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We know that the two functions $\{\;\cos (ax),\;2\cos (b x)\;\}$ where $\frac ab \notin \mathbb{Q}$ are independently positive (and negative) over $\frac 12$ of the domain.

Is it possible to estimate the fraction of the domain in which $\;\cos (ax)+2\cos (b x) \;$ is positive (in this case, the function is not periodic)?

In other words, can we estimate $$ \lim_{m\to\infty} \frac 1m \int_{0}^{m} \Big(\cos (ax)+2\cos (b x) >0\Big) \,dx?$$

Any hints and comments are greatly appreciated.

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    $\begingroup$ I found "the ratio of the domain" confusing; I think you meant something like "the fraction of the domain". I have edited accordingly, hopefully without changing the meaning. $\endgroup$ – LSpice May 12 at 1:28
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    $\begingroup$ @LSpice Yes, that is true, thanks. $\endgroup$ – user215601 May 12 at 11:03
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    $\begingroup$ Interesting question. If the amplitudes were equal, i.e. the function was $\cos(ax)+\cos(bx)$, it could be rewritten as a product $2\cos(Ax)\cos(Bx)$ with $A=(a+b)/2$, $B=(a-b)/2$, and one would only have to worry about the signs of the two cosines. But with the different amplitudes it "beats" me. (Pun intended, en.wikipedia.org/wiki/Beat_(acoustics) ) $\endgroup$ – Jukka Kohonen May 13 at 16:36
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    $\begingroup$ I guess the answer is of the form “mumble ergodic mumble fraction of $(u,v) \in (\mathbb{R}/2\pi\mathbb{Z})^2$ such that $\cos(u) + 2\cos(v) > 0$, which is equal to mumble”, but I don't know the magic words with which to replace the first two “mumble” and I don't have the patience to compute the value of the last “mumble”. 😔 $\endgroup$ – Gro-Tsen May 13 at 18:04
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    $\begingroup$ PS: The mean motion theorem and the references given in this question is relevant to a closely related question (computing the number of zeros of this function in a given interval). $\endgroup$ – Gro-Tsen May 13 at 18:07
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Since $a,b$ are incommensurable, $(ax,bx)$ is asymptotically equidistributed in the torus $({\bf R} / 2\pi{\bf Z})^2$. [One proof is via a continuous version of Weyl's equidistribution criterion: for any integers $r,s$ with $(r,s) \neq (0,0)$ we have $$ \frac1m \int_0^m e^{i(rax+sbx)}\, dx = O_{r,s}(1/m) \to 0 $$ as $m \to \infty$.] Therefore $\{x > 0 \mid \cos ax + 2 \cos bx > 0 \}$ has the same density in the positive reals as $\{ (\theta,\phi) \in ({\bf R} / 2\pi{\bf Z})^2 \mid \cos \theta + 2 \cos \phi > 0\}$, which is $1/2$ by symmetry.

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Let $r$ be an irrational real number. For real $x>0$, let $U_x$ be a random variable (r.v.) uniformly distributed on the interval $[0,x]$, and then let $$C_x:=\cos rU_x+2\cos U_x.$$ Then the problem can be restated as follows: Is it true that \begin{equation*} P(C_x>0)\to1/2\,\text{?} \tag{1} \end{equation*} Everywhere here, the limits are taken for $x\to\infty$.

The answer to this question is yes. Indeed, for each $(k,n)\in\{0,1,\dots\}^2$, \begin{align*} &E\cos^k rU_x\,\cos^n U_x \\ &=2^{-k-n}\,\frac1x\,\int_0^x du\,(e^{iru}+e^{-iru})^k \,(e^{iu}+e^{-iu})^n \\ &=2^{-k-n}\sum_{p=0}^k\sum_{q=0}^n\binom kp\binom nq \frac1x\,\int_0^x du\, \exp\{iu[(2p-k)r+2q-n]\} \\ &\to2^{-k-n}\sum_{p=0}^k\sum_{q=0}^n\binom kp\binom nq 1(2p=k,2q=n), \end{align*} since $r$ is irrational. So, \begin{align*} &E\cos^k rU_x\,\cos^n U_x\to m_k m_n, \end{align*} where \begin{equation*} m_k:=1(k\text{ is even})2^{-k}\binom k{k/2}=E\cos^k U_{2\pi}. \end{equation*} So, by dominated convergence, for the joint characteristic function (c.f.) $f_{r,x}$ of the pair $(\cos rU_x,\cos U_x)$ of r.v's and all real $s,t$ we have \begin{align*} f_{r,x}(s,t)&=E\exp\{i(s\cos rU_x+t\cos U_x)\} \\ &=\sum_{n=0}^\infty \frac{i^n}{n!}\,E(s\cos rU_x+t\cos U_x)^n \\ &=\sum_{n=0}^\infty \frac{i^n}{n!}\, \sum_{k=0}^n\binom nk s^k t^{n-k}E\cos^k rU_x\,\cos^{n-k} U_x \\ &\to\sum_{n=0}^\infty \frac{i^n}{n!}\, \sum_{k=0}^n\binom nk s^k t^{n-k}m_k m_{n-k} \\ &=h(s)h(t), \end{align*} where \begin{equation*} h(s):=\sum_{k=0}^\infty \frac{i^k s^k m_k}{k!}=E\exp\{is\cos U_{2\pi}\}, \end{equation*} so that $h$ is the c.f. of the (symmetric absolutely continuous) r.v. $\cos U_{2\pi}$.

So, the pair $(\cos rU_x,\cos U_x)$ of r.v's converges in distribution to a pair $(A,B)$ of independent copies of the r.v. $\cos U_{2\pi}$. So, for any real $b$, the r.v. $\cos rU_x+b\cos U_x$ converges in distribution to the symmetric absolutely continuous r.v. $A+bB$.

Thus, (1) follows.

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  • $\begingroup$ Thank you very much for the answer. $\endgroup$ – user215601 May 13 at 19:31

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