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I've been conducting some simulations to explore the distribution properties of sequences in the unit interval under modular arithmetic, specifically modulo $2\pi$. The sequence in question includes the natural numbers and prime numbers. Preliminary results suggest that these sequences might be uniformly distributed modulo $2\pi$, but I'm looking for a more rigorous mathematical foundation or a strategy to prove or disprove this. How can one rigorously prove that a sequence like the natural numbers or the primes is uniformly distributed modulo $2\pi$? What mathematical tools and approaches are typically employed in such investigations? Any references to theorems or prior research that addresses similar questions would be especially helpful.

AgnostMystic
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2 Answers2

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It's more traditional to study equidistribution mod $1$ rather than $2\pi$, but of course they are related: $a_n \mod 1 = (2 \pi a_n \mod 2\pi)/(2\pi)$.See Equidistributed sequence. Equidistribution of the naturals mod $2\pi$, or equivalently the multiples of $1/(2\pi)$ mod $1$, is easy from the fact that $1/(2\pi)$ is irrational. Equidistribution of the primes mod an irrational, or equivalently the prime multiples of an irrational mod $1$, is a theorem of Vinogradov from 1948.

Robert Israel
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The standard method for proving equidistribution is Weyl's criterion which states that a sequence $a_r$ is equiditributed modulo $1$ if and only if for all non-zero integers $k$,

$$ \lim_{n \to \infty}\frac{1}{n}\sum_{r=1}^{n}e^{2\pi i k a_r} = 0 $$

Practical test using Riemann integration: Let your sequence be $a_1,a_2,\ldots,a_n$ is ascending order. Take any function which is Riemann integrable in $(0,1)$. If

$$ \lim_{n \to \infty}\frac{1}{n}\sum_{r=1}^{n}f\left(\frac{a_r}{a_n}\right) = \int_0^1 f(x)dx $$

Then the sequence of ratios $\displaystyle \frac{a_1}{a_n}, \frac{a_2}{a_n},\ldots, \frac{a_n}{a_n}$ are uniformly distributed in $(0,1)$. For simplicity, you can take $f(x) = x$ and check if

$$ \lim_{n \to \infty}\frac{1}{n}\sum_{r=1}^{n}\frac{a_1+a_2+\cdots+a_n}{a_n} = \frac{1}{2}. $$

Nilotpal Sinha
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  • Lots of thanks for the above two helpful answers. – AgnostMystic Apr 15 '24 at 04:46
  • ,what if the sequence is not in ascending order .for example $sin(n).$ is it uniform in [-1,1] – AgnostMystic Apr 15 '24 at 06:30
  • To use the Riemann integral equivalence, notice that $0 \le x \le 1$ and $x$ increases in $(0,1)$. So if your original sequence is not in ascending, then first sort it in ascending order and define a new sequence whose $n$-th term is the $n$-th term of the original sequence. If the original sequence is uniformly distributed, the new sequence will also be uniformly distributed. Again, if your sequence has negative values say the values are in $(-c,c)$ then you can simply create a new sequence $b_i = a_i + c$ and this sequence will be non negative in $(0,2c)$. – Nilotpal Sinha Apr 15 '24 at 09:08
  • @Nilopal Sinha.can I take $a_k=2 \pi p_k,$ where $p_k$ is the kth prime?In that case the last sum should be $\lim_{n \to \infty} \frac1n \sum_{i=1}^{n} \frac{p_k}{p_n}=1/2$.Or ,do you mean by $a_k \mod 1$ instead of $a_k?$ Kindly clarify – AgnostMystic Apr 15 '24 at 09:55
  • Yes, it must be $1/2$. In fact for primes, its is a co-incidence that I had used the uniform distribution of the seaquence $p_k/p_n$ is one of my previous answers. Check this out: https://math.stackexchange.com/questions/2917887/do-arithmetic-mean-and-geometric-mean-of-prime-numbers-converge/2919825#2919825 – Nilotpal Sinha Apr 15 '24 at 09:59