我使用以下代码将特定段落分成两列:
\documentclass[12pt]{extreport}
\usepackage[left = 0.5in, right = 0.5in, top = 0.5in, bottom = 0.5in]{geometry}
\usepackage{multicol}
\setlength{\columnseprule}{1pt}
\begin{document}
\begin{multicols}{2}
\textbf{Theorem 1} \quad The distribution of $$\mathcal F_n: = \sup_{x \in \mathcal X} |F_n (x) - F(x)|$$ is known and the same for any DGP whose $F$ is \textbf{continuous}.
\textbf{Theorem 2} \quad If $F$ is \textbf{continuous} then for all $t > 0$, we have $$\mathbb P \left [\sqrt{n} \mathcal F_n \le t \right ] \longrightarrow H(t)= 1-2 \sum_{j=1}^{\infty}(-1)^{j-1} e^{-2 j^{2} t^2}$$ where $H$ is the c.d.f of K-S distribution.
\end{multicols}
\end{document}
和结果
如果我可以将第一列的宽度减少 2 厘米,从而将第二列的宽度增加 2 厘米,那么从我的角度来看,该段落会更加平衡。
您能告诉我如何实现这个目标吗?非常感谢!
答案1
这是一个使用tabularx
具有两个可变宽度类型列的环境的解决方案X
。要求是各个相对宽度(此处为:0.75 和 1.25)的总和等于类型列的数量X
(此处为 2)。(如果您希望强制通常的宽度比率为 1:2,请使用 0.6667 和 1.3333 作为 X 列的相对宽度。)
\documentclass[12pt]{extreport}
\usepackage[margin = 0.5in]{geometry}
\usepackage{mathtools,amssymb,tabularx,ragged2e}
\DeclarePairedDelimiter\abs\lvert\rvert
\newcolumntype{L}[1]{>{\RaggedRight\arraybackslash%
\hsize=#1\hsize\linewidth=\hsize}X}
\begin{document}
\noindent
\begin{tabularx}{\textwidth}{@{} L{0.75}|L{1.25} @{}} % 0.75+1.25=2.0
\textbf{Theorem 1} \quad The distribution of
\[
\mathcal{F}_n \coloneqq \sup_{x\in\mathcal X}\abs{F_n(x)-F(x)}
\]
is known and the same for any DGP whose $F$ is \textbf{continuous}.
&
\textbf{Theorem 2} \quad If $F$ is \textbf{continuous} then for all $t > 0$, we have
\[
\mathbb{P} [\sqrt{n} \mathcal F_n \le t ] \longrightarrow H(t)
= 1- 2\sum_{j=1}^{\infty} (-1)^{j-1} e^{-2 j^2 t^2}
\]
where $H$ is the cdf of the K-S distribution.
\end{tabularx}
\end{document}
答案2
另一种可能性是使用paracol
包和真实定理环境:
\documentclass[12pt]{extreport}
\usepackage[left = 0.5in, right = 0.5in, top = 0.5in, bottom = 0.5in]{geometry}
\usepackage{multicol}
\usepackage{paracol}
\usepackage{amsmath}
\usepackage{amsthm, amsfonts}
\newtheorem{theorem}{Theorem}
\setlength{\columnseprule}{1pt}
\begin{document}
\columnratio{0.4}
\begin{paracol}{2}
\begin{theorem}
The distribution of
\[ \mathcal F_n: = \sup_{x \in \mathcal X} |F_n (x) - F(x)| \]
is known and the same for any DGP whose $F$ is \textbf{continuous}.
\end{theorem}
\switchcolumn \stepcounter{theorem}
\begin{theorem}
If $F$ is \textbf{continuous} then for all $t > 0$, we have
\[ \mathbb P \left [\sqrt{n} \mathcal F_n \le t \right ] \longrightarrow H(t)= 1-2\smash[t]{ \sum_{j=1}^{\infty}}(-1)^{j-1} e^{-2 j^{2} t^2}, \]
where $H$ is the c.d.f of K-S distribution.
\end{theorem}
\end{paracol}
\end{document}
答案3
您可能需要 vwcol 包。texdoc vwcol
详情请参阅。不幸的是,此包不跨页
\documentclass[12pt]{extreport}
\usepackage[left = 0.5in, right = 0.5in, top = 0.5in, bottom = 0.5in]{geometry}
\usepackage{vwcol,amsmath,amssymb}
\begin{document}
\begin{vwcol}[widths={5.5cm,9.5cm}]
\textbf{Theorem 1} \quad The distribution of $$\mathcal F_n: = \sup_{x \in \mathcal X} |F_n (x) - F(x)|$$ is known and the same for any DGP whose $F$ is \textbf{continuous}.
\textbf{Theorem 2} \quad If $F$ is \textbf{continuous} then for all $t > 0$, we have $$\mathbb P \left [\sqrt{n} \mathcal F_n \le t \right ] \longrightarrow H(t)= 1-2 \sum_{j=1}^{\infty}(-1)^{j-1} e^{-2 j^{2} t^2}$$ where $H$ is the c.d.f of K-S distribution.
\end{vwcol}
\end{document}