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The Mandelbrot Set is the set of all complex values $c$ that cause the recurrence relation $z_n = z_{n-1} + c$ to remain bounded within a circle of radius 2 when iterated from $z_0 = 0$. In general, there is no way to prove that a given value of $c$ is a member of the set. Instead, we iterate the recurrence relation some large number of times until we either find some value of $n$ where $|z_n| > 2$ (proving that the current value of $c$ is not in the set) or give up and declare that the current value of $c$ is probably part of the set (or at least very close to it).

When the value of $c$ is far from the set, $z_n$ quickly escapes the bounding circle. As $c$ gets closer to the set, the number of iterations $n$ required to exceed the bounding circle grows without bound. If we look at values of $c$ on the positive real axis only and plot how many iterations it takes to exceed the bounding circle, we get a graph like this:

Graph showing how quickly the orbit of z escapes the bounding circle for real values of c greater than 0.25

If $c$ is greater than two, it only takes one iteration to escape the circle because $z_0$ is always zero and $z_1$ is always $c$. When $1 < c ≤ 2$ it takes two iterations. The next twelve transition values are approximately 0.68233, 0.5352, 0.4537, 0.4034, 0.36999, 0.34659, 0.32952, 0.31666, 0.30672, 0.29887, 0.29256, and 0.28741 — that's as far as I can get with WolframAlpha. Positive real values of $c$ less than or equal to 0.25 are part of the Mandelbrot Set and will never escape even after infinitely many iterations.

To me, that looks very much like a graph of $1/x$ with some scaling and translating (for example, the vertical asymptote is at $x = 0.25$), but I can't find an equation that fits perfectly.

Is there a formula that fits these points exactly?

UPDATE based on responses from PM 2Ring and Claude

As the distance $\epsilon$ from the boundary of the set tends towards zero on the positive real axis, the number of iterations required for escape tends towards $\frac{π} {\sqrt{ϵ}}$. As $ϵ$ increases to 1.75 along the positive real axis, the relationship changes to $\frac{2\sqrt{1.75}} {\sqrt{ϵ}}$. It seems like I need some function $v(ϵ)$ such that $\frac{v(ϵ)} {\sqrt{ϵ}}$ fits my points exactly. Based on the points I have, it looks like this:

Graph showing the value that is divided by to fit the points from the previous graph.

UPDATE 2

I calculated the iteration counts for $c = -0.75 + ϵ i$ and discovered that approaching the boundary of the set from this direction is more closely approximated by $\frac{π}{ϵ}$ than by $\frac{π}{\sqrt{ϵ}}$. As in the first approach, the actual iteration count starts out far from the function and approaches it as $ϵ$ goes to zero, though it wobbles around more than it does when approaching from the positive real axis. It seems my hope for a single function to calculate iteration counts may go unfulfilled.

Two graphs showing a comparison of approaching the boundary of the set from along the positive real axis, in blue, and from the positive imaginary direction towards the root of the main bulb, in red.

Lawton
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  • Related: https://en.wikipedia.org/wiki/Feigenbaum_constants Here's a fun little exercise. How wide is the "neck" of the Mandelbrot set? That is, what's the largest $k$ such that $-0.75\pm ki$ is in the M set? – PM 2Ring Mar 10 '21 at 20:14
  • @PM2Ring The Feigenbaum constant is related to the period-doubling region, which I don't think applies here. As for the width of the connection between the main cardioid and the main bulb, I believe that is a singular point and any nonzero $k$ will cause the orbit of $z$ to escape the bounding circle. Plotting the value of $k$ against the number of iterations required to escape produces a similar curve to the one in my question. – Lawton Mar 10 '21 at 20:59
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    Sure, the Feigenbaum constant is only marginally related to your question, which is why I posted that link as a comment, not as part of an actual answer. And of course, period doubling comes up all over the place in the M set. OTOH, there's other good info in that article which is more closely related to your question. You are correct regarding my little puzzle: the neck narrows down to the real line. And there's a nice relationship between $k$ and the number of iterations to escape which involves pi, as mentioned in that Feigenbaum constant article. – PM 2Ring Mar 10 '21 at 21:27
  • @PM2Ring I assume you're talking about the relationship described here: https://en.wikipedia.org/wiki/Mandelbrot_set#Pi_in_the_Mandelbrot_set Though I don't see a mention of that on the page for the Feigenbaum constant. That relationship does support my idea that $1/x$ is involved, though! – Lawton Mar 11 '21 at 00:40
  • Oop! You're right. Sorry about that. (I got a bit confused about which Wiki page I was on because I'm browsing with my phone). The Feigenbaum page has that cute anim of the self similarity at the neck / head. – PM 2Ring Mar 11 '21 at 14:02

1 Answers1

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Mark McClure's comment references a nice historical overview by G. Edgar.

The $c = -\frac{3}{4} \pm \epsilon i$ and $c = \frac{1}{4} + \epsilon$ relationships were found by D. Boll.

The $c = \frac{1}{4} + \epsilon$ relationship is proven by A. Klebanoff:

The $\pi$-Theorem

Choose $\epsilon>0$ and let $N(\epsilon)$ be the number of iterations required for the orbit of zero, under the map $Q_{\frac{1}{4}+\epsilon}(x)=x^2+\frac{1}{4}+\epsilon$, to exceed $2$, i.e. $$N(\epsilon) = \min_n Q_{\frac{1}{4}+\epsilon}(0)>2$$ Then $$\lim_{\epsilon \to 0^+} \sqrt{\epsilon}N(\epsilon) = \pi$$

-- "$\pi$ in the Mandelbrot set" by Aaron Klebanoff, Fractals, Vol. 9, No. 4 (2001) 393-402 https://doi.org/10.1142/S0218348X01000828

A. Klebanoff's paper also mentions Jay Hill's $\pi$-result for $-\frac{5}{4}-\epsilon^2 \pm \epsilon i$. G. Edgar's page says Jay Hill also mentioned a $\pi$-result for $-\frac{7}{4} + f(\epsilon)$ for some unspecified $f$. Another point to investigate with an easy route is $-2 - \epsilon$, though I suspect the relationship here will not involve $\pi$.

Tabulating:

$c(\epsilon)$ $N(\epsilon)$
$\frac{1}{4} + \epsilon$ $\frac{\pi}{\sqrt{\epsilon}}$
$-\frac{3}{4} \pm \epsilon i$ $\frac{\pi}{\epsilon}$
$-\frac{5}{4} - \epsilon^2 \pm \epsilon i$ $\frac{\pi}{2 \epsilon}$
$-\frac{7}{4} + f(\epsilon)$ ?
$-2 - \epsilon$ ?

All of these are parabolic apart from the last, which is a Misiurewicz point with aymptotically geometric scaling.

Claude
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  • Rearranging the limit equation, I see that $\lim_{\epsilon \to 0^+} N(\epsilon) = \pi / \sqrt{\epsilon}$. This definition of $N(\epsilon)$ fits the points I have better than any of the expressions I tried previously, but it doesn't fit perfectly. As $\epsilon$ gets bigger, the expression tends towards about $2.6 / \sqrt{\epsilon}$. To make this fit, I need to find some expression $f(\epsilon)$ that equals $\pi$ when $\epsilon = 0$ and $2 \sqrt{1.75}$ when $\epsilon = 1.75$, so that $N(\epsilon)$ is exactly $v(\epsilon) / \sqrt{\epsilon}$... – Lawton Mar 12 '21 at 20:49
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    @Lawton This relationship is also described on this site here and there's a fairly simple heuristic explanation on this web page. – Mark McClure Mar 14 '21 at 02:11