       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>

```

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