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...