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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.

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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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@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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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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2 Answers

up vote 105 down vote accepted

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 $z_n=n$.

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@Didier: Nice observation +1 –  user9413 Jun 14 '11 at 10:57
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@Luboš: What 20-minute older answer? What are you talking about? Wow... –  Did Jun 14 '11 at 11:33
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@Luboš: Compared to this answer yours is very inaccurate and more reasoning than a rigorous proof. For example Didier doesn't try to argue that the numbers are "random" which is simply false. Neither does he always use expressions like "obvious" or "its easy to see" or resort to giving examples.. –  Listing Jun 14 '11 at 11:41
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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... :-) en.wikipedia.org/wiki/Squeeze_theorem –  Did Jun 14 '11 at 13:25
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@Didier: +1, this answer is perfect. Need I say more? I will: For what its worth, I have had Luboš act in a very similar way to me when I posted a different style of answer to another question. He continued to claim it was "wrong" and "misleading" and said that he hoped everyone else would downvote it as well... the answer was not wrong (unlike that $\sinh(x)$ thing you pointed out for me!) but rather it was an elementary way to get a non optimal bound. (Which is still useful!!) I think he just likes arguing. –  Eric Naslund Jun 14 '11 at 14:13
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The limit is zero. The expression $\sin(n)$ for a large integer $n$ is a "random" number between $-1$ and $+1$. You might be afraid that by applying $\sin$ many times, it will be even more random.

But it's not the case because $\sin x$ for any $x$ may be Taylor-expanded $$\sin x = x - \frac{x^3}{3!} + \frac {x^5}{5!} - \dots $$ and for small $|x|$, only the first few terms are important, so the effect of computing the sines is to do almost nothing, except for a small reduction of the absolute value of $|x|$.

It is enough to notice that $|\sin n| < 1$ for all positive integers $n$; the inequality is strict because $\pi$ is irrational. We may rewrite the Taylor expansion above as an inequality $$|\sin x| \leq |x| - \frac{|x|^3}{7} $$ for any $|x|<1$. To prove the inequality above, check that it holds for $|x|=1$, and it consequently has to hold for $|x|<1$ because for ever smaller $|x|$, the adjustments from $x^5/5!$ are ever smaller. The inequality is sharp unless $x=0$.

So we're led to compute the $(n-1)$ times "embedded sine" of a number between $-1$ and $+1$. Each embedding subtracts at least $|x|^3/7$ from the value of $|x|$. It is easy to see that this sequence is decreasing and positive, so it must have a non-negative limit. And the limit can't be positive, $L$, because one more iteration brings $L$ to a strictly smaller number which cannot converge back to $L$ because of the monotonic character of the sequence.

So the limit has to be zero.

By the way, for large enough $n$, the member of the sequence becomes independent of small variations of $n$ (except for the sign). So for example, for all $n$ around $10,000$, the composite sine is very close to $\pm 0.017$.

The absolute value of the $n$-th element of the sequence actually goes like $1.7/\sqrt{n}$ where the value $1.7$ is approximate. One may actually show why this decrease of $a_n$ is right. It is because $$\frac{1.7}{\sqrt{n}} - \frac{1}{3!} \left( \frac{1.7}{\sqrt{n}} \right)^3 = \frac{1.7}{\sqrt{n+1}} + O(n^{-5/2}) $$ which is true and easily provable if you Taylor-expand this expression: $$ 1.7 (n+1)^{-1/2} = 1.7 n^{-1/2} + 1.7 (-1/2) n^{-3/2} - \dots $$ Well, you can see that $$1.7 \times (1/2) \approx 1.7^3/6$$ and it is actually possible to get the exact value of the constant $1.7$: $$1.7 \approx \sqrt{3}$$ So for very large $n$, $$ |a_n| = \frac{\sqrt{3}}{\sqrt{n}} + O(n^{-3/2}).$$

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@Lubos: Well, this is hardly more than a heuristic... "Each embedding subtracts at least |x|^3/7 from the value of |x|". Note that x changes with n, but you seem to be assuming that we can treat this x to be fixed and then only have to apply the sine (n-1) times - this is not the case. –  Sam Jun 14 '11 at 10:57
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I do not understand the proof of the asymptotics. First, assume that a given $n$ approaches a multiple of $\pi$ extremely well, then $|\sin(n)|$ will be extremely small and $|a_n|$ even more so. Second, as Sam noticed, the reasoning with limited expansions seems to assume that one applies $n$ times the sine function to a fixed starting point. (But, Luboš, to prevent misunderstandings, I mention that I did not downvote your answer.) –  Did Jun 14 '11 at 11:10
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If we iterate your inequality, we get that $|\sin^(n)x|\le |x|-\frac{1}{7}(|x|^3+|\sin x|^3 +|\sin\sin x |^3+ \dots+|\sin^(n-1)x|^3), which contradicts your claim that each embedding subtracts at least |x|^3/7. (I also did not downvote) –  mpiktas Jun 14 '11 at 11:22
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Actually, the only trouble I have with the answer is that @Luboš nowhere pointed out that $\sin([-1,1]) \subset [-\sin(1),\sin(1)]$, so that one indeed can restrict oneself to only considering the case $|x| = 1$ and the argument goes through. (btw. I did not downvote your answer, I don't think "more intuitive" arguments are necessarily bad habit, but I wanted to point out that I don't think the argument is complete as it stands. –  Sam Jun 14 '11 at 11:35
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@Sam Here are few of the things I have a hard time digesting. The reasoning to deduce that $|\sin(x)|<|x| -\frac{|x|^3}{7}$ is wrong (even if the conclusion is right). If the terms of an alternating series decrease it doesn't imply that the sum of the series decrease... The part about each embedding subtracting $|x|^3/7$ is again wrong, since $x$ changes with the embedding.... And the last part of the argument, namely deciding that the limit must be $0$, is explained very badly... What does "one more iteration cannot converge back" even mean? –  N. S. Apr 25 '13 at 14:13
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