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:

g(x) = 1/x when x in (0, 1) uu (-oo, -1)

g(x) = -x when x in (-1, 0) uu (1, oo)

works, but is not as simple as f(x) = 1/(1-x)

Explanation:

We can split RR \ { -1, 0, 1 } into four open intervals (-oo, -1), (-1, 0), (0, 1) and (1, oo) and define g(x) to map between the intervals cyclically.

This is a solution, but are there any simpler ones?