I need to prove that the sequence $\{a \log n \}$ is NOT equidistributed for any $a \in \mathbb{R}$.

Now, I think that Weyl's criterion will be a good idea to use here. So, I need to show that $$\frac{1}{N} \sum_{n=1}^{N} e^{2 \pi i k (a \log n)} \rightarrow 0 $$ as $N \rightarrow \infty$ for any $k \in \mathbb{Z} - \{0\}$ doesn't hold.

I have two problems here. Firstly, I don't know whether the base of $\log$ is $10$ or $e$. So, this can be tried for both bases.

Secondly, if I assume that the base is $e$, then I have $$ e^{\log n^{a 2 \pi i k }}$$ inside the summation. This is equal to $n^{a 2 \pi i k}. $ So, I am left with $$\frac{1}{N} \sum_{n=1}^{N} n^{ 2 \pi i k a} $$ Now suppose I take a negative $k$ and positive $a$ or vice-versa, then wouldn't this series converges to $0$ as $N \rightarrow \infty$ ?

  • $\begingroup$ You can prove this without exponential sums. The idea is that the function grows really, really slowly. You can assume the sequence is equidistributed and use this to get a contradiction. $\endgroup$ – Vik78 Oct 2 '16 at 4:45
  • $\begingroup$ Sorry but I don't get the idea of this function being really slow? You are talking about the whole summation function, right? $\endgroup$ – Dark_Knight Oct 2 '16 at 4:54
  • $\begingroup$ No, i'm talking about the logarithm function. Just say you look at the elements of the sequence out to index $n$, such that the proportion of elements up to $n$ within an interval $I \subset [0, 1)$ is within $\epsilon$ of the length of $I$, $a_n$ is not in $I$ and $a_{n+1}$ is. If the sequence next exits $I$ up at index $m$, you should be able to choose $n$ large enough such that the proportion of elements in $I$ at index $m$ to the total is no longer within $\epsilon$ of the length of $I$. This is because the log grows so slowly: you can write log $n \le$ log$(n + k) \le$ log $n$ $+ k/n$. $\endgroup$ – Vik78 Oct 2 '16 at 8:28
  • $\begingroup$ Indeed you can see from the above bounds that $m$ should be asymptotically at least $n |I|$. $\endgroup$ – Vik78 Oct 2 '16 at 8:42
  • 1
    $\begingroup$ I don't even know the equidistribution criterion, this is just how I proved it. If $r(k)$ is the number of times $(a_n)$ falls in $I$ before the $k$th index, find $k$ such that $r(k)/k$ is within $\epsilon$ of the length $|I|$ of $I$ and $a_k$ is in $I$. Let $m$ be the next index such that $a_m$ is not in $I$. As I explained above, as $k \to \infty$ we have $m \ge k + |I|k$. Therefore $r(m)/m \ge \frac{r(k) + |I|k}{k + |I|k} = \frac{r(k)}{k} + \frac{|I|}{1 + |I|}$. If you choose $|I|$ and $\epsilon$ small enough this is about $2|I|$, so $r(n)/n$ cannot converge to the length of $I$. $\endgroup$ – Vik78 Oct 3 '16 at 21:30

Apply the Euler–Maclaurin summation formula. For $f\in C^1\big([1,N]\big)$, it reads $$\sum_{n=1}^N f(n)=\int_1^N f(x)\,dx+\frac{f(1)+f(N)}{2}+\int_1^N\left(\{x\}-\frac12\right)f'(x)\,dx,$$ where $\{x\}=x-\lfloor x\rfloor$ is the fractional part of $x$.

For $f(x)=e^{ic\log x}$ with $c=2\pi ka$, we have $f'(x)=icf(x)/x$ and $$W_N:=\frac1N\sum_{n=1}^N e^{ic\log N}=I_N+F_N+R_N, \\I_N=\frac1N\left.\frac{e^{(1+ic)\log x}}{1+ic}\right|_{x=1}^{x=N}=\frac{e^{ic\log N}-1/N}{1+ic}, \\F_N=\frac{1+e^{ic\log N}}{2N},\quad|R_N|\leqslant\frac{|c|}{N}\int_1^N\frac12\frac{dx}{x}=\frac{|c|}{2}\frac{\log N}{N}.$$

As $N\to\infty$, clearly $F_N\to 0$ and $R_N\to 0$ but $I_N\not\to 0$. Hence, $W_N\not\to 0$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.