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