Show that the limit of the resulting sequence is $\sqrt{2}$. Consider the sequence of rational numbers defined recursively by the following formula: $$(x,y) \mapsto (x^2 + 2y^2, 2xy)$$ and starting at the point $(x,y) = (2,1)$. Show that the limit of the ratio of the resulting sequence is $\sqrt{2}$.
Is there any easy way to look it down and solve it.
I was writting in the form $$(x,y) \mapsto (x^2 + 2y^2, 2xy)= ((x+y)^2+x^2, (x-y)^2+x^2)$$
Still it is not showing any easy calculation later.
 A: The ratio is
$$\frac{x^2+2y^2}{2xy}=\frac{\dfrac xy+2\dfrac yx}2$$
and by setting 
$$t:=\frac xy,$$
$$\frac{x^2+2y^2}{2xy}=\frac{t+\dfrac2t}2.$$
You should recognize Henon's formula for the square root, which is a case of Newton's method and is known to converge from any nonzero starting value.
It is easy to see that convergence is to one of $\pm\sqrt2$ by solving
$$t=\frac{t+\dfrac2t}2.$$
A: Define $x_n = x_{n-1}^2+2y_{n-1}^2$, $y_n =2x_{n-1}y_{n-1}$ and $x_0 = 2$, $y_0=1$. It is obvious that $x_n$ and $y_n$ are positive for all $n$.
Assume that $\frac{x_n}{y_n}$ converges and denote it's limit by $L$. Then we have $$\frac{x_n}{y_n} = \frac{x_{n-1}^2+2y_{n-1}^2}{2x_{n-1}y_{n-1}} = \frac{\frac{x_{n-1}^2}{y_{n-1}^2}+2}{2\frac{x_{n-1}}{y_{n-1}}}$$ and by taking limit we get $$L = \frac{L^2+2}{2L} \implies 2L^2 = L^2 + 2 \implies L^2 = 2\implies L=\pm \sqrt 2.$$
Since, $\frac{x_n}{y_n}>0$ for all $n$, $L\geq 0$, and thus $L = \sqrt 2$.
To prove that $\frac{x_n}{y_n}$ converges, we can prove that it is decreasing and bounded bellow. Note that
$$\frac{x_n}{y_n}\geq \frac{x_{n+1}}{y_{n+1}} \iff \frac{x_n}{y_n}\geq \frac{x_n^2+2y_n^2}{2x_ny_n}\iff 2x_n^2\geq x_n^2+2y_n^2 \iff \frac{x_n}{y_n}\geq \sqrt 2,$$
so we can get both that the sequence is decreasing and is bounded bellow in one shot. You can prove that $\frac{x_n}{y_n}\geq 2$ by induction.
A: As per Yves' response, but a more general approach for $$(x,y) \mapsto (x^2+ay^2, 2xy), a >0$$
we have 
$$\frac{x^2+ay^2}{2xy}= \frac{1}{2}\left(\frac{x}{y}+a\frac{y}{x}\right)$$
which is equivalent to
$$s_{n+1}=\frac{1}{2}\left(s_{n}+\frac{a}{s_{n}}\right)$$
by noting $s_n=\frac{x_n}{y_n}$, with a proof here of the fact that this sequence has $\sqrt{a}$ as the limit.
