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comment Intuitive and/or philosophical explanation for set theory paradoxes
Interesting. "Empty" is not the same thing as "vicious" indeed.
Feb
4
accepted Every positive polynomial with rational coefficients is above a completely Q-factorized nonnegative polynomial ?
Feb
4
comment Every positive polynomial with rational coefficients is above a completely Q-factorized nonnegative polynomial ?
(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
comment Every positive polynomial with rational coefficients is above a completely Q-factorized nonnegative polynomial ?
@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
revised Every positive polynomial with rational coefficients is above a completely Q-factorized nonnegative polynomial ?
corrected spelling in title
Feb
3
comment Every positive polynomial with rational coefficients is above a completely Q-factorized nonnegative polynomial ?
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
comment Every positive polynomial with rational coefficients is above a completely Q-factorized nonnegative polynomial ?
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
comment Every positive polynomial with rational coefficients is above a completely Q-factorized nonnegative polynomial ?
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 $x$.