The MWE below produces a warning  LaTeX Font Warning: Font shape `T2A/ntxtlf/m/n' undefined (Font)              using `T2A/cmr/m/n' instead on input line 4.   I tried to locally change the font to CMR but no success. So, how to fix it?   MWE  \documentclass[11pt,a4paper]{report} \usepackage[utf8]{inputenc} \usepackage[T1,T2A]{fontenc} \usepackage{newtxtext,newtxmath} \usepackage[russian,english]{babel} \newcommand{\ru}{\selectlanguage{russian}}  \begin{document} \ru Інститут математики \end{document}
I am writing an article in English, with some pages in Russian.  \documentclass[12pt]{article}  \usepackage[T1,T2A]{fontenc}  \usepackage[utf8]{inputenc}  \usepackage[russian,english]{babel}  \usepackage{newtxmath} \usepackage{newtxtext} \begin{document}  hello \textbf{hello}  \selectlanguage{russian} привет \textbf{привет}  \end{document}   Output:  hello hello привет привет  I am using \selectlanguage{} to switch between languages when needed.  However, I noticed that when I try to use \textbf{} when in Russian language mode, it does not work because of newtxtext, which I am using for math formulas.  Does anyone know how to solve this problem?  Thank you.
Let $f : \mathbb{R}^n → \mathbb{R}_∞$ be a function and let $C ⊂ dom f$ be a convex set.  $$**Part I**$$ Prove that $f$ is a convex function if and only if $f$ is convex over every line $L_{v,x_0}$ where $L_{v,x_0} := \{x_0 + tv : t \in \mathbb{R}\}$.  i.e. f is convex iff when restricted to any line, it be convex on that certain line.  So essentially to handle this proof I need to show that convexity for $f$ holds when its convex over any arbitrary line, and on the converse it does not hold when $f$ is not convex over that arbitrary line. If $f$ is convex then,  $$\forall x,y \in C, a \in [0,1]$$ it holds that $$f(ax+(1-a)y) \leq af(x) + (1-a)f(y) \space \space\space\space\space\space\space\space\space\space\space\space (1)$$  EDIT: $$**Part II**$$  Let $f : \mathbb{R}^n \to \mathbb{R}_∞$ be a function.   I want to prove that $f$ is convex over the line $L_{v,x_0}$ iff $\psi : \mathbb{R} \to \mathbb{R}_∞$  $\psi(t) := f (x_0 + tv)$, is convex over $I (x_0, v )$.  where $L_{v,x_0}:= \{x_0 + tv : t ∈ \mathbb{R}\}$, and $I(x_0,v):=\{t \in \mathbb{R} : x_0 + tv \in C\}$  Any help is very much appreciated Why is it bad to log in as root? I've often come across posts on forums or other websites where you see people joking in such a manner about running/logging in as root as if it's something awful and everyone ought to know about it. However, there isn't much that a search reveals on the matter.   It may be widely known to Linux experts, but I really don't know why. I remember always running as root when I first tried Linux years ago (Redhat and Mandrake) and don't remember running into any problems because of that.   There are actually some distros that have a bright red background with alert signs all over it as wallpaper for the root user (SuSe?). I still use the "Administrator" account for regular use on my Windows installation and haven't ever run into any problems there either. Why are my textures pink? We had in our lecture on numerical analysis the following: Let $\mathrm{Lin}(X,Y)$ be the set of all linear maps $X\rightarrow Y$. Let $A\in\mathrm{Lin}(\mathbb R^l,\mathbb R^n)$ and $B\in\mathrm{Lin}(\mathbb R^n,\mathbb R^m)$ and $\|C\|_{op}:=\max_{\|x\|\leq1}\|C(x)\|$. Then our lecturer followed $\|A\circ B\|_{op}\leq\|A\|_{op}\cdot\|B\|_{op}$. So he didn't prove it and so I've tried it by my own. My attempt: $$\begin{aligned} \|A\circ B\|_{\mathrm{op}} &amp;= \max_{\|x\| &lt; 1}\|(A \circ B)(x)\| \\ &amp; \leq \max_{\|x\| &lt; 1} \|A\|_{\mathrm{op}} \|B(x)\| \\ &amp;= \|A\|_{\mathrm{op}} \max_{\|x\|\leq1}\|B(x)\|\\ &amp;= \|A\|_{\mathrm{op}}\|B\|_{\mathrm{op}} \end{aligned}$$ But this seems too easy. I am really interested in a nice proof so anybody could help? Thanks a lot! I see that one time out of the twenty total tests they run, $p &lt; 0.05$, so they wrongly assume that during one of the twenty tests, the result is significant ($0.05 = 1/20$).      Title: Significant Hover text: "'So, uh, we did the green study again and got no link. It was probably a--' 'RESEARCH CONFLICTED ON GREEN JELLY BEAN/ACNE LINK; MORE STUDY RECOMMENDED!'" Is $x^x=y$ solvable for $x$? Given that   $x^x = y$; and  given some value for $y$   is there a way to expressly solve that equation for $x$? Can I get UAC prompt for user from batch file? Can I get UAC prompt for user from batch file?  For example I need put some keys to registry under CurrentControlSet from batch file by     reg import path\to\my.reg   PS. runas is not a solution as my customers uses different Windows localisation and I can not guess which name used for Administrator account (this can be Администратор)...  And script don't know administrator password.  And UAC dialog box does not require password entering... A hard problem on exponential integration Suppose $a : [0 , 1] \to \Bbb R$ is an infinitely smooth function. For $\lambda\ge1$, define $$F(\lambda) := \lambda \int_0^1 e^{\lambda t} a(t) \, dt.$$ If $\sup_{\lambda\ge1}|F(\lambda)|\lt\infty$, then $a$ is the identically zero function. Below are the some results I have derived:   Derivatives of any order of $a$ vanishes at $t = 1.$ For any $\delta\lt1$, $a(t)$ has a zero in the open interval $(\delta , 1).$  The same is true for all the derivatives of $a(t)$. This tells us that the $n^\text{th}$ derivative of $a$ has infinitely many distinct zeros for all natural $n$. If $a$ is analytic, then I can show that $a(t)\equiv 0.$ If $a(t)\ge0$ on $[0 , 1]$, then it is obvious that $a\equiv 0.$   I would appreciate any hint.  Actually $(1)$ and $(3)$ follows from $(2)$. For $(2)$, suppose that $a(1)\gt0$ (the case that $a(1)\lt0$ can be argued in the same way as the following), then there exists $\delta\lt1$ such that $a(t)$ is strictly positive on $I = [1-\delta , 1]$. Then write $F(\lambda) = \lambda\int_0^{1-\delta}e^{\lambda t}a(t)dt + \lambda\int_{1-\delta}^1e^{\lambda t}a(t)dt$. Now there exists $c\gt0$ such that $a(t)&gt;c$ on $I$ since $I$ is compact, so that the second term is bounded below by $c (e^{\lambda}-e^{\lambda(1-\delta)})$. But the first term is only $O(e^{\lambda(1-\delta)})$, so this contradicts the fact that $F(\lambda)$ is bounded. Morever, using the same idea and integration by parts, one can see that $n^{th}$ derivative of $a$ must vanish at $1$. In the Android app, left sidebar, there is an editable list of "Your Sites". In the editing pane, the sites are re-orderable (although awkwardly), and the order one places the sites in there is preserved, even after quitting the app and returning to this pane. One expects that order to be reflected in the "Your Sites" portion of the left sidebar when the editing pane is closed, but it is not.  I'm not sure what is determining the order of the sites in the left sidebar, it doesn't seem to be by date they were "pinned" into the list one wants to show, either. Proving $\left(1+\dfrac{1}{1^3}\right)\left(1+\dfrac{1}{2^3}\right)\cdots\left(1+\dfrac{1}{n^3}\right)<3$ for all positive integers $n$ Prove that $\left(1+\dfrac{1}{1^3}\right)\left(1+\dfrac{1}{2^3}\right)\cdots\left(1+\dfrac{1}{n^3}\right)&lt;3$   for all positive integers $n$   This problem is copied from Math Olympiad Treasures by Titu Andreescu and Bogdan Enescu.They start by stating that induction wouldn't directly work here since the right hand side stays constant while the left increases.They get rid of this problem by strengthening the hypothesis.$$\left(1+\dfrac{1}{1^3}\right)\left(1+\dfrac{1}{2^3}\right)\cdots\left(1+\dfrac{1}{n^3}\right)\le3-\dfrac{1}{n}$$ and then proceed by induction.  The problem is that I can't find a motivation for the above change.I mean,we could have subtracted a lot of things from the RHS but what should nudge us to try $\dfrac{1}{n}$?The rest of the proof is quite standard,but I can't see how I am supposed to have thought of it.Is it just experience?Or is it a standard technique?A little guidance and motivation will be appreciated. Find the value of $\sum_0^n \binom{n}{k} (-1)^k \frac{1}{k+1}$ Find the value of $\sum_0^n \binom{n}{k} (-1)^k \frac{1}{k+1}$.  Writing out several terms, I think the answer is $1/(n+1)$, but I'm struggling to prove this. I would greatly appreciate it if someone could show how to prove this. In other words, how to prove:  For all real constants $a$ and $b$ such that $a &gt; 1$, $$\lim_{n\to\infty}\frac{n^b}{a^n} = 0$$  I know the definition of limit but I feel that it's not enough to prove this theorem. I've created a script that runs every night on my Linux server that uses mysqldump to back up each of my MySQL databases to .sql files and packages them together as a compressed .tar file.  The next step I want to accomplish is to send that tar file through email to a remote email server for safekeeping.  I've been able to send the raw script in the body an email by piping the backup text file to mailx like so:  $ cat mysqldbbackup.sql | mailx backup@email.com   cat echoes the backup file's text which is piped into the mailx program with the recipient's email address passed as an argument.  While this accomplishes what I need, I think it could be one step better, Is there any way, using shell scripts or otherwise, to send the compressed .tar file to an outgoing email message as an attachment?  This would beat having to deal with very long email messages which contain header data and often have word-wrapping issues etc. Is the git storage model wasteful? Wolfram alpha says there are no solutions. Is this because of the following?  $$\tan(z)=i \Rightarrow \frac{e^{iz}-e^{-iz}}{i(e^{iz}+e^{-iz})}=i \Rightarrow \frac{e^{iz}-e^{-iz}}{(e^{iz}+e^{-iz})}=-1$$ Let $x=e^{iz}$. Then,  $$\frac{x-1/x}{x+1/x}=-1 \Rightarrow x-1/x=-x-1/x \Rightarrow 2x=0$$  Then, $$x=e^{iz} \Rightarrow \ln(0)=iz$$ but $\ln(0)$ does not exist so we have no solutions? How to prove that $\lim_{n\rightarrow \infty}\frac {n^{2}}{2^{n}}=0$ & $\lim_{n\rightarrow \infty}\frac {n^{2}}{n!}=0$? Prove that:  $$\lim_{n\rightarrow \infty}\dfrac {n^{2}}{2^{n}}=0$$  I think I need to show $2^n \geq n^3$ $\forall n \geq10$. Is it true?  What about: $$\lim_{n\rightarrow \infty}\dfrac {n^{2}}{n!}=0$$  What to do with $n!$? The phrase for when: working in a group to divide responsibility and reduce malicious intent Is the following correct, or is there more to it?  "I forgot his name" — I knew his name, but I forgot it.  "I forget his name" — I keep forgetting his name. Where using "forget" basically means that you tried to remember this information before as well, but you couldn't then either. Wouldn't it be a good idea to be able to vote for questions to be moved to any Stack Exchange site? For example, I just voted  as off topic, but that was all I could do. I wish I could've voted it be moved to .  It's OK to have the predefined list of sites that is now when voting to move, but it would be nice to have a more link, where we could pick any Stack Exchange site.  Here's a quick mockup of how I see this implemented:    When the more link is clicked, a list of all the Stack Exchange sites should appear. Sort Order on the User Filter is Surprising To Me When filling in the "Type to Find Users" , it seems to sort by exact match and then by reputation when I am on the reputation tab.   For example, please consider the following search results:    This threw me off since I thought that the sort order is reputation. I guess this often might give people what they want, but when picking a popular name is shows a lot of 0 rep users. Maybe adding a filter for low rep to the exact match might give more expected results, or just sorting on Rep?