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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. |