I have a problem with this exercise.
Let a function $f∈C^{p+1}(\mathbb{R})$ be given. let f have a p-fold root in $x^∗$, $p ∈ N$.
Now consider the following variant of Newton's method to determine $x^∗$:
$x_{n+1}=x_{n}-p·\frac{f(x_{n})}{f'(x_{n})} \text{ for } n=0,1,2,...$.
Determine the local order of convergence of the iteration procedure.
I really don't know how to proceed. Can someone help me?
You basically need to show that $g'(z)=0$, where $g(x)= x- p\cdot \frac{f(x)}{f'(x)}$ and $z$ is the solution of the equation. Since $z$ is a root with order $p$, we know that $f(x) = (x-z)^p \phi(x)$, where $\phi(z)\ne 0$. Use this to show that $\lim_{x\to z} g'(x)=0$.
If the sequence $x_{n+1} = g(x_n)$ converges to $z$ and $$ g'(z)=g''(z)=\cdots =g^{(p-1)}(z), \quad g^{(p)}(z)\ne 0 $$ you can conclude that it converges with order $p$. This is a direct consequence of Taylor's formula: $$ g(x_n) = g(z) + g'(z)(x_n-z) + \cdots + \frac{g^{(p-1)}(z)}{(p-1)!}(x_n-z)^{p-1} + \frac{g^{(p)}(\xi_n)}{p!}(x_n-z)^p. $$
Noting that $g(z)=z$, $x_{n+1} = g(x_n)$, and that the derivatives of $g$ at $z$ are zero up to the order $p-1$, the previous expression becomes $$ |x_{n+1}-z| = \dfrac{|g^{(p)}(\xi_n)|}{p!} |x_n -z|^p, $$
which establishes convergence with order $p$.