One says that a sequence $(s_n)_{n \in \mathbb{N}}$ is equidistributed modulo 1 if for every $0 \leq a < b < 1$, one has that $$\dfrac{1}{N}\#\{ 1 \leq n \leq N \ : \ a \leq \{s_n\} \leq b\} \to b-a $$ as $N \to \infty$. Here $\{x\} := x - \lfloor x \rfloor$ denotes the fractional part of $x \in \mathbb{R}$.

Weyl a long time ago came up with a popular criterion, which states that $(s_n)$ is equidistributed mod 1 iff for every fixed non-zero integer $h$, $$ \dfrac{1}{N}\sum_{n=1}^N e(hs_n) \to 0 $$ as $N \to \infty$. Here $e(x) := e^{2\pi i x}$. Using this he for instance proved that the sequence $(\alpha n)_{n \in \mathbb{N}}$ is equidistributed mod 1 for any fixed irrational $\alpha$.

I was wondering if it is known that, given a fixed integer $k \geq 2$ and an irrational number $\alpha$, the sequence $(\alpha k^n)_{n \in \mathbb{N}}$ is equidistributed mod 1? Thanks.


1 Answer 1


From the Wikipedia page on normal numbers we learn that the sequence $\{\alpha k^n\}$ is equidistributed modulo 1 if and only if $\alpha$ is a normal number to base $k$. In general, it is very difficult to show that a given real number is normal, and essentially the only explicit examples of normal real numbers are artificial constructions like Champernowne's constant.

If you are interested in exponential sums $$\sum_{n \leq N} e(\alpha k^n)$$ where $\alpha$ is a rational number, there are techniques for estimating such sums. This recent preprint might be a good place to start.

  • $\begingroup$ Thanks a lot. I wasn't aware of normal numbers and how little is known about them. Interesting. Also a little intriguing given that $k^x$ is a smooth function of $x$... I imagine, in the Euler-Maclaurin formula, that the integral term which has the derivative of $\alpha k^x$ as part of the integrand, will yield a large error with the typical methods non? $\endgroup$
    – user152169
    Commented Jan 10, 2017 at 3:18
  • 1
    $\begingroup$ Yes. Euler-Maclaurin summation is really only useful for functions with small derivatives. $\endgroup$
    – user397701
    Commented Jan 10, 2017 at 3:45

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .