0

I have been asked about solving (EDIT: find explicit $a_n$) the recurrence relation containing square root, but I have tried days without success.

\begin{align} a_{n+1}=\sqrt{2+a_n}, \end{align} with $a_1=\sqrt 2$. I tried to use the ODE equivalence for the homogeneous equation $L=(u')^2-u=0$ like this \begin{align} L'=u'(2u''-1)=0, \end{align} so $u=n^2/4+c_1n+c_2$ where $c_1$, $c_2$ are constants. Nothing can be obtained further...

Will it be the characteristic equation $x^2-x-2=(x+1)(x-2)=0$? Then $a_n$ would be related to $2^n$?

What techniques or strategies can be used to solve them?

MathArt
  • 1,053
  • 1
    Another one: https://math.stackexchange.com/q/4117644. – Martin R Jul 22 '21 at 12:49
  • By "solve" do you mean a closed formula for $a_n$? Or just information about the limit? – lulu Jul 22 '21 at 12:50
  • I was trying to find the closed expression. – MathArt Jul 22 '21 at 13:49
  • @Martin R: Thanks for the link! Can this $\underbrace{\sqrt{2+\sqrt{2+\sqrt{2+\cdots}}}}_{ n ~ \text{terms},~n≥1}=2\cos \frac{\pi}{2^{n+1}}$ be derived? – MathArt Jul 22 '21 at 13:56
  • 1
    @MathArt: See for example https://math.stackexchange.com/a/2053715. – All these (and more) related questions can be found with a search on https://approach0.xyz/. – Martin R Jul 22 '21 at 14:06

0 Answers0