Re: Hypergeometric1F1 polynomial

*To*: mathgroup at smc.vnet.net*Subject*: [mg91451] Re: Hypergeometric1F1 polynomial*From*: m.r at inbox.ru*Date*: Fri, 22 Aug 2008 03:12:25 -0400 (EDT)*References*: <g8je5u$a4n$1@smc.vnet.net>

Alec Mihailovs wrote: > Mathematica gives the wrong answer to the following sum, > > In[1]:= Sum[Binomial[n, k]/Binomial[2 n, k]/k! (2 x)^k, {k, 0, n}] > > Out[1]= 2^(-(1/2) - n) E^x x^(1/2 + n) > BesselI[1/2 (-1 - 2 n), x] Gamma[1/2 - n] > > The correct answer is 1 for n=0 and Hypergeometric1F1[-n, -2 n, 2 x] for > integer n>0, which would be equal to the expression given by Mathematica if > n was not a positive integer. > > Another form of the correct answer is > > (2 x)^(n+1/2) E^x BesselK[n+1/2,x] n!/(2 n)!/Sqrt[Pi] > > Is there a way to apply some assumptions to get the correct answer? > > Alec Here's one way to obtain the correct answer. The intermediate steps are only formally correct but in the end the singularities cancel out: In[1]:= Sum[ Binomial[n, k] (2 x)^k/(Binomial[2 n, k] k!) // FunctionExpand // # /. Gamma[a_ - k] :> (-1)^k Pi Csc[a Pi]/Gamma[1 - a + k]&, {k, 0, n}] // FullSimplify // Simplify[#, Element[n, Integers]]& Out[1]= 1/Gamma[1/2 + n] 2^-n ((-2)^n E^x Pi Hypergeometric0F1Regularized[1/2 - n, x^2/4] + Sqrt[Pi] (-x)^(1 + n) HypergeometricPFQRegularized[{1, 1}, {1 - n, 2 + n}, 2 x]) Maxim Rytin m.r at inbox.ru