Timeline for Estimating the volume of a region bounded by polynomial inequalities
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 22, 2017 at 10:48 | vote | accept | Stanley Yao Xiao | ||
| Sep 14, 2017 at 1:52 | answer | added | fedja | timeline score: 3 | |
| Sep 13, 2017 at 9:05 | comment | added | fedja | Then $\rho=2$ (easy Analysis) and, if you invoke Arithmetic (a harder subject, so I may have a mistake here), then you get $\frac 89\sqrt X\log^2 X(1+o(1))$ as $X\to+\infty$. My general advice is to never even think of using Calculus. It is an impossibly hard subject; even the simplest Calculus exercises like finding $\int e^{x^2}\,dx$ or $\min_{x} (x^2+e^{-x})$ are as hard as the three famous problems of antiquity.together. I wonder why we still torture young minds with it instead of teaching elementary Analysis. Anyway, I'll post the details when I finally wake up :-) | |
| Sep 13, 2017 at 6:19 | comment | added | Stanley Yao Xiao | @fedja I am specifically interested in the case when $Q(x,y,z) = z^2 - 4xy$. | |
| Sep 13, 2017 at 2:40 | comment | added | fedja | Still not enough information for a unique answer. It is, indeed, of the kind you guessed but $\rho$ can be both $1$ and $2$ depending on the form. If you have some particular (class of) forms in mind, you'd better tell us what exactly it is. | |
| Sep 11, 2017 at 18:00 | comment | added | David G. Stork | If you merely want to "estimate the order of magnitude" of the bounded region (and not, say, derive a formula), then I recommend you do a simple stochastic simulation, choosing random points in a volume and testing whether they satisfy the inequalities. | |
| Sep 11, 2017 at 16:29 | history | edited | Stanley Yao Xiao | CC BY-SA 3.0 |
added 51 characters in body
|
| Sep 11, 2017 at 16:28 | comment | added | Stanley Yao Xiao | I have not considered this type of degenerate case, but in the cases I am interested in $z^2$ appears with non-zero coefficient, so this type of degeneracy is avoide. I will add that assumption to the question | |
| Sep 11, 2017 at 16:20 | comment | added | Fedor Petrov | Ok, what about $x^2+z(x+y)$? If $x=2,y=-2$ and large $z$ it is bounded. | |
| Sep 11, 2017 at 15:56 | comment | added | Stanley Yao Xiao | @Fedor Petrov importantly $x,y$ are bounded away from zero; so the $yz$ term cannot be eliminated. Now suppose $z$ is arbitrarily large. Then since $x^2 + yz$ is small, it follows that $x^2$ has to cancel $yz$ (which is at least as big as $|z|$ in absolute value)... but $x^2$ and s bounded, so this leads to a contradiction. | |
| Sep 11, 2017 at 15:53 | comment | added | Fedor Petrov | Ok, then why if $x,y$ and $x^2+yz$ are bounded, so is $z$? I do not think so. | |
| Sep 11, 2017 at 15:24 | comment | added | Stanley Yao Xiao | @Fedor Petrov it just means that it is irreducible over $\mathbb{C}$. In fact one can assume in addition that $Q$ is also indefinite. | |
| Sep 11, 2017 at 15:21 | comment | added | Fedor Petrov | What is geonetrically irreducible quadratic form? Say, is $x^2+yz$ such? | |
| Sep 11, 2017 at 13:32 | history | asked | Stanley Yao Xiao | CC BY-SA 3.0 |