Change font size in the middle of a document or for individual included .tex file
Switching to different font size(s) locally
How do I make my program watch for file modification in C++? Let $G$ be a graph of minimum degree $k&gt;1$. Show that $G$ has a cycle of length at least $k+1$.  How would I show this? I understand that a cycle is a sequence of non-repeated vertices and the degree of a graph is the number of neighbors the vertex has. The most common way to refer to non-immediate family? What is the difference between using a colon and a semi-colon to join two sentences? I've seen a few questions on this site relating to semi-colons, which I believe I correctly understand, but what I'm not as clear about is colons.  For example:     The man ate the apple; it tasted good.   Is this not correctly also written as:     The man ate the apple: it tasted good.   In this case, what is the difference between the two sentences? What does the comma operator , do? Prove that for every $a, b\in R\setminus\{0\}$ is correct this inequality:  $$\frac{|a+b|}{1+|a+b|}&lt;\frac{|a|}{1+|a|}+\frac{|b|}{1+|b|}.$$ "Can/may/will you help me with this?" Which word to use when we ask for help?  Some conditions:     We know that the person asked is able to do it. We don't know if the person asked is able to do it. I need to simulate the warning:  @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ @ WARNING: REMOTE HOST IDENTIFICATION HAS CHANGED!  @ @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@  IT IS POSSIBLE THAT SOMEONE IS DOING SOMETHING NASTY!   I have two Linux machines and when I do ssh from Linux1 to Linux2, ssh accepted and reports no problem.  But because some of testing, I want to simulate the - Identification Has Changed error  So what I need to edit on the remote server in order to create the problem?  what I did until now is to change the knpwn_hosts on both machine but its not creating the problem.  I get this:   key_read: uudecode AAAAB3NzaC1y3jrzG62xyHrb3qxc9eYpg1Cp1PrRElTFfYbEHsRfTZU21S8kO5/Tq+ED8=  failed  key_read: uudecode AAAAB3NzaC1y3jrzG62xyHrb3qxc9eYpg1Cp1PrRElTFfYbEHsRfTZU21S8kO5/Tq+ED8=  failed  The authenticity of host 'ste1 (10.106.164.111)' can't be established.  RSA key fingerprint is 84:82:2b:80:ca:3f:57:c7:8c:14:44:dd:b9:70:e5:d8.  Are you sure you want to continue connecting (yes/no)? Nowhere monotonic continuous function Does there exist a nowhere monotonic continuous function from some open subset of $\mathbb{R}$ to $\mathbb{R}$? Some nowhere differentiable function sort of object? I use to work at night and it is eye-hurting to watch at lightbulb-bright screens (I've got two - laptop's built-in panel and an external 18-inch CRT) in a dark environment. So I adjust my screens to be darker.  Can I do this a software way instead of pressing monitors' buttons that many times every time? How are the Tenchi timelines related? Between Muyo, Universe, GXP, and Ryo-ohki, I'm thoroughly confused. Sometimes Tenchi series seem to be related (Ryo-ohki and Universe) and sometimes not (GXP). What are all the Tenchi series, are they related, and if they are, how? Case insensitive regular expression without re.compile? In Python, I can compile a regular expression to be case-insensitive using re.compile:  &gt;&gt;&gt; s = 'TeSt' &gt;&gt;&gt; casesensitive = re.compile('test') &gt;&gt;&gt; ignorecase = re.compile('test', re.IGNORECASE) &gt;&gt;&gt;  &gt;&gt;&gt; print casesensitive.match(s) None &gt;&gt;&gt; print ignorecase.match(s) &lt;_sre.SRE_Match object at 0x02F0B608&gt;   Is there a way to do the same, but without using re.compile. I can't find anything like Perl's i suffix (e.g. m/test/i) in the documentation. Display error message if same record exist with combination of some fields? biblatex and autonum don't work together biblatex and autonum don't seem to work together in the newest version. Consider the following MWE:  \documentclass{article} \usepackage[utf8]{inputenc} \usepackage[backend=biber]{biblatex} \usepackage{autonum}  \begin{document}  \end{document}   I get the following error messages:  0 File ended while scanning use of \ettl@for@do1 0 pdflatex: \pdfmatch: repetition-operator operand invalid pdfTeX warning: pdflatex: \pdfmatch: repetition-operator operand invalid)Runaway argument?{{\blx@bibdata@i {\blx@tempb }}\blx@bibfiles \eappto \blx@tempa {\blx@xml@comme   After commenting out either biblatex or autonum, the errors disappear. Let $b>0.$ Evaluate $\lim_{n \to \infty} \int^{b}_{0} \frac{\sin nx}{nx}dx$ I'm stumped at the last part of this problem. Could anyone advise, please? Thank you.  Here is  my  proof.   Define  $f: \mathbb{R} \to \mathbb{R}$ by $$ f(x) = \left\{\begin{aligned} &amp;\frac{\sin nx}{nx}&amp;&amp;, x \neq0 \\ &amp;1 &amp;&amp;, x=0 \end{aligned} \right.$$   Hence, $$f(x) = \sum^{\infty}_{k=0}(-1)^k\frac{(nx)^{2k}}{(2k+1)!} , \forall x \in \mathbb{R} \mbox{ and } f \in {\mathscr R[0,b]}.$$  $$\int^{b}_{0} \frac{\sin nx}{nx}dx = \int^{b}_{0}f(x)dx=\sum^{\infty}_{k=0}\int^{b}_{0}(-1)^k\frac{(nx)^{2k}}{(2k+1)!}dx= \sum^{\infty}_{n=0}(-1)^k\frac{(n)^{2k}b^{2k+1}}{(2k+1)(2k+1)!}$$  $$\lim_{n \to \infty} \sum^{\infty}_{k=0}(-1)^k\frac{(n)^{2k}b^{2k+1}}{(2k+1)(2k+1)!}=  ?$$ This is the first part of the first half of the a proof in the Axiom of Choice. The proof that the Axiom of Choice is equivalent to the choice function. REsources: Keith Devlin's Joy of Sets: pg. 58. F.R. Drake and Singh Intermediate Set Theory: pg. 110).   I am confused on proving disjointness and will ask further questions of proof once I get this first part v. clear for myself.   AC1: Let $\mathcal F$ be a family of pairwise disjoint, non-empty sets. Then there is a set $\mathcal M$ that consists of precised one element from each member of $\mathcal F$  AC2: For every family $\mathcal F$ of non-empty sets, there is a choice function $\mathcal f$. (I.e. $\mathcal f(X)$ is a member of X, for each X in $\mathcal F$.   Disjointness: Two sets are disjoin if Q $\cap$ R = $\emptyset$. Meaning that they have zero elements in common. These sets do not overlap.   Show: AC1 $\leftrightarrow$ AC2   (1) ($\rightarrow$) Let $\mathcal F$ be a set of nonempty sets.   (2)For each X $\in$ $\mathcal F$, let $X^*$ = (X $\times$ {X}) = {(x, X) | x $\in$ X})  (3) Show Disjointness.  An outline of how book proves  disjointness:  (a) if u = {(X $\times$ {X}) | x $\in$ X} then u is of the form $\lt  z, t \gt$ where z = X and t=x, i.e. x $\in$ X.   (b) then if u $\in$ ({X} $\times$ x) $\cap$ ({Y} $\times$ Y) $\rightarrow$ u = $\lt z, t \gt$ with z = X and z=Y, i.e. X=Y. Which means ({X} $\times$ x) = ({Y} $\times$ y).  Questions: (1) Apparently we cannot prove disjointness in $\mathcal F$ but must prove in a different set. If so, why is that?  (2) why is the case that if u $\in$ ({X} $\times$ X) $\cap$ ({Y} $\times$ Y) then z =X and z=Y? I am not following that step at all.   (3) why does the axiom of choice have to do with ordered pairs? Why are we setting X $\in$ $\mathcal F$, let $X^*$ = (X $\times$ {X})?    Thank you! Can I open Microsoft Word and Excel files? Can I open Microsoft Word and Excel files with Ubuntu? Calculation of $\int_0^{\pi} \frac{\sin^2 x}{a^2+b^2-2ab \cos x}\mathrm dx$ Calculate the definite integral      $$ I=\int_0^{\pi} \frac{\sin^2 x}{a^2+b^2-2ab \cos x}\;\mathrm dx $$      given that $a&gt;b&gt;0$.   My Attempt:  If we replace $x$ by $C$, then  $$ I = \int_{0}^{\pi}\frac{\sin^2 C}{a^2+b^2-2ab\cos C}\;\mathrm dC $$  Now we can use the Cosine Formula ($A+B+C=\pi$). Applying the formula gives  $$ \begin{align} \cos C &amp;= \frac{a^2+b^2-c^2}{2ab}\\ a^2+b^2-2ab\cos C &amp;= c^2 \end{align} $$  From here we can use the formula $\dfrac{\sin A}{a} = \dfrac{\sin B}{b} = \dfrac{\sin C}{c}$ to transform the integral to  $$ \begin{align} I &amp;= \int_{0}^{\pi}\frac{\sin^2 C}{c^2}\;\mathrm dC\\ &amp;= \int_{0}^{\pi}\frac{\sin^2A}{a^2}\;\mathrm dC\\ &amp;= \int_{0}^{\pi}\frac{\sin^2 B}{b^2}\;\mathrm dC \end{align} $$  Is my process right? If not, how can I calculate the above integral? How do I show a function on 2-adic units is continuous (using the 2-adic metric)?   I'd be happy to learn the general rule or definition. But in particular I need to show that $f(x)=\dfrac{3x+1}{2^{v_2(3x+1)}}$ is continuous at all odd numbers.  Since the function is isometric in $\lvert\cdot\rvert_2$, i.e. since $\lvert f(x)\rvert_2=\lvert x\rvert_2$ every orbit of the function is duplicated, multiplied by any power of $2$.  This can therefore be formulated in various ways; as a function though the odd numbers, through the dyadic rationals, but the following seems good to work with.  If we let $n$ be a positive integer and $k$ be the power of $2$ we can define:  $f(2^k(2n+1))=(3(2n+1)+1)\cdot2^k\cdot\lvert3(2n+1)+1\rvert_2$  So $f(2^k(2n+1)=(3n+2)\times2^k\times\lvert3n+2\rvert_2$  Here is what I have so far:  I think to prove continuity in the 2-adic metric I need $\lvert x_n-x\rvert_2=0\implies\lvert f(x_n)-f(x)\rvert_2=0$  I think $k$ and $n$ are independent so I think I can examine them independently.  Taking $k$ to infinity brings both $x$ and $f(x)$ to zero so that seems to satisfies the continuity requirement.  Moving on to $n$; this seems to be an exercise in proving convergence within odd integers which are in a sense a subset of the 2-adic units.  I know all Cauchy sequences in these converge to 2-adic units but not much more than that.  However I do have a little insight into this particular function.  For example if we examine the inputs $x$ which map to any given output of $f(x)$, it can fairly easily be shown that these $x$'s take the form of a set $\left\{4^mp+\dfrac{4^m-1}{3}:m,n\in\mathbb{N}\right\}$ so, at least for any given output $f(x)$ the inputs always converge to $x=\frac{-1}{3}$ as the intervals between them become large powers of $2$.  In fact the orbit of the function $a(2^kx)=2^k(4x+1)$ on variation of $x$ and holding $k$ fixed is in a sense orthogonal to $f(x)$; which is a restatement of the above except not in closed form. Name finding problem with comment notification feature I just received the following in my Response pane (from ), but you can see it was meant for another user.  The @Robert line seems to have found 'Robert' in my name first, instead of Robert C. Cartaino.  I think the comment notification feature should match based on the closest match from the first character in a name (in posting order), and not just the first match found in posting order.