TITLE: A Recurrence Relation Involving a Square Root
QUESTION [5 upvotes]: Consider the recurrence relation:
$a_{n+1} = \sqrt{a_n^2 -k},$
where $k>0$, $n\in\{0,1,n:a_n^2\geq k\}$, and $a_0>0$ is known.
Is it possible to obtain an expression for $a_n$ in terms of $n$?

REPLY [9 votes]: Let $v_n=a^2_n$ then we have $v_{n+1}=v_n-k$: arithmetic sequence, so 
$v_n=v_0-nk$ and then 
$$a_n=\sqrt{a^2_0-nk}.$$

REPLY [3 votes]: Let $b_n=a_n^2$
$b_{n+1}=b_n -k$ 
Which is easy to solve.
Note that this will eventually become negative, so you'll have to restrict n or allow the number's to be complex.