proving the convergence of an iterative method

Question

I was reading an optimization book where for the following recursion $x_{k+1}=e^{-x_k}$ the book claims that $|z-x_{k+1}|\le r|z-x_k|$, where $r=\max(e^{-x_1},e^{-x_2})$ for $k=3,4,5,\ldots$., As a hint the book says that one can use $ 1-e^{-x} \le x$ identity to easily show it. I tried to use it but couldn't able to prove it. So any help would be appreciated.

Answer 1

Since the function $\mathrm e^{-t}$ is decreasing, you can show it has a unique fixed point, $z$, and supposing $x_1>z$, that the odd terms $x_1, x_3, x_5, \dots$ are decreasing, whereas the even terms, $x_2, x_4, x_6, \dots$, are increasing,and that, furthermore, $$x_2Mean value theorem.

Some details:

$z-x_{k+1}=\mathrm e^{-z}-\mathrm e^{-x_k}$, since $z$ is a fixed point of $\mathrm e^{-t}$, hence, by the Mean value theorem, there exists $\xi$ between $z$ and $x_k$ such that $$z-x_{k+1}=-\mathrm e^{-\xi}(z-x_k).$$ As $\xi $ lies between $z$ and $x_k$, it a fortiori lies between $x_1$ and $x_2$, so $\;\mathrm e^{-\xi}\le\max(\mathrm e^{-x_1},\mathrm e^{-x_2})$, and $$\lvert z-x_{k+1}\rvert=\mathrm e^{-\xi}\lvert z-x_k\rvert\le\max(\mathrm e^{-x_1},\mathrm e^{-x_2})\lvert z-x_k\rvert. $$