WARNING: This blog post involves mathematics.

Hello, Robloxians. You may remember me for making a three part blog post series (proving that in game mechanics in Robloxia are unrealistic)and one thread on Dizzy that proves that Robloxia is unrealistic. Today, we take a break from all those subjects and talk about the Mandelbrot Set. Specifically, the one topic in mathematics that I researched outside of Roblox Cities Wiki. Here is a summary of my research:

Note: All differential equations mentioned were solved using this defined constant: ${\displaystyle c_{1}=1}$

This page is a WIP. As more discoveries are found, this page will update.

Newton Fractal: Just a Julia Set?

The Newton Fractal is essentially a Julia Set. Specifically, the Newton fractal of ${\displaystyle f \left( z \right)}$ is similar to the Julia Set of ${\displaystyle z-\frac{f\left(z\right)}{f'\left(z\right)}}$. Now, one might wonder if it is possible to have the Julia Set of ${\displaystyle g\left(z\right)}$ and convert to a Newton fractal of some function. Turns out it is possible, but it is complicated, and requires solving the differential equation ${\displaystyle z-\frac{f\left(z\right)}{f'\left(z\right)}=g\left(z\right)}$.

A specific ${\displaystyle g\left(z\right)}$

Now we are curious to see what happens if we let ${\displaystyle g\left(z\right)=z^{2}+c}$, where ${\displaystyle c}$is an arbitrary mathematical constant that can be made complex if desired. Finding ${\displaystyle f \left( z \right)}$ is equivalent to solving the differential equation ${\displaystyle z-\frac{f\left(z\right)}{f'\left(z\right)}=z^{2}+c}$ for ${\displaystyle f \left( z \right)}$. We will spare all the steps to you that ${\displaystyle f\left(z\right)=\mathrm{e}^{-\frac{2\left(\arctan\left(\frac{2z-1}{\sqrt{4c-1}}\right)+\arctan\left(\frac{1}{\sqrt{4c-1}}\right)\right)}{\sqrt{4c-1}}}}$.

Why ${\displaystyle f \left( z \right)}$?

The function ${\displaystyle f \left( z \right)}$ has special branch points, namely two of them: ${\displaystyle \frac{1-\mathrm{i}\sqrt{4c-1}}{2}}$ and ${\displaystyle \frac{\mathrm{i}\sqrt{4c-1}+1}{2}}$. Only the former one interests us. Twice the former one, or

${\displaystyle \mathrm{i}\sqrt{4c-1}+1}$, is the length of the Julia Set (this is primarily due to the Julia Set of ${\displaystyle g\left(z\right)=z^{2}+c}$ being symmetric on the complex axis). This function has domain ${\displaystyle -2\le c\le\frac{1}{4}}$(complex values of ${\displaystyle c}$ tend to make this complex, something we will not even consider in this blog post) , which is coincidentally when the Julia Set of ${\displaystyle g\left(z\right)=z^{2}+c}$ is connected. Let's define the length function of the Julia Set to be ${\displaystyle L\left(c\right)=\sqrt{1-4c}+1}$.

We are also interested on which Julia sets have integral lengths. To do so, we calculate the inverse of ${\displaystyle L}$ to be ${\displaystyle L^{-1}\left(c\right)=\frac{1}{4}-\frac{\left(c-1\right)^{2}}{4}}$. This function links length of the Julia Set to the c. Doing so, we find out that the lengths of the Julia Set can be ${\displaystyle 1,2,3,4}$, corresponding one-to-one with ${\displaystyle c=\frac{1}{4},0,-\frac{3}{4},-2}$, respectively. The first one in the set corresponds to the main Mandelbrot set's cusp. The second refers to the origin. The third refers to the main Mandelbrot's seahorse valley, and the fourth refers to the tip of the Mandelbrot Set. This appears to be a ${\displaystyle 5:3:1}$ ratio.

Notable lengths

${\displaystyle c}$	${\displaystyle L\left(c\right)}$
${\displaystyle -2}$(tip)	${\displaystyle 4}$
${\displaystyle \frac{-\frac{\sqrt[3]{25-3\sqrt{69}}}{\sqrt[3]{2}}-\frac{\sqrt[3]{2}}{\sqrt[3]{25-3\sqrt{69}}}-2}{3}}$(nucleus of largest minibrot)	${\displaystyle \sqrt{1-\frac{4\left(-\frac{\sqrt[3]{25-3\sqrt{69}}}{\sqrt[3]{2}}-\frac{\sqrt[3]{2}}{\sqrt[3]{25-3\sqrt{69}}}-2\right)}{3}}+1}$
${\displaystyle -\frac{7}{4}}$(cusp of largest minibrot)	${\displaystyle 2^{\frac{3}{2}}+1}$
${\displaystyle -\frac{5}{4}}$(border between period 4 - period 2 )	${\displaystyle \sqrt{6}+1}$
${\displaystyle -1}$(between period 2 and period 1 )	${\displaystyle \sqrt{5}+1}$
${\displaystyle -\frac{3}{4}}$(seahorse)	${\displaystyle 3}$
${\displaystyle 0}$(origin)	${\displaystyle 2}$
${\displaystyle \frac{1}{4}}$(cusp)	${\displaystyle 1}$