TITLE: Newton's method for roots of multiplicity $> 1$
QUESTION [10 upvotes]: I'm studying numerical analysis and I read in Wikipedia that "If the root being sought has multiplicity greater than one, the convergence rate is merely linear".
The article in Wikipedia also mentions that "if the multiplicity $m$ of the root is known, one can use the following modified algorithm that preserves the quadratic convergence rate:
$$x_{n+1} = x_n -m \bigl[f(x_n)/f'(x_n)\bigr].$$
I tried proving this and I couldn't do it (I only managed to prove that convergence rate is at least linear).
Any ideas on how can this be proven ?

REPLY [4 votes]: Suppose that $x$ is a m-multiple root.
then you can have $f(x_k)=(x_k-x)^mf^{(m)}(x)/m!+O((x_k-x)^{m+1})$, derivate it 
$f'(x_k)=m(x_k-x)^{m-1}f^{(m)}(x)/m!+O((x_k-x)^m)$
$$x_{k+1}-x=x_k-f(x_k)/f'(x_k)-x$$
$$=x_k-x-(x_k-x)/m+O((x_k-x)^2)$$
you see this is linear convergence.
But if you change here $x_{k+1}=x_k-m[f(x_k)/f'(x_k)]$
then you can get this result $x_k-x-m(x_k-x)/m+O((x_k-x)^2)=O((x_k-x)^2)$