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

*To*: mathgroup at smc.vnet.net*Subject*: [mg102657] Re: [mg102734] Re: [mg102717] Re: [mg102710] Incongruence? hmm...*From*: Filippo Miatto <miatto at gmail.com>*Date*: Sun, 23 Aug 2009 05:33:43 -0400 (EDT)*References*: <200908200856.EAA05738@smc.vnet.net>

I see. So, since the solution that Mathematica finds is valid between 0 and 2Pi, what if I take the result (8 \[Pi]^4 - 60 \[Pi]^2 x^2 + 60 \[Pi]x^3 - 15 x^4)/720 and substitute Mod[x,2Pi] in place of all x? That would make it valid for every x. I don't get why Mathematica doesn't find the right answer for x in [-2pi,2pi], though.. Either using Abs[x] for the 3rd power term, or at least using the right sign in the interval [-2pi,0]. is it a bug? 2009/8/22, DrMajorBob <btreat1 at austin.rr.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 >

**Follow-Ups**:**Re: Incongruence? hmm...***From:*"Tony Harker" <a.harker@ucl.ac.uk>

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

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

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