Timeline for Resultant of $f(x)$ and $f(-x)$
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
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| Dec 3 at 16:37 | history | edited | darij grinberg | CC BY-SA 4.0 |
fix typo and remove unnecessary assumptions
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| Jan 29, 2021 at 15:25 | vote | accept | Anton | ||
| Jan 29, 2021 at 15:10 | answer | added | Richard Stanley | timeline score: 12 | |
| Jan 29, 2021 at 12:47 | comment | added | Anton | @NoamD.Elkies, thanks! I think your observation regarding the connection between $b_n$ and the remainder of $f(x)$ mod $x^2 - \beta$ is actually what I was looking for. | |
| Jan 29, 2021 at 4:20 | comment | added | Noam D. Elkies | @Anton That follows from the product formula $\prod_{i,j=1}^n (\alpha_i + \alpha_j)$. The $i=j$ factors give $2^n \prod_i \alpha_i = (-2)^n a_n$, and each of the $i\neq j$ factors appears twice. (Their product, which you call $b_n$, is $\pm$ the resultant w.r.t. $\beta$ of the linear and constant coefficients of the remainder of $f(x) \bmod x^2-\beta$.) | |
| Jan 29, 2021 at 4:10 | history | edited | Anton | CC BY-SA 4.0 |
edited body
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| Jan 29, 2021 at 4:09 | answer | added | Anton | timeline score: 0 | |
| Jan 29, 2021 at 4:04 | comment | added | Anton | Thank you, @JoeSilverman. Looks like I got my answer: this may not be the right question to ask and I need to look closely at the properties of polynomials that I am working with instead. Curious fact though: if $f(x) = x^n + \cdots + a_{n - 1}x + a_n$, then it looks like $\operatorname{Res}(f(x), f(-x)) = 2^na_nb_n^2$ for some integer $b_n$. | |
| Jan 29, 2021 at 2:04 | comment | added | Joe Silverman | It's obviously a symmetric function in the roots of $f(x)$ (explicitly described in Will's answer), so it's a polynomial in the coefficients of $f(x)$. Have you tried writing it out for $n=2,3,4$, say, and seeing what that polynomial looks like. For $f(x)=x^2+ax+b$, I get $4a^2b$, which doesn't seem related to much of anything. | |
| Jan 29, 2021 at 1:39 | history | edited | Anton | CC BY-SA 4.0 |
added 373 characters in body
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| Jan 29, 2021 at 1:25 | comment | added | Anton | @WillSawin thanks, but I am trying to avoid expressing this quantity through its roots, and use things like discriminant / degree / constant coefficient of f(x) instead. Perhaps, this is impossible, or maybe it is possible with the specific case which I am dealing with, where the Galois group of f(x) is Abelian. | |
| Jan 29, 2021 at 1:22 | comment | added | Will Sawin | Up to sign it's $\prod_{i=1}^n \prod_{j=1}^n (\alpha_i + \alpha_j)$. | |
| Jan 29, 2021 at 1:16 | comment | added | Anton | @Wojowu, I see, that makes sense. | |
| Jan 29, 2021 at 0:52 | comment | added | Wojowu | Just $n$ and the discriminant of $f$ are definitely not enough to determine the resultant you want - there will be polynomials of same degree and same discriminant but different resultants. | |
| Jan 29, 2021 at 0:49 | history | asked | Anton | CC BY-SA 4.0 |