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Offline claude

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(Question) Re: iterating corners to find period
« Reply #30 on: November 19, 2020, 08:07:34 AM »
You need to iterate the derivative in full precision for Newton's method at least.  Indeed Newton's method doubles number of accurate bits every iteration (provided you are close enough to start with), quadratic convergence, and some other methods have higher order.  Benchmarks would be interesting, there's probably a threshold precision higher than which the higher order methods are beneficial.  The higher derivatives could maybe be calculated in more threads, so the wall-clock time might not be much slower if you have enough cores? (But more threads means higher cost to synchronize each iterations, hm)

Offline marcm200

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Re: iterating corners to find period
« Reply #31 on: December 15, 2020, 09:17:16 AM »
Another hypothetical collection of ideas that might lead to area by the corner iteration (not thought through):

1. Pick 4 complex values c1-c4 in the 2-square.
2. Corner-iterate them until they capture the origin, say p times.
3. If p is not a prime >= 3, discard the points, otherwise
4. They lie in (different) period-p components or the main cardioid (wobbly)
5. Test if any one of c1-c4 lies in the main cardioid (https://en.wikibooks.org/wiki/Fractals/Iterations in the complex plane/Mandelbrot set/mandelbrot#Cardioid and period2 checking.), if so,discard the 4 points, otherwise
6. Define the number \( d_p :=inf \{ |a-b| \} \) where a,b are complex numbers in different hyperbolic components of period p. If d_p turns out to be > 0 (I think it will be, otherwise two period p components need to share a boundary point), proceed,
7. If all pairs of distances |c_i - c_j| are below d_p, then the 4 points lie in the same hyperbolic component, otherwise discard them.
8. Use pauldelbrot's derivative idea (forum link) to check whether their connecting edges lie in "the same curvature ", so not straddling an indentation or a cusp. If so, the interior of the polygon generated by c1-c4 lies fully inside one component and the Mset.

As for the d_p value: If there was a reliable upper estimate on the sizes of period-p components, one could use this to get a lower bound on d_p. Any other ideas?

I wonder: Does there exist a universal real-valued constant D such that 0 < D < d_p for all prime p >= 3 - supposing d_p > 0 is true (or maybe even all periods?). There are ever more but ever smaller components for growing p, so there must be a lot of space between them, but maybe there are always 2 that group ever closer together?

Offline pauldelbrot

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Re: iterating corners to find period
« Reply #32 on: December 15, 2020, 03:29:44 PM »
Quote
I think it will be, otherwise two period p components need to share a boundary point

Whereas this seems not to occur for zn + c, it can occur for generic parameter spaces, and even for unicritical maps. Proof: consider z2 + c2 + a. For generic a, the c-parameter plane contains two copies of the Mandelbrot set, centered on ±i√a. The point midway between them corresponds to the point at a in the normal Mandelbrot set's space. If a is the coordinate of a point on the Mandelbrot set's perimeter, the two copies will be tangent at that point (and highly warped to avoid any overlaps or additional points of contact), and if a lies in the regular M-set's interior the two copies will be conjoined.

If a is on the periphery of a period p component, the two M-set copies will touch there, and there will therefore be two period p components tangent to one another.

The logistic map rz(1 - z) furnishes a more restricted example: two tangent period-1 components. The point of tangency is at r = 1 and both components are unit-radius disks, one centered on the origin and one centered on 2.

Offline gerrit

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Re: iterating corners to find period
« Reply #33 on: December 15, 2020, 05:36:43 PM »
Another hypothetical collection of ideas that might lead to area by the corner iteration (not thought through):
AFAIK the "corner method" is just a heuristics which works most of the time.

Offline hobold

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Re: iterating corners to find period
« Reply #34 on: December 15, 2020, 07:53:39 PM »
AFAIK the "corner method" is just a heuristics which works most of the time.
I think it is slightly more. Let's see if I can cobble some line of reasoning together ...

The mapping z -> z*z wraps the complex plane around the origin twice. To illustrate this, imagine a point p = (p_x + i*p_y) on the unit circle, moving around to do one full rotation. Then the point q := p*p performs two full rotations as p is revolving once. So this mapping will keep a quadrangle intact, if the origin is not inside its convex hull. That is, if the quadrangle is far away from the origin compared to the diameter of the quadrangle, then the "wrapping around" does move (and rotate) more than it bends.

The mapping x -> x + c is just a translation of all points x. It will always leave a quadrangle intact.

Once the origin is inside the quadrangle, all four corners will (eventually) get mixed up by the "wrapping around", because the bending aspect dominates.

So, in other words, as long as the four quadrangle corners get to keep their relative topology - because the origin was outside the quadrangle up to this point - they all share essentially the same "fate" with respect to chaotic bifurcations and the like.

Hmm, okay, but what would this prove? I guess the other direction makes more sense: if the original four corners all started in the same disk/cardioid, then one can conclude that all points of that disk/cardioid share the same fate, i.e. they all have the same periodicity as the four corners. Hmm, that's not what I tried to prove. :)

Hmm, I'm sorry, this is not leading anywhere. But there is a little more to the four corner rule than just heuristics.

Offline marcm200

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Re: iterating corners to find period
« Reply #35 on: December 16, 2020, 03:49:47 PM »
@pauldelbrot: I used your constructive proof in my other thread (forum link). I had a proof for a very specific period-1 combination, but not for the general case, so thanks for that!

@gerrit, hobold: What types of errors by the corner method have you encountered? The ones I saw (finding a multiple of the period at capture incident - all 4 or only a fraction thereof), those will be counteracted by only using prime iterations and the period-1 additional test. Is there an example for e.g. 4 points in a period-11 (random prime) component(s) and the corner iteration returns 7 for the capture event - or 13 e.g?

Offline gerrit

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Re: iterating corners to find period
« Reply #36 on: December 16, 2020, 05:50:33 PM »
@pauldelbrot: I used your constructive proof in my other thread (forum link). I had a proof for a very specific period-1 combination, but not for the general case, so thanks for that!

@gerrit, hobold: What types of errors by the corner method have you encountered? The ones I saw (finding a multiple of the period at capture incident - all 4 or only a fraction thereof), those will be counteracted by only using prime iterations and the period-1 additional test. Is there an example for e.g. 4 points in a period-11 (random prime) component(s) and the corner iteration returns 7 for the capture event - or 13 e.g?
I used it a bit years ago, you can try in old KF versions, sometimes it doesn't find any perdiod.
Of course there's nothing special about 4, it works equally good with 3, but the best with 0 using the ball method.
I call it heuristic because you can make it plausible but no rigorous proof AFAIK.

Offline marcm200

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Re: iterating corners to find period
« Reply #37 on: January 18, 2021, 01:58:59 PM »
I call it heuristic because you can make it plausible but no rigorous proof AFAIK.

Some thoughts on that using interval arithemtics:

Suppose we start with 4 complex points A,B,C,D in a period p component, forming an axis-parallel rectangle that contains the hyperbolic center of the component.

This rectangle ABCD is used as a complex interval seed c.

If one performs p iterations using IA, starting with the origin, f[p](0) will contain the origin again due to interval arithmetics' inclusion property: the true orbit of 0 under the hyperbolic center seed is contained in all f[k](0), 0 <= k <= p. And as the origin is periodic for that one specific seed in the used complex interval seed c, this particular orbit ends again with 0.

Hence the origin is definitely captured at the component's period.

However, as the true image of an axis-parallel rectangle under z^2+c is a tilted 4-object with arcs as edges and IA always returns an axis-parallel rectangle as a superset of the true image, I cannot rule out for now, that the origin might be captured earlier, misindicating the period. But it feels like I'm on the right track.

Below is an example for a period-11 component.

The hyperbolic center is contained in
Code: [Select]
[-0.697845458984375..-0.6978302001953125] + i*[-0.2793121337890625..-0.279296875]

which was rigourously proven using IA, subdivision root finding and outward rounding in (forum link); a double-precision outward-rounded enclosement of root #688 was used here).

The 4 points A-D were the corners of the rectangle:
Code: [Select]
real: -0.701995849609375 .. -0.69366455078125
imag: -0.283538818359375 .. -0.27520751953125

which contain the entire hyperbolic center region from above.

Periods for points A-D and the hc region were confirmed to be 11 using the TSA cell-mapping with a sufficient number-type.

Iterations in the capturing routine were performed with the kv IA-library and outward rounding.

A-D and hc were simpley visually judged to lie in the same hyperbolic component due to a point-sampled,non-rounding controlled image (upper left).



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