我正在努力创建两列、全页表。目前我编写了以下代码：
\documentclass{article}
\usepackage{booktabs}
\usepackage[margin=13mm,paper=a4paper]{geometry}
\usepackage{graphics}
\begin{document}
\begin{table}
\centering
\small
\resizebox{\columnwidth}{!}{%
\begin{tabular}{l c}
\toprule
y & y' \\
\midrule
as & well \\
using & the \\
as & well \\
using & the \\
as & well \\
using & the \\
as & well \\
using & the \\
as & well \\
using & the \\
as & well \\
using & the \\
as & well \\
using & the \\
\bottomrule
\end{tabular}
}
\end{table}
\end{document}
但不知为何，它的字体太大了，我需要大约 20 行小数学表达式。我是 Latex 新手，所以可能缺少一些简单的东西。
这是输出：
答案1
这里有几种可能的解决方案（在评论中讨论）。
第一个是
\documentclass{article}
\usepackage{booktabs}
\usepackage[margin=13mm,paper=a4paper]{geometry}
\usepackage{graphics}
\begin{document}
\begin{table}
\centering
\small
\begin{tabular*}{\linewidth}{@{\extracolsep{\fill}}l c}
\toprule
y & y' \\
\midrule
as & well \\
using & the \\
as & well \\
using & the \\
as & well \\
using & the \\
as & well \\
using & the \\
as & well \\
using & the \\
as & well \\
using & the \\
as & well \\
using & the \\
\bottomrule
\end{tabular*}
\end{table}
\end{document}
结果是
第二个是
\documentclass{article}
\usepackage{booktabs}
\usepackage[margin=13mm,paper=a4paper]{geometry}
\usepackage{graphics}
\begin{document}
\begin{table}
\centering
\small
\begin{tabular}{l c}
\toprule
y & y' \\
\midrule
as & well \\
using & the \\
as & well \\
using & the \\
as & well \\
using & the \\
as & well \\
using & the \\
as & well \\
using & the \\
as & well \\
using & the \\
as & well \\
using & the \\
\bottomrule
\end{tabular}
\end{table}
\end{document}
结果是
第三个是
\documentclass{article}
\usepackage{booktabs}
\usepackage[margin=13mm,paper=a4paper]{geometry}
\usepackage{graphics}
\begin{document}
\begin{table}
\centering
\small
\resizebox{!}{391pt}{
\begin{tabular}{l c}
\toprule
y & y' \\
\midrule
as & well \\
using & the \\
as & well \\
using & the \\
as & well \\
using & the \\
as & well \\
using & the \\
as & well \\
using & the \\
as & well \\
using & the \\
as & well \\
using & the \\
\bottomrule
\end{tabular}}
\end{table}
\end{document}
结果是
（从图片上很难看出来，但它填满了整个页面并且没有溢出。）
第四个是
\documentclass{article}
\usepackage{booktabs}
\usepackage[margin=13mm,paper=a4paper]{geometry}
\usepackage{graphics}
\usepackage{tabularx}
% ...
\begin{document}
\begin{table}
\centering
\small
\begin{tabularx}{\textwidth}{ XXXXXXX }
\toprule
y & y' \\
\midrule
as & well \\
using & the \\
as & well \\
using & the \\
as & well \\
using & the \\
as & well \\
using & the \\
as & well \\
using & the \\
as & well \\
using & the \\
as & well \\
using & the \\
\bottomrule
\end{tabularx}
\end{table}
\end{document}
这使
我可能错误地安排了tabularx选项的间距，但即便如此，我个人认为第三个选项是最好的选择。
感谢所有评论者（Fran、David Carlisle、Johannes_B 和 Heiko Oberdiek）的建议。
最后，根据你的评论，我正在努力为你真正想要创造的东西找到解决方案。到目前为止，我已经
\documentclass{article}
\usepackage{booktabs}
\usepackage[margin=13mm,paper=a4paper]{geometry}
\usepackage{graphics}
% ...
\begin{document}
\begin{table}
\centering
\small
\resizebox{520pt}{320pt}{%
\begin{tabular}{l l}
Derivatives and Integrals \\
\\
\toprule
Basic Differentiation Rules \\
\midrule
\\
1. $\frac {d}{dx} [cu] = cu'$ & 2. $\frac{d}{dx} [u \pm v] = u' \pm
v'$ \\
3. $\frac{d}{dx} [uv] = uv' + vu'$ & 4. $\frac{d}{dx} [\frac{u}{v}]
= \frac{vu' - uv'}{v^2}$ \\
5. $\frac{d}{dx} [c] = 0$ & 6. $\frac{d}{dx} [u^n] = nu^{n-1} \quad
u'$ \\
7. $\frac{d}{dx} [x] = 1$ & 8. $\frac{d}{dx} [\mid u \mid] =
\frac{u}{\mid u \mid} (u'), \quad u \neq 0$ \\
9. $\frac{d}{dx} [ln \quad u] = \frac{u'}{u}$ & 10. $\frac{d}{dx}
[e^u] = e^u \quad u'$ \\
11. $\frac{d}{dx} [sin \quad u] = (cos \quad u) u'$ & 12. $\frac{d}
{dx} [cos \quad u] = -(sin \quad u) u'$ \\
13. $\frac{d}{dx} [tan \quad u] = (sec^2 \quad u)u'$ & 14.
$\frac{d}{dx} [cot \quad u] = -(csc^2 \quad u) u'$ \\
15. $\frac{d}{dx} [sec \quad u] = (sec \quad u \quad tan \quad u)
u'$ & 16. $\frac{d}{dx} [csc \quad u] = -(csc \quad u \quad cot
\quad u) u'$ \\
17. $\frac{d}{dx} [arcsin \quad u] = \frac{u'}{\sqrt{-1 - u^2}}$ &
18. $\frac{d}{dx} [arccos \quad u] = \frac{-u'}{\sqrt{1-u^2}}$ \\
19. $\frac{d}{dx} [arctan \quad u] = \frac{u'}{1 + u^2}$ & 20.
$\frac{d}{dx} [arccot \quad u] = \frac{-u'}{1 + u^2}$ \\
21. $\frac{d}{dx} [arcsec \quad u] = \frac{u'}{\mid u \mid
\sqrt{u^2 - 1}}$ & 22. $\frac{d}{dx} [arcsec \quad u] = \frac{-u'}
{\mid u \mid \sqrt{u^2 - 1}}$ \\
\\
Basic Integration Formulas \\
\\
1. $\int k \quad f(u) \quad d u = k \int f(u) \quad du$ & 2. $\int
\quad [f(u) \pm g (u)] \quad du = \int f(u) \quad du \pm \int g(u)
\quad du$ \\
3. $\int d u = u + C$ & 4. $\int u^n d u = \frac{u^{n+1}}{n + 1} +
C, \quad n \neq -1$ \\
5. $\int \frac{d}{u} = ln \mid u \mid + C$ & 6. $\int e^u d u = e^u
+ C$ \\
7. $\int sin \quad u \quad du = -cos u + C$ & 8. $\int cos \quad u
\quad d u = sin \quad u + C$ \\
9. $\int tan \quad u \quad du = -ln \mid cos \quad u \mid + C$ &
10. $\int cot \quad u \quad du = ln \mid sin \quad u \mid + C$ \\
11. $\int sec \quad u \quad du = ln \mid sec \quad u + tan \quad du
\mid + C$ & 12. $\int csc \quad u \quad du = -ln \mid csc \quad u +
cot \quad u \mid + C$ \\
13. $\int sec^2 u \quad du = tan \quad u + C$ & 14. $\int csc^2
\quad u \quad du = -cot \quad u + C$ \\
15. $\int sec \quad u \quad tan \quad u \quad du = sec \quad u + C$
& 16. $\int csc \quad u \quad cot \quad du = -csc \quad u + C$ \\
17. $\int \frac{du}{\sqrt{a^2 - u^2}} = arcsin \frac{u}{a} + C$ &
18. $\int \frac{du}{a^2 + u^2} = \frac{1}{a} arctan \frac{u}{a} +
C$ \\
19. $\int \frac{du}{u \sqrt{u^2 - a^2}} = \frac{1}{a} arcsec
\frac{\mid u \mid}{a} + C$ \\
\bottomrule
\end{tabular}}
\end{table}
\end{document}
这使
我正在努力实现这个目标：
有些线条没有居中，调整大小也不完美，但您可以调整大小。我会尝试找出如何正确居中线条。
希望这可以帮助！