I'm using XeLaTeX to build my homework but the commas and the parentheses are all messed up. Could someone explain this?  Source code:  4). $(1,5,6,2,3,10,4,7,1)$ is a cycle of length 8.  5). Since $\{3,4\}$ is not an edge and $(3,10,4)$ is a path, the length of a shortest path from 3 to 4 is 2.  6). The vertices you can travel to from 8 within 2 edges are 2, 3, 9, 10. Since $(8,3,2,7)$ is a path, the length of a shortest path from 8 to 7 is 3.  7). $(4,7,1,5,2,6)$ is such a path.
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I just updated my TeX live installation (tlmgr update --all) and now I get some wrong symbols in math mode if I use XeLaTeX or LuaTeX. This MWE for example  \documentclass{standalone} \begin{document} $. + , - / $ \end{document}   produces the following output:    I suppose it is a bug in the new Latex2e version from 2016/02/01. Any ideas how to fix this?    EDIT:  Adding \usepackage{fontspec} produces a different output. But it is still not perfect.
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I solved this problem by manually adding parentheses and counting them, and got correct answer of 32. Is there a simple to find the answer? Thanks.     The value of the expression $1÷2÷3÷5÷7÷11÷13$ can be altered by including parentheses. If we are allowed to place as many parentheses as we want, how many distinct values can be obtained for this expression?
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Where should the period go when using parentheses? For example:     In sentence one, I use this example (which has a parenthesis at the end.)   Should the period be inside, or outside of the parentheses? What about if the entire sentence is a parenthetical (as below)?     (Where does the terminal punctuation go here?)   Is there a hard and fast rule?
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I have used the next code to get a commutative diagram:  \documentclass[10pt,a4paper]{article}  \usepackage[spanish]{babel} \usepackage{amsmath} \usepackage{tikz-cd}% diagramas  \begin{document}  %-----Diagrama------- \deactivatequoting \[ \begin{tikzpicture} \matrix(m)[matrix of math nodes, row sep=4em, column sep=3em, text height=1.5ex, text depth=0.25ex] { M   &amp;   TN \\     &amp;    N \\ }; \path[-&gt;] (m-1-1) edge node[auto] {$\psi\circ Y$} (m-1-2); \path[-&gt;] (m-1-1) edge node[left] {$\psi$} (m-2-2); \path[-&gt;] (m-1-2) edge node[right] {$\pi$} (m-2-2); \path[-&gt;] (m-2-2) edge node[left] {$Y$} (m-1-2); \end{tikzpicture} \] \activatequoting %-------------------  \end{document}   However, I would like two vertical parallel arrows and not a "leftrightarrow".  What is the solution?
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I'm not sure how to go about this. One of my thoughts was to take the number of edges in a complete graph, and subtract all of the edges that would be needed to make a smaller cycle.   If you have $C_5$, there is only one way to make a $5$-cycle on $5$ vertices, but there's $10$ ways to make a $3$-cycle, and...  I have no idea how to do this without counting things multiple times.
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Consider the red path from A that zigzags to B, which takes $n$ even steps of length $w$. The path length of the route $P_n$ will be equal to:  $ P_n = P_x + P_y = \frac{n}{2}\times w + \frac{n}{2}\times w = n \times w $  But $\frac{n}{2}\times w = 1$ beacuse it is the length of one of the sides of the triangle so:  $P_n = 2$  Which will be true no matter how many steps you take. However in the limit $n \to \infty, w \to 0$ the parth length $P_\infty$ suddenly becomes:  $P_\infty = \sqrt{1^2 + 1^2} =  \sqrt{2}$  Due to Pythagoras. Why is this the case? It seems the path length suddenly decreases by 0.59!
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It's always drilled into us that we shouldn't use comma splices. They're bad, to be avoided and poor form.  But why? What makes the comma a faulty choice in something like this:     I never liked school, it was boring and repetitive most of the time.   It could be replaced with a semicolon, dash, period or colon and be "correct." Yet to a fair amount of people, the comma seems fine. If it weren't, it wouldn't be found so often in online forums, text messages, emails and so on.  What makes the comma between two sentences an error?
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If I type it in between these parentheses -> () the backward slash inside the parentheses is removed.  As you see it is not there.  However if I surround it with spaces ( \ ), then it is rendered correctly.
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This is what I have come up with so far, but I am rather lost at lines 3-4...  $$\begin{align} (a,b)\in A\times (B\setminus C)\iff&amp; a\in A \land b\in (B\setminus C)\\ \iff&amp; a\in A \land (b\in B \land \lnot(b\in C))\\ \iff&amp; (a\in A \land b\in B) \land (a\in A \land b\notin C)\\ \iff&amp; (a\in A \land b\in B)\setminus (a\in A \land b\in C)\\ \iff&amp; (a,b)\in (A\times B)\setminus ((a,b)\in A\times C)\\ \iff&amp; (a,b)\in (A\times B)\setminus (A\times C) \end{align}$$
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I have  \begin{tikzpicture}[-&gt;,auto,node distance=3cm,   thick,main node/.style={draw,font=\bfseries}]   \node[main node] (111) at (0,1) {$X_{1,1}^{(1)}$};   \node[main node] (121) at (0,2) {$X_{1,2}^{(1)}$};   \node[main node] (1n1) at (0,4) {$X_{1,n}^{(1)}$};   \node[main node] (211) at (0,6) {$X_{2,1}^{(1)}$};   \node[main node] (221) at (0,7) {$X_{2,2}^{(1)}$};   \node[main node] (2n1) at (0,9) {$X_{2,n}^{(1)}$};   \node[main node] (n11) at (0,11) {$X_{n,1}^{(1)}$};   \node[main node] (n21) at (0,12) {$X_{n,2}^{(1)}$};   \node[main node] (nn1) at (0,14) {$X_{n,n}^{(1)}$};   \node[main node] (112) at (2.5,1) {$X_{1,1}^{(2)}$};   \node[main node] (122) at (2.5,2) {$X_{1,2}^{(2)}$};   \node[main node] (1n2) at (2.5,4) {$X_{1,n}^{(2)}$};   \node[main node] (212) at (2.5,6) {$X_{2,1}^{(2)}$};   \node[main node] (222) at (2.5,7) {$X_{2,2}^{(2)}$};   \node[main node] (2n2) at (2.5,9) {$X_{2,n}^{(2)}$};   \node[main node] (n12) at (2.5,11) {$X_{n,1}^{(2)}$};   \node[main node] (n22) at (2.5,12) {$X_{n,2}^{(2)}$};   \node[main node] (nn2) at (2.5,14) {$X_{n,n}^{(2)}$};   \node[main node] (11k) at (6,1) {$X_{1,1}^{(k)}$};   \node[main node] (12k) at (6,2) {$X_{1,2}^{(k)}$};   \node[main node] (1nk) at (6,4) {$X_{1,n}^{(k)}$};   \node[main node] (21k) at (6,6) {$X_{2,1}^{(k)}$};   \node[main node] (22k) at (6,7) {$X_{2,2}^{(k)}$};   \node[main node] (2nk) at (6,9) {$X_{2,n}^{(k)}$};   \node[main node] (n1k) at (6,11) {$X_{n,1}^{(k)}$};   \node[main node] (n2k) at (6,12) {$X_{n,2}^{(k)}$};   \node[main node] (nnk) at (6,14) {$X_{n,n}^{(k)}$};   \node[main node] (11n2) at (9.5,1) {$X_{1,1}^{(n-2)}$};   \node[main node] (12n2) at (9.5,2) {$X_{1,2}^{(n-2)}$};   \node[main node] (1nn2) at (9.5,4) {$X_{1,n}^{(n-2)}$};   \node[main node] (21n2) at (9.5,6) {$X_{2,1}^{(n-2)}$};   \node[main node] (22n2) at (9.5,7) {$X_{2,2}^{(n-2)}$};   \node[main node] (2nn2) at (9.5,9) {$X_{2,n}^{(n-2)}$};   \node[main node] (n1n2) at (9.5,11) {$X_{n,1}^{(n-2)}$};   \node[main node] (n2n2) at (9.5,12) {$X_{n,2}^{(n-2)}$};   \node[main node] (nnn2) at (9.5,14) {$X_{n,n}^{(n-2)}$};   \node[main node] (11n1) at (12,1) {$X_{1,1}^{(n-1)}$};   \node[main node] (12n1) at (12,2) {$X_{1,2}^{(n-1)}$};   \node[main node] (1nn1) at (12,4) {$X_{1,n}^{(n-1)}$};   \node[main node] (21n1) at (12,6) {$X_{2,1}^{(n-1)}$};   \node[main node] (22n1) at (12,7) {$X_{2,2}^{(n-1)}$};   \node[main node] (2nn1) at (12,9) {$X_{2,n}^{(n-1)}$};   \node[main node] (n1n1) at (12,11) {$X_{n,1}^{(n-1)}$};   \node[main node] (n2n1) at (12,12) {$X_{n,2}^{(n-1)}$};   \node[main node] (nnn1) at (12,14) {$X_{n,n}^{(n-1)}$};   \path     (111) edge (112)         edge (122)         edge (1n2)     (121) edge (112)         edge (122)         edge (1n2)     (1n1) edge (112)         edge (122)         edge (1n2)     (121) edge[--,dashed] (1n1); \end{tikzpicture}   I just want simple lines, no arrows. How to manage with edge to achieve that? Thanks
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Note that a product   may be parenthesized in two different ways:   and . Similarly, there are several different ways to parenthesize . Two such ways are  and . Let  be the number of different ways to parenthesize the product . Show that if , then      for all integers .  Keep in mind that the sequence  is the same as the sequence of Catalan numbers.   I am entirely lost. Please can someone offer some hint or direction of where to begin? I humbly thank you so very much.
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Original  The remainder provided by polynom.sty is NOT complete (as shown in the following figure).      My Objective  When I was in senior high school, my teacher made the diagram like below. :-)  Thus the final step must show the TOTAL remainder and mark it with L-shaped curve.  NOTE  I am sorry. The previous post I forgot to put back -1/4 in the original position. It have made you confused. Now there is no mistake anymore. :-)      Minimal Code  \documentclass{article} \usepackage{polynom,array} \usepackage[table]{xcolor}   \makeatletter \def\pld@ArrangeResult#1{%     \ifx\pld@remainder\@empty         \@tempcnta\pld@maxcol\relax         \pld@InsertItems@do\pld@lastline             {\pld@firsttrue\pld@PLD{\pld@R{0}{1}}}%     \fi     \ifnum\pld@currstage&gt;\z@         \pld@Extend\pld@allines{\pld@lastline\cr}%     \else         \pld@InsertFake\pld@lastline     \fi     \pld@iftopresult         \def\pld@lastline{\pld@PrintPoly\pld@divisor%         %====================================================================================         \quad\smash{{\color{red}\rule[-6pt]{\arrayrulewidth}{17pt}}}\kern-\arrayrulewidth&amp;}%         %====================================================================================     \else         \let\pld@lastline\@empty         \ifx B\pld@style\else             \def\pld@lastline{\pld@leftdelim\strut\pld@rightxdelim&amp;}%         \fi     \fi     \expandafter\pld@AR@col\expandafter\pld@PLD                            \expandafter\pld@lastline#1+\relax+%     \pld@SplitQuotient     \pld@iftopresult         \let\pld@currentline\@empty         \expandafter\pld@AR@col\expandafter\pld@PLD                                \expandafter\pld@currentline                                            \pld@quotient+\relax+%         \expandafter\pld@AR@col\expandafter\pld@XPLD                                \expandafter\pld@currentline                                            \pld@shadow+\relax+%         \edef\pld@subline{%             \noexpand\cline{\tw@-\pld@maxcol}%             \noalign{\vskip\jot}}%         \pld@Extend\pld@currentline{\expandafter\cr\pld@subline}%     \else         \@tempcnta-\@tempcnta         \advance\@tempcnta\pld@maxcol\relax \advance\@tempcnta\@ne         \edef\pld@span{\the\@tempcnta}%         \ifx B\pld@style           \pld@AddTo\pld@lastline{%             &amp;\multispan\pld@span${}=%             \pld@PrintPolyWithDelims\pld@divisor             \expandafter\pld@IfSum\expandafter{\pld@divisor}{}{\cdot}%             \expandafter\pld@IfSum\expandafter{\pld@quotient}\pld@true                                                              \pld@false             \pld@if \pld@leftdelim                     \pld@PrintPolyShadow                     \pld@rightdelim               \else \pld@PrintPolyShadow \fi             \pld@firstfalse             \expandafter\pld@PrintRemain\expandafter{\pld@remainder}$}%         \else           \pld@AddTo\pld@lastline{%             &amp;\multispan\pld@span$\pld@leftxdelim\strut\pld@rightdelim             \pld@div             \pld@PrintPolyWithDelims\pld@divisor=             \pld@PrintPolyShadow             \ifx\pld@remainder\@empty\else                 +{}%                 \setbox\z@=\hbox{$\displaystyle                   \frac{\let\strut\@empty\pld@firsttrue \expandafter                         \pld@PrintRemain\expandafter{\pld@remainder}}%                        {\let\strut\@empty\pld@PrintPoly\pld@divisor}$}%                 \dp\z@=\z@\box\z@             \fi             $}%         \fi     \fi \expandafter\pld@AR@\pld@allines\relax} \makeatother   \arrayrulecolor{red} \arrayrulewidth=0.8pt \def\strut{\rule[-6pt]{0pt}{12pt}}  \begin{document} \polylongdiv[style=A]{\frac{3}{7}x^9+x^2-\frac{1}{4}}{\frac{9}{5}x^4-1}  \vspace{1cm} \polylongdiv[style=A]{x^4-1}{x^2-1} \end{document}
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I often get stuck when forming or formatting a sentence with one or more parentheses. For example, I wrote an answer on another Stack Exchange site, in which I wrote:     ...whereas, my phone(xperia x10 mini)   comes with several widgets...   which, afterwards, I thought should be:     ...whereas, my phone (xperia x10 mini)   comes with several widgets...   But in some cases I feel the former way is correct, like:     ... phones which have a camera include   IPhone(5MP), Nokia N8(12MP), Nokia X6(3MP)....   Is there any rule for the placement of space after and before parentheses?
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I've always heard this reasoning, and it makes obvious sense, but how do you actually show it for some arbitrary product? Would it be something like this?  $$(a(b(cd)))e=((ab)(cd))e=(((ab)c)d)e=abcde?$$  Do you just say that the grouping of the parentheses now corresponds to just multiplying straight through? Thanks.
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What is the difference between a[bc]d and a{b,c}d? Why do people use a{b,c}d when there is already a[bc]d?
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This question has bugged me for a while. I want to write a nested function, f(g(x)). If I naively enter it:  \begin{align}     f(g(x))=y \end{align}     The problem is that each and every parenthesis is the same size as all of the others. Shouldn't the outer parentheses be larger? But if I do this:  \begin{align}     f\left(g(x)\right)=y \end{align}     Now it looks even less like the functions are nested. What's the proper way to do this?
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LaTeX commands like this:  Here again, four different simulations are run on networks of size  $C^{G} = 25 \times\ 25 = 625$, $C^{G} = 50 \times\ 50 = 2500$, $C^{G} = 250 \times\ 250 = 62500$, and $C^{G} = 500 \times\ 500 = 250000$...   When compiled generate the following:    The spacing in math mode really irritates me. There's too much white space around the equal sign, and the multiplication sign has more right padding than left padding. Is there any way to reduce this spacing?