Showing point superattractive using Taylor expansion

Question

$\newcommand{\ee}{ \varepsilon }$

The Newton-Raphson function $N:\mathbb{R}\to\mathbb{R}$ associated with a function $f:\mathbb{R}\to \mathbb{R}$ is defined as

$$N(x)=x-\frac{f(x)}{f'(x)}$$

Using a Taylor series expansion of $N(x)$ about the point $x=p$, show that such a fixed point is superattractive, i.e.

$$|N(x)-p|\le K|x-p|^2,$$

for some $K>0$ and all $x$ in some neighbourhood $I$ of $p$.

Here's what approach to this problem I had:

$$N''(p) = \frac{f''(p)}{[f'(p)]}$$ $$N(x) = N(p)+N'(p)(x-p)+N''(p)(x-p)^2 + \mathcal{O}[(x-p)^3]=p+N''(p)(x-p)^2+\mathcal{O}[(x-p)^3]$$ $$=p+\frac{f''(p)}{f'(p)}(x-p)^2+\mathcal{O}[(x-p)^3]$$ $$\implies |N(x)-p|=\left| \frac{f''(p)}{f'(p)} (x-p)^2 + \mathcal{O}[(x-p)^3] \right|\le \left| \frac{f''(p)}{f'(p)}\right||x-p|^2+\left|\mathcal{O}[(x-p)^3]\right|$$ Let $|x-p|=\ee >0$, such that $\ee << 1$. Then $|N(x)-p|\le K\ee^2 + \underbrace{|O(\ee^3)|}_\text{$:=\ee'<\ee$}$. As $\ee$ tends to $0$, $\ee'$ tends to $0$ faster, so that

$$|N(x)-p|\le K\ee^2 = K|x-p|^2$$

Unfortunately, the proof is not very rigorous, since I didn't exactly get rid of the error term. I'm wondering how can one get rid of this term using the Taylor expansion approach?

Answer 1

By definition $\mathcal{O}[(x-p)^3]=B(x)(x-p)^3$ where $B$ is a bounded function near $x=p$. Then you can write $$ |N(x)-p|=\left| \frac{f''(p)}{f'(p)} (x-p)^2 + B(x)(x-p)^3 \right|=\Bigg|\underbrace{\frac{f''(p)}{f'(p)} + B(x)(x-p)}_{\le K\text{ near }x=p}\Bigg|\cdot |x-p|^2. $$

Answer 2

(I write $O_{f,p}$ to denote the dependency of the implicit constant on $f$ and $p$.)

If you know that $f'(p)\neq 0$ and that $$N(x)=p+\frac{f''(p)}{f'(p)}(x-p)^2+O_{f,p}[(x-p)^3]$$ then you know that $$N(x)=p+O_{f,p}[(x-p)^2]$$ which is precisely what the question is asking you to prove. This is just unwrapping the big O notation.