Timeline for Every positive polynomial with rational coefficients is above a completely Q-factorized nonnegative polynomial ?
Current License: CC BY-SA 3.0
18 events
| when toggle format | what | by | license | comment | |
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| Feb 5, 2012 at 4:16 | comment | added | Brendan McKay | (some editing overlap there ;) | |
| Feb 5, 2012 at 4:16 | comment | added | Brendan McKay | To answer my own objection, the coefficient of $x^{2d-2}$ in $P-\prod (x-a_i)^2$ is precisely $\sum b_i$, which is positive. | |
| Feb 5, 2012 at 4:13 | comment | added | Will Sawin | The difference you're forming is $\prod ((x-a_i)^2+b_i)-\prod (x-a_i)^2$, so the coefficient is $\sum b_i>0$. | |
| Feb 5, 2012 at 4:00 | comment | added | Brendan McKay | I think there's a gap in the proof. The minimum of $(P-\prod (x-a_i)^2)/(1+x^{2d-2})$ might be 0, occurring at $x\to\infty$, so there might be no space for an open set where it is positive. I think it will be ok if the coefficient of $x^{2d-2}$ in $P-\prod (x-a_i)^2$ is not zero, but is it? | |
| Feb 4, 2012 at 22:14 | vote | accept | Ewan Delanoy | ||
| Feb 4, 2012 at 22:14 | vote | accept | Ewan Delanoy | ||
| Feb 4, 2012 at 22:14 | |||||
| Feb 4, 2012 at 22:11 | vote | accept | Ewan Delanoy | ||
| Feb 4, 2012 at 22:13 | |||||
| Feb 4, 2012 at 20:03 | comment | added | Will Sawin | I edited it to include a uniformity argument. | |
| Feb 4, 2012 at 20:02 | history | edited | Will Sawin | CC BY-SA 3.0 |
added 497 characters in body
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| Feb 4, 2012 at 16:17 | comment | added | Ewan Delanoy | (precedent comment, continued) This does not happen because, roughly speaking, polynomials are very regular objects and we have uniform bounds for everything. I'm still thinking about how to show this rigorously with a minimum of notation. | |
| Feb 4, 2012 at 16:16 | comment | added | Ewan Delanoy | @Will : I am now convinced that your construction works, but I still think that your proof that it works is incomplete because there's a "uniformity" argument lacking. What I mean is that we might have $P-(\prod_{k}(x-d_k))^2$ nonnegative everywhere except on an interval $[N,N+1]$, say, with $N$ tending to infinity as the $d_k$'s gets closer to $a_k$'s. So that the construction would work both for small x and for large x, but not in-between. | |
| Feb 3, 2012 at 23:41 | comment | added | Will Sawin | When you have a bit of breathing room, my two constructions do not interfere. The small-$X$ construction is a ball around the irrational points, the large-$x$ is a hyperplane through them. The ball and the hyperplane must have nontrivial rational intersection. | |
| Feb 3, 2012 at 23:39 | comment | added | Will Sawin | You're misinterpreting "small" and "large" here. In that case, since $P(x)$ is not positive, there is no construction that works for small $x$ because small $x$ means $x$ near $\sqrt{2}$. Similarly, there is no construction that works for large $x$ because large $x$ means large $|x|$. | |
| Feb 3, 2012 at 21:19 | comment | added | Ewan Delanoy | Of course, $P(x)=(x-\sqrt{2})^2$ does not satisfy my hypotheses, but it illustrates my point : you can find a $q_1$ such that $|x-\sqrt{2}| \geq |x-q_1|$ for all small $x$, and another $q_1$ for large $x$. But one cannot find a $q_1$ that works for all $x$ at once. | |
| Feb 3, 2012 at 21:17 | comment | added | Ewan Delanoy | Your construction is essentially a local one - you can obtain a set of roots that work for all small $x$, and another that work for all large $x$. But those two constructions get into each other's way. | |
| Feb 3, 2012 at 19:29 | comment | added | Will Sawin | That polynomial does not satisfy the axiom $P(x)>0$. Getting it to hold for small $x$ is easy - if you don't change the roots very much, you don't change the polynomial very much, and since there must be a buffer of positive size between $P$ and the polynomial with roots, which will be enough for all small $x$. | |
| Feb 3, 2012 at 19:13 | comment | added | Ewan Delanoy | I already thought of your idea, but unfortunately it does not work. The problem is that your construction will yield a case where () holds for "almost all" x (i.e. when $x$ is large enough), but not all x, and mathematicians do mind this ... Also, note that () does not hold when $P$ has an irrational root : e.g. if $P=(x-\sqrt{2})^2$, there is no $q_1$ such that $P \geq (x-q_1)^2$ for all $x$. | |
| Feb 3, 2012 at 18:38 | history | answered | Will Sawin | CC BY-SA 3.0 |