# # Nucleus

The nucleus $$c$$ of a hyperbolic component of period $$p$$ satisfies:

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

Applying Newton’s method in one complex variable:

$c_{m+1} = c_m - \frac{F^p(0, c_m)}{\frac{\partial}{\partial c}F^p(0, c_m)}$

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

$$c_0 = -1.8$$, $$p = 3$$: \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}

$$c_0 = i$$, $$p = 4$$: \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}