On Mar 25, 5:29 pm, hru...@[EMAIL PROTECTED]
(Herman Rubin) wrote:
> In article
<aa9545df-690b-4565-8497-ef2e97c78...@[EMAIL PROTECTED]
>,
> Vladimir Bondarenko <v...@[EMAIL PROTECTED]
> wrote:
>
>
>
> >An interesting catch of yours!
> >Indeed,
> >NSum[Cos[n]^2/n^2, {n,1,Infinity}, Method -> "AlternatingSigns"]
> >NSum[Cos[n]^2/n^2, {n,1,Infinity}, Method -> "WynnEpsilon"]
> >NSum[Cos[n]^2/n^2, {n,1,Infinity}, Method -> "EulerMaclaurin"]
> >0.543756
> >0.553446
> >0.574138
> >So Mathematica 6.0.2 *had* a clear chance to produce a
> >nice answer (0.574138)... but it missed it...
> >Maybe someone from Wolfram Research could enlighten us
> >the customers why such a misfire happened?
> >On Mar 25, 12:55=A0am, SzH <szhor...@[EMAIL PROTECTED]
> wrote:
> >> On Mar 25, 5:28 am, Vladimir Bondarenko <v...@[EMAIL PROTECTED]
> wrote:
> >> > NSum[(Cos[n]/n)^1, {n, 1, Infinity}] =A0 =A0 slow convergence
> >> > NSum[(Cos[n]/n)^2, {n, 1, Infinity}] =A0 =A0 fast convergence
> >> > 0.0420195 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0RIGHT
> >> > 0.572578 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 WRONG
> >> > Any comments?
> >> Interestingly, the setting Method -> "EulerMaclaurin" gives a much
> >> more precise answer (0.574138), however, no explicit setting for
> >> Method can reproduce the answer that we get with Method -> Automatic.
>
> All of this is not surprising. "Automatic" manages to
> find an analytic function it knows how to compute.
>
Do you mean that in the first case Mathematica computes the result
analytically, and then applies N to get a numerical answer, while in
the second case it is unable do this? This cannot be the explanation
because Mma can compute the exact result in both cases:
Sum[(Cos[n]/n)^1, {n, 1, Infinity}]
(1/2)*(-Log[1 - E^(-I)] - Log[1 - E^I])
N[%]
0.04201950582536906 + 0.*I
Sum[(Cos[n]/n)^2, {n, 1, Infinity}]
(1/12)*(Pi^2 + 3*PolyLog[2, E^(-2*I)] + 3*PolyLog[2, E^(2*I)])
N[%]
0.5741377400533296 + 0.*I
|
