Re: plotting even function

[Mike Miller]

On Sun, 20 Mar 2005, Geraint Paul Bevan wrote:

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> John B. Thoo wrote:
>
> > Also, why is  (x^(1/3))^2  not even?  If  f(x) = (x^(1/3))^2,  then isn't
> >   f(-x) = ((-x)^(1/3))^2 = (- x^(1/3))^2 = (x^(1/3))^2 = f(x)
>
>
> It is not true to say that (-x)^(1/3) is the same as -(x^(1/3)). That is 
> one possibility but there are two more. If n is an integer, x^n has a 
> unique value but x^(1/n) has n possible values i.e. the square root 
> x^(1/2) has two possible values, the cube root x^(1/3) has three 
> possible values, etc.
>
> Consider that (-x)^(1/n) can be re-written as ((-1)*(+x))^(1/n) which is 
> ((-1)^(1/n))*((+x)^(1/n))
>
> For symmetry about the y-axis, you therefore require that:
> (-1)^(1/n) = -1.
>
> However, for n=3, y=(-1)^(1/n) has three solutions:
> y = -1
> y = 0.5+i*sqrt(3)/2
> y = 0.5-i*sqrt(3)/2
>
> You obviously want to use the first of these solutions to get symmetry 
> about the y-axis but Octave doesn't know that unless you tell it.


I am not a mathematician, but I believe that you are missing something. 
The equation y^3 + 1 = 0 has three solutions, as noted above.  The 
equation y^3 - 1 = 0 also has three solutions:

y = 1
y = -0.5-i*sqrt(3)/2
y = -0.5+i*sqrt(3)/2

How does Octave "know what we want" (1) in the latter case, but it does 
not know what we want (-1) in the former case?  I think the answer has to 
do with numerical analysis.  The binary representation of 1/3 is never
exactly 1/3, but it is only at the exact value of 1/3 that (-1)^(1/3) 
equals -1.  This is not a problem for (+1)^(1/3).

For (-1)^(333/1000), Octave gives a very similar answer to what it gives 
for (-1)^(1/3), which seems appropriate.  Octave doesn't seem to know or 
care that there are 1000 possible solutions in the complex plane!

Mike



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