Justify: if $x\gt 0$, $\;\lim_{n\to\infty} \sqrt{n}\cdot{\overbrace{\sin\sin\cdots\sin}^{n\space\text{sines}}(x)}=\sqrt{3}$

Question

I believe that I have managed to show that (if $x\gt 0$) $$\lim_{n\to\infty} \sqrt{n}\cdot{\overbrace{\sin\sin\cdots\sin}^{n\space\text{sines}}(x)}=\sqrt{3}$$ I did this by defining a sequence as $a_0=x$ and the recursion $$a_{n+1}=\sin a_n$$ I then approximated the recursion with the first two nonzero terms of the Maclaurin series for sine, giving me $$\Delta a_n=-\frac{x^3}{6}$$ I then approximated this with a differential equation $$y'=-\frac{y^3}{6}$$ Which I then easily solved... the answer follows from here.

Question: How can this be made more rigorous? I don't know how to justify that my approximations are good enough for th error to vanish under the limit. What theorems are generally used to justify approximations of discrete recursions with differential equations? I think I know how to justify the approximation of some with its Maclaurin series using the Lagrange error bound.

Answer 1

This may not be the answer you wanted, cuz it may not be generalized.

Claim: $\lim n a_n^2 = 3$.

Proof. $\blacktriangleleft$ Note that not matter which value $x$ is, $a_1 = \sin (x) \in (-1,1)$, so it suffice to consider if $x \in (0, \pi/2)$. Now $a_1 > 0$ and $a_2 < a_1$. By induction it would be easy to verify that $0 < a_n< a_{n-1}$. Hence $a_n \searrow$ and $a_n > 0$. Therefore the limit $A = \lim a_n$ exists. Solve $\sin (A) = A$ in $[0, \pi/2]$ gives that $A=0$, i.e. $a_n = o(1)$.

Now compute the limit: \begin{align*} \lim n a_n^2 &= \lim \frac n {a_n^{-2}} \\ &= \lim \frac 1 {a_{n+1}^{-2} - a_n^{-2}}\quad [\text {Stolz formula}]\\ &= \lim \frac {a_n^2 a_{n+1}^2} {a_n^2 - a_{n+1}^2} \\ &= \lim \frac {a_n^2 \sin(a_n^2)} {(a_n - \sin(a_n) )(a_n + \sin (a_n))} \\ &= \lim \frac {a_n^4} {a_n^3/6 \cdot 2a_n} \quad [\sin(a_n) \sim a_n; a_n - \sin (a_n) \sim a_n^3/6]\\ &= 3. \end{align*} Hence $\lim a_n \sqrt n = \sqrt 3. \blacktriangleright$

Answer 2

I'll assume $0<x<1$, so that your sequence decreases to zero.

Write $$b_n=a_n^2=\sin^2(\sin^{[n-1]}(x))$$ where I'll use $\sin^{[m]}(x)$ for the $m$-fold composite of $\sin$. We'd like to prove $b_n\sim 3/n$. But $b_{n+1}=\phi(b_n)$ where $$\phi(t)=\sin^2\sqrt t=t-\frac{t^2}3+\frac{t^3}{12}+\cdots.$$ Then $$\frac1{b_{n+1}}=\frac1{b_n}\left(1+b_n+O(b_n^2)\right) =\frac1{b_n}+\frac13+O(b_n).$$ As $b_n\to0$ then $1/b_n$ increases by at least $1/4$ eventually, so that $1/b_n<n/4+C$ for some constant $C$, and so $b_n=O(1/n)$. Then$$\frac1{b_n}=\frac n3+O(\ln n)$$ which is good enough.

Answer 3

We have

$$ \sqrt n \sin(\cdots \sin x) = a_n\\ \sqrt{n+1}\sin(\sin(\cdots \sin x)) = a_{n+1} $$

hence

$$ \sin\left(\frac{a_n}{\sqrt n}\right) = \frac{a_{n+1}}{\sqrt{n+1}} $$

Analyzing now the iterative procedure

$$ u_{n+1} = \sin u_n $$

with $u_0 = \frac{\pi}{2}$ the sequence

$$ \sqrt k u_k $$

is convergent and has a behavior shown in the attached plot

(in red $\sqrt 3$ and in blue $\sqrt k u_k$)

Answer 4

Since you asked for a reference, here is one that I have enjoyed for many years:

Asymptotic Methods in Analysis by N. G. de Bruijn, available quite inexpensively at Dover and Amazon.

Chapter 8 has an extensive discussion of iterated functions which includes most of the methods in the answers to this question.