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MathGroup Archive 2006

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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
> 


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