Timeline for Bound for some trigonometric polynomials
Current License: CC BY-SA 4.0
4 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 30, 2023 at 10:16 | comment | added | Itachi | Using cyclotomic polynomials is very clever! However the bounds then start getting more complicated because not neccesarily all coefficients are $\pm 1$ or $0$. Also I don't know how to deal with the case when $1+t^{10} \Phi_{33}\Phi_{99}=0$, one root is for instance $t=i$, and then the reason $g_3=0$ is because of the term $t^{100}-1$ (contrary to what I thought it would have something to do with $t^{909}-1$) | |
| Mar 29, 2023 at 18:43 | comment | added | Pavel Gubkin | Oh, not divisible. The correct claim is: if $t$ is a root of $t\Phi_3\Phi_9(1 + t^{10}\Phi_{33}\Phi_{99})$ with $|t | = 1$ then it is also a root of $g_3$. | |
| Mar 29, 2023 at 18:27 | comment | added | Pavel Gubkin | To show the claim for $f_1 + f_2 + f_3$ you have to show that $$g_3 = \frac{t^{1010} - t^{101}}{t^{101 - 1}}\cdot \frac{t^{100} - 1}{t^{10} - 1}$$ is divisible by $g_1 + g_2 = t\Phi_3\Phi_9(1 + t^{10}\Phi_{33}\Phi_{99})$ | |
| Mar 29, 2023 at 18:18 | history | answered | Pavel Gubkin | CC BY-SA 4.0 |