Re: Re: Re: Incongruence? hmm...

*To*: mathgroup at smc.vnet.net*Subject*: [mg102656] Re: [mg102734] Re: [mg102717] Re: [mg102710] Incongruence? hmm...*From*: DrMajorBob <btreat1 at austin.rr.com>*Date*: Sun, 23 Aug 2009 05:33:32 -0400 (EDT)*References*: <200908200856.EAA05738@smc.vnet.net>*Reply-to*: drmajorbob at bigfoot.com

Yes, Cos is an even function, so the two results should agree at 3 and -3, for instance: one = Sum[Cos[m x]/m^4, {m, 1, \[Infinity]}, Assumptions -> {0 < x < 2 Pi}] one3 = one /. x -> 3; 1/720 (8 \[Pi]^4 - 60 \[Pi]^2 x^2 + 60 \[Pi] x^3 - 15 x^4) two = Sum[Cos[m x]/m^4, {m, 1, \[Infinity]}, Assumptions -> {-2 Pi < x < 0}] two3 = two /. x -> -3; 1/720 (8 \[Pi]^4 - 60 \[Pi]^2 x^2 + 60 \[Pi] x^3 - 15 x^4) two3 - one3 // Expand -((9 \[Pi])/2) But they're not the same, so one of those results is wrong. As Tony explained, the sum MUST be between + and - Pi^4/40 for ALL x, and the second result fails that test: limit = Pi^4/90. 1.08232 two /. x -> -Pi // N -17.1819 Plot[{limit, -limit, two}, {x, -Pi, 0}, PlotRange -> All] An expression that's correct (or even and properly bounded, at least) over both intervals is three = 1/ 720 (8 \[Pi]^4 - 60 \[Pi]^2 x^2 + 60 \[Pi] Abs[x]^3 - 15 x^4); Plot[{limit, -limit, three}, {x, -2 Pi, 2 Pi}] This expression agrees with the first result on 0 to 2Pi, as well. Plot[{one, three}, {x, 0, 2 Pi}, PlotRange -> All] Bobby On Sat, 22 Aug 2009 06:41:33 -0500, Filippo Miatto <miatto at gmail.com> wrote: > 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 >> >> -- DrMajorBob at bigfoot.com

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