I recently ran into this series: $$\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^3}$$  Of course this is just a special case of the Beta Dirichlet Function , for $s=3$.   I had given the following solution: $$\begin{aligned}  1-\frac{1}{3^3}+\frac{1}{5^3}-\cdots &amp;=\sum_{n=0}^{\infty}\frac{(-1)^n}{\left ( 2n+1 \right )^3} \\    &amp;\overset{(*)}{=} \left ( 1+\frac{1}{5^3}+\frac{1}{9^3}+\cdots \right )-\left ( \frac{1}{3^3}+\frac{1}{7^3}+\frac{1}{11^3}+\cdots \right )\\    &amp;=\sum_{n=0}^{\infty}\frac{1}{\left ( 4n+1 \right )^3} \; -\sum_{n=0}^{\infty}\frac{1}{\left ( 4n+3 \right )^3} \\    &amp;= -\frac{1}{2\cdot 4^3}\psi^{(2)}\left ( \frac{1}{4} \right )+\frac{1}{2\cdot 4^3}\psi^{(2)}\left ( \frac{3}{4} \right )=\frac{1}{2\cdot 4^3}\left [ \psi^{(2)}\left ( 1-\frac{1}{4} \right )-\psi^{(2)}\left ( \frac{1}{4} \right ) \right ]\\   &amp;=\frac{1}{2\cdot 4^3}\left [ 2\pi^3 \cot \frac{\pi}{4} \csc^2 \frac{\pi}{4}  \right ] \\   &amp;=\frac{\pi^3 \cot \frac{\pi}{4}\csc^2 \frac{\pi}{4}}{4^3}=\frac{\pi^3}{32}  \end{aligned}$$  where I used polygamma identities and made use of the absolute convergence of the series at $(*)$ in order to re-arrange the terms.   Any other approach using Fourier Series, or contour integration around a square, if that is possible?
Prove that : $$\sum_{k=0}^{\infty} \frac{(-1)^{k}}{(2k+1)^3}=\frac{\pi^3}{32}.$$    I think this is known (see ), I appreciate any hint or link for the solution (or the full solution).
Identifying Coordinate System of Shapefile when Unknown? Let $k^p = k$, $K = k(x)$ and $F = k(u)$ for some $u \in K$. Show that $K/F$ is separable $\iff u \not\in K^p$. Let $K$ be the rational function field $k(x)$ over a perfect field $k$ of   characteristic $p &gt; 0$. Let $F = k(u)$ for some $u\in K$, and write   $u = f(x)/g(x)$ with $f$ and $g$ relatively prime. Show that $K/F$ is   a separable extension if and only if $u\notin K^p$.   There is already a post on this but it has not yet been answered :   The truth is that I am very confused with this problem, I would like someone please help me to understand what happens with this exercise and explain to me how it could be done by giving me a help or something. I already have an approach for the left-to-right implication, but I do not know very well how good it is. Could someone help me with the other implication please? Thank you very much. Transparent model when exporting fbx file to Unity So I have this shark model which I animated and ready to export it on fbx and transport it to Unity.  But When I put it on Unity, It looks transparent like this.    Many of the solutions online say that the problem is because of my faces, and they told me to recalculate it or something, but normals are just fine. Is there any other solution?  here's the blend file: Integration by partial fractions; how and why does it work? Could someone take me through the steps of decomposing  $$\frac{2x^2+11x}{x^2+11x+30}$$ into partial fractions?   More generally, how does one use partial fractions to compute integrals $$\int\frac{P(x)}{Q(x)}\,dx$$ of rational functions ($P(x)$ and $Q(x)$ are polynomials) ?    This question is asked in an effort to cut down on duplicates. See .  Also see . How would I remove the border from an iframe embedded in my web app? An example of the iframe is:  &lt;iframe src="myURL" width="300" height="300"&gt;Browser not compatible.&lt;/iframe&gt;   I would like the transition from the content on my page to the contents of the iframe to be seamless, assuming the background colors are consistent. The target browser is IE6 only and unfortunately solutions for others will not help. How to prove that if $\lim_{n \rightarrow \infty}a_n=A$, then $\lim_{n \rightarrow \infty}\frac{a_1+...+a_n}{n}=A$ I'm wondering if I have a sufficient proof of the following:   If $(a_n)$ is a sequence such that $\lim_{n \rightarrow \infty}a_n=A$, then $\lim_{n \rightarrow \infty}\frac{a_1+...+a_n}{n}=A$.   My approach:  For all $\varepsilon &gt; 0$, there exists $N$ such that for all $k&gt;N$, $|a_k -A|&lt;\varepsilon$. So we can break the limit up as follows $$\lim_{n\rightarrow \infty} \frac{a_1+...+a_k}{n} + \lim_{n \rightarrow \infty}\frac{a_{k+1}+...+a_n}{n} \overset{\epsilon \rightarrow 0}{=} 0 + \lim_{n \rightarrow \infty}\frac{nA}{n}=A$$  Is this on the right track, or am I missing something about breaking up the limit in the way I have? I am creating a controller for a 12 V system using an Arduino Uno microcrontroller. For the outputs I am using a relay shield to switch the 12 V components. I have a 12 V toggle switch that turns on some 12 V components in the system and I want to use a trigger signal off of this same switch to send to an Arduino digital input. I know that the Arduino can only handle 5 V max. What would be the best way to step down the 12 V coming off of the switch to the 5 V for the input?  EDIT: The system is for use in a car. Would the amperage of the car battery need to be lowered somehow as to not blow up the components? Is the total energy of the universe zero? In popular science books and articles, I keep running into the claim that the total energy of the Universe is zero, "because the positive energy of matter is cancelled out by the negative energy of the gravitational field".  But I can't find anything concrete to substantiate this claim. As a first check, I did a calculation to compute the gravitational potential energy of a sphere of uniform density of radius $R$ using Newton's Laws and threw in $E=mc^2$ for energy of the sphere, and it was by no means obvious that the answer is zero!  So, my questions:   What is the basis for the claim &ndash; does one require General Relativity, or can one get it from Newtonian gravity? What conditions do you require in the model, in order for this to work? Could someone please refer me to a good paper about this? Hello Physics StackExchange!  I will put the TL;DR in the beginning: I need a self contained, relatively hand-holding sequence of textbooks that covers up from the end of Griffith's Intro to QM to relativistic QM/particle physics to QFT.    This will be my first post. I'm a medical physics grad student who majored in biomedical engineering for my undergrad, with serious regrets about not doing math/physics double major in hindsight. Oh well. I've made it my goal to eventually understand QFT.  My background in relevant subjects is Calc 1 to 3 and Diff EQ, 2 part freshman physics, an E&amp;M course that used Griffiths Electrodynamics, a summer of self studying Griffith's QM (got most of the way through). I will also be officially taking a class in QM using Griffith's textbook this next spring. I am aware of the Lagrangian and Hamiltonian approaches to mechanics as well. In addition I have familiarity with special and basic general relativity using tensors and all this. What would be the best way for me to get from where I am now to a strong understanding of QFT autodidactically? I have no aversion to studying pure maths where I need to.  EDIT: My question is much different than simply asking for a handholding explanation of QFT. Generally these textbooks assume a certain amount of physics/math background, and I was asking for a sequence of textbooks leading up to QFT from the end of a specific textbook, Griffith's QM. Did you even read my question? Did you read the thing you linked me to? I'm an Indian National. I have a Type C Schengen Business Visa which is valid from 26-11-2015 to 26-11-2020. I received it when I intended to travel to Netherlands, although I have not used it.   When I applied for a spouse endorsement, the passport on which the visa is stamped was cancelled and a new passport was issued.  Now, I plan to travel to Italy on business from July 8th to 16th.    Can travel with the same visa in old passport, or should I cancel the existing visa and apply for a new one?  I contacted the Netherlands Embassy call center (not VFS Netherlands) which responded that I can use it without any issues. However, they have not confirmed it by email.   I asked about any immigration issues in Italy and was told me that, if I have a valid itinerary and invitation letter from the office, there should be no problem. I was also told that Italian immigration authorities are able to contact the Dutch, if necessary.   I emailed the Italian Embassy in Mumbai (visa.mumbai@esteri.it) and await its response. Proof that the Euler characteristic is additive I'm reading through a set of notes which assumes that the Euler characteristic is additive, but doesn't give a proof, so I would like to understand why this is.  Let $A_n$ be a finitely generated Abelian group, non-zero for $0&lt;n&lt;k$. Consider the long exact sequence  $ \cdots \rightarrow A_n \rightarrow B_n \rightarrow C_n \rightarrow A_{n-1} \rightarrow B_{n-1} \rightarrow C_{n-1} \rightarrow \cdots $  where $B_n$, $C_n$ defined similarly to $A_n$.   I know that the Euler characteristic of $A$ is given by $\chi(A) = \sum_0^k (-1)^n \text{rank}(A_n)$, with similar definitions for $B$ and $C$. I want to prove that $\chi(B) = \chi(A) + \chi(C)$.  The notes say that for an exact sequence  $0 \rightarrow X \rightarrow Y \rightarrow Z \rightarrow 0$  of finitely generated Abelian groups $X,Y,Z$, we have the result rank($Y$) $=$ rank($X$) +  rank($Z$), and that the additivity of the Euler characteristic follows from this fact, but I can't see how. If $\mathrm{Tr}(A)=0$ then $T=R^{-1}AR$ has all entries on its main diagonal equal to $0$ gzip: stdin: not in gzip format When I run:  tar -zxvf john-1.7.0.2.tar.gz   I get this:  gzip: stdin: not in gzip format tar: Child returned status 1 tar: Error is not recoverable: exiting now   It should start unpacking.  Some people told me not to use the z but I get the same error.  It's a file that I downloaded this way:  wget http://openwall.com/john/f/john-1.7.0.2.tar.gz Let's stop displaying a user's accept rate My feature request (or anti-feature-request perhaps) is the following: Let's stop displaying a user's accept rate.  For those of you who know "Fawlty Towers", the whole issue of a visible accept rate and our behavior towards it starts to feel like .   We generally seem to agree that harassing a user about it is not appropriate and that our decision to answer a question should not depend on it, yet we display it in varying terrorist-threat-levels of color. Yes dear users, please don't pay any attention to the brightly orange colored percentage you see there. And whatever you do, don't bring it up.  But as long as you're displaying the accept rate, people will comment on it, criticize users for it and possibly reconsider answering because of it. And while I'm of the opinion that a low accept rate should not stop you from answering (we're here to make the internet better after all) and that the possible reputation gain (or lack thereof) should not factor in either, I can't blame users for considering it. It's in their face after all. And we're only people.   So let's take the issue off the table. Or at least partially. Information about a user's acceptance behavior can still always be gathered from the profile. But at least that would be less in-your-face.   "But it helps me to see if a user is a help vampire!" I'm still wondering if those are of the sparkly variety as well, but I digress. So what? If the question is bad, downvote it. Evaluate a question on its individual value. There is no need to take a user's history into account. There might be a correlation between the quality of a question and a user's accept rate, but there doesn't have to be. And if there is, the bigger problem is the question's quality.  "Exactly! A user with a low accept rate might indicate there's a problem!" I'm not arguing we should get rid of the accept-rate value within SO/SE. After all, this Q&amp;A works based on up and downvotes, as well as the acceptance of an answer. But why not use it behind the scenes to (very occasionally) nudge the user with a statement like:      "You have received several answers to your questions, yet have not accepted an answer as correct or helpful in a while. You might want to evaluate them and where possible ".    After all, in the various questions with regards to comments on low accept rates, if any comments are encouraged at all, then they are of this soft variety. So why not let the system take care of that?  I feel this would take the whole issue of accept rate off the table, while still providing the user with the information he might have missed or needs. (As a bonus, it would reduce the frequency with which the topic is brought up here on Meta) And those users who would ignore this information are most likely not convinced by a displayed percentage either. Uncountable $\sigma$-algebra I'm stuck on the following problem (Source: Real Analysis for Graduate Students; Exercise 2.6; ):     Suppose $\mathcal A$ is a $\sigma$-algebra with the property that whenever $A \in \mathcal A$, there exists $B, C \in \mathcal A$ with $B \cap C = \emptyset$, $B \cup C = A$, and neither $B$ nor $C$ is empty. Prove that $\mathcal A$ is uncountable.   I think there is the added assumption that this is only true for $A \in \mathcal A$ having at least two elements, so it doesn't hold for singletons and the emptyset?  I tried to show this by way of contradiction and say that $\mathcal A = \{A_k\}$ is countable, but I didn't see how this could get me to a contradiction.  The other approach I tried was to look at $X = B_1 \cup C_1$ where $B_1 \cap C_1 = \emptyset$, then look at $B_1 = B_2 \cup C_2$ where $B_2 \cap C_2 = \emptyset$ and so on. If I look at the $C_k$'s they are pairwise disjoint and I either have a finite number of them or an infinite number of them in which case I've created a countable sequence of pairwise disjoint nonempty elements of $\mathcal A$. This path looked promising, but I couldn't see what to do next. Any ideas?  As a secondary question: I think this result is supposed to be used to show that if $\mathcal A$ is a $\sigma$-algebra with infinitely many elements, then it's uncountable, but I wasn't able to show that the property mentioned above (the one I'm trying to show) was satisfied in this case. Upgrade to the latest PHP version in Ubuntu I am using  in my server and NextCloud informs me that:     You are currently running PHP 7.0.32-0ubuntu0.16.04.1. Upgrade your   PHP version to take advantage of performance and security updates   provided by the PHP Group as soon as your distribution supports it.   So, I tried to upgrade me OS (Ubuntu 16.04), by executing:  apt-get update  apt-get dist-update  Then I restarted the server and I re-executed the above commands. After all these, when I connect from terminal I am getting:  Welcome to Ubuntu 16.04.5 LTS (GNU/Linux 4.4.0-83-generic x86_64)   * Documentation:  https://help.ubuntu.com  * Management:     https://landscape.canonical.com  * Support:        https://ubuntu.com/advantage  0 packages can be updated. 0 updates are security updates.   and my PHP version is still PHP 7.0.32-0ubuntu0.16.04.1 (cli) ( NTS ).  This means that PHP 7.0.32 is the latest possible version that I can install in ubuntu 16.04; If $F_n$ is the $n$-th fibonacci number, then prove that, $$F_n={n-1 \choose 0 }+{n-2 \choose 1 }+{n-3 \choose 2 }+\ldots$$  I tried the idea of using Pascal's triangle, but it seems to need some adjustment, so I am stuck, please help.    Any combinatorial argument would be of greater help than rigorous proof.  Thank you. Answering a phone call with "Yes, XX, tell me" I have heard a lot of people pick their phone and go "Yes, XX, tell me" (highly used in India).  I think the right way should be "Hey, XX, what's up?" or "Hey, XX, what's going on"? But does this work in a professional environment? Let $U\subset \mathbb{C}$ be a nonempty connected open set such that for every $z\in U$, $\overline z\in U$.  Let $f$ be analytic on $U$. Suppose $f(x)\in\mathbb R$ for every $x\in U\cap\mathbb R$. Prove that $f(\overline{z})=\overline{f(z)}$ for any $z \in U$.    By definition, I know that $f$ analytic on $U$ means that for every $z_0 \in U$, there exists $r&gt;0$ and a sequence of complex numbers $\left(a_n\right)_{n=0}^\infty$ such that $f(z)=\sum_{n=0}^\infty a_n\left(z-z_0\right)^n$ on the disc $D(z_0,r)\subseteq U$. I see that $f$ takes points without imaginary components to other points without imaginary components. But I don't see how this implies a symmetry that $f(\overline{z})=\overline{f(z)}$ for $z \in \mathbb C \setminus \mathbb R$. It seems like a very strong conclusion and I'm not sure how I would prove it.  I have also shown in the preceding question that $U\cap\mathbb R$ contains an open interval, however I am not sure if that detail is meant to be helpful to this question. How to create individual chapter PDFs from included TeXs? My document structure has   one file with preamble (i.e. MainTex) and around 75 individual .tex files (i.e. Chapter_01, Chapter_02, ..., Chapter_75) representing each chapter.   I am using \include command to create a one big PDF file, the output is MainTex.pdf. Life is all good up to this point.   If I have to create individual PDF for each chapter then do I need to run 75 times with each chapter name and rename the PDF manually to match to chapter name. Because if I include only Chapter_01 and run it creates PDF named as MainTex.pdf.  Is there a way I can automate this workflow?