Suppose $p$ is a simple root of $f(x)=0$ and $f^{(5)}(x)$ is continuous in the neighborhood of $p$. Define $$g(x)=x-{f(x) \over f'(x)}-{f''(x)\over {2f'(x)}}\left({f(x)\over f'(x)}\right)^2.$$ Show that the convergence rate of the fixed point iteration $p_n=g(p_{n-1})$ is at least cubic if $p_n\to p$.
The theorem says that if $g'(p)=g''(p)=g'''(p)=\dotsb=g^{(m-1)}(p), g^{(m)}(p) \neq 0$ for some $m \geq 2$ and $g^{(m)}(x)$ is continuous in a neighborhood of $p$, then
$$\lim_{n \to \infty}\left|{{p-p_n}\over{(p-p_{n-1})^m}}\right|={|g^{(m)}(p)|\over m!}>0$$ and the rate of convergence is $m$. I am getting some nasty derivative and expressions. Maybe I am doing something wrong.
I proved that the first two derivatives of $g(x)$ at p are zero. However, I have the problem with the last derivative which has to be non zero. I got $$g'''(p)=3({f''(p)\over f'(p)})^2-{f'''(p)\over f'(p)}\neq 0$$ How I show that it is indeed not equal to zero?