Re: Showing that a hypergeometric expression is 0?

*To*: mathgroup at smc.vnet.net*Subject*: [mg73282] Re: Showing that a hypergeometric expression is 0?*From*: Paul Abbott <paul at physics.uwa.edu.au>*Date*: Fri, 9 Feb 2007 02:28:51 -0500 (EST)*References*: <20070208190342.879$d6@newsreader.com>

>[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. Cheers, Paul