Perturbation of Mandelbrot set fractal

Question

I recently discovered very clever technique how co compute deep zooms of the Mandelbrot set using Perturbation and I understand the idea very well but when I try to do the math by myself I never got the right answer.

I am referring to original PDF by K.I. Martin but I will put the necessary equations below.

Theory

Mandelbrot set is defined as $X_{n+1} = X_n^2 + X_0$.

Where the complex number $X_0$ is in the Mandelbrot set if $|X_n| \leq 2$ for all n. Otherwise we assign a color based on $n$ where $|X_n| > 2$.

Now consider another point $Y_0$ that gives us $Y_{n+1} = Y_n^2 + Y_0$.

Let $\Delta_n = Y_n - X_n$, Then

$\Delta_{n+1} = Y_{n+1} - X_{n+1} = 2X_n \Delta_n + \Delta_n^2 + \Delta_0$

So far this is crystal clear to me. But now we want to compute $\Delta_n$ directly from $\Delta_0$ using pre-computed coefficients of the recursive equation.

The author continues:

Let $\delta = \Delta_0$

$\Delta_1 = 2X_0\delta + \delta^2+\delta = (2X_0+1)\delta + \delta^2\\ \Delta_2 = (4X_1X_0 - 2X_1-1)\delta + ((X_0-1)^2+2X_1)\delta^2 + (4X_0-2)\delta^3 + o(\delta^4)$

Let $\Delta_n=A_n\delta+B_n\delta^2+C_n\delta^3+o(\delta^4)$

Then

$A_{n+1} = 2 X_n A_n + 1\\ B_{n+1} = 2 X_n B_n + A_n^2\\ C_{n+1} = 2 X_n C_n + 2 A_n B_n$

Knowing all $X_n$ we can pre-compute $A_n$, $B_n$, and $C_n$ and given new point $Z_0$ we can compute $\delta_z$ and searching for $|Z_n| > 2$ is just binary search that is O(log n).

Question

My question is how to compute the equations for $A_n$, $B_n$, and $C_n$? I tried to "check" the equations by applying the $\Delta$ recurrence but I obtained:

$\Delta_2 = 2 X_n \Delta_1 + \Delta_1^2+\Delta_0 =\\ =(4X_1 X_0 + 2 X_1 + 1) \delta + (2X_1 + (2X_0 + 1)^2)\delta^2 + (4X_0+2)\delta^3+\delta^4$

Which does not match author's $\Delta_2$.

I have also tried to apply given formulas for $A_n$, $B_n$, and $C_n$ to compute forst few $\Delta$'s but they matched my $\Delta_2$, not authors (for example $C_2 = 4 X_0 + 2).$

What am I doing wrong? Is it something with complex numbers?

Bonus

There is probably a general formula for $\Delta_n$, can you help me to find it? Something like $\Delta_n=\sum_{i=0}^\infty C_n^{(i)} \delta^i$.

Edit

Anybody? The "Tumbleweed" badge for this question is cool but I thought this should be rather "simple" problem. The solution should probably involve Taylor series, I just need to point out to the right direction. Thanks!

Answer 1

Your computation is correct:

$\Delta_2 = (4X_1 X_0 + 2 X_1 + 1) \delta + (2X_1 + (2X_0 + 1)^2)\delta^2 + (4X_0+2)\delta^3+\delta^4$

Their computation for the induction is correct:

$ A_{n+1} = 2 X_n A_n + 1 \\ B_{n+1} = 2 X_n B_n + A_n^2 \\ C_{n+1} = 2 X_n C_n + 2 A_n B_n $

Initialise $A_0 = 1, B_0 = 0, C_0 = 0$ and use their computation for the induction.