Cube Roots

*To*: mathgroup at yoda.physics.unc.edu*Subject*: Cube Roots*From*: Dana_Scott at pop.cs.cmu.edu*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: In[81]:= N[(-1 )^(1/3)] Out[81]= 0.5 + 0.866025 I In[82]:= %^3 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 again? 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).