we have fixedpoint iteration $x_{k+1} = 1 + e^{-x_k} $. I need to prove that that the iteration converges for any starting point $x_0\in [1,2]$ and determine its convergence rate and estimate the number of iterations it will take to achieve $10^{-5}$.
Let $g(x) = 1 + e^{-x}$ so we have scheme $x_{k+1} = g(x_k)$ We have $g'(x) = -e^{-x}$ and on $[1,2]$ $|g'(x^*)| < e^{-1} < 1 $ and so the iteration converges in that interval. Im stuck on the second part. By definition, the convergent rate of an interation is given by
$$ \lim \frac{ |E_{k+1}| }{|E_k|^r } = C $$
where $E$ is the error and $C> 0$ is constant. In our case,
$$ |E_{k+1}| =| x_{k+1} - x^*| = |g(x_k) - g(x^*)| < g'(\xi )|x_k-x^*|<|E_k| $$
since $g'(anything) < 1$ So
$$ \frac{ |E_{k+1} |}{|E_k|} < 1 $$
So $r=C=1$ can we conclude then that the rate of convergence is linear? how do we find the number of iteration to achieve $10^{-5}$?
The rate of convergence is certainly linear because $g(x)$ can be approximated by a linear function when $x$ is very close to $x^*$. You have already written that $|E_{n+1}|=|x_{n+1}-x^*|=|g'(\xi)||x_{n}-x^*|$, where $\xi\in[x_n,x^*]$ or $\xi\in[x^*,x_n]$. Rate of convergence can be calculated in the following way: $$ \lim_{n\to \infty}\frac{|E_{n+1}|}{|E_{n}|}=\lim_{\xi\to x^*}{|g'(\xi)|}={x^*-1}. $$ (Note that $g'(x^*)=-e^{-x^*}=1-x^*$)
The number of iteration needed to reach an error of $10^{-5}$ depends on the starting point(if you start with $x^*$ then NO iteration is needed!). But let's suppose that $x_0=2$. Then $|E_0|$ is less than $1$. We can check that $x^*<1.28$. So $\lim_{n\to \infty}\frac{|E_{n+1}|}{|E_{n}|}<0.28$ for large $n$ in general. So we want to have $0.28^n<10^{-5}$. This means $n>9$. So about a dosen iterations are needed.