Prove $\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\frac{m^2-n^2}{(m^2+n^2)^2}=\frac{\pi}{4}$ can anyone show me how to prove this since it seems to not be able to swap the summation because $\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}|\frac{m^2-n^2}{(m^2+n^2)^2}| $may not converge.
I have been trying around different ranges of summations:  $$ \sum_{(m,n) \in \mathbb{Z}^2} \frac{m^2 - n^2}{(m^2 + n^2)^2} = 0$$  That's not any good.  What about if we restrict to $m, n \in \mathbb{Z}$ as positive integers.    $$ \sum_{m &gt; 0, n &gt; 0} \frac{m^2 - n^2}{(m^2 + n^2)^2} = 0$$  Now here's an anti-symmetry.  I do not like taking out the symmetry, but perhaps I can ask about:   $$ \sum_{m &gt;  n &gt; 0} \frac{m^2 - n^2}{(m^2 + n^2)^2}  = \; ? \tag{$*$} $$  There doesn't seem to be a change of variables that can work.  And we've used symmetry about as much as we can.  This looks related to:  $$  \sum \frac{1}{n^2} = \frac{\pi^2}{6}$$  Perhaps this other series ($*$) also has a special value.
Question Prove that for every integer k≥1, exists a self-complementary graph with 4k vertices half of which are of degree 2k-1 and the other half of degree 2k.  My approach So, I think the easiest way to prove what is asked is by induction. I can prove the base case for k=1, but as far as the inductive step is concerned, I cannot seem to reach a conclusion as to why if the statement holds for n, then it holds for n+1.  Any help would be appreciated. general solution sum of two uniform random variables aY+bX=Z? is there a general solution to that? I have seen simple examples for Y+X=Z but I was wondering how this would be with rescaling? Prove every element of $G$ has finite order. This is a homework problem and I'm having trouble just getting a basic understanding.  I understand that $S_6$ is a symmetric group of degree $6$.  I'm not sure how to start looking for elements. Which tag is used in the title of a question?   It does not seem to be the most popular It does not seem to be the first in the tagging order:   This question   is prefixed with NGINX (which is actually completely fine)  And it is tagged with "php nginx fastcgi magento" in the order of popularity.  Can we choose the tag to use in the title manually? I'm trying to solve hard combinatorics that involve complicated factorials with large values.  In a simple case such as $8Pr = 336$, find the value of $r$, it is easy to say it equals to this: $$\frac{8!}{(8-r)!} = 336.$$  Then $(8-r)! = 336$ and by inspection, clearly $8-r = 5$ and $r = 3$.  Now this is all and good and I know an inverse function to a factorial doesn't exist as there is for functions like sin, cos and tan etc. but how would you possibly solve an equation that involves very large values compared to the above problem without the tedious guess and checking for right values.  Edit: For e.g. if you wanted to calculate a problem like this (it's simple I know but a good starting out problem) Let's say 10 colored marbles are placed in a row, what is the minimum number of colors needed to guarantee at least $10000$ different patterns? WITHOUT GUESS AND CHECKING  Any method or explanation is appreciated! Is "Smaug" pronounced "Smog"? Did Tolkien the linguist intend for the name of the dragon Smaug to be pronounced "Smog" [IPA: /smɒɡ/ or /smɔːɡ/]? The trailer for the second installment of The Hobbit trilogy pronounces the word (pronounced 'sma-ow-g' [IPA: /smaʊɡ/]) entirely differently than the way I have been reading it for all these many years. Proving that $n$-cycle has order $n$ Express symmetric polynomial $\prod_{i < j} (X_i+X_j)$ in terms of elementary symmetric functions Exercise: Define a polynomial $\Sigma(X_1,\ldots,X_n)$ as \begin{align*} \Sigma(X_1,\ldots,X_n) = \prod_{i &lt; j} (X_i+X_j) \end{align*} This is a symmetric polynomial, quite clearly. I want to express $\Sigma$ as a polynomial in the elementary symmetric functions, for $n=2,3,4$. Case $n=2$ is trivial: $\Sigma(X_1,X_2)= \sigma_1 = X_1+X_2$. Case $n=3$ is easy. I write $\Sigma$ as follows: \begin{align*} \Sigma(X_1,X_2,X_3) = (\sigma_1 - X_1)(\sigma_1 - X_2)(\sigma_1 - X_3), \end{align*} then apply  formally, obtaining \begin{align*} \Sigma(X_1,X_2,X_3) = \sigma_1^3 - \sigma_1 \sigma_1^2 + \sigma_2 \sigma_1 - \sigma_3 = \sigma_2 \sigma_1 - \sigma_3. \end{align*} The case $n=4$ seems harder to me, and I'm unable to work it out. I hope there is some trick to avoid unnecessary computations. My attempt is to write down $\Sigma$ completely: \begin{align*} \Sigma = (X_1+X_2)(X_3+X_4)(X_1+X_4)(X_2+X_3)(X_2+X_4)(X_1+X_3) \end{align*} and then multiply factors in groups of two, obtaining: \begin{align*} \Sigma = (\sigma_2 - X_1X_2-X_3X_4)(\sigma_2 - X_1X_4 - X_2X_3)(\sigma_2 - X_2X_4-X_1X_3), \end{align*} but it doesn't seem to help much. How can I solve the exercise? Does anyone have a simple work-flow for drawing cross sections of earth features in QGIS?   My thing is that QGIS is being used to present the plan view, but we have a little interest in some elevations as well. All the data is one set and is simply an elevation value in the attribute table, I don't want to have to move all that to CAD.  One logical workaround is to present the z value as the y.  I am a dunce in paper space in CAD, and would be more comfortable with the QGIS Composer. "Not a COM32R image" error when trying to install from a USB key Not a COM32R image boot   is a message that the console infinitely keeps spitting out when I try to install Ubuntu 10.04 from a USB. I created the USB using the StartupDiskCreator in 11.04.  What is the reason for this behavior? What do I do to set it right?   I have followed the steps perfectly, as enlisted in the . Travel to Cuba as dual citizen (USA+other)? I have a map a at the link below:    I added one of my GeoServer WMS layers to the map, now I try to add the attributes window pop up on click. I found to examples on the OpenLayers 3 homepage, but I cannot put together a working code...      Is there any chance to have a working sample somewhere? give space above and below a DIV How did nobody at Hogwarts (especially Dumbledore) figure out what killed Myrtle? The information is all clearly there, since Harry and Ron, two 12-year-olds, were able to figure it out (with a bit of help from a petrified Hermione), thought at the time Aragog was much younger and smaller and wouldn't have been as easy to find, so maybe the "spiders flee before it" clue would have been lost, although to be honest it's not exactly as vital a piece of the puzzle as the others.  Dumbledore himself is obviously a very capable and intelligent Wizard, and after Myrtle died she came back as a ghost and was able to provide (admittedly limited) details of her own death.  I don't know about you, but I'd have thought at least someone would have tried to piece it together at the time, and it wouldn't have taken them long to figure out "Oh, she saw a pair of big yellow eyes after hearing a boy speak a funny language in the bathroom and dropped dead? Sounds like she might have heard somebody order a Basilisk to look her in the eyes in parseltongue", especially if the investigator in question was someone like Dumbledore.  Do we have any info on why nobody seemed to have found out anything in 50 years prior to Harry Potter (and, probably more importantly, Hermione Granger) turning up at Hogwarts? Does $\{f_ng_n\}\to fg$ uniformly? How to parse a string to an int in C++? What's the C++ way of parsing a string (given as char *) into an int? Robust and clear error handling is a plus (instead of ). Does D&amp;D 5E have a mechanic similar to  or  as it exists in D&amp;D 3.5?  To put it simply, in D&amp;D 3.5, touch AC excludes armor bonuses when armor can't help against an attack, and flatfooted AC excludes your Dexterity bonus when you haven't acted yet in combat. This user is only displaying 56 rep.   But surely (s)he has much more than that.        The user has also never given out any bounties.  Also, badges are missing: $\mathbb{Q}(\sqrt{n}) \cong \mathbb{Q}(\sqrt{m})$ iff $n=m$ Page 105 of D. Burton's A First Course in Rings and Ideals reads     It is not difficult to show that if (I'll call them $n$ and $m$ instead of $n_1$ and $n_2$) $n$, $m$ are square free integers, then $\mathbb{Q}(\sqrt{n}) \cong \mathbb{Q}(\sqrt{m})$ if and only if $n=m$.   Well, it is getting difficult for me, any help or would be appreciated.    My progress so far: Suppose $\phi :\mathbb{Q}(\sqrt{n})\to\mathbb{Q}(\sqrt{m})$ is an isomorphism. Then $\phi(u)=u$ for every $u \in \mathbb{Q}$, so that $\phi(\sqrt{n})^2=\phi(n)=n$. If $\phi(\sqrt{n})=a+b\sqrt{m}$ for some $a,b\in\mathbb{Q}$, then $(a+b\sqrt{m})^2=a^2+b^2m + 2ab\sqrt{m}=n$ implies $$a^2+b^2m = n$$ and $$2ab=0.$$  Then $b=0$ cannot happen since that would imply that $\sqrt{n}\in\mathbb{Q}$, where $n$ is a square-free integer. Also $a=b=0$ cannot happen. Hence $a=0$ and we have $$b^2m=n.$$ If $\psi:\mathbb{Q}(\sqrt{n})\to\mathbb{Q}(\sqrt{m})$ were an isomorphism then there would exist some $s\in\mathbb{Q}$ such that $$s^2m=n,$$ then $s=b$ or $s=-b$. Therefore the only isomorphisms from $\mathbb{Q}(\sqrt{n})$ to $\mathbb{Q}(\sqrt{n})$ are $\phi(u+v\sqrt{n})=u+vs\sqrt{m}$ and $\psi(u+v\sqrt{n})=u-vs\sqrt{m}$.    I don't know if there's an easier way and I don't know if I'm in the correct way here. Thanks beforehand :)