We have observed that cycles are born in the Feigenbaum Plot in one of two ways: bifurcations of lower cycles and spontaneously out of chaos. Yesterday we understood how these cycles are spontaneously born and how this phenomenon coupled with the self-similarity of the Feigenbaum Plot suggests an order to cycles. What it also gives us is a way of counting how many cycles of each type there are.
We know there are two fixed points on the original f (x) = ax(1 – x). One is attracting for a < 3, the other (at zero) is always repelling.
For 2-cycles, we look at f (f (x)). We see as many as four fixed points on this graph. But two are already members of 1-cycles and must be eliminated. This leaves only two points eligible to be members of 2-cycles, meaning we only have one 2-cycle, which we see bifurcating at a = 3.
To count 3-cycles, we observe that f (f (f (x))) has as many as eight fixed points. A 3-cycle pattern n, ___, ____, n could be explained two ways: as a 1-cycle (n, n, n, n) or as a real 3-cycle (n, o, p, n). So again, we remove the two 1-cycle points and are left with six points eligible to be members of 3-cycles, suggesting two 3-cycles. We see one in the Feigenbaum plot at a ≈ 3.83, but that’s an attracting fixed point. Where’s the other one? It turns out that for every cycle born out of chaos, there is an “evil twin” repelling cycle born as well. As a result, there are actually two 3-cycles born at a ≈ 3.83: one attracting and one repelling.
To count 4-cycles, we observe that f(f(f(f(x)))) has as many as 16 fixed points. But if we see a pattern of n, ___, ____, ____, n, we could explain this by:
- A 1-cycle: n, n, n, n, n
- A 2-cycle: n, o, n, o, n
- A real 4-cycle: n, o, p, q, n
So we eliminate the two 1-cycle points and the two 2-cycle points. This leaves 12 points eligible to be members of 4-cycles, suggesting three 4-cycles. One is the bifurcating cycle we clearly see in the Feigenbaum plot, the second and third are an attracting/repelling 4-cycle pair found at a ≈ 3.96.
To count 5-cycles, we start with the 32 fixed points, eliminate the two that are members of 1-cycles, and observe that the 30 remaining points must create six 5-cycles. One attracting/repelling pair we see at a ≈ 3.74, but the other two are harder to find (one is at a ≈ 3.906, the other pair at a ≈ 3.99028).
Your homework for this weekend: continue this train of thought and identify how many 6-, 7-, and 8-cycles there are. If you’re feeling ambitious, try to locate all of them!