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$ a(n+1) = a(n) + \frac {1}{a(n)}, a(1) = 1 $

What is the function that generates all the values of $a(n)$?

Upon first inspection, this function appears to lie somewhere between a fractional power of $k$ and a logarithmic function of $k$. My reason for suspecting logarithms is that the recurrence relation is harmonic. However, this is just a contemplation, and my other suspect possibility is a Taylor Series of some kind.

Edit: Echo provided me a link to OEIS, so now I will reveal the problem that generated this curiosity of mine.

Given $ a(n+1) = a(n) + \frac {1}{a(n)}, a(1) = 1 $

Prove $ a(100)>14 $

Jack Tiger Lam
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  • You attempt to find the function(or formula) like $f: \mathbb N \rightarrow \mathbb R$ such that f(n)=a(n)? –  Jun 15 '15 at 07:49
  • The reverse is also acceptable. Map the rational values of a(n) to n. The problem is that the values are harmonic relative to each other, which is why I suspect a logarithmic/exponential solution is the function i am searching for. – Jack Tiger Lam Jun 15 '15 at 08:46
  • $\sqrt{2x}+\frac {1} {\sqrt{x}}$ improves on 2000's answer by 1 s.f. – wlad Jun 15 '15 at 11:30
  • You may refer to https://oeis.org/A073833 – PSPACEhard Jun 15 '15 at 12:14
  • echo, it's a bit difficult to separate the text, mentally, on OEIS, but i'll look at it. – Jack Tiger Lam Jun 15 '15 at 12:17
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    I edited my answer, that it show $a(100)>14$. –  Jun 15 '15 at 12:43
  • It might also make sense to state the error bounds of the approximation, and also why it is a justified approximation. I'm unable to see how you derived the approximation, so that would be useful. Also, this was a problem in a mathematics competition, so I think a more abstract approach to the size of a(100) is what the answer should be. – Jack Tiger Lam Jun 15 '15 at 13:48

2 Answers2

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I guess this idea is useful:

We know that $\sqrt {x+1} \simeq \sqrt {x} + \frac {1}{2 \sqrt {x}}$ and now if we set $f(x)= \sqrt{2x}$ then for large $x$ we have: $f(x+1) \simeq f(x)+\frac{1}{f(x)}$

For example: $a(100)=14.284...$ and $f(100)=\sqrt{200}=14.142...$
And it's not hard to show $\mid a(n)-f(n)\mid<1 \quad \forall n \in \mathbb N$ and for $n>2$ we have: $a(n)>f(n)$ (because $\sqrt {x} + \frac {1}{2 \sqrt {x}}> \sqrt {x+1})$

  • Yeah, I though of an asymptotic solution, but I am on the look for an exact solution... Although it is possible one does not exist. – Jack Tiger Lam Jun 15 '15 at 08:47
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I don't think you can find an easy explicit formula for your recurrence relation. But there is an easy solution to you problem.

Consider the sequence $b_n = a_n^2$. Then we observe that $b_{n+1} = b_n + \frac{1}{b_n}+2>b_n+2$. So $b_n > 2n-1$ for all $n$. In particular, $$b_{100} > 199 > 196 = 14^2$$

Jef
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