Closed form of the sequence $a_{n+1}=a_n^2+1$

If $$a_{n+1}=a_n^2+1,$$ with first $a_1=\frac{1}{2}$. Just how to address this series trouble, i.e., how to stand for $a_n$ in shut kind?

9
2022-07-25 17:47:13
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Answers: 2

Perhaps allow $a_n=\tan \theta_n$. After that $\tan \theta_{n+1}=\sec^2 \theta_n$. So, $\tan \theta_2=\sec^2 \theta_1$, $\tan \theta_3=1+\sec^4 \theta_1$, etc I am not exactly sure if this procedure will certainly generate a shut kind.

1
2022-07-25 19:22:12
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The reappearance $a_{n+1} = a_n^2+c$ has actually a (recognized) shut kind if and also just if $c=0$ or $c=-2$. See for even more description.

3
2022-07-25 19:19:54
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