   ISSN 1088-6850(online) ISSN 0002-9947(print)

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