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
36 events
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| S Aug 27, 2013 at 21:52 | history | bounty ended | Vesselin Dimitrov | ||
| S Aug 27, 2013 at 21:52 | history | notice removed | Vesselin Dimitrov | ||
| Aug 27, 2013 at 15:02 | comment | added | Vesselin Dimitrov | @ Noam Elkies: Thank you! It was of indeed the example of $\binom{n}{k}$ which led me to ask question 2. (Otherwise I was considering rather different kinds of stuff, more in line with the first question.) | |
| Aug 27, 2013 at 14:44 | comment | added | Noam D. Elkies | @Vesselin Dimitrov: Your contribution is not minimal; not only is asking a good question valuable in itself, but you also gave the example showing $A \leq 2$ which suggested that this "Question 2" has quite a different flavor from "Question 1" on ${\bf Z}[x]$. | |
| Aug 27, 2013 at 10:16 | comment | added | Vesselin Dimitrov | (cont.) This generalizes in an obvious way to global fields and to algebraic curves of higher genus. It would thus be a sharp algebraicity criterion, of which I can only prove a weaker version involving an inequality stronger than 1). It was this type of problems which motivated me to ask question 1. | |
| Aug 27, 2013 at 10:12 | comment | added | Vesselin Dimitrov | (cont.) is a rational fraction of degree $\leq h_p$. Then: 1) If $\log{\rho} + \liminf_n \frac{1}{n} \sum_{p : h_p < n} \log{p} > 0$, the set $S(G,h)$ should only contain rational functions in $\mathbb{Z}[[x]]$. 2) If however the inequality in 1) is not satisfied, the set $S(G,h)$ should be uncountable. | |
| Aug 27, 2013 at 9:55 | comment | added | Vesselin Dimitrov | Here is a generalization of Ruzsa's conjecture which, though not involving polynomial coefficients anymore, is nonetheless in the spirit of the two questions. It is suggested to me by the connection with Fekete's theorem (and by that of Polya-Carlson-Bertrandias). Let $G \ni 0$ be a pointed (say) simply connected domain, and $\rho$ its conformal mapping radius. Let $h : P \to \mathbb{N}_0 \cup \{\infty\}$ be a function on the set $P$ of primes, and consider the set $S(G,h)$ of those $f \in \mathbb{Z}[[x]]$ which are meromorphic on $G$ and whose reduction $f \mod{p}$ at each prime $p$ (cont.) | |
| Aug 27, 2013 at 9:35 | comment | added | Vesselin Dimitrov | @ Noam Elkies: I think so; the relation to Fekete's theorem which you and David discovered seems particularly interesting to me. My contribution here, though, is minimal and reduces to having merely asked the question. | |
| Aug 26, 2013 at 23:20 | comment | added | Noam D. Elkies | Looking beyond Mathoverflow, is it worth writing this up for publication as a joint paper? | |
| Aug 26, 2013 at 23:20 | comment | added | Noam D. Elkies | Ah, I misunderstood. I'm entirely fine with David getting double my bounty, since his answer is definitive (besides being a double answer - both an upper and a lower bound which "happen" to coincide!). | |
| Aug 26, 2013 at 22:10 | comment | added | Vesselin Dimitrov | Well, actually, he would have to donate it to you - for I already awarded you the old 100 bounty. Then I intended to set a new 100 bounty for him, but it turned out this wasn't an option, so I went for 200. (The latter hasn't yet been awarded, so if I award it to you instead, what you say could be an option if you set a 150-point bounty, if that's possible, for his answer :). | |
| Aug 26, 2013 at 21:59 | comment | added | Noam D. Elkies | Hm, does this mean I should somehow donate half of the 200-point bounty to David Speyer? I suppose I can achieve this purpose by posting a new question "What's the optimal $A$?", putting a 100-point bounty on it, and accepting David's answer which would be basically a link or pointer here... | |
| S Aug 26, 2013 at 21:50 | history | bounty started | Vesselin Dimitrov | ||
| S Aug 26, 2013 at 21:50 | history | notice added | Vesselin Dimitrov | Reward existing answer | |
| S Aug 26, 2013 at 21:42 | history | bounty ended | Vesselin Dimitrov | ||
| S Aug 26, 2013 at 21:42 | history | notice removed | Vesselin Dimitrov | ||
| Aug 26, 2013 at 16:23 | answer | added | David E Speyer | timeline score: 45 | |
| Aug 26, 2013 at 4:05 | vote | accept | Vesselin Dimitrov | ||
| Aug 26, 2013 at 1:54 | answer | added | Noam D. Elkies | timeline score: 47 | |
| Aug 26, 2013 at 1:22 | comment | added | Michael Zieve | Also, there are only finitely many integer-valued $f$'s of any fixed degree $k$ which satisfy $|f(n)|<2^n$ for all $n$, since if we write $f(x)=\sum_{i=0}^k a_i \binom{x}i$ with $a_i\in\mathbf{Z}$ then $|f(0)|<2^0$ bounds $|a_0|$, after which $|f(1)|<2^1$ bounds $|a_1|$, and so on. So Question 2 is really asking whether there exist $f$'s of arbitrarily large degree. | |
| Aug 26, 2013 at 1:02 | comment | added | Michael Zieve | Regarding question 2: when $k$ is sufficiently large we can't take $f(x)=\binom{x}k$, since $\binom{2k}k$ is asymptotic to $2^{2k}/\sqrt{\pi k}$ as $k\to\infty$. In general if $f(x)$ is integer-valued of degree $k$, with $k$ sufficiently large, then the inequalities $|f(n)|<A^n$ with $n$ near $2k$ will impose huge constraints on $f$. Maybe somehow one can use them to show that there are no $f$'s of suff.large degree? | |
| Aug 25, 2013 at 15:06 | history | edited | Vesselin Dimitrov | CC BY-SA 3.0 |
deleted 510 characters in body; edited title
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| S Aug 25, 2013 at 15:01 | history | bounty started | Vesselin Dimitrov | ||
| S Aug 25, 2013 at 15:01 | history | notice added | Vesselin Dimitrov | Improve details | |
| Aug 12, 2013 at 21:15 | history | edited | Vesselin Dimitrov | CC BY-SA 3.0 |
Last edit: Gamma(t) had to be changed by n!, as I had neglected that Gamma(n) = (n-1)! rather than n!.
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| Aug 12, 2013 at 21:00 | history | edited | Vesselin Dimitrov | CC BY-SA 3.0 |
In the follow-up question, "inf" had to be replaced with "sup." My apology for being careless here.
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| Aug 12, 2013 at 20:39 | history | edited | Vesselin Dimitrov | CC BY-SA 3.0 |
Correction: ...does *not* appear to be easy to construct...
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| Aug 12, 2013 at 20:11 | history | edited | Vesselin Dimitrov | CC BY-SA 3.0 |
A follow-up on David Speyer's solution.
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| Aug 12, 2013 at 20:04 | history | edited | Vesselin Dimitrov | CC BY-SA 3.0 |
A follow-up on David Speyer's solution.
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| Aug 12, 2013 at 18:57 | comment | added | Vesselin Dimitrov | Indeed, and perhaps the title should now be changed to: "Are there infinitely many integer-valued polynomials dominated by $1.9^n$ on all of $\mathbb{N}$?" | |
| Aug 12, 2013 at 18:35 | comment | added | Harry Altman | It's probably worth noting here the fact that any integer-valued polynomial is an integral linear combination of the polynomials $\binom{x}{k}$. | |
| Aug 12, 2013 at 17:28 | answer | added | David E Speyer | timeline score: 55 | |
| Aug 11, 2013 at 13:10 | comment | added | Paul Siegel | Oops! Groggy mistake indeed... | |
| Aug 11, 2013 at 13:06 | comment | added | Vesselin Dimitrov | It is not smaller in absolute value, though. Your polynomial has $|f_k(1)| = m^k - 1 > e$. | |
| Aug 11, 2013 at 12:53 | comment | added | Paul Siegel | For any fixed $k$, $n^k < e^n$ for all $n$ sufficiently large - say, larger than $m$. It follows that $f_k(n) := n^k - m^k$ is smaller than $e^n$ for every $n$ and $k$. Unless I'm making a groggy mistake... | |
| Aug 11, 2013 at 8:47 | history | asked | Vesselin Dimitrov | CC BY-SA 3.0 |