Determining the order of convergence of $ X_{n+1} = \frac{(X^3_n + 3aX_n)}{(3X^2_n + \alpha)} $

Solution 1:

You are studying a special case of the fixpoint iteration $x_{n+1}=g(x_n)$. If your case $g : \mathbb{R} \rightarrow \mathbb{R}$ is given by \begin{equation} g(x) = \frac{x^3+6x}{3x^2+2}. \end{equation} You are computing the derivative of $g$ in an attempt to determine if the restriction of $g$ to some closed interval $I$ is a contraction, which maps $I$ into itself. This is frequently an excellent strategy, but when the derivative(s) are complicated, alternatives should be explored.

From your equation it is clear that if the iteration converges to $X$, then $X \in \{-\sqrt{2},0,\sqrt{2}\}$. Since $g(x) > 0$ for all $x>0$, and $x_0 = 1 >0$, we see that $X = \sqrt{2}$ is the only viable option. We now find that \begin{equation} x_{n+1} - \sqrt{2} = \frac{x_n^3+6x_n-\sqrt{2}(3x_n^2+2)}{3x_n^2 +2}=\frac{x_n^3+3(\sqrt{2})^2x_n^2-3\sqrt{2}x_n^2-(\sqrt{2})^3}{3x_n^2+2}=\frac{(x_n-\sqrt{2})^3}{3x_n^2+2}. \end{equation} Now, if $e_n = \left|x_n - \sqrt{2} \right|$ denotes the absolute value of the error at the $n$th stage, then \begin{equation} e_{n+1} = \frac{e_n^3}{3x_n^2 +2} \leq \frac{1}{2}e_n^3. \end{equation} Since $e_0 = \left|\sqrt{2}-x_0\right|=\sqrt{2}-1 < 1$, it is clear that $e_n \rightarrow 0$ or equivalently that $x_n \rightarrow \sqrt{2}$. As for the exact order of the method, we observe that \begin{equation} \frac{e_{n+1} }{e_n^3} = \frac{1}{3x_n^2 + 2} \rightarrow \frac{1}{8} \not = 0, \quad n \rightarrow \infty, \quad n \in \mathbb{N}. \end{equation} By definition, this shows that the order is $p=3$.