大公式中的换行符和对齐中的错误消息？

Question

我有一个很长的align公式，我不知道如何以最佳方式（看起来不错并且符合 LaTeX 规则）换行。目前我有

\begin{align*}
\mathrm{Var}(\alpha)&\approx (1+\pi^{2})\exp(-2\delta^{*}+2\zeta^{*})\cdot \mathrm{Var}(\zeta^{*})+\frac{\pi^{2}\exp(-2\delta^{*}+2\zeta^{*})}{1+\pi^{2}}\cdot  \mathrm{Var}(\pi)\\ &+ (1+\pi^{2})\exp(-2\delta^{*}+2\zeta^{*})\cdot \mathrm{Var}(\delta^{*}) + \\
2&\times \left[ \pi\exp(-2\delta^{*}+2\zeta^{*})\cdot \mathrm{Cov}(\pi,\zeta^{*}) - (1+\pi^{2})\exp(-2\delta^{*}+2\zeta^{*})\cdot \mathrm{Cov}(\delta^{*},\zeta^{*})  - \pi\exp(-2\delta^{*}+2\zeta^{*})\cdot \mathrm{Cov}(\delta^{*},\pi)\right]
\end{align*}

我试过

\begin{align*}
\mathrm{Var}(\alpha)&\approx (1+\pi^{2})\exp(-2\delta^{*}+2\zeta^{*})\cdot \mathrm{Var}(\zeta^{*})+\frac{\pi^{2}\exp(-2\delta^{*}+2\zeta^{*})}{1+\pi^{2}}\cdot  \mathrm{Var}(\pi)\\ &+ (1+\pi^{2})\exp(-2\delta^{*}+2\zeta^{*})\cdot \mathrm{Var}(\delta^{*}) + \\
2&\times \left[ \pi\exp(-2\delta^{*}+2\zeta^{*})\cdot \mathrm{Cov}(\pi,\zeta^{*}) - (1+\pi^{2})\exp(-2\delta^{*}+2\zeta^{*})\cdot \mathrm{Cov}(\delta^{*},\zeta^{*}) \\ & - \pi\exp(-2\delta^{*}+2\zeta^{*})\cdot \mathrm{Cov}(\delta^{*},\pi)\right]
\end{align*}

但我收到错误信息：

! Extra }, or forgotten \right.
<template> }
            $}\ifmeasuring@ \savefieldlength@ \fi \set@field \hfil \endtempl...
l.1533 \end{align*}

Answer 1

您可以做的一件事是使用以下命令\hphantom：

\documentclass{article}

\usepackage{mathtools}

\DeclareMathOperator*\var{Var}
\DeclareMathOperator*\cov{Cov}

\begin{document}

\begin{align*}
  \var(\alpha)
  &\approx (1 + \pi^{2}) \exp(-2\delta^{*} + 2\zeta^{*}) \cdot \var(\zeta^{*})\\
  &\hphantom{{}\approx} + \frac{\pi^{2} \exp(-2\delta^{*} + 2\zeta^{*})}{1 + \pi^{2}} \cdot \var(\pi)\\
  &\hphantom{{}\approx} + (1 + \pi^{2}) \exp(-2\delta^{*} + 2\zeta^{*}) \cdot \var(\delta^{*})\\
  &\hphantom{{}\approx} + 2 \times \bigl[\pi \exp(-2\delta^{*} + 2\zeta^{*}) \cdot \cov(\pi, \zeta^{*})\\
  &\hphantom{{}\approx + 2 \times \bigl[} - (1 + \pi^{2}) \exp(-2\delta^{*} + 2\zeta^{*}) \cdot \cov(\delta^{*}, \zeta^{*})\\
  &\hphantom{{}\approx + 2 \times \bigl[} - \pi \exp(-2\delta^{*} + 2\zeta^{*}) \cdot \cov(\delta^{*}, \pi)\bigr]
\end{align*}

\end{document}

关于芭芭拉的注释：egreg 解释了为什么需要\bigland 而\bigr不是\leftand \right。

评论

如果您想节省一些击键时间，可以使用以下命令：

\documentclass{article}

\usepackage{mathtools}

\DeclareMathOperator*\var{Var}
\DeclareMathOperator*\cov{Cov}

\begin{document}

\begin{align*}
  \var(\alpha)
  &\approx (1 + \pi^2) \exp(-2\delta^* + 2\zeta^*) \cdot \var(\zeta^*)\\
  &\hphantom{{}\approx} + \frac{\pi^2 \exp(-2\delta^* + 2\zeta^*)}{1 + \pi^2} \cdot \var(\pi)\\
  &\hphantom{{}\approx} + (1 + \pi^2) \exp(-2\delta^* + 2\zeta^*) \cdot \var(\delta^*)\\
  &\hphantom{{}\approx} + 2 \times \bigl[\pi \exp(-2\delta^* + 2\zeta^*) \cdot \cov(\pi, \zeta^*)\\
  &\hphantom{{}\approx + 2 \times \bigl[} - (1 + \pi^2) \exp(-2\delta^* + 2\zeta^*) \cdot \cov(\delta^*, \zeta^*)\\
  &\hphantom{{}\approx + 2 \times \bigl[} - \pi \exp(-2\delta^* + 2\zeta^*) \cdot \cov(\delta^*, \pi)\bigr]
\end{align*}

\end{document}

Answer 2

您可以通过这种方式减少方程的大小和一些重复的输入：

\documentclass{article}

\usepackage{mathtools}

\DeclareMathOperator*\var{Var}
\DeclareMathOperator*\cov{Cov}

\begin{document}

\begin{align*}
  \var(\alpha)
  &\approx (1+\pi^{2})A\cdot \var(\zeta^{*})
   + \frac{\pi^{2}A}{1+\pi^{2}}\cdot \var(\pi)
   + (1+\pi^{2})A\cdot \var(\delta^{*})\\
  &\hphantom{{}\approx} + 2 \times \bigl[ \pi A\cdot \cov(\pi,\zeta^{*})
   - (1+\pi^{2})A\cdot \cov(\delta^{*},\zeta^{*})
   - \pi A\cdot \cov(\delta^{*},\pi)\bigr]
\end{align*}
where 
\[
    A = \exp(-2\delta^{*}+2\zeta^{*})
\]

\end{document}