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 88133
Prime numbers, diophantine equations, diophantine approximations, analytic or algebraic number theory, arithmetic geometry, Galois theory, transcendental number theory, continued fractions
4
votes
Complexity of a polynomial
Let $k_1,\dotsc,k_r$ be the even integers with even $2$-adic valuation in the range you want. For any polynomial $p$ meeting your condition, $p(x)-2^{n-1}$ has a root between $k_i-2$ and $k_i$, and a …
- 6,237
1
vote
Accepted
Conjectures inspired in the context of Casas-Alvero conjecture, via the logarithmic derivati...
I'm probably making a stupid mistake, but is Conjecture 1 possibly rather easy? At $\ell=0$ we're assuming
$$ p(x) = \frac{a_n}{n!} \left( \frac{n}{\frac{d}{dx} \log p(x)} \right)^n $$
Write $\frac{d} …
- 6,237
3
votes
There is a nice theory of quadratic forms. How about cubic forms, quartic forms, quintic for...
Here is an additional comment. Every homogeneous polynomial can be regarded as a symmetric tensor. Quadrics correspond to symmetric matrices; cubics correspond to order $3$ tensors; in general, a homo …
- 6,237