Timeline for Partitions of $\mathbb{R}^d$ by implicit polynomial equations
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Sep 7, 2012 at 11:52 | comment | added | Joseph O'Rourke | (And, as Aaron pointed out, $d=10$, not $6$ as I claimed above.) | |
| Sep 7, 2012 at 8:01 | comment | added | Pietro Majer | @Matthew: isn't your latter bound shifted by one? For a 2-variables real polynomial of degree $d$, I'd say at least $d(d-1)/2+1$ components in the real projective plane (and $d(d+1)/2+1$ in the affine plane), which is reached if the polynomial is a product of linear terms (see my comment above). | |
| Sep 7, 2012 at 3:32 | comment | added | Matthew Badger | @Joseph: Oops! I confused components of the curve and components of the complement of the curve. I've updated the answer to fix this. | |
| Sep 7, 2012 at 3:28 | history | edited | Matthew Badger | CC BY-SA 3.0 |
Earlier post confused components of the curve and components of the complement of the curve. Now corrected.; added 5 characters in body; deleted 16 characters in body
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| Sep 7, 2012 at 1:28 | comment | added | Joseph O'Rourke | @Matthew: Thanks! I didn't know this theorem. But I am confused because the Chebyshev example I posted has $d=6$, and so should have at most $11$ regions, but I count $24$ bounded regions... | |
| Sep 7, 2012 at 0:21 | history | edited | Matthew Badger | CC BY-SA 3.0 |
added 24 characters in body; edited body; deleted 4 characters in body
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| Sep 7, 2012 at 0:15 | history | answered | Matthew Badger | CC BY-SA 3.0 |