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Re: silly questions?
Thanks Andrzej and others for your informative replies on LeafCount regarding my simplification query. I see how it all makes sense regarding the Cos simplification but I have one little nagging issue left as the following example shows: Factor[x^4 - 16] (-2 + x)*(2 + x)*(4 + x^2) FullSimplify[(x^4-16)/(-2+x)] (2 + x)*(4 + x^2) FullSimplify[(x^4-16)/(2+x)] (-16 + x^4)/(2 + x) LeafCount[(-16+x^4)/(2+x)] 11 LeafCount[(-16+x^4)/(-2+x)] 11 LeafCount[(2+x)*(4+x^2)] 9 LeafCount[(-2+x)*(4+x^2)] 9 The question to you gifted individuals is why doesn't FullSimplify[(x^4-16)/(2+x)] give (-2 + x)*(4 + x^2) which has a lower leaf count? On Thu, 28 Jul 2005 07:55:58 +0100, Andrzej Kozlowski <akoz at mimuw.edu.pl> 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 > > >