$ z^2+c+s_i$
s_i = -1, 1, -1, 1, …
Center: C1.00: 0.500000, 1.322876, Zoom = 0.8

The helper functions will be redefined for the new sequence. (Notice the sign flip in key places.)
f₀(z,c) = z²+c-1
f₁(z,c) = z²+c+1
f_c(z) = f(z,c) =f₁(f₀(z,c),c) = (z²+c-1)²+c+1.
Derivative: f_c'(z) = 4z(z²+c-1)

z₀ = 0 is a critical point and will again be the default starting point. The other two critical points will be ignored without guilt. They just bring us back to the sequence starting 1,-1…, about which we already know more than we ever wanted to know.

Fixed points are found when f(0,c) = 0. so c = $ \frac{1\pm\sqrt{-7}}{2}$ = 0.5 ± 1.322876i.

  2 fixed points
C1.00:   0.500000,  1.322876
C1.01:   0.500000, -1.322876

I will start naming named points with a leading letter to avoid confusion with decimal numbers.

$ z^2+c+s_i$
s_i = -1, 1, -1, 1, …
Center: C2.00: 0.053427, 0.781326, Zoom = 16

First of three pairs of two-cycles, C2.00. Below and to the left of C1.00 in Sequence Fractals Part I #21

  6 two cycles
C2.00:   0.053427,  0.781326
C2.01:   0.053427, -0.781326
C2.02:   0.633096,  1.288065
C2.03:   0.633096, -1.288065
C2.04:   0.813477,  1.815442
C2.05:   0.813477, -1.815442

Everything is reflected across the real axis. C2.01 is exactly the same, just flipped vertically. So I will ignore everything going on in the bottom half of the complex plane.

z²+c+s_i
s_i = -1, 1, -1, 1, …
Center: C2.00: 0.053427, 0.781326, Zoom = 40

C2.00, 5x zoom, 3x colors compared to yesterday.

z²+c+s_i
s_i = -1, 1, -1, 1, …
Center:C2.04 0.633096, 1.288065, Zoom = 16

C2.04, up and right of C1.00, Sequence Fractals Part I #21. The other two cycle, C2.02 is the period two bulb in the mini associated with the fixed point host C1.00, also found in Sequence Fractals Part I #21

z²+c+s_i
s_i = -1, 1, -1, 1, …
Center:C2.04 (0.633096, 1.288065), Zoom = 16

C2.04 again, 20x yesterday's zoom.

z²+c+s_i
s_i = -1, 1, -1, 1, …
Center:C3.00 -0.028857, 0.822581, Zoom = 100

C3.00 can be found in the upper left corner of the C2.00 picture Sequence Fractals Part I #22

Three cycles can be found by solving f(f(f(0,c),c),c) = 0. The degree 32 polynomial equation gives 32 roots, then take away the two fixed points, leaving 30 three cycles

  30 three cycles
C3.00:  -0.028857,  0.822581
C3.01:  -0.028857, -0.822581
C3.02:   0.000323,  0.777540
C3.03:   0.000323, -0.777540
C3.04:   0.123345,  0.786543
C3.05:   0.123345, -0.786543
C3.06:   0.123768,  0.684849
C3.07:   0.123768, -0.684849
C3.08:   0.256240,  1.261825
C3.09:   0.256240, -1.261825
C3.10:   0.456809,  1.544145
C3.11:   0.456809, -1.544145
C3.12:   0.475692,  1.211907
C3.13:   0.475692, -1.211907
C3.14:   0.543537,  1.408400
C3.15:   0.543537, -1.408400
C3.16:   0.729946,  1.274941
C3.17:   0.729946, -1.274941
C3.18:   0.731182,  1.053796
C3.19:   0.731182, -1.053796
C3.20:   0.774380,  1.836743
C3.21:   0.774380, -1.836743
C3.22:   0.802629,  1.829248
C3.23:   0.802629, -1.829248
C3.24:   0.827269,  1.802481
C3.25:   0.827269, -1.802481
C3.26:   0.828125,  1.397028
C3.27:   0.828125, -1.397028
C3.28:   0.855612,  1.812967
C3.29:   0.855612, -1.812967

z²+c+s_i
s_i = -1, 1, -1, 1, …
Center:C3.00 -0.028857, 0.822581, Zoom = 5000

C3.00 with 50x zoom and 2x color compared to yesterday. Sequence Fractals Part I #26

z²+c+s_i
s_i = -1, 1, -1, 1, …
C3.02: 0.000323, 0.777540, Zoom = 40

C3.02 is between C2.00 and C3.00 in the C2.00 picture Sequence Fractals Part I #22.

z²+c+s_i
s_i = -1, 1, -1, 1, …
C3.02: 0.000323, 0.777540, Zoom = 5000

C3.02 with 50x zoom, 2x color

I am not going to show all three-cycle hosts. They all look similar. They are found in an interesting pattern. Three are connected to C1.00 in the usual Mandelbrot way, top and bottom bulbs, and largest mini off the period 2 bulb. Each of lower cycle hosts, C1.00, C2.00, and C2.04 are surrounded by four satellite three cycles.

 15 +i half plane 3 cycles
C1.00:   0.500000,  1.322876  - connected   Sequence Fractals Part I #21
 C3.12:   0.475692,  1.211907 
 C3.14:   0.543537,  1.408400
 C3.16:   0.729946,  1.274941
C1.00:   0.500000,  1.322876  - satellites  Sequence Fractals Part I #21
 C3.08:   0.256240,  1.261825
 C3.10:   0.456809,  1.544145 
 C3.18:   0.731182,  1.053796 
 C3.26:   0.828125,  1.397028 
C2.00:   0.053427,  0.781326  - satellites  Sequence Fractals Part I #22
 C3.00:  -0.028857,  0.822581 
 C3.02:   0.000323,  0.777540
 C3.04:   0.123345,  0.786543 
 C3.06:   0.123768,  0.684849
C2.04:   0.813477,  1.815442  - satellites  Sequence Fractals Part I #24
 C3.20:   0.774380,  1.836743
 C3.22:   0.802629,  1.829248 
 C3.24:   0.827269,  1.802481 
 C3.28:   0.855612,  1.812967

z²+c+s_i
s_i = -1, 1, -1, 1, …
M1,1.00: 0.142733, 0.691465, Zoom = 50

  4 1/1 preperiodic points, with derivatives
M1,1.00:    0.143,   0.691  f'=   3.432,  -5.565, |f'|=  6.539
M1,1.01:    0.143,  -0.691  f'=   3.432,   5.565, |f'|=  6.539
M1,1.02:    0.857,   1.810  f'=   8.568,  10.873, |f'|= 13.843
M1,1.03:    0.857,  -1.810  f'=   8.568, -10.873, |f'|= 13.843

Here is the first M_1,1 Misiurewicz point.

Back in Sequence Fractals Part I #16 I remarked how the derivative near a Misiurewicz point describes the self-similarity. This is well-known in the fractal circles, however I could find only one sentence on my go-to reference, Wikipedia. Here is a link to the original proof by Tan Lei. similarityMJ.pdf

If c is a M_m,n, and λ is the derivative of the n-cycle, λ = (f_cⁿ)'(0), then for arbitrary small ε there is a radius r such that |x- λ(f_cⁿ)(x)| < ε whenever |x-c| < r.

Or more simply, if M is the set of non-escaping parameter space points (the picture), then M ≈ λ(f_cⁿ)(M) near c.

z²+c+s_i
s_i = -1, 1, -1, 1, …
M1,1.00: 0.142733, 0.691465, Zoom = 0.5
Julia, variable plane, view.

Julia view of M1,1.00. Tan Lei's paper similarityMJ.pdf also shows how the Julia set is self-similar, this time without the λ factor. It also describes the similarity between the Julia view and the parameter space non escaping set. The discussion and examples in the paper are about the Mandelbrot set. The theorems are stated more generally for rational functions, and so applies to our situation. (The two-step function is a degree 4 polynomial.)

z²+c+s_i
s_i = -1, 1, -1, 1, …
M2,1.00: -0.028482, 0.832244, Zoom = 5000

And here is the first (left most) M_2,1 Misiurewicz point.

    26 2/1 preperiodic points
M2,1.00:   -0.028,   0.832  f'=   5.429,   8.069, |f'|=  9.725
M2,1.01:   -0.028,  -0.832  f'=   5.429,  -8.069, |f'|=  9.725
M2,1.02:   -0.004,   0.767  f'=   3.550,  -4.296, |f'|=  5.573
M2,1.03:   -0.004,  -0.767  f'=   3.550,   4.296, |f'|=  5.573
M2,1.04:    0.108,   0.771  f'=  -1.304,   2.085, |f'|=  2.459
M2,1.05:    0.108,  -0.771  f'=  -1.304,  -2.085, |f'|=  2.459
M2,1.06:    0.143,   0.691  f'=   3.432,  -5.565, |f'|=  6.539
M2,1.07:    0.143,  -0.691  f'=   3.432,   5.565, |f'|=  6.539
M2,1.08:    0.252,   1.245  f'=   3.815,  -8.645, |f'|=  9.449
M2,1.09:    0.252,  -1.245  f'=   3.815,   8.645, |f'|=  9.449
M2,1.10:    0.464,   1.552  f'=   7.181,  10.114, |f'|= 12.404
M2,1.11:    0.464,  -1.552  f'=   7.181, -10.114, |f'|= 12.404
M2,1.12:    0.727,   1.039  f'=   7.840,   8.341, |f'|= 11.447
M2,1.13:    0.727,  -1.039  f'=   7.840,  -8.341, |f'|= 11.447
M2,1.14:    0.756,   1.270  f'=   4.497,   0.392, |f'|=  4.514
M2,1.15:    0.756,  -1.270  f'=   4.497,  -0.392, |f'|=  4.514
M2,1.16:    0.772,   1.834  f'=   2.074, -12.196, |f'|= 12.371
M2,1.17:    0.772,  -1.834  f'=   2.074,  12.196, |f'|= 12.371
M2,1.18:    0.804,   1.833  f'=   4.556,   3.965, |f'|=  6.040
M2,1.19:    0.804,  -1.833  f'=   4.556,  -3.965, |f'|=  6.040
M2,1.20:    0.823,   1.408  f'=   1.284, -10.258, |f'|= 10.338
M2,1.21:    0.823,  -1.408  f'=   1.284,  10.258, |f'|= 10.338
M2,1.22:    0.826,   1.806  f'=  -2.921,  -2.545, |f'|=  3.874
M2,1.23:    0.826,  -1.806  f'=  -2.921,   2.545, |f'|=  3.874
M2,1.24:    0.857,   1.810  f'=   8.568,  10.873, |f'|= 13.843
M2,1.25:    0.857,  -1.810  f'=   8.568, -10.873, |f'|= 13.843

z²+c+s_i
s_i = -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: f_c(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.

z²+c+s_i
s_i = -1, 1, -1, 1, …
Center: 0.00555195607569483, 0.796788357586998, Zoom = 8388608000

Summary so far: Comparing to z²+c, the escape set picture for this formula looks surprising normal. There are two main, separated, cardioids. They are surrounded by many little satellites. We have found more and "thinner" Misiurewicz points. However, despite the sparseness, every neighborhood of every M point contains infinitely many minis. (Conjecture supported by many pictures.)

When I first started looking at sequence fractals, my approach was quite random and the early results were nothing like these pictures. I had to dial back to the simplest cases to find familiar territory. We have been exploring in almost familiar territory, in the suburbs just outside of Mandelbrot city. Next let's take a cautious step into the jungle.