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