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Mar 30, 2018 at 9:50 comment added Fedor Petrov @Jean you mean equidistributed modulo 1? Of course no, take $x_n=n\sqrt{2}, y_n=[x_n]$.
Mar 30, 2018 at 9:03 comment added Jean @GregMartin I mean $x_n/y_n\to 1$ as $n\to \infty$.
Mar 30, 2018 at 1:45 comment added Greg Martin If by "asymptotic equivalent" you mean that $x_n-y_n$ tends to $0$ as $n\to\infty$, then yes, it is an easy consequence of the definition of equidistribution that $\{x_n\}$ is equidistributed if and only if $\{y_n\}$ is.
Mar 29, 2018 at 22:22 comment added Jean Thanks, I didn't see that. But for example, if I have two asymptotic equivalent sequences $x_n$ and $y_n$. If $x_n$ is equidistributed then so $y_n$?
Mar 29, 2018 at 21:13 history answered Fedor Petrov CC BY-SA 3.0