Re: sum problem with infinity
- To: mathgroup at smc.vnet.net
- Subject: [mg33010] Re: sum problem with infinity
- From: Jens-Peer Kuska <kuska at informatik.uni-leipzig.de>
- Date: Tue, 26 Feb 2002 04:34:50 -0500 (EST)
- Organization: Universitaet Leipzig
- References: <a5clnv$9ip$1@smc.vnet.net>
- Reply-to: kuska at informatik.uni-leipzig.de
- Sender: owner-wri-mathgroup at wolfram.com
Hi,
Sum[] has the attribute HoldAll and the D[] operation is
not evaluated. So You keep Sum[D[something,y],__] and replace
y by a numerical value and D[] say that D[something,0.9] is
not possible. So, you must force to evaluate the argument of Sum[]
with
dsum = D[\[Psi][x, y], y] /.
HoldPattern[Sum[b_.*a_D, ll_]] :> Sum[Evaluate[b*a], ll];
and now you can replace x and y
dsum /. y -> 0.9 /. x -> 0.8
Regards
Jens
Doron wrote:
>
> hello,
> I am a problem :
> I defined ->
> \!\(\[Psi][x_, y_] := \(1\/4\) a\^2 -
> x\^2 - \(\(8
> a\^2\)\/\[Pi]\^3\) \(\[Sum]\+\(n = \
> 0\)\%\[Infinity]\(\((\(-1\))\)\^n\/\((2 n + 1)\)\^3\) \(Cosh[k[n]*y]\ \
> Cos[k[n]*x]\)\/Cosh[\(1\/2\) k[n]\ b]\)\)
> and tried to compute ->
> D[\[Psi][x, y], y] /. y -> 0.9 /. x -> 0.8
> the answer I got was ->
> General::"ivar": "\!\(0.9`\) is not a valid variable."
>
> General::"stop": "Further output of \!\(General :: \"ivar\"\) will be \
> suppressed during this calculation."
>
> \!\(\(-\(\(32\ \(\[Sum]\+\(n = \
> 0\)\%\[Infinity]\[PartialD]\_0.9`\(\(\((\(-1\))\)\^n\ \((Cosh[
> k[n]\ 0.9`]\ Cos[
> k[n]\ 0.8`])\)\)\/\(\((2\ n + 1)\)\^3\ Cosh[
> 1\/2\ k[n]\ b]\)\)\)\)\/\[Pi]\^3\)\)\)
> But when I replaced the sums limit from infinity to 99
> there was no problem and I got a numerical answer.
> Any ideas why ?
> Thank you , Doron .