# Is there a definition of a “pseudo period” for $f(x)=\sin(3x)+\sin(\pi x)$?

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 $$f(x) = \sin(3x) + \sin(\pi x)$$ 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.

• How about: a function is pseudoperiodic if it is the sum of finitely many periodic functions? – Mees de Vries Jan 11 '17 at 16:09
• 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 $b$. – John Wayland Bales Jan 11 '17 at 16:39
• However, I see a 'period' of $\frac{1}{4}\cdot\dfrac{2\pi}{\frac{\pi}{3}-1}$ for $\sin\left(\dfrac{\pi}{3} x\right)+\sin(x)$. – John Wayland Bales Jan 11 '17 at 16:51
• The point is that one can come up with a local approximation of a "period" but this approximation will vanish on bigger scales. – marco trevi Jan 11 '17 at 16:59
• 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? – daniel Jan 13 '17 at 10:16

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 $$|f(x\pm kT)-f(x)| for every $$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 $$|f(x±kT)-f(x)| for every $$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 $$|f(x±kT)-f(x)| for every $$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.