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对齐多行方程的右侧

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对齐多行方程的右侧

我想对齐多线方程的右侧并为整个方程分配一个编号。

\begin{multline*}
\frac{\partial P(y_{i}\succ0)} {\partial z_{ij}}=\tfrac {a(1-\tau^{2})^{1/2}}{(2 \pi)^{1/2}} e^{-\tfrac{1}{2}a^{2}(1-\tau^{2})^{2}\left(z_{1i}\prime\gamma_{1j}\right)^{2}}\\
\Phi\left(b\tfrac{z_{2i}\prime\gamma_{2j}}{\sigma}-a\tau \left(z_{1i}\prime\gamma_{1j}\right)\right)\gamma_{1j}+\tfrac {b(1-\tau^{2})^{1/2}}{(2 \pi)^{1/2}} e^{-\tfrac{1}{2}b^{2j}(1-\tau^{2})^{2}\left(\tfrac{z_{2i}\prime\gamma_{2j}}{\sigma}\right)^{2}}\\
\Phi\left(a\left(z_{1i}\prime\gamma_{1j}\right)-b\tau \tfrac{z_{2i}\prime\gamma_{2j}}{\sigma}\right)\tfrac{\gamma_{2j}}{\tau}\\
\end{multline*}

\begin{multline*}
=\phi\left(z_{1i}\prime\gamma_{1}\right)\Phi\left(b\tfrac{z_{2i}\prime\gamma_{2j}}{\sigma}-a\tau z_{1i}\prime\gamma_{1j}\right)\gamma_{1j}\\
+\phi\left(\tfrac{z_{2i}\prime\gamma_{2}}{\sigma}\right)\Phi\left(a\left(z_{1i}\prime\gamma_{1j}\right)-b\tau \tfrac{z_{2i}\prime\gamma_{2j}}{\sigma}\right)\tfrac{\gamma_{2j}}{\tau}
\end{multline*}

\begin{multline*}
=\phi\left(z_{1i}\prime\gamma_{1}\right)\Phi\left(\tfrac{\tfrac{z_{2i}\prime\gamma_{2j}}{\sigma}-\tau z_{1i}\prime\gamma_{1}}{(1-\tau^{2})^{1/2}}\right)\gamma_{1j}\\
+\phi\left(\tfrac{z_{2i}\prime\gamma_{2}}{\sigma}\right)\Phi\left(\tfrac{z_{1i}\prime\gamma_{1j}-\tau \tfrac{z_{2i}\prime\gamma_{2j}}{\sigma}}{(1-\tau^{2})^{1/2}}\right)\tfrac{\gamma_{2j}}{\tau}
\end{multline*}

我该怎么做？我无法使用align

答案1

\documentclass[preview,border=12pt]{standalone}
\usepackage[a4paper,margin=1cm]{geometry}
\usepackage{mathtools}
\begin{document}
\abovedisplayskip=0pt\relax% don't use this line in your production (egreg does not like this)
\begin{equation}
\begin{split}
\frac{\partial P(y_{i}\succ0)} {\partial z_{ij}}
&=
\!
\begin{multlined}[t][10cm]
\tfrac{a(1-\tau^{2})^{1/2}}{(2 \pi)^{1/2}} e^{-\tfrac{1}{2}a^{2}(1-\tau^{2})^{2}\left(z_{1i}\prime\gamma_{1j}\right)^{2}}
\Phi\left(b\tfrac{z_{2i}\prime\gamma_{2j}}{\sigma}
-a\tau \left(z_{1i}\prime\gamma_{1j}\right)\right)
\gamma_{1j}\\
+\tfrac {b(1-\tau^{2})^{1/2}}{(2 \pi)^{1/2}} e^{-\tfrac{1}{2}b^{2j}(1-\tau^{2})^{2}\left(\tfrac{z_{2i}\prime\gamma_{2j}}{\sigma}\right)^{2}}\\
\Phi\left(a\left(z_{1i}\prime\gamma_{1j}\right)-b\tau \tfrac{z_{2i}\prime\gamma_{2j}}{\sigma}\right)\tfrac{\gamma_{2j}}{\tau}
\end{multlined}\\
&=
\!
\begin{multlined}[t][10cm]
\phi\left(z_{1i}\prime\gamma_{1}\right)\Phi\left(b\tfrac{z_{2i}\prime\gamma_{2j}}{\sigma}-a\tau z_{1i}\prime\gamma_{1j}\right)\gamma_{1j}\\
+\phi\left(\tfrac{z_{2i}\prime\gamma_{2}}{\sigma}\right)\Phi\left(a\left(z_{1i}\prime\gamma_{1j}\right)-b\tau \tfrac{z_{2i}\prime\gamma_{2j}}{\sigma}\right)\tfrac{\gamma_{2j}}{\tau}
\end{multlined}\\
&=
\!
\begin{multlined}[t][10cm]
\phi\left(z_{1i}\prime\gamma_{1}\right)\Phi\left(\tfrac{\tfrac{z_{2i}\prime\gamma_{2j}}{\sigma}-\tau z_{1i}\prime\gamma_{1}}{(1-\tau^{2})^{1/2}}\right)\gamma_{1j}\\
+\phi\left(\tfrac{z_{2i}\prime\gamma_{2}}{\sigma}\right)\Phi\left(\tfrac{z_{1i}\prime\gamma_{1j}-\tau \tfrac{z_{2i}\prime\gamma_{2j}}{\sigma}}{(1-\tau^{2})^{1/2}}\right)\tfrac{\gamma_{2j}}{\tau}
\end{multlined}
\end{split}
\end{equation}
\end{document}

在此处输入图片描述

为了获得更好的分辨率，请单击此处

答案2

据我了解你的意图：

\documentclass{article}

\usepackage{amsmath}

\begin{document}

\begin{equation}
\begin{aligned}
 \frac{\partial P(y_{i}\succ0)} {\partial z_{ij}}&=\tfrac {a(1-\tau^{2})^{1/2}}{(2 \pi)^{1/2}} e^{-\tfrac{1}{2}a^{2}(1\tau^{2})^{2}\left(z_{1i}\prime\gamma_{1j}\right)^{2}}\times\\
  &\qquad\Phi\left(b\tfrac{z_{2i}\prime\gamma_{2j}}{\sigma}-a\tau \left(z_{1i}\prime\gamma_{1j}\right)\right)\gamma_{1j}\\
  &\qquad{}+\tfrac {b(1-\tau^{2})^{1/2}}{(2 \pi)^{1/2}} e^{-\tfrac{1}{2}b^{2j}(1-\tau^{2})^{2}\left(\tfrac{z_{2i}\prime\gamma_{2j}}{\sigma}\right)^{2}}\\
  &\qquad\Phi\left(a\left(z_{1i}\prime\gamma_{1j}\right)-b\tau \tfrac{z_{2i}\prime\gamma_{2j}}{\sigma}\right)\tfrac{\gamma_{2j}}{\tau}\\
  &=\phi\left(z_{1i}\prime\gamma_{1}\right)\Phi\left(b\tfrac{z_{2i}\prime\gamma_{2j}}{\sigma}-a\tau z_{1i}\prime\gamma_{1j}\right)\gamma_{1j}\\
  &\qquad{}+\phi\left(\tfrac{z_{2i}\prime\gamma_{2}}{\sigma}\right)\Phi\left(a\left(z_{1i}\prime\gamma_{1j}\right)-b\tau \tfrac{z_{2i}\prime\gamma_{2j}}{\sigma}\right)\tfrac{\gamma_{2j}}{\tau}\\
  &=\phi\left(z_{1i}\prime\gamma_{1}\right)\Phi\left(\tfrac{\tfrac{z_{2i}\prime\gamma_{2j}}{\sigma}-\tau z_{1i}\prime\gamma_{1}}{(1-\tau^{2})^{1/2}}\right)\gamma_{1j}\\
  &\qquad{}+\phi\left(\tfrac{z_{2i}\prime\gamma_{2}}{\sigma}\right)\Phi\left(\tfrac{z_{1i}\prime\gamma_{1j}-\tau \tfrac{z_{2i}\prime\gamma_{2j}}{\sigma}}{(1-\tau^{2})^{1/2}}\right)\tfrac{\gamma_{2j}}{\tau}
\end{aligned}
\end{equation}

\end{document}

在此处输入图片描述

仍需要进行一些更改。特别是，\prime看起来非常奇怪。

答案3

这是一个尝试性的解决方案，它使用split环境中的环境equation来确保整个构造只分配一个“数字”。

请注意，我使用\qquad来缩进与前面内容相乘相关的术语，以及\quad与前面内容相加相关的术语。我还用 替换了各种\prime语句'；如果您愿意\prime，我相信它应该排版在上标位置，因此应该输入为^{\prime}。最后，在第三个等号后面的行中，我用 and 替换了包含大术语的\left(and语句，因为从视觉上讲，括号在其他方面完全占主导地位。\right)\Phi\biggl(\biggr)

在此处输入图片描述

\documentclass{article}
\usepackage{amsmath}
\begin{document}
\begin{equation} 
\begin{split}
\frac{\partial P(y_{i}\succ0)} {\partial z_{ij}}
&=\tfrac {a(1-\tau^{2})^{1/2}}{(2 \pi)^{1/2}} 
e^{-\tfrac{1}{2}a^{2}(1-\tau^{2})^{2} \left(z_{1i}'\gamma_{1j}\right)^{2}}\\
&\qquad\times\Phi\left(b\tfrac{z_{2i}'\gamma_{2j}}{\sigma}-a\tau \left(z_{1i}'\gamma_{1j}\right)\right)\gamma_{1j}\\ 
&\quad+\tfrac {b(1-\tau^{2})^{1/2}}{(2 \pi)^{1/2}} 
e^{-\tfrac{1}{2}b^{2j}(1-\tau^{2})^{2}\left(\tfrac{z_{2i}'\gamma_{2j}}{\sigma}\right)^{2}}\\
&\qquad \times \Phi\left(a\left(z_{1i}'\gamma_{1j}\right)- b\tau \tfrac{z_{2i}'\gamma_{2j}}{\sigma}\right)\tfrac{\gamma_{2j}}{\tau}\\
&=\phi\left(z_{1i}'\gamma_{1}\right)\Phi\left(b\tfrac{z_{2i}'\gamma_{2j}}{\sigma}-a\tau z_{1i}'\gamma_{1j}\right)\gamma_{1j}\\
&\quad+\phi\left(\tfrac{z_{2i}'\gamma_{2}}{\sigma}\right)\Phi\left(a\left(z_{1i}'\gamma_{1j}\right)-b\tau \tfrac{z_{2i}'\gamma_{2j}}{\sigma}\right)\tfrac{\gamma_{2j}}{\tau}\\
&=\phi\left(z_{1i}'\gamma_{1}\right)
\Phi\biggl(\tfrac{\tfrac{z_{2i}'\gamma_{2j}}{\sigma}-\tau z_{1i}'\gamma_{1}}{(1-\tau^{2})^{1/2}}\biggr)\gamma_{1j}\\
&\quad+\phi\left(\tfrac{z_{2i}'\gamma_{2}}{\sigma}\right)
\Phi\biggl(\tfrac{z_{1i}'\gamma_{1j}-\tau \tfrac{z_{2i}'\gamma_{2j}}{\sigma}}{(1-\tau^{2})^{1/2}}\biggr)\tfrac{\gamma_{2j}}{\tau}
\end{split}
\end{equation}
\end{document}

附录：假设您的文本块足够宽，整个方程式结构可以排版成四行而不是八行；但是，我认为这种节省垂直空间的尝试不会提高可理解性：

在此处输入图片描述

\documentclass{article}
\usepackage[margin=1in]{geometry}
\usepackage{amsmath}
\begin{document}
\begin{equation} 
\begin{split}
\frac{\partial P(y_{i}\succ0)} {\partial z_{ij}}
&=\tfrac {a(1-\tau^{2})^{1/2}}{(2 \pi)^{1/2}} 
e^{-\tfrac{1}{2}a^{2}(1-\tau^{2})^{2} \left(z_{1i}'\gamma_{1j}\right)^{2}}
\Phi\left(b\tfrac{z_{2i}'\gamma_{2j}}{\sigma}-a\tau \left(z_{1i}'\gamma_{1j}\right)\right)\gamma_{1j}\\ 
&\quad+\tfrac {b(1-\tau^{2})^{1/2}}{(2 \pi)^{1/2}} 
e^{-\tfrac{1}{2}b^{2j}(1-\tau^{2})^{2}\left(\tfrac{z_{2i}'\gamma_{2j}}{\sigma}\right)^{2}}
\Phi\left(a\left(z_{1i}'\gamma_{1j}\right)- b\tau \tfrac{z_{2i}'\gamma_{2j}}{\sigma}\right)\tfrac{\gamma_{2j}}{\tau}\\
&=\phi\left(z_{1i}'\gamma_{1}\right)\Phi\left(b\tfrac{z_{2i}'\gamma_{2j}}{\sigma}-a\tau z_{1i}'\gamma_{1j}\right)\gamma_{1j}
+\phi\left(\tfrac{z_{2i}'\gamma_{2}}{\sigma}\right)\Phi\left(a\left(z_{1i}'\gamma_{1j}\right)-b\tau \tfrac{z_{2i}'\gamma_{2j}}{\sigma}\right)\tfrac{\gamma_{2j}}{\tau}\\
&=\phi\left(z_{1i}'\gamma_{1}\right)
\Phi\biggl(\tfrac{\tfrac{z_{2i}'\gamma_{2j}}{\sigma}-\tau z_{1i}'\gamma_{1}}{(1-\tau^{2})^{1/2}}\biggr)\gamma_{1j}
+\phi\left(\tfrac{z_{2i}'\gamma_{2}}{\sigma}\right)
\Phi\biggl(\tfrac{z_{1i}'\gamma_{1j}-\tau \tfrac{z_{2i}'\gamma_{2j}}{\sigma}}{(1-\tau^{2})^{1/2}}\biggr)\tfrac{\gamma_{2j}}{\tau}
\end{split}
\end{equation}
\end{document}

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