Sines and cosines (part 2)

In my previous post I asked the following question.

What $x$-values satisfy the equation

$\sin(\sin(\sin(\sin(x))))=\cos(\cos(\cos(\cos(x))))$?

More generally, let $s(x)=\sin(x)$ and $c(x)=\cos(x)$. Let $s^n=s\circ\cdots\circ s$ be the composition of $s$ with itself $n$ times. Similarly, let $c^m$ be the composition of $c$ with itself $m$ times.

What are the solutions to $s^n(x)=c^m(x)$?

First, let us look at some properties of the iterates of $s(x)$ and $c(x)$.

$s^n(x)$ is periodic with period $2\pi$. The maxima occur at $\pi(4k+1)/2$ and minima occur at $\pi(4k-1)/2$. (The first few iterates are shown below.)

When $n>1$$c^n(x)$ is periodic with period $\pi$. When $n$ is odd the maxima occur at $k\pi$ and the minima occur at $\pi (2k+1)/2$. When $n$ is even the minima occur at $k\pi$ and the maxima occur at $\pi (2k+1)/2$. (The first few iterates are shown below.)

Now we will address the question at hand by examining some specific cases.

m=2 and n=2

We will show that there are no solutions. In particular, we will show that $c^2(x)-s^2(x)>0$ and hence $c^2(x)>s^2(x)$ for all $x$.

First notice that $\big|\dfrac{\cos(x)\pm\sin(x)}{2}\big|\le\dfrac{\sqrt{2}}{2}<\dfrac{\pi}{4}$. Then see that

$\begin{array}{rl}c^2(x)-s^2(x)&=\cos(\cos(x))-\sin(\sin(x))\\&=\sin\big(\dfrac{\pi}{2}-\cos(x)\big)-\cos\big(\cos(x))\big)\\&=2\cos\big(\dfrac{\pi}{4}-\dfrac{\cos(x)}{2}+\dfrac{\sin(x)}{2}\big)\sin\big(\dfrac{\pi}{4}-\dfrac{\cos(x)}{2}-\dfrac{\sin(x)}{2}\big)\\&>2\cos\big(\dfrac{\pi}{4}+\dfrac{\pi}{4}\big)\sin\big(\dfrac{\pi}{4}-\dfrac{\pi}{4}\big)\\&=0\end{array}$

Here are the graphs of $y=s^2(x)$ and $y=c^2(x)$.

m=3 and n=3

As we see in the graph below, $y=s^3(x)$ and $y=c^3(x)$ cross twice in each interval of length $2\pi$.

m=4 and n=4 (the original question)

Above we showed that $c^2(x)>s^2(x)$ for all $x$. Since $\sin(x),s^2(x),c^2(x)\in[-1,1]$ and $\sin(x)$ is strictly increasing in $[-1,1]$, $s^2(x)$ is increasing in $[-1,1]$. So,

$s^4(x)=s^2(s^2(x)) for all $x$.

Thus there are no solutions.

Here are the graphs of $y=s^4(x)$ and $y=c^4(x)$.

m=5 and n=5

The maximum value of $s^5$ is $s^5(\pi/2)=0.627571832\ldots$ and the minimum value of $c^5$ is $c^5(\pi/2)=0.65428979\ldots$ Thus the two graphs are disjoint. We see the graphs of $y=s^5(x)$ and $y=c^5(x)$ below.

m>5 and n>5

It appears that the graphs of $y=c^n(x)$ and $y=s^n(x)$ flatten out as $n$ gets larger. Indeed this is the case. It turns out that as $n\to\infty$, the graph of $y=c^n(x)$ limits on the line $y=.739085...$, and the graph of $y=s^n(x)$ limits on the line $y=0$. The distances to those lines decrease with each iteration, thus the two graphs never cross again. We justify this below.

A value $x\in\mathbb{R}$ is a fixed point of a function $f$ if $f(x)=x$. Graphically, we can identify fixed points by finding the points of intersection of the line $y=x$ and the graph $y=f(x)$. Our functions have one fixed point each: $s(x)$ has a fixed point at $x=0$ and $c(x)$ has a fixed point at $x=.739085...$ (i.e., the unique solution to $\cos(x)=x$).

A fixed point $x$ is attracting if, whenever $x'$ is close to $x$ the sequence (or orbit, using the terminology of dynamical systems) $\{f^n(x')\}$ limits upon $x$; that is, $\displaystyle\lim_{n\to\infty}f^n(x')=x$. It is globally attracting if the orbit of every point limits upon $x$.

As we can see in the cobweb plots below, $0$ is an attracting fixed point for $s(x)$; we see that it attracts all points in the interval $[-1,1]$, and since $-1\le s(x)\le 1$, $0$ is a global attractor.

Similarly, as we see below, the fixed point for $c(x)$ is a global attractor.

Thus, since $\displaystyle\lim_{n\to\infty}s^n(x)=0$ for all $x\in\mathbb{R}$ and $\displaystyle\lim_{n\to\infty}c^n(x)=.739085...$ for all $x\in\mathbb{R}$, the graphs of $y=c^n(x)$ and $y=s^n(x)$ limit on the lines $y=.739085...$ and $y=0$, respectively.

Other cases

For all the other cases I used graphing software to find the number of points of intersection of $y=s^n(x)$ and $y=c^m(x)$. The values along the top correspond to $n$ and the values along the side correspond to $m$. The values in the table are the number of points of intersection in each interval of length $2\pi$ (the arrows mean that the last given value repeats indefinitely).

$\begin{array}{|c||c|c|c|c|c|c|c|c|}\hline & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8\\\hline\hline 1 & 2 & 2 & 2 & 2 & 2 & 2 & 2 & \rightarrow \\\hline 2 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & \rightarrow \\\hline 3 & 2 & 2 & 2 & 2 & 2 & 2 & 0 & \rightarrow \\\hline 4 & 2 & 0 & 0 & 0 & 0 & 0 & 0 & \rightarrow \\\hline 5 & 2 & 2 & 2 & 0 & 0 & 0 & 0 & \rightarrow \\\hline 6 & 2 & 2 & 0 & 0 & 0 & 0 & 0 & \rightarrow \\\hline 7 & 2 & 2 & 0 & 0 & 0 & 0 & 0 & \rightarrow \\\hline 8 & \downarrow & \downarrow & \downarrow & \downarrow & \downarrow & \downarrow & \downarrow & \searrow \\\hline \end{array}$

[I found the $n=2,m=2$ and $n=4,m=4$ proofs here.]