Newtons Modified Method

Question

In class we derived the convergence rate for Newton's method for a function such that the root was not simple. And we found it to be linear convergent However our professor then went on to a modified version of Newtons method for roots of multiplicity $m$: $x_{n+1} = x_n - m \frac{ f(x_n)}{f'(x_n)}$. However he just mentioned that assuming $x_n \rightarrow \alpha$ then the modified method is quadratically convergent. But I can't see why?

Answer 1

Consider $g(x)=\sqrt[m]{|f(x)|}$. Away from the roots of $f$ this function is differentiable, left and right of the root of multiplicity $m$ of $f$ it behaves like having a simple root.

Since $g'(x)=\frac1m sign(f(x))f'(x)|f(x)|^{1/m-1}$ one gets $$ \frac{g(x)}{g'(x)}=m\frac{f(x)}{f'(x)} $$ and since the convergence to the simple root of $g$ is quadratic, this trivially holds also true for the (same) m-fold root of $f$ with the modified formula.