Zeroes of the third derivative of an iterated sine.

Solution 1:

It is possible to explain the location of the knee in the graphs above without taking derivatives of $f_n$ or $g_n$.

Suppose that, for $x\in(0,\pi/2]$, we set $x_0:=x$, $x_{n+1}:=\sin x_n$. For all $n\ge 0$, we then have $x_n\in(0,\pi/2]$, so we can set $x_n:=\sqrt{3/t_n}$ for some $t_n=3/x_n^2$ in $[1.2,\infty)$. Expanding the Taylor series for $\sin x$ around $0$, we then have $$x_{n+1}=x_n-\frac{x_n^3}{6}+\theta_n \frac{x_n^5}{120}, \qquad 0\le \theta_n\le 1, $$ so $$t_{n+1}=t_n (1-\frac{1}{2 t_n} +\theta_n \frac{3}{40 t_n^2})^{-2}=t_n+1+O(t_n^{-1})$$ and with some computation we can find the bounds $$0\le t_{n+1}-(t_n+1)\le \frac{2}{t_n},$$ so, in particular, $t_n\ge n+1$ for all $n$ (since $t_0\ge 1.2)$, and $$0\le t_n-(t_0+n)\le \frac{2}{t_0}+\cdots+\frac{2}{t_{n-1}}\le 2 H_n,\qquad (*)$$ where $H_n$ is the $n$th harmonic number, $$H_n=1+\frac12+\cdots+\frac1n=O(\log n).$$

Now, from (*), we can see that if we iterate $\sin x$ $n$ times, starting from $x_0=\pi/2$, we will reach the value $t_n=n+O(\log n)$, or $M_n:=x_n=(1+o(1))\sqrt{3/n}$. If, as is done in the graphs above, we normalize $x_n$ by dividing by $M_n$, setting $\bar x_n:=x_n/M_n$, we get $$\bar x_n=\sqrt{\frac{n+O(\log n)}{t_n}}=(1+\frac{t_0}{n})^{-1/2} (1+O(\frac{\log n}{n})).\qquad (**)$$ Therefore, we will have $\bar x_n\approx 1$ for $t_0=o(n)$. On the other hand, if $t_0/n\to\infty$, then from (*), $t_n/t_0=1+o(1)$, so $x_n=x_0(1+o(1))$ and $\bar x_n\approx x_0/ M_n \approx x_0 \sqrt{n/3}$. The knee in the graph is located between these two regimes, which is where $t_0$ is on the order of $n$, i.e., $x_0$ is on the order of $\sqrt{3/n}$. For example, if we set $x_0=\sqrt{3/n}$ and $t_0=n$, we see from (**) that $\bar x_n\to 2^{-1/2}$ as $n\to\infty$.

Solution 2:

David Moews has given a very precise analysis of the observed phenomenon. On the other hand it is very easy to prove that $$\lim_{n\to\infty} g_n(x)=1\qquad\Bigl(0<x\leq{\pi\over2}\Bigr)\ .$$ Put $\alpha_n:=f_n\bigl({\pi\over2}\bigr)$. The $\alpha_n$ converge monotonically to zero, whereby $$\lim_{n\to\infty}{\alpha_{n+1}\over\alpha_n}=\lim_{n\to\infty}{\sin\alpha_n\over\alpha_n}=1\ .$$ Given any $x\in\ ]0,1]$ there is an $r\geq 0$ with $\alpha_{r+1}<x\leq\alpha_r$. It follows that $$\alpha_{n+r+1}=f_n(\alpha_{r+1})<f_n(x)\leq f_n(\alpha_r)=\alpha_{n+r}\ .$$ Therefore we have $${\alpha_{n+r+1}\over\alpha_n}<g_n(x)\leq{\alpha_{n+r}\over\alpha_n}\ ,$$ and letting $n\to\infty$ the claim follows (note that $r$ is fixed here).