Is there a way to make it so that equations and mathematical type looks formatted on mobile (iOS) devices on the main page (where the questions are)? Thanks!
TeX support works fine in question and answer bodies, and comments can be rendered manually by tapping and selecting Render MathJax. See related question .  However, the titles are not rendered which makes the browsing experience less than perfect. See screenshot below.   Are there any plans to implement MathJax support in titles?  (I'm running Stack Exhange iOS App Version 1.0.0 on an iPhone 4S with iOS 7.1.1)    EDIT: I was just informed about  asking for the same functionality for the Android app. I guess I shouldn't hope too much then...
Would the Animate Objects spell end an ongoing spell affecting the targeted object(s)? Change logo url link I'm embarking on a new web map project that seeks to display simple geometries (lines, points, polygons) and rasters/basemaps. Ideally, the map will also allow authorized users to add/remove/edit geometries and their attributes.  What are the available options for the storage of data (e.g. SQL Server Spatial)?  What are the available options for the presentation of data (e.g. ArcGIS Server)?  I'm new to web mapping and am attempting to build a solid understanding of the available options and their pros/cons. What is the word that describes the absence of bureaucracy? What is the word which describes the absence of formless, of long procedure with bureaucrats? I was thinking about formsless but it sounds weird... Why is $\mathbb{Z}[\sqrt{-n}], n\ge 3$ not a UFD? I'm considering the ring $\mathbb{Z}[\sqrt{-n}]$, where $n\ge 3$ and square free. I want to see why it's not a UFD.   I defined a norm for the ring by $|a+b\sqrt{-n}|=a^2+nb^2$. Using this I was able to show that $2$, $\sqrt{-n}$ and $1+\sqrt{-n}$ are all irreducible. Is there someway to conclude that $\mathbb{Z}[\sqrt{-n}]$ is not a UFD based on this? Thanks. Saving memoization to disk On Windows 10 Professional, with the November update (Build 10586, Version 1511) , is there any possibility to completely remove Windows Defender ?  I don't want to just disable Windows Defender.  I want to completely uninstall from my system. Java rounding to nearest 0.05 I own a lot of games on Steam, and I'm overall a big fan of the service.  However, one thing about it really bothers me, and I'm wondering if anything can be done about it.  Let's say I have two different online, multiplayer games, both of which I legally own.   If I bought these games normally, I could be playing one of them online, while my fiance could play the other one online on another computer.     However, with Steam, I can only be signed in on one computer at a time.  So, despite the fact that I have bought about 40 games, if I'm playing one of them, someone else in my house is not able to play any of the others.  Is there a way to get around this?  I don't think this is illegal or immoral as I own both of the games in question, and only one person is playing the game at a given time (same IP address and the same household, so it should be easy for Steam to verify that I am not sharing my games unfairly).    If I had bought simple boxed copies, this wouldn't be an issue, but since I became such a fan of Steam, I can't play any of those games at the same time as my fiance.    Is there anything that can be done? Changing website favicon dynamically I have a web application that's branded according to the user that's currently logged in. I'd like to change the favicon of the page to be the logo of the private label, but I'm unable to find any code or any examples of how to do this. Has anybody successfully done this before?  I'm picturing having a dozen icons in a folder, and the reference to which favicon.ico file to use is just generated dynamically along with the HTML page. Thoughts? Prove the series $ \sum_{n=1}^\infty \frac{1}{(n!)^2}$ converges to an irrational number How can one prove that the series $\displaystyle \sum\limits_{n=1}^\infty \frac{1}{(n!)^2}$ converges to an irrational number? There's no need to use Taylor expansion, integrals or any advanced/professional techniques. It can be proved using only basic techniques. My first attempt was to assume that the series does converge to a rational number $ \frac{p}{q}$ and then it follows by definition that for every positive epsilon there exists an integer k such that $ {\sum_{n=1}^{k} \frac{1}{(n!)^2} -\frac{p}{q}}&lt;=\epsilon$ I tried to get a contradiction but I failed. Integral closure of $\mathbb{Z}$ in $\mathbb{C}$ is not finitely generated as a $\mathbb{Z}$-module? Let     $$ \mathbb{Z}^{'}_{\mathbb{C}}=\{ z \in \mathbb{C} | \exists f \in \mathbb{Z}[X] \text{ monic such that } f(z)=0\} $$   be the integral closure of $ \mathbb{Z} $ in $ \mathbb{C} $. Prove that $ \mathbb{Z}^{'}_{\mathbb{C}} $ is not finitely generated as a $ \mathbb{Z} $-module.  My attempt: Suppose $ \mathbb{Z}^{'}_{\mathbb{C}}=\mathbb{Z}\alpha_{1}+...+\mathbb{Z}\alpha_{n} $ for some $ \{ \alpha_{i}\}_{i=\overline{1,n}} \subset \mathbb{Z}^{'}_{\mathbb{C}} $. I know that given $ \alpha , \beta \in \mathbb{Z}^{'}_{\mathbb{C}} $ with minimal polynomials of degrees $ m $ and $ n $ respectively, then we can construct a monic polynomial with integral coefficients $ P(z)=\prod(z-(\alpha_{i}+\beta_{j})) $ where $ \alpha_{i},\beta_{j} $ are the conjugates of $\alpha $ , $\beta $ and therefore the minimal polynomial of $ \alpha + \beta $ has degree $ \leq mn $.   I was thinking that then the degrees of the minimal polynomials of the elements of $ \mathbb{Z}\alpha_{1}+...\mathbb{Z}\alpha_{n} $ could be somehow bounded by a certain constant, whereas $ \mathbb{Z}^{'}_{\mathbb{C}}$ contains elements whose degrees of the minimal polynomials are arbitrarily large, take $ 2^{\frac{1}{n}} $ for example,$ n\geq 2$. I think this works when we are asked to prove that the set of all algebraic numbers $ \overline{\mathbb{Q}} $ is not a finitely dimensional $\mathbb{Q}$-vector space, but I don't know how to proceed in my case.  I would really appreciate any ideas, hints or solutions. Thank you very much! Bug: biblatex can not deal name prefix like "van der ", Miktex updated on 2016.01.16 Firstly, hello everyone. I am new here : )  I am writing my phd thesis using MikTeX + TeXnicCenter and everything goes well, until I have updated MikTeX today.  My Problem:  The program gives an error message to the "same old" command:  \printbibliography[prefixnumbers={},notkeyword={Eigene}, notkeyword={Datenblatt}, title={Fremdliteratur}, heading=subbibliography]   The error message looks like this:  Undefined control sequence. &lt;argument&gt; ...nameprefix {van~der}\isdot }\ifprefchar    Investigation  I have found out that this error comes from the name prefix "van der". If I write them together without space in *.bib file and compile it, it works again.  What's the best way to fix this or work around it? Is it a bug? We have noticed that when using nginx internal 301 and 302 handling, nginx will serve a small document body with the appropriate Location: ... header.  Something along the lines of (in html): 301 redirect - nginx.  As appropriate in the above behaviour, a content-type text/html and content-length header is also sent.  We do a lot of 302 and some 301 redirects, the above behaviour is wasted bandwidth in our opinion.  Any way to disable this behaviour?  One idea that crossed our mind was to set error_page 301 302 to an empty text file. We have not tested this yet, but I am assuming even with the above, the content-type and content-length (0) headers will be sent.  So, is there a clean way to send a "body-less" 301/302 redirect with nginx? Accessing Email logs using REST or SOQL Center image element in parent div Matrix Group induction proof and order of elements question $H$ is the set of the matrices $A$ of the form $$A= \begin{pmatrix} \cos(\theta) &amp; \sin (\theta) \\ -\sin (\theta) &amp; \cos (\theta)\end{pmatrix}$$  where $0\leq \theta &lt; 2\pi$  is a group with respect to matrix multiplication.  Question part (1) Prove by induction that  for any $\theta$ and any positive integer $n$,  $$\begin{pmatrix} \cos(\theta) &amp; \sin (\theta) \\ -\sin (\theta) &amp; \cos (\theta)\end{pmatrix}^{n} = \begin{pmatrix} \cos(n\theta) &amp; \sin (n\theta) \\ -\sin (n\theta) &amp; \cos (n\theta)\end{pmatrix}$$   When $n =1$ we have $$\begin{pmatrix} \cos(\theta) &amp; \sin (\theta) \\ -\sin (\theta) &amp; \cos (\theta)\end{pmatrix}^{1} = \begin{pmatrix} \cos(\theta) &amp; \sin (\theta) \\ -\sin (\theta) &amp; \cos (\theta)\end{pmatrix}$$   and clearly this holds for the base case.   Now I'm going to assume true when $n=k$ and try and prove for $k+1$. So we have   $$\begin{pmatrix} \cos(\theta) &amp; \sin (\theta) \\ -\sin (\theta) &amp; \cos (\theta)\end{pmatrix}^{k}\begin{pmatrix} \cos(\theta) &amp; \sin (\theta) \\ -\sin (\theta) &amp; \cos (\theta)\end{pmatrix}^{1} = \begin{pmatrix} \cos(k\theta + \theta) &amp; \sin (k\theta + \theta) \\ -\sin (k\theta + \theta) &amp; \cos (k\theta + \theta)\end{pmatrix}$$   Now I'm sure I can cheese this, but is there some clever way to pull this apart? I want to just be able to say something like the determinant of the left side is $1^{k+1}$ and its the same on the right ...  Question part (2) Find an element of $H$ of order 5.  Here I'm totally lost, I seem to think that every element is of order 1 cause the determinant is 1.  Question part (3) Find an element of $H$ of infinite order.  Again I'm lost; I want to somehow make the determinant 0 for this but it can never be 0 for any $\theta$. After a company has issued shares, why should it care if the value of the shares fluctuate? Since the shares are now publicly traded, if the value fluctuates, that is strictly a transaction between buyer and seller, none of whom are directly related to the company. E.g. if I sell X an iPhone for $100, why should I care if X is selling the iPhone to Y for 1 dollar or 1000 dollars? I have got my $100, so I need not bother. No? What is the command line way of sending files to the recycle bin? Is there a command line program that can send files to the recycle bin? This is on XP and Vista. Do I leave a revised question closed even though it falls out of scope for a different reason than the original closure reason?