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 | |
Results tagged with riemann-zeta-function
Search options answers only
not deleted
user 10811
The Riemann zeta function is the function of one complex variable $s$ defined by the series $\zeta(s) = \sum_{n \geq 1} \frac{1}{n^s}$ when $\operatorname{Re}(s)>1$. It admits a meromorphic continuation to $\mathbb{C}$ with only a simple pole at $1$. This function satisfies a functional equation relating the values at $s$ and $1-s$. This is the most simple example of an $L$-function and a central object of number theory.
5
votes
Accepted
On the critical line $ \Re \zeta'(s)/\zeta(s) =? 1/2 \log(\pi) - 1/2 \Re \psi(s/2)$ ?
Yes, your formula is true (and no RH is needed),
Your symmetrized zeta-function is symmetrized such that the functional equation
$$
\zeta^* (s)=\zeta^* (1-s)
$$
holds. Thus we have that $\zeta^* …
- 2,913
7
votes
Accepted
What are the fallacies that this RH inequality may fail at most finitely often?
I suspect the mistake is in relying on Sage or Maple in the last step. Instead use the asymptotic expansion of li(x) ( http://en.wikipedia.org/wiki/Logarithmic_integral_function )
$$\operatorname{li}( …
- 2,913
18
votes
Accepted
A note by N. A. Carella on zero-free regions
Well,
In this case it is a short elementary paper (6 pages) and it is easy to verify, although just the length of the paper and the importance of the result and some other aspects of the paper, such …
- 2,913
10
votes
Accepted
lower bound for $\Re\zeta(1+it)$
There is no lower bound for Re$(\zeta(1+it))$. This can be found in Lamzouri's paper http://www.math.uiuc.edu/~lamzouri/distribzeta.pdf (see e.g. page 3, formula (5)). (Choose $\tau$ large enough, $\ …
- 2,913
6
votes
non-trivial zeros of partial zeta functions
Micah's answer answers your Q1. My reply gives some information on your Q2. Let us assume that $1 \leq a\leq N$ for simplicity (this condition can be removed). Your zeta-functions $\zeta$ and $\Psi$ …
- 2,913
5
votes
Accepted
Pair correlation for the Riemann zeros and $(\zeta^\prime(s)/\zeta(s))^\prime$
I believe the comments of joro and Carlo Benakker right under your question is to the point. Since the zeroes close to $s$ will be the ones that contributes in the sum, the zeroes close to s must be c …
- 2,913