Re: Re: Incongruence? hmm...
- To: mathgroup at smc.vnet.net
- Subject: [mg102734] Re: [mg102717] Re: [mg102710] Incongruence? hmm...
- From: DrMajorBob <btreat1 at austin.rr.com>
- Date: Sat, 22 Aug 2009 03:38:23 -0400 (EDT)
- References: <200908200856.EAA05738@smc.vnet.net>
- Reply-to: drmajorbob at bigfoot.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
- References:
- Incongruence? hmm...
- From: Filippo Miatto <miatto@gmail.com>
- Incongruence? hmm...