A Recurrence Relation Involving a Square Root

Question

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 $n$?

Answer 1

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 $$a_n=\sqrt{a^2_0-nk}.$$

Answer 2

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.