Re: Incongruence? hmm...
- To: mathgroup at smc.vnet.net
- Subject: [mg102735] Re: [mg102710] Incongruence? hmm...
- From: "Tony Harker" <a.harker at ucl.ac.uk>
- Date: Mon, 24 Aug 2009 07:55:15 -0400 (EDT)
- References: <200908200856.EAA05738@smc.vnet.net> <200908230933.FAA02257@smc.vnet.net>
Bug? Oversight? I don't know how to classify it. Actually, Mathematica makes a bit of a meal of sums of this sort: compare its results for Sum[Cos[m x]/m^2, {m, 1, Infinity}, GenerateConditions -> True] Sum[Sin[m x]/m^3, {m, 1, Infinity}, GenerateConditions -> True] Sum[Sin[m x]/m^5, {m, 1, Infinity}, GenerateConditions -> True] Sum[Cos[m x]/m^2, {m, 1, Infinity}] Series[%, {x, 0, 5}] Sum[Sin[m x]/m^3, {m, 1, Infinity}] Series[%, {x, 0, 5}] Sum[Sin[m x]/m^5, {m, 1, Infinity}] Series[%, {x, 0, 5}] with Gradshteyn and Ryzhik's results under item 1.443. I find it particularly strange that including GenerateConditions -> True seems to wrong-foot Mathematica on the first of these. Tony Harker ]-> -----Original Message----- ]-> From: Filippo Miatto [mailto:miatto at gmail.com] ]-> Sent: 23 August 2009 10:34 ]-> To: mathgroup at smc.vnet.net ]-> Subject: [mg102657] Re: [mg102734] Re: [mg102717] Re: ]-> [mg102710] Incongruence? hmm... ]-> ]-> 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 ]-> > ]-> ]->
- References:
- Incongruence? hmm...
- From: Filippo Miatto <miatto@gmail.com>
- Re: Re: Re: Incongruence? hmm...
- From: Filippo Miatto <miatto@gmail.com>
- Incongruence? hmm...