Benchmarks

Table 1: Newton maps $N\_\{\backslash frac\{z^4\}\{p(z)\}\}$ with an exceptional point

1. Let $p$ be generic. If $p$ is linear, then it follows from the Scaling property that $N\_R(z)$ is conjugate to $N\_\{\backslash frac\{z\}\{z-1\}\}$ .

Let $p$ be non-linear. As $p$ is generic and $N\_\{\backslash frac\{z^d\}\{p(z)\}\}$ is a polynomial, we have $zp\text{'}(z)\; -\; dp(z)\; =\; \backslash alpha$ for some non-zero $\backslash alpha$ . Letting $y\; =\; p(z)$ , we have the first-order linear differential equation $y\text{'}\; -\; \backslash frac\{d\}\{z\}y\; =\; \backslash frac\{\backslash alpha\}\{z\}$ . The solution is

$$y\; \backslash cdot\; \backslash frac\{1\}\{z^d\}\; =\; \backslash int\; \backslash frac\{\backslash alpha\}\{z\}\; \backslash cdot\; \backslash frac\{1\}\{z^d\}\; dz\; +\; \backslash beta\; =\; -\backslash frac\{\backslash alpha\}\{dz^d\}\; +\; \backslash beta,$$

for an arbitrary constant $\backslash beta$ . Therefore, $p(z)\; =\; \backslash beta\; z^d\; -\; \backslash frac\{\backslash alpha\}\{d\}$ and $R(z)\; =\; \backslash frac\{z^d\}\{\backslash beta\; z^d\; -\; \backslash frac\{\backslash alpha\}\{d\}\}$ . Consider $c$ such that $c^d\; =\; \backslash frac\{\backslash alpha\}\{d\backslash beta\}$ and use the Scaling property to see that $N\_R$ is conjugate to $N\_\{\backslash beta\; R(cz)\}$ . We are done since $\backslash beta\; R(cz)\; =\; \backslash frac\{z^d\}\{z^d\; -\; 1\}$ and the resulting Newton map is $\backslash frac\{z^\{d+1\}\; +\; (d-1)z\}\{d\}$ .

2. Let $n$ denote the number of distinct roots of $p$ . Using the Scaling property, we assume without loss of generality that 1 is a multiple root of $p$ whenever $p$ is not generic. Along with this, what is going to be repeatedly used in all the following cases is that $g(z)$ is a non-zero constant (see Equation (3.1)).

(a) Let $\backslash deg(p)\; =\; 3$ . Then there are three cases depending on the values of $n$ . If $n\; =\; 3$ , then $p$ is generic, and from the first part of this theorem, it follows that $p(z)\; =\; z^3\; -\; 1$ , and hence $N\_R(z)\; =\; \backslash frac\{1\}\{3\}z(z^3\; +\; 2)$ . If $n\; =\; 2$ , then $p$ has a root with multiplicity 2 and therefore $p(z)\; =\; (z-1)^2(z-a)$ , where $a\; \backslash neq\; 0,\; 1$ . In this case, $g(z)\; =\; -(a+2)z\; +\; 3a$ and therefore $a\; =\; -2$ . Thus, $N\_R(z)\; =\; \backslash frac\{z\}\{6\}(z^2\; +\; z\; +\; 4)$ . If $n\; =\; 1$ , then $p(z)\; =\; (z-1)^3$ , and we get $N\_R(z)\; =\; \backslash frac\{1\}\{3\}z(z+2)$ .

(b) Let $\backslash deg(p)\; =\; 4$ . All possible cases of $p$ and the resulting Newton maps are given in Table 1.

(c) For $\backslash deg(p)\; =\; 5$ , all the possible forms of $p$ and resulting Newton maps are given in Table 2.

10