Linear convergence with Newton method multiple root

Question

I've a function $f \in C^{3}(\mathbb{R})$ with $f(\alpha)=f'(\alpha)=0$ and $f''(\alpha) \neq 0 $ and I'm trying to proove that the Newton's method has a linear convergence and that my sequence is well defined. Since $f$ has a multiple root, I can write $f(x)=(x-\alpha)^{2}g(x).$ We have for the Newton method $$x_{n+1}= x_n-\frac{f(x_n)}{f'(x_n)}.$$ Hence $$ |x_{n+1}-\alpha| =\left| \frac{g(x_n)+(x_n-\alpha)g'(x_n)}{2g(x_n)+(x_n-\alpha)g'(x_n)} \right||x_n-\alpha|$$ I want to find $k<1$ (where $k$ does not depend on $n$), with $$ |x_{n+1}-\alpha|\leq k|x_n-\alpha|.$$ How can I do it?

Answer 1

As $g(α)\ne 0$ you can find an interval around $α$ where $$ \left|\frac{g'(x)(x-α)}{g(x)}\right|<\frac13 $$ Then on this interval the fraction that has to be bounded is smaller than $$ \frac{1+\frac13}{2-\frac13}=\frac45. $$ By making the first bound smaller towards $0$, you decrease also the second towards $\frac12$.