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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.

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  • $\begingroup$ 35 minutes. // Are you sure that This result is in paper X is a better answer than an actual (basic and complete) proof? $\endgroup$
    – Did
    Commented Jan 12, 2013 at 10:11
  • $\begingroup$ @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). $\endgroup$
    – Belgi
    Commented Jan 12, 2013 at 19:13
  • $\begingroup$ @did - I don't know about that, after all - it is possible to change the accepted answer $\endgroup$
    – Belgi
    Commented Jan 12, 2013 at 19:31
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    $\begingroup$ possible duplicate of Convergence of $\sqrt{n}x\_{n}$ where $x\_{n+1} = \sin(x\_{n})$ $\endgroup$ Commented Jan 12, 2013 at 21:00
  • 2
    $\begingroup$ Possible duplicate of math.stackexchange.com/questions/3215/… $\endgroup$ Commented Jan 12, 2013 at 21:00

2 Answers 2

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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}$.

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  • $\begingroup$ (+1) Excellent answer! You may wish to correct a typo in the bottom line: the $x_1$ should be squared. $\endgroup$ Commented Jan 12, 2013 at 10:14
  • $\begingroup$ Thanks, I corrected the answer. $\endgroup$ Commented Jan 12, 2013 at 15:41
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    $\begingroup$ Great proof. See this for a generalization. $\endgroup$
    – Pedro
    Commented Jun 23, 2013 at 1:13
  • $\begingroup$ Just noting that the convergence will be very slow, given the fact that the derivative of the iteration map is equal to $1$ at the fixed point. $\endgroup$ Commented Apr 30, 2019 at 13:48
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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 ($k \in \mathbb{Z}$):

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

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  • $\begingroup$ Thanks for the answer, for what values of $x$ is this valid ? for example for $x=0$ it is not valid $\endgroup$
    – Belgi
    Commented Jan 12, 2013 at 1:01
  • $\begingroup$ @Belgi $x= \pi/2$ it is valid. I would assume $x \in (0, \pi)$. $\endgroup$
    – user17762
    Commented Jan 12, 2013 at 1:02
  • $\begingroup$ From the text it seems that it is valid for $0<x<\pi$. What do we do for $\pi<x<2\pi$ ? $\endgroup$
    – Belgi
    Commented Jan 12, 2013 at 1:04
  • $\begingroup$ @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. $\endgroup$
    – user17762
    Commented Jan 12, 2013 at 1:05
  • $\begingroup$ @Marvis - but since $\sin(-x)=-\sin(x)$ I would actually say it doesn't converge.What do you think ? $\endgroup$
    – Belgi
    Commented Jan 12, 2013 at 1:09

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