33
votes
Accepted
Distribution of square roots mod 1
You can certainly use Vinogradov's method to show that $\sqrt{p}$ is equidistributed $\pmod 1$. I haven't thought about more subtle properties, such as the gap spacing considered by Elkies and ...
24
votes
Accepted
Anti Arzela-Ascoli
Under the stated conditions, there always exists a subsequence that Cesaro converges almost everywhere. This was a question of Steinhaus, solved by Revesz [1].
More generally, it suffices that the ...
18
votes
Distribution of square roots mod 1
Here is an insight on what happens for the bin $[0.5,0.501]$
$n^2+n+\frac14=(n+\frac12)^2$ and $(n+\frac12+\frac3{8n+5})^2=n^2+n+1-\frac{3(8n+7)}{4(8n+5)^2}.$
So, until $\frac3{8n+5} \lt \frac1{1000}...
17
votes
Accepted
Is the divisor counting function equidistributed mod $p$?
$\newcommand{\Y}{\mathfrak{X}_p(X)}$I haven't checked all the details on the application of Selberg–Delange below, so it's possible something I am saying is nonsense, but even after accounting for the ...
16
votes
Digits in an algebraic irrational number
What is known is that every real irrational has a $0$ in its $g$-ary expansion for infinitely many $g$. WLOG take $0 < x < 1$.
Taking an even-numbered convergent of the continued fraction of $...
14
votes
Accepted
Equidistribution of $\{\alpha p\}$ for $p$ in an arithmetic progression
I think the sought result follows from Vinogradov's theorem. By Weyl's criterion and the orthogonality of Dirichlet characters, the sought result can be reformulated as follows. For every nonzero ...
14
votes
Behavior of $\sum_{n=1}^{\infty} (\{n \xi \} - \frac{1}{2})$ for irrational number $\xi$
This is studied in detail in Sums of Fractional Parts of Integer Multiples of an Irrational
The
sum
$$C(N)=\sum_{n=1}^{N} \bigl( \{ n \xi \} - \tfrac{1}{2}\bigr)$$
satisfies $|C(N)|>c\log N$ for ...
14
votes
Accepted
Is the product of two equidistributed power series equidistributed?
No. Let $f(T)$ be chosen uniformly at random from all power series with constant term nonzero and let $g(T) = 1/ f(T) $. Then $g(T)$ is also chosen uniformly at random from all power series with ...
13
votes
Uniform distribution of points on Riemannian manifolds
Let $\mu=(\delta_A+\delta_B)/2$. Then the claim of Arnold - Krylov is the weak convergence of the convolutions $\mu^{*n}*\delta_x$ to the rotation invariant probability measure on the sphere (where $\...
13
votes
Accepted
Equidistribution of $\{p_n^2α\}$
Yes - this follows from a general theorem of Bergelson, Kolesnik, Madritsch, Son, and Tichy (Theorem 2.1 in https://people.math.osu.edu/bergelson.1/BKMS_PrimePowers.pdf):
Let $\xi(x)=\sum_{j=1}^m\...
13
votes
Accepted
Continuous variant of the Chinese remainder theorem
Your guess is correct. For $\alpha = (\alpha_1, \alpha_2, \ldots, \alpha_d) \in \mathbb{R}^d$, the values of $m \alpha \bmod 1$ are equidistributed (and in particular dense) in $(\mathbb{R}/\mathbb{Z})...
11
votes
Accepted
Upper bound for maximum modulus of polynomial on unit circle in term of the distribution of its roots
Let me first start with the other side: does the maximum being small guarantee that the roots are equidistributed? This is indeed the case, and is a beautiful theorem of Erdos and Turan. For a ...
11
votes
Accepted
Equidistribution of CM points in the principal genus
This is known, and follows from Theorem 2 in Harcos and Michel's paper The subconvexity problem for Rankin-Selberg $L$-functions and equidistribution of Heegner points. II (Invent. math., vol. 163, ...
10
votes
Equidistribution of CM points in the principal genus
Vesselin Dimitrov gave a nice answer, but let me point out that for the OP's question one does not need the result of Harcos-Michel (2006). Instead, the original work of Duke (1988) or alternatively ...
10
votes
Accepted
Cancellation in a very rapidly oscillating exponential sum
I doubt that one is able to get as far as $T = \exp(\log^{2-\varepsilon} x)$ with Weyl differencing. Standard Weyl differencing arguments, such as that in Theorem 8.4 of
Iwaniec, Henryk; Kowalski, ...
9
votes
Accepted
Let $(a_n)_{n\in N}=(1,2,3,4,6,8,9,12,\cdots)$ list the set$\{2^n3^m\mid m,n\in N\}$. Find $α$ such that $(a_n)\alpha\pmod1$ is not equidistributed
$\alpha =\sum_{n=1}^{\infty} \frac{1}{ 6^{100^n}}$ should do the trick. A positive proportion of numbers on your list are of the form $2^a 3^b$ for $a,b$ within a reasonable constant factor of each ...
9
votes
Must bounded sequences be well-distributed to most *composite* moduli?
This question is related to the results in
Bergelson, Vitaly; Richter, Florian K., Dynamical generalizations of the prime number theorem and disjointness of additive and multiplicative semigroup ...
8
votes
Distribution of square roots mod 1
For the spikes, note that in your picture, the predicted number of points in the bin around 0 is 100, but there are 300 squares in your range. Similarly, there are 75 quarter-squares (where by quarter ...
8
votes
Accepted
$\{(\log n)^\alpha\}$ not equidistributed if $0<\alpha\leq 1$, so how is it distributed?
There is no limit distribution at all for $a:=\alpha\in(0,1)$: for each $x\in(0,1)$, the relative frequency
\begin{equation*}
f_n:=f_n(x):=\frac1n\,\sum_{j=1}^{n-1}1(x_j<x)
\end{equation*}
will ...
8
votes
Reference request - Pillai-Selberg Theorem
The original references are:
S. Selberg [Math. Z. 44 (1939), 306–318; zbMATH:0019.39308]
S. S. Pillai [Proc. Indian Acad. Sci. Sect. A. 11 (1940), 13–20; zbMATH:66.0168.01, MR0001761].
Interestingly ...
8
votes
Accepted
Equidistribution on $\mathrm{SU}_2$
In the article "On the spectral gap for finitely-generated subgroups of SU(2)" by Jean Bourgain and Alex Gamburd (Invent. Math. 171, No. 1, 83-121 (2008)), they show that free subgroups of $...
6
votes
Accepted
Sequences similar to $\{n\alpha\}$ that are both equidistributed and truly random-like
For any $p>0$ the sequence of fractional parts $x_n=\{n^p\alpha\}$ cannot be random-like in the sense defined in the appendix. The case of integer $p$ was already discussed in the comment by ...
6
votes
Elementary proof of the equidistribution theorem
An elementary proof was given by Alberto Zorzi in
An Elementary Proof for the Equidistribution Theorem
The Mathematical Intelligencer
September 2015, Volume 37, Issue 3, pp 1–2
Unfortunately the ...
6
votes
Accepted
Convolution sum of divisor functions
Too long for a comment
Maybe this helps though it has $\sigma(n)=\sigma_1(n)$ not $\sigma_0(n)$. It may be that a similar identity may hold for $\sigma_0(n).$ In a long paper available here we find
...
6
votes
for which $A\subset \mathbb{N}$ does there exist irrational $\alpha$ for which $\alpha\cdot A \pmod 1$ is not uniformly distributed?
Consider a random set $A$ where $n\in A$ with probability $f(n)/n$ for $f$ a function going to $\infty$ arbitrarily slowly. Then I think $A$ has $\alpha \cdot A$ uniformly distributed for all nonzero $...
6
votes
Must bounded sequences be well-distributed to most *composite* moduli?
The answer to both questions is negative. I will now try to be careful about constants.
Let $\omega(n)$ be the number of prime factors of $n$, let $$a_n = \begin{cases} 1 & \omega(n) < \log \...
5
votes
Accepted
The lonely runner conjecture and equidistribution on tori
A corrected version of this argument is contained in Section 4 of Six Lonely Runners by Bohman, Holzman, and Kleitman.
Their argument shows, using equidistribution, that the case of the lonely runner ...
5
votes
Accepted
Duke and Schulze-Pillot condition for equidistribution
I recommend that you study the paper by Duke and Schulze-Pillot, Representation of integers by positive ternary quadratic forms and equidistribution of lattice points on ellipsoids (Inventiones, 1990)....
5
votes
Accepted
Equidistribution of distances of integer points to a circle
Yes. What you are looking for follows from the known error bounds in the Gauss Circle Problem. In particular, in the notation of that Wiki article, $\text{card}(A_r\cap [a,b])=N(r+b)-N(r+a^-)$, while $...
4
votes
Accepted
Is this a criterion for uniform distribution modulo one?
I provide a sequence $(x_k)_k$ that is not uniformly distributed but satisfies $\frac{1}{N}\sum_{k \le N} a_k^m e^{2\pi i x_k m} \to 0$ as $N \to \infty$ for each $m \ge 1$, where $a_k = 1$ if $k$ is ...
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