Timeline for Real roots for polynomials
Current License: CC BY-SA 3.0
11 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Jul 18, 2011 at 16:03 | comment | added | mathphysicist | @Yemon: provided one considers $P$ as map from $\mathbb{R}$ to $\mathbb{C}$ (rather than from $\mathbb{C}$ to $\mathbb{C}$), of course. But in the case under study we are interested in real roots only anyway. | |
| Jul 18, 2011 at 15:19 | comment | added | mathphysicist | @Yemon: That's one way to look at it, yes. | |
| Jul 18, 2011 at 15:03 | comment | added | Yemon Choi | Put it this way: P "is" an ordered tuple of complex numbers. Out of those, one wishes to define a new ordered tuple Q of real numbers, such that when you interpret P as a polynomial (over the complex field) then its zeros are precisely those of Q as a polynomial (with real coefficients but allowing complex roots). Is that something like what you meant? | |
| Jul 18, 2011 at 15:01 | comment | added | Yemon Choi | MP: OK, but then I prefer Denis' formulation. (The polynomials have complex coefficients and complex roots, I wasn't keen on your elision between the variable and the roots. So your answer is what Denis suggested, is that correct?) | |
| Jul 18, 2011 at 15:00 | comment | added | mathphysicist | @Yemon: of course I meant for real $x$. | |
| Jul 18, 2011 at 14:38 | comment | added | mathphysicist | @Yemon: In fact $Q(x)=P(x)\bar P(x)$ even for complex $x$ (sorry for a poor formulation in previous comment), but, and that's the main point, for real $X$ $P(x)\bar P(x)$ is clearly a polynomial in $x$. | |
| Jul 18, 2011 at 13:39 | comment | added | mathphysicist | @Yemon: $x$ is assumed to be real, and then my $Q(x)$ equals $P(x)\overline{P(x)}$ (cf. Denis' answer). | |
| Jul 18, 2011 at 12:59 | comment | added | Yemon Choi | Specifically, take $P(z)=z-1$ | |
| Jul 18, 2011 at 12:58 | comment | added | Yemon Choi | Why is $Q(z)$ a polynomial in $z$? | |
| Jul 18, 2011 at 11:51 | history | answered | mathphysicist | CC BY-SA 3.0 |