Can anybody give me a closed form for the (limit of the) recurrence relation $a_0 = 0$, $a_{n+1} = \frac12\cdot\big(1 + a_n^2\big)$?
And more general: Can anybody give me a closed form for the (limit of the) recurrence relation $b_0 = 0$, $b_{n+1} = \frac12\cdot\big(1 + b_n^m\big)$?
Hint: For $m = 3$ it seems to be $\lim\limits_{n\to\infty} b_n = \frac{\sqrt5-1}{2}$. For $m = 2$ I have no clue.
On
As others have said, the limit for $m=2$ is $x=1$.
Let $x_n=1-\epsilon_n$.
$$x_{n+1}=\frac12(1+x_n^2)=\frac12(2-2\epsilon_n+\epsilon_n^2)$$
$$\epsilon_{n+1}=\epsilon_n-\frac12\epsilon_n^2$$
$$\frac1{\epsilon_{n+1}}=\frac1{\epsilon_n}(1-\epsilon_n/2)^{-1}$$
$$\frac1{\epsilon_{n+1}}=\frac1{\epsilon_n}(1+\frac{\epsilon_n}2+\frac{\epsilon_n^2}4+...)$$
$$\frac1{\epsilon_{n+1}}=\frac1{\epsilon_n}+\frac12+\frac{\epsilon_n}4+...$$
So $1/\epsilon$ increases by at least 1/2 each step, and $\epsilon\to0$
Assuming that the sequence has a limit, it is a root of the equation $$x=\frac12(1+x^m)$$