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 |
Questions in which polynomials (single or several variables) play a key role. It is typically important that this tag is combined with other tags; polynomials appear in very different contexts. Please, use at least one of the top-level tags, such as nt.number-theory, co.combinatorics, ac.commutative-algebra, in addition to it. Also, note the more specific tags for some special types of polynomials, e.g., orthogonal-polynomials, symmetric-polynomials.
15
votes
0
answers
717
views
Solving polynomial systems with homotopy. Where is the bottleneck?
I have a polynomial system with $n+k$ unknowns ($n+k$ can be greater than 8), that is known to have a limited number of isolated solutions.
I want to solve this system numerically, but if I plug it …
7
votes
0
answers
2k
views
Why is solving polynomial systems NP hard?
My interest is in the case of systems of multivariate polynomials over the real field. … Intuitively, I can imagine that part of the problem consists in the counting of the roots of the system: Bezout bound states that the number of roots is bounded by the product of the degrees of the polynomials …