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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.

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    $\begingroup$ 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
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    $\begingroup$ @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
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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.

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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 $994.2$

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  • $\begingroup$ 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

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