Solution set of $\cos(\cos(\cos(\cos(x)))) = \sin(\sin(\sin(\sin(x))))$

(I'm new here, so I hope this question hasn't come up before)

A bit of motivation for the problem:
It is well-known that the equations $\cos(x) = \sin(x)$, $\cos(\cos(x)) = \sin(\sin(x))$, and $\cos(\cos(\cos(x))) = \sin(\sin(\sin(x)))$ all have infinitely many solutions in $\mathbb{C}$ (the first and third have infinitely many solutions in $\mathbb{R}$ as well). The proofs in these cases are elementary, but break down when applying it to further (lacking a better word) iterations. Using the fourth to illustrate:

Consider the two functions $\cos(\cos(\cos(\cos(z))))$ and $\sin(\sin(\sin(\sin(z))))$, and for convenience, let
$H(z) = \cos(\cos(\cos(\cos(z)))) - \sin(\sin(\sin(\sin(z))))$.
Also, let $V = \{z \in \mathbb{C} \, | \, H(z) = 0\}$.

The original question (while not phrased in this manner) was:
Find $V$.

It is not difficult to verify that $\not\exists z\in V$ such that $\Im(z) = 0$. To prove this, locate local extrema of the function $H(x)$ (where in an abuse of notation, I use $H(x)$ to denote the restriction of $H$ to the real numbers), and one will find that all (relative) maximum and minimum values of the function $H(x)$ are strictly positive. In fact, one can prove that $H(x) \geq \frac{1}{10} > 0$, $\forall x \in \mathbb{R}$. There is a sharper estimate, but this is sufficient for our purposes, and proves that there are no real solutions to the equation $H(x) = 0$.

Having proved that there exist no real solutions, the question now becomes:
Is it possible to analytically (i.e. without numerical methods) prove that $V$ is non-empty?

My first instinct was to try to use the Argument Principle, as applied either to a ball of radius $n\in\mathbb{N}$ centered at $0$, or a rectangle of side-length $n$ centered at $0$, but I'm not sure if those integrals can be computed explicitly (even as contour integrals).

Note: It would be elementary to write an algorithm based on Newton's Method or some improvement thereof and attempt to find roots. But stability is an issue if you are far from a root.

• Sorry...To clarify, it should have said "infinitely many solutions", instead of "infinitely many real solutions", as Colin pointed out. It is actually an isolated instance that in the first and third cases, there are infinitely many real solutions, as for any other iterations, there are no real solutions. – Nicholas Stull Apr 13 '12 at 16:43
• Since your function $H$ is entire, this is essentially just showing that its 'omitted value' isn't $0$ (or equivalently, that it's not of the form $e^{f(z)}$ for some $f$), but unfortunately my analysis background isn't strong enough to go much past Picard's theorem for this... – Steven Stadnicki Apr 13 '12 at 17:23
• @Steven That was something I had thought of. $H$ is indeed analytic and entire, and has an essential singularity at the point at infinity. If we express this as $H(z) = e^{f(z)}$ for some $f$, I believe we would need to ensure that $f$ is analytic. Thanks for the suggestion. I'll update the question if I make any progress. – Nicholas Stull Apr 13 '12 at 23:06
• One can sharpen the condition (without analytical methods). Consider the vanishing of $H^2$ too to see, that its necessary for $cos(cos(cos(cos(z))))$ and $sin(sin(sin(sin(z))))$ to coincide, to have the same value $\pm i\sqrt{1/2}$, i.e. $V = \{z|cos(cos(cos(cos(z)))) = \pm i\sqrt{1/2} = sin(sin(sin(sin(z))))\}$. But I don't know how to go further. – Ben Apr 14 '12 at 11:42
• It appears that there are infinitely many roots on the line $\operatorname{Re} z=\pi/2$. In this case the function $H(z)=\cos^4 z −\sin^4 z$ becomes $H(y)=\cos^3(\sinh y)−\sin^3(\cosh y)$, where the exponent denotes repeated application of the function. – Antonio Vargas Apr 14 '12 at 16:06

There was a proof that $\cos^{(3)}\sinh x=\sin^{(3)}\cosh x$ has infinitely many solutions in a previous version of this answer, but it turns out this is irrelevant to the question.

We will try to solve $\cos^{(n)}z=\sin^{(n)}z$ using Rouche's theorem. We begin by inverting the equation to (almost) solve for $z$ on the left. Let $\arccos z$ be the standard inverse cosine function with branch cuts for $z\ge 1$ and $z\le -1$, and let $\arccos_2 z=\arccos z$ if $\Im[z]\le 0$ and $\arccos_2 z=-\arccos z$ otherwise, which has a branch cut for $z\le 1$. Now $\arccos z$, $-\arccos z$, and $\arccos_2 z$ are all inverse cosines of $z$, so if we let $w=f(z):=(-\arccos)^{(n-2)}(\arccos_2 z)$ then $\cos^{(n-1)} w=z$, and we are left to solve $\cos z=\sin^{(n)}f(z)$.

Observe that $f$ has no branch cuts on the half-plane $\Re[z]>\pi$. The branch cut of $\arccos_2 z$ is $z\le 1$, and other branch cuts arise if $(-\arccos)^{(i)}(\arccos_2 z)$ is a branch cut point of $\arccos$, for some $i<n$. That is, $(-\arccos)^{(i)}(\arccos_2 z)$ is real of magnitude at least 1. But if $\Re[z]>\pi$ then $\arccos_2 z$ lies in the rectangle $|\Re[z]|<\pi/2$, $\Im[z]>0$, and $-\arccos z$ maps the region $|\Re[z]|<\pi$, $\Im[z]>0$ to a subset of itself, namely $-\pi<\Re[z]<0$, $\Im[z]>0$, so it can never be a real number.

As $\Re[z]$ becomes large, $f(z)$ behaves roughly as $\log^nz$. So it does not ever stabilize completely, but it does grow slowly enough that it is effectively constant across each period of $\cos z$. Here is a plot of $\cos z-\sin^{(n)} f(z)$:

Near $0$ we have the crazy behavior of $\sin^{(n)}$ visible, and the branch cut at negative reals is also visible, but on the right, we have very regular behavior, essentially the same as $\cos z$ plus a slowly varying part. We now show that for some sufficiently large (but not too large - see below) real $x$, there are at least $2$ solutions of $\cos z=\sin^{(n)} f(z)$ with $|\Re[z]-x|\le 2\pi$. It follows that $\cos^{(n)}z=\sin^{(n)}z$ has infinitely many solutions.

Let $g(z)=\sin^{(n)} f(z)$, and suppose we have chosen $x$ sufficiently large. We will approximate $g(z)$ by $g(x)$ in a nearby region, and use Rouche's theorem on the function $\cos z-g(x)$, with $g(x)-g(z)$ as the perturbation. The region of interest is a rectangle bounded by $|\Im[z]|\le R$ and either $2\pi n\le\Re[z]\le 2\pi n+2\pi$, or $2\pi n-\pi\le\Re[z]\le 2\pi n+\pi$, where $n$ is chosen so that the rectangle includes $x$. If $\Re[g(x)]$ is negative, then $\cos z-g(x)$ takes a minimum value at least $1$ on the vertical lines $\Re[z]=2\pi n$, so we use the region $2\pi n\le\Re[z]\le 2\pi n+2\pi$, and if it is positive then $\cos z-g(x)$ takes a minimum value at least $1$ on vertical lines $\Re[z]=2\pi n+\pi$, so we use the other region.

It suffices to show $|g'(z)|\le\frac1{2\pi}$ for all $z$ in the rectangle, because then $$|g(x)-g(z)|\le \frac1{2\pi}|z-x|\le \frac1{2\pi}\sqrt{\Im[z]^2+(2\pi)^2}\le \cosh \Im[z]\le \left|\cos z\right|$$ so Rouche's theorem applies.

It turns out that it is surprisingly hard to show that $g(z)$ is slowly varying. The problem is that $\sin^{(n)} z$ has the potential to blow up the value, but it doesn't in this case, and the reason is more subtle than a crude estimation will give you.

Below is a plot of $\sin^{(100)}z$. We can see the fractal nature of $\sin^{(n)}z$ here; the red and blue basins are the regions where the sequence $(\sin^{(n)}z)_n$ converges (slowly) to zero more or less along the real line near $\sin 1$ (red) or $-\sin 1$ (blue). The white regions are where $\sin^{(n)}$ blows up (this process is extremely nonuniform, since the points are sent all over the plane and may hit a convergent region again).

The orange point marked in both graphs is one of the fixed points of $-\arccos z$, approximately $z^*\approx -2.487 + 1.809 i$. As you can see this is within a basin of attraction of $\sin^{(n)} z$, so eventually it will converge to zero as $n$ grows. The derivatives $\frac{d}{dz}\sin^{(n)} z\large|_{z=z^*}$ bounce around for small $n$ but eventually converge to zero; in particular one can verify that $|\frac{d}{dz}\sin^{(n)} z\large|\le 120$, and $|\sin^{(n)} z\large|\le 6$, for all $n$ and all $z$ within $0.002$ of $z^*$.

Every point in $\Re[z]>\pi$ converges to $z^*$ under the sequence $z_n=(-\arccos)^{(n)}(\arccos_2z)$, and $|\frac{d}{dz}(-\arccos z)\large|_{z=z^*}|\approx 0.32$ so convergence is quite rapid; we have $|z_{n+1}-z^*|\le \frac12|z_n-z^*|$ for every $n$ and initial point $z$ with $\Re[z]>\pi$.

Now, suppose $\Re[z]>\pi$; then $$z_0\in S'=\{z\mid -\pi<\Re[z]<\pi\wedge \Im[z]\ge 3/2\},$$ and if we define $$S=\{z\mid -\pi<\Re[z]<0\wedge\Im[z]>0\wedge |z|>3/2\},$$ then $z\in S'\cup S\to-\arccos z\in S$, so $z_n\in S$ for all $n>0$.

On the region $S$, the derivative of $-\arccos$ has magnitude at most $2/\sqrt 5\approx 0.89$. Thus $|\frac{dz_n}{dz_0}|\le (2/\sqrt 5)^n$ for all $n$ and all initial points $z_0\in S'$.

Returning to our Rouche's theorem application, we need $R$ to be large enough to enclose a root of $\cos z-g(x)$, meaning that $\Im[\arccos g(x)]<R$; since $g$ will be bounded by $6$ in the region of interest it suffices to take $R=3$. In this case the entire rectangle lies within $\sqrt{(x+2\pi)^2+3^2}\le x+7$ of the origin. If we take the smallest $x$ that produces a box which does not touch the branch point, namely $x=2\pi$, then we find that $z_7$ is the first iterate which is guaranteed to lie within $0.002$ of the fixed point. Additionally, using that the derivative of $\arccos_2 z$ is bounded by $1/\sqrt{35}$, we find that $|g'(z)|\le 120\cdot \frac1{\sqrt{35}}\cdot (2/\sqrt 5)^{n-2}\le \frac 1{2\pi}$ for $n\ge 46$ (or $n\ge 20$, with the better estimate $|\frac{d}{dz}\sin^{(n)} z\large|\le 6$ for $n\ge 20$).

Thus we have solved the equation for all $n\ge 20$. Unfortunately this leaves quite a few cases to check, including the case $n=4$ in this question! We can still use the Rouche method for smaller $n$, but we must be more careful about the derivation that $g'(z)$ is small.

There is a much simpler method for $n=4$ specifically, along the lines of the elementary solutions for $n=1,2,3$. Let $z=\pi/2+i\operatorname{arcsinh} x$, so that we must solve $g(x)=\cos^{(2)}\cosh x-\sin^{(3)}\sqrt{x^2+1}=0$. Here's a plot:

By some easy bounds we can show that $g(0)=\cos^{(2)}1-\sin^{(3)}1>0$ and $g(5/3)<0$, so there is a root by the IVT.

• That first plot has some of the most bizarre anything I've ever seen anywhere. – The Count Aug 13 '18 at 22:11
• @TheCount It's even crazier when you realize, as I did, that the regular behavior on the right doesn't continue forever, but rather it starts going crazy again around $e^{e^{e^e}}$. – Mario Carneiro Aug 14 '18 at 5:06
• Do you mind if I save this as an example of "not at all nicely predicatable" behavior? – The Count Aug 14 '18 at 17:58