格式化方程式

Question

\begin{align} \frac{1}{2} (1-\varepsilon^2)\left(\partial_\tau \phi\right)^2
= \frac{1}{2} \left(1-\epsilon ^2\right) \left(
 -\frac{S \epsilon \sin\tau }{\sqrt{\lambda }}
 -\frac{\text{g2} S^2 \epsilon ^2\sin2\tau }{3 \lambda }
 +\epsilon ^3 \left(-\frac{\left(-\frac{1}{54} \text{g2}^2 S  
 \left(32+19 S^2\right) \lambda +Z \left( 
 \frac{35 \text{g2}^4}{27}
 -\frac{7 \text{g2} \text{g4}}{4}+\frac{5 \text{g5}}{8}
 -\frac{\text{g2}^2 \lambda }{6} 
 +\frac{\lambda ^2}{24}\right)\right) \sin\tau }{\lambda ^{5/2}}
 -\frac{S^3 \left(4 \text{g2}^2-3 \lambda \right) \sin3\tau }
       {24 \lambda ^{3/2}}\right)\right)^2 
\end{align}

我想拆分 latex 文件中的方程式，我已经使用过， & \qquad但由于括号问题，它不起作用。我还能做什么？

Answer 1

在换行之前\left(用 an关闭打开的文本，然后用\right.\left.

答案如下：

\begin{align*} \frac{1}{2} (1-\varepsilon^2)\left(\partial_\tau \phi\right)^2
&= \frac{1}{2} \left(1-\epsilon ^2\right)\left(
 -\frac{S \epsilon \sin\tau }{\sqrt{\lambda }}
 -\frac{\text{g2} S^2 \epsilon ^2\sin2\tau }{3 \lambda }\right.\\
&\quad\left. +\epsilon ^3 \left(-\frac{\left(-\frac{1}{54} \text{g2}^2 S  
 \left(32+19 S^2\right) \lambda +Z \left( 
 \frac{35 \text{g2}^4}{27}
 -\frac{7 \text{g2} \text{g4}}{4}+\frac{5 \text{g5}}{8}
 -\frac{\text{g2}^2 \lambda }{6} 
 +\frac{\lambda ^2}{24}\right)\right) \sin\tau }
 {\lambda ^{5/2}}\right.\right.\\
 &\quad\left.\left.-\frac{S^3 \left(4 \text{g2}^2-3 \lambda \right) \sin3\tau }
       {24 \lambda ^{3/2}}\right)\right)^2 
\end{align*}

Answer 2

\left( \right)用（比如说）替换每一个\Bigl( \Bigr) 如果您以这种方式手动选择大小，那么您可以将括号拆分到行和对齐单元格上。

此外，请始终发布完整的文档，而不仅仅是片段：

\documentclass{article}
\usepackage{amsmath}
\begin{document}

\begin{align*} \frac{1}{2} (1-\varepsilon^2)(\partial_\tau \phi)^2
&= \frac{1}{2} (1-\epsilon ^2) \Biggl(
 -\frac{S \epsilon \sin\tau }{\sqrt{\lambda }}
 -\frac{\text{g2} S^2 \epsilon ^2\sin2\tau }{3 \lambda }\\
&\quad +\epsilon ^3 \Bigl(-\frac{\Bigl(-\frac{1}{54} \text{g2}^2 S  
 (32+19 S^2) \lambda +Z \Bigl( 
 \frac{35 \text{g2}^4}{27}
 -\frac{7 \text{g2} \text{g4}}{4}+\frac{5 \text{g5}}{8}
 -\frac{\text{g2}^2 \lambda }{6} 
 +\frac{\lambda ^2}{24}\Bigr)\Bigr) \sin\tau }{\lambda ^{5/2}}\\
&\qquad -\frac{S^3 (4 \text{g2}^2-3 \lambda) \sin3\tau }
       {24 \lambda ^{3/2}}\Bigr)\Biggr)^2 
\end{align*}

\end{document}

Answer 3

\left由于某种原因，您想将三条线居中。然后您可以执行以下操作（观察带有和 的注释版本\right）：

\documentclass{article}
\usepackage{amsmath}

\begin{document}

\[
\begin{array}{c}
\displaystyle
 \frac{1}{2} (1-\varepsilon^2)
\left(
\partial_\tau \phi\right)^2
= \frac{1}{2} \left(1-\epsilon ^2\right) 
%\left(
\Biggl(
 -\frac{S \epsilon \sin\tau }{\sqrt{\lambda }}
 -\frac{\text{g2} S^2 \epsilon ^2\sin2\tau }{3 \lambda }
%\right.
\\
\displaystyle
 +\epsilon ^3
% \left(
\Biggl(
-\frac{\left(-\frac{1}{54} \text{g2}^2 S  
 \left(32+19 S^2\right) \lambda +Z \left( 
 \frac{35 \text{g2}^4}{27}
 -\frac{7 \text{g2} \text{g4}}{4}+\frac{5 \text{g5}}{8}
 -\frac{\text{g2}^2 \lambda }{6} 
 +\frac{\lambda ^2}{24}\right)\right) \sin\tau }{\lambda ^{5/2}}
%\right.
\\
\displaystyle
%\left.\left.
 -\frac{S^3 \left(4 \text{g2}^2-3 \lambda \right) \sin3\tau }
       {24 \lambda ^{3/2}}
%\right)\right)^2 
\Biggr)\Biggr)^2
\end{array}
\]

\end{document}