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