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$\qquad \qquad \bbox[15px,border:2px solid red] { x_n:=\text{\{α$\cdot$ log(n)\}}_{n\in \mathbb N}}$

I want to show that the sequence $x_n$ is not uniformly distributed mod1 in $[0, 1]$ for any $α\in \mathbb R$.

Note:

1)$\qquad \qquad \qquad \qquad$ Euler summation Formula:

$ \qquad \qquad \bbox[15px,border:2px solid red] { \sum_{n=1}^Nf(n)=\int_1^Nf(t)dt+\frac{1}{2}(f(1)+f(N)) +\int_1^N(\text{\{t\}-$\frac{1}{2})$ }f'(t)dt }$

2) $\qquad \qquad \qquad \qquad$ Weyl's equidistributed criterion:

$\qquad \qquad \qquad \qquad \quad \qquad$ The following are equivalent
$\qquad \qquad \qquad \qquad \qquad \quad \bbox[15px,border:2px solid blue] {x_n \quad \text{is equivalent modulo 1} }$ $\qquad \qquad \qquad \qquad \qquad \quad \bbox[15px,border:2px solid blue] {\forall \text{continuous & 1-peridic f:}\quad \frac{1}{N}\sum_{n=1}^Nf(x_n)\rightarrow\int_0^1f }$ $\qquad \qquad \qquad \qquad \qquad \quad \bbox[15px,border:2px solid blue] {\forall k\in \mathbb Z^*:\quad \frac{1}{N}\sum_{n=1}^Ne^{2πikx_n}\rightarrow 0 }$

I've already proved it by using (1) & (2) , is there any other way to approach this problem?

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  • $\begingroup$ How did you use the Euler summation formula to show $x_n$ is not uniformly distributed? $\endgroup$ Aug 4, 2020 at 19:11
  • $\begingroup$ Do you mean $\{x_n\}$ is equidistributed in $[0,1]$? The adjective Uniformly distributed means a different thing, that is $X:(\Omega,\mathscr{A},\mathbb{P})\rightarrow[0,1]$ is uniformly distributed if $\mathbb{P}[X\in A]=\lambda(A)$ for all measurable set $A\subset[0,1]$, where $\lambda$ is the Lebesgue measure. $\endgroup$
    – Mittens
    Aug 4, 2020 at 19:13
  • $\begingroup$ @Verun Vejalla : math.rice.edu/~michael/teaching/426_Spr14/UDmod1A.pdf page 8 EXAMPLE 2.4. $\endgroup$ Aug 4, 2020 at 19:15
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    $\begingroup$ @JohnMars: Weyl's equidistribute theorem is the natural thing to use. Whether one uses Euler summation, Abel summation or Sonin's summation is just a tool of calculation. $\endgroup$
    – Mittens
    Aug 4, 2020 at 20:06
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    $\begingroup$ Another way would be directly from dynamics. If this sequence is uniformly distributed, it is recurrent around 0 (namely has syndetic set of return times). Pick M large so that $\alpha\cdot\log(M)$ is near $0$, one can now estimate that for many consecutive numbers n after $M$, $\alpha\cdot\log n$ is small, by some Taylor expansion or so. Hence having large gaps. $\endgroup$
    – Asaf
    Aug 5, 2020 at 17:32

1 Answer 1

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Yea, just think about what's going on. For $n \approx e^k$, $\alpha\log(n) \approx k\alpha$ and for $n \approx e^{k+\frac{1}{2\alpha}}$, $\alpha\log n \approx k\alpha+\frac{1}{2}$. So $\{\alpha \log n\}$ is in a particular interval of size $\frac{1}{2}$ for $n$ between $e^k$ and $e^{k+\frac{1}{2\alpha}}$. And for large $k$ ($\alpha$ is fixed), nearly all positive integers less than $e^{k+\frac{1}{2\alpha}}$ are greater than $e^k$. We conclude that nearly all $n \le e^{k+\frac{1}{2\alpha}}$ (you can put a floor function if you want) have $\{\alpha \log(n)\} \in (k\alpha,k\alpha+\frac{1}{2}) \pmod{1}$, clearly violating uniform distribution.

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    $\begingroup$ I think the last part is missing something: the proportion of integers between $e^k$ and $e^{k+u}$ (with $u=1/2\alpha$) for large $k$ among integers less than $e^{k+u}$ is $1-e^{-u}$. So you only get a contradiction if $1-e^{-u} > 1/2$ ie $e^{-u} < 1/2$, ie if $\alpha$ is small… by taking variable length intervals I think we can reach a contradiction if $\alpha < 1$. In general we should be able compute the exact limit densities and deduce a contradiction for (almost?) every $\alpha$. $\endgroup$
    – Aphelli
    Nov 25, 2021 at 19:57

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