prove that $x_{n+1}=\frac{x_n(x_n^2+15)}{3x_n^2+5}$ is cubic order of convergence near $x_0=\sqrt{5}$

Question

To solve the equation $$x^2-5=0$$ There exitsts a iteration method $$x_{n+1}=\frac{x_n(x_n^2+15)}{3x_n^2+5}$$ I know that it is cubic convergence but I don't know how to prove it.

I have tried the following $$ \begin{align} e_{k+1}&=x_{n+1}-\sqrt{5} \\ &=\frac{x_n(x_n^2+15)}{3x_n^2+5}-\sqrt{5} \\ &=\frac{\frac{1}{3}x_n(3x_n^2+5+40)}{3x_n^2+5}-\sqrt{5} \\ &=\frac{1}{3}x_n(1+\frac{40}{3x_n^2+5})-\sqrt{5} \\ &=\frac{\frac{40}{3}x_n}{3x_n^2+5}+\frac{1}{3}x_n-\sqrt{5} \end{align} $$ I'm trying to extract $e_n=x_n-\sqrt{5}$ from $e_{n+1}$ in an effort of proving $lim \frac{e_{n+1}}{e_n^3}\rightarrow C$ . But it seems that I'm not on the right way.

Do you have any idea? Any discussion is appreciated. Thanks in advance.

Answer 1

In general, Halley's method for the square root reads as $$ x_{n+1}=x_n·\frac{x_n^2+3a}{3x_n^2+a} $$ The difference to the square root develops then as $$ x_{n+1}-\sqrt{a}=\frac{x_n^3+3x_n\sqrt a^2 - 3x_n^2\sqrt a-\sqrt a^3}{3x_n^2+a} =\frac{(x_n-\sqrt a)^3}{3x_n^2+a} $$ which gives you the third order convergence.

Answer 2

Hint: Define $f(x):=\frac{x(x^2+15)}{3x^2+5}$, and Taylor expand it around $x=\sqrt{5}$.

If you want to continue with the method you're currently working with, then you'll want to replace then $x_n$ terms by $e_n+\sqrt{5}$.