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 nt.number-theory
Search options not deleted
user 94470
Prime numbers, diophantine equations, diophantine approximations, analytic or algebraic number theory, arithmetic geometry, Galois theory, transcendental number theory, continued fractions
4
votes
Statements going against the grain of Riemann Hypothesis (R.H.)
If you have $M(N) > N^{0.99}$ for infinitely many $N$ then this means there is a an exceptional zero with real part $\geq 0.99$, that therefore invalidates RH.
As to the order of growth of $M(x)$ a …
- 104
2
votes
Accepted
new proof of Halasz inequality
Heuristically (and in fact rigorously with a bit of work) we have
$$
\sum_{p \leq x} \frac{1 + \Re(p^{it})}{p} = \log\log x + \Re \log \zeta(1 + \frac{1}{\log x} + it) + O(1)
$$
Therefore if $\zeta$ h …
- 104
1
vote
Higher dimensional analogs of logarithmic density
The natural $k$-dimensional analogue of logarithmic density is
$$
\lim_{x \rightarrow \infty} \frac{1}{(\log x)^{k}} \sum_{\substack{n_1, \ldots, n_k \leq x \\ (n_1, \ldots, n_k) \in S}} \prod_{i = 1} …
- 104