Attractiver vs super attractive Newton's method

mark

Let [imath]k[/imath] be an integer and let [imath]f(x)=k x + x^2[/imath].

• Find the Newton's method iteration function [imath]N[/imath] for [imath]f[/imath] and simplify it.
• Show that the origin is super attractive if [imath]k\neq 0[/imath] but that it is just attractive if [imath]k=0[/imath].

SomeCallMeTim

Let [imath]k \in \mathbb{Z} [/imath], and let [imath]f(x) = kx + x^2 [/imath].

Then the Newton's method iteration function [imath] N(x)[/imath] for [imath]f(x)[/imath] is [dmath]N(x)=x-\frac{f(x)}{f'(x)}=x-\frac{kx+x^2}{2x+k} = \frac{(2x+k)x-(kx+x^2)}{2x+k}[/dmath]
[dmath]= \frac{2x^2+kx -kx -x^2}{2x+k}=\frac{x^2}{2x+k}.[/dmath]

Let us assume that [imath]k \neq 0[/imath]. Then note by the quotient rule [dmath]N'(x)=\frac{(2x)(2x+k)-2x^2}{(2x+k)^2}= \frac{2x^2+2kx}{4x^2+4kx+k^2}.[/dmath]
If we then evaluate the value of this derivative at [imath]0[/imath], we have: [dmath]N'(0)=\frac{2(0)^2+2k(0)}{4(0)^2+4k(0)+k^2}=\frac{0}{k^2}=0.[/dmath]
Accordingly, for [imath] k \neq 0, 0[/imath] is a super attractive fixed point for [imath]N(x)[/imath].

Now let us return to our formula [imath]N(x)=\frac{x^2}{2x+k}[/imath], but let us assume that [imath] k=0 [/imath]. Then note, [dmath]N(x)=\frac{x^2}{2x+0}=\frac{x^2}{2x}=\frac{x}{2} \Rightarrow N'(x) = \frac{1}{2} = N'(0).[/dmath]

Since [imath]|N'(0)|=\frac{1}{2}[/imath] whenever [imath] k=0, 0 [/imath] is an attractive, but not super attractive, fixed point of [imath]N(x)[/imath] in this case.