Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Notice: We are no longer accepting new posts, but the forums will continue to be readable.

Topic: How to simplify hypergeometrics
Replies: 4   Last Post: Oct 16, 2012 10:50 PM

 Messages: [ Previous | Next ]
 Dr. Wolfgang Hintze Posts: 195 Registered: 12/8/04
How to simplify hypergeometrics
Posted: Oct 11, 2012 2:14 AM

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

Date Subject Author
10/11/12 Dr. Wolfgang Hintze
10/12/12 Roland Franzius
10/13/12 Dr. Wolfgang Hintze
10/14/12 Roland Franzius
10/16/12 Dr. Wolfgang Hintze