Iterated function ($F^n$)

For a map $F:X\to X$, its iterates are defined by $F^0$ equal to the identity map and $F^{n+1}=F\circ F^n$.

For a partial function $f:D\to X$ with $D\subseteq X$, its $n$th iterated function is defined only when each intermediate application is defined. Put $D_0=X$, $f^0={\mathrm{id}}_X$, and use the displayed recursion for the domain of a function. Simplifying an algebraic expression for $f^n$ does not restore points excluded by an intermediate application. For example, $f(x)=1/x$ has $f^2(x)=x$ on $\mathbb{R}\setminus \{0\}$; the simplified expression $x$ has a larger natural domain of a function.

The sine fixed-point iteration converges to zero from every real starting value. After one step, $|x_n|\le 1$, and $|x_{n+1}|=\sin |x_n|$. Since $0<\sin t<t$ for $0<t\le 1$, the absolute values form a bounded monotone sequence. Its limit $a$ satisfies $a=\sin a$ by continuity, forcing $a=0$. This proves convergence even though the sine is not a contraction mapping on any interval containing zero in its interior.

A map on $(0,1)$ can satisfy $0<f(x)<x$ and still have no orbit tending to zero. Partition the domain into $I_j=(1/(j+1),1/j]\cap (0,1)$ and define $f(x)=(x+1/(j+1))/2$ on $I_j$. Every interval is invariant under the iteration of a map, and $f^n(x)=1/(j+1)+2^{-n}(x-1/(j+1))$ has a positive limit. The discontinuities at interval boundaries prevent that limit from being a fixed point. In contrast, continuity of a self-map satisfying $f(x)<x$ permits passage to a positive orbit limit and would force an impossible fixed point.