# Julia Size

The central feature of a Julia set associated with a baby Mandelbrot set is a circle.

Let f0=0,f1=c,f2=c2+c,f_0 = 0, f_1 = c, f_2 = c^2 + c, \ldots be the orbit of the origin.

Consider z0=λz_0 = \lambda on the boundary of the circle.

Then z1=c+λ2=f1+hz_1 = c + \lambda^2 = f_1 + h.

Write the Taylor expansion about ff:

zp=fp+dfpdf1h+z_p = f_p + \frac{\mathrm{d} f_p}{\mathrm{d} f_1} h + \ldots.

If cc is periodic with period pp, then fp=0f_p = 0 and you can pick z0z_0 on the boundary of the circle such that z0=zpz_0 = z_p.

This gives λ=dfpdf1λ2\lambda = \frac{\mathrm{d} f_p}{\mathrm{d} f_1} \lambda^2, that is,

λ=1/dfpdf1 \lambda = 1 / \frac{\mathrm{d} f_p}{\mathrm{d} f_1}

# 1 C99 Code

#include <complex.h>

double _Complex m_julia_size(double _Complex c, int p)
{
    double _Complex z = c;
    double _Complex dz = 1;
    for (int q = 1; q < p; ++q)
    {
        dz = 2 * z * dz;
        z = z * z + c;
    }
    return 1 / dz;
}

# 2 Examples

Circle: c=0c = 0, p=1p = 1: λ=1\lambda = 1.

Airplane: c=1.7548c = -1.7548\ldots, p=3p = 3: λ=0.10753\lambda = -0.10753\ldots.

Kokopelli: c=0.15652+i1.0322c= -0.15652\ldots + i 1.0322\ldots, p=4p = 4: λ=0.074812+i0.038616\lambda = -0.074812\ldots + i 0.038616\ldots, |λ|=0.084191|\lambda| = 0.084191\ldots.