Re: Newbie question: equations with sums.
- To: mathgroup at smc.vnet.net
- Subject: [mg92871] Re: Newbie question: equations with sums.
- From: dh <dh at metrohm.ch>
- Date: Thu, 16 Oct 2008 05:05:22 -0400 (EDT)
- References: <gd4dlf$bgb$1@smc.vnet.net>
Hi Vend,
Mathematica is a pattern matcher. Thta means it tries to transform both side of
the expression to some standard form and compares them. Obviously it
does not change dummy variables to some standard variable. But we may
tell Mathematica to do so. I assume that the variable i is the standard dummy
variable and not used elsewhere. Then we can define a transformation
function and tell Simplyfy to use it:
uniqueDummy[e_]:= e/. Sum[x1_,{x2_,x3_,x4_}]:>Sum[x1/.x2->i,{i,x2,x3}];
Simplify[Sum[f[i],{i,0,n}]==Sum[f[j],{j,0,n}],TransformationFunctions->{uniqueDummy}]
hope this helps, Daniel
Vend wrote:
> Hello.
>
> FullSimplify[\!\(
> \*UnderoverscriptBox[\(\[Sum]\), \(i = 0\), \(\(-1\) + n\)]i\ f[
> i]\) == \!\(
> \*UnderoverscriptBox[\(\[Sum]\), \(k = 0\), \(\(-1\) + n\)]k\ f[k]\)]
>
> Evaluates to:
>
> \!\(
> \*UnderoverscriptBox[\(\[Sum]\), \(i = 0\), \(\(-1\) + n\)]i\ f[
> i]\) == \!\(
> \*UnderoverscriptBox[\(\[Sum]\), \(k = 0\), \(\(-1\) + n\)]k\ f[k]\)
>
> Why doesn't it evaluate to true? Is there a way to make the system
> solve this kind of equations?
>
--
Daniel Huber
Metrohm Ltd.
Oberdorfstr. 68
CH-9100 Herisau
Tel. +41 71 353 8585, Fax +41 71 353 8907
E-Mail:<mailto:dh at metrohm.com>
Internet:<http://www.metrohm.com>