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

Re: silly questions?

• To: mathgroup at smc.vnet.net
• Subject: [mg59114] Re: silly questions?
• From: Peter Pein <petsie at dordos.net>
• Date: Fri, 29 Jul 2005 00:41:51 -0400 (EDT)
• References: <dc77me$k74$1@smc.vnet.net> <dc9van$cka$1@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

snoofly schrieb:
> Not silly at all.
> I was going to say this could be a case of one mans simple is another mans 
> complex.
> One may think (x^5-32)/(x-2) is simpler than 16 + 8*x + 4*x^2 + 2*x^3 + x^4 
> due to the lower number of components but experimenting below I can see that 
> my theory is not the case.
> 
> Messing around, I was suprised to find:
> Factor[x^4 - 16]
> 
> (-2 + x)*(2 + x)*(4 + x^2)
> 
> FullSimplify[(x^4 - 16)/(2 + x)]
> 
> (-16 + x^4)/(2 + x)
> 
> FullSimplify[(x^4 - 16)/(-2 + x)]
> 
> (2 + x)*(4 + x^2)
> 
> Why does the first FullSimplify not return (-2 + x)*(4 + x^2) in similar 
> style to the second? Clearly this must be simpler than the first (which 
> involves same # components but at higher orders). Additionally, if the 
> second can be done, why is that the algorithm misses on the first almost 
> identical case?
> 
...

> 
> "Kent Holing" <KHO at statoil.com> wrote in message 
> news:dc77me$k74$1 at smc.vnet.net...
> 
>>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
>>
> 
> 
Hi,

 if you simply try Factor[(x^4-16)/(x-#)]&/@{-2,2}, Mathematica returns
the canceled polynomials, you want. I _guess_ Factor is not among the
TransformationFunctions (see help), because it /can/ become very time
consuming.

-- 
Peter Pein
Berlin