Given the function $f(x) = (e^x - 1)^2$, we can use a fixed-point iteration to approximate the root.
$$x_{n+1} = x_n - \frac{(e^{x_n} - 1)^2}{2e^{x_n}(e^{x_n}-1)}$$
This gives the following iterations after an initial guess $x_0 = 1$:
$$x_1 = 0.6839$$ $$x_2 = 0.4363$$ $$x_3 = 0.2595$$$$x_4=0.1452$$ And so on. The error $E$ for each iteration is just the value of the iteration itself, given that the exact solution is $0$. My question is: How does one find both the rate and order of convergence, given these iterations? Is there a specific formula or does one try to find a pattern from the ratio of consecutive errors? Then, can you prove these claims using Taylor series about the root? Any explanations would be brilliant.
If the sequence is converging with order $p$, you have that $$ \lim_{n \to \infty} \dfrac{|z-x_{n+1}|}{|z-x_n|^p} = K_{\infty}^{[p]} $$
Imagining that $n$ is large enough (and using $z=0$), you would expect $|x_{n+1}| \approx K |x_n|^p$. In particular, $$ \frac{|x_{n+1}|}{|x_n|} \approx \frac{K|x_n|^p}{K|x_{n-1}|^p} = \left(\frac{|x_n|}{|x_{n-1}|}\right)^p. $$
From this relation you can estimate $$ p = \frac{\log(|x_{n+1}|/|x_n|)}{\log(|x_n|/|x_{n-1}|)} $$
In this situation, we have
$$ p \approx \frac{\log(|x_4/x_3|))}{\log(|x_3/x_2|)}\approx 1.17 $$
which suggests linear convergence, as expected.
We could have guessed this right from the start... The iteration process is $x_{n+1}= \underbrace{x_n+\frac 12 e^{-x_n}-\frac 12}_{g(x_n)}$ Using Taylor's formula you get
\begin{align*} |x_{n+1} - z| = & |g(x_n)-z|=|g(z) + g'(\xi)(x_n -z)|, \xi \in (z,x_n)\\ = & |g'(\xi)| |x_n-z| \end{align*}
So, when $x_n$ is close to $z$, the constant in front of $|x_n-z|$ is close to $|g'(0)| = \frac 12$.