MathGroup Archive 1994

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

Search the Archive

Cube Roots

  • To: mathgroup at
  • Subject: Cube Roots
  • From: Dana_Scott at
  • Date: Thu, 10 Feb 94 10:36:37 EST

I am trying to tell my class about roots of polynomials.  I was
surprised that I could not evaluate numerically the radical expression
for the real root of a cubic (which had two other complex roots) in
what I thought was the expected way.  After trying to isolate the
problem it came down to this:

	N[(-1 )^(1/3)]

	0.5 + 0.866025 I


 Out[82]=      	        -19
	-1. + 1.89735 10    I

I agree that -1 has three cube roots (two complex), but this is a
really dumb convention to choose for N when a real root is available.
What do people think?  Is there a way to work around this "counter-
clockwise" feature of Mathematica?  Or does the Robotic Mind triumph

By the way, the actual expression I was trying to evaluate was:

	2^(1/3)/(-27 + 3*69^(1/2))^(1/3) + 
	  (-27 + 3*69^(1/2))^(1/3)/(3*2^(1/3))

Not so pretty, but it is real (~ -1.324717957244746026, I think).

  • Prev by Date: Sum
  • Next by Date: ComplexExpand
  • Previous by thread: Sum
  • Next by thread: ComplexExpand