Timeline for Convex polynomial homogenization and convexity
Current License: CC BY-SA 2.5
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 26, 2011 at 14:57 | vote | accept | Will | ||
| May 26, 2011 at 14:57 | answer | added | Will | timeline score: 1 | |
| May 26, 2011 at 14:53 | vote | accept | Will | ||
| May 26, 2011 at 14:57 | |||||
| May 9, 2011 at 17:30 | answer | added | Amir Ali Ahmadi | timeline score: 6 | |
| Mar 3, 2011 at 18:11 | comment | added | Deane Yang | Will, I agree that there should be a direct proof but I don't know it. With the Hessian, I don't recall all the details offhand, but I believe that you can show that the last diagonal term dominates the two terms arising from the last row and column. | |
| Mar 3, 2011 at 16:36 | comment | added | Will | Hi Deane, thanks for your response. I would like to prove convexity directly by invoking a result (that I presume to exist somewhere) about convexity of homogenized convex polynomials. This would be the most elegant way to show it I think. But I have also tried to show that x'Hx is >= 0 for the homogeneous function, which of course can be decomposed into a sum involving the Hessian of the inhomogeneous function plus the column/row for the variable a, but I was not successful in doing that, because it was not clear that the inner products involving x and that column+row were > 0. | |
| Mar 3, 2011 at 16:25 | comment | added | Deane Yang | But are you trying to prove convexity directly or by computing the Hessian of the homogeneous function in terms of the Hessian of the inhomogeneous function. It seems to me that the latter is a straightforward computation. | |
| Mar 3, 2011 at 14:52 | comment | added | Deane Yang | Keep trying. It should work. | |
| Mar 3, 2011 at 14:19 | history | edited | Will | CC BY-SA 2.5 |
added 560 characters in body; edited body
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| Mar 3, 2011 at 13:32 | history | asked | Will | CC BY-SA 2.5 |