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
> >>
> >
> >
>
>