Re: silly questions?
- To: mathgroup at smc.vnet.net
- Subject: [mg59111] Re: silly questions?
- From: David Bailey <dave at Remove_Thisdbailey.co.uk>
- Date: Fri, 29 Jul 2005 00:41:49 -0400 (EDT)
- References: <200507270526.BAA20063@smc.vnet.net> <firstname.lastname@example.org>
- Sender: owner-wri-mathgroup at wolfram.com
Andrzej Kozlowski wrote: > On 27 Jul 2005, at 07:26, Kent Holing wrote: > > >>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 >> >> > > > Very "simple". What makes you think the cancelled out form is "simpler"? > > > LeafCount[(x^5-32)/(x-2)] > > > 11 > > while > > > LeafCount[Cancel[(x^5-32)/(x-2)]] > > > 18 > > The cancelled form is much more "complicated", at least as measured > by LeafCount (and Mathematica's default complexity function). > > So if you want your answer it is better to make ask Mathematica to > make the expression more "complex": > > > Simplify[(x^5 - 32)/(x - 2), ComplexityFunction -> > (1/LeafCount[#1] & )] > > > x^4 + 2*x^3 + 4*x^2 + 8*x + 16 > > But it is of course much more sensible to just use Cancel. > > Andrzej Kozlowski > > > Maybe we need two new Mathematica operations - Complicate and FullComplicate - the trouble is, I guess they would always hang trying to return an infinitely large answer! More seriously, a function that presented a range of possible alternatives (together with a command that could achieve the each result directly) might be quite useful. David Bailey http://www.dbaileyconsultancy.co.uk
- silly questions?
- From: Kent Holing <KHO@statoil.com>
- silly questions?