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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.

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