Newton's method yields super-attractive fixed points

mark
Mon, 11 Sep 2017 14:38:34 +0000
Problems

Let [imath]f:\mathbb R \to \mathbb R[/imath] be continuously differentiable. We say that [imath]x_0[/imath] is a simple root of [imath]f[/imath] if [imath]f(x_0)=0[/imath] and [imath]f'(x_0)\neq 0[/imath]. Show that if [imath]x_0[/imath] is a simple root of [imath]f[/imath], then [imath]x_0[/imath] is a super-attracting fixed point of the Newton's method iteration function [imath]N[/imath] for [imath]f[/imath].