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MathGroup Archive 2005

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Re: silly questions?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg59083] Re: silly questions?
  • From: "snoofly" <snoofly at snoofly.com>
  • Date: Thu, 28 Jul 2005 02:27:21 -0400 (EDT)
  • References: <dc77me$k74$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

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?

Another situation that I spotted in a similar vein where a small change in a 
constant multiplier blows the algorithm:

Simplify[Cos[1 m Pi], Assumptions -> m \[Element] Integers] // InputForm

(-1)^m

Simplify[Cos[2 m Pi], Assumptions -> m \[Element] Integers] // InputForm

1

Simplify[Cos[3 m Pi], Assumptions -> m \[Element] Integers] // InputForm

Cos[3*m*Pi]

Simplify[Cos[4 m Pi], Assumptions -> m \[Element] Integers] // InputForm

1

I'm curious that if the 1,2,4 multiplier cases can be deduced, the third 
ideally giving (-1)^m causes a problem.


"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
> 


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