Iterated function

Remarks

By induction, $f^m \circ f^n = f^{m + n}$ and $(f^m)^n = f^{mn}$ for all $m, n \in \N$. If $f$ is a bijection, one also sets $f^{-n} = (f^{-1})^n$, and the rule $f^m \circ f^n = f^{m+n}$ then holds for all $m, n \in \Z$.

Example: for $f(x) = x^2$ on $\R$, one has $f^2(x) = (x^2)^2 = x^4$ and in general $f^n(x) = x^{2^n}$. The check is the induction step $f^{n+1}(x) = f(x^{2^n}) = x^{2^{n+1}}$.

Non-example: for $g(x) = x + 1$ on $\R$, the pointwise square $x \mapsto g(x)^2 = (x+1)^2$ is not the second iterate, which is $g^2(x) = x + 2$. The notation $f^n$ can mean either object, so texts state which one is meant; for example $\sin^2 x$ usually means $(\sin x)^2$.