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Re: silly questions?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg59111] Re: silly questions?
  • From: David Bailey <dave at Remove_Thisdbailey.co.uk>
  • Date: Fri, 29 Jul 2005 00:41:49 -0400 (EDT)
  • References: <200507270526.BAA20063@smc.vnet.net> <dc9vhu$cnk$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Andrzej Kozlowski wrote:
> On 27 Jul 2005, at 07:26, Kent Holing wrote:
> 
> 
>>Why does not (x^5-32)/(x-2)//FullSimplify in Mathematica  work?
>>Compare with Factor[x^5-32]//InputForm which returns (-2 + x)*(16 +  
>>8*x + 4*x^2 + 2*x^3 + x^4).
>>So why does not the first command just return 16 + 8*x + 4*x^2 +  
>>2*x^3 + x^4?
>>As in a factorization above, how is the easiest way to pick  
>>automatically (by a function) the factors of say degree >=2,  if any ?
>>
>>Kent Holing
>>
>>
> 
> 
> Very "simple". What makes you think the cancelled out form is "simpler"?
> 
> 
> LeafCount[(x^5-32)/(x-2)]
> 
> 
> 11
> 
> while
> 
> 
> LeafCount[Cancel[(x^5-32)/(x-2)]]
> 
> 
> 18
> 
> The cancelled form is much more "complicated", at least as measured  
> by LeafCount (and Mathematica's default complexity function).
> 
> So if you want your answer it is better to make ask Mathematica to  
> make the expression more "complex":
> 
> 
> Simplify[(x^5 - 32)/(x - 2), ComplexityFunction ->
>     (1/LeafCount[#1] & )]
> 
> 
> x^4 + 2*x^3 + 4*x^2 + 8*x + 16
> 
> But it is of course much more sensible to just use Cancel.
> 
> Andrzej Kozlowski
> 
> 
> 
Maybe we need two new Mathematica operations - Complicate and 
FullComplicate - the trouble is, I guess they would always hang trying 
to return an infinitely large answer!

More seriously, a function that presented a range of possible 
alternatives (together with a command that could achieve the each result 
directly) might be quite useful.

David Bailey
http://www.dbaileyconsultancy.co.uk


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