I've run into a scenario where it would be extremely useful to know if, for $x$ irrational, the sequence $\{(2^nx)\}$ is equidistributed on in $[0,1]$, where $(\cdot)$ denotes the ``fractional part" of a number. I've tried to show that this is true using Weyl's criterion to no avail. I am not a number theorist in any way, and I am curious if anybody here knows how to show this, or if anybody here is aware of some tools that could be useful.
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1$\begingroup$ This is equivalent to $x$ being normal in base $2$. Almost all, but not all, irrational numbers have this property. $\endgroup$– WojowuCommented Mar 12, 2023 at 0:27
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Not always. The sum
$x=2^{-1}+2^{-2}+2^{-4}+2^{-8}+...$
with the exponent doubling on each successive term is irrational, but $(2^nx)$ is not uniform as $n\to\infty$. It clusters around each negative-integer power of $2$ and around $0$.
The problem of identifying normal numbers in any base is highly non-trivial.
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$\begingroup$ Thank you! I greatly appreciate it. $\endgroup$ Commented Mar 12, 2023 at 0:58
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$\begingroup$ I meant the measure of $\{x: (2^nx) \text{is equidistributed}\}$. But I now realize that the wikipedia article you linked answers this question. $\endgroup$ Commented Mar 12, 2023 at 1:25