Re: Incongruence? hmm...

*To*: mathgroup at smc.vnet.net*Subject*: [mg102712] Re: [mg102710] Incongruence? hmm...*From*: DrMajorBob <btreat1 at austin.rr.com>*Date*: Fri, 21 Aug 2009 04:42:04 -0400 (EDT)*References*: <200908200856.EAA05738@smc.vnet.net>*Reply-to*: drmajorbob at bigfoot.com

Your alternative calculation goes something like this: (* the term you defined: *) term[m_, n_] = ((m x)^(2 n) (-1)^n)/((2 n)! m^4) ((-1)^n (m x)^(2 n))/(m^4 (2 n)!) (* six terms added: *) Sum[term[m, n], {n, 0, 5}] 1/m^4 - x^2/(2 m^2) + x^4/24 - (m^2 x^6)/720 + (m^4 x^8)/40320 - ( m^6 x^10)/3628800 (* and that matches: *) Normal@Series[Cos[m x]/m^4, {x, 0, 10}] 1/m^4 - x^2/(2 m^2) + x^4/24 - (m^2 x^6)/720 + (m^4 x^8)/40320 - ( m^6 x^10)/3628800 So far, so good! (* summing first on m: *) term[n_] = Sum[term[m, n], {m, 1, Infinity}] ((-1)^n x^(2 n) Zeta[-2 (-2 + n)])/(2 n)! (* most terms are zero: *) term /@ Range[0, 10] {\[Pi]^4/90, -(1/12) \[Pi]^2 x^2, -(x^4/48), 0, 0, 0, 0, 0, 0, 0, 0} (* adding the non-zero ones: *) f3[x_] = Sum[term[n], {n, 0, 2}] \[Pi]^4/90 - (\[Pi]^2 x^2)/12 - x^4/48 (Same as you posted.) But the calculation for term[n] was spurious (a bug?), since term[m, 2] x^4/24 can't be summed over m: Sum[term[m, 2], {m, 1, Infinity}] Sum::div: Sum does not converge. >> \!\( \*UnderoverscriptBox[\(\[Sum]\), \(m = 1\), \(\[Infinity]\)] \*FractionBox[ SuperscriptBox[\(x\), \(4\)], \(24\)]\) Thus, term[2] is undefined, and we can't sum over all n. Bobby On Thu, 20 Aug 2009 03:56:35 -0500, Filippo Miatto <miatto at gmail.com> 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)! > > 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

**References**:**Incongruence? hmm...***From:*Filippo Miatto <miatto@gmail.com>