# is the Buddhabrot well-defined?

Define the Mandelbrot set $M = \{ c \in \mathbb{C} : P_c^n(0) \not\to \infty \text{ as } n \to \infty \}$ where $P_c(z) = z^2 + c$.

Define the complement of the Mandelbrot set $\overline{M} = \mathbb{C} - M$.

Define the escape time $N : \overline{M} \to \mathbb{N}$ by $N(c) = \min\{ n \in \mathbb{N} : |P_c^{n+1}(0)| > R \ge |P_c^{n}(0)| \}$ where $R = 2$.

Define level sets (dwell bands) $L_n = \{ c \in \overline{M} : N(c) = n \}$, call the area of each $A_n$.

Then the area of $M$ is $A = \pi R^2 - \sum_{n=1}^\infty A_n$, so $A_n \to 0$ as $n \to \infty$.

The Buddhabrot visualization accumulates $N(c)$ points for each $c \in \overline{M}$. Weighting points plotted by the area they represent would be an appropriate normalization for taking the limit as the number of samples and iteration count goes to infinity.

Claim: the Buddhabrot is well-defined only if $\sum_{n=1}^\infty n A_n$ is well-defined.

Question: is my claim valid?

Answer: no, as pointed out to me on IRC:

Even if $\sum_n n A_n$ diverges, it's plausible that the (normalized) distribution corresponding to the Buddhabrot where you limit to $N$ iterations converges to some distribution as $N \to \infty$. Moreover, even the unnormalized distribution could maybe plausibly converge to some interesting measure, the fact that $\sum_n n A_n$ diverges just means that the measure of the whole picture is infinite; maybe the infinite bits are "localized" somehow. Here's a maybe useful rephrasing of your question: picking $c$ outside the Mandelbrot set uniformly at random, does the escape time corresponding to $c$ have finite expectation?

Question: is the Buddhabrot well-defined? (in the sense of a convergent distribution / finite expectation)

My hunch says "yes", as exhibited by this image showing the dwindling contribution between increasing $2^n$ iteration thresholds: