In the FPI method, we are trying to come up with some $g(x)$ such that $f(x)$ can represented as:
$f(x) = x - g(x)$, correct?
What I don't understand is how these translate to: $$ x_{k+1} = g(x_k)? $$ Can someone provide a more intuitive explanation for this?
On
If the sequence defined inductively by
$$x_{k+1}=g(x_k)$$ converges, then it converges to a value $x_\infty$ such that
$$x_\infty=g(x_\infty),$$
which is a root of $f$ (because$f(x_\infty)=x_\infty-g(x_\infty)=0$).
Establishing convergence is another story. But an example shows that it can work. Consider e.g. $x=\cos x$, starting the iterations from $x=1$:
$$1.0 \\ 0.5403023058681398 \\ 0.8575532158463933 \\ 0.6542897904977792 \\ 0.7934803587425655 \\ 0.7013687736227566 \\ 0.7639596829006542 \\ 0.7221024250267077 \\ 0.7504177617637605 \\ 0.7314040424225098 \\ 0.7442373549005569 \\ 0.7356047404363473 \\ 0.7414250866101093 \\ 0.7375068905132428 \\ 0.7401473355678757 \\ 0.7383692041223232 \\ 0.739567202212256 \\ 0.7387603198742114 \\ 0.7393038923969057\\ \cdots$$
Convergence is slow, though. It takes $90$ iterations to get to the final value
$$0.7390851332151607$$
You have to consider this the other way around. You want some process $x_{k+1}=g(x_k)$ that in each iteration step gets closer to the root of $f$. "Getting closer" includes that the sequence converges, the limit is a fixed point $x^*=g(x^*)$.
Now you need that $x-g(x)=0$ is equivalent to or at least implies that $f(x)=0$, which would be trivially true if $x-g(x)=f(x)$. But usually one takes $g(x)=x-h(x)f(x)$ where $h(x)$ is non-zero in the region under consideration where the root of $f$ is expected. $h$ can be selected to ensure that $g$ is contracting.