Timeline for Are there infinitely many integer-valued polynomials dominated by $1.9^n$ on all of $\mathbb{N}$?
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 26, 2013 at 21:42 | history | bounty ended | Vesselin Dimitrov | ||
| Aug 26, 2013 at 19:55 | comment | added | Noam D. Elkies | For $m$ large, the average of $|f_m(z)|$ over $|z|=r$ is $o(1)$ because $|f_m|$ behaves like the $m$-th power of a function that's at most $1$ but attains that value at only finitely many points. (Most likely $O(m^{-1/2})$ unless for some reason $1 - \lim_{m\rightarrow\infty} |f_m|^{1/m}$ has a quadruple or higher-order zero at some point of $|z|=r$.) | |
| Aug 26, 2013 at 19:44 | comment | added | David E Speyer | Question: This constructs a constant $C$ and infinitely many integer values polynomials with $|f_m(n)|<C r^{-n}$. Clearly, at the cost of increasing $r$ infinitesimally, we can find an $N$ so that we get $|f_m(n)| < r^{-n}$ for $n>N$. But how do we guarantee ourselves $|f_m(n)|<r^{-n}$ for all $n$? | |
| Aug 26, 2013 at 18:41 | comment | added | David E Speyer | @FelipeVoloch It looks like $|f(2n)|$ is $(6n)! (2n)!/((4 n)! (3n)! n!)$, see oeis.org/A211419 . | |
| Aug 26, 2013 at 18:12 | comment | added | Felipe Voloch | The integers $f(n)$, when $n = 3m$ seem to be divisible by all primes between $2m$ and $4m$, which leads to a Chebyshev estimate slightly better than the one from the middle binomial coefficient. I haven't looked at David's new example but this suggests that getting $A$ down to $1$ might involve the PNT. | |
| Aug 26, 2013 at 15:22 | comment | added | David E Speyer | This is the opposite of mathoverflow.net/questions/32961, which aims to find monic polynomials $f$ and $g$ so that $\{ |f(z)|=|g(z)| \}$ stays as close to $0$ as possible. | |
| Aug 26, 2013 at 14:46 | comment | added | David E Speyer | Let me spell out what I think Noam is implying. Let $p(x)$ be a polynomial with integer coefficients and degree $d$. Let $f_m(n)$ be the coefficient of $x^n$ in $p(x)^m/(1-x)^{dm+1}$. Then $f_m$ is an integer valued polynomial. If $|p(x)/(1-x)^d| \leq 1$ on the circle $|x| = r$, then integrating on this contour gives a bound $|f_m(n)| \leq C r^{-n}$. So our goal is to find polynomials $p$ which can push the radius $r$ as far out as possible. | |
| Aug 26, 2013 at 6:23 | comment | added | Vesselin Dimitrov | (And I meant, after all, to write $\mathbb{Z}[1/(1-x)] \subset \mathbb{Z}[[x]]$ and $f \in \mathbb{Z}[[x]]$ in the above comment. Sorry about this, as well as for the switch of notation, $f$ now denoting the generating function.) | |
| Aug 26, 2013 at 5:40 | comment | added | Vesselin Dimitrov | (continued.) And a closer look at their argument shows that not only the order of the differential equation can be taken to depend only on the radius of convergence, but that if the coefficients satisfy $|a(n)| < A^n$ for all $n \geq n_0$ and $A < e$, then in fact $f$ satisfies one of a finite set (depending on $A$ and $n_0$) of linear homogeneous differential equations. This implies that the degree of a polynomial in Question 1 is bounded for $A < e$. | |
| Aug 26, 2013 at 5:36 | comment | added | Vesselin Dimitrov | Not an application, but it came up in relation to an old conjecture of Ruzsa characterizing the subring $\mathbb{Z}[1/(1-x)] \subset \mathbb{Q}[[x]]$ by the two properties: 1) radius of convergence (and I would add: meromorphy) is strictly $> 1/e$; and 2) for a set of primes $p$ of full density, the mod $p$ reduction is $A_p(x)/(1-x)^p$, with $A_p$ a polynomial of degree $< p$. Perelli and Zannier have shown that such an $f \in \mathbb{Q}[[x]]$ is at least $D$-finite. (continued.) | |
| Aug 26, 2013 at 4:51 | comment | added | Noam D. Elkies | Was there a motivating application for either question? (It's a natural problem in any case but it would be nice if the answer has some further use.) | |
| Aug 26, 2013 at 4:47 | comment | added | Noam D. Elkies | Thanks. Since $25 \geq 2 \cdot 11$ (and $11 \geq 10$) this also gives David Speyer the rare "Populist" badge, which is his third gold and which David may actually prefer to the bounty points :-) | |
| Aug 26, 2013 at 4:30 | comment | added | Vesselin Dimitrov | Thanks also to David Speyer for his solution to Question 1. But the bounty was offered for Question 2, so I'm marking this as the accepted answer. | |
| Aug 26, 2013 at 4:29 | comment | added | Vesselin Dimitrov | So I was wrong on both questions :-). Since you believe the limit infimum [of values of $A$] to be nonetheless strictly $> 1$, interesting what the actual value would be. Regarding integer polynomials the right question would have been: what is the limit infimum $\delta$ of the values of $c > 0$ such that there are infinitely many $f \in \mathbb{Z}[t]$ with $|f(n)|/n! < c$ for all $n \in \mathbb{N}$? We know from David Speyer's solution that $\delta \in [1/2,1]$. | |
| Aug 26, 2013 at 4:05 | vote | accept | Vesselin Dimitrov | ||
| Aug 26, 2013 at 1:54 | history | answered | Noam D. Elkies | CC BY-SA 3.0 |