我正在用 IEEEtrans 模板写一篇日志。我有多行方程，无法放在一列中。因此，我做了以下事情：
\documentclass[final]{IEEEtran}
\IEEEoverridecommandlockouts
% \overrideIEEEmargins
\setlength{\textheight}{237mm}
\usepackage{multirow}
\usepackage{amssymb,amsmath,amsthm}
\usepackage{color}
\usepackage{graphicx}
\usepackage{tikz}
\usepackage{epstopdf}
\usepackage{romannum}
\usepackage{algorithm}
\usepackage{subcaption}
\usepackage{graphicx}
\usepackage{float}
\usepackage{multicol}
\usepackage[noend]{algpseudocode}
\usepackage{romannum}
\def\m{\tilde{m}}
\begin{document}
\begin{align}
m & = 2^{\m} - 1 = 2^4 ( 2^{\m/2 - 2})^2 - 1 = 16 k^2 -1, \label{eq:92} \\
n & = (2^{2\m - 1} + 2^{3\m/2 -2} - 2^{\m - 2} + 2^{\m/2} + 1) (2^{\m/2} - 1) \nonumber \\
\begin{split}
& = \Big( 2^7 (2^{\m/2 - 2})^4 + 2^4 (2^{\m/2 - 2} )^3 - 2^2 (2^{\m/2 - 2})^2 \\
& \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad + 2^2 \cdot 2^{\m/2 - 2} + 1 \Big)\\
& \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \times (2^2 \cdot 2^{\m/2 - 2} - 1)\\
& = (2^7 k^4 + 2^4 k^3 - 2^2 k^2 + 2^2 k + 1) (2^2 k - 1) \\
& = 2^9 k^5 - 2^6 k^4 - 2^5 k^3 + 20 k^2 - 1.
\end{split}
\end{align}
\end{document}
但正如您在图片中看到的，我的结果看起来不太好。此外，最后一个方程的方程编号 (2) 远低于最后一个方程。有人能帮我找到输入方程的正确方法吗？
答案1
也许
\documentclass[final]{IEEEtran}
\IEEEoverridecommandlockouts
% \overrideIEEEmargins
\setlength{\textheight}{237mm}
\usepackage{amssymb,mathtools,amsthm}
\def\m{\tilde{m}}
\begin{document}
\begin{align}
m & = 2^{\m} - 1 = 2^4 ( 2^{\m/2 - 2})^2 - 1 = 16 k^2 -1, \label{eq:92} \\
n & = (2^{2\m - 1} + 2^{3\m/2 -2} - 2^{\m - 2} + 2^{\m/2} + 1) (2^{\m/2} - 1) \nonumber \\
&=
\begin{multlined}[t]
\Bigl( 2^7 (2^{\m/2 - 2})^4 + 2^4 (2^{\m/2 - 2} )^3 - 2^2 (2^{\m/2 - 2})^2 \\
+ 2^2 \cdot 2^{\m/2 - 2} + 1 \Bigr) (2^2 \cdot 2^{\m/2 - 2} - 1)
\end{multlined}\nonumber\\
& = (2^7 k^4 + 2^4 k^3 - 2^2 k^2 + 2^2 k + 1) (2^2 k - 1) \nonumber\\
& = 2^9 k^5 - 2^6 k^4 - 2^5 k^3 + 20 k^2 - 1.
\end{align}
\end{document}
我认为将显式\times与+内部项对齐没有帮助，不这样做也可以节省一行。另请\Big[lr]注意\Big
答案2
我不认为它“小”。
下面我添加了lipsum仅用于显示被文本包围的公式。
其理念是， 中的第一列aligned与右对齐，这样我们就可以避免手动添加空格。aligned是“顶部对齐”，因此其第一行与等号齐平。
还要注意\bigl(和\bigr)。你可能想要\Big大小（我不会），但它们应该是\Bigl(和\Bigr)。
\documentclass[final]{IEEEtran}
\IEEEoverridecommandlockouts
% \overrideIEEEmargins
%\setlength{\textheight}{237mm}% don't change the class defaults
\usepackage{multirow}
\usepackage{amssymb,amsmath,amsthm}
\usepackage{color}
\usepackage{graphicx}
\usepackage{tikz}
%\usepackage{epstopdf}% not needed
\usepackage{romannum}
\usepackage{algorithm}
\usepackage{subcaption}
\usepackage{graphicx}
\usepackage{float}
\usepackage{multicol}
\usepackage[noend]{algpseudocode}
%\usepackage{romannum}% twice?
\usepackage{lipsum}% to see in context
\newcommand{\m}{\tilde{m}}% not \def
\begin{document}
\lipsum*[1][1-3]
\begin{align}
m & = 2^{\m} - 1 = 2^4 ( 2^{\m/2 - 2})^2 - 1 = 16 k^2 -1, \label{eq:92} \\
n & = (2^{2\m - 1} + 2^{3\m/2 -2} - 2^{\m - 2} + 2^{\m/2} + 1) (2^{\m/2} - 1) \nonumber \\
& = \begin{aligned}[t]
\bigl( 2^7 (2^{\m/2 - 2})^4 + 2^4 (2^{\m/2 - 2} )^3 - 2^2 (2^{\m/2 - 2})^2 \\
{}+ 2^2 \cdot 2^{\m/2 - 2} + 1 \bigr)\\
{} \times (2^2 \cdot 2^{\m/2 - 2} - 1)
\end{aligned} \nonumber \\
& = (2^7 k^4 + 2^4 k^3 - 2^2 k^2 + 2^2 k + 1) (2^2 k - 1) \nonumber \\
& = 2^9 k^5 - 2^6 k^4 - 2^5 k^3 + 20 k^2 - 1.
\end{align}
\lipsum
\end{document}
如果我添加
\usepackage{newtxtext,newtxmath}
（您需要删除amssymb，因为newtx包无论如何都会覆盖符号集）。我建议这样做，以便您的数学公式具有与文本相同的字体系列。
\documentclass[final]{IEEEtran}
\IEEEoverridecommandlockouts
% \overrideIEEEmargins
%\setlength{\textheight}{237mm}% don't change the class defaults
\usepackage{multirow}
\usepackage{amsmath,amsthm}
\usepackage{newtxtext,newtxmath}
\usepackage{color}
\usepackage{graphicx}
\usepackage{tikz}
%\usepackage{epstopdf}% not needed
\usepackage{romannum}
\usepackage{algorithm}
\usepackage{subcaption}
\usepackage{graphicx}
\usepackage{float}
\usepackage{multicol}
\usepackage[noend]{algpseudocode}
%\usepackage{romannum}% twice?
\usepackage{lipsum}% to see in context
\newcommand{\m}{\tilde{m}}% not \def
\begin{document}
\lipsum*[1][1-3]
\begin{align}
m & = 2^{\m} - 1 = 2^4 ( 2^{\m/2 - 2})^2 - 1 = 16 k^2 -1, \label{eq:92} \\
n & = (2^{2\m - 1} + 2^{3\m/2 -2} - 2^{\m - 2} + 2^{\m/2} + 1) (2^{\m/2} - 1) \nonumber \\
& = \begin{aligned}[t]
\bigl( 2^7 (2^{\m/2 - 2})^4 + 2^4 (2^{\m/2 - 2} )^3 - 2^2 (2^{\m/2 - 2})^2 \\
{}+ 2^2 \cdot 2^{\m/2 - 2} + 1 \bigr)\\
{} \times (2^2 \cdot 2^{\m/2 - 2} - 1)
\end{aligned} \nonumber \\
& = (2^7 k^4 + 2^4 k^3 - 2^2 k^2 + 2^2 k + 1) (2^2 k - 1) \nonumber \\
& = 2^9 k^5 - 2^6 k^4 - 2^5 k^3 + 20 k^2 - 1.
\end{align}
\lipsum
\end{document}
