Mathematica 9 is now available
Services & Resources / Wolfram Forums
-----
 /
MathGroup Archive
2005
*January
*February
*March
*April
*May
*June
*July
*August
*September
*October
*November
*December
*Archive Index
*Ask about this page
*Print this page
*Give us feedback
*Sign up for the Wolfram Insider

MathGroup Archive 2005

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: silly questions?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg59088] Re: [mg59052] silly questions?
  • From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
  • Date: Thu, 28 Jul 2005 02:27:42 -0400 (EDT)
  • References: <200507270526.BAA20063@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com


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




  • Prev by Date: Re: "Substract one and add one" algorithm
  • Next by Date: Re: Showing Mathematica notebooks within Firefox browser (like Adobe PDF's)
  • Previous by thread: silly questions?
  • Next by thread: Re: silly questions?