Limit of infinite loops of sin x as n tends to infinity

Question

Show that $$lim_{n\to\infty} \text {sin sin ... sin x} = 0 $$ for all x.

Note that the n here refers to the number of sin in the expression above.

Answer 1

This is less of an answer and more of a mental note, and is certainly not a rigorous proof. But I think the idea is the same.

We know that $\sin x$ is bounded above and below - i.e. no matter what $x$ is, $-1 \le \sin x \le 1$.

We also know that $-x \le \sin x \le x$ because $-1 \le \cos x \le 1$ (just take the derivative of the whole inequality).

Ergo, with each iteration of $\sin$, the domain decreases, and eventually reaches $0$.

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For reference, here are the values of possible values of the iterated $\sin$s.

$$n = 0 \rightarrow x \in (-\infty, \infty)$$

$$n = 1 \rightarrow \sin x \in [-1,1]$$

$$n = 2 \rightarrow \sin \sin x \in [-a_2,a_2] \text{ where } a_2 \approx 0.84147$$

$$n = 3 \rightarrow \sin \sin \sin x \in [-a_3,a_3] \text{ where } a_3 \approx 0.74562$$

$$n = 4 \rightarrow \sin \sin \sin \sin x \in [-a_4,a_4] \text{ where } a_4 \approx 0.67843$$

$$n = 5 \rightarrow \sin \sin \sin \sin \sin x \in [-a_5,a_5] \text{ where } a_5 \approx 0.62757$$

Answer 2

Let $x \in \mathbb{R}$ and let $(a_n)$ be as follow

$a_1 = sin(x)$

$a_n = sin(a_{n-1})$

Note that, as $Sin(0)=0$ and $Sin'(x) = Cos(x) \le 1 = x'$ o that $Sin(x) < x$ for all positive x, and so we know for sure that ($|a_n|)_{n\in \mathbb{N}}$ is decreasing (because $sin$ is an odd function). It remains to prove that $a_n \to 0$

Let $|a_n| \to k$, then:

$k = lim_{n \to \infty} |a_n| = lim_{n \to \infty} sin(|a_n|) = sin( lim_{n \to \infty} |a_n|) = sin(k)$

So $k\ge0$ and is a fixed point of $sin$, so that $k=0$.

Answer 3

This is equivalent to $\sin y=y$, whose only solution is $y=0$.