Posted by: Dave Richeson | March 9, 2009

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.)

picture-5

When n>1c^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.)

picture-4

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).

picture-6

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.

picture-7

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))<s^2(c^2(x))<c^2(c^2(x))=c^4(x) for all x.

Thus there are no solutions.

Here are the graphs of y=s^4(x) and y=c^4(x).picture-8

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.

picture-10

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.

picture-21picture-3

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

picture-1

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.]

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