       How to simplify hypergeometrics

• To: mathgroup at smc.vnet.net
• Subject: [mg128379] How to simplify hypergeometrics
• From: "Dr. Wolfgang Hintze" <weh at snafu.de>
• Date: Thu, 11 Oct 2012 02:08:47 -0400 (EDT)
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• Delivered-to: l-mathgroup@wolfram.com
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```Consider the probability function

p[n_, a_, b_, k_] :=
Binomial[k - 1, a - 1]*(Binomial[n - k, b - a]/Binomial[n, b]) /; {0
<= a <=
b, n >= a}

In:= p[n, a, b, k]

Out= (Binomial[-1 + k, -1 + a]*Binomial[-k + n, -a + b])/
Binomial[n, b]

Now let's look for the zeroeth moment k^0 (for the higher ones the
situation is similar)

In:= k0 = Sum[p[n, a, b, k], {k, 1, n}]

Out= ((-Binomial[-1, -a + b])*Binomial[n, -1 + a]*
Hypergeometric2F1[1 - a + b, 1 + n, 2 - a + n, 1] +
Binomial[0, -1 + a]*Binomial[-1 + n, -a + b]*
HypergeometricPFQ[{1, 1, 1 - a + b - n}, {2 - a, 1 - n}, 1])/
Binomial[n, b]

This should give 1, but it looks clumsy.

Simplifying simply gives

In:= FullSimplify[k0]

Out= ComplexInfinity

but also a qualified Simplify does not help, because in this case
Mathematica makes slight cosmetic changes but the result is still far
from being recognized as 1:

In:= FullSimplify[k0, {Element[{a, b, n}, Integers], 0 <= a <= b,
n >= a}]

Out= ((-(-1)^(a + b))*Binomial[n, -1 + a]*
Hypergeometric2F1[1 - a + b, 1 + n, 2 - a + n, 1] +
Binomial[0, -1 + a]*Binomial[-1 + n, -a + b]*
HypergeometricPFQ[{1, 1, 1 - a + b - n}, {2 - a, 1 - n}, 1])/
Binomial[n, b]

Any help would be greatly appreciated.

Best regards,
Wolfgang

```

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