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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.

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    $\begingroup$ Assuming $a_n$ converges to some $k$, for $m=2$, you just have to solve $k=\frac{1}{2} (1+ k^2)$ $\endgroup$ – Nicholas Oct 30 '15 at 15:04
  • $\begingroup$ Why does this suffice? $\endgroup$ – tender Oct 30 '15 at 15:13
  • $\begingroup$ If $a_n$ converges to some $k$ then as $n \rightarrow \infty$, $a_n \rightarrow k, a_{n+1} \rightarrow k$. Put this back into the recurrence and you get the equation. $\endgroup$ – Nicholas Oct 30 '15 at 15:16
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Assuming that the sequence has a limit, it is a root of the equation $$x=\frac12(1+x^m)$$

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  • $\begingroup$ Any insight why this is the solution? $\endgroup$ – tender Oct 30 '15 at 15:13
  • $\begingroup$ @ajotatxe You might want to specify which root is that. (For example, 1 is a root too, but that's the wrong one, it works only for $m\le2$). $\endgroup$ – Ivan Neretin Jun 19 '16 at 20:10
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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 $\epsilon\to0$

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