The identity as a third iterate

mark

Here's a fun problem that you can solve by applying Sharkovsky's theorem.

Suppose that [imath]f:\mathbb R \to \mathbb R[/imath] is a continuous function such that
[dmath]f\circ f \circ f(x) = x[/dmath]
for all [imath]x[/imath]. That is, [imath]f^3[/imath] is the identity function. Prove that [imath]f(x)=x[/imath].

Rex

A New Challenger appears!

I'll Give this a shot.

The Sharkovsky theorem tells us that

$$Odds > Odds * 2 > odds * 2^2 > odds * 2^3, ..., 2^3 * 1 > 2^2 * 1 > 2*1 > 1$$

so we've got $f$ and $f^3$, and we know that $f^3 =x $.

We know that f has a period of 3, and Ol' Sharky tells us that if a function has a period of 3, then it has a period of every number under the sun.

That's all I got for now. I haven't quite figured this out yet.