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We are interested in the boundary of the Mandelbrot set. It is closed, so does have a boundary. It seems a good idea to increase the boundary by making holes in the set. Which is easy to do. We then have not lost anything and have gained additional boundary.

The Mandelbrot set is defined by repeated application of the function $ f(z) = z^2 + c $ starting from z=0. The set is then those c for which z does not tend to infinity. Of these about a third (mostly near the origin) tend to a point, and some cycle. For example if c=-2. z=0 goes to -2, then to 2 and stays at 2. If c = i, z cycles between -1 + i and -i.

We can make holes by excluding c for which z either goes to a point or to a cycle. But I am not able to see in any detail what the resulting Mandelbrot set looks like. So this question is not theoretical. It is a request for a diagram so I can see what the Mandelbrot set looks like with these holes. I am hoping that this is sufficiently interesting to make it worthwhile for someone to produce this diagram.

There is a similar situation on the inside to the outside where we ask does z go to infinity? For any c we ask whether z has stopped moving or is cycling. We can give values of c different colours for z cycles of different lengths. There are certainly enough cycles, at least of length 2, to make this worthwhile. If a point c results in z cycling, a point nearby is also likely to result in z cycling with the same cycle length. This implies we have pools of colours.

I am putting some findings in my comments about the answer to this question.

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  • $\begingroup$ what do you mean by the perimeter of the Mandelbrot set? do you mean the perimeter of its boundary? in any case this will not be finite $\endgroup$ – Glougloubarbaki Feb 9 '18 at 21:51
  • $\begingroup$ "$c$ goes to a point" happens precisely for the interior of the main cardioid. If by "vibrate" you mean that the sequence is (eventually) periodic, this removes "only" countably many points (some of them "centres"of "bulbs" of $M$, e.g., $-1$ is the centre of the disk-shaped head). $\endgroup$ – Hagen von Eitzen Feb 9 '18 at 22:02
  • $\begingroup$ Does the "vibration" between two values for some points have any relevance except for the general case when the function does not escape within a certain number of iterations? For what proportion of points would ma "vibration" test be worth inplementing? $\endgroup$ – Weather Vane Feb 9 '18 at 22:05
  • $\begingroup$ @WeatherVane if the sequence is preperiodic, then $c$ is at a "tip" of the Mandelbrot set. if the sequence is periodic, is is a center of a component of the interior of $M$ $\endgroup$ – Glougloubarbaki Feb 9 '18 at 22:09
  • $\begingroup$ @Glougloubarbaki pls see my (with typo) edit to my previous comment. $\endgroup$ – Weather Vane Feb 9 '18 at 22:10
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It is not completely clear to me what is being asked here, but here goes.

If $c$ is in the Mandelbrot set, the sequence $z_{n+1}=z_n^2+c$ is bounded. A special case of that instance is when the sequence $z_n$ converges to a cycle, meaning that there is a periodic sequence $u_n$ such that $\lim z_n - u_n=0$. I guess this is what you mean by vibrating. An even more particular case is when the sequence $u_n$ is periodic of period 1 (in other words, constant), in which case $c$ belongs to the heart-shaped part of the Mandelbrot set (called the main cardioid). There are other kinds of points in the Mandelbrot set, where the sequence $(z_n)$ has more complicated behaviour.

The set of those $c$ such that $z_n$ converges to a cycle contains an union of infinitely (countably) many "bulbs". A central conjecture in complex dynamics is that that those bulbs (called hyperbolic components) are exactly the interior of the Mandelbrot set, that is any component of the interior of the Mandelbrot set is made of $c$ such that the corresponding $z_n$ converges to a periodic point.

So it is probable (but not known) that in some sense most points in the Mandelbrot set do satisfy this property.

However if you restrict your attention to parameters $c$ such that the corresponding sequence $(z_n)$ is exactly periodic, then as mentionned in the comments you will get only countably many points, so very few in some sense. However these points will be well distributed in the boundary of the Mandelbrot set, so you will get a somewhat good picture.

But the most efficient way to draw the Mandelbrot set is probably still the divergence test, modulo a few numerical tricks.

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  • $\begingroup$ 1 1 ¯1 121 3 2 ¯1 138 6 1 ¯1 80 4 5 ¯1 122 2 8 4 1 2 8 4 1 2 43 ¯1 332 1 1 2 2 1 1 2 3 1 1 2 2 1 1 2 2 1 2 2 2 1 2 2 2 1 2 2 1 1 2 2 2 1 2 2 2 1 3 2 1 1 3 2 1 1 3 2 2 1 2 2 2 1 3 2 1 1 4 2 1 1 3 2 1 1 4 2 1 1 4 2 1 1 8 2 1 1 4 2 1 1 4 2 1 1 4 2 1 1 4 2 1 1 13 2 1 1 112 ¯1 105 0 305 $\endgroup$ – Stephen Wynn Feb 14 '18 at 13:40
  • $\begingroup$ My last post was accidental. But I can explain. This is d $\endgroup$ – Stephen Wynn Feb 14 '18 at 14:20
  • $\begingroup$ This is going from -2 in a positive direction with steps of 0.001. 1 1 means that c=-2 is stationary. -1 121 means that the next 121 points do not move. $\endgroup$ – Stephen Wynn Feb 14 '18 at 14:51
  • $\begingroup$ -2 is not stationnary. it is preperiodic/prefixed: $-2 \mapsto 2 \mapsto 2 \mapsto \ldots$ $\endgroup$ – Glougloubarbaki Feb 14 '18 at 14:56
  • $\begingroup$ I mean of course stationary after it has settled down. All these points wander around before cycling. -1 means it does not cycle. 0 means it goes to infinity. An interesting aspect of this output is that the -1 points occur in relatively large groups. $\endgroup$ – Stephen Wynn Feb 14 '18 at 15:14
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We have to check that removing cycles of length 1 2 3 ... does not significantly effect the boundary. We do this in order starting with length 1. Removing cycles of length 1 removes the whole of the central cardioid. But this has little if any effect on the boundary which comes almost entirely from the bulbs.

Moving on to cycles of length 2. We have to check that removing cycles of length 2 does not effect the boundary of the first bulb centre x = -1 radius .25. The output of my program running along the line x = -1 in the y direction from 0 to .25i intervals 1/10,000 is:

¯2 1885 ¯99 1 ¯2 10 ¯99 1 ¯2 4 ¯8 1 ¯2 4 ¯99 1 ¯2 3 ¯99 1 ¯2 8 ¯99 1 ¯2 5 ¯8 1 ¯2 3 ¯99 2 ¯2 3 ¯99 1 ¯2 1 ¯99 1 ¯8 2 ¯2 2 ¯99 1 ¯8 2 ¯2 2 ¯8 1 ¯2 6 ¯8 2 ¯99 2 ¯2 2 ¯99 1 ¯2 2 ¯8 1 ¯2 1 ¯99 1 ¯2 1 ¯99 1 ¯2 1 ¯8 1 ¯2 1 ¯99 1 ¯8 2 ¯2 2 ¯99 5 ¯8 1 ¯2 1 ¯99 1 ¯8 1 ¯2 1 ¯8 4 ¯2 1 ¯8 6 ¯99 1 ¯8 6 ¯2 1 ¯8 15 ¯2 1 ¯8 346 ¯99 135

There are 1885 cycles of length 2 at the centre. I am using -99 for c points for which z does not cycle. So there are 135 consecutive points on the boundary which do not cycle. So removing 2 cycles does not effect the boundary.

There is a similar situation with 3 cycles with x = -.125. So my conclusion is that removing cycles of length 1 2 3 ... does indeed add additional boundary. It removes the central cardioid, and makes a large hole in the bulbs surrounded by small holes.

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