Limit of infinite composition of sin(x) I was playing around on desmos the other day, and noticed that $\sin\left(\sin\left(x\right)\right)$ is basically a version of sin with a lower amplitude (which makes intuitive sense). To me, it seems intuitve that this curve, when composed infinite times, becomes a straight line, as the values at $x=\frac{\pi}{2}\mathbb{Z}$ would move (slowly) towards the values at $x=\pi\mathbb{Z}$ by virtue of them moving away from the peaks, but is there a way to go about properly proving this?
 A: This is straightforward to prove, but needs attention to some detail.
Suppose we pick some $x_0$ and let $x_{n+1} = \sin x_n$. The idea is to show that $x_n \to 0$.
If $x_0 = 0$ then $x_n = 0 $ for all $n$, so suppose $x_0 \neq 0$. Note that $|x_n| \le 1$ for $n \ge 1$, so we might as well suppose $|x_0| \le 1$.
The mean value theorem shows that for for $x \neq 0$ and $|x| \le 1$ that
$\sin x-0 = (\cos \xi) x$ for some $\xi \in (0, x)$ (or $(x,0)$ if $x <0$), and since
$\cos \xi \in (\cos 1,1)$ (using $x\neq0$ and $|x| \le 1$ here) we see that $|\sin x| = \sin |x| < |x|$.
In particular, $|x_n| $ is non increasing and so $|x_n| \to t^*$ for some $t^*$. By continuity we have $\sin t^* = t^*$ and from the previous paragraph we see that we must have $t^* = 0$. Hence $x_n \to 0$ always.
A: Good question. The straight line you are talking about is actually the line $y=0$.
One way to see this is to start at any point $x_0\in\mathbb{R}$ and apply a fixed point iteration with the formula $$x_{n+1} = \sin(x_n).$$
It's almost as if the maths Gods knew you were going to ask this question, because Wikipedia even has a nice diagram of what's going on in your exact example!

The diagram shows what happens with $x_0 = 2$. You get a staircase diagram converging towards the only root of the equation $x = \sin(x),\ $ namely $x = 0.$
But this was with a starting point of $x_0 = 2.$ What about other starting points? Well, we know that, for all $x \in \mathbb{R},\ -1 \leq \sin(x) \leq 1.$ If $x_0$ is such that $0\leq x_1=\sin(x_0)\leq 1,\ $ then we get a convergence similar to that in the diagram above (staircase). And if $x_0$ is such that $-1\leq x_1=\sin(x_0)\leq 0,\ $ then we also get something similar, but happening below the $x-$axis, but still towards the root $x=0$.
It follows that the curve that $\sin\left(\sin\left(\sin\left(\ldots x \right)\right)\right)$ converges to is in fact the $x$-axis, ie. the line $y=0.$

Why/when does this occur / how can we prove that we get convergence via
this "fixed-point iteration" process?

It can be shown that a fixed-point iteration $x_{n+1} = g(x_n)$:

*

*converges if $\ \vert g'(x)\vert < 1\ $ near the root and $x_1$ is sufficiently close to the root

*diverges if $\ \vert g'(x) > 1\ $ near the root.

There's lots of material out there if you want to look up proofs of these things, but the above is a summary of when convergence happens.
In the $x_{n+1} = \sin(x_n)\ $ example, we have $g(x)=\sin(x).\ $ We know that $g'(0) = 1$, and $g'(x)<1$ for $x$ close to but not equal to the root. We have also seen that $x_1\in [-1,1]\ $ and is sufficiently close to the root. Therefore we get convergence towards $0$ for any starting value $x_0.$
A: We know that $\sin(\mathbb{R})=[-1,1]$  Since the sine function is odd, it will be enough to consider the interval $[0,1]$.  The sine function is increasing on this interval so $\sin([0,1])=[0,\sin(1)]$  Since $|sin(x)|<|x|$ this will be a shorter interval than $[0,1]$.  We can apply the process repeated getting a succession of smaller intervals, which must either converge to the origin, or to an interval $[0,y]$ with $y=\sin y$.  But the only such $y$ is $y=0$, and the intervals shrink to the origin.
Thus, for any $\varepsilon>0$ there is $N$ such for $n>N,\ |\sin^{(n)}x|<\varepsilon$ for every $x\in \mathbb{R}$.  Thus, you are correct in saying that the graph will look like a straight line, in fact the $x$-axis.
