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 >