MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

What can be said about the limiting distribution of the sequence of fractional parts of $\{n^{a},n>0\}$ for $a\in(1,2)$. I ran a computer experiment for $n\sqrt{n}$ and it looks like uniformly distributed. Is there a simple proof?

share|cite|improve this question
Can you explain what is "discrepancy theory"? I hope that "number theory" will be more appropriate. – Wadim Zudilin Aug 2 '10 at 3:12
@Wadim, I've taken the liberty of adding the Number Theory tag. – Gerry Myerson Aug 2 '10 at 4:08
@Wadim: I added the "discrepancy theory" tag (see ) under the impression that it was standard, if slightly recent, terminology. My impression was that it might be relevant. – Yemon Choi Aug 2 '10 at 5:10
@Gerry, thanks for your correction but also for your very clear answer. @Yemon, the page you send me says "This article may require cleanup to meet Wikipedia's quality standards." I found no relation of that theory to this particular problem! I'd justify your tag by your personal love to this theory, is it allright? ;-) – Wadim Zudilin Aug 2 '10 at 5:50
"Discrepancy" is a well-established term in the theory of uniform distribution - indeed, a sequence is uniformly distributed if and only if the discrepancy of its first $n$ terms goes to zero as $n$ goes to infinity. But "discrepancy theory" I've only seen in a more combinatorial context, where (e.g.) you 2-color some finite set and ask how far you are from using the 2 colors equally often. I can't get too worked up either way about using it as a tag for this problem. "Distribution-of-sequences" might be a better tag. – Gerry Myerson Aug 2 '10 at 6:34
up vote 7 down vote accepted

Exercise 2.23 in Kuipers and Niederreiter, Uniform Distribution Of Sequences: Use Theorem 2.7 to show that the sequence $(\alpha n^{\sigma})$, $n=1,2,\dots$, $\alpha\ne0$, $1\lt\sigma\lt2$, is u.d. mod 1.

They are using $(x)$ for the fractional part. Theorem 2.7 is Let $a$ and $b$ be integers with $a\lt b$, and let $f$ be twice-differentiable on $[a,b]$ with $f''(x)\ge\rho\gt0$ or $f''(x)\le-\rho\lt0$ for $x\in[a,b]$. Then $$\left|\sum_{n=a}^be^{2\pi if(n)}\right|\le(|f'(b)-f'(a)|+2)\left({4\over\sqrt\rho}+3\right).$$ Theorem 2.7 is attributed to van der Corput, Zahlentheoretische Abschatzungen, Math. Ann. 84 (1921) 53-79.

share|cite|improve this answer
Thank you! Do they prove the theorem in the book? If you know the proof, how hard is it? – Zarathustra Aug 2 '10 at 17:42
The proof can be found in Montgomery's 10 lecture book. The problem with this proof is that it does not generalize to $x^{\rho}$ for $\rho > 2$ not an integer, but the statement remains valid. See M. Boshernitzan, Uniform distribution and Hardy fields, J. Anal. Math. 62, 225-240 (1994) – Helge Aug 2 '10 at 18:56
@Zarathustra, yes, a proof is given in Kuipers and Niederreiter. It's only half a page, but it refers to a previous lemma, and I haven't tracked it back to see how long it is if you unwind the whole thing. – Gerry Myerson Aug 2 '10 at 23:16

I think you can try Weyl's criterion on this.

share|cite|improve this answer
If I recall correctly, Weyl's criterion works when we take fractional parts of integer multiples of a fixed irrational number. What makes you think it will work for a sequence of the form mentioned in the present question? – Yemon Choi Aug 2 '10 at 1:35
@Yemon, can you follow the link? Weyl's criterion is for any sequence of real numbers, you speak about one of its application. @Pencil, I don't understand why people downvote, your link is definitely a standard one for this type of problems, but probably has to go as comment, because it's quite helpless in this particular problem. The resulting exponential sums are very hard... – Wadim Zudilin Aug 2 '10 at 3:49
Hi, I believe "answers" seems more appropriate for solved answers. Next time I will write such thing in comments... – Kerry Aug 2 '10 at 4:17
@Wadim: yes, I followed the link, and once in a former mathematical life read a little about Weyl's criterion. My point was that an answer of the form "hey this result I've heard of might help" seemed a bit ill-judged in this context. As you say, the exponential sums that would need to be estimated here in order to appeal to Weyl's criterion look hard to my inexpert eye. (For the record, I didn't downvote Pencil's answer, although I didn't upvote it either.) – Yemon Choi Aug 2 '10 at 5:13
@Wadim (cont.) I admit that my initial comment was poorly phrased. What I meant to say was that Weyl's criterion gives an effective method of proving equidistribution of integer multiples modulo 1, but that I didn't see why Pencil thought it might help here. – Yemon Choi Aug 2 '10 at 5:14

Here's how to carry out direct proof:

By Weyl's criterion it suffices to show $$ S_N = \frac{1}{N} \sum_{n=1}^{N} e(k n^{\rho}) \to 0 $$ for $k \in \mathbb{Z} \setminus \{0\}$ and $\rho \in (1,2)$. Now $$ |S_N|^2 = \frac{1}{N^2} \sum_{m=1}^{N} \sum_{n=1}^{N} e(k (n^{\rho} - m^{\rho})) $$ Write $n = m + h$. Then by Taylor's theorem $(m+h)^{\rho} - m^{\rho} = \rho h \cdot m^{\rho - 1} + \frac{\rho(\rho - 1)h^2 }{2 (m + \xi)^{2 - \rho}}$ for some $|\xi| \leq h$. Hence $$ |S_N|^2 \leq \frac{1}{N^2} \sum_{m=1}^{N} \left|\sum_{h} e(k \rho h \cdot m^{\rho - 1} + \dots) \right| $$ here one needs to figure out the limit of $h$ and how to get rid of the $\dots$ term. This trick is called Weyl differencing (e.g. how you show the claim for the sequence $\alpha n^2$). The conclusion is that $|S_N|^2 \leq N$, which suffices to deduce the claim.

share|cite|improve this answer
Thank you. You forgot 1/N and rho must be between 1 and 2. Also I don't understand your formula for the difference (m+h)^-m^, it's wrong. Also, you forgot k in the last formula. Anyhow, if you can elaborate that'd be great. – Zarathustra Aug 2 '10 at 19:47
Thank you, I wasn't able to completely carry out the proof, but it looks like something doable. – Zarathustra Aug 5 '10 at 6:04

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.