FW: FW: Hypergeometric functions

["Dan Lozier" <lozier@nist.gov>]

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From: opsftalk@nist.gov [mailto:opsftalk@nist.gov]On Behalf
Of Christian
Krattenthaler
Sent: Tuesday, October 21, 2003 3:21 AM
To: dlozier@nist.gov
Subject: Re: FW: Hypergeometric functions


This message was submitted by Christian Krattenthaler
<kratt@igd.univ-lyon1.fr>
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Sender: Christian Krattenthaler <kratt@igd.univ-lyon1.fr>
Subject: Re: FW: Hypergeometric functions


>
> Eleutherius wrote:
>
> in my research work on the Poisson integral for a ball in
> non-euclidean spaces I have obtained the following
> identities for the
> hypergeometric function:
>
> ... and
> $$
> \frac{1}{2}+\sum_{k=1}^\infty \frac{{2k-1 \choose
> k-1}{{\frac{n}{2}
> +k-2}
> \choose k-1}}{{\frac{n}{2}+2k-2 \choose k-1}}F\left(k,
>
\frac{n-1}{2}+k;\frac{n}{2}+2k;x\right)\left(-\frac{x}{4}\ri
> ght)^k
> \equiv\frac{\sqrt{1-x}}{2}\,,
> $$

This identity (the second) is easy to verify:
Take the coefficient of z^n on the left-hand side.
It is a single sum, and in hypergeometric notation it is

                    n  3   n         3  n
              [ 1 + -, - + -, 1 - n, -, -    ]
              |     2  2   4         2  2    |  1   n
         -( F |                          ; 1 | (-
     )
           5 4| 1   n      3 n  1   n        |  2   2 -1 + n
              [ - + -, 1 + ---, - + -, 2     ]
                2   4       2   2   2
         ---------------------------------------------------
-
                                  n
                           4 (2 + -)
                                  2 -1 + n

This 5F4-series is very-well-poised and can therefore be
summed by
means of (Slater, Appendix (III.12))

                             a
            [         a, 1 + -, b, c, d             ]
            |                2                      |
          F |                                   ; 1 | ==
         5 4| a                                     |
            [ -, 1 + a - b, 1 + a - c, 1 + a - d    ]
              2

     Ga(1 + a - b) Ga(1 + a - c) Ga(1 + a - d) Ga(1 + a -
b - c - d)
>    -------------------------------------------------------
--------
     Ga(1 + a) Ga(1 + a - b - c) Ga(1 + a - b - d) Ga(1 +
a - c - d)


   With best wishes,

     Christian Krattenthaler