I'm curious, in the Mandelbrot set, why is the escape radius $2$? I've seen few proofs of that on the internet, but i can't understand them enough.
Why is the bailout value of the Mandelbrot set 2?
Mandelbrot sets and radius of convergence
https://mrob.com/pub/muency/escaperadius.html
Some of the statements in them seem "out of the blue" for me.
For example, in the second in-site link I gave above: $ |c|≤2 \Rightarrow|z_n+1|≥|z_n|2−|c|>2|z_n|−2$
Where does $2|z_n|−2$ come from?
From the third link I interpeted the proof like this :
What we want to proof is the criteria $|z_n| \le 2$.
Let $|z_n|>2$ and $|z_n|>|c|$. This is done to create a situation where the ratio from $|z_{n+1}|$ to $|z_n|$ is always greater 1 so the next term in the sequence always gets bigger than the one before (sequence is unbound) : $\frac{|z_{n+1}|}{|z_n|}>1$.
So we now have to prove $\frac{|z_{n+1}|}{|z_n|}>1$ :
In the linked article they first use the triangle inequality on the term (https://en.wikipedia.org/wiki/Triangle_inequality) :
$\frac{|z_{n+1}|}{|z_n|} = \frac{|z_n^2+c|}{|z_n|} \ge \frac{|z_n|^2-|c|}{|z_n|} = |z_n|-\frac{|c|}{|z_n|}$
From $|z_n|>|c|$ we get $\frac{|c|}{|z_n|}<1$ so :
$|z_n|-\frac{|c|}{|z_n|} > |z_n|-1$
From $|z_n|>2$ we get :
$|z_n|-1 > 1$ , so all in all we have proven $\frac{|z_{n+1}|}{|z_n|}>1$
This inequality is now true (and the sequence unbound) for all $|c| \le 2 < |z_n|$.
But what if $|c|>2$ (second case)? Then we can show that the sequence always "escapes" at least after 2 iterations : $z_{n=2}=|c^2+c| \ge |c|^2-|c| > 2$.
If we put this all togehter we get the criteria for the sequence to be unbound if $|z_n| > 2$ , so the criteria for the sequence to be bound (number in the mandelbrot set) is : $|z_n| \le 2$.
PS : I'm a student for myself and not an expert. As I mentioned above this is just my interpretation of the proof in the linked article. If some expert comes along this thread pls have a quick look over my explaination and if everything is correct.
PPS : Sorry for my English im from Germany xD