The Mandelbrot set $M$ is defined by iterations of $z \to z^2 + c$: it is the set of $c \in \mathbb{C}$ for which the orbit starting from $0$ remains bounded. Most(1) points in $M$ tend towards a $c$-specific limit cycle (the interior of the hyperbolic components), almost all points on $\partial M$ are aperiodic (pre-periodic Misiurewicz points are only countably infinite, as are the parabolic bond (and cusp) points with rational multiplier). (1) depends on whether $\partial M$ has an area, which is an open problem as far as I know.
The Burning Ship is defined similarly, by iterations of $x + i y \to (|x| + i |y|)^2 + c$. Again, many points that remain bounded tend to limit cycles, but there seems to be a larger proportion of points whose orbits are aperiodic. Key to the image (plotted with the imaginary axis increasing downwards):
- white through blue: unbounded, coloured according to exterior distance estimate
- cyan: unbounded, but the derivative w.r.t $c$ overflowed so exterior distance estimate could not be computed
- green through black: converged to limit cycle (derivative w.r.t. $(x,y)$ became very small), coloured according to multiplier
- red: maximum iteration count reached without any conclusions
Conjecture 1: aperiodic orbits are dense in the red region
Conjecture 2: there are parts of the red region without any green regions inside (no matter how small). Perhaps this can be phrased as the red region having non-empty interior?
Conjecture 3: the Hausdorff dimension of the red region is $2$
Conjecture 4: the red region has positive area
For example, plotting the first $100000$ points of the orbit of $c = \frac{1 - 3 i}{4}$ gives this shape:
And the same number of points of the orbit of $c = \frac{-3 - 1 i}{4}$ gives this shape:
Question: how can I show that the two example orbits are aperiodic?
It is possible to show that they are bounded by adapting the exact interval arithmetic techniques from:
Images of Julia sets that you can trust Luiz Henrique de Figueiredo, Diego Nehab, Jorge Stolfi, Joao Batista Oliveira http://webdoc.sub.gwdg.de/ebook/serien/e/IMPA_A/721.pdf
Ideas for resolving my conjectures would be welcome too.
Update:
I will show that the first example orbit with $c = \frac{1 - 3 i}{4} = x_1 + y_1 i$ is not (pre-)periodic. The first iterates are
$$\frac{1 - 3 i}{4}, \frac{-2 - 3i}{8}, \frac{11 - 36i}{64}, \frac{-151 - 2280i}{4096}, \frac{-981295 - 11894352i}{16777216}, \ldots$$
Suppose the iterates for $k \ge K$ for some $K \in \mathbb{N}$ are of the form $$x_k + y_k i = \frac{(4a_k + 1) + 2^{c_k}b_k i}{2^{m_k}}$$ where $x_k, y_k \in \mathbb{Q}, a_k, b_k, c_k, m_k \in \mathbb{Z}, m_k > 2, c_k > 2$ and $b_k$ is odd. Algebra gives: $$ \begin{align} x_{k+1} &= \frac{4(16a_k^2+8a_k+1-2^{2c_k}b_k^2 + 2^{2m_k-2})}{4(2^{2m_k})} \\ a_{k+1} &= 4 a_k^2 + 2 a_k - 2^{2c_k - 2} b_k^2 + 2^{2m_k-4} \\ m_{k+1} &= 2m_k \\ y_{k+1} &= \frac{4\left|2(4a_k+1)2^{c_k}b_k\right| - 3 \cdot 2^{2m_k}}{4\cdot 2^{2m_k}} \\ &= \frac{\left|(4a_k+1)2^{c_k+1}b_k\right| - 3 \cdot 2^{2m_k - 2}}{2^{2m_k}} \\ c_{k+1} &= c_k + 1 \\ b_{k+1} &= |(4a_k+1)b_k| - 3\cdot 2^{2m_k-2-(c_k+1)} \end{align} $$ where $b_{k+1}$ is odd as the sum of an odd and an even, assuming $2m_k > c_k + 3$. Tabulating values calculated from the iterates shows these assumptions are met for $K \ge 4$. As $q_k = 2^{m_k - c_k} \to \infty$, strictly increasing as $k \to \infty$, and $b_k$ is odd for all $k \ge K$, then the sequence $y_k = \frac{b_k}{q_k}$ (with $\gcd(b_k, q_k) = 1$) has every value distinct.
| $k$ | $m_k$ | $c_k$ |
|---|---|---|
| 4 | 12 | 3 |
| 5 | 24 | 4 |
To show that it does not converge to a periodic limit cycle will take more work - any ideas?