Question

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?

Answer 1

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.

Answer 2

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

Answer 3

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.