Distribution of fractional parts of n^{3/2} - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T08:01:58Z http://mathoverflow.net/feeds/question/34188 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/34188/distribution-of-fractional-parts-of-n3-2 Distribution of fractional parts of n^{3/2} Zarathustra 2010-08-02T01:23:10Z 2010-08-02T20:23:10Z <p>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?</p> http://mathoverflow.net/questions/34188/distribution-of-fractional-parts-of-n3-2/34191#34191 Answer by Changwei Zhou for Distribution of fractional parts of n^{3/2} Changwei Zhou 2010-08-02T01:31:45Z 2010-08-02T01:31:45Z <p>I think you can try <a href="http://en.wikipedia.org/wiki/Weyl%27s_criterion" rel="nofollow">Weyl's criterio</a>n on this. </p> http://mathoverflow.net/questions/34188/distribution-of-fractional-parts-of-n3-2/34202#34202 Answer by Gerry Myerson for Distribution of fractional parts of n^{3/2} Gerry Myerson 2010-08-02T04:05:40Z 2010-08-02T04:05:40Z <p>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. </p> <p>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. </p> http://mathoverflow.net/questions/34188/distribution-of-fractional-parts-of-n3-2/34293#34293 Answer by Helge for Distribution of fractional parts of n^{3/2} Helge 2010-08-02T19:07:51Z 2010-08-02T20:23:10Z <p>Here's how to carry out direct proof:</p> <p>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.</p>