MathGroup Archive 1998

[Date Index] [Thread Index] [Author Index]

Search the Archive

RE: How Mathematica select one




 S. Molyavko  wrote:
|
|  I have some question about Mathematica 3.0. |
|  When Mathematica 3.0 calculate
|  (-1.)^(1/3)
|  it generate complex value result
|  0.5 + 0.8660254037844385*I.
|
|  But when I try to Solve equation
|  x^3+1==0
|  it generates three different roots. |
|  So, here is the question:
|
|  "How Mathematica 3.0 select one of the roots in the first example?" |

Suppose you are doing (-2.0)^(1/3) . Then  -2.0=2.0 E^(Pi*I)     (*
where I=Sqrt[-1] *) so (-2.0)^(1/3) gives (2.0^(1/3))*E^(Pi*I*(1/3))
You then get (2.0^(1/3)) (Cos[Pi/3] + I*Sin[Pi/3])

The root above is called the principal root. The other roots are:
(2.0^(1/3)) (Cos[2*Pi/3] + I*Sin[2*Pi/3]) (2.0^(1/3)) (Cos[3*Pi/3] +
I*Sin[3*Pi/3])

This works for any (x^a) when (x<0); (-1<a<1).

Ted Ersek




  • Prev by Date: Announcing Java-Interface to MathLink
  • Next by Date: Doing a table of contents ; also : footnotes ?
  • Prev by thread: Announcing Java-Interface to MathLink
  • Next by thread: Doing a table of contents ; also : footnotes ?