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MathGroup Archive 2005

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Re: Adding new rules to Simplify

  • To: mathgroup at smc.vnet.net
  • Subject: [mg56387] Re: Adding new rules to Simplify
  • From: Peter Pein <petsie at arcor.de>
  • Date: Sat, 23 Apr 2005 01:16:23 -0400 (EDT)
  • References: <27457333.1114080708731.JavaMail.jakarta@nitrogen.mathforum.org> <d4al4s$iv0$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

John Billingham wrote:
> Here's another one!
> 
> r1[x_ /; ! FreeQ[x, f]] := x /. f[y_] -> a[y]/b[y]; 
> r1[x_] := x;
> 
> r2[x_ /; ! FreeQ[x, g]] := x /. g[y_] -> b[y]/a[y]; 
> r2[x_] := x;
> 
> Simplify[f[X] g[X], TransformationFunctions -> {r1, r2}]
> 
> gives 
> 
> f(X) g(X)
> 
> 
> I would have hoped to get 1 here, particularly since
> 
> r1[r2[f[X]g[X]]]
> 
> gives me 1 as expected! 
> 
> So I tell it to Simplify using just two rules, the successive application of which gives 1, but Mathematica fails to simplify at all.
> 
> What's going on here??
> 
> John
> 
Hi John,

Simplify doesn't apply r1 and r2 simultaneous.
If Simplify applies e.g. r1 to f[x]g[x], this product becomes
a[x]g[x]/b[x] which is assumed (correctly) to be more complex.

 You either have to use an own ComplexityFunction, or you use only one
replacement r3:

In[6]:=
Simplify[h, TransformationFunctions -> {r1, r2},
  ComplexityFunction ->
   (LeafCount[#1] + 10^6*Count[#1, f | g, Heads -> True] & )]
Out[6]=
1

In[7]:=
r3[expr:(p_.)*(f | g)[y_] + (q_.)] :=
  expr /. {f -> (a[#1]/b[#1] & ), g -> (b[#1]/a[#1] & )}
In[8]:=
Simplify[h, TransformationFunctions -> {r3}]
Out[8]=
1

-- 
Peter Pein
Berlin


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