Notation of iterated composition of functions

Question

Let $f$ be a function from $A$ to $B$ such that the image $f(A)\subset A$. Is there a widely accepted notation for the expression $$f\circ\left(f\circ\cdots(f\circ f)\right),$$ where $f$ composite with itself $n$ times? I failed to find a natural way to include the information $n$ in the notation.

Answer 1

You wrote $f(B)\subset A$ but probably meant to write something like $B\subseteq A$ (or $f(A)\subseteq A$).

I would plead for presenting $f$ not as a function $A\to B$ but as a function $A\to A$.

This because $f\circ f$ is only properly defined if the codomain of $f$ coincides with the domain of $f$.

The collection $M$ of functions $f:A\to A$ can be looked at as a monoid with the map prescribed by $a\mapsto a$ as identity and composition of functions as (associative) multiplication.

In a monoid $\langle M,\circ\rangle$ it is quite common to write $f^n$ for expression $f\circ\cdots\circ f$ containing $f$ exactly $n$ times.

So this is a nice justification for that notation.

Answer 2

A cleaner way to define the function above and the notation is presented below.

Let $f: A \longrightarrow A$ and for $n \in \mathbb{N}$, let $$ f^n(x) := \begin{cases} f(x), & n=1 \\ f(f^{n-1}(x)), & n>1. \end{cases} $$

A straightforward proof by induction shows that $f(f^n(x)) = f^n(f(x))$ for every natural number $n$.

Some have suggested the notation $f^{(n)}$ but, as noted in other comments, this is usually reserved for the $n^\text{th}$-derivative of $f$, which is also defined recursively.