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
your solution.
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)"
>
> (* your expression *)
> 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