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](, specifically $$s_n\sim\sqrt{\frac{3}{n}}+O\left(\frac{\ln n}{n^{3/2}}\right).$$
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