       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).
>
> 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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