Re: Re: Incongruence? hmm...-a followup- correction

*To*: mathgroup at smc.vnet.net*Subject*: [mg102731] Re: [mg102713] Re: Incongruence? hmm...-a followup- correction*From*: Filippo Miatto <miatto at gmail.com>*Date*: Sat, 22 Aug 2009 03:37:50 -0400 (EDT)*References*: <34c814850908210636u4f9ce335ic47ffb739f8182da@mail.gmail.com>

Thank you very much, this was really helpful! I'll also have a look at the Matsubara technique, I think it will come handy in other occasions. Filippo 2009/8/21, Leonid Shifrin <lshifr at gmail.com>: > Sorry, in my step #4 (in the follow-up post) the rules for coefficients > should read > > 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] > > (the sign of c2 is negative - due to differentiating Cos twice), and then > the final result > also has a minus in front of x^2-term: > > 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] > > > 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 >> >> >> > >> > >> >> >