       Re: Re: Incongruence? hmm...-a followup- correction

• To: mathgroup at smc.vnet.net
• Subject: [mg102731] Re: [mg102713] Re: Incongruence? hmm...-a followup- correction
• From: Filippo Miatto <miatto at gmail.com>
• Date: Sat, 22 Aug 2009 03:37:50 -0400 (EDT)
• References: <34c814850908210636u4f9ce335ic47ffb739f8182da@mail.gmail.com>

```Thank you very much, this was really helpful!
I'll also have a look at the Matsubara technique, I think it will come
handy in other occasions.
Filippo

2009/8/21, Leonid Shifrin <lshifr at gmail.com>:
> Sorry, in my step #4 (in the follow-up post) the rules for coefficients
>
> In =
> result = result/.{c1 -> 0, c2 -> - Sum[1/m^2, {m, 1, Infinity}], c3 -> 0,
>  c4 -> Sum[1/m^4, {m, 1, Infinity}]}
>
> Out = \[Pi]^4/90 - (\[Pi]^2 x^2)/12 - x^4/48 +  1/6 \[Pi] x^3
> HeavisideTheta[x]
>
> (the sign of c2 is negative - due to differentiating Cos twice), and then
> the final result
> also has a minus in front of x^2-term:
>
> In =
> result = result/.HeavisideTheta[x] :> 1/2 Sign[x]
>
> Out =\[Pi]^4/90 - (\[Pi]^2 x^2)/12 - x^4/48 + 1/12 \[Pi] x^3 Sign[x]
>
>
> Regards,
> Leonid
>
>
>
> On Fri, Aug 21, 2009 at 12:42 PM, 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
>> >>
>> >
>> >
>>
>>
>

```

• Prev by Date: Re: Re: Incongruence? hmm...
• Next by Date: Re: Re: Incongruence? hmm...
• Previous by thread: Re: Re: Incongruence? hmm...-a followup- correction
• Next by thread: Re: Re: Incongruence? hmm... - a follow-up