This is a follow-up on my previous question that turned out to be almost-trivial.

Let $\varphi(t)=\sin(t)+\sin(t\sqrt{2})+\sin(t\sqrt{3})$. Such function is not periodic, but it is bounded, Lipshitz-continuous and with mean zero, i.e. $\lim_{b\to +\infty}\frac{1}{b-a}\int_{a}^{b}\varphi(t)\,dt = 0$. Its real zeroes are simple, hence by denoting as $\zeta_0<\zeta_1<\zeta_2<\zeta_3<\ldots$ the real positive zeroes we have that $$ E = \sup_{n\in\mathbb{N}}\left(\zeta_{n+1}-\zeta_n\right) < +\infty. $$

Now my actual question:

Q: How can we improve the previous inequality and show, for instance, that $E\leq 2\pi$?

My thoughts:

  1. If one is able to produce accurate bounds for $\frac{1}{2\pi i}\oint_{\gamma}\frac{\varphi'(t)}{\varphi(t)}\,dt$, with $\gamma$ being the boundary of a thin rectangle in the complex plane enclosing the real interval $[a,b]$, is also able to estimate the density of real zeroes;
  2. If for some non-negative function $\psi(t)$ over the interval $[a,b]$ the integrals $\int_{a}^{\frac{a+b}{2}}\varphi(t)\psi(t)\,dt $ and $\int_{\frac{a+b}{2}}^{b}\varphi(t)\psi(t)\,dt $ have opposite signs, $\varphi(t)$ has a zero in $[a,b]$. But what is an efficient way for constructing such weigth functions $\psi$? Can we exploit the convergents of the continued fractions of $\sqrt{2}$ and/or $\sqrt{3}$?
  3. It might by practical to consider the winding number of the curve $\gamma:[0,T]\to \mathbb{C}$ given by $\gamma(t) = e^{it}+e^{it\sqrt{2}}+e^{it\sqrt{3}}$.

Addendum: an explicit proof of $E<+\infty$ through Diophantine Approximation. Let $R\subset\mathbb{R}^+$ the set of real numbers such that $r,r\sqrt{2},r\sqrt{3}$ are simultaneously close to integer multiples of $\pi$. For any $r\in R$ we have that $\varphi(r)$ is close to zero: since $\varphi$ is bounded and Lipschitz-continuous, it is enough to show that $R$ is syndetic to have that $E$ is finite. If we consider a cube with side length $\varepsilon>0$ centered at $\frac{m}{\pi}\left(1,\sqrt{2},\sqrt{3}\right)\pmod{1}$ we easily get than for some integer $m\leq\frac{1}{\varepsilon^3}$ the numbers $m,m\sqrt{2},m\sqrt{3}$ are simultaneously at most $\pi\varepsilon$-apart from an integer multiple of $\pi$. By choosing $\varepsilon=\frac{1}{3\pi}$ the ridiculous bound $E\leq 6+27\pi^3$ can be easily derived.

The inequality $E\leq 12$ can be deduced from my approach below, however the optimal bound for $E$ seems to be around $4.5$, so there still is some work to be done.

  • 1
    $\begingroup$ Sorry, if this is trivial, but how did you obtain, that the zeros are simple and that the supremum over the differeneces of the zeros is finite? $\endgroup$ – Severin Schraven Aug 5 '17 at 22:21
  • 1
    $\begingroup$ @SeverinSchraven: assuming that $\varphi(t)$ and $\varphi'(t)$ vanish at the same point one gets a contradiction through the Cauchy-Schwarz inequality. About the supremum being finite, it is enough to apply the approach in 1. with horribly crude bounds. $\endgroup$ – Jack D'Aurizio Aug 5 '17 at 22:41
  • $\begingroup$ Thank you, for explaining it :) I like your question very much $\endgroup$ – Severin Schraven Aug 5 '17 at 23:17
  • $\begingroup$ Numerically zeros of $\phi t$ are well distributed on the $x$ axis in the sense that if in the interval $(0,22)$ we find $10$ zeroes then in the interval $(0,220)$ we can expect $100/quad$. The average distance between $\zeta_n$ and $\zeta_{n+1}$ is about $2.23$ and the maximum is about $4.5$ $\endgroup$ – Raffaele Aug 6 '17 at 16:01
  • $\begingroup$ It might be useful to mention an element of $R$. By choosing $n=821$, both $n\sqrt{2}$ and $n\sqrt{3}$ are at most $\frac{1}{14}$-apart from an integer. Netwon's method then gives that a root of $\varphi$ is close to $821\pi$, namely in a small neighbourhood of $2579$. $\endgroup$ – Jack D'Aurizio Aug 6 '17 at 17:10

Here there are some extra thoughts. It is not difficult to show by the sum formulas that the roots of $\sin(t)+\sin(t\sqrt{2})=2\sin\left(t\frac{\sqrt{2}+1}{2}\right)\cos\left(t\frac{\sqrt{2}-1}{2}\right)$ are located at $$ 2\pi(\sqrt{2}-1)\mathbb{Z} \cup \left(\pi(\sqrt{2}+1) + 2\pi(\sqrt{2}+1)\mathbb{Z}\right) $$

hence $\sin(t)+\sin(t\sqrt{2})$ has a sign change on any interval $[a,b]\subset\mathbb{R}^+$ whose length exceeds $2\pi(\sqrt{2}-1)$. By a similar argument, $\sin(t\sqrt{2})+\sin(t\sqrt{3})$ has a sign change on any interval $[a,b]\subset\mathbb{R}^+$ whose length exceeds $2\pi(\sqrt{3}-\sqrt{2})\leq 2$.
Let $g(t)=\sin(t\sqrt{2})+\sin(t\sqrt{3})$ and $h(t)=g(t)g(t+2\pi(\sqrt{3}-\sqrt{2}))$.
$h$ can be factored through the sum formulas and has the following behaviour: enter image description here

Let us consider the set $H=\{t\in\mathbb{R}^+: h(t)\leq -2\}$. Since $|g(t)|\leq 2$, for any $t\in H$ we have that $g(t)$ and $g(t+2\pi(\sqrt{3}-\sqrt{2}))$ have opposite signs and absolute values $\geq 1$. In particular:

For any $t\in H$ there is a root of $\varphi$ in the interval $[t,t+2\pi(\sqrt{3}-\sqrt{2})]$.

and $E$ is bounded by the length of the largest interval over which $h(t)\geq -2$.

This proves $\color{blue}{E\leq 12}$, roughly.

And that can be probably improved up to $E\leq 10$ by directly considering the largest interval over which $g(t)^2\leq 1$, since $h(t)$ is rarely positive.


I have estimated analytically the integral in your first thought

$$g(t)=\frac{\cos (t)+\sqrt{2} \cos \left(\sqrt{2} t\right)+\sqrt{3} \cos \left(\sqrt{3} t\right)}{\sin (t)+\sin \left(\sqrt{2} t\right)+\sin \left(\sqrt{3} t\right)}$$

I considered a very thin isosceles trapezoid having height the interval $[0;\;22]$ and larger basis on the imaginary axis from $-0.01i$ to $0.01i$ and smaller basis from $22-0.05i$ to $22+0.05i$ and I got

$$\dfrac{1}{2\pi i}\oint_{\gamma}\dfrac{g'(t)}{g(t)}\,dt\approx 9.5$$

The actual number of root is $10$

I repeated on $[0,\;220]$ and I got $99.3$ while the actual number of roots is $99$ and on $[0,\;2200]$ it gives $994.2$

  • $\begingroup$ Were you able to produce lower bounds for the number of roots in any (or in any large enough) interval $[a,b]\subset\mathbb{R}^+$? $\endgroup$ – Jack D'Aurizio Aug 6 '17 at 17:15
  • $\begingroup$ I used numerical methods, you know the integral is intractable. $\endgroup$ – Raffaele Aug 6 '17 at 17:55
  • $\begingroup$ Despite that, it is possible to prove that $E\leq 12$, for instance. Please have a look at the approach I have just outlined. $\endgroup$ – Jack D'Aurizio Aug 6 '17 at 18:30

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.