How to find $x_n$ from $x_{n+1} = \frac{x_n}{1-a+a x_n}$? For $n\geq 0$ let $x_{n+1} = \frac{x_n}{1-a+a x_n}$, where $a\in (0,1)$.
I would like to know if it is possible to express $x_n$ as a function of $a$ and $x_0$ for all $n\geq0$.

Note that $$\frac1{x_{n+1}}=\frac{1-a}{x_n}+a$$ which implies that $$\frac1{x_{n+1}}-1=(1-a)\left(\frac1{x_n}-1\right)$$ hence $$\frac1{x_n}-1=(1-a)^n\left(\frac1{x_0}-1\right)$$ from which an explicit formula for $x_n$ in terms of $n$, $a$ and $x_0$ follows.
(Of course, there is some well known theory behind all this...)
