Timeline for Reference request for anti-palindromic polynomials.
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 2, 2018 at 21:44 | comment | added | Student | @GerryMyerson, I made another post about this here: mathoverflow.net/questions/314436/…. | |
| Oct 30, 2018 at 7:01 | comment | added | François Brunault | @SuperMario If $f$ has integral coefficients then there will be of course additional conditions on the roots. If you are interested with the roots on the unit circle, I encourage you to ask a new question (but make it precise enough). | |
| Oct 30, 2018 at 4:15 | comment | added | Gerry Myerson | Coefficients even or odd? Were we meant to assume the coefficients are integers? This was never stated. | |
| Oct 29, 2018 at 23:08 | comment | added | Student | Do you think if the coefficients are even or odd the roots the number of roots lying on the unit circle will change? | |
| Oct 29, 2018 at 22:46 | comment | added | François Brunault | @SuperMario By $\mathbb{C}^\times$ I mean the multiplicative group of all nonzero complex numbers. I should also have mentioned that in the case $h=0$ (so $n \equiv 0 \textrm{ mod } 4$), the polynomial $g(x)=f(ix)$ is reciprocal, and there are many results on roots of such polynomials (lying within, on, or outside the unit circle). | |
| Oct 29, 2018 at 21:35 | comment | added | Student | $n$ stands for the degree. By $\mathbb{C}^{\times}$ you mean non-zero complex number or set of all unit elements? | |
| Oct 29, 2018 at 19:41 | vote | accept | Student | ||
| Oct 29, 2018 at 18:56 | history | edited | François Brunault | CC BY-SA 4.0 |
Corrected the argument
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| Oct 29, 2018 at 18:25 | history | answered | François Brunault | CC BY-SA 4.0 |