How can find the values of $\alpha$ and $\beta$, such that $x_{n+1}=\alpha x_n\left(3-\dfrac{x_n^2}{a}\right)+\beta x_n\left(1+\dfrac{a}{x_n^2}\right)$ has $3^{\text{rd}}$ order convergence to $\sqrt a$ ?
Set $x_{n+1}=f(x_n)=\alpha x_n(3-\frac{x_n^2}{a})+\beta x_n(1+\frac{a^2}{x_n^2})$ and denote $z=\sqrt{a}$. Replace $x_n$ by $x$.
Using Taylor's development we find:
$$f(x)=f(z)+f'(z)(x-z)+\frac{f''(z)}{2}(x-z)^2+\frac{f'''(z)}{6}(x-z)^3+\frac{f''''(\xi)}{24}(x-z)^4$$
for some $\xi\in(x,z)$. Since $f(z)=z$ and dividing the previous equation by $|x-z|^3$ we have:
$$\frac{f(x)-f(z)}{|x-z|^3}=\frac{f'(z)}{|x-z|^2}+\frac{f''(z)}{2|x-z|}+\frac{f'''(z)}{6}+\frac{f''''(\xi)}{24}|x-z|$$
So in order to converge to the $3^{rd}$ order we have to impose both $f'(z)=0$ and $f''(z)=0$. This give raise to the following system of equations:
$$ \begin{align} \beta(1-a)&=0\\ -3\alpha+\beta&=0 \end{align} $$
and hence
$$ \begin{align} \beta(1-a)&=0\\ \beta&=3\alpha& \end{align} $$
Looking at the proposed solutions we have to choose $\color{red}{\boxed{\color{black}{(2)}}}$
Remark: solution $(4)$ seems to be a duplicate of $(2)$.
(2) $\alpha=\frac{1}{8}$ and $\beta=\frac{3}{8}$
(3) $\alpha=\frac{2}{8}$ and $\beta=\frac{2}{8}$
(4) $\alpha=\frac{1}{8}$ and $\beta=\frac{3}{8}$
– Surya Deep Mishra Feb 21 '17 at 15:48