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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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