我有以下等式:
\begin{equation}
\sum^{n}_{\stackrel{i=1}{(j,i)\in Q}} \left\{ w^{1}_{ijR}+\sum^{m=1}_{\stackrel{(m,i)\in Q}{m\neq j}} \left[ (w^{2}_{ijmR}+w^{2}_{ijm,R-1}) + \sum^{l=1}_{\stackrel{(l,i)\in Q}{i\neq j\neq m}} (w^{3}_{ijmlR}+w^{3}_{ijml,R-1}+w^{3}_{ijml,R-2}) \right] \right\}+ \sum^{n}_{\stackrel{j=1}{(l,i)\in Q}} \left\{ w^{1}_{jiR}+ \sum^{m=1}_{\stackrel{(m,j)\in Q}{m\neq i}} \left[ (w^{2}_{jimR}+w^{2}_{jmi,R-1}) +\sum^{l=1}_{\stackrel{(l,j)\in Q}{m\neq i\neq l}}(w^{3}_{jimlR}+w^{3}_{jmil,R-1}+w^{3}_{jmli,R-2}) \right] \right\}=1 ~ ~ ~ i=1,...,n ~ ~ ~ R=1,...,n-1
\end{equation}
我怎样才能将其分成三行?
答案1
如果您不需要边注,这里有一个解决方案,第二个等式有两种变体。一个小注释:对于多行索引,您应该使用\substack
,而不是\stackrel
,后者旨在在 上添加一些内容relation symbol
,以便两行使用相同的字体大小。我还使用了较小的分隔符。
\documentclass{article}
\usepackage{geometry}
\usepackage{mathtools}
\begin{document}
\begin{multline}
\begin{aligned}
&\smashoperator{ \sumⁿ_{\substack{i=1\\(j,i) ∈ Q}}}\, \Biggl\{ w^{1}_{ijR}+\sum^{m=1}_{\substack{(m,i) ∈ Q\\m ≠ j}} \Biggl[ (w^{2}_{ijmR}+w^{2}_{ijm,R-1}) + \smashoperator{\sum^{l=1}_{\substack{(l,i) ∈ Q\\i ≠ j ≠ m}}} (w^{3}_{ijmlR}+w^{3}_{ijml,R-1}+w^{3}_{ijml,R-2}) \Biggr] \Biggr\} \\
+ & \smashoperator{\sumⁿ_{\substack{j=1\\(l,i) ∈ Q}}} \Biggl\{ w^{1}_{jiR}+ \sum^{m=1}_{\substack{(m,j) ∈ Q\\ m ≠ i}} \Biggl[ (w^{2}_{jimR}+w^{2}_{jmi,R-1}) + \smashoperator{\sum^{l=1}_{\substack{(l,j) ∈ Q\\ m ≠ i ≠ l}}}(w^{3}_{jimlR}+w^{3}_{jmil,R-1}+w^{3}_{jmli,R-2}) \Biggr] \Biggr\}
\end{aligned} \\
=1 ,\qquad i=1,...,n,\quad R=1,...,n-1
\end{multline}\\
\begin{multline} \smashoperator{\sum^{n}_{\substack{j=1\\ (j,i)\in Q}}} \Biggl\{w^{1}_{ijR}+w^{1}_{ij,R+1}+\smashoperator{\sum^{n}_{\substack{m=1 \\ (m,i)\in Q \\ m\neq j}}} \,\biggl[w^{2}_{ijm,R-1}+w^{2}_{ijmR}+w^{2}_{ijm,R+1} \\[-6ex]
+ \smashoperator{\sum^{n}_{\substack{l=1 \\ (l,i)\in Q \\ i\neq j\neq m}}}(w^{3}_{ijml,R-2}+w^{3}_{ijmlR}+w^{3}_{ijml,R+2}) \biggr] \Biggr\} \leq 1
\end{multline} \\
\begin{equation}
\begin{aligned}[b]
\smashoperator{\sum^{n}_{\substack{j=1\\ (j,i)\in Q}}} \Biggl\{w^{1}_{ijR}+w^{1}_{ij,R+1}+\smashoperator{\sum^{n}_{\substack{m=1 \\ (m,i)\in Q \\ m\neq j}}} \,\biggl[w^{2}_{ijm,R-1} & + w^{2}_{ijmR}+w^{2}_{ijm,R+1} \\[-5.5ex]
& + \smashoperator{\sum^{n}_{\substack{l=1 \\ (l,i)\in Q \\ i\neq j\neq m}}}(w^{3}_{ijml,R-2}+w^{3}_{ijmlR}+w^{3}_{ijml,R+2}) \biggr] \Biggr\} \leq 1
\end{aligned}
\end{equation}
\end{document}
答案2
- 假设页面边框宽度为 25 毫米
- 使用包
multlined
中的数学环境mathtools
:
(红线表示文字边界)
\documentclass{article}
\usepackage[margin=25mm]{geometry}
\usepackage{mathtools}
%---------------- show page layout. don't use in a real document!
\usepackage{showframe}
\renewcommand\ShowFrameLinethickness{0.15pt}
\renewcommand*\ShowFrameColor{\color{red}}
%---------------------------------------------------------------%
\begin{document}
\begin{equation}
\begin{multlined}[0.95\hsize]
\sum^{n}_{\stackrel{i=1}{(j,i)\in Q}}
\left\{ w^{1}_{ijR}+\smashoperator[l]{\sum^{m=1}_{\stackrel{(m,i)\in Q}{m\neq j}}}
\left[ \Bigl(w^{2}_{ijmR}+w^{2}_{ijm,R-1}\Bigr) +
\smashoperator[l]{\sum^{l=1}_{\stackrel{(l,i)\in Q}{i\neq j\neq m}}}
\Bigl(w^{3}_{ijmlR}+w^{3}_{ijml,R-1}+w^{3}_{ijml,R-2} \Bigr)
\right]
\right\}+ \\
\sum^{n}_{\stackrel{j=1}{(l,i)\in Q}}
\left\{ w^{1}_{jiR}+ \smashoperator[l]{\sum^{m=1}_{\stackrel{(m,j)\in Q}{m\neq i}}}
\left[ \Bigl(w^{2}_{jimR}+w^{2}_{jmi,R-1}\Bigr) +
\smashoperator[l]{\sum^{l=1}_{\stackrel{(l,j)\in Q}{m\neq i\neq l}}}
\Bigl(w^{3}_{jimlR}+w^{3}_{jmil,R-1}+w^{3}_{jmli,R-2}\Bigr)
\right]
\right\}=1,\\
i=1,...,n ;\ R=1,...,n-1
\end{multlined}
\end{equation}
\end{document}