Solving a recurrence relation $x_{n+1}^2 =2 + x_n$?

Asked Jun 02 '18 at 05:19

Active Jun 02 '18 at 06:12

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How to show rigorously that the recurrence relation $x_{n+1}^2 =2 + x_{n}$, the limit tends to $2$ and can we get a closed form of sequence $x_{n}$ depending on $n$? with $x_{0}=2$.

I was tryin to observe $|x_{n+1} -x_{n}| = |\sqrt{2+x_{n}} - x_{n}| = |\sqrt{2+x_{n}} - x_{n}| \frac{\sqrt{2+x_{n}} + x_{n}}{\sqrt{2+x_{n}} + x_{n}} = \frac{2+x_{n}-x_{n}^2}{\sqrt{2+x_{n}}+x_{n}}$

I could do the numerical computation of a few iterates through which I see it converges to 2.

edited Jun 02 '18 at 06:12

Martin Sleziak

53,687

asked Jun 02 '18 at 05:19

BAYMAX

4,972

Definitely has been answered here already, but if you already know it converges to say $L$, then you have two sequences on both sides, both having same limit, and so by continuity of $x^2$ and $2+x$, the $L^2=2+L$ must hold... – Sil Jun 02 '18 at 05:23
Yes that is nice! but can we get a closed form expression for $x_{n}$ depending on $n$ ? – BAYMAX Jun 02 '18 at 05:26
3

Look at Proof of an equality involving cosine $\sqrt{2 + \sqrt{2 + \cdots + \sqrt{2 + \sqrt{2}}}}\ =\ 2\cos (\pi/2^{n+1})$ . – Sil Jun 02 '18 at 05:43
@Sil: Your comment should have been an answer :) – Saša Jun 02 '18 at 05:52
2

By the way, your condition $x_0=2$ makes the problem trivial, since then $x_1=\sqrt{2+2}=2$, and so $x_n=2$. I guess you meant $x_0=0$... – Sil Jun 02 '18 at 05:52
@Oldboy Maybe, but I prefer if an answer gives some additional info, instead of just merely refering to other question. For these we have a duplicate flag... – Sil Jun 02 '18 at 06:06

Solving a recurrence relation $x_{n+1}^2 =2 + x_n$?

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