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If $$a_{n+1}=a_n^2+1,$$ with initial $a_1=\frac{1}{2}$. How to solve this sequence problem, i.e., how to represent $a_n$ in closed form?

Yimin
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  • What is the question? – MJD May 16 '12 at 15:39
  • By "solve," I assume you want to find a closed form for $a_n.$ Right? –  May 16 '12 at 15:42
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    Yes, a closed form for $a_n$. – Yimin May 16 '12 at 15:43
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    Let's not jump into voting to close, before we ask OP for clarifications. Especially if OP is willing to promptly reply and edit the question within few minutes of "clarification request" comments. It's annoying. –  May 16 '12 at 15:51
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    It is unlikely that there is a closed form. Recursions of the form $a_{n+1}=a_n^2+c$ are very complicated. The Mandelbrot set reflects this. – lhf May 16 '12 at 16:04
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    My first though is to represent $a_n=\frac{b_n}{c_n}$ in lowest terms. It becomes $c_n=2^{2^{(n-1)}}$, $b_{n+1}=b_n^2+2^{2^{(n-1)}}$, $b_1=1$, which gets you into $\mathbb N$. But I'm not sure it helps any. – Ross Millikan May 16 '12 at 16:04
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    Why do you think there is a closed form? What would you do if you had one? Perhaps you're trying to solve a problem related to this sequence that does not need a closed form. – lhf May 16 '12 at 16:06
  • @RossMillikan I have tried that, it does help much. I also tried to use sinh and cosh functions to replace $a_n$, but it did not provide any convenience either. – Yimin May 16 '12 at 16:16
  • See http://applied-discrete-structures.wiki.uml.edu/Chapter+8 – leo May 16 '12 at 16:44
  • @J.D.Thanks for the correction. I immediately regretted my vote to close, but the software does not allow me to retract it. Fortunately there is no harm done. – MJD May 16 '12 at 19:44

3 Answers3

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EDIT: The answer below assumes $a_1$ is an integer, and as such is not directly applicable to your question, but I am leaving it here in case people can make use of the results mentioned in the paper below.

Aho and Sloan proved that for sequences like yours, there is a constant $k$ such that

$$a_n = \lfloor k^{2^n} \rfloor$$

for sufficiently large $n$. $k$ can be defined as a limit of a sequence using $a_n$ itself. If you include $k$ as one of your closed form constants, you are done!

See their paper for details: http://www.fq.math.ca/Scanned/11-4/aho-a.pdf

Of course, for your special case, one might still be able to find a different "closed form" which might be more appealing to you.

Aryabhata
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The recurrence $a_{n+1} = a_n^2+c$ has a (known) closed form if and only if $c=0$ or $c=-2$. See this answer for more explanation.

Community
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GEdgar
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Perhaps let $a_n=\tan \theta_n$. Then $\tan \theta_{n+1}=\sec^2 \theta_n$. So, $\tan \theta_2=\sec^2 \theta_1$, $\tan \theta_3=1+\sec^4 \theta_1$, etc. I am not sure if this procedure will produce a closed form.

PAD
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