# Nucleus

The nucleus cc of a hyperbolic component of period pp satisfies:

Fp(0,c)=0F^p(0,c) = 0

Applying Newton’s method in one complex variable:

cm+1=cm−Fp(0,cm)∂∂cFp(0,cm)c_{m+1} = c_m - \frac{F^p(0, c_m)}{\frac{\partial}{\partial c}F^p(0, c_m)}

A reasonable starting guess for Newton’s method is within the atom domain of the component: Mandelbrot set Newton basins (2012). Note however, that even points in the atom domain might not converge to the desired root.

# 1 C99 Code

#include <complex.h>

double _Complex m_nucleus
    (double _Complex c0, int p, int n)
{
    double _Complex c = c0;
    for (int m = 0; m < n, ++m)
    {
        double _Complex z = 0;
        double _Complex dc = 0;
        for (int i = 0; i < p; ++i)
        {
            dc = 2 * z * dc + 1;
            z = z * z + c;
        }
        c = c - z / dc;
    }
    return c;
}

# 2 Examples

c0=−1.8c_0 = -1.8, p=3p = 3: c0=−1.8c1=−1.75(78298397040690…)c2=−1.7548(913878280483…)c3=−1.754877666(5449095…)c4=−1.7548776662466929(…)\begin{aligned} c_0 &= -1.8 \\ c_1 &= -1.75(78298397040690\ldots) \\ c_2 &= -1.7548(913878280483\ldots) \\ c_3 &= -1.754877666(5449095\ldots) \\ c_4 &= -1.7548776662466929(\ldots) \end{aligned}

c0=ic_0 = i, p=4p = 4: c0=ic1=−0.(11926605504587157…)+0.(9357798165137615…)ic2=−0.(23004008246918839…)+1.0(769584254327267…)ic3=−0.1(8143326956347502…)+1.0(265172581955131…)ic4=−0.15(552525333120717…)+1.0(295018249244792…)ic5=−0.1565(4521565515017…)+1.0322(758207945100…)ic6=−0.1565201(7318905109…)+1.03224710(96777757…)ic7=−0.1565201668337550(8…)+1.032247108922831(6…)ic8=−0.1565201668337550(5…)+1.0322471089228318(…)ic9=−0.1565201668337550(3…)+1.0322471089228318(…)ic10=−0.15652016683375508(…)+1.0322471089228318(…)i\begin{aligned} c_0 &= & i \\ c_1 &= -0.(11926605504587157\ldots) &+ 0.(9357798165137615\ldots) i \\ c_2 &= -0.(23004008246918839\ldots) &+ 1.0(769584254327267\ldots) i \\ c_3 &= -0.1(8143326956347502\ldots) &+ 1.0(265172581955131\ldots) i \\ c_4 &= -0.15(552525333120717\ldots) &+ 1.0(295018249244792\ldots) i \\ c_5 &= -0.1565(4521565515017\ldots) &+ 1.0322(758207945100\ldots) i \\ c_6 &= -0.1565201(7318905109\ldots) &+ 1.03224710(96777757\ldots) i \\ c_7 &= -0.1565201668337550(8\ldots) &+ 1.032247108922831(6\ldots) i \\ c_8 &= -0.1565201668337550(5\ldots) &+ 1.0322471089228318(\ldots) i \\ c_9 &= -0.1565201668337550(3\ldots) &+ 1.0322471089228318(\ldots) i \\ c_{10} &= -0.15652016683375508(\ldots) &+ 1.0322471089228318(\ldots) i \end{aligned}