4
$\begingroup$

There have been a number of discussions of zeros of random polynomials here (the most recent being: Why do roots of polynomials tend to have absolute value close to 1?).

Here is a closely related question: consider $N$ polynomials in $\mathbb{R}[x_1, \dotsc, x_N]$ whose coefficients are independent centered Gaussians (not identical, necessarily). These define a variety which is almost surely zero-dimensional. Now, the question is: what is the distribution of the number of its real points (or, at least, what can be said about its mean and variance)?

Improve this question
$\endgroup$
5
  • 2
    $\begingroup$ Are there bounds on the degrees of these polynomials? $\endgroup$
    – Per Alexandersson
    Oct 21, 2014 at 12:33
  • $\begingroup$ @PerAlexandersson typically, we would want the degrees to go to infinity, and the results we get would be asymptotic in these (all degrees the same might be easiest, but certainly not the only interesting case...) $\endgroup$
    – Igor Rivin
    Oct 21, 2014 at 13:04
  • $\begingroup$ Even if all of them are linear they will have one solution in general I believe. $\endgroup$
    – joro
    Oct 21, 2014 at 13:44
  • $\begingroup$ @joro true, and in fact exactly one (almost surely). $\endgroup$
    – Igor Rivin
    Oct 21, 2014 at 15:47
  • $\begingroup$ I believe it is $\sqrt{d}$, where $d$ is the product of degrees = number of complex solutions, but I do not remember the reference in higher dimension. $\endgroup$
    – Alexandre Eremenko
    Oct 21, 2014 at 21:53

1 Answer 1

Reset to default
2
$\begingroup$

Under certain conditions on the variance of the coefficients it is $\sqrt{d}$ where $d$ is the product of degrees, MR2182444 .

Improve this answer
$\endgroup$
2
  • $\begingroup$ Yes, that is true, but that is only true with certain variance assumptions. $\endgroup$
    – Igor Rivin
    Oct 21, 2014 at 23:14
  • $\begingroup$ Yes, with different assumptions the result is different. As you can see from the one polynomial in one variable case. $\endgroup$
    – Alexandre Eremenko
    Oct 21, 2014 at 23:17

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.