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I need your help with evaluating this limit:

$$ \lim_{n \to \infty }\underbrace{\sin \sin \dots\sin}_{\text{$n$ compositions}}\,n,$$

i.e. we apply the $\sin$ function $n$ times.

Thank you.

Thomas Andrews
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    and why Did I get the votedown? –  Jun 14 '11 at 10:41
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    I got two downvotes for my answer, even though my proof and answer are obviously correct. By the way, I just upvoted your question because it's a clever limit. – Luboš Motl Jun 14 '11 at 10:42
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    Well, I can't speak for the people who voted (I didn't vote -- as a good Swiss citizen I remained neutral), but I suspect that you got the down vote because you have this habit of just asking questions without exhibiting the least work of your own. As for the votes on Luboš's answer I think this has to do with the fact that he has the habit of being rather verbose and intuitive but I for one wouldn't want my students to hand in such solutions. – t.b. Jun 14 '11 at 11:05
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    @Theo: You can go through my questions and notice that when I have any clue or when I am close to solutions, I definitely write it as part of my question.Those guys don't own the site, and if they get annoyed by me they can simply ignore my question. You know, without us, the less talanted Mathematicians this site wouldn't exist. –  Jun 14 '11 at 11:10
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    @Nir: I must object in the strongest sense of the term to your characterization of the position expressed by Theo in his comment, as an opposition between less talanted [sic] Mathematicians (whatever that means) and other ones. This simply makes no sense. – Did Jun 14 '11 at 11:21
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    @Theo: and regarding Lubos's answers- He has the full right to write what he wants as along as he thinks it correct. this is what the site for- Discussions, attempts to help,and if someone does not agree with the answer he can write down his opinion and not hide behind the downvote buttons. We should let people here write and express their opinions as long it belongs to math, relevant and properly written. Talking about habits- a lot of people here have the hebit to be cynical and sarcastic. –  Jun 14 '11 at 11:22
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    Dear Nir, I do not think there is any problem with you "being less talented" (in fact, I think you are talented!). The point is that it would be better for you to show some attempt(s) at your own question. "Attempt" can (and should) be interpreted in a broad sense. For example, what do you think the limit is? Why do you think it is this value? If you simply have no idea as to what the limit is, why do you think this is? In particular, what property of the sine function is difficult for you? In any case, one way to at least guess the limit is to "plug in" some values of $n$ (such as $1$). – Amitesh Datta Jun 14 '11 at 11:25
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    Dear @Theo, it's mutual, I would prefer not to have students or teachers who don't want to understand things and who are proud of having a limited intuition so that they always look for ways how to do things in mechanical ways that don't require intuition. – Luboš Motl Jun 14 '11 at 11:28
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    I do not want to argue because there's no point in that. You asked for an explanation, I gave the one I find plausible, so don't complain now. If you insist on the possibilities that this site offers, you also have to live with the fact that people can vote without giving explanations. – t.b. Jun 14 '11 at 11:35
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    I don't have problem with that, I asked that because I was afraid that I might be wrong with the question itself, perhaps something was wrong with the way I wrote it. –  Jun 14 '11 at 12:53
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    The problem is routine, anyone with experience can solve it in a few minutes. To me, the debate has to do with pedagogic style. There are strong arguments in either direction. Luckily, there are practitioners of each style present here, so students are being exposed to each style. – André Nicolas Jun 14 '11 at 16:35
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    @Nir: You seem to have a mistaken concept of what this site is about. This site is not about discussions. It is about specific answers to specific questions and the voting is to separate the wheat from the chaff. There are certain rules one is expected to follow when using this site. Please read the FAQ. btw, if you hover over the downvote arrow, it clearly mentions "prior research". – Aryabhata Jun 14 '11 at 17:17
  • @Aryabhata: Does my question look like a chaff? It did get 10 votes and tagged as a "nice question". I believe that the site is for talking about math through helping other people, mostly for users who love math. This discussion went over proporation, I don't have nothing against downvotes when you deserve them, but for this kind of a question, without explaining- so yes, I asked the person to reason his downvoting- and as I said JUST BECAUSE I was afraid that I wrote something wrong, this reputation game is not intresing me at all, I just want clever answeres from the clever people here. –  Jun 14 '11 at 17:49
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    @Nir: Sorry to say this, but your beliefs about what the site is, are immaterial. Why don't you read the FAQ to find out what this site is really about? Also, did you miss the part of the comment about "prior research"? The site rules specifically encourage one to downvote questions which show no prior research. And I am only saying this for your benefit. By following the rules of the site, you will get a more positive response. How hard would it have been to show what you tried, or even mentioned that you have no clue where to start? (FWIW, I haven't downvoted your question). – Aryabhata Jun 14 '11 at 17:56
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    @Arybhata: As I said, I usually write what I think or what I did try to do, here I didn't have anything smart to say, So I just asked. So sure, there are rules about the site, and again, cause you having a hard time to understand that, I don't care AT ALL about the votedowns, Just wanted to know if I wrote something wrong. Please check my previous questions and see that I did write what I think and I tried. this is my last comment about this, Have agreat night\day. –  Jun 14 '11 at 20:09

1 Answers1

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The first sine is in $I_1=[-1,1]$ hence the $n$th term of the sequence is in the interval $I_n$ defined recursively by $I_1=[-1,1]$ and $I_{n+1}=\sin(I_n)$. One sees that $I_n=[-x_n,x_n]$ where $x_1=1$ and $x_{n+1}=\sin(x_n)$. The sine function is such that $0\le\sin(x)\le x$ for every nonnegative $x$ hence $(x_n)$ is nonincreasing and bounded below by zero hence it converges to a limit $\ell$. The sine function is continuous hence $\ell=\sin(\ell)$. The only fixed point of the sine function is zero hence $\ell=0$. This proves that $x_n\to0$, that the sequence $(I_n)$ is nonincreasing and that its intersection is reduced to the point zero and finally, that the sequence considered in the post converges to zero.

Edit: The argument above shows that for every sequence $(z_n)$, the sequence $(s_n)$ defined by $s_n=\sin\sin\cdots\sin(z_n)$ (the sine function being iterated $n$ times to define $s_n$) converges to zero. In other words, there is nothing particular about the choice $z_n=n$.

Did
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    The only problem with this is that the initial argument is $n$, so the series is not uniformly decreasing in $n$, even in absolute value. In particular, $I_3\approx 0.14, I_4\approx -0.59$. But the limit is $0$. – Ross Millikan Jun 14 '11 at 13:13
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    @Ross: Once again, the only decreasing sequence in the picture is $(I_n)$, a sequence of intervals. You are right to point out that the sequence whose $n$th term is $\sin\sin\cdots\sin n$ may, and probably does, behave in a quite erratic way. But, surprise, this is not relevant! All we need to know is that the $n$th term must be somewhere in $I_n$ and that $I_n$ shrinks to the point zero. O squeeze lemma, so useful and yet so despised... :-) http://en.wikipedia.org/wiki/Squeeze_theorem – Did Jun 14 '11 at 13:25
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    Nice solution. You can also rewrite the solution without intervals by observing that $|\sin^{(n)} (z_n)| \leq |\sin^{(n-1)}(1) |$ where by power I mean composition. The sequence $|\sin^{(n-1)}(1) |$ is nonnegative and decreasing, thus convergent, and the limit must be $0$. – N. S. Apr 25 '13 at 13:59
  • @N.S. Looks like paraphrasing to me, but thanks for the comment. – Did Apr 25 '13 at 14:15
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    @Did It is paraphrasing, but IMO the students are more familiar with convergence of sequences than with convergence of compact sets, so the paraphrasing might make the solution readable to more people :) Also, your solution probably works for any function for which $|f(x)| < |x| ,\forall |x| >0$ :) – N. S. Apr 25 '13 at 14:20
  • I really liked this approach. (How) will it be different for $cos$ and the fact that $cos(0) = 1$? – Michael Aug 03 '14 at 12:35
  • @Michael The approach can be adapted to the cosine function. The identity $\cos(0)=1$ is irrelevant. The limit is the unique root of $\cos\ell=\ell$, approximately $\ell=.739$. – Did Aug 03 '14 at 13:10
  • @Did Thank you for the response, @Did! I was so far as well and what I am curious about whether we can find out the exact result :) – Michael Aug 03 '14 at 13:44
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    @Michael "The exact result" is the unique root of $\cos\ell=\ell$. – Did Aug 03 '14 at 13:49
  • I'm trying work through your proof. Here is where I'm stuck: to show that $s_n \in I_n$ for all $n$, I use the fact that $sinx$ is increasing on $[-1,1]$. Thus, $sinz_n \leq 1$ implies that $sin^{(2)}z_n < sin(1)$ by $sinx$ increasing. Applying this idea $n$ times, $sin^{(n)}z_n < sin^{(n-1)}(1)$. So, $s_n \in I_{n}$. Do we need to rely on the increasing nature of $sinx$ on $[-1,1]$ to assert the general sequence $s_n$ is in the nested intervals, or is there another way to prove this? I ask because the construction you used for $x_{n+1} = sin(x_n)$ didn't rely on this: only that $sinx \leq x$. – tmastny Dec 23 '14 at 17:02
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    @tmastny The monotonicity of the function sine is necessary to assert that $\sin([-x,x])=[-\sin x,\sin x]$ for every suitable $x$. The inequality $0\leqslant\sin x\leqslant x$ for every suitable $x$ guarantees that $(x_n)$ is nonincreasing and bounded below by $0$. – Did Dec 23 '14 at 21:20
  • Right. I understand the construction of $x_n$ and why it converges to zero. However, I'm confused about the general sequence $s_n$ defined in your edit. We need to show that $|\sin^{(n)}z_n| \leq \sin^{(n-1)}(1)$ for $s_n$ to be in $I_n$. But by monotonicity, we can only show that $\sin^{(n)}z_n \leq \sin^{(n-1)}z_n$, because monotonicity requires that the right hand side be the argument of the left. But we need that $s_n$ is less than $\sin^{(n-1)}(1)$ or $\sin(x_{n-1})$. The only way I understand how to do that is by using the method of an increasing function I described in my previous post. – tmastny Dec 23 '14 at 21:46
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    @tmastny Sorry? Actually, we need to know (and we do know) that $|\sin^{(n)}z_n|\leqslant x_{n-1}$, which is obvious since $\pm x_{n-1}=\sin^{(n-1)}(\pm1)$, $-1\leqslant\sin z_n\leqslant1$ and $\sin^{(n-1)}$ is nondecreasing on $[-1,1]$. – Did Dec 23 '14 at 22:00
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    Okay thanks for the clarification. You didn't mention it explicitly, so I thought I was doing something wrong when I had to use the fact $\sin$ is non-decreasing on $[-1,1]$. – tmastny Dec 23 '14 at 22:13