Re: Why is 2/3 not seen as rational? [was "plotting even function"]

[Paul Kienzle]

On Mar 20, 2005, at 11:00 AM, John B. Thoo wrote:

> On March 19, 2005 4:20:04 PM PST I wrote:
>
> > Hi.  I hope that I'm not embarrassing myself by asking the following.
> >
> > I believe that
> >
> >        3       2/3    1
> >   y = --- (2 x)    + ---
> >        4              2
> >
> > is an even function
>
> And on Mar 19, 2005, at 12:36 PM, Thomas Shores replied:
>
> > Here's the problem: how do you know that the exponent is a
> > rational fraction, so that you can reinterpret it?  Octave
> > rightfully doesn't.
>
> I don't understand why  2/3  is not seen as rational.  At first I 
> thought it may be because division by  3  results in a nonterminating 
> decimal, but then I found that plotting  x^(2/5)  gives the same 
> (formerly) unexpected result as plotting  x^(2/3)  (not symmetrical 
> about the y-axix).  So, why is  m/n  not seen as rational?

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

And it is indeed a root:

  octave> ((-8).^(1/3)).^3
  ans = -8.0000e+00 + 2.2204e-15i

This is the root used when computing -8^2/3

  octave> (-8).^(2/3)
  ans = -2.0000 + 3.4641i

  octave> ((-8).^(1/3)).^2
  ans = -2.0000 + 3.4641i

And it is a correct solution, with (-8^2/3)^3 == -8^2

  octave> ((-8).^(2/3)).^3
  ans = 6.4000e+01 - 2.6042e-14i


- Paul



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