MathGroup Archive 2009

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: Incongruence? hmm...

  • To: mathgroup at
  • Subject: [mg102711] Re: [mg102710] Incongruence? hmm...
  • From: Leonid Shifrin <lshifr at>
  • Date: Fri, 21 Aug 2009 04:41:53 -0400 (EDT)
  • References: <>


First, the correct answer: Mathematica is right, given that 0<=x<=2Pi. Take
what Mathematica
gives  if you only need positive x, or use a Sign[x] function in front of
cubic term, if you also need the interval -2Pi<=x<=0:

1/720 (8 \[Pi]^4 - 60 \[Pi]^2 x^2 + Sign[x]*60 \[Pi] x^3 - 15 x^4)

You should not have expanded the Cos, since then your sum over m

Sum[((m x)^(2n) (-1)^n)/((2n)!m^4),{m,1,Infinity}]

is not convergent for n>2. I am actually surprised that Mathematica did not
issue a warning when performing this sum. But in fact, it gave you a hint
the behavior changes qualitatively at n=2 - the argument of Zeta changes
So, the values of Zeta are indeed zero for n>2, but for n>2 this is no
a correct answer - this is where the cubic term is coming from. The fact
there is a Sign function in front means that there is no way you could have
obtained this term through expansion in x -  it is "non-perturbative". It is
way of saying that you should not have expanded.

One other way to obtain a correct answer valid also for negative x
is to ask Mathematica to compute

Sum[Sin[m x]/m^5, {m, 1, \[Infinity]}],

then differentiate over x and simplify the result under an assumption that x
is real.
Eventually, you will get the second result I mentioned. Needless to say, the
result is only
correct for  -2Pi<=x<=2Pi.


On Thu, Aug 20, 2009 at 12:56 PM, Filippo Miatto <miatto at> 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)!
> 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: Incongruence? hmm...
  • Next by Date: Strange behavior: random and plot
  • Previous by thread: Re: Incongruence? hmm...
  • Next by thread: Re: Re: Incongruence? hmm...