# Closed form for the sequence defined by $a_0=1$ and $a_{n+1} = a_n + a_n^{-1}$

Today, we had a math class, where we had to show, that $a_{100} > 14$ for

$$a_0 = 1;\qquad a_{n+1} = a_n + a_n^{-1}$$

Apart from this task, I asked myself: Is there a closed form for this sequence? Since I didn't find an answer by myself, can somebody tell me, whether such a closed form exists, and if yes what it is?

-
A little Googling leads to mathworld.wolfram.com/MycielskiGraph.html (scroll to the bottom) – Byron Schmuland Mar 29 '11 at 19:18

I agree, a closed form is very unlikely. As for more precise asymptotics, I think $a_n = \sqrt{2n} + 1/8\,{\frac {\sqrt {2}\ln \left( n \right) }{\sqrt {n}}}-{\frac {1}{ 128}}\,{\frac {\sqrt {2} \left( \ln \left( n \right) -2 \right) ^{2} + o(1)} {{n}^{3/2}}}$

-
+1: The first two terms seem to match what I have :-) Can you please elaborate on how you got this? Seems like the $C$ in my answer can probably be given in "closed form" using your answer. – Aryabhata Mar 29 '11 at 20:03
It seems you are also this guy. Mysterious stranger. :) – Wok Mar 29 '11 at 21:23
@Robert: You can ask a moderator to merge your two accounts. – Bill Dubuque Mar 30 '11 at 19:41
@Bill: thanks for the headsup. – Willie Wong Mar 30 '11 at 20:31
@Robert: since I'm pretty sure you are who you are, I've taken the liberty to merge your old unregistered account into your new one. – Willie Wong Mar 30 '11 at 20:40

A closed form I doubt there is. But asymptotics are easy: $$a_{n+1}^2=a_n^2+2+1/a_n^2,$$ hence, for every $n\ge1$, $$a_n^2=2n+1+\sum_{k=0}^{n-1}\frac1{a_k^2}.\qquad\qquad\qquad\qquad (*)$$ This shows that $a_n^2\ge2n+2$ for every $n\ge1$, for example $a_{100}\ge\sqrt{202}>10\sqrt{2}>14$. In particular, $a_n\to+\infty$. Plugging this into $(*)$ yields $a_n^2=2n+1+o(n)$ hence $$\sqrt{2n}\le a_n\le\sqrt{2n}+o(\sqrt{n}).$$ At this point, we know that $a_n^2\ge2n+2$ for every $n\ge1$. Using $(*)$ again, one sees that, for every $n\ge1$, $$a_n^2\le2n+2+\sum_{k=1}^{n-1}\frac1{2+2k}\le2n+2+\frac12\log(n).$$ Plugging this upper bound of $a_n^2$ into $(*)$ would yield a refined lower bound of $a_n^2$. And one could then plug this refined lower bound into $(*)$ again to get a refined upper bound. And so on, back and forth between better and better upper bounds and better and better lower bounds. (No more asymptotics here.)

-
I just tried to understand this answer again, but I didn't understood, why $2n+1+\sum_{0\le k<n}a_k^{-2}=2n+1+o(n)$ – FUZxxl May 10 '11 at 20:15
The result is quite general: for every nonnegative sequence $(x_n)$ such that $x_n=o(1)$, $\sum_{k\le n}x_k=o(n)$. You could try to prove this by the usual epsilon-delta method. – Did May 11 '11 at 5:30
@FUZxxl: what's more, since $a_k>\sqrt{2k+1}$, we have that $\sum_{0\le k<n}a_k^{-2}<\frac{1}{2}\gamma+\log(2)+\frac{1}{2}\log(n)+\frac{1}{48n^2}$ – robjohn Aug 16 '11 at 12:07

To elaborate on how I got my answer: I started with Didier's $a(n) \approx \sqrt{2n}$ and looked for a next term. $a(n) = \sqrt{2n}$ would make $a(n+1) - (a(n) + a(n)^{-1}) = \sqrt {2\,n+2}-\sqrt {2}\sqrt {n}-1/2\,{\frac {\sqrt {2}}{\sqrt {n}}} = - \frac{\sqrt{2}}{8} n^{-3/2} + O(n^{-5/2})$. With $a(n) = \sqrt{2n} + c n^{-1/2}$ I don't get a change in the $n^{-3/2}$ term, so I tried $a(n) = \sqrt{2n} + c \ln(n) n^{-1/2}$ and got $a(n+1) - (a(n) + a(n)^{-1}) = (-\frac{2}{\sqrt{8}} + c) n^{-3/2} + \ldots$. So to get rid of the $n^{-3/2}$ term I want $c = \frac{2}{\sqrt{8}}$. Then look at the leading term for $a(n) = \sqrt{2n} + \frac{2}{\sqrt{8}} \ln(n) n^{-1/2}$ and continue in that vein...

-
Not sure this try-and-guess procedure constitutes a proof. The argument in my post proves that $a_n=\sqrt{2n}+o(\sqrt{n})$. – Did Apr 1 '11 at 6:36
@Didier: Do you disagree with my answer? – Aryabhata Apr 1 '11 at 14:35
You should edit your other answer, rather than adding a new one. – Aryabhata Apr 1 '11 at 14:36

Reposting here, as it is kind of lost in that thread and this thread is more suitable for it.

Note: I have no clue if a closed form exists, but here is an asymptotic estimate...

I think we can show that $$\displaystyle a_{n}^2 \sim 2n + \dfrac{\log n}{2} - C$$ for some constant $\displaystyle C \gt 0$

By $\displaystyle x_n \sim y_n$ I mean $\displaystyle \lim_{n \to \infty} (x_n - y_n) = 0$

Consider $b_n = a_{n}^2 - 2n$

Then we have that $\displaystyle b_{n+1} = b_n + \dfrac{1}{b_n + 2n}$

Notice that $b_0 \gt 0$ and thus $\displaystyle b_n \gt 0$.

(Note that the other thread linked above starts with $a_1 = 1$ and not $a_0 = 1$.)

We can easily show that $b_n \lt 2 + \log n$, as

$b_{n+1} - b_n = \dfrac{1}{b_n + 2n} \lt \dfrac{1}{2n}$

Adding up gives us the easy upper bound. Note, even though we can give tighter bounds, this is sufficient for our purposes.

Now we have that, for sufficiently large $\displaystyle m,n$

$\displaystyle b_{m+1} - b_n = \sum_{k=n}^{m} \dfrac{1}{b_k + 2k}$

we have that

$\displaystyle \sum_{k=n}^{m} \dfrac{1}{2k} \gt b_{m+1} - b_n \gt \sum_{k=n}^{m} \dfrac{1}{2k}(1- \dfrac{b_k}{2k})$

(Here we used $\displaystyle \dfrac{1}{1+x} \gt \ \ 1-x, 1 \gt x \gt 0$)

Now Since $b_k \lt 2 + \log k$, we have that

$\displaystyle \sum_{k=n}^{m} \dfrac{1}{2k} \gt b_{m+1} - b_n \gt \sum_{k=n}^{m} \dfrac{1}{2k} - \sum_{k=n}^{m} \dfrac{2 + \log k }{4k^2}$

Using the fact that $\displaystyle H_m - H_n = \log(\dfrac{m+1}{n}) + O(\dfrac{1}{n}) + O(\dfrac{1}{n} - \dfrac{1}{m})$, where $\displaystyle H_n = \sum_{k=1}^{n} \dfrac{1}{k}$ is the $\displaystyle n^{th}$ harmonic number.

We see that,

if $c_n = b_n - \dfrac{\log n}{2}$, then

$\displaystyle O(\dfrac{1}{n} -\dfrac{1}{m}) + O(\dfrac{1}{n}) \gt c_{m+1} - c_n \gt O(\dfrac{1}{n} -\dfrac{1}{m}) + O(\dfrac{1}{n}) -\sum_{k=n}^{m} \dfrac{2 + \log k }{4k^2}$

Now $\displaystyle \sum_{k=1}^{\infty} \dfrac{2 + \log k}{k^2}$ is convergent and so by the Cauchy convergence criteria, we have that $\displaystyle c_n$ is convergent.

Thus the sequence $\displaystyle a_{n}^2 - 2n - \dfrac{\log n}{2}$ converges and hence, for some $\displaystyle C$ we have that

$$\displaystyle a_{n}^2 \sim 2n + \dfrac{\log n}{2} - C$$

or in other words

$$\displaystyle a_{n} \sim \sqrt{2n + \dfrac{\log n}{2} - C}$$

A quick (possibly incorrect) computer simulation seems to show a very slow convergence to $\displaystyle C = 1.47812676429749\dots$

Note: Didier suggested an alternate proof in the comments below, which might simpler.

-
That's still only an asymptotic. – FUZxxl Mar 29 '11 at 19:33
@Fuz: Yes, I doubt if there is a known closed form for it. But being able to peg down the difference from $2n + \log n/2$ to be a constant should be reasonably useful, I would say. – Aryabhata Mar 29 '11 at 19:42
@Moron This is to answer to the question you asked in a comment on Robert's post. First, do we agree that you ask me to check the mathematical accuracy of your answer? And that I only do that and state publicly the result of the checking because you asked me to? If I misunderstood you, please say so and I will delete my comment right away. So... here we go. To summarize, the general line of your proof is correct but some details annoy me, most often because you make some unnecessary détours where a more direct path exists. .../... – Did Apr 2 '11 at 10:04
.../... For example: (1) $b_n>0$ for every $n\ge0$ and not only for $n\ge3$. (2) The easy argument I can think of to bound $b_n$ is based on the inequality $b_n+2n\ge1$ for every $n\ge0$, it leads to $b_n\le n+1$ and not to $b_n<2n$. Comment on (2): as a consequence, when I read that you will use $b_n<2n$ and that there is an easy argument to prove it, I am puzzled and worried that I missed something or that my argument leading to $b_n\le n+1$ is wrong, so, ultimately, I am distracted. And this is bad. :-) .../... – Did Apr 2 '11 at 10:05
.../... (3) The easy argument I can think of to bound $b_n$ by a $\log$ is based on he inequality $b_n+2n\ge2n$ for every $n\ge1$ and leads to an upper bound of leading term $\frac12\log(n)$ and not $\log(n)$. So the comment on (2) applies here as well, mutatis mutandis. (3') You should expand this step. (4) The bound $b_k<\log(k)$ is not useful the first time you invoke it (but it is useful the second time). (5) I would postpone the introduction of $\log(n)$ to as late a point as possible. If you define $c_n$ by $c_n=b_n-\frac12H_n$, $(c_n)$ is nonincreasing and bounded below by .../... – Did Apr 2 '11 at 10:06

Let us consider the functional formulation. Given $y(0)=1$, $y' = \frac{1}{y}$ yields $y(x) = \sqrt{2x+1}$.

I am not saying $a(n)=y(n)$. Yet there is a link between the two approaches (finite difference).

-
What are you saying? That $a_n=y(n)$? Clearly not, since $a_n$ is always rational and $y(n)$ is typically irrational. There's no reason why replacing a difference by a derivative should work. – joriki Mar 29 '11 at 19:28
@joriki: Actually, it could be useful. If I recollect correctly, Donald J Newman recommends this as a rough method to guess the asymptotic behaviour of certain sequences, in one of his books. – Aryabhata Mar 29 '11 at 19:39
@joriki Indeed one can compare the solution $y$ of the ODE $y'=\varphi(y)$ with the sequence $(a_n)$ defined by $a_{n+1}=a_n+\varphi(a_n)$. For $\varphi(y)=-2y$ the result is interesting. – Did Apr 4 '11 at 17:51
@Moron See comment above. – Did Apr 4 '11 at 17:52
@Didier: Thanks for the example :-) – Aryabhata Apr 4 '11 at 17:58