Newton's method yields super-attractive fixed points

mark

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].

SomeCallMeTim

Let $f$ be a function as described above. Then note, the newton's method defined for the function is [dmath] N(x) = x - \frac{f(x)}{f'(x)}. [/dmath]

Let us consider the derivative of this function, [imath]N'(x)[/imath]. Using the quotient rule, we calculate [dmath] N'(x) = 1 - \frac{f'(x)f'(x)-f''(x)f(x)}{f'(x)f'(x)} .[/dmath] We know that each of these terms exist, because $f$ is continuously differentiable.

Now let us consider [imath] N'(x_0)[/imath]. We know that [imath] f(x_0)=0, f'(x_0)\neq 0[/imath], so we have [dmath]N'(x_0)=1 - \frac{f'(x_0)^2 - f''(x_0)\cdot f(x_0)}{f'(x_0)^2} = 1- \frac{f'(x_0)^2 - f''(x_0)\cdot 0}{f'(x_0)^2} = 1-1 = 0.[/dmath]

Since [imath]N'(x_0)=0, x_0[/imath] is a super-attractive fixed point for [imath]N(x)[/imath].