From owner-help-octave at bevo dot che dot wisc dot edu  Tue Nov 12 12:53:59 1996
Subject: Re:  fractional powers
From: Marvin Vis <mvis at advtech dot uswest dot com>
To: heberf at u dot washington dot edu, help-octave@bevo.che.wisc.edu
Date: Tue, 12 Nov 96 11:53:37 MST

> I'm somewhat disturbed by the following behaviour on octave.  Perhaps it
> is standard and I shouldn't be worried but it surprised me.  Consider the
> following question: what is the cube root of -1?  Clearly the answer
> should be -1.  Now ask octave
> 
> octave:1> x = (-1)^(1/3)
> x = 0.50000 + 0.86603i
> 
> it gets wierder if you now cube that number
> 
> octave:2> x^3
> ans = -1.0000e+00 + 1.2246e-16i
>  
> This is pretty close to the truth but still strange to my way of thinking.
> Similar wierdness shows up with other fractional powers: 1/5, 1/7, etc.
> 
> Any thoughts?

There are actually 3 cube roots of -1 (here, j = sqrt(-1)):

	(-1)^(1/3) = {exp(j*pi/3), exp(-j*pi/3), -1}

Most rooting algo's will find the n^th root of a number as follows:

	Starting with a number z = |z|*exp(j*theta),

		z^(1/n) = |z|^(1/n) * exp(j*theta/n)

I'd call this the primary root.  If you want to have a special case for
negative (real) z and odd n, you could use a routine that does:

	z^(1/n) = - |z|^(1/n)

This would give you -1 as the cube root of -1.

M.