The function f(x) = 1/(1-x) on RR \ {0, 1} has the (rather nice) property that f(f(f(x))) = x. Is there a simple example of a function g(x) such that g(g(g(g(x)))) = x but g(g(x)) != x?
1 Answer
Jul 10, 2015
The function:
works, but is not as simple as
Explanation:
We can split
This is a solution, but are there any simpler ones?