Uniform partitions of an interval

Author:
Vladimir Drobot

Journal:
Trans. Amer. Math. Soc. **268** (1981), 151-160

MSC:
Primary 10K30; Secondary 10K05

DOI:
https://doi.org/10.1090/S0002-9947-1981-0628451-3

MathSciNet review:
628451

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Abstract: Let $\{ {x_n}\}$ be a sequence of numbers in $[0, 1]$; for each $n$ let ${u_0}(n), \ldots , {u_n}(n)$ be the lengths of the intervals resulting from partitioning of $[0, 1]$ by $\{ {x_1}, {x_2}, \ldots , {x_n}\}$. For $p > 1$ put ${A^{(p)}}(n) = {(n + 1)^{p - 1}}\sum \nolimits _0^n {{{[{u_j}(n)]}^p}}$; the paper investigates the behavior of ${A^{(p)}}(n)$ as $n \to \infty$ for various sequences $\{ {x_n}\}$. Theorem 1. *If* ${x_n} = n\theta (\bmod 1)$ *for an irrational* $\theta > 0$, *then* $\lim \inf {A^{(p)}}(n) < \infty$. *However* $\lim \sup {A^{(p)}} < \infty$ *if and only if the partial quotients of* $\theta$ *are bounded* (*in the continued fraction expansion of* $\theta$). Theorem 2 gives the exact values for $\lim \inf$ and $\lim \sup$ when $\theta = \tfrac {1} {2}(1 + \sqrt 5 )$. Theorem 3. *If* ${x_n}{\text {’}}s$ *are random variables, uniformly distributed on* $[0, 1]$, *then* $\lim {A^{(p)}}(n) = \Gamma (p + 1)$ *almost surely*.

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Article copyright:
© Copyright 1981
American Mathematical Society