Re: Re: Incongruence? hmm... - a follow-up

*To*: mathgroup at smc.vnet.net*Subject*: [mg102729] Re: [mg102713] Re: Incongruence? hmm... - a follow-up*From*: Leonid Shifrin <lshifr at gmail.com>*Date*: Sat, 22 Aug 2009 03:37:28 -0400 (EDT)

Hi Filippo, There is a relatively simple way to compute your sum: 1. Diffrerentiate it 4 times with respect to x, to get Sum[Cos[m*x],{m,1,Infinity}] 2. Use identity Sum[Cos[m*x],{m,1,Infinity}] = 1/2 (2 Pi DiracDelta[x] - 1) 3. Integrate it over x 4 times, as follows: In[1] = Clear[result,c1,c2,c3,c4]; result = Fold[Integrate[#1, x] + #2 &, 1/2 (2 Pi DiracDelta[x] - 1), {c1, c2, c3, c4}] Out[1] = c4+c3 x+(c2 x^2)/2+(c1 x^3)/6-x^4/48+1/6 \[Pi] x^3 HeavisideTheta[x] 4. Substitute constants (found by differentiation and then setting x to 0): In[2] = result = result/.{c1 -> 0, c2 -> Sum[1/m^2, {m, 1, Infinity}], c3 -> 0, c4 -> Sum[1/m^4, {m, 1, Infinity}]} Out[2] = \[Pi]^4/90 + (\[Pi]^2 x^2)/12 - x^4/48 + 1/6 \[Pi] x^3 HeavisideTheta[x] 5. Finally, symmetrize the cubic term correctly: In[3] = result = result/.HeavisideTheta[x] :> 1/2 Sign[x] Out[3] =\[Pi]^4/90 + (\[Pi]^2 x^2)/12 - x^4/48 + 1/12 \[Pi] x^3 Sign[x] This is the same result that I posted in my previous post, and I think this method is simpler than Matsubara summation technique I suggested earlier. 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 > >> > > > > > >