Timeline for Possible application of Rouche's theorem to aproblem of complex roots of polynomials
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
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| Sep 26, 2011 at 20:09 | comment | added | Luis H Gallardo | dear amit: can you eventually ellaborate on your comment (line 3 of it mainly) (e.g. transforming it with more work in an answer) this will be useful in order to compare with robert's solution. | |
| Sep 10, 2011 at 1:01 | comment | added | Yemon Choi | [previous comment, making risibly wrong statement -- as pointed out by Robert Israel -- deleted] | |
| Sep 9, 2011 at 2:06 | comment | added | Robert Israel | How does it follow from Gauss-Lucas? There is a third root of $P$ that may be outside the region $R$. Actually, though, it follows from the fact that $P$ is of the form $P(x) = (a x + b )(x^2 - 1)$ that the product of the roots of $P'$ is $-1/3$. So at least one of those roots has absolute value $\le 1/\sqrt{3}$. | |
| Sep 8, 2011 at 23:06 | comment | added | Amit Kumar Gupta | What would a "no" answer to this question look like? The only thing I can think of would be to do some reverse math and find a reasonable theory where Rouch$\acute{e}$'s theorem is provable but this statement is not. However this statement can be proved easily by integrating the real part of $P′(x)$ and Rouch$\acute{e}$'s theorem is normally proved with some integration (correct?) so I suspect it would be hard to separate the two in this way. | |
| Sep 8, 2011 at 22:23 | history | asked | Luis H Gallardo | CC BY-SA 3.0 |