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Re: Incongruence? hmm...
*To*: mathgroup at smc.vnet.net
*Subject*: [mg102737] Re: [mg102710] Incongruence? hmm...
*From*: DrMajorBob <btreat1 at austin.rr.com>
*Date*: Mon, 24 Aug 2009 07:55:37 -0400 (EDT)
*References*: <200908200856.EAA05738@smc.vnet.net>
*Reply-to*: drmajorbob at bigfoot.com
That should work for all x, yes.
Why can't Mathematica solve every problem you give it, always correctly?
Because humans can't write such programs and, it's been proven, we never
will.
Bobby
On Sun, 23 Aug 2009 04:33:43 -0500, Filippo Miatto <miatto at gmail.com>
wrote:
> 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
>>
>
--
DrMajorBob at bigfoot.com
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