Re: Re: Incongruence? hmm...

*To*: mathgroup at smc.vnet.net*Subject*: [mg102721] Re: [mg102713] Re: Incongruence? hmm...*From*: Leonid Shifrin <lshifr at gmail.com>*Date*: Sat, 22 Aug 2009 03:36:00 -0400 (EDT)*References*: <h6j33u$5j4$1@smc.vnet.net> <4A8D74DC.5050805@gmail.com>

Hi Filippo, >but still, since they alternate sign for n even or odd, couldn't it be >that they sort of 'cancel out' giving that x^3 term in the end? >if this is the case, how can i check it with mathematica? The term you have is not x^3, but Abs[x]^3 - there is a difference. This sum is not analytic in the compex plane (taking x complex), therefore you can not compute it in any way by expanding around x = 0 - full stop here. If you really want to compute this sum all by yourself analytically, read about Matsubara technique (used in statistical physics and quantum field theory) - it uses residues and methods of complex variables to compute this sort of sums - and apply it to this sum - this is a nice exercise, should not be too hard. The simplest example is Sum[1/(n^2+a^2),{n,-Infinity,Infinity}] - computation of this should be easy to find online (related to harmonic oscillator at finite temperature, for instance), and it will get you started. Regards, Leonid On Fri, Aug 21, 2009 at 12:42 PM, Filippo Miatto <miatto at gmail.com> wrote: > Thank you all for contributing, I didn't see that the sums do not > converge for n>2, what a mistake... > but still, since they alternate sign for n even or odd, couldn't it be > that they sort of 'cancel out' giving that x^3 term in the end? > if this is the case, how can i check it with mathematica? > Filippo > > > > 2009/8/20, Szabolcs Horv=E1t <szhorvat at gmail.com>: > > On 2009.08.20. 10:56, Filippo Miatto 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)! > > > > Hello Filippo, > > > > I believe the result above to be valid only for n=0 and n=1. For oth= > er > > values of n the series will not be covergent. > > > >> > >> 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 > >> > > > > > >