What is the solution to the functional equation $f(f(f(x)))=x$?

Question

Is there a solution to the function $f^3(x)=x$ where $f^n(x)$ denote function composition?

I'm trying to find a function analogous to the reciprocal function - which satisfies the functional equation $f^2(x)=x$ - but with period of three instead of two.

Answer 1

Yep. Try $f(x) = 1 - \frac{1}{x}$.

Answer 2

There are actually infinitely many such functions. As the_fox pointed out, $f(x) = 1 - 1/x$ works, and is probably what you're looking for. But you can get more general: let $g(x)$ be any invertible function on $\mathbb{R}$, and let $f$ be a solution to $f^3(x) = x$. Then $h(x) = g(f(g^{-1}(x)))$ also works (you can check this).

Answer 3

Hint.-the element $x$ is not necessarily a number. For example a rotation of $120^{\circ}$ is a linear function of $\mathbb R\times\mathbb R$ on itself. So its matrix respect to all base must verify

$$\begin{bmatrix} a&b \\ c&d\end{bmatrix}*\begin{bmatrix} a&b \\ c&d\end{bmatrix}*\begin{bmatrix} a&b \\ c&d\end{bmatrix}=\begin{bmatrix} 1&0 \\ 0&1\end{bmatrix}$$ because $120+120+120=360$. And you know that there are infinitely many bases for that rotation.......

Answer 4

The (canonical) solution is $$f(x)=\frac{1}{2} \left(-i \sqrt{3} \sqrt{\left(x-c_3\right){}^2-4 c_1 c_2}+3 c_3-x\right)$$

Basically, the equation can be reduced to difference equation of 3rd order $F(x+3)=F(x)$, where $F(x)$ is the flow of the original function. It has the solution $F(x)=c_1 e^{-\frac{2}{3} i \pi x}+c_2 e^{\frac{2}{3} i \pi x}+c_3$. Now apply the anti-flow operator.

The correctness of the solution may depend on the choice of the branches of the radicals though, which I did not check.