Recursive sequence…

I have a problem that I definitely should be able to solve but I can't seem to do some simple algebra at the end...

The problem gives a recursive sequence defined as $x_1= \sqrt2$, and $x_{n+1} = \sqrt{2+\sqrt{x_n}}$

I proved that it has a limit, but when I go to solve $L=\sqrt{2+\sqrt L}$, I can't seem to get anywhere past squaring both sides.

• To get $x_{n+1}$, use curly braces around the subscript: x_{n+1}. Similarly to get the nested square roots: \sqrt{2+\sqrt{x_n}}. – Brian M. Scott Aug 25 '13 at 20:56
• How can you solve that using induction...I just want to know the algebra that will let me solve the last "L" equation. – Johnny Apple Aug 25 '13 at 20:58
• Are you sure that it is $\sqrt{x_n}$ instead of just $x_n$ in the formula for $x_{n+1}$? – user84413 Aug 25 '13 at 21:24

If $L = \sqrt{2 + \sqrt{L}}$, square both sides to get

$$L^{2} = 2 + \sqrt{L}$$

Rearrange and square again:

$$L^{2} - 2 = \sqrt{L} \implies L^{4} - 4L^2 + 4 = L$$

So we know have a fourth degree equation

$$f(L) = L^4 - 4L^2 - L + 4 = 0$$

By inspection, this has a root at $L = 1$, and another root at about 1.83. It also has two complex roots. Note that the solution $L = 1$ was introduced as a result of the final squaring.

Since the sequence terms are all greater than $\sqrt{2}$, we see that the limit is the larger of the two roots. The exact form of this root could be obtained using the quartic formula, but it's not pretty.

• I don't think it has a root at 1... – Johnny Apple Aug 25 '13 at 21:01
• "this can be seen by noting that the cubic f(L)/(L−1) is strictly increasing"... It is not (but has a unique real root nevertheless). – Did Aug 25 '13 at 21:04
• @AnthonyVasaturo The root at 1 is due to the last squaring. – Did Aug 25 '13 at 21:05
• @AnthonyVasaturo Note that if $L = 1$, then $L^2 - 2 = -1$; so this extra solution was introduced by the final squaring operation. – user61527 Aug 25 '13 at 21:08