TITLE: Convergence rate of Newton's method
QUESTION [15 upvotes]: Let $f(x)$ be a polynomial in one variable $x$ and let $\alpha$ be its $\delta$-multiple root 
($\delta\ge2$). 
Show that in the Newton's $x_{k+1}=x_k-f(x_k)/f'(x_k)$, the rate of convergence to $\alpha$ is not quadratic.
My solution: Suppose that $\alpha$ is one regular root of equation.Then 
$$x_{k+1}=x_k-\frac{f(x_k)}{f'(x_k)}=\phi(x_k)$$
convergence rate to $\alpha$:
$$x_k-\alpha=\phi(x_k)-\phi(\alpha)=(x_k-\frac{f(x_k)}{f'(x_k)})-(x-\frac{f(x)}{f'(x)})$$
$$(x_k-\alpha)=(f'(x_k))^{-1}(f(x_k)-f(x))=(x_k-\alpha)+(f'(x_k))^{-1}\{ f'(x_k)(x_k-\alpha)+O((x_k-\alpha)^2)\}=f'(x_k)^{-1}O((x_k-\alpha)^2)) 
\tag{1}$$
so the convergence rate to $\alpha$ is quadratic.
But if $\alpha$ is not regular root, then $(f'(x))^{-1}$ has no meaning. We need do slightly change in $(1)$,
$$(x_k-f'(x_k)^{-1}f(x_k))-(x-f^{(\delta)}(x)\delta!f(x))=(x_k-x)-f'(x_k)(f(x_k)-f(x))$$
$$=(x_k-x)-f'(x_k)^{-1}(\frac{f^{(\delta)}(x_k) (x_k-x)^{\delta}}{\delta!}+O((x_k-x)^{\delta+1}))$$
convergence rate to non regular root $\alpha$ is one.
Is my solution correct?
Then I do same thing to next nonlinear equation:
$$f(x)=x^2(x-1) $$, m is integer.
so the newton's formula is above, and how about convergence rate to $0,1$? 
I think convergence to 1 is one, absolutely convergence to 0 is quadratic.
when convergence rate is 1, the how about the convergence rate?

REPLY [3 votes]: You are making it too hard.
If
$x_{k+1}=x_k-af(x_k)/f'(x_k)
$
and
$f(x_k)$ is small
and
$f'(x_k)$ is bounded away from zero,
then,
since
$f(x+h)
=f(x)+hf'(x)+O(h^2)
$,
$\begin{array}\\
f(x_{k+1})
&=f(x_k-af(x_k)/f'(x_k))\\
&=f(x_k)-(af(x_k)/f'(x_k))f'(x_k)+O((af(x_k)/f'(x_k))^2\\
&=f(x_k)-af(x_k)+O((f(x_k)/f'(x_k))^2)\\
&=f(x_k)(1-a)+O((f(x_k)/f'(x_k))^2)\\
\end{array}\\
$
If
$a=1$
then
$f(x_{k+1})
=O((f(x_k)/f'(x_k))^2)
$
and the convergence is quadratic.
If
$a\ne 1$
then
$f(x_{k+1})
=f(x_k)(1-a)+O((f(x_k)/f'(x_k))^2)
$
and the convergence is linear.