Re: Re: Incongruence? hmm...
- To: mathgroup at smc.vnet.net
- Subject: [mg102732] Re: [mg102713] Re: Incongruence? hmm...
- From: DrMajorBob <btreat1 at austin.rr.com>
- Date: Sat, 22 Aug 2009 03:38:01 -0400 (EDT)
- References: <h6j33u$5j4$1@smc.vnet.net> <4A8D74DC.5050805@gmail.com>
- Reply-to: drmajorbob at bigfoot.com
You already checked, and it didn't work.
Bobby
On Fri, 21 Aug 2009 03:42:15 -0500, Filippo Miatto <miatto at gmail.com>
wrote:
> Thank you all for contributing, I didn't see that the sums do not
> converge for n>2, what a mistake...
> but still, since they alternate sign for n even or odd, couldn't it be
> that they sort of 'cancel out' giving that x^3 term in the end?
> if this is the case, how can i check it with mathematica?
> Filippo
>
>
>
> 2009/8/20, Szabolcs Horv=E1t <szhorvat at gmail.com>:
>> On 2009.08.20. 10:56, Filippo Miatto wrote:
>>> 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)!
>>
>> Hello Filippo,
>>
>> I believe the result above to be valid only for n=0 and n=1. For oth=
> er
>> values of n the series will not be covergent.
>>
>>>
>>> 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