Problem:
With an initial guess of $x_0$, the fixed point iteration is given by $$x_{k+1} = e^{-x_k}, \mbox{ for } k=0,1,2,...$$ If $x^*$ is the exact solution, then the approximation error is $$e_n = x_n - x^*, \mbox{ for } n=0,1,2,...$$ Show that for a choice of $x_0$ close enough to $x^*$, we have $$e_{n+1}\approx -e^{-x^*}e_n, \mbox{ for } n=0,1,2,...$$
Thoughts:
I feel like I'm missing something really obvious. I know that the fixed point iteration is solving the equation $x=e^{-x}$. If $x^*$ is an exact solution to the equation, and we have $x_0\approx x^*$, then $x_0 \approx e^{-x^*}$. How would I proceed?
Simple Hint: What can you say about $|x_n -x^*|$ and $|x_{n+1} - x^*|$ ? (Mean Value Theorem!)