Re: plotting even function

[Thomas Shores]

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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