Re: Re: Re: Incongruence? hmm...
- To: mathgroup at smc.vnet.net
- Subject: [mg102651] Re: [mg102734] Re: [mg102717] Re: [mg102710] Incongruence? hmm...
- From: Filippo Miatto <miatto at gmail.com>
- Date: Sun, 23 Aug 2009 05:32:37 -0400 (EDT)
- References: <200908200856.EAA05738@smc.vnet.net>
But the cosine is an even function, why should the result not be even in x? I mean, the assumption 0<x<2pi should be equivalent to the assumption -2pi<x<0 since every term of the sum does not change value. F 2009/8/22, DrMajorBob <btreat1 at austin.rr.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>
- Incongruence? hmm...