Timeline for Geometry of Level sets of elliptic polynomials in two real variables
Current License: CC BY-SA 4.0
42 events
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| Mar 27, 2020 at 11:25 | vote | accept | Ali Taghavi | ||
| Mar 25, 2020 at 17:30 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Mar 25, 2020 at 14:27 | history | bounty ended | Ali Taghavi | ||
| Mar 24, 2020 at 23:51 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Mar 24, 2020 at 22:17 | comment | added | Iosif Pinelis | @AliTaghavi : Also, in general the level sets $P^{-1}((-\infty,c])$ will not be convex, even if $P$ is an elliptic homogeneous polynomial. E.g., take $P(x,y)=(x - y)^2 (x + y)^2 + (x^4 + y^4)/10$. | |
| Mar 24, 2020 at 21:34 | comment | added | Iosif Pinelis | @AliTaghavi : I did use an additional convexity assumption in an earlier version of the answer, but later was able to remove it. Indeed, no such assumption is needed to obtain the limit of the centroid. | |
| Mar 24, 2020 at 21:31 | comment | added | Iosif Pinelis | @AliTaghavi : I have now added details on why without loss of generality $p'(x_\pm)\ne0$ for all $p\in\mathcal P_{n,d_*,D}$ and all real $c\ge c_*(n,d_*,D)$ -- this statement is now formula (1.5). | |
| Mar 24, 2020 at 21:26 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Mar 24, 2020 at 2:23 | comment | added | Ali Taghavi | BTW do we realy need the convexity? And is it realy the case?Each line passing irigin intersect the level set at exactly 2 points but it does not necessarily imply that it is convex, right? So is the theorem which I wrote in the first paragraph completly true? Should not we remive the word convex? | |
| Mar 24, 2020 at 1:52 | comment | added | Ali Taghavi | I mean is it obvious that we may assume that c is regular value UNIFORMLY independent of choosing coefficient in $\mathcal{p} (n,d_*, D)$? | |
| Mar 24, 2020 at 1:31 | comment | added | Ali Taghavi | Thank you for your comments. Is there an easy argument for "wlog we may assume "c" is a regular value that is $p'(r^+)$ and $p'(r^-)$ are non zero. You uses this fact in 2 variable setting.(Sòrry if my question is elementary). | |
| Mar 24, 2020 at 0:05 | comment | added | Iosif Pinelis | @AliTaghavi : I am glad you liked the answer. Please let me know if more clarifications are needed in some places. | |
| Mar 23, 2020 at 15:12 | comment | added | Ali Taghavi | I appreciate again your very onteresting answer. I read it several times I am finalizong my full understanding. | |
| Mar 8, 2020 at 12:27 | comment | added | Ali Taghavi | Thank you so much! | |
| Mar 8, 2020 at 3:12 | comment | added | Iosif Pinelis | @AliTaghavi : I have added details on why the $c$-level curve for an elliptic polynomial is necessarily simple and closed for all large enough $c>0$. I have not stated or used any convexity properties. | |
| Mar 8, 2020 at 3:07 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Mar 6, 2020 at 21:49 | comment | added | Ali Taghavi | Can you please give a reason? The cordinate (sin t, Cost) implies the convexity? Does not? Just other question why it is not possible that large level sets would be open curves? From Morse theorem and also Bezout theorem we know that for a generic polynomial p, all large level sets are homeomorphic to each other but it does not guarantee that they are closed curv, yes? | |
| Mar 5, 2020 at 19:00 | comment | added | Iosif Pinelis | @AliTaghavi : For all large enough $c$, the $c$-level curve of $P$ is $\{R(t)(\cos t,\sin t)\colon t\in[0,2\pi)\}$, where $R(t):=r_+(t)$ for $t\in[0,\pi)$ and $R(t):=r_-(t-\pi)$ for $t\in[\pi,2\pi)$. So, the level curve is closed and simple, and there is no need to additionally assume that. As for the centroid, this is the way I understand it; is there any other way? | |
| Mar 5, 2020 at 15:16 | comment | added | Ali Taghavi | I really wish to understand your answer but I admit it is a little complicated. BTW what (equivalent formulation of centroid you are using? $\iint (x,y)dxdy$ divided by area? | |
| Mar 5, 2020 at 15:14 | comment | added | Ali Taghavi | Yes I see. Please give me a few days to underestand your interesting answer. But just a question: Do you think my assumption that for c sufficiently large the preimage of c is simple clised curve is redandant? | |
| Mar 5, 2020 at 3:31 | comment | added | Iosif Pinelis | @AliTaghavi : The convexity is not needed because we now have the uniformity: (1) holds uniformly over all polynomials in $\mathcal P_{n,d_*,D}$ if $c\ge c_*(n,d_*,D)$. So, if $c>0$ is large enough, then for all $t\in[0,2\pi)$ at once the polynomials $P(r\cos t,r\sin t)-c$ in $r$ have exactly two roots, $r_\pm(t)$, satisfying condition (2a). I have now inserted the previously missing qualification "for all large enough $c>0$" into the sentence "The main idea for the elliptic polynomial case is ...". | |
| Mar 5, 2020 at 3:27 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Mar 5, 2020 at 0:14 | comment | added | Ali Taghavi | To be honnest i have some question about your answer. But this is my first question: in this new version of your answer, how do not you need the convexity assumption? Is not possible that a ray intersect our closed curve in several points? | |
| Mar 1, 2020 at 17:35 | comment | added | Iosif Pinelis | Please let me know where details are needed the most. Since the answer is already very long, I felt it necessary to omit some of the details. | |
| Mar 1, 2020 at 13:53 | comment | added | Ali Taghavi | Once again I thank you very much for your answer I try to understand its details. | |
| Mar 1, 2020 at 13:34 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Mar 1, 2020 at 11:37 | history | bounty ended | Ali Taghavi | ||
| Mar 1, 2020 at 1:32 | comment | added | Iosif Pinelis | I have removed the "eventual" convexity condition (which was previously imposed for elliptic polynomials). | |
| Mar 1, 2020 at 1:31 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Feb 28, 2020 at 22:44 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Feb 28, 2020 at 21:00 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Feb 28, 2020 at 20:53 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Feb 28, 2020 at 20:38 | comment | added | Iosif Pinelis | I have added the case of elliptic polynomials. | |
| Feb 28, 2020 at 20:37 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Feb 28, 2020 at 14:44 | comment | added | Ali Taghavi | It would be interesting if we could prove that the centroid of level set of a (somewhat) elliptic polynomial is convergent and the cordinates of limiting centroid can be expreced in terms of coefficients of polynomial. (May be in terms of coefficients of degree 2n and 2n-1 and not lower degree. This is really the case for one variable) | |
| Feb 28, 2020 at 14:24 | comment | added | Ali Taghavi | They are not equivalent but what I was thinking about initially is the first one. | |
| Feb 28, 2020 at 14:22 | comment | added | Ali Taghavi | mathoverflow.net/questions/117437/… | |
| Feb 28, 2020 at 14:20 | comment | added | Ali Taghavi | I am not sure my terminoligy is good.(please see the next comment too. But what was in mind is the following: The last homogeneous part is positive (or negative) on punctured plane. Example $x^4+y^4+ \text{lower terms}$. In fact every polynomial whose corresponding pde is elliptic. | |
| Feb 28, 2020 at 13:34 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Feb 28, 2020 at 13:30 | comment | added | Iosif Pinelis | @AliTaghavi : How do you define an elliptic polynomial? | |
| Feb 28, 2020 at 9:58 | comment | added | Ali Taghavi | Thanks for your answer. What about an elliptic polynomial? | |
| Feb 28, 2020 at 5:46 | history | answered | Iosif Pinelis | CC BY-SA 4.0 |