# Distribution of fractional parts of n^{3/2}

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?

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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 en.wikipedia.org/wiki/Discrepancy_theory ) 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

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.

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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

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.

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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