       • 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)

)

the following expression can be reduced to 1

z = (2 + Sqrt)^(1/3) + (2 - Sqrt)^(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)^(1/3), (2 + Sqrt)^(1/3)}
{0.30901699437494756 + 0.535233134659635*I, 1.618033988749895}

However defining

mycuberoot[x_] := Block[{w}, w = w /. Solve[w^3 == 1][]; If[Re[x] <
0, w*x^(1/3), x^(1/3)]]

Then

{2 - Sqrt, 2 + Sqrt}
mycuberoot /@ %
FullSimplify[%]
Together[Plus @@ %]

{2 - Sqrt, 2 + Sqrt}
{(-1)^(2/3)*(2 - Sqrt)^(1/3), (2 + Sqrt)^(1/3)}
{(1/2)*(1 - Sqrt), (1/2)*(1 + Sqrt)}
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)^(1/3) + (10 - 6*Sqrt)^(1/3)

is equal to 2. Indeed

{10 + 6*Sqrt, 10 - 6*Sqrt}
mycuberoot /@ %
FullSimplify[%]
Together[Plus @@ %]

{10 + 6*Sqrt, 10 - 6*Sqrt}
{(10 + 6*Sqrt)^(1/3), (-1)^(2/3)*(10 - 6*Sqrt)^(1/3)}
{1 + Sqrt, 1 - Sqrt}
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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