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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).