Re: plotting even function

[Thomas Shores]

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:
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> >
> > 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
> >
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> >
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