if f(x) = 2x^2 - 1,
then the nth iteration of f(x), i.e. f(f(f(...x))) = cos (2 ^ n *
arccos(x)).
This is easy to see from the identity:
cos(2x) = 2 * cos(x)^2 - 1,
cos(4x) = 2 * cos(2 * cos(x)^2 - 1)^2 - 1
etc.
In general, if F(x) = the inverse of f(x), the nth iteration of f(g(F(x)))
is equal to
f( the_nth_iteration_of_g(F(x))), where f(x)=cos(x), F(x)=arccos(x), and
g(x)=2x.
Interestingly, the general formula works for non-integer values of n,
conveniently
defining the value of the iterated function over *all* numbers.
At the time, I was working with the Mandelbrot set, trying to find a
closed-form
solution for f(x) = x^2 + a, for any value of a. Though I was utterly
unable to
generalize the above solution for 2x^2-1 (which can easily be transformed
to
x^2-1/2), I have always been curious as to whether a closed-form solution
does indeed exist. Any takers?
-Tim Sweeney