I was recently reminded of the following cute fact which I will state as a proposition to fix notation:

Proposition Given $\epsilon > 0$, let $c = -3/4 + \epsilon i \in \mathbb{C}$ and $q_c(z) = z^2 + c$. Define the sequence of polynomials $q_c^n$ inductively by $q_c^0(z) = z$ and $$ q_c^{n+1}(z) = q_c(q_c^n(z)). $$ Since $q_c^n(0) \to \infty$ as $n \to \infty$, $$ N(\epsilon) = \min\{n ~:~ |q_c^n(0)| > 2 \} $$ is well-defined. The cute fact is that: $$ \epsilon N(\epsilon) \to \pi, $$ as $\epsilon \to 0$.

The next-simplest example of this phenomenon uses $c = 1/4 + \epsilon$. If we define $N(\epsilon)$ similarly then this time it turns out that: $$ \sqrt{\epsilon} N(\epsilon) \to \pi, $$ as $\epsilon \to 0$.

The points $-3/4$ and $1/4$ are the "neck" and "butt", respectively, of the Mandelbrot set and there are various other examples of these "$\pi$ paths": they are suitably-parameterised curves simultaneously tangent to two bulbs of the Mandelbrot set where they meet it. In fact I'd guess any part of the boundary where a bulb bubbles off works and so there are infinitely many such paths. External rays seem the obvious paths and they have a natural parameterisation. (Note that although $\epsilon \mapsto -3/4 + \epsilon i$ is not an external ray, it is asymptotic to one, as least set-wise, which is really all that matters.)

If true, I'm sure this is all in the literature (perhaps implicitly) but I don't know the field and after of searching through some lovely papers I couldn't find what I wanted. A naive guess is that something like the following might be true:

Half-serious conjecture Let $M \subset \mathbb{C}$ be the Mandelbrot set, $\overline D \subset \mathbb{C}$ the closed unit disc and $\Phi : \mathbb{C} - M \to \mathbb{C} - \overline D$ the unique conformal isomorphism such that $\Phi(c) \sim c$ as $c \to \infty$. Let $e^{i\theta} \in S^1$ (for $\theta$ rational) and let: $$ \alpha : (1, \infty) \to \mathbb{C} - M\\ r \mapsto \Phi^{-1}(re^{i\theta}) $$ Let $c = \alpha(r)$ and define $N(r)$ as above, then: $$ N(r) \sim \pi / f_\theta(r) $$ for some function $f_\theta$, such that $f_\theta(r) \to 0$ as $r \to 1$ and in particular $f_0(r) \sim \sqrt{\alpha(r)-1/4}$ and $f_{1/3} \sim -i(\alpha(r)+3/4)$.

Here then, at last, is my question:

Question Is some suitably-modified form of the above conjecture correct and is there a proof in the literature toward which somebody could direct me?

Aside from general curiosity my motivation stems from the fact that it's not too hard to prove the stated results for the $c = -3/4 + \epsilon i$ and $c = 1/4 + \epsilon$ paths separately but I'd like to have a unified proof. E.g., to prove the result for the (easier) $c = 1/4 + \epsilon$ path we consider: $$ z_{n+1} = z_n^2 + 1/4 + \epsilon\\ \Rightarrow z_{n+1} - z_n = (z_n - 1/2)^2 + \epsilon $$ Then approximate: $$ \frac{dz}{dn} \simeq (z-1/2)^2 + \epsilon\\ \Rightarrow z(n) = 1/2 + \sqrt{\epsilon}\tan(\sqrt{\epsilon}n)\\ \Rightarrow \sqrt{\epsilon}N(\epsilon) \simeq \tan^{-1}(\frac{3}{2\sqrt{\epsilon}} ) - \tan^{-1}(-\frac{1}{2\sqrt{\epsilon}}) $$ And so provided we can justify the approximation is accurate as $\epsilon \to 0$ (which is tricky but possible) the result follows.

  • 2
    $\begingroup$ This doesn't answer your very interesting question (and surely you are already aware of what follows), but a good reference for this $\pi$ phenomenon is the paper " $\pi$ in the Mandelbrot set" by Klebanoff. In the conclusion, Klebanoff conjectures that there are infinitely many routes at each of the infinitely many pinches of the Mandelbrot set which lead to $\pi$ in this way. $\endgroup$ Aug 19, 2015 at 19:29
  • $\begingroup$ Thank you @MalikYounsi! I was indeed aware of Klebanoff's paper but I had not looked at it in years and had forgotten his concluding remarks about infinitely-many such paths. Indeed I see he even gives a parameterisation of a path to -5/4. $\endgroup$ Aug 19, 2015 at 20:13
  • $\begingroup$ Thinking about this a tiny bit more, I could believe the right conjecture is $N(r) \sim \pi / |\alpha(r) - \alpha(1)|$ (the square root for $c=1/4$ being absorbed by parameterisation). Note $\alpha(1)$ exists because rational rays land. This should be easy to decide one way or another by numerical calculation, if all else fails. $\endgroup$ Aug 20, 2015 at 1:05

1 Answer 1


These results (which are indeed cute - I hadn't seen them before) are well-explained by the theory of parabolic explosion, which is by now classical. Indeed, for the Mandelbrot set, I think that the relevant statements were known to the experts already in the 1980s; they may already be contained implicitly in the Orsay notes by Douady and Hubbard. For an introduction, see Shishikura, Bifurcation of parabolic fixed points, in The Mandelbrot set, Theme and Variations.

Let me focus on the case of "primitive" parabolics, i.e. the "cusps" of Mandelbrot copies, with $c=1/4$ being the simplest case.

Then (see Theorem 3.2.2 of Shishikura's article) - under perturbations as you describe - the number of iterates to escape through the "eggbeater" dynamics into a prescribed part of the basin of infinity is essentially $$N(c) \approx \frac{1}{|\alpha(c)|},$$ where $\lambda(c) = e^{2\pi i \alpha(c)}$ is the multiplier of one of the orbits bifurcating from the parabolic fixed point. (This multiplier is a holomorphic function defined on a double cover around the parabolic; hence the square root in the expression.)

So, for perturbations $c=1/4+\epsilon$, the repelling fixed point has multiplier $$\lambda(c) = 1 + 2\sqrt{-\epsilon}$$ by an elementary and well-known calculation. So $$N(c) \approx \frac{2\pi}{|\log \lambda(c)|}\approx \frac{2\pi}{2\sqrt{\epsilon}}=\frac{\pi}{\sqrt{\epsilon}}.$$

Hence indeed $$ N(c)\cdot \sqrt{\epsilon}\to\pi.$$ For other parameters, you will get similar results, which will however depend on the derivative of the multiplier map with respect to the parameter in a suitable sense. Note that the above formulation (in terms of repelling orbits) is, in any case, the natural one, since it is independent of parameterisation.

For the satellite case, there are similar results that you can find in the literature. Again, for the bifurcation at -3/4, you get an explicit formula because you can compute the multipliers of the orbits in question directly.

  • $\begingroup$ Marvellous, thank you. Most satisfying to get an idea of what's going on at last. I look forward to reading Shishikura's article in more detail over lunch. $\endgroup$ Aug 22, 2015 at 12:10

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