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Let $S_1$ be a subset of $[0,1]$ with consecutive elements separated by a distance of at most $\epsilon/2\pi$ and let $S_2$ be a subset of $[0,1]$ with consecutive elements separated by at most $\epsilon / (2 \pi m)$. We can take $|S_1| < 2\pi\epsilon^{-1} +1$ and $|S_2| < 2\pi m \epsilon^{-1} +1$. Let $\gamma \in S_1$ be the closest point to $\alpha ... 2 Let me provide a slightly different proof to that of David Loeffler, using only that$\zeta_K^\ast(0)<0$for any number field$K$. A real valued linear character factors through$\mathbb{Z}/2\mathbb{Z}$, corresponding to a quadratic extension$L/K$, so we can suppose that$G=\mathbb{Z}/2\mathbb{Z}$; denoting$\chi$its non trivial character we have the ... 3 If$\chi$is real-valued, then the question makes sense. Using the functional equation, it reduces to computing the sign of the non-zero real number$L(\chi, 1)$if$\chi$is non-trivial, or the residue of$L(\chi, s) = \zeta_K(s)$at$s = 1$if$\chi$is trivial. In either case,$L(\chi, s)$tends to$+1$for$s$large and real, and it cannot vanish on$Re(...

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For $k\in\mathbb{N}$ one has the inverse Mellin transform $$\frac{1}{2\pi i}\int\limits_{a-i\infty}^{a+i \infty}\varGamma^k(z) u^{-z}dz=G_{0,k}^{k,0}\left(u\left| \begin{array}{c} 0^{\otimes k} \\ \end{array} \right.\right),$$ with $G$ the Meijer-G function and $0^{\otimes k}$ is the string $0,0,0\ldots 0$ of length $k$. For $k=1$ this is the exponential $e^... 1 Montgomery (1) gives a list of 40 exponent pairs$(\kappa,\lambda)$which can be plugged into Iwaniec's formula $$\eta(\theta)=\left(1+\frac{1-\lambda+2\kappa}{3-\lambda-\kappa/2}\right)\theta - \frac{\kappa}{3-\lambda-\kappa/2}$$ to yield bounds for$0<\theta\le1/2.$Of these, 36 are optimal in some interval; adding the zeta function value for$\theta&...

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This problem has been addressed in a paper of Friedlander and Iwaniec in Acta Mathematica 1978, called Quadratic polynomials and quadratic forms. Under general conditions they count the number of integers $n\le x$ for which the values $g(n) = an^2+ bn+c$ (with $a>0$, and $a$, $b$, $c$ integers) may be represented by a given quadratic form $\phi(u,v) = Au^... 3 The function $$f(s):=\sin(\tfrac{\pi}{2}(s+1)) \zeta(s)+\sin(\tfrac{\pi s}{2}) L(\tfrac{s+1}{2})\tag{1}$$ where$L(s)$is the second Dirichlet series $$L(s)=\sum\limits_{n=1}^{\infty } \frac{(-1)^{n-1}}{(2 n-1)^{2 s-1}}\tag{2}$$ which is valid for$\Re(s)>1$but which can be analytically extended using the Hurwitz zeta function as $$L(s)=\frac{\zeta \left(... 9 No. Such a bound would imply a similar bound on$$\displaystyle \left \lvert \sum_{N(z) = M} \left(\frac{z}{w} \right)_3 \right \rvert.$$If M is a product of distinct primes p_1,\dots p_n congruent to 1 mod 3, the norms of primes \pi_1,\dots,\pi_n then$$\sum_{N(z) = M} \left(\frac{z}{w} \right)_3 = \left( \left(\frac{1}{w} \right)_3 + \left(\... 3 One can do a bit better. For simpler presentation assume that we instead consider the function$b'$that is the indicator function of integers all of whose prime divisors are$1 \mod 4$. We have$b'\leq b$but$b'$and$b$agree on square-free odd integers and any proof for$b'$can be adapted to$b$. Furthermore, we have$\sum_{p\leq n} b'(n-p)\ll \sum_{d\...

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If the $x_j$'s are distinct modulo $1$ (which is the natural assumption), then $$\lim_{N\to\infty}\frac{1}{N}\sum_{n=1}^{N}\left|\sum_{j}c_j e^{2\pi i n x_j}\right|^2=\sum_{j}|c_j|^2.$$ Indeed, let us assume (without loss of generality) that the $x_j$'s lie in $[0,1]$. Then $$\sum_{n=1}^{N}\left|\sum_{j}c_j e^{2\pi i n x_j}\right|^2=\sum_{j,k}c_j\overline{... 5 I found one paper that improves on the quoted result (in a narrow range):$$\eta(\theta )=\frac{100\theta -45}{11}\qquad\text{is admissible for}\qquad \frac{6}{11}<\theta \le \frac{11}{20}.$$See Lou-Qi: Upper bounds for primes in intervals (Chinese), Chinese Ann. Math. Ser. A 10 (1989), 255-262. 0 From the announcement of the Séminaire Bourbaki du vendredi on 28/1/2022 : 13 This answer is based on Lucia's remark, and is included for completeness. By (8.111) in Ivić's book "The theory of the Riemann zeta function with applications", we have$$\int_T^{2T}|\zeta(\sigma+it)|\,dt\asymp_\sigma T,\qquad T\geq 1,\quad 1/2<\sigma<1.$$Hence, by the functional equation for \zeta(s) and Stirling's approximation, we also ... 3 Here are two equivalences. Theorem 1. For each m \geq 1, the following are equivalent. a) For all nontrivial Dirichlet characters \chi \bmod m, L(1,\chi) \not= 0. b) For all a \in (\mathbf Z/m\mathbf Z)^\times, the set of primes p \equiv a \bmod m has Dirichlet density 1/\varphi(m). Proof of Theorem 1. We will will compute the Dirichlet density ... 3 I assume that \mathbb Zp means the field with p elements, and denote it by \mathbb F_p. Since \lambda= (\lambda X)/X, the fraction field of R is equal to the fraction field of F[[X]], which is the ring of Laurent power series F((X)). Let f=\sum_{n=m}^{\infty} a_n X^n \in F((X)) with a_m\neq 0. After possibly replacing f by its inverse we ... 10 To add to GH from MO's answer, Chapter 10 of Iwaniec and Kowalski's "Analytic Number Theory" is another good reference. A few of the classical papers are Ingham's "On the Estimation of N(\sigma,T)," Montgomery's "Mean and Large Values of Dirichlet Polynomials" and "Zeros of L-functions," and Huxley's "On the ... 18 There are, provably, very few zeros with real part close to 1 (or bigger than 0.51 for that matter). These theorems go under the name of "zero density estimates", and they have a vast literature. See Chapter 11 in Ivić: The Riemann zeta function (1985). The book was reprinted in 2013 by Dover Publications. 2 Assuming the strengthened version of Hardy-Littlewood conjecture I discuss here (which follows from Dickson's conjecture), the following much stronger result holds: let a_0,\dots,a_m be any sequence of natural numbers with a_0\geq 1 and a_{i+1}\geq a_i+2 for i<m. Then there are infinitely many integers n such that k_i(n)=a_i for i\leq m. Let ... 1 This is not mean to be a full answer, but one which illustrates how one can prove an estimate like the one in my comment through a rather brute force'' approach. To illustrate the idea of the computation, let's deal with the case when p=2 and q is some odd prime number, so$$ S(2,q) = \sum_{1 \leq y < q} \frac{1}{\frac{1}{2} \left|\left| \frac{y}{q}\...

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If you just want $o()$, the story is rather simple. Let $a_{xy}$ be the remainder of $qx+py\mod pq$ where all remainders modulo $P$ are assumed to be between $-P/2$ and $P/2$. Note that all $a_{xy}$ are distinct, so if we have any set $Z$ of pairs $(x,y)$, then $\sum_{(x,y)\in Z}\frac 1{a_{xy}}\le 2(1+\log|Z|)$. What we want to show is just  \sum_{0<|x|&...

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