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)*References*: <20070208190342.879$d6@newsreader.com> <eqh8r2$iii$1@smc.vnet.net>

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

**Follow-Ups**:**Re: Re: Showing that a hypergeometric expression is 0?***From:*Andrzej Kozlowski <akoz@mimuw.edu.pl>