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如何处理乳胶中的这个长片段

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如何处理乳胶中的这个长片段 
$$  \pi(\tau|x)= \frac{ B(S_{\tau}+a,b) \,_2F_1(S_{\tau}+\tau r, S_{\tau}+a, S_{\tau}+a+b; -\frac{1}{r})}
{\sum_{\tau=1}^{n} B(S_{\tau}+a,b) \,_2F_1(S_{\tau}+\tau r, S_{\tau}+a, S_{\tau}+a+b; -\frac{1}{r})}  $$

\begin{equation}
  \times \frac{B(S_{n-\tau}+a_1,b_1) \,_2F_1(S_{n-\tau}+(n-\tau) r, S_{n-\tau}+a_1, S_{n-\tau}+a_1+b_1;  -\frac{1}{r}) }{B(S_{n-\tau}+a_1,b_1) \,_2F_1(S_{n-\tau}+(n-\tau) r, S_{n-\tau}+a_1, S_{n-\tau}+a_1+b_1;  -\frac{1}{r})}
\end{equation}

答案1

我假设您希望将两个方程合并为一个两行方程。在这种情况下，使用包amsmath或mathtools（的改进版本amsmath）及其数学环境（如等multline）align非常方便：

在此处输入图片描述

\documentclass{article}
\usepackage[margin=30mm]{geometry}
\usepackage{mathtools}

\usepackage{showframe}% for show page layout, in real use had to be omitted
\renewcommand*\ShowFrameColor{\color{red}}

\begin{document}
\begin{multline}  
\pi(\tau|x) = 
    \frac{B(S_{\tau}+a,b)_2 F_1(S_{\tau}+\tau r, S_{\tau}+a, S_{\tau}+a+b; -\frac{1}{r})}
          {\sum\limits_{\tau=1}^{n} B(S_{\tau}+a,b)_2 
                             F_1(S_{\tau}+\tau r, S_{\tau}+a, S_{\tau}+a+b; -\frac{1}{r})} 
          \times\\
    \frac{B(S_{n-\tau}+a_1,b_1)_2F_1(S_{n-\tau}+(n-\tau)r, S_{n-\tau}+a_1, S_{n-\tau}+a_1+b_1;  -\frac{1}{r}) }
        {B(S_{n-\tau}+a_1,b_1)_2F_1(S_{n-\tau}+(n-\tau)r, S_{n-\tau}+a_1, S_{n-\tau}+a_1+b_1;  -\frac{1}{r})}
\end{multline}
or
\begin{equation}
\pi(\tau|x) =
    \frac{B(S_{\tau}+a,b)_2 F_1(S,\tau,r)}
          {\sum\limits_{\tau=1}^{n} B(S_{\tau}+a,b)_2 F_1(S,\tau,r)} \times
    \frac{B(S_{n-\tau}+a_1,b_1)_2F_1(S,n,\tau,r)}
         {B(S_{n-\tau}+a_1,b_1)_2F_1(S,n,\tau,r)}
\end{equation}
where
    \begin{align*}
F_1(S,\tau,r)   & = F_1(S_{\tau}+\tau r,      S_{\tau}+a,     S_{\tau}+a+b;       -\frac{1}{r})\\
F_1(S,n,\tau,r) & = F_1(S_{n-\tau}+(n-\tau)r, S_{n-\tau}+a_1, S_{n-\tau}+a_1+b_1; -\frac{1}{r})     
    \end{align*}
\end{document}

请注意，等式的第二部分等于 1:)（或者其中存在一些错误）。

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