Re: Re: Showing that a hypergeometric expression is 0?

*To*: mathgroup at smc.vnet.net*Subject*: [mg73480] Re: [mg73435] Re: Showing that a hypergeometric expression is 0?*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>*Date*: Mon, 19 Feb 2007 01:29:15 -0500 (EST)*References*: <20070208190342.879$d6@newsreader.com> <eqh8r2$iii$1@smc.vnet.net> <200702160603.BAA15760@smc.vnet.net>

On 16 Feb 2007, at 07:03, David W.Cantrell wrote: > 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 > Here is an informal argument which, while of course not a proof, indicates that the result ought to be true and perhaps a formal proof = could be constructed along these lines. We define f[x_] := n Hypergeometric2F1Regularized[k, 1 - n, 1 + k - n, x] + Hypergeometric2F1Regularized[k, -n, k - n, x] + k n Hypergeometric2F1Regularized[1 + k, 1 - n, 2 + k - n, x] - k Hypergeometric2F1Regularized[1 + k, -n, 1 + k - n, x] and simply compute Table[FullSimplify[Limit[Normal[f[x] + O[x]^p], x -> -1], (n | k) =E2=88=88 Integers], {p, 1, 5}] {(k*n)/Gamma[k - n + 2], (k*(k + 1)*(n - 1)*n)/ Gamma[k - n + 3], (k*(k + 1)*(k + 2)*(n - 2)*(n - 1)* n)/(2*Gamma[k - n + 4]), (k*(k + 1)*(k + 2)*(k + 3)*(n - 3)*(n - 2)*(n - 1)*n)/ (6*Gamma[k - n + 5]), (k*(k + 1)*(k + 2)*(k + 3)* (k + 4)*(n - 4)*(n - 3)*(n - 2)*(n - 1)*n)/ (24*Gamma[k - n + 6])} The answers strongly suggest that if p>n the Taylor series truncated at p is actually 0 and hence so is the full Taylor series. (This is because the numerators contain products of the form (n-p+1) ... (n - 4)*(n - 3)*(n - 2)*(n - 1). While this is, of course, not a proof it suggest a possible proof, by induction on p. Andrzej Kozlowski

**References**:**Re: Showing that a hypergeometric expression is 0?***From:*"David W.Cantrell" <DWCantrell@sigmaxi.net>