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I was asked today by a friend to calculate a limit and I am having trouble with the question.

Denote $\sin_{1}:=\sin$ and for $n>1$ define $\sin_{n}=\sin(\sin_{n-1})$. Calculate $\lim_{n\to\infty}\sqrt{n}\sin_{n}(x)$ for $x\in\mathbb{R}$ (the answer should be a function of $x$ ).

My thoughts:

It is sufficient to find the limit for $x\in[0,2\pi]$ , and it is easy to find the limit at $0,2\pi$ so we need to find the limit for $x\in(0,2\pi)$.

If $[a,b]\subset(0,\pi)$ or $[a,b]\subset(\pi,2\pi)$ we have it that then $$\max_{x\in[a,b]}|\sin'(x)|=\max_{x\in[a,b]}|\cos(x)|<\lambda\leq1$$ hence the map $\sin(x)$ is a contracting map.

We know there is a unique fixed-point but since $0$ is such a point I deduce that for any $x\in(0,2\pi)$ s.t $x\neq\pi$ we have it that $$\lim_{n\to\infty}\sin_{n}(x)=0$$

So I have a limit of the form "$0\cdot\infty$" and I can't figure out any way on how to tackle it.

Can someone please suggest a way to find that limit ?

Note: I am unsure about the tags, please change them if you see fit.

S L
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Belgi
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  • 35 minutes. // Are you sure that This result is in paper X is a better answer than an actual (basic and complete) proof? – Did Jan 12 '13 at 10:11
  • @did - Having looked at the proof at the book I thought there is no "easy" solution for my question and since the answer is complete I accepted it (I accepted it before there were any other answers and I saw no reason to delay on accepting the answer). I have changed the accepted answer since the other answer given is more basic and complete, as you stated (and I agree). – Belgi Jan 12 '13 at 19:13
  • @did - I don't know about that, after all - it is possible to change the accepted answer – Belgi Jan 12 '13 at 19:31
  • Do you realize that MSE people are located all around the Globe? Hence, accepting an answer half an hour after you posted the question effectively prevents people not in your time zone (or close to it) to answer. This is one reason amongst many. – Did Jan 12 '13 at 19:34
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    possible duplicate of Convergence of $\sqrt{n}x_{n}$ where $x_{n+1} = \sin(x_{n})$ – Martin Argerami Jan 12 '13 at 21:00
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    Possible duplicate of http://math.stackexchange.com/questions/3215/convergence-of-sqrtnx-n-where-x-n1-sinx-n – Eric Naslund Jan 12 '13 at 21:00

2 Answers2

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I will deal with the case when $x_0 \in (0,\pi)$ If $x_0 \in (0,\pi)$ and $x_{n+1} = \sin x_n $, for $ n \geq 0$ then $x_1 \in (0,1] \subseteq (0,\pi/2)$, and it is easy to see that from that point onwards, $0<x_{n+1}<x_{n}$ and hence $x_n$ converges to a fixed point of $\sin$ which has to be $0$.

We have $$ \dfrac{1}{\sin^2 x} - \dfrac{1}{x^2} = \dfrac{x-\sin x}{x^3} \times \dfrac{x}{\sin x} \times \left(\dfrac{x}{\sin x} + 1\right) \to \dfrac{1}{3}$$ as $x \to 0$.

This implies, putting $x = x_n$ $$ \dfrac{1}{x_{n+1}^2} - \dfrac{1}{x_n^2} \to \dfrac{1}{3}.$$

The Ceasaro mean of above, $$ \dfrac{1}{n}\sum_{i=0}^{n-1}\left(\dfrac{1}{x_{i+1}^2} - \dfrac{1}{x_i^2}\right) = \dfrac{1}{n}\left(\dfrac{1}{x^2_{n}} -\dfrac{1}{x^2_0}\right)$$ must also converge to $\dfrac{1}{3}$ and since $x_n > 0$, $ \sqrt{n} x_n \to \sqrt{3}$.

Gary
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Arin Chaudhuri
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De Bruijn proves this asymptotic for the sine's iterates:

$$ \sin_n x \thicksim \sqrt{\frac{3}{n}} $$

Now we have:

$$ \lim_{n\to\infty} \sqrt{n} \sin_{n}{x} $$

We have $n\to\infty$.

$$ \lim_{n\to\infty} \sqrt{n} \sqrt{\frac{3}{n}} $$

$$ \lim_{n\to\infty} \sqrt{3} = \sqrt{3} $$

It is interesting to note that this result is independent of $x$. (As De Bruijn notes, G. Polya and G. Szegu prove a weaker result, namely, exactly this limit.)

This is only true for $x \in \left(0, \pi\right)$. For $x = 0$, the limit is $0$. For $x = \pi$, the limit is likewise, $0$.

For $\sin x$ negative, the limit goes to $-\sqrt{3}$. A proof follows. Note that the sine function is odd, that is:

$$ \sin_n (-x) = -\sin_n x$$

Now, we have:

$$ \lim_{n\to\infty} \sqrt{n} \sin_n (-x) $$

Or:

$$ -\lim_{n\to\infty} \sqrt{n} \sin_n (x) $$

Which we know to be $\sqrt{3}$, so:

$$ -\sqrt{3} $$

As a final summary ($k \in \mathbb{Z}$):

$$ \begin{cases} 0 & \mbox{if } x = k\pi \\ -\sqrt{3} & \mbox{if } x \in (2 \pi k - \pi, 2 \pi k) \\ \sqrt{3} & \mbox{if } x \in (2 \pi k, \pi + 2 \pi k) \\ \end{cases} $$

George V. Williams
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  • Thanks for the answer, for what values of $x$ is this valid ? for example for $x=0$ it is not valid – Belgi Jan 12 '13 at 01:01
  • @Belgi $x= \pi/2$ it is valid. I would assume $x \in (0, \pi)$. –  Jan 12 '13 at 01:02
  • From the text it seems that it is valid for $0<x<\pi$. What do we do for $\pi<x<2\pi$ ? – Belgi Jan 12 '13 at 01:04
  • @Belgi I would assume it would be $-\sqrt{3}$. Since the first term will be negative, which you can pull out through all the sine's all the way out. –  Jan 12 '13 at 01:05
  • @Marvis - but since $\sin(-x)=-\sin(x)$ I would actually say it doesn't converge.What do you think ? – Belgi Jan 12 '13 at 01:09
  • @Marvis - I was wrong, the minus sign remains throwout and this proves the answer is indeed $-\sqrt{3}$ as you say – Belgi Jan 12 '13 at 01:11
  • @Belgi, while you guys were talking about that, I was actually adding that to my post. ;) – George V. Williams Jan 12 '13 at 01:13
  • Thanks George! Did you really mean it is valid for $x\in (0,\pi/2)$ or should it be $(0,\pi)$ ? – Belgi Jan 12 '13 at 01:16
  • @Belgi, I meant for $x \in (0, \pi)$, my bad. – George V. Williams Jan 12 '13 at 01:18
  • Hi, George. I'm quite interested in the approximation of iterate series, say $\sin_nx$. I know that $\sin_nx\sim\sqrt{3/n}$ could be obtained from Stolz-Cesàro theorem, but I don't know how to obtain more accurate results. The book you've mentioned is unavailable here. Is there any systematical way to deal with the asymptotic of iterate series? – Yai0Phah Jan 12 '13 at 04:08
  • @FrankScience, the method de Bruijn uses is basically outlined as such: 1. Prove that the limit converges to something. 2. Take the Taylor Expansion of an iterate (i.e., $\sin_{n+1} = \sin_{n} - \frac{1}{6} \sin_{n}^3 + \cdots$). 3. Make a series of substitutions which preserve the limit, namely, $\sin_{n} \to \sin_{n}^2 \to \frac{1}{3} \sin_{n}^2$, and then letting the polynomial approach zero. 4. We then have $\frac{1}{3}\sin_{n}^2 = n^{-1} O(n^{-2} \log n)$, and returning to $\sin_{n}$ we have $3^{.5}n^{.5} (1 + O(n^{-1} \log n))$. I am unaware if this method can be extended in general. – George V. Williams Jan 12 '13 at 17:41
  • @GeorgeV.Williams $\newcommand\abs[1]{\left|#1\right|}$ used some tricks to read that book. It seems that the last way is elementary, i.e., prove $\abs{x_n-1/n}<Kn^{-2}\log n$ inductively. Is there any advanced approach, say using calculus? – Yai0Phah Jan 14 '13 at 06:35
  • @GeorgeV.Williams For example, when we prove $nx_n\sim1$, we use Stolz-Cesàro theorem, which doesn't seem able to be applied in the more accurate asymptotic. – Yai0Phah Jan 14 '13 at 06:38
  • Hi George -- since there's no mechanism to send personal messages, I picked this comment thread that's already long anyway to thank you for the extra bounty on the tacoid answer :-) I think that's the first time I ever got an extra bounty post factum! I've been thinking recently that I should start awarding bounties; you've certainly encouraged that train of thought :-) – joriki Feb 03 '13 at 18:47
  • @joriki, No problem! I always enjoy giving out bounties to answers which I feel are exemplary, and your answer certainly fits that criteria. – George V. Williams Feb 03 '13 at 18:56