The first zero-crossing of a combination of sines

Question

Let $\{c_i\}_{i=1}^n$ be a sequence of real numbers such that $c_i \geq 0$ for each $i$ and $\sum_{i=1}^n c_i = 1$. Let $\omega_i \in [\delta, \Delta]$ for each $i$, where $\delta$ and $\Delta$ are strictly positive reals. Define $$f(t) = \sum_{i=1}^n c_i \sin(\omega_i t)$$ for $t \in [0, \infty)$, and let $t'$ be the smallest non-zero value of $t$ for which $f(t) = 0$.

• How can we show that there exists a constant $T > 0$ such that $t' \leq T$ for all possible choices of $c_i$ and $\omega_i$?
• Can we give upper/lower bounds on $T$ in terms of any of the other variables?
• In particular, how can we determine what, if any, is the dependence of $T$ on $n$?

Answer 1

That one cannot guarantee a zero on $(0,\pi/\delta]$ or on any interval $(0,c]$ with $c$ independent on $n$ follows from the example of B. Logan, (Theorem 5.5.1 of his thesis Properties of high-pass signals, Columbia Univ., 1965). He constructs a bounded $L^2$ function whose Fourier transform is supported on any given set $[-b,-a]\cup [a,b]$ with $0<a<b$, and which is positive on any given interval $(0,c)$. This function can be approximated by functions of your class with large $n$.

Logan's example is reproduced as Example 1 here.

Here is the solution suggested by user Fedja.

Theorem. $f$ must have a zero on the interval $(0,\pi n/\delta)$. (This estimate is unlikely to be exact for all $n$).

Proof. Define the linear operators $f\mapsto A_kf$, where $(A_kf)(t)=f(t)+f(t+\pi/\omega_k)$. When we apply $A_k$ to our $f$ it kills all summands of the form $\sin\omega_k(t+\alpha)$. So $$A_nA_{n-1}\ldots A_1f=0.$$ On the other hand $$(A_nA_{n-1}\ldots A_1f)(t)=f(t)+f(t+\pi/\omega_1)+f(t+\pi/\omega_2)+f(t+\pi/\omega_1+\pi/\omega_2)+\ldots,$$ all summands are of the form $f(t+\alpha)$ where $\alpha$ is a sum of some $\pi/\omega_j$, with at most $n$ such summands, so $\alpha\leq \pi n/\delta$. Thus if our $f(t)>0$ on $(0,\pi n/\delta+\epsilon)$ this sum will be positive on $(0, \epsilon)$ - contradiction.

Notice that the estimate does not depend on $\Delta$.

EDIT. For $f(x)=(2/3)\sin x+(1/3)\sin(2.5 x)$ the smallest positive zero is about $3.48>\pi$,

Answer 2

The upper bound $\frac{\pi n}{\delta}$, with $\delta = \min_i(\omega_i)$, can be improved when $n = 2$.

Claim. Let $f(t) = c_1 \sin(\omega_1 t) + c_2 \sin(\omega_2 t)$, with $c_i, \omega_i > 0$ for $i = 1, 2$. Then there is $0 < \theta < \frac{3\pi}{2 \delta}$ such that $f(\theta) = 0$.

Proof. If $\omega_1 = \omega_2$, set $\theta = \frac{\pi}{\omega_1}$. Swapping $\omega_1$ and $\omega_2$ if needed, we can hence assume, without loss of generality, that $\omega_1 < \omega_2$. If $\frac{\pi}{\omega_2} \ge \frac{\pi}{2\omega_1}$, then we have $f(\frac{\pi}{\omega_1}) = c_2 \sin(\frac{\omega_2}{\omega_1} \pi) \le 0$. Since $f(\epsilon) = (c_1 \omega_1 + c_2 \omega_2) \epsilon + o(\epsilon^2)$ is positive for all sufficiently small $\epsilon > 0$, the claim follows from the intermediate value theorem. Let us assume now that $\frac{\pi}{\omega_2} < \frac{\pi}{2\omega_1}$. As $\sin(\omega_2 t)$ cannot be positive on any interval of length larger than $\frac{\pi}{\omega_2}$, it takes a negative value at some $\theta$ in $(\frac{\pi}{\omega_1}, \frac{3\pi}{2\omega_1})$, an interval on which $\sin(\omega_1 t)$ is negative. Hence $f(\theta) < 0$, which completes the proof.

Answer 3

Below is a contour plot (courtesy of Wolfram Alpha) of sin t + sin ct. A better plot would be of sin t + A(c)*sin ct to maximize the light regions. This plot confirms the $3\pi/2$ bound for the case of equal amplitudes. Maybe someone could make a better plot with different amplitudes?

Gerhard "Use Pretty Pictures To Bound" Paseman, 2019.04.19.