枚舉方程式列表

Question

考慮以下 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}

如何得到與方程式在同一行的枚舉數（i、ii 等）？

Answer 1

我在您的問題中沒有看到任何無法使用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 放置標籤

Answer 2

具有枚舉環境的解決方案：

\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} 

Answer 3

解決方案基於技術員的評論。

\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}

參考：

在同一文件中（amsmath 的）leqno 和 reqno 選項之間切換

如何在對齊環境中將數字變更為羅馬數字