Re: Substitute expressions with FullSimplify
- To: mathgroup at smc.vnet.net
- Subject: [mg108687] Re: Substitute expressions with FullSimplify
- From: "Nasser M. Abbasi" <nma at 12000.org>
- Date: Sat, 27 Mar 2010 05:11:26 -0500 (EST)
- References: <hoi2n9$qj7$1@smc.vnet.net>
"Guido Walter Pettinari" <coccoinomane at gmail.com> wrote in message news:hoi2n9$qj7$1 at smc.vnet.net... > Hello world! > > This is my first post in this group, but it has been a while since I > started reading it. I always found it quite useful, therefore I wish > to thank everibody for their contributions! > > Here is my problem. Let's say I have an expression. I would like to > substitute all the occurences of a given subexpression in this > expression with a symbol. I want to do it in an intelligent way, i.e. > by using FullSimplify instead of ReplaceAll. > > If my expression is: > > x^2 + y^2 > > I know that: > > FullSimplify [ x^2 + y^2, x^2 + y^2 == r ] > > will produce 'r' as a result, which is what I want. > > However, if my expression is > > x + y , > > then > > FullSimplify [ x + y, x + y == r ] > > produces 'x + y' and not 'r' ! I tried to use > > FullSimplify [ x + y, x + y == r, ComplexityFunction -> LeafCount ] > > but I still get 'x+y' as a result. > > Do you have any idea on how to substitute x+y with r in an expression? > Good question. I do not know myself, may be a Mathematica expert can tell us. But I am just curious, why do you consider Simplify [ x + y, x + y == r ] more "intelligent" than In[39]:= x + y /. {x + y -> r} Out[39]= r ? --Nasser