2
$\begingroup$

Take a polynomial $p \in \mathbb{Z}[x_1, \dotsc, x_k],$ and consider the values of $p(a_1, \dotsc, a_k)$ as $\mathbf{a}$ (the vector of the $a_i$) ranges over, say, the integer lattice points in the ball of radius $R.$ What is known about how the values of $p$ are distributed, under some reasonable assumption (for example, if $p$ is irreducible)?

$\endgroup$
  • $\begingroup$ What do you mean by how the variables are distributed? Presumably we know the distribution well on a large scale but not so well on a short scale. $\endgroup$ – Will Sawin Jan 15 '17 at 14:36
  • $\begingroup$ @WillSawin I was intentionally vague. For example, a simple question would be whether the map is onto some interval $[-N, N].$ This is not true if the polynomial is, say a square, and presumably not if it's reducible (you are not going to hit many primes), but otherwise?! This is the "short scale" question, I assume. A simpler (I assume) question is: how many distinct values are assumed, and the easiest (as you say) is what does the probability distribution is like when you rescale (but I don't know the answer to even that). $\endgroup$ – Igor Rivin Jan 15 '17 at 15:53
  • $\begingroup$ At the highest scale, I think one can show the value distribution, rescaled by an appropriate power of $R$, converges as $R$ goes to $\infty$ to the value distribution of the leading term of $p$ on a ball of radius $1$. There shouldn't be any simpler characterization of this measure. $\endgroup$ – Will Sawin Jan 15 '17 at 20:19
  • $\begingroup$ @WillSawin That will be the law of large numbers, but there should be a limit theorem, as well... $\endgroup$ – Igor Rivin Jan 15 '17 at 21:10
  • $\begingroup$ I don't understand the distinction you're drawing between a law of large numbers and a limit theorem. One wants to subtract off the main term I've just described and measure the error term, I guess. But here the main term is a limiting probability distribution, not a limiting value as in the law of large numbers. So the error term could be the error in this probability distribution, but then it would not itself be a probability distribution. $\endgroup$ – Will Sawin Jan 15 '17 at 22:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.