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Re: Re: Simplifying in Mathematica

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  • Subject: [mg71247] Re: [mg71200] Re: Simplifying in Mathematica
  • From: Daniel Lichtblau <danl at>
  • Date: Sat, 11 Nov 2006 03:39:27 -0500 (EST)
  • References: <eish5b$nq1$> <>

pierodancona at wrote:
> 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 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!

Random point evaluation, and various other heuristics, are what lie 
behind the Mathematica predicate Developer`ZeroQ.

Daniel Lichtblau
Wolfram Research

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