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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
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