Mandelbrot Images generated with “optimized_Fractal_inner_estimate.py”
$r(N)$ calculated with “rN_List.py”

Introduction

The Mandelbrot set $M_0$ is the set of complex numbers such that for $f(z,c)=z^2+c$ the sequence $z_0(c)=f(0,c),z_1(c)=f(f(0,c),c),\dots,z_k(c)=f(z_{k-1},c),\dots$ doesn’t diverge to infinity.
A starting value $c$ is not in $M_0$ if $|z_k(c)|>2$ for some $k\in \mathbb N_0$.
When approximating this set using a computer, we give each Pixel a coordinate in the complex numbers and calculate the sequence up to some Iteration $N$ and look at all numbers $c$ such that $\forall k\in \mathbb{N}_0$ we have $|z_k(c)|\le 2$.

Since $|z_k(c)|>2\Rightarrow|z_{k+1}(c)|>2$ it suffices to discard any points $c$ at iteration $0\le n\le N$ for which $|z_n(c)|>2$ holds.
Therefore the following definitions are justified:

where $M_0\subset M_0(N+1)\subset M_0(N)\subseteq \overline{B}_2(0)$.

We know, that $M_0$ is compact and connected (Douady and Hubbard, or Topological proof by Jeremy Kahn), and simply connected. We here assume that $M_0(N)$ is also compact, connected (which follows from the proof of Jeremy Kahn) and simply connected.
For our (bad) Optimization on finding points inside the set we define

and $r(N):=r^{M_0}(N):=\inf R_N$. We note, that $\forall N\in\mathbb N_0$ we get $0<r(N)\le 2$
If $r(N)=0$ for some $N$, then $M_0(N)=M_0$. For a variable $x$ $z_k(x)$ is a Polynomial and therefore holomorphic (and continuous).

Lemma 1.
$\partial M_0 \subset M_0\subsetneq M_0(N+1)\subsetneq M_0(N)\subseteq \overline{B}_2(0)$

Proof.
We know $\{z_k(c):k\in [N]\}\subset \{z_k(c):k\in [N+1]\}\subset {z_k(c):k\in \mathbb N_0}$ and $M_{0}(0)=\overline{B}_2(0)$.

$\partial M_{0}\subset M_{0}$ follows from $M_{0}$ being compact.

Now, to show that $M_0(N+1)\subsetneq M_0(N)$, we can construct a sequence of points such that $x_k\in M_0(k)\land x_k\not\in M_0(k+1)$ as follows:
We use the fact that since $z_k(x)$ is a polynomial, so holomorphic it is continuous and therefore for every iteration and any point $x_1\in M_0\cap\mathbb R_+$, for example 0 or 0.5 and $x_2\not\in M_0\cap\mathbb R_+$, for example 4, we have some point $y\in [x_1,x_2]$ such that $|z_k(y)|=z_k(y)=2$.
Now since $|y|>0$ we get, that $z_{k+1}(y)=z_k(y)^2+y=4+y>2$, so $y\in M_0(k)$ and $y\not\in M_0(k+1)$. $\square$

Lemma 2. Let $N\in\mathbb N_0$, then $R_N\ne\emptyset$

Proof. Let $r>2$ and $c\in \mathbb C$, then $B_r(c)\cap (\overline{B}_2(0)^c)\ne \emptyset$, because else $B_r(c)\subseteq \overline{B}_2(0)$.
Therefore $B_r(c)\not\subset M_0(N)$, so $(B_r(c)\subset M_0(N)\Rightarrow c\in M_0)$ holds trivially.  $\square$

This also shows, that $r(N)$ is well-defined and $0\le r(N)\le 2$

Lemma 3. $\partial M_0(N)\cap\partial M_0(N+1)={-2}$

Proof. Let $c:=r_c e^{i\varphi_c}\in \partial M_0(N)\cap \partial M_0(N+1)$ and $z_k(c):= re^{i\varphi}$.
So $|z_k(c)|=r=2=|z_{k+1}(c)|=|z_k(c)^2+c|=|r^2e^{i\varphi}+r_ce^{i\varphi_c}|=|4e^{i\varphi}+r_ce^{i\varphi_c}|\ge 4-r_c$.
And therefore $r_c=2$. Since $c\in M_0(N)$ $\forall N\in [k+1]$, so $c\in M_0(1)$ we get $c=-2$

Lemma 4. Let $N\in\mathbb N_0$, then $M_0(N)\setminus M_0\ne \emptyset$

Proof. Assume $M_0(N)\setminus M_0 = \emptyset$, then $M_0\subseteq M_0(N+1)\subseteq M_0(N)\subseteq M_0$, which is a contradiction to $\partial M_0(N)\cap\partial M_0(N+1)={-2}$. $\square$

Lemma 5. Let $N\in\mathbb N_0$, then $0<r(N)$

Proof. Assume $r(N)=0$ for some $N$, we want to show $\partial M_0(N)\cap \partial M_0(N+1)\supsetneq \{-2\}$ which is a contradiction.
By definition $\forall r>0,\ c\in M_0(N):\ \overline{B}_r(c)\subset M_0(N)\Rightarrow c\in M_0$.
So $\forall c\in M_0(N)\setminus M_0,\ r>0$ we know, that $\overline{B}_r(c)\not\subset M_0(N)$, so $c\in \partial M_0(N)$.
So $M_0(N)\setminus M_0 \subset \partial M_0(N)$.
Let $z\in M_0(N+1)\setminus M_0\ne \emptyset$, then $z\in \partial M_0(N+1)\subset M_0(N+1)\subset M_0(N)$, so $z\in M_0(N)\setminus M_0\subset \partial M_0(N)$, so $z\in \partial M_0(N)\cap \partial M_0(N+1)$, but $z\ne -2$ which is a contradiction to the boundaries only intersecting in $-2$. $\square$

Lemma 6. Let $N\in\mathbb N_0$, then $r(N)>r(N+1)$

Proof. $r(N)\ge r(N+1)$ as $M_0(N)\supset M_0(N+1)$.
Assume $r(N)=r(N+1)$, we want to show $r(N)=0$ which is a contradiction.
Then there is a $c\in \partial M_0$ such that $B:=\overline{B}_{r(N)}(c)\subset M_0(N+1)$ and $\exists z\in \partial M_0(N+1)\cap B$ with $B\subset M_0(N)$ and $z\in \partial M_0(N)\cap B$.
We get, that $z\in \partial M_0(N+1)\cap \partial M_0(N)$, so $z=-2$ and therefore $r(N)=0$, which is a contradiction to $r(N)>0$. $\square$

Lemma 7. $r(N)\overset{N\to\infty}{\longrightarrow}0$

Proof. Since $r(N)$ is strictly decreasing and at least $0$, we know, that it must converge.
Assume there is some $\varepsilon>0$ such that $R(N)\to \varepsilon$, so $R(N)>\varepsilon$ for all $N\in\mathbb N_0$.
Then there exists some $c\in M_0^c$ such that $d(c,M_0)=\varepsilon$ and $c\in M_0(N)$ for all $N\in\mathbb N_0$, which is a contradiction. $\square$

Conjecture 8. Let $N\in \mathbb N_0$, then $-0.75\in {c\in \partial M_0: \overline{B}_{r(N)}(c)\subset M_0(N)}$

The goal of this project is to find upper and lower bounds for $r(N)$.
Upper bounds are useful, because with them, we know, that any point that hasn’t escaped at iteration N is within at most $r(N)$ distance from $M_0$, and assuming infinite Resolution, we can take any point inside $M_0(N)$ with a distance to $\partial M_0(N)$ of at least $r(N)$ (or at least our upper bound) and know they are in the set $M_0$, therefore leaving them out of any further computing. Of course we do not have infinite Resolution, but it still gives us a good approximation and optimization. (Which is worse then border tracing, as border tracing doesn’t check any points of $M_0$ apart from the border)

Conjecture 9. For $N\ge 12$ it holds, that $\frac{3}{N}\le r(N)\le \frac{\pi}{N}$
a more precice, though still rough upper bound would be:

While trying to prove the theorems, and optimizing the code, I encountered the following equivalent formulations/characterizations of $r(N)$. I use the next characterization in my code because it is easier to work with, though I couldn’t prove it yet.

Here is the direction that I could prove:

Proof. Assume the set is empty, then $\forall \varepsilon>0, \forall c\in \partial M_0$ we get, that

$\overline{B}_\varepsilon(c)\not\subset M_0(N)$, since $M_0(N)$ is compact we would get $\partial M_0\subseteq \partial M_0(N)$ which is a contradiction to $\partial M_0(N)\ne \{-2\}$:

Let $c\in\partial M_0\setminus\{-2\}$, then $c\in M_0(N)$, we show $c\in\partial M_0(N)$ by constructing a sequence $x_m$ of elements in $M_0(N)^c$ that converge to $c$.

We know, that $\forall m\ge 1, \overline{B}_{\frac{1}{m}}(c)\cap M_0(N)^c\ne \emptyset$.

So let $x_m\in \overline{B}_{\frac{1}{m}}(c)\cap M_0(N)^c$, then since $\frac{1}{m}\overset{m\to\infty}{\longrightarrow} 0$, that $x_m\overset{m\to\infty}{\longrightarrow} c$, therefore $c\in\partial M_0(N)$, but $c\ne -2$ which is a contradiction.

Let $r:=\max \{\varepsilon>0:\exists c\in \partial M_0\text{ with }\overline{B}_\varepsilon(c)\subset M_0(N)\}$. To show: $r\le r(N)$: Assume $r> r(N)$.

Then there exists some $c\in\partial M_0 \text{ s.t. } \overline{B}_r(c)\subset M_0(N)$ and some $c’\in M^c_0\cap \overline{B}_{r-r(N)}(c)$

(because $c\in\partial M_0\text{, so } \forall \varepsilon>0\ \exists c’\in M_0^c\cap B_\varepsilon(c)$),

but $\overline{B}_{r(N)}(c’)\subset \overline{B}_r(c)\subset M_0(N)$, so $c’\in M_0$ which is a contradiction. $\square$

The direction we would still need to show is $r\ge r(N)$. or equivalently

To give some experimental “evidence” for Conjecture 8 using code: “Dist_local_maxima.py”.
In the code, we look at the distance of any Point in $M_0(N)$ to $\partial M_0(N)$. (More accurately, we look at the distance to $\partial M_0(N)\cap Im$, where $Im$ is the Box of our Image).
We then take locally maximal Points in $M_0\setminus int(M_0)$ and $M_0$ and colour them accordingly.
(for $M_0(N)\setminus int(M_0)$ we colour them orange to red increasingly, for $M_0$, we colour them blue to violet).
Also visible in the following Images is the violetly coloured circle (that is maximal in $M_0(N)$) around the mouse-cursor, here positioned on $-0.75$.

Looking at the images, we can see, that any point that connects two sections (so a point, such that removing it disconnects the set) is a local maximizer of the distance function. The “larger” the “shapes”, the greater its distance value. $-0.75$ is the point that connects the largest two shapes, the main cardioid, and $1/4$ sphere around $-1$.

Now to give some experimental “evidence” for Conjecture 9 using code: “rN_List.py”.
The algorithm without optimizations works like this:
given the Iteration $N$, we want to approximately calculate $r(N)$.
We do a Binary search of the radius between $r_l,\ r_h$ Given, that the Conjecture 8 holds, and our characterization $r_2(N)$ does too, we check if $\partial B_r(-0.75)\subset M_0(N)$ for a radius $r:r_l \le r \le r_h$.

If it is, we try a bigger radius, if it is not, we try a smaller radius (using Binary Search, we half our step each time until it is smaller then our threshold $\varepsilon$).
Since we can only calculate finite Points on $\partial B_r(-0.75)$, we have a step_amount, and take step_amount many evenly spaced samples of $(\alpha_l, \alpha_h)$.

In the following graphs, we can see, that $\frac{1}{r(N)}-\frac{1}{r^{upp}(N)}\ge 0$, so therefore we know, that $r^{upp}(N)-r(N)\ge 0$, so $r^{upp} (N)$ is an upper bound for $r(N)$, and that it goes towards $0$.