Nesting the Sine Function

## Problem
#### Let $x\in(0, \pi)$. Consider the sequence of real numbers given by $s_1=\sin x$ and $s_{n+1}=\sin s_n$ for $n\in\{2,3,\ldots\}$. Thus we consider iterations of sine nesting:
$$s_n=\sin\Big(\sin\big(\cdots\sin (x)\big)\Big).$$
#### (a) Prove that $\{s_n\}$ is a convergent sequence.
#### (b) Determine its leading-order asymptotic behavior for large $n$.
## Solution
**(a)** The restriction of $x$ to $(0,\pi)$ makes it so that $s_1=\sin x>0$. We would like to show that $\{s_n\}$ is decreasing and bounded below by zero. We may do it by induction: suppose $s_k>0$ so that, because $0<\sin u< u$ for $u>0$, we have $s_k>s_{k+1}=\sin s_k>0$. This is exactly the desired statement. Therefore the sequence is convergent by the monotone convergence theorem, and we may define
$$s=\lim_{n\to\infty} s_n.$$
Now to get the value of the limit $s$, we write
$$0=\lim_{n\to\infty}\Big(s_{n+1}-\sin s_n\Big)=s-\sin\left(\lim_{n\to\infty}s_n\right)=s-\sin s.$$
This equation is solved by $s=0$, and therefore
$$s=\lim_{n\to\infty}s_n=0,$$
independent of the value of $x$.
**(b)** For large $n$, $s_n$ is small and positive. We postulate an ansatz of algebraic decay,
$$s_n=\alpha n^{-r}.$$
Now, the recursive definition of $s_n$ gives
$$s_{n+1}=\sin s_n=s_n-\tfrac{1}{6}s_n^3+O\left(s_n^5\right),$$
and, plugging in the ansatz, we find
$$\alpha n^{-r}\left(1+\frac{1}{n}\right)^{-r}=\alpha n^{-r}\Big(1-\tfrac{1}{6}\alpha^2 n^{-2r}+O\left(n^{-5r}\right)\Big).$$
The left hand side is expanded according to
$$\left(1+\frac{1}{n}\right)^{-r}=1-\frac{r}{n}+O\left(\frac{1}{n^2}\right).$$
Forcing agreement to first order, we have
$$
-\frac{r}{n}=-\frac{\alpha^2}{6n^{2r}}.
$$
In order to match powers of $n$, we need $r=\frac{1}{2}$ and subsequently $\alpha=\sqrt{3}$. Therefore, the leading order asymptotic behavior is
$$s_n\sim\sqrt{\frac{3}{n}},$$
or equivalently
$$\lim_{n\to\infty} \sqrt{n}\sin^{(n)}(x)=\sqrt{3},\qquad x\in(0,\pi)$$
where the notation $\sin^{(n)}$ refers to iterating rather than differentiating or exponentiating.
## Continuation
If we make another ansatz,
$$s_n=\sqrt{\frac{3}{n}}+\beta n^{-s},$$
we quickly find that it gives us no information. This failure indicates that the subsequent correction to the asymptotic form is not algebraic. In this case, [the next term in the expansion turns out to be logarithmic](http://books.google.com/books?hl=en&id=_tnwmvHmVwMC&pg=PA157), specifically
$$s_n\sim\sqrt{\frac{3}{n}}+O\left(\frac{\ln n}{n^{3/2}}\right).$$