Let $\alpha$ and $\beta$ be incommensurate real numbers.

Consider the function $f(x)={\rm cos}(x)+{\rm cos}(\alpha x)+{\rm cos}(\beta x)$ and its positive zeros $x_k(\alpha,\beta)$.

Fix $\alpha$ and $\beta$, and consider the distances between successive zeros $z_k=x_{k+1}-x_k$.

The concrete distribution of the zeros depends on $\alpha$ and $\beta$, but the distribution always shows singularities. Here are some examples for the first 100k zeros for various $\alpha$ and $\beta$:

distribution of zero distances

The position of the singularities does obviously depend on the continued fraction representation of $\alpha$ and $\beta$, as higher rational approximations to $\alpha$ and $\beta$ will display similar distributions.

My question is: Can one write down explicit approximations for the positions of these singularities in these distributions of the distances between the zeros in terms of the continued fraction expansions of $\alpha$ and $\beta$?

(The distances $z_k$ and $z_{k+1}$ are strongly correlated.)

The function $f(x)$ is almost periodic. In the literature about almost periodic functions I could not locate information about the distribution of the distances between the zeros.

Mathematica code to generate the images: https://drive.google.com/open?id=0B649LNvIOdYnaDctRElFYjQ2RlU


2 Answers 2


The peaks arise from envelopes that arise when the function value nearly repeats. The peaks are at the roots of the implicit equation $\pm {\rm cos}(x)\pm {\rm cos}(\alpha x)\pm {\rm cos}(\beta x)=0$ for the case of the function $f(x)= {\rm cos}(x)+ {\rm cos}(\alpha x)+ {\rm cos}(\beta x)=0$. More details can be found here.

  • $\begingroup$ How does this even address the question? $\endgroup$
    – Alex M.
    Apr 24, 2018 at 17:57

Here is how I think of this problem from a ergodic theory point of view. Because the system is ergodic for long time the point $(t [2\pi] , \alpha t [2\pi], \beta t [2\pi] )$ should span $[2\pi]^3 $ with equidistribution . Therefore the distribution of the distance $t$ between two consecutive zeros should be given by the measure of the set defined by $$\begin{cases} \cos(x)+\cos(y)+\cos(z)=0 \\ \cos(x+t)+\cos(y+\alpha t)+\cos(z+\beta t)=0\end{cases} $$ Let $(u,v,w)$ an orthogonal basis with $u=(1,\alpha,\beta)$ and let $(a,b,c)$ the coordinate of $(x,y,z)$ in this basis. The previous equations give us $$\begin{cases} a_1=g_1(b,c) \\ a_2=g_2(b,c) \\ t=a_2-a_1\end{cases} $$We can think of $a_1$ and $a_2$ as the fist and second time we cross the set $\cos(x) +\cos(y)+\cos(z) =0 $ starting from the point $(0,b,c)$.

On the $\nabla (g_2-g_1)\neq 0$, the solution is a smooth curve,therefore the singalarities can only appear when $\nabla (g_2-g_1) = 0$. As a conclusion, the singularities should only exists for $t$ such that exists $(x,y,z)$ solution of $$\begin{cases} \cos(x)+\cos(y)+\cos(z)=0 \\ \cos(x+t)+\cos(y+\alpha t)+\cos(z+\beta t)=0 \\ \nabla(g_2-g_1)=0 \end{cases}$$ This explain the singularity at the boundaries of the set on your first diagram (extremum $\Rightarrow \nabla (g_2 -g_1)=0$ ).

I try to solve the system numerically for $\alpha=\sqrt{2},\beta=\sqrt{3}$ and it gives me $t \approx 2.275$. Does it make sense with your last diagram?


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