Re: Re: Incongruence? hmm...

*To*: mathgroup at smc.vnet.net*Subject*: [mg102732] Re: [mg102713] Re: Incongruence? hmm...*From*: DrMajorBob <btreat1 at austin.rr.com>*Date*: Sat, 22 Aug 2009 03:38:01 -0400 (EDT)*References*: <h6j33u$5j4$1@smc.vnet.net> <4A8D74DC.5050805@gmail.com>*Reply-to*: drmajorbob at bigfoot.com

You already checked, and it didn't work. Bobby On Fri, 21 Aug 2009 03:42:15 -0500, 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 >>> >> >> > -- DrMajorBob at bigfoot.com