QUESTION: What is the average distance between the consecutive real zeroes of the function $$f(x)=\sin(x)+\sin(x\sqrt{2})+\sin(x\sqrt{3})$$ or, more specifically, if $z(x)$ is defined as the number of zeroes $\zeta$ satisfying $|\zeta|<x$, what is the value of $$\lim_{x\to\infty} \frac{2x}{z(x)}=?$$

Here’s some context. I’ve been studying sums of sinusoids with “mutually irrational” periods, such that the sum of the sinusoids is not actually a periodic function. For example, the function $$\sin(x)+\sin(x\sqrt{2})$$ is not periodic, because $\sqrt{2}$ is irrational. In particular, I’ve been looking at the asymptotic distribution of solutions $x$ to equations in the form $$\sin(x)+\sin(\tau x)=\alpha$$ where $\tau \notin \mathbb Q$ and $|\alpha|<2$. I’ve actually come up with a formula for the average distance between the solutions to the above equation along the real line, but it’s messy so I won’t type it out unless someone cares enough to ask for it. The case of $\alpha = 0$ is almost trivial though, and can be figured out with an easy trig identity.

However, when dealing with three summed sinusoids the case of $\alpha = 0$ is no longer trivial. For two sinusoids, $$\sin(x)+\sin(\tau x)=2\sin\bigg(\frac{\tau+1}{2}x\bigg)\cos\bigg(\frac{\tau - 1}{2}x\bigg)$$ so we can easily calculate the actual explicit values of the zeroes. But for three sinusoids with mutually irrational periods so that $\tau_1, \tau_2, \tau_1/\tau_2 \notin\mathbb Q$, $$\sin(x)+\sin(\tau_1 x)+\sin(\tau_2 x)$$ I haven’t been able to come up with any explicit formulas for zeroes, or even an asymptotic density of/average distance between zeroes.

Can anyone figure out how to work out this problem for the specific case of $\tau_1 = \sqrt{2}$, $\tau_2 =\sqrt{3}$?

  • $\begingroup$ If we let $u=\sin\frac{x(1+\sqrt2)}2$ then the zeroes satisfy the equation $$2u(-\sqrt{1-u^2}+2\cos\frac x2\cos\frac x{\sqrt2})=-\sin(x\sqrt 3).$$ If only $\sqrt3$ became $\sqrt4$... $\endgroup$
    – TheSimpliFire
    May 25, 2019 at 18:11
  • $\begingroup$ Interesting question. So what is the average distance between zeroes in the case of 2? ($\alpha = 0$) $\endgroup$
    – Paul
    May 26, 2019 at 10:25
  • 1
    $\begingroup$ @Paul The average distance between real zeroes of $\sin(x)+\sin(\tau x)$ is $\pi/\tau$ for $\tau > 1$. Look at the factorization in the body of my question. Note that $\sin(\frac{\tau + 1}{2}x)$ and $\cos(\frac{\tau - 1}{2}x)$ have no zeroes in common. The average number of zeroes per unit length on the real line of these two sinusoids are $\frac{\tau + 1}{2\pi}$ and $\frac{\tau - 1}{2\pi}$ (you can calculate the actual zeroes explicitly) so the average number of zeroes per unit length of the product is their sum $\tau/\pi$. The average distance between zeroes is the reciprocal of this. $\endgroup$ May 26, 2019 at 12:23
  • 1
    $\begingroup$ @Paul It gets a lot messier when $\alpha\ne 0$. If this is the case, the formula is $$\frac{2\pi^2}{(1+\tau)\pi-2\arcsin(\frac{\sqrt{\tau^4+(\alpha^2-2)\tau^2+1}-\alpha}{\tau^2-1})-2\tau\arcsin(\frac{\alpha\tau^2-\sqrt{\tau^4+(\alpha^2-2)\tau^2+1}}{\tau^2-1})}$$ for $\tau > 1$. The proof of this is a lot trickier. $\endgroup$ May 26, 2019 at 12:24
  • 4
    $\begingroup$ The main result of math.purdue.edu/~eremenko/dvi/novik1011.pdf shows that the zeroes have density at least $1/\pi$ (since the Fourier transform is supported at $\{ \pm 1, \pm \sqrt{2}, \pm \sqrt{3} \}$). The mean motion theorem mathoverflow.net/questions/278996 , which shows that the winding number of $e^{i t} + e^{i \sqrt{2} t} + e^{i \sqrt{3} t}$ for $|t|<T$ grows like $cT$, also seems relevant. But, in general, I would say this seems hard! $\endgroup$ May 28, 2019 at 18:31

1 Answer 1


Partial answer

I thought of a way to transform the problem into a double integral. I don't prove every step, so I can't say I'm a 100% sure this is right. I'm pretty confident that this approach works, but let me know if I made a mistake.

I'm gonna add cosines together instead of sines. It's the same thing, but cosine is a little easier to work with because it's an even function.

Let $n \ge 2$ be the number of cosine functions we're adding together and let $\tau$ be an $n$-dimensional vector containing the rationally independent positive coefficients. We define: $$ \begin{align} C &= [-\pi, \pi]^n && \text{($n$ dimensional hypercube)} \\ S &= \left\{x \in C\ \middle|\ \sum_{i=1}^n \cos(x_i) = 0\right\} && \text{($n{-}1$ dimensional surface}) \\ g(x) &= \sum_{i=1}^n \cos(\tau_i x) \\ l_i(x) &= ((\tau_i x + \pi) \operatorname{mod} 2\pi) - \pi && \text{(line through $C$)} \end{align} $$ For $n=2$ and $n=3$, $S$ looks like this:

Surface when n=2 and n=3

The function $l(x)$ is a line that starts at the origin and goes in direction $\tau$. Whenever it reaches an edge of $C$, it comes out at the edge on the other side.

Now $g(x) = 0$ whenever $l(x) \in S$. So to count the zeroes we can follow the line $l(x)$ and see how often it crosses the surface $S$.

Because of the rational independence, it seems intuitive that the line will travel through each part of $C$ equally often. Therefore we can integrate over the surface $S$ to calculate how often $S$ is crossed.

I figured out the following formula for calculating the frequency $f$. The average distance between zeroes is $1/f$. The function $p(x)$ gives one of the two possible unit normal vectors on the surface $S$ at $x$. The dot represents the dot product of two vectors. $$ f = \frac{1}{(2\pi)^n} \int_S |p(x) \cdot\tau |\ \mathrm{d}x \label{surfaceint}\tag{1} $$ This gives for $n=2$: $$ \begin{align} f_2 &= \frac{1}{(2\pi)^2} \cdot 2 \pi \sqrt{2} \cdot \left(\left| \left[\begin{smallmatrix}\frac12 \sqrt{2} \\ \frac12 \sqrt{2}\end{smallmatrix}\right] \cdot \tau \right| + \left| \left[\begin{smallmatrix}\frac12 \sqrt{2} \\ -\frac12 \sqrt{2}\end{smallmatrix}\right] \cdot \tau \right| \right) \\ &= \frac{1}{(2\pi)^2} \cdot 2 \pi \cdot (|\tau_1 + \tau_2| + |\tau_1 - \tau_2|) \\ &= \max(\tau_1, \tau_2) / \pi \end{align} $$ For higher $n$, the surface $S$ is more complex and this isn't as easy. Because $S$ is mirror symmetric, we can make it ourselves easier by only integrating over the positive part of $S$. But we do have to take the different normals into account.

$$ \begin{align} R &= \{x \in S\ |\ \forall_i\ x_i \ge 0\} \\ I(x) &= \sum_{d \in \{-1, 1\}^n} \left| \sum_{i=1}^n d_i \cdot p_i(x) \cdot \tau_i \right| \\ f &= \frac{1}{(2\pi)^n} \int_R I(x)\ \mathrm{d}x \end{align} $$ For the normal vector $p(x)$ we can use the normalized gradient of $\sum_{i=1}^n \cos(x_i)$. I multiplied the whole thing by $-1$ to get a positive normal. $$ p_i(x) = \sin(x_i) / \sqrt{\sum_{j=1}^n \sin(x_j)^2} $$

The case $n=3$

Let $u$ be a vector of 3 non-negative elements. The following equation holds: $$ \sum_{d \in \{-1, 1\}^3} |d \cdot u| = 4 \max (2u_1, 2u_2, 2u_3, u_1+u_2+u_3) $$ If we combine that equation with the equation for the normal, we get: $$ I(x) = \frac{ 4 \max \left( \begin{array}{} 2\sin(x_1) \tau_1, \\ 2\sin(x_2) \tau_2, \\ 2\sin(x_3) \tau_3, \\ \sin(x_1)\tau_1+\sin(x_2) \tau_2+ \sin(x_3) \tau_3 \end{array} \right) } { \sqrt{\sin(x_1)^2 + \sin(x_2)^2 + \sin(x_3)^2} } $$ To rewrite the equation to a normal two dimensional integral, instead of a surface integral, we first replace $x_3$. $$ \begin{align} x_3 &= \arccos(-\cos(x_1)-\cos(x_2)) \\ \sin(x_3) &= \sqrt{1 - (\cos(x_1)+\cos(x_2))^2} \\ I(x) &= \frac{ 4 \max \left( \begin{array}{} 2\sin(x_1) \tau_1, \\ 2\sin(x_2) \tau_2, \\ 2 \tau_3\sqrt{1 - (\cos(x_1)+\cos(x_2))^2} , \\ \sin(x_1)\tau_1+\sin(x_2) \tau_2+ \tau_3\sqrt{1 - (\cos(x_1)+\cos(x_2))^2} \end{array} \right) } { \sqrt{\sin(x_1)^2 + \sin(x_2)^2 + 1 - (\cos(x_1)+\cos(x_2))^2} } \end{align} $$ Now we use the equation: $$ \begin{align} \int_R I(x)\ \mathrm{d}x = \int_A I(x)J(x)\ \mathrm{d}x \end{align} $$ Where: $$ \begin{align} A &= \left\{ \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0\end{bmatrix}x\ \middle|\ x \in R \right\} \\ J(x) &= \sqrt{\left(\frac{\partial x_3}{\partial x_1}\right)^2 + \left(\frac{\partial x_3}{\partial x_2}\right)^2 + 1} \\ &= \sqrt{\frac{\sin(x_1)^2+\sin(x_2)^2+1-(\cos(x_1)+\cos(x_2))^2}{1-(\cos(x_1)+\cos(x_2))^2}} \end{align} $$ Combining $I$ and $J$, we get: $$ I(x)J(x) = \frac{ 4 \max \left( \begin{array}{} 2\sin(x_1) \tau_1, \\ 2\sin(x_2) \tau_2, \\ 2 \tau_3\sqrt{1 - (\cos(x_1)+\cos(x_2))^2} , \\ \sin(x_1)\tau_1+\sin(x_2) \tau_2+ \tau_3\sqrt{1 - (\cos(x_1)+\cos(x_2))^2} \end{array} \right) } { \sqrt{1-(\cos(x_1)+\cos(x_2))^2} } $$ So our new integral becomes: $$ \begin{align} f_3 &= \frac{1}{(2\pi)^3} \int_A I(x)J(x)\ \mathrm{d}x \\ &= \frac{1}{(2\pi)^3} \left(\int_0^{\frac12\pi} \int_{\arccos(1-\cos(x_1))}^\pi I(x)J(x)\ \mathrm{d}x_2\ \mathrm{d}x_1 + \int_{\frac12\pi}^\pi \int_0^{\arccos(-1-\cos(x_1))} I(x)J(x)\ \mathrm{d}x_2\ \mathrm{d}x_1 \right) \\ &= \frac{1}{4\pi^3} \int_0^{\frac12\pi} \int_{\arccos(1-\cos(x_1))}^\pi I(x)J(x)\ \mathrm{d}x_2\ \mathrm{d}x_1 \end{align} $$ We can get rid of those nasty sines and cosines by using integration by substitution. Replacing $x_2$ with $\arccos(v)$ gives: $$ \begin{align} H(x_1, v) &= \frac{ \max \left( \begin{array}{} 2\sin(x_1) \tau_1, \\ 2\tau_2\sqrt{1-v^2}, \\ 2 \tau_3\sqrt{1 - (\cos(x_1)+v)^2} , \\ \sin(x_1)\tau_1+\tau_2\sqrt{1-v^2} + \tau_3\sqrt{1 - (\cos(x_1)+v)^2} \end{array} \right) } { \sqrt{1-v^2} \cdot \sqrt{1-(\cos(x_1)+v)^2} } \\ f_3 &= \frac{1}{\pi^3} \int_0^{\frac12\pi} \int_{-1}^{1-\cos(x_1)} H(x_1, v) \ \mathrm{d}v\ \mathrm{d}x_1 \end{align} $$ Then replacing $x_1$ with $\arccos(u)$ gives: $$ \begin{align} G(u, v) &= \frac{ \max \left( \begin{array}{} 2\tau_1\sqrt{1-u^2} , \\ 2\tau_2\sqrt{1-v^2}, \\ 2 \tau_3\sqrt{1 - (u+v)^2} , \\ \tau_1\sqrt{1-u^2}+\tau_2\sqrt{1-v^2} + \tau_3\sqrt{1 - (u+v)^2} \end{array} \right) } { \sqrt{1-u^2} \cdot \sqrt{1-v^2} \cdot \sqrt{1-(u+v)^2} } \\ f_3 &= \frac{1}{\pi^3} \int_0^1 \int_{-1}^{1-u} G(u, v) \ \mathrm{d}v\ \mathrm{d}u \end{align} $$ One possible way to continue solving this integral, is splitting it into four integrals, one for each of the arguments of $\max$. To do this we need to find the values for $u$ and $v$ in which these arguments are the maximum. The conditions can be reduced to: $$ \begin{align} 1\colon&&\ \tau_1 \sqrt{1-u^2} &> \tau_2 \sqrt{1-v^2} + \tau_3 \sqrt{1-(u+v)^2} \\ 2\colon&&\ \tau_2 \sqrt{1-v^2} &> \tau_1 \sqrt{1-u^2} + \tau_3 \sqrt{1-(u+v)^2} \\ 3\colon&&\ \tau_3 \sqrt{1-(u+v)^2} &> \tau_1 \sqrt{1-u^2} + \tau_2 \sqrt{1-v^2} \\ 4\colon&&\ \text{otherwise} \end{align} $$

Numerical approximation

The following Mathematica code calculates an approximation to $\ref{surfaceint}$. I tried to write it to work in any number of dimensions, but it only seems to work in exactly 3 dimensions (Mathematica 11.2).

frequency[t_] := Module[{n, vars, x, r},
   n = Length[t];
   vars = Table[x[i], {i, n}];
   r = ImplicitRegion[Total[Cos[vars]] == 0, Evaluate[{#, -Pi, Pi}& /@ vars]];
   NIntegrate[Abs[t.Normalize[Sin[vars]]], vars \[Element] r] / (2Pi)^n
Print["Average distance between zeroes: ", 1 / frequency[{1, Sqrt[2], Sqrt[3]}]];

The code outputs $2.22465$. I don't know how many digits of that are right.

  • 1
    $\begingroup$ Cool! Yep, that formula works for $n=2$... But does it facilitate the calculation for $n=3$? $\endgroup$ May 27, 2019 at 14:11
  • $\begingroup$ Amazing, that’s great. While it appears unlikely at this point that we will find a “nice” expression for the answer, your approach seems like it would yield the answer to anyone with enough patience. Bravo. $\endgroup$ May 30, 2019 at 20:02
  • $\begingroup$ @Frpzzd Thanks! Maybe I'll see if I can improve it further. $\endgroup$
    – Paul
    May 31, 2019 at 15:25
  • $\begingroup$ Hi Paul: If you can take 50 points more of my reputation (like Frpzzd did) I autorize you to take these (I don't not know how to do). $\endgroup$
    – Piquito
    Jun 1, 2019 at 0:30
  • $\begingroup$ @Piquito Thanks! I'm not sure if it's possible since Frpzzd already did that, but the thing is called bounty, $\endgroup$
    – Paul
    Jun 1, 2019 at 9:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.