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-hypothesis
Search options not deleted
user 26522
Questions about the famous conjecture from Riemann saying that the non-trivial zeroes of the Riemann Zeta function all lie on the so-called critical line $\Re(s)=\dfrac{1}{2}$, its various generalizations and the different approaches towards its solution.
9
votes
1
answer
493
views
Subspaces of $L^2(0,1)$ dense on every truncation $L^2(c,1)$
It may be better to move this to a separate question.
Let me call a linear subspace $V \subset L^2(0,1)$ to be tame if, for every linear subspace $W \subset V$, either $W$ is dense in $L^2(0,1)$, or …
- 13.8k
10
votes
1
answer
581
views
Are the polynomials in $\{1/t\}$ dense in $L^2(0,1)$?
Added. My question in the title was solved (in the negative) by Nik Weaver (in the answer below) and Mateusz Kwaśnicki (in the comments). In both solutions, the reason is that the $L^2$ density fails …
- 13.8k
6
votes
Accepted
Lexicographic distribution of irreducible polynomials
This is true.
By Gauss's theorem (the inclusion-exclusion formula for the number of irreducibles of a given degree), we may restrict to polynomials of a fixed degree $r$. A moment of reflection then …
- 13.8k