Timeline for Partitions of $\mathbb{R}^d$ by implicit polynomial equations
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
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| S Apr 24, 2017 at 12:46 | history | suggested | Rodrigo de Azevedo |
Added tag to question
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| Apr 24, 2017 at 12:36 | review | Suggested edits | |||
| S Apr 24, 2017 at 12:46 | |||||
| Apr 24, 2017 at 12:22 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
Image links broken; now fixed.
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| Sep 9, 2012 at 1:47 | answer | added | Saugata Basu | timeline score: 7 | |
| Sep 8, 2012 at 15:40 | comment | added | Igor Rivin | See the edit to my response for a sharper (in fact, I am pretty sure it is sharp) bound. | |
| Sep 8, 2012 at 10:19 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
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| Sep 8, 2012 at 3:17 | comment | added | Aaron Meyerowitz | I'm pretty sure it is $56$ and not $58$ regions. I may have heard once that this is optimal. You lose something in not having curved borders which can multiply intersect but gain more in having lots of linear borders heading off in many directions. Look at the quintic factor $y-T_5(x).$ Wouldn't you rather straighten the rounded turns and replace them with $5$ lines going off to infinity? The $\pm \epsilon$ adjustment gives a prettier picture but then the number of regions is $n^2+O(n).$ | |
| Sep 7, 2012 at 11:59 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
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| Sep 7, 2012 at 7:53 | answer | added | Aaron Meyerowitz | timeline score: 1 | |
| Sep 7, 2012 at 2:46 | answer | added | Igor Rivin | timeline score: 7 | |
| Sep 7, 2012 at 0:42 | comment | added | Pietro Majer | If $p$ is a product of $k$ affine functions in $\mathbb{R}^d$, its zero-set is the union of $k$ affine hyperplanes. If these are in generic position, the number of components of the co-set is the well-known $\sum_{j=0}^d\binom kj$. I'd conjecture this is not worse than any polynomial of degree $k$. Note that $2n$ lines disconnect the plane into $2n^2+n+1\ge n^2+3$ components. | |
| Sep 7, 2012 at 0:15 | answer | added | Matthew Badger | timeline score: 7 | |
| Sep 6, 2012 at 23:53 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
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| Sep 6, 2012 at 23:15 | comment | added | Greg Martin | An observation, which can probably be generalized: let $T_n$ be the $n$th Chebyshev polynomial of the first kind. For $n$ odd, consider the polynomial $(y-T_n(x))(x-T_n(y))$. The graph of the zero locus of this degree-$2n$ polynomial splits the plane into $n^2+3$ pieces (4 unbounded pieces and $n^2-1$ bounded pieces). The bounded pieces come up because the graph of $y-T_n(x)=0$ oscillates up and down $n$ times inside the unit box $[-1,1]^2$, while $x-T_n(x)$ oscillates $n$ times side to side. See: en.wikipedia.org/wiki/Chebyshev_polynomials#Roots_and_extrema | |
| Sep 6, 2012 at 21:10 | history | asked | Joseph O'Rourke | CC BY-SA 3.0 |