Why is integer approximation of a function interesting?

Solution 1:

Because after a long time the question received a number of upvotes, but no answer was given, I will try to write up whatever justification I have managed to come up with until now.

$\newcommand{\bbR}{\mathbb{R}}$ $\newcommand{\bbZ}{\mathbb{Z}}$

Contrast with equidistribution. It is a classical result that if $g(x) \in \bbR[x]$ is a polynomial with at least one irrational coefficient, then the sequence $\left( g(n) \bmod{1} \right)_{n \in \bbZ}$ is equdistributed. This means that if you consider any interval $I \subset [0,1)$, and look at the set $A_I := \{ n \in \bbZ \ : \ g(n) \bmod{1} \in I \}$, then $A_I$ is a set of integers with density: $$ \operatorname{dens} A_I = \lim_{N-M \to \infty} \frac{A_I \cap [M,N)}{N-M} = |I| $$ The limit is to be understood in the sense that we have convergence $\lim_{i \to \infty} \frac{A_I \cap [M_i,N_i)}{N_i-M_i}$ whenever $\lim_{i \to \infty}(N_i - M_i) = +\infty$. It is not clear a priori that this limit should exist, but the theorem say that it indeed exists, and is equal to the length of $I$, which is what one would expect for a random sequence. This particular result is due to Weyl, and can be found on Wikipedia, along with other results in the similar spirit. It would seem that there has been significant interest in these problems, for instance Vinogradov theorem falls under this heading.

On the other hand, the theorem I mentioned says basically that in the same setup, if $I = (-\varepsilon,\varepsilon)$, then $A_I$ is $IP^*$. (that's a bit of cheating, since I said that $I \subset [0,1)$, but since things are happening modulo $1$, I hope it is clear what is meant). A "random" set is not $IP^*$, and a set has to be comparatively structured or large to be such. For instance, an $IP^*$ has be an $IP$ set, meaning that it contains a pattern $FS(a) := \{ \sum_{i \in I} a_i \ : \ I \subset \mathbb{N},\ \text{finite} \}$ for some increasing sequence $a_i$. Generically speaking, a translate of an $IP^*$ set is not $IP^*$. So, it turns out that $A_I$ is combinatorially rich (in agreement with Weyl's equidistribution), and is even richer than first intuition might suggest (in contrast with the equidistribution).