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

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Offline lkmitch

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« Reply #15 on: November 29, 2018, 05:19:39 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 admit that I'm coming at this discussion from the viewpoint of an engineer rather than a mathematician, but could this process be implemented using the boundary tracing method? It basically treats the Mandelbrot set as a very concave polygon and marches around the boundary. A double limit of step size and iteration count might conceivably converge on the functional change of dimension with resolution.

Offline FractalDave

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« Reply #16 on: November 29, 2018, 05:33:05 PM »
I admit that I'm coming at this discussion from the viewpoint of an engineer rather than a mathematician, but could this process be implemented using the boundary tracing method? It basically treats the Mandelbrot set as a very concave polygon and marches around the boundary. A double limit of step size and iteration count might conceivably converge on the functional change of dimension with resolution.

Yes (that's just the yardstick method) and at each resolution you'd need to use a high enough iteration count and bailout such that increasing either for that resolution (step-size) has no effect on the results.
Also for resolutions below a given amount the results may give a variation in dimension (even a consistently drifting one with resolution) that could simply be due to statistical errors i.e. lack of "sample size", I'd guess that anything less than around a million steps wouldn't really be accurate enough.
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Offline gerrit

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« Reply #17 on: November 29, 2018, 05:55:08 PM »
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).
This is equivalent to proving the set is "locally connected"; most introductions on M-math mention this open problem.

Ratio of 2 boundary segment's lengths should be well defined as long as the segments have the same fractal dimension, just the box counting length ratio limit.

Offline claude

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« Reply #18 on: November 29, 2018, 06:11:40 PM »
thanks, I didn't realize it was equivalent (I knew MLC was open).

boundary from the interior has dim 1, smooth disc-like/cardioid-like shapes, from the exterior dim 2 (because minibrots are dense in the boundary so there are hairy Feigenbaum points everywhere)

Offline TGlad

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« Reply #19 on: November 30, 2018, 12:12:32 AM »
Quote
Ratio of 2 boundary segment's lengths should be well defined as long as the segments have the same fractal dimension, just the box counting length ratio limit.

Ah, this is very clever if you mean what I think you mean.
Take the ratio of the boundary length between the two segments, with yardstick length l. Now find the limit as l-->0
Equivalently, count the pixels on the boundary for pixel width l and take the ratio of this between the two segments. Now find the limit of the ratio as l-->0

It means we don't have to find the growth function, we just assume the growth functions tend to the same shape with increasing resolution.

So we could answer the question of which border is bigger. We could even compare the lengths of the multibrot sets... does the Mandelbrot set have the 'longest' border of all of them?

Does anyone know of a Mandelbrot border pixel counting (or yardstick length measuring) tool? Let's try it.

Offline fractower

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« Reply #20 on: November 30, 2018, 08:09:12 AM »
How about combining re-normalization with the boundary scan.

For any finite iteration method the boundary is defined by iteration count and escape limit.
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1. Assume that the escape limit is second order error.
2. Assume there is a Nyquist limit like relationship to iteration(frequency) and sampling such that a valid boundary length (including end points) can be defined for every iteration count. 
3. Magic happens. At some iteration count a (power, poly, etc.) law emerges for sufficiently separated end points such that the ratios of lengths becomes meaningful.

I know how to do this computationally and will probably try. I am not sure how to do it analytically however.
« Last Edit: November 30, 2018, 07:40:20 PM by fractower »

Offline fractower

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« Reply #21 on: November 30, 2018, 07:15:07 PM »
Perhaps I should expand on the Magic happens step.

For some segment x and iteration n assume the length is X(n) and X(n+1) can be written as:

X(n+1) = X(n) * S(n)

Where S(n) is some number >1 (divergent series) but probably less than 2. If for large n, S(n) converges to a constant then:

X(n_large+m) ~= X(n_large) * S(inf)^m

If S(inf) is independent of chosen segment then the ratio of two segments length X(n_large+m)/Y(n_large+m) will be constant as m approaches infinity.

For this to work antenna and seahorses must have the same S(inf). They look pretty different but I am not sure how to know without trying.

If it turns out that S(inf) is larger for antenna than seahorses this would imply that the boundary length for any segment of antenna is infinitely longer than any segment of seahorse.
« Last Edit: November 30, 2018, 07:39:12 PM by fractower »

Offline gerrit

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« Reply #22 on: November 30, 2018, 08:49:20 PM »
Take the ratio of the boundary length between the two segments, with yardstick length l. Now find the limit as l-->0
Here some brute force result for power 2/power 3 boundary length. Seems to be about 0.6.

Offline TGlad

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« Reply #23 on: December 01, 2018, 12:48:26 AM »
Awesome!  :thumbs:  :thumbs:
So the power 3 multibrot border is about two Mandelbrots long  :)
I guessed it would be shorter, since the shape tends to a circle at higher power minibrots.

Offline FractalDave

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« Reply #24 on: December 01, 2018, 01:58:49 AM »
Awesome!  :thumbs:  :thumbs:
So the power 3 multibrot border is about two Mandelbrots long  :)
I guessed it would be shorter, since the shape tends to a circle at higher power minibrots.

I'm guessing that the result is what it is because the higher power tends to the limit dimension 2 faster. This checks out against the difference in rate of change of the widths of bailouts at a given iteration for different powers. (I mean the iteration bands)
« Last Edit: December 01, 2018, 11:34:29 AM by FractalDave »

Offline FractalDave

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« Reply #25 on: December 01, 2018, 11:33:49 AM »
Here some brute force result for power 2/power 3 boundary length. Seems to be about 0.6.

Hmmm, I have an idea - but first what do you get for power 2/power 4 ?

Offline TGlad

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« Reply #26 on: December 01, 2018, 12:20:16 PM »
I don't think the growth curve shape is different, so I don't think 'the higher power tends to the limit dimension 2 faster', if it did then the ratio would tend to 0.

Offline FractalDave

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« Reply #27 on: December 01, 2018, 01:59:24 PM »
I don't think the growth curve shape is different, so I don't think 'the higher power tends to the limit dimension 2 faster', if it did then the ratio would tend to 0.

Not necessarily - due to the nature of infinity the ratio can remain constant at 0.6 or whatever forever ! i.e. for any finite resolution.
« Last Edit: December 01, 2018, 02:21:29 PM by FractalDave »

Offline FractalDave

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« Reply #28 on: December 01, 2018, 02:12:31 PM »
Awesome!  :thumbs:  :thumbs:
So the power 3 multibrot border is about two Mandelbrots long  :)
I guessed it would be shorter, since the shape tends to a circle at higher power minibrots.

The boundary tends to a circle but with some area, I guess a torus ;) The inside tends to a circle.

Having said that I guess that the infinite limit boundary of z^infinty+c may be a plain 1D circle...anyone ? But anything less than z^infinity and the infinite limit boundary will be an area.

Offline FractalDave

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« Reply #29 on: December 01, 2018, 02:33:24 PM »
I'm guessing that the result is what it is because the higher power tends to the limit dimension 2 faster. This checks out against the difference in rate of change of the widths of bailouts at a given iteration for different powers. (I mean the iteration bands)

Of course that doesn't allow for more complex polynomials e.g. for a given resolution iterating (z^2+c)^2+c should have exactly the same length as iterating z^2+c but of course the first is using non-linear space due to the use of the constant (pixel) being non-linear ;)