Re: HypergeometricPFQ - simplification

• To: mathgroup at smc.vnet.net
• Subject: [mg90808] Re: HypergeometricPFQ - simplification
• From: janos <janostothmeister at gmail.com>
• Date: Fri, 25 Jul 2008 06:12:39 -0400 (EDT)
• References: <g1gqir\$1fn\$1@smc.vnet.net> <g27r4b\$d6g\$1@smc.vnet.net>

```Thanks, Oleksander,

I could only now return to the problem, but I appreciate very much
However, I keep  saying,  that a symbol manipulation system
is the least appropriate to manipulate symbols :)
Or, these kinds of problems are much more complicated than
to solve a system of nonlinear PDEs.

Best regards,

Janos

On j=FAn. 5, 06:47, sashap <pav... at gmail.com> wrote:
> On May 27, 2:16 pm, janos <janostothmeis... at gmail.com> wrote:
>
>
>
> > Could anyone show using Mathematica that the expression
>
> > -2/27 + (253*HypergeometricPFQ[{-1/2, -1/6, 1/6}, {4/3, 5/3}, -1/729])/
> > 108 - (911*HypergeometricPFQ[{1/2, 5/6, 7/6}, {7/3, 8/3}, -1/729])/
> > 6298560 +
> >  (73*HypergeometricPFQ[{3/2, 11/6, 13/6}, {10/3, 11/3}, -1/729])/
> > 9795520512
>
> > is equal to
>
> > (-2*(-1 + 10*Sqrt[10]))/27? (Using, say, TraceInternal, to see what is
> > going on.)
>
> > Thank you,
>
> > Janos
>
> Dear Janos,
>
> A way to prove this in Mathematica  takes couple of lines of code:
>
> In[1]:= \$Version
>
> Out[1]= "6.0 for Microsoft Windows (32-bit) (March 13, 2008)"
>
> In[2]:= expr = -(2/27) +
>    253/108 HypergeometricPFQ[{-(1/2), -(1/6), 1/6}, {4/3, 5/3}, -(1/
>       729)] - (
>    911 HypergeometricPFQ[{1/2, 5/6, 7/6}, {7/3, 8/3}, -(1/729)])/
>    6298560 + (
>    73 HypergeometricPFQ[{3/2, 11/6, 13/6}, {10/3, 11/3}, -(1/729)])/
>    9795520512;
>
> Now define
>
> In[3]:= e1 = ({D[
>       HypergeometricPFQ[{-(1/2), -(1/6), 1/6 + a} - 1, {4/3, 5/3} -
>         1, -z^3], z],
>      D[HypergeometricPFQ[{-(1/2), -(1/6), 1/6 + a} - 1, {4/3, 5/3} -
>         1, -z^3], z, z],
>      D[HypergeometricPFQ[{-(1/2), -(1/6), 1/6 + a} - 1, {4/3, 5/3} -
>         1, -z^3], z, z, z]} /. {a -> 0, z -> 1/9} // Expand)
>
> Out[3]= {(35/144)*
>   HypergeometricPFQ[{-(1/2), -(1/6), 1/6}, {4/3, 5/3}, -(1/729)],
>    (35/8)*
>    HypergeometricPFQ[{-(1/2), -(1/6), 1/6}, {4/3,
>      5/3}, -(1/729)] - (7*
>      HypergeometricPFQ[{1/2, 5/6, 7/6}, {7/3, 8/3}, -(1/729)])/
>        124416, (315/8)*
>    HypergeometricPFQ[{-(1/2), -(1/6), 1/6}, {4/3, 5/3}, -(1/729)] -
>      (7*HypergeometricPFQ[{1/2, 5/6, 7/6}, {7/3, 8/3}, -(1/729)])/
>    2304 +
>      (35*HypergeometricPFQ[{3/2, 11/6, 13/6}, {10/3,
>        11/3}, -(1/729)])/214990848}
>
> Now repeat the same with variable a set to 0 from the outset:
>
> In[4]:= e2 =
>  FullSimplify[
>   With[{a = 0}, {D[
>       HypergeometricPFQ[{-(1/2), -(1/6), 1/6 + a} - 1, {4/3, 5/3} -
>         1, -z^3], z],
>      D[HypergeometricPFQ[{-(1/2), -(1/6), 1/6 + a} - 1, {4/3, 5/3} -
>         1, -z^3], z, z],
>      D[HypergeometricPFQ[{-(1/2), -(1/6), 1/6 + a} - 1, {4/3, 5/3} -
>         1, -z^3], z, z, z]}] /. z -> 1/9]
>
> Out[4]= {Sqrt[
>  11236809 + 778034 Sqrt[73] -
>   2000 Sqrt[12368090 + 7780340 Sqrt[73]]]/1458,
>  7/324 Sqrt[
>   21873 + 10658 Sqrt[73] - 200 Sqrt[-781270 + 106580 Sqrt[73]]],
>  35/36 Sqrt[73/2 (15 + Sqrt[73]) + 10 Sqrt[5 (595 + 73 Sqrt[73])]]}
>
> Now solve for hypergeometric functions:
>
> In[7]:= ru =
>   Thread[Union[Cases[e1, _HypergeometricPFQ, Infinity]] -> {h1, h2,
>      h3}];
>
> and substitute back into your original expression:
>
> In[8]:= FullSimplify[
>  expr /. (Solve[((e1 - e2) /. ru) == 0, {h1, h2, h3}] /. (Reverse /@
>       ru))]
>
> Out[8]= {2/27 (-1 + 10 Sqrt[10])}
>
> Hope this helps,
> Oleksandr Pavlyk
> Wolfram Research

```

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