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Re: Simplifying complicated expressions

• To: mathgroup at smc.vnet.net
• Subject: [mg110446] Re: Simplifying complicated expressions
• From: Bob Hanlon <hanlonr at cox.net>
• Date: Fri, 18 Jun 2010 07:44:23 -0400 (EDT)
• Reply-to: hanlonr at cox.net

Keep the LHS of the replacement rule simple and the match will be robust:

2 x^2 + y^3 - x^2 y^2 + Sqrt[z3 + y2] /.
 
 Sqrt[z3 + y2] -> flxyz - (x^2 + y^3 - x^2 y^2)


flxyz + x^2


Bob Hanlon

---- "Sjoerd C. de Vries" <sjoerd.c.devries at gmail.com> wrote: 

=============
Mathematica does not replace the expression if I put a simple 2 in
front of the first term, whereas most any human sees the simple
reduction that's possible:

In: 2 x^2 + y^3 - x^2 y^2 + Sqrt[z3 + y2] /.
 x^2 + y^3 - x^2 y^2 + Sqrt[z3 + y2] -> flxyz

Out: 2 x^2 - x^2 y^2 + y^3 + Sqrt[y2 + z3]

could be easily reduced to x^2 +flxyz

Cheers -- Sjoerd

On Jun 6, 12:42 pm, "David Park" <djmp... at comcast.net> wrote:
> "The normal
> /. and -> substitutions and patterns are not adequate for this."
>
> That sounds like a completely unfounded statement so why don't you
> demonstrate it?
>
> David Park
> djmp... at comcast.nethttp://home.comcast.net/~djmpark/ 
>
> From: S. B. Gray [mailto:stev... at ROADRUNNER.COM]
>
> Suppose I  have a long complex expression in which terms like
> (x^2+y^3-x^2y^2+Sqrt[z3+y2]) (for a simple example) appear many times
> along with various powers and the reciprocals of it, etc. To make the
> expression comprehensible and to make the computation faster, I would
> like to substitute say "f1xyz" for it everywhere it appears. The normal
> /. and -> substitutions and patterns are not adequate for this. Of
> course at evaluation time I want to compute f1xyz only once and not have
> the final formula revert to the original variables. How do I prevent that?
>
> Also a welcome addition to Mathematica would be the ability to find these
> repeated expressions automatically and put them in, because doing it
> manually is very error-prone and slow.
>
> Tips will be appreciated!
>
> Steve Gray