## Fractal Christmas Cards and More

For Christmas, I was handing out fractal Christmas Cards (they had images made in gnofract4d with the equation beneath… I gave my mom a handbound book with pictures of fractals made in sage in it since I didn’t get the latest revision of the math textbook done in time to get that). Thus, I made many fractals. At the end, I noticed something.

First I’ll make a (IMHO, reasonable) conjecture:

$\forall n \in \mathbb{R}, n > 2;$ $k \in \{x|x\in \mathbb{C}; \lim_{N\to\infty} (z\to z^n+x)^N(x) \neq \pm \infty\}$

$\implies$

$\{x|x\in \mathbb{C}; \lim_{N\to\infty} (z\to z^n+k)^N(x) \neq \pm \infty\} \in \{\text{connected}\}$

Now, it’s often interesting to look at julia sets when they fracture, thus they make good pictures. And it is easy to find the point of fracture on the positive real number line.

The $(z\to z^n+x)-$orbits of these points exhibit a rather simple behaviour: $z$ converges to a value that satisfies $z=z^n+x$ if it is possible.

So we want to find the maximum value of $z-z^n$ (ie. $M\cdot (z\to z-z^n)^{\to}(\mathbb{R}^+)$ where $M$ is the supreumum).

This is trivial to find: $z|_{\frac{d}{dz}z-z^n=0}=\sqrt[n-1]{\frac{1}{n}}$ and if we plug that into $z-z^n$ we get $\sqrt[n-1]{\frac{1}{n}} -(\sqrt[n-1]{\frac{1}{n}})^n$