3
$\begingroup$

I'm studying certain polynomials of 2 complex variables, say x and y. These polynomials have roots at the non-negative integers, that is both $x$ and $y$ have to be

$x,y \in \mathbb{N}$

simultaneously with the restriction that

$x+y \leq A$

for some non-negative integer A. I'd like to know the basis for these polynomials of minimal total degree (sum of degrees in $x$ and $y$), up to overal constants. I think this basis is:

$\left(x \atop n\right)\left(y \atop A+1-n\right)$

for integer $n$ with $0 \leq n \leq A+1$. The brackets are falling factorials:

$\left(x \atop n\right) = x (x-1)....(x-n-1)$

with $\left(x \atop 0\right)\equiv 1$

At the moment I have only a proof-by-lack-of-imagination argument that this is the full set. Of course, I'd also like to know how to generalize to multiple complex variables :)

Improve this question
$\endgroup$
8
  • $\begingroup$ What do you mean exactly by "total degree" of a finite set of polynomials? The maximum of their (total) degrees, the sum of their degrees... $\endgroup$
    – Federico Poloni
    Aug 30, 2013 at 9:53
  • $\begingroup$ the sum of $i$ and $j$ in term of the type $x^i y^j$ $\endgroup$
    – twistorial
    Aug 30, 2013 at 10:00
  • $\begingroup$ No, I mean, you want to compare different sets of generators; how should we define the "total degree" of such a set? I know what is the "total degree" of a polynomial, but here you're asking how to compare sets of them if I am understanding the question correctly. $\endgroup$
    – Federico Poloni
    Aug 30, 2013 at 16:26
  • $\begingroup$ The question is about the "set of these (polynomials of minimal total degree)", not "(set of these polynomials) of minimal total degree". $\endgroup$
    – user39445
    Sep 2, 2013 at 9:09
  • $\begingroup$ edited for clarification... apologies for not thinking like a mathematician :) $\endgroup$
    – twistorial
    Sep 2, 2013 at 9:25

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.