![枚举方程列表](https://linux22.com/image/353496/%E6%9E%9A%E4%B8%BE%E6%96%B9%E7%A8%8B%E5%88%97%E8%A1%A8.png)
考虑以下 MWE:
\documentclass[12pt,a4paper]{article}
\usepackage[fleqn]{amsmath}
\usepackage{amssymb}
\usepackage{amsthm}
\usepackage{enumitem}
\newcommand{\C}{\mathbb{C}}
\begin{document}
\begin{enumerate}[label=(\roman*)]
\itemsep -0.2em
\item \begin{equation*}
e^z = \sum_{n=0} z^n / n!
\end{equation*}
\item \begin{equation*}
\sin z = \sum_{n=0} (-1)^n z^{2n+1}/ (2n+1)!
\end{equation*}
\item \begin{equation*}
\cos z = \sum_{n=0} (-1)^n z^{2n}/ 2n! \\
\text{which converges } \forall \ z \in \C
\end{equation*}
\item \begin{equation*}
\log_e (1 - z) = - \sum_{n=1} z^n/n
\end{equation*}
\item \begin{equation*}
(1 - z)^{-1} = \sum_{n=0} z^n \\
\text{which converges for } |z| < 1
\end{equation*}
\end{enumerate}
\end{document}
我怎样才能使枚举数(i、ii 等)与方程式位于同一行?
答案1
我没有在您的问题中看到任何无法使用 完成的事情equation
。
\documentclass[12pt,a4paper]{article}
\usepackage[fleqn]{amsmath}
\usepackage{amssymb}
\usepackage{amsthm}
\newcommand{\C}{\mathbb{C}}
\newcounter{dummy}
\makeatletter
\newcommand{\leqnomode}{\tagsleft@true}
\makeatother
\begin{document}
\begin{equation}
normalequation
\end{equation}
\bgroup
\leqnomode
\setcounter{dummy}{\theequation}
\renewcommand{\theequation}{\roman{equation}}%
\begin{equation}
e^z = \sum_{n=0} z^n / n!
\end{equation}
\begin{equation}
\sin z = \sum_{n=0} (-1)^n z^{2n+1}/ (2n+1)!
\end{equation}
\begin{equation}
\cos z = \sum_{n=0} (-1)^n z^{2n}/ 2n! \\
\text{which converges } \forall \ z \in \C
\end{equation}
\begin{equation}
\log_e (1 - z) = - \sum_{n=1} z^n/n
\end{equation}
\begin{equation}
(1 - z)^{-1} = \sum_{n=0} z^n \\
\text{which converges for } |z| < 1
\end{equation}
\setcounter{equation}{\thedummy}
\egroup
\bgroup
\leqnomode
\setcounter{dummy}{\theequation}
\renewcommand{\theequation}{\roman{equation}}%
\begin{equation}
e^z = \sum_{n=0} z^n / n!
\end{equation}
\begin{equation}
\sin z = \sum_{n=0} (-1)^n z^{2n+1}/ (2n+1)!
\end{equation}
\begin{equation}
\cos z = \sum_{n=0} (-1)^n z^{2n}/ 2n! \\
\text{which converges } \forall \ z \in \C
\end{equation}
\begin{equation}
\log_e (1 - z) = - \sum_{n=1} z^n/n
\end{equation}
\begin{equation}
(1 - z)^{-1} = \sum_{n=0} z^n \\
\text{which converges for } |z| < 1
\end{equation}
\setcounter{equation}{\thedummy}
\egroup
\begin{equation}
normal equation
\end{equation}
\end{document}
\leqnomode
借自使用 amsmath 放置标签
答案2
使用枚举环境的解决方案:
\documentclass[12pt, a4paper]{article}
\usepackage[fleqn]{mathtools}
\usepackage{amssymb}
\usepackage{amsthm}
\usepackage{enumitem}
\newcommand{\C}{\mathbb{C}}
\begin{document}
\begin{enumerate}[label=(\roman*), leftmargin=*, itemsep=0.4ex, before={\everymath{\displaystyle}}]%
\item $ e^z = \sum_{n=0} z^n / n! $ \label{eq-1}
\item $ \sin z = \sum_{n=0} (-1)^n z^{2n+1}/ (2n+1)! $\label{eq-2}
\item $ \begin{aligned}[t]
& \cos z = \sum_{n=0} (-1)^n z^{2n}/ 2n!\\
& \text{which converges $ \forall \ z \in \C $} \label{eq-3}
\end{aligned}
$
\item $ \log_e (1 - z) = - \sum_{n=1} z^n/n $ \label{eq-4}
\item $\begin{aligned}[t] & (1 - z)^{-1} = \sum_{n=0} z^n ,\\
&\text{which converges for $ |z| < 1 $} \end{aligned}$ \label{eq-5}
\end{enumerate}
It is easy to prove eq. \ref{eq-3}.
\end{document}
答案3
基于的解决方案技术员的评论。
\documentclass[12pt,a4paper]{article}
\usepackage[leqno, fleqn]{amsmath}
\usepackage{amssymb}
\usepackage{ mathtools }
\newcommand{\C}{\mathbb{C}}
\makeatletter
\newcommand{\leqnomode}{\tagsleft@true}
\newcommand{\reqnomode}{\tagsleft@false}
\makeatother
\newtagform{Alph}[\renewcommand{\theequation}{\Alph{equation}}]()
\newtagform{roman}[\renewcommand{\theequation}{\roman{equation}}]()
\newtagform{scroman}[\renewcommand{\theequation}{\scshape\roman{equation}}]
[]
\begin{document}
\usetagform{roman}
\begin{align}
&e^z = \sum_{n=0} z^n / n! \\
&\sin z = \sum_{n=0} (-1)^n z^{2n+1}/ (2n+1)! \\
&\cos z = \sum_{n=0} (-1)^n z^{2n}/ 2n! \text{ which converges }
\forall \ z \in \C \\
&\log_e (1 - z) = - \sum_{n=1} z^n/n \\
&(1 - z)^{-1} = \sum_{n=0} z^n \text{ which converges for } |z| < 1
\end{align}
%\setcounter{equation}{0}
\end{document}
参考: