Ich habe mich gefragt, ob es möglich ist, ein Array auf zwei Seiten aufzuteilen, da ich eine riesige Formel für mein Array habe. Ich habe die Formel bereits gedreht und die Größe reduziert, aber ich denke, ich muss sie trotzdem auf zwei Seiten aufteilen, damit sie lesbar ist. Ich habe die folgenden Pakete und den folgenden Code verwendet (tut mir leid, ich weiß nicht, wie ich meinen Code anzeigen soll, außer die Latex-Gleichungen zu schreiben):
\documentclass[a4paper,twoside,12pt,openany]{report}
\begin{document}
\usepackage{mathtools}
\usepackage{rotating}
\usepackage{graphicx}
\usepackage[T1]{fontenc}
\usepackage{placeins}
\usepackage{amssymb}
\usepackage{amsmath,scalefnt}
Eigenvector (D12): \scalefont{0.3}{\begin{equation} \rotatebox{-90}{$\left(\begin{array}{ccc} \frac{6\,\left(d-2\right)\,\left(\frac{d}{3}-\frac{2^{1/3}\,{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{1/3}}{6}+\frac{2^{2/3}\,\left(d^2-1\right)}{3\,{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{1/3}}-\frac{1}{3}\right)}{d\,\left(d+2\right)}-\frac{{\left(2\,d-2^{1/3}\,{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{1/3}+\frac{2\,2^{2/3}\,\left(d^2-1\right)}{{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{1/3}}-2\right)}^2}{3\,d\,\left(d+2\right)}-\frac{4\,\left(d-1\right)}{d+2} & -\frac{4\,\left(d-1\right)}{d+2}-\frac{{\left(4\,d+2^{1/3}\,{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{1/3}-\frac{2\,2^{2/3}\,\left(d^2-1\right)}{{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{1/3}}+\frac{2\,2^{2/3}\,\sqrt{3}\,\left(\frac{2^{2/3}\,{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{2/3}\,1{}\mathrm{i}}{4}+d^2\,1{}\mathrm{i}-\mathrm{i}\right)}{{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{1/3}}-4\right)}^2}{12\,d\,\left(d+2\right)}+\frac{6\,\left(d-2\right)\,\left(\frac{d}{3}+\frac{2^{1/3}\,{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{1/3}}{12}-\frac{2^{2/3}\,\left(d^2-1\right)}{6\,{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{1/3}}+\frac{2^{2/3}\,\sqrt{3}\,\left(\frac{2^{2/3}\,{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{2/3}\,1{}\mathrm{i}}{4}+d^2\,1{}\mathrm{i}-\mathrm{i}\right)}{6\,{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{1/3}}-\frac{1}{3}\right)}{d\,\left(d+2\right)} & -\frac{4\,\left(d-1\right)}{d+2}-\frac{{\left(\frac{2\,2^{2/3}\,\left(d^2-1\right)}{{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{1/3}}-2^{1/3}\,{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{1/3}-4\,d+\frac{2\,2^{2/3}\,\sqrt{3}\,\left(\frac{2^{2/3}\,{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{2/3}\,1{}\mathrm{i}}{4}+d^2\,1{}\mathrm{i}-\mathrm{i}\right)}{{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{1/3}}+4\right)}^2}{12\,d\,\left(d+2\right)}-\frac{6\,\left(d-2\right)\,\left(\frac{2^{2/3}\,\left(d^2-1\right)}{6\,{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{1/3}}-\frac{2^{1/3}\,{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{1/3}}{12}-\frac{d}{3}+\frac{2^{2/3}\,\sqrt{3}\,\left(\frac{2^{2/3}\,{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{2/3}\,1{}\mathrm{i}}{4}+d^2\,1{}\mathrm{i}-\mathrm{i}\right)}{6\,{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{1/3}}+\frac{1}{3}\right)}{d\,\left(d+2\right)}\\ \frac{6\,\left(\frac{d}{3}-\frac{2^{1/3}\,{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{1/3}}{6}+\frac{2^{2/3}\,\left(d^2-1\right)}{3\,{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{1/3}}-\frac{1}{3}\right)}{d}-\frac{{\left(2\,d-2^{1/3}\,{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{1/3}+\frac{2\,2^{2/3}\,\left(d^2-1\right)}{{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{1/3}}-2\right)}^2}{6\,d\,\left(d-1\right)}-3 & -\frac{{\left(4\,d+2^{1/3}\,{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{1/3}-\frac{2\,2^{2/3}\,\left(d^2-1\right)}{{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{1/3}}+\frac{2\,2^{2/3}\,\sqrt{3}\,\left(\frac{2^{2/3}\,{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{2/3}\,1{}\mathrm{i}}{4}+d^2\,1{}\mathrm{i}-\mathrm{i}\right)}{{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{1/3}}-4\right)}^2}{24\,d\,\left(d-1\right)}+\frac{2^{2/3}\,\left(\frac{2^{2/3}\,{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{2/3}}{4}-2^{1/3}\,{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{1/3}-d^2+2^{1/3}\,d\,{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{1/3}+1-\sqrt{3}\,1{}\mathrm{i}+\sqrt{3}\,d^2\,1{}\mathrm{i}+\frac{2^{2/3}\,\sqrt{3}\,{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{2/3}\,1{}\mathrm{i}}{4}\right)}{d\,{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{1/3}}-3 & -\frac{{\left(\frac{2\,2^{2/3}\,\left(d^2-1\right)}{{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{1/3}}-2^{1/3}\,{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{1/3}-4\,d+\frac{2\,2^{2/3}\,\sqrt{3}\,\left(\frac{2^{2/3}\,{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{2/3}\,1{}\mathrm{i}}{4}+d^2\,1{}\mathrm{i}-\mathrm{i}\right)}{{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{1/3}}+4\right)}^2}{24\,d\,\left(d-1\right)}+\frac{2^{2/3}\,\left(\frac{2^{2/3}\,{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{2/3}}{4}-2^{1/3}\,{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{1/3}-d^2+2^{1/3}\,d\,{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{1/3}+1+\sqrt{3}\,1{}\mathrm{i}-\sqrt{3}\,d^2\,1{}\mathrm{i}-\frac{2^{2/3}\,\sqrt{3}\,{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{2/3}\,1{}\mathrm{i}}{4}\right)}{d\,{\left(\sqrt{d^2\,{\left(d-1\right)}^2\,\left(17\,d^2+40\,d+24\right)}-d^3-3\,d^2+4\right)}^{1/3}}-3\\ 1 & 1 & 1 \end{array}\right) $} \end{equation}}
\end{document}
Antwort1
Wenn Sie möchten, dass Ihre Leser überhaupt eine Chance haben, zu verstehen, was vor sich geht, müssen Sie die Elemente der Eigenvektoren anhand ihrer gemeinsamen Faktoren neu ausdrücken, möglicherweise in etwa wie folgt (keine Garantie, dass ich es richtig gemacht habe!):
\documentclass[a4paper,twoside,12pt,openany]{report}
\usepackage{mathtools}
\usepackage[T1]{fontenc}
\providecommand\dsq{d^{\mkern1mu 2}}
\begin{document}
The eigenvector matrix $D$ is given by
\[
D=\begin{pmatrix}
d_{11} & d_{12} & d_{13} \\
d_{21} & d_{22} & d_{23} \\
1 & 1 & 1
\end{pmatrix}
\]
where \allowdisplaybreaks
\begin{align*}
d_{11}
&= \frac{6(d-2)V}{d(d+2)}
-\frac{{6V}^2}{3d(d+2)}
-\frac{4(d-1)}{d+2} \\
d_{12}
&=-\frac{(W+X)^2}{12d(d+2)}
+\frac{6(d-2)(Y+Z)}{d(d+2)}
-\frac{4(d-1)}{d+2} \\
d_{13}
&=-\frac{(W-X)^2}{12d(d+2)}
-\frac{6(d-2)(Y-Z)}{d(d+2)}
-\frac{4(d-1)}{d+2} \\
d_{21}
&=\frac{6V}{d}
-\frac{(6V)^2}{6d(d-1)}
-3 \\
d_{22}
&=-\frac{(W+X)^2}{24d(d-1)}
+\frac{2^{2/3}(K-L)}{d\,U}
-3
\\
d_{23}
&= -\frac{(W-X)^2}{24d(d-1)}
+\frac{2^{2/3}\left(K+L\right)}{d\,U}
-3
\\
\shortintertext{and}
U&= (\sqrt{\dsq (d-1)^2(17\dsq +40d+24)}-d^3-3\dsq +4)^{1/3}\\
V&= \frac{d}{3}-\frac{2^{1/3}U}{6}+\frac{2^{2/3}(\dsq -1)}{3U}-\frac{1}{3}\\
W&= 2\cdot2^{2/3}\sqrt{3}(2^{2/3}U^2 \mathrm{i}/4+(\dsq -1) \mathrm{i})/U \\
X&= 4d+2^{1/3}U-2\cdot2^{2/3}(\dsq -1)/U-4 \\
Y&= 2^{2/3}\sqrt{3}(2^{2/3}U^2 \mathrm{i}/4+(\dsq -1) \mathrm{i})/(6U) \\
Z&= \frac{d}{3}+\frac{2^{1/3}U}{12}-\frac{2^{2/3}(\dsq -1)}{6U}-\frac{1}{3} \\
K&= 2^{2/3}U^2/4-2^{1/3}U-\dsq +2^{1/3}d\,U+1 \\
L&= \sqrt{3} \mathrm{i}-\sqrt{3}\dsq \mathrm{i}-2^{2/3}\sqrt{3}U^2 \mathrm{i}/4 \,.
\end{align*}
\end{document}