# 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?

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.$$

• This is a clever way to formalize this. Thanks! May 7, 2021 at 16:10
• @Tuatarian It's pretty much the first example you will see in a dynamical systems class (cosine is a little more interesting). May 7, 2021 at 18:53

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.

• +1. I like how concise and clear this answer is. May 7, 2021 at 5:36
• This is not a really conclusive argument. $y = \sin y$ is besides the point, because $\sin 0$ never actually turns up in the iteration process. The important thing is that there is a Lipschitz cone and the function stays strictly below it in an open region around 0. May 7, 2021 at 7:53
• @leftaroundabout As I said, it's enough to consider the interval $[0,1]$, and $0$ certainly is one of the endpoints. May 7, 2021 at 11:17
• This is the cleverest answer imo Aug 17, 2021 at 20:44

Evaluation of the Limit

Let $$a_{n+1}=\sin(a_n)$$ and suppose $$a_0\gt0$$.

As shown in this answer, for $$x\gt0$$, $$0\lt\sin(x)\lt x$$. Thus, $$a_n$$ is a decreasing sequence bounded below by zero. Therefore, $$\lim\limits_{n\to\infty}a_n$$ exists. Since $$\sin(x)$$ is continuous, \begin{align} \lim_{n\to\infty}a_n &=\lim_{n\to\infty}\sin(a_n)\tag{1a}\\ &=\sin\left(\lim_{n\to\infty}a_n\right)\tag{1b} \end{align} Since $$\sin(x)=x$$ only at $$x=0$$, $$(1)$$ says that $$\bbox[5px,border:2px solid #C0A000]{\lim_{n\to\infty}a_n=0}\tag2$$ Since $$\sin(x)$$ is an odd function, $$(2)$$ holds for $$a_0\lt0$$.

Of course, if $$a_0=0$$, then $$a_n=0$$ for all $$n\ge0$$, and $$(2)$$ holds for $$a_0=0$$.

Asymptotics

As shown in this answer $$\lim\limits_{x\to0}\frac{x-\sin(x)}{x^3}=\frac16\tag3$$ If $$a_0\ne0$$, since $$a_{n+1}=\sin(a_n)$$, $$(2)$$ and $$(3)$$ say $$\lim_{n\to\infty}\frac{a_n-a_{n+1}}{a_n^3}=\frac16\tag4$$ An immediate consequence of $$(2)$$ and $$(4)$$ is \begin{align} \lim_{n\to\infty}\frac{a_{n+1}}{a_n} &=\lim_{n\to\infty}\left(1-a_n^2\frac{a_n-a_{n+1}}{a_n^3}\right)\tag{5a}\\ &=1-0^2\cdot\frac16\tag{5b}\\[6pt] &=1\tag{5c} \end{align} Thus, $$(4)$$ and $$(5)$$ give \begin{align} \lim_{n\to\infty}\left(\frac1{a_{n+1}^2}-\frac1{a_n^2}\right) &=\lim_{n\to\infty}\frac{a_n^2-a_{n+1}^2}{a_n^2a_{n+1}^2}\tag{6a}\\ &=\lim_{n\to\infty}\frac{a_n-a_{n+1}}{a_n^3}\lim_{n\to\infty}\frac{a_n+a_{n+1}}{a_{n+1}}\lim_{n\to\infty}\frac{a_n}{a_{n+1}}\tag{6b}\\ &=\frac16\cdot2\cdot1\tag{6c}\\[3pt] &=\frac13\tag{6d} \end{align} Finally, Stolz-Cesàro and $$(6)$$ say that $$\lim\limits_{n\to\infty}\frac1{na_n^2}=\frac13$$; that is, $$\bbox[5px,border:2px solid #C0A000]{\lim_{n\to\infty}na_n^2=3}\tag7$$ In fact, starting with $$a_0=1$$, we get $$\begin{array}{r|l} n&na_n^2\\\hline 1&0.7080734183\\ 10&2.1433001582\\ 100&2.8511162950\\ 1000&2.9803925383\\ 10000&2.9976147637\\ 100000&2.9997198782\\ 1000000&2.9999678410 \end{array}$$

Bounds

Substituting $$x\mapsto x/\pi$$ in $$(7)$$ from this answer, we get $$\cot(x)=\sum_{k\in\mathbb{Z}}\frac1{k\pi+x}\tag8$$ Subtracting $$(8)$$ from $$\frac1x$$ and taking the derivative, we get $$\frac1{\sin^2(x)}-\frac1{x^2}=\sum_{\substack{k\in\mathbb{Z}\\k\ne0}}\frac1{(k\pi+x)^2}\tag9$$ Evaluating $$(9)$$ at $$x=0$$ gives $$\frac2{\pi^2}\zeta(2)=\frac13$$, which agrees with $$(6)$$.

Taking two derivatives of $$(9)$$ gives \begin{align} \frac{\mathrm{d}^2}{\mathrm{d}x^2}\left(\frac1{\sin^2(x)}-\frac1{x^2}\right) &=\sum_{\substack{k\in\mathbb{Z}\\k\ne0}}\frac6{(k\pi+x)^4}\\ &\ge0\tag{10} \end{align} Thus, $$\frac1{\sin^2(x)}-\frac1{x^2}$$ is an even convex function with a minimum of $$\frac13$$ at $$x=0$$.

Since $$a_k^2$$ is decreasing, $$\frac13\le\frac1{a_{k+1}^2}-\frac1{a_k^2}\le\frac1{\sin^2(a_0)}-\frac1{a_0^2}\tag{11}$$ Summing $$(11)$$ gives $$\frac n3\le\frac1{a_n^2}-\frac1{a_0^2}\le n\left(\frac1{\sin^2(a_0)}-\frac1{a_0^2}\right)\tag{12}$$ Since $$\sin^2(x)\le1$$, assume that $$a_0^2\le1$$, whence $$\frac1{\sin^2(a_0)}-\frac1{a_0^2}\le\cot^2(1)$$.

Solving $$(12)$$ for $$a_n^2$$ yields $$\frac{\tau a_0^2}{\tau+na_0^2}\le a_n^2\le\frac{3a_0^2}{3+na_0^2}\tag{13}$$ where $$\tau=\tan^2(1)\approx2.42551882$$.

Graphical Verification

Limit $$(7)$$ holds for any $$a_0\ne0$$. Since $$\sin(x)$$ is an odd function, changing the sign of $$a_0$$, changes the sign of all $$a_n$$. Thus, we have that $$\newcommand{\sgn}{\operatorname{sgn}} \lim_{n\to\infty}\sqrt{n}\,a_n=\sqrt3\sgn(a_0)\tag{14}$$ Therefore, if we define $$\sin^{\ast n}(x)$$ to be the $$n$$-fold composition of $$\sin(x)$$, we get $$\bbox[5px,border:2px solid #C0A000]{\lim_{n\to\infty}\sqrt{n}\sin^{\ast n}(x)=\sqrt3\sgn(\sin(x))}\tag{15}$$ Here is graphical verification of $$(15)$$: Asymptotically, $$(15)$$ says that $$\bbox[5px,border:2px solid #C0A000]{\sin^{\ast n}(x)\sim\sqrt{\frac3n}\sgn(\sin(x))}\tag{16}$$ which would be a more precise way of saying what you were trying to say.

• This is also quite clever, +1 May 8, 2021 at 19:59
• Thanks. I have expanded the answer to include the functions you mentioned, although scaled so that we can better see what they look like.
– robjohn
May 8, 2021 at 22:35
• – Gary
May 9, 2021 at 11:13

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.