Re: Re: Incongruence? hmm...

*To*: mathgroup at smc.vnet.net*Subject*: [mg102734] Re: [mg102717] Re: [mg102710] Incongruence? hmm...*From*: DrMajorBob <btreat1 at austin.rr.com>*Date*: Sat, 22 Aug 2009 03:38:23 -0400 (EDT)*References*: <200908200856.EAA05738@smc.vnet.net>*Reply-to*: drmajorbob at bigfoot.com

So, of these, Mathematica gets the second one wrong: Sum[Cos[m x]/m^4, {m, 1, \[Infinity]}, Assumptions -> {0 < x < 2 Pi}] 1/720 (8 \[Pi]^4 - 60 \[Pi]^2 x^2 + 60 \[Pi] x^3 - 15 x^4) Sum[Cos[m x]/m^4, {m, 1, \[Infinity]}, Assumptions -> {-Pi < x < 0}] 1/720 (8 \[Pi]^4 - 60 \[Pi]^2 x^2 + 60 \[Pi] x^3 - 15 x^4) Bobby On Fri, 21 Aug 2009 03:43:01 -0500, Tony Harker <a.harker at ucl.ac.uk> wrote: > > I don't see how either expression can be correct for all x. The 'direct' > form is clearly wrong, as every term in the summation is an even > function of > x. However, the modulus of each term is less than or equal to 1/m^4, so > the > sum must be bounded above and below by plus and minus Pi^4/90, which > neither > result is. > > The indirect approach obviously has a limited radius of convergence for > the sum over m, and GenerateConditions->True will show that, so that's > what > goes wrong there. The direct approach does not generate any conditions, > so > seems to be just plain wrong. > > The key to sorting this out is to note that the original expression is > a > Fourier series, so any polynomial form can only be valid over an interval > such as -Pi to Pi, and must then repeat periodically. > > In fact, the result from the direct sum is correct for 0<x<Pi, and for > -Pi<x<0 the expression is similar, but the sign of the coefficient of > x^3 is > changed. Basically, the original sum is a polynomial in (Pi-x) which has > been made symmetrical about x=0. > > Tony Harker > > > ]-> To: mathgroup at smc.vnet.net > ]-> Subject: [mg102710] Incongruence? hmm... > ]-> > ]-> 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>