Nonlinear reccurence relation: $a^{2}_{n} - 2a_{n-1} = 0, a_{0} = 2$ How would one go about solving such recurrence relation? I tried writting it as $$a_{n} = \sqrt{2a_{n-1}}$$
and after inputing some numbers it is obvious that $$a_n = 2 $$
But is this correct?
 A: Let $f (x)=\sqrt {2x}$
the fixed points are $0$ and $2$.

if $a_0=0$ then $\forall n  \,a_n=0$
if $a_0=2$ then $\forall n  \;a_n =2$

if $a_0 <0$ the sequence is not defined.
suppose $a_0>0.$
then $\forall n \;\; a_n>0$.
$f'(x)=\frac {1}{\sqrt {2x}}>0$
$\implies f $ is increasing at $[0,+\infty) $
$\implies (a_n) $ is monotonic.
on the other hand
$f (x)-x=\sqrt {x}(\sqrt {2}-\sqrt {x}). $


*

*first case :$0 <a_0 <2$

then by induction  $0 <a_n<2$  and
$a_{n+1}-a_n=f (a_n)-a_n>0$
increasing and bounded, the sequence goes to $2$.


*

*second case for you.


A: From 
$$a_n^2=2a_{n-1}$$ you draw $$a_n=\pm\sqrt{2a_{n-1}}.$$ Obviously, the negative solution must be rejected as it causes the next iterates to be undefined. Also note that $a_0=0$ makes the sequence all zeroes.
Otherwise, take the logarithm and set $b:=\log a$. The recurrence is
$$b_n=\frac{b_{n-1}+\log2}2.$$
Then by induction,
$$b_n=2^{-k}b_{n-k}+(1-2^{-k})\log 2=2^{-n} b_0+(1-2^{-n})\log 2$$
and
$$a_n=2\left(\frac{a_0}2\right)^{(2^{-n})}.$$
For any $a_0>0$, the sequence quickly converges to $2\left(\dfrac{a_0}2\right)^0=2$.
