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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?

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    $\begingroup$ For what it's worth, I put a bounty on one of the questions you linked. Hopefully that gets a couple of answers that explain this is a way that doesn't require as much background knowledge. $\endgroup$
    – Mr. Brooks
    Commented Feb 12, 2020 at 20:57

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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

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  • $\begingroup$ Maybe I’m missing something, but why does the triangle inequality step end with: |z|^2 - |c|, doesn’t the triangle inequality imply |z^2 + c| <= |z^2| + |c| $\endgroup$
    – user668074
    Commented Dec 6, 2021 at 11:33

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