Third order convergence of an iteration scheme Consider the iteration scheme 
$x_{n+1}=\alpha x_n(3-\frac{x_n^2}{a})+\beta x_n(1+\frac{a}{x_n^2})$
For third order convergence to $\sqrt 2$, the values of $\alpha$ and $\beta$ are ......
I tried it by plugging $x_{n+1}=x_n=\sqrt a$ as $n\rightarrow\infty$ and got $\alpha +\beta =1$. 
After this I tried to put it in the form 
$Lim_{n\rightarrow\infty}$$\frac{|x_{n+1}-\sqrt a|}{|x_n-\sqrt a|^3}=M$ where $M\ne 0$ 
but couldn't succeed. Any suggestion and hints please!
The given answer is $\alpha=1/8$,$\beta=3/8$.
 A: Let $x_n = \sqrt{2}+\epsilon$ (and of course $a=2$).  Then 
$$
x_{n+1} = \alpha (\sqrt{2}+\epsilon)(3-\frac12(\sqrt{2}+\epsilon)^2
+\beta (\sqrt{2}+\epsilon) \left( 1+\frac{2}{(\sqrt{2}+\epsilon)^2}\right)
$$
Expanding in $\epsilon$, to one order more than we need in the first line,
$$
x_{n+1} = (\sqrt{2}+\epsilon) \left( \alpha (2-\sqrt{2}\epsilon-\epsilon^2/2)+\beta\left(1+1-\sqrt{2}\epsilon +\frac32 \epsilon^2 -\sqrt{2} \epsilon^3 + O(\epsilon^4)\right) \right)
\\
x_{n+1} = (2\sqrt{2}\alpha + 2\sqrt{2}\beta) + \left( -2\alpha +2\alpha -2\beta + 2\beta\right) \epsilon \\+ \left(-\frac12\sqrt{2}\alpha-\sqrt{2}\alpha +\frac32\sqrt{2}\beta-\sqrt{2}\beta\right)\epsilon^2+O(\epsilon^3)
$$
and since this needs to equal $\sqrt{2} + O(\epsilon^3)$ we have 
$$
\alpha+\beta = \frac12\\ \frac12 \beta - \frac32 \alpha = 0
$$
the solution to which is $\alpha = \frac18, \beta = \frac38$.
A: We start by considering two sequences - 
$$ A \rightarrow x_{n+1} = \frac{x_{n}}{2}(1 + \frac{a}{x_{n}^2})$$ 
and 
$$ B \rightarrow x_{n+1} = \frac{x_{n}}{2}(3 - \frac{x_{n}^2}{a}) $$
We can observe that by letting $\lim_{n \rightarrow \infty} x_{n} = \xi$ and $\lim_{n \rightarrow \infty} x_{n+1} = \xi$ , $\xi^2 = a$ where $\xi$ is the exact solution.
Next we substitute $x_{n} = \xi + \epsilon_{n}$ and $x_{n+1} = \xi + \epsilon_{n+1}$ and $a = \xi^2$ to both $A$ and $B$ and we obtain the following equations respectively.
In case of 
$$A : \epsilon_{n+1} = \frac{\epsilon_{n}^2}{2 \xi} + O(\epsilon^3)$$
which implies that iti s 2nd oreder convergent with error constant $\frac{1}{2\xi}$.
for the case 
$$B :  \epsilon_{n+1} = \frac{-3}{2\xi} (\epsilon^{n})^2 + O(\epsilon_{n}^2)$$
We next do this operation - adding three times equation $A$ to equation $B$,in order to get third order convergence as after this operation $\epsilon_{n+1} = O(\epsilon_{n}^3)$.
so after the above operation we get the following equation 
$$x_{n+1}  = \frac{1}{8}x_{n}(6 + 3\frac{a}{x_{n}^2} - \frac{x_{n}^2}{a})$$
and comparing to your given formula
we get the coefficients $\alpha = \frac{1}{8} , \beta = \frac{3}{8}.$
