Fixed point method: order of convergence

Question

Consider the following fixed point method to evaluate the cubic root of a number $N$: $$ x(n+1)=(2x(n)/3)+(N/3x(n)^2) $$ Let $g(x)=(2x/3)+(N/3x^2)$

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Theoretically, the order of convergence is exactly 2 since $g'(N^1/3)=0$ and $g''(N^1/3)≠0$

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After this algorithm has been applied for the calculation of the cubic root of 100 starting with $x(0)=5$, the values of $e(n+1)/e(n)$ converges to $0$ and the values of $e(n+1)/e(n)^2$ converges to $0.21544$.

This numerical result seems not to confirm the theoretical order of convergence.

Any interpretation???

Answer 1

The computation does confirm the order.

If the order is $k$, then $\lim_{n \to \infty} \dfrac{e_{n+1}}{e_n^k} $ exists and $\lim_{n \to \infty} \dfrac{e_{n+1}}{e_n^j} =0 $ for $j < k$.

Your data matches this.

Why do you think otherwise?