Reduction of Radicals

*To*: mathgroup at smc.vnet.net*Subject*: [mg71902] Reduction of Radicals*From*: "dimitris" <dimmechan at yahoo.com>*Date*: Sun, 3 Dec 2006 06:26:16 -0500 (EST)

Based on this reference Cardan Polynomials and the Reduction of Radicals (by T. Osler) (see also references therein) (you can download the paper here: http://www.jstor.org/view/0025570x/di021218/02p0059q/0?currentResult=0025570x%2bdi021218%2b02p0059q%2b0%2c03&searchUrl=http%3A%2F%2Fwww.jstor.org%2Fsearch%2FBasicResults%3Fhp%3D25%26so%3DNewestFirst%26si%3D1%26Query%3DOsler ) the following expression can be reduced to 1 z = (2 + Sqrt[5])^(1/3) + (2 - Sqrt[5])^(1/3) Mathematica gives N[%] 1.9270509831248424 + 0.535233134659635*I This is because by default it returns a complex number for the cube root of a negative number List @@ z N[%] {(2 - Sqrt[5])^(1/3), (2 + Sqrt[5])^(1/3)} {0.30901699437494756 + 0.535233134659635*I, 1.618033988749895} However defining mycuberoot[x_] := Block[{w}, w = w /. Solve[w^3 == 1][[3]]; If[Re[x] < 0, w*x^(1/3), x^(1/3)]] Then {2 - Sqrt[5], 2 + Sqrt[5]} mycuberoot /@ % FullSimplify[%] Together[Plus @@ %] {2 - Sqrt[5], 2 + Sqrt[5]} {(-1)^(2/3)*(2 - Sqrt[5])^(1/3), (2 + Sqrt[5])^(1/3)} {(1/2)*(1 - Sqrt[5]), (1/2)*(1 + Sqrt[5])} 1 Is there a particular reason why by default Mathematicas returns a complex number for the cube root of a negative number or it is a matter of choise? Following the same procedure I prove that (10 + 6*Sqrt[3])^(1/3) + (10 - 6*Sqrt[3])^(1/3) is equal to 2. Indeed {10 + 6*Sqrt[3], 10 - 6*Sqrt[3]} mycuberoot /@ % FullSimplify[%] Together[Plus @@ %] {10 + 6*Sqrt[3], 10 - 6*Sqrt[3]} {(10 + 6*Sqrt[3])^(1/3), (-1)^(2/3)*(10 - 6*Sqrt[3])^(1/3)} {1 + Sqrt[3], 1 - Sqrt[3]} 2 This behavior of Mathematica does not affect simplifications by e.g. RootReduce? I must admit that I have gaps on my knowledge in these symbolic aspects (I start to be interested in after I try to solve the the secular Rayleigh equation) so more experienced members of the forum may forgive any possible mistakes of mine! Anyway I don't understand this difference in treating nested radicals between literature and Mathematica. I really appreciate any kind of insight/guideness/comments. Regards Dimitris

**Follow-Ups**:**Re: Reduction of Radicals***From:*Daniel Lichtblau <danl@wolfram.com>

**Re: Reduction of Radicals***From:*Andrzej Kozlowski <akoz@mimuw.edu.pl>