What is the length of a section of the Mandelbrot set boundary?

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Offline TGlad

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« on: November 28, 2018, 12:50:10 PM »
I have another maths question.

I have heard that the boundary of the Mandelbrot set has dimension 2, but that the boundary has no area. I assume this means that the Hausdorff measure of the boundary is zero.

If this is the case then how can anyone compare the size of two different sections of the boundary? So no-one can answer questions like "How much longer is the first bud's boundary to the cardoid's boundary" or "which boundary is longer, the Mandelbrot set or the power 3 multibrot set?"

Is there are way to quantify the 'size/length' of the boundary?

Offline hobold

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« Reply #1 on: November 28, 2018, 01:34:22 PM »
The boundary of the Mandelbrot set consists of a) outlines of disks and cadioids, and b) antennas. There may be further types, but these two suffice to illustrate the following argument.

As you zoom in closer to the boundary, the apparent shape of (connected) disks and cardioids does not change in fundamental ways. That is, the relative size and number of disks/cardioids does not vary based on scale. Or in other words: the shape of minibrots does not fundamentally change as you keep zooming in.

However, antennas get relatively longer and more numerous as you zoom deeper. The outlines of disks/cardioids become furry, and their pelts grow longer and denser the deeper you zoom. In that sense, the boundary of the Mandelbrot set "gradually" fills the plane as you zoom in.

Maybe one could quantify the growth of fur in relation to zoom factor? The traditional way only looks at the ("infinitely zoomed") limit and sees dimension 2, but maybe there are differences in "convergence speed"?

Another way to reason about such fuzziness is to look at intentionally blurred variants. Instead of plotting the (infinitely many) points of the boundary, plot small disks centered around each point, all with the same radius (the specific radius is a tuning parameter, or a resolution of choice). The union of all those disks has a nonzero finite surface area that can be measured. That surface area can be compared between different images, as a proxy for length at a specific resolution (disk radius).

Or one could look at convergence speed to surface area zero as disk radius approaches zero?

Offline FractalDave

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« Reply #2 on: November 28, 2018, 02:42:54 PM »
<snip>
Is there are way to quantify the 'size/length' of the boundary?

Do you mean analytically ?

One slightly related question I have - is it possible that taking any given point on the (boundary) Mandelbrot Set at some viewing resolution that point will be on the edge of a visible minibrot ? Further - if infinite zooming were in fact possible, would every such point actually be on the boundary of infinite such minibrots ?
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Offline hobold

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« Reply #3 on: November 28, 2018, 03:34:45 PM »
is it possible that taking any given point on the (boundary) Mandelbrot Set at some viewing resolution that point will be on the edge of a visible minibrot ?
Not in a strict sense. Tips of antennas and centers of spirals are not on minibrot edges.

Now I wonder how many different types of points there are on the Mandelbrot boundary. Besides tips and centers there are touch points between two disks or between a disk and a cardioid. Is there such a thing as an "ordinary" boundary point at all? Hmm, there are "inverse tips" at the cusps of cardioids. What else?

Offline pauldelbrot

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« Reply #4 on: November 28, 2018, 03:38:09 PM »
Every point on the boundary of the M-set is, I think, either on a disk boundary, on a cardioid boundary, or a limit point of an infinite sequence of those. The points on filaments in particular are in the latter category. That every filament boundary point is a limit of a sequence of disk/cardioid boundary points is the same as saying any image rectangle containing a filament boundary point contains a minibrot (which might require zooming in further to see).

Offline lkmitch

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« Reply #5 on: November 28, 2018, 04:54:45 PM »
I think Misiurewicz points (typically, tips of filaments and branching points) are not on the boundary of any cardioid or disk. If they were, there should be some sort of neighborhood analysis that could show that a point "right beside" the given point is inside a periodic component. That may exist, but I'm not aware of it.

As to "ordinary" boundary points, using the Mandelbrot main cardioid as an example, the boundary of the cardioid is parameterized by the polar angle theta. If theta is a fraction of 2pi, then the boundary point is a tangent point between the cardioid and a disk. For example, the large disk on the top of the cardioid is at an angle theta = 1/3 * 2pi. The uncountably infinite number of boundary points where theta is not a fraction of 2pi (e.g., theta = 1 radian) are not tangent points. Maybe these are the "ordinary" boundary points.

Offline Adam Majewski

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« Reply #6 on: November 28, 2018, 08:07:20 PM »
I think Misiurewicz points (typically, tips of filaments and branching points) are not on the boundary of any cardioid or disk. If they were, there should be some sort of neighborhood analysis that could show that a point "right beside" the given point is inside a periodic component. That may exist, but I'm not aware of it.

It seems that it is a good question for math.stackexchange.com

Offline gerrit

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« Reply #7 on: November 28, 2018, 08:48:04 PM »
Every point on the boundary of the M-set is, I think, either on a disk boundary, on a cardioid boundary, or a limit point of an infinite sequence of those. The points on filaments in particular are in the latter category. That every filament boundary point is a limit of a sequence of disk/cardioid boundary points is the same as saying any image rectangle containing a filament boundary point contains a minibrot (which might require zooming in further to see).
True for open rectangles, but false for closed ones I think.

Offline FractalDave

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« Reply #8 on: November 28, 2018, 09:05:30 PM »
I realised about filament tips after posing the question about minibrots - but hadn't considered spiral centres.
To be honest (yet again) I was really making the (math) rookie mistake of considering that infinity contains infinite possibilites, which it doesn't of course - otherwise for example infinite places of pi could contain an infinite copy of a recurring fraction, which it can't.

Offline TGlad

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« Reply #9 on: November 29, 2018, 04:33:46 AM »
Going back to Hobold's answer about quantifying the boundary length, I think that is a very good description of the problem.

We know the complexity/fuzziness/tentacles increase as you zoom in, and this isn't captured in the basic notions of Hausdorff dimension or Hausdorff measure. Does it all grow in complexity at the same rate? If not then measuring the length at any particular scale could be very different to the actual size of the set.

If it does grow at the same rate, then the coefficient on that function would be a measure of the boundaries size, and would allow different segments to be compared.

Let's make the problem more concrete (for someone in 1000 years to solve):

what is the larger part of the Mandelbrot set: the boundary around the main bud or the boundary around the left side of the cardoid (excluding the main bud)?

Offline FractalDave

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« Reply #10 on: November 29, 2018, 05:06:45 AM »
<snip>
what is the larger part of the Mandelbrot set: the boundary around the main bud or the boundary around the left side of the cardoid (excluding the main bud)?

In the infinite limit any given section would be infinitely long, also IMO it's quite possible that which is the longer at a given resolution could change at a different resolution - so what resolution are you asking about ?

Offline TGlad

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« Reply #11 on: November 29, 2018, 05:25:28 AM »
I don't mean length in metres. I mean some measure to compare the size of the two boundary sections.
The Koch curve is infinitely long in metres but in metres^1.26 it has a 'length' or rather a Hausdorff measure, so you can quantify and compare sections.
You can't do that with the Mandelbrot set because its Hausdorff measure (in m^2) is 0. That's the crux of the puzzle.

Offline claude

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« Reply #12 on: November 29, 2018, 06:54:46 AM »
Is it really known that the 2-dimensional Hausdorff measure of the boundary of the Mandelbrot set is 0?  Links to references would be appreciated.  I think I read once that it was an open problem, alas I have no reference for that either.

One way to quantify parts of the boundary could be to use external rays - compute the dimension (and d-dimensional measure for disambiguation) of the set of external angles whose rays land on the boundary of the part in question.  But I think I read (no reference, sorry) that it is also an open problem whether all rays (including irrational rays) do actually land on the boundary.  (Rational rays do land).

Offline TGlad

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« Reply #13 on: November 29, 2018, 11:16:44 AM »
You're right, it doesn't seem to be proven that the measure is zero, but it seems like mathematicians think it probably is zero, and it sounds like numerical computations back that up.

So I guess this puzzle can't be solved until it is known for sure.

Assuming it is zero I think what we're seeing is a set which is not a typical fractal, it looks like the local fractal dimension (local in scale) increases with zoom depth. I'd like to see a log-log plot of the boundary length (box counting or yardstick), if it does get steeper with detail level then that suggests the linear 'fractal' description is incomplete, a non-linear model would be needed to give it a finite measure.

I like the idea of this.

Offline FractalDave

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« Reply #14 on: November 29, 2018, 12:13:37 PM »
Assuming it is zero I think what we're seeing is a set which is not a typical fractal, it looks like the local fractal dimension (local in scale) increases with zoom depth. I'd like to see a log-log plot of the boundary length (box counting or yardstick), if it does get steeper with detail level then that suggests the linear 'fractal' description is incomplete, a non-linear model would be needed to give it a finite measure.

I like the idea of this.

When I posted I was thinking you couldn't mean "metres" ;) And when asking did you mean "analytical" I was thinking that the only method would be relative "box-counting or "yardstick" :)
I'm guessing you're correct about increasing dimensionality with resolution and I still think it's possible that such may not be uniform in different areas so that the relative "lengths" of any two sections may in fact also change with resolution....
Also I'd guess anything involving c*z or z^c or z^z will complicate things exponentially ;)
I say c*z due to the issues involved in simply getting an all-over smooth iteration fraction.