frostfern
Forums Newbie
Posts: 4
|
 |
« on: April 04, 2014, 12:19:21 AM » |
|
I've been thinking about this lately just for fun. I wouldn't call myself a mathematician, but I've taken plenty of courses and consider math a hobby of sorts.
Staying strictly on the real line (i.e. Im(c) = 0) there seems to be a relatively straightforward pattern. In general the number of solutions to P^n(x) = x (where P^n(x) = (...(((x^2+c)^2+c)^2+c)...)^2+c i.e. n iterations of x^2+c) increases from zero when 0.25 < c to 2^n when c < 2.0. The roots either appear as sets of attracting-repelling pairs with a multiplicity that is some factor of n, or from period doublings of already existing attracting cycles.
Attracting cycles that form cardioids always appear via the birth of attracting-repelling pairs. They occur when one or more hills and/or troughs of the polynomial intersect the line y=x. The derivative must be exactly one when the growing hills/troughs touch the y=x line. As a hill/trough grows across the y=x line, the single tangent point roots splits into two roots. For a growing trough, the root on the left is initially an attracting cycle while the root on the right is forever a repelling cycle. For a hill it's just reversed, the right root is an attracting cycle while the left root is a repelling cycle.
Attracting cycles that form bulbs always appear via the splitting of an attracting cycle into two, leaving a repelling cycle in the middle. This happens when an inflection point forms on the intersection of P^n(x) with y=x. The twisting of the inflection point from a slope less than one to a slope greater than one always creates two new attracting cycles via the birth of a new hill/trough pair. The divisions can happen either from the main cardioid or from a subsequent mini-brot cardioid.
You can basically go through a process starting with the first attracting-repelling pair (which appears at the cusp point c = 0.25), checking off any subsequent bulbs of period 2^k where 2^k is a factor of n. Once you get rid of those you go on to the odd factors of n. For each new cardioid you also check off bulbs that divide off it in factors of 2^k. Just keep going until you've checked off all 2^n/2 pairs.
For n = 6 there are 2^6 there are 64 total roots, 32 of which become attracting at some point via the formation of new pairs. The main cardioid and it's attached period 2 bulb account for 2 of the 32 pairs. That leaves 30 pairs left.
The next factor is 3, which corresponds to the birth of a period 3 minibrot. The period 3 cardioid and it's period 6 bulb account for another 6 pairs. This leaves 24 more pairs to use
To see that there can only be one period 3 minibrot you have to look at the n=3 case (i.e. you have 8 roots, 4 of which become attracting, and 1 is associated with the main cardioid, leaving the remaining three for a single period 3 cardioid). Because there can't be any more period 3 minibrots, the remaining pairs must be associated with period 6 minibrots, of which there must be exactly 4.
Along the real line I've computed that there's 1 period 1 cardioid, 0 period 2 cardioids, 1 period 3 cardioid, 1 period 4 cardioid, 3 period 5 cardioids, and 4 period 6 cardioids.
I'm wondering if there's a way to do this for all cardioids, not just the ones that lie along the real axis? Is there a general way to determine the number of period n cardioids located anywhere in the complex plane. I'm guessing the number cardioids or bulbs for any given period is finite (I'm not 100% certain though).
|
March 15, 2009, 09:06:57 PM
Dilber MC Theme by HarzeM