9
$\begingroup$

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?

$\endgroup$
10
  • 1
    $\begingroup$ Yes, a closed form for $a_n$. $\endgroup$
    – Yimin
    May 16, 2012 at 15:43
  • 11
    $\begingroup$ 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. $\endgroup$
    – user2468
    May 16, 2012 at 15:51
  • 4
    $\begingroup$ 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. $\endgroup$
    – lhf
    May 16, 2012 at 16:04
  • 1
    $\begingroup$ 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. $\endgroup$ May 16, 2012 at 16:04
  • 2
    $\begingroup$ 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. $\endgroup$
    – lhf
    May 16, 2012 at 16:06

3 Answers 3

12
$\begingroup$

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.

$\endgroup$
4
  • 1
    $\begingroup$ But $a_n$ is never an integer, is it? $\endgroup$
    – Did
    May 16, 2012 at 17:47
  • 1
    $\begingroup$ @Didier: Yes, it isn't. The paper deals with that too. I read the question too quickly and thought $a_1 = 1$. Will update. $\endgroup$
    – Aryabhata
    May 16, 2012 at 18:00
  • $\begingroup$ @anon: No, it is my fault. $\endgroup$
    – Aryabhata
    May 16, 2012 at 18:00
  • $\begingroup$ @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. $\endgroup$
    – Aryabhata
    May 16, 2012 at 20:17
3
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .