# Solving $\scriptsize a_n=\sqrt{a_{n-1}+\sqrt{a_{n-2}+\sqrt{a_{n-3}+\ldots}}}$ with range transformation

This is a practice problem provided by a textbook on recurrences.

Solve using range transformation:

$a_n=\sqrt{a_{n-1}+\sqrt{a_{n-2}+\sqrt{a_{n-3}+...}}}$, where $a_0$ =4

The hint is to view the first few terms to find a pattern. So what I've got: $a_0=4$, $a_1=\sqrt{a_0}$ $= 2$, $a_2=\sqrt{a_1+\sqrt{a_0}}$ $=2$, $a_3=\sqrt{a_2 +\sqrt{a_1+\sqrt{a_0}}}$ $=2$

My guess is the range transformation needs to be based on $n^k$ where k is the degree of the root, but the degree of the root is increasing with the recurrence.

• Is $a_0 = 0$ or $a_0 = 4$? – JimmyK4542 Apr 30 '15 at 1:14
• It's given as 4 in the problem, I'll edit the first few terms to express this. – Whitesizzle Apr 30 '15 at 1:17

HINT: If $a_n = \sqrt{a_{n-1}+\sqrt{a_{n-2}+\sqrt{a_{n-3}+\cdots}}}$ for all positive integers $n$, then we have $a_{n+1} = \sqrt{a_{n}+\underbrace{\sqrt{a_{n-1}+\sqrt{a_{n-2}+\sqrt{a_{n-3}+\cdots}}}}_{a_n}} = \sqrt{a_n+a_n} = \sqrt{2a_n}$ for all $n \ge 2$.