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Re: Replacement rule limitations

  • To: mathgroup at smc.vnet.net
  • Subject: [mg110576] Re: Replacement rule limitations
  • From: Peter Pein <petsie at dordos.net>
  • Date: Sat, 26 Jun 2010 03:11:07 -0400 (EDT)
  • References: <i023o1$r6u$1@smc.vnet.net>

Am Fri, 25 Jun 2010 11:25:53 +0000 (UTC)
schrieb "S. B. Gray" <stevebg at ROADRUNNER.COM>:

> Several responders have told me that replacement rules are the way to 
> simplify complex expressions and to reduce redundant computations (?).
> 
> So let's try an arbitrary expression as a very simple example of 
> something that could be much more complicated:
> 
> exp = 1/Sqrt[
>     x^2+y^2+z^2] - (x^2+y^2+z^2) + (x^2+y^2+z^2)^(
>     1/3) /. x^2+y^2+z^2->dd               This gives
> 
> 1/Sqrt[dd] + dd^(1/3)-x^2-y^2-z^2        which is not that useful.
> 
> But if I introduce a superfluous multiplier "s":
> 
> exp = 1/Sqrt[x^2+y^2+z^2] -
>     s(x^2+y^2+z^2) + (x^2+y^2+z^2)^(
>     1/3) /. {x^2+y^2+z^2->dd, s->1}     I get
> 
> 1/Sqrt[dd] + dd^(1/3) - dd             which is better. Asking for
> 
> exp^2                                  gives, as desired,
> 
> (1/Sqrt[dd] + dd^(1/3) - dd)^2 .       But trying to proceed as if
> this were regular algebra where cascaded substitutions are routine, I
> try:
> 
> exp/.{x->a^2, y->3b, z->Sqrt[d + e]} , I get the useless result
> 
> 1/Sqrt[dd] + dd^(1/3) - dd .
> 
> Unless I am missing something important (it wouldn't be the first 
> time!), replacement rules are not a good substitute for real 
> intermediate variables. This does not even address a feature I'd like
> to see in Mathematica in which it would figure out what
> subexpressions appear repeatedly and make up its own simplifying
> intermediate variables. This could be incorporated into FullSimplify.
> 
> Comments will be greatly appreciated.
> 
> Steve Gray
> 

Hi,

rules are one way constructs. Once you eliminated x, y and z by
your multiplier-method, exp is free of these variables. You'll have to
do a backsubstitution first:

exp/.{dd->x^2+y^2+z^2,x->a^2, y->3b, z->Sqrt[d + e]}

should give what you want.

Peter

P.S.: Did I hear a flat hand slapping against someone's forehead? ;-)




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