Re: difficulty in calculating first two terms of an

*To*: mathgroup at smc.vnet.net*Subject*: [mg110137] Re: difficulty in calculating first two terms of an*From*: Bob Hanlon <hanlonr at cox.net>*Date*: Thu, 3 Jun 2010 05:40:38 -0400 (EDT)

Clear[p]; p[x_, n_] = (-1 + x)^(n + 1)* Sum[Binomial[n + 2*k, 2*k]*x^k, {k, 0, Infinity}] // FullSimplify ((1/2)*(-(1 - Sqrt[x])^(n + 1) - (Sqrt[x] + 1)^(n + 1))* (x - 1)^n)/(1 - x)^n Simplify[(1 - x)^-n (x - 1)^n == (-1)^n, Element[n, Integers]] True p[x_, n_Integer] = ((1/2)*(-(1 - Sqrt[x])^(n + 1) - (Sqrt[x] + 1)^(n + 1)))* (-1)^n; Bob Hanlon ---- Roger Bagula <roger.bagula at gmail.com> wrote: ============= The zero and one terms for the binomial version exist, but this Narayana number analog has difficulty in Mathematica on the n=1 term : Clear[p, x] p[x_, n_] = (1 - x)^(2*n - 1)*Sum[(Binomial[2*k + n, 2*k] Binomial[2*k + n, 1 + 2*k]/(2*k + n))*x^k, {k, 0, Infinity}] Table[FullSimplify[ExpandAll[p[x, n]]], {n, 0, 10}] Table[CoefficientList[FullSimplify[ExpandAll[p[x, n]]], x], {n, 0, 10}] It is probably, like the n equal zero term, one. Try : Clear[p, x] p[x_, n_] = (-1 + x)^( n + 1)*Sum[Binomial[n + 2*k, 2*k]*x^k, {k, 0, Infinity}] Table[FullSimplify[ExpandAll[p[x, n]]], {n, 0, 10}] Table[CoefficientList[FullSimplify[ExpandAll[p[x, n]]], x], {n, 0, 10}] ( see if you can find the really simple Sqrt[x] form for this!) I wouldn't have any chance doing these without Mathematica. It also fails several ways on the Eulerian numbers analogs of these infinite sums which are interesting in what they might look like. Respectfully, Roger L. Bagula 11759 Waterhill Road, Lakeside,Ca 92040-2905,tel: 619-5610814 : http://www.google.com/profiles/Roger.Bagula alternative email: roger.bagula at gmail.com