Re: Incongruence? hmm...

*To*: mathgroup at smc.vnet.net*Subject*: [mg102711] Re: [mg102710] Incongruence? hmm...*From*: Leonid Shifrin <lshifr at gmail.com>*Date*: Fri, 21 Aug 2009 04:41:53 -0400 (EDT)*References*: <200908200856.EAA05738@smc.vnet.net>

Filippo, First, the correct answer: Mathematica is right, given that 0<=x<=2Pi. Take what Mathematica gives if you only need positive x, or use a Sign[x] function in front of cubic term, if you also need the interval -2Pi<=x<=0: 1/720 (8 \[Pi]^4 - 60 \[Pi]^2 x^2 + Sign[x]*60 \[Pi] x^3 - 15 x^4) You should not have expanded the Cos, since then your sum over m Sum[((m x)^(2n) (-1)^n)/((2n)!m^4),{m,1,Infinity}] is not convergent for n>2. I am actually surprised that Mathematica did not issue a warning when performing this sum. But in fact, it gave you a hint that the behavior changes qualitatively at n=2 - the argument of Zeta changes sign. So, the values of Zeta are indeed zero for n>2, but for n>2 this is no longer a correct answer - this is where the cubic term is coming from. The fact that there is a Sign function in front means that there is no way you could have obtained this term through expansion in x - it is "non-perturbative". It is another way of saying that you should not have expanded. One other way to obtain a correct answer valid also for negative x (-2Pi<=x<=0) is to ask Mathematica to compute Sum[Sin[m x]/m^5, {m, 1, \[Infinity]}], then differentiate over x and simplify the result under an assumption that x is real. Eventually, you will get the second result I mentioned. Needless to say, the result is only correct for -2Pi<=x<=2Pi. Regards, Leonid On Thu, Aug 20, 2009 at 12:56 PM, Filippo Miatto <miatto at gmail.com> wrote: > Dear all, > I'm calculating the sum > > Sum[Cos[m x]/m^4, {m, 1, \[Infinity]}] > > in two different ways that do not coincide in result. > If i expand the cosine in power series > > ((m x)^(2n) (-1)^n)/((2n)!m^4) > > and sum first on m i obtain > > ((-1)^n x^(2n) Zeta[4-2n])/(2n)! > > then I have to sum this result on n from 0 to infinity, but Zeta[4-2n] > is different from 0 only for n=0,1,2 and the result is > > \[Pi]^4/90 - (\[Pi]^2 x^2)/12 - x^4/48 > > Three terms, one independent on x, with x^2, one with x^4. > > however if I perform the sum straightforwardly (specifying that > 0<x<2pi) the result that Mathematica gives me is > > \[Pi]^4/90 - (\[Pi]^2 x^2)/12 + (\[Pi] x^3)/12 - x^4/48 > > with the extra term (\[Pi] x^3)/12. Any idea on where it comes from?? > Thank you in advance, > Filippo > >

**References**:**Incongruence? hmm...***From:*Filippo Miatto <miatto@gmail.com>