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

  • To: mathgroup at smc.vnet.net
  • Subject: [mg102651] Re: [mg102734] Re: [mg102717] Re: [mg102710] Incongruence? hmm...
  • From: Filippo Miatto <miatto at gmail.com>
  • Date: Sun, 23 Aug 2009 05:32:37 -0400 (EDT)
  • References: <200908200856.EAA05738@smc.vnet.net>

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
>
>


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