How to find the almost period of an exponential polynomial - MathOverflow

How to find the almost period of an exponential polynomial

2010-09-10T13:00:18Z

Let $u(t) = \Sigma_{k=1}^n c_k e^{i \lambda_k t} (c_k \in \mathbb C, \lambda_k \in \mathbb R) $ be an exponential polynomial of order $n$ with purely imaginary exponents. We can assume that the exponents are $\delta$-separated.

Define $E_n$ to be the collection of all exponential polynomial of order $n$. i.e.,

$$ E_n:= \{ u : u(t) = \sum_{k=1}^n c_k e^{i \lambda_k t}, c_k \in \mathbb C, \lambda_k \in \mathbb R, and |\lambda_i - \lambda_j| > \delta\ when\ i \neq j \}. $$

Of course $u$ is an almost periodic function i.e., given $\epsilon >0$ there exists $T_\epsilon$ such that every interval of length $T_\epsilon$ contains an almost period of $u$. i.e., $\forall x \in \mathbb R,$ $\exists \tau \in (x,x+T_\epsilon)$ such that

$$ \sup_{t\in \mathbb R} |u(t)-u(t+\tau)| < \epsilon$$

Is it possible to find a bound on the $T_\epsilon$ for such an exponential polynomial, I added the separation condition hoping that it would lead to an affirmative answer.

I have this suspicion that the almost period may have to do something with the number theoretical properties of the set of exponents. After all if all the exponents are in a lattice then the function is periodic.

and may I even dare hope to find a bound which will work for the entire class? Some nice subclass maybe ?

If that is too much to ask then what is a good question to ask ?

Has some one studied related questions or variations of it ? I would be glad to know.

Answer by fedja for How to find the almost period of an exponential polynomial

2010-09-10T13:25:48Z

Unfortunately, not. Take $\lambda_1=\delta$, $\lambda_2=2\delta+\gamma$ with very small $\gamma$ that is not a rational multiple of $\delta$. Then we can choose $t$ such that $e^{\lambda_1 t}\approx 1$ and $e^{\lambda_2t}\approx -1$. Now, there will be no $\varepsilon$-almost period in $[t,t+0.1\gamma^{-1}]$ (for a polynomial, an almost-period should be an almost-period of each exponent).

Answer by alext87 for How to find the almost period of an exponential polynomial

2010-09-10T13:27:44Z

Here is a first attempt:

For each $\lambda_k$ find a rational number $\frac{p_k}{q_k}$ such that $\lambda_k\approx\frac{p_k}{q_k}$.

Then an approximation pestimistic bound for $T_{\epsilon}$ would be $(q_1\ldots q_n)2\pi$ since $\frac{p_k}{q_k}(q_1\ldots q_n)2\pi$ is a multiple of $2\pi$ for every $k$.

Answer by Nick S for How to find the almost period of an exponential polynomial

2011-03-26T01:30:53Z

You are basically interested in what is called $\epsilon$-dual Characters.

For a set $\Lambda \subset \R^d$ we define

$$\Lambda^\epsilon := { t \in \R^d | \left| e^{2 \pi x \cdot t} -1 \right| < \epsilon \, \forall x \in \Lambda }\,.$$

In your case $\Lambda := { \lambda_1, .., \lambda_n }$ and for all $t \in \Lambda^\epsilon$ you get

$$\| T_tu - u\|_{\infty} < (|c_1|+...+|c_n|) \epsilon \,.$$

Here is a short review of what is known about $\Lambda^\epsilon$:

1) If $\Lambda \subset \R^d$ is finite, then $\Lambda^\epsilon$ is relatively dense for all $0< \epsilon <2$. , that is $\Lambda^\epsilon +K_\epsilon = \R^d$ for some compact $K_\epsilon$. How to finding this $K$ is exactly waht you need. I will address this question below (*).

2) If $\Lambda \subset \R^d$ is relatively dense and $\Lambda-\Lambda := { x-y | x,y \in \R^d }$ is uniformly discrete (i.e. $|z_1-z_2| > \delta$, for all $z_1, z_2 \in \Lambda- \Lambda$) then $\Lambda^\epsilon$ is relatively dense for all $0< \epsilon <2$. Such a set is called a {\bf Meyer set}.

(*) How do we actually find $K$ so that $\Lambda^\epsilon +K= \R^d$? This is covered by Meyer in the book mentioned below, when he studies the connection between $\Lambda^\epsilon $ relatively dense and $\Lambda$ being harmnious.

I will explain directly what happens in your case, since in general things are a little more complicated.

First lets observe that if $t= \frac{2pi n}{ \lambda}\, n \in \Z$ then $e^{2 \pi \lambda \cdot t} =1 $, which is the reson the problem is easy for latices.

If we only have 2 $\lambda$'s, here is what we need to do: First if $\lambda_1/\lambda_2$ is rational, then the intersection of the latices $ \frac{2pi}{ \lambda_1}\Z \cap \frac{2pi}{ \lambda_2}\Z$ is non-empty and any $t$ in here will do.

If $\lambda_1/\lambda_2$ is irrational, then the latices $ \frac{2pi}{ \lambda_1}\Z$ and $\frac{2pi}{ \lambda_2}\Z$ come arbitrarily close within a fixed gap, and that will do.

More exactly, fix a $\alpha$ so that if $|x-y| < \alpha $ then $ \left| e^{2 \pi x \cdot \lambda_i} -e^{2 \pi y \cdot \lambda_i} \right| < \epsilon$, and then the dirichclet theorem tells you that there exists number $N$ so that you can find integers $m,n$, with $m$ in any interval of length $N$ so that:

$$ | m\frac{2pi}{ \lambda_1} - n\frac{2pi}{ \lambda_2} | < \alpha \,.$$

Then $m\frac{2pi}{ \lambda_1}$.

If you have more than two $\lambda$'s, exactly as above all you have to do is prove that there exists a number $N$ so that within any interval of lenght $N$ there exists a $t$ so that for all $i$ we have:

$$ | t- n_i\frac{2pi}{ \lambda_i} | < \alpha \,.$$

This is what Meyer calls a Harmnious set, and a stronger version of Dirischlet Theorem proves it.

I recommend to you the following book(s):

-Y. Meyer, Algebraic numbers and Harmonic Analysis.

-R. V. Moody - Meyer Sets and their duals.