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怎样在乳胶中输入这个长方程式？

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怎样在乳胶中输入这个长方程式？ 
\documentclass[a4paper,12pt]{report}
\usepackage[latin1]{inputenc}
\usepackage[T1]{fontenc}
\usepackage[round]{natbib}
\usepackage{pgf}
\usepackage{url}
\usepackage[english]{babel}
\usepackage{multirow}
\usepackage{fancyhdr}
\usepackage{anysize}
\usepackage{amsmath, mathtools}

\begin{document}
\begin{equation}\label{eq:20}
\begin{split} \
\MoveEqLeft[3] \{T_{1} \in (x_{1},x_{1}+h_{1}],\dots,T_{n} \in (x_{n},x_{n}+h_{n}], N(t)=n\} = \\
&\{N(0,x_{1}]=0, N(x_{1},x_{1}+h_{1}] = 1, N(x_{1}+h_{1},x_{2}]=0, \\
& N(x_{2}, x_{2}+h_{2}]=1,\dots, N(x_{n-1}+h_{n-1},x_{n}]=0, \\
& N(x_{n},x_{n}+h_{n}]=1, N(x_{n}+h_{n},t]=0\}
\end{split}
\end{equation}

Taking probabilities on both sides and using the property of independen increments of the Poisson Process N, we obtain:

\begin{equation} \label{eq:21}
\begin{split} 
\MoveEqLeft[6]  P(T_{1} \in (x_{1},x_{1}+h_{1}], \dots, T_{n} \in (x_{n},x_{n}+h_{n}], N(t)=n) \\ P(N(0,x_{1}]=0)P(N(x_{1},x_{1}+h_{1}]=1)P(N(x_{1}+h_{1},x_{2}]=0)= \\
& P(N(x_{2},x_{2}+h{2}]=1) \dots P(N(x_{n-1}+h_{n-1},x_{n}]=0) \\
&P(N(x_{n}, x_{n}+h_{n}]=1) P(N(x_{n}+h_{n},t]=0)= \\
&e^{-\mu(x_{1}}[\mu(x_{1},x_{1}+h_{1}]e^{-\mu(x_{1},x_{1}+h_{1}]}]e^{-\mu(x_{1}+h_{1},x_{2}]} \\
&[\mu(x_{2},x_{2}+h_{2}]e^{-\mu(x_{2}, x_{2}+h_{2}]}] \dots e^{-\mu(x_{n-1}+h_{n-1},x_{n}]} \\
&[\mu(x_{n}, x_{n}+h_{n}]e^{-\mu(x_{n}, x_{n}+h_{n}]]e^{\mu(x_{n}+h_{n},t]}= \\
&e^{-\mu(t)} \mu(x_{1}, x_{1}+h_{1}] \dots \mu(x_{n}, x_{n}+h_{n}] 
\end{split}
\end{equation}
\end{document}

我用 LaTeX 写这个但出现了错误。

因为我之前有过其他拆分，所以出现错误？

可以帮我解方程吗？

答案1

正如 David Carlisle 所指出的，第二个equation环境中语法错误的直接来源是 a}被错误地写成]。

此外，似乎还有一些数学错误内容，例如方括号放置不正确。我已尽力清理这两个方程的外观。希望您可以使用此代码作为清理剩余问题的中途点。

请注意，我用（“点二进制”）替换了一些\dots指令\dotsb，因为它们（即印刷省略号）似乎表示乘法省略。

在此处输入图片描述

\documentclass[a4paper,12pt]{report} 
\usepackage[latin1]{inputenc} 
\usepackage[T1]{fontenc} 
%% Commented out unneeded \usepackage statements:
%\usepackage[round]{natbib} 
%\usepackage{pgf} 
%\usepackage{url} 
%\usepackage[english]{babel} 
%\usepackage{multirow} 
%\usepackage{fancyhdr} 
%\usepackage{anysize} 
\usepackage{mathtools} % 'mathtools' loads 'amsmath' automatically
\begin{document}
\setcounter{equation}{19} % just for this example

\begin{equation}\label{eq:20}
\begin{split}
\MoveEqLeft[3] 
\bigl\{T_{1} \in (x_{1},x_{1}+h_{1}],\dots,T_{n} \in (x_{n},x_{n}+h_{n}], N(t)=n \bigr\} \\
{}=\bigl\{&N(0,x_{1}]=0, N(x_{1},x_{1}+h_{1}] = 1, N(x_{1}+h_{1},x_{2}]=0, \\
  & N(x_{2}, x_{2}+h_{2}]=1,\dots, N(x_{n-1}+h_{n-1},x_{n}]=0, \\
  & N(x_{n},x_{n}+h_{n}]=1, N(x_{n}+h_{n},t]=0 \bigr\}
\end{split}
\end{equation}

Taking probabilities on both sides and using the property of independent increments of the Poisson Process $N$, we obtain:
\begin{equation} \label{eq:21}
\begin{split} 
\MoveEqLeft[3]  
P\bigl(T_{1} \in (x_{1},x_{1}+h_{1}], \dots, T_{n} \in (x_{n},x_{n}+h_{n}], N(t)=n\bigr) \\ 
{}={}&P(N(0,x_{1}]=0)\,
      P(N(x_{1},x_{1}+h_{1}]=1)\,
      P(N(x_{1}+h_{1},x_{2}]=0) \\
     &P(N(x_{2},x_{2}+h_{2}]=1) \dotsb 
      P(N(x_{n-1}+h_{n-1},x_{n}]=0) \\
     &P(N(x_{n}, x_{n}+h_{n}]=1)\,
      P(N(x_{n}+h_{n},t]=0) \\
{}={}&e^{-\mu(x_{1}}[\mu(x_{1},x_{1}+h_{1}]\,
      e^{-\mu(x_{1},x_{1}+h_{1}]}]\,
      e^{-\mu(x_{1}+h_{1},x_{2}]}[\mu(x_{2},x_{2}+h_{2}] \\
     &e^{-\mu(x_{2}, x_{2}+h_{2}]}] \dotsm 
      e^{-\mu(x_{n-1}+h_{n-1},x_{n}]}[\mu(x_{n}, x_{n}+h_{n}] \\
     &e^{-\mu(x_{n}, x_{n}+h_{n}}]\,
      e^{\mu(x_{n}+h_{n},t]} \\
{}={}&e^{-\mu(t)} \mu(x_{1}, x_{1}+h_{1}] \dotsb \mu(x_{n}, x_{n}+h_{n}] 
\end{split}
\end{equation}
\end{document}

答案2

该行&[\mu(x_{n}, x_{n}+h_{n}]e^{-\mu(x_{n}, x_{n}+h_{n}]]e^{\mu(x_{n}+h_{n},t]}=缺少一行\}（第二段倒数第二行）。

这是可以工作的 LaTeX，但要注意，它看起来很糟糕：

\begin{equation}\label{eq:20}
\begin{split} \
\MoveEqLeft[3] \{T_{1} \in (x_{1},x_{1}+h_{1}],\dots,T_{n} \in (x_{n},x_{n}+h_{n}], N(t)=n\} = \\
&\{N(0,x_{1}]=0, N(x_{1},x_{1}+h_{1}] = 1, N(x_{1}+h_{1},x_{2}]=0, \\
& N(x_{2}, x_{2}+h_{2}]=1,\dots, N(x_{n-1}+h_{n-1},x_{n}]=0, \\
& N(x_{n},x_{n}+h_{n}]=1, N(x_{n}+h_{n},t]=0\}
\end{split}
\end{equation}

Taking probabilities on both sides and using the property of independen increments of the Poisson Process N, we obtain:

\begin{equation} \label{eq:21}
\begin{split} 
\MoveEqLeft[6]  P(T_{1} \in (x_{1},x_{1}+h_{1}], \dots, T_{n} \in (x_{n},x_{n}+h_{n}], N(t)=n) 
\\ 
P(N(0,x_{1}]=0)P(N(x_{1},x_{1}+h_{1}]=1)P(N(x_{1}+h_{1},x_{2}]=0)= 
\\
& P(N(x_{2},x_{2}+h{2}]=1) \dots P(N(x_{n-1}+h_{n-1},x_{n}]=0) 
\\
&P(N(x_{n}, x_{n}+h_{n}]=1) P(N(x_{n}+h_{n},t]=0)= 
\\
&e^{-\mu(x_{1}}[\mu(x_{1},x_{1}+h_{1}]e^{-\mu(x_{1},x_{1}+h_{1}]}]e^{-\mu(x_{1}+h_{1},x_{2}]} 
\\
&[\mu(x_{2},x_{2}+h_{2}]e^{-\mu(x_{2}, x_{2}+h_{2}]}] \dots e^{-\mu(x_{n-1}+h_{n-1},x_{n}]} 
\\
&[\mu(x_{n}, x_{n}+h_{n}]e^{-\mu(x_{n}, x_{n}+h_{n}]]e^{\mu(x_{n}+h_{n},t]}}= 
\\
&e^{-\mu(t)} \mu(x_{1}, x_{1}+h_{1}] \dots \mu(x_{n}, x_{n}+h_{n}] 
\end{split}
\end{equation}

但我必须建议您使用\begin{align}或\begin{align*}代替，\begin{equation}\begin{split}因为它现在已成为常态。

以下是一个例子：

\begin{align*}
    &P(T_{1} \in (x_{1},x_{1}+h_{1}], \dots, T_{n} \in (x_{n},x_{n}+h_{n}], N(t)=n) = 
    \\
    & \hspace{1cm} \{N(0,x_{1}]=0, N(x_{1},x_{1}+h_{1}] = 1, N(x_{1}+h_{1},x_{2}]=0, 
    \\  
    & N(x_{2}, x_{2}+h_{2}]=1,\dots, N(x_{n-1}+h_{n-1},x_{n}]=0, 
    \\
    & N(x_{n},x_{n}+h_{n}]=1, N(x_{n}+h_{n},t]=0\}
\end{align*}

请注意\hspace使行稍晚一些开始，以使其看起来像换行符而不是新行。

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