Let's say I want to find a fun way to write a number $l$. I can procede by doing so: (as I saw here https://math.stackexchange.com/questions/400535/can-we-get-just-3-from-pi/400538)
Let $f(x) = \sqrt{2lx - l^2}$. The only fixed point is $f(x) = x \Rightarrow (x-l)^2 = 0 \Rightarrow x = l$.
Now, if we show that $f(x)$ is a contraction over $\mathbb{R}$, (with the usual euclidean distance), then (using the Banach-Caccioppoli theorem) we can start with any value (say $\pi$) and write
$$...\sqrt{2l\sqrt{2l\sqrt{2l\sqrt{2l\pi - l^2} - l^2}-l^2}-l^2}... = l$$
But I'm having trouble showing that $f(x)$ is a contraction.
Also,are there some nice generalization and/or related concepts?
It isn't. Take $l=1$ as an example.
$$f(x)=\sqrt{2x - 1}$$
Then $|f(1)-f(\frac{1}{2})|=1>|1-\frac{1}{2}|$.
This doesn't rule out that it may be a contraction on certain parts of its domain, of course.
Indeed, let
$$f(x)=\sqrt{2lx-l^2}$$
First of all, look at the graph of $f$ (just use some example value for $l$). You can see that after a while, $f$ is no longer very steep. It would make sense for a function with low slope to be contracting. Specifically, if $f'<1$ between $x$ and $y$, then $f(y)-f(x)=\int^y_xf'(t)dt\leq\int^y_x1dt=y-x$, so $f$ is indeed contracting between $x$ and $y$. So the question is: when does $f'(x)<1$?
This is a simple calculus exercice so I won't bore you, but the answer turns out to be that $f'(x)\leq1$ when $x\geq l$ (assuming $l>0$), and furthermore $f$ maps $[l, +\infty[$ into itself. So $f$ is a contraction on $[l, +\infty[$ and you're fine as long as you pick a starting point greater than $l$.