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# Asymptotic and partial equidistribution modulo one

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1. Introduction

A sequence of real numbers $a_n$ is equidistributed on a given interval if the probability of finding $a_n$ in any subinterval is proportional to the subinterval length. In particular, if $\{a_n\}$ denotes the fractional part of $a_n$ by then a sequence $a_n$ is said to be equidistributed modulo 1 or uniformly distributed modulo 1 if , for all $0 \le b < c \le 1$,

$\displaystyle{\lim_{N \rightarrow \infty}\frac{\#\{r \le N : b < \{a_n\} \le c\}}{N} = c - b.}$

Often problems related to equidistribution modulo 1 can be established using a powerful theorem called Weyl’s criterion which gives the necessary and sufficient conditions for a sequence to be equidistributed mod 1. Weyl’s criterion can be stated in the following equivalent forms.

Theorem 1. (Weyl’s Criterion) Let $a_n$ be a sequence of real numbers. The following are equivalent:

1. The sequence $a_n$ is equidistributed mod 1.
2. For each nonzero integer $k$, we have $\displaystyle{\lim_{n \rightarrow \infty}\frac{1}{n}\sum_{r
3. For each Riemann-integrable $f:[0,1] \rightarrow \mathbb{C}$ we have

$\displaystyle{\lim_{n \rightarrow \infty}\frac{1}{n}\sum_{r \le n}f(\{a_r\}) = \int_{0}^{1} f(x)dx}.$

2. Asymptotic equidistribution mod 1

Lemma 2.1. If $u_n$ is a bounded sequence of real numbers such that for every fixed $\epsilon$, $0 < \epsilon \le 1$$\lim_{n \rightarrow \infty}u_{[n\epsilon]} = \epsilon$ then $u_n$ is equidistributed mod 1.

Example. Taking $u_r = r/n$ in Lemma 2.1, we have

$\displaystyle{\lim_{n \rightarrow \infty}\frac{1}{n}\sum_{r \le n}f\Big(\frac{r}{n}\Big) = \int_{0}^{1} f(x)dx}.$

Recall that the above example, also known as the rectangle formula, is a standard method in elementary calculus for evaluating the limit of a sum using definite integral. Since the rectangle formula holds for all functions $f$ Riemann integrable in $(0,1)$ it implies that the sequence $r/n$ approaches equidistribution mod 1 as $n \rightarrow \infty$. This leads us to the concept of asymptotic equidistribution mod 1 which is defined as follows.

Definition 2.2. A sequence of positive real numbers $s_n$ is said to be asymptotically equidistributed modulo one if $s_n$ is increasing and the sequence of ratios

$\displaystyle{\frac{s_1}{s_n},\frac{s_2}{s_n},...,\frac{s_n}{s_n}}$

approach uniform distribution modulo one as $n \rightarrow \infty$.

Theorem 2.3. The sequence $s_n = a n(\ln (b n))^{c}$, where $a, b$ and $c$ are positive constants, is asymptotically equidistributed mod 1 and therefore by Weyl’s Criterion we have

$\displaystyle{\lim_{n \rightarrow \infty}\frac{1}{n}\sum_{r \le n}f\Big(\frac{s_r}{s_n}\Big) = \int_{0}^{1} f(x)dx}.$

Thus the sequence of natural numbers, the sequence of primes $p_n$, the sequence of composite numbers $c_n$ and the sequence of the imaginary parts of the non-trivial zeroes of the Riemann zeta function $\gamma _n$ are asymptotically equidistributed mod 1.

Example.

$\displaystyle{\lim_{n \rightarrow \infty} \frac{1}{n} \sum_{r \le n}\frac{p_n ^{2}}{p_n ^{2}+ p_r ^{2}}=\lim_{n \rightarrow \infty} \frac{1}{n} \sum_{r \le n}\frac{\gamma _n ^{2}}{\gamma _n ^{2}+ \gamma _r ^{2}} = \frac{\pi}{4}.}$

Trivially if $a_n$ equidistributed modulo 1 then $a_n$ is also asymptotically equidistributed modulo one. The converse of this however is highly non trivial and to appreciate this fact we shall look at three celebrated results of analytic number theory in the field of equidistribution mod 1.

Let $\alpha$ be any irrational number and $\beta$ be any real number. We have

1. H. Wely: The sequence $\alpha n$ is equidistributed mod 1.
2. I. Vinogradov: The sequence $\alpha p_n$ is equidistributed mod 1
3. E. Hlawka: The sequence $\beta \gamma_n$ is equidistributed mod 1.

This three results would follow as special cases if the following conjecture on asymptotic equidistribution were true.

Conjecture 2.4. If $a_n$ is also asymptotically equidistributed modulo 1 and $\alpha$ is any irrational number then the sequence $\alpha a_n$ is equidistributed modulo 1.

A proof or disproof of this conjecture has eluded me so far.

3. Partial equidistribution modulo 1

Consider any the set $S_0 = {a_1, a_2, ... , a_n}$ where $a_n$ is any sequence equidistributed modulo 1. Now from the set $S_0$ let us remove all elements  that satisfy the condition $0.2 \le a_i < 0.3$ to obtain a new set $S_1$. Clearly no element of $S_1$ will have the digit 2 in the first place after the decimal point and therefore the set $S_1$ no longer equidistributed modulo 1 because no element of $S_0$ will lie in the interval $(0.2,0.3[$. In general we can remove from the set $S_0$ all the elements which begins with a given sequence of digits after the decimal point. The remaining elements of $S_1$ will form a partially equidistributed sequence.We call the set $S_1$ to be partially equidistributed modulo 1 because as we shall see below that the set $S_1$ shows properties analogous to the set $S_0$ which is equidistributed modulo 1.

If numbers are represented in decimal nation, the first digit after the decimal point can have 10 different values, the first two digits after the decimal point can have 100 different values and in general, if we consider the first $m$ digits after the decimal point, we can have $10^m$ different possibilities.  From these $10^m$ different possibilities let us form a sequence $b_n$ partially equidistributed modulo 1 such that the first $m$ digits of $b_r$ after the decimal point can take $k$ different values from the set $D=\{d_1, d_2, ... , d_k\}$.

Theorem 3.1. If $b_r$ is partially equidistributed modulo 1 and $D=\{d_1, d_2, ... , d_k\}$ is the set of allowed values of the first $m$ digits after the decimal point then

$\displaystyle{\lim_{n \rightarrow \infty}\frac{1}{n}\sum_{r=1}^{n}f(b_r) = \frac{10^m}{k} \sum_{d \in D}\int_{\frac{d}{10^m}}^{\frac{d+1}{10^m}}f(x)dx}.$

Theorem 3.1 is a generalization of (3) of Theorem 1. If $k=10^m$ then set $D$ will contain all possible combinations of $m$ digits. Hence $b_r$ will be equidistributed modulo 1 and RHS of Theorem 3.1 reduces to

$\displaystyle{\int_{0}^{1} f(x)dx}.$

To be contd.