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.


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


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

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).


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?


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


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!


1 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.

  • $\begingroup$ Thanks for your answer. If my computation as well as their induction are correct then why am I getting different Delta2 than the author started with? Also, do you know what technique is used to obtain the A, B, and C equations? Taylor? $\endgroup$
    – NightElfik
    Dec 19, 2014 at 6:22
  • $\begingroup$ The author's computation for $\Delta_2$ is incorrect, as you showed. The equations for $A_n$, $B_n$, and $C_n$ come from plugging $\Delta_n = A_n \delta + B_n \delta^2 + C_n \delta^3 + o(\delta^4)$ into $\Delta_{n+1} = 2 X_n \Delta_n + \Delta_n^2 + \Delta_0$. $\endgroup$ Dec 19, 2014 at 12:10
  • $\begingroup$ How do you know what to initialize the variables to? $\endgroup$
    – Byte11
    Nov 27, 2017 at 23:24
  • $\begingroup$ Can this be used for the Julia fractal too (I don't see any reason why not)? Julia replaces $X2n+X0$ by $X2n+c$. $\endgroup$
    – Bim
    Nov 15, 2018 at 19:40
  • $\begingroup$ @Byte11 because $\Delta_0 = \delta = 1 \delta + 0 \delta^2 + 0 \delta^3 + \ldots$ $\endgroup$
    – Claude
    Nov 18, 2018 at 15:51

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