Services & Resources / Wolfram Forums / MathGroup Archive
-----

MathGroup Archive 2012

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: Bug in infinite sum

  • To: mathgroup at smc.vnet.net
  • Subject: [mg127403] Re: Bug in infinite sum
  • From: Peter Pein <petsie at dordos.net>
  • Date: Sun, 22 Jul 2012 04:32:46 -0400 (EDT)
  • Delivered-to: l-mathgroup@mail-archive0.wolfram.com
  • Delivered-to: mathgroup-newout@smc.vnet.net
  • Delivered-to: mathgroup-newsend@smc.vnet.net
  • References: <jud8jf$j2m$1@smc.vnet.net>

Am 21.07.2012 05:44, schrieb Dr. Wolfgang Hintze:
> Consider this nice sum
>
> s[x_] = Sum[Binomial[x, k], {k, 0, Infinity}]
> 2^x
>
> and that, (with the odd terms only)
>
> u[x_] = Sum[Binomial[x, 2*k + 1], {k, 0, Infinity}]
> 2^(-1 + x)
>
> But now the even terms only ... and the surprise:
> g[x_] = Sum[Binomial[x, 2*k], {k, 0, Infinity}]
>
> (Sqrt[Pi]*Gamma[1 + x]*GegenbauerC[x, 1/2 - x, 1])/
>    (2^x*Gamma[1/2 + x])
>
> looks complicated, but let's see
> FullSimplify[g[x], x > 0]
>
> 2^x*Cos[Pi*x]
>
> much simpler, but definitely wrong (giving e.g. 0 for x=1/2)
> Of course g should be s - u = 2^x -1/2 2^s = 1/2 2^s.
>
> Best regards,
> Wolfgang
>

Hmm on version 8 (Windows 64-bit) I get:

  Sum[Binomial[x, 2*k], {k, 0, Infinity}]
    2^(-1+x)

and the same result for

  Sum[Binomial[x, k], {k, 0, Infinity, 2}]

as it should be.

Peter



  • Prev by Date: Re: How to Scale and vary plot for a Differential Equation
  • Next by Date: Re: How to Scale and vary plot for a Differential Equation
  • Previous by thread: Re: Bug in infinite sum
  • Next by thread: Re: Using Mathematica in VB.NET