Timeline for A discrete operator begets even/odd polynomials
Current License: CC BY-SA 4.0
20 events
| when toggle format | what | by | license | comment | |
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| Apr 7, 2019 at 3:42 | vote | accept | T. Amdeberhan | ||
| Apr 7, 2019 at 3:42 | |||||
| Mar 11, 2019 at 6:27 | vote | accept | T. Amdeberhan | ||
| Mar 11, 2019 at 6:28 | |||||
| Mar 11, 2019 at 6:27 | vote | accept | T. Amdeberhan | ||
| Mar 11, 2019 at 6:27 | |||||
| May 13, 2018 at 14:41 | comment | added | Richard Stanley | How is this question different from your mathoverflow.net/questions/299981? | |
| May 13, 2018 at 1:10 | comment | added | Matemáticos Chibchas | What "$\lambda\vdash n$" means? | |
| May 12, 2018 at 23:18 | history | edited | darij grinberg | CC BY-SA 4.0 |
An empty product would not be divisible by $\delta$
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| May 25, 2017 at 15:05 | comment | added | T. Amdeberhan | @darijgrinberg: I appreciate your meticulous care to help everyone understand the subtle points in the operators actions. Thank you. | |
| May 25, 2017 at 13:27 | comment | added | darij grinberg | Ah, I see. The two interpretations do give rise to the same $L_\lambda\left(x\right)_n$, because the $L_\lambda\left(x\right)_n$ computed using Interpretation 2 has constant term $0$. (Why it has constant term $0$ is not completely obvious; it basically comes from observing that $E^k \left(x\right)_n$ has constant term $0$ for all $k \in \left\{0,1,\ldots,n-1\right\}$.) | |
| May 25, 2017 at 13:20 | comment | added | darij grinberg | ... give rise to the same $L_\lambda$ (indeed, for any given polynomial $f$, the $L_\lambda f$'s defined using the two interpretations can have different constant terms). Do they nevertheless give rise to the same $L_\lambda \left(x\right)_n$ ? (Your own answer relies on Interpretation 1, as far as I understand it.) | |
| May 25, 2017 at 13:19 | comment | added | darij grinberg | ... Interpretation 2 is to understand the fraction in the definition of $L_\lambda$ as being a formal fraction which has to be evaluated as a polynomial in $E$ (using $\delta = E - 1$) before the actual operator $E$ is substituted into it. Thus, $L_\lambda$ is redefined as follows: The polynomial $\left(x^{\lambda_1} - 1\right) \cdots \left(x^{\lambda_k} - 1\right)$ is divisible by $x-1$ as long as $k \geq 1$. Denote the quotient by $q$, and define $L_\lambda$ to be $q\left(E\right)$. These two interpretations both make sense, but they do not ... | |
| May 25, 2017 at 13:17 | comment | added | darij grinberg | There is an ambiguity here: The operator $\delta$ is not invertible (in fact, it is not even injective), so how do you divide by $\delta$ ? In light of this, I see two reasonable interpretations of the definition of $L_\lambda$. Interpretation 1 is to redefine $L_\lambda$ as $L_\lambda = \delta' \left(E^{\lambda_1} - 1\right) \cdots \left(E^{\lambda_k} - 1\right)$, where $\delta'$ is the linear operator on $\mathbb{Q}\left[x\right]$ defined as follows: For each polynomial $f$, we let $\delta' f$ be the unique polynomial $g$ with constant term $0$ satisfying $\delta g = f$. Meanwhile, ... | |
| May 25, 2017 at 4:57 | history | edited | T. Amdeberhan | CC BY-SA 3.0 |
edited title
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| May 24, 2017 at 16:14 | answer | added | T. Amdeberhan | timeline score: 2 | |
| May 23, 2017 at 5:35 | comment | added | T. Amdeberhan | It's a composition. Example: $(E^2-1)(E-1)f(x)=(E^3-E^2-E+1)f(x)=f(x+3)-f(x+2)-f(x+1)+f(x)$. | |
| May 23, 2017 at 5:25 | comment | added | Włodzimierz Holsztyński | Is the numerator of $L_{\lambda}$ a product or a composition? | |
| May 23, 2017 at 5:10 | history | edited | T. Amdeberhan | CC BY-SA 3.0 |
added 126 characters in body
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| May 23, 2017 at 0:32 | comment | added | T. Amdeberhan | Just $\delta$. It is not a mistake. Higher powers are not giving the same result. | |
| May 23, 2017 at 0:31 | comment | added | LSpice | In the definition of $L_\lambda$, do you mean to divide by $\delta^{\lvert\lambda\rvert}$, or really just by $\delta$? | |
| May 22, 2017 at 19:54 | history | edited | T. Amdeberhan | CC BY-SA 3.0 |
added 29 characters in body
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| May 22, 2017 at 19:40 | history | asked | T. Amdeberhan | CC BY-SA 3.0 |