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Sums of trigonometric functions may or may not be periodic functions; in particular, $\sin(ax)+\sin(bx)$ is periodic if $a/b$ is rational.

If we consider the function \begin{equation} f(x) = \sin(3x) + \sin(\pi x) \end{equation} it surely looks periodic, even if it's not; to me it feels like the period itself is somewhat periodic (or is the result of a kind of "period cascade").

My question is: is there some way to capture this "quasiperiodic" nature of this kind of functions, i.e. does there exist a measure of how "repetitive" a function is even if it is not a periodic function?

To narrow the scope of the question, I'm trying at the moment to figure out the case of the above sum of sines.

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    $\begingroup$ How about: a function is pseudoperiodic if it is the sum of finitely many periodic functions? $\endgroup$ – Mees de Vries Jan 11 '17 at 16:09
  • $\begingroup$ Just an observation in the case of your specific example. The longer of the two periods, between the places where the sum almost vanishes, is close to the value of $\dfrac{2\pi}{\pi-3}$. I tried one other, $\sin(2x)+\sin\left(\frac{\pi}{2}\right)$ and the longer period is close to $\dfrac{2\pi}{2-\frac{\pi}{2}}$. Perhaps it would be worthwhile to check out this relationship $\dfrac{2\pi}{\vert b-a\vert}$ with other values of $a$ and $b$. $\endgroup$ – John Wayland Bales Jan 11 '17 at 16:39
  • $\begingroup$ However, I see a 'period' of $\frac{1}{4}\cdot\dfrac{2\pi}{\frac{\pi}{3}-1}$ for $\sin\left(\dfrac{\pi}{3} x\right)+\sin(x)$. $\endgroup$ – John Wayland Bales Jan 11 '17 at 16:51
  • $\begingroup$ The point is that one can come up with a local approximation of a "period" but this approximation will vanish on bigger scales. $\endgroup$ – marco trevi Jan 11 '17 at 16:59
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    $\begingroup$ In the Bohr-Weyl development of almost-periodic functions, the criterion is $|f(t+T)-f(t)|< \epsilon.$ I wonder if this is related to your question? $\endgroup$ – daniel Jan 13 '17 at 10:16
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I will try to propose some definitions, with my main intention being not necessarily to provide a complete answer, but to put some options on the table.

$\varepsilon-$periodicity

As mentioned by @daniel, according to Bohr-Weyl, we may define:

A function $f$ is said to be $\varepsilon-$periodic for some constant number $\varepsilon>0$ if there exists some $T>0$ such that, for every $x\in\mathbb{R}$: $$|f(x+T)-f(x)|<\varepsilon.$$

Under this definition, $f(x)=\sin(3x)+\sin(\pi x)$ is clearly $5-$periodic with any period, since: $$|f(x+T)-f(x)|\leq|f(x+T)|+|f(x)|\leq2+2<5.$$

What would be interesting is to define some notion of principal precision for $\varepsilon-$periodicity, for instance, such as:

Given an $\varepsilon-$periodic function $f$ we define as principal precision of $f$ with respect to $T>0$ the non-negative number: $$\varepsilon_0:=\inf\{\varepsilon>0\mid|f(x+T)-f(x)|<\varepsilon,x\in\mathbb{R}\}.$$

Note that $\varepsilon_0$ depends, of course, on $f$, but also on $T$. In our case, note that: $$\begin{align} |f(x+T)-f(x)|&=|\sin(3x+3T)+\sin(πx+πT)-\sin(3x)-\sin(\pi x)|=\\ &=2\left|\sin\frac{3T}{2}\cos\frac{6x+3T}{2}+\sin\frac{πT}{2}\cos\frac{2πx+πT}{2}\right|\leq\\ &\leq2\left|\sin\frac{3T}{2}\right|+2\left|\sin\frac{\pi T}{2}\right|. \end{align}$$ Seeing this, we may change a little bit our $\varepsilon_0$ definition and drop dependence on $T$ in the following way:

Given an $\varepsilon-$periodic function $f$ we define as global principal precision of $f$ the non-negative number: $$\delta_0:=\inf_{T>0}\inf\{\varepsilon>0\mid|f(x+T)-f(x)|<\varepsilon,x\in\mathbb{R}\}=\inf_{T>0}\varepsilon_0(T).$$

Under this definition, it would be useful to find a minimizer of $$g(T)=\left|\sin\frac{3T}{2}\right|+\left|\sin\frac{\pi T}{2}\right|.$$ Plotting the graph of $g$, I suspect that $\delta_0=0$. After taking a closer look to the graph of $g$, one observes that, if $T=2k$, where $k=1,2,\ldots$, we get: $$g(2k)=|\sin(3k)|+|\sin(k\pi)|=|\sin(3k)|.$$ Now, since $|sin(3k)|$ is a dense subset of $[0,1]$, we can find $k_n$ increasing w.r.t. $n$ such that $$|\sin k_n|\to0.$$ Hence, $\delta_0(f)=0$. In our case, this implies that choosing arbitrarily large periods $T_n=2k_n$ (as above), makes our function even more "periodic" - that is, its principal precision, as defined above, vanishes when $n\to\infty$.

Lastly, observe that if we define as $P_ε$ the set of all $ε-$periodic functions, then we have the following properties for two functions $f,g$ with $f∈ P_ε$ and $g\in P_δ$ and $\lambda\in\mathbb{R}$:

  1. $f+g\in P_{ε+δ}$.
  2. $0f∈ P_ε$ for every $ε>0$.
  3. $\lambda f∈ P_{\varepsilon/|\lambda|}$, for $λ\neq0$.

Thus, the set: $$\mathcal{P}:=\bigcup_{ε>0}P_ε,$$ is a vector space. Noteably, $\mathcal{P}$ is very "large", in the sense that any bounded function belongs to it.

$(a_k)-$periodicity

A drawback of the above definition is that it does not give us control of the function's behaviour for large values of $|x|$. So, we may seek an alternative definition for our purposes. For instance, consider the following definition:

Attempt 1

A function $f$ is said to be $(a_k)-$periodic, w.r.t. to some positive Cesaro summable sequence $(a_k)$ if $$|f(x\pm kT)-f(x)|<a_k,\ x\in[0,T),$$ for every $k=1,2,\ldots$, where: $$T=\lim_{n\to\infty}\frac{a_1+a_2+\ldots+a_n}{n}>0.$$ We will call $T$ the average period of $f$.

Note that we demand the Cesaro sum to be positive as well as some kind of symmetry for such a function - well, periodicity implies such a symmetry as well.

At first, note that any periodic function is also trivially $(a_k)-$periodic, with $a_k=T$. However, a problem occurs at this point. If we let $T>1$, $a_k=T+(-1)^k$, ad also demand that $|f(x)|<\frac{T-1}{2}$ - for instance, $f(x)=\frac{1}{100}\sin x$ -,then we have: $$\frac{a_1+\ldots+a_n}{n}=T+\frac{\delta_n}{n}\to T+0=T,$$ where $\delta_n=-\frac{1}{2}+\frac{(-1)^n}{2}$. But, this behaviour of the definition we gave on trivial cases - i.e. normally periodic functions - seems quite absurd, so, we would like to avoid it.

An idea would be to enforce some additional normality in $a_k$, so we may try the following:

Attempt 2

A function $f$ is said to be $(a_k)-$periodic, w.r.t. to some positive and monotonous Cesaro summable sequence $(a_k)$ if $$|f(x±kT)-f(x)|<a_k,\ x\in[0,T),$$ for every $k=1,2,\ldots$, where: $$T=\lim_{n\to\infty}\frac{a_1+a_2+\ldots+a_n}{n}>0.$$ We will call $T$ the average period of $f$.

The additional feature we have added is that $a_k$ is monotonous, so as to avoid cases such as such the previous one.

At this time, it is useful to see what our definition is intuitively representing. According to it, we measure how much a function differs from its restriction on the principal interval $[0,T]$.

Now, observe that the definition of $(a_k)-$periodicity implies that $a_k\to a$, for some $a\in[0,+\infty]$. Let: $$s_n:=\sum_{k=1}^na_k.$$

Cesaro-Stolz Lemma gives us, for any monotonous and unbounded sequence $b_n$: $$\liminf\frac{a_{n+1}-a_n}{b_{n+1}-b_n}\leq\liminf\frac{a_n}{b_n}\leq\limsup\frac{a_n}{b_n}\leq\limsup\frac{a_{n+1}-a_n}{b_{n+1}-b_n}.$$ Let $a_n=s_n$ and $b_n$, which yields: $$\liminf{a_n}\leq\liminf\frac{s_n}{n}\leq\limsup\frac{s_n}{n}\leq\limsup{a_n},$$ or, equivalently: $$\liminf{a_n}\leq T\leq\limsup{a_n}.$$ Since $a_n$ is convergent, we have: $$\liminf{a_n}=\limsup{a_n},$$ so: $$\liminf{a_n}=\limsup{a_n}=T\Leftrightarrow\lim_{n\to\infty}a_n=T.$$

At this point, since we have proved that, under the above assumptions, $a_k$ is convergent and since convergence of $a_k$ implies Cesaro summability, we may refine our definition as follows:

Attempt 3

A function $f$ is said to be $(a_k)-$periodic, w.r.t. to some positive and monotonically convergent sequence $(a_k)$ if $$|f(x±kT)-f(x)|<a_k,\ x\in[0,T),$$ for every $k=1,2,\ldots$, where: $$T=\lim_{n\to\infty}a_n>0.$$ We will call $T$ the asymptotic period of $f$.

I will keep this post constantly updated for the next days.

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