# number of zeros of the superposition/interference of sine oscillations

There is a tricky problem to solve and we ask for your kind help.

In a rather simple version of the problem, we have a set of oscillations of type $f(x)=\sin (\omega x + \phi)$ with different phases and frequencies. They all start at $f(x_0=0)=0$.

Is there a way to estimate or precise calculate the number of zeros of the superposition of these oscillations: $$F(x_0)= \sum_{i=1}^{m}{\sin (\omega_i x + \phi_i)}=0$$ up to $m$. To emphasise: it is about finding a counting function $c(m)$ that gives us an estimate or precise number of zeros of the interference/superposition function; what we do NOT look for is to calculate first the roots and then count them.

As this problem resembles near to spectral theory and Fourier, any hint from other research in other disciplines or mathemtical areas that would cross with our problem would be very welcome as well. This includes also any stochastic approaches on how to determine the statistical distribution of the zeros.

Thanks

• Note that $f(0)=0$ implies $\phi=n\pi$ with $n\in\mathbb N$. Commented May 1, 2013 at 8:24

[EDIT 2 this is actually an answer, see below]. Not a complete answer but it's a start. First note that, depending on the properties of the frequencies $\omega_j$ your function $F(x)$ has a very different behavior. When the frequencies are rational numbers, the function $F(x)$ is periodic. To find the number of zeros in an interval $[0,X]$ you will need the explicit form of the frequencies. If the frequencies are rationally independent (i.e. linearly independent over the set of rationals) your function is almost periodic. You can find some information in this wikipedia article.

Now note that you can write the number of zeros of $F(x)$ in $[0,X]$ as

$$N(X) = \int_0^X \delta(F(x)) |F'(x)| dx \tag{1}$$

where $\delta()$ is Dirac delta function. Now, with the assumption of rational independence, you can use the theorem of the average of Weyl:

$$\lim_{X\to\infty} \frac{1}{X} \int_0^X f(x) dx = \langle f \rangle$$

where $\langle f \rangle$ is the "phase-space" average over an $m$ dimensional torus ($m$ is the number of frequencies). More precisely

$$\langle f \rangle = \left ( \prod_{j=1}^{m} \int_0^{2\pi} \frac{d\theta_j}{2\pi} \right ) f(\theta_1,\ldots,\theta_m) \tag{2}$$

Now, for large $X$ you can write $$N(X) \simeq X \langle N \rangle$$ where the average $\langle N \rangle$ can be computed as a phase space average. Now typically phase space average are much easier to compute than "time average" (your $x$) and you should be able to estimate $\langle G \rangle$.

In any case your question is so natural that I would guess that somebody might have already computed the answer.

There is a book by Besicovitch on almost periodic functions that may contain some information. Not surprisingly the title is "Almost periodic functions". If you google "theorem of average", "Weyl", you should get more information on what I just wrote above.

Hope this helps.

EDIT

I guess it should be quite clear that in general your function has an infinite number of zeros, therefore I interpreted your question as: find the number of zeros in a given interval $[0,X]$ (possibly with $X$ large)

It turns out that Eq. (1) is basically Rice's formula for counting zeros, and it's a big thing in signal processing (are you an engineer?). You can find some more information in the bible here. There you will basically also find the answer, at least for the case of rationally independent frequencies and large $m$ as in this case your function is a Gaussian random process. In this case the average number of zeros in an interval $[0,T]$ (sorry I switched to time notation, it just makes much more sense) is given by

$$\langle N(T) \rangle = T \frac{1}{\pi} \sqrt{-\frac{\psi''(0)}{\psi(0)}}$$

where $\psi(\tau)$ is the correlation function:

$$\psi(\tau) = \lim_{T\to\infty} \frac{1}{T} \int_0^T F(t) F(t+\tau) dt$$

Now you should be able to do the final steps and get the answer.

Hint. If your frequencies $\omega_j$ are reasonably distributed $N(T)$ is basically independent of $m$. For instance, if $\omega_j$ are uniformly distributed in $[0,1]$ then

$$N(T) \approx \frac{T}{\pi \sqrt{3}}$$

• thanks lev. this helps us a step forward. we will read through the links and also appreciate your suggested Ansatz. Thanks for taking time. Commented May 2, 2013 at 8:54
• Hi lev - fantastic extension. I just got back to this today early in the morning while seeking for understanding my autocorrelation function of the series. It looks like a mircale. Indeed I will need to digest all and understand all details but your help was absolutely crucial. Thanks Commented May 13, 2013 at 7:39