作为我两天前问题的后续如何从 Mathematica 渲染可读的大型 TeXForm 表达式
我也想类似地融入以下两种表达方式：
1.
\frac{2 (\alpha +3) (2 \alpha +5) (10 \alpha +17) (10 \alpha +19) (10 \alpha +21) (10
\alpha +23) (\alpha (\alpha (\alpha (20 \alpha (5 \alpha (740 \alpha
+5783)+91541)+3013197)+2724024)+1284280)+246960) \, _7F_6\left(1,\alpha
+\frac{4}{5},\alpha +\frac{5}{6},\alpha +\frac{7}{6},\alpha +\frac{6}{5},\alpha
+\frac{7}{5},\alpha +\frac{8}{5};\alpha +\frac{17}{10},\alpha +\frac{19}{10},\alpha
+\frac{21}{10},\alpha +\frac{23}{10},\alpha +\frac{5}{2},\alpha
+3;\frac{27}{64}\right)+16875 \left(\alpha +\frac{4}{5}\right) \left(\alpha
+\frac{5}{6}\right) \left(\alpha +\frac{7}{6}\right) \left(\alpha +\frac{6}{5}\right)
\left(\alpha +\frac{7}{5}\right) \left(\alpha +\frac{8}{5}\right) \left((\alpha
(\alpha (20 \alpha (25 \alpha (888 \alpha +5783)+366164)+9039591)+5448048)+1284280)
\, _7F_6\left(2,\alpha +\frac{9}{5},\alpha +\frac{11}{6},\alpha +\frac{13}{6},\alpha
+\frac{11}{5},\alpha +\frac{12}{5},\alpha +\frac{13}{5};\alpha +\frac{27}{10},\alpha
+\frac{29}{10},\alpha +\frac{31}{10},\alpha +\frac{33}{10},\alpha +\frac{7}{2},\alpha
+4;\frac{27}{64}\right)+(\alpha (40 \alpha (25 \alpha (1110 \alpha
+5783)+274623)+9039591)+2724024) \, _8F_7\left(2,2,\alpha +\frac{9}{5},\alpha
+\frac{11}{6},\alpha +\frac{13}{6},\alpha +\frac{11}{5},\alpha +\frac{12}{5},\alpha
+\frac{13}{5};1,\alpha +\frac{27}{10},\alpha +\frac{29}{10},\alpha
+\frac{31}{10},\alpha +\frac{33}{10},\alpha +\frac{7}{2},\alpha
+4;\frac{27}{64}\right)+(40 \alpha (25 \alpha (1480 \alpha +5783)+183082)+3013197)
\, _9F_8\left(2,2,2,\alpha +\frac{9}{5},\alpha +\frac{11}{6},\alpha
+\frac{13}{6},\alpha +\frac{11}{5},\alpha +\frac{12}{5},\alpha
+\frac{13}{5};1,1,\alpha +\frac{27}{10},\alpha +\frac{29}{10},\alpha
+\frac{31}{10},\alpha +\frac{33}{10},\alpha +\frac{7}{2},\alpha
+4;\frac{27}{64}\right)+20 (25 \alpha (2220 \alpha +5783)+91541) \,
_{10}F_9\left(2,2,2,2,\alpha +\frac{9}{5},\alpha +\frac{11}{6},\alpha
+\frac{13}{6},\alpha +\frac{11}{5},\alpha +\frac{12}{5},\alpha
+\frac{13}{5};1,1,1,\alpha +\frac{27}{10},\alpha +\frac{29}{10},\alpha
+\frac{31}{10},\alpha +\frac{33}{10},\alpha +\frac{7}{2},\alpha
+4;\frac{27}{64}\right)+100 (4440 \alpha +5783) \, _{11}F_{10}\left(2,2,2,2,2,\alpha
+\frac{9}{5},\alpha +\frac{11}{6},\alpha +\frac{13}{6},\alpha +\frac{11}{5},\alpha
+\frac{12}{5},\alpha +\frac{13}{5};1,1,1,1,\alpha +\frac{27}{10},\alpha
+\frac{29}{10},\alpha +\frac{31}{10},\alpha +\frac{33}{10},\alpha +\frac{7}{2},\alpha
+4;\frac{27}{64}\right)+74000 \, _{12}F_{11}\left(2,2,2,2,2,2,\alpha
+\frac{9}{5},\alpha +\frac{11}{6},\alpha +\frac{13}{6},\alpha +\frac{11}{5},\alpha
+\frac{12}{5},\alpha +\frac{13}{5};1,1,1,1,1,\alpha +\frac{27}{10},\alpha
+\frac{29}{10},\alpha +\frac{31}{10},\alpha +\frac{33}{10},\alpha +\frac{7}{2},\alpha
+4;\frac{27}{64}\right)\right)}{4288785520939920}
和
2.
\text{DifferenceRoot}\left[\left\{G_2^1,\alpha \right\}\unicode{f4a1}\left\{(\alpha +3)
(\alpha +4) (2 \alpha +5) (2 \alpha +7) (10 \alpha +17) (10 \alpha +19) (10 \alpha
+21) (10 \alpha +23) (10 \alpha +27) (10 \alpha +29) (10 \alpha +31) (10 \alpha +33)
(\alpha (\alpha (\alpha (20 \alpha (5 \alpha (740 \alpha
+5783)+91541)+3013197)+2724024)+1284280)+246960)+804147285176235 (\alpha +1) (5 \alpha
+4) (5 \alpha +6) (5 \alpha +7) (5 \alpha +8) (6 \alpha +5) (6 \alpha +7) G_2^1+\alpha
(-2144392760469960 (\alpha +4) (2 \alpha +7) (10 \alpha +27) (10 \alpha +29) (10
\alpha +31) (10 \alpha +33)) G_2^1=0,1 G_2^1=\frac{45}{286}\right\}\right]
答案1
确实，这是同一个问题：
\documentclass[preprint,showpacs,preprintnumbers,amsmath,amssymb]{revtex4}
\usepackage{amsmath}
\newcommand\numberthis{\addtocounter{equation}{1}\tag{\theequation}}
\begin{document}
\begin{center}
$\let\left\relax\let\right\relax
\def\frac#1#2{((#1)/(#2))}
\frac{2 (\alpha +3) (2 \alpha +5) (10 \alpha +17) (10 \alpha +19) (10 \alpha +21) (10
\alpha +23) (\alpha (\alpha (\alpha (20 \alpha (5 \alpha (740 \alpha
+5783)+91541)+3013197)+2724024)+1284280)+246960) \, _7F_6\left(1,\alpha
+\frac{4}{5},\alpha +\frac{5}{6},\alpha +\frac{7}{6},\alpha +\frac{6}{5},\alpha
+\frac{7}{5},\alpha +\frac{8}{5};\alpha +\frac{17}{10},\alpha +\frac{19}{10},\alpha
+\frac{21}{10},\alpha +\frac{23}{10},\alpha +\frac{5}{2},\alpha
+3;\frac{27}{64}\right)+16875 \left(\alpha +\frac{4}{5}\right) \left(\alpha
+\frac{5}{6}\right) \left(\alpha +\frac{7}{6}\right) \left(\alpha +\frac{6}{5}\right)
\left(\alpha +\frac{7}{5}\right) \left(\alpha +\frac{8}{5}\right) \left((\alpha
(\alpha (20 \alpha (25 \alpha (888 \alpha +5783)+366164)+9039591)+5448048)+1284280)
\, _7F_6\left(2,\alpha +\frac{9}{5},\alpha +\frac{11}{6},\alpha +\frac{13}{6},\alpha
+\frac{11}{5},\alpha +\frac{12}{5},\alpha +\frac{13}{5};\alpha +\frac{27}{10},\alpha
+\frac{29}{10},\alpha +\frac{31}{10},\alpha +\frac{33}{10},\alpha +\frac{7}{2},\alpha
+4;\frac{27}{64}\right)+(\alpha (40 \alpha (25 \alpha (1110 \alpha
+5783)+274623)+9039591)+2724024) \, _8F_7\left(2,2,\alpha +\frac{9}{5},\alpha
+\frac{11}{6},\alpha +\frac{13}{6},\alpha +\frac{11}{5},\alpha +\frac{12}{5},\alpha
+\frac{13}{5};1,\alpha +\frac{27}{10},\alpha +\frac{29}{10},\alpha
+\frac{31}{10},\alpha +\frac{33}{10},\alpha +\frac{7}{2},\alpha
+4;\frac{27}{64}\right)+(40 \alpha (25 \alpha (1480 \alpha +5783)+183082)+3013197)
\, _9F_8\left(2,2,2,\alpha +\frac{9}{5},\alpha +\frac{11}{6},\alpha
+\frac{13}{6},\alpha +\frac{11}{5},\alpha +\frac{12}{5},\alpha
+\frac{13}{5};1,1,\alpha +\frac{27}{10},\alpha +\frac{29}{10},\alpha
+\frac{31}{10},\alpha +\frac{33}{10},\alpha +\frac{7}{2},\alpha
+4;\frac{27}{64}\right)+20 (25 \alpha (2220 \alpha +5783)+91541) \,
_{10}F_9\left(2,2,2,2,\alpha +\frac{9}{5},\alpha +\frac{11}{6},\alpha
+\frac{13}{6},\alpha +\frac{11}{5},\alpha +\frac{12}{5},\alpha
+\frac{13}{5};1,1,1,\alpha +\frac{27}{10},\alpha +\frac{29}{10},\alpha
+\frac{31}{10},\alpha +\frac{33}{10},\alpha +\frac{7}{2},\alpha
+4;\frac{27}{64}\right)+100 (4440 \alpha +5783) \, _{11}F_{10}\left(2,2,2,2,2,\alpha
+\frac{9}{5},\alpha +\frac{11}{6},\alpha +\frac{13}{6},\alpha +\frac{11}{5},\alpha
+\frac{12}{5},\alpha +\frac{13}{5};1,1,1,1,\alpha +\frac{27}{10},\alpha
+\frac{29}{10},\alpha +\frac{31}{10},\alpha +\frac{33}{10},\alpha +\frac{7}{2},\alpha
+4;\frac{27}{64}\right)+74000 \, _{12}F_{11}\left(2,2,2,2,2,2,\alpha
+\frac{9}{5},\alpha +\frac{11}{6},\alpha +\frac{13}{6},\alpha +\frac{11}{5},\alpha
+\frac{12}{5},\alpha +\frac{13}{5};1,1,1,1,1,\alpha +\frac{27}{10},\alpha
+\frac{29}{10},\alpha +\frac{31}{10},\alpha +\frac{33}{10},\alpha +\frac{7}{2},\alpha
+4;\frac{27}{64}\right)\right)}{4288785520939920}
$
\end{center}
\end{document}