## Powers of a mapping

Let $$f(x)=(1-x)^{-1}$$. Determine $$f\circ f$$ a $$f\circ f\circ f$$.
$$\displaystyle (f\circ f)(x)=f(f(x))=\frac{1}{1-f(x)}=\frac{1}{1-\frac{1}{1-x}}=\frac{x-1}{x}=1-\frac1x$$
$$\displaystyle (f\circ f\circ f)(x)=f((f\circ f)(x))=\frac{1}{1-\left[1-\frac1x\right]}=x$$
Iterating the function yields $$(f\circ f)(x)=1-\frac1x$$ and $$(f\circ f\circ f)(x)=x$$.