If I'm given the following parameters,
Let the function $f(x)$ be four times continuously differentiable and have a simple zero ξ. Successive approximation $x_n, n = 1,2,...$ to ξ are computed from $x_{n+1} = {(x^′_{n+1}+x^{′′}_{n+1})\over2}$, where $x^{'}_{n+1} =x_n − {f(x_n) \over f^{′}(x_n)}$ $x^{′′}_{n+1} =x_n − {u(x_n) \over u^{′}(x_n)}$ $u = { f(x) \over f^{′}(x)}$
How can I prove that if the sequence ${x_n}$ converges to ξ, then the convergence is cubic.