Re: question

*To*: mathgroup at smc.vnet.net*Subject*: [mg75661] Re: question*From*: Peter Pein <petsie at dordos.net>*Date*: Tue, 8 May 2007 05:48:45 -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. > > In[36]:= > Plot[Chop[x/Pi - (1/2)*x*(StruveH[-1, x] + StruveH[1, x])], {x, -2, > 2}, Axes -> False, Frame -> True] > > with Mathematica. > > Simplify, FullSimplify, FunctionExpand fail to do this task. Even > replacing x by specific value does > not yield better results. > > In[40]:= > FullSimplify[3/Pi - (1/2)*3*(StruveH[-1, 3] + StruveH[1, 3])] > > Out[40]= > 3/Pi - (3/2)*(StruveH[-1, 3] + StruveH[1, 3]) > > On the contrary, in another CAS I took > > simplify(3/Pi - (1/2)*3*(StruveH(-1, 3) + StruveH(1, 3))); > 0 > > 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 > > Hi Dimitris, I don't get it in the usual way (showing, that the derivative is identical to zero _and_ expression at x==0 is zero (the latter is easy)). My way is similar to your approach via integration: ee = x/Pi - (1/2)*x*(StruveH[-1, x] + StruveH[1, x]); InverseLaplaceTransform[LaplaceTransform[ee, x, s], s, x] --> 0 and calculations like Series[ee, {x, 0, 50}] --> O[x]^51 are no proof. Playing around with SeriesTerm[] leads to a funny result: e2 = Sum[SeriesTerm[ee, {x, 0, k}]x^(k)*k!^(3/2), {k, 0, â??}] gives two warnings "Sum does not converge" and the "result" x/Pi. But really astonishing is: e3 = Sum[SeriesTerm[ee, {x, 0, k}]*(x^k/k), {k, 0, Infinity}] --> x/Pi - (1/(9*Pi))*(x^3*HypergeometricPFQ[{1}, {5/2, 5/2}, -(x^2/4)]) -(1/2)*StruveH[0, x] ZeroQ[e3] --> False *but:* FullSimplify[e3] --> 0 I understood ZeroQ as if it tries transformations, which FullSimplify uses too?? But using SeriesTerm, we get a proof: First FullSimplify the even terms, then the odd terms with k>=3 and at last the coefficient with k==1: ak = SeriesTerm[ee, {x, 0, k}]; FullSimplify[ak /. k -> 2*j, j \[Element] Integers && j >= 0] --> 0 FullSimplify[ak /. k -> 2*j + 1, j \[Element] Integers && j > 0] --> 0 and finally FullSimplify[ak /. k -> 1] --> 0 Cheers, Peter