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MathGroup Archive 2007

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Re: question

  • To: mathgroup at smc.vnet.net
  • Subject: [mg75664] Re: question
  • From: Peter Pein <petsie at dordos.net>
  • Date: Tue, 8 May 2007 05:50:25 -0400 (EDT)
  • References: <f1mrii$rb4$1@smc.vnet.net>

dimitris schrieb:
> Hi.
> 
> I want to show that
> 
> x/Pi - (1/2)*x*(StruveH[-1, x] + StruveH[1, x])
> 
> is equal to zero.
...

> Any ideas?
> 
> A quick way to show that x/Pi - (1/2)*x*(StruveH[-1, x] + StruveH[1,
> x]) is indeed
> zero is
> 
> In[41]:=
> Integrate[x/Pi - (1/2)*x*(StruveH[-1, x] + StruveH[1, x]), x]
> 
> Out[41]=
> 0
> 
> but I would really appreciate any other suggestions.
> 
> Thanks
> 
> 

Hello again,

there is another way using Prof. Wolfram Koepf's Package Specialfunctions.m.
You can find it as supplement to his book Computeralgebra (in german;
http://www.mathematik.uni-kassel.de/~koepf/CA/index.html) and the zip-archive
for Mathematica
(http://www.mathematik.uni-kassel.de/~koepf/CA/Sitzungen/Mathematica.zip)
contains SpecialFunctions.m. It fails with Mathematica 5.1 to find the
powerseries directly:
<< SpecialFunctions`
ee = x/Pi - (1/2)*x*(StruveH[-1, x] + StruveH[1, x]);
PS[ee, x, 0]
--> SpecialFunctions`Private`df[5]
(don't ask me what this means)

but trying to find and solve a recurrence for the coefficients a[k] gives:

RSolve[Simplify[{SimpleRE[ee, x, a, k], a[0] == ee /. x -> 0,
  a[1] == Limit[D[ee, x], x -> 0]}], a, k]
--> {{a -> Function[{k}, 0]}}

P²


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