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我在用
\documentclass[preprint,showpacs,preprintnumbers,amsmath,amssymb]{revtex4}
\usepackage{amsmath}
\newcommand\numberthis{\addtocounter{equation}{1}\tag{\theequation}}
我想知道如何将 Mathematica 生成的以下两种形式(传统形式和标准形式)以某种可读形式(不会丢失材料 - 如溢出的线条)合并到我的文档中
Mathematica 为我提供了(使用 TradtionalForm)代码
\frac{1}{250} \left(50 (2 k+7)+\sqrt{50-10 \sqrt{5}} \left(-3 \sin \left(\frac{2}{5} \pi
(1-2 k)\right)+2 \sin \left(\frac{4 \pi k}{5}\right)-2 \sin \left(\frac{6 \pi
k}{5}\right)-3 \sin \left(\frac{1}{5} (\pi -6 \pi k)\right)+2 \sin \left(\frac{1}{5}
(\pi -4 \pi k)\right)-3 \left(\sin \left(\frac{2}{5} (3 \pi k+\pi )\right)+\sin
\left(\frac{1}{5} (4 \pi k+\pi )\right)\right)+2 \sin \left(\frac{1}{5} (6 \pi k+\pi
)\right)\right)+\sqrt{10 \left(5+\sqrt{5}\right)} \left(-2 \sin \left(\frac{2}{5} \pi
(1-4 k)\right)-2 \sin \left(\frac{2 \pi k}{5}\right)+2 \sin \left(\frac{8 \pi
k}{5}\right)-2 \sin \left(\frac{2}{5} \pi (k+1)\right)-3 \left(\sin \left(\frac{1}{5}
(\pi -2 \pi k)\right)-\sin \left(\frac{2}{5} (4 \pi k+\pi )\right)+\sin
\left(\frac{1}{5} (8 \pi k+\pi )\right)\right)+3 \cos \left(\frac{1}{10} (4 \pi
k+\pi )\right)\right)\right)
或(使用 StandardForm)
\frac{1}{250} \left(50 (7+2 k)+\sqrt{50-10 \sqrt{5}} \left(-3 \text{Sin}\left[\frac{2}{5}
(1-2 k) \pi \right]+2 \text{Sin}\left[\frac{4 k \pi }{5}\right]-2
\text{Sin}\left[\frac{6 k \pi }{5}\right]-3 \text{Sin}\left[\frac{1}{5} (\pi -6 k \pi
)\right]+2 \text{Sin}\left[\frac{1}{5} (\pi -4 k \pi )\right]-3
\left(\text{Sin}\left[\frac{2}{5} (\pi +3 k \pi )\right]+\text{Sin}\left[\frac{1}{5}
(\pi +4 k \pi )\right]\right)+2 \text{Sin}\left[\frac{1}{5} (\pi +6 k \pi
)\right]\right)+\sqrt{10 \left(5+\sqrt{5}\right)} \left(3 \text{Cos}\left[\frac{1}{10}
(\pi +4 k \pi )\right]-2 \text{Sin}\left[\frac{2}{5} (1-4 k) \pi \right]-2
\text{Sin}\left[\frac{2 k \pi }{5}\right]+2 \text{Sin}\left[\frac{8 k \pi
}{5}\right]-2 \text{Sin}\left[\frac{2}{5} (1+k) \pi \right]-3
\left(\text{Sin}\left[\frac{1}{5} (\pi -2 k \pi )\right]-\text{Sin}\left[\frac{2}{5}
(\pi +4 k \pi )\right]+\text{Sin}\left[\frac{1}{5} (\pi +8 k \pi
)\right]\right)\right)\right)
答案1
\documentclass[preprint,showpacs,preprintnumbers,amsmath,amssymb]{revtex4}
\usepackage{amsmath}
\newcommand\numberthis{\addtocounter{equation}{1}\tag{\theequation}}
\begin{document}
$\let\left\relax\let\right\relax
\frac{1}{250} \left(50 (2 k+7)+\sqrt{50-10 \sqrt{5}} \left(-3 \sin \left(\frac{2}{5} \pi
(1-2 k)\right)+2 \sin \left(\frac{4 \pi k}{5}\right)-2 \sin \left(\frac{6 \pi
k}{5}\right)-3 \sin \left(\frac{1}{5} (\pi -6 \pi k)\right)+2 \sin \left(\frac{1}{5}
(\pi -4 \pi k)\right)-3 \left(\sin \left(\frac{2}{5} (3 \pi k+\pi )\right)+\sin
\left(\frac{1}{5} (4 \pi k+\pi )\right)\right)+2 \sin \left(\frac{1}{5} (6 \pi k+\pi
)\right)\right)+\sqrt{10 \left(5+\sqrt{5}\right)} \left(-2 \sin \left(\frac{2}{5} \pi
(1-4 k)\right)-2 \sin \left(\frac{2 \pi k}{5}\right)+2 \sin \left(\frac{8 \pi
k}{5}\right)-2 \sin \left(\frac{2}{5} \pi (k+1)\right)-3 \left(\sin \left(\frac{1}{5}
(\pi -2 \pi k)\right)-\sin \left(\frac{2}{5} (4 \pi k+\pi )\right)+\sin
\left(\frac{1}{5} (8 \pi k+\pi )\right)\right)+3 \cos \left(\frac{1}{10} (4 \pi
k+\pi )\right)\right)\right)
$
\end{document}
让它适合后,您当然可以修改布局,您可能更喜欢\displaystyle
将它放入或通过结束等来begin{center}..\end{center}
伪造方程式编号。..$\hfill\refstepcounter{equation}(\theequation)