<s>[INST] 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}$$\frac{|x_{n+1}-\sqrt a|}{|x_n-\sqrt a|^3}=M$ where $M\ne 0$ 
but couldn't succeed. Any suggestion and hints please!
The given answer is $\alpha=1/8$,$\beta=3/8$.
 [/INST] Let $x_n = \sqrt{2}+\epsilon$ (and of course $a=2$).  Then 
$$
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 $\alpha = \frac18, \beta = \frac38$.
</s>