我正在用 IEEEtrans 範本寫一篇日記。我有多行方程，無法容納在一列中。因此，執行了以下操作：
\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}
但正如你在圖片中看到的，我的結果看起來不太好。除此之外，最後一個方程式的方程式編號 (2) 遠低於最後一個方程式。有人可以幫我找出輸入方程式的正確方法嗎？
答案1
也許
\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}
我不認為將顯式內容\times與+內部術語保持一致有幫助，不這樣做也可以節省一行。另請注意\Big[lr]不\Big
答案2
我不認為這是“小”。
在下面我添加lipsum只是為了顯示文字包圍的公式。
這個想法是 an 中的第一列aligned右對齊，因此我們可以避免手動添加空間。是aligned“頂部對齊”，因此它的第一行與等號齊平。
注也\bigl(和\bigr)。你可能想要\Bigsize （我不會），但它們應該是\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}
如果我添加也一樣
\usepackage{newtxtext,newtxmath}
（您需要刪除amssymb，newtx軟體包無論如何都會覆蓋符號集）。我推薦它，以便您的數學公式具有與文字相同的字體系列。
\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}
