Re: Showing that a hypergeometric expression is 0?

• To: mathgroup at smc.vnet.net
• Subject: [mg73435] Re: Showing that a hypergeometric expression is 0?
• From: "David W.Cantrell" <DWCantrell at sigmaxi.net>
• Date: Fri, 16 Feb 2007 01:03:36 -0500 (EST)

```Paul Abbott <paul at physics.uwa.edu.au> wrote:
> >[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.

Many, many thanks for your response, Paul!

BTW, if anyone has other ways of showing that the expression is 0, I'd love
to see them.

David Cantrell

```

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