Timeline for bounding roots of a polynomial with Rouche's Theorem
Current License: CC BY-SA 2.5
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 24, 2010 at 22:00 | comment | added | Harald Hanche-Olsen | @FC: Yes, I realized that after asking. See my answer. (I am going to delete my above comments now, since they have more or less become parts of the answer. Late comers to the discussion may notice that FC did indeed answer a question of mine, now gone.) | |
| Feb 24, 2010 at 21:34 | answer | added | Harald Hanche-Olsen | timeline score: 4 | |
| Feb 24, 2010 at 16:10 | comment | added | Michael Lugo | Note that f(z)*(z-1) = z^{n+1} - (1+k) z^n + k. This polynomial has the same roots as f(z), plus a root at z = 1. This might be a better function to look at. | |
| Feb 24, 2010 at 15:34 | answer | added | Gabriel Benamy | timeline score: 0 | |
| Feb 24, 2010 at 15:33 | history | edited | Steve Huntsman |
retag
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| Feb 24, 2010 at 15:24 | history | asked | Josh | CC BY-SA 2.5 |