Re: Reduction of Radicals

*To*: mathgroup at smc.vnet.net*Subject*: [mg71935] Re: [mg71902] Reduction of Radicals*From*: Bob Hanlon <hanlonr at cox.net>*Date*: Mon, 4 Dec 2006 06:39:26 -0500 (EST)*Reply-to*: hanlonr at cox.net

To: mathgroup at smc.vnet.net Search the archives for "branch cut" and/or "principal value" z==(2 + Sqrt[5])^(1/3) + (2 - Sqrt[5])^(1/3); #^3&/@(#-(2 + Sqrt[5])^(1/3)&/@%) (-(2 + Sqrt[5])^(1/3) + z)^3 == 2 - Sqrt[5] Reduce[%,z,Reals]//ToRules {z -> 1} z==(10 + 6*Sqrt[3])^(1/3) + (10 - 6*Sqrt[3])^(1/3) z == (10 - 6*Sqrt[3])^(1/3) + (10 + 6*Sqrt[3])^(1/3) #^3&/@(#-(10 + 6*Sqrt[3])^(1/3)&/@%) (-(10 + 6*Sqrt[3])^(1/3) + z)^3 == 10 - 6*Sqrt[3] Reduce[%,z,Reals]//ToRules {z -> 2} Bob Hanlon ---- dimitris <dimmechan at yahoo.com> wrote: > 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 >