Search Results
Search type | Search syntax |
---|---|
Tags | [tag] |
Exact | "words here" |
Author |
user:1234 user:me (yours) |
Score |
score:3 (3+) score:0 (none) |
Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
Views | views:250 |
Code | code:"if (foo != bar)" |
Sections |
title:apples body:"apples oranges" |
URL | url:"*.example.com" |
Saves | in:saves |
Status |
closed:yes duplicate:no migrated:no wiki:no |
Types |
is:question is:answer |
Exclude |
-[tag] -apples |
For more details on advanced search visit our help page |
Prime numbers, diophantine equations, diophantine approximations, analytic or algebraic number theory, arithmetic geometry, Galois theory, transcendental number theory, continued fractions
2
votes
Accepted
Error term in França-LeClair approximation of zeta zeros
There is no simple equation that given $n$ gives us $\gamma_n$, with sufficient approximation. Given a constant $c>0$, the error in both equations are at times greater than $c/\log n$ for any given c …
3
votes
Odlyzko's reformulation of Montgomery's pair correlation conjecture
I will say that the natural normalisation of the zeros of zeta is
$$\tilde\gamma=\frac{1}{\pi}\vartheta(\gamma)$$
where
$$\vartheta(t)=\Im(\log\Gamma(\frac14+\frac{it}{2}))-\frac{t}{2}\log\pi=\frac{t} …
3
votes
Is there a way to tie up even and "newly suggested odd" Riemann zeta values?
The function
$$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 …
1
vote
Zeroes of real and imaginary parts of $\zeta(1+it)$ separately (if any for $t>1$ say)
The solutions of $\Re\zeta(1+it)=0$ are scarce. They limit small intervals where
$\Re\zeta(1+it)<0$. The probability in the sense of the limit of the
quotient of the measure of the
set $\{0<t<T: \Re\ …
3
votes
Explicit estimates for $N(T,\chi)$ (not $N(T,\chi)+N(T,\overline{\chi})$)
For the first question about $2\log\zeta(\sigma_0)|$, I think the reasoning is this:
For $\sigma>2$ we have
$$|L(s,\chi)-1|\le \sum_{n=2}^\infty\frac{1}{n^\sigma}=\zeta(\sigma)-1<1.$$
Hence $\log L(s, …
2
votes
Are Li's numbers $\lambda_n$ absolutely convergent for $n>1$?
Assuming the Riemann hypothesis $\rho=\frac12+i\gamma$, then
$$1-\frac{1}{\rho}=-\frac{\frac12-i\gamma}{\frac12+i\gamma}=e^{2i\theta},\qquad
\theta=\arctan\frac{1}{2\gamma}.$$
$$\sum_\rho\Bigl[1-\Big …
1
vote
Riemann–Von Mangoldt formula
You have also Guinand formula for $N(T)$, see, for example, in this answer
https://mathoverflow.net/a/104570/7402
13
votes
For $x$ irrational, is $a_{n} =\sum_{k=1}^{n}(-1)^{⌊kx⌋}$ unbounded?
We have
Theorem. Let $\psi(x)$ and $\varphi(x)$ be positive increasing functions such that
$$\int_1^\infty \frac{dx}{\psi(x)}=+\infty,\qquad \int_1^\infty \frac{dx}{\varphi(x)}<+\infty.$$
Then for a …
2
votes
Enquiry on an equality involving the Riemann zeta function
Your question is a little imprecise, because you use $\arg(\zeta(1/2+it))$ without explaining what value to take. I assume that you refer to $\pi S(t)$ as defined in the book of Titchmarsh (section 9. …
7
votes
$\pi((n+1)^2)-\pi(n^2) \le \pi(n)$ for all $n \ge 370$?
We know is that the difference between
$\pi(x)$ y $\textrm{Li}(x)$ is less than $xe^{-c\sqrt{\log x}}$, and we have the asymptotic expansion
$$\textrm{Li}(x)-\textrm{Li}((x+1)^2)+\textrm{Li}(x^2)=$ …
12
votes
Accepted
Is this differential equation for zeta on the critical line? One can compute it from its der...
The zeta function can be written $\zeta(1/2+it)=Z(t)e^{-i\vartheta(t)}$ where
$Z(t)$ are real analytic and $\vartheta(t)$ is a simple function related to
$\Gamma(s)$. The relation between $\zeta(1/2 …
14
votes
Accepted
A conjecture about certain values of the Fabius function
I have posted in arXiv:1702.05442 the English translation of a paper about Fabius
function that I published in Spanish in 1982 (we will refer to it as (A)).
With the Theorems in this paper
the ques …
2
votes
Do the complex zeros of $\big(\zeta(s-1) -\zeta(s)\big) \pm \big(\zeta(1-s) - \zeta(2-s)\big...
I do not think this is so difficult as the Riemann hypothesis, I will only
explain why this is so without giving complete proof.
First on the line $s=1+it$ the functions are
$$(\zeta(it)-\zeta(1+it)) …
4
votes
Accepted
References on Taylor series expansion of Riemann xi function
In the paper:
M. W. Coffey, "Asymptotic estimation of $\xi^{(2n)}(1/2)$: On a conjecture
of Farmer and Rhoades", Mathematics of Computation, {\bf 78} (2009) 1147--1154
you may find the first terms o …
8
votes
Accepted
Fourier transform of the critical line of zeta?
If $\varphi$ is in the class of Schwartz
we have
$$\int_{-\infty}^{+\infty}\varphi(t)\zeta(\frac12+it)\,dt=
\sum_{n=0}^\infty\Bigl\{
\frac{1}{\sqrt{n+1}}\widehat{\varphi}\Bigl(\frac{1}{2\pi}\log
(n+1) …