Nesting the Sine Function

2012-10-05 math asymptotics

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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 $n\in \{2,3,\dots \}$. Thus we consider iterations of sine nesting:

$$s_n=\sin (\sin (\cdots \sin (x)\left)\right)\mathrm{.}$$

(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 $$s=\underset{n\to \mathrm{\infty }}{\lim }{s}_n\mathrm{.}$$

Now to get the value of the limit $s$, we write $$0=\underset{n\to \mathrm{\infty }}{\lim }(s_{n+1}-\sin s_n)=s-\sin \left(\underset{n\to \mathrm{\infty }}{\lim }{s}_n\right)=s-\sin s\mathrm{.}$$

This equation is solved by $s=0$, and therefore $$s=\underset{n\to \mathrm{\infty }}{\lim }{s}_n=0,$$

independent of the value of $x$.

(b) For large $n$, $s_n$ is small and positive. We postulate an ansatz of algebraic decay, $$s_n=\alpha n^{-r}\mathrm{.}$$

Now, the recursive definition of $s_n$ gives $$s_{n+1}=\sin s_n=s_n-\frac{1}{6}{s}_n^3+O\left(s_n^5\right),$$

and, plugging in the ansatz, we find $$\alpha n^{-r}{(1+\frac{1}{n})}^{-r}=\alpha n^{-r}(1-\frac{1}{6}{\alpha }^2{n}^{-2r}+O\left(n^{-5r}\right))\mathrm{.}$$

The left hand side is expanded according to $${(1+\frac{1}{n})}^{-r}=1-\frac{r}{n}+O\left(\frac{1}{n^2}\right)\mathrm{.}$$

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 $$\underset{n\to \mathrm{\infty }}{\lim }\sqrt{n}{\sin }^{(n)}(x)=\sqrt{3},x\in (0,\pi )$$

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