centers of circles in mandelbrot set ?

[cKleinhuis]

arrh, i know this has been discussed often, but i just can not find it right now

for my show i plan the next issue on julia/mandelbrot fractals, for that i want to use 5 different julia visualisations,
i have the seeds (0,0) and (-1,0) wich are perfect centers of the circles, and i want to have the exact centers of 3 more
circles, the one would be the biggest up, and then one from the -1 up and one on the left big circle side

how are these calculated ? root/zero points finding for higher polynomals ?! just expanding the formula for a few iterations and take the zero locations of that graph

[Pauldelbrot]

arrh, i know this has been discussed often, but i just can not find it right now

for my show i plan the next issue on julia/mandelbrot fractals, for that i want to use 5 different julia visualisations,
i have the seeds (0,0) and (-1,0) wich are perfect centers of the circles, and i want to have the exact centers of 3 more
circles, the one would be the biggest up, and then one from the -1 up and one on the left big circle side

how are these calculated ? root/zero points finding for higher polynomals ?!

Yep. The polynomials are (((((c² + c)² + c)² + c)² + c)...)² + c = 0.

The solutions to the ith polynomial are the centers of the components of periods dividing i. So the center 0 of the component of period 1 is always a root -- and indeed the polynomials are all divisible by c. After c, the second polynomial is c² + c, which is divisible by c and leaves c + 1 when this is factored out -- so, the only period 2 component is at -1. So far, so consistent, right?

Now if you look at (c² + c)² + c you can again factor out a c to remove the period-1 component root, giving (after multiplying out) c³ + 2c² + c + 1 = 0, a cubic whose three roots are the center of the upper bud with a lightning bolt attached, the center of the lower bud with a lightning bolt attached, and the center of the largest spike minibrot.

The next polynomial has degree 8! But it factors by c² + c, leaving a degree 6 polynomial for the roots of period-4 components. These are in conjugate pairs, so I could advise you to symbolically manipulate the thing for half an hour into a real and imaginary part and then use the fact that the real part polynomial will contain a perfect square factor of degree 4 (two lightning bolt minibrots and two buds between there and Elephant Valley) and a quadratic factor (a bud and a minibrot at the spike) which can be factored out of the original, complex polynomial, but it's undoubtedly easier to just start using Newton's Method at this point.

Besides, the next one will be degree-16 and divisible only by c among the polynomials that came before it, leaving a degree-15 polynomial with 14 conjugate pairs of roots and one on the axis, leaving you to try to find that one and then take the square root of a degree-14 polynomial. Ouch.

(Incidentally, the Newton Julia fractals for these polynomials are pretty, and for higher orders resemble a shattered and "flooded" Mandelbrot set, as most of the roots begin to cluster near the M-set boundary for higher periods for obvious reasons.)

So, tl;dr: use Newton's Method on the lower-degree polynomials in the set (((((c² + c)² + c)² + c)² + c)...)² + c = 0.

The specific ones you noted are as follows.

• Biggest up is period 3, so the root of (c² + c)² + c with the largest imaginary component.
• Biggest up from -1 is period 6, so the root with positive imaginary part nearest -1 of the degree-28 polynomial you get by taking the degree-32 polynomial (the sixth) in the original sequence and then dividing by the period 3 polynomial above.
• The next one left from -1 is period 4, so a root of ((c² + c)² + c)² + c/(c² + c) on the real axis. There will be two such roots; the one you want is on the right. (The left is way out near -2 and is a minibrot center.)

HTH.

[cKleinhuis]

arrh, isnt there somewhere just a list for it ? lol, no head right now to solve it, can i put it into wolfram alpha?

[lkmitch]

In Wolfram Alpha, you can build up the list iteratively.  Start with "c^2 + c = 0" in the search box, and Alpha will return the real and complex solutions.  Then, wrap the lefthand side of the equation in parentheses, add a "^2 + c" (so you have "(c^2 + c)^2 + c = 0") and run it.  Lather, rinse, repeat.

[lkmitch]

I have a list of the complex roots up to the 9th iteration in an Excel file, if anyone wants it.

[cKleinhuis]

can you send it to me ? i am using such locations to show the development of a corresponding julia set, because those locations do have the "clear loops" with exactly 1,2,3,4,5...n points that are repeated all over the time, and thus these locations show very nicely the behaviour of the julia sets, you can send it directly to my email, pm me yours

[claude]

Some time ago I made a database of all islands up to period 16, found by tracing external rays:

http://mathr.co.uk/mandelbrot/feature-database.csv.bz2 (2.5MB, decompresses to 8.7MB)

It has lines like:

3,True,1 2 3,3,7,4,7,3.141592653589793,1.8201981627989673e-2,-1.754877666,0

meaning:

period, islandhood, angled internal address, lower external angle numerator, denominator, upper numerator, denominator, orientation, size, centre realpart, imagpart

Unfortunately it only has complete data for islands, not the circles you are looking for

But, given a component (island or bulb) and an internal angle for a child bulb, you can use Newton's method to find the bond (tangency point) and there's also a formula for approximate bulb size which you can use to find an approximate centre, then you can use Newton's method again to find the centre more precisely.  Repeat as necessary to descend down a chain of bulbs...


http://www.math.nagoya-u.ac.jp/~kawahira/programs/mandel-exray.pdf  describes external ray tracing
http://arxiv.org/abs/math/9411238 describes angled internal addresses

[knighty]

@cKleinhuis:
http://www.fractalforums.com/theory/the-mandelbrot-polynomial-roots-challenge/

[cKleinhuis]

Lkmitch gave me his excel table thanks lkmitch