## New answers tagged analytic-number-theory

0

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(...

1

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^...

answered Jan 20 at 13:06

Carlo Beenakker

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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&...

6

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\...

5

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 :

answered Jan 15 at 5:10

Chandan Singh Dalawat

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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}\...

8

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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