I want to add this equation to my letter, but it doesn't fit the page: $$ I_{GRV3}=\int_{\Sigma_{i}}\left[4 \pi S_{i}^{i}-v^{\mid i} v_{\mid i}+\frac{1}{4} h^{i j}\left(\Delta_{i m}^{l} \Delta_{j l}^{m}-\Delta_{i m}^{l} \Delta_{i j}^{m}\right)+  \frac{3}{4}\left(K_{i j} K^{i j}-K^{2}\right)-\frac{K}{N} N^{i} \nu_{i i}\right] \sqrt{h} \mathrm{d}^{3} x=0 $$  The problem is that I tried to use an array method to separate them, but I get an error cause I'm separating the \right] from the \left[.
5909
I want to align equations (using the align environment from the amsmath package) inside matching \left( and \right) commands. For example:  \documentclass[a4paper,12pt]{article}  \usepackage{amsmath}  \begin{document}  \begin{align}     a \left( b &amp;= c \right)\\     d \left( e &amp;= f \right) \end{align}  \end{document}   But that code gives this error:     Extra }, or forgotten \right.   The error disappears when I remove the delimiters. Is there a workaround for this?
12561
\begin{align} \begin{split} \pd{EU}{\alpha} &amp;= d(\theta) U'(\pi_{1}) [G_{\alpha}(\alpha, \theta, \epsilon) - \widebar{P} C_{\alpha}(\alpha, \theta)] \\                      &amp;+ (1-d(\theta)) U'(\pi_{0}) [G_{\alpha}(\alpha, \theta, 0) -  C_{\alpha}(\alpha, \theta)] = 0,\\ \pd{EU}{\theta} &amp;= h'(\theta)[U(\pi_{0}) - U(\pi_{1})] \\                 &amp;+ d(\theta) U'(\pi_{1})[G_{\theta}(\alpha, \theta, \epsilon) - \widebar{P} C_{\theta}(\alpha, \theta)] \\                 &amp;+(1-d(\theta))U'(\pi_{0})[G_{\theta}(\alpha, \theta, 0) -  1] = 0.\\ \end{split} \end{align}     I want to label the equations 4 and 5 and then reference them. Here I only have the two equations labelled as one.   EDIT:  From the answer below, I was able to label the equations:  \begin{align} \begin{split} \pd{EU}{\alpha} &amp;= d(\theta) U'(\pi_{1}) [G_{\alpha}(\alpha,   \theta, \epsilon) - \widebar{P} C_{\alpha}(\alpha, \theta)] \\                 &amp;+ (1-d(\theta)) U'(\pi_{0}) [G_{\alpha}(\alpha, \theta, 0) - C_{\alpha}(\alpha, \theta)]=0 \end{split}\label{eqn:4}\\ \begin{split} \pd{EU}{\theta} &amp;= h'(\theta)[U(\pi_{0}) - U(\pi_{1})] \\                 &amp;+ d(\theta) U'(\pi_{1})[G_{\theta}(\alpha, \theta, \epsilon) - \widebar{P} C_{\theta}(\alpha, \theta)] \\                 &amp;+(1-d(\theta))U'(\pi_{0})[G_{\theta}(\alpha, \theta, 0) -  1] = 0.\\ \end{split}\label{eqn:5}\ \end{align}     However, now the two equations are not aligned with each other, and I want them to be aligned. Any suggestions?
11193
I know this is a very common corollary of the class equation. And I know how to do it by using class equation.   But can you do it bu using group action, maybe find a nice set for G to act on?
15307
Given $[a_{ij}]_{10\times10}=\left[ \begin{array}{ccc}   1&amp;2&amp;4&amp;7&amp;...&amp;...&amp;...\\   3&amp;5&amp;8&amp;...&amp;...&amp;...&amp;...\\   6&amp;9&amp;...&amp;...&amp;...&amp;...&amp;...\\   10&amp;...&amp;...&amp;...&amp;...&amp;...&amp;...&amp;\\   ...&amp;...&amp;...&amp;...&amp;...&amp;...&amp;...\\   ...&amp;...&amp;...&amp;...&amp;...&amp;...&amp;98\\   ...&amp;...&amp;...&amp;...&amp;...&amp;99&amp;100\\ \end{array} \right]$, what's $a_{7,3}+a_{7,7}$?  Is there a better way than write down all the elements? It's not easy to find the formula of $a_{i,j}$
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I have a group of equations, such as  /begin{gather} Eq1 // Eq2 // Eq3 // Eq4 /end{gather}   Unfortunately, some of the equations in the group are too long to fit on the page.  For equations that are only slightly too long, is there a way to slightly decrease the font size of that equation only?  For equations that are much too long, is there any automated way to split the equation (where the computer decides where and how to split it to fit the page width), or would this split need to be defined by the user?  Thank you, Kevin  P.S.  Here is my code where the matrix is too wide and goes off the end of the page:  \documentclass{article} \begin{document} \[ \mathbf{L} = \left[ \begin{array}{*{6}{c}} {{L_S}}&amp;{ - {L_{SS}}}&amp;{ - {L_{SS}}}&amp;{M\cos \theta }&amp;{M\cos \left( {\theta  +       \frac{{2\pi }}{3}} \right)}&amp;{M\cos \left( {\theta  - \frac{{2\pi }}{3}}   \right)}\\ { - {L_{SS}}}&amp;{{L_S}}&amp;{ - {L_{SS}}}&amp;{M\cos \left( {\theta  - \frac{{2\pi }}{3}} \right)}&amp;{M\cos \theta }&amp;{M\cos \left( {\theta  + \frac{{2\pi }}{3}} \right)}\\ { - {L_{SS}}}&amp;{ - {L_{SS}}}&amp;{{L_S}}&amp;{M\cos \left( {\theta  + \frac{{2\pi }}{3}} \right)}&amp;{M\cos \left( {\theta  - \frac{{2\pi }}{3}} \right)}&amp;{M\cos \theta }\\ {M\cos \theta }&amp;{M\cos \left( {\theta  - \frac{{2\pi }}{3}} \right)}&amp;{M\cos \left( {\theta  + \frac{{2\pi }}{3}} \right)}&amp;{{L_R}}&amp;{ - {L_{RR}}}&amp;{ - {L_{RR}}}\\ {M\cos \left( {{\theta} + \frac{{2\pi }}{3}} \right)}&amp;{M\cos \theta }&amp;{M\cos \left( {\theta  - \frac{{2\pi }}{3}} \right)}&amp;{ - {L_{RR}}}&amp;{{L_R}}&amp;{ -     {L_{RR}}}\\ {M\cos \left( {{\theta _2} - \frac{{2\pi }}{3}} \right)}&amp;{M\cos \left( {{\theta _2} + \frac{{2\pi }}{3}} \right)}&amp;{M\cos \theta }&amp;{ - {L_{RR}}}&amp;{ - {L_{RR}}}&amp;{{L_R}} \end{array}\right] \] \end{document}
93368
For context, this equation is used in the following (from Schwartz's QFT 3.44)  $$\partial_{\mu} \frac{\partial(\partial_{\beta}A_{\gamma})^2}{\partial(\partial_{\mu}A_{\nu})} = \partial_{\mu}\left[2(\partial_{\alpha}A_{\alpha})\frac{\partial(\partial_{\beta}A_{\gamma})}{\partial(\partial_{\mu}A_{\nu})} g_{\beta\gamma}\right] = \partial_{\mu}[2(\partial_{\alpha}A_{\alpha})g_{\beta\mu}g_{\gamma\nu}g_{\beta\gamma}]=2\partial_{\nu}(\partial_{\alpha}A_{\alpha}).\tag{3.44}$$  But doesn't $$\frac{\partial(\partial_{\beta}A_{\gamma})}{\partial(\partial_{\mu}A_{\nu})} = g_{\beta\mu}g_{\gamma\nu}$$ imply, for instance, that  $$\frac{\partial(\partial_0 A_3)}{\partial(\partial_0 A_3)} = g_{00}g_{33} = -1$$ instead of $=1$? Shouldn't we instead have $\frac{\partial(\partial_{\beta}A_{\gamma})}{\partial(\partial_{\mu}A_{\nu})} = \delta_{\beta\mu}\delta_{\gamma\nu}$?
28429
so I'm trying to use the equation: $Var(X) = E[X^2] - (E[x])^2$, And for the $E[X^2]$ part, I'm trying to use the method of indicators... However, when I do that, I get the same value as with $E[X]$... Is it wrong to try to use the method of indicators for this case? Basically, I end up with $Var(X) = np(1-np)$,  when it should be $Var(X) = np(1-p)$.
4489
Expanding the equation out gives  $(3^{2n}\times3)+(2^n\times2^2) \equiv 0\pmod{7}$   Is this correct? I'm a little hazy on my index laws.  Not sure if this is what I need to do? Am I on the right track?
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\documentclass{article} \usepackage{amsmath} \begin{document} \begin{equation*}   X_i =    \begin{cases}     \left[\begin{smallmatrix}       0 \\ 1 \\ 2     \end{smallmatrix}\right]     &amp;     \text{if $i$ is even;}     \\     \left[\begin{smallmatrix}       2 \\ 1 \\ 0     \end{smallmatrix}\right]     &amp;     \text{otherwise.}   \end{cases} \end{equation*} \end{document}   This is the LaTeX code that I wrote based on a cases environment. There's no space between the matrices on the right. What's the best way to add some vertical spaces to separate them?    p.s. This is a simplified form of my original writing, so I cannot just accept suggestions like 'use a matrix transpose to transform the column vector into a row vector.'
28865
When I used this piece down, I got.  \documentclass[a4paper,12pt]{report} \usepackage[inner=0.75in,outer=0.65in,top=0.75in,bottom=0.75in]{geometry} \usepackage{graphicx} \usepackage{url} \usepackage[sc]{mathpazo} %\usepackage{mathptmx} \usepackage{url} \usepackage[onehalfspacing]{setspace} \usepackage{parskip} \usepackage{paralist} \usepackage{multirow} \begin{document} \begin{table}[!htb] \centering \caption{Present Strength-Production} \begin{tabular}{|c|l|l|c|c|c|c|c|c|} \hline \multirow{2}{*}{Sl.No} &amp;  \multicolumn{2}{c|}{\multirow{2}{*}{\textbf{Designation}}} &amp;  \multicolumn{4}{c|}{\textbf{Shifts}} &amp;\textbf{Leave/Off} &amp;  \multirow{2}{*}{\textbf{Total}}\\ \cline{4-7} &amp; &amp; &amp;\textbf{A} &amp; \textbf{B} &amp; \textbf{C}&amp; \textbf{G}&amp;\textbf{Reserve}&amp;\\ \hline 1 &amp; \multicolumn{2}{l|}{Plant Engineer} &amp; &amp; &amp; &amp; 1&amp; &amp; 1\\ \hline \multicolumn{3}{|l|}{\textbf{Sub Total}} &amp; &amp; &amp; &amp; \textbf{1} &amp; &amp; \textbf{1}\\ \hline 2&amp;\multirow{2}{*}{Operator}&amp;Work related to water regeneration&amp;1&amp;1&amp;1&amp;&amp;1&amp;  4\\ \cline{3-9} 3&amp; &amp; Work related to valve control&amp; 1&amp;1&amp;1&amp; &amp;1&amp;4\\ \hline \multicolumn{3}{|l|}{\textbf{Sub Total}}&amp;\textbf{2}&amp;\textbf{2}&amp;\textbf{2}&amp; &amp;\textbf{2} &amp; \textbf{8}\\ \hline 1 &amp; \multicolumn{2}{l|}{Worker} &amp;2 &amp; 2&amp; 2&amp; &amp; 2&amp; 8\\ \hline \multicolumn{3}{|l|}{\textbf{Sub Total}} &amp;\textbf{2}&amp;\textbf{2}&amp;\textbf{2}&amp; &amp;\textbf{2} &amp; \textbf{8}\\ \hline \hline \multicolumn{3}{|l|}{\textbf{Total}} &amp;\textbf{4} &amp;\textbf{4} &amp;\textbf{4}  &amp;\textbf{1}&amp;\textbf{4} &amp; \textbf{17}\\ \hline \end{tabular} \label{tab:pro_pro} \end{table} \end{document}     Can anyone show me the right path?  Certain important things are   The starting place of Sub total and Total should be from the second column The vertical line in the Designation box Horizontal line between 2 and 3
14940
$$\int \frac{\mathrm{d}x}{x}$$ If I integrate this by parts ($\displaystyle u=\frac{1}{x}, \mathrm{d}u = -\frac{\mathrm{d}x}{x^2}, \mathrm{d}v= \mathrm{d}x, v = x$), then why does this happen?  $$\int \frac{1}{x} \: \mathrm{d}x = u v - \int v \:\mathrm{d}u = \frac{x}{x} - \int x\left(\frac{-1}{x^2}\right) \:\mathrm{d}x=1+\int \frac{1}{x}\:\mathrm{d}x$$  So $\displaystyle\int \frac{1}{x}\:\mathrm{d}x = 1 + \int \frac{1}{x} \:\mathrm{d}x$ and cancel out both sides of $\displaystyle\int \frac{1}{x} \:\mathrm{d}x$ and I get $0=1$. How can I find the error in the process because right now I cant see it.
90273
Can someone tell me what's wrong with this? because i can't figure out where the missing \right( is.  \begin{equation}  \begin{split} Μ=&amp;-\pi \rho \frac{c^{2}}{4}    \left[c\left(\frac{α}{c}-\frac{1}{2}\right)\ddot{h}+Uc\left(\frac{3}{4}-     \frac{α}{c}\right)\dot{\alpha} +\frac{c^{2}}{4}\left(\frac{9}{8}+      \frac{4α^{2}}{c^{2}}-\frac{4α}{c}\right)\ddot{\alpha} \\  &amp;-\left(\frac{4α}{c}-1\right)U C(k) \left(-\dot{h} +U\alpha + c \left(\frac{3}{4}-\frac{α}{c}\right)\dot{\alpha}\right)\right]  \end{split} \end{equation}
7909
Can someone kindly please help and explain how can obtain this equation?  $$ \Gamma^i =\gamma^{jk} \Gamma^i_{jk}= - \frac{1}{\sqrt{|\gamma}|}\partial_l \Big(\sqrt{|\gamma|}~ \gamma^{il} \Big) $$  where $\Gamma^i = ~^{(3)}\Gamma^i$ are the 3D Christoffel symbols and $\gamma_{ij}$ is the spatial metric tensor. The source is the book "Numerical Relativity" (Baumgarte, Shapiro), Eq. 4.40. (in this book the equation uses the 4D metric and the 4D Christoffel symbols, but in the Baumgarte video lectures he uses the above equation.) Here BSSN is the Baumgarte-Shapiro-Shibata-Nakamura formulation.  Thanks
59966
Here is my code:   \begin{align*} (a-bu)e^{-\frac{1}{4}(v^2+u^2)(1+c)+\frac{1}{2}(1-c)u|v|}  &amp;=(a-bx-by)e^{-\frac{1}{4}((x-y)^2+(x+y)^2)(1+c)+\frac{1}{2}(1-c)(x+y)(y-x)}\\ &amp;=(a-bx-by)e^{-(x^2+cy^2)} \\ &amp;=f(x,y). \end{align*}   I have a long equation. My question is as follows: I want to isolate the expression   (a-bu)e^{-\frac{1}{4}(v^2+u^2)(1+c)+\frac{1}{2}(1-c)u|v|}   on the first line so that I don't have an equation on the first line and I want this expression left justified with the other lines (that have the equal sign). How can I do this?
88180
I'd like to know if it is possible to create a equation like the one shown in the figure below. And if it is possible, how can I do it?  Desired characteristics:   Side comments aligned between equations Curved arrows pointing from one equation to another   Can someone help me?
101373
If I have a document like this:  \documentclass[landscape, 12pt]{report}  \usepackage[landscape]{geometry}  \usepackage[utf8x]{inputenc} \usepackage[T1]{fontenc}  \usepackage{amsmath, amssymb, graphics, setspace} \newcommand{\mathsym}[1]{{}} \newcommand{\unicode}[1]{{}} \newcounter{mathematicapage}  \begin{document}    \begin{equation}\label{Equation:Naive_Bayes_Classifier}     P\left(H_h|E_1,E_2,\ldots ,E_e,\ldots    E_{\mathbb{E}}\right)=\frac{P\left(H_h\right) P\left(E_1|H_h\right)    P\left(E_2|H_h,E_1\right) \text{$\ldots $P}     \left(E_e|H_h,E_1,E_2,\ldots ,E_{e-1},E_{e+1} \ldots, E_{\mathbb{E}}\right) \text{$\ldots    $P}     \left(E_{\mathbb{E}}|H_h,E_1,E_2,\ldots ,E_{\mathbb{E}-1}\right)}{ P\left(E_1,E_2,\ldots,E_e, \ldots    ,E_{\mathbb{E}}\right)} \end{equation} \end{document}   How can I auto adjust this equation to appear in the entire page, using all horizontal space? (Without badboxes or smaller than the width of the page?)  If is there a kind of scale to use in equations, giving explicity how greater the equation should become, this solver my problem too.
61397
I have a equation that needs to be left aligned within the cases environment. Here's my code so far, which does not work:  \documentclass{article}  \usepackage{amsmath}  \newcommand{\diff}{\mathop{}\!d} \begin{document}  \begin{equation}\label{eq:algin}  \begin{aligned} \begin{cases} u_i^{k+1}(x)=\arg \min_{u} \Bigg\{ {}&amp; \lambda\sum_{i=1}^{N}\int_{\Omega} r_i(x)u_i(x)\diff x  + \frac{\gamma}{2} \sum_{i=1}^{N}\int_{\Omega} \left( u_i(x)-m(x) \right)^2\diff x  + \frac{\mu}{2}\int_{\Omega} \left( u_i(x)-m(x) \right)^2\diff x \Bigg\}, \\ f'(x_{0})=0 \text{ : } z=f(x_{0}) \text{ is de horizontale raaklijn.} \\ \lim_{x\to x_{0}} f'(x)=\pm\infty \text{ : } x=x_{0} \text{ is de verticale raaklijn.} \\ g(x_{0}){R}_{0} \text{ : } z-f(x_{0})=f'(x_{0})\cdot(x-x_{0})   \text{ is de raaklijn.} \end{cases} \end{aligned} \end{equation}  \end{document}    My wrong output is:   I'm trying to get output like this:    This is my expected result    Thanks egreg. I found that when equation is so long (we cut it before), the equation number will automatically locate in bottom. Could we can relocate it in middle?  Code  \documentclass{article}  \usepackage{amsmath,mathtools}  \newcommand{\diff}{\mathop{}\!d} \DeclareMathOperator*{\argmin}{arg\,min}  \begin{document}  \begin{equation} \begin{dcases} u_i^{k+1}(x) \!\begin{multlined}[t] =\frac{-\lambda r_i(x)+\gamma \bigl( r_i(x)+b_i(x)\bigr)+\gamma \bigl( \alpha-\beta\bigr)}{\theta}\\  -\frac{\sigma \bigl( -\lambda\int_{i=1}^N r_i(x) \diff x + \sigma\sum_{i=1}^N \bigl( r_i(x)-b_i(x)\bigr) +\gamma N \bigl( 1-r_i(x)\bigr)\bigr)}{\gamma\bigl( \sigma N -1\bigr)}, \end{multlined} \\[1ex] f'(x_{0})=0 : z=f(x_{0}) \text{ is de horizontale raaklijn.} \\[1ex] \lim_{x\to x_{0}} f'(x)=\pm\infty : x=x_{0} \text{ is de verticale raaklijn.} \\[1ex] g(x_{0}){R}_{0} : z-f(x_{0})=f'(x_{0})\cdot(x-x_{0})   \text{ is de raaklijn.} \end{dcases} \end{equation} This is text \end{document}