我最近在发数学讲义，想在其中添加一个包含练习的小节。讲义的编号是 (sec.eq)，但对于练习，我希望编号为 (sec.ex.eq)。我不知道如何实现这一点。
(2017 年 9 月 6 日) 感谢您迄今为止的评论。提供一份最简短的文档：
\documentclass{article}
\usepackage{amsmath,amsthm,amsfonts,amssymb}
\usepackage{hyperref}
\newtheorem{Exercise}{Exercise}[section]
\begin{document}
\section{sec1}
Here is section 1.
\subsection{ssec1.1}
Here I have a subsection.
\subsection{ssec1.2}
Here I have another subsection.
\subsection{exercises1}
Here I have the exercises (using the above defined theroem enviroment).
\section{sec2}
Here the same.
\subsection{ssec2.1}
Here the same.
\subsection{ssec2.2}
Here the same.
\subsection{exercises2}
And again exercises.
\end{document}
编号是关于方程式编号的，我在那里遇到了我的问题。
答案1
的输出可以在通过以下方式定义的环境\theequation中本地更改：Exercise\newtheorem
\AtBeginEnvironment{Exercise}{%
\renewcommand{\theequation}{\theExercise.\arabic{equation}}
}
由于这发生在环境组内，因此重新定义不会影响\theequation使用\begin{equation}...\end{equation}。
由于Exercise定义为在内计算section，\theExercise因此已\thesection.\arabic{Exercise}按设计使用。
请注意，这里方程计数器不会每次都重置Exercise！
\documentclass{article}
\usepackage{amsmath,amsthm,amsfonts,amssymb}
\usepackage{etoolbox}
\usepackage{hyperref}
\newtheorem{Exercise}{Exercise}[section]
\AtBeginEnvironment{Exercise}{%
\renewcommand{\theequation}{\theExercise.\arabic{equation}}
}
\begin{document}
\section{sec1}
Here is section 1.
% Regular equation environment, using a different counter.
\begin{equation}
E = mc^{2}
\end{equation}
\subsection{ssec1.1}
Here I have a subsection.
\subsection{ssec1.2}
Here I have another subsection.
\subsection{exercises1}
Here I have the exercises (using the above defined theroem enviroment).
\begin{Exercise}
\begin{equation}
E = mc^{2}
\end{equation}
And other equations:
\begin{align}
E^{2} &= (p c)^{2} + (m_{0} c^{2})^{2} \\
c^{2} &= a^{2} + b^{2}
\end{align}
\end{Exercise}
\section{sec2}
Here the same.
\subsection{ssec2.1}
Here the same.
\subsection{ssec2.2}
Here the same.
\subsection{exercises2}
And again exercises.
\begin{Exercise}
\begin{equation}
E = mc^{2}
\end{equation}
And other equations:
\begin{align}
E^{2} &= (p c)^{2} + (m_{0} c^{2})^{2} \\
c^{2} &= a^{2} + b^{2}
\end{align}
\end{Exercise}
\end{document}
