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From: Vladimir Bondarenko on 23 Jul 2010 22:49
Hello,

Mathematica:

(5 Hypergeometric2F1[-1/2, 1, 1/4, -1] -
4 Hypergeometric2F1[ 1/2, 1, 1/4, -1])/
(1 + 2^(7/4) EllipticF[ArcCsc[2^(1/4)], -1])

Maple:

(5*hypergeom([-1/2, 1], [1/4], -1)-
4*hypergeom([ 1/2, 1], [1/4], -1))/
(1+2*2^(3/4)*EllipticF(1/2^(1/4),I))

?

Cheers,

Vladimir Bondarenko

Co-founder, CEO, Mathematical Director

http://www.cybertester.com/ Cyber Tester Ltd.

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"...... the challenges imply that a solution is built within the
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From: Axel Vogt on 24 Jul 2010 16:53
Vladimir Bondarenko wrote:
> Hello,
>
> Mathematica:
>
> (5 Hypergeometric2F1[-1/2, 1, 1/4, -1] -
> 4 Hypergeometric2F1[ 1/2, 1, 1/4, -1])/
> (1 + 2^(7/4) EllipticF[ArcCsc[2^(1/4)], -1])
>
> Maple:
>
> (5*hypergeom([-1/2, 1], [1/4], -1)-
> 4*hypergeom([ 1/2, 1], [1/4], -1))/
> (1+2*2^(3/4)*EllipticF(1/2^(1/4),I))

Numerical it is 3 (as a trivial observation).

It would be more interesting to look at transformations
for hypergeometrics like the reformulation

5*hypergeom([-1/2, 1],[1/4],-x)-
4*hypergeom([1/2, 1],[1/4],-x)-
3*4^(7/8)*hypergeom([1/4, 3/4],[5/4],-x) = 3 for x=1

Not that I have that. But googling for Vidunas, arxiv
should give plenty of identities not being covered by
MMA or Maple.
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