From help-request at octave dot org Mon Mar 21 08:51:27 2005
Subject: Re: Why is 2/3 not seen as rational? [was "plotting even function"]
From: Mike Miller <mbmiller at taxa dot epi dot umn dot edu>
To: Paul Kienzle <pkienzle at users dot sourceforge dot net>
cc: "John B. Thoo" <jb2 at ms dot yuba dot cc dot ca dot us>,         Octave_post mailing list <help@octave.org>
Date: Mon, 21 Mar 2005 08:51:53 -0600 (CST)

On Mon, 21 Mar 2005, Paul Kienzle wrote:

> The cubed root function is multi-valued and Octave is choosing a 
> different root than you expect.  Look at -8 for example:
>
> x^3 + 8 has three roots:
>
>  octave> roots([1,0,0,8])
>  ans =
>
>  -2.0000 + 0.0000i
>   1.0000 + 1.7321i
>   1.0000 - 1.7321i
>
> Octave chooses one of them:
>
>  octave> (-8).^(1/3)
>  ans = 1.0000 + 1.7321i


It seems to 'choose' using De Moivre's theorem.

For x > 0 and integer n != 0:

-x = x * (cos(pi) + sin(pi)*i)

(-x)^(1/n) = x^(1/n)*(cos(pi/n)+sin(pi/n)*i)

As someone else pointed out, the abs function will force the answer to be 
a real integer:

  - Mapping Function:  abs (Z)
      Compute the magnitude of Z, defined as |Z| = `sqrt (x^2 + y^2)'.

      For example,

           abs (3 + 4i)
                => 5

So, for x > 0 and integer n != 0,

-abs((-x)^(1/n))

seems to give the desired answer when n is odd, but that isn't very 
helpful because -(x)^(1/n) gives the same answer when n is odd.

On the other hand, this seems to do what the guy originally wanted (for 
scalar integer values of a and b and any real-valued x vector.):

abs(rem(a,2))*abs(rem(b,2))*sign(x).*(abs(x).^(a/b)) + (1-abs(rem(a,2)))*abs(x).^(a/b) + abs(rem(a,2))*(1-abs(rem(b,2)))*x.^(a/b)

Mike



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