Re: Incongruence? hmm...

• To: mathgroup at smc.vnet.net
• Subject: [mg102712] Re: [mg102710] Incongruence? hmm...
• From: DrMajorBob <btreat1 at austin.rr.com>
• Date: Fri, 21 Aug 2009 04:42:04 -0400 (EDT)
• References: <200908200856.EAA05738@smc.vnet.net>

```Your alternative calculation goes something like this:

(* the term you defined: *)

term[m_, n_] = ((m x)^(2 n) (-1)^n)/((2 n)! m^4)

((-1)^n (m x)^(2 n))/(m^4 (2 n)!)

Sum[term[m, n], {n, 0, 5}]

1/m^4 - x^2/(2 m^2) + x^4/24 - (m^2 x^6)/720 + (m^4 x^8)/40320 - (
m^6 x^10)/3628800

(* and that matches: *)

Normal@Series[Cos[m x]/m^4, {x, 0, 10}]

1/m^4 - x^2/(2 m^2) + x^4/24 - (m^2 x^6)/720 + (m^4 x^8)/40320 - (
m^6 x^10)/3628800

So far, so good!

(* summing first on m: *)

term[n_] = Sum[term[m, n], {m, 1, Infinity}]

((-1)^n x^(2 n) Zeta[-2 (-2 + n)])/(2 n)!

(* most terms are zero: *)

term /@ Range[0, 10]

{\[Pi]^4/90, -(1/12) \[Pi]^2 x^2, -(x^4/48), 0, 0, 0, 0, 0, 0, 0, 0}

(* adding the non-zero ones: *)

f3[x_] = Sum[term[n], {n, 0, 2}]

\[Pi]^4/90 - (\[Pi]^2 x^2)/12 - x^4/48

(Same as you posted.)

But the calculation for term[n] was spurious (a bug?), since

term[m, 2]

x^4/24

can't be summed over m:

Sum[term[m, 2], {m, 1, Infinity}]

Sum::div: Sum does not converge. >>

\!\(
\*UnderoverscriptBox[\(\[Sum]\), \(m = 1\), \(\[Infinity]\)]
\*FractionBox[
SuperscriptBox[\(x\), \(4\)], \(24\)]\)

Thus, term[2] is undefined, and we can't sum over all n.

Bobby

On Thu, 20 Aug 2009 03:56:35 -0500, Filippo Miatto <miatto at gmail.com>
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??
> Filippo
>

--
DrMajorBob at bigfoot.com

```

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