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MathGroup Archive 2006

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Re: Simplifying in Mathematica

  • To: mathgroup at smc.vnet.net
  • Subject: [mg71200] Re: Simplifying in Mathematica
  • From: pierodancona at gmail.com
  • Date: Fri, 10 Nov 2006 06:38:01 -0500 (EST)
  • References: <eish5b$nq1$1@smc.vnet.net>

Define a function (depending on the variables B,b,A,a,c,d...etc,
all of them) equal to your expression, and compute it at many points.
If the expression is 0, the result should always be 0. Notice that if
your expression is a polynomial or rational and you choose your
points in a "not too special" way, this gives actually a rigorous
"proof"
that the expression is 0. In general, this is an extrremely reliable
heuristic (and of course a "proof" if you do not get 0 for some
values).

I encounter the same problem frequently, and this method works
very well, also to test if two complicated expressions are the same
or not (guess how :)

Piero



330006 at gmail.com wrote:
> I have a function which is a sum of many terms which look like this:
>
> (2*(B-b)^2 - 2*(A-a)*c*d^2)/(4*b^2*(1-c*2)*d^2)
>
> I think the function is actually equal to 0, but I have a hard time in
> trying to simplify it in Mathematica. Any ideas or commands I should
> try? Any suggestions in general about simplifying formulas will also
> be greatly appreciated!
> 
> Thanks a lot!


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