$A,B$ coefficients for maximal convergence

Question

In order to approximate the solution to the equation $f(x) = x^2-2$ we use the following iteration: $$x_{n+1} = x_n + A\frac{x_n^2 -2}{x^n} + B \frac{x_n^2-2}{x_n^3}$$ Find $A,B$ for maximal rate of convergence.

So first we can easily see that it is a proper function since if we apply $\sqrt 2$ for $x_n$ we indeed get $x_{n+1} = \sqrt 2$.

Now, I've understood that in order to get a maximal rate of convergence one should check when the derivatives are vanished. Is that the case here? What exactly the criteria?

I'd be glad for guidance.

Thanks!

Answer 1

By varying the coefficients $A$ and $B$, we can achieve some mathematical convenience, this is called method of undetermined coefficients.

Let $x_{n}=\sqrt{2}+e_{n}$, then

\begin{align*} x_{n+1} &= x_{n}+A\frac{x_{n}^{2}-2}{x_{n}}+B\frac{x_{n}^{2}-2}{x_{n}^{3}} \\ &= \sqrt{2}+e_{n}+A\frac{(\sqrt{2}+e_{n})^{2}-2}{\sqrt{2}+e_{n}}+ B\frac{(\sqrt{2}+e_{n})^2-2}{(\sqrt{2}+e_{n})^3} \\ &= \sqrt{2}+e_{n}+A(\sqrt{2}+e_{n})+\frac{B-2A}{\sqrt{2}+e_{n}}- \frac{2B}{(\sqrt{2}+e_{n})^{3}} \\ &= \sqrt{2}+e_{n}+A(\sqrt{2}+e_{n})+(B-2A) \left( \frac{1}{\sqrt{2}}-\frac{e_{n}}{2}+ \frac{e_{n}^{2}}{2\sqrt{2}}-\frac{e_{n}^{3}}{4}+\ldots \right) \\ & \quad -2B \left( \frac{1}{2\sqrt{2}}-\frac{3e_{n}}{4}+ \frac{3e_{n}^{2}}{2\sqrt{2}}-\frac{5e_{n}^{3}}{4}+\ldots \right) \\ &= \sqrt{2}+(1+2A+B)e_{n}+\frac{2A+5B}{2\sqrt{2}}e_{n}^{2}+ \frac{2A+9B}{4}e_{n}^{3}+\ldots \end{align*}

To optimize the rate of convergence, we make the linear and quadratic terms in $e_{n}$ vanished:

$$ \left \{ \begin{align*} 1+2A+B &= 0 \\ 2A+5B &= 0 \end{align*} \right.$$

On solving, we have $$(A,B)=\left( -\frac{5}{8},\frac{1}{4} \right)$$

Substitute back into $x_{n+1}$, we obtain cubic convergence: $$e_{n+1} \approx \frac{e_{n}^3}{4}$$