Re: plotting even function

[John B. Thoo]

First, thanks to Steve T, Henry M, Geraint B, and Thomas S for their 
replies.

Thomas, your plotting example

octave:5> y = 3/4*((2*x).^2).^(1/3)+1/2;
octave:6> plot(x,y)

gives what I expected.

I learnt something tonight.  When I teach (high school) algebra, I tell 
my students that for rational  m/n  in lowest terms,

  a^(m/n) = (a^m)^(1/n) = ( a^(1/n) )^m

as long as  a^(1/n)  is real.  And, for  n  odd, there is no reason at 
this level to think that  a^(1/n)  is not real.

In my example,  2/3  is clearly(?) a rational exponent with  n=3  odd, 
so I never expected that Octave would treat  a^(1/3)  as complex.  
Certainly food for thought for me.

So, thanks again to all for your help.

Oh, in case you're interested in how this came up for me, I was 
plotting normals to the curve  y = x^2,

> x = -1:0.04:1;  t = -1:1:1;
> for i = 1:51
> plot (t, -t ./ (2 * x (i)) + (1 + 2 * x (i).^2) ./ 2, "-b")
> endfor

and the envelope of the normals (the evolute of the parabola?) is given 
by the curve I had asked about,

  y = (3/4) (2x)^(2/3) + 1/2.

Anyhow, thanks again.

---John.


On Mar 19, 2005, at 12:48 PM, Thomas Shores wrote:

> Umm... make that
> y = 3/4*((2*x)^2)^(1/3) + 1/2
>
> So add to my last plotting suggestions
>
> octave:5> y = 3/4*((2*x).^2).^(1/3)+1/2;
> octave:6> plot(x,y)
>
> Tom Shores
>
> On Saturday 19 March 2005 08:36 pm, Thomas Shores wrote:
> > Hmmm...
> > Well, actually the function really *is* even if you think of
> > it this way:
> >
> > y  = 3/4*((2*x)^2)^3 + 1/2
> >
> > 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.  It evidently uses complex analysis to
> > find non-integral powers.  Try this
> >
> > octave:16> x = (-1)^(1/3)
> > x = 0.50000 + 0.86603i
> > octave:17> x^3
> > ans = -1.0000e+00 + 3.6370e-16i
> >
> > So octave calculates a complex third root of unity, rather
> > than the real root x = -1 that might spring to mind.  Makes
> > sense, since you can generate the other roots from the
> > complex primitive root of unity.
> >
> > Tom Shores
> >
> > On Sunday 20 March 2005 01:20 am, Geraint Paul Bevan wrote:
> > > -----BEGIN PGP SIGNED MESSAGE-----
> > > Hash: SHA1
> > >
> > > John B. Thoo 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, yet
> > > |
> > > | x = -1:0.04:1;
> > > | plot (x, 0.75 * (2 .* x).^(2 / 3) + 0.5)
> > > |
> > > | is not symmetrical about the y-axis.  What is wrong with
> > > | my thinking?
> > > |
> > > | TIA.
> > > | ---John.
> > >
> > > x^(2/3) is not an even function,
> > >
> > > octave:1> [ (+8)^(2/3) ; (-8)^(2/3) ]
> > > ans =
> > >
> > > ~    4.0000 + 0.0000i
> > > ~   -2.0000 + 3.4641i
> > >
> > > so nor is the function which you are plotting.
> > >
> > > - --
> > > Geraint Bevan
> > > http://homepage.ntlworld.com/geraint.bevan
> > >
> > > -----BEGIN PGP SIGNATURE-----
> > > Version: GnuPG v1.2.4 (GNU/Linux)
> > >
> > > iEYEARECAAYFAkI8z/sACgkQcXV3N50QmNOi1QCdEvZ44CO0bbl0h+TQvgF
> > > pD 9uy YQgAn2a4A1x4m7ZdST8vG9IBRgBu1lCW
> > > =7F+W
> > > -----END PGP SIGNATURE-----
> > >
> > >
> > >
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