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Transactions of the American Mathematical Society

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
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.


References [Enhancements On Off] (What's this?)

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DOI: http://dx.doi.org/10.1090/S0002-9947-1981-0628451-3
Article copyright: © Copyright 1981 American Mathematical Society