• 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)

```To: mathgroup at smc.vnet.net

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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)
>
>
> )
>
> 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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