# Julia Set Distance Estimate

# 2.1 Problem

There is a problem with the third formula when implemented on a computer: “spots” at pre-images of the attracting orbit, worse for strong (but not super) attraction.

See my decade-old math.stackexchange question (accessed 2025-02-20).

For example, the Julia set of $z \to z^2 + 0.01$:

# 2.2 Derivation Using Koenigs Linearisation

See Chapter 6 of:

Dynamics in one complex variable: introductory lectures, John W. Milnor, 1990, https://arxiv.org/abs/math/9201272

Corollary 6.4 introduces $\phi(z) = \lim_{n \to \infty} (z_n - \hat z) / \lambda^n$ as linearisation near an attractive fixed point with multiplier $\lambda$.

When using this for distance estimation, one calculates $\phi / \phi'$, and the $\lambda^n$ factor cancels.

However, as Milnor points out,

Note that $\phi$ has zeros at all iterated preimages of $\hat z$, and critical points at all iterated preimages of the critical point $\omega$. The function $z\mapsto\phi(z)$ is unbounded, and oscillates wildly as $z$ tends to $J = \delta\Omega$.

This means that $\phi / \phi'$ is zero and infinite at various points inside the attractive basin $\Omega$, and thus it isn’t really useful as a distance estimate.

# 2.3 Doomed Attempt Using L’Hospital’s Rule

L’Hospital’s Rule for limits says that (under some conditions) if $\lim\frac{f}{g}$ cannot be evaluated as $\frac{\lim f}{\lim g}$, then it might be evaluated as $\lim \frac{f'}{g'}$.

Applying L’Hospital’s Rule to the third formula gives a sequence of distance estimates:

$$\begin{aligned} \delta_0(z) &= \lim_{n \to \infty} \frac{|z_{pn} - z_*|}{\left|\frac{\partial}{\partial z} z_{pn} \right|} \\ \delta_1(z) &= \lim_{n \to \infty} \frac{\left|\frac{\partial}{\partial z} z_{pn} \right|}{\left|\frac{\partial^2}{\partial z^2} z_{pn} \right|} \\ \delta_2(z) &= \lim_{n \to \infty} \frac{\left|\frac{\partial^2}{\partial z^2} z_{pn} \right|}{\left|\frac{\partial^3}{\partial z^3} z_{pn} \right|} \\ &\vdots \end{aligned}$$

Then define a combined distance estimate as the maximum of all of these

$$\delta(z) = \max\{ \delta_i(z) : i = 0, 1, 2, \ldots \}$$

In practice, only a finite number can be calculated.

This is unprincipled as the pre-conditions for l’Hospital’s rule are not checked, and it’s not obvious that taking the maximum is always the right thing to do.

Anyway it doesn’t really work: the problem is not the evaluation of the limit, but that the value of the limit is bad as a distance estimate.