z2+c+si
si = -1, 1, -1, 1, …
Center: 0.82813995, 1.39705173, Zoom = 5120000
This is just a random zoom. It has nothing to do with today’s topic. I just feel obligated to always attach a new picture to each post.
Recall that we are actually looking at pictures
generated by two alternating functions, see Sequence Fractals Part I #19
f0(z,c) = z2+c-1
f1(z,c) = z2+c+1
Two step function: fc(z) = f(z,c) =f1(f0(z,c),c)
Let’s go back to the original definition today. The fixed points and the n-cycles for the two-step function are actually 2-cycles and 2n-cycles in the original setup. A natural question is “Are there any fixed points?”.
The answer is no. A fixed point would require f0(0,c) = 0 and f1(0,c) = 0. Observe that f0(0,c) = f1(0,c)-2, . So the simultaneous equalities cannot be solved.
It is essentially impossible for any odd cycle to exist. We need to solve two equations with one unknown. An old cycle gives rise to two polynomials
g0 = f0f1…f0
g1 = f1f0…f1
Please forgive the abuse of notation, this is function composition not multiplication (don’t make me write all of the parenthesis). An odd cycle requires
g0(0,c) = 0
g1(0,c) = 0
Except in very rare, carefully designed situations, two equations and one unknown has no solutions.
Side note: I tried to add another degree of freedom so that the system of equations could be solved. I was able to find a sequence which had an odd cycle, but the picture did not look interesting. So that search in on the back burner now.