It is an elementary exercise to show that for irrational $\alpha \in \mathbb{R}$, the sequence $\{ \alpha n \mod 1 \}_{n \in \mathbb{N}}$ is dense in $T := \mathbb{R}/\mathbb{Z}$. With more work, it can be shown that the sequence is in fact equidistributed.

This result has two natural generalizations.

The first of them concerns the higher dimensions: suppose that $1,\alpha_1,\alpha_2,\dots,\alpha_d$ are linearly independent over $\mathbb{Q}$ (in other words, no nontrivial combination of $\alpha_i$ with rational coefficients is a rational number). Then the sequence $\{ (\alpha_i n)_{i=1}^d \mod 1 \}_{n \in \mathbb{N}}$ is equidistributed in $T^d$.

The second concerns more general functions; this result is due to Weyl, I think. Suppose that $p(n) = \sum_{k=1}^r a_k n^k$ is a polynomial with $0$ constant term and at least one irrational coefficient. Then, the sequence $\{ p(n) \mod 1 \}_{n \in \mathbb{N}}$ is equidistributed in $T$.

The following common generalisation comes to mind: Suppose that $p_1(n), p_2(n), \dots, p_d(n)$ are polynomials with $p_i(0) = 0$ such that any linear combination of $p_i$ with rational coefficients is a polynomial with at least one irrational coefficients not counting the constant term. (Equivalently, the polynomials $p_1(n), p_2(n), \dots, p_d(n), n,n^2,\dots$ are linearly independent over $\mathbb{Q}$). Then, the sequence $\{ (p_i(n))_{i=1}^d \mod 1 \}_{n \in \mathbb{N}}$ is equidistributed in $T^d$.

Question: Is this more general result true? Also, is it known?

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    $\begingroup$ You are familiar with the Weyl criterion? The multi-dimensional version? If you apply that, and the known results that you cite, do you get the result you want? $\endgroup$ Commented May 9, 2013 at 12:03
  • $\begingroup$ I used to be familiar; now I would call it more of a nodding acquaintance. I will refresh my memory and see if the answer follows. $\endgroup$ Commented May 9, 2013 at 12:06
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    $\begingroup$ small comment : the polynomial you suggest in weyl's result , i think result should be atleast one $a_n$ is irrational for n > 0 $\endgroup$
    – rohit
    Commented May 9, 2013 at 12:28
  • $\begingroup$ @rohit: Thank you, I forgot that the constant term doesn't count. I just corrected it. $\endgroup$ Commented May 9, 2013 at 12:30

1 Answer 1


Many thanks to Gerry Myerson for pointing me in the right direction. I am citing some results from lecture notes by Terence Tao. I would still appreciate a check from the community.

Weyl's Equidistribution Theorem says the following: Let $x_n \in T^d$ be a sequence on the $d$-dimensional torus. Then, the following conditions are equivalent:

($1.$) $\qquad$ $\{x_n\}_n$ is equidistributed. ($2.$) $\qquad$ for any $k \in \mathbb{Z}^d$, $k \neq 0$, it holds that $$\lim_{N\to\infty} \frac{1}{N} \sum_{n=1}^N e^{2\pi i k \cdot x_n} = 0$$ where $k \cdot x_n$ is the standard scalar product: $k_1 x_{n,1} + \dots k_d x_{n,d}$.

It is a corollary of the above result that a sequence $\{x_n\}_n \subset T^d$ is equidistributed if and only if:

($\ast$) $\qquad$ $\{x_n \cdot k\}_n \subset T$ is equidistributed for any $k \in \mathbb{Z}^d,\ k\neq 0$.

Indeed, applying Weyls criterion in one dimension for a fixed $k$, the condition ($\ast$) is equivalent to the condition:

($2.'$) for any $k \in \mathbb{Z}^d$, $k \neq 0$, for any $l \in \mathbb{Z}$, $k \neq 0$ $$\lim_{N\to\infty} \frac{1}{N} \sum_{n=1}^N e^{2\pi i l (k \cdot x_n)} = 0$$

Now, the conditions ($2.$) and ($2.'$) are clearly equivalent, so $(1) \equiv (2) \equiv (2') \equiv (\ast)$.

The result asked for in the question is the immediate application of the above corollary.

Indeed, the required linear independence of $p_i$ says in particular that for $k \in \mathbb{Z}^d$, $k \neq 0$, the polynomial $q(n) = \sum_{i=1}^d k_i p_i(n)$ has an irrational coefficient. The result on equidistribution for single polynomials now says that $q(n) \mod{1}$ is equidistributed. Because $k$ was arbitrary, it follows that $(p_i(n))_{i=1}^d$ is equidistributed by the virtue of the Corollary.


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