Show $x_k = F^k(x_0)$, with any seed $x_0$, satisfies $\lim_{k\to\infty} x_k=\bar{x}$

Question

Consider the dynamical system $F(x) = ax+b$ where $a \neq 1$ and $b\in \mathbb{R}$. Show $x_k = F^k(x_0)$, with any seed $x_0$, satisfies $\lim_{k\to\infty} x_k=\bar{x}$

I found $\bar{x}$ to be $\frac{b}{1-a}$

But how does one prove as $k\to\infty$ that $x_k\to\bar{x}$?

Answer 1

By induction we see \begin{gather}F^k(x_0)=a^kx_o+b\sum_{i=0}^{k-1} a^{i}. \end{gather}So for $|a|<1$ the left summand converges to $0$ and the right summand converges to $b\frac1{1-a}=\frac b{1-a}$, as $\sum_{i=0}^k a^{i}$ is a geometric series.

Answer 2

Need $|a| < 1$, otherwise, take b = 0, a = -2, there is no limit for $x+k$ since it goes to $+/- ]infty$ alternatively . When $|a| <1$, it comes from fixed point theorem since $|F(x) - F(y)| < a |x -y| $.