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 open-problems
Search options not deleted
user 1631
If it turns out that a problem is equivalent to a known open problem, then the open-problem tag is added. After that, the question essentially becomes, "What is known about this problem? What are some possible ways to approach this problem? What are some ways that people have tried to attack it before, and with what results?"
19
votes
4
answers
2k
views
Irreducible polynomials with constrained coefficients
Over at the Cafe, after reading about TWF 285, I asked more-or-less
About how many polynomials with coefficients in $\{\pm 1\}$ and of degree $d$ are irreducible?
and that's what I want to ask …
- 1,987
-1
votes
When shorter means smaller?
Another tact; consider the set of foldings of our convex shape. Since the fold axis partitions the perimeter into two parts, and by convexity both have finite length, so one of them is at least half …
- 1,987
1
vote
When shorter means smaller?
An incomplete answer; but perhaps it helps to rephrase the problem as below. The reason the round circle does have this property is that without loss of generality, the map $f$ fixes the origin; and …
- 1,987