☰

当方程式被打破时添加标签

tag-icon • tag-icon equations labels
当方程式被打破时添加标签 
\begin{align}
\begin{split}
\pd{EU}{\alpha} &= d(\theta) U'(\pi_{1}) [G_{\alpha}(\alpha, \theta, \epsilon) - \widebar{P} C_{\alpha}(\alpha, \theta)] \\     
                &+ (1-d(\theta)) U'(\pi_{0}) [G_{\alpha}(\alpha, \theta, 0) -  C_{\alpha}(\alpha, \theta)] = 0,\\
\pd{EU}{\theta} &= h'(\theta)[U(\pi_{0}) - U(\pi_{1})] \\
                &+ d(\theta) U'(\pi_{1})[G_{\theta}(\alpha, \theta, \epsilon) - \widebar{P} C_{\theta}(\alpha, \theta)] \\
                &+(1-d(\theta))U'(\pi_{0})[G_{\theta}(\alpha, \theta, 0) -  1] = 0.\\
\end{split}
\end{align}

在此处输入图片描述

我想标记方程 4 和方程 5，然后引用它们。这里我仅将两个方程标记为一个。

编辑：

根据以下答案，我能够标记方程式：

\begin{align}
\begin{split}
\pd{EU}{\alpha} &= d(\theta) U'(\pi_{1}) [G_{\alpha}(\alpha,   \theta, \epsilon) - \widebar{P} C_{\alpha}(\alpha, \theta)] \\
                &+ (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} &= h'(\theta)[U(\pi_{0}) - U(\pi_{1})] \\
                &+ d(\theta) U'(\pi_{1})[G_{\theta}(\alpha, \theta, \epsilon) - \widebar{P} C_{\theta}(\alpha, \theta)] \\
                &+(1-d(\theta))U'(\pi_{0})[G_{\theta}(\alpha, \theta, 0) -  1] = 0.\\
\end{split}\label{eqn:5}\
\end{align}

在此处输入图片描述

但是，现在这两个方程式彼此不一致，我希望它们对齐。有什么建议吗？

答案1

使用\nonumber(或\notag) 避免对 内的特定方程式进行编号align，否则它将被编号（您可以\label-\ref它）：

在此处输入图片描述

\documentclass{article}
\usepackage{amsmath,mathabx}
\newcommand{\pd}[2]{\frac{\partial #1}{\partial #2}}
\begin{document}

\begin{align}
  \pd{EU}{\alpha} &= d(\theta) U'(\pi_{1}) \bigl[G_{\alpha}(\alpha, \theta, \epsilon) - \widebar{P} C_{\alpha}(\alpha, \theta)\bigr] \nonumber \\
                  &\hphantom{=} + (1-d(\theta)) U'(\pi_{0}) [G_{\alpha}(\alpha, \theta, 0) -  C_{\alpha}(\alpha, \theta)] = 0, \label{eq:first} \\
  \pd{EU}{\theta} &= h'(\theta)\bigl[ U(\pi_{0}) - U(\pi_{1})\bigr] \nonumber \\
                  &\hphantom{=} + d(\theta) U'(\pi_{1})\bigl[G_{\theta}(\alpha, \theta, \epsilon) - \widebar{P} C_{\theta}(\alpha, \theta)\bigr] \nonumber \\
                  &\hphantom{=} + (1-d(\theta))U'(\pi_{0})\bigl[G_{\theta}(\alpha, \theta, 0) -  1\bigr] = 0. \label{eq:second}
\end{align}
Consider reviewing~\eqref{eq:first} and~\eqref{eq:second}.

\end{document}

答案2

这里有一种方法可以给你你想要的结果。

\documentclass{article}
\usepackage{amsmath}
\newcommand{\pd}[2]{\frac{\partial {#1}}{\partial {#2}}}
\begin{document}
\setcounter{equation}{3}
\begin{align} 
\begin{split}\smash[b]{\pd{EU}{\alpha}}
 &= d(\theta) U'(\pi_{1})
    [G_{\alpha}(\alpha, \theta, \epsilon) - \bar{P} C_{\alpha}(\alpha, \theta)] \\
 & \quad+ (1-d(\theta)) U'(\pi_{0})
    [G_{\alpha}(\alpha, \theta, 0) - C_{\alpha}(\alpha, \theta)] = 0,
\end{split}\label{aa}\\
\begin{split}\smash[b]{\pd{EU}{\theta}}
 &= h'(\theta)[U(\pi_{0}) - U(\pi_{1})] \\
 & \quad+ d(\theta) U'(\pi_{1})
    [G_{\theta}(\alpha, \theta, \epsilon) -  \bar{P} C_{\theta}(\alpha, \theta)] \\
 & \quad+(1-d(\theta))U'(\pi_{0})
    [G_{\theta}(\alpha, \theta, 0) - 1] = 0.
\end{split}\label{bb}
\end{align}

some text \eqref{aa} some more text \eqref{bb}

\end{document}

示例代码的输出

有几件事需要注意：

  • 分数会分散每个“分割”组的第一行和第二行之间的空间，因此我应用了\smash 底部[b]；两个组之间更宽的空间是合适的，但由于右侧没有任何高度和深度超过单行的东西，所以结果（在我看来）这样看起来更好。

  • 分数是每组中位于符号左边的唯一部分=。所有后续行均缩进，\quad以使结构更清晰。

答案3

这是你想要的吗：

\documentclass{article}
\usepackage{amsmath}
\newcommand{\pd}[2]{\frac{\partial {#1}}{\partial {#2}}}
\begin{document}

\begin{align} 
\begin{split} \pd{EU}{\alpha} &= d(\theta) U'(\pi_{1}) [G_{\alpha}(\alpha,   \theta, \epsilon) - \bar{P} C_{\alpha}(\alpha, \theta)] \\
&\quad+ (1-d(\theta)) U'(\pi_{0}) [G_{\alpha}(\alpha, \theta, 0) - C_{\alpha}(\alpha, \theta)] = 0,\end{split}\label{aa}\\
\begin{split}\pd{EU}{\theta} &= h'(\theta)[U(\pi_{0}) - U(\pi_{1})] \\
&\quad+ d(\theta) U'(\pi_{1})[G_{\theta}(\alpha, \theta, \epsilon) -  \bar{P} C_{\theta}(\alpha, \theta)] \\ &\quad+(1-d(\theta))U'(\pi_{0})[G_{\theta}(\alpha, \theta, 0) - 1] = 0. \end{split}\label{bb}
\end{align}

some text \eqref{aa} some more text \eqref{bb}

\end{document}

相关内容