TITLE: Alternative unconditional form of $\sqrt{n -\sqrt{n -\sqrt{n -\cdots}}}$?
QUESTION [5 upvotes]: Consider $a_n$, where
$$\begin{align} a_n &=\small{\sqrt{n -\!\!\!\sqrt{n -\!\!\!\sqrt{n -\!\!\sqrt{n -\!\!\sqrt{n -\!\!\sqrt{n -\!\sqrt{n - \cdots}}}}}}}}\end{align}$$
Using a recursive solution, such that:
$$a_n = f(n) = \sqrt{n - f(n)}$$
is too slow, while an iterated form don't fit my usage.
Is there a unconditional form of $a_n$ which don't rely as heavily on self-reference or recursion? Maybe an approximation?

REPLY [4 votes]: $a_n =\small{\sqrt{n -\!\!\!\sqrt{n -\!\!\!\sqrt{n -\!\!\sqrt{n -\!\!\sqrt{n -\!\!\sqrt{n -\!\sqrt{n - \cdots}}}}}}}}$
$a_n^2 =\small{n -\!\!\!\sqrt{n -\!\!\!\sqrt{n -\!\!\sqrt{n -\!\!\sqrt{n -\!\!\sqrt{n -\!\sqrt{n - \cdots}}}}}}}$
$a_n^2 -n =\small{ -\!\!\!\sqrt{n -\!\!\!\sqrt{n -\!\!\sqrt{n -\!\!\sqrt{n -\!\!\sqrt{n -\!\sqrt{n - \cdots}}}}}}}$
$a_n^2 -n =-a_n$
$a_n^2+a_n -n =0$
Using quadratic formula for positive root.
$a_n=\dfrac{-1+\sqrt{1+4n}}{2}$

REPLY [3 votes]: $$a_n=\sqrt{n-a_n}\implies a_n^2 = n-a_n \implies a_n^2+a_n-n = 0 \implies a_n\in\left\{\frac12\left(-1\pm\sqrt{1+4n}\right)\right\}$$
Since $a_n$ is positive, we therefore have
$$a_n =\frac{\sqrt{4n+1}-1}{2}$$