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Re: Incongruence? hmm...
*To*: mathgroup at smc.vnet.net
*Subject*: [mg102713] Re: Incongruence? hmm...
*From*: Filippo Miatto <miatto at gmail.com>
*Date*: Fri, 21 Aug 2009 04:42:15 -0400 (EDT)
*References*: <h6j33u$5j4$1@smc.vnet.net> <4A8D74DC.5050805@gmail.com>
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
>>
>
>
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