Re: plotting even function

[Geraint Paul Bevan]

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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.


- --
Geraint Bevan
http://homepage.ntlworld.com/geraint.bevan

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