Weyl's Theorem says that if $p(x)$ is a polynomial with at least one of the coefficients (non-constant) is irrational then the sequence $\{p(n)\}$ is equidistributed in $\mathbb T$ (Torus of $1$ dimension)

I understand the case if the leading coefficient is irrational then we can use induction and the fact that if $m^{th}$ difference of a sequence is equidistributed for all $m\in\mathbb N$ then the sequence itself is equidistributed.

The exercise in the notes I'm following says that the general case (described in 1st para) can be easily deduced from this special case. However, I do not see any obvious jumps. I did try proving sum of an equidistributed sequence and a periodic sequence is equidistributed but unfortunately found a counterexample to that. Any clue on what "easy conclusion" I'm missing?


1 Answer 1


It's because each of the cosets of the period is equidistributed. For instance, if $p(n) = \frac{1}{2}n^2+\pi n$, then both $(p(2n))_{n \ge 1}$ and $p((2n+1))_{n \ge 1}$ are equidistributed.

Exercise: If $(x_n)_n = (p_n+y_n)_n$ where $p_n$ is periodic with period $p$ and $(y_{pk+j})_{k \ge 1}$ is uniformly distributed for each $0 \le j \le p-1$, then $(x_n)_n$ is uniformly distributed.


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