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[68]:= p[n, a, b, k]
Out[68]= (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[69]:= k0 = Sum[p[n, a, b, k], {k, 1, n}]
Out[69]= ((-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[71]:= FullSimplify[k0]
Out[71]= 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[73]:= FullSimplify[k0, {Element[{a, b, n}, Integers], 0 <= a <= b,
n >= a}]
Out[73]= ((-(-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