Re: Re: Simplifying in Mathematica

*To*: mathgroup at smc.vnet.net*Subject*: [mg71247] Re: [mg71200] Re: Simplifying in Mathematica*From*: Daniel Lichtblau <danl at wolfram.com>*Date*: Sat, 11 Nov 2006 03:39:27 -0500 (EST)*References*: <eish5b$nq1$1@smc.vnet.net> <200611101138.GAA13711@smc.vnet.net>

pierodancona at gmail.com 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 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! Random point evaluation, and various other heuristics, are what lie behind the Mathematica predicate Developer`ZeroQ. Daniel Lichtblau Wolfram Research

**Follow-Ups**:**Re: Re: Re: Simplifying in Mathematica***From:*Andrzej Kozlowski <akoz@mimuw.edu.pl>

**References**:**Re: Simplifying in Mathematica***From:*pierodancona@gmail.com