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MathematicaBeatifier.tex
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\documentclass[10pt,a4paper]{jsarticle}
\usepackage[utf8]{inputenc}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{graphicx}
\setcounter{MaxMatrixCols}{16}
\author{SUGIMOTO Kentaro}
\title{Mathematica整形用}
\begin{document}
\begin{align}
\hat{M}^{(2,2)}(\beta)=\tilde{C}^{(2,2)}_{0} + \mathrm{e}^{-4\beta}\tilde{C}^{(2,2)}_{4} + \mathrm{e}^{-8\beta}\tilde{C}^{(2,2)}_{8}\\
\tilde{C}^{(2,2)}_{0} = \begin{bmatrix}
1 & \frac{1}{4} & \frac{1}{4} & 0 & \frac{1}{4} & 0 & 0 & 0 & \frac{1}{4} & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & \frac{1}{4} & 0 & 0 & 0 & \frac{1}{4} & 0 & 0 & 0 & \frac{1}{4} & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & \frac{1}{4} & 0 & 0 & 0 & \frac{1}{4} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & \frac{1}{4} & \frac{1}{4} & \frac{1}{2} & 0 & 0 & 0 & \frac{1}{4} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & \frac{1}{2} & \frac{1}{4} & \frac{1}{4} & 0 & 0 & 0 & 0 & 0 & \frac{1}{4} & 0 & 0 & 0 \\
0 & \frac{1}{4} & 0 & 0 & \frac{1}{4} & \frac{1}{4} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \frac{1}{4} & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & \frac{1}{4} & 0 & \frac{1}{4} & \frac{1}{4} & \frac{1}{2} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \frac{1}{4} & \frac{1}{4} & 0 & 0 & \frac{1}{4} & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & \frac{1}{4} & 0 & 0 & 0 & 0 & 0 & \frac{1}{4} & 0 & \frac{1}{2} & 0 & 0 & 0 & \frac{1}{4} & 0 \\
0 & 0 & 0 & \frac{1}{4} & 0 & 0 & 0 & 0 & 0 & \frac{1}{4} & \frac{1}{4} & \frac{3}{4} & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \frac{1}{4} & 0 & 0 & 0 & 0 & \frac{1}{4} & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \frac{1}{4} & 0 & 0 & \frac{1}{4} & \frac{1}{4} & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & \frac{1}{4} & 0 & 0 & 0 & \frac{1}{4} & 0 & \frac{1}{4} & 0 & \frac{1}{2} & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & \frac{1}{4} & 0 & 0 & 0 & \frac{1}{4} & 0 & \frac{1}{4} & \frac{1}{4} & 1 \\
\end{bmatrix}\\
\frac{1}{256}\begin{bmatrix}
-779 & -201 & -169 & 76 & -302 & -100 & -70 & 35 & -199 & -66 & 78 & 0 & -74 & -16 & 48 & 0 \\
100 & 39 & 44 & 33 & 25 & 17 & 22 & 33 & 38 & 13 & 0 & 0 & 10 & 4 & 3 & 0 \\
85 & 44 & 51 & 22 & 47 & 30 & 36 & 18 & 46 & 20 & 27 & 0 & 28 & 11 & 26 & 0 \\
76 & -21 & 8 & -25 & 17 & -31 & -8 & -32 & 34 & 20 & 58 & 85 & 16 & 3 & 37 & 45 \\
144 & 29 & 51 & 13 & -15 & -14 & 10 & 21 & 37 & 0 & 17 & 0 & -8 & -19 & 32 & 0 \\
68 & 20 & 30 & 34 & -14 & 2 & 12 & 39 & 14 & 4 & 5 & 0 & -2 & -13 & 14 & 0 \\
0 & 4 & 6 & 0 & 13 & 9 & 12 & 0 & 11 & 7 & 13 & 0 & 16 & 10 & 14 & 0 \\
35 & -25 & -1 & -21 & -2 & -22 & 8 & 9 & 16 & 13 & 43 & 90 & 16 & 31 & 67 & 130 \\
91 & 38 & 42 & 0 & 50 & 25 & 28 & 0 & 45 & 15 & 31 & 0 & 25 & 12 & 29 & 0 \\
0 & 4 & 0 & 0 & 4 & 5 & 1 & 0 & 5 & 9 & 0 & 0 & 8 & 8 & 0 & 0 \\
74 & 34 & 7 & 54 & 40 & 15 & -14 & 16 & -21 & 18 & -33 & 74 & -18 & -5 & -59 & 0 \\
40 & -13 & -60 & -92 & 12 & 3 & -3 & 18 & -11 & -18 & -93 & -60 & 14 & 67 & 13 & 186 \\
23 & 17 & 15 & 0 & 30 & 22 & 20 & 0 & 18 & 14 & 25 & 0 & 26 & 10 & 25 & 0 \\
6 & 17 & 3 & 0 & 26 & 23 & 10 & 0 & 8 & 5 & 9 & 0 & 10 & 12 & 11 & 0 \\
25 & 11 & -14 & 16 & 36 & 14 & -16 & 2 & -32 & -4 & -64 & 34 & -25 & -4 & -76 & 0 \\
12 & 3 & -13 & -110 & 33 & 2 & -48 & -159 & -9 & -50 & -116 & -223 & -42 & -111 & -184 & -361 \\
\end{bmatrix}
\end{align}
\begin{align}
m_{\rm b,eq}(z,y)=\frac{z^{-1}-z}{z/y-y/z}\left[\frac{b^{2}}{2\pi}\mathrm{K}(16w^{2})+\frac{b^{2}}{4\pi w}\frac{(a+y^{2}/z)^{2}}{1-by^{2}/c^{2}z}\mathrm{\Pi}\left(\frac{(1-by^{2}/z)}{1-by^{2}/c^{2}z},16w^{2}\right) + \frac{Y^{1/2}-Y^{-1/2}}{2(z^{-1}-z)}-\frac{1}{4}\right]\\
z:=\tanh K\\
y:=\tanh h_{\rm b}\\
\chi_{\rm b, eq}(z)=(z^{-1}-1)\times\left[(1+2w-8w^{2})\frac{\mathrm{K}(16w^{2})}{4\pi w}-\frac{\mathrm{E}(16w^{2})}{4\pi w} - \frac{1}{4}\right]\\
w:=\frac{z(1-z^{2})}{(1+z^{2})^{2}}\\
a:=\frac{1-2z-z^{2}}{1+z^{2}}\\
b:=\frac{1+2z-z^{2}}{1+z^{2}}\\
c:=\frac{2z}{1+z^{2}}\\
Y:=\frac{az/c^{2}y^{2}+1}{by^{2}/c^{2}z-1}\\
\mathrm{K}(k)=\int_{0}^{\pi/2}\frac{d\theta}{\sqrt{1-k^{2}\sin^{2}\theta}}\\
\mathrm{E}(k)=\int_{0}^{\pi/2}d\theta\;\sqrt{1-k^{2}\sin^{2}\theta}\\
\mathrm{\Pi}(n,k)=\int_{0}^{\pi/2}\frac{d\theta}{(1-n\sin^{2}\theta)\sqrt{1-k^{2}\sin^{2}\theta}}
\end{align}
\end{document}