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For a $a_n$ defined recursively by $a_{n+1} = a_n - \frac{1}{a_n}$,$a_0 = k >0$. Prove that if the first $n$ such that $a_n \leq 0$, then $n \in O(k^2)$.

I ran a computer simulation, and it seems true.

Moreover, these questions appear similar to this question: Closed form for the sequence defined by $a_0=1$ and $a_{n+1} = a_n + a_n^{-1}$ .

However, I have no idea how to prove that. Do you have any ideas?

tovdan
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    It is not clear what are you asking. What is $a$? Is it a fixed number or the variable of $f$? – Marcos Mar 22 '24 at 10:27
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    did you mean maybe that there is a limit for $f(a)$ instead of a limit for $a$? – pygri Mar 22 '24 at 10:32
  • thanks i edit it – tovdan Mar 22 '24 at 10:34
  • Now it is clear? – tovdan Mar 22 '24 at 10:41
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    You've still got to be a bit more precise. It's possible that $a_0 > 0$ is small, that (say) $a_{100} < 0$, but that $a_{200} > 0$. Maybe a better framing would be to introduce $N = N(a_0)$ as $n-1$, where $n$ is the first index such that $a_n \leq 0$. – D. Thomine Mar 22 '24 at 12:26
  • You are right.!. – tovdan Mar 22 '24 at 12:28
  • Is it now clear? – tovdan Mar 22 '24 at 12:34
  • I would start by examining: what can we say about $k$ if $a_2 > 0$? If $a_2$ and $a_3$ are both positive? If $a_2, a_3, a_4$ are all positive? We might need to find relaxed bounds rather than strict bounds, but hopefully some pattern easy to prove by induction will show up. – aschepler Mar 22 '24 at 12:50
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    I don't know if this is helpful. If $(a^k_n)_n$ is the sequence defined by given recursion with $a^k_0=k$ and $f(k) = \min{n\in\mathbb N\mid a^k_n \leq 0 }$, then using computer I got $f(1) = 1$, $f(2) = 3$, $f(3) = 6$, $f(4) = 10$, $f(5) = 14$, $f(6) = 20$, $f(7) = 26$, which suggests $f(k) \leq f(k-1) + k$ which would give $f(k) \leq k(k+1)/2$ by telescoping. – Ennar Mar 22 '24 at 13:18
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    Think about how much you're subtracting off with $ - 1/a_n$, and how that relates to $ \lfloor a_n \rfloor$. In particular, what's the minimal $m$ such that $a_m < k-1$? $\quad$ Using that, show that we need at least $ 0 + 1 + \ldots + k-1$ steps, and at most $ 1 + 2 + \ldots + k$ steps to get under 0, hence the result holds. – Calvin Lin Mar 22 '24 at 13:53

1 Answers1

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Show the following for $ a_0 = k$. If you're stuck, explain what you've tried.

  1. $a_{k-1} > k-1 $
  2. $a_{k-1 + k-2} > k-2 $
  3. $a_{k-1 + k-2 + k-3 } > k-3$
  4. $ a_{k-1 + k-2 + k-2 + \ldots + 1 } > 0 $
  5. $a_k < k-1 $
  6. $a_{k + k-1} < k-2 $
  7. $a_{k + k-1 + k-2 } < k-3 $
  8. $a_{k + k-1 + k-2 + \ldots + 1 } < 0 $
  9. Hence $ \frac{k^2 - k } { 2 } < n_k \leq \frac{ k^2 + k}{2}$.

Hence, the result follows.

Calvin Lin
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    Thanks you! it make sense now, not sure i will have time to post the full answer here, If someone has time I would really appreciate it if they would do it, this is an interesting question and an interesting answer. In addition it will be interesting to compute the exact $n_k$ for integer $k$, i think it is possible do so. – tovdan Mar 22 '24 at 15:23