Timeline for Cyclotomic polynomial written as $x^d g\left(x + \frac{1}{x}\right)$
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| Oct 20, 2020 at 7:17 | comment | added | borntomath | Thanks for setting me on the right track! I found chapter 3 of Antonio Cafure, & Eda Cesaratto. (2017). Irreducibility Criteria for Reciprocal Polynomials and Applications. The American Mathematical Monthly, 124(1), 37-53 quite helpful. The functions $f_k$ there are more or less Chebyshev polynomials. | |
| Sep 16, 2020 at 0:01 | comment | added | Gerry Myerson | $g$ is just the minimal polynomial for $2\cos(2\pi/n)$, no? | |
| Sep 15, 2020 at 17:10 | review | Close votes | |||
| Oct 1, 2020 at 3:08 | |||||
| Sep 15, 2020 at 16:55 | comment | added | François Brunault | @JoeSilverman Right, let me try again: starting from the polynomial $x^{2n}+1$, you get essentially the Chebyshev polynomials. Now you can write $x^{2n}+1$ as a product of cyclotomic polynomials, and try to apply Möbius inversion. | |
| Sep 15, 2020 at 16:54 | comment | added | Joe Silverman | @FrançoisBrunault I think that would be correct if $f(x)$ were $x^n-1$, but in this case $f(x)$ is the cyclotomic polynomial $\prod_{d\mid n}(x^d-1)^{\mu(n/d)}$. So as you say, the answer is closely related to the Chebyshev polynomials (maybe a product of them), but they're not equal to the Chebyshev polynomials. | |
| Sep 15, 2020 at 14:56 | history | asked | borntomath | CC BY-SA 4.0 |