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MathGroup Archive 2007

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Re: Showing that a hypergeometric expression is 0?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg73282] Re: Showing that a hypergeometric expression is 0?
  • From: Paul Abbott <paul at physics.uwa.edu.au>
  • Date: Fri, 9 Feb 2007 02:28:51 -0500 (EST)
  • References: <20070208190342.879$d6@newsreader.com>

>[Message also posted to: comp.soft-sys.math.mathematica]
>
>How can I use Mathematica to show that
>
>n Hypergeometric2F1Regularized[k, 1 - n, 1 + k - n, -1] +
>Hypergeometric2F1Regularized[k, -n, k - n, -1] +
>k n Hypergeometric2F1Regularized[1 + k, 1 - n, 2 + k - n, -1] -
>k Hypergeometric2F1Regularized[1 + k, -n, 1 + k - n, -1]
>
>is identically 0?
>
>I presume that I could do that by hand, perhaps without much trouble. But I
>also have far, far messier expressions involving hypergeometric functions
>which I need to show to be identically 0, and I really don't want to have
>to do so by hand!

One approach is to use FunctionExpand to convert 
Hypergeometric2F1Regularized ->
Hypergeometric2F1 and then apply

   http://functions.wolfram.com/07.24.17.0039.01

in the form of a generic rule:

   Hypergeometric2F1[b_, (m_.) - n, c_, -1] :>
    (n! Hypergeometric2F1[-n, 1 - b, 1, 2])/((-1)^n Pochhammer[1 - b, n]) /;
       c == b + m - n

If you FunctionExpand and then FullSimplify the result you obtain

  (Gamma[1 - k] Gamma[1 + n] (k (1 + k - n) Hypergeometric2F1[1 - k, -n, 1, 2] -
   (1 + k - 2 n) (k - n) Hypergeometric2F1[-k, -n, 1, 2]) Sin[(k - n) Pi])/
    ((-1)^n (k - n) (1 + k - n) Pi)

Simplification of this result for Element[{k,n}, Integers] yields 0. 
Observe that

  Simplify[Gamma[1 - k] Sin[(k - n) Pi], Element[{k,n}, Integers]]

  0

Another approach might be to show that (z+1) is a factor of

n Hypergeometric2F1Regularized[k, 1 - n, 1 + k - n, z] +
Hypergeometric2F1Regularized[k, -n, k - n, z] +
k n Hypergeometric2F1Regularized[1 + k, 1 - n, 2 + k - n, z] -
k Hypergeometric2F1Regularized[1 + k, -n, 1 + k - n, z]

for arbitrary z.

Cheers,
Paul


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