density is too low to make up for that. If you think any of my approximations don't apply (personally, I'm not too sure about the particle density) then please correct me in a comment! - Your error lies in the fact you assume the ionization to start from the fundamental n=1 state. In a flamme, electrons first climb by collision the energy ladder up to some excited states. When their bounding energy is close to the thermal energy they get ionized. – Shaktyai Sep 4 '12 at 13:23 I see what you mean, but I don't think you're quite right. The vast majority of atoms in a plasma are either ionized or in the ground state. This is because the continuum has a huge statistical weight compared to the bound states (the density of states is large). You can actually use (a slightly more general form of) the Saha equation to find out how many atoms are in the excited states in a Hydrogen plasma; it turns out it's safe to ignore all except the ground state. – poorsod Sep 4 '12 at 14:31 I did run a kinetic (Fokker Planck) atomic code for fusion plasmas at the edge of Tokamaks. At 0.1 eV all the gas is ionized. The energy difference between the n=6 and n=7 is already of the order of 0.1 eV. And collisions pump up level by level electrons from ground state to the continuum. The problem is extremely complex because the system is not in thermodynamical equilibrium (because of radiations)and one needs to introduce different temperatures for the upper levels, the lower ones and the free electrons. – Shaktyai Sep 4 '12 at 16:07 2 Flames can have non-negligible ionization due to chemical reactions involving free radicals (starting, in hydrocarbon flames, with ${\rm CH+O\rightarrow CHO^{+}}+e^{-}$). Check this old survey about the topic for more information. – mmc Sep 5 '12 at 2:40 Fire is a plasma. There are two kinds of plasmas: hot plasmas relevant to astrophysics or fusion are indeed a mixture of totally ionized gas. In cold plasmas ( northen lights, Neon tubes,flamme) the ionization degree is less than one but the mixture typically exhibit collective behaviour and a zoo of waves one do not encounter in gases. The most famous is the plasma oscillation and the Alfven wave but they are many others. poorsod's calculus assume the ionization takes place between n=1 and n=infinity. In reality, the atoms are first excited by collisions, their electrons jump on higher n until their bounding energy is lower than thermal energy of free electrons. For 0.1 eV more than 99% of the atoms are ionized (I did work on a computer model to analyze this problem). Though the equilibrium Saha approach is known to be false (the electron distribution function is not Maxwellian), you can get a preety good idea of the problem if you split your neutral atoms population into atoms in the fundamental, n=2, n=3, etc.. and use Saha equation for each population. - interesting, can other people please confirm this too? – daniel.sedlacek Sep 7 '12 at 1:35 1 I am sorry, as mentioned by @Mitchell Fire is generally not plasma. it rare for fire to be plasma. A single electron or ion or few of then dose not qualify for being plasma, it has to gather in sufficient amount so that it can show collective behavior. – Samir Chauhan Jan 4 at 8:33 fire is not a plasma if heated with high temperature it can become plasma - 1 Welcome! Typically one-line answers should be avoided and more information provided. – tpg2114 Dec 9 '12 at 8:16 There's this very interesting demonstration on "1veritaserum" channel on youtube involving a candle, two large metal paddles and 1000 volts of electricity. check it out:	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 10, "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": 0.3, "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": 0.932265043258667, "perplexity_flag": "middle"}
http://nrich.maths.org/5338/index?nomenu=1	## 'Odd Stones' printed from http://nrich.maths.org/ ### Show menu #### $27$ stones are distributed between $3$ circles On a "move" a stone is removed from two of the circles and placed in the third circle. So, in the illustration, if a stone is removed from the $4$ and the $10$ circles and added to the $13$ circle, the new distribution would be $3$ - $9$ - $15$ #### Check you can turn $2$ - $8$ - $17$ into $3$ - $9$ - $15$ in two "moves" Here are five of the ways that $27$ stones could be distributed between the three circles : $6$ - $9$ - $12$ $3$ - $9$ - $15$ $4$ - $10$ - $13$ $4$ - $9$ - $14$ $2$ - $8$ - $17$ There is always some sequence of "moves" that will turn each distribution into any of the others - apart from one. Identify the distribution that does not belong with the other four. Can you be certain that this is actually impossible rather than just hard and so far unsuccessful?	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 30, "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": 0.9518617987632751, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/tagged/matrices	Tagged Questions 0answers 28 views Gershgorin interval of an eigenvalue and the largest coordinate of the corresponding eigenvector Let $A=(a_{ij})$ be a $n\times n$ -- symmetric matrix with positive diagonal entries. The smallest eigenvalue, $\lambda_1$, is simple, and the corresponding unit eigenvector has … 2answers 198 views spectral radius monotonicity I encountered an inequality when reading a paper. Can someone help to show how to prove it? Let be the spectral radius of matrix $A$ or \$\rho(A)=\max{\|\lambda\|, \lambda \text{ are … 2answers 167 views Dimension of incomplete matrix over finite fields. Hi, Suppose one has an incompletely specified $2^n \times 2^n$ matrix over some fixed finite field $\mathbb{F}_{p^k}$. In fact, one knows that the diagonal entries are zero and al … 1answer 45 views On solution of a class of discrete-time Lyapunov equation Hello members, let's consider the following equation $$X=F_{1}XF_{1}^{T}+...+F_{p}XF_{p}^{T}+C$$ where $p$ is an positive integer and $C$ is a known positive semidefinite matrix. I … 2answers 224 views tracial triples Say that a triple of real numbers $(a,b,c)$ is a realizable triple if there are matrices $A,B\in SL_2(\mathbb{R})$ such that $tr (A)=a$, $tr (B)=b$, and $tr (AB)=c$. Question: what … 1answer 71 views On solution of a discrete-time equation Hello, members. I have a problem for the following problem when I derive an optimization algorithm for stochastic singular systems S(k+1)=A(k)S(k)A^{T}(k)+R(k)+F(k)S(k+1)F^{T}( … 1answer 70 views An Interesting variant of Rayleigh Quotient Let $A$ and $B$ be two given hermitian positive semi-definite matrices, then what is the solution for \begin{align} \max_{x\neq 0}\frac{x^HAx}{x^HBx+1}. \end{align} I am looking f … 1answer 74 views On solution of a recursion with rectangular matrices Greetings to members here. The question is how to calculate the solution $S(k)$ of the following recursive equation $$J(k)S(k+1)J^{T}(k)=A(k)S(k)A^{T}(k)+R(k)$$ where $J$ and $A$ … 1answer 72 views Schur product, partial order Let $A, B$ be positive definite matrices. Then $A^r\circ B^r \le (A\circ B)^r$ for $0\le r\le 1$, where $\circ$ is Schur product. Here the inequality is in the sense of Loewner par … 1answer 33 views Feasibility of a given set of homogenuous nonconvex quadratic inequality constraints Let $C_1$,$C_2$,...$C_N$ be $M \times M$ indefinite hermitian matrices. What can we say about the following quadratic constriants \begin{align} w^{H}C_1w>0 \\ w^{H}C_2w>0 \\ ...~~~ … 2answers 191 views Efficient computation of Markov chain transition probability matrix Consider a continuous Markov chain $X = (X_t)$ on a finite state space and let $Q$ be the (given) transition rate matrix. This matrix is very sparse, with non-zero values on 3 diag … 0answers 87 views Examples of functions from matrices to real numbers with certain properties Let $M(\mathbb{R})$ be the set of all matrices (of any size) over $\mathbb{R}$. Let $v : M(\mathbb{R}) \rightarrow \mathbb{R}$ be a function which satisfies the following 5 proper … 2answers 73 views Strictly positive definite autocovariance function of fGn Hi, let $\gamma(k) = 1/2 (\|k+1\|^{2H} + \|k-1\|^{2H}-2\|k\|^{2H}),k\in\mathbb{Z},$ be autocovariance function of fractional Gaussian noise where $H\in(0,1)$ is parameter. I want to sh … 0answers 101 views Products of matrices of a certain form Are $n \times n$ matrices of the form $$\pmatrix{1&1&1&1 \cr x&1&1&1 \cr x&x&1&1 \cr x&x&x&1}$$ studied anywhere? I am interested in … 0answers 238 views Noncommutative arithmetic mean geometric mean inequality and symmetric polynomials While analyzing convergence speed of stochastic-gradient methods for convex optimization problems, Recht et al (2011) posed a tantalizing conjecture. It seems quite tricky, so afte … 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": 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": 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": 0.8032317757606506, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/70016/voronoi-region-as-a-polyhedron	# Voronoi region as a polyhedron I have been asked to prove that a Voronoi region $V=\{x \in \mathbb{R}^n : \\|x-x_0\\| \le \\|x-x_i\\|, i=1,\dots,k\}$ around $x_0$ with respect to $x_1,\dots,x_k$ is a polyhedron. My idea is to find the set of hyperplanes $h_i : a_i^Tx=b_i$ defined by the pairs $(x_0,x_i)$, such as they divide the entire space in halfspaces that fulfill $a_i^Tx \le b_i \Leftrightarrow \\|x-x_0\\| \le \\|x-x_i\\|$. Intuitively, I can see that the hyperplanes I am looking for are symmetric with respect to each pair of points, so $a_i = x_i-x_0$ and $b_i = (x_i-x_0)^T (\frac{x_i+x_0}{2})$ could be a valid solution. As always, I am able to imagine it in a visual way, but I think I have no idea of how could I try to prove it in a more formal way, that is, to derive the valid $a_i$ and $b_i$ values without drawing. Added: Following the Boyd and Vanderberghe book on convex optimization, the definition of polyhedron I am using is that of a intersection of a finite number of halfspaces and hyperplanes. Could anybody please give me a hint on whether this is possible or not? Could a graphical proof like this just do its job? Thanks in advance. - 1 You are not asked to compute the relevant data of $V$ but to prove that $V$ is a polyhedron. This proof depends on your definition of a polyhedron and general theorems you have at your disposal. – Christian Blatter Oct 5 '11 at 8:00 1 I'm a little bit curious about the following: A polyhedron apparently doesn't need to be bounded here? Say, if $x_0$ is not 'completely surrounded' by the other points, then the Voronoi region may 'escape to the infinity'. This is irrelevant to the question. I have just seen Voronoi regions of (full rank) lattices, where this is never an issue. – Jyrki Lahtonen Oct 5 '11 at 8:23 @Jyrki, in the Boyd book, which is the one I am using, first defines polyhedron as the intersection of a finite number of halfspaces and hyperplanes, then states that it is convex, and finally talks about the difference between bounded and unbounded ones, specially about its nomenclature when mixing the concepts of polyhedra and polytopes. – Fernandez Oct 5 '11 at 9:03 @ChristianBlatter, I'll edit the question to include the definition of polyhedron. Thank you. – Fernandez Oct 5 '11 at 9:04 ## 2 Answers Your hyperplane is correct. If you square the right-hand inequality of your equivalence and cancel the terms $x^Tx$ on both sides, you're left with a linear equation for $x$ that you can bring into the form of the left-hand inequality with the values for $a_i$ and $b_i$ that you gave. Given the definition you cite, that completes the proof. Note, however, that there are other definitions of polyhedra which imply that polyhedra are bounded. The Wikipedia article states that polyhedra are bounded but has a section discussing the relationship to the definition you cite. - Thank you, Joriki. From the Boyd book, I have the definition of a polyhedron as the intersection of a finite number of halfspaces and hyperplanes. I was trying to figure the nature of every halfspace generated for each of the hyperplanes I talked about in my question. Then, if I am able to characterize those hyperplanes and associated halfspaces, I thought I could be able to argue that, given a point in the Voronoi region, it would be in one of those halfspaces, and viceversa. – Fernandez Oct 5 '11 at 8:59 1 @Fernandez: OK; in that case, you're done. I've edited my answer in response to your comment and edit. – joriki Oct 5 '11 at 9:49 I think that you are on the right track. AFAICT your equation defines the hyperplane that bisects the line segment from $x_0$ to $x_i$, and also gives you the constraint for a point to belong to the Voronoi region. You can argue as follows. Let $x$ be an arbitrary point. We can write $x-x_0=x'+ka_i$ for some constant $k$ and some vector $x'\perp a_i$. Then $$\\|x-x_0\\|^2=\\|x'\\|^2+k^2\\|a_i\\|^2$$ by Pythagorean theorem. But here $x-x_i=x-x_0-a_i=x'+(k-1)a_i$, and therefore $$\\|x-x_i\\|^2=\\|x'\\|^2+(k-1)^2\\|a_i\\|^2.$$ We can thus conclude that $$\\|x-x_0\\|\le \\|x-x_i\\|\Leftrightarrow k^2\le (k-1)^2\Leftrightarrow k\le\frac12.$$ This means exactly that $x$ is on the same side of the hyperplane $h_i$ as $x$. In other words, the vector $ka_i$ is the orthogonal projection of $x-x_0$ onto the vector $a_i$, and if the projection is on the nearer side of the midway point, we have the desired inequality between the distances. - 1 That's rather a roundabout way of doing it :-) – joriki Oct 5 '11 at 8:24 1 This calculation is meant to be a substitute to the visually obvious fact that this hyperplane splits the space into two sets: one consisting of points closer to $x_0$ the other of points closer to $x_i$. If you wanted something else, then I apologize. – Jyrki Lahtonen Oct 5 '11 at 8:27 @joriki Yeah, I may have misunderstood, which step the OP had problems with. Leaving this here for now, and wait for comments from Fernandez. Prepared to delete :-) – Jyrki Lahtonen Oct 5 '11 at 8:29 I didn't mean to substitute it by the visually obvious fact; I meant that the argument is quite a bit more complicated than the direct transformation I sketched in my answer. – joriki Oct 5 '11 at 8:30 1 @Fernandez Thanks for the kind comment. Of course, joriki's approach is cleaner, and should get the preference. I also had a picture in my mind. In that picture the 'coordinate' difference from $x$ to $x_0$ vs $x_i$ agree at $n-1$ coordinates, and disagree at only one. I was just porting that image to a calculation :-) – Jyrki Lahtonen Oct 5 '11 at 10:12 show 1 more comment	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 36, "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": 0.9523949027061462, "perplexity_flag": "head"}
http://terrytao.wordpress.com/2009/06/page/2/	What’s new Updates on my research and expository papers, discussion of open problems, and other maths-related topics. By Terence Tao # Monthly Archive You are currently browsing the monthly archive for June 2009. ## Random matrices: universality of local eigenvalue statistics 3 June, 2009 in math.PR, paper \| Tags: eigenvalues, Lindeberg replacement trick, random matrices, universality, Van Vu, Wigner matrices \| by Terence Tao \| 21 comments Van Vu and I have just uploaded to the arXiv our paper “Random matrices: universality of local eigenvalue statistics“, submitted to Acta Math.. This paper concerns the eigenvalues $\lambda_1(M_n) \leq \ldots \leq \lambda_n(M_n)$ of a Wigner matrix $M_n = (\zeta_{ij})_{1 \leq i,j \leq n}$, which we define to be a random Hermitian $n \times n$ matrix whose upper-triangular entries $\zeta_{ij}, 1 \leq i \leq j \leq n$ are independent (and whose strictly upper-triangular entries $\zeta_{ij}, 1 \leq i < j \leq n$ are also identically distributed). [The lower-triangular entries are of course determined from the upper-triangular ones by the Hermitian property.] We normalise the matrices so that all the entries have mean zero and variance 1. Basic examples of Wigner Hermitian matrices include 1. The Gaussian Unitary Ensemble (GUE), in which the upper-triangular entries $\zeta_{ij}, i<j$ are complex gaussian, and the diagonal entries $\zeta_{ii}$ are real gaussians; 2. The Gaussian Orthogonal Ensemble (GOE), in which all entries are real gaussian; 3. The Bernoulli Ensemble, in which all entries take values $\pm 1$ (with equal probability of each). We will make a further distinction into Wigner real symmetric matrices (which are Wigner matrices with real coefficients, such as GOE and the Bernoulli ensemble) and Wigner Hermitian matrices (which are Wigner matrices whose upper-triangular coefficients have real and imaginary parts iid, such as GUE). The GUE and GOE ensembles have a rich algebraic structure (for instance, the GUE distribution is invariant under conjugation by unitary matrices, while the GOE distribution is similarly invariant under conjugation by orthogonal matrices, hence the terminology), and as a consequence their eigenvalue distribution can be computed explicitly. For instance, the joint distribution of the eigenvalues $\lambda_1(M_n),\ldots,\lambda_n(M_n)$ for GUE is given by the explicit formula $\displaystyle C_n \prod_{1 \leq i<j \leq n} \|\lambda_i-\lambda_j\|^2 \exp( - \frac{1}{2n} (\lambda_1^2+\ldots+\lambda_n^2))\ d\lambda_1 \ldots d\lambda_n$ (0) for some explicitly computable constant $C_n$ on the orthant $\{ \lambda_1 \leq \ldots \leq \lambda_n\}$ (a result first established by Ginibre). (A similar formula exists for GOE, but for simplicity we will just discuss GUE here.) Using this explicit formula one can compute a wide variety of asymptotic eigenvalue statistics. For instance, the (bulk) empirical spectral distribution (ESD) measure $\frac{1}{n} \sum_{i=1}^n \delta_{\lambda_i(M_n)/\sqrt{n}}$ for GUE (and indeed for all Wigner matrices, see below) is known to converge (in the vague sense) to the Wigner semicircular law $\displaystyle \frac{1}{2\pi} (4-x^2)_+^{1/2}\ dx =: \rho_{sc}(x)\ dx$ (1) as $n \to \infty$. Actually, more precise statements are known for GUE; for instance, for $1 \leq i \leq n$, the $i^{th}$ eigenvalue $\lambda_i(M_n)$ is known to equal $\displaystyle \lambda_i(M_n) = \sqrt{n} t(\frac{i}{n}) + O( \frac{\log n}{n} )$ (2) with probability $1-o(1)$, where $t(a) \in [-2,2]$ is the inverse cumulative distribution function of the semicircular law, thus $\displaystyle a = \int_{-2}^{t(a)} \rho_{sc}(x)\ dx$. Furthermore, the distribution of the normalised eigenvalue spacing $\sqrt{n} \rho_{sc}(\frac{i}{n}) (\lambda_{i+1}(M_n) - \lambda_i(M_n))$ is known; in the bulk region $\varepsilon n \leq i \leq 1-\varepsilon n$ for fixed $\varepsilon > 0$, it converges as $n \to \infty$ to the Gaudin distribution, which can be described explicitly in terms of determinants of the Dyson sine kernel $K(x,y) := \frac{\sin \pi(x-y)}{\pi(x-y)}$. Many further local statistics of the eigenvalues of GUE are in fact governed by this sine kernel, a result usually proven using the asymptotics of orthogonal polynomials (and specifically, the Hermite polynomials). (At the edge of the spectrum, say $i = n-O(1)$, the asymptotic distribution is a bit different, being governed instead by the Tracy-Widom law.) It has been widely believed that these GUE facts enjoy a universality property, in the sense that they should also hold for wide classes of other matrix models. In particular, Wigner matrices should enjoy the same bulk distribution (1), the same asymptotic law (2) for individual eigenvalues, and the same sine kernel statistics as GUE. (The statistics for Wigner symmetric matrices are slightly different, and should obey GOE statistics rather than GUE ones.) There has been a fair body of evidence to support this belief. The bulk distribution (1) is in fact valid for all Wigner matrices (a result of Pastur, building on the original work of Wigner of course). The Tracy-Widom statistics on the edge were established for all Wigner Hermitian matrices (assuming that the coefficients had a distribution which was symmetric and decayed exponentially) by Soshnikov (with some further refinements by Soshnikov and Peche). Soshnikov’s arguments were based on an advanced version of the moment method. The sine kernel statistics were established by Johansson for Wigner Hermitian matrices which were gaussian divisible, which means that they could be expressed as a non-trivial linear combination of another Wigner Hermitian matrix and an independent GUE. (Basically, this means that distribution of the coefficients is a convolution of some other distribution with a gaussian. There were some additional technical decay conditions in Johansson’s work which were removed in subsequent work of Ben Arous and Peche.) Johansson’s work was based on an explicit formula for the joint distribution for gauss divisible matrices that generalises (0) (but is significantly more complicated). Just last week, Erdos, Ramirez, Schlein, and Yau established sine kernel statistics for Wigner Hermitian matrices with exponential decay and a high degree of smoothness (roughly speaking, they require control of up to six derivatives of the Radon-Nikodym derivative of the distribution). Their method is based on an analysis of the dynamics of the eigenvalues under a smooth transition from a general Wigner Hermitian matrix to GUE, essentially a matrix version of the Ornstein-Uhlenbeck process, whose eigenvalue dynamics are governed by Dyson Brownian motion. In my paper with Van, we establish similar results to that of Erdos et al. under slightly different hypotheses, and by a somewhat different method. Informally, our main result is as follows: Theorem 1. (Informal version) Suppose $M_n$ is a Wigner Hermitian matrix whose coefficients have an exponentially decaying distribution, and whose real and imaginary parts are supported on at least three points (basically, this excludes Bernoulli-type distributions only) and have vanishing third moment (which is for instance the case for symmetric distributions). Then one has the local statistics (2) (but with an error term of $O(n^{-1+\delta})$ for any $\delta>0$ rather than $O(\log n/n)$) and the sine kernel statistics for individual eigenvalue spacings $\sqrt{n} \rho_{sc}(\frac{i}{n}) (\lambda_{i+1}(M_n) - \lambda_i(M_n))$ (as well as higher order correlations) in the bulk. If one removes the vanishing third moment hypothesis, one still has the sine kernel statistics provided one averages over all i. There are analogous results for Wigner real symmetric matrices (see paper for details). There are also some related results, such as a universal distribution for the least singular value of matrices of the form in Theorem 1, and a crude asymptotic for the determinant (in particular, $\log \|\det M_n\| = (1+o(1)) \log \sqrt{n!}$ with probability $1-o(1)$). The arguments are based primarily on the Lindeberg replacement strategy, which Van and I also used to obtain universality for the circular law for iid matrices, and for the least singular value for iid matrices, but also rely on other tools, such as some recent arguments of Erdos, Schlein, and Yau, as well as a very useful concentration inequality of Talagrand which lets us tackle both discrete and continuous matrix ensembles in a unified manner. (I plan to talk about Talagrand’s inequality in my next blog post.) Read the rest of this entry » ### Recent Comments Sandeep Murthy on An elementary non-commutative… Luqing Ye on 245A, Notes 2: The Lebesgue… Frank on Soft analysis, hard analysis,… andrescaicedo on Soft analysis, hard analysis,… Richard Palais on Pythagoras’ theorem The Coffee Stains in… on Does one have to be a genius t… Benoît Régent-Kloeck… on (Ingrid Daubechies) Planning f… Luqing Ye on 245B, Notes 7: Well-ordered se… Luqing Ye on 245B, Notes 7: Well-ordered se… Arjun Jain on 245B, Notes 7: Well-ordered se… %anchor_text% on Books Luqing Ye on 245B, Notes 7: Well-ordered se… Arjun Jain on 245B, Notes 7: Well-ordered se… Luqing Ye on 245A, Notes 2: The Lebesgue…	{"extraction_info": {"found_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": 35, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8935421109199524, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/13270?sort=votes	## orthogonality relation for quadratic Dirichlet characters ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Hello, I've been working deriving the orthogonality relation for quadratic Dirichlet characters $\chi_d(n)$ (or real primitive characters). The statement I'm trying to prove is $$\lim_{X \rightarrow \infty} \frac{1}{D} \sum_{0 < \|d\| \leq X} \chi_d(n)= \begin{cases} \prod_{p\|n} \left(1 + \frac{1}{p}\right)^{-1} \quad &\text{if $m$ is a perfect square,} \newline 0 \quad &\text{otherwise,} \end{cases}$$ where the sum ranges over fundamental discriminants $d$ and $D$ is the number of terms appearing in the sum. You could recast this as $$\lim_{X \rightarrow \infty} \frac{1}{D} \sum_{0 < \|d\| \leq X} \chi_d(n)\chi_d(m)= \begin{cases} \prod_{p\|nm} \left(1 + \frac{1}{p}\right)^{-1} \quad &\text{if $mn$ is a perfect square,} \newline 0 \quad &\text{otherwise,} \end{cases}$$ which has the standard form of an orthogonality relation for characters. In the first case, $\chi_d(n) = 1$ unless $\gcd(d,n) > 1$, so essentially i'm trying to count fundamental discriminants. But this still seems a bit tricky to me. My approach is to try and count fundamental discriminants by using the generating function for squarefree numbers, i.e. $$\frac{\zeta(s)}{\zeta(2s)} = \sum_{n=1}^\infty \frac{\|\mu(n)\|}{n^s}.$$ I know that Jutila proves this in his paper On the Mean Values of $L(1/2,\chi_d)$ for Real Characters, but I would like to prove this lemma equation using analytic methods. Can anyone help me. - 1 Jutila is an analytic number theorist, and I guess that his arguments are very likely analytic. What do you mean when you say that you want an analytic proof? Did he prove it algebraically? I am unable to check the paper now. – Anweshi Jan 28 2010 at 18:54 I'm sorry, a bit confusing. Jutila's is analytic. However, his arguments are more elementary in nature. I want to incorporate $\zeta(s)/\zeta(2s)$ into my arguments somehow, which he does not. – Matthew Alderson Jan 28 2010 at 19:25 Does the case where $mn$ is not a perfect square simply follow from the symmetry of the roots of unity? – Matthew Alderson Jan 28 2010 at 20:22 ## 2 Answers So I think I solved half of the problem. Suppose that $n$ is a perfect square. Then $\chi_d(n) = 1$ unless $\gcd(d,n) > 1$, in which case its $0$. So, for $\gcd{d,n} = 1$, we are simply pulling out the subset of fundamental discriminants having no common divisor with $n$. To quantify the size of this subset, we must first count fundamental discriminants. The set of fundamental discriminants consists of all square-free integers congruent to $1$ modulo $4$ (i.e. odd fundamental discriminants) and all such numbers multiplied by $-4$ and $\pm 8$ (i.e. even fundamental discriminants). The odd fundamental discriminants may be counted by considering the series $$\sum_{\text{$d$ odd}} \frac{1}{\|d\|^s}.$$ In fact, this is a Dirichlet series. For observe that $$\sum_{\text{$d$ odd}} \frac{1}{\|d\|^s} = 1 + \frac{1}{3^s} + \frac{1}{5^s} + \frac{1}{7^s} + \cdots = \prod_{p>2} \left(1 + \frac{1}{p^s}\right) = \frac{\zeta(s)}{\zeta(2s)} \left(1 + \frac{1}{2^s}\right)^{-1},$$ where $$\frac{\zeta(s)}{\zeta(2s)} = \sum_{n=1}^\infty \frac{\|\mu(n)\|}{n^s},$$ is the Dirichlet series which generates the square-free numbers. So, by following the definition of fundamental discriminants given above, we can count fundamental discriminants by using the Dirichlet series $$\left(1 + \frac{1}{4^s} + \frac{2}{8^s}\right) \frac{\zeta(s)}{\zeta(2s)} \left(1 +\frac{1}{2^s}\right)^{-1} = \left(1 + \frac{1}{4^s} + \frac{2}{8^s}\right) \underbrace{\prod_{p>2} \left(1 + \frac{1}{p^s}\right)}_{l_p(s)}.$$ Now, to omit those discriminants with $\gcd(d,m) > 1$, we just omit the corresponding factors in $l_p(s)$. What's missing is $$\prod_{\substack{p>2 \ p\|m}} \left(1 + \frac{1}{p^s}\right),$$ so the relative density (compared to all fundamental discriminants $d$) is can be quantified by the expression $$\frac{1}{D} \cdot \prod_{\substack{p>2 \ p\nmid m}} \left(1 + \frac{1}{p^s}\right) = \prod_{\substack{p>2 \ p\|m}} \left(1 + \frac{1}{p^s}\right)^{-1}.$$ As in the proof of the prime number theorem, the main contribution here comes from the simple pole at $s=1$ (of $\zeta(s)$). In fact, $p=2$ also fits at $s=1$ since $1 + \frac{1}{4} + \frac{2}{8} = 1 + \frac{1}{2}$. Thus, in the end we obtain $$\lim_{X\rightarrow \infty} \frac{1}{D} \sum_{0 < \|d\| \leq X} \chi_d(m) = \prod_{p\|m} \left(1 + \frac{1}{p}\right)^{-1},$$ as desired. Now, does anyone know how to explain the other case. I assume it just follows from the symmetry of the roots of unity. But can anyone validate and perhaps clarify this. Thank you. - ### 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'm not 100% sure I know what you mean by an "analytic proof" but I'll try to give an argument similar in style to what you already worked out. In particular I won't go into any detail on the correspondence between average values of arithmetical functions, and the analytic behavior of the relevant generating Dirichlet series. One annoying issue with fundamental discriminants is that they come in a few families and it is difficult to avoid case-by-case analysis. For simplicity let's consider the case where $n$ is odd and squarefree, and $d$ runs over fundamental discriminants of the form $8c$ where $c$ is positive, odd, and squarefree. The other cases are similar. The goal is to show $$(1) \qquad \sum_{c \leq x}' (\frac{c}{n}) = o(x),$$ with the prime indicating $c$ is odd and squarefree, and where $(\frac{c}{n})$ is the Jacobi symbol. The Dirichlet series $A_n(s) = \sum_{c}' (\frac{c}{n}) c^{-s}$ has the Euler product $A_n(s) = \prod_{p > 2}(1 + (\frac{p}{n}) p^{-s})$. A simple calculation then shows $A_n(s) = L(s, \chi_n) \prod_{p \nmid 2n} (1-p^{-2s})$. Here $L(s, \chi_n)$ is a primitive Dirichlet $L$-function that is entire. In particular, there is no pole at $s=1$ so that (1) holds. - Thank you for the response. Greatly appreciated. – Matthew Alderson Feb 2 2010 at 14:50	{"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": 38, "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": 0.9348211884498596, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/22276/the-limit-of-z-cdot-overlinez-as-z-to-i?answertab=votes	# The limit of $z\cdot\overline{z}$ as $z\to i$ How can I compute this limit: $$\lim_{z\rightarrow i} (z\cdot \overline{z})$$ (where $\overline{z}$ is the conjugate of $z$)? - Please make your titles informative, and please make the body of your post self-contained. In particular, since you are asking a question, the body of the message should actually contain a question. – Arturo Magidin Feb 16 '11 at 3:52 Sory, I'll try that next time. – Tomas Sironi Feb 16 '11 at 3:54 ## 2 Answers As $\mathbb{C}$ comes nicely equipped with a metric, limits behave as nice as ever. Therefore, the limit of the product is the product of the limits. And the limit of the conjugate is the conjugate of the limit (this can be proved using a standard epsilon-delta argument). Hence, in your example: $$\lim_{z \to i} (z \cdot \overline{z})=\lim_{z \to i}(z) \lim_{z \to i}(\overline{z})= i \cdot(-i) = 1.$$ - Thanks, I should revise the properties of the limit in the complex plane – Tomas Sironi Feb 16 '11 at 11:27 Hint: Is $z'$ the conjugate of $z$ and the lower dot a multiplication? If so, what is the problem? You might try expressing $z=a+bi$, with $a,b$ real and trying a real limit. - It is a conjugate, but I don't remember the symbol in LaTeX so I used that instead. Anyway, I'll try that. Thanks! – Tomas Sironi Feb 16 '11 at 3:55 I am used to $\overline{z}$ but as people are used to different things it is worth defining. – Ross Millikan Feb 16 '11 at 3:59 I'm also used to that, but again, I didn't remember what's the command therefore I used $'$. – Tomas Sironi Feb 16 '11 at 4:01 `\overline`, just got it! – Tomas Sironi Feb 16 '11 at 4:02	{"extraction_info": {"found_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": 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": 0.9129254817962646, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/tagged/differential-geometry+thermodynamics	# Tagged Questions 5answers 574 views ### Introduction to differential forms in thermodynamics I've studied differential geometry just enough to be confident with differential forms. Now I want to see application of this formalism in thermodynamics. I'm looking for a small reference, to learn ... 1answer 151 views ### How to express the heat capacity in terms of heat? The first law of thermodynamics divides the internal energy change into contributions of heat and work. $$\text dU=\omega_Q-\omega_W,$$ Here I chose the notation to emphasise that the two parts are ... 4answers 318 views ### Discontinuities and nondifferentiability in thermodynamics In physics and engineering sources, calculus-based formalisms - whether differential forms on a manifold, or "differentials" of functions of several variables - are presented as a way of modeling and ...	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9279465079307556, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/27681/which-symmetric-pure-qudit-states-can-be-reached-within-local-operations/27682	# Which symmetric pure qudit states can be reached within local operations? There are two pure symmetric states $\|\psi\rangle$ and $\|\phi\rangle$ of $n$ qudits. Is there any known set of invariants $\{I_i:i\in\{1,\ldots,k\}\}$ which is equal for both states iff $\|\phi\rangle=U^{\otimes n}\|\psi\rangle$ for a $U\in\text{SU}(d)$? There is a theorem by Hilbert saying that for a compact group acting on a linear space there is a finite number of polynomial invariants characterizing orbits. However, (as far as I know) it does not provide and explicit construction of the invariants. The problem is easy when $n=2$ (Schmidt decomposition) and $d=2$ (Majorana representation). Partial solutions, and solutions with changed assumptions (e.g. linear operators instead of unitary operations) are welcome as well. - ## 5 Answers This is an algorithm for the computation of the homogeneous polynomial invariants in the general case, however without any attempt to reduce the algorithm's complexity. The basic needed ingredient of the algorithm is the ability to average over the Haar measure of the group of the local transformations. In the qubit case it is just an integration over copies of SU(2). But even in the more general cases of compact Lie groups, this task is possible but cumbersome, please, see for example the following parameterization of SU(N) by: Bertini, Cacciatori, Cerchiai that can be used for the integration. The procedure is as follows: First one computes the Molien function of the group of local transformations (The Wikipedia page describes the finite group case which can be generalized to compact groups): $M(t) = \int \frac{d_H(g)}{det(1-tg)}$ The coefficient of $t^n$ in the Taylor expansion of $M(t)$ is the number of linearly independent homogeneous invariants of degree n. Since the integrand is a class function the integration can be performed on the maximal torus by the Weyl integration formula. Now, for each degree, one constructs all possible combinations of lower degree inavriants and if this number doesn't exhaust the required number from the Molien series, then the additional invariants are computed by averaging monomials of the required degree. This operation involves the integration overthe Haar measure which cannot be reduced to an integration on the maximal torus, which is the most complex step in the algorithm. This operation is repeated until a sufficient number of linearly independent polynomials is produced, then one passes to the next degree. This process is continued until the total number of invariants reaches the difference between the dimension of the vector space and the local symmetry group. More practical variations of this procedure were used to construct polynomial invariants for special cases of entanglement problems, please, see for example the following work by: Grassl, Rotteler and Beth. - Thanks David, it looks nice. And thanks for the references. I will see if it is practical numerically. However, as far as I understand, the procedure is longer than the straightforward optimization of $\|\langle \phi \| U^{\otimes} \|\psi \rangle\|^2$ (unless a low degree invariant suffices to distinguish two states). – Piotr Migdal Oct 6 '11 at 17:31 It seems to me that the stabilizer formalism provides an answer to your question (see section 3.1 of quant-ph/0603226 for an introduction to the formalism). Given the two states, you simply take the stabilizer group for that 2-dimensional subspace of the total Hilbert space, and they will give you a such a set of invariants. However, this of course has nothing to do with the globally symmetric operations you consider, and could be done with any pair of states. However, given that you are kept within the symmetric subspace, this insures that these stabilizers will always have a description at worst polynomial in the number of local systems. If you wish to go further, and try to find some set of invariants which will uniquely identify states which are equivalent up to this form of globally symmetric local operation, then you are out of luck. This is because the set of product states which is globally symmetric spans the space of symmetric states, and hence any globally symmetric state can be written as a superposition of globally symmetric separable states. That is to say, without fixing the states, it is impossible to produce an observable which is invariant between the two states but varies across the space of symmetric states. Thus the only invariants which exists depend only on the symmetry of the states. UPDATE: I notice Norbert and I have interpreted the question somewhat differently. I have focused on the existence of observables which have the same value for the LU equivalent states. This is effectively answering the question of whether there is a measurement which distinguishes these states from other symmetric states. The paper Norbert links to is about the mathematical structure of the states, and is not testable with a single copy of the pair of states. I have no idea which setting Piotr had in mind (I had originally thought it was this one, but Norbert's answer has caused me to rethink that position). - Of course, it's important that not only does the symmetric subspace have polynomial dimension, but it has a basis which is easy to describe (in fact, it has a basis of product states); you can then compute the restriction of operators to the symmetric subspace efficiently. – Niel de Beaudrap Oct 1 '11 at 14:29 @NieldeBeaudrap: Yes. In fact the basis of product states is exactly the reason why there cannot be some observable which has a different expectation for some entangled state on this space if it's expectation value is constant across all product states. – Joe Fitzsimons Oct 1 '11 at 14:33 @JoeFitzsimons: I have in mind the mathematical structure of states, and invariants that can be computed numerically. For this question I'm not that interested if invariants are directly related to any measurements. – Piotr Migdal Oct 3 '11 at 12:10 @Piotr: I posted a new answer which hopefully addresses your question as I now understand it. I'll leave this answer, as I think it is interesting in its own right, but is entirely different from the new answer. – Joe Fitzsimons Oct 3 '11 at 16:22 It seems from Piotr's comments on my other answer, that he is looking for an invariant of the mathematical representation of the state, rather than an observable that remains unchanged. In this case the answer is very different, and hence I am posting a new answer, rather than replacing the old one (since it would mean entirely rewriting it, and the current version may be of interest to some people). Any density matrix can be written as $\rho = \sum_k \alpha_k \sigma_k$, where $\sigma_k=\bigotimes_i \sigma_{k_i}$ and $\{\sigma_{k_i}\}$ form an orthonormal basis for Hermitian matrices corresponding to the subsystem dimensionality and includes the identity matrix. When you apply local unitaries you obtain $\rho' = \big(\bigotimes_i U_i \big) \rho \big(\bigotimes_i U_i^\dagger \big)$. Now if you consider what happens term by term, you will notice that each operator $\sigma_k$ is mapped only to operators of the same weight (i.e. operating non-trivially on the same number of subsystems). I'll take $w_k = w(\sigma_k)$ to be the weight function for each operator $\sigma_k$. So, we have $w(\sigma_k) = w\big(\big(\bigotimes_i U_i \big) \sigma_k \big(\bigotimes_i U_i^\dagger \big)\big)$. This is trivially true, since for every subsystem where $\sigma_k$ in $\rho$ acts as the identity (i.e. for every $i$ such that $\sigma_{k_i}=\mathbb{I}$) $U_i$ and $U_i^\dagger$ cancel, and so the transformed operator also acts as the identity on that subsystem. Conversely, if $\sigma_{k_i}\neq \mathbb{I}$ then $U_i\sigma_{k_i}U_i^{\dagger}\neq \mathbb{I}$. There is a pretty intuitive reason for this: local operations should not create non-local correlations. Now, from this, it should be clear that $\{\beta_w=\sum_{i:w(\sigma_i)=w} \|\alpha_i \|^2\}_{w=0}^n$ is invariant, since $U \sigma_k U^\dagger = \sum_j \gamma_{jk} \sigma_j$ such that $\sum_j \|\gamma_{jk}\|^2 = 1$. This is a conserved quantity independent of the symmetry, and depends only on the fact that all applied unitaries are local, however I believe this is the kind of thing you want. Once you impose the criteria that the states and operations are symmetric, you have the additional criteria that $\alpha_i = \alpha_j$ if $\sigma_i$ can be obtained from $\sigma_j$ by permutation of the qudits, and hence $\|\alpha_i-\alpha_j\|$ is also invariant for all such pairs. - Thanks Joe - actually I was looking at something similar. Or rather - with more details (e.g. diagonalization of the matrix for $w=(\sigma_i)2$ (for symmetric states there are only 9 entries)). And do you have any idea which states have the same $\beta_w$s? – Piotr Migdal Oct 6 '11 at 17:12 It seems that this question is addressed in this paper: http://arxiv.org/abs/1011.5229 (Edit: I just noticed this is seems to be restricted to qubits, so it probably does not answer your question ... ) - Welcome to TP.SE! I wasn't aware of that paper, but it seems an interesting observation, and looks like it might generalize beyond qubits. – Joe Fitzsimons Oct 1 '11 at 7:10 This answer is incomplete, but it should provide an answer for almost all symmetric states (i.e. it suffices for all but a set of symmetric states having measure zero). The symmetric subspace is spanned by product states. We may then consider different ways in which a particular symmetric state decomposes into symmetric products; in particular, if any choice of decomposition naturally gives rise to an invariant. A greedy way to go about decomposing a symmetric $\def\ket#1{\|#1\rangle}\ket\psi$ state into symmetric products would be to simply look for the symmetric product $\ket\phi\ket\phi\cdots\ket\phi$ with which $\ket\psi$ has the overlap of the largest magnitude. Let $\ket{\phi_0}$ be the single-spin state satisfying this, and $$\def\bra#1{\langle#1\|}\alpha_0 = \Bigl[\bra{\phi_0}\cdots\bra{\phi_0}\Bigr]\ket\psi$$ which without loss of generality is positive. With probability 1, the state $\ket{\phi_0}$ is unique (in that the set of symmetric states for which it is not unique has measure zero). Let $\ket{\psi_1}$ be the projection of $\ket{\psi}$ into the orthocomplement of $\ket{\phi_0}^{\otimes n}$: this is another symmetric state. So, we define $\ket{\phi_1}$ to be the (again with probability 1, unique) single spin state such that $\ket{\phi_1}^{\otimes n}$ has maximal overlap with $\ket{\psi_1}$; we let $\alpha_1$ be that overlap; and we define $\ket{\psi_2}$ to be the projection of $\ket{\psi}$ onto the orthocomplement of $\mathrm{span}\{\ket{\phi_0}^{\otimes n}\!\!,\;\; \ket{\phi_1}^{\otimes n}\}$. And so on. By continually projecting $\ket{\psi}$ onto the orthocomplement of spans of larger and larger sets of symmetric products, we ensure that the resulting projections $\ket{\psi_j}$ will not have maximal overlap with any product state that came before, or more generally which can be spanned by the preceding symmetric products. So at each iteration we obtain a single-spin state $\ket{\phi_j}$ such that $\mathrm{span}\{\ket{\phi_0}^{\otimes n}\!\!,\;\ldots\,,\; \ket{\phi_j}^{\otimes n}\}$ has dimension one larger than in the previous iteration. In the end, we will obtain a collection of symmetric products which, if they don't span the symmetric subspace, at least contain $\ket{\psi}$ in their span. So we obtain a decomposition $$\ket{\psi} \;=\; \sum_{j = 0}^\ell \alpha_j \ket{\phi_j}^{\otimes n}$$ where the sequence $\alpha_0, \alpha_1, \ldots$ is strictly decreasing. Let us call this the symmetric product decomposition of $\ket{\psi}$. (It wouldn't take too much to generalize this representation to one where the states $\ket{\phi_0}$, $\ket{\phi_1}$, etc. are non-unique; but for what comes next uniqueness will be important.) Given the symmetric product decomposition of $\ket{\psi}$, it is trivial to describe the corresponding representation for $U^{\otimes n} \ket{\psi}$: just multiply each of the $\ket{\phi_j}$ by $U$. And in fact, if you computed $U\ket{\psi}$ and then determined its symmetric product decomposition, the decomposition $$U^{\otimes n}\ket{\psi} = \sum_{j=0}^\ell \alpha_j \Bigl[ U\ket{\phi_j} \Bigr]^{\otimes n}$$ is exactly what you would find: $\ket{\phi_0}^{\otimes n}$ has maximal overlap with $\ket{\psi}$ if and only if $[ U \ket{\phi_0} ]^{\otimes n}$ has maximal overlap with $U^{\otimes n} \ket\psi$, and so on. So to show that two symmetric states are LU-equivalent, it suffices to show that the sequence of amplitudes $\alpha_j$ are the same, and that the sequence of single-spin states $\ket{\phi_j}$ are related by a common unitary. The last part can be most easily done by finding a normal form for the states which are equal, for sequences of single-spin states which are related by a common single-spin unitary. We can do this by finding a unitary $T$ which 1. maps $\ket{\phi_0}$ to $\ket{0}$, 2. maps $\ket{\phi_1}$ to a state $\ket{\beta_1}$ in the span of $\ket{0}$ and $\ket{1}$, with $\bra{1}\beta_1\rangle \geqslant 0$, 3. and for each subsequent $j > 1$, maps $\ket{\phi_j}$ to some state $\ket{\beta_j}$ which is in the span of standard basis states $\ket{0}, \ldots, \ket{b_j}$ for $b_j$ as small as possible, with $\bra{b_j}\beta_j\rangle \geqslant 0$ if possible. (For any state $\ket{\phi_j}$ which is in the span of preceding states $\ket{\phi_h}$, the state $\ket{\beta_j}$ will similarly be determined by the states $\ket{\beta_h}$ for $h < j$, in which case we do not have a choice of the value of $\bra{b_j}\beta_j\rangle$.) We then have a unitary such that $T \ket{\phi_j} = \ket{\beta_j}$; and for any two sequences of states $\ket{\phi_j}$ and $U\ket{\phi_j}$, we should obtain the same sequence of states $\ket{\beta_j}$. You can then determine that two symmetric states are equivalent if they give rise to the same sequence of amplitudes $\alpha_j$ and the same "normal form" single-spin states $\ket{\beta_j}$. In the case that there is not a unique state $\ket{\phi_j}^{\otimes n}$ which has maximal overlap with $\ket{\psi_j}$ in the construction of the symmetric product decomposition, the problem is then in defining the normal form states $\ket{\beta_j}$. However, so long as the states $\ket{\phi_j}$ are unique, which happens with measure 1, you should have a polynomial-size invariant (up to precision limitations) for determining if two symmetric states are the same. - Thanks Niel. I like the approach, but (unfortunately) it has one drawback - interesting states tend to have a lot of symmetries and are often in the zero-measure set of all permutation symmetric states. E.g. for a Bell state $(\|00\rangle+\| 00\rangle)/\sqrt{2}$ the maximum is achieved on entire equator (i.e. separable states with $\|\psi\rangle = \cos(\alpha)\|0\rangle+\sin(\alpha)\|1\rangle$) so your decomposition is not unique. – Piotr Migdal Oct 6 '11 at 17:21 @PiotrMigdal: Fair enough; I'm too aware of the bitter-sweet irony that the nicest structures to study tend to be examples from a set of measure zero. Of course, if you have enough symmetry between a (discrete) set of maximizing products, such as with a Bell state, this would simplify the task of obtaining a normal form; perhaps one could formulate particular symmetry conditions for which a useful normal form is possible. – Niel de Beaudrap Oct 6 '11 at 17:51	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 104, "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": 0.9510730504989624, "perplexity_flag": "head"}
http://mathhelpforum.com/differential-geometry/122338-basic-topology-closed-open-sets.html	Thread: 1. Basic Topology, closed and open sets Consider the metric space ([0, 1], \| · \|).In this space, no number smaller than 0 or more than 1 exists. a) Show that any interval in the form (r,s), where 0 < r < s < 1 is open in this space. b)Show that any interval in the form [0, s), where s < 1, is open in this space. c) Show that any interval in the form (r, 1], where r > 0, is open in this space. 2. Hello, there may be a simpler way, but this proved it for me: a) Let $y\in (r,s)$. Then, for $\delta =\min \{d(r,y),d(s,y)\},$ we have that $B_d(y,\delta )\subset (r,s)$, hence $(r,s)$ is open. For b) and c), adapt the same argument accordingly.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.3, "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": 0.8636229038238525, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/26155?sort=oldest	## Finite number of minimal ideals ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) What is the necessary condition on a ring that guarantees the number of minimal non-zero ideals to be finite? Neither Noetherian or Artinian condition seems sufficient, and the ring being semisimple seems too strong. - 1 Noetherian is sufficient (although obviously not necessary). Just take a primary decomposition of the ideal (0). – JT May 27 2010 at 15:08 3 I believe kwan is asking for finitely many minimal ideals, not minimal prime ideals. If so, then every ring has a unique minimal ideal: $(0)$. If instead you're asking for minimal non-zero ideals, it's much harder. Can you give us an idea of why you're asking? – Graham Leuschke May 27 2010 at 15:22 Thanks for point that out. Yes, I meant non-zero ideals, not necessarily prime. I was reading about the annihilator of a maximal ideal in a ring being a direct sum of non-zero minimal ideals (socle), and wondered if it's common for a ring to have infinitely many non-zero minimal ideals. – ashpool May 27 2010 at 20:31 9 Here are a couple things to notice. (1) if $R$ is noetherian local and reduced, then it has no minimal non-zero ideals, since the square of any ideal is a strictly smaller one. So that's an example of what you're asking about, but not a very satisfying one. (I needed $R$ to be local to rule out things like $I=I^2$.) (2) Let $R = k[x,y]/(x^2,xy,y^2)$. Then the non-maximal non-zero ideals of $R$ are all of the form $(\alpha x - \beta y)$ for field elements $\alpha, \beta$. Thus $R$ has your property if and only if $k$ is a finite field, which again is not a very satisfying answer. – Graham Leuschke May 27 2010 at 21:33 1 @kwan, for your recent question: If $\text{dim}_kI$ is at least $2$, then the set of lines will not be finite unless $k$ is finite. By the way, I just notice you have not accepted any of my answers and many others, for example the one by BCrnd on flatness. If you think the answers are correct, please accept them. – Hailong Dao Aug 16 2010 at 2:29 show 2 more comments ## 3 Answers (Inspired by Graham's comment) For simplicity I will consider the case $(R,m)$ is a Noetherian, local ring. A non-zero minimal ideal better be principal. Also, if $(x)$ is such ideal, then for any $y\in m$, the ideal $(xy)$ has to be $0$, so $mx=0$. Let $I= \{x\in R\|mx=0\}$, the socle of $R$. Since $Im=0$, $I$ is a vector space over $k=R/m$. You want to know when the set of $1$-dimensional subspaces of $I$ is finite. This happens if and only if $\dim_kI\leq 1$ or $k$ is a finite field. If $R$ is also Artinian (i.e. $\dim R=0$), then $\dim_kI\leq 1$ means precisely that $R$ is Gorenstein. So in this case (Noetherian, local of dimension $0$) one has a particularly nice answer: the set of nonzero minimal ideals is finite iff $R$ is Gorenstein or $k$ is finite. Note that if $\dim R\geq 1$, then $I=0$ iff $\text{depth}\ R\geq 1$. In general one can localize to get at least necessary conditions. For example, if the height of all maximal ideals is at least $1$, then $R$ satisfying Serre's condition $(S_1)$ is certainly sufficient, since the socle when you localize at any maximal ideal will then be $0$, so there is no non-zero minimal ideals. - I'd like to be credited with an assist on this answer :) . – Graham Leuschke Jun 3 2010 at 12:38 @Graham: Done, thanks! – Hailong Dao Jun 4 2010 at 19:55 Thanks for your answer. How can I see that $2\leq \mbox{dim }_{k}I<\infty$ is impossible? – ashpool Aug 15 2010 at 21:12 ### 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. Please clarify: are you asking about commutative rings or noncommutative rings? (The tag "ac.commutative-algebra" suggests the former; your reference to semisimplicity suggests the latter.) If the question is about noncommutative rings, then presumably by "ideals" you mean two-sided ideals. - Sorry, I should have mentioned it was commutative ring. – ashpool Jun 3 2010 at 11:37 I'm not sure how to answer your exact question ("the necessary condition that guarantees..."?), but here are a few minor observations. A direct product of commutative rings has only finitely many minimal (by definition nonzero) ideals if and only if each component ring has only finitely many and all but finitely many of the component rings have none. So it suffices to consider indecomposable commutative rings with only finitely many ideals. There are all sorts of indecomposable commutative rings with no minimal ideals. Now suppose the indecomposable commutative ring has a positive finite number of minimal ideals. The socle of such a ring has square zero; thus, the socle is a nonunital subring with the structure of an additive abelian group with zero multiplication on it (of course, this additive abelian group need not have only finitely many minimal subgroups). One can construct all sorts of examples of this sort. For instance, let $A$ be an indecomposable commutative ring with no minimal ideals, and let $M$ be an $A$-module with only finitely many simple submodules. (For example, one might take $M$ to be a uniserial $A$-module.) Let $R = A \oplus M$ as an additive group, with multiplication given by $(a_1, m_1) (a_2, m_2) = (a_1 a_2, a_1 m_2 + a_2 m_1)$. Then $R$ is an indecomposable commutative ring with only finitely many minimal ideals. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 51, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": 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": 0.9214548468589783, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/tagged/type-theory?sort=votes&pagesize=15	# Tagged Questions The type-theory tag has no wiki summary. 6answers 2k views ### Learning Lambda Calculus What are some good online/free resources (tutorials, guides, exercises, and the like) for learning Lambda Calculus? Specifically, I am interested in the following areas: Untyped lambda calculus ... 1answer 468 views ### If $f(x)=g(x)$ for all $x:A$, why is it not true that $\lambda x{.}f(x)=\lambda x{.}g(x)$? There's something about lambda calculus that keeps me puzzled. Suppose we have $x:A\vdash f(x):P(x)$ and $x:A\vdash g(x):P(x)$ for some dependent type $P$ over a type $A$. Then it is not necessarily ... 0answers 111 views ### Hao Wang's $\mathfrak S$ system: a “transfinite type” theory? Long ago, while I was reading a book ($$) about the various way to build set theories (Zermelo-Freankel, Von Neumann–Bernays–Gödel, and type theories), I read about a variant of type theory with ... 2answers 335 views ### Classic type theory textbooks There are many classic textbooks in set and category theory (as possible foundations of mathematics), among many others Jech's, Kunen's, and Awodey's. Are there comparable classic textbooks in ... 1answer 277 views ### Intuitionistic Banach-Tarski Paradox While the Banach-Tarski paradox is a counter-intuitive result which requires the Axiom of Choice, leading some people to argue specifically against Choice, and others to argue for constructive ... 0answers 167 views ### Looking for an approach to mathematical notation wherein the universe is divided into disjoint worlds. Is there a rigorous approach to mathematical notation wherein the "universe" is divided into disjoint "worlds," and the meaning of notation is world-dependent? This would solve a few pesky problems. ... 2answers 335 views ### What's the meaning of algebraic data type? I'm reading a book about Haskell, a programming language, and I came across a construct defined "algebraic data type" that looks like ... 2answers 63 views ### Is there a theory of extensible definitions? We can define $+$ as a function $\mathbb{N}^2 \rightarrow \mathbb{N}$, and then prove the theorem: Thm. The range of $+$ is $\mathbb{N}$. If we later wish to extend $+$ to a function \$\mathbb{Z}^2 ... 2answers 69 views ### Eilenberg Moore category I've been trying to code up the Eilenberg-Moore category for a monad in Haskell. As I understand it, given a category $C$ and a monad $(T,\eta,\mu)$ on $C$ we build the Eilenberg-Moore category $C^T$ ... 2answers 118 views ### Peano axioms expressed in type theory I have a very strong understanding of 1st order logic and am trying to lean type theory as an alternative. Could someone express the Peano axioms with type-theory? I am especially interested to see ... 2answers 271 views ### Curry-Howard correspondence I read that the Curry-Howard correspondence introduces an isomorphism between typed functions and logical statements. For example, supposedly the function \begin{array}{l} I : \forall a. a \to a\\ ... 0answers 41 views ### Are large cardinals bi-interperable with type theory? "Sets in Types, Types in Sets" establishes that the calculus of constructions is bi-interpretable with ZFC + infinitely many inaccessibles. Are there any more results like this higher up the large ... 2answers 212 views ### Has a Dependent Type always a Type? I am experimenting with dependent types. Lets assume the following short notation: ... 1answer 216 views ### Constructing dependent product (right adjoint to pullback) in a locally cartesian closed category I've been trying to find a proof that the pullback functors in a locally cartesian closed category have right adjoints (used to model the notion of indexed product inside a category (rather than ... 0answers 60 views ### Introductory text about different stratification methods in higher-order logic and set theory Could someone recommend me a good overview text about stratification of predicates, comprehension axioms, and other methods of avoiding the paradoxes in untyped or only loosely/relatively typed ... 2answers 243 views ### formal rules for avoiding bound/unbound variable problems in lambda calculus I have been interested in learning formal math formally enough so that I could write a proof assistant using some simple parsing tools, and explain to someone else with little math knowledge how to ... 2answers 132 views ### Accessible formal specification and explanation of First Order Logic? I am trying to get good at proofs by working through How To Prove It. Unfortunately I am very bothered by the fact that I do not understand all the formalities in First Order Logic + set theory ... 1answer 28 views ### Variables in Types in type theory I'm slowly grasping this, though the different formulations of type theory make it difficult. In http://imps.mcmaster.ca/doc/seven-virtues.pdf types can only be formed from , i, and a->b when a and ... 1answer 27 views ### in type theory does (x:A) imply ((x:A):A) In the formulation of type theory I'm reading, (x:A) is an expression of type A. This would seem to imply ((x:A):A) and (((x:A):A):A)... Is this a common feature of type theories? Or am I reading too ... 2answers 28 views ### mapping simple first oreder problems to type theory Since Peano axioms expressed in type theory doesn't seem to be going anywhere, here is a simpler question: How would I map simple first order systems to an equivalent type theoretic notation. ... 0answers 39 views ### How does type theory handle division by zero and such? Say I have a program that needs to not divide by zero: f(x): if nonzero(x): return sin(x)/x else: return 1 If we divide by zero, we get ... 0answers 26 views ### Enumeration of symbols in grammatical expressions or vertices in tree graphs I have expressions (type of a function) like e.g. $$f:(A\to B)\to C \to (D\to E)\to F.$$ (Where I understand $A\to B\to C$ as $A\to (B\to C)$, in case that is relevant.) There might be information ... 1answer 56 views ### Natural numbers as types. My question could be simple but I haven't found an answer for it: could we define a type theory where the types are the natural numbers themselves (not the set of the natural numbers)? Thanks. 1answer 166 views ### How to understand inductive definitions of recursive data types? The problem was encountered when learning Computability Notes by Roberto Zunino (link), page 9 and 11. It seems as if it is a well-known issue and I don't need to specify the "canonical" meaning of ...	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 21, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9054675102233887, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/92192/hot-topics-in-error-correcting-coding-related-to-interesting-math/118041	## Hot-topics in error correcting coding related to interesting math. ? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) What are topics in error-correcting coding which are related to interesting math. ? I am primarely interested in nowdays hot topics, but old days topics are also welcome. Let me try to mention what I heard about. 1) Hot topic in error-correction is finding LDPC codes with very low "error-floor" for code lengths dozens thoursands bits, this might be useful for optic transmission. However it is not clear for me what kind of math playing role here ? ("Error-floor" is related with codewords with small Hamming weight. So the code might be quite good - means majority of codewords have big Hamming weight, so in most case code performs well, but very small number having small Hamming weight will cause small number of errors - it can be seen on the BER/SNR plot as a "floor".) 2) There is certain number of papers applying number theory (lattices in algebraic number fields) to consruct good codes. One may see papers by F. Oggier, G. Rekaya-Ben Othman, J.-C. Belfiore, E. Viterbo: e.g. this one : http://arxiv.org/abs/cs/0604093. I am not aware how "hot" is this topic and how far it is from practical applications... 3) Polar codes is a hot topic. What kind of math is playing role here ? 4) Probably most classical example is the Golay code (1948) and sporadic simple Mathieu groups. Let me quote Wikipedia: http://en.wikipedia.org/wiki/Binary_Golay_code : "The automorphism group of the binary Golay code is the Mathieu group . The automorphism group of the extended binary Golay code is the Mathieu group . The other Mathieu groups occur as stabilizers of one or several elements of W." By the way - is it occasional coincidence of there is something behind it ? - There is certainly some "recent" mathematical research going into ECC as applications. The first example that comes into mind are expanders graphs, and the construction of LDPC codes by Margulis (see for example here - nd.edu/~rosen/Paper/margulis_8.pdf). – Asaf Mar 25 2012 at 19:50 arxiv.org/abs/1209.3460 Expander-like Codes based on Finite Projective Geometry Swadesh Choudhary, Hrishikesh Sharma, B. S. Adiga, Sachin Patkar (Submitted on 16 Sep 2012) We present a novel error correcting code and decoding algorithm which have construction similar to expander codes. The code is based on a bipartite graph derived from the subsumption relations of finite projective geometry, and Reed-Solomon codes as component codes. We use a modified version of well-known Zemor's decoding algorithm for expander codes, for decoding our codes. ........... – Alexander Chervov Sep 18 at 6:25 books.google.com/… Coding theory and algebraic curves over finite fields G VAN DER GEER - … of Algebraic Geometry to Coding Theory, …, 2001 ... tautological classes on the moduli space and the not less spectacular proof by Kontsevich of this ... Deligne and Lusztig showed that irreducible representations of finite Lie groups can be found in a ... For example, the bounds on the number of rational points on cu – Alexander Chervov Sep 22 at 17:16 arxiv.org/abs/1210.0083 Decoding a Class of Affine Variety Codes with Fast DFT Hajime Matsui An efficient procedure for error-value calculations based on fast discrete Fourier transforms (DFT) in conjunction with Berlekamp-Massey-Sakata algorithm for a class of affine variety codes is proposed. Our procedure is achieved by multidimensional DFT and linear recurrence relations from Grobner basis and is applied to erasure-and-erro ... A motivating example of our algorithm in case of a Reed-Solomon code and a numerical example of our algorithm in case of a Hermitian code are also described. – Alexander Chervov Oct 2 at 5:08 arxiv.org/abs/1210.0140 Polycyclic codes over Galois rings with applications to repeated-root constacyclic codes Cyclic, negacyclic and constacyclic codes are part of a larger class of codes called polycyclic codes; namely, those codes which can be viewed as ideals of a factor ring of a polynomial ring. The structure of the ambient ring of polycyclic codes over GR(p^a,m) and generating sets for its ideals are considered. Along with some structure details of the ambient ring, the existance of a certain type of generating set for an ideal is proven. – Alexander Chervov Oct 2 at 5:12 show 10 more comments ## 2 Answers Your list certainly has many nice topics. 1) Yup. This would be nice to have. In practical applications we can get rid of the error-floor by concatenating a decent LDPC with a good high rate algebraic code such as a BCH-code that can then correct the residual errors (the one application I know about is the second generation standard for European digital video broadcast, aka digi-TV, their the code length is 64800 or 16200 bits). What makes this challenging is that designing a good LDPC-code requires familiarity with some tools from stochastics (lost me at that point), but those tools don't say anything about the minimum Hamming distance. Many a standard (IIRC in addition to European DVB also MediaFlo, a US standard for something similar) uses families of LDPC-codes designed around a specific decoding circuitry. This is more or less necessary, because otherwise the problem of routing the messages generated by the belief propagation algorithm becomes prohibitive. An exception to this rule is the Chinese video broadcast standard. At least the parts of that standard that I have seen describe the LDPC-codes in such a way that no structure is apparent. They may be protecting their intellectual property :-) So a breakthru in this area would probably have to also keep this in mind in order to end up in future applications. Hopefully more knowledgable people can comment. I do expect something to happen here in years to come, but the existing LDPC codes already work quite well. 2) This was a relatively hot topic a few years. I am a bit hesitant to call it coding theory - calling it multiantenna signal constellation design might be more fitting, but whatever :-). By using basic facts of global class field theory my graduate students managed to "improve" upon the Golden code (by Oggier et al). I put the "improve" in quotes, because the improvement is somewhat theoretical. A more precise way of stating their result is that if you carve a given number of multiantenna signals from their lattice (representing a maximal order of a division algebra), you are less likely to make an error at the receiving end than what would happen, if you carve your signal set from one of the codes proposed by Oggier et al. However, that's not the end of the story. If you combine that multiantenna constellation with, for example, an LDPC code, our construction loses its theoretical advantage, because an LDPC-decoder wants to have reliability information about individual bits. When you pack several bits worth of information into a selection of a single multiantenna signal, our method creates more dependencies among those reliability figures, and that makes things worse in the end. Anyway, the math in the construction of my students is fun, and they all graduated, so... As the number of antennas increases, the computational complexity grows really badly. Some codes suffer more from this than others. A relatively recent idea (B.S. Rajan and his students, couldn't find a proper reference, sorry) is to use representations of Clifford algebras with a view of reducing this complexity. That is a promising idea. All of the above constructions depend on the receiver knowing the channel state. From some point on you need to allocate too large fraction of the bandwidth to pilot symbols to make that assumption true. So another thread in this area has been to use differential modulation (=use the preceding signal as a pilot for the next) or Grassmannian codes (=the signal is to an extent its own pilot). A lot of fun math going on there, but don't know whether they will stay. Another theoretically interesting thread within this topic is: "How will the rules change, when two or more independent users transmit simultaneously?" A beginning graduate student in our group has come up with some number theoretic constructions. As a new tool he needed some facts from Diophantine approximation. The information theory in that thread is, I'm sad to say, over my head. I am willing to also bet that this question on MO derives its motivation from this problem area :-) This question hit too close. It is not entirely clear that I managed to be objective. The Golden Code (Oggier et al) is in a hyperWLAN standard. Don't know how widely that part of the standard is used. Multiantenna coding in cellular applications goes largely by different rules. This is because there is a feedback channel there, so the transmitter also has an idea of the channel state, and can take advantage. The math becomes easier then (so I've been told). This is not possible in a broadcast application, because there may be millions of receivers, and knowing their channel states is A) impossible, B) useless because you can't optimize the transmission for all of them simultaneously. 3) The Polar codes were a big surprise to me. I can't comment on them for lack of familiarity. Leave this for someone else to answer. 4) The Golay codes have been around. They are a rich source of algebraic and combinatorial miracles - a lot of fun! The codes are way too short to be useful in transmitting bulk data, but do make an appearance in other applications. In their book (SPLAG) Conway & Sloane study these in great detail. Probably the most investigated error-correcting codes of all times! 5) I want to add network coding as a hot topic. It has certainly received a lot of attention lately. It is not clear how deep math they end up using. Sometimes it looks like it is just Grassmannians over a finite field. - Wow ! Thank you very much for this detailed answer ! – Alexander Chervov Mar 27 2012 at 6:23 ### 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. Edit: since I only implicitly answered your question in the title (i.e., hot topics in error coding codes related to interesting math), I think one of the current hottest ones is quantum error-correcting codes. Since the discovery by P. Shor of the fact that we can correct quantum errors, the field of quantum information science exploded and made a quantum leap towards realization of large-scale quantum computation and reliable quantum communication. The list of the branches of mathematics used in the field is endless and growing as we speak. [\end Edit] Jyrki already gave a great answer to many of your questions. So, here's to complement it by addressing a couple points he (she?) didn't cover: 1) Hot topic in error-correction is finding LDPC codes with very low "error-floor" for code lengths dozens thoursands bits, this might be useful for optic transmission. However it is not clear for me what kind of math playing role here ? That depends on the channel (and decoding method) you have in mind. In what follows, we only consider a binary code (because I don't know much about the non-binary case!). The simplest case is BEC (the binary erasure channel), where stopping sets determine the characteristic of the BER of an LDPC code. Small stopping sets are like small weight codewords in a traditional code; the more your code has small stopping sets, the worse its error correction performance gets in general. Stopping sets can be described in a completely combinatorial way. A stopping set of a parity-check matrix $H$ is a set of columns in which every row has at least two $1$'s (within the columns). You can rephrase the definition in the language of set systems by regarding $H$ as an incidence matrix; it's equivalent to a full configuration. Or you can reword it by using terms like check nodes and variable nodes as is standard in the LDPC code literature. Anyway, the smallest stopping sets dominate at sufficiently high SNR. So improving BER directly corresponds to an avoidance problem (or equivalently a forbidden configuration problem) in a binary matrix. So, it's a combinatorial problem that studies a set system avoiding full configurations that gives the desired code length, rate, other additional properties you want such as a particular automorphism. So it's pretty much the same as good ol' algebraic coding theory; avoiding low weight codewords (or equivalently achieving larger minimum distances) is now replaced by avoiding small stopping sets. In fact, the size of a smallest stopping set is called the stopping distance of $H$. One thing to note is that the notion of stopping distance is defined for a parity-check matrix. Since one same linear code has various different parity-check matrices, you don't say the stopping distance of a code. Other than this, stopping distance is more or less the analogue of minimum distance when it comes to an LDPC code over BEC. Things are much more complicated for other well-studied channels like everyone's favorite, the AWGN channel, and the coding theory 101 channel, the binary symmetric channel. Some say that in general better stopping distances tend to lead to better BER or at least it's not a bad sign. But this isn't always the case. The kind of sub-structure in $H$ that screws up your decoding algorithm at high SNR for channels other than BEC is typically not easy to describe by simple combinatorics. If you're curious, searching IEEE Xplore with keywords like "trapping sets" and "pseudo-codewords" should direct you to the right papers. You can also correct quantum errors by taking advantage of the theory of LDPC codes. But I haven't seen a paper that directly studies the analogue of stopping sets, trapping sets, etc. for quantum channels. The main difficulty in the quantum domain is that not every $H$ defines a quantum LDPC code; the rows should correspond to the generators of a stabilizer that is an abelian subgroup of the Pauli group or otherwise you need either a certain number of qubits in the Bell states shared between sender and receiver or assume a very reliable auxiliary quantum channel to "force" your $H$ to define a stabilizer of a larger group. So, you have more things to consider before you can translate the error floor problem of LDPC coding for quantum channels into the language of mathematics. Or you can say the kind of math playing a role for this particular subfield is the ones you need for quantum information on top of the usual math for LDPC codes such as combinatorics, information theory, and probability. Since you mentioned polar codes in your third question, spatially-coupled LDPC codes are among the hottest classes of LDPC codes that are competing with polar codes. The noise threshold of spatially-coupled ensamble under iterative decoding matches the MAP decoding's threshold over BEC. The quantum analogues of spatially-coupled LDPC codes and polar codes have been/are being studied, too, albeit with a limited success compared to the remarkable progress for the classical case. 4) Probably most classical example is the Golay code (1948) and sporadic simple Mathieu groups... [Quote from Wikipedia about how Mathieu groups are automorphism groups of the Golay codes, extended Golay codes, etc. goes here] ...Is this a coincidence, or is there something behind it? Yes. There's something behind it. For instance, the Mathieu group $M_{24}$ is by definition the automorphism group of the extended binary Golay code, or equivalently, of the Steiner $5$-design of order $24$ and block size $8$, which is called the Witt design in combinatorial design theory. Here is a lecture note on this by A. E. Brouwer. This is one of those interesting interactions between finite simple groups, designs, and codes. I think this answer is getting too long for a forum post, so for the rest I'll just recommend you search the internet with google or your favorite searching engine by keywords like Golay codes, Steiner designs, Witt designs, simple groups, and the like. That way, I won't upset my previous and current posdoc mentors and Ph.D. supervisor at the same time by writing something stupid about what I should know better. - Thank you very much!!! – Alexander Chervov Jan 4 at 11:42	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 11, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_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": 0.9280902743339539, "perplexity_flag": "middle"}
http://divisbyzero.com/2010/09/08/irrational-rotations-of-the-circle-and-benfords-law/?like=1&source=post_flair&_wpnonce=8149a33672	# Division by Zero A blog about math, puzzles, teaching, and academic technology Posted by: Dave Richeson \| September 8, 2010 ## Irrational rotations of the circle and Benford’s law Take a collection of real-world data such as the lengths of all rivers in the world, the populations of counties in the United States, the net worths of American corporations, or the street addresses of all residents of Detroit. Strip away all the information except the leading digits. What percentage of these digits do you expect to be 1′s? 2′s? 9′s? Surely the answer is the same for all digits: 11% (1/9), right? It turns out that in many cases, the leading digits are not uniformly distributed, but obey Benford’s law: the leading digit ${d}$ occurs with frequency ${\log_{10}(1+1/d)}$. [See Terence Tao's post for details on when Benford's law applies, but roughly speaking the data must be very spread out (e.g., spanning several orders of magnitude), so data sets like zip codes in the US or heights of adult males won't follow Benford's law.] Thus we expect the leading digits to occur with the following frequencies: $\displaystyle \begin{array}{cc} \text{Leading digit} &\text{Benford's prediction}\\ 1 & 30.10\%\\ 2 & 17.61\%\\ 3 & 12.49\%\\ 4 & 9.69\%\\ 5 & 7.92\%\\ 6 & 6.69\%\\ 7 & 5.80\%\\ 8 & 5.11\%\\ 9 & 4.58\% \end{array}$ Let us look at a concrete example: powers of 2. The first 20 powers of 2 are: 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768, 65536, 131072, 262144, and 524288. The leading digit is 1 six times, that’s 30%, it is 2 four times (20%), and 9 zero times (umm,… 0%). I computed the first 100 and first 500 powers of 2 and found that the leading digits have percentages: $\displaystyle \begin{array}{cccc} \text{Leading digit} & \text{First 100 powers} &\text{First 500 powers} & \text{Benford's prediction}\\ 1 & 30\% &30.2\%& 30.10\%\\ 2 & 17\% &17.6\%& 17.61\%\\ 3 & 13\% &12.4\%& 12.49\%\\ 4 & 10\% &9.6\%& 9.69\%\\ 5 & 7\% &7.8\%& 7.92\%\\ 6 & 7\% &6.8\%& 6.69\%\\ 7 & 6\% &5.6\%& 5.80\%\\ 8 & 5\% &5.0\%& 5.11\%\\ 9 & 5\%& 4.6\%&4.58\% \end{array}$ Amazing! Benford’s law has been used to detect fraud in tax returns and in election results—the numbers that people make up typically do not follow Benford’s law. For more information you may want to read other accounts of Benford’s law on the web. In this blog post I will give a proof just for the powers of 2. It is a clever proof that uses a theorem about rotations of the circle that I wrote about last summer. For simplicity we’ll consider a circle of circumference 1; equivalently we may think of the circle as ${[0,1]}$ with 0 and 1 glued together or as ${\mathbb{R}/\mathbb{Z}}$, the set of real numbers, modulo 1. Let ${\{x\}}$ denote the fractional part of the real number ${x}$. Then ${\{3.25\}=\{239.25\}=\{-1.75\}=0.25}$ all represent the same point on the circle. Let ${\alpha}$ be any real number. We can think of the sequence ${0,\{\alpha\},\{2\alpha\},\{3\alpha\},\ldots}$ as the orbit of 0 under repeated rotations of the circle by ${\alpha}$. We begin with the point 0 on the circle, rotate by ${\alpha}$ to obtain the point ${\{\alpha\}}$, rotate by ${\alpha}$ again to obtain ${\{2\alpha\}}$, etc. In my earlier blog post I stated the following theorem. Theorem. Suppose ${\alpha}$ is an irrational number. 1. Then ${\big\{0,\{\alpha\},\{2\alpha\},\{3\alpha\},\ldots\big\}}$ is a dense subset of the circle. 2. Moreover, if ${I}$ is an interval in the circle of length ${l(I)}$ and there are ${n(k)}$ elements of ${\big\{0,\{\alpha\},\{2\alpha\},\{3\alpha\},\ldots,\{(k-1)\alpha\}\big\}}$ in ${I}$, then ${\displaystyle\lim_{k\rightarrow\infty}\frac{n(k)}{k}=l(I)}$. Part 1 of the theorem states that the orbit of the point 0 under an irrational rotation by ${\alpha}$ “fills up” the entire circle; that is, every open interval on the circle, regardless of how small, intersects this set of points. Part 2 says something even stronger. It says that the orbit fills up the circle in a uniform way—asymptotically the amount of time the orbit spends in an interval is equal to the length of the interval (we often say that “the time average equals the space average”). So what does this have to do with Benford’s law? The leading digit of ${2^{m}}$ is ${d}$ provided there exists a nonnegative integer ${n}$ such that $\displaystyle d10^{n}\le 2^{m}< (d+1)10^{n}.$ Taking the logarithm (base 10) of the above inequality we obtain $\displaystyle n+\log_{10}(d)\le m\log_{10}2<n+\log_{10}(d+1).$ Equivalently, $\displaystyle m\log_{10}2\in[n+\log_{10}(d),n+\log_{10}(d+1)).$ Notice that for ${d=1,\ldots 9}$, ${\log_{10}(d)}$ and ${\log_{10}(d+1)}$ are between 0 and 1. Thus we conclude that the leading digit of ${2^{m}}$ is ${d}$ provided $\displaystyle \{m\log_{10}2\}\in[\log_{10}(d),\log_{10}(d+1)).$ But ${\log_{10}(2)}$ is an irrational number. By the second part of our theorem we know that the percentage of the set ${\big\{0,\{\log_{10} 2\},\{2\log_{10} 2\},\{3\log_{10} 2\},\ldots\big\}}$ that intersects the interval ${I=[\log_{10}(d),\log_{10}(d+1))}$ is precisely the length of the interval ${I}$, ${l(I)=\log_{10}(d+1)-\log_{10}(d)=\log_{10}(1+1/d)}$, which is what Benford’s law says. ### Like this: Posted in Math \| Tags: Benford's law, circle, dynamical systems, irrational rotation, logarithms, statistics ## Responses 1. Dave, In my blog on Benford’s law I point out that, “As you might have guessed, someone else did it earlier; a half century earlier. In 1881 a note to the American Journal of Mathematics by an American astronomer named Simon Newcomb described an unusual observation. He had noticed that the tables of logarithms that were in common use back then by astronomers, always had the pages of the lower numbers more dog-eared than the pages of the higher numbers. He suggested that natural observations tend to start with the number one more often than with an eight or nine. For some reason, the observation went without much comment. ” For my students, I point out that before they think Newcomb was TOOOO bright, I point out that in 1903 (only months before the Wright Bros. flight at Kittyhawk) he declared boldly, “”Aerial flight is one of that class of problems with which man cannot cope”. By: Pat Ballew on September 8, 2010 at 11:59 am • Thanks, Pat. I hadn’t seen your earlier post. I like that story about Newcomb. I’ll have to see if I can find a book of logarithms in our college’s archives so that I can see the page usage for myself. Also thanks for giving me the publication year of his note. I was able to find it in our online holdings in an instant. By: Dave Richeson on September 8, 2010 at 3:12 pm 2. [...] strikes again September 8, 2010 mcxxiii Leave a comment Go to comments Dave Richeson has a post about Benford’s law up which looks really nice. It includes an example using the powers of 2, and a neat proof. I [...] By: Benford strikes again « 11:23 on September 8, 2010 at 2:32 pm 3. [...] Irrational rotations of the circle and Benford’s law. Benford’s law is amazing. It predicts, for instance, that in a large list of spread out [...] By: Benford’s Law & Infinite Primes « Random Walks on September 9, 2010 at 1:00 pm 4. There is a misspelling in the last formula. should be “log(1 + 1/d)”; there is sign mismatch. Feel free to remove this comment) By: Ivan on September 10, 2010 at 4:15 am • Thanks. It is fixed now. By: Dave Richeson on September 10, 2010 at 6:52 am 5. I think it’s useful to imagine a counter. If you have a lot of numbers you’re counting, the dials at the lower end would move quickly; the ones at the higher end (that represent millions, or hundreds of thousands) would move slowly. Those high numbers wouldn’t flip over very often as the smaller numbers kept turning through all the digits. How often do you hear of 900 million anyway. It’s usually 1.2 million or some lower number, especially with \$\$\$. Thanks for the post. By: Rick Meese on September 11, 2010 at 10:04 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": 0, "wp_latex": 42, "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": 0.9338980317115784, "perplexity_flag": "middle"}
http://quant.stackexchange.com/questions/978/formal-proof-for-risk-neutral-pricing-formula?answertab=oldest	# Formal proof for risk-neutral pricing formula As you know, the key equation of risk neutral pricing is the following: $\exp^{-rt} S_t = E_Q[\exp^{-rT} S_T \| \mathcal{F}_t]$ That is, discounted prices are Q-martingales. It makes real-sense for me from an economic point of view, but is there any "proof" of that? I'm not sure my question makes real sense, and an answer could be "there is no need to prove anything, we create the RN measure such that this property holds"... Is this sufficient to prove that, within this model, the risk-neutral measure exists? EDIT: Some answers might have been misled by my notation. Here is a new one: $\exp^{-rt} X_t = E_Q[\exp^{-rT} X_T \| \mathcal{F}_t]$ where $X_t$ can be any financial asset. For example, a binary option on an underlying stock $S$. In order to price the option, you would start with this equation and develop the right-hand side to finally solve for $X_t$. My key question: what allows me to write the initial equation assuming I have no information about the dynamics of the option or its underlying. - assume Laplace. like they said in the 90s "don't be normal". will the ideal probability distribution please stand up? – user3232 Feb 15 at 5:58 ## 5 Answers Note first that this key equation is only assumed to hold true under some extra assumptions. Typically those assumptions are taken to be about absence of arbitrage, though it is possible to weaken them somewhat if you are willing to consider portfolio arguments or collectively agreeable objective function. Anyway, the argument is this: if all the risk can be arbitraged away, then the price of any contingent claim should be equal to its price under the risk-neutral measure Q. The mathematical proof can be grasped most easily by the old-school arguments where one shows delta-hedges eliminating stochastic terms from the SDE. More mathematically elegant arguments involving the Girsanov theorem and Feynman-Kac formula are less intuitive. - Thanks for the answer, I updated the question, because I still can't see how to conclude.... – SRKX♦ Apr 14 '11 at 21:02 2 If you have absolutely no information, then the conditions of the Girsanov theorem fail to apply and you cannot write the equation. – Brian B Aug 12 '11 at 16:58 Since I did not get any comments to my latest update, and since I find it quite convincing, I hereby post my solution as an answer. maybe I can prove that Q exists assuming a lognormal distribution of $S_t$. Assuming $dS_t = \mu S_t dt + \sigma S_t dW_t$ By Itô, $d(e^{-rt} S_t) = -r e^{-rt} S_t dt + e^{-rt} S_t dS_t$. Replacing with the definition of $dS_t$, I get: $d(e^{-rt} S_t) = e^{-rt} S_t [(\mu - r)dt + \sigma dW_t] = e^{-rt} S_t \sigma [\theta dt + dW_t]$. (with $\theta=\frac{\mu-r}{\sigma}$) By Girsanov, we know that there exists a measure s.t $dW_t = dW_t^* - \theta dt$ We thus get $d(e^{-rt} S_t) = e^{-rt} S_t \sigma [\theta dt + dW_t] = e^{-rt} S_t \sigma dW_t^$ which means that $e^{-rt} S_t$ is a Q-martingale. If you think it's incorrect, feel free to comment, or even better, to correct! The thing that worries me is that I need to know the dynamics of the asset, and I think I shouldn't have to... - 1 The price to set up a dynamic hedge portfolio absolutely depends on the dynamics of the asset's price process through the quadratic variance. If you change your assumption about the dynamics of the price process, say by using a jump-diffusion or variance gamma process instead of geometric Brownian motion, you change the value of the option. – user3296 Jul 8 '11 at 6:44 You can find a simple proof in the discrete time case at http://kalx.net/ftapd.pdf. I'm not sure what you are trying to derive with your Ito calculus, but here is a rigourous derivation of the Black-Sholes/Merton PDE: http://kalx.net/dsS2011/bms.pdf. The Black-Scholes '73 derivation is not mathematically correct. The modern approach does not use so called real-world measures. - 1 Ok I'll have a look a it. Regarding my Ito calculus, I'm just trying to prove that the equation in my question holds. I can't find a way to prove that it holds regardless the dynamics of the asset. – SRKX♦ Apr 18 '11 at 14:35 The first think you have to ask is ¿¿What price??? Monetary price or equity price?? All answers,the ones I read, related to monetary price, but are equity price really risk free???? One of the biggest problem with Black Scholes (personal opinion) is that they consider the behave of equity price as monetary price: Solve this ODE: S(t)'/dt= rS(0), this tell you how money change in a deposit account (monetary price). All finance theory comes form this simple implementation. Now check Hull, and you will see that they use this expression to get the final result en the E(S(T)). Nice question, and I dont pretent to solve it, cause risk neutral in equity price does not exist........., is better to understand the model, limitation, etc....... - 1 I am sure you mean something interesting but what you are saying is far from obvious. – nicolas Jul 9 '11 at 18:46 1 option theory does not say that equity prices are risk free. it says that you can offset the risk from the stock in the option using a correct amount of stocks. – nicolas Jul 9 '11 at 18:47 "Monetary price or equity price?": not sure what you mean... – SRKX♦ Jul 13 '11 at 9:31 The only requirement if you are risk neutral is the property of martingale on your discounted stock price $M_t=e^{-rt} S_t$. But if you apply Itô $d( S_t\cdot e^{-rt} e^{rt})=d(M_t\cdot e^{rt})=r_tM_te^{rt}dt + ..dW_t=r_tS_tdt+..dW_t$ you see see that under the risk free probability, the asset price must have $r_t$ as yield and to answer to your question, why the GB is so popular? simply because $dS_t=S_t(r dt+ \sigma dW_t)$ is the most simple SDE that hold this property. - I'm not sure if my question was clear enough. I edited it because I don't think this is the answer I'm looking for. – SRKX♦ Jul 13 '11 at 9:43	{"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": 18, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9312777519226074, "perplexity_flag": "middle"}
http://mathoverflow.net/revisions/68360/list	## Return to Question 2 added 25 characters in body Let $C$ be a smooth projective connected curve of genus $g$ over $\bar{\mathbf{Q}}$. Fix a finite non-empty (Edit) set of closed points $S$ in $C$ and let $U$ be the complement of $S$ in $C$. Q1. (Algebraic formulation) Does there exist a finite (surjective) morphism `$\pi:C\longrightarrow \mathbf{P}^1_{\bar{\mathbf{Q}}}$` such that $\pi\|_{U}$ is etale? Equivalently, let $X$ be a compact connected Riemann surface of genus $g$ which can be defined over $\bar{\mathbf{Q}}$ and let $B$ be a finite set of of closed points in $X$ with complement $Y$. Q1. (Analytic formulation ) Does there exist a finite topological cover `$Y\longrightarrow \mathbf{P}^1(\mathbf{C})-\{0,1,\infty\}$` ? The equivalence of these two questions follows from the proof of Belyi's theorem and Riemann's existence Theorem. If the answer to Question 1 is positive, I would be very interested in knowing if the degree of $\pi$ can be bounded effectively. Q2. Does there exist a finite (surjective) morphism $\pi:C\longrightarrow \mathbf{P}^1$ such that $\pi\|_{U}$ is etale and $\deg \pi \leq c$, where $c$ is a constant depending only on $S$ and $g$? Example. Suppose that $g=0$. Then, following Belyi's proof of his theorem, the answer to Question 1 is yes. The answer to Question 2 is also positive and an explicit upper bound for such a rational function is given by Khadjavi in An effective version of Belyi's Theorem. I don't expect the answer to Question 1 to be easy. In fact, what I'm asking is to prove the existence of a Belyi morphism `$\pi:C\longrightarrow \mathbf{P}^1_{\bar{\mathbf{Q}}}$` with prescribed ramification. Now, that's probably very hard but definitely very interesting to find out. Trivial Remark. Suppose that $g>1$. Then the automorphism group of $C$ is finite. Choose a Belyi morphism `$\pi:C\longrightarrow \mathbf{P}^1_{\bar{\mathbf{Q}}}$` and let $U_0\subset C$ be the complement of the ramification points of $\pi$. Then we see that Question 1 has a positive answer if we take $U$ to be $\sigma(U_0)$ with $\sigma$ an automorphism of $C$. But that's only finitely many examples. 1 # Can we construct rational functions with prescribed ramification on an algebraic curve over \Qbar Let $C$ be a smooth projective connected curve of genus $g$ over $\bar{\mathbf{Q}}$. Fix a finite set of closed points $S$ in $C$ and let $U$ be the complement of $S$ in $C$. Q1. (Algebraic formulation) Does there exist a finite (surjective) morphism `$\pi:C\longrightarrow \mathbf{P}^1_{\bar{\mathbf{Q}}}$` such that $\pi\|_{U}$ is etale? Equivalently, let $X$ be a compact connected Riemann surface of genus $g$ which can be defined over $\bar{\mathbf{Q}}$ and let $B$ be a finite set of of closed points in $X$ with complement $Y$. Q1. (Analytic formulation ) Does there exist a finite topological cover `$Y\longrightarrow \mathbf{P}^1(\mathbf{C})-\{0,1,\infty\}$` ? The equivalence of these two questions follows from the proof of Belyi's theorem and Riemann's existence Theorem. If the answer to Question 1 is positive, I would be very interested in knowing if the degree of $\pi$ can be bounded effectively. Q2. Does there exist a finite (surjective) morphism $\pi:C\longrightarrow \mathbf{P}^1$ such that $\pi\|_{U}$ is etale and $\deg \pi \leq c$, where $c$ is a constant depending only on $S$ and $g$? Example. Suppose that $g=0$. Then, following Belyi's proof of his theorem, the answer to Question 1 is yes. The answer to Question 2 is also positive and an explicit upper bound for such a rational function is given by Khadjavi in An effective version of Belyi's Theorem. I don't expect the answer to Question 1 to be easy. In fact, what I'm asking is to prove the existence of a Belyi morphism `$\pi:C\longrightarrow \mathbf{P}^1_{\bar{\mathbf{Q}}}$` with prescribed ramification. Now, that's probably very hard but definitely very interesting to find out. Trivial Remark. Suppose that $g>1$. Then the automorphism group of $C$ is finite. Choose a Belyi morphism `$\pi:C\longrightarrow \mathbf{P}^1_{\bar{\mathbf{Q}}}$` and let $U_0\subset C$ be the complement of the ramification points of $\pi$. Then we see that Question 1 has a positive answer if we take $U$ to be $\sigma(U_0)$ with $\sigma$ an automorphism of $C$. But that's only finitely many examples.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 62, "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": 0.9561601877212524, "perplexity_flag": "head"}
http://mathoverflow.net/questions/67763?sort=votes	## Separability of continuous functions with compact support [closed] ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Hi, is the space $C_0(\mathbb{R}^m)$, $m \in \mathbb{N}$ of continuous functions with compact support separable? If yes: where can I find a proof for that? Please note: this is not a duplicate of this question (click), since I asked for compact support. The reason I'm asking is: here (click) the author (L.C. Evans) seems to use the fact on page 18 (page 20 of the pdf) first line without mentioning the difficulty, that $\mathbb{R}$ isn't compact. The entire proof is senseless without this fact. Your help is very much appreciated! Thanks! - 2 Isn't $C_0(\mathbb{R}^m)$ usually continuous functions that vanish at infinity? – Keenan Kidwell Jun 14 2011 at 14:15 1 This is not a research-level question. – Nate Eldredge Jun 14 2011 at 14:29 Oh... he doesn't clarify his notation. But that could very well be. But doesn't that make it even worse? – fjodor_d Jun 14 2011 at 14:30 3 The notation for $C_0(U)$ is indeed explained at page 7, line 18 (functions with compact support, as you said). It is not stated if a topology is chosen. In any case, in the proof at page 18 just uniform convergence is used. So the separability you need comes e.g. from the separability of the Banach space $C(S)$ where $S$ is the one-point compactification of $\mathbb{R}^m$ (you can see $C_0(\mathbb{R}^m)$ as a subset of $C(S)$). – Pietro Majer Jun 14 2011 at 15:12 ## 1 Answer Each $C([-N,N]^m)$ is separable, and so is the subspace $X_N\subseteq C([-N,N]^m)$ consisting of functions which vanish on the boundary $\partial[-N,N]^m$. Take a countable dense subset of each $X_N$, and take the union over all integers $N$. This is then a countable dense subset of $C_0(\mathbb R^m)$. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 16, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9340361952781677, "perplexity_flag": "head"}
http://mathhelpforum.com/differential-equations/168534-finite-difference-method-boundary-value-problem.html	# Thread: 1. ## Finite difference method on boundary value problem I want to solve boundary value problem in n points, $x_{n}\in(0,1)$: $-u'(x)(ku'(x))=12x^{2}$ where: $u(x)\in[0,1]$ $u(0)=0$ $ku'(1)+hu'(1)=hu_{z}$ $x\in R$ with finite difference method instead of finite element method. I substituted finite difference: $u'(x)=\frac{u(x_{i-1})-u(x_{i+1})}{2h}$ $h=\frac{1}{n}$ and obtained: $-kn^{2}[u(x_{i-1})-u(x_{i+1})]^{2}=3x_{i}^{2}$ $i=1,...,n$ And I don't know how to turn that into the matrix equation. Is there a way to do it?	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 10, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "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": 0.8999195098876953, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/42851?sort=votes	## Can we define geometric morphisms (between ETCS categories) elementarily? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) The ETCS axioms give conditions on a category for it to be a category of sets. These axioms can be written out in first order language, resulting in a finite axiomatisation of the category of sets. Given a model of ZFC one can form the associated category of sets and this will satisfy the axioms. On the other hand, the axioms do not require a priori defined sets. (The existence of a model of the ETCS axioms then is probably on par with the assuption there exists a model of the ZFC axioms) The appropriate definition of map between ETCS categories is a geometric morphism. Is it possible to define geometric morphisms elementarily? In first order language? Then one can define the category of ETCS categories and consider the relations between models. This is related to my previous question. - ## 1 Answer Yes, it is possible. Precisely, we can write down a first-order theory for which a model is a pair of ETCS-models and a geometric morphism between them (am I right in thinking this is what you're asking for?). To do this, on top of axiomatising “a pair of models of ETCS”, you add some extra function symbols for the adjunction. The conditions of functoriality, etc. are easily written algebraically; the adjunction can be expressed in various ways, of which the simplest to write down is probably the triangle-inequalities form. “Preserving finite limits”, when you write it out, is also just a scheme of first-order conditions; if you want to reduce it to a finite axiomatisation, note that it's enough to ask for preservation of finite products and equalisers (by the usual proof that all finite limits can be constructed from these). This said, I disagree somewhat with an implicit premise of your question. You say: “Then one can define the category of ETCS categories…” But to do this, you don't need to show that geometric morphisms can be defined in first-order terms. To talk about “the category of ETCS categories”, you already need to be working in a meta-theory with some sort of notion of set or similar (eg types, etc.); and so don't need the definitions of the morphisms to be first-order. The foundational advantage of a first-order axiomatisation of widgets is that you can then study a single widget without needing any meta-theory. But to study the collection of all widgets (as a category or whatever else), you still need a meta-theory. - Is NBG a sufficient meta-theory to study the category of all models of ETCS+(some large cardinal axiom)? – Harry Gindi Oct 20 2010 at 2:22 1 Absolutely! But tbh, it's probably overkill. To do basic model theory of any first-order logical system, assuming you want the usual completeness theorem and things, your minimal metatheory is probably something like ZC (I don't believe you need replacement, though I wouldn't swear to this); a reverse-mathematician could surely sharpen this up a bit. If you want to prove existence of models, then replacement is enough: sets of cardinality bounded by any strong limit cardinal (eg $2^{2^{\ddots ^ \omega}}$) give a model of bounded Zermelo set theory, and hence of ETCS. (cont’d) – Peter LeFanu Lumsdaine Oct 20 2010 at 2:41 1 On the other hand, for large cardinal axioms, as usual, you get what you pay for: the more you put into your meta-theory, the more you'll get back in existence of models. I'm not sure if the distinction between ZFC-style and NBG-style is particularly relevant here: the advantage one might expect in NBG, that it might allow one to talk about class models, is (if I'm not mistaken) illusory: how do you define satisfaction of formulas in putative class models? (Satisfaction is a recursive definition, and one can't generally recurse over classes, thanks to the restriction on comprehension) – Peter LeFanu Lumsdaine Oct 20 2010 at 2:53 So a theory can only model theories weaker than it? – Harry Gindi Oct 20 2010 at 3:58	{"extraction_info": {"found_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": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9172727465629578, "perplexity_flag": "head"}
http://www.citizendia.org/Nyquist_rate	Not to be confused with Nyquist frequency. The Nyquist frequency, named after the Swedish-American engineer Harry Nyquist or the Nyquist–Shannon sampling theorem, is half the Sampling frequency Spectrum of a bandlimited signal as a function of frequency In signal processing, the Nyquist rate is two times the bandwidth of a bandlimited signal or a bandlimited channel. A bandlimited signal is a Deterministic or Stochastic signal whose Fourier transform or Power spectral density is zero above a certain finite Signal processing is the analysis interpretation and manipulation of signals Signals of interest include sound, images, biological signals such as Bandwidth is the difference between the upper and lower Cutoff frequencies of for example a filter, a Communication channel, or a Signal spectrum A bandlimited signal is a Deterministic or Stochastic signal whose Fourier transform or Power spectral density is zero above a certain finite This term is used to mean two different things under two different circumstances: as a lower bound for the sample rate for alias-free signal sampling,[1] and as an upper bound for the signaling rate across a bandwidth-limited channel such as a telegraph line. [2] ## Nyquist rate relative to sampling The Nyquist rate is the minimum sampling rate required to avoid aliasing, equal to twice the highest frequency contained within the signal. Sampling theorem The Nyquist–Shannon sampling theorem states that perfect reconstruction This article applies to signal processing including computer graphics $f_N \ \stackrel{\mathrm{def}}{=}\ 2 B\,$ where $B\,$ is the highest frequency at which the signal can have nonzero energy. Frequency is a measure of the number of occurrences of a repeating event per unit Time. To avoid aliasing, the sampling rate must exceed the Nyquist rate: $f_S > f_N\,$. To avoid aliasing, the bandwidth must be considered to be the upper frequency limit of a baseband signal. In Signal processing, baseband is an adjective that describes signals and systems whose range of Frequencies is measured from zero to a maximum bandwidth Bandpass sampling signals must be sampled at at least twice the frequency of the highest frequency component of the bandpass signal in order to avoid aliasing. In Signal processing, sampling is the reduction of a Continuous signal to a Discrete signal. However, it is typical to use aliasing to advantage, to allow sampling of bandpass signals at rates as low as 2B, where B is the bandwidth of the bandpass signal. An alternative is to mix (heterodyne) the bandpass signals down to baseband, and sample there in the usual way; in this case, the baseband bandwidth can be as low as B/2 in the case of symmetric signals such as amplitude modulation, so the sampling rate can be as low as B in such cases. In Radio and Signal processing, heterodyning is the generation of new frequencies by mixing or multiplying two Oscillating waveforms Amplitude modulation ( AM) is a technique used in electronic communication most commonly for transmitting information via a Radio Carrier wave ## Nyquist rate relative to signaling Long before Harry Nyquist had his name associated with sampling, the term Nyquist rate was used differently, with a meaning closer to what Nyquist actually studied. Harry Nyquist ( né Harry Theodor Nyqvist pron, not as often pronounced ( February 7, 1889 – April 4, 1976) was an important Quoting Harold S. Black's 1953 book Modulation Theory, in the section Nyquist Interval of the opening chapter Historical Background: "If the essential frequency range is limited to B cycles per second, 2B was given by Nyquist as the maximum number of code elements per second that could be unambiguously resolved, assuming the peak interference is less half a quantum step. Harold Stephen Black (1898-1983 was an American Electrical engineer, who revolutionized the field of applied electronics by inventing the Negative feedback amplifier This rate is generally referred to as signaling at the Nyquist rate and 1/(2B) has been termed a Nyquist interval. " (bold added for emphasis; italics as in the original) According to the OED, this may be the origin of the term Nyquist rate. The Oxford English Dictionary ( OED) published by the Oxford University Press (OUP is a comprehensive Dictionary of the English [3] Nyquist's famous 1928 paper was a study on how many pulses (code elements) could be transmitted per second, and recovered, through a channel of limited bandwidth. Signaling at the Nyquist rate meant putting as many code pulses through a telegraph channel as its bandwidth would allow. Shannon used Nyquist's approach when he proved the sampling theorem in 1948, but Nyquist did not work on sampling per se. The Nyquist–Shannon sampling theorem is a fundamental result in the field of Information theory, in particular Telecommunications and Signal processing Black's later chapter on "The Sampling Principle" does give Nyquist some of the credit for some relevant math: "Nyquist (1928) pointed out that, if the function is substantially limited to the time interval T, 2BT values are sufficient to specify the function, basing his conclusions on a Fourier series representation of the function over the time interval T. " ## See also • Harry Nyquist • Nyquist–Shannon sampling theorem • Sampling frequency • Nyquist frequency — The Nyquist rate is defined differently from the Nyquist frequency, which is the frequency equal to half the sampling rate of a sampling system, and is not a property of a signal. Harry Nyquist ( né Harry Theodor Nyqvist pron, not as often pronounced ( February 7, 1889 – April 4, 1976) was an important The Nyquist–Shannon sampling theorem is a fundamental result in the field of Information theory, in particular Telecommunications and Signal processing Sampling theorem The Nyquist–Shannon sampling theorem states that perfect reconstruction The Nyquist frequency, named after the Swedish-American engineer Harry Nyquist or the Nyquist–Shannon sampling theorem, is half the Sampling frequency • Nyquist ISI criterion ## References 1. ^ Yves Geerts, Michiel Steyaert, and Willy Sansen (2002). In communications the Nyquist ISI criterion describes the conditions which when satisfied by a communication channel, result in no Intersymbol interference or ISI Design of multi-bit delta-sigma A/D converters. Springer. 2. ^ Roger L. Freeman (2004). Telecommunication System Engineering. John Wiley & Sons, 399. 3. ^ Black, H. S., Modulation Theory, v. Harold Stephen Black (1898-1983 was an American Electrical engineer, who revolutionized the field of applied electronics by inventing the Negative feedback amplifier 65, 1953, cited in OED The Oxford English Dictionary ( OED) published by the Oxford University Press (OUP is a comprehensive Dictionary of the English	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 3, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9237139821052551, "perplexity_flag": "middle"}
http://terrytao.wordpress.com/tag/singular-values/	What’s new Updates on my research and expository papers, discussion of open problems, and other maths-related topics. By Terence Tao # Tag Archive You are currently browsing the tag archive for the ‘singular values’ tag. ## From the Littlewood-Offord problem to the Circular Law: universality of the spectral distribution of random matrices 18 October, 2008 in math.CO, math.PR, math.SP, paper \| Tags: circular law, negative second moment, random matrices, second moment method, singular values, universality \| by Terence Tao \| 8 comments Van Vu and I have just uploaded to the arXiv our survey paper “From the Littlewood-Offord problem to the Circular Law: universality of the spectral distribution of random matrices“, submitted to Bull. Amer. Math. Soc.. This survey recaps (avoiding most of the technical details) the recent work of ourselves and others that exploits the inverse theory for the Littlewood-Offord problem (which, roughly speaking, amounts to figuring out what types of random walks exhibit concentration at any given point), and how this leads to bounds on condition numbers, least singular values, and resolvents of random matrices; and then how the latter then leads to universality of the empirical spectral distributions (ESDs) of random matrices, and in particular to the circular law for the ESDs for iid random matrices with zero mean and unit variance (see my previous blog post on this topic, or my Lewis lectures). We conclude by mentioning a few open problems in the subject. While this subject does unfortunately contain a large amount of technical theory and detail, every so often we find a very elementary observation that simplifies the work required significantly. One such observation is an identity which we call the negative second moment identity, which I would like to discuss here. Let A be an $n \times n$ matrix; for simplicity we assume that the entries are real-valued. Denote the n rows of A by $X_1,\ldots,X_n$, which we view as vectors in ${\Bbb R}^n$. Let $\sigma_1(A) \geq \ldots \geq \sigma_n(A) \geq 0$ be the singular values of A. In our applications, the vectors $X_j$ are easily described (e.g. they might be randomly distributed on the discrete cube $\{-1,1\}^n$), but the distribution of the singular values $\sigma_j(A)$ is much more mysterious, and understanding this distribution is a key objective in this entire theory. Read the rest of this entry » ## Milliman Lecture II: Additive combinatorics and random matrices 5 December, 2007 in math.CO, math.PR, math.SP, talk \| Tags: additive combinatorics, condition number, eigenvalues, Littlewood-Offord problem, Milliman lecture, random matrices, singular values \| by Terence Tao \| 14 comments This is my second Milliman lecture, in which I talk about recent applications of ideas from additive combinatorics (and in particular, from the inverse Littlewood-Offord problem) to the theory of discrete random matrices. Read the rest of this entry » ## Random matrices: the circular law 23 August, 2007 in math.CO, math.PR, math.SP, paper \| Tags: additive combinatorics, condition number, eigenvalues, Littlewood-Offord problem, random matrices, singular values \| by Terence Tao \| 28 comments Van Vu and I have recently uploaded our joint paper, “Random matrices: the circular law“, submitted to Contributions to Discrete Mathematics. In this paper we come close to fully resolving the circular law conjecture regarding the eigenvalue distribution of random matrices, for arbitrary choices of coefficient distribution. More precisely, suppose we have an $n \times n$ matrix $N_n$ for some large n, where each coefficient $x_{ij}$ of $N_n$ is an independent identically distributed copy of a single random variable x (possibly complex-valued). x could be continuous (e.g. a Gaussian) or discrete (e.g. a Bernoulli random variable, taking values +1 and -1 with equal probability). For simplicity, let us normalise x to have mean 0 and variance 1 (in particular, the second moment is finite). This matrix will not be self-adjoint or normal, but we still expect it to be diagonalisable, with n complex eigenvalues. Heuristic arguments suggest that these eigenvalues should mostly have magnitude $O(\sqrt{n})$; for instance, one can see this by observing that the Hilbert-Schmidt norm (a.k.a. the Frobenius norm) $\hbox{tr} N_n^* N_n$, which can be shown to dominate the sum of squares of the magnitudes of the eigenvalues, is of size comparable to $n^2$ on the average. Because of this, it is customary to normalise the matrix by $1/\sqrt{n}$; thus let $\lambda_1,\ldots,\lambda_n$ be the n complex eigenvalues of $\frac{1}{\sqrt{n}} N_n$, arranged in any order. Numerical evidence (as seen for instance here) soon reveals that these n eigenvalues appear to distribute themselves uniformly in the unit circle $\{ z \in {\Bbb C}: \|z\| \leq 1 \}$ in the limit $n \to \infty$. This phenomenon is known as the circular law. It can be made more precise; if we define the empirical spectral distribution $\mu_n: {\Bbb R}^2 \to [0,1]$ to be the function $\mu_n(s,t) := \frac{1}{n} \# \{ 1 \leq k \leq n: \hbox{Re}(\lambda_k) \leq s; \hbox{Im}(\lambda_k) \leq t \}$ then with probability 1, $\mu_n$ should converge uniformly to the uniform distribution $\mu_\infty$ of the unit circle, defined as $\mu_\infty(s,t) := \frac{1}{\pi} \hbox{mes}(\{ (x,y): \|x\|^2 + \|y\|^2 \leq 1; x \leq s; y \leq t \}.$ This statement is known as the circular law conjecture. In the case when x is a complex Gaussian, this law was verified by Mehta (using an explicit formula of Ginibre for the joint density function of the eigenvalues in this case). A strategy for attacking the general case was then formulated by Girko, although a fully rigorous execution of that strategy was first achieved by Bai (and then improved slightly by Bai and Silverstein). They established the circular law under the assumption that x had slightly better than bounded second moment (i.e. ${\Bbb E}\|x\|^{2+\delta} < \infty$ for some $\delta > 0$), but more importantly that the probability density function of x in the complex plane was bounded (in particular, this ruled out all discrete random variables, such as the Bernoulli random variable). The reason for this latter restriction was in order to control the event that the matrix $N_n$ (or more precisely $\frac{1}{\sqrt{n}} N_n - z I$ for various complex numbers z) becomes too ill-conditioned by having a very small least singular value. In the last few years, work of Rudelson, of myself with Van Vu, and of Rudelson-Vershynin (building upon earlier work of Kahn, Komlos, and Szemerédi, and of Van and myself), have opened the way to control the condition number of random matrices even when the matrices are discrete, and so there have been a recent string of results using these techniques to extend the circular law to discrete settings. In particular, Gotze and Tikhomirov established the circular law for discrete random variables which were sub-Gaussian, which was then relaxed by Pan and Zhou to an assumption of bounded fourth moment. In our paper, we get very close to the ideal assumption of bounded second moment; we need ${\Bbb E} \|x\|^2 \log^C(1+\|x\|) < \infty$ for some $C > 16$. (The power of 16 in the logarithm can certainly be improved, though our methods do not allow the logarithm to be removed entirely.) The main new difficulty that arises when relaxing the moment condition so close to the optimal one is that one begins to lose control on the largest singular value of $\frac{1}{\sqrt{n}} N_n$, i.e. on the operator norm of $\frac{1}{\sqrt{n}} N_n$. Under high moment assumptions (e.g. fourth moment) one can keep this operator norm bounded with reasonable probability (especially after truncating away some exceptionally large elements), but when the moment conditions are loosened, one can only bound this operator norm by a quantity bounded polynomially in n, even after truncation. This in turn causes certain metric entropy computations to become significantly more delicate, as one has to reduce the scale $\epsilon$ of the net below what one would ordinarily like to have. Read the rest of this entry »	{"extraction_info": {"found_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": 33, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9105526208877563, "perplexity_flag": "head"}
http://mathhelpforum.com/geometry/128318-rhombus.html	# Thread: 1. ## rhombus a rhombus has sides of length 10 inches, and the lengths of its diagonals difference by 4 inches . what is the area of the rhoumbus in square inches ? 2. Originally Posted by sri340 a rhombus has sides of length 10 inches, and the lengths of its diagonals difference by 4 inches . what is the area of the rhoumbus in square inches ? Area = half of the product of the diagonals. To get the length x of the shorter diagonal, note that the diagonals of a rhombus are perpendicular to each other. Therefore: $x^2 + (x - 2)^2 = 10^2$.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 1, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9295526146888733, "perplexity_flag": "middle"}
http://www.r-bloggers.com/functional-anova-using-inla/	## R-bloggers R news and tutorials contributed by (452) R bloggers # Functional ANOVA using INLA January 13, 2012 By simonbarthelme (This article was first published on dahtah » R, and kindly contributed to R-bloggers) [Update alert: INLA author Håvard Rue found a problem with the code below. See here] Ramsay and Silverman’s Functional Data Analysis is a tremendously useful book that deserves to be more widely known. It’s full of ideas of neat things one can do when part of a dataset can be viewed as a set of curves – which is quite often. One of the methods they’ve developed is called Functional ANOVA. It can be understood I think as a way of formulation a hierarchical prior, but the way they introduce it is more as a way of finding and visualising patterns of variation in a bunch of curves. Ramsay and Silverman use classical penalised likelihood estimation techniques, but I thought it’d be useful to explore functional ANOVA in a Bayesian framework, so here’s a quick explanation of how you can do that armed with R and INLA (Rue et al., 2009). The quickest way to understand what fANOVA is about is to go through Ramsay and Silverman’s analysis of Canadian temperature profiles (a summary of their analysis is available here). The dataset is the daily temperature average in 35 Canadian locations, for the period 1960-1994. Here’s a summary plot: Each curve represents the smoothed temperature profile at a location, and the fuzzy points around the curve are the raw averages (not much noise here). Note to potential American readers who can’t figure out the y-axis: those temperatures range from “Cool” down to “Not appreciably above absolute zero”. Importantly, the 35 locations are grouped into 4 regions: Atlantic, Continental, Pacific and Arctic. The profiles seem to differ across regions, but it’s hard to immediately say how from this graph, especially since, in a not totally unexpected fashion, there’s a big Summer-Winter difference in there (spot it?) Enter fANOVA: we will attempt to describe the variability in these curves in terms of the combined effects of variation due to Season and Region and ascribe whatever variability remains to individual Locations (and noise). The model used by Ramsay & Silverman is $\displaystyle y_{ij}(t)=s(t)+r_{j}\left(t\right)+\epsilon_{ij}\left(t\right)$ ${y_{ij}\left(t\right)}$ is the temperature at location ${i}$ in region ${j}$ at time ${t}$. Every location shares the same basic profile ${s(t)}$, then there’s some extra variation ${r_{j}\left(t\right)}$ that depends on the region, and everything else gets lumped into a term ${\epsilon_{ij}\left(t\right)}$ that is specific to a particular location. Ramsay & Silverman fit this model and obtain the following regional effects ${r_{j}\left(t\right)}$: (Figure nicked from their website). What’s really nice is that this is immediately interpretable: for example, compared to everybody else, the Pacific region gets a large bonus in the winter, but a smaller one in the summer, while the Atlantic region gets the same bonus year-round. The Artic sucks, but significantly less so in the summer. To fit the model Ramsay and Silverman use a penalty technique (and some orthogonality constraints I won’t get into), essentially, to estimate ${s(t),r_{j}\left(t\right),\ldots}$ they maximise the sum of the log-likelihood plus a penalty on the wiggliness of these functions, to get smooth curves as a result. In a Bayesian framework one could do something in the same spirit by putting priors on the functions ${s(t),r_{j}\left(t\right),\ldots}$. Inference would be done as usual by looking for example at the posterior ${p\left(s(t)\|\mathbf{y}\right)}$ to draw conclusions about ${s(t)}$. We want to set up the priors so that: 1. The constant null function ${f(t)=0}$ is a priori the most likely, and everything that’s not the null function is less likely than the null. 2. Smooth functions are more likely than non-smooth functions (the smoother the better). (NB: if the prior is proper than this rids us of the need for pesky orthogonality constraints) Why insist that the prior impose conditions (1) and (2)? Condition (2) makes sense among other things because we don’t want our functions to fit noise in the data, i.e. we don’t want to interpolate exactly the signal. More importantly, conditions (1) and (2) are needed because we want variation to be attributed to a global effect or inter-regional differences whenever it makes sense to do so. Going back to the model equation: $\displaystyle y_{ij}(t)=s(t)+r_{j}\left(t\right)+\epsilon_{ij}\left(t\right)$ it’s obvious that this is as such unidentifiable – we could stick all seasonal variation in the regional terms ${r_{1},\ldots,r_{4}}$ and the predictions would be the same. Explicitly if we change the variables so that ${s'(t)=0}$, ${r_{j}'\left(t\right)=s(t)+r_{j}\left(t\right)}$, than the likelihood has not changed but all the global seasonal variability has now moved to the intra-regional level. This is of course something we’d like to avoid. Imposing conditions (1) and (2) lets us do that. It’s useful at this point to think of the prior as imposing a cost: the further from 0 the function, and the less smooth, the higher the cost. When we shift variability away from the global level ${s(t)}$ to the regional level ${r_{j}\left(t\right)}$, with the change ${s'(t)=0,\, r_{j}'\left(t\right)=s(t)+r_{j}\left(t\right)}$ then: • The cost for ${s(t)}$ has now been reduced. • However we have now four functions ${r_{j}'\left(t\right)=s(t)+r_{j}\left(t\right)}$ that are now in all likelihood wigglier and further from 0 than they used to be. In total the cost will have risen – the prior puts a price on coincidence (four regions having the same global pattern by chance). There’s admittedly a lot of hand-waving here, but this is the rough shape of it. One way to impose conditions (1) and (2) is to put Gaussian Process priors on our latent functions and use MCMC, but we’d have a posterior distribution with ${365\times\left(35+4+1\right)=14,600}$ dimensions, and sampling from it would be rather slow. The INLA package in R lets us do the same thing, much, much more efficiently. INLA is very fast for two reasons. One, it prefers Gauss-Markov Processes to Gaussian processes – Gauss-Markov Processes more or less approximate GPs, but have sparse inverse covariance matrices, which speeds up inference a lot. Second, the philosophy of INLA is to not even attempt to capture the joint posterior distribution, but only approximate uni-dimensional marginals – for example ${p(s(23)\|\mathbf{y})}$, the posterior distribution of the seasonal effect at day 23. It does so using a variant of the Laplace approximation of Tierney & Kadane (note to ML folks: it’s related to, but not what you probably think is a Laplace approximation). This means that MCMC is not used in INLA, but fast optimisation algorithms instead. The R INLA package has an interface that’s not completely unlike that of MGCV (itself similar to lm and glm), although they’re very different behind the scenes. You specify a model using the formula interface, e.g.: ``` inla(y ~ x,data=dat,family=''gaussian'') ``` will perform inference on the linear regression model: $\displaystyle \begin{array}{rcl} y_{i} & = & \beta x_{i}+\beta_{0}+\epsilon\\ \epsilon & \sim & \mathcal{N}\left(0,\tau^{-1}\right) \end{array}$ This means INLA will return marginal distributions for ${\beta}$, ${\beta_{0}}$, integrated over the hyperparameter ${\tau}$ (unknown noise precision). A nonlinear regression $\displaystyle y_{i}=f(x_{i})+\beta_{0}+\epsilon$ can be done using: ```res <- inla(y ~ f(x,model="rw2",diagonal=1e-5),dat=df,family="gaussian") ``` • model=”rw2” specifies a 2nd-order random walk model, which is for all instance and purposes similar to a spline penalty (i.e., the estimated ${f(x)}$ will be smooth). • diagonal = 1e-5 makes the prior proper by adding a small diagonal component to the precision matrix of the Gaussian prior. This is not always necessary but stabilises INLA, which sometimes won’t work when two values of ${x}$ are too close together. This is due to an ill-conditioned prior precision matrix (INLA is great but still has some rough edges, and it helps to know the theory when trying to figure out why something is not working). Every term that comes enclosed in a ${f()}$ in the formula, INLA calls a random effect. Everything else is called a fixed effect. The different priors are called models. This clashes a bit with traditional terminology, but one gets used to it. In the case of the Canadian temperature data, I found after some tweaking that the formula below works well: ``` temp~f(day,model="rw2",cyclic=T,diagonal=.0001) +f(day.region,model="rw2",replicate=region.ind,cyclic=T,diagonal=.01) +f(day.place,model="rw2",cyclic=T,diagonal=.01,replicate=place.ind) +region ``` • temp is the temperature • day is the time index (1 to 365). Several functions will depend on the time index, which INLA doesn’t like, so we duplicate it artificially (day.region, day.place). • compared to Ramsay & Silverman’s specification, ${\epsilon_{ij}\left(t\right)}$ is decomposed into a smooth component (the third in the formula), and measurement noise (because we use Gaussian likelihood) • the regional and place effects are “replicated”, which means in the case of the regional effects that INLA will consider that ${r_{1}\left(t\right),\ldots,r_{4}\left(t\right)}$ have the same hyperparameters (here this means the same level of smoothness). • the “region” factor comes in as a linear effect – effectively this decomposes the regional effect into a constant shift and a smooth part. • We set cyclic=T because the functions we are trying to infer are periodic (over the year). • Roughly speaking, the diagonal component imposes a penalty on ${\int\left\|f(t)\right\|^{2}dt}$. We want the global component to be larger than the regional ones, so it gets a smaller diagonal penalty. The data and formula can now be fed into INLA. As a first check, we can plot the “fitted” model: The smooth lines are the posterior expected value of the linear predictor. We do not seem to be doing anything terribly wrong so far. We can also plot the estimated regional effects (again, we plot posterior expected values): Compared to the effects estimated by Ramsay&Silverstein, we have a lot more wiggly high-frequency stuff going on. Closer examination shows that the wiggly bits actually seem to really be in the data, and not just made up by the procedure. Since I’m no meteorologist I have no idea what causes them. Smoothing our curves (using MGCV) recovers essentially the curves R&S had originally inferred (INLA first, R&S second): What about the global, seasonal component? Here’s another plot modelled on one by R&S, showing the global component (dashed gray line), and the sum of the global component and each seasonal effect (coloured lines) Here’s R&S’s original plot: The code to reproduce the examples is given below. You’ll need to download INLA from r-inla.org, it’s not on CRAN. ## References Ramsay, J. and Silverman, B. W. (2005). Functional Data Analysis (Springer Series in Statistics). Springer, 2nd edition. Rue, H., Martino, S., and Chopin, N. (2009). Approximate bayesian inference for latent gaussian models by using integrated nested laplace approximations. Journal of the Royal Statistical Society: Series B (Statistical Methodology), 71(2):319-392. ```</div> <div>#INLA for functional ANOVA on the Canadian temperature dataset #Usage: #Load the data #cw = load.data() #Run INLA #res <- inla.fanova.temperature(cw) # #Plot "fitted" model #plot.fitted(cw,res) # #etc. require(fda) require(INLA) require(ggplot2) require(stringr) require(mgcv) require(directlabels) #Use the same colour scheme Ramsay&Silverman did col.scale <- c('red','blue','darkgreen','black') load.data <- function() { cw <- with(CanadianWeather,ldply(1:length(place),function(ind) data.frame(temp=dailyAv[,ind,1],place=place[ind],region=region[ind],day=1:365))) #We need to create multiple copies of the time index because we need multiple functions of time cw <- mutate(cw,day.region=day,day.place=day, region.ind=as.numeric(region), place.ind=as.numeric(place)) #INLA apparently doesn't like the original factor levels, we modify them levels(cw$place) <- str_replace(levels(cw$place),'. ','_') cw } inla.fanova.temperature <- function(cw) { #The formula for the model formula <- temp ~ f(day.region,model="rw2",replicate=region.ind,cyclic=T,diagonal=.01)+f(day,model="rw2",cyclic=T,diagonal=.0001)+f(day.place,model="rw2",cyclic=T,diagonal=.01,replicate=place.ind) + region #Note that 'region' comes in as a factor, and inla treats factors #in the same way as the 'lm' function, i.e., using contrasts #This is not strictly necessary in a Bayesian analysis and here complicates things a bit #The default "treatment" contrasts used here mean that the "place" factor has the Pacific region as the default level to which other regions are compared #Call inla #We use control.fixed to impose a proper gaussian prior on the fixed effects #and control.predictor to make INLA compute marginal distributions for each value of the linear predictor #The call takes 160 sec. on my machine inla(formula,data=cw,family="gaussian", control.predictor=list(compute=T), control.fixed=list(prec=list(default=0.01,prec.intercept = 0.001)),verbose=T) } plot.raw.data <- function(cw) { p <- ggplot(cw,aes(day,temp,group=place,colour=region))+geom_point(alpha=.1)+geom_smooth(method="gam",form = y ~ s(x))+labs(x='Day of the year',y='Temp. (° C)')+scale_colour_manual(values=col.scale)+facet_wrap(~ region) p + theme_bw() + opts(legend.position="none") } plot.fitted <- function(cw,res) { cw$fitted <- (res$summary.fitted.values$mean) p <- ggplot(cw,aes(day,temp,group=place,colour=region))+geom_point(alpha=.1)+facet_wrap(~ region)+scale_colour_manual(values=col.scale)+geom_path(aes(y=fitted)) p + theme_bw() + opts(legend.position="none") } extract.regional.effects <- function(cw,res) { reg.effect <- reff(res)$day.reg names(reg.effect)[1] <- 'day' reg.effect$region <- gl(4,365,lab=levels(cw$region)) #Total regional effect is equal to smooth component + intercept + regional effect feff.region <- feff(res)[str_detect(attributes(feff(res))$row.names,"region"),] feff.inter <- feff(res)[str_detect(attributes(feff(res))$row.names,"Inter"),] offsets <- c(feff.inter$mean,feff.inter$mean+feff.region$mean) reg.effect$total.effect <- reg.effect$mean+rep(offsets,each=365) reg.effect } plot.regional.effects <- function(cw,res,smooth=F) { reg.effect <- extract.regional.effects(cw,res) p <- ggplot(reg.effect,aes(day,total.effect,colour=region))+geom_line()+scale_colour_manual(values=c('red','blue','darkgreen','black'))+scale_y_continuous(lim=c(-18,15)) if (smooth) p <- p + geom_smooth(method="gam",form = y ~ s(x),lty=2) p <- p + geom_abline(slope=0,intercept=0,lty=3,col="lightblue") p <- p + geom_dl(aes(label=region),method="top.qp") p + theme_bw() +opts(panel.grid.minor = theme_blank(),panel.grid.major = theme_blank(),legend.position="none") + labs(x='\n Day of the year',y='Temp. (° C)') } plot.regional.avg <- function(cw,res) { reg.effect <- extract.regional.effects(cw,res) regionavg <- data.frame(day=1:365, glob.effect=reff(res)$day$mean, reg.avg=reff(res)$day$mean+reg.effect$total.effect, region=reg.effect$region ) p <- ggplot(regionavg,aes(day,reg.avg,colour=region))+geom_line()+facet_wrap(~ region)+scale_colour_manual(values=col.scale)+geom_line(aes(y=glob.effect),col="darkgrey",lty=2) p <- p + geom_abline(slope=0,intercept=0,lty=3,col="lightblue") p+ theme_bw() +opts(panel.grid.minor = theme_blank(),panel.grid.major = theme_blank(),legend.position="none") + labs(x='\n Day of the year',y='Temp. (° C)') } #Extract "random effects" summary from an inla object reff <- function(res.inla) { res.inla$summary.random } #Extract "fixed effects" feff <- function(res.inla) { as.data.frame(res.inla$summary.fixed) } ``` R-bloggers.com offers daily e-mail updates about R news and tutorials on topics such as: visualization (ggplot2, Boxplots, maps, animation), programming (RStudio, Sweave, LaTeX, SQL, Eclipse, git, hadoop, Web Scraping) statistics (regression, PCA, time series,ecdf, trading) and more... Tags: Functional Data Analysis, INLA, meteorology, R, stats Comments are closed.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 36, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 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": 0.9099155068397522, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/tagged/measure-theory+elementary-set-theory	# Tagged Questions 1answer 39 views ### Showing that a union of the subsets of two $\sigma$-algebras is a $\sigma$-algebra. I got back an assignment for a first course in analysis and I have made a very basic error, and I'm having a lot of trouble pinpointing exactly what piece of information I'm missing. You have two ... 1answer 18 views ### why S' is closed with respect to the formation of finite unions? $Proposition$ : Let S be a semiring of subsets of a set X. Define S' to be the collection of unions of finite disjoint collections of sets in S. Then S' is closed with respect to the formation of ... 2answers 57 views ### The cardinality of Lebesgue sets Suppose $A=\{S\;\|\;S \subset \mathbb R^n, S\text{ is Lebesgue measurable}\}$. What is the cardinality of $A$? Is it the same as the cardinality of all of the real numbers? 1answer 47 views ### what does $(\Omega^T,\mathcal{A}^T)$ mean? Let $(\Omega_t,\mathcal{A}_t), t\in T$ be a collection of measurable spaces. What does the notation mean? $(\Omega^T,\mathcal{A}^T)$ 1answer 52 views ### Additive set function properties I am reading an introduction to measure theory, which starts by defining $\sigma$-rings then additive set functions and their properties which are given without proof. I was able to prove two of them ... 0answers 28 views ### What does the notation mean? $A_t\times \boldsymbol\times_{s\neq t}\Omega_s$ Let $(\Omega_t,\mathcal{A}_t), \;t\in T$ be a collection of measurable spaces. What does the following notation mean? $$A_t\times \boldsymbol\times_{s\neq t}\Omega_s$$ where $t\in T$ and \$A_t\in ... 1answer 37 views ### Why is $\{x:f(x)\ge a\}=\bigcap_{n=1}^\infty\{x: f(x)\gt a-\frac{1}{n}\}$ and $\{x: f(x)\gt a\}=\bigcup_{n=1}^\infty\{x: f(x)\ge a+\frac{1}{n}\}$? It is known from measure theory that for all $a \in \mathbb{R}$, the following are equivalent: $f$ is measurable. The set $\{x : f(x) \ge a\}$ is measurable. The set $\{x : f(x) \gt a\}$ is ... 0answers 38 views ### Question on integrable set What is an integrable set? "If $T$ is a Hausdorff topological space and $\mu$ is a positive Radon measure on $T$, and if $F$ is a measurable multifunction, then for each integrable set ... 3answers 71 views ### How to derive a union of sets as a disjoint union? $$\bigcup_{n=1}^\infty A_n = \bigcup_{n=1}^\infty (A_{1}^c \cap\cdots\cap A_{n-1}^c \cap A_n)$$ The results is obvious enough, but how to prove this 1answer 47 views ### simple question from set theory/measure theory This is a simple question. On pages 5-6 of Measure Theory,Vol 1, Vladimir Bogachev he writes that: for $E=(A\cap S)\cup (B\cap (X-S))$ Now, he writes that: \$X-E = ((X-A)\cap S) \cup ((X-B)\cap ... 1answer 44 views ### Halmos Measure Theory section 5 Th5.B E is a class of sets. $R_o (E)$ is the ring generated by E. Any set in $R_o (E)$ can be covered by a finite union of sets in E. The proof: The class of all sets that can be covered by a finite ... 2answers 73 views ### Proof that the pre-image is a sigma algebra We shall prove that for sets X,Y and a map f: X $\rightarrow$ Y, if B is a $\sigma$-algebra on Y, then $\{f^{-1}(E) ; E \in B\}$ is a $\sigma$-algebra on X. I have shown all the properties except ... 0answers 79 views ### $\left(0,1\right]\neq\biguplus_{k=1}^{n}\bigcap_{j=1}^{k}G_{j},$ Proving elegantly How can I show (without making many distinctions by cases) that the equality $$\left(0,1\right]=\biguplus_{k=1}^{n}\bigcap_{j=1}^{k}G_{j},$$ can't hold, if \$G_{j}\in\mathcal{A}\cup\left\{ ... 2answers 92 views ### What does the completed graph of a function mean zab said: the Levy metric between two distribution functions $F$ and $G$ is simply the Hausdorff distance $d_C$ between the closures of the completed graphs of $F$ and $G$. I have difficulty in ... 1answer 43 views ### Finite Partitions of the Unit Interval Does the unit interval have a finite partition $P$ such that no element of $P$ contains an open interval? I would think that the answer is no, because each element of $P$ would have Lebesgue measure ... 0answers 63 views ### How are algebras and rings of subsets generated in this paragraph? From ncatlab What is missing is a simple description of the σ-algebra generated by ℬ. For a mere algebra, this is easy; any ℬ can be taken as a subbase of an algebra, the symmetric unions ... 1answer 146 views ### Making sense out of “field”, “algebra”, “ring” and “semi-ring” in names of set systems There are some set systems with algebraic titles, such as "field", "algebra", "ring" and "semi-ring" (and possibly other titles), in their names. Examples are a sigma field (aka sigma algebra, ... 1answer 221 views ### Bolzano-Weierstrass and measures Let $\{\mathcal{A}_n\}$ be an infinite sequence of sets with $\mathcal{A}_n \subset \mathcal{M}$, where $\mathcal{M}$ is a bounded subset of $\mathbb{R}$ (for simplicity). Is there a "nice" limit ... 1answer 159 views ### When does it make sense to define a base of a set system? In a topology, a base is defined to be a class of subsets such that every open set is the union of some members of it. In a convexity structure, a base is defined to be a class of subsets ... 2answers 85 views ### When does it make sense to define a generator of a set system? In a set system, such as a topology, sigma algebra or convexity structure, a generator is defined to be a class of subsets such that the given set system is the coarsest such set system ... 2answers 266 views ### Hyperreal measure? If AC be accepted, then there exists a Lebesgue unmeasurable set called Vitali Set. However, I'm curious about measure valued in hyperreal numbers. Argument in disproof of unmeasurability of Vitali ... 1answer 65 views ### Difference between “measure on” and “measure over” I want to make sure I understand the difference between the terms "measure on" and "measure over," assuming there is one. Is a measure on the set $X$ the same as a measure over its power set ... 1answer 60 views ### Generating a Sigma Algebra from an Algebra Let $\mathscr{A}$ be an algebra of sets over $X$. Let $\mathscr{A}_\sigma$ contain $\mathscr{A}$ as well as any countable union of any sequence of members of $\mathscr{A}$. I'm trying to figure out ... 1answer 57 views ### Algebra generated by countable family of sets is countable? If I have a countable family of sets $\mathcal{A}=\{A_1,A_2,...\}$ and construct the Algebra generated by $\mathcal{A}$. Will it also be countable? My intuition screams YES, but I cannot seem to ... 4answers 221 views ### Cardinality of Vitali sets: countably or uncountably infinite? I am a bit confused about the cardinality of the Vitali sets. Just a quick background on what I gather about their construction so far: We divide the real interval $[0,1]$ into an uncountable number ... 2answers 192 views ### Example of strictly subadditive lebesgue outer measure One of the properties of the Lebesgue outer measure is that it is subadditive and not countably additive. In fact, I have read that even when the sets A_i are disjoint, there is still generally ... 2answers 102 views ### What does the supremum of a sequence of sets represent? I'm trying to understand more about the limits of sequences of sets in Measure Theory. Given a sequence of sets $\{A_n\}_{n\in \mathbb{N}} = \{ A_1,A_2, \ldots \}$, what does $\sup_n \{ A_n \}$ ... 2answers 93 views ### Is there an efficient Hausdorff Distance algorithm? Two sided Hausdorff distance is calculated as $$H(r_1,r_2)=\max\{h(r_1,r_2),h(r_2,r_1)\}$$ where $$h(r_1,r_2)=\max_{a \in r_1}\min_{b\in r_2}\\|r_1-r_2\\|$$ and vice-verse $r_1$ and $r_2$ are two ... 2answers 260 views ### Closed under countable union I am reading a tutorial on measure theory and it states: "Given an interval $E = [a, b]$ and a set $S$ of subsets of $E$ which is closed under countable unions, we deﬁne the following..." I was ... 1answer 201 views ### Limit Inf/Sup of Sequence of Set Example In "A Probability Path", they have an example that states that the lim inf and lim sup of [0,n/(n+1)) is equal to [0,1). I guess I don't see how [0,1) is in all the sets except a finite number of ties ... 1answer 226 views ### Why is the Vitali set not necessarily equal to the interval e.g. [0,1]? I don't seem to get the special properties of Vitali sets which makes them different from the intervals, e.g. [0,1]. 4answers 582 views ### Interpretation of limsup-liminf of sets What is an intuitive interpretation of the 'events' $$\limsup A_n:=\bigcap_{n=0}^{\infty}\bigcup_{k=n}^{\infty}A_k$$ and $$\liminf A_n:=\bigcup_{n=0}^{\infty}\bigcap_{k=n}^{\infty}A_k$$ when $A_n$ are ... 1answer 123 views ### Finding an irrational not covered in standard proof that $\mu(\mathbb{Q} \cap [0,1]) = 0$ [duplicate] Possible Duplicate: How would one go about proving that the rationals are not the countable intersection of open sets? Constructing a number not in \$\bigcup\limits_{k=1}^{\infty} ... 2answers 86 views ### Generated $\sigma(X)$ where $\Omega \neq \mathbb{R}$ Simple question here. I am trying to enumerate the sigma field generated by the random variable: $$X(\omega)=2+1_{\left\{a,b\right\}}(\omega)$$ where $\Omega=\left\{a,b,c,d\right\}$. I think what is ... 1answer 617 views ### limsup liminf of sequence of sets Following up from the discussion here: Liminf and Limsup of a sequence of sets I wanted to confirm my understanding of these concepts with another example. Suppose we have: $a_n>0$, $b_n >1$ ... 1answer 346 views ### Number of $\sigma$ -Algebra on the finite set Let $X$ is a nonempty set with $m$ members . How many $\sigma$ -algebra can we make on this set? 3answers 188 views ### A problem in Sigma algebra. How do I conceptualise this expression : Let {$A_n$}$^{n=\infty}_{n=1}$ belong to sigma algebra $A$. Define, $\limsup\{A_{n}\}=\bigcap_{n=1}^{\infty}\{\bigcup_{m=n}^{\infty}A_{n}\}$ and similarly ... 1answer 248 views ### “Converse” to composition of measurable functions is measurable Here is a restatement of a problem in a textbook I encountered. I'm well beyond the age of doing homework and this is purely for self-study. Exercise: Let $f : (X,\Sigma_1) \to (Y, \Sigma_2)$ and ... 1answer 224 views ### $A$ uncountable thus $\mu(A)>0$ I was thinking if it is possible to come up with a $\sigma$-finite measure on $\mathbb R$ which is positive on any uncountable set. I think that I have a proof that there is no such measure - but I am ... 1answer 310 views ### Sigma algebra example I am having a bit of trouble understanding this definition. "A collection $\Sigma$ of subsets of S is called a $\sigma$-algebra on S if $\Sigma$ is an algebra on S such that whenever \$F_n \in\Sigma (n ... 1answer 271 views ### Outer measure defined on sets of $\mathbb R$ - help with showing if outer measure or not I have a long list of definitions of outer measures that I am trying to (a) show IF they are outer measures, and (b) if it is, determine its outer measurable subset of $\mathbb{R}$. My books ... 0answers 92 views ### Semi-partition or pre-partition For a given space $X$ the partition is usually defined as a collection of sets $E_i$ such that $E_i\cap E_j = \emptyset$ for $j\neq i$ and $X = \bigcup\limits_i E_i$. Does anybody met the name for a ... 0answers 103 views ### algebra of sets [closed] 1)prove that an algebra of sets is closed under finite union and intersections. 2)prove that for a semiring S,countable unions and finite intersections of sigma sets are sigma sets. 3answers 214 views ### Intersection of two 'huge' sets in the plane Consider two sets on the plane $A=\mathbb{Q}\times \mathbb{R}$ and $B=\mathbb{R}\times \mathbb{Q}$. We know that $A\cap B=\mathbb{Q}\times \mathbb{Q}\neq\emptyset$. What about the general cases? That ... 2answers 771 views ### Infinite product of measurable spaces Suppose there is a family (can be infinite) of measurable spaces. What are the usual ways to define a sigma algebra on their Cartesian product? There is one way in the context of defining product ... 2answers 748 views ### Finite additivity follows from countable additivity! How to prove that finite additivity follows from countable additivity!!? 1answer 122 views ### What's a closed countably infinite intersection? When operating with sigma algebras, what does it mean when we talk about a countably infinite set? And, what are then closed countably infinite interesections? 3answers 867 views ### Preimage of generated $\sigma$-algebra For some collection of sets $A$, let $\sigma(A)$ denote the $\sigma$-algebra generated by $A$. Let $C$ be some collection of subsets of a set $Y$, and let $f$ be a function from some set $X$ to $Y$. ... 1answer 527 views ### Field of sets and Sigma algebra of sets (1). According to Wikipedia, a field of subsets of X is defined to be a non-empty subset of the power set of X closed under the intersection and union of pairs of sets and under complements of ... 1answer 821 views ### limit superior and limit inferior of the given sequence of sets A sequence of sets is defined as $A_n=\{x \in [0,1] : \|\sum_{i=0}^{n-1} 1_{[\frac{i}{2n},\frac{2i+1}{4n})} - 1_{[\frac{2i+1}{4n},\frac{i+1}{2n})}\| \geq p\}$ for some positive $p\geq0$. What 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": 103, "mathjax_display_tex": 8, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9422239661216736, "perplexity_flag": "head"}
http://mathoverflow.net/questions/62528?sort=oldest	## Twisted forms and $\check{H}^1$ ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I am reading Milne's Étale cohomology, III.4. A twisted form of an object $Y$ (a scheme, a sheaf of modules, of algebras...) over a scheme $X$ is an object $Y'$ such that there exists a covering in some topology (let's say étale to fix the ideas), $(U_i \to X)$ such that $Y \times_X U_i \cong Y' \times_X U_i$ for all $i$. Then, the book says, any twisted form of $Y$ defines a cocycle in $\check{H}^1(X,\mathrm{Aut}(Y))$ as follows: let $f_i$ be the isomorpisms $Y \times_X U_i \to Y' \times_X U_i$, then the cocycle is given by $(\alpha_{ij}): Y \times_X U_i \times_X U_j \to Y \times_X U_i \times_X U_j$ where $\alpha_{ij} = f_i^{-1} \circ f_j$, which, in turn, constitutes the descent data that eventually allows to recover $Y'$. As far as I understand, the descent data does not have to be always effective in general, but in some cases it is. The book gives two examples when this is the case: Severi-Brauer varieties and vector bundles. I would like to understand when the Cech cohomology classes are in bijective correspondnence with the (isomorphism classes of) twisted forms. So I have two questions: 1. what are the general criteria that the descent data as described above is effective? 2. what happens if the trivialising cover of a twisted form $Y'$ contains just one étale morphism $U_0 \to X$? It seems like the cocycle defined by $Y'$ is always trivial then ($\alpha_{00} = f_0^{-1} \circ f_0$), yet the form $Y'$ might be not isomorphic to $Y$. - ## 2 Answers For your question 2, note that the fiber product $U_0\times_XU_0$ can be non-trivial (unlike the case of an open sub-set $U_0\subset X$, where it would be $U_0$ again) with two different projections $p_1$ and $p_2$ onto $U_0$ giving two different structures to it as a $U_0$-scheme. The isomorphism $\alpha_{0,0}$ is an isomorphism between these two different $U_0$-structures on $Y\times_XU_0\times_XU_0$ and is also a non-trivial piece of information As an example, if $X=Spec\;k$ and $U_0=Spec\;K$ where $K/k$ is a Galois extension with Galois group $G$, then we get an isomorphism $G\times U_0\rightarrow U_0\times_XU_0$ using the group action. The two structures of $G\times U_0$ as a $U_0$-scheme correspond to the two maps $(g,u)\mapsto u$ and $(g,u)\mapsto gu$. The isomorphism $\alpha_{0,0}$ satisfying the co-cycle condition now equates precisely to giving an action of $G$ on $Y'$ compatible with its action on $U_0$. This is Galois descent. See Serre's Local fields, for example. EDIT: It occurs to me that I didn't directly talk about twisted objects. The idea is the same. Let us start with an object $Y'$ over $U_0$ equipped with an isomorphism $f_0:Y'\rightarrow Y\times_XU_0$. This isomorphism has to satisfy the co-cycle condition, which means the following: We have two projections $p_1,p_2:U_0\times_XU_0\rightarrow U_0$. So we have two different ways to pull-back $Y'$ to over $U_0\times_XU_0$, giving us $p_1^Y'$ and $p_2^Y'$. The two pull-backs of $Y\times_XU_0$ under these projections are canonically isomorphic, since the two projections $p_1,p_2$, when composed with the map $U_0\rightarrow X$ agree with the structure map for $U_0\times_XU_0\rightarrow X$. So pulling back $f_0$ gives us isomorphisms $$p_1^Y'\rightarrow Y\times_X(U_0\times_XU_0)\rightarrow p_2^Y'.$$ This is your $\alpha_{0,0}$; if it satisfies the co-cycle condition--this amounts to the required compatibility between the pull-backs of $\alpha_{0,0}$ to $U_0\times_{X}U_0\times_{X}U_0$ under the three different projections to $U_0\times_XU_0$--then it gives you descent data for $Y'$. In the Galois setting, if you use $f_0$ to identify $Y'$ with $Y\times_XU_0$, then $\alpha_{0,0}$ is giving you a `twisted' action of $G$ on $Y\times_XU_0$. This data is not always effective. I would highly recommend the chapter on descent in Bosch-Lutkebohmmert-Raynaud's `Neron Models' for an explanation of all these things. - Dear Keerthi, thank you for you answer. I will look at the chapter in "Neron models". Is it easy to answer from what general theorem does it follow that in the case of vector bundles or Severi-Brauer varieties the descent data is always effective? In the chapter of Milne's book I have mentioned, SGA1.I.VIII.7.8 is referenced, but so far I couldn't extract the precise statement that would imply the bijection between isomorphism types of twisted forms and cocycle classes. – Dima Sustretov Apr 21 2011 at 19:19 Dear Dmitry--Yes, I believe BLR has a resume of effectivity results. For vector bundles, it's the fact that descent data for quasi-coherent sheaves are always effective. For Severi-Brauer varieties, it's the fact that varieties with an ample line bundle always have effective descent data. I don't have BLR at hand, so I can't give you references for the precise statements there. Once you have effectivity, the bijection you need follows formally. Again, I don't know of a good reference where this is laid out cleanly, though I can say that SGA1 is quite clearly written in general. – Keerthi Madapusi Pera Apr 21 2011 at 20:43 ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. If your $Y$ is an object over $X$ of a stack in the étale topology, then you can use the cocycles as descent data to get a twisted form (because by definition of a stack all descent data are effective). Some examples other than the ones you mentioned are quasi-coherent sheaves and affine morphisms of schemes (i.e. quasi-coherent sheaves of algebras). This is far from a necessary condition on $Y$, I don't know if there is a sharper characterization. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 70, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9329719543457031, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/27660/solution-to-the-schrodinger-equation-for-periodically-time-dependent-hamiltonian	# Solution to the Schrodinger equation for periodically time dependent Hamiltonians I have a Hamiltonian which is time dependent but possesses periodic symmetry: $H(t+t_0)=H(t)$. Is there any clever techniques to exploit this? Edit: In particular, I would like a practical method to simulate the dynamics of such a system (as opposed to naive time-slicing). - 1 Do you want dynamics or average ground state? – Joe Fitzsimons Jan 25 '12 at 19:04 @JoeFitzsimons -- good question. This was an example question I used at a StackExchange participation drive, so I unfortunately didn't give it much thought. I will make it clearer now. – Chris Ferrie Jan 26 '12 at 0:36 ## 1 Answer I would suggest looking at the formalism of Floquet space. The basic idea is that one uses a time-independent but infinite dimensional Hamiltonian to simulate evolution under a time-dependent but finite dimensional Hamiltonian by using a new index to label terms in a Fourier series. A good, short introduction can be found in Levante et al. For more details, Leskes et al provides a very through review. Finally, a simple example of an application of Floquet theory is given by Bain and Dumont. - Chris, can you add some comments about truncating the series in order to implement this method? – Chris Ferrie Jan 26 '12 at 0:40 @ChrisFerrie Another paper about Floquet theory: Shirley JH. Solutions of the Schrodinger equation with a Hamiltonian periodic in time. Phys Rev 1965; 138:B979-B987. – user6048 Apr 2 at 22:51	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.8991726636886597, "perplexity_flag": "middle"}
http://unapologetic.wordpress.com/2009/04/09/decomposing-real-linear-transformations/?like=1&source=post_flair&_wpnonce=bcc517a348	# The Unapologetic Mathematician ## Decomposing Real Linear Transformations Finally, we come to the analogue of Jordan normal form over the real numbers. Given a linear transformation $T:V\rightarrow V$ on a real vector space $V$ of dimension $d$, we can find its characteristic polynomial. We can factor a real polynomial into the product of linear terms $(X-\lambda_i)$ and irreducible quadratic terms $(X^2-\tau_jX+\delta_j)$ with $\tau_j^2<4\delta$. These give us a list of eigenvalues and eigenpairs for $T$. For each distinct eigenvalue $\lambda_i$ we get a subspace $U_i=\mathrm{Ker}\left((T-\lambda_iI_V)^d\right)\subseteq V$ of generalized eigenvectors, with $m$ distinct eigenvalues in total. Similarly, for each distinct eigenpair $(\tau_j,\delta_j)$ we get a subspace $V_j=\mathrm{Ker}\left((T^2-\tau_jT+\delta_jI_V)^d\right)\subseteq V$ of generalized eigenvectors, with $n$ distinct eigenpairs in total. We know that these subspaces are mutually disjoint. We also know that the dimension of $U_i$ is equal to the multiplicity of $\lambda_i$, which is the number of factors of $(X-\lambda_i)$ in the characteristic polynomial. Similarly, the dimension of $V_j$ is twice the multiplicity of $(\tau_j,\delta_j)$, which is the number of factors of $(X^2-\tau_jX+\delta)$ in the characteristic polynomial. Since each linear factor contributes ${1}$ to the degree of the polynomial, while each irreducible quadratic contributes ${2}$, we can see that the sum of the dimensions of the $U_i$ and $V_j$ is equal to the degree of the characteristic polynomial, which is the dimension of $V$ itself. That is, we have a decomposition of $V$ as a direct sum of invariant subspaces $\displaystyle V=\bigoplus\limits_{i=1}^mU_i\oplus\bigoplus\limits_{j=1}^nV_j$ Further, we know that the restrictions $(T-\lambda_iI_V)\|_{U_i}$ and $(T^2-\tau_jT+\delta_jI_V)\|_{V_j}$ are nilpotent transformations. I’ll leave it to you to work out what this last property implies for the matrix on the generalized eigenspace of an eigenpair, in analogy with a Jordan block for an eigenvalue. ### Like this: Posted by John Armstrong \| Algebra, Linear Algebra ## 3 Comments » 1. The formula after “For each distinct eigenvalue” does not parse. Comment by jr \| April 9, 2009 \| Reply 2. Thanks. Not sure when the d in lambda disappeared. Comment by \| April 9, 2009 \| Reply 3. [...] Real Inner Products Now that we’ve got bilinear forms, let’s focus in on when the base field is . We’ll also add the requirement that our bilinear forms be symmetric. As we saw, a bilinear form corresponds to a linear transformation . Since is symmetric, the matrix of must itself be symmetric with respect to any basis. So let’s try to put it into a canonical form! [...] Pingback by \| April 15, 2009 \| 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": 28, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8825870156288147, "perplexity_flag": "head"}
http://en.wikipedia.org/wiki/Transpose	# Transpose Note that this article assumes that matrices are taken over a commutative ring. These results may not hold in the non-commutative case. The transpose AT of a matrix A can be obtained by reflecting the elements along its main diagonal. Repeating the process on the transposed matrix returns the elements to their original position. In linear algebra, the transpose of a matrix A is another matrix AT (also written A′, Atr,tA or At) created by any one of the following equivalent actions: • reflect A over its main diagonal (which runs from top-left to bottom-right) to obtain AT • write the rows of A as the columns of AT • write the columns of A as the rows of AT Formally, the i th row, j th column element of AT is the j th row, i th column element of A: $[\mathbf{A}^\mathrm{T}]_{ij} = [\mathbf{A}]_{ji}$ If A is an m × n matrix then AT is an n × m matrix. The transpose of a matrix was introduced in 1858 by the British mathematician Arthur Cayley.[1] ## Examples • $\begin{bmatrix} 1 & 2 \end{bmatrix}^{\mathrm{T}} = \, \begin{bmatrix} 1 \\ 2 \end{bmatrix}$ • $\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}^{\mathrm{T}} = \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix}$ • $\begin{bmatrix} 1 & 2 \\ 3 & 4 \\ 5 & 6 \end{bmatrix}^{\mathrm{T}} = \begin{bmatrix} 1 & 3 & 5\\ 2 & 4 & 6 \end{bmatrix}$ ## Properties For matrices A, B and scalar c we have the following properties of transpose: 1. $( \mathbf{A}^\mathrm{T} ) ^\mathrm{T} = \mathbf{A} \quad \,$ Taking the transpose is an involution (self-inverse). 2. $(\mathbf{A}+\mathbf{B}) ^\mathrm{T} = \mathbf{A}^\mathrm{T} + \mathbf{B}^\mathrm{T} \,$ The transpose respects addition. 3. $\left( \mathbf{A B} \right) ^\mathrm{T} = \mathbf{B}^\mathrm{T} \mathbf{A}^\mathrm{T} \,$ Note that the order of the factors reverses. From this one can deduce that a square matrix A is invertible if and only if AT is invertible, and in this case we have (A−1)T = (AT)−1. By induction this result extends to the general case of multiple matrices, where we find that (A1A2...Ak-1Ak)T = AkTAk-1T... A2TA1T. 4. $(c \mathbf{A})^\mathrm{T} = c \mathbf{A}^\mathrm{T} \,$ The transpose of a scalar is the same scalar. Together with (2), this states that the transpose is a linear map from the space of m × n matrices to the space of all n × m matrices. 5. $\det(\mathbf{A}^\mathrm{T}) = \det(\mathbf{A}) \,$ The determinant of a square matrix is the same as that of its transpose. 6. The dot product of two column vectors a and b can be computed as $\mathbf{a} \cdot \mathbf{b} = \mathbf{a}^{\mathrm{T}} \mathbf{b},$ which is written as ai bi in Einstein notation. 7. If A has only real entries, then ATA is a positive-semidefinite matrix. 8. $(\mathbf{A}^\mathrm{T})^{-1} = (\mathbf{A}^{-1})^\mathrm{T} \,$ The transpose of an invertible matrix is also invertible, and its inverse is the transpose of the inverse of the original matrix. The notation A−T is often used to represent either of these equivalent expressions. 9. If A is a square matrix, then its eigenvalues are equal to the eigenvalues of its transpose. ## Special transpose matrices A square matrix whose transpose is equal to itself is called a symmetric matrix; that is, A is symmetric if $\mathbf{A}^{\mathrm{T}} = \mathbf{A}$ A square matrix whose transpose is equal to its negative is called skew-symmetric matrix; that is, A is skew-symmetric if $\mathbf{A}^{\mathrm{T}} = -\mathbf{A}$ The conjugate transpose of the complex matrix A, written as A∗, is obtained by taking the transpose of A and the complex conjugate of each entry: $\mathbf{A}^* = (\overline{\mathbf{A}})^{\mathrm{T}} = \overline{(\mathbf{A}^{\mathrm{T}})}$ A square matrix whose transpose is also its inverse is called an orthogonal matrix; that is, G is orthogonal if $\mathbf{G G}^\mathrm{T} = \mathbf{G}^\mathrm{T} \mathbf{G} = \mathbf{I}_n , \,$ the identity matrix, i.e. GT = G−1. ## Transpose of linear maps Main article: Dual space#Transpose of a linear map Main article: Hermitian adjoint If f : V → W is a linear map between vector spaces V and W with nondegenerate bilinear forms, we define the transpose of f to be the linear map tf : W → V, determined by $B_V(v,{}^tf(w))=B_W(f(v),w) \quad \forall\ v \in V, w \in W$ Here, BV and BW are the bilinear forms on V and W respectively. The matrix of the transpose of a map is the transposed matrix only if the bases are orthonormal with respect to their bilinear forms. Over a complex vector space, one often works with sesquilinear forms instead of bilinear (conjugate-linear in one argument). The transpose of a map between such spaces is defined similarly, and the matrix of the transpose map is given by the conjugate transpose matrix if the bases are orthonormal. In this case, the transpose is also called the Hermitian adjoint. If V and W do not have bilinear forms, then the transpose of an F-linear map f : V → W is only defined as a linear map tf : W∗ → V∗ between the dual spaces of W and V. This means that the transpose (and even the orthogonal group) can be defined abstractly, and completely without reference to matrices (nor the components thereof). If f : V → W then for any o : W → F (that is, any o belonging to W∗), if Tf(o) is defined as o composed with f then it will map V → F (that is, Tf will map W∗ to V∗). If the vector spaces have metrics then V∗ can be uniquely mapped to V, etc., such that we can immediately consider whether or not fT : W → V is equal to f −1 : W → V. ### Dual basis Main article: Dual basis The transpose of a vector is generalized by the dual basis. Given a basis B for a vector space V, indexed by a set I, one has a dual basis B* for the dual space V, with the same index set I. This gives an isomorphism between vectors (elements of V) and dual vectors (elements of V) which corresponds to transpose, using B and B* to give coordinates on V and V*. ### As a shorthand for contraction with the metric tensor Introductory linear algebra generally does not distinguish between the notion of a vector and a dual vector. Once that distinction is made, many common expressions seem to be freely transposing vectors to create dual vectors, in seeming disregard for the distinction. For example, this is the case in defining the inner product as $u\cdot v \equiv u^\mathrm{T} v$ Here $u^\mathrm{T}$ is a notational shortcut for tensor contraction with the metric tensor. Using the Einstein summation convention, with regular (contravariant) vectors having upper indices, this is computing $u\cdot v \equiv g_{ij} u^i v^j$ with the metric tensor for the Euclidean metric being the Kronecker delta. In other words, the notation $u^\mathrm{T}$ to create a dual vector is really shorthand: $\,(u^\mathrm{T})_j = g_{ij} u^i$ with the assumption that $g_{ij}=\delta_{ij}$. ## Implementation of matrix transposition on computers On a computer, one can often avoid explicitly transposing a matrix in memory by simply accessing the same data in a different order. For example, software libraries for linear algebra, such as BLAS, typically provide options to specify that certain matrices are to be interpreted in transposed order to avoid the necessity of data movement. However, there remain a number of circumstances in which it is necessary or desirable to physically reorder a matrix in memory to its transposed ordering. For example, with a matrix stored in row-major order, the rows of the matrix are contiguous in memory and the columns are discontiguous. If repeated operations need to be performed on the columns, for example in a fast Fourier transform algorithm, transposing the matrix in memory (to make the columns contiguous) may improve performance by increasing memory locality. Main article: In-place matrix transposition Ideally, one might hope to transpose a matrix with minimal additional storage. This leads to the problem of transposing an n × m matrix in-place, with O(1) additional storage or at most storage much less than mn. For n ≠ m, this involves a complicated permutation of the data elements that is non-trivial to implement in-place. Therefore efficient in-place matrix transposition has been the subject of numerous research publications in computer science, starting in the late 1950s, and several algorithms have been developed. ## References 1. Arthur Cayley (1858) "A memoir on the theory of matrices," Philosophical Transactions of the Royal Society of London, 148 : 17-37. The transpose (or "transposition") is defined on page 31.	{"extraction_info": {"found_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": 0.3, "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": 0.9068298935890198, "perplexity_flag": "head"}
http://mathoverflow.net/questions/106877/line-bundles-connections-and-covariantly-holomorphic-sections	## Line bundles, connections, and covariantly holomorphic sections ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I have a confusion regarding the line bundles arising in Kahler quantization for the torus. I know of course that the space of holomorphic sections should be isomorphic to a space of theta functions. However, there is a gap in my understanding of these bundles. Although a bit pedantic let me first specify the construction of these bundles which I am thinking of. We may regard the torus as the quotient $\mathbb{C}/\Lambda$, where $\Lambda$ is a lattice generated by two complex numbers, and $\mathbb{C}$ has the standard symplectic form $dx\wedge dy$. Let us for concreteness take $\Lambda$ to be generated by 1 and $i$. Any bundle over the torus can be regarded as a quotient of $\mathbb{C}\times\mathbb{C}$ by an action of $\Lambda$, with $\lambda(z,s)=(z+\lambda,g(\lambda, z)s)$ where $g$ is a cocycle. The line bundles over the torus are classified by their Chern class, and hence are classified by a level $k$. To obtain the bundle $L_k$ for level $k$ we take $g(\lambda)=e^{\pi ik\omega(\lambda, z)}$, where the hermitian form $h$ on $\mathbb{C}\times\mathbb{C}$ is given by $h((z,s),(z,s'))=s\overline{s'}$. Clearly the cocycle $g$ preserves the hermitian form. A connection $D$ on $\mathbb{C}\times\mathbb{C}$ with curvature $-2\pi ik\omega$ may be taken to be of the form $D=d-2\pi ikzd\overline{z}$. Now the covariantly constant sections are represented by functions $f$ obeying $\partial_\overline{z}f=2\pi ikzf$. This has solutions $f(z,\overline{z})=\phi(z)e^{2\pi i kz\overline{z}}$ where $\phi$ is holomorphic. However, for $\|f\|$ to be invariant under $\Lambda$, we must have $\phi$ invariant under $\Lambda$, since $\|f\|=\|\phi\|$. This is impossible unless $\phi$ is constant, as may be argued using the maximum modulus principle, for example. Therefore the space of covariantly holomorphic functions which descend to sections on $L_k$ appears to be 1-dimensional. Obviously there must be a mistake somewhere - one should get something related to theta functions. But it is not clear to me why on these bundles $L_k$ constructed as above the space of covariantly holomorphic sections is one dimensional? What is the problem with this approach, and what is the correct way to approach this? - The connection you wrote down is not invariant under $\Lambda$. Since $\omega$ contains both $z$ and $\overline{z}$, the gauge transformed connection $g^{-1}Dg$ will have both $dz$ and $d\overline{z}$, while your $D$ has only $d\overline{z}$. If, however, $\omega$ only contained $z$, you could not have $\|g\|=1$. – Pavel Safronov Sep 11 at 5:32	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 46, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9489868879318237, "perplexity_flag": "head"}
http://mathhelpforum.com/algebra/46479-solved-domain-range.html	# Thread: 1. ## [SOLVED] Domain and range Describe the domain range of f and g. form the function f(g(x)) and describe the domain and range of this composed function. F(x)=sqrtX G(x) = x+5 i really need help with this i dont really understand domain and range 2. Originally Posted by john doe Describe the domain range of f and g. form the function f(g(x)) and describe the domain and range of this composed function. F(x)=sqrtX G(x) = x+5 i really need help with this i dont really understand domain and range Keep in mind that the domain and range of $\sqrt{x}$ is $\left[0,\infty\right)$ The domain and range of $x+5$ is $\left(-\infty,\infty\right)$ If $f(x)=\sqrt{x}$ and $g(x)=x+5$, then $f(g(x))=\sqrt{x+5}$ Keep in mind that $\sqrt{x+5}$ exists when $x+5>0$. Solve this for x, and this will give you the domain [convert it to interval notation, though] The range would be all values of $y>0$. No value of x will cause this square root function to be negative. Rewrite this in interval notation as well. I hope this makes sense! --Chris 3. I understand the domain and range. What i don't get is, the product of squaring any number would give me two answer so f(g(x)) wouldn't be a function. ex: if i square 4 wouldn't i get + or - 2. 4. Originally Posted by john doe I understand the domain and range. What i don't get is, the product of squaring any number would give me two answer so f(g(x)) wouldn't be a function. ex: if i square 4 wouldn't i get + or - 2. no, if you square 4 you get 16. what you are getting at is the square-root of 4. in that case the answer is 2. only 2, -2 is not $\sqrt{4}$. since $\sqrt{x} \ge 0$ for all $x \in \mathbb{R}$ definitions that always seem to help me: the domain is the set of input values (in this case, x values) for which a function is defined, or has an output value. the range is the set of output values (in this case, y-values) so just figure out what x-values work in your function, and that's your domain. it is often easier to find what does not work, and say the domain is the set of all values but those that don't work. the range is the set of outputs you can obtain by plugging in any one of these domain values hope that helps	{"extraction_info": {"found_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": 0.9343252778053284, "perplexity_flag": "head"}
http://crypto.stackexchange.com/questions/1401/how-many-possible-combinations-of-input-files-are-possible-using-a-63-bit-length/1402	# How many possible combinations of input files are possible using a 63-bit length? I'm trying to calculate how many possible combinations of files there can be using a signed 64-bit file length, but can't seem to find a formula (or I'm using the wrong keywords). For example, the number of unique files that have a length of 256 bytes is $2^{256*8}$ = $2^{2048}$. I assume the total number of combinations across all file lengths will be some kind of geometric sequence. The reason I want to calculate it, is because I'm curious how it relates to the number of possible SHA-256 hashes ($2^{256}$) of all these input files. Also, if I then calculate the SHA-256 hash of the SHA-256 hash of the file contents, how much have I reduced the entropy (if that's the correct term)? Thanks for your help! - It looks like you did not work on that one hard enough. Some hints follow. You will find the restrictions on the input of SHA-256 in the standard. That does not match the question and title, any way I read length. Worse, there's a clash in the usual meaning of the term. You need to define what length is in your context. – fgrieu Dec 6 '11 at 12:50 ## 1 Answer For a given file length L bytes the combinations for that length is 256^L. The total combination is the sum of combinations for all file lengths L from zero to (2^63)-1. Just substitute parameters in the formula for summing the n first terms of a geometric series. The rest is left as an exercise to the reader :-) If all you need is a rough order-of-magnitude estimate it is really sufficient to just look at the highest order term, which provides a quick lower-bound estimate and which will already tells you the ratio is a very large number (very large for cryptographic purposes), which means the factor you have "reduced entropy" is a number very close to zero (once again, for cryptographic purposes). -	{"extraction_info": {"found_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": 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": 0.9183147549629211, "perplexity_flag": "head"}
http://www.maths.usyd.edu.au/u/UG/JM/MATH1903/Quizzes/quiz4.html	# The University of Sydney-School of Mathematics and Statistics • Mathematics & Statistics • University Home • Science Faculty # Quiz 4: Integration by Parts Question ## Question 1 Recall that integration by parts is a technique to re-express the integral of a product of two functions $u$ and $\frac{dv}{dx}$ in a form which allows it to be more easily evaluated. The formula is $\int \phantom{\rule{1em}{0ex}}u\phantom{\rule{2.77695pt}{0ex}}\frac{dv}{dx}\phantom{\rule{2.77695pt}{0ex}}dx=uv-\int \phantom{\rule{1em}{0ex}}v\phantom{\rule{2.77695pt}{0ex}}\frac{du}{dx}\phantom{\rule{2.77695pt}{0ex}}dx$. When applying the method of integration by parts to find $\int \phantom{\rule{1em}{0ex}}x{e}^{x}\phantom{\rule{1em}{0ex}}dx$, the best choice of $u$ and $\frac{dv}{dx}$ is \| \| \| \| \| \|----\|----------------------------------------------------------------------------\|----\|----------------------------------------------------------------------------\| \| a) \| $u={e}^{x},\phantom{\rule{1em}{0ex}}\frac{dv}{dx}=x$ \| b) \| $u=x{e}^{x},\phantom{\rule{1em}{0ex}}\frac{dv}{dx}=1$ \| \| c) \| $u=1,\phantom{\rule{1em}{0ex}}\frac{dv}{dx}=x{e}^{x}$ \| d) \| $u=x,\phantom{\rule{1em}{0ex}}\frac{dv}{dx}={e}^{x}$ \| \| e) \| None of the above \| \| \| \| \| \| \| \| Not correct. Choice (a) is false. This expresses the original integral in terms of a new integral which is even harder! After applying integration by parts as suggested, we have $\int \phantom{\rule{1em}{0ex}}x{e}^{x}\phantom{\rule{1em}{0ex}}dx=\frac{1}{2}{x}^{2}{e}^{x}-\int \phantom{\rule{2.77695pt}{0ex}}\frac{1}{2}{x}^{2}{e}^{x}\phantom{\rule{2.77695pt}{0ex}}dx$. Not correct. Choice (b) is false. This expresses the original integral in terms of a new integral which is even harder! After applying integration by parts as suggested, we have $\int \phantom{\rule{1em}{0ex}}x{e}^{x}\phantom{\rule{1em}{0ex}}dx={x}^{2}{e}^{x}-\int \phantom{\rule{2.77695pt}{0ex}}\left({x}^{2}{e}^{x}+x{e}^{x}\right)\phantom{\rule{2.77695pt}{0ex}}dx$. Not correct. Choice (c) is false. The problem of finding $v$ if $dv=x{e}^{x}\phantom{\rule{2.77695pt}{0ex}}dx$ is exactly the integral we started with! So no progress in this case. This gives genuine simplification and results in an easy integral. We obtain $\int \phantom{\rule{2.77695pt}{0ex}}x{e}^{x}\phantom{\rule{2.77695pt}{0ex}}dx=x{e}^{x}-\int \phantom{\rule{2.77695pt}{0ex}}{e}^{x}\phantom{\rule{2.77695pt}{0ex}}dx$. Not correct. Choice (e) is false. ## Question 2 Find ${\int }_{0}^{1}\phantom{\rule{1em}{0ex}}x{e}^{x}\phantom{\rule{1em}{0ex}}dx$ using integration by parts, and enter your answer. Choosing $u=x,\phantom{\rule{1em}{0ex}}\frac{dv}{dx}={e}^{x}$ gives ${\int }_{0}^{1}\phantom{\rule{1em}{0ex}}x{e}^{x}\phantom{\rule{1em}{0ex}}dx={\left[x{e}^{x}\right]}_{0}^{1}-{\int }_{0}^{1}\phantom{\rule{2.77695pt}{0ex}}{e}^{x}\phantom{\rule{2.77695pt}{0ex}}dx={\left[{e}^{x}\left(x-1\right)\right]}_{0}^{1}=1.$ Not correct. You may try again. Try choosing $u=x,\phantom{\rule{1em}{0ex}}\frac{dv}{dx}={e}^{x}$. ## Question 3 The reduction formula for ${I}_{n}=\int \phantom{\rule{2.77695pt}{0ex}}x{\left(lnx\right)}^{n}\phantom{\rule{2.77695pt}{0ex}}dx$ is ${I}_{n}=\frac{1}{2}{x}^{2}{\left(lnx\right)}^{n}-\frac{n}{2}{I}_{n-1}$. Given that ${I}_{0}=\frac{1}{2}{x}^{2}+C$, find ${\int }_{1}^{e}\phantom{\rule{2.77695pt}{0ex}}x\left(lnx\right)\phantom{\rule{2.77695pt}{0ex}}dx$. Give your answer correct to three decimal places. The integral ${I}_{1}$ is $\frac{1}{2}{x}^{2}lnx-\frac{1}{4}{x}^{2}+C$, using the reduction formula with $n=1$. Evaluating this between $1$ and $e$ gives 2.097 to three decimal places. Not correct. You may try again. Substitute $n=1$ into the reduction formula to obtain ${I}_{1}$. ## Question 4 Which option is an antiderivative of ${x}^{4}{e}^{x}$ ? Use the reduction formula ${I}_{n}=\int \phantom{\rule{2.77695pt}{0ex}}{x}^{n}{e}^{x}\phantom{\rule{2.77695pt}{0ex}}dx={x}^{n}{e}^{x}-n{I}_{n-1}$ to help answer this question. \| \| \| \| \| \|----\|-----------------------------------------------------------------------------\|----\|-----------------------------------------------------------------------------\| \| a) \| ${e}^{x}{x}^{4}+4{x}^{3}{e}^{x}-12{x}^{2}{e}^{x}+24x{e}^{x}-24{e}^{x}$ \| b) \| ${e}^{x}\left({x}^{4}-4{x}^{3}-12{x}^{2}-24x+24\right)$ \| \| c) \| ${e}^{x}{x}^{4}-4{x}^{3}{e}^{x}+12{x}^{2}{e}^{x}-24x{e}^{x}+24{e}^{x}+C{e}^{x}$ \| d) \| ${e}^{x}\left({x}^{4}-4{x}^{3}+12{x}^{2}-24x+24\right)+15$ \| \| e) \| None of the above \| \| \| \| \| \| \| \| Not correct. Choice (a) is false. Not correct. Choice (b) is false. Not correct. Choice (c) is false. Not correct. Choice (e) is false. ## Question 5 Which option equals ${\int }_{0}^{1}\phantom{\rule{2.77695pt}{0ex}}x{tan}^{-1}x\phantom{\rule{2.77695pt}{0ex}}dx$ ? (Hint: use integration by parts with $u={tan}^{-1}x$ and $\frac{dv}{dx}=x$.) \| \| \| \| \| \|----\|-----------------------------------------------------------------------------\|----\|-----------------------------------------------------------------------------\| \| a) \| $\frac{\pi }{4}-\frac{1}{2}$ \| b) \| $-\frac{1}{2}$ \| \| c) \| $\frac{\pi }{8}-\frac{1}{2}$ \| d) \| $\frac{1}{2}\left(\frac{\pi }{4}+1\right)$ \| \| e) \| $\frac{\pi }{4}$ \| \| \| \| \| \| \| \| The indefinite integral is $\frac{1}{2}{x}^{2}{tan}^{-1}x-\frac{1}{2}x+\frac{1}{2}{tan}^{-1}x+C$, which, when evaluated between 0 and 1, matches this option. Not correct. Choice (b) is false. Not correct. Choice (c) is false. Not correct. Choice (d) is false. Not correct. Choice (e) is false. ## Question 6 Which option equals ${\int }_{0}^{\frac{1}{2}}\phantom{\rule{2.77695pt}{0ex}}{sin}^{-1}x\phantom{\rule{2.77695pt}{0ex}}dx$ ? (Hint: use integration by parts with $u={sin}^{-1}x$ and $\frac{dv}{dx}=1$.) \| \| \| \| \| \|----\|-----------------------------------------------------------------------------\|----\|-----------------------------------------------------------------------------\| \| a) \| $\frac{\pi }{12}-\frac{\sqrt{3}}{2}+1$ \| b) \| $\frac{\pi }{4}+\frac{\sqrt{3}}{2}$ \| \| c) \| $\frac{\pi }{12}+\frac{\sqrt{3}}{2}-1$ \| d) \| $\frac{\pi }{3}-1$ \| \| e) \| $\frac{\pi }{12}-\frac{\sqrt{3}}{2}-1$ \| \| \| \| \| \| \| \| Not correct. Choice (a) is false. Not correct. Choice (b) is false. The indefinite integral is $x{sin}^{-1}x+\sqrt{1-{x}^{2}}+C$, which, when evaluated between $0$ and $\frac{1}{2}$, matches this option. Not correct. Choice (d) is false. Not correct. Choice (e) is false. ## Question 7 The finite area bounded by the curve $y=lnx$, the line $y=1$ and the tangent line to $y=lnx$ at $x=1$ is given as an integral with respect to $x$ by ${\int }_{1}^{2}\phantom{\rule{2.77695pt}{0ex}}\left(x-1-lnx\right)\phantom{\rule{2.77695pt}{0ex}}dx+{\int }_{2}^{e}\phantom{\rule{2.77695pt}{0ex}}\left(1-lnx\right)\phantom{\rule{2.77695pt}{0ex}}dx$. Which option equals the same area given as an integral with respect to $y$ ? (You must draw a sketch to help you with this question.) \| \| \| \| \| \|----\|-----------------------------------------------------------------------------\|----\|-----------------------------------------------------------------------------\| \| a) \| ${\int }_{1}^{e}\phantom{\rule{2.77695pt}{0ex}}{e}^{y}-\left(y+1\right)\phantom{\rule{2.77695pt}{0ex}}dy$ \| b) \| ${\int }_{0}^{1}\phantom{\rule{2.77695pt}{0ex}}lny-y+1\phantom{\rule{2.77695pt}{0ex}}dy$ \| \| c) \| ${\int }_{0}^{e}\phantom{\rule{2.77695pt}{0ex}}{e}^{y}-y+1\phantom{\rule{2.77695pt}{0ex}}dy$ \| d) \| ${\int }_{0}^{1}\phantom{\rule{2.77695pt}{0ex}}{e}^{y}-\left(y+1\right)\phantom{\rule{2.77695pt}{0ex}}dy$ \| \| e) \| None of the above \| \| \| \| \| \| \| \| Not correct. Choice (a) is false. Not correct. Choice (b) is false. Not correct. Choice (c) is false. Not correct. Choice (e) is false. ## Question 8 Which option equals $\int \phantom{\rule{2.77695pt}{0ex}}x{sec}^{2}x\phantom{\rule{2.77695pt}{0ex}}dx$ ? \| \| \| \| \| \|----\|-----------------------------------------------------------------------------\|----\|-----------------------------------------------------------------------------\| \| a) \| $xsecxtanx-ln\left(cosx\right)+C$ \| b) \| $xtanx+ln\|cosx\|+C$ \| \| c) \| $x{tan}^{2}x-ln\|cosx\|+C$ \| d) \| $xtanx-ln\left(cosx\right)+C$ \| \| e) \| None of the above \| \| \| \| \| \| \| \| Not correct. Choice (a) is false. Not correct. Choice (c) is false. Not correct. Choice (d) is false. Not correct. Choice (e) is false. ## Question 9 In some problems you need to apply the integration by parts method twice in order to obtain the required answer. The integral $\int \phantom{\rule{2.77695pt}{0ex}}sin\left(lnx\right)\phantom{\rule{2.77695pt}{0ex}}dx$ is one such problem. Which of the following options gives the expression obtained after one application of integration by parts? \| \| \| \| \| \|----\|-----------------------------------------------------------------------------\|----\|-----------------------------------------------------------------------------\| \| a) \| $xsin\phantom{\rule{0.3em}{0ex}}\left(lnx\right)-\int \phantom{\rule{2.77695pt}{0ex}}cos\phantom{\rule{0.3em}{0ex}}\left(lnx\right)\phantom{\rule{2.77695pt}{0ex}}dx$ \| b) \| $-cosxlnx+\int \phantom{\rule{2.77695pt}{0ex}}\frac{cosx}{x}\phantom{\rule{2.77695pt}{0ex}}dx$ \| \| c) \| $-xsin\phantom{\rule{0.3em}{0ex}}\left(lnx\right)+\int \phantom{\rule{2.77695pt}{0ex}}xcos\phantom{\rule{0.3em}{0ex}}\left(lnx\right)\phantom{\rule{2.77695pt}{0ex}}dx$ \| d) \| $xsin\phantom{\rule{0.3em}{0ex}}\left(lnx\right)-\int \phantom{\rule{2.77695pt}{0ex}}\frac{sin\phantom{\rule{0.3em}{0ex}}\left(lnx\right)}{x}\phantom{\rule{2.77695pt}{0ex}}dx$ \| \| e) \| $cosxlnx+\int \phantom{\rule{2.77695pt}{0ex}}\frac{cosx}{x}\phantom{\rule{2.77695pt}{0ex}}dx$ \| \| \| \| \| \| \| \| Not correct. Choice (b) is false. Try $u=sin\phantom{\rule{0.3em}{0ex}}\left(lnx\right)$ and $\frac{dv}{dx}=1$ in the integration by parts formula. Not correct. Choice (c) is false. Not correct. Choice (d) is false. Not correct. Choice (e) is false. ## Question 10 Which option equals ${\int }_{1}^{e}\phantom{\rule{2.77695pt}{0ex}}sin\left(lnx\right)\phantom{\rule{2.77695pt}{0ex}}dx$ ? \| \| \| \| \| \|----\|-----------------------------------------------------------------------------\|----\|-----------------------------------------------------------------------------\| \| a) \| $\frac{1}{2}\left(esin1+ecos1\right)$ \| b) \| $sin1-cos1$ \| \| c) \| $\frac{1}{2}\left(esin1-ecos1+1\right)$ \| d) \| $\frac{1}{2}sin1-2ecos1+1$ \| \| e) \| $esin1-ecos1$ \| \| \| \| \| \| \| \| Not correct. Choice (a) is false. Not correct. Choice (b) is false. Not correct. Choice (d) is false. Not correct. Choice (e) is false. © 2002-09 The University of Sydney. Last updated: Thursday 16 February 2012 ABN: 15 211 513 464. CRICOS number: 00026A. Phone: +61 2 9351 2222. Authorised by: Head, School of Mathematics and Statistics. Contact the University \| Disclaimer \| Privacy \| Accessibility	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 86, "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": 0.8260449767112732, "perplexity_flag": "middle"}
http://mathhelpforum.com/advanced-applied-math/104663-statics-problem.html	# Thread: 1. ## Statics problem The uniform ladder in Fig. 2 weighs 0.6kN. If x = 4m, how large must the coefficient of friction against the ground be for the ladder not to slip down? If the coefficient of friction is m = 0.5, show that the ladder will only fall when the man climbs up to x = 2.8m or a height of 3.5m above the ground! Now I think I'm correct in saying, at A, there's a vertical reaction pointing upwards and a horizontal reaction pointing to the right? At B there's a horizontal reaction pointing left? I worked out the reactions and got FAy=1.6Kn, FAx=1.04Kn, FBx=1.04Kn. Just got stuck at the last part where its asking for the coefficient of friction. Could anyone explain how to answer the last part of the question? Cheers. 2. Hello Haris Originally Posted by Haris The uniform ladder in Fig. 2 weighs 0.6kN. If x = 4m, how large must the coefficient of friction against the ground be for the ladder not to slip down? If the coefficient of friction is m = 0.5, show that the ladder will only fall when the man climbs up to x = 2.8m or a height of 3.5m above the ground! Now I think I'm correct in saying, at A, there's a vertical reaction pointing upwards and a horizontal reaction pointing to the right? At B there's a horizontal reaction pointing left? I worked out the reactions and got FAy=1.6Kn, FAx=1.04Kn, FBx=1.04Kn. Just got stuck at the last part where its asking for the coefficient of friction. Could anyone explain how to answer the last part of the question? Cheers. I take it you're referring to the last part of the first question, which asks for the coefficient of friction when $x = 4$; i.e. when the man is at the top of the ladder. I agree with your results, which are obtained by resolving horizontally, vertically and taking moments (about A, for instance). Then you simply say that if the ladder is on the point of slipping at A, then $F_{Ax} = \mu F_{Ay}$ $\Rightarrow \mu = \frac{1.04}{1.6}=0.65$ Grandad 3. I understand the first part of the last question. Could you please explain how to calculate the last part now? Cheers. 4. Hello Haris Originally Posted by Haris I understand the first part of the last question. Could you please explain how to calculate the last part now? Cheers. In the final part, $\mu = 0.5$ and the ladder is on the point of slipping. So, now we get: Resolve vertically: $F_{Ay}=1.6$ (as in part 1) Resolve horizontally: $F_{Ax} = F_{Bx}$ (also as in part 1) But now: $F_{Ax} = \mu F_{Ay} = 0.8 \Rightarrow F_{Bx} = 0.8$ Take moments about A: $0.6\times2+1\times x = F_{Bx}\times 5=0.8\times 5 = 4$ $\Rightarrow x = 4-1.2=2.8$ So, by similar triangles, the man's height above the ground is $\frac{2.8\times5}{4}=3.5$ m Grandad	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 10, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9316149950027466, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/120799/manifolds-admitting-cw-structure-with-single-n-cell	## Manifolds admitting CW-structure with single n-cell ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Let $M$ be a topological $n$-manifold, closed and connected (not necessarily oriented): When does $M$ not admit (up to homotopy-type) a CW-structure with a single $n$-cell? By classification of surfaces we assume $n>2$. By existence of smooth structures we assume $n>3$. In particular, if $M$ is smoothable then Morse theory provides us the desired structure. []: To put this question into context, we have various ways of showing that $H_{n-1}(M)$ has either $0$ or $\mathbb{Z}_2$ as its torsion subgroup depending on orientability. One way, when $M$ is such a CW-complex, is to quickly look at the chain-complex differential $d:C_n(M)\cong\mathbb{Z}\to C_{n-1}(M)$ and note that $H_n(M)\cong\mathbb{Z}$ for $M$ orientable and $H_n(M;\mathbb{Z}_2)\cong\mathbb{Z}_2$ otherwise. So I would like to see how large of a class of manifolds this argument holds for. [[Addendum]]: After chatting with Allen Hatcher and Rob Kirby, who reaffirm the comments below, here are their resulting thoughts: 1) We should be careful with the Kirby theorem of $M$ being homotopy-equivalent to a finite complex, because this complex is obtained by first embedding $M$ into $\mathbb{R}^N$ and then wiggling the boundary of a tubular neighborhood ($M\times D^{N-n}$) of $M$ to be PL, and so the resulting complex could have $i$-cells with $i>n$. 2) When $\dim M\ne 4$ there is a handlebody-decomposition, and this can be arranged to have a single 0-handle (canceling the other 0-handles with available 1-handles -- we can do this because there are no smoothing obstructions in a neighborhood of the 3-skeleton). Taking the dual handlebody, we have a decomposition with a single n-handle. Passing from the handlebody-decomposition to the CW-decomposition (shrinking everything to their cores), we obtain the desired CW-complex with a single n-cell. 3) When $\dim M=4$ then a handlebody-decomposition exists if and only if $M$ is smoothable. So when $M$ is smoothable we can apply the argument in (2). 4) But even when $M$ is not smooth we get some positive results, in particular for the $E_8$ manifold. We build $E_8$ using Kirby calculus on an 8-link diagram, giving a decomposition of $E_8$ into a 0-handle plus eight 2-handles plus a contractible piece (without the contractible piece we get a space with boundary being a homology 3-sphere, namely the Poincare-sphere $S^3/G$ with $G=$ binary icosahedral group). In particular, flipping this structure over we see that $E_8$ is homotopy-equivalent to a CW-complex with a single 4-cell. Furthermore, Lennart Meier's remark gets us all other simply-connected 4-manifolds. We are thus left with the scenario that $M$ (up to homotopy) is a closed connected non-simply-connected non-smoothable 4-manifold. (which the comments below assert) - 2 Since not all topological manifolds admit CW structures, your question must be about homotopy type. So any counter-examples would have to fail to be homotopy equivalent to a smooth manifold. The examples here might be candidates: mathoverflow.net/questions/34848/… – Mark Grant Feb 4 at 21:52 Isn't the $E_8$ manifold homotopy equivalent to a CW complex with one 0-cell, 8 2-cells, 0 3-cells, and 1 4-cell? – Will Sawin Feb 4 at 22:28 2 I just want to remark: If you care only about homotopy type, then every simply-connected closed, connected manifold has a CW-structure with just one 1-cell. This follows from Proposition 4C.1 of Hatcher's Algebraic Topology (about minimal cell structures) and Poincare duality. This does not exclude the aspherical manifold mentioned by Misha in the other Question. – Lennart Meier Feb 4 at 22:35 3 @Mark: Nevertheless, every compact manifold of dimension other than four admits a CW-structure (see mathoverflow.net/questions/36838). The question posed by Chris Gerig then makes sense precisely as stated. In fact, it is still interesting in dimension four, even if some 4-manifolds may not admit a CW-structure. Note: the question of CW-structures on 4-manifolds seems to be fairly open (see mathoverflow.net/questions/73428). – Ricardo Andrade Feb 4 at 23:17 1 See also mathoverflow.net/questions/42234/rugged-manifold (especially Greg's answer to a similar question; it works in dimension 4 as well). – Misha Feb 8 at 21:12 show 1 more comment	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 35, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8980233669281006, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/36155/linear-diophantine-equation	# Linear diophantine equation a) $$130x + 143y = 5957$$ b) $$44x + 19y = 75$$ the theorem says ax + by = c has solution if and only if d \| c however I work out both question with no solution as a) $$143 = 130 . 1 + 13$$ $$130 = 13 . 10 + 0$$ and 5957 is unable to be divided by 13 therefore no solution exists. b) $$44 = 19 . 2 + 6$$ $$19 = 6 . 3 + 1$$ $$6 = 3 . 2 + 0$$ for b) I am unsure if there are any solution exists? - Can you try to clean up your question please? It is hard for me to tell exactly what you are doing and what you are asking, without simply assuming that you are trying to apply Bezout's identity and getting stuck somewhere. – Alex Becker May 1 '11 at 6:31 ## 1 Answer Your second one, however, is not. When doing the Euclidean Algorithm, you mess up between your second and third steps. From $19 = 63 + 1$, you should go on the next next step $6 = 16 + 0$, indicating that 1 is the gcd of 19 and 44. Let's also consider for a moment what your solution said: it said that either 2 or 3 divided both 44 and 19 (neither of which are correct). We also know that 19 is prime, so unless 44 is a multiple of 19 (which it isn't), they share no factors. With that, it turns out that there are infinitely many solutions. - ops. mistake in b) for my euclidean algorithm. if 1 is the common divisor which means there are no common divisor? – optimus May 1 '11 at 6:38 @liangteh: No, it means that 1 is the common divisor. – mixedmath♦ May 1 '11 at 6:43 in this case, for gcd = 1 how do I derive the general solution and particular solution? – optimus May 1 '11 at 6:46 1 @Liangteh: I will not do your homework for you, but I will do a quick example. Consider the numbers 3 and 8. $8 = 32 + 2$, and $3 = 2 1 + 1$. So 1 is the gcd. But then, by reversing my equations, I get that $1 = 3 - 2(1) = 3 - (8 - 3(2)) = 3(3) - 8(1)$. So I have a solution. – mixedmath♦ May 1 '11 at 6:52	{"extraction_info": {"found_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": 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": 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": 0.9551032185554504, "perplexity_flag": "head"}
http://mathhelpforum.com/calculus/182771-inequality.html	# Thread: 1. ## inequality Solve the inequality $-0.02\leqslant -(-\frac{2}{3})^n\leqslant 0.02$ 2. Originally Posted by Punch Solve the inequality $-0.02\leqslant -(-\frac{2}{3})^n\leqslant 0.02$ I am going to assume that n is an integer otherwise the power may give complex values and they are not ordered. To get a ruff Idea of the bound solve $\left(\frac{2}{3} \right)^n=0.02 \iff n=\frac{\ln(0.02)}{\ln\left( \frac{2}{3}\right)} \approx 9.68...$ So for $n=10$ the inequality will hold 3. Originally Posted by TheEmptySet I am going to assume that n is an integer otherwise the power may give complex values and they are not ordered. To get a ruff Idea of the bound solve $\left(\frac{2}{3} \right)^n=0.02 \iff n=\frac{\ln(0.02)}{\ln\left( \frac{2}{3}\right)} \approx 9.68...$ So for $n=10$ the inequality will hold sorry but i dont understand the rationale behind the workings	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 6, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8967061042785645, "perplexity_flag": "middle"}
http://www.reference.com/browse/Vacuum_solution	Definitions # Vacuum solution A vacuum solution is a solution of a field equation in which the sources of the field are taken to be identically zero. For example, in Maxwell's theory of electromagnetism, a vacuum solution would represent the electromagnetic field in a region of space where there are no electromagnetic sources (charges and electric currents), i.e. where the current 4-vector vanishes: $J^a=0$ Another example is Einstein's theory of general relativity where a vacuum solution would represent the gravitational field in a region of spacetime where there are no gravitational sources (masses), i.e. where the energy-momentum tensor vanishes: $T_\left\{ab\right\}=0$	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 2, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.9081533551216125, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/219069/showing-that-if-a-subgroup-is-normal-its-homomorphic-image-is-normal	# Showing that if a subgroup is normal, it's homomorphic image is normal $\alpha : G \to H$ is a surjective homomorphism. And $U \subset G$ is a subgroup of $G$. Verfiy the claim - The image of $U$, ie $\alpha(U)$, is a subgroup of $H$, and if $U$ is normal in $G$, then $\alpha(U)$ is normal in $H$. Answer: Firstly, do I have to show $\alpha(U)$, is a subgroup of $H$ or is that statement just a statement of fact as part of the question? Anyway here is what I have done..taking it as a given that $\alpha(U)$, is a subgroup of $H$ - As $U$ is normal we have $U = gUg^{-1}$ $\alpha(U) = \alpha(gUg^1) =$ {applying homomorphic mapping into H} = $\alpha(g)\alpha(U)\alpha(g^{-1})$ Is that correct? I have a feeling I should take the $-1$ exponent outside the bracket as an extra final step or is that superfluous? - 1 The hypotheses should include the statement that $U$ is a subgroup of $G$. You must verify (1) that $\alpha[U]$ is a subgroup of $H$, and (2) that if $U$ is normal in $G$, than $\alpha[U]$ is normal in $H$. Yes, you need to finish the proof of (2) by pulling the inverse outside, so that the last factor is $(\alpha(g))^{-1}$. – Brian M. Scott Oct 23 '12 at 0:04 1. Yes, if you are uncertain, try to prove it. 2. Yes, final step, inverse out. – Berci Oct 23 '12 at 0:05 @BrianM.Scott: Yes I have edited in that it $U$ is a subgroup of $G$. – Jim_CS Oct 23 '12 at 0:06 ## 2 Answers To show that $\alpha(U)$ is a normal subgroup, you need to prove that $\alpha(U) = x \alpha(U) x^{-1}$ for all $x \in H$. But any $x \in H$ can be written in the form $\alpha(g)$ for $g \in G$ since $\alpha$ is surjective. Thus, you need only prove that $\alpha(U) = \alpha(g) \alpha(U) \alpha(g)^{-1}$ for all $g \in G$. Note the $-1$ exponent is "on the outside", so you really do need to take that last step as you suspected. And yes, as written it is meant that you should prove that $\alpha(U)$ is a subgroup. The proof requires you to carefully work through the definition/criterion for being a subgroup, but nothing beyond that. Also, +1 for showing your reasoning and clearly indicating what you are unsure of. - I think you must have meant $\,\alpha(U)=x\alpha(U)x^{-1}\,$ for all $\,x\in H\,$ ... – DonAntonio Oct 23 '12 at 2:53 Yes, thanks. :) I've corrected the mistake. – Michael Joyce Oct 23 '12 at 3:35 You have to prove that $\alpha(U)$ is normal subgroup in $H$, so for all $x\in H$, one has $x\cdot \alpha(U)\cdot x^{-1} \subseteq \alpha(U)$. -	{"extraction_info": {"found_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": 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": 0.9657390713691711, "perplexity_flag": "head"}
http://stats.stackexchange.com/questions/11634/how-do-i-calculate-error-propagation-with-different-measures-of-error	# How do I calculate error propagation with different measures of error? I am calculating the age of lake sediments at the base of a sediment core by dividing the total sediment mass of the core ($\mathrm{mg} \ \mathrm{cm}^{-2}$) by the sediment accumulation rate ($\mathrm{mg} \ \mathrm{cm}^{-2}\ \mathrm{y}^{-1}$). Both the sediment mass and the accumulation rate have variation associated with them. The sediment mass is a mean of 3 samples and the accumulation rate is reported as $\pm 10\%$. It is my understanding that I can calculate the error associated with age as: $\sqrt{\left(\frac{\delta x}{x}\right)^2 + \left(\frac{\delta y}{y}\right)^2}$ where $\delta x$ and $\delta y$ are the relative error of the measurements being divided (i.e., $x$ and $y$). The error associated with the sedimentation rate is $\pm \ 10%$ but the error associated with the sediment mass in a standard deviation. Are these forms of error compatible within the above error propagation formula? Thank you. - 1 – whuber♦ Jun 9 '11 at 19:48 @whuber - thanks for suggesting that search. I did try that but unless I am missing it there isn't anything that answers how to propagate the error when one measure is relative error ($\pm$ 10%) and the other is a standard deviation. Perhaps my question isn't clear. – KennyPeanuts Jun 9 '11 at 20:51 ## 2 Answers In some sense this depends on what you mean by $x$ and $\delta x$. Usually people mean that they are modeling $X$ as a random variable with mean $x$ and variance $(\delta x)^2$. Sometimes they mean the stronger condition that $X$ is actually Gaussian, and sometimes they have a broader meaning that $x$ and $\delta x$ can possible be other measures of the center and the spread. A bit of calculus and handwaving shows that for small variations that are also approximatable as Gaussian, and $X$ and $Y$ independent, $f(X, Y)$ can be approximately described as having mean $f(x, y)$, and $(\delta f)^2 = (\delta x)^2 (\frac{\partial f}{\partial x})^2 + (\delta y)^2 (\frac{\partial f}{\partial y})^2$. We can do the same thing for $a(m, r) = m /r$, where $a$ is the calculated age, $m$ is the mass, and $r$ is the rate. $$\begin{align} (\delta a)^2 &= (\delta m)^2 / r^2 + (\delta r)^2 m^2 / r^4 \\ a^2 &= m^2 / r^2 \\ (\delta a)^2/a^2 &= (\delta m)^2/m^2 + (\delta r)^2 / r^2 \\ (\delta a)/a &= \sqrt{(\delta m)^2/m^2 + (\delta r)^2 / r^2} \\ \end{align}$$ This matches the formula you have. You just have to convert between absolute errors and relative errors to be able to use it. EDITed to add (incorporating comments): To convert the sedimentation rate to relative error, just use $(\delta r)/r = 10\% = 0.1$. You need to find the $\delta m$ = standard error for the mean. It's not clear whether you have $\delta m_i$ for each individual sediment core measurements. If you do, you want to find $m$ with a weighted mean and calculating the standard error is a bit tricky, but the prescription given above for general $f$ expands fine to three arguments. If it's not, the standard mean can be used and the variance in the sample can be used to calculate the standard error of the mean. The relative error is of course just $(\delta m)/m$. - thanks for your answer. I guess my real question is what you allude to in your last sentence. How do I convert the error from each term (\$\pm 10%) and standard deviation into compatible forms so they can be used in those formulas? Thanks again. – KennyPeanuts Jun 9 '11 at 20:53 – wnoise Jun 9 '11 at 22:01 awesome! thanks. I am assuming you mean standard error for $\delta m$ since thats what you linked to and it seems to make more sense. I edited my question to make it what I was asking a bit more clear and I am accepting your answer. If you are so inclined you may want to add what you have in the comment to the answer itself just to have it all in the the answer. Thanks again! – KennyPeanuts Jun 10 '11 at 12:36 See my book: "Propagation of Errors" by Mike Peralta at amazon.com It treats the seldom covered topic of Second Order "Propagation of Errors" - 3 – gung Feb 27 at 21:37 2 I beg to differ a little, @gung. We always appreciate experts who weigh in and it's natural that when their work is relevant to a question that it should be referenced. But unfortunately a mere reference is inadequate as a stand-alone answer, as I'm sure you will appreciate, Mike. If you could amplify this answer to explain how your treatment answers the present question, I am sure many readers--now and future--would be delighted. – whuber♦ Feb 28 at 6:50	{"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": 34, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9458121061325073, "perplexity_flag": "head"}
http://mathoverflow.net/revisions/57546/list	## Return to Answer 6 added 427 characters in body Mirror symmetry gives some remarkable connections between certain varieties. The first step in this connection is that certain homology groups have the same rank. An explicit case for mirror symmetry duals is the case coming from toric varieties. In this case, the dual objects comes from duality of polytopes. So duality of polytopes: associating the octahedron to the cube and the icosahederon to the dodecahederon is related to Mirror symmetry. Perhaps the very first facts about polytopes which demonstrates unexpected equalities for certain homologies can be described as follows: For 2-dimensional polytopes this is the following numerical fact: A polygon has the same number of edges as the dual. (Well, this is not so unexpected.) For 4 dimensional polytope P it is the following numerical fact. Start with a 4-polytope with n vertices and e edges. Triangulate every 2-face by non crossing diagonals. Let $e^+$ be the number of edges including the added diagonals. Consider the quantity $$\gamma (P) = e^+ - 4n .$$ It is true that for every dual pair of 4-polytopes $P$ and $P^$, $$\gamma (P^)=\gamma(P).$$ This is more surprising. For example, let P be the 4-dimensional cross polytope and Q be the 4-dimensional cube. P has 8 bertices 24 edges and all the 2-faces are triangles so $\gamma (P)=24~-~4\cdot 8~=~-8$. The 4-cube Q has 16 vertices, and 32 edges and it has also 24 2-faces which are squares, so $e^+(Q)=56$. $\gamma (Q)=56-64 = -8$. Voila! This reflects some properties of toric varieties (unexpected equalities between Hodge numbers) which express (sort of the 0-th step of) mirror symmetry. Related papers: V. Batyrev and L. Borisov, Mirror duality and string-theoretic Hodge numbers; V. Batyrev and B. Nill, Combinatorial aspect of mirror symmetry. Here is a lecture by B. Nill. Another manifestation of mirror symmetry of combinatorial nature, that can be formulated in simple words, is in terms of typical shape of various classes of partitions. I mentioned it in a remark above and let me quote a description taken from my adventure book. A partition is just a way to write a number as a sum of other numbers. Like 9=4+2+1+1+1. Partitions have attracted mathematicians for centuries. Among others, the famous Indian mathematician Ramanujan was well known for his identities regarding partitions. And now enters another idea, baring the names of Ulam, Vershik, Kerov, Shepp and others who studied partitions as stochastic objects. In particular, it was discovered that "most" partitions, say of a number n, come in a "typical shape". The emergent picture drawn by Okounkov and his coauthors goes very roughly like this: an "algebraic variety" (a manifold of some sort) that takes part in a certain string theory is related to a class of partitions, and when we consider the typical shape of a partition in the class this gives us another algebraic variety, and - lo and behold - the typical shape IS the mirror image of the original one. The mirror relations translate to asymptotic results on the number of partitions, somewhat in the spirit of the famous asymptotic formulas of the mathematicians Hardy and Ramanujan for p(n)- the total number of partitions for the number n. As mentioned in the comments I am not sure about good references to this connection between mirror symmetry and limit shapes of classes of partitions. The 2003 paper Quantum Calabi-Yau and Classical Crystals by Andrei Okounkov, Nikolai Reshetikhin, and Cumrun Vafa describes this connection in Section 2.3 called "mirror symmetry and the limit shape". 5 added 1621 characters in body; deleted 37 characters in body Another manifestation of mirror symmetry of combinatorial nature, that can be formulated in simple words, is in terms of typical shape of various classes of partitions. I mentioned it in a remark above and let me quote a description taken from my adventure book. A partition is just a way to write a number as a sum of other numbers. Like 9=4+2+1+1+1. Partitions have attracted mathematicians for centuries. Among others, the famous Indian mathematician Ramanujan was well known for his identities regarding partitions. And now enters another idea, baring the names of Ulam, Vershik, Kerov, Shepp and others who studied partitions as stochastic objects. In particular, it was discovered that "most" partitions, say of a number n, come in a "typical shape". The emergent picture drawn by Okounkov and his coauthors goes very roughly like this: an "algebraic variety" (a manifold of some sort) that takes part in a certain string theory is related to a class of partitions, and when we consider the typical shape of a partition in the class this gives us another algebraic variety, and - lo and behold - the typical shape IS the mirror image of the original one. The mirror relations translate to asymptotic results on the number of partitions, somewhat in the spirit of the famous asymptotic formulas of the mathematicians Hardy and Ramanujan for p(n)- the total number of partitions for the number n. 4 added 93 characters in body Mirror symmetry gives some remarkable connections between certain varieties. The first step in this connection is that certain homology groups have the same rank. An explicit case for mirror symmetry duals is the case coming from toric varieties. In this case, the dual objects comes from duality of polytopes. So duality of polytopes: associating the octahedron to the cube and the icosahederon to the dodecahederon is related to Mirror symmetry. Perhaps the very first facts about polytopes which demonstrates unexpected equalities for certain homologies can be described as follows: For 2-dimensional polytopes this is the following numerical fact: A polygon has the same number of edges as the dual. (Well, this is not so unexpected.) For 4 dimensional polytope P it is the following numerical fact. Start with a 4-polytope with n vertices and e edges. Triangulate every 2-face by non crossing diagonals. Let $e^+$ be the number of edges including the added diagonals. Consider the quantity $$\gamma (P) = e^+ - 4n .$$ It is true that for every dual pair of 4-polytopes $P$ and $P^$, $$\gamma (P^)=\gamma(P).$$ This is more surprising. For example, let P be the 4-dimensional cross polytope and Q be the 4-dimensional cube. P has 8 bertices 24 edges and all the 2-faces are triangles so $\gamma (P)=24~-~4\cdot 8~=~-8$. The 4-cube Q has 16 vertices, and 32 edges and it has also 24 2-faces which are squares, so $e^+(Q)=56$. $\gamma (Q)=56-64 = -8$. Voila! This reflects some properties of toric varieties (unexpected equalities between Hodge numbers) which express (sort of the 0-th step of) mirror symmetry. Related papers: V. Batyrev and L. Borisov, Mirror duality and string-theoretic Hodge numbers; V. Batyrev and B. Nill, Combinatorial aspect of mirror symmetry. Here is a lecture by B. Nill. 3 added 461 characters in body; deleted 14 characters in body Mirror symmetry gives some remarkable connections between certain varieties. The first step in this connection is that certain homology groups have the same rank. An explicit case for mirror symmetry duals is the case coming from toric varieties. In this case, the dual objects comes from duality of polytopes. So duality of polytopes: associating the Octaheder octahedron to the cube and the icosaheder icosahederon to the dodecaheder dodecahederon is related to Mirror symmetry. Perhaps the very first facts about polytopes which demonstrates unexpected equalities for certain homologies can be described as follows: For 2-dimensional polytopes this is the following numerical factthat a : A polygon has the same number of edges as the dual. (Well, this is not so unexpected.) For 4 dimensional polytope P it is the following numerical fact. Start with a 4-polytope with n vertices and e edges. Triangulate every 2-face by non crossing diagonals. Let $e^+$ be the number of edges including the added diagonals. Consider the quantity $$\gamma (P) = e^+ - 4n +10. .$$ It is true that for every dual pair of 4-polytopes $P$ and $P^$, $$\gamma (P^)=\gamma(P).$$ This is more surprising. For example, let P be the 4-dimensional cross polytope and Q be the 4-dimensional cube. P has 8 bertices 24 edges and all the 2-faces are triangles so $\gamma (P)=24~-~4\cdot 8~=~-8$. The 4-cube Q has 16 vertices, and 32 edges and it has also 24 2-faces which are squares, so $e^+(Q)=56$. $\gamma (Q)=56-64 = -8$. Voila! This reflects some properties of toric varieties (unexpected equalities between Hodge numbers) which express (sort of the 0-th step of) mirror symmetry. Related papers: V. Batyrev and L. Borisov, Mirror duality and string-theoretic Hodge numbers; V. Batyrev and B. Nill, Combinatorial aspect of mirror symmetry. 2 added 290 characters in body Mirror symmetry gives some remarkable connections between certain varieties. The first step in this connection is that certain homology groups have the same rank. An explicit case for mirror symmetry duals is the case coming from toric varieties. In this case, the dual objects comes from duality of polytopes. So duality of polytopes: associating the Octaheder to the cube and the icosaheder to the dodecaheder is related to Mirror symmetry. Perhaps the very first facts about polytopes which demonstrates unexpected equalities for certain homologies can be described as follows: For 2-dimensional polytopes this is the fact that a polygon has the same number of edges as the dual. (Well, this is not so unexpected.) For 4 dimensional polytope P it is the following numerical fact. Start with a 4-polytope with n vertices and e edges. Triangulate every 2-face by non crossing diagonals. Let $e^+$ be the number of edges including the added diagonals. Consider the quantity $$\gamma (P) = e^+ - 4n +10.$$ It is true that for every dual pair of 4-polytopes $P$ and $P^$, $$\gamma (P^)=\gamma(P).$$ This reflects some properties of toric varieties which express mirror symmetry. Related papers: V. Batyrev and L. Borisov, Mirror duality and string-theoretic Hodge numbers; V. Batyrev and B. Nill, Combinatorial aspect of mirror symmetry. 1 Mirror symmetry gives some remarkable connections between certain varieties. The first step in this connection is that certain homology groups have the same rank. An explicit case for mirror symmetry duals is the case coming from toric varieties. In this case, the dual objects comes from duality of polytopes. So duality of polytopes: associating the Octaheder to the cube and the icosaheder to the dodecaheder is related to Mirror symmetry. Perhaps the very first facts about polytopes which demonstrates unexpected equalities for certain homologies can be described as follows: For 2-dimensional polytopes this is the fact that a polygon has the same number of edges as the dual. (Well, this is not so unexpected.) For 4 dimensional polytope P it is the following numerical fact. Start with a 4-polytope with n vertices and e edges. Triangulate every 2-face by non crossing diagonals. Let $e^+$ be the number of edges including the added diagonals. Consider the quantity $$\gamma (P) = e^+ - 4n +10.$$ It is true that for every dual pair of 4-polytopes $P$ and $P^$, $$\gamma (P^)=\gamma(P).$$ This reflects some properties of toric varieties which express mirror symmetry.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 24, "mathjax_display_tex": 10, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9301404356956482, "perplexity_flag": "middle"}
http://mathhelpforum.com/pre-calculus/76823-inequality-graph.html	# Thread: 1. ## Inequality graph -x + y > 3 please list 4 points on a graph for x & y thanks!!! 2. It's not clear what you asking for. Please state the full problem. Are you asking for 4 points in the region? Are you asking for 4 points on the graph of the line? Does X and Y have to be integer valued? 3. ## It's a simple line, (I think) Hi, The equation is the area above a line. $y>=x+3$ consists of all the points including and above $y=x+3$ which is a line with gradient 1, x intercept $(-3,0)$ and y intercept $(0,3)$ 4. Yes it is a region bounded by a simple line. I can name lots of points that have something to do with the line and the region. It is not clear what the question is asking. 5. Originally Posted by craigmain Hi, The equation is the area above a line. $y>=x+3$ consists of all the points including and above $y=x+3$ which is a line with gradient 1, x intercept $(-3,0)$ and y intercept $(0,3)$ I think the term "gradient" is too advanced for pre-calculus, although it's certainly correct and might be understood. Slope seems to be more commonly because early algebra only teaches about lines with constant gradients. I'm not correcting you, I just want to point out to take in mind the level of the poster. Originally Posted by meymathis Yes it is a region bounded by a simple line. I can name lots of points that have something to do with the line and the region. It is not clear what the question is asking. Agreed. Badly written question. "on a graph" could mean so many things.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 8, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "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": 0.9626309275627136, "perplexity_flag": "head"}
http://mathoverflow.net/questions/102879?sort=oldest	## Bijection between irreducible representations and conjugacy classes of finite groups ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Is there some natural bijection between irreducible representations and conjugacy classes of finite groups (as in case of $S_n$)? - 2 I do not see how it is natural for $S_n$, $n>3$? Surely, you can use either a partition or its dual partition for a representation. Is there a mathematical reason to choose one over another or is it purely historical? – Bugs Bunny Jul 23 at 9:37 @Bugs Bunny: maybe one can argue that there are two natural choices in this case? It all boils down to interpreting the word "natural", of course. – Vladimir Dotsenko Jul 23 at 13:14 1 @Bugs It's fairly natural. For any partition $\lambda$ of $n$, let $S(\lambda)$ be the subgroup $\prod S_{\lambda_i}$ of $S_n$. Let $H(\lambda)$ be the induction of the trivial rep from $S(\lambda)$ to $S_n$. The characters of the $H(\lambda)$ are upper triangular in the characters of the irreps; the conjugacy classes pairs upper triangularly against the $H(\lambda)$. The pairing is forced by wanting these matrices to be upper triangular rather than a permutation of upper triangular. I'm not sure that there is no other pairing which achieves this, but most of them won't work. – David Speyer Jul 23 at 13:30 Is there any other case in math where bijection exists, but there is no natural one ? – Alexander Chervov Jul 24 at 7:30 ## 6 Answers This is a different take on Steven Landsburg's answer. The short version is that conjugacy classes and irreducible representations should be thought of as being dual to each other. Fix an algebraically closed field $k$ of characteristic not dividing the order of our finite group $G$. The group algebra $k[G]$ is a finite-dimensional Hopf algebra, so its dual is also a finite-dimensional Hopf algebra of the same dimension; it is the Hopf algebra of functions $G \to k$, which I will denote by $C(G)$. (The former is cocommutative but not commutative in general, while the latter is commutative but not cocommutative in general.) The dual pairing $$k[G] \times C(G) \to k$$ is equivariant with respect to the action of $G$ by conjugation, and it restricts to a dual pairing $$Z(k[G]) \times C_{\text{cl}}(G) \to k$$ on the subalgebras fixed by conjugation; $Z(k[G])$ is the center of $k[G]$ and $C_{\text{cl}}(G)$ is the space of class functions $G \to k$. Now: The maximal spectrum of $Z(k[G])$ can be canonically identified with the irreducible representations of $G$, and the maximal spectrum of $C_{\text{cl}}(G)$ can be canonically identified with the conjugacy classes of $G$. The second identification should be clear; the first comes from considering the central character of an irreducible representation. Now, the pairing above is nondegenerate, so to every point of the maximal spectrum of $Z(k[G])$ we can canonically associate an element of $C_{\text{cl}}(G)$ (the corresponding irreducible character) and to every point of the maximal spectrum of $C_{\text{cl}}(G)$ we can canonically associate an element of $Z(k[G])$ (the corresponding sum over a conjugacy class divided by its size). - @Qiaochu: This is a nice abstract way to re-focus rigorously the original somewhat fuzzy question. But I still feel unable to treat concrete cases involving finite groups of Lie type in any explicit way: e.g., given the family `$\mathrm{SL}_2(\mathbb{F}_p)$`, how to prescribe an actual bijection (uniformly for all odd primes) and pass to the quotient group by the center as well? Even for `$S_n$` I'm reminded that the original Springer Correspondence assigned characters/partitions to cohomology degrees of the flag variety in one way which later got dualized/transposed. – Jim Humphreys Jul 24 at 20:47 I'm not sure I understand your question. I'm not claiming that a bijection between irreducible representations and conjugacy classes exist. What this argument does is exhibit irreducible representations and conjugacy classes as canonical bases of two vector spaces which are canonically dual to each other. But these bases are not dual to each other, so I don't get a bijection this way. – Qiaochu Yuan Jul 24 at 22:39 1 @Qiaochu: I guess I was expecting a more explicit answer to the original question, such as "no" (?). My concern beyond that is whether your interesting higher level viewpoint adds any concrete detail about actual finite groups of interest, starting with symmetric groups. – Jim Humphreys Aug 2 at 18:08 @Jim: yes, my answer is essentially "no." That is, I'm suggesting that there shouldn't be a nice bijection for arbitrary finite groups. I think one way to see this is to note that there is no reasonable sense in which this is true for infinite groups (e.g. $\text{SU}(2)$ has countably many irreducible representations but uncountably many conjugacy classes). The correct generalization of the duality above is that by Peter-Weyl, for a compact Hausdorff group $G$ the characters of irreducible representations are dense in the space of class functions. – Qiaochu Yuan Aug 2 at 18:35 ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. In general there is no natural bijection between conjugacy classes and irreducible representations of a finite group. To see this think of abelian groups for example. The conjugacy classes are the elements of the group, while the irreducible representations are elements of the dual group. These are isomorphic, via the Fourier transform, but not canonically. - Nevetheless among all set-theoretic bijections one has subclass of "good" bijections which are isomorphisms of the groups G and and G^dual. – Alexander Chervov Sep 16 at 14:42 Let $k$ be an algebraically closed field whose characteristic is either zero or prime to the order of $G$. Then the center of the group ring $kG$ has one basis in natural bijective correspondence with the set of irreducible representations of $G$ over $k$, and another basis in natural bijective correspondence with the conjugacy classes of $G$. Namely: 1) $kG$ is semisimple (this is called Maschke's Theorem) and Artinian, so it is a direct sum of matrix rings over division rings, hence (because $k$ is algebraically closed) a direct sum of matrix rings over $k$. There is (up to isomorphism) one irreducible representation for each of these matrix rings. Those representations are therefore in natural one-one correspondence with the central idempotents that generate those matrix rings, and these form a basis for the center. 2) For each conjugacy class, we can form the sum of all elements in that conjugacy class. The resulting elements of $kG$ form a basis for the center. This gives a (non-natural) bijection between irreducible representations and conjugacy classes, because there is a (non-natural) bijection between any two bases for a given finite-dimensional $k$-vector space. I do not see any way you can make this natural. - This only works when $k$ is algebraically closed, and the result itself also only works in that generality. For example, over $\mathbb{R}$, the cyclic group of order $3$ has two irreducible representations but still has three conjugacy classes. The error in the argument is that, for a division ring $\Delta$ occuring in the decomposition of $k[G]$, the contribution of $M_n(\Delta)$ to $Z(k[G])$ is $Z(\Delta)$, which may be larger than $k$. – David Speyer Jul 22 at 22:47 David Speyer: Point taken; I've inserted the words "algebraically closed" where I ought to have inserted them in the first place. – Steven Landsburg Jul 22 at 23:30 May I suggest that you then change "matrix rings over division rings" to "matrix rings over $k$"? Since an algebraically closed field has nonnontrivial finite division ring extensions. – David Speyer Jul 23 at 10:08 David Speyer: Done. Thanks. – Steven Landsburg Jul 23 at 13:00 Steven's and Gjergji answers points that there is no bijection, however possibly this idea should not be put into the rubbish completely. Ideologically conjugacy classes and irreducible representations are somewhat dual to each other. The other instances of this "duality" is Kirillov's orbit method - this is "infinitesimal version" of the duality: orbits in Lie algebra are infinitesimal versions of the conjugacy classes. But pay attention orbits are taken not in Lie algebra, but in the dual space g^. This again manifests that there irreps and conj. classes are dual to each other. However think of semi-simple Lie algebra - then g^ and g can be canonically identified... Another instance is Langlands parametrization of the unitary irreducible representations of the real Lie group G. They are parametrized by conjugacy classes in Langlands dual group G^L. Again here are conjugacy classes in G^L, not in G itself. However for example GL=GL^L... So it might be one should ask the question what are the groups such that conjugacy classes and irreps are in some natural bijection or something like this ? PS Here is some natural map conjugacy classes -> representations. But it does not maps to irreducible ones, and far from being bijection in general. A colleague of mine suggested the following - take vector space of functions on a group which are equal to zero everywhere except given conjugacy class "C". We can act on these functions by $f \to g f g^{-1}$ - such action will preserve this class. So we get some representation. In the case of abelian group this gives trivial representation, however in general, it might be non-trivial. It always has trivial component - the function which is constant on "C". I have not thought yet how this representation can be further decomposed, may be it is well-known ? - Take a finite abelian group $G$ and fix a non-degenerate pairing $G \times G \to \mathbb{C}^{\times}$. Unlike in the case of semisimple Lie algebras I do not see a canonical way to pick such a pairing. – Qiaochu Yuan Jul 22 at 19:21 @Qiaochu Yuan I agree. I did not pretend that there always exists some natural bijection. Just wanted to softly point out that "not giving up tooo easily", as Paul Garrett wrote in his comment above. E.g. Take G=Z/2Z in this case we may say there is natural bijection :) – Alexander Chervov Jul 23 at 7:39 @Geoff Sorry may be I misunderstanding however it seems my question was different. I do not take "class functions" (functions constant on conj. class (is it correct?)) but take NON constant on conj. class. Pay attention I am NOT acting by G in standadrd way - but act by conjugation f-> g f g^-1 hence non-constant functions are preserved by this action. This is a representation which clear have trivial one as a submodule (class function is trivial submodule). How this repr. is decomposed ? – Alexander Chervov Jul 24 at 9:50 E.g. take S_3 then conj. class corresponding to transpositions (just 3 elements) generaters standard 3D representation. We know it is trivial + 2D irreducible representation – Alexander Chervov Jul 24 at 9:54 OK Alexander. I did not read carefully enough. – Geoff Robinson Jul 24 at 10:31 I would suggest that no such general natural bijection has been found to date. I am not sure how one would prove" that such a natural bijection could not be found, notwithstanding Gjergi's answer. I take the view that the equality between numbers of irreducible characters (over an algebraically closed field of characteristic zero) and the number of conjugacy classes is most naturally obtained by counting the dimension of the center of the group algebra in two different categorical settings: from the group theoretic perspective, the natural distinguished basis for the group algebra (the group elements) makes it clear that the dimension of the center is the number of conjugacy classes. On the other hand, from a ring-theoretic perspective, the structure of semi-simple algebras makes it clear that the dimension of the center of the group algebra is the number of isomorphism types of simple modules, that is, the number of irreducible characters. Moving to prime characteristic (still over an algebraically closed field, now of characteristic $p$, say), it is rather more difficult to prove, as R. Brauer did, that the number of isomorphism types of simple modules is the number of conjugacy classes of group elements of order prime to $p.$ However, there are contemporary conjectures in modular representation theory which suggest that there may one day be a different explanation for this equality. In particular, Alperin's weight conjecture suggests counting the number of (isomorphism types of) absolutely irreducible modules in characteristic $p$ in quite a different way, but one which still degenerates to the usual "non-natural" count when the characteristic $p$ does not divide the group order, which is essentially the same as the characteristic zero case. No general conceptual explanation for the conjectural count of Alperin has been found to date, though a number of approaches have been suggested, including a 2-category perspective. But it is not impossible that such an explanation could one day be found, and such an explanation might shed light even on the "easy" characteristic zero situation. Later edit: In view of some of the comments below on the action of the automorphism group on irreducible characters and on conjugacy classes (which is really an action of the outer automorphism group, since inner automorphisms act trivially in each case), I make some comments on (well-known) properties of these actions, which while not identical, have many compatible features. Brauer's permutation lemma states that for any automorphism $a$ of the finite group $G,$ the number of $a$-stable complex irreducible characters of $G$ is the same as the number of $a$-stable conjugacy classses. Hence any subgroup of ${\rm Aut}(G)$ has the same number of orbits on irreducible characters as it does on conjugacy classes. The Glauberman correspondence goes further with a group of automorphisms $A$ of order coprime to $\|G\|$. In that case the $A$-actions on the irreducible characters of $G$ and on the conjugacy classes of $G$ are permutation isomorphic. While the actions of a general subgroup of the automorphism group are not always as strongly compatible as in the coprime case, various conjectures from modular representation theory suggest that it might be possible to have more compatibilty when dealing with complexes of modules than with individual modules. As a matter of speculation, I have sometimes wondered whether there might be some analogue of Glauberman correspondence in the non-coprime situation for actions on suitable complexes, although I have no idea for a precise formulation at present. Since the dimension of the center of an algebra is invariant under derived equivalence, this is one reason why I do not dismiss the idea of a more subtle explanation for numerical equalities. - 2 If I read the discussion at mathoverflow.net/questions/46900/… correctly, no bijection between the conjugacy classes and the irreducible representations respects the action of outer automorphisms. That seems like pretty convincing evidence to me. – Qiaochu Yuan Jul 22 at 21:53 mathoverflow.net/questions/21606/… seems to say the same thing. – Qiaochu Yuan Jul 22 at 21:54 1 @Qiaochu: It's a matter of opinion to some extent. That is some evidence, but I wouldn't consider it conclusive myself. – Geoff Robinson Jul 22 at 22:06 2 In my opinion, before I'd call a bijection natural, I'd want it to be invariant under isomorphisms of groups. That is, it should depend only on the group structure, not on what the specific elements of the group are. In particular, then, it should be invariant under arbitrary automorphisms, including outer ones. – Andreas Blass Jul 22 at 23:22 3 For what it's worth, even if the original straightforward version of the question has a somewhat negative answer (instead, duality...), as a general methodological attitude I endorse "not giving up tooo easily". Thus, I like Geoff R.'s points. Yet, dangit, there's that pesky factoid about (outer) automorphisms...?!? Ok, well, I myself "concede" that automorphisms' action on repns is natural in all ways that likely concern me (tho' I'm open to persuasion otherwise). So then the scope of that negative result is relevant. Still, "the orbit method"... and all that. Fun stuff... – paul garrett Jul 23 at 0:39 show 1 more comment Expanding slightly on the other answers: To ask for a "natural" bijection is presumably to ask for a natural isomorphism between two functors from the category of finite groups to the category of sets. First, we have the contravariant functor $S$ that associates to each $G$ the set of isomorphism classes of irreducible representations. Then we have the covariant "functor" $T$ that associates to each $G$ the set of its conjugacy classes. The first problem is that $T$ is not in fact functorial, because the image of a conjugacy class might not be a conjugacy class. So at the very least we should restrict to some subcategory on which $T$ is functorial, e.g. finite groups and surjective morphisms. But the key problem still remains: There is no good way to define a natural transfomation between two functors of opposite variances. So when I said in my earlier answer that "I do not see any way you can make this natural" I might better have said "This is not a situation in which the notion of naturality makes sense". All of this, of course, is really just an expansion of Gjergji's and Qiaochu's observations. - Representations can be induced (loosing irreducibleiity) so it also can thought as contr variant functor. There are many problems still exist but ..... – Alexander Chervov Sep 8 at 9:25 There is no hope for unique bijection in general but there are some "good" bijectionS in many examples although I do not know definitive characterization.... – Alexander Chervov Sep 8 at 9:32	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 78, "mathjax_display_tex": 2, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9266514778137207, "perplexity_flag": "head"}
http://mathhelpforum.com/calculus/183971-integral-area.html	# Thread: 1. ## Integral Area Hello fellow math members! I can't find this area of this two function intersecion: Y1= $x^3-x$ Y2=sin(pi*x) For $-1\leq x\leq 1$ From what I know, I need to do y1=y2 initially, right? But even that I did't get to resolve =/ Any help? ^^ Here is the answer: Spoiler: Area=(8+pi)/pi Thanks! 2. ## Re: Integral Area A graphical representation of the intersections of the two functions would be sufficient in my opinion, but if you want to be slightly more rigorous: When x = -1, 0, 1 then both $x^3-x = x(x-1)(x+1)$ and $sin([pi x)$ = 0 In the interval (-1,0), $sin([pi x) > 0$ and $x^3-x < 0$ so the two functions cannot be equal. In the interval (0,1), "" < 0 and "" > 0 so the two functions cannot be equal. Hence the only intersection points are -1, 0, 1. then set up your integrals. one will be from -1 to 0 with $sin(pi x) - (x^3-x)$ and the other will be from 0 to 1 with $(x^3-x) - sin(pi x)$. After solving both, add them to get your total area. 3. ## Re: Integral Area Thanks a lot mate! Just did it!	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 8, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8863198161125183, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/76186/definition-of-injective-function/76189	# Definition of injective function From wikipedia I obtain the following definition of an injective function : Let $f$ be a function whose domain is a set $A$. The function $f$ is injective if for all $a$ and $b$ in $A$, if $f(a) = f(b)$, then $a = b$; that is, $f(a) = f(b)$ implies $a = b$. From this I conclude that a function $f$ is injective if the below statement is true for all $a,b \in A$: $$f(a)=f(b) \implies a=b$$ My question is: Can I re-formulate the above statement as $f(a)=f(b) \iff a=b$ ? - 4 $a = b \implies f(a) = f(b)$ is already part of the definition of a function since a function assigns exactly one output to each input. – user17762 Oct 26 '11 at 21:10 Thus: Yes, your reformulation will also work. – Henning Makholm Oct 26 '11 at 21:12 1 Sivaram: that's not really part of the definition of functions... If $a$ is equal to $b$, then how could possibly $f(a)$ be not equal to $f(b)$? That's waaaay below, in the rules of manipulation of equality. (notice that for all this to make sense one has to define what $f(a)$ is, but even with one-to-many 'functions' it is true that $a=b\implies f(a)=f(b)$!) – Mariano Suárez-Alvarez♦ Oct 26 '11 at 21:15 Part of the definition of a function is $(a,b) \in f$ and $(a,c) \in f$ then $b=c$. This is exactly the condition $a=b$ implies $f(a)=f(b)$. – Bill Cook Oct 26 '11 at 21:27 1 @Arturo: but that is a different statement alltogether! :D The meaning of the notation $f(a)$ cannot depend on whether the relation $f$ is or not a function, if the implication $a=b\implies f(a)=f(b)$ is to have sense as part of the definition of what it means for a relation to be a function. I claim that for any definition of what the notation $f(a)$ means for a relation $f$ and an element $a$, the implication will hold for all relations. – Mariano Suárez-Alvarez♦ Oct 26 '11 at 21:31 show 12 more comments ## 4 Answers Yes, but the implication $a=b\Rightarrow f(a)=f(b)$ holds because you have a function. So in a sense it is redundant. That is, "$f$ is a function" implies "$a=b\Rightarrow f(a)=f(b)$". So $$\begin{align} f\text{ is an injective function} &\text{is equivalent to } f\text{ is a function and f is injective}\\ &\text{is equivalent to } f\text{ is a function and } f(a)=f(b)\Rightarrow a=b\\ &\text{implies }\Bigl( (a=b\Rightarrow f(a)=f(b))\text{ and }(f(a)=f(b)\Rightarrow a=b)\Bigr)\\ &\text{is equivalent to } f(a)=f(b)\Leftrightarrow a=b. \end{align}$$ Conversely, since "$f$ is a function and $a=b \Rightarrow f(a)=f(b)$" is equivalent to "$f$ is a function", you also have the implication going the other way provided you know that $f$ is a function. - Yes. Let us see why: $f$ is a function if: 1. $f\subseteq A\times B$ for some sets $A, B$. That is the elements of $f$ are ordered pairs. 2. For all $a\in A$ there exists $b\in B$ such that $(a,b)\in f$. 3. For every $a\in A$ if $b,c\in B$ such that $(a,b)\in f$ and $(a,c)\in f$ then $b=c$. We then denote this unique $b$ as $f(a)$. This means that if $x=y$ then $f(x)=f(y)$ by the definition of a function. Therefore for injective functions, it is enough to require $f(x)=f(y)\Rightarrow x=y$, since by the definition of $f$ as a function we have the other direction. - Yes, you can. You can formally prove that if a=b, then f(a)=f(b), where f denotes any unary predicate, as follows: ````1 \|a=b hypothesis 2 \|f(a)=f(a) equality (identity) introduction 3 \|f(a)=f(b) equality elimination 1, 2, or replacing "a" on the right by "b" 4 If a=b, then f(a)=f(b) 1-3 conditional introduction ```` So, if f also denotes a function, then a=b implies f(a)=f(b). Thus, f(a)=f(b)⟹a=b can get reformulated as f(a)=f(b)⟺a=b - Yes. $a=b$ $\Longrightarrow$ $f(a)=f(b)$ means the function $f$ is well-defined. By definition every function is well-defined. Example: $f:\mathbb{Q} \to \mathbb{Z}$ "defined" by $f(a/b)=a$ (the numerator "function") is not well-defined since $4/2=6/3$ but $f(4/2)=4$ and $f(6/3)=6$. The same input gave two different outputs. Thus $f$ is not well-defined. It's not a function! - The sentence "$f$ is well-defined" makes no sense: if something is not well-defined, it cannot even be the subject of a sentence. What one really means is that "the relation $f$ which we have defined is a function". – Mariano Suárez-Alvarez♦ Oct 26 '11 at 21:18 1 And notice that if $f\subseteq X\times Y$ is a any relation whatsoever from $X$ to $Y$ and you define, as usual, for each $x\in X$ that $f(x)=\{y\in I:(x,y)\in R\}$, then it is also true that $x=x'\implies f(x)=f(x')$... so that implication is not characteristic of functions, really. – Mariano Suárez-Alvarez♦ Oct 26 '11 at 21:21 I stand behind the term "well-defined" -- this is common usage. $f$ and the relation defining $f$ are commonly identified. – Bill Cook Oct 26 '11 at 21:21 @MarianoSuárez-Alvarez I don't follow your second comment. Do you mean that the implication fails to make sure that $f$ is defined for all elements of the domain? I wasn't claiming that "well-defined" is synonymous with "$f$ is a function". It's just part of the definition. – Bill Cook Oct 26 '11 at 21:24 I am saying that for the implication $x=y\implies f(x)=f(y)$ to be part of the definition of a relation being a function, then the symbol $f(x)$ has to be defined for all relations, not just functions (for otherwise the definition wouldbe circular) I claim that for any definition you give of what $f(x)$ means for a relation $f\subseteq X\times Y$ and an element $x\in X$, the implication will hold independently of whether the relation is or not a function. – Mariano Suárez-Alvarez♦ Oct 26 '11 at 21:26 show 3 more comments	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 83, "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": 0.9192168116569519, "perplexity_flag": "head"}
http://mathhelpforum.com/differential-equations/83152-solve-following-differential-equation.html	Thread: 1. solve the following differential equation I am supposed to solve the following zy'' + (z-2)y' -3y = 0 and y=z^3 is a solution. I have a feeling I should use the Wronskian but I don't know how because my text book is horrible for explaining anything. 2. Originally Posted by sarah232 I am supposed to solve the following zy'' + (z-2)y' -3y = 0 and y=z^3 is a solution. I have a feeling I should use the Wronskian but I don't know how because my text book is horrible for explaining anything. You can but an easy way is let $y = z^3 u$. So $y' = z^3 u' + 3z^2 u$ and $y'' = z^3u'' + 6z^2u' + 6z u$. Now substitute into your ODE and it becomes $z(z^3u'' + 6z^2u' + 6z u)+(z-2)(z^3 u' + 3z^2 u) - 3 z^3 u = 0$ which becomes, after some simplification $z u'' + (z+4)u' = 0$ If you let $v = u'$ then we obatin $z v' + (z+4)v = 0$ (separable). Integrate, put back in $u'$ and integrate again to get u. Then $y = z^3 u$. 3. Do you mind showing me the Wronskian method? I believe I will need to know this for my exam. Thanks! 4. Originally Posted by sarah232 Do you mind showing me the Wronskian method? I believe I will need to know this for my exam. Thanks! Here a post from this forum on that http://www.mathhelpforum.com/math-he...e-problem.html	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 9, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": 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": 0.9169216752052307, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/18702/why-was-the-universe-in-a-extraordinarily-low-entropy-state-right-after-the-big/18743	Why was the universe in a extraordinarily low-entropy state right after the big bang? Let me start by saying that I have no scientific background whatsoever. I am very interested in science though and I'm currently enjoying Brian Greene's The Fabric of the Cosmos. I'm at chapter 7 and so far I understand most of general ideas he has talked about. Or at least, I think I understand them :-) There is however one part, at the end of chapter 6 that I can't grasp. It is about entropy and the state of the universe a few minutes after the big bang. On page 171 he says: Our most refined theories of the origin of the universe -our most refined cosmological theories- tell us that by the time the universe was a couple of minutes old, it was filled with a nearly uniform hot gas composed of roughly 75 percent hydrogen, 23 percent helium, and small amounts of deuterium and lithium. The essential point is that this gas filling the universe had extraordinarily low entropy. And on page 173-174: We have now come to the place where the buck finally stops. The ultimate source of order, of low entropy, must be the big bang itself. In its earliest moments, rather than being filled with gargantuan containers of entropy such as black holes, as we would expect from probabilistic considerations, for some reason the nascent universe was filled with a hot, uniform, gaseous mixture of hydrogen and helium. Although this configuration has high entropy when densities are so low that we can ignore gravity, the situation is otherwise when gravity can't be ignored; then, such a uniform gas has extremely low entropy. In comparison with black holes, the diffuse, nearly uniform gas was in an extraordinarily low-entropy state. In the first part of the chapter Brian Greene explains the concept of entropy with tossing the 693 pages of War and Peace in the air. At first, the pages are ordered. The specific order they are in make sense and are required to recognize the pages as a readable book called War and Peace. This is low entropy. It is very highly ordered and there is no chaos. Now, when you throw the pages in the air, let them fall and then pick them up one by one and put them on top of each other, the chances you get the exact same order as the initial state are extremely small. The chance you get another order (no matter what order, just not the one from the beginning) is extremely big. When the pages are in the wrong order, there is high entropy and a high amount of chaos. The pages are not ordered and when they are not ordered you would not notice the difference between one unordered state and another one. However, should you swap two pages in the ordered, low entropy version, you would notice the difference. So I understand low entropy as a highly ordered state with low chaos in which a reordering of the elements would be noticeable. I hope I'm still correct here :-) Now, what I don't understand is how a uniform mixture of hydrogen and helium can by highly ordered? I'd say you wouldn't notice it if some particles traded places. I'd say that a uniform mixture is actually in a state of high entropy because you wouldn't notice it if you swapped some hydrogen atoms. Brian Greene explains this would indeed be the case when gravity plays no important role, but that things change when gravity does play a role; and in the universe right after the big bang, gravity plays a big role. Is that because a reordering of the particles would change the effects of gravity? Or is there something else that I'm missing here? - Hi Kristof, this site is devoted to research related questions by professional physicists. I will redirect your question to physics.se where it hopefully get some good answers. – user566 Dec 24 '11 at 7:53 – Qmechanic♦ Dec 24 '11 at 9:34 5 Answers Your specific question is about why uniform gas is a low entropy state for the universe. The reason is that you make entropy by allowing the gas to self-gravitate and compress, releasing heat to the environment in the process. The end result is a black hole where the gas is compressed maximally, and these are the maximum entropy gravitational states. But the uniform gas comes from a nearly uniform inflaton field shaking over all space after inflation. This inflaton produces uniform denisty of matter, which then becomes uniform baryons and hydrogen. Ultimately, it is the uniformity of the energy density in the inflaton field which is responsible for the low entropy of the initial conditions, and this is linked to the dynamics of inflation. The dynamics of inflation produce low entropy initial conditions without fine tuning. This seems like a paradox, because low entropy is fine tuning by definition, don't you need to choose a special state to have low entropy? The answer in inflation is that the state is only special in that there is a large positive cosmological constant, but it is otherwise generic, in that it is a maximum entropy state given the large cosmological constant. The theory of inflation explains the specialness of the initial conditions completely. This was proposed by Davies in 1983, but it is rejected by cosmologists. The rest of this answer discusses arguments that support Davies' position. deSitter space If you consider a deSitter space with some mass density added, and you look in a causal patch (meaning what one observer can see), the mass density gives an additional curvature without (significant) pressure and turns deSitter into more like a sphere. There is a continuous deformation of deSitter space into the Einstein static universe, which is obtained by making the density of matter as large as possible. Any matter you add reduces the horizon area of the cosmological horizon, and this is true for black holes as well. If you consider the ds-Schwartschild exact solution, for example, you can have an isolated black hole in deSitter space: $$ds^2 = - f(r) dt^2 + {dr^2\over f(r) } + r^2 d\Omega^2$$ $$f(r) = 1 - {2m\over r} - {\Lambda r^2\over 3}$$ but there are two horizons, and the causal patch is the region between the black hole and the cosmological horizon. It is easy to check that the total horizon area, cosmological plus black-hole is maximum for m=0. It is also easy to check that there is a certain value of m where the black hole radius and the cosmological radius degenerate. At this degeneration, the distance between the black hole and cosmological horizon stays constant, they do not collide except in the bad r coordinate, and the space turns into AdS_2 x S_2. Nariai dynamics Imagine starting near a Nariai solution with additional matter between the two horizons. These are both still black holes, neither is a cosmological horizon, as you can see by adding more matter with a uniform density, until you approach the limit of the Einstein static universe with two antipodal black holes. This is a physical configuration of the static cosmology. So you can start with an Einstein static universe, and evolve it forward in time, you will produce black holes, and they will merge and grow. If you take all the matter in the static universe and push it into one of the black holes, this black hole area will increase past the Nariai limit and it will become the cosmological horizon. At this point, the singularity runs away to infinity. If you push the matter into another black hole, the other black hole will be the cosmological horizon. It's up to you. So if you start with the Einstein static universe, the black holes compete for mattter, until eventually the biggest black hole will surround all the others, and become the cosmological horizon. The lessons are the following: • Cosmological horizons are the same stuff as black hole horizons. Their other side is described by black hole complementarity, just as for black holes. It is wrong to think of the universe in a global picture. • deSitter space is the maximum entropy configuration of a positive cosmological constant universe, everything else eventually thermalizes into deSitter space. • The global picture of black holes is not particularly physical, because the singularity of the Nariai solution runs away to infinity in the Nariai limit. There are cases where black hole interior structure degenerates. Inflation Produces Low Entropy Initial Conditions The second point answers your question, because the early universe is in a deSitter phase. So given a large positive value of the cosmological constant in the early universe, the maximum entropy state is a deSitter space with a cosmological horizon of small area, and this is necessarily a low entropy initial condition for later times, during which the cosmological horizon grows. There is no further explanation required for the low-entropy initial conditions. This is the same explanation as for all the other miracles of inflation, the killing of fluctuations, the flatness condition, the monopole problem. The whole point of inflation is to produce a theory of low entropy initial conditions, including gravity, and it does so naturally, because deSitter space is the only low entropy maximal entropy gravitational state. This answer was first given by Davies, and it is just plain correct. This plain-as-the-nose-on-your-face idea is not accepted despite the nearly thirty years since Davies' paper. I should add that Tom Banks and Leonard Susskind both now say similar things, although I don't want to put words in their mouths. - To say that the universe had a very low entropy just after the big bang is a bit misleading without saying what you're comparing it with. We know (well we're pretty sure) that black holes have an entropy, and that this entropy is enormous. Have a look at http://en.wikipedia.org/wiki/Black_hole_thermodynamics for how to calculate the entropy of a black hole. If you take some region of a uniform hot gas and you compress it into a black hole then the entropy goes up enormously, so the entropy of the universe just after the big bang was much much lower as a hot gas than as a assortment of black holes. This isn't the same as saying the entropy was low in any absolute sense. It's certainly true that the entropy of for example a given (small) mass of ice is a lot lower than the entropy of the same mass of steam, so you'd expect that if you take some mass of the early universe and allow it to condense into e.g. a planet, then the entropy should go down. Brian Greene's point is that this is only true if you ignore the fact that by concentrating the mass into a small area you are increasing the gravitational potential in that area. The increased gravitational potential makes another contribution to the entropy that you need to include when working out the total entropy. It's entirely reasonable to ask what is the physical mechanism for gravity's contribution to the energy, but no-one knows. In both string theory and loop quantum gravity you can construct models for the entropy of a black hole and come up with the correct answer, but of course the physical origins of the entropy are different in these two cases, so which do you believe (if either)? - This is not correct at all. If you consider the inflation state and let the inflaton be constant (whcih is approximately true in the earliest stages) you have deSitter space. You just cannot put a black hole into a deSitter space of horizon area A without decreasing the cosmological horizon area by more than A. Further, if you try to pack the contents of the early universe into a black hole, there is an upper limit to the black hole size, the Nariai solution is the maximum area. The deSitter thermal state is both the maximal entropy at fixed inflaton, and the universe's initial condition. – Ron Maimon Dec 24 '11 at 13:37 I don't think he's correct about the entropy being low in this case. If the gravity were distributed unevenly, that could be different, but with the uniform gas distribution, gravity must have been uniform as well. I'm not even sure that we could talk about this weird world we had in the beginning of times in nowadays terms. After all, "order" is a human concept (War and Peace is readable only by humans, "objectively" the "ordered" pages are just as good as "unordered"). - I think the following definition of entropy might be enlightening in the statistical thermodynamics paragraph: $$S=k_b\ln(\Omega)$$ $S$, the entropy, is proportional to the natural logarithm of the number of microstates, $\Omega$. You could think of it this way: At the Big Bang there is only one microstate, since everything is a point in space and time. At some time after the Bang microstates appear. In the model you are discussing these are the number of states that the hydrogen and helium atoms can have. The complexity of these states, and therefore the number of possible microstates, is much smaller than if the full nuclear spectrum were available, for example, or the present extent in space time . The more phase space the more microstates are available to be counted in $\Omega$. Now the comment you quote: Although this configuration has high entropy when densities are so low that we can ignore gravity, the situation is otherwise when gravity can't be ignored; then, such a uniform gas has extremely low entropy. In comparison with black holes, the diffuse, nearly uniform gas was in an extraordinarily low-entropy state. is cryptic for me. Certainly with respect to black holes a uniform gas has lower entropy, and certainly the existence of a strong gravitational field will constrain and stratify a diffuse gas, and stratification reduces the number of available microstates, so that is what I would keep from this quote. If you keep in mind that order means less available microstates because of the constraints order introduces, and disorder more available microstates because constraints are loosened you cannot go wrong. - The above answers are very interesting and informative, but not all perhaps ideally suited to someone with, by their own admission, no scientific background. Kristof, you can think of entropy as a measure of the "typicalness" of a system's arrangement, be it a pack of cards or a bottle (or universe) full of gas molecules. There are a relatively small number of "special" arrangements, such as cards ascending order, and many orders of magnitude more typical or random arrangements. In the absence of gravity, the typical arrangement of gas after the Big Bang is a fairly uniform distribution, as it would be in a bottle in thermal equilibrium. But introduce gravity and the tables are turned - What was high entropy, or a typical distribution, suddenly becomes anything but! The reason of course is that with gravity the typical distribution is more clumpy, and in an equilibrium state comprises merely black holes. -	{"extraction_info": {"found_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": 3, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9464859366416931, "perplexity_flag": "head"}
http://www.physicsforums.com/showthread.php?s=7fc32e8edb5b013530451976eb57decb&p=4288780	Physics Forums ## a charged segment of a ring? Hi everyone, i have been looking at this problem for a few days now and i am pretty stuck, ill explain why first the problem : a charge Q is distributed over a ring section of radius r, between angles 90 and 180. find the V and E at the origin. Now the problem im having ais everything that is kind of similar to this problem or about charged rings is only for uniformly charged rings and i cant find anything about just a segment, and in those cases i think E at the center is 0. Also every where i look im seeing a different formula for E of a charged ring. I am pretty confused 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 i also dont understand how only one segment and in this case the 2nd quadrant can be charged.. $$\vec{E}=\int\frac{\lambda\hat{r}}{r^2}ds$$ Where $\lambda$ merely represents the charge distribution $Q$, and $\hat{r}$ represents a unit vector perpendicular to each segment pointing towards the origin. You want to take the line integral on the interval $(\pi/2, \pi)$. Normally, the electric field would be zero in the center of a circle with uniform charge distribution. However, it is not in this instance. ## a charged segment of a ring? thank you, what source did you get that equation from id like to read more about it i see 1/4pieo and to the power of 3/2 in alot of ring equations thats was confusing me is the difference between those and this one Not entirely sure what level you are at, but this was in both my vector calculus and electrostatics textbook. You may find variations depending on how the charge is distributed (line, surface, etc). College, and the one in my Electromagnetics book is E=$\hat{z}$[h/(4$\pi$ε0(b^2+h^2)^3/2)]Q where b is the radius and h is the height, but like i said its for a uniform charge Recognitions: Gold Member Science Advisor Quote by Octavius1287 College, and the one in my Electromagnetics book is E=$\hat{z}$[h/(4$\pi$ε0(b^2+h^2)^3/2)]Q where b is the radius and h is the height, but like i said its for a uniform charge This is probably what it's referring to. I don't know if this will be helpful, but take a look. http://hyperphysics.phy-astr.gsu.edu...ic/elelin.html Regards ya all those make sense to me, but i read that page and once again it said it was only for a uniform charge Recognitions: Gold Member Science Advisor Quote by Octavius1287 ... i read that page and once again it said it was only for a uniform charge Yes. Quote by Octavius1287 ...problem : a charge Q is distributed over a ring section of radius r, between angles 90 and 180. This doesn't mean it is NON-uniform for the rest of the ring. Aren't they looking for the field from the contribution of charge on just this piece of the ring? If so, Quote by sandy.bridge ...You want to take the line integral on the interval $(\pi/2, \pi)$. Thread Tools \| \| \| \| \|---------------------------------------------------\|-------------------------------\|---------\| \| Similar Threads for: a charged segment of a ring? \| \| \| \| Thread \| Forum \| Replies \| \| \| Advanced Physics Homework \| 1 \| \| \| Introductory Physics Homework \| 1 \| \| \| Introductory Physics Homework \| 11 \| \| \| Mechanical Engineering \| 7 \| \| \| Advanced Physics Homework \| 1 \|	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 9, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9404860734939575, "perplexity_flag": "middle"}
http://mathoverflow.net/revisions/67272/list	## Return to Answer 3 added 9 characters in body If the proof system is recursively axiomatizable, this situation cannot occur. If there exists a proof of $\Theta$, there exists an algorithm to find that proof. Namely, search the recursively enumerable set of deductions until you find a proof of $\Theta$. This must terminate, as we have proved that $\Theta$ is provable. If the proof system is NOT recursive, then this may be possible. Consider the following set of axioms $\Sigma$ in the signature of arithmetic. Let $A$ be an infinite set which does not contain any infinite r.e. set. Define $$\Sigma = \{ (\bar k = \bar k) \wedge \sigma \mid k \in A, k > \ulcorner \sigma \urcorner, \mathfrak N \models \sigma \}$$ Now, note that any sentence provable in $Th(\cal N)$ is provable is in $\Sigma$. However, there is no algorithm to transform produce such proofs from $\Sigma$. To do so would require enumerating arbitrarily large elements of $A$, which is impossible 2 added 574 characters in body; added 9 characters in body; added 16 characters in body If the proof system is recursively axiomatizable, this situation cannot occur. If there exists a proof of $\Theta$, there exists an algorithm to find that proof. Namely, search the recursively enumerable set of deductions until you find a proof of $\Theta$. This must terminate, as we have proved that $\Theta$ is provable. If the proof system is NOT recursive, then this may be possible. Consider the following set of axioms $\Sigma$ in the signature of arithmetic. Let $A$ be an infinite set which does not contain any r.e. set. Define $$\Sigma = \{ (\bar k = \bar k) \wedge \sigma \mid k \in A, k > \ulcorner \sigma \urcorner, \mathfrak N \models \sigma \}$$ Now, note that any sentence provable in $Th(\cal N)$ is provable is in $\Sigma$. However, there is no algorithm to transform produce such proofs from $\Sigma$. To do so would require enumerating arbitrarily large elements of $A$, which is impossible 1 If the proof system is recursively axiomatizable, this situation cannot occur. If there exists a proof of $\Theta$, there exists an algorithm to find that proof. Namely, search the recursively enumerable set of deductions until you find a proof of $\Theta$. This must terminate, as we have proved that $\Theta$ is provable.	{"extraction_info": {"found_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": 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": 0.9428767561912537, "perplexity_flag": "head"}
http://johncarlosbaez.wordpress.com/2012/07/28/the-noisy-channel-coding-theorem/	# Azimuth ## The Noisy Channel Coding Theorem Here’s a charming, easily readable tale of Claude Shannon and how he came up with information theory: • Erico Guizzo, The Essential Message: Claude Shannon and the Making of Information Theory. I hadn’t known his PhD thesis was on genetics! His master’s thesis introduced Boolean logic to circuit design. And as a kid, he once set up a telegraph line to a friend’s house half a mile away. So, he was perfectly placed to turn information into a mathematical quantity, deeply related to entropy, and prove some now-famous theorems about it. These theorems set limits on how much information we can transmit through a noisy channel. More excitingly, they say we can cook up coding schemes that let us come as close as we want to this limit, with an arbitrarily low probability of error. As Erico Guizzo points out, these results are fundamental to the ‘information age’ we live in today: Can we transmit, say, a high-resolution picture over a telephone line? How long will that take? Is there a best way to do it? Before Shannon, engineers had no clear answers to these questions. At that time, a wild zoo of technologies was in operation, each with a life of its own—telephone, telegraph, radio, television, radar, and a number of other systems developed during the war. Shannon came up with a unifying, general theory of communication. It didn’t matter whether you transmitted signals using a copper wire, an optical fiber, or a parabolic dish. It didn’t matter if you were transmitting text, voice, or images. Shannon envisioned communication in abstract, mathematical terms; he defined what the once fuzzy concept of “information” meant for communication engineers and proposed a precise way to quantify it. According to him, the information content of any kind of message could be measured in binary digits, or just bits—a name suggested by a colleague at Bell Labs. Shannon took the bit as the fundamental unit in information theory. It was the first time that the term appeared in print. So, I want to understand Shannon’s theorems and their proofs—especially because they clarify the relation between information and entropy, two concepts I’d like to be an expert on. It’s sort of embarrassing that I don’t already know this stuff! But I thought I’d post some preliminary remarks anyway, in case you too are trying to learn this stuff, or in case you can help me. There are various different theorems I should learn. For example: • The source coding theorem says it’s impossible to compress a stream of data to make the average number of bits per symbol in the compressed data less than the Shannon information of the source, without some of the data almost certainly getting lost. However, you can make the number of bits per symbols arbitrarily close to the Shannon entropy with a probability of error as small as you like. • the noisy channel coding theorem is a generalization to data sent over a noisy channel. The proof of the noisy channel coding theorem seems not so bad—there’s a sketch of a proof in the Wikipedia article on this theorem. But many theorems have a hard lemma at their heart, and for this one it’s a result in probability theory called the asymptotic equipartition property. You should not try to dodge the hard lemma at the heart of the theorem you’re trying to understand: there’s a reason it’s there. So what’s the asymptotic equipartition property? Here’s a somewhat watered-down statement that gets the basic idea across. Suppose you have a method of randomly generating letters—for example, a probability distribution on the set of letters. Suppose you randomly generate a string of $n$ letters, and compute $-(1/n)$ times the logarithm of the probability that you got that string. Then as $n \to \infty$ this number ‘almost surely’ approaches some number $S.$ What’s this number $S$? It’s the entropy of the probability distribution you used to generate those letters! (Almost surely is probability jargon for ‘with probability 100%’, which is not the same as ‘always’.) This result is really cool—definitely worth understanding in its own right! It says that while many strings are possible, the ones you’re most likely to see lie in a certain ‘typical set’. The ‘typical’ strings are the ones where when you compute $-(1/n)$ times the log of their probability, the result is close to $S.$ How close? Well, you get to pick that. The typical strings are not individually the most probable strings! But if you randomly generate a string, it’s very probable that it lies in the typical set. That sounds a bit paradoxical, but if you think about it, you’ll see it’s not. Think of repeatedly flipping a coin that has a 90% chance of landing heads up. The most probable single outcome is that it lands heads up every time. But the typical outcome is that it lands up close to 90% of the time. And, there are lots of ways this can happen. So, if you flip the coin a bunch of times, there’s a very high chance that the outcome is typical. It’s easy to see how this result is the key to the noisy channel coding theorem. The ‘typical set’ has few elements compared to the whole set of strings with $n$ letters. So, you can make short codes for the strings in this set, and compress your message that way, and this works almost all the time. How much you can compress your message depends on the entropy $S$. So, we’re seeing the link between information and entropy! The actual coding schemes that people use are a lot trickier than the simple scheme I’m hinting at here. When you read about them, you see scary things like this: But presumably they’re faster to implement, hence more practical. The first coding schemes that come really close to the Shannon limit are the turbo codes. Surprisingly, these codes were developed only in 1993! They’re used in 3G mobile communications and deep space satellite communications. One key trick is to use, not one decoder, but two. These two decoders keep communicating with each other and improving their guesses about the signal they’re received, until they agree: This iterative process continues until the two decoders come up with the same hypothesis for the m-bit pattern of the payload, typically in 15 to 18 cycles. An analogy can be drawn between this process and that of solving cross-reference puzzles like crossword or sudoku. Consider a partially completed, possibly garbled crossword puzzle. Two puzzle solvers (decoders) are trying to solve it: one possessing only the “down” clues (parity bits), and the other possessing only the “across” clues. To start, both solvers guess the answers (hypotheses) to their own clues, noting down how confident they are in each letter (payload bit). Then, they compare notes, by exchanging answers and confidence ratings with each other, noticing where and how they differ. Based on this new knowledge, they both come up with updated answers and confidence ratings, repeating the whole process until they converge to the same solution. This can be seen as “an instance of loopy belief propagation in Bayesian networks.” By the way, the picture I showed you above is a flowchart of the decoding scheme for a simple turbo code. You can see the two decoders, and maybe the loop where data gets fed back to the decoders. While I said this picture is “scary”, I actually like it because it’s an example of network theory applied to real-life problems. This entry was posted on Saturday, July 28th, 2012 at 2:26 am and is filed under information and entropy, mathematics, probability. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site. ### 15 Responses to The Noisy Channel Coding Theorem 1. davidtweed says: What you’re looking at here is (AIUI) general codes where one is receives all the transmitted “symbols” but there is generally some garbling going on. Another model is the “erasure codes”, where it’s assumed you can “check” (with eg a checksum) each symbol that you receive, so you KNOW if it’s correct or not but you may not receive all the symbols. (Of course these are the ends of a spectrum, eg, erasure could be said to be the case where garbled symbols are “completely” garbled.) In both cases you want to reconstruct an original message from a “redundancy via coding” message that’s gone over the unreliable channel. The interesting thing is that slightly different mathematics appears to apply best in the different cases, eg, consider “fountain codes” such as described in the last chapter of McKay’s book. There it considers using random graphs to encode, then reconstruct, a message from an arbitrary set of packets that’s just bigger than the minimal amount needed.It would be interesting to see what you could figure out about the representing a “set” of messages (eg, a set of viable genomes) via one of these kind of processes since it might be related to “useful” within species diversity. (It might be useful to have some different genes within a population, but for many — although not all — genetic ailments there is no benefit associated with it. So you want some set, but not all, variants to be “recoverable” from the population of animals in a given species.) • John Baez says: David wrote: What you’re looking at here is (AIUI) general codes where one is receives all the transmitted “symbols” but there is generally some garbling going on. That’s right—that’s the model here. Another model is the “erasure codes”, where it’s assumed you can “check” (with eg a checksum) each symbol that you receive, so you KNOW if it’s correct or not but you may not receive all the symbols. Hmm! Now you’re making me wonder what’s the most general model of ‘garbled’, ‘partially erased’, or otherwise screwed-up data. Category-theoretic computer scientists and logicians have extremely general theories of data types, which I sort of understand, but I’ve never tried to imagine an extremely general theory of ‘screwed-up’ or ‘damaged’ data. It seems information theorists are often content to think of data as a string of symbols drawn from a pre-established alphabet. Clearly anything discrete can be encoded in this way, but is this model sufficiently general for thinking about the ways data can be damaged? It’s easy to imagine a string being ‘garbled’ with some symbols replaced by others, and it’s easy to imagine it being ‘partially deleted’. As you note, the latter can be conceived as a special case of the former if we allow a symbol ‘blank’ and think of deletion as replacing some other symbol by a blank. But are there other fundamentally different ways to damage data… or is there some theorem saying that all other ways can be reduced to ‘garbling’? And when I say “information theorists are often content to think of data as a string of symbols drawn from a pre-established alphabet” maybe I’m being old-fashioned. Maybe they do more general things I don’t know about. For example, maybe I’m stuck back in the era where data was sent along a fixed ‘channel’, and now I should be thinking about data moving around in ‘packets’… so that one form of ‘garbling’ could be getting a packet that should have gone to someone else, or getting a packet really late, or something like that! Examples chosen from biology could force us to think about information transmission in even more general ways. Just because you can model data as a string of symbols in some alphabet, doesn’t mean you should. There’s information in the cytoplasm of a human egg cell, which is important for the development of the fetus. But modeling it as a string of symbols seems awkward (while for the genome, it’s not quite so bad). • Tobias Fritz says: John said: Maybe they do more general things I don’t know about. Having a single communication channel is very simplistic: there is one sender and one receiver which share a channel. For our deeply connected world, this scenario is highly unrealistic! Nowadays, communication is organized in big networks formed by lots of nodes and connections between these. Such a network typically have many senders and many receivers. In such situations, capacity ceases to be a single number; rather, it becomes an entire region in the space of matrices $R_{ij}$, the entries of which stand for the amount of information transmitted from sender $i$ to receiver $j$ when using a certain coding scheme. It’s called the ‘rate region’. The butterfly network is a beautiful example of how coding in a network can work — and why the internet, with its routing-based approach, is likely to be in a pretty mediocre part of the rate region. I’m mentioning all this because it’s obviously related to Bayesian networks, and maybe even to other aspects of your work on network theory, I don’t know. • John Baez says: Thanks! More to learn and think about. I’d like to combine network theory with information theory, since green mathematics, if such a thing exists, should involve both. • Robert says: DNA copying can also be thought of as data transmission. Besides point transcription and deletion errors, corresponding to those already described, chunks of DNA can get duplicated, or moved around. – ‘genes 1,2,3′ becoming ‘genes 1,2,2,3′ or ‘genes `1,3,2′. I don’t think there’s any equivalent for messages sent over serial channels. Tangentially, the types of errors you see can indicate how the data is being stored. If ‘riddle’ is misspelt ‘ridlle’, that suggests an underlying representation as r-i-double-d-l-e, with a single error producing r-i-d-double-l-e. • Jamie Vicary says: In quantum error correction, people often consider two primary types of errors: bit flips, where $\| 0 \rangle$ and $\| 1 \rangle$ switch; and sign flips, where $\| 0 \rangle$ is unchanged but $\| 1 \rangle$ is multiplied by ${-}1$. These are conveniently represented by the Pauli $X$ and $Z$ matrices. There are results saying that as long as you can correct adequately for these discrete sorts of errors, you can correct for all errors in some sense. I wonder if there are classical analogues of these sorts of result. Tobias, I agree, it’s really exciting to think about how all this stuff generalizes to topologically interesting situations. It’s entrancing to see linear algebra coming into play in the butterfly network. • John Baez says: Let’s keep talking about all this stuff. By the way, Jamie, speaking of error correction… Everyone please remember: on all WordPress blogs, LaTeX is done like this: \$latex E = mc^2\$ with the word ‘latex’ directly following the first dollar sign, no space. Double dollar signs don’t work here. 2. romain says: Noteworthy, the asymptotic equipartition property (AEP) shows that the plug-in estimate of entropy converges to the true entropy in the large sampling limit. This is interesting with respect to the post “Mathematics of biodiversity 7” to understand the whole concept of bias correction of entropy estimates in the case of a limited sample. Consider the quantity $\displaystyle{H_n=-\frac{1}{n} \log p(X_{i_j}^n)}$ where $X_{i_j}^n$ is a sequence of $n$ letters $X_i$ occurring with probabilities $p_i$. And let’s assume there are $m$ different letters.. In the case where the letters are i.i.d., then $\displaystyle{p(X_{i_j}^n)=\prod_j^n p(X_{i_j})}$ Now, the same $X_i$ may appear many times, say $n_i$, so this formula can be rewritten $\displaystyle{p(X_i^n)=\prod_i^m p(X_i)^{n_i}}$ so that: $\displaystyle{H_n=-\sum_i^m \frac{n_i}{n} \log p(X_i)}$ When $N$ is large, then $\displaystyle{\frac{n_i}{n} \rightarrow p(X_i)}$ QED. • John Baez says: That’s nice! Thanks! I hope I have enough energy to say more about the asymptotic equipartition property. 3. [...] So I’m boning up on the subject by reading Khinchin’s great classic, Mathematical Foundations of Information Theory. In the future I’ll share some findings here, but in the mean time, follow the link to Azimuth and read Baez’s great discussion of the Noisy Channel Coding Theorem. [...] 4. arch1 says: John, My quick skim of your summary of the hard lemma left me wondering why it is a hard lemma. Isn’t it just a baby step away from the fact that as n-> infinity, samples of size n look more and more like the underlying distribution from which they are drawn? That in itself is progress, since my quick skim of Wikipedia’s summary of the hard lemma left me almost clueless as to the content of the hard lemma (it struck me as very vague:-) • John Baez says: I didn’t mean to say the asymptotic equipartition property is extremely hard. However, the rest of the proof looks easy in comparison, so one is inclined to look at this part and say “yuck, that’s some technical fact I’d rather take on faith”. It seems like the pit in the peach. But I was trying to convince everyone that unlike the pit in the peach, it’s highly nutritious, and tasty in its own way. I stated a watered-down version of the asymptotic equipartition theorem: just for purposes of exposition, I assumed that each letter in the string was drawn independently from the same probability distribution on letters. In other words, I was assuming that they’re ‘independent identically distributed’ random variables. This is clearly too restrictive—it sure ain’t true for English text! The statement and proof gets a bit harder when we do the full-fledged thing: you can see a proof here for the i.i.d. case and here for the more general case. The more general case uses Lévy’s martingale convergence theorem, Markov’s inequality and Borel-Cantelli lemma. For some people that would be very scary, while others eat such stuff for breakfast. But anyway, regardless of whether the proof is hard, the result seems both interesting and highly believable: not shocking. • Florifulgurator says: Sounds like yummy breakfast. But no Cramer large deviation stuff? Last century I ran a seminar on Cramer’s theorem, iterated logarithm, arcsin law etc. Now suddenly it seems Shannon is yummy, too. • arch1 says: Thanks for the extra context John. 5. blake561able says: It is interesting that Weyl was Shannon’s adviser at IAS. The communication between those two must have come close to exceeding these limits on information!	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 32, "mimetex.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": 0.934399425983429, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/36545/which-simple-puzzles-have-fooled-professional-mathematicians	# Which simple puzzles have fooled professional mathematicians? Although I'm not a professional mathematician by training, I felt I should have easily been able to answer straight away the following puzzle: Three men go to a shop to buy a TV and the only one they can afford is £30 so they all chip in £10. Just as they are leaving, the manager comes back and tells the assisitant that the TV was only £25. The assistant thinks quickly and decides to make a quick profit, realising that he can give them all £1 back and keep £2. So the question is this: If he gives them all £1 back which means that they all paid £9 each and he kept £2, wheres the missing £1? 3 x £9 = £27 + £2 = £29...?? Well, it took me over an hour of thinking before I finally knew what the correct answer to this puzzle was and, I'm embarrassed. It reminds me of the embarrassement some professional mathematicians must have felt in not being able to give the correct answer to the famous Monty Hall problem answered by Marilyn Vos Savant: http://www.marilynvossavant.com/articles/gameshow.html Suppose you're on a game show, and you're given the choice of three doors. Behind one door is a car, behind the others, goats. You pick a door, say #1, and the host, who knows what's behind the doors, opens another door, say #3, which has a goat. He says to you, "Do you want to pick door #2?" Is it to your advantage to switch your choice of doors? Yes; you should switch. It's also mentioned in the book: The Man Who Only loved Numbers, that Paul Erdos was not convinced the first time either when presented by his friend with the solution to the Monty Hall problem. So what other simple puzzles are there which the general public can understand yet can fool professional mathematicians? - 18 Well, I am pretty easily fooled… The list would be too long! – Mariano Suárez-Alvarez♦ May 3 '11 at 0:27 5 I don't worry to much about being wrong becuase I'm focused on learning and really have no one I want to impress. – a little don May 3 '11 at 0:31 6 – Mitch May 3 '11 at 0:45 8 While this is nitpickery, I should note that the Monty Hall problem was posed and answered long before vos Savant came along, and the confusion and logical debates it engenders have lingered well after her 'answering' it - I'm not sure she did anything for the problem except making it marginally more well-known. – Steven Stadnicki May 3 '11 at 2:13 8 It should also be noted that the Monty Hall problem is notorious in part because slight variations change the correct answer, and vos Savant's original description of the problem was ambiguous enough to possibly include several variants with different possible solutions (e.g., not specifying clearly the host always opens a door he knows does not contain a car, always offers the choice to switch). And, how did Erdos get dragged into this? I was not aware he was part of that little imbroglio. – Arturo Magidin May 3 '11 at 3:49 show 13 more comments ## 8 Answers How about the Two envelopes problem? - Along the same lines as the Monty Hall Problem is the following (lifted from Devlin's Angle on MAA and quickly amended): I have two children, and (at least) one of them is a boy born on a Tuesday. What is the probability that I have two boys? Read a fuller analysis here. - 6 – ShreevatsaR May 3 '11 at 5:45 Actually, perhaps, accepting the argument for "13/27" without careful thought is the real mistake made by many mathematicians. :-) – ShreevatsaR May 3 '11 at 6:05 2 ShreevatsaR, suppose there are two mathematicians, and at least one of them made this mistake, and also gave it careful thought. – Dan Brumleve May 3 '11 at 8:11 @ShreevatsaR, I opted for the short version instead of getting bogged down in semantics. To me, this conveys the idea of the problem enough so that people understand what is being asked. That being said, this is the more rigorous version: You take a poll of $N$ families and discard all but families with exactly 2 children and that have at least one boy born on Tuesday. Of the restricted set, what is the proportion of families that have both children as boys as $\lim N \to \infty$? (and, of course, 1/2 prob. of a child being a boy with 1/7 prob. being born on a given day of the week) – user4143 May 3 '11 at 10:05 – ShreevatsaR May 3 '11 at 11:21 I guess this well known von Neumann anecdote fits the description of the question: The following problem can be solved either the easy way or the hard way. Two trains 200 miles apart are moving toward each other; each one is going at a speed of 50 miles per hour. A fly starting on the front of one of them flies back and forth between them at a rate of 75 miles per hour. It does this until the trains collide and crush the fly to death. What is the total distance the fly has flown? The fly actually hits each train an infinite number of times before it gets crushed, and one could solve the problem the hard way with pencil and paper by summing an infinite series of distances. The easy way is as follows: Since the trains are 200 miles apart and each train is going 50 miles an hour, it takes 2 hours for the trains to collide. Therefore the fly was flying for two hours. Since the fly was flying at a rate of 75 miles per hour, the fly must have flown 150 miles. That's all there is to it. When this problem was posed to John von Neumann, he immediately replied, "150 miles." "It is very strange," said the poser, "but nearly everyone tries to sum the infinite series." "What do you mean, strange?" asked Von Neumann. "That's how I did it!" - Does this demonstrate the calculating brilliance of Von Neuman or him lacking a creative intuition to see the beautiful, elegant solution? I also solved it as he did, but in a few minutes, only to be utterly humiliated by the true solution. Moral of the story - always check for the elegant solution first, before someone else humiliates you with it. – John McVirgo Jul 4 '12 at 20:35 1 I'd say it was that Von Neumann saw summing the infinite series as the beautiful, elegant solution! – Joe Z. Mar 13 at 14:53 I'm not sure if this qualifies as I don't have any reports of this actually tricking any mathematician, but it's a good problem. So, at the risk of violating the criteria, the following is Robert Connelly's "Say Red" (taken from Gardner's "Fractal Music, Hypercards and more", Chapter 14): The banker shuffles a standard deck of 52 cards and slowly deals them face up. The dealt cards are left in full view where they can be inspected at any time by the player. Whenever the player wants, he may say "Red." If the next card is red, he wins the game, otherwise he loses. He must call red before the deal ends, even if he waits to call on the last card. What odds should the banker give to make it a fair game, assuming that the player adopts his best strategy on the basis of feedback form the dealt cards? The player must announce the size of his bet before each game begins. - Weird! I did the maths, and it seems that your probability of winning is 1/2, whatever strategy you choose. But I can't quite believe it. – TonyK May 5 '11 at 19:06 @TonyK, that's right. That's what makes this such a wonderful puzzle. – user4143 May 6 '11 at 7:42 5 @TonyK, The other way to see it is to consider flipping over the bottom card when the player says 'red'. The probability is exactly the same as flipping the top card in the remaining pile, but obviously only 1/2. – user4143 May 10 '11 at 0:55 Perhaps not what you were aiming at, but have a look at this. Fabio Massacci provides a counterexample for a lower bound proved by Cook and Reckhow (that's the same Cook from Cook's theorem), and also referred to in several other papers (Section 5 of the paper). - "the same Cook from Cook's theorem" -- I'm afraid that doesn't help me out any. – Pete L. Clark May 10 '11 at 20:24 – Yuval Filmus May 10 '11 at 20:54 thanks. – Pete L. Clark May 10 '11 at 22:49 I don't know if this qualifies as simple. A train line from an airport to the Cantor Hotel operates in the following manner. There is a station at each ordinal number, and every station is assigned a unique ordinal. The train stops at each station, in order. At each station people disembark and board, in order, as follows: i) if any passengers are on the train, exactly $1$ disembarks, and then ii) $\aleph_0$ passengers board the train. Station $0$ is at the airport, and the Cantor Hotel is at station $\omega_1$, the first uncountable cardinal. The train starts its journey empty. $\aleph_0$ passengers board the train to the Cantor Hotel at the airport (station $0$), and off it goes. When the train pulls into the Cantor Hotel at station $\omega_1$, how many passengers are on it? - 1 Surely it depends on which passenger disembarks at which station? I believe I can arrange it so that the train has anywhere between 0 and $\aleph_1$ passengers. – Nate Eldredge Jul 4 '12 at 19:13 It's already "hard to tell" how many are on board at station $\omega$ if at each finite station two get on and one gets off the train. If odd-numbered people never leave, there are $\aleph_0$ people on on borad; but if people leave in th esame order they enter, nobody is left. – Hagen von Eitzen Feb 21 at 20:58 OK, here's one which had me stumped and when I knew the answer, I was pretty embarrassed. Even more so when I told my mother and she immediately saw the solution. And she's no mathematician. A music lover lives in a town which has two concerts every weekend. One is in the North of the town, the other in the South. The music lover lives in the centre of town and decides to take the bus to the concert. There's a bus to the North and one to the South every hour. The music lover just decides to take the first bus he sees and go to that concert. At the end of the year, he notices that he only went to the concert in the South. How come? Here's the answer: The clue is that there's a bus every hour, but I didn't specify at what time in the hour the buses leave for the North and South. Imagine that the bus for the South leaves only 5 minutes before the one for the North. Then, it is more likely that if the music lover arrives at a random time, he will take the bus to the South. The odds are 1 to 11 that he takes the one to the North. If they leave even closer to one another, the probability will be even smaller. - – Willie Wong♦ May 3 '11 at 11:11 Right, forgot about that. Also realised I made a mistake in the solution. And then I wonder why such simple stuff can stump me when I even fail at simple arithmetic. :P – Raskolnikov May 3 '11 at 11:20 I would have said that the bus to the North remains at the stop in the center longer than for the bus for the south. – user10389 May 5 '11 at 17:51 Erm... I don't even understand the solution, but there is this one: An ant starts to crawl along a taut rubber rope 1 km long at a speed of 1 cm per second (relative to the rubber it is crawling on). At the same time, the rope starts to stretch by 1 km per second (so that after 1 second it is 2 km long, after 2 seconds it is 3 km long, etc). Will the ant ever reach the end of the rope? -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 12, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9718485474586487, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/37231/are-all-parametrizations-via-polynomials-algebraic-varieties/37271	## Are all parametrizations via polynomials algebraic varieties? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Suppose that we have a parametrization via polynomials as follows: $$t\longrightarrow (f_1(t),\ldots,f_n(t)),$$ where $t$ is a vector in $\mathbb{C}^r$ and $f_i$ are polynomials of arbitrary degree. Can we always find equations such that the image is an affine algebraic variety? The question is motivated by Exercise 1.11 in Hartshorne: Let $Y\subseteq A^3$ be the curve given parametrically by $x = t^3, y= t^4, z = t^5$. Show that $I(Y)$ is a prime ideal of height 2 in $k[x,y,z]$ which cannot be generated by 2 elements. I am not interested in the exercise in particular. Finding the variety is easy sometimes, for instance $t\rightarrow (t^2,t^3)$ is given by $I(x^3-y^2)$. I am looking for a result which says that the image is always an affine algebraic variety AND a procedure to find the ideal. - 3 As others note in answers, the image is not necessarily a subvariety if you have more than one source variable---but the image is always constructible. You can calculate the Zariski closure easily by elimination (in theory---may take a long time): for Hartshorne exercise, ask a CAS to eliminate t from the list of defining parametric equations. When you are parameterizing by A^1, the image is always closed, since A^1 --> A^r is a finite map if nontrivial. In Hartshorne exercise, for example, t satisfies the monic polynomial T^3 - x over k[x,y,z]. – james-parson Aug 31 2010 at 16:15 ## 4 Answers I can't comment (b/c I'm not a registered user) but let me add: in case the dimension of the domain is 1 (as in your motivating example) the image is in fact an affine variety. To see this, note that the map can always be extended to a map from the projective line to projective space by homogenizing things (compare with Dan's example---if you tried homogenizing his map, you'd get $[x:y:z] \mapsto [xy:yz:z^2]$ which is not defined if $x=z=0$ or $y=z=0$), and use the fact that the image of a projective variety by a regular map is closed. Finally, observe that the image of the affine line you started with is precisely the intersection of the image of the homogenized map with the affine coordinate patch determined by your homogenization. Therefore the image of the map you started with is an affine variety. So Hartshorne's example is not an accident. - ### 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. Consider $f \colon {\mathbb C}^2 \to {\mathbb C}^2$ given by $(x,y) \mapsto (xy,y)$. The image consists of ${\mathbb C}^2$ minus the subset $y = 0, x \neq 0$. Since the image is not closed, it is not a variety. The notion of a constructible subset was invented to deal with questions like this. A constructible subset is one which can be constructed from subvarieties using "boolean operations". Equivalently it is a subset defined by polynomial equations and inequations. It is true that the image of a constructible subset is again constructible (Chevalley's theorem). - If I change the question slightly the answer is yes'. Changed question: ` Is the Zariski closure of the image of a polynomial map always an (affine) algebraic variety''. Explicitly finding this variety (i.e the ideal defining it) is the subject of ````elimination theory''. See ch. 4 of the book```Introduction to Algebraic Geometry' by Brendan Hasset. - 7 While what you write is correct, it's not what you intend, I think. The Zariski closure is an affine variey by definition. But the stronger fact, mentioned by Dan Petersen above, is the constructibility of the image. So the image is Zariski open inside its Zariski closure. This implies, in particular, that the 'usual' closure (with respect to the complex topology) is an affine variety. Meanwhile, all these closures are indeed computable using elimination theory. – Minhyong Kim Aug 31 2010 at 16:07 I think this is ok,. Suppose you have $t\longrightarrow (f_1(t),\ldots,f_n(t)).$ Let's call Y the image of that. Then to see if $Y$ is a variety, you can see that $I(Y)$ is prime. But let $\phi: k[X_1, \ldots, X_n] \longrightarrow k[T]$ be given by $X_i \mapsto f_i(T)$. Then $$Ker \phi = \{f: \phi(f) = 0\} = \{f: f(f_1(T), \ldots, f_n(T)) = 0\} = I(Y).$$ So $I(Y)$ is prime because it is the kernel of a map whose image is an integral ring, and then $Y$ is a variety. - 1 How do you know that the zero set of $I(Y)$ is the image of the map (and not something bigger)? – Sheikraisinrollbank Sep 22 2010 at 14:18 1 I think you have just proved that the Zariski closure of the image is irreducible, which is certainly true if the source is irreducible. But this is not what was asked; e.g. $(x,y) \mapsto (x,xy)$ does not have Zariski closed, or even locally closed, image. – Emerton Sep 22 2010 at 15:37	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 25, "mathjax_display_tex": 2, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9432336688041687, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/180429/local-coordinates-for-a-tranversal-intersection-of-a-curve-with-a-coordinate-axi?answertab=votes	# Local coordinates for a tranversal intersection of a curve with a coordinate axis Suppose I have a smooth curve $\gamma: \mathbb{R} \rightarrow \mathbb{R}^2$ in the $xy$-plane given by $t \mapsto \gamma(t)=(\gamma_1(t),\gamma_2(t))$ which intersects the $x$-axis transversely. Is it then possible to locally express $\gamma$ in terms of $\gamma_2$? I have not been able to construct a counter example yet but I have not been able to come up with a proof either. Any help is welcome. - 1 The implicit function theorem should help you out here. – Kris Aug 8 '12 at 21:57 ## 2 Answers To flesh out my comment into an answer: the condition that $\gamma$ intersects the $x$-axis transversely is equivalent to stating that $\frac{\partial \gamma}{\partial \gamma_1}$ is non-zero, and so by the implicit function theorem we can write $\gamma_1$ as a function of $\gamma_2$ and hence $\gamma$ as a function of $\gamma_2$. - Notice that formally $\partial\gamma/\partial\gamma_1=(1,0) \ne 0$ for any curve $\gamma=(\gamma_1,\gamma_2)$, therefore you are saying that every curve $\gamma=(\gamma_1,\gamma_2)$ intersects transversally the $x$-axis, which of course is not true. – Mercy Aug 8 '12 at 22:33 Guess you mean $\frac{\partial \gamma_2}{\partial \gamma_1}$ is non-zero. However, my problem is that the latter can be undefined. For example consider $\gamma=(t,t^2)$ at $t=0$. Does the implicit function theorem in these cases still hold? – user29751 Aug 9 '12 at 18:14 Let $t_0 \in \mathbb{R}$ be such that $\gamma_2(t_0)=0$. Since $\gamma$ intersects transversally the $x$-axis, we have $\dot{\gamma}_2(t_0) \ne 0$. Thanks to the inverse function theorem, there exist $\epsilon,\delta>0$ such that $\gamma_2: (t_0-\epsilon,t_0+\epsilon)\to (-\delta,\delta)$ is a diffeomorphism. Hence $\gamma(\gamma_2^{-1}(\tau))=(\gamma_1\circ\gamma_2^{-1}(\tau),\tau) \quad \forall\ \tau \in (-\delta,\delta)$. - Suppose I have the curve $\gamma:t \mapsto (t^3,t^3)$ then this curve intersects the $x$-axis transversely but $\dot{\gamma}_2(0)=0$! – user29751 Aug 9 '12 at 17:59 It's just a matter of parametrization. The curve $\gamma: t \mapsto (t^3,t^3)$ can be reparametrize as $s \mapsto (s,s)$. – Mercy Aug 9 '12 at 19:15	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 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": 0.9295934438705444, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/11293/determining-wave-function-for-term-symbol-1d	# Determining wave function for term symbol 1D I am trying to follow a book (Introduction to Ligand Field Theory by Ballhausen in 1962 on pg 15), but it isn't clear how they make a particular leap. Background I want to find the wave function for the term $^1 D$. We know that $\psi(L,M_L,S,M_S) = \psi (2,2,0,0)$ is made up of three micro states: $(2^+,0^-)$, $(2^-,0^+)$, $(1^+,1^-)$. Thus, a linear combination must be taken of these micro states. Now in the book they say they must be orthogonal to $\psi(3,2,1,0)$ and $\psi(4,2,0,0)$. Presumably these are picked because we just determined those via lowering operators in the previous section. I don't take it that there is another reason those two wave functions are mentioned. $(2^+,0^−)$ means that electron 1 has $m_s=+1/2$ and $m_l=2$ and that electron 2 has $m_s=−1/2$ and $m_l=0$. We write that $\Psi=\|(\psi^+_1)(\psi^−_2)\|$ where we have written the short form of the determinantal antisymmetrized normalized wave function, which comes from the diagonal element in the Slater determinant AND separated the orbital- and spin-dependent parts of the wave function (spin is denoted by super plus or minus sign). Problem Anyway, I understand that $\psi(2,2,0,0) = a (2^+,0^-) + b (2^-,0^+) + c(1^+,1^-)$. Then they write "and we get $a \sqrt{3} - b \sqrt{3} + c \sqrt{8} = 0$ and $a + b = 0$." I'm totally lost how they made this leap and why these must be equal to zero. Does this have to do with the orthogonal wave functions just mentioned? Where did the numbers in the sqrt come from? And the minus sign! Greatly appreciated. - Which book is it? Your question seems to be missing something as is, so it would help if we could check the book to fill in any gaps. – David Zaslavsky♦ Jun 20 '11 at 6:13 it's "Introduction to Ligand Field Theory" by Ballhausen (1962) on page 14. The point of this is that we "must develop a technique for picking out suitable linear combinations of these micro states in order to form eigenfunctions for $\hat{L}$ and $\hat{S}$." – Chris Jun 20 '11 at 6:38 1 I presume this problem is about expressing one state in some other basis. If so, we can't really help you until you tell us what that basis is. I can only guess that this is a system of two spin-$1/2$ particles and you are trying to express the common wavefunction as the wavefunction of individual particles. But if so, I don't understand why there are only three microstates and not $(0^+,2^-)$ etc. Please, fill in the details. – Marek Jun 20 '11 at 9:39 Or is $(2^+,0^-)$ supposed to represent already antisymmetrized product? – Marek Jun 20 '11 at 9:40 $(2^+,0^-)$ means that electron 1 has $m_s=+1/2$ and $m_l=2$ and that electron 2 has $m_s = -1/2$ and $m_l=1$. We write that $\Psi = \| (\psi_1^+) (\psi_2^-) \|$ where we have written the short form of the determinantal antisymmetrized normalized wave function, which comes from the diagonal element in the Slater determinant AND separated the orbital- and spin-dependent parts of the wave function. In the complete expanded wave function, we have all possible permutations of electrons. – Chris Jun 20 '11 at 17:50 show 1 more comment ## 1 Answer Apparently I figured it out. ANSWER Previously on pg. 14 we determined using ladder operators that $\psi(4,2,0,0) = \sqrt{3/14} (2^+, 0^-) + \sqrt{8/14} (1^+,1^-) - \sqrt{3/14}(2^-,0^+)$ and $\psi(3,2,1,0) = sqrt{1/2} [(2^+,0^-) + (2^-,0^+)]$. For the singlet D term, we have $\psi(2,2,0,0) = a (2^+,0^-) + b(2^-,0^+) + c(1^+,1^-)$. And if we take $\psi(2,2,0,0) \cdot \psi(4,2,0,0) = \sqrt{3/14}a + \sqrt{8/14}c - \sqrt{3/14}b = 0$ because he has said that these micro states must be orthogonal. Multiply by $\sqrt{14}$ to obtain his result. It also follows that $\psi(2,2,0,0) \cdot \psi(3,2,1,0) = \sqrt{1/2} a + \sqrt{1/2} b = 0$. Multiplication by $\sqrt{2}$ will give his result of $a + b = 0$. Also realize that $a^2 + b^2 + c^2 = 1$ must be true as a normalizing condition. Now we have three equations and three unknowns, so solving it is simple. $\psi(3,2,1,0)$ and $\psi(4,2,0,0)$ were picked in the first place because we had three unknowns and needed three equations to solve it. These wave functions must be orthogonal in the first place, therefore it was a matter of convenience AND that they were made up of the corresponding micro states in the first place. -	{"extraction_info": {"found_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": 0.9595993757247925, "perplexity_flag": "head"}
http://en.wikipedia.org/wiki/Preorder	# Preorder In mathematics, especially in order theory, a preorder or quasiorder is a binary relation that is reflexive and transitive. All partial orders and equivalence relations are preorders, but preorders are more general. The name 'preorder' comes from the idea that preorders are 'almost' (partial) orders, but not quite; they're neither anti-symmetric nor symmetric. Because a preorder is a binary relation, the symbol ≤ can be used as the notational device for the relation. However, because they are not anti-symmetric, some of the ordinary intuition associated to the symbol ≤ may not apply. On the other hand, a pre-order can be used, in a straightforward fashion, to define a partial order and an equivalence relation. Doing so, however, is not always useful or worthwhile, depending on the problem domain being studied. In words, when a ≤ b, one may say that b covers a or that b precedes a, or that b reduces to a. Occasionally, the notation ← or $\lesssim$ is used instead of ≤. To every preorder, there corresponds a directed graph, with elements of the set corresponding to vertices, and the order relation between pairs of elements corresponding to the directed edges between vertices. The converse is not true: most directed graphs are neither reflexive nor transitive. Note that, in general, the corresponding graphs may be cyclic graphs: preorders may have cycles in them. A preorder that is antisymmetric no longer has cycles; it is a partial order, and corresponds to a directed acyclic graph. A preorder that is symmetric is an equivalence relation; it can be thought of as having lost the direction markers on the edges of the graph. In general, a preorder may have many disconnected components. The diamond lemma is an important result for certain kinds of preorders. Many order theoretical definitions for partially ordered sets can be generalized to preorders, but the extra effort of generalization is rarely needed.[citation needed] ## Formal definition Consider some set P and a binary relation ≤ on P. Then ≤ is a preorder, or quasiorder, if it is reflexive and transitive, i.e., for all a, b and c in P, we have that: a ≤ a (reflexivity) if a ≤ b and b ≤ c then a ≤ c (transitivity) Note that an alternate definition of preorder requires the relation to be irreflexive. However, as this article is examining preorders as a logical extension of non-strict partial orders, the current definition is more intuitive. A set that is equipped with a preorder is called a preordered set (or proset[citation needed]). If a preorder is also antisymmetric, that is, a ≤ b and b ≤ a implies a = b, then it is a partial order. On the other hand, if it is symmetric, that is, if a ≤ b implies b ≤ a, then it is an equivalence relation. A preorder which is preserved in all contexts (i.e. respected by all functions on P) is called a precongruence. A precongruence which is also symmetric (i.e. is an equivalence relation) is a congruence relation. Equivalently, a preordered set P can be defined as a category with objects the elements of P, and each hom-set having at most one element (one for objects which are related, zero otherwise). Alternately, a preordered set can be understood as an enriched category, enriched over the category 2 = (0→1). A preordered class is a class equipped with a preorder. Every set is a class and so every preordered set is a preordered class. Preordered classes can be defined as thin categories, i.e. as categories with at most one morphism from an object to another. ## Examples • The reachability relationship in any directed graph (possibly containing cycles) gives rise to a preorder, where x ≤ y in the preorder if and only if there is a path from x to y in the directed graph. Conversely, every preorder is the reachability relationship of a directed graph (for instance, the graph that has an edge from x to y for every pair (x, y) with x ≤ y). However, many different graphs may have the same reachability preorder as each other. In the same way, reachability of directed acyclic graphs, directed graphs with no cycles, gives rise to partially ordered sets (preorders satisfying an additional anti-symmetry property). • Every finite topological space gives rise to a preorder on its points, in which x ≤ y if and only if x belongs to every neighborhood of y, and every finite preorder can be formed as the specialization preorder of a topological space in this way. That is, there is a 1-to-1 correspondence between finite topologies and finite preorders. However, the relation between infinite topological spaces and their specialization preorders is not 1-to-1. • A net is a directed preorder, that is, each pair of elements has an upper bound. The definition of convergence via nets is important in topology, where preorders cannot be replaced by partially ordered sets without losing important features. • The relation defined by $x \le y$ iff $f(x) \le f(y)$, where f is a function into some preorder. • The relation defined by $x \le y$ iff there exists some injection from x to y. Injection may be replaced by surjection, or any type of structure-preserving function, such as ring homomorphism, or permutation. • The embedding relation for countable total orderings. • The graph-minor relation in graph theory. • A category with at most one morphism from any object $x$ to any other object $y$ is a preorder. Such categories are called thin. In this sense, categories "generalize" preorders by allowing more than one relation between objects: each morphism is a distinct (named) preorder relation. In computer science, one can find examples of the following preorders. • The subtyping relations are usually preorders. • Simulation preorders are preorders (hence the name). • Reduction relations in abstract rewriting systems. Example of a total preorder: • Preference, according to common models. ## Uses Preorders play a pivotal role in several situations: • Every preorder can be given a topology, the Alexandrov topology; and indeed, every preorder on a set is in one-to-one correspondence with an Alexandrov topology on that set. • Preorders may be used to define interior algebras. • Preorders provide the Kripke semantics for certain types of modal logic. ## Constructions Every binary relation R on a set S can be extended to a preorder on S by taking the transitive closure and reflexive closure, R+=. The transitive closure indicates path connection in R: x R+ y if and only if there is an R-path from x to y. Given a preorder $\lesssim$ on S one may define an equivalence relation ~ on S such that a ~ b if and only if a $\lesssim$ b and b $\lesssim$ a. (The resulting relation is reflexive since a preorder is reflexive, transitive by applying transitivity of the preorder twice, and symmetric by definition.) Using this relation, it is possible to construct a partial order on the quotient set of the equivalence, S / ~, the set of all equivalence classes of ~. Note that if the preorder is R+=, S / ~ is the set of R-cycle equivalence classes: x ∈ [y] if and only if x = y or x is in an R-cycle with y. In any case, on S / ~ we can define [x] ≤ [y] if and only if x $\lesssim$ y. By the construction of ~, this definition is independent of the chosen representatives and the corresponding relation is indeed well-defined. It is readily verified that this yields a partially ordered set. Conversely, from a partial order on a partition of a set S one can construct a preorder on S. There is a 1-to-1 correspondence between preorders and pairs (partition, partial order). For a preorder "$\lesssim$", a relation "<" can be defined as a < b if and only if (a $\lesssim$ b and not b $\lesssim$ a), or equivalently, using the equivalence relation introduced above, (a $\lesssim$ b and not a ~ b). It is a strict partial order; every strict partial order can be the result of such a construction. If the preorder is anti-symmetric, hence a partial order "≤", the equivalence is equality, so the relation "<" can also be defined as a < b if and only if (a ≤ b and a ≠ b). (Alternatively, for a preorder "$\lesssim$", a relation "<" can be defined as a < b if and only if (a $\lesssim$ b and a ≠ b). The result is the reflexive reduction of the preorder. However, if the preorder is not anti-symmetric the result is not transitive, and if it is, as we have seen, it is the same as before.) Conversely we have a $\lesssim$ b if and only if a < b or a ~ b. This is the reason for using the notation "$\lesssim$"; "≤" can be confusing for a preorder that is not anti-symmetric, it may suggest that a ≤ b implies that a < b or a = b. Note that with this construction multiple preorders "$\lesssim$" can give the same relation "<", so without more information, such as the equivalence relation, "$\lesssim$" cannot be reconstructed from "<". Possible preorders include the following: • Define a ≤ b as a < b or a = b (i.e., take the reflexive closure of the relation). This gives the partial order associated with the strict partial order "<" through reflexive closure; in this case the equivalence is equality, so we don't need the notations $\lesssim$ and ~. • Define a $\lesssim$ b as "not b < a" (i.e., take the inverse complement of the relation), which corresponds to defining a ~ b as "neither a < b nor b < a"; these relations $\lesssim$ and ~ are in general not transitive; however, if they are, ~ is an equivalence; in that case "<" is a strict weak order. The resulting preorder is total, that is, a total preorder. ## Number of preorders Number of n-element binary relations of different types n all transitive reflexive preorder partial order total preorder total order equivalence relation 0 1 1 1 1 1 1 1 1 1 2 2 1 1 1 1 1 1 2 16 13 4 4 3 3 2 2 3 512 171 64 29 19 13 6 5 4 65536 3994 4096 355 219 75 24 15 OEIS A002416 A006905 A053763 A000798 A001035 A000670 A000142 A000110 As explained above, there is a 1-to-1 correspondence between preorders and pairs (partition, partial order). Thus the number of preorders is the sum of the number of partial orders on every partition. For example: • for n=3: • 1 partition of 3, giving 1 preorder • 3 partitions of 2+1, giving 3 × 3 = 9 preorders • 1 partition of 1+1+1, giving 19 preorders i.e. together 29 preorders. • for n=4: • 1 partition of 4, giving 1 preorder • 7 partitions with two classes (4 of 3+1 and 3 of 2+2), giving 7 × 3 = 21 preorders • 6 partitions of 2+1+1, giving 6 × 19 = 114 preorders • 1 partition of 1+1+1+1, giving 219 preorders i.e. together 355 preorders. ## Interval For a $\lesssim$ b, the interval [a,b] is the set of points x satisfying a $\lesssim$ x and x $\lesssim$ b, also written a $\lesssim$ x $\lesssim$ b. It contains at least the points a and b. One may choose to extend the definition to all pairs (a,b). The extra intervals are all empty. Using the corresponding strict relation "<", one can also define the interval (a,b) as the set of points x satisfying a < x and x < b, also written a < x < b. An open interval may be empty even if a < b. Also [a,b) and (a,b] can be defined similarly. ## See also • partial order - preorder that is antisymmetric • equivalence relation - preorder that is symmetric • total preorder - preorder that is total • total order - preorder that is antisymmetric and total • directed set • category of preordered sets • prewellordering • Well-quasi-ordering • Newman's lemma ## References • Schröder, Bernd S. W. (2002), Ordered Sets: An Introduction, Boston: Birkhäuser, ISBN 0-8176-4128-9	{"extraction_info": {"found_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": 28, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9102530479431152, "perplexity_flag": "head"}
http://sbseminar.wordpress.com/2008/08/21/this-paper-was-written-for-our-blog/	## This paper was written for our blog August 21, 2008 Posted by David Speyer in Algebraic Geometry, mathematical physics, Number theory, Paper Advertisement, papers, representation theory. trackback I’ve recently been reading a paper which ties together a number of this blog’s themes: Canonical Quantization of Symplectic Vector Spaces over Finite Fields by Gurevich and Hadani. I’m going to try to write an introduction to this paper, in order to motivate you all to look at it. It really has something for everyone: symplectic vector spaces, analogies to physics, Fourier transforms, representation theory of finite groups, gauss sums, perverse sheaves and, yes, $\theta$ functions. In a later paper, together with Roger Howe, the authors use these methods to prove the law of quadratic reciprocity and to compute the sign of the Gauss sum. For the experts, Gurevich and Hadani’s result can be summarized as follows: they provide a conceptual explanation of why there is no analgoue of the metaplectic group over a finite field. Not an expert? Keep reading! Let’s start with some simple “physics”. Let $H$ be the vector space of complex-valued functions on the real line. (We probably want to impose some conditions on these functions, but this is (1) a blog post (2) “physics” and (3) only for motivation. I can think of no better time to ignore analytic issues.) There are two important operators on $H$: the operator $x$, which sends $f(x)$ to $x f(x)$ and the operator $D$ which sends $f(x)$ to $i \partial f/ \partial x$. They obey the relation $D \circ x - x \circ D = i \mathrm{Id}$. Basically all of the math in an introductory quantum mechanics course is playing with these operators. For our purposes, we would rather exponentiate these operators. The exponential $e^{i a x}$ is the operator which takes $f(x)$ to $e^{i a x} f(x)$ and the exponential $e^{i b D}$ is the operator which takes $f(x)$ to $f(x-b)$. (If you haven’t seen this before, it is worth thinking about.) These exponentiated operators obey the relation $e^{i b D} \circ e^{i a x} = e^{i ab} \cdot e^{i a x} \circ e^{i b D}.$ It is nice to build an abstract group to capture the formal properties of the above relation. The Heisenberg group is the group of formal symbols $z e^{i a x + i b D}$ where $z$ is a complex number of norm ${1}$ and $a$ and $b$ are real numbers. Multiplication is determined by the following relations: For any $z$ in $S^1$, the element $z$ is central. $e^{i a x} e^{i a' x} = e^{i (a + a') x}$ $e^{i b D} e^{i b' D} = e^{i (b + b') D}$ $e^{i a x} e^{i b D} = e^{(1/2) i a b} e^{i a x+ i b D}$ $e^{i b D} e^{i a x} = e^{-(1/2) i a b} e^{i a x+ i b D}$. So $H$ is a representation of the Heisenberg group. The Stone-von Neumann Theorem roughly states that $H$ is, up to isomorphism, the only irreducible representation of the Heisenberg group where the symbol $z$ acts by multiplication by the scalar $z$. Let $V$ be the vector space of formal linear combinations $a x + b D$, with the skew symmetric bilinear form $\omega(a_1 x + b_1 D, a_2 x + b_2 D) = a_1 b_2 - a_2 b_1$. Then we can more concisely define the Heisenberg group by saying that it is the group of symbols $z e^{i v}$, with $z \in S^1$ and $v \in V$, given the relations that $z$ is central and $e^{i v_1} e^{i v_2} = e^{(i/2) \omega(v_1, v_2)} e^{i (v_1+v_2)}$. Mathematical physicists like to work with $V$, without choosing a splitting of $V$ into transverse subspaces $\mathbb{R} x$ and $\mathbb{R} D$. Note that the definition of the Heisenberg group respects that aesthetic, but the definition of $H$ does not. For future reference, we set $L_1 = \mathbb{R} x$ and $L_2 = \mathbb{R} D$. An interruption: all of this works equally well for functions on a vector space of dimension larger than ${1}$. I find that right about now is the transition point where that added generality becomes illuminating, rather than confusing. You might want to go back and rework everything for functions on a $g$ dimensional space. Note that, if you do things correctly, $L_1$ and $L_2$ will naturally be dual vector spaces. I’ll keep writing in the one dimensional case, then switch to the general case when I talk about what is new in Gurevich and Hadani. Here is a definition of the Heisenberg group which uses the subspace $L_2$ but not the subspace $L_1$. Let $S^1 e^{i L_2}$ denote the abelian subgroup of the Heisenberg group consisting of expressions $z e^{i \ell}$ where $\ell \in L_2$. The quotient of the Heisenberg group by $S^1 e^{i L_2}$ is canonically isomorphic to $L_1$. Rather than realize $H$ as functions on $L_1$, we will realize it as functions on the quotient of the Heisenberg group by $S^1 e^{i L_2}$. The language of induced representations lets us do this in a very concise way: $H$ is (canonically isomorphic to) the induction from $S^1 e^{i L_2}$ of the obvious one dimensional representation of $z e^{i \ell}$. By “the obvious one dimensional representation”, I mean $z e^{i \ell} \mapsto z$. For any one dimensional subspace $L$ of $V$, we can define the group $S^1 e^{i L}$. Let $H(L)$ be the induction to the Heisenberg group of the character $z e^{i \ell} \mapsto z$ of $S^1 e^{i L}$. (If you are doing the $g$-dimensional generalization, then $L$ should be $g$ dimensional and $\omega\|_L$ should be zero, i.e., $L$ should be Lagrangian.) All of the representations $H(L)$ are isomorphic, but not canonically so. To get some intuition, let’s look at $H(L_2)$ and $H(L_1)$. To find an isomorphism, we must find a way to translate functions on $L_1$ to functions on the dual space $L_2$. The Fourier transform, $\hat{f}(y) = C \int_{x \in M_1} e^{i \omega(y,x)} f(x) \ dx$ is such an isomorphism, and the standard fact that Fourier transform switches $x$ and $D$ turns into the fact that this is a map of representations. That $C$ is the constant I always forget, you know, the one that is something like $1/\sqrt{2 \pi}$. The issue of this constant is actually very important for us. For almost any two subspaces $M_1$ and $M_2$, we can similarly define a map $H(M_2)$ to $H(M_1)$ by $\hat{f}(y) = C(M_1, M_2) \int_{x \in M_1} e^{i \omega(y,x)} f(x) dx$ where we use the canonical identifications between $H(M_i)$ and functions on $M_{3-i}$. (Why “almost any”? Stay tuned!) We hit a technical point: what do I mean by $dx$? I need to choose a volume form on $M_1$ in order to take the integral. So, properly, we should work not with subspaces $M_1$ and $M_2$ but with subspaces equipped with a nondegenerate volume form. But now something strange happens! Even once the issue about volume forms is straightened out, there is no compatible continuous way to choose the constants $C(M_1, M_2)$. The space of one dimensional spaces of $V$, equipped with volume forms, is a Mobius strip. Requiring that form to be nondegenerate cuts the strip down the center, leaving a cylinder. If you travel around that cylinder and come back to where you started, you’ll find you’ve picked up a sign error! I’m sure there is a lot to say here about connections, topology and so forth. But I want to explain what happens in Gurevich and Hadani’s work. Instead of working with $\mathbb{R}$, they work over the finite field $\mathbb{F} := \mathbb{Z}/p$ for $p$ an odd prime. Fix a nontrivial $p^{\textrm{th}}$ root of unity $\zeta$. Let $V$ be a $\mathbb{F}$-vector space of dimension $2g$, equipped with a perfect skew symmetric form $\omega$. Define the discrete Heinsenberg group to be the group of formal symbols $\zeta^k e^{v}$, with $v \in V$, subject to the relations that $\zeta$ is central and $e^{v_1} e^{v_1} = \zeta^{(1/2) \omega(v_1, v_2)} e^{v_1 + v_2}$. For any Lagrangian subspace $L$ of $V$, we can define $H(L)$ as before. If $L'$ is any lagrangian subspace transverse to $L$, we can identify $H(L)$ with complex-valued functions on $L'$. If $M_1$ and $M_2$ are transverse, there is an isomorphism $H(M_2) \to H(M_1)$ by $\hat{f}(y) = C(M_1, M_2) \sum_{x \in M_1} \zeta^{\omega(y,x)} f(x).$ (If $M_1$ and $M_2$ are not transverse, we can’t identify $H(M_1)$ with $\mathbb{C}^{M_2}$. This is why I said “almost any” two subspaces above. We’ll talk more about this issue later.) Now, once again, we need to think carefully about volume forms. (And this time, it is less obvious why, because there is a standard counting measure on $M_1$.) However, once that issue is straightened out, a miracle happens: this time, we can choose the constants $C(M_1, M_2)$ so that everything is compatible. In particular, the group $Sp_{2g}(\mathbb{F})$ of automorphisms of $V$ preserving $\omega$ acts on $H$, even though the analgous statement for $Sp_{2g}(\mathbb{R})$ is false. My understanding is that the above was known before Gurevich and Hadani, but their explanation, and in particular an explicit recipe for choosing the constants $C(M_1, M_2)$, was not. One of the innovations of their paper is that, using perverse sheaves, they extend the Fourier transform to the case where $M_1$ and $M_2$ are not transverse. Also, while I do not follow their proof yet, it seems to involve a number of identities between Gauss sums, so the number theorists should like this too. A few questions that came to mind reading their paper: (1) Does this result say that the space of lagrangian subspaces of $\mathbb{F}_p^{2g}$ is simply connected? (A statement which is false for $\mathbb{R}^{2g}$.) If not, is there some similar statement about characteristic $p$ topology which is true and give a conceptual explanation for this? (2) Gurevich and Hadani work very hard to handle the nontransverse case. Is this necessary? Explicitly, consider this simplicial complex whose vertices are lagrangian subspaces equipped with a non-degnerate volume form and whose faces are pairwise transverse collections of lagrangians. Is this complex connected and simply connected? If so, then we could give an alternate definition of $H(M_2) \to H(M_1)$ by composing the definition for the transverse case. Once we checked that going around a triangle in both ways gave the same result, we would know that our map was well defined. This certainly works when $g=1$! ## Comments» 1. Scott Carnahan - August 21, 2008 Shamgar has been a postdoc at Berkeley for the past couple years, and gave several talks about his work with Hadani. It was apparently inspired by a 1982 letter from Deligne to Kazhdan that did not receive much attention for a while. There is some similar work by Lysenko (not the “biologist”) and V. Lafforgue, related to geometric Langlands and sheafy theta correspondence. 2. Scott Carnahan - August 21, 2008 For question number one, there is a rather unenlightening answer. Symplectic groups over finite fields don’t have nontrivial central extensions, except for some small exceptions (I’m afraid I’ve never seen a proof). The space of Lagrangians is a homogeneous space with connected stabilizer (I think), making the quotient simply connected. This is notably false for nonarchimedean local fields. Our own Cal Moore did some work on central extensions of these groups in the middle of the last century, and there is exactly one nontrivial isomorphism type in each case. 3. Kea - August 21, 2008 Wow, thanks! So cool to learn about Deligne’s ideas for the quantum (or discrete if you like) Fourier transform and its ‘phase space’ or ‘quantum torus’ (for the physicists who don’t like the word symplectic). You know, such Hilbert spaces are well studied in quantum computation and the field known as Quantum Foundations, but I haven’t come across such a mathematical connection before. Very nice, and very close to some interesting physics. 4. Kea - August 21, 2008 …and I suspect that number theorists will be more intrigued by their remark that the proof will be close in spirit to the procedure of analytic continuation … coupled with the name of Weil. 5. Andy P. - August 24, 2008 Scott — The result you are quoting is due to Steinberg, and is contained in his paper “Générateurs, relations et revêtements de groupes algébriques”. It is also (I think) contained in his “Lectures on Chevalley groups”. 6. Ben Wieland - August 24, 2008 The space of Lagrangians is a homogeneous space with connected stabilizer That is a hard statement to make sense of for finite fields. The usual way of making sense of it really involves passing to algebraically closed fields, but then the argument would apply just as well to R. 7. Emmanuel Kowalski - August 24, 2008 One has to be careful with things like “simply connected” for algebraic groups: the notion bearing this name in the category of algebraic groups is not the same as the corresponding one when other natural topologies are involved (e.g., SL(n) is simply connected as an algebraic group, but SL(2,R) is not with the usual topology — one reason the metaplectic groups, which are not algebraic groups, come up). (In a probably different direction, having to do with Comment 6, it may be interesting to note that Steinberg, among others, discusses the philosophy that, over finite fields, a notion analogue to “connectedness” is “generated by unipotents” — I don’t remember the precise reference however). 8. Scott Carnahan - August 25, 2008 I’m not sure what I was thinking when I wrote the bit about “simply connected,” but it only seems to bear a superficial resemblance to actual mathematics. If I were pressed, I’d probably wave my hands and mumble something about spherical buildings, but I don’t know if they’re really useful here. %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": 143, "mimetex.cgi": 0, "/images/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": 0.9315115809440613, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/195189/is-there-a-lower-dimensional-vector-operator-analogous-to-curl-or-div?answertab=active	# Is there a lower dimensional vector operator analogous to Curl or Div? The Curl vector operator can be defined as the limit of an infinitesimal closed line integral over a surface divided by the \|area\| as it approaches zero. Likewise for Div as the limit of an infinitesimal surface integral over a volume divided by the \|volume\| as it approaches zero. Taking the analogy to a lower dimension: Is there a lower dimensional vector operator analogous to Curl and Div? Is there a theorem associated with it analogous to Stokes' for Curl or Ostrogradsky's for Div? - By "lower dimensional" do you mean: "for scalar and vector fields in plane domains"? – Christian Blatter Sep 13 '12 at 13:33 – user8268 Sep 13 '12 at 15:01 @ChristianBlatter I see Curl as being a lower dimensional vector operator compared to Div and I'm looking for a lower dimensional operator compared to Curl. I'm open to suggestions on how to better express this in my question. – user10389 Sep 13 '12 at 15:05 1 – Jonathan Sep 13 '12 at 16:36 ## 1 Answer Note that all three operators: ${\rm div}$, ${\rm curl}$, and $\nabla$ are first order operators, only the Laplace operator $\Delta:={\rm div}\circ\nabla$ is of second order. Your question then concerns only the dimension of the domains over which one integrates. The operator ${\rm div}$ acts on flow fields ${\bf v}$, and is implicitly defined by $$\int_{\partial B} {\bf v}\cdot{\rm d}{\omega}\ \doteq\ {\rm div}\,{\bf v}({\bf p})\ {\rm vol}(B)$$ for small three-dimensional balls $B:=B_\epsilon({\bf p})$ with center ${\bf p}$. The operator ${\rm curl}$ acts on force fields ${\bf F}$, and is implicitly defined by $$\int_{\partial D} {\bf F}\cdot d{\bf x}\ \doteq\ {\rm curl}\,{\bf F}({\bf p})\cdot {\bf n}\ \ {\rm area}(D)$$ for small oriented two-dimensional disks $D\subset{\mathbb R}^3$ with center ${\bf p}$ and normal ${\bf n}$. Note that the dimension of the considered "balls" has decreased from $3$ to $2$, and the dimension of their boundaries (spheres) has decreased from $2$ to $1$. Now you want to go one step further down. This is indeed possible, and results in the following sentence, formulated in the same way as the first two: The operator $\nabla$ acts on scalar fields $f$, and is implicitly defined by $$\int_{\partial S} f\ dp\doteq\ \nabla f({\bf p})\cdot {\bf u}\ \ {\rm length}(S)$$ for small oriented segments $S\subset{\mathbb R}^3$ with center ${\bf p}$ and direction ${\bf u}$. But what is $\partial S$ in this case? It is the endpoint ${\bf b}$ of $S$ with weight $+1$ and the initial point of $S$ with weight $-1$. Therefore the last formula should be read as $$f({\bf b})-f({\bf a})\ \doteq\ \nabla f({\bf p})\cdot ({\bf b}-{\bf a})\ .$$ This formula can be viewed as "implicit" definition of the gradient at ${\bf p}$. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 29, "mathjax_display_tex": 4, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9331651329994202, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/265881/show-the-derivative-of-an-activation-function?answertab=votes	# Show the derivative of an activation function I am learning about neural networks and am using the sigmoid activation function $$q(z)=\frac{1}{1+e^{-z}}.$$ The problem is that I need to use its derivative $q^{\prime}(z)$. Would anyone have any hints as to how I would go about calculating this? - ## 1 Answer Hint: $$\left(\frac 1f\right)^{\prime}=-\frac{f^{\prime}}{f^2}$$ when $f$ is non zero and differentiable. This means $$q^{\prime}(z)=\left(\frac{1}{1+e^{-z}}\right)^{\prime}=-\frac{(1+e^{-z})^{\prime}}{(1+e^{-z})^2}$$ - Mhh, sorry, I don't quite get it...one more hint? :) Is it just q'(z) = q(z)^2 ? – user1796218 Dec 27 '12 at 12:19 Ahh, ok, that makes sense. Thank you!! So, q'(z) = q(z)(1-q(z)). – user1796218 Dec 27 '12 at 12:35	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9322623610496521, "perplexity_flag": "middle"}
http://mathhelpforum.com/trigonometry/123816-trig-identities-problem.html	# Thread: 1. ## Trig Identities Problem cos(arcTan (5/12) - arcCos(2/3)) so basically I set up the two triangles and get the hypotenuse 13 and the side length 2sqrt5. I'm supposed to evaluate without using a calculator, what do I do next? 2. First - recognize that what you have are two angles - call them alpha and beta, where: $<br /> \alpha = tan ^{-1} (\frac 5 {12}), \ \ \beta = cos ^{-1} (\frac 2 3 )<br />$ You are aked to find $cos(\alpha - \beta)$. So use the trig identity: $cos(\alpha - \beta) = cos \alpha \cdot \cos \beta + sin \alpha \cdot sin \beta$ You can figure the values of these sines and cosines from the information you've been given. For example, $cos \alpha = \frac {12} {13}$. BTW - the side legth of the second triangle is $\sqrt { 3^2 - 2^2} = \sqrt 5$ 3. Hello, hellojellojw! Smplify: . $\cos\left(\arctan\tfrac{5}{12} - \arccos\tfrac{2}{3}\right)$ .[1] Let: . $\alpha \:=\: \arctan\tfrac{5}{12} \quad\Rightarrow\quad \tan\alpha \:=\:\tfrac{5}{12} \:=\:\tfrac{opp}{hyp}$ We have: . $opp = 5,\;adj = 12 \quad\Rightarrow\quad hyp = 13$ . . Hence: . $\sin\alpha \,=\,\tfrac{5}{13},\;\cos\alpha \,=\,\tfrac{12}{13}$ .[2] Let: . $\beta \,=\,\arccos\tfrac{2}{3} \quad\Rightarrow\quad \cos\beta \:=\:\tfrac{2}{3} \:=\:\tfrac{adj}{hyp}$ We have: . $adj = 2,\;hyp = 3 \quad\Rightarrow\quad opp = \sqrt{5}$ . . Hence: . $\sin\beta \,=\,\tfrac{\sqrt{5}}{3},\;\cos\beta \,=\,\tfrac{2}{3}$ .[3] From [1], we have: . $\cos(\alpha - \beta) \;=\;\cos\alpha\cos\beta + \sin\alpha\sin\beta$ Substitute [2] and [3]: . $\cos(\alpha + \beta) \;=\;\left(\tfrac{12}{13}\right)\left(\tfrac{2}{3} \right) + \left(\tfrac{5}{13}\right)\left(\tfrac{\sqrt{5}}{3 }\right) \;=\;\frac{24+5\sqrt{5}}{39}$ 4. why does arctan being 5/12 and arccos being 2/3 imply that the tangent and cosine are also those values? 5. Originally Posted by BugzLooney why does arctan being 5/12 and arccos being 2/3 imply that the tangent and cosine are also those values? Definition of the arctan: it's the angle whose tangent is the given value. Think of arctan as the inverse function of tan. Hence the tangent of the arctangent of x is simply x. You can see this with a simple example: consider the angle whose tangent = 1. You can probably see right away that the angle is $\pi/4$. Now find the tangent of $\pi/4$ - it's 1. So you see that tan(arctan(x)) = x.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 16, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8821921944618225, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/35524/interference-photons-phase-and-the-hilbert-space	# Interference, photon's phase, and the Hilbert space Dirac said that a photon can only interfere with itself. This is consistent with the tensor product of two photon spaces representation. On the other hand, it is known that there is interference between distinct sources, between two photons, between two photons again, and between two electrons. This is related to this SE question, and this one. Let $\frac 1 {\sqrt 2}\left(\|1\rangle\|2\rangle + \|2\rangle\|1\rangle\right)$ be a two-photon state. Let's consider interference between the two photons. If one "rotates" one of the wavefunctions with a phase factor, one expects that this will affect the interference, and the result will depend on the phase factor. So, the interference between $\|1\rangle$ and $\|2\rangle$ should be different than the interference between $\|1\rangle$ and $e^{i\varphi}\|2\rangle$, and that between $e^{i\varphi}\|1\rangle$ and $\|2\rangle$ (where $\varphi$ is an arbitrary overall phase). This can be checked experimentally by using a phase shifter. Suppose that there is an algorithm to calculate the interference between the two photons, and that this algorithm receives as input the two-photon state. The state $\frac 1 {\sqrt 2}\left(\|1\rangle\|2\rangle + \|2\rangle\|1\rangle\right)$ is equivalent to $\frac 1 {\sqrt 2}\left(\|1\rangle\|2'\rangle + \|2'\rangle\|1\rangle\right)$, where $\|2'\rangle=e^{i\varphi}\|2\rangle$. If the algorithm which gives us the interference depends only on the two-photon state, then it apparently gives the same result, even when we phase-shift the second photon. Apparently, the interference between photons depends on the way we decompose the two-photon state in two photons: in terms of $\|1\rangle,\|2\rangle$ we obtain different interference than in terms of $\|1\rangle,e^{i\varphi}\|2\rangle$, respectively $e^{i\varphi}\|1\rangle,\|2\rangle$. So, either the Hilbert space does not contain all the needed information, or I am missing something (which is by far more likely). Could you please explain me where I am wrong, and what is the correct description of interference using the two-photon state? P.S. Let me emphasize that I am fully aware that QM has been tested successfully by so many experiments, and that most likely I am missing something. I just don't know what. - ## 2 Answers Two photons do not interfere with each other. One photon interferes with itself. You cannot add together the quantum states belonging to different Hilbert space. Ok, two photons can have interference, because they are indistinguishable, and thus should always be described in whole. You can't use $\|1\rangle+\|2\rangle$ to describe the state of two photons, because that is not a symmetric state. It is not even a state in 2-photon space. You should add some labels to clarify your thinking. For example, $\|1\rangle_1+\|2\rangle_2$. - Thank you. I updated the question with some references which seem to prove that there may be interference between different sources, different photons and different electrons. – Cristi Stoica Sep 3 '12 at 16:09 Thanks for the update. I can't see how labeling the photons allows me to avoid what I wrote in my second update. Namely, that the state obtained by shifting the phase of one of the photons is equivalent, in the two-state Fock space, with the original one. So I can't see where the information about that phase shift is in the vector state from the two-photon space. Yet, it has physical effects. – Cristi Stoica Sep 4 '12 at 4:42 @CristiStoica: I don't understand you. You have specify the state of the two photons. What next? – C.R. Sep 4 '12 at 6:12 The state of two photons remains unchanged if we shift the phase of one of them. But the interference pattern changes. So apparently the same two-photon state yields distinct interference patterns, depending on how we choose to write it in terms of two photon states. What I am missing? If I am wrong, then what is the correct algorithm to get the interference pattern solely from the two-photon state? – Cristi Stoica Sep 4 '12 at 6:24 @CristiStoica: I don't know where to begin. You should read those papers you cite, and understand what exactly two photon interference means. It does not in any way mean that you add the quantum states of two different photons together; it means you superimpose two different states in the two-photon system's space. – C.R. Sep 4 '12 at 8:41 show 2 more comments $\newcommand{\ket}[1]{\|#1\rangle}$The states $$\ket{\phi^+}=\frac{1}{\sqrt{2}}\left(\ket{1}\ket{2}+\ket{2}\ket{1}\right)\textrm{ and }\ket{+}=\frac{1}{\sqrt{2}}\left(\ket{1}+\ket{2}\right)$$ don't really have much to do with each other, and from their structure you can see that they are very different. If you want to obtain the state $\ket{\varphi}=\frac{1}{\sqrt{2}}\left(\ket{1}+e^{i\varphi}\ket{2}\right)$ in port 1 of your system, starting from the common, entangled state $\ket{\phi^+}$, then you'd have to do a projective measurement on to $\ket{\varphi}$ on port 2, which leaves the desired state in port 1. (Check it!). Of course, this has the attendant risk that the measurement will yield $\ket{\varphi}$'s orthogonal complement, $\ket{\overline{\varphi}}=\frac{1}{\sqrt{2}}\left(\ket{1}- e^{i\varphi}\ket{2}\right)$, on port 2, thus collapsing port 1 onto $\ket{\overline{\varphi}}$. Thus measurements on port 1 done irrespectively of coincidence counts with port 2 will not yield any information; this reflects the fact that the reduced density matrix for either port in the state $\ket{\phi^+}$ is a completely mixed state. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 24, "mathjax_display_tex": 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": 0.9438645243644714, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/34028/what-is-the-quantity-in-physics-that-is-advanced-from-speed/34031	# What is the quantity in physics that is advanced from speed? [closed] What is the quantity in physics that is advance from speed? i know one is velocity, one is speed, one is acceleration. - 4 It seems logical that speed is related to speed... – Fabian Aug 12 '12 at 18:15 2 I am at a loss to determine what the question here is. In particular I don't understand "is advanced from" as it is used here. Can someone clarify the question? – dmckee♦ Aug 12 '12 at 21:38 ## closed as not a real question by Qmechanic♦, Ron Maimon, Colin K, dmckee♦Aug 14 '12 at 17:00 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. ## 1 Answer To give an overview of the different measurements encountered in kinematics (as there doesn't seem to be a real, precise question asked here). The most familiar quantity we encounter in kinematics is speed. Speed is abstractly how fast something is going. It does not take into account direction, and is therefore known as a scalar quantity. It is what we measure on the speedometer of a car. Speed is often denoted by $v$ in physics. Velocity takes into account both speed and direction. It is a vector quantity, and is often denoted by $\vec{v}$ or $\mathbf v$ in physics. Velocity is the rate of change of distance with respect to time, which in calculus notation is written: $$\vec{v}=\mathbf{v}=\frac{d\vec{s}}{dt}$$ It is also useful to note that if we know the velocity, it is easy to find the speed, as we have the simple relationship: $$v=\\|\vec{v}\\|=\\|\mathbf{v}\\|$$ Where $\\|\cdot\\|$ denotes the euclidean norm. Acceleration formally is defined as the rate of change with velocity with respect to time, that is, it's a measurement of how fast the velocity is changing, so it is a vector quantity. That is, in calculus notation: $$\vec{a}=\mathbf{a}=\frac{d\vec{v}}{dt}=\frac{d^{2}\vec{s}}{dt^{2}}$$ However, it is also sometimes used to refer to the norm of that quantity. So you may also see: $$a=\left\\|\frac{d\vec{v}}{dt}\right\\|=\\|\vec{a}\\|=\\|\mathbf{a}\\|$$ Bear in mind, there are also various higher derivatives (rates of change) which are occasionally used in kinematics, such as the derivative of acceleration with respect to time, called jerk and the derivative of that with respect to time, called jounce. I hope this helps clear things up a bit, if you still have questions, feel free to ask them in the comments. - 2 As far as I know, the norm of $\mathbf{a}$ isn't the same as the derivative of the norm of $\mathbf{v}$. The latter is what is usually called tangential acceleration. – Javier Badia Aug 12 '12 at 19:18 @JavierBadia: as written, they have a = the norm of the derivative, not the other way around. – Jerry Schirmer Aug 13 '12 at 3:58	{"extraction_info": {"found_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": 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": 0.9796347618103027, "perplexity_flag": "head"}
http://www.physicsforums.com/showthread.php?s=0373a0abf5938a477ee428caac7aab23&p=3860242	Physics Forums ## Virtual photon-antiphoton pairs? I have heard of the phenomenon of virtual particle-antiparticle pairs popping out of the vacuum and then back into it within a time $\Delta t \approx \hbar / \Delta E$. Do virtual photon-antiphoton pairs pop out of the vacuum in a similar way? I understand that antiphotons are the "same" as photons - is that right? In that case how would the two photons annihilate? Does one have positive energy and the other negative energy? John PhysOrg.com physics news on PhysOrg.com >> Promising doped zirconia>> New X-ray method shows how frog embryos could help thwart disease>> Bringing life into focus No, not at all. When we say that a photon and an anti-photon are the same, we really mean it. There is no such thing as an anti-photon, because if you apply the transformations on it that would in other particles yield their counterparts, you get exactly the same you started with. What is to say, an anti-photon is just a regular photon like any other. They wouldn't annihilate, they would just interfere with each other and then go on their ways. And if you do assume they annihilate... annihilation produces only more photons. So you would have two photons colliding, which would produce two more photons that would just go their merry way, without minding each other too much. There are no anti-photons. All photons have positive energy. Two photons can not annihilate without creating a new particle/antiparticle pair, because of the conservation of energy. If they have enough energy (more than 1.022 GeV in total), they can annihilate and create an electron-positron pair. If they have more energy they could also create other particles. The electron is the lightest (=least energy) massive particle, so that is the first process that can happen. ## Virtual photon-antiphoton pairs? Quote by M Quack If they have enough energy (more than 1.022 GeV in total), they can annihilate and create an electron-positron pair. Wait, I didn't know that. So all annihilation processes can be time-reversed? Yes. It is one of the main mechanisms of how very hard gamma rays get absorbed - the second photon is usually a virtual photon of the electric field near a nucleus. Look up "pair production" (Wiki entry is wrong because it give the impression that this can happen with a single photon) http://www.icecube.wisc.edu/~tmontaruli/801/lect9.pdf Interesting. But I think it's misleading of you, or it may cause some confusion, to say that they 'annihilate' because, at least as far as I know, the concept is linked to the idea of a particle/anti-particle pair generating energy when in contact, and not the other way around :P Well, after the process you don't have any photons any more... Can you suggest a better word? As far as "generating energy" goes, rest mass may be converted into energy (or the other way around). There is no energy generated in the strict sense. Total energy and momentum are conserved, along with charge, etc. True enough, I suppose. Coming back to the original question: Yes, a pair of virtual photons can pop into existence for a short time and then disappear again. This can happen in vacuum and has even been measured in the Casimir effect. http://en.wikipedia.org/wiki/Casimir_effect Mentor Here we go again. There are at least three threads here where the argument that "Casmir proves virtual photons" is rebutted. One can calculate the Casimir effect without invoking virtual particles at all; therefore one cannot logically say it tells you anything about them. Recognitions: Science Advisor Quote by M Quack If they have enough energy (more than 1.022 GeV in total), they can annihilate and create an electron-positron pair. I wouldn't call that annihilation, though. They get absorbed by the created particles. In annihilation process, you are typically looking at a single world line that does a U-turn. And yes, there has to be a virtual photon involved. Two light-cone photons aren't going to just interact in empty space to give you particle-anti-particle pairs. Quote by Vanadium 50 Here we go again. There are at least three threads here where the argument that "Casmir proves virtual photons" is rebutted. One can calculate the Casimir effect without invoking virtual particles at all; therefore one cannot logically say it tells you anything about them. I'll leave that discussion to the experts then. Can you give a better example/proof of virtual particles from vacuum fluctuations? The Cotton-Mouton effect has not yet been observed in vacuum, I believe. And I am not sure if it would prove the eistence of virtual particles from vacuum. But then again I am not an expert, as you already know. Mentor If you want to discuss that, I suggest you continue on one of the Casimir threads. Quote by K^2 I wouldn't call that annihilation, though. They get absorbed by the created particles. In annihilation process, you are typically looking at a single world line that does a U-turn. And yes, there has to be a virtual photon involved. Two light-cone photons aren't going to just interact in empty space to give you particle-anti-particle pairs. So there are no "real" antiphotons. Real photons travel on light-cones and don't "experience" time. Therefore there can't be any negative-energy photons traveling backwards in time to act like "real" antiphotons. But if a photon is virtual then it can travel on a path off a light-cone. If a negative-energy virtual photon travels on the time-reversed path would it then be a virtual antiphoton? If virtual antiphotons can exist could virtual photon-antiphoton pairs be created from the vacuum? I presume that at the edge of a black hole the gravitational field is strong enough to prevent the virtual photon-antiphoton pairs from annihilating so that one "real" photon can escape as Hawking radiation while the other "real" photon falls down the black hole. Most elementary discussions of Hawking radiation describe particle-antiparticle pairs being pulled apart by the gravitational field. But the wavelength of Hawking radiation is of the same size as the black hole itself so it has to be of the form of photons and not massive particle/antiparticles like electron/positrons. Recognitions: Science Advisor Quote by Vanadium 50 Here we go again. There are at least three threads here where the argument that "Casmir proves virtual photons" is rebutted. One can calculate the Casimir effect without invoking virtual particles at all; therefore one cannot logically say it tells you anything about them. The Casimir effect can conveniently be calclulated using the concept of "virtual photons". I am not surprised that there are other formulations. I can also do quantum mechanics without using imaginary numbers. Nevertheless they are a useful concept, not more and not less. Recognitions: Science Advisor This particle-anti-particle concept has less to do with particles travelling backwards in time (a lousy trick of Feynman to draw funny pictures) but with the concept of charge. Photons don't carry charge, so they can can pop out in any reaction in any amount. You don't need to create them together with a particle carrying opposite charge. Do virtual particles and antiparticles annihilate because their wavefunctions are mirrors of each other in every respect and therefore exactly cancel? If an electron/positron pair pops out of the vacuum I guess they have opposite spin as well as opposite charge. If one of the spins was altered somehow would that stop them annihilating completely if they collided? Thread Tools \| \| \| \| \|-------------------------------------------------------\|-------------------\|---------\| \| Similar Threads for: Virtual photon-antiphoton pairs? \| \| \| \| Thread \| Forum \| Replies \| \| \| Quantum Physics \| 3 \| \| \| Quantum Physics \| 7 \| \| \| General Astronomy \| 4 \| \| \| Quantum Physics \| 17 \| \| \| General Physics \| 5 \|	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 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": 0.9296937584877014, "perplexity_flag": "middle"}
http://nrich.maths.org/4759/index?nomenu=1	## 'Von Koch Curve' printed from http://nrich.maths.org/ ### Show menu The Von Koch curve is a fractal. The rule for generating this curve is to start with an equilateral triangle and to replace each line segment by a zig-zag curve (a generator) made up of $4$ copies of the line segment it replaces, each reduced to one third of the original length. Click on the red button below to see the first six stages of the infinite process for generating the Von Koch curve. This text is usually replaced by the Flash movie. The original equilateral triangle has side $1$ unit. Work out the length of this curve in the first few stages and the length of the fractal curve formed when the process goes on indefinitely. Now suppose you made a poster for your classroom with coloured paper by drawing an equilateral triangle for Stage 0 and then cutting smaller equilateral triangles and sticking them on the edge. What is the total area of all the triangles you would stick on one edge if you could continue the process indefinitely to make the Von Koch curve? So what is the area inside the Von Koch curve? Find the dimension of the Von Koch curve using the formula $n=m^d$, where where $n$ is the number of self similar pieces in the generator and $m$ is the magnification factor. The diagrams below show Stages 0 to 5 in the evolution of the Von Koch curve. The Logo program for drawing this fractal is given in the Notes. See First Forward for a ten part series giving an introduction to Logo programming for beginners.	{"extraction_info": {"found_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": 0.9085591435432434, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/295233/simple-module-over-a-basic-algebra-is-one-dimensional/295253	# Simple module over a basic algebra is one-dimensional May I refer you to page $32$ of: http://www.staff.city.ac.uk/a.g.cox/LTCC/Week3.pdf Part (b) of proposition $3.2.4$. Why the fact that $A/\operatorname{rad}(A)$ is isomorphic to $k^m$ for some $m$ implies that the dimension of any simple $A$-module is $1$? - ## 1 Answer There are two claims: 1. Every simple $A$-module is a simple $A/\mathrm{rad}A$-module. 2. Every simple $k^m$-module is $1$-dimensional. First note that if $I \subseteq A$ is an ideal of any ring $A$ and $M$ is an $A$-module such that $IM = 0$, then $M$ is an $A/I$-module. Moreover, we haven't changed the submodules of $M$, i.e., any $N \subseteq M$ is a submodule for the $A$-action on $M$ if and only if it is a submodule for the $A/I$-action on $M$. (If you don't already know this you should stop and figure out why this is true, it's an important point) Now, for (1) you use the Nakayama lemma which says that for any finitely generated module $M$ the submodule $(\mathrm{rad}A)M$ is proper. If $M$ is simple this forces $(\mathrm{rad}A)M = 0$ so that $M$ is an $A/\mathrm{rad}A$-module. It is still simple because we haven't changed the submodules of $M$. For (2) let $e_1, \ldots, e_m$ be the standard basis for $k^m$. Then given any nonzero $k^m$-module $M$ we can form the submodule $e_iM$. As the unit $1 \in k^m$ is $1 = e_1 + e_2 + \cdots + e_m$ it cannot be the case that $e_iM = 0$ for all $i$ (otherwise we'd have $1M = 0$). So fix $i$ such that $e_iM \neq 0$. If $M$ is simple we must have $e_iM = M$ (there are no proper non-zero submodules). Note $e_ie_j = 0$ so $e_jM = e_je_iM = 0$ if $i \neq j$. Thus the ideal generated by $e_j$ with $i \neq j$ annihilates $M$. When we factor the action again we get that $M$ is a simple $k$-module. Now module = vector space and submodule = subspace. That $M$ is a simple module means it is a nonzero vector space with no proper non-zero subspace. Hence $M$ must be a $1$-dimensional vector space. - thanks! so the idea of the proof is to show that every simple $k^{m}$ module is also a simple $k$-module? is that the point? – user10 Feb 6 at 1:43 Yes, in fact the idea is to show that every simple $A$-module is a simple $k$-module. – Jim Feb 6 at 2:05	{"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": 60, "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": 0.8554667234420776, "perplexity_flag": "head"}
http://mathoverflow.net/questions/94480/generalization-of-hilbert-94-and-capitulation	## Generalization of Hilbert 94 and capitulation ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Let $L/K$ be a finite, cyclic extension of number fields, say with $\mathrm{Gal}(L/K)=G$. In my context $G$ is actually of order $p$, an odd prime number, but let me state my question for every cyclic $G$. Hilbert theorem 94 says that if $L/K$ is everywhere unramified (hence contained in the Hilbert class field $H$ of $K$), then the ''capitulation kernel'', namely the kernel of the natural map $\iota:Cl_K\to Cl_L$, has order divisibile by $\vert G\vert$. Some (but, to my knowledge, not many) generalizations have been proven, mainly removing the cyclicity assumption although losing something. My question goes into another direction: if $L/K$ is allowed to ramify and is contained in the ray class field $H(\mathfrak{f})$ modulo some conductor $\mathfrak{f}$, what can we say about the capitulation kernel $\iota:Cl_K(\mathfrak{f})\to Cl_L(\mathfrak{F})$, where these groups are now the ray class groups modulo the respective conductors (and $\mathfrak{F}=\mathfrak{f}\mathcal{O}_L)$)? Can it be trivial? I must confess my ignorance with respect to an even more basic question, namely: what is the state of the art concerning the Principal Ideal Theorem for ray class fields? Is it known – or false, or trivial, or... – that all classes in $Cl_K(\mathfrak{f})$ become principal (perhaps, only $\mathfrak{f}$-principal?) in $H(\mathfrak{f})$? - 2 Your second question is indeed well-known: the principal ideal theorem remains true with appropriate modification ($\mathfrak f$ principality) in the ray class group. This is a classical theorem of Iyanaga whose proof is identical to the principal ideal theorem (both reduce to the computation of the transfer map towards the derived subgroup). Then again, Grothendieck was famously stumped by this question. See also mathoverflow.net/questions/63465/… – Olivier Apr 19 2012 at 7:37 Thanks! I did not think, when asking, that the proof is only group-theoretic. Still, I would be interested in the H94 stuff... – Filippo Alberto Edoardo Apr 19 2012 at 12:34 1 Iyanaga's article is in German: Ueber den allgemeinen Hauptidealsatz, Japanese Journ. of Math. 7 (1934), 315-333 – Franz Lemmermeyer Apr 19 2012 at 12:38 Kein Problem! Und jetzt bin ich in Japan... ;-) Danke, Filippo – Filippo Alberto Edoardo Apr 19 2012 at 13:31 ## 1 Answer The answer to both my question is that "adding conductors does not change anything". Olivier has already discussed this for the Principal Ideal Theorem, and for Hilbert 94 this is proven by Suzuki in http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?view=body&id=pdf_1&handle=euclid.nmj/1118782786 The whole point is that Hilbert 94 follows from a generalization of Furtwängler theorem, saying that the kernel of the transfer map $$\mathrm{ver}:G^\text{ab}\to N^\text{ab}$$ has order divisible by $[G:N]$, and this for all normal subgroups $N$ containing the commutators $[G,G]$ (or, equivalently, such that the quotient by them is abelian). The case $N=[G,G]$ is Furtwangler's theorem and applying it in my setting with $N=\mathrm{Gal}(H_L(\mathfrak{F})/K)$ in the notations of the question (now $H_L(-)$ are ray class fields of $L$) shows that $\iota$ is never invective. -	{"extraction_info": {"found_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": 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": 0.9043717980384827, "perplexity_flag": "middle"}
http://mathoverflow.net/revisions/20373/list	## Return to Question 3 deleted 6 characters in body I want to know about people in researching complex (maybe differential) geometry are careing about what currently ? For example ,$L^2$ estimate inspired by Lars Hormander is a very useful tool,and how does this theory be developed currently ? As myself , i like this method very much ,but i don't know which is the next important problem be solved by this method . How far will this method go ? As well , just like the holomorphic morse inequalities , when it is proved by Demailly in 1985,twenty years passed , it seems that during thest twenty years there are no important results comes out in complex geometry ? I'm a beginner in complex geometry, i think that i can't scratch the direction of complex geometry ? I don't know people are careing about what in complex geometry ?Only doing some little questions or leave this field? So this is just the purpose of this question i asked , i want help , help me see the possible direction of complex geometry ?to communicate with all who are interested in this field. 2 added 866 characters in body I want to know about people in researching complex (maybe differential) geometry are careing about what currently ? For example ,$L^2$ estimate inspired by Lars Hormander is a very useful tool,and how does this theory be developed currently ? As myself , i like this method very much ,but i don't know which is the next important problem be solved by this method . How far will this method go ? As well , just like the holomorphic morse inequalities , when it is proved by Demailly in 1985,twenty years passed , it seems that during thest twenty years there are no important results comes out in complex geometry ? I'm a beginner in complex geometry, i think that i can't scratch the direction of complex geometry ? I don't know people are careing about what in complex geometry ?Only doing some little questions or leave this field? So this is just the purpose of this question i asked , i want help , help me see the possible direction of complex geometry ? 1 # the central issues in complex geometry I want to know about people in researching complex geometry are careing about what currently ?	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9688708782196045, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/21685?sort=oldest	## What is the “correct” category of multisets ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) During seminar the other day, when speaking about subobject classifiers, I asked if the subobject classifier for the category of multisets would have integer truth values, corresponding to the number of times and element is in the set. We attempted to show this, but quickly realized that we were not even sure of the "correct" category for multisets. To clarify, when I say correct I want my category to 1. Have objects identified by multisets 2. Have maps between the multisets be on the level of elements in the multiset, and forget the order of those elements, e.g. there is only one map {111223}->{55} 3. The subobject classifier will behave as I had hoped, with {1} having truth value 3 in {111} My question is; Can you construct a category satisfying these properties? Thanks in advance! EDIT: First, sorry about not checking nLab, I forget about that site far too often. Second, I should say that I have a little bit of motivation for my property two. So let me clarify what I meant in property two. Given a multiset, it can be thought of as a pair $S\times\mathbb{N}$ for a set $S$. Now, when considering morphisms between multisets I want the maps $f,g:\lbrace 1122\rbrace\rightarrow\lbrace34\rbrace$ such that $f$ sends $\begin{eqnarray} 1&\mapsto& 3,\ 1&\mapsto& 4,\ 2&\mapsto& 3,\ 2&\mapsto& 4\ \end{eqnarray}$ and $g$ sends $\begin{eqnarray} 1&\mapsto& 4,\ 1&\mapsto& 3,\ 2&\mapsto& 4,\ 2&\mapsto& 3\ \end{eqnarray}$ to be the same morphism. But if $h$ sends $\begin{eqnarray} 1&\mapsto& 4,\ 1&\mapsto& 4,\ 2&\mapsto& 4,\ 2&\mapsto& 3\ \end{eqnarray}$ then $h$ is not the same as $g$ or $f$. Further I would like it such that $\lbrace 112\rbrace$ is not a subobject of $\lbrace 12\rbrace$ but it is a subobject of $\lbrace 11122\rbrace$. Hopefully this will clear it up. - 1 Interesting question! I would personally benefit from a little more information on your condition #2 above regarding maps. What does "on the level of elements" mean, and how does this imply that there is only one map {111223} -> {55} in the category? Does it mean that 1, 2, and 3 must all map to 5, forgetting their multiplicity, and that 5 then "becomes doubled"? – Manny Reyes Apr 17 2010 at 20:24 2 Just wanted to note that there is a discussion in nLab regarding the "right" definition of the category of multisets (ncatlab.org/nlab/show/multiset). AFAICT, subobject classifiers are not considered, but there is some discussion (and references) regarding the "right" definition of arrows in that category (if you are willing to consider alternatives to your (2)). – unknown (google) Apr 17 2010 at 21:34 +1 good questions – Shizhuo Zhang Apr 17 2010 at 21:51 Am I correct in assuming you want {1,1} $\not\cong$ {1,2}, and {1,1} $\not\cong$ {1}? – S. Carnahan♦ Apr 17 2010 at 22:57 2 It looks to me like the problems are arising from a sort of inconsistency in how you're thinking about the elements with multiplicity. For {1122}, the two 1s are different enough that you can send one to a 3 and the other to a 4. However, the two 5s in {55} are the same, so you can't tell the difference between any two maps from {111223} to it. (From there it's just a short hop to {55} being a terminal object, hence isomorphic to every other terminal object, including {5}.) I advise you think very carefully about what sort of morphisms a particular application calls for, and go from there. – Owen Biesel Apr 18 2010 at 4:15 show 7 more comments ## 3 Answers According to the way I understand your conditions, I think the answer is No. In particular, condition 2 seems to suggest that there should be unique maps {1}->{111} and {111}->{1}, and also that those maps be inverses of each other (since there is only one map {1}->{1} and only one {111}->{111}). Hence the map {1}->{111} is an isomorphism, so its truth value is "true" regardless of whether the latter multiset has three or any other number of 1s. Edited to add: To me a very natural candidate for the category of multisets would be the category of sets equipped with an equivalence relation, whose morphisms are functions on the underlying set that preserve the equivalence relation. In other words, the category whose objects are surjections A->A' and whose morphisms from (A->A') to (B->B') are pairs of maps A->B and A'->B' making the square commute. The idea is that for a multiset like {1122}, the set A has four elements (like the multiset should) and the set A' only has two elements (like the underlying set {12} does), and the surjection A->A' tells you which elements of A are "the same" and which are different. The commuting square condition tells you that if two elements are equal, so are their images under any map. (So there's no map from {55} to {12} sending one 5 to 1 and the other to 2. However, there are two distinct maps from {5} to {55}.) This category does have small limits, and the monomorphisms from (A->A') to (B->B') are the ones whose underlying map A->B is injective. However, I don't know whether this category has a subobject classifier, or what it might look like if it exists. - Moreover, it seems that your condition 2 suggests that from the Category's point of view, you can't tell a multiset from its underlying set (throw away multiplicities). I think better is to have some sort of "category enriched in multisets", but that begs the question. – Theo Johnson-Freyd Apr 17 2010 at 22:20 Yes Owen this looks correct :/ Which is disappointing. I would like to preserve property two or something similar for the reason that I added. @Theo, any idea how to make that clear? – B. Bischof Apr 18 2010 at 4:00 I will not accept this answer just yet, in hopes of somebody fixing this issue and producing a category that is as close as possible to what I want. If nobody can come up with something, I will accept this, since you have really answered my original question. – B. Bischof Apr 18 2010 at 4:08 @Owen Biesel:Your suggested category of multisets is a subcategory of $\mathbf{Sets}^{\mathbf{2}}$, where $\mathbf{2}$ is the category of 2 distinct objects and one arrow between them (you consider only surjective functions, hence a subcategory). Now, $\mathbf{Sets}^{\mathbf{2}}$ has a subobject classifier (see pp. 105--106 of Mac Lane). I wonder whether this can be used somehow for your subcategory. [Ref to Mac Lane:books.google.com/… – unknown (google) Apr 18 2010 at 13:33 @unknown(google): That category is just the category $Arr(Sets)$. Unless the inclusion functor admits a lex left-adjoint (thereby making it a topos as well), I don't see how having the subobject classifier in the big category would help. – Harry Gindi May 6 2011 at 13:35 ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. This is all just off the cuff, so it might not work at all, and even if it it does, it's probably too complicated. I think Owen Biesel is right to say that the answer is no as stated, so I think condition 2 ought to be dropped. Assuming that a multiset can only have finitely many copies of each element (as your desire for integer truth values seems to suggest) I think you could have the objects be sets of pairs (x,n) where x is a set and n is a nonnegative integer (so kind of like the elements of a tagged disjoint union) with the property that, if (x,n) is in the set, then so is (x,m) for any m≤n, and for which only finitely many "copies" of the same element appear. A morphism $f:S\to T$ is an equivalence class of set functions with the following property: take $x\in S$, and consider the sequence of elements $f(x,0),f(x,1),\ldots$. This sequence should have the property that, for each $y\in Y$, the appearances of y in the sequence occur in nondecreasing order with no gaps. Two functions are the same if they differ by permuting the indices in the domain. For example, if this definition does what I hope it does, there are two functions from {x,x} to itself, namely the identity and the one which sends (x,0) and (x,1) to (x,0). I think this at least solves the isomorphism problem; I'm pretty sure that the only isomorphic multisets will be the ones that "ought" to be isomorphic. I don't think this makes the subobject classifier behave like you'd want, but I'm not sure you'd actually want it to. The integers seem like a bad candidate, because there's nothing "special" about the element 79 as opposed to the element 3; the "truth" map can only pick out one element. I think the way to make it work might be to drop the finiteness assumption and use {0,1,1,...} as the classifier. EDIT: I'm pretty sure this actually doesn't work; see the comments to the question. - This is one definition that we tried, but if I understand what you are saying, this is a slice category of sets, which give the same subobject classifier. In response to the integers not being a good subobject classifier; what about the time till truth Subobject classifier? This is very interesting based on the interpretation. I want to see what kind of topoi we get from certain SO classifiers like the integers or the reals. I think this is very strange. I would also like to see an SO classifier even more bizzare than just a well-ordered set. Especially because it is not just time till truth. – B. Bischof Apr 18 2010 at 4:06 Are you thinking this is just Sets/$\mathbb{N}$? I think the maps are different. – Nicolas Ford Apr 18 2010 at 4:41 For purposes not obviously related to the question we (arXiv:math/0412287) considered the following definition of maps of (finite) multisets: In the case of sets a map $S\rightarrow T$ is a subset $\Gamma\subseteq S\times T$ such that the projection on the first factor $\Gamma\rightarrow S$ is a bijection. Let us say that a map $f\colon S\rightarrow T$ of the sets underlying two multisets is a multijection if $\mu(t)=\sum_{f(s)=t}\mu(s)$ for all $t\in T$ (a bijection is a map such that the cardinality of a fibre over $t$ is equal to the cardinality of ${t}$). Composites of multijections are then multijections and they form a category (not a groupoid though). We can then define a multimap $S \rightarrow T$ to be a submultiset $\Gamma$ of the multiset product $S\times T$ such that the projection on the first factor is a multijection. This seems to fit with the examples given in the question. The problem is that composition of multimaps becomes multivalued as there will be ambiguities. One can solve this by looking at all possible composites making the composite a multiset of multimaps (technically we did this rather by constructing a category enriched in abelian monoids whose monoid of morphisms $S \rightarrow T$ was the free abelian monoid on the multimaps, but elements in this free monoid corresponds exactly to multisets of multimaps and the basis of a free abelian monoid can be canonically recovered from the monoid). If this approach is to be used to deal with the question one would have to set up a theory of categories with multivalued composition. I haven't thought at all about the problem of putting this on an abstract enough foundation so that one can even speak of subobject classifiers (multitoposes anyone?). -	{"extraction_info": {"found_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": 0.9472121596336365, "perplexity_flag": "head"}
http://mathematica.stackexchange.com/questions/8654/creating-diagrams-for-category-theory/8682	# Creating diagrams for category theory Lately I've been doing algebra and I have found myself drawing a bunch of diagrams when I attempt to solve a problem. Most of the diagrams are very simple so I thought, I bet I can do this in Mathematica and include these pictures in a nice document. What I would like to do is create diagrams like the ones found in category theory: Here is an attempt based on a solution provided by Mr.Wizard ````vertex = { {White, Disk[#2, 0.04]}, Text[#, #2] } &; GraphPlot[{ {"A" -> "B", "f"}, {"A" -> "C", "h"}, {"B" -> "C", "g"} }, DirectedEdges -> True, VertexLabeling -> True ] /. {Text -> vertex, Framed -> (# &)} ```` I like the idea of how in `GraphPlot` you can define the vertices and the labels. Unfortunately I'm still not very savvy on controlling many aspects of the `Graphics`. Does anyone know how to create diagrams like the ones I'm attempting to draw? It would be awesome if the font size could be specified. ## Edit: As belisarius pointed out: Text length and font size should be considered with care As requested by Mr.Wizard, here is are some interesting diagrams: http://scientopia.org/blogs/goodmath/category/good-math/category-theory/ The second picture can be found here A quick search on google images for "category theory diagrams" will give you some other diagrams including the ones I have put in here. It seems based on the figures that most of them involve simple rectangles with diagonal lines. In some cases some lines might be curved but this can all be done with straight lines. Since the arrangement of the lines is up to personal choice it would be good to have a solution in which we can move the vertices locations, a similar functionality should be done with the edge labels. I picture an edge label as an image that can rotate around a point in a line. That way we can select a position in the line and the position angle. In most cases this angle is multiples of 45 degrees. - 3 – belisarius Jul 24 '12 at 5:08 Can you provide a reference to a range of these diagrams? For example I am wondering of they are always a rectangle or a diagonal across a rectangle as shown? That would be pretty easy to write a specific function for. – Mr.Wizard♦ Jul 24 '12 at 6:20 Use `VertexCoordinates` or `GraphLayout` to "move the vertices locations". Use `Placed`, `Rotate` and even `EdgeShapeFunction' to play with edge labels placement. – Vitaliy Kaurov Jul 24 '12 at 7:06 – Szabolcs Jul 24 '12 at 7:20 3 Indeed , as @Szabolcs mentioned, WildCats is a category theory package for Mathematica. I am the developer. WildCats can plot commutative (and non-commutative) categorical diagrams. But it can do much more. It can do (some) calculations in category theory, both symbolically and -when appropriate- visually using diagrams. I am preparing a more complete answer with examples, shortly – magma Jul 24 '12 at 8:25 show 5 more comments ## 3 Answers WildCats is a category theory package for Mathematica. It is still under development. Current version is 0.51.0 I am the developer. WildCats can plot commutative (and non-commutative) categorical diagrams. But it can do much more. It can do (some) calculations in category theory, both symbolically and - when appropriate - visually, using diagrams. This is because, in WildCats, diagrams are not just pretty pictures, but retain most of their mathematical semantic. So it is possible to input a diagram to a functor (which is an operator between categories) and obtain a new diagram. Functors are operators which preserve the topology of diagrams (that means: it transforms vertices and arrows and an arrow between 2 vertices is transformed into an arrow between the transformed 2 vertices). Let me show some of the current diagram-drawing capabilities in WildCats and give some flavour of category theory along the way. The following example is taken from the "Displaying diagrams" tutorial. We are in the category (cat in some popular textbook) Grp, containing groups and their group-homomorphisms. Wildcats knows something about 27 well-known cats, but you can build your own cats too and derive new cats from old cats...that's why they can get wild pretty soon :-). Grp is the category of groups. We define 4 groups (`g1`...`g4`) and 3 group-homomorphisms (`f1`...`f3`) between them. They are the objects and morphisms in Grp. The following code defines the groups and morphisms. When we define a morphism (an arrow) in Category theory, we must always specify where it starts and where it ends. So here `f1` goes from groups `g1` to `g2`. WildCats at this point knows that we can compose the various morphisms and where the composite morphisms should start and end. Composite morphisms are represented as `Op[f2,f1,Grp]` or, even better, as since WildCats uses custom notation for almost everything. We can visualize all this with a `CategoryPlot` The `DC` functions find the Domain and Codomain of an arrow (its starting and ending vertex) As you see I just tell `CategoryPlot` which arrows I want to draw and it does the rest by itself (with just some help for the coordinates of the vertices). The above diagram visually says that the composition in Grp is associative, that is something that WildCats knows well. Another example might be It visually represents the relations describing the fact that `Id` (identity morphisms) are bilateral identities in the morphism-composition. Notice that the same vertex (eg. `g1`) appears to be displayed in 2 different locations. This is not easily possible in standard Mathematica. This is a visual depiction of the structure of (some of) the mathematical knowledge of WildCats (some standard categories and their relations). The tantalizing thing in WildCats is that you can take a diagram, give it to a functor and get another diagram! This is an example taken from the Tutorial "Functors". I won't go into the mathematical details too deeply. Suppose you have some groups and homomorphims in a diagram `cp` Now we tell WildCats that `g4` is the Quaternions group (a group with 8 elements) and apply to the above diagram the forgetful functor U[Grp , Set] which forgets the group structure and gives the underlying set of a group As you see the groups `IntegersPlus` and `RealsPlus` (groups of numbers with the Plus operation) have become the simple sets `Z` and `R` of integers and reals (without any specific operation),`Ab0`, the trivial group, has become the `Singleton` set, while the g4 (the quaternions group) has become a simple set with 8 elements. Also notice that the composition in Grp is now a composition in Set the category of sets and functions. In the future releases of WildCats, I would like to use some Dynamic features to bring life (and even more intuition) to all these diagram drawing. Anyone interested in testing/using WildCats, can contact me at maderri2 at gmail dot com - Hey magma, this is pretty interesting. But I cannot follow it. How do you get the $\mathbb{Grp}$? Could you perhaps include a screenshot of the things that you click on the palette? – jmlopez Jul 24 '12 at 14:29 there is no palette at the moment. Will be made soon. For now, just write it with doublescripted characters or copy/paste it from one of the tutorials. Watching a webseminar from WRI now, I will be back in 2 hours. Where is the popcorn ? :-) – magma Jul 24 '12 at 15:02 @jmlopez Here's a palette button: `PasteButton[\[DoubleStruckCapitalG]\[DoubleStruckR]\[DoubleStruckP]]` – Szabolcs Jul 24 '12 at 18:01 I meant: doublestruck characters. You can find them in the Special characters palette. But the fastest way is probably to evaluate `StandardCategories` and then copy/paste from its output. – magma Jul 24 '12 at 18:01 Or type `ESC` `dsG` `ESC`. – Szabolcs Jul 25 '12 at 6:45 show 1 more comment I don't like the new `Graph` functionality, but in your case it might be easier to use for label styling etc. ````ef1[el_, ___] := Arrow[el, 0.1] vertexLabels = MapIndexed[#2[[1]] -> Placed[Style[#, Opacity[1], Background -> White], {1/2, 1/2}] &, {F[X], F[Y], G[Y], G[X]}]; Graph[{Property[1 \[DirectedEdge] 2, EdgeLabels -> F[f]], Property[2 \[DirectedEdge] 3, EdgeLabels -> \[Eta]\[Gamma]], Property[1 \[DirectedEdge] 4, EdgeLabels -> \[Eta]x], Property[4 \[DirectedEdge] 3, EdgeLabels -> G[f]]}, VertexLabels -> vertexLabels, ImagePadding -> Scaled[.1], VertexSize -> 0, VertexLabelStyle -> Directive[Italic, 20], EdgeLabelStyle -> Directive[Italic, 20], EdgeStyle -> Black, EdgeShapeFunction -> ef1, VertexStyle -> None, GraphLayout -> "SpringEmbedding" (Thanx Vitaliy)] ```` - Option `GraphLayout -> "SpringEmbedding"` will make it "more square" ;-) – Vitaliy Kaurov Jul 24 '12 at 5:38 I tried a few of the layout alogorithms using the context menu (right mouse click) but for some reason I forgot `SpringEmbedding`, thank you @VitaliyKaurov – phantomas1234 Jul 24 '12 at 5:52 2 +1 I am glad you used `Graph` and not `GraphPlot`. BTW I do not think it is necessary, but if OP doesn't like edge crossing over label `Placed` is the way to deal with this: reference.wolfram.com/mathematica/ref/EdgeLabels.html#72169267 – Vitaliy Kaurov Jul 24 '12 at 6:42 After exploring phantomas1234 answer and the useful comments from Vitaliy Kaurov I have made a simple module which does what I want. ````PlotDiagram[vertex_List, edge_List, args___] := Module[ { v, vl, vp, e, el, es }, v = Range[Length@vertex]; vl = Table[vertex[[i, 1]], {i, 1, Length@vertex}]; vp = Table[vertex[[i, 2]], {i, 1, Length@vertex}]; e = Table[edge[[i, 1]], {i, 1, Length@edge}]; el = Table[edge[[i, 2]], {i, 1, Length@edge}]; es = Table[edge[[i, 3]], {i, 1, Length@edge}]; Graph[e, VertexCoordinates -> vp, VertexLabels -> Table[v[[i]] -> Placed[vl[[i]], {0.5, 0.5}], {i, 1, Length[v]}], EdgeLabels -> Table[e[[i]] -> el[[i]], {i, 1, Length[e]}], EdgeShapeFunction -> Table[e[[i]] -> es[[i]], {i, 1, Length[e]}], args, VertexLabelStyle -> Directive[Italic, 18], EdgeLabelStyle -> Directive[Italic, 15], VertexSize -> 0, VertexStyle -> Directive[EdgeForm[], White] ] ] ```` The default behavior is to make the diagram with vertex labels of font size 18 and the edge labels of font size 15. The first argument of the function has be a list containing the vertex labels and the positions of the vertices. The second argument is a list containing the edges. Each entry in edges is a list of 3 elements. The first one specifies the edge, the second specifies the label and the third specifies the arrow to be used. After that you may use any option you want for `Graph`. For a simple example I will make the figure that I showed in the post. ````arrowShape = ({Black, Arrowheads[0.1], Arrow[#, {2, 2}]} &); vertex = { {"X", {0, 10}}, {"Y", {10, 10}}, {"Z", {10, 0}} }; edge = { {1 \[DirectedEdge] 2, Placed["f", {.5, {.5, -0.2}}], arrowShape}, {2 \[DirectedEdge] 3, Placed[ "g", {.5, {-1.2, .5}}], arrowShape}, {1 \[DirectedEdge] 3, Placed[ "g\[SmallCircle]f", {.5, {1.2, 1.2}}], arrowShape} }; graph = PlotDiagram[ vertex, edge, AspectRatio -> 1, ImageSize -> 272, ImagePadding -> {{15, 15}, {5, 20}} ] Export["graph1.png", graph] ```` ````vertex = { {"T", {0, 10}}, {Superscript["T", 2], {10, 10}}, {"T", {20, 10}}, {"T", {10, 0}} }; edge = { {1 \[DirectedEdge] 2, Placed["T\[Eta]", {.5, {.5, -.2}}], ({Black, Arrowheads[0.05], Arrow[#, {2, 2}]} &)}, {3 \[DirectedEdge] 2, Placed[ "\[Eta]T", {.5, {.5, -.2}}], ({Black, Arrowheads[0.05], Arrow[#, {2, 2}]} &)}, {2 \[DirectedEdge] 4, Placed[ "\[Mu]", {.5, {-1.2, -.5}}], ({Black, Dashed, Arrowheads[0.05], Arrow[#, {1, 1}]} &)}, {1 \[DirectedEdge] 4, Placed[ "id", {.5, {2, 1.2}}], ({Black, Arrowheads[0.05], Arrow[#, {2, 2}]} &)}, {3 \[DirectedEdge] 4, Placed[ "id", {.5, {-1.2, 1.2}}], ({Black, Arrowheads[0.05], Arrow[#, {2, 2}]} &)} }; graph = PlotDiagram[ vertex, edge, AspectRatio -> 1/GoldenRatio, ImageSize -> 472, ImagePadding -> {{15, 15}, {5, 20}} ] Export["graph2.png", graph] ```` Notice how I was able to change the style of one of the edges. We can even change the tip of the arrow by using something similar as in one of the examples at Wolfram website. - 1 – rm -rf♦ Jul 24 '12 at 18:16 Those diagrams look really good. – murray Jul 26 '12 at 19:36 lang-mma	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9048752784729004, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/187750/edp-with-complementary-terms?answertab=oldest	# EDP with complementary terms I am considering a problem of a two-dimensional ODE involving Karush, Kuhn and Tucker conditions on one of the unknowns. After a few algebraic manipulations, I end up having to solve the following PDEs: $$\begin{aligned}&Q(x_2,y_2)=\frac{\partial P(x_2,y_2)}{\partial x_2}y_2+\frac{\partial P(x_2,y_2)}{\partial y_2}(f_2(x_2,y_2)+\lambda)\\ &f_1(x_2,y_2)=\frac{\partial Q(x_2,y_2)}{\partial x_2}y_2+\frac{\partial Q(x_2,y_2)}{\partial y_2}(f_2(x_2,y_2)+\lambda)\end{aligned}$$ with: $$x_2-d\leq 0 \quad;\quad \lambda\geq 0 \quad;\quad (x_2-d)\lambda=0$$ In these equations, $d$, $f_1(x_2,y_2)$, and $f_2(x_2,y_2)$ are known; $P(x_2,y_2)$, $Q(x_2,y_2)$, and $\lambda$ should be found. I am not detailing the boundary conditions. Have you ever seen PDEs of this type (\$\lambda is problematic)? If yes, are you aware of a numerical strategy that would solve them? -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 6, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9553136229515076, "perplexity_flag": "head"}
http://mathoverflow.net/questions/43815?sort=votes	Leading eigenvalues Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) If I know about the leading eigenvalues and the eigenfunctions of two operators, is there any result about the leading eigenvalue of the sum of the two operators? - I asked my question in a very general way. In fact, what I have is: My operator is the transfer operator P on L1 functions defined on compact X. It is the pre-dual of the operator U:L∞ \rightlarrow L∞ defined by U(ϕ)=ϕ \circ f for a fixed map f on X. I have PU=Id and UP is the projection. Now my specific question is, if P1(h)=h and P2(g)=g for g,h∈L1 and if 1 is the leading simple isolated eigenvalue for both P1 and P2, then does (P1+P2)/2 have 1 as a leading eigenvalue and what about the corresponding eigenfunction? – filiz Oct 27 2010 at 20:48 ask this as a separate question? – ohai Oct 27 2010 at 20:56 3 Answers You need to add some assumption, otherwise $\begin{pmatrix} 1 & n \\ 0 & 1 \end{pmatrix}$ and $\begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}$ add to a matrix with eigenvalues $1 \pm \sqrt{n}$. Maybe your operators are self-adjoint? - You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. If $A$ is self adjoint, and $\lambda$ its leading eigenvalue, then `$\lambda = \mathrm{sup}_{\langle u,u \rangle =1} \langle u, Au \rangle$`. If $A$ and $B$ are self-adjoint, we have the obvious consequence `$$\mathrm{sup}_{\langle u,u \rangle =1} \langle u, (A+B) u \rangle = \mathrm{sup}_{\langle u,u \rangle =1} \left( \langle u, A u \rangle + \langle u, B u \rangle \right) \leq \mathrm{sup}_{\langle u,u \rangle =1} \langle u, Au \rangle + \mathrm{sup}_{\langle u,u \rangle =1} \langle u, Bu \rangle$$` so the leading eigenvalue of the sum is bounded by the sum of the leading eigenvalues. - I would guess that the magnitude of the leading eigenvalue of the sum is at most the sum of the magnitudes of the leading eigenvalues of the two operators, because the size of the leading eigenvalue is like a norm and the norms have the triangle inequality. Added: I guess this assumes the operators are self adjoint. - "size of the leading eigenvalue is like a norm" is only true if you make additional assumptions. See Homology's example. – Willie Wong Oct 27 2010 at 15:55 @ohai. The modulus of the leading eigenvalue is called spectral radius. It is not a norm at all, since it vanishes on non-zero nilpotent matrices. It does not satisfy a triangle inequality either. Take the sum of the matrices $$\begin{pmatrix} 0 & 1 \\\\ 0 & 0 \end{pmatrix},\qquad\begin{pmatrix} 0 & 0 \\\\ 1 & 0 \end{pmatrix}.$$ – Denis Serre Oct 27 2010 at 16:23 However, the spectral radius equals the operator norm when the matrix is Hermitian. – Denis Serre Oct 27 2010 at 16:24	{"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": 7, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.903828501701355, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/10870/calculating-position-in-space-assuming-general-relativity/10888	# Calculating position in space assuming general relativity Suppose two pointed masses are given in space. Suppose further that one of the masses has a given velocity at (local) time 0. Is there a way to compute its position in a future time? Neglecting general relativity, I will simply compute an integral, but with general relativity, we see that the metric of the space changes with time, so I need to compute an integral with respect to a measure that changes along time. Can this be done? If so, how? Thank you! - ## 1 Answer If the moving mass is small enough, you can do this using the geodesic equation, $$\frac{\mathrm{d}^2x^\lambda}{\mathrm{d}t^2} + \Gamma^{\lambda}_{\mu\nu}\frac{\mathrm{d}x^\mu}{\mathrm{d}t}\frac{\mathrm{d}x^\nu}{\mathrm{d}t} = 0$$ This is essentially the general relativistic equivalent of Newton's second law: it's a differential equation that governs how a test particle's position changes in time. The connection coefficients $\Gamma^{\lambda}_{\mu\nu}$ (also called Christoffel symbols) can be calculated from the metric. So if the metric is known, even if it's time-dependent, you can calculate the connection coefficients in your desired reference frame, plug them in, and find a solution to the geodesic equation that tells you how the particle will move. There may not be an analytic solution, but in all but the most extreme cases you can either solve the equation numerically, or make some approximation that might make it analytically solvable. If the moving particle is not small enough to be considered a test particle, then the situation becomes more complicated because the metric, and thus the connection coefficients, will depend on the motion of the test particle itself. So you wind up with a coupled system of three equations: the geodesic equation for the moving particle, the geodesic equation for the other particle, and the Einstein equation $$G^{\mu\nu} = 8\pi T^{\mu\nu}$$ which tells you how the metric changes in response to the motion of the two particles. In this case it's highly unlikely that you could find an analytic solution, but you could potentially still use a numeric differential equation solver, at least for some range of time. (All the equations are nonlinear so it's likely that your solution would lose accuracy quickly.) -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9413105845451355, "perplexity_flag": "head"}
http://mathoverflow.net/revisions/10833/list	## Return to Answer 2 added 288 characters in body From the recent discussion in the question you linked to, it seems that even (some? most?) finite-dimensional Banach spaces do not have categorical duals in the precise sense you describe. In fact it's clear what goes wrong here with the construction used in the algebraic setting of finite-dimensional vector spaces: the composition of the two maps you describe would be multiplication on $I$ by the dimension of $A$, but that is not a contraction unless $\dim A \le 1$! You might have more luck with a category of Banach spaces and all continuous linear maps, although I do not know whether that also has a closed symmetric monoidal structure. 1 From the recent discussion in the question you linked to, it seems that even (some? most?) finite-dimensional Banach spaces do not have categorical duals in the precise sense you describe. You might have more luck with a category of Banach spaces and all continuous linear maps, although I do not know whether that also has a closed symmetric monoidal structure.	{"extraction_info": {"found_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": 0.9360896944999695, "perplexity_flag": "head"}
http://mathoverflow.net/questions/12232/functional-calculus-for-direct-integrals/12320	## Functional calculus for direct integrals ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Suppose I have a direct integral of Hilbert spaces $H = \int^\oplus H_x dx$, and suppose I have an operator $T: H \to H$ which is decomposable, and so it can be written as $T = \int^\oplus T_x$ for some measurable field of operators $T_x$. Suppose furthermore that every $T_x$ is self-adjoint (and so also $T$ is self-adjoint), and let $f$ be a bounded measurable function on $\mathbb R$. Under what conditions $f(T)$ is decomposable (I guess always) and equal to the integral of the field $f(T_x)$ ? One paper which says something about this problem is Chow, Gilfeather, "Functions of direct integrals of operators". It actually states that the only necessary condition is that $T_x$ are contractions. But unfortunately I don't understand this paper, since it doesn't state its assumptions very precisely - for example, it doesn't seem to be assumed that the operator $T$ (or operators $T_x$) is (are) normal, and so I don't what kind of functional calculus is considered. - (1) I guess you mean every T_x is self-adjoint. (2) The Chow and Gilfeather paper tells us that they are using the functional calculus developed by Schreiber in "A functional calculus for general operators in Hilbert space." The rough idea appears to be compression of functional calculus for unitary dilations of contractions. – Jonas Meyer Jan 18 2010 at 21:39 Normal generalizes self-adjoint. en.wikipedia.org/wiki/Normal_operator – Steve Huntsman Jan 18 2010 at 21:45 I assumed Lukasz already knows that. The typo I am referring to is in the last sentence of the first paragraph "...T_x is normal (and so also T is self-adjoint)...". Since f is defined on R, it is clear that self-adjoint was intended. – Jonas Meyer Jan 18 2010 at 21:55 Jonas: (1) I edited the question accordingly, (2) Thanks for this clarification. I will look in the paper of Schreiber then, but probably this isn't what I'm looking for, since, for example, Chow-Gilfeather deal with contractions which I don't know how could be relevant to the question I asked. – Łukasz Grabowski Jan 19 2010 at 16:10 ## 3 Answers Your guess that it is always decomposable is correct. Here is a way to see this without verifying the expected formula: Borel functional calculus keeps you inside the von Neumann algebra generated by $T$, and the set of decomposable operators on $H$ is the von Neumann algebra of operators that commute with the diagonal operators on $H$ (Kadison-Ringrose 5.2.8, Takesaki 8.16; see also K-R 14.1.10 which has no Google preview). (However, I don't know a reference (or have a proof) that the expected formula is correct. I think it should follow by Fubination once the case of characteristic functions is known.) - what do you mean by fubination? – Martin Brandenburg Jan 20 2010 at 1:13 I mean using Fubini's theorem. There are 2 integrals going on in the Borel functional calculus of a decomposable operator: the decomposition as mentioned in the question; and the functional calculus itself, which uses the spectral decomposition of a normal operator. The case of characteristic functions is just making sure that the spectral decompositions of the T_x's behave as one would want, and my feeling was that if this were established the rest would follow by a form (or analogue) of Fubini's Theorem. It was left vague here and above because I have not put much thought into it yet. – Jonas Meyer Jan 20 2010 at 2:27 ### 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 wanted to make it a comment to Jonas' answer, but the system didn't allow me (because it's too long?) I might be forced to write down my own proof of the expected formula. What do you think about the following sketch? The statement is clear for polynomials. Then take a sequence of polynomials $p_n$ converging somehow to $f$ (How?). This should imply that $p_n(T)$ converges weakly to $f(T)$ and similarly for $p_n(T_x)$ for a.e. $x$. Now one needs to check that $f(T_x)$ is a measurable field, i.e. whether $lim_n (p_n(T_x)v_x,w_x)$ converges to a measurable function whenever $v_x, w_x$ are measurable vecotr fields. But this limit is the same as $lim_n ([p_n(T)]_xv_x,w_x) = ([f(T)]_xv_x,w_x)$, because by your argument we know that $f(T)$ is decomposable (there is some argument needed here). The expected formula then holds because of the same reasoning and uniqueness of the weak limit. - If you want just to give a brief argument with possible references to known results, you can proceed in the following way: one picks a sequence $p_n$ of polynomials converging to $f$ in the weak-measure topology on the Borel functions; then $p_n(T)$ converges to $f(T)$ even strongly (see e.g. Helemski. Lectures and exercises on functional analysis, p. 388). As $p_n(T)$ commuted with every diagonal operator, $f(T)$ does commute as well, and therefore is decomposable (Dixmier. Les algèbres d'opérateurs dans l'espace hilbertien, Thm. II.2.5.1), say, as $\int^\oplus S_x d\nu(x)$. Now, there is a subsequence $p_{n_k}$ such that $p_{n_k}(T_x)$ converges strongly to $S_x$ $x$-almost everywhere (Dixmier, Prop. II.2.3.4), so $S_x=f(T_x)$ almost everywhere. - How do you know that there exists a sequence of polynomials converging to $f$ in the weak-measure topology? You can certainly find a sequence of polynomials for a given operator $T$ such that $p_n(T)$ converges to $f(T)$, but I'm not sure how to get such convergence for all operators $T$. For example, you can't just take a sequence $p_n$ which converges to $f$ point-wise because such a sequence might not exist. – Łukasz Grabowski Feb 10 2010 at 23:35	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 46, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "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": 0.9401957988739014, "perplexity_flag": "head"}
http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.jsl/1183745472	### The Algebraic Sum of Sets of Real Numbers with Strong Measure Zero Sets Andrej Nowik, Marion Scheepers, and Tomasz Weiss Source: J. Symbolic Logic Volume 63, Issue 1 (1998), 301-324. #### Abstract We prove the following theorems: (1) If X has strong measure zero and if Y has strong first category, then their algebraic sum has property s$_0$. (2) If X has Hurewicz's covering property, then it has strong measure zero if, and only if, its algebraic sum with any first category set is a first category set. (3) If X has strong measure zero and Hurewicz's covering property then its algebraic sum with any set in $\mathcal{APC}$ ' is a set in $\mathcal{APC}$ '. ($\mathcal{APC}$ ' is included in the class of sets always of first category, and includes the class of strong first category sets.) These results extend: Fremlin and Miller's theorem that strong measure zero sets having Hurewicz's property have Rothberger's property, Galvin and Miller's theorem that the algebraic sum of a set with the $\gamma$-property and of a first category set is a first category set, and Bartoszynski and Judah's characterization of SR$^\mathcal{M}$-sets. They also characterize the property (*) introduced by Gerlits and Nagy in terms of older concepts. Primary Subjects: 03E20 Secondary Subjects: 28E15, 54F65, 54G99 Full-text: Remote access If you are a member of the ASL, log in to Euclid for access. Full-text is available via JSTOR, for JSTOR subscribers. Go to this article in JSTOR. Permanent link to this document: http://projecteuclid.org/euclid.jsl/1183745472	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 6, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9077861905097961, "perplexity_flag": "middle"}
http://mathematica.stackexchange.com/questions/tagged/numerics+integral-equation	Tagged Questions 1answer 396 views how to solve an implicit integral equation? (iterate to a functional fixed point?) [closed] I reduced a (special case) of my problem to the following code. Even though in this special case all related functions are analytical, DSolve is not the tool for this, though I am indeed looking for a ... 3answers 1k views Solving a Volterra integral equation numerically I would like to solve for $P(t)$, in Mathematica, a Volterra integral equation of the 2nd kind. It is: $$P(t) = R_0(t) + \int_0^t P(t') R_0(t-t')dt'$$ I know the function $R_0$ and would ...	{"extraction_info": {"found_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": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.916664719581604, "perplexity_flag": "middle"}
http://www.reference.com/browse/intergalactic+space	Related Searches Definitions # Intergalactic space Intergalactic space is the physical space between galaxies. Generally free of dust and debris, intergalactic space is very close to a total vacuum. Some theories put the average density of the Universe as the equivalent of one hydrogen atom per cubic meter. The density of the Universe, however, is clearly not uniform; it ranges from relatively high density in galaxies (including very high density in structures within galaxies, such as planets, stars, and black holes) to conditions in vast voids that have much lower density than the Universe's average. The temperature is only 2.73 Kelvin. NASA's COBE mission (COsmic Background Explorer) measured the temperature as 2.725 +/- 0.002 K Surrounding and stretching between galaxies, there is a rarefied plasma that is thought to possess a cosmic filamentary structure and that is slightly denser than the average density in the Universe. This material is called the intergalactic medium (IGM) and is mostly ionized hydrogen, i.e. a plasma consisting of equal numbers of electrons and protons. The IGM is thought to exist at a density of 10 to 100 times the average density of the Universe (10 to 100 hydrogen atoms per cubic meter). It reaches densities as high as 1000 times the average density of the Universe in rich clusters of galaxies. The reason the IGM is thought to be mostly ionized gas is that its temperature is thought to be quite high by terrestrial standards (though some parts of it are only "warm" by astrophysical standards). As gas falls into the Intergalactic Medium from the voids, it heats up to temperatures of $10^5$ K to $10^7$ K, which is high enough for the bound electrons to escape from the hydrogen nuclei upon collisions. At these temperatures, it is called the Warm-Hot Intergalactic Medium (WHIM). Computer simulations indicate that on the order of half the atomic matter in the universe might exist in this warm-hot, rarefied state. When gas falls from the filamentary structures of the WHIM into the galaxy clusters at the intersections of the cosmic filaments, it can heat up even more, reaching temperatures of $10^8$ K and above. ## External links • Intergalactic Space, Natural History, Feb 1998	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 3, "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": 0.9278699159622192, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/36350?sort=newest	## Differential equation for a ratio of consecutive Bessel functions ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) My attempts to search via Google seem to be failing, so I thought of asking here. All the derivatives of the function `$r_n(z):=\frac{J_n(z)}{J_{n-1}(z)}$` where `$J_n(z)$` is the Bessel function of the first kind are expressible in terms of `$r_n(z)$`, for instance `$\frac{\mathrm{d}}{\mathrm{d}z}r_n(z)=r_n(z)^2-\frac{2n-1}{z}r_n(z)+1$` . I've been trying to derive a (linear?) differential equation (hopefully just second-order) that might be satisfied by `$r_n(z)$`, but my manipulative ability does not seem to be up to snuff. Probably my problem can be resolved in two ways: 1. Are there any papers where generating functions/differential equations of ratios of Bessel functions have been studied? ; or 2. How can I derive a differential equation for `$r_n(z)$` with the knowledge that all higher derivatives are expressible in terms of `$r_n(z)$`? (On the other hand, the difference equation for `$r_n(z)$` (and thus its continued fraction representation) is easily derived, so no problem for me there.) I will be interested in any input. Thanks! EDIT: Per Pietro's request, I now tip my hand and reveal my reason for interest: I saw this paper many years ago on a neat method for computing the first few roots of the Bessel function of the first kind. Some time later, I came across J.F. Traub's "Iterative Methods for the Solution of Nonlinear Equations", where he shows the construction of iteration functions involving derivatives and can be constructed to have quadratic, cubic... convergence. (Newton's method is but the first member of this family). I also came across this short note by D.J. Hofsommer on how one might profitably exploit the methods derived by Traub if the function of interest satisfies a simple differential equation (Essentially, one just constructs the Newton correction `$u=\frac{f(z)}{f^{\prime}(z)}$`, and the high-order iteration functions are merely a series in powers of `$u$`). That got me wondering on how one might recursively generate iteration functions with increasing order of convergence for the case of finding the roots of the Bessel function. (On another note, I was able to successfully use the ideas of Traub and Hofsommer for the generation of Gaussian quadrature rules, e.g. Legendre, Lobatto, Radau, and was hoping things might be just as successful for Bessel function root-finding). - A note: I am not of course counting "trivial" differential equations like the derivative formula I gave in the post. So probably something like a relation between `$r_n(z)$` and its first few derivatives. – J. M. Aug 22 2010 at 3:46 1 Would you clarify why the Riccati equation you wrote is not satisfacory for you? – Pietro Majer Aug 22 2010 at 8:05 Pietro: I suppose I should've inserted the adjective "linear" somewhere... :) I'll edit my post and detail my motivation further. – J. M. Aug 22 2010 at 9:36 ## 3 Answers No, you will not find a linear ordinary differential equation (with polynomial coefficients) for $r_n$. This is because $J_{n-1}$ has infinitely many zeroes which are not cancelled out by the zeroes of $J_n$, so that $r_n$ has infinitely many poles. Holonomic functions, aka solutions of LODEs with polynomial coefficients, can only have a finite number of singularities. The transformation between LODEs and Riccati equations involves transformations between an equation for a quantity $y(z)$ [your $r_n$] and an LODE for $-u'/u$. So what you'll get is indeed an LODE for a single Bessel $J$ function! - Thank you Jacques! I suppose I can't do better than ELF and GNOME then. :) – J. M. Aug 22 2010 at 13:39 ### 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. I do not know how helpful this would be to you however it was very helpful to understand the physics and numerics of Bessel. if you are studying elastic wave propagation. The solution of the differential equations of potential wave is the cylindrical Bessel: $r^2 \frac{d^2 R}{dr^2} + r \frac{dR}{dr} + (r^2 - \alpha^2)R = f(r)$ for an arbitrary real integer number α (the '''order''' of the Bessel function). In solving problems in cylindrical coordinate systems, Bessel functions are of integer order (α = ''n''). Since this is a second-order differential equation, there must be two [[linearly independent]] solutions. My solutions use Bessel J(n,.) and Hankel H(n,.) (as previously mentioned) The potential is assumed for each media to be: $\phi=\left(a_{1}J_{n}(Kr)+a_{2}H_{n}(Kr)\right)e^{in\theta} ,$ $\psi(t) = \left(a_{3}J_{n}(kr)+a_{4}H_{n}(kr)\right)e^{in\theta} \,$ However for numerical stability: They (ref. 1) normalize the potential for each layer and for each nth iteration The potential will have Hankel function equal to 1 at the inner radius, while Bessel J will be multiplied by Hankel at outer radius. $\phi=\left(a_{1}J_{n}(Kr)H_{n}(Kr_{out})+a_{2}\frac{H_{n}(Kr)}{H_{n}(Kr_{in})}\right)e^{in\theta},$ $\psi(t)=\left(a_{3}J_{n}(kr)H_{n}(kr_{out})+a_{4}\frac{H_{n}(kr)}{H_{n}(Kr_{in})}\right)e^{in\theta},$ For more details: http://mathoverflow.net/questions/82303/bessel-functions-in-wave-propagation-and-scattering Reference: David C. Ricks and Henrik Schmidt, "A numerically stable global matrix method for cylindrically layered shells excited by ring forces" 1994 - What you wrote does not appear to be related to answering the question. In particular, at no point do you do even mention the function $r_n$. The part of your first sentence where you said "it was very helpful" does not seem to make any sense in the context of this web page. – S. Carnahan♦ Dec 2 2011 at 9:53 1 @ S. Carnahan, the use of Bessel ratios prove to be very useful. I was sharing my own experience that the ratio could be used to reduce the overflow. At higher order n and small arguments, very small real and large imaginary. This resulted in numerical instability. In other words the ratio can be used to normalize the potentials as: $<math>\phi=\left(a_{1}J_{n}(Kr)H_{n}(Kr_{out})+a_{2}\frac{H_{n}(Kr)}{H_{n}(Kr_{in})}\right)e^{in\theta},</math>$ $<math>\psi(t)=\left(a_{3}J_{n}(kr)H_{n}(kr_{out})+a_{4}\frac{H_{n}(kr)}{H_{n}(Kr_{in})}\right)e^{in\theta},</math>$ – Chad Dec 3 2011 at 3:49 As noted by Pietro, the nonlinear differential equation you gave is a Riccati equation. Next, there is a standard method (due to Ince?) to convert a Riccati equation into a second-order linear equation. http://en.wikipedia.org/wiki/Riccati_equation Try it, and see if that is what you want. - The equation I got after trying that transformation out (I came upon this earlier after Pietro pointed it out to me) has the regular solution `$\frac{J_{n-1}(x)}{x^{n-1}}$` ... using that would require me to reckon the Bessel functions themselves from the ratio (resulting in more lines of code, sigh...). Am I to imply that I have to use the Bessel functions themselves, and not the ratios, to fully exploit Hofsommer's observation? – J. M. Aug 22 2010 at 13:13 ...and Jacques's answer makes the answer to my question in the previous comment "yes". – J. M. Aug 22 2010 at 13:45	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 16, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9373842477798462, "perplexity_flag": "middle"}
http://en.wikipedia.org/wiki/Bipolar_coordinates	# Bipolar coordinates See also: two-center bipolar coordinates Bipolar coordinate system Bipolar coordinates are a two-dimensional orthogonal coordinate system. There are two commonly defined types of bipolar coordinates.[1] The other system is two-center bipolar coordinates. There is also a third coordinate system that is based on two poles (biangular coordinates). The first is based on the Apollonian circles. The curves of constant σ and of τ are circles that intersect at right angles. The coordinates have two foci F1 and F2, which are generally taken to be fixed at (−a, 0) and (a, 0), respectively, on the x-axis of a Cartesian coordinate system. Bipolar coordinates form the basis for several sets of three-dimensional orthogonal coordinates. The bipolar cylindrical coordinates are produced by projecting in the z-direction. The bispherical coordinates are produced by rotating the bipolar coordinates about the $x$-axis, i.e., the axis connecting the foci, whereas the toroidal coordinates are produced by rotating the bipolar coordinates about the y-axis, i.e., the axis separating the foci. The classic applications of bipolar coordinates are in solving partial differential equations, e.g., Laplace's equation or the Helmholtz equation, for which bipolar coordinates allow a separation of variables. A typical example would be the electric field surrounding two parallel cylindrical conductors. The term "bipolar" is sometimes used to describe other curves having two singular points (foci), such as ellipses, hyperbolas, and Cassini ovals. However, the term bipolar coordinates is reserved for the coordinates described here, and never used to describe coordinates associated with those other curves, such as elliptic coordinates. Geometric interpretation of the bipolar coordinates. The angle σ is formed by the two foci and the point P, whereas τ is the logarithm of the ratio of distances to the foci. The corresponding circles of constant σ and τ are shown in red and blue, respectively, and meet at right angles (magenta box); they are orthogonal. ## Definition The most common definition of bipolar coordinates (σ, τ) is $x = a \ \frac{\sinh \tau}{\cosh \tau - \cos \sigma}$ $y = a \ \frac{\sin \sigma}{\cosh \tau - \cos \sigma}$ where the σ-coordinate of a point P equals the angle F1 P F2 and the τ-coordinate equals the natural logarithm of the ratio of the distances d1 and d2 to the foci $\tau = \ln \frac{d_1}{d_2}$ (Recall that F1 and F2 are located at (−a, 0) and (a, 0), respectively.) Equivalently $x + i y = a i \cot\left( \frac{\sigma + i \tau}{2}\right)$[2][3] ## Curves of constant σ and τ The curves of constant σ correspond to non-concentric circles $x^2 + \left( y - a \cot \sigma \right)^2 = \frac{a^{2}}{\sin^2 \sigma}$ that intersect at the two foci. The centers of the constant-σ circles lie on the y-axis. Circles of positive σ are centered above the x-axis, whereas those of negative σ lie below the axis. As the magnitude \|σ\| increases, the radius of the circles decreases and the center approaches the origin (0, 0), which is reached when \|σ\| = π/2, its maximum value. The curves of constant $\tau$ are non-intersecting circles of different radii $y^2 + \left( x - a \coth \tau \right)^2 = \frac{a^2}{\sinh^2 \tau}$ that surround the foci but again are not concentric. The centers of the constant-τ circles lie on the x-axis. The circles of positive τ lie in the right-hand side of the plane (x > 0), whereas the circles of negative τ lie in the left-hand side of the plane (x < 0). The τ = 0 curve corresponds to the y-axis (x = 0). As the magnitude of τ increases, the radius of the circles decreases and their centers approach the foci. ## Scale factors The scale factors for the bipolar coordinates (σ, τ) are equal $h_\sigma = h_\tau = \frac{a}{\cosh \tau - \cos\sigma}$ Thus, the infinitesimal area element equals $dA = \frac{a^2}{\left( \cosh \tau - \cos\sigma \right)^2} \, d\sigma\, d\tau$ and the Laplacian is given by $\nabla^2 \Phi = \frac{1}{a^2} \left( \cosh \tau - \cos\sigma \right)^2 \left( \frac{\partial^2 \Phi}{\partial \sigma^2} + \frac{\partial^2 \Phi}{\partial \tau^2} \right)$ Other differential operators such as $\nabla \cdot \mathbf{F}$ and $\nabla \times \mathbf{F}$ can be expressed in the coordinates (σ, τ) by substituting the scale factors into the general formulae found in orthogonal coordinates. ## References 1. Polyanin, Andrei Dmitrievich (2002). Handbook of linear partial differential equations for engineers and scientists. CRC Press. p. 476. ISBN 1-58488-299-9. 2. Happel, John; Brenner, Howard (1983). Low Reynolds number hydrodynamics: with special applications to particulate media. Mechanics of fluids and transport processes 1. Springer. p. 497. ISBN 978-90-247-2877-0. • H. Bateman "Spheroidal and bipolar coordinates", Duke Mathematical Journal 4 (1938), no. 1, 39–50 • Lockwood, E. H. "Bipolar Coordinates." Chapter 25 in A Book of Curves. Cambridge, England: Cambridge University Press, pp. 186–190, 1967. • Korn GA and Korn TM. (1961) Mathematical Handbook for Scientists and Engineers, McGraw-Hill.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 13, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8288360834121704, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/53627/finding-the-electric-field/53644	# Finding the electric field The charge per unit length on a long, straight filament is -91.9 µC/m. (a) Find the electric field 10.0 cm from the filament, where distances are measured perpendicular to the length of the filament. (Take radially inward toward the filament as the positive direction.) MN/C (b) Find the electric field 50.0 cm from the filament, where distances are measured perpendicular to the length of the filament. MN/C (c) Find the electric field 150 cm from the filament, where distances are measured perpendicular to the length of the filament. $λ_q=\frac{dq}{dl}→dq=λ_qdl$ $\vec{E} =k_e\frac{q}{r^2}\hat{r}$ $\|\|\vec{E}\|\|=\frac{λ_qdl}{r^2}$ How can I solve this without knowing the length of the filament? Supposedly this is the suggested way of solving this problem--and I'd like to understand this method very much. Could someone help me? Also, in addition to solving it by this expedient, is it possible to solve this problem by employing a Gaussian surface? - ## 1 Answer You have basically indicated the two natural ways to solve this problem: 1. Integrate the electric field due to small segments along the filament to find the total electric field at a specified point. 2. Use Gauss's law with an appropriate gaussian surface. For method 1, you are basically almost there given your manipulations; you simply need to integrate $$dE = k_e\frac{\lambda\,d\ell}{r^2}$$ But what are the limits of integration if we don't know the length of the filament? Well, the problem states that the wire is "long," which is just physics-speak for infinitely long. So you need to integrate from $-\infty$ to $\infty$. Just make sure to be careful that $r$ denotes the distance from the point where you want to find the field, to the point where $dq$ is located on the filament, so you need to use the Pythagorean theorem to related $r$ to the perpendicular distance to the filament, say $d$, and the length along the filament, say $\ell$. For method 2, you simply pick a cylindrical Gaussian cylinder whose axis of symmetry is along the filament. I'll leave out the details for this method and let you try it. Let me know if you need more help, but this is something you can easily find by googling. Hope that helps! Physics Rocks. - Actually, I am having difficulty applying Gauss's law to a cylinder. – Mack Feb 11 at 21:50 1 – joshphysics Feb 11 at 22:36	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 10, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9412431120872498, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/22157/connection-between-poisson-brackets-and-symplectic-form/22160	# Connection between Poisson Brackets and Symplectic Form Jose and Saletan say the matrix elements of the Poisson Brackets (PB) in the ${q,p}$ basis are the same as those of the inverse of the symplectic matrix $\Omega^{-1}$, whereas the matrix elements of the symplectic form are equal to those of the symplectic matrix $\Omega$ . I have no problem with the latter statement, but I do with the first. That's because the PBs are introducing by writing Hamilton's equation as $\dot{\xi^j} = \omega^{jk}{\frac{\partial H }{\partial \xi^k}}$, where $\omega^{jk}$ are the elements of the $\Omega$, and then taking the Lie derivative of a dynamical variable $f$ along the dynamical vector field, which gives $L_{\Delta}f = (\partial_j f) \omega^{jk}(\partial_k H)+ \partial_{t}f$. It is later said that the term containing the $\omega^{jk}$ is the PB $\left \{ f,H \right \}$, which I have no problem at all, as it gives the right expression for the PBs of the canonical coordinates when the $\omega^{jk}$ are the elements of the symplectic matrix $\Omega$, i.e., in the way it was first introduced through Hamilton's equations. However, as I have mentioned, in a later consideration they say that that the $\omega^{jk}$ of the PBs are the elements of $\Omega^{-1}$, which made me confused, especially because at several times in the book they use $\omega^{jk}$ as components of $\Omega$, and $\Omega_{jk}$ as components of $\Omega^{-1}$, at various derivations. However, I do not believe the statement on the book about the elements of the PBs being those of $\Omega^{-1}$ is wrong, because this is used in the derivation of the preservation of the symplectic form under canonical transformations. Therefore, I think there is a misconception on my behalf somewhere, which I do not know where it is, and I would be thankful if anyone could shed some light on this. - ## 4 Answers It seems that OP just wants to trace a sign convention in a specific book (Ref. [JS]). To answer this most convincingly, we should document where we read what. • In the case of canonical coordinates (also known as Darboux coordinates), [JS] uses a convention where the positions $q^i$ are ordered before the momenta $p_i$, $$\xi^i~=~(q^1, \ldots, q^n,p_1, \ldots p_n),$$ cf. e.g. p. 215 in [JS]. • On top of p. 230 in [JS] is written: The elements $\omega_{jk}$ of the symplectic form and the $\omega^{jk}$ of the Poisson bracket [...] are the matrix elements of $\Omega$ and $\Omega^{-1}$, respectively. Since we are only after a sign convention, let us for simplicity restrict to canonical/Darboux coordinates. Then $\Omega^2=-1_{2n}$, and hence the difference between $\Omega$ and $\Omega^{-1}$ boils down to a sign. • On p. 216 eqs. (5.42b) and (5.43) read $$\tag{5.42b} \omega_{\ell j} \dot{\xi}^j~=~ \partial_{\ell} H,$$ $$\tag{5.43} \Omega ~=~\begin{bmatrix} 0_n & -I_n \cr I_n & 0_n \end{bmatrix}.$$ • On p. 218 eq. (5.47) defines the Poisson bracket $$\tag{5.47} \{f,g\} ~\equiv~ (\partial_jf) \omega^{jk} (\partial_kg) ~\equiv~ \frac{\partial f}{\partial q^{\alpha}}\frac{\partial g}{\partial p_{\alpha}}-\frac{\partial f}{\partial p_{\alpha}}\frac{\partial g}{\partial q^{\alpha}},$$ leading to Hamilton's equation of motion (5.50), $$\tag{5.50} \dot{\xi}^j~=~ \{\xi^j , H\}.$$ We conclude that [JS] has the convention that $$\Omega^{-1} ~=~\begin{bmatrix} 0_n & I_n \cr -I_n & 0_n \end{bmatrix}.$$ So far so good. • On p.228 is written $$\tag{5.75} \omega ~=~ dq^{\alpha} \wedge dp_{\alpha}.$$ Comparing with $\Omega$ and $\omega_{ij}$, we conclude that [JS] has the convention that $$\omega ~=~ \frac{1}{2} \omega_{ij} d\xi^j \wedge d\xi^i ~=~ -\frac{1}{2} \omega_{ij} d\xi^i \wedge d\xi^j .$$ Note the opposite ordering of $i$ and $j$! This is probably the point where OP and many others instead would have liked to defined it oppositely as $$\tag{Opposite JS} \omega ~=~ \frac{1}{2} \omega_{ij} d\xi^i \wedge d\xi^j,$$ and $$\tag{Opposite JS} \omega ~=~ dp_{\alpha} \wedge dq^{\alpha}.$$ References: [JS] Jose and Saletan, Classical Dynamics: A contemporary Approach, 1998. - Oh, so they did do an index swap. This is a very wrong convention--- it conflicts with what you would do with a metric manifold. They should have defined one of the two with a minus sign. OP was right to get confused. Thanks for tracking it down, +1. – Ron Maimon May 14 '12 at 1:42 Thanks for the clarification. It still doesn't make any sense to me why they chose such confusing way to present this, but at least I understand what happened now. I do have to say though that in my version of the book (also from 1998) the matrix you have in eq. 5.43 is for $\Omega ^{-1}$ not $\Omega$, and as a result your matrix after "we conclude..." should be $\Omega$. It doesn't change anything else you pointed out, but could create confusion for someone else. – Raphael R. Jun 11 '12 at 9:29 I have now checked again my 1998 version of [JS]. Its eq. (5.43) is for $\Omega$ not $\Omega^{-1}$. – Qmechanic♦ Jun 28 '12 at 21:06 Let's see: A symplectic form on a manifold $P$ (the phasespace) is a nondegenerate closed two-form $\omega$. This gives you for every function $f \colon P \to \mathbf{R}$ a vectorfield $\xi_f$ defined by $$i_{\xi_f}\omega = -df$$, where $i$ denotes the interior product. Then for two functions $f,g \colon P \to \mathbf{R}$ the Poissonbracket is $$\{f,g\} = \xi_f g = \omega(\xi_f,\xi_g) = - \{g,f\}$$ Now in local coordinates $\xi_f = \xi^i_f \partial_i$, so you get $$i_{\xi_f}\omega(\partial_j) = \omega(\xi^i_f \partial_i,\partial_j) = \xi^i_f \omega(\partial_i,\partial_j) = \xi^i_f \omega_{ij} = -df(\partial_j) = -\partial_j f$$ Since $\omega$ is invertible this means $\xi^i_f = -\omega^{ij}\partial_j f$, hence $$\{f,g\} = \omega^{ij}\partial_i f \partial_j g$$ - I do not see how you answered the confusion exposed on my post. I know the matrices representing the PB and the symplectic form are inverse to one another, the problem lies on their representation, i.e., in the (q,p) representation w^{ij} have to be the matrix elements of the symplectic matrix so we get the right expression for the PB (if the order of the \xi are q1,q2,p1,p2, for example). Whereas, if I use w_(ij) as the matrix elements of the inverse of omega to obtain the symplectic form I get (again in the q,p rep), w = - dq ^ dp instead of w = dq ^ dp. – Raphael R. Mar 10 '12 at 5:05 I don't have Jose and Saletan, but see Theorem 18.1.3, in particular (18.6), in my book Classical and Quantum Mechanics via Lie algebras - The issue here is the raising and lowering of indices. The form $\omega_{ij}$ with lower indices is not the same as the two-tensor $\omega^{ij}$, although with a flat metric (as your examples have), the two are the same (since raising and lowering an index makes no difference). But there is a sign issue--- the two tensors are antisymmetric, and you could define raising the index to swap the index position too. In this case, you get an extra minus sign. The issue is important, because the symplectic matrix has one down and one up index, and squares to -1: $$\omega^i_j \omega ^j_k = -\delta^i_k$$ (assuming no swap definition). Then lowering and raising with the metric, $$\omega^{ij} = g^{jk}\omega^i_k$$ $$\omega_{ij} = g_{ik}\omega^k_i$$ so that if you multiply the entries as matrices, $$\omega^{ij}\omega_{jk} = \omega^i_k g^{kj}g_{jl}\omega^l_k = -\delta^i_k$$ since the g's are inverse to each other, so you get cancellation in the middle. The result is that the entries of the upper index $\omega$ are the inverse matrix (up to a sign) of the lower index $\omega$, and this is what the authors are trying to say. They are either sloppy about the sign, or have an index flipping convention that fixes it up, I don't know. But the sign on the inverse is the reason for your confusion. I wouldn't use their terminology--- I would say that the upper index $\omega$ has entries which are the negative inverse of the lower index $\omega$'s, but they probably fix up any signs using their experience and intuition, so that the formulas end up correct in the end. ### EDIT: Qmechanic found the reference It's a swap convention. They flip the indices for the tensor vs. the form, absorbing the minus sign. This is not a great convention, but it's what they do. Thanks Qmechanic for figuring it out. - Comment to the answer(v1): The notions of a symplectic 2-form $\omega_{ij}$ and a Poisson bi-vector field $\omega^{ij}$ can (and should) be defined without relying on the existence of an arbitrary choice of a metric tensor field $g_{ij}$. – Qmechanic♦ May 11 '12 at 10:20 @RonMaimon: bedroom, haha. – Nick Kidman May 11 '12 at 14:46	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 55, "mathjax_display_tex": 18, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9437106847763062, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/23901/spin-matrices-in-dirac-equation	# Spin matrices in Dirac equation Why in every textbook when deriving Dirac's equation the smallest possible matrices ($2 \times 2$) are used? I wonder why one couldn't use spin 1 matrices ($3 \times 3$) and get relativistic equation for spin 1 particle? - ## 2 Answers One can write down equations for spin-one particles but they won't be the Dirac equation because the Dirac equation, by definition, describes spin-1/2 particles (i.e. fermions such as leptons and quarks). The equations for free fields associated with spin-one particles are the Maxwell equations (for photons etc.) and/or Proca equations (for W-bosons or Z-bosons). - but i think you would agree that its still an interesting exercise to try to obtain the bulk maxwell equations from such a field equation – lurscher Apr 17 '12 at 14:18 @Luboš Motl "by definition, describes spin-1/2" - could you elaborate? Wherever I see Dirac's equation derivation there is only considered the case of 2x2 matrices which obviously leads to spin-1/2 but what about other cases? – John Apr 17 '12 at 17:53 Dear @lurscher, the spin-one equation is Maxwell's equation. The work needed to get from one to another is zero. Well, at most, you may change a basis or something like that. John: electron has $j=1/2$ because the eigenvalue of $J_z=J_{12}=\gamma_1\gamma_2/2$ acting on the electron state is $1/2$ times the original state. Photon has $J=1$ because when you act with $J_{12}$ on the photon state moving in the z-direction, you get $1$ times the original state. One-half is something else than one. For example, if people only eat 1/2 of a bread, 1 bread may feed 2 of them, not one. – Luboš Motl Apr 18 '12 at 4:57 Particles in the standard model are in principle massless and acquire mass through interactions via the Lagrangian. This poses a problem if you want to use the 3x3 matrix generators of angular momentum because massless particles have different lie group representations as particles with mass. The 3x3 generators for particles of spin 1 acts on three states, -1, 0 +1. In the massless case only two states remain. With fermions and the 2x2 SU(2) matrices this issue doesn't exist. The two chiral components $\psi_L$ and $\psi_R$ propagate lightlike and the two states of each chiral component correspond with spin up and down where the spin is parallel or anti parallel to the direction of propagation. Hans - so long story short if I get you right: when it's spin-1/2 both $m=\pm 1/2$ values are ok for particles wheather or not they has a mass, while for spin-1 particles there's a problem with massless particles $m \neq 0$? – John Apr 17 '12 at 17:58	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 12, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9151040315628052, "perplexity_flag": "middle"}
http://mathhelpforum.com/algebra/171803-matrix-transformations.html	# Thread: 1. ## Matrix Transformations With transformations of [3,2] and [2,-2], i am to find the image of the following equations(noting that each of the above transformations are treated separately for each equation: a)y=\|x\| b)y=x^2-3x 2. Originally Posted by johnsy123 With transformations of [3,2] and [2,-2], i am to find the image of the following equations(noting that each of the above transformations are treated separately for each equation: a)y=\|x\| b)y=x^2-3x First get the transformation matrix. Can you do this? Please show what you've tried and say where you get stuck. 3. i am critically most stuck on applying the transformations(they're are horizontal and vertical) of [3,2] and [2,-2] to the equation x^2-3x. i factorized x^2-3x which came out to be x(x-3) and then i applied the transformations seperately, which is what is asked, but then it got a lil messy and wrong. 4. Originally Posted by johnsy123 With transformations of [3,2] and [2,-2], i am to find the image of the following equations(noting that each of the above transformations are treated separately for each equation: a)y=\|x\| b)y=x^2-3x Here's how to do part b) i), you follow the same procedure to do part b) ii)... If you remember back to when you studied quadratics, the quadratic equation $\displaystyle y = a(x - h) + k$ represents a translation of $\displaystyle (h, k)$ to the equation $\displaystyle y = ax^2$. So for your quadratic $\displaystyle y = x^2 - 3x$ $\displaystyle = x^2 - 3x + \left(-\frac{3}{2}\right)^2 - \left(-\frac{3}{2}\right)^2$ $\displaystyle = \left(x - \frac{3}{2}\right)^2 - \frac{9}{4}$, applying the transformation $\displaystyle (3, 2)$ gives $\displaystyle y = \left(x - \frac{3}{2} - 3\right)^2 - \frac{9}{4} + 2$ $\displaystyle = \left(x - \frac{9}{2}\right)^2 - \frac{1}{4}$ $\displaystyle = x^2 - 9x + \frac{81}{4} - \frac{1}{4}$ $\displaystyle = x^2 - 9x + 20$.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 11, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9013824462890625, "perplexity_flag": "middle"}
http://en.wikipedia.org/wiki/Multinomial_theorem	# Multinomial theorem In mathematics, the multinomial theorem says how to expand a power of a sum in terms of powers of the terms in that sum. It is the generalization of the binomial theorem to polynomials. ## Theorem For any positive integer m and any nonnegative integer n, the multinomial formula tells us how a sum with m terms expands when raised to an arbitrary power n: $(x_1 + x_2 + \cdots + x_m)^n = \sum_{k_1+k_2+\cdots+k_m=n} {n \choose k_1, k_2, \ldots, k_m} \prod_{1\le t\le m}x_{t}^{k_{t}}\,,$ where ${n \choose k_1, k_2, \ldots, k_m} = \frac{n!}{k_1!\, k_2! \cdots k_m!}$ is a multinomial coefficient. The sum is taken over all combinations of nonnegative integer indices k1 through km such that the sum of all ki is n. That is, for each term in the expansion, the exponents of the xi must add up to n. Also, as with the binomial theorem, quantities of the form x0 that appear are taken to equal 1 (even when x equals zero). In the case m = 2, this statement reduces to that of the binomial theorem. ### Example The third power of the trinomial a + b + c is given by $(a+b+c)^3 = a^3 + b^3 + c^3 + 3 a^2 b + 3 a^2 c + 3 b^2 a + 3 b^2 c + 3 c^2 a + 3 c^2 b + 6 a b c.$ This can be computed by hand using the distributive property of multiplication over addition, but it can also be done (perhaps more easily) with the multinomial theorem, which gives us a simple formula for any coefficient we might want. It is possible to "read off" the multinomial coefficients from the terms by using the multinomial coefficient formula. For example: $a^2 b^0 c^1$ has the coefficient ${3 \choose 2, 0, 1} = \frac{3!}{2!\cdot 0!\cdot 1!} = \frac{6}{2 \cdot 1 \cdot 1} = 3$ $a^1 b^1 c^1$ has the coefficient ${3 \choose 1, 1, 1} = \frac{3!}{1!\cdot 1!\cdot 1!} = \frac{6}{1 \cdot 1 \cdot 1} = 6$. ### Alternate expression The statement of the theorem can be written concisely using multiindices: $(x_1+\cdots+x_m)^n = \sum_{\|\alpha\|=n}{n \choose \alpha}x^\alpha$ where α = (α1,α2,…,αm) and xα = x1α1x2α2⋯xmαm. ### Proof This proof of the multinomial theorem uses the binomial theorem and induction on m. First, for m = 1, both sides equal x1n since there is only one term k1 = n in the sum. For the induction step, suppose the multinomial theorem holds for m. Then $(x_1+x_2+\cdots+x_m+x_{m+1})^n = (x_1+x_2+\cdots+(x_m+x_{m+1}))^n$ $= \sum_{k_1+k_2+\cdots+k_{m-1}+K=n}{n\choose k_1,k_2,\ldots,k_{m-1},K} x_1^{k_1}x_2^{k_2}\cdots x_{m-1}^{k_{m-1}}(x_m+x_{m+1})^K$ by the induction hypothesis. Applying the binomial theorem to the last factor, $= \sum_{k_1+k_2+\cdots+k_{m-1}+K=n}{n\choose k_1,k_2,\ldots,k_{m-1},K} x_1^{k_1}x_2^{k_2}\cdots x_{m-1}^{k_{m-1}}\sum_{k_m+k_{m+1}=K}{K\choose k_m,k_{m+1}}x_m^{k_m}x_{m+1}^{k_{m+1}}$ $= \sum_{k_1+k_2+\cdots+k_{m-1}+k_m+k_{m+1}=n}{n\choose k_1,k_2,\ldots,k_{m-1},k_m,k_{m+1}} x_1^{k_1}x_2^{k_2}\cdots x_{m-1}^{k_{m-1}}x_m^{k_m}x_{m+1}^{k_{m+1}}$ which completes the induction. The last step follows because ${n\choose k_1,k_2,\ldots,k_{m-1},K}{K\choose k_m,k_{m+1}} = {n\choose k_1,k_2,\ldots,k_{m-1},k_m,k_{m+1}},$ as can easily be seen by writing the three coefficients using factorials as follows: $\frac{n!}{k_1! k_2! \cdots k_{m-1}!K!} \frac{K!}{k_m! k_{m+1}!}=\frac{n!}{k_1! k_2! \cdots k_{m+1}!}.$ ## Multinomial coefficients The numbers ${n \choose k_1, k_2, \ldots, k_m} = \frac{n!}{k_1!\, k_2! \cdots k_m!},$ which can also be written as $= {k_1\choose k_1}{k_1+k_2\choose k_2}\cdots{k_1+k_2+\cdots+k_m\choose k_m} = \prod_{i=1}^m {\sum_{j=1}^i k_j \choose k_i}$ are the multinomial coefficients. Just like "n choose k" are the coefficients when you raise a binomial to the nth power (e.g. the coefficients are 1,3,3,1 for (a + b)3, where n = 3), the multinomial coefficients appear when one raises a multinomial to the nth power (e.g. (a + b + c)3) ### Sum of all multinomial coefficients The substitution of xi = 1 for all i into: $\sum_{k_1+k_2+\cdots+k_m=n} {n \choose k_1, k_2, \ldots, k_m} x_1^{k_1} x_2^{k_2} \cdots x_m^{k_m} = (x_1 + x_2 + \cdots + x_m)^n\,,$ gives immediately that $\sum_{k_1+k_2+\cdots+k_m=n} {n \choose k_1, k_2, \ldots, k_m} = m^n\,.$ ### Number of multinomial coefficients The number of terms in multinomial sum, #n,m, is equal to the number of monomials of degree n on the variables x1, …, xm: $\#_{n,m} = {n+m-1 \choose m-1} = {n+m-1 \choose n}\,.$ The count can be performed easily using the method of stars and bars. ### Central multinomial coefficients All of the multinomial coefficients for which the following holds true: $\left\lfloor\frac{n}{m}\right\rfloor \le k_i \le \left\lceil\frac{n}{m}\right\rceil,\ \sum_{i=1}^m{k_i} = n,$ are central multinomial coefficients: the greatest ones and all of equal size. A special case for m = 2 is central binomial coefficient. ## Interpretations ### Ways to put objects into boxes The multinomial coefficients have a direct combinatorial interpretation, as the number of ways of depositing n distinct objects into m distinct bins, with k1 objects in the first bin, k2 objects in the second bin, and so on.[1] ### Number of ways to select according to a distribution In statistical mechanics and combinatorics if one has a number distribution of labels then the multinomial coefficients naturally arise from the binomial coefficients. Given a number distribution {ni} on a set of N total items, ni represents the number of items to be given the label i. (In statistical mechanics i is the label of the energy state.) The number of arrangements is found by • Choosing n1 of the total N to be labeled 1. This can be done $N\choose n_1$ ways. • From the remaining N − n1 items choose n2 to label 2. This can be done $N-n_1 \choose n_2$ ways. • From the remaining N − n1 − n2 items choose n3 to label 3. Again, this can be done $N-n_1-n_2 \choose n_3$ ways. Multiplying the number of choices at each step results in: ${N \choose n_1}{N-n_1\choose n_2}{N-n_1-n_2\choose n_3}...=\frac{N!}{(N-n_1)!n_1!}\frac{(N-n_1)!}{(N-n_1-n_2)!n_2!}\frac{(N-n_1-n_2)!}{(N-n_1-n_2-n_3)!n_3!}....$ Upon cancellation, we arrive at the formula given in the introduction. ### Number of unique permutations of words In addition, the multinomial coefficient is also the number of distinct ways to permute a multiset of n elements, and ki are the multiplicities of each of the distinct elements. For example, the number of distinct permutations of the letters of the word MISSISSIPPI, which has 1 M, 4 Is, 4 Ss, and 2 Ps is ${11 \choose 1, 4, 4, 2} = \frac{11!}{1!\, 4!\, 4!\, 2!} = 34650.$ (This is just like saying that there are 11! ways to permute the letters—the common interpretation of factorial as the number of unique permutations. However, we created duplicate permutations, due to the fact that some letters are the same, and must divide to correct our answer.) ### Generalized Pascal's triangle One can use the multinomial theorem to generalize Pascal's triangle or Pascal's pyramid to Pascal's simplex. This provides a quick way to generate a lookup table for multinomial coefficients. The case of n = 3 can be easily drawn by hand. The case of n = 4 can be drawn with effort as a series of growing pyramids. ## References 1. National Institute of Standards and Technology (May, 11, 2010). "NIST Digital Library of Mathematical Functions". Section 26.4. Retrieved August 30, 2010.	{"extraction_info": {"found_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": 25, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_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": 0.851693868637085, "perplexity_flag": "head"}
http://cs.stackexchange.com/questions/9265/uniform-cost-search-problem	# Uniform-cost Search Problem Suppose that we take an initial search problem and we add $c > 0$ to the costs on all edges. Will uniform-cost search return the same answer as in the initial search problem? Definitions: Uniform-cost search is also known as lowest cost first. Initial search problem can be any graph with a start and a goal state. You just apply the uniform cost search algorithm on the graph. - 2 What is an initial search problem? What is uniform-cost search? Where do the edges come in? – Yuval Filmus Jan 29 at 2:55 @Yuval: Uniform-cost search is also known as lowest cost first. Initial search problem can be any graph with a start and a goal state. You just apply the algorithm on the graph. – Dave Jan 29 at 3:27 ## 2 Answers The shortest path may indeed change. This is not because of some property of the uniform cost search, but rather, the property of the graph itself. Note that adding a constant positive cost to each edge affects more severely the paths with more edges. Here is an example, where the shortest path has cost $5$: Adding a cost of $1$ to each edge changes the shortest path in the graph as: The original shortest path has a new cost of $10$, whereas the other path has a cost of only $9$. Therefore, any optimal shortest path algorithm, such as Dijkstra's or uniform cost search, will find a different shortest path. The shortest path will not change if it also has the least number of paths among all paths from source to destination. If the shortest distance is $d_{min}$ and $k_1$ edges, and there is some other path with distance $d_{min} + \Delta$ with $k_2$ edges ($k_2 < k_1$), it can be shown that adding any $c > \frac{\Delta}{k_1 - k_2}$ to all the edges will change the shortest path, and therefore your result. Scaling the edges by a constant positive factor (multiplying all edges by $c>0$) will not change the shortest path. - The answer is no, solutions may differ. If suppose your optimal solution in a graph was $s \in \mathbb{R}$ for a path of length $\ell$. If you add $c > 0$ to all edges, this new path will get cost $s + \ell c$. If there is a different (worse) solution of cost $s - \epsilon$, where $0 < \epsilon < c$, and with a path length $< \ell$, you will now get an alternate solution of cost $\leq s + \epsilon + (\ell - 1) c$ which is less then the optimal solution before adding $c$ to every edge. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 21, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9308444857597351, "perplexity_flag": "head"}
http://mathoverflow.net/questions/19084/equivalences-in-model-categories	## Equivalences in Model Categories ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) If $\mathcal M$ is a model category and I know that $A$ and $B$ are isomorphic in $\mathrm{Ho}(\mathcal M)$, is it guaranteed that there is a zig-zag of weak-equivalences in $\mathcal M$ connecting $A$ and $B$? - In particular, you could have searched google.com/… . If you click the top link, it has the answer. – Harry Gindi Mar 23 2010 at 5:06 ## 1 Answer Yes. The isomorphism in $\mathrm{Ho}(\mathcal{M})$ is represented by a morphism in $\mathcal{M}$ from a cofibrant replacement for $A$ to a fibrant replacement for $B$. The "converse to the Whitehead lemma" states that a map in a model category is a weak equivalence iff its image in the homotopy category is an isomorphism. Combining this with the definition of (co)fibrant replacement, we see that $A$ and $B$ are connected by a 3-step zig-zag of weak equivalences. -	{"extraction_info": {"found_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": 0.9231072664260864, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/127517/probability-of-full-rank-of-a-random-matrix/127520	Probability of full rank of a random matrix. Suppose, $G$ is a $k \times n$ binary matrix with $\operatorname{rank}(G) = k$. The first $k$ columns of $G$ are linearly independent and the next $n-k$ columns are linear combinations of the first $k$ columns. $G$ matrix can be a generator matrix of a linear code where the fist $k$ columns are the $k \times k$ identity matrix and the next $n-k$ column contain the parity bits and these columns are linear combinations if the first $k$ columns. I am choosing $u$ number of columns randomly from $G$ to form the matrix $G_u$, where $u<=k$. I want to find the probability that $\operatorname{rank}(G_u) = u$ for any random sub-matrix. Really appreciate any suggestions to solve this problem. Thanks in advance. - I don't think there's enough information to give an answer. For example, if the last $n-k$ columns are all zeros then the probability is very small; for a different matrix of the same size, the probability could be 1. – Gerry Myerson Apr 3 '12 at 6:02 I am actually talking about a generator matrix of a linear channel code in which the first k columns are the identity matrix and the next n-k columns contain the parity bits. G = [ I \| P ]. I have updated my question as well. Thanks for your comment. – anne Apr 3 '12 at 6:08 1 I don't know what restrictions there are on $P$, but I suspect it's still the case that even if you fix $k,n,u$ the answer will depend on $P$. Take the case where $n-k=1$; the answer will (for some values of $u$) depend on how many ones there are in $P$. – Gerry Myerson Apr 3 '12 at 6:23 Let's say P does not contain any all zero column vectors. Each column of P will be a linear combination of any of the k columns. I do not understand how the answer will depend on the number of ones in P. Thanks again... – anne Apr 3 '12 at 6:29 1 Answer As an example showing, why the answer depends on $P$, consider the matrices ($n=4$, $k=u=2$) $$G_1=\pmatrix{1&0&1&1\cr0&1&0&0\cr},\quad G_2=\pmatrix{1&0&1&0\cr0&1&0&1},\quad G_3=\pmatrix{1&0&1&1\cr0&1&0&1}.$$ Two columns of $G_1$ will be linearly independent, iff the second column is among them. There the probability of a full rank submatrix is $3/6=1/2$. OTOH two columns of $G_2$ will be linearly independent unless you happened to pick 1st and 3rd or 2nd and 4th, so the probability of full rank is $4/6=2/3$. As the last case, two columns of $G_3$ will be linearly independent iff they are distinct, so the only pair violating that condition is 1st and 3rd. Therefore the probability of full rank has gone up to $5/6$. It is easy to prove that with this set of parameters: $N=4, k=u=2$ it is impossible that the probability of a full rank submatrix would be equal to $1.$ Probability of full rank can be $1$ only, if the minimum Hamming distance of the dual code is $>u$ (and this is not possible for many choices of $n,k,u$). The way to see this is to view $G$ as a parity check matrix. If some $u$ columns are linearly dependent, then some subset of those columns sums to zero. Putting 1s in the positions of that subset we get a non-zero word of the dual code that has weight $\le u$. If the minimum Hamming distance of the dual code is $>u$, such sets do not exist, and the probability of full rank is equal to $1$. So it all comes down to finding the minimum Hamming distance of the dual code. - I understand above. Thanks. The probability of rank does depend on P. But for large k and n, when I chose u columns randomly then there is a fair possibility of having Pr(rank(Gu)=u) = 1. I want to find a way to prove this analytically. My simulation results show that this is true. – anne Apr 3 '12 at 6:40 My simulation results show that for N=100, k =50 and if I run for u = 1 to 50 Pr(rank(Gu)=u)=1 for all u<k/2. Is it possible to explain this with the minimum hamming distance of the dual code property ? – anne Apr 3 '12 at 6:57 @anne, I doubt it very much. The dual code also has parameters $n=100,k=50$. I don't remember the bounds on the minimum Hamming distance of such codes, but I very much doubt that it could be as high as $25$. And that would likely require a very specific choice of $G$! But surely you haven't done an exhaustive search of the set of $u$ columns either! The low weight words of the dual code may be relatively rare, and if your simulation has just been lucky. It is very hard to prove by simulation that the probability is exactly $1$ :-) – Jyrki Lahtonen Apr 3 '12 at 7:05 I select the u number of columns over 100 trials and in each trials the column selection is different. If I select 25 column randomly during each trials and check the rank of that matrix, then the matrix has rank = 25. This is true for any random matrix created in Matlab. When I think of the hamming distance property you state, the generator matrix of the dual code will be sparse and therefore the minimum distance will not be > u for sure. – anne Apr 3 '12 at 7:25 1 Let's take your $n=100$, $k=50$ case. Your first 50 columns form an identity matrix. Suppose the 51st column just has two ones in it. Say $u=20$. If your 20 columns include the 51st and include the two columns of the first 50 that have their ones in the same places as the 51st column, you have linear dependence, and rank less than 20. The probability of 20 chosen columns including those three particular ones is low, but it isn't zero. – Gerry Myerson Apr 3 '12 at 7:51 show 4 more comments	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 51, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": 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": 0.9303361773490906, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/109537/product-of-a-principal-proper-ideal-by-itself?answertab=active	# Product of a principal proper ideal by itself Let $P$ be a principal proper ideal in an integral domain. Is it $P^2 \subset P$ in general? If yes, how to prove it? For example, if you look at the ideal $(3)=3\mathbb{Z}$ in $\mathbb{Z}$, it is quite simple to show that $3$ is not in $(3)^2$, but how to prove a similar property in a more general context? Now suppose that we have a Dedekind domain. If $P$ is a principal prime ideal, does $P^2 \neq P$ (and so $P^2 \subset P$) follow from the fact that there is a sort of unique factorization of proper ideals in terms of prime ideals? - 4 By $\subset$ you mean $\subsetneq$ (gotten from `\subsetneq`). It is usually a good idea to be emphatic about this in notation. – Mariano Suárez-Alvarez♦ Feb 15 '12 at 6:15 Hint: contains = divides for principal ideals and ideals in Dedekind domains. – Math Gems Feb 15 '12 at 6:27 ## 1 Answer If $P=(x)$, then $P^2=(x^2)$. If $P\subseteq P^2$, there is an $a$ such that $x=ax^2$, so $x(1-ax)=0$. If $x\neq0$, then $1-ax=0$ and $x$ is a unit. Therefore $P=P^2$ implies either $P=0$ or $P=A$, the whole ring. - Nice one. And if we replace "principal" with "prime", does the property still hold? – Oo3 Feb 15 '12 at 18:44	{"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": 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": 0.8961783051490784, "perplexity_flag": "head"}
http://mathhelpforum.com/differential-geometry/170142-proof-critique.html	# Thread: 1. ## Proof critique Hello, I am trying to prove that given a proper integral, $\displaystyle \int_a^b \sum_{n=0}^\infty f_n(x) dx = \sum_{n=0}^\infty \int_a^b f_n(x) dx$, assuming both converge. Here's my attempted proof: $\displaystyle \sum_{n=0}^\infty f_n(x) = \sum_{n=0}^N f_n(x) + R_N(x)$, where $\forall x, \; R_N(x) \to 0$ as $N \to \infty$. Thus $\displaystyle \int_a^b \sum_{n=0}^\infty f_n(x) dx = \int_a^b \sum_{n=0}^N f_n(x) dx + \int_a^b R_N(x) dx = \sum_{n=0}^N \int_a^b f_n(x) dx + \int_a^b R_N(x) dx$. But $\displaystyle 0 \leq \left\| \int_a^b R_N(x) dx \right\| \leq \int_a^b \left\|R_N(x)\right\| dx \leq (a-b)\cdot\max_{a\leq x\leq b}\bigg\{\|R_n(x)\|\bigg\} \to 0$ as $N \to \infty$. Hence $\displaystyle \int_a^b \sum_{n=0}^\infty f_n(x) dx = \sum_{n=0}^\infty \int_a^b f_n(x) dx$. Does this look valid? 2. Originally Posted by mathman88 $(a-b)\cdot\max_{a\leq x\leq b}\bigg\{\|R_n(x)\|\bigg\} \to 0$ as $N \to \infty$. There's a mistake in this step, since we don't know that we can bound $R_N(x)$ uniformly on the interval. That said what you're trying to prove is false as stated (indeed, if it were true, theorems like dominated and monotone convergence would be unecessary), there is a sequence of functions such that all terms have the same integral and the sequence converges pointwise to zero.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 11, "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": 0.9608001112937927, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/59752/how-to-find-onto-one-to-one-and-everywhere-defined-from-a-formula/59755	# How to find onto, one to one and everywhere defined from a formula? I was reading `functions`, I came across this question, Next, the author has given an exercise to find out 3 things from the example, ```` 1. Onto 2. Everywhere defined 3. One to one ```` I am stuck with how do I come to know if it has these there qualities? I mean if I had values I could have come up with an answer easily but with just a function formula, I don't get how to proceed. Please guide me. thanks - I see that the domain is fully occupied so everywhere defined is true. I am stuck with the rest – Fahad Uddin Aug 25 '11 at 19:30 Well, the "one-to-one" part is easily demonstrated... you know what that means, right? – J. M. Aug 25 '11 at 19:30 @J.M: Yes, that I get it now – Fahad Uddin Aug 25 '11 at 19:32 ## 2 Answers A function is "onto" if for every element $c$ in the range there is an element $b$ in the domain such that $f(b)=c$. So can you determine whether this is true? If I give you a number $c$, can you find an appropriate $b$? A function is "one to one" if $f(c)=f(d)$ implies $c=d$. $g(x)=x^2$ is not one to one because $g(2)=g(-2)$. What happens here? Added: If the sets are finite, you can draw a diagram. You have some points that are set A and some points that are set B. If you draw a line from each point in A to its image in B, • Onto means that every point in B is the image of at least one point in A. So there must be a line to all the points of B • Everywhere defined means there is a line from each point in A. Sometimes people require that a function be everywhere defined, reducing the set A as necessary so every element has a function value. • One to one means that each point in B is connected to at most one point in A. - Sure, for every c, I can come up with a b. I did not got your one to one part/ – Fahad Uddin Aug 25 '11 at 19:46 1 A function $f$ is one-to-one if $f(c) = f(d)$ implies $c = d$. Therefore, assume that $f(c) = f(d)$, and see if you can prove that $c = d$; if so, this means that $f$ is one-to-one. – Tanner Swett Aug 25 '11 at 19:54 I would be very thankful if anyone of you could give me a virtual representation through the diagram. – Fahad Uddin Sep 11 '11 at 15:15 @fahad: see my addition. – Ross Millikan Sep 11 '11 at 16:12 @Ross: Thanks a lot for that :) – Fahad Uddin Sep 11 '11 at 16:18 Presumably when the problem says $A=B=Z$ it means that $A$ and $B$ are each the set of (positive, zero and negative) integers. So you now have the values and can test each of the three properties for this function. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 18, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9464110136032104, "perplexity_flag": "head"}
http://mathoverflow.net/questions/tagged/mathematical-finance	## Tagged Questions 1answer 266 views ### Mathematical properties of financial prices Prices of financial assets (stock-market prices or currency exchange rates) obviously resemble trajectories of stochastic processes. What is known about their mathematical prope … 0answers 193 views ### American put option pricing by “binomial trees” Dear MO World, I'm teaching a financial mathematics course and have found a fascinating (to me) numerical phenomenon and wonder if anyone has studied it, or knows anything similar … 1answer 221 views ### Trajectorial version of Doob’s $L^2$ inequality In the paper http://www.mat.univie.ac.at/~schachermayer/pubs/preprnts/prpr0154.pdf you can find a trajectorial version of Doob's inequality. It is given by: \bar{s}^2_T+4\su … 10answers 2k views ### Expected value as decision criterion in the context of rare events I have often seen discussions of what actions to take in the context of rare events in terms of expected value. For example, if a lottery has a 1 in 100 million chance of winning, … 1answer 220 views ### Solving an Ornstein-Uhlenbeck-like SDE $y(t,T)=H_t + \mathbb{E}[\int_t^T y(s-,T)dX_s\|\mathcal{F}_t]$ I have asked a similar question involving some finance background some time ago here math.stackexchange, however no really good answer came up. I was able to find a solution at lea … 2answers 526 views ### Compactness of the set of densities of equivalent martingale measures Consider an incomplete market $(\Omega,\mathcal F,\mathbb P)$ driven by a semimartingale $S=(S_t)_{t\in[0,T]}$. Under the no free lunch under vanishing risk (NFLVR) assumption, the … 10answers 886 views ### Is there any straightforward way to substitute for Gaussian/Brownian assumptions in financial mathematics? A huge amount of financial mathematics assumes Gaussian distributions of risks and Brownian movement of prices. What efforts have there been to replace these with heavy-tailed dist … 2answers 149 views ### market completion in stochastic volatility model Hi all, Consider a stochastic volatility model. As there are two sources of risk and one asset only, this is an imcomplete market. One can complete the market by considering a der … 1answer 204 views ### Arbitrage free price of a derivative when the price is collected over the lifetime of the derivative [closed] Let $X_t$ be an american style financial derivative with random exercise time $T$ where $t$ and $T$ belongs to some finite set $A$. Buying this derivative requires the buyer to pay … 0answers 73 views ### stochastic volatility valuation equation I'm trying to derive the valuation equation under a general stochastic volatility model. What one can read in the litterature is the following reasonning: One consider a replicat … 5answers 2k views ### Discrete version of Ito’s lemma Could anyone give me some references where I could find (a) discrete version(s) of Ito's lemma (b) a proof how it converges to the continuous form in the limit (c) its usage within … 3answers 4k views ### Visualisation of Riemann-Stieltjes-Integrals The Riemann-Stieltjes-Integral $\int_a^bf(x)dg(x)$ is a generalization of the Riemann Integral. It is e.g. heavily used as a starting point for stochastic integration. The approxim … 3answers 1k views ### Rigorous definition, detection and test for trending vs. mean-reverting behaviour of stochastic processes This is a question that has haunted me for some time. In the domain of time series you always talk about trends and mean reversion. But at least to me these concepts are either def … 1answer 347 views ### Better understanding of the Datar Mathews Method - Real Option Pricing [closed] Hi all, in their paper "European Real Options: An intuitive algorithm for the Black and Scholes Formula" Datar and Mathews provide a proof in the appendix on page 50, which is not … 1answer 9k views ### Covariance and standard deviation relationship I would like to know if an increase in the covariance between two variables would imply that the standard deviation for one of the variables has increased? This is assuming that t … 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": 11, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9044950008392334, "perplexity_flag": "middle"}
http://nrich.maths.org/325/index?nomenu=1	## 'Picture Story' printed from http://nrich.maths.org/ ### Show menu The picture illustrates the formula for the sum of the first six cube numbers: $1^3 + 2^3 + 3^3 + ... + 6^3 = ( 1+ 2 + 3 + ... + 6)^2$ Can you see which parts of the picture represent each part of the formula? Could you draw a similar picture to represent the sum of the first seven cube numbers? What about other sums of cubes? Suggest a formula for the sum of the first $n$ cube numbers. Can you prove that your formula works, using diagrams and explanations? Send us your thoughts!	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8774117231369019, "perplexity_flag": "middle"}
http://www.physicsforums.com/showthread.php?p=3054894	Physics Forums Electric Potential Difference on a Cone 1. The problem statement, all variables and given/known data I'm working out of Griffith's "Intro to Electrodynamics" and the problem states: "A conical surface (an empty ice-cream cone) carries a surface charge $$\sigma$$. The height of the cone is h as is the radius of the top. Find the potential difference between points a (the vertex) and b (the center of the top). 2. Relevant equations and Attempt at a solution So, since this is the chapter that I'm in, I'm going to use $$\[V(R)=\frac{\sigma }{4\pi \varepsilon _{0}}\int _{S}\frac{da{}'}{R}\]$$. Now since a is at the vertex I chose $$\[\vec{a}=0\]$$ and $$\[\vec{b}=h\hat{z}\]$$. Thus the equation would become $$\[V(\mathbf{b})-V(\mathbf{a})=\frac{\sigma }{4\pi \varepsilon _{0}}\int _{S}\left [ \frac{{da}'}{\sqrt{(h-{z}')^2+{s}'^2}} -\frac{{da}'}{\sqrt{{z}'^2+{s}'^2}}\right ]\]$$ Now da' is what I was having a little trouble attaining, so I thought the best place to start would be with the surface area of the cone: $$\[a'=\pi s\sqrt{s^2+z^2}\]$$ but since the radius s is equal to the height z in our case the formula becomes $$\[a'=\pi s\sqrt{s^2+s^2}=\sqrt{2}\pi s^2\]$$. Now since fractions of this area can be represented by multiplying in terms of the angle that determines the fraction of area, $$\[\frac{\theta }{2\pi }\]$$. Thus $$\[a'=(\sqrt{2}\pi s^2)\cdot (\frac{\theta }{2\pi })=\frac{\sqrt{2}}{2}s^2\theta \]$$ and if I consider the angle to be small $$\[a'=\frac{\sqrt{2}}{2}s^2d\theta \]$$. Now to find the differential area I should subtract to get [tex]$da'=\frac{\sqrt{2}}{2}(s+ds)^2d\theta -\frac{\sqrt{2}}{2}s^2d\theta=\frac{\sqrt{2}}{2}d\theta(s^2+sds+ds^2-s^2)=\frac{\sqrt{2}}{2}sdsd\theta$ [/tex] since ds^2 is to small to matter. The main equation then becomes: $$\[V(\mathbf{b})-V(\mathbf{a})=\frac{\sigma }{4\pi \varepsilon _{0}}\int _{S}\left [ \frac{{da}'}{\sqrt{(h-{z}')^2+{s}'^2}} -\frac{{da}'}{\sqrt{{z}'^2+{s}'^2}}\right ]=\frac{\sigma }{4\pi \varepsilon _{0}}\int _{S}\left [ \frac{1}{\sqrt{(h-{s}')^2+{s}'^2}} -\frac{1}{\sqrt{{s}'^2+{s}'^2}}\right ](\frac{\sqrt{2}}{2}{s}'{ds}'{d\theta}' )\]$$ $$\[=\frac{\sqrt{2}\sigma }{8\pi \varepsilon _{0}}\int_{0}^{2\pi }\int_{0}^{h}\left ( \frac{{s}'}{\sqrt{(h-{s}')^2+{s}'^2}}-\frac{\sqrt{2}}{2} \right ){ds}'{d\theta}' \]$$ $$\[=\frac{\sqrt{2}\sigma }{4\varepsilon _{0}} [(-hln({s}'-h)+\frac{{s}'^3}{3}+{s}')\|_{0}^{h}-\frac{\sqrt{2}}{2}h]\]$$ but the above does not converge when evaluated so I'm at a loss. This isn't for a class or anything, I'm just self studying so answer at your convenience. PhysOrg.com science news on PhysOrg.com >> Front-row seats to climate change>> Attacking MRSA with metals from antibacterial clays>> New formula invented for microscope viewing, substitutes for federally controlled drug Recognitions: Homework Help Science Advisor Try slicing the cone along the vertical axis into rings of area $dA = 2\pi s dz$ where s = radius of the ring at height z, which is a linear function of z. So each ring carries a charge that is proportional to z. That should be easy to integrate. AM Tags cone, difference, electric, potential Thread Tools \| \| \| \| \|--------------------------------------------------------------\|-------------------------------\|---------\| \| Similar Threads for: Electric Potential Difference on a Cone \| \| \| \| Thread \| Forum \| Replies \| \| \| Advanced Physics Homework \| 12 \| \| \| Advanced Physics Homework \| 2 \| \| \| Advanced Physics Homework \| 3 \| \| \| Introductory Physics Homework \| 1 \| \| \| Introductory Physics Homework \| 5 \|	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 13, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9156445860862732, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/304771/bookshelf-problem/304778	# Bookshelf problem There is this thought problem I've been trying to solve, it goes as follows Imagine a bookshelf with a finite number of books in it, to which a finite number of people have access. Each person has a paper and pen where they can keep track of what books they have already read. the books have no covers, so in order to find out if you have read it you must take it out and open it. What is the most effective method each person can use to find a book they haven't read before? conditions: • You cannot change the order of the books in the bookshelf (you can however order the bookshelf into blocks/sections) • The bookshelf - and its order - is the same for all people edit: • The bookshelf can also have new books added, so the system must be able to adapt to that, • books can get stolen too. so a missing index can appear. For example we can let each person take a book at random, then check on their list if they have read this book before, this will work. but once you reach the 900th/1000 book, you need to take 10 books in order to be able to read one. Another example is where everyone writes down what they read, but after a few 100 books this gets rather slow. since you need to keep checking a list of 500 before you can pick one. - 2 Everybody walks up to the bookshelf, and writes each book's title on the spine of said book. – anorton Feb 15 at 12:06 heh, smart but that would be avoiding the problem – Sam Feb 15 at 12:11 2 Life is about avoiding problems. – Gerry Myerson Feb 15 at 12:19 This question is extremely underspecified, to the point where there is not going to be any "correct" answer. I am going to remove the "logic" and "algorithm" tags because this is not about mathematical logic nor about "algorithms" in the formal sense. It's just a puzzle. I added the "modeling" tag because the question is so underspecified that the main difficulty is modeling the situation that is described. – Carl Mummert Feb 15 at 12:52 ## 2 Answers I would say that writting on the spine solves the problem, not avoids it. However, given your attitude, let me ask this question: is it possible to make an index? That is, if it is not allowed to write on spines, you could write on paper, everybody can count books and bookshelves (and if you can't write on paper you could do a "spoken index"). If yes, then just go through the index (make many copies) and the problem is solved (indirect references are a common way in computer science). If no, then the readers are independent of each other but for making some books unavailable to others at some particular time. Then, let $n$ be the number of readers. Split the books into $n$ shelves and make a rule that reader $i$ at time $t$ can read only from shelf $(i+t) \bmod n$ (so there are no collisions). Then for every reader assign a random permutation of books "to read", and at any given time the reader would be to read first available book from the list (and of course, remove it from the list afterwards). After any new book has been added, put it into the remaining list of all the readers and reshuffle them all (lists, not readers). Is this close to what you want? - sounds perfect! can't see anything wrong with it. thanks a ton – Sam Feb 15 at 12:46 1 and my attitude, I may come across wrong, sorry if that is the case I do really appreciate all the answers and help you guys are giving! – Sam Feb 15 at 12:49 All jokes aside (see comment on OP)... Let the people be numbered $P_1, P_2, P_3,\ldots$ and the books $B_1, B_2, B_3, \ldots$ If we have $P_n$ start on book $B_n$, the next book they have not yet read is book $B_{n+1}$. If this book does not exist, wrap around to the start of the shelf. Essentially, have each person start with a different book, and move down the shelf one-by-one. No unnecessary books are opened, so this is optimal. - You are assuming everyone reads at the same speed, no? – user1729 Feb 15 at 12:16 Great answer, however this would mean some people are reading all the old books, if a new book is added(B1001) P1 would take ages to get to it. I forgot to mention this in the question. – Sam Feb 15 at 12:18 @user1729 yes everyone is a speed reader, takes them about 2 seconds :) speed is not an issue here. also books can magically be read by 2 people at the same time – Sam Feb 15 at 12:19 You should edit the body of the question so it asks what you actually want to ask. But first, think about what you want to ask, so we don't wind up with $17$ edits. – Gerry Myerson Feb 15 at 12:21 1 @Sam: that is not a limitation according to the question that you actually asked. And, of course, once enough books are out then everyone will be reading at the same time. It's only at the beginning and end that some people have to wait. – Carl Mummert Feb 15 at 12:53 show 4 more comments	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 11, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9644126892089844, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/tagged/complex-analysis?sort=active&pagesize=50	# Tagged Questions The theory of functions of one complex variable with an emphasis on the theory of complex analytic (or holomorphic) functions of one complex variable. Typical topics include: Cauchy's integral formula, singularities, poles, meromorphic functions, Laurent and Taylor series, maximum modulus principle, ... 1answer 29 views ### Integrate: $\int_0^{\infty}\frac{\sinh (ax)}{\sinh x} \cos (bx) dx$ Q: If $\|a\|< 1$ and $b>0$, show that $$\int_0^{\infty}\frac{\sinh (ax)}{\sinh x} \cos (bx) dx = \frac{\pi \sin (\pi a)}{2 (\cos (\pi a)+\cosh (\pi b))}$$ I need to evaluate the above ... 3answers 37 views ### Harmonic Function bounded by a linear function Let $u$ be a harmonic function on $\mathbb C$. Suppose that for each $\epsilon > 0$, there is a constant $C_\epsilon$ such that $$u(z) \leq C_\epsilon + \epsilon \|z\| .$$ I am trying to show that ... 0answers 28 views ### Conformal mapping from exterior of closed unit disk onto exterior of horizontal interval. This is a problem from Bak-Newman's "Complex Analysis", #4 from Chapter 14 "The Riemann Mapping Theorem". The question is this: Verify directly that $F(z) = z + \frac{1}{z}$ is the unique conformal ... 0answers 20 views ### Series expansion of a meromorphic function in a theoretical physics book, the author makes the following claim: $$\frac{1}{e^z + 1} = \frac{1}{2} + \sum_{n=-\infty}^\infty \frac{1}{(2n+1) i\pi - z}$$ and justifies this as These series ... 1answer 59 views ### Integrate: $\int_{a - i\infty}^{a + i\infty} \frac{e^{tz}}{z^2 + p^2}dz$ Q. Show that : $$\int_{a - i\infty}^{a + i\infty} \frac{e^{tz}}{z^2 + p^2}dz = \frac{\sin pt}{p}$$ I considered the following contour \int_\Gamma \frac{e^{tz}}{z^2 + p^2}dz + \int_{a - ... 1answer 217 views ### interpolating the primorial $p_{n}\#$ The primorial $p_{n}\#$ is given by the product $p_n\# = \prod_{k=1}^n p_k$ (where $p_{k}$ is the $k$th prime) -- is there a natural (a la the gamma function $\Gamma(z)$) way of interpolating it for ... 2answers 31 views ### Does having multiple limit values at a point imply essential discontinuity? In Complex Analysis, do "jump discontinuities" exist? If I find that a function of $z$ approaches two different values as z is approached from two different directions, can I immediately conclude ... 2answers 73 views ### limit of $e^z$ at $\infty$ What's the limit of $e^z$ as $z$ approaches infinity? I am given that the answer is "There is no such limit." Is this correct, and if so, am I correct to demonstrate this by showing that as $y$ ... 2answers 206 views ### Möbius transform which completely preserves circles (how to map a circle?) (remmert theory of complex function) I am trying to solve this exercise, however it seems impossible because I don't know how to map a circle, and I will be very thankful if somebody points out to ... 1answer 20 views ### What is the inverse z transform of 1/(z-1)^2? I'd like to know how to calculate the inverse z transform of $\frac{1}{(z-1)^2}$ and the general case $\frac{1}{(z-a)^2}$ 1answer 123 views ### Graphing the complex function I'm looking for some software that my help me to graph some complex functions on unit circle. I.e. let say if I have $\ f(z)=1/(1-z)$ I want to see to give an input an image with unit circle and want ... 3answers 92 views ### Finding the Fourier Series of $\sin(x)^2\cos(x)^3$ I'm currently struggling at calculation the Fourier series of the given function $$\sin(x)^2 \cos(x)^3$$ Given Euler's identity, I thought that using the exponential approach would be the easiest ... 1answer 29 views ### Complex Analysis problem related to T/F statements I am stuck on the following problem: For each of the statements below, indicate whether they are true or false. If true, give a proof. If false, give a counter example or explain why it cannot ... 1answer 85 views ### Integrating: $\int_0^\infty \frac{\sin (ax)}{e^x + 1}dx$ I am trying to evaluate the following integral using the method of contour which I am not being able to. Can anyone point out what mistake I am making? $$\int_0^\infty \frac{\sin ax}{e^x + 1}dx$$ I ... 1answer 49 views ### Show that F is a normal family G is a domain in $C$, where $C$ is the complex plane and $G \neq C$ and $M>0$ we have that: $F=\{f\in H(G): \int_{G}\|f(z)\|^2 dm_2(z)\leq M\}$ How do I show that F is a normal family? - Some of the ... 2answers 13 views ### Show coth is a conformal mapping of the horizontal strip I want to show that $\coth=\frac{e^{2z}+1}{e^{2z}-1}$ is a conformal mapping of the horizontal strip $S=\{z\in C: \pi/4<\text{Im}(z)<3\pi/4\}$ onto the unit disc U, but I can't seem to get the ... 2answers 21 views ### Problem involving the computation of the following integral I was solving the past exam papers and stuck on the following problem: Compute the integral $\displaystyle \oint_{C_1(0)} {e^{1/z}\over z} dz$,where $C_1(0)$ is the circle of radius $1$ around ... 2answers 32 views ### Computing real integrals using the Residue Theorem where singularities are on the real line How would you compute, for $a>0$ the integral $$\int_0^\infty \frac{\sin x}{x(x^2 + a^2)} dx \, \, ?$$ I've computed the residues of the function $$f(z) = \frac{e^{iz}}{z(z^2 + a^2)}$$ which I ... 1answer 55 views ### Does there exist $g$ s.t $g'=f$? I have the following homework question: Let G be the bounded open set shown in gray in this picture, whose boundary consists of eight line segments. The endpoints of those segments are, as ... 1answer 98 views ### An issue with approximations of a recurrence sequence By trying to give an approximation to a given recurrence sequence I encountered a problem. To be more precise I have a method but it fails if the right condition is not met and I wonder how I should ... 1answer 329 views ### Why is every holomorphic bijection of the Riemann sphere a Möbius transformation? Just based on some reading, I know that every Möbius transformation is a bijection from the Riemann sphere to itself. I'm curious about the converse. For any holomorphic bijection on the sphere, why ... 3answers 36 views ### Finding the Taylor series of $f(z)=\frac{1}{z}$ around $z=z_{0}$ I was asked the following (homework) question: For each (different but constant) $z_{0}\in G:=\{z\in\mathbb{C}:\, z\neq0$} find a power series $\sum_{n=0}^{\infty}a_{n}(z-z_{0})^{n}$ whose sum ... 1answer 72 views ### Problem with calculating a winding number I have a problem with calculating the winding number $n\left ( \gamma ,\frac{1}{3} \right )$ of the curve $\gamma :\left [ 0,2\pi \right ]\rightarrow \mathbb{C}, t \mapsto \sin(2t)+i\sin(3t)$. ... 1answer 26 views ### Evaluate the following contour integral I was solving old exam papers and I am stuck on the following question: Evaluate the contour integral $\displaystyle \oint_{C} \frac{dz}{(\bar z-1)^2}$ where $C$ is the semi-circle \$\|z-1\|=1, \Im ... 1answer 120 views ### Lipschitz Domain Let $U$ denote the open unit ball in the complex plane and $G=${$(z(1)+z(2),z(1)z(2))$ belongs to $C^2:z(1),z(2)$ belongs to U}.Is the boundary of $G$ Lipschitz?Justify. Thanks for any help. 1answer 388 views ### What is wrong with this fake proof $e^i = 1$? $$e^{i} = e^{i2\pi/2\pi} = (e^{2\pi i})^{1/(2\pi)} = 1^{1/(2\pi )} = 1$$ Obviously, one of my algebraic manipulations is not valid. 2answers 65 views ### If $\mathbb f$ is analytic and bounded on the unit disc with zeros $a_n$ then $\sum_{n=1}^\infty \left(1-\lvert a_n\rvert\right) \lt \infty$ I'm going over old exam problems and I got stuck on this one. Suppose that $\mathbb{f}\colon \mathbb{D} \to \mathbb{C}$ is analytic and bounded. Let $\{a_n\}_{n=1}^\infty$ be the non-zero zeros of ... 3answers 199 views ### Riemann sphere and Maps Could somebody please clarify the following for me? I am not too clear about the relationship between the Riemann sphere and Mobius maps. I know that we can through projection make some Mobius maps ... 1answer 33 views ### A sequence of analytic functions with converging path integrals converges how? Suppose you have a sequence $\{f_n\}_{n \in \mathbb Z^+}$ of analytic functions on a compact set $K$ such that for every smooth curve $\gamma$ in $K$ the sequence \$\{\int_\gamma f_n\}_{n \in \mathbb ... 2answers 138 views +50 ### Integrate: $\int_0^{\infty} \frac{\sin (ax)}{e^{\pi x} \sinh(\pi x)}dx$ How to evaluate the following $$\int_0^{\infty} \frac{\sin (ax)}{e^{\pi x} \sinh(\pi x)} dx$$ Given hints says to construct a rectangle $0\to R\to R+i\to i \to 0$ and consider \$\displaystyle ... 2answers 37 views ### Largest disc around which this complex function is one-to-one? How would you determine explicitly the largest disc round the origin on which the function $f(z) = z^2 +z$ is one-to-one? Is there a general method to do this for functions of this type? 2answers 46 views ### Finding all $z$ such that the modulus of $f(z)=e^{(z+1)/(z-1)}$ is equal to/at most $1$ I was solving a previous exam paper and there I got stuck on the following problem: Let $f(z)=e^{\frac{z+1}{z-1}}$. Then find all $z \in \Bbb C$ for which $\|f(z)\|=1$, \$\|f(z)\|\le ... 1answer 137 views ### Sequence of entire functions with some conditions Let $\{f_{n}(z)\}$ be a sequence of analytic functions in the upper half plane (in a Hilbert space $H$) and continuous on the real axis, such that (1) $0<\|f_{n}(x)\|\leq 1$ for all \$x\in \mathbb ... 0answers 78 views ### Why is Every Elliptic Function of Order $2$ the Möbius Tranform of a $\wp$-function? I'm trying to prove that every elliptic function of order $2$ has the form $$f(z)=\frac{a\wp(z-z_0)+b}{c\wp(z-z_0)+d}$$ I've got the following so far. Let $f$ be an elliptic function of order 2. ... 2answers 63 views ### Calculate the following integral I was thinking about the following problem: Calculate $\displaystyle \oint_{C}(\bar z)^2 \mathrm{dz}$ where $C:\|z-1\|=1$ is oriented counter clock-wise. My Attempt: I take $z-1=e^{i\theta}$ ... 4answers 125 views ### Why is it meaningless for a closed set to be polygonal path connected? My textbook (Complex Analysis by Saff & Snider) defines connectedness for open sets; the given definition of a connected open set is: a set in which every pair of points can be joined by a ... 2answers 81 views ### Laurent series, complex analysis. I am working on a complex analysis exercice. I need to find the Laurent series of the function: $$f(z) = \frac{e^z}{z + 1}$$ about $z = −1$. I know that the result is ... 3answers 36 views ### When speaking of neighbourhoods in complex analysis, are we always referring to circular neighbourhoods? In Complex Analysis, does "neighbourhood" automatically mean "circular neighbourhood", or do non-circular ones exist? 1answer 152 views ### Solution of a differential matrix equation Given a differential matrix equation, ie $X'=A(z)X+B(z)$ where both $A$ and $B$ are matrix of size $n\times n$ with coefficients that are holomorfic functions in a convex open set $\Omega$ and ... 4answers 83 views ### what is the relation of smooth compact supported funtions and real analytic function? What is the major difference between real analytic and test function (smooth compact supported functions). Can we find a real analytic function $f$ on $R^n$ which is also smooth compact supported? If ... 1answer 20 views ### Extension of real functions to Riemann surface Let $f:\mathbb{R}^*_+\to \mathbb{R}$ be a function that is locally the restriction of an holomorphic function. Notating $R$ the Riemann surface of the complex logarithm, with coordinates ... 1answer 34 views ### Show $5z^n=e^z$ has a finite number of zero in $\{a<\Im z < v\}$ and $\{a < \Re z < b \}$ Show that the number of roots $N$ of the function $$h(z)= f(z)-g(z)=5z^n - e^z, n\ge 1,$$ is at most finite in any horizontal strip $$\alpha=\{z:a<\Im z<b\},$$ and any vertical strip, ... 1answer 62 views ### To show that $\sqrt{z^2-1}=\exp(\frac{1}{2} \log(z^2-1))$ is analytic in the plane minus [-1,1] By the definition of logarithm branch, we set that $\sqrt{z^2-1}=\exp(\frac{1}{2} \log(z^2-1))$ However to show that the $\sqrt{z^2-1}$ is analytic in the entire plane minus the interval [-1,1], it ... 1answer 25 views ### Impossibility of polynomial approximation This is exercise 12.6 in David Ullrich's Complex Made Simple. He has discussed many ways to prove the existence of polynomial approximations to functions in the complex plane, but not how to show such ... 3answers 69 views ### A boundary version of Cauchy's theorem I am looking for a reference for the following theorem (or something like it) that is not Kodaira's book. Let $D$ be a domain and $\overline{D}$ be it's closure. Suppose that \$f:\overline{D} ... 2answers 39 views ### Contour Integral: $\int^{\infty}_{0}(1+z^n)^{-1}dz$ I'm working through Priestley's Complex Analysis (really good book by the way) and this Ex 20.2: Evaluate $\int^{\infty}_{0}(1+z^n)^{-1}dz$ round a suitable sector of angle $\frac{2\pi}{n}$ for ... 1answer 38 views ### Dual of holomorphic functions (with the $L^1$ topology) Let $\Omega$ be a connected domain of the complex plane, and let $E$ be the vector space of integrable holomorphic functions on $\Omega$. Then it can be checked that $E$ is a closed subspace of ... 1answer 24 views ### Conformal Map from Vertical Strip to Unit Disc I haven't found a similar question on here, though I suspect the question may be rather well-covered. I want to find a conformal map from the vertical strip $\{z:-1<Re(z)<1\}$ onto the unit ... 2answers 249 views ### True/False Questions for Complex Analysis I am studying for my (introductory) complex analysis final exam tomorrow. I am practicing an old final exam, which unfortunately has no answer key. Here is a link: ... 2answers 39 views ### Application of Open Mapping Theorem This was stated without proof in the complex analysis text I am reading (Complex Made Simple by Ulrich, page 107). I'm sure it's easy, but I'm tired and need a little help. Let $f$ be nonconstant and ...	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 112, "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": 0.9217082858085632, "perplexity_flag": "head"}
http://mathoverflow.net/questions/54975?sort=newest	## When is a finitely generated group finitely presented? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I think the question is very general and hard to answer. However I've seen a paper by Baumslag ("Wreath products and finitely presented groups", 1961) showing, as a particular case, that the lamplighter group is not finitely presented. To prove this, he gives conditions to say if a wreath product of groups is finitely presented. The question is: which ways (techniques, invariants, etc) are available to determine whether a finitely generated group is also finitely presented? For instance, is there another way to show that fact about the lamplighter group? Thanks in advance for references and comments. - 2 This question seems like it's asking for a list. If it has no single correct answer, it should probably be Community Wiki. – HW Feb 10 2011 at 2:41 1 Maybe this is too basic, but algebraic (ex. nilpotent) and geometric (ex. hyperbolic) properties can give f.p. automatically. – Steve D Feb 10 2011 at 4:48 ## 2 Answers An often-used method is to compute $H_2$. If the group is finitely presentable then $H_2$ is of finite rank with any coefficients. For instance, you can use this technique to show that if $q:F\to\mathbb{Z}$ is the map from the free group of rank two that sends both generators to one then the fibre product $H\subseteq F\times F$, ie $(q\times q)^{-1}$ of the diagonal, is infinitely presented. A famous theorem of Bestvina and Brady shows that this doesn't always work: they give a similar example which is infinitely presented but has finite-rank $H_2$. A related technique shows that this question is indeed `very hard'. Grunewald showed that the fibre product coming from a surjection $f:F\to Q$ is finitely presented if and only $Q$ is finite. It follows that you cannot in general tell if a recursively presented group is (in)finitely presented. - ...with any small choice of coefficients. – Mariano Suárez-Alvarez Feb 10 2011 at 3:40 Mariano - right. I suppose I want the coefficient module itself to be finitely generated. – HW Feb 10 2011 at 14:56 ### 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 general method is to consider an infinite presentation of the group, and then show that every finite subset of the set of relations defines a group with clearly different property. for example, the lamplighter group has the presentation $\langle \ldots a_{-n}, \ldots, a_1, a_2, \ldots, a_n,\ldots,t \mid a_0^2=1, [a_i,a_j]=1, ta_it^{-1}=a_{i+1}\rangle$. Every finite subpresentation defines a group that has as a quotient one of the following groups $H_n=\langle a_{-n}, \ldots, a_1, a_2, \ldots, a_n,t \mid a_0^2=1, [a_i,a_j]=1, ta_it^{-1}=a_{i+1}\rangle$ for some $n$. The group $H_n$ is an HNN extension of a finite Abelian group $\langle a_{-n},\ldots, a_n\rangle$ with the free letter $t$. Hence $H_n$ is a virtually free group, in particular, $H_n$ contains a non-Abelian free subgroup. Therefore every finite subpresentation defines a group containing a free non-Abelian subgroup, while the Lamplighter group is solvable and thus cannot contain a free non-Abelian subgroup. Similarly lacunary hyperbolic but not hyperbolic groups given by presentations satisfying small cancelation conditions or their generalizations are infinitely presented since every finite subpresentation of their presentation defines a hyperbolic group. - Can this type of argument be used to show that the group has no finite presentation? (Or does it only show what it seems to show, namely that the group has no finite sub-presentation of a given infinite presentation.) – aaron Feb 10 2011 at 13:26 1 If a finitely generated group $G$ is finitely presented with finite set of relators $Q$, then every infinite presentation $R$ has a finite subset that is also a presentation. Indeed, consider any proof of $Q$ using $R$. It involves only finite subset $R'$ of $R$. The relations from $R'$ define the group $G$. Indeed, let $G'$ be the group defined by $R'$. Then the identity map on the generating set induces a hom. $\phi: G'\to G$. All relations of $Q$ hold in $G'$, so the kernel of $\phi$ is trivial, and $G'=G$. – Mark Sapir Feb 10 2011 at 13:36 ... of course, you need to use the fact that if $G$ has finite presentation with one generating set, then for any other finite generating set, it also has finite presentation (just rewrite all relations in the new generating set). – Mark Sapir Feb 10 2011 at 14:27 @Mark: Very nice. Thanks. – aaron Feb 10 2011 at 21:27	{"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": 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": 0.9334697723388672, "perplexity_flag": "head"}
http://www.impan.pl/cgi-bin/dict?probability	## probability Define $a_k$ to be the probability that exactly $k$ out of the $2n$ values $X_i$ exceed $T$, conditional on $X_0>T$. Go to the list of words starting with: a b c d e f g h i j k l m n o p q r s t u v w y z	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 6, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.6908437013626099, "perplexity_flag": "head"}
http://cms.math.ca/Reunions/hiver12/abs/cm	Réunion d'hiver SMC 2012 Fairmont Le Reine Elizabeth (Montréal), 7 - 10 décembre 2012 Mécanique céleste Org: Florin Diacu (Victoria), Manuele Santoprete (Wilfrid Laurier) et Cristina Stoica (Wilfrid Laurier) [PDF] LENNARD BAKKER, Brigham Young University The Rhomboidal Symmetric-Mass Four-Body Problem. [PDF] We consider the existence and stability of periodic solutions with regularizable collisions in the rhomboidal symmetric-mass four-body problem. In the two degrees of freedom setting, where the analytic existence of the periodic solutions is given by a variational method, we show that the periodic solutions are numerically linearly stable for most of the values of the mass parameter. In the four degrees of freedom setting, we establish the analytic existence of the periodic solutions and numerically investigate their linear stability. FLORIN DIACU, University of Victoria Rotopulsators of the curved N-body problem [PDF] We consider the $N$-body problem in spaces of constant curvature and study its rotopulsators, i.e.\ solutions for which the configuration of the bodies rotates and, usually, changes size during the motion. Rotopulsators fall naturally into five groups: positive elliptic, positive elliptic-elliptic, negative elliptic, negative hyperbolic, and negative elliptic-hyperbolic, depending on the nature and number of their rotations and on whether they occur in spaces of positive or negative curvature. After obtaining existence criteria for each type of rotopulsator, we derive their conservation laws. We further deal with the existence and uniqueness of some classes of rotopulsators in the 3-body case and prove two general results about the qualitative behaviour of these orbits. An interesting finding is that of a class of rotopulsators that behave like relative equilibria, i.e.\ maintain constant mutual distances during the motion, but cannot be generated from any single element of the underlying subgroup $SO(2)\times SO(2)$ of the Lie group $SO(4)$. TOSHIAKI FUJIWARA, Kitasato University Saari's homographic conjecture for the planar equal-mass three-body problem under the Newton potential and a strong force potent [PDF] The Saari's homographic conjecture for $N$-body problem is the following: For the $N$-body problem under the homogenious potential $U=\sum m_i m_j/r_{ij}^\alpha$, the configurational measure $I^{\alpha/2}U$ is constant if and only if the motion is homographic. Here, $m_i$ $(i=1,2,\dots N)$ is mass for the body $i$, $r_{ij}$ is the mutual distance between the body $i$ and $j$ and $I=\sum m_i m_j r_{ij}^2$ is the moment of inertia. In this year, Fukuda, Ozaki, Taniguchi and the present author proved this conjecture for the planar equal-mass three-body problem under the Newton potential($\alpha=1$) and the strong force potential($\alpha=2$). In this talk, I will review our work. This is a joint work with Hiroshi Fukuda, Hiroshi Ozaki and Tetsuya Taniguchi. JUAN LUIS GARCIA GUIRAO, Universidad Politécnica de Cartagena, SPAIN Periodic orbits and $\mathcal{C}^1$-integrability in the planar Stark-Zeeman problem [PDF] The aim of the present talk is to study the periodic orbits of a hydrogen atom under the effects of a circularly polarized microwave field and a static magnetic field orthogonal to the plane of polarization of the magnetic field via averaging theory. Moreover, the technique used for proving the existence of isolated periodic orbits allows us to provide information on the $\mathcal{C}^1$–integrability of this mechanic–chemical system. JEFFREY LAWSON, Western Carolina University Saari's Conjecture on simple mechanical systems with symmetry [PDF] In 1970 Donald Saari famously conjectured that if a Newtonian $N$-body system has a constant moment of inertia then it is in relative equilibrium (i.e., it is in rigid rotation with a constant angular velocity). In 2002 Jerrold Marsden hypothesized that the conjecture may be generalized to a simple mechanical system that admits a Lie symmetry. The aim of this talk is to explore a geometric approach to Saari's Conjecture. In particular we may pose the problem on a Manakov rigid body and prove a refined statement of the conjecture. (The refinement is essentially necessary to handle higher dimensional symmetry.) By employing a Palais slice decomposition, the question may be further enlarged to simple mechanical systems in which the group action has no points of isotropy. We will conclude with a brief discussion on handling points of isotropy by using a blowup technique. ABDALLA MANSUR, Queen's University The Maslov index and Stability of Periodic Solutions [PDF] We employ techniques from symplectic geometry and specifically a variant of the Maslov index for curves of Lagrange planes along action-minimizing solutions to develop conditions which preclude eigenvalues of the monodromy matrix on the unit circle. This analytical method of proving stability is demonstrated in the context of selected special cases of the n-body problem, namely rhombus and parellelogram solutions of the four-body problem and hip-hop solutions of the 2n-body problem. DANIEL OFFIN, Queen's University Dynamics, stability and symmetric minimization [PDF] We discuss the connection between minimization of action and hyperbolic structure of invariant sets. The notion of absolute minimization always leads to hyperbolic behavior. The weaker notion of symmetric minimization over fundamental time domain, or equivalently minimization over a symmetry class may involve hyperbolic behavior. We give examples from N-body dynamics and outline the proof of hyberbolicity in the reduced space for the periodic hip hop family of the 2N-body problem with equal mass. Finally we state a condition for periodic symmetric minimizing orbits to be absolutely minimizing over arbitrary compact time intervals. This condition concerns the representation of the reversing subgroup of the spatio-temporal group for the given periodic solution. GARETH ROBERTS, College of the Holy Cross Stability of Relative Equilibria in the N-Vortex Problem [PDF] In the weather research and forecasting models of certain hurricanes, vortex crystals'' are found within a polygonal-shaped eyewall. These special configurations can be interpreted as relative equilibria (rigidly rotating solutions) of the point vortex problem introduced by Helmholtz. Their stability is thus of considerable importance. Adapting an approach of Moeckel's for the companion problem in celestial mechanics, we present some useful theory for studying the linear stability of relative equilibria in the N-vortex problem. The analysis is developed in a rotating coordinate frame and special properties of the Hamiltonian play a key role. For example, we show that in the case of equal strength vortices, a relative equilibrium is linearly stable if and only if it is a minimum of the Hamiltonian restricted to a level surface of the moment of inertia. Some symmetric examples will be presented, including a linearly stable family of rhombii in the four-vortex problem. MANUELE SANTOPRETE, Wilfrid Laurier University Relative equilibria in the four-vortex problem with two pairs of equal vorticities [PDF] The N-vortex problem concerns the dynamics of N point vortices moving in the plane. Of particular interest in this problem are solutions that appear fixed when viewed in an uniformly rotating frame. Such solutions are called relative equilibria. In this talk I will give a complete classification of the relative equilibria in the four vortex problem with two pairs of equal vorticities. I will also describe some of the methods used to obtain such results. CRISTINA STOICA, Wilfrid Laurier University Relative equilibria in symmetric $(2N+1)$-body problems [PDF] Consider the Newtonian $(2N+1)$-body problem where $2N$ of the bodies have unit mass and at any time form two regular $N$-gons with a common centre, and where an additional mass $m$ is centrally situated. It is known that in this context there is a $m_0>0$ so that the number of central configurations is three for $m< m_0$ and one if $m>m_0.$ Also, it is known that there is a $m_c>0$, $m_c\neq m_0,$ so that the regular $2N$-gon with central mass is linearly stable if $m>m_c$. Using the discrete and rotational symmetries, we reduce the problem to a three degrees of freedom Hamiltonian system. In this setting, we show that the central configurations mentioned above are in fact relative equilibria and that $m_0$ marks a pitchfork/steady-state bifurcation. The value $m_c$ marks a Hamiltonian-Hopf bifurcation (i.e., a 1:-1 resonance). J. GUADALUPE REYES VICTORIA, UAM	{"extraction_info": {"found_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": 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": 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": 0.8472930788993835, "perplexity_flag": "middle"}
http://mathoverflow.net/revisions/40914/list	## Return to Answer 1 [made Community Wiki] I recall that the following simple "proof" of $\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$ is attributed to Euler: Begin with the fact that for a polynomial $a_0 + a_1 x + \cdots + a_N x^N$ the sum of the inverses of the roots is given by $\sum_{n=1}^N \frac{1}{x_n} = -\frac{a_1}{a_0}$. (If you only remember the formula for the sum of the roots just make a change of variable $y=1/x$). Now consider the "polynomial" $$\frac{\sin\sqrt{x}}{\sqrt{x}} = 1 - \frac{x}{3!} + \cdots$$ whose roots are $x_n = (n\pi)^2$ for $n\in N$. By applying the aforementioned fact the desired result is immediate.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 6, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": 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": 0.919714629650116, "perplexity_flag": "head"}
http://mathoverflow.net/revisions/11789/list	Return to Answer Post Made Community Wiki by Anton Geraschenko♦♦ 1 This is maybe stretching it a little bit, but Tim Gowers' quasi-random groups describes and references some extremal combinatorial properties of graphs constructed from the groups $PSL_2(\mathbb{F}_q)$ which ultimately rely on the fact that they have no non-trivial low-dimensional irreducible representations.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8909623026847839, "perplexity_flag": "middle"}

Subsets and Splits

No community queries yet

The top public SQL queries from the community will appear here once available.