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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?

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    What is the question?2012-05-16
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    By "solve," I assume you want to find a closed form for $a_n.$ Right?2012-05-16
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    Yes, a closed form for $a_n$.2012-05-16
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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.**2012-05-16
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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.2012-05-16
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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.2012-05-16
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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.2012-05-16
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    @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.2012-05-16
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    See http://applied-discrete-structures.wiki.uml.edu/Chapter+82012-05-16
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    @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.2012-05-16

3 Answers 3

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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.

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    But $a_n$ is never an integer, is it?2012-05-16
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    @Didier: Yes, it isn't. The paper deals with that too. I read the question too quickly and thought $a_1 = 1$. Will update.2012-05-16
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    @anon: No, it is my fault.2012-05-16
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    @Didier: I added a disclaimer! It might need some work to get those results to work for this problem (if at all they work). But I have left the answer, in case someone wants to try.2012-05-16
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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.

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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.