Does a gap of 10 months after Master's degree affect my chances for a future academic career?
Will taking a gap year after my Master's degree improve or degrade my potential of doing a PhD?
How to install Google Chrome Solution of limit $\lim\limits_{x\to 0} \frac {e^{-1/x^2}}{x} $ Small question, I'm trying to solve this limit but I just can't wrap my head around this problem.  $$\lim_{x\to 0} \frac {e^{-1/x^2}}{x} $$  L'Hopital just seems to make it messier.   It's probably pretty simple - I'd like to hear what I'm missing. Looking for an appropriate English name to replace my Chinese name Background: I am Chinese, and after living abroad from China for several years, I realize that I need an English name anyway. I’ve seen too many confused faces looking at my Chinese name. My Chinese Name: I am male, my Chinese first name is (IPA): /ʃəʊ/-/dəʊ/, and my family name is: /dʒi/. (Check ipa-reader.xyz if not clear.) Any advice about what could be a good idea of a first name, if the last name sounds like &quot;G&quot;? Something about myself: I'm a introvert, nerdy person. I don't like social, small talk, public area. I enjoy programming, writing, arguing. Physically I am short, slim, with a huge glasses. Inequality with complex numbers Consider the following problem.  Fix $n \in \mathbb N$. Prove that for every set of complex numbers $\{z_i\}_{1\le i \le n}$, there is a subset $J\subset \{1,\dots , n\}$ such that   $$\left|\sum_{j\in J} z_j\right|\ge \frac{1}{4\sqrt 2} \sum_{k=1}^n |z_k|.$$  I believe I proven have a stronger statement. Is this proof correct, and if so, what is the optimal constant?  My proof. Consider all the $z_i$ with positive real part. Call the real part of the sum of these numbers $X^+$. In a similar way, form $X^-$, $Y^+$, and $Y^-$. Without loss of generality, let $X^+$ have the greatest magnitude of these.  Note that because $|\operatorname{Re}(z)|+|\operatorname{Im}(z)|\ge |z|$, we have  $$ \left(\sum_{k=1}^n  |\operatorname{Re}(z_k)|+|\operatorname{Im}(z_k)| \right) \ge \sum_{k=1}^n |z_k|.$$  But note that $\sum \limits_{k=1}^n  |\operatorname{Re}(z_k)|+|\operatorname{Im}(z_k)| = X^+ + |X^-|+ Y^+ +|Y^-|$, so we have  $$ 4X^+ \ge \sum_{k=1}^n |z_k|.$$ By choosing $J$ to be the set of complex number with positive real part, this proves a stronger statement, because the factor of $1/\sqrt 2$ isn't needed. Browse for Servers does not display local instance How can I run a command after boot? I would like to run the simple shell command echo 1 &gt; /proc/sys/kernel/sysrq at each boot, to enable the sysrq keys. When in boot should this be done, and where should I put a script to do it? Problem with literate and breaklines=true in listings package As the title says, the option breaklines=true seems to have an undesired interaction when using literate; in the following example the closing parenthesis doesn't get colorized   \documentclass{article} \usepackage{xcolor} \usepackage{listings}  \lstset{   literate=     {)}{{\textcolor{red}{)}}}{1}     {(}{{\textcolor{red}{(}}}{1},   breaklines=true, }  \begin{document}  \begin{lstlisting} () \end{lstlisting}  \end{document}     Commenting out the breaklines=true option produces the desired result. What is producing this odd behaviour and how can it be prevented? Verify the identity: $\tan^{-1} x +\tan^{-1} (1/x) = \pi /2$ Verify the identity: $\tan^{-1} x + \tan^{-1} (1/x) = \frac\pi 2, x &gt; 0$  $$\alpha= \tan^{-1} x$$  $$\beta = \tan^{-1} (1/x)$$  $$\tan \alpha = x$$  $$\tan \beta = 1/x$$  $$\tan^{-1}[\tan(\alpha + \beta)]$$  $$\tan^{-1}\left [{\tan\alpha + \tan\beta\over 1 - \tan\alpha \tan\beta} \right]$$  $$\tan^{-1}\left[ {x + 1/x\over 1- x/x }\right]$$  $$\tan^{-1}\left[{x + (1/x)\over 0} \right]$$  I can't find out what I'm doing wrong.. Is it purely a 'gotta-have-them-all' mindset, or do they actually serve some purpose?  I notice that you are able to sell them on a community market. For a test I listed one, and someone bought it for 50p. Not exactly a huge amount, but I don't understand why someone would spend real money on them.  Am I missing something?   Are they redeemable for games or money off vouchers or something 'tangible' that I'm missing? Find the limit of $\log\Big(\frac{(n^2+n)!}{n^2!n^{2n}}\Big)$ I have to prove that      $$\Big[\log\Big(1+\frac 1{n^2}\Big)+ \log\Big(1+\frac 2{n^2}\Big)+...+ \log\Big(1+\frac n{n^2}\Big)\Big] \to \frac 12$$   I know that $$\log\Big(1+\frac 1{n^2}\Big)+ \log\Big(1+\frac 2{n^2}\Big)+...+ \log\Big(1+\frac n{n^2}\Big)= \\ =\log\Big(\frac {n^2+1}{n^2}\Big)+ \log\Big(\frac {n^2+2}{n^2}\Big)+...+ \log\Big(\frac {n^2+n}{n^2}\Big)=\\ =\log\Big(n^2+1\Big)+ \log\Big(n^2+2\Big)+...+ \log\Big(n^2+n\Big)-2n\log n$$  Doing some manipulation we get:           $$\lim_{n \to +\infty}\log\Big(\frac{(n^2+n)!}{n^2!n^{2n}}\Big)=\frac 12$$      How can I prove this limit? Android app "Unknown error" when uploading photo Spent a half hour on this. Android 5.1.1 latest version SE app (Play Store won't let me update so latest).   Tried to upload photo into an answer. "Something went wrong".  Did tricks to reduce photo size to 275k -- same.    45k rep on the stack.  Here is the photo..... oh wait.  How do I solve? What equipment and settings do I use for panning shots of moving objects? How do I use the Chinese remainder theorem to find all the square roots of 11 in $\mathbb Z_{35}$? The iOS-app should provide (at least) the same features as the site and the ability to edit your profile is not so unimportant.  In my opinion, the user should be able to change the every single info-field in his profile (including, but not limited to nickname, description, profile picture) by pressing an "Edit" button which should be located somewhere on his or her profile view. "Speed" of Gravity and Speed of Light Some threads here touching speed of gravity made me think about that. This lead to some questions.    The speed of gravity was not measured until today (at least there are no undebated papers to that effect). It seems common knowledge/belief among physicists that the speed of gravity is the same as the speed of light.    And this is my question: Is that speed of light = speed of gravity a result of GR or is that fed into the theory? Or is there some evidence from other sources than GR for this? Does the same speed imply some deep-lying relation between gravity and electromagnetism? Converting JSON data to Java object If $X$ and $Y$ are sets, their cartesian product $X\times Y$ is a set. In ZFC set theory, every element object is a set. We know or feel that if $X$ and $Y$ are sets, their cartesian product $X\times Y$ is a set.  My question is: Why if $X$ and $Y$ are sets, their cartesian product $X\times Y$ is a set? Finitely generated monoids- axiomatization- compactness theorem. Protect email addresses in the user description from bots I've noticed that emails are not protected in the user description. For example  entered his email and when viewing the source code of the page it is plain text that malicious bots could harvest and exploit to spam. It would be nice if email addresses were automatically encoded to prevent this. The limit of a convergent Gaussian random variable sequence is still a Gaussian random variable I'm trying to prove this conclusion but have some problems with one of the steps.  Assume $X_1,\ldots,X_n,\ldots$ is a sequence of Gaussian random variables, converging almost surely to $X$, prove that $X$ is Gaussian.  We use characteristics function here. Since $|\phi_{X_n}(t)|\leq 1$, by dominated convergent theorem, we have for any $t$  $$ \lim_{n\rightarrow\infty}e^{it\mu_n-t^2\sigma_n^2/2}=\lim_{n\rightarrow \infty}\phi_{X_n}(t) = \lim_{n\rightarrow \infty}\mathbb{E}\left[e^{itX_n}\right] = \mathbb{E}\left[e^{itX}\right] = \phi_X(t) $$  this is the step that I cannot figure out: $e^{it\mu_n-t^2\sigma_n^2/2}$ converges for any $t$ if and only if $\mu_n$ and $\sigma_n$ converges.   Let $\mu=\lim_n \mu_n$, and $\sigma=\lim_n\sigma_n$, then $\phi_X(t)=e^{it\mu-t^2\sigma^2/2}$, which proves that $X$ is a Gaussian random variable.  Why can we get that $\mu_n$ and $\sigma_n$ converge? This looks intuitive for me, but I cannot make a rigorous prove.