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As a follow-up to my question of two days ago How to render readable a large TeXForm expression from Mathematica

I'd like to also similarly incorporate the following two expressions:

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}

and

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]
  • 1
    why doesn't the solution provided by David Carlisle in the linked question work here? What have you tried to make it work? – Runar Jun 24 '16 at 15:54
  • @runartrollet what do you mean, "egreg" :-) – David Carlisle Jun 24 '16 at 15:57
  • @DavidCarlisle I am so sorry, must have been sitting here too long :P – Runar Jun 24 '16 at 15:59
  • The original David Carlisle commands--though they worked in the previous case--don't succeed here, apparently, with just a first line appearing and nothing more. So, almost all the output is not shown. So, I guess the original command was "custom-made" for that particular output. – Paul B. Slater Jun 24 '16 at 16:14
  • @user40881 not really, you just need to look at the expression and define linear forms of terms, that time it was just left.right that was stopping lien breaking (and were not needed anyway) here the entire term is a fraction so write fractions as a/b so they can break, you similarly need to add a suitable mathematically equivalent form for any other constructs. – David Carlisle Jun 24 '16 at 16:50
2

Really it is the same question:

\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}
  • OK, thanks--mea culpa, apparently. The Carlisle suggestion does work--though not highly esthetic, but guess I should be satisfied with getting the entire expression presented. Actually, to deal with this type of (large, cumbersome) output, I have been creating pdf's of the Mathematica output and including it in my documents as Figures. (For example, Fig. 3 in arxiv.org/pdf/1301.6617.pdf, and Fig. 3 in arxiv.org/pdf/1504.04555.pdf.) A little unconventional--and maybe referees are not too happy--but it accomplishes its purpose. – Paul B. Slater Jun 24 '16 at 17:01
  • Thanks again! So, is there a good source of documentation that helps to explain these issues? (maybe not) It's a little hard "coming in from the cold" to all these what seem highly technical points. – Paul B. Slater Jun 24 '16 at 19:16
  • 1
    @user40881 it's not really very technical, I just ran your document looked at the output and it was clearly a massive fraction and a displayed fraction can't break so I made them all inline. – David Carlisle Jun 24 '16 at 19:21

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