Timeline for Proof of a combinatorial equation
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 23, 2018 at 0:35 | comment | added | MTyson | @darijgrinberg Thank you for writing this! This point is much more clear to me now. | |
| Aug 23, 2018 at 0:28 | comment | added | darij grinberg | This is a nice proof (though I think the steps you and I have filled in here in the comments are not less difficult than the steps you have shown in your answer). Reminds me of some of the ways partial derivatives are used around Capelli identities. | |
| Aug 23, 2018 at 0:24 | comment | added | darij grinberg | ... the polynomial $q_{m,k}\left(s\right)$ equals $s^m$ when $k = m$ but is a linear combination of smaller powers of $s$ when $k \neq m$. (This fact, in turn, can be proven by induction on $m$ using the above recurrence.) | |
| Aug 23, 2018 at 0:20 | comment | added | darij grinberg | ... recursive equation (along with the fact that $q_{0,k}$ is the Kronecker delta $\delta_{0,k}$ for each $k \geq 0$) determines the polynomials $q_{m,k}$ uniquely. Now, set $q_m = \sum\limits_{k=0}^{\infty} q_{m,k}$ for each $m \geq 0$. It is then easy to see that $\left(\partial_t^m e^{se^t}\right)\mid_{t=0} = q_m\left(s\right)$. So we need to show that the leading term of $q_m\left(s\right)$ is $s^m$. But this easily follows from the fact that ... | |
| Aug 23, 2018 at 0:15 | comment | added | darij grinberg | Ah. Since I can't quite follow your "$\partial_t$ pulled out an $s$" argument, let me re-argue your first claim in my language: For each $m \geq 0$, there is a sequence $\left(q_{m,0}, q_{m,1}, q_{m,2}, \ldots\right)$ of univariate polynomials such that $\partial^m_t e^{se^t} = \sum\limits_{k=0}^{\infty} q_{m,k}\left(s\right) e^{kt+se^t}$, and such that $q_{m,k} = 0$ for all $k > m$. This is proven by induction on $m$, and this argument also shows that $q_{m,k} = k q_{m-1,k} + s q_{m-1,k-1}$ for all $m$ and $k$, where $q_{m-1,k-1}$ is understood to be $0$ if $k = 0$. This ... | |
| Aug 22, 2018 at 22:23 | comment | added | MTyson | @darijgrinberg By induction, $\partial_t^m e^{se^t}$ is a sum of terms of the form $q(s)e^{kt+se^t}$ for all $m$. There's exactly one term where the $\partial_t$ pulled out an $s$ each time, so $p(s)$ has degree $n-k$ and leading coefficient $1$. For the latter point, note that $(\partial_s-1)(p(s)e^s)=p'(s)e^s$. | |
| Aug 22, 2018 at 22:03 | comment | added | darij grinberg | How do you get $p\left(s\right)$ and its initial term? And why does $\left(\partial_s-1\right)^{n-k}$ take $p\left(s\right)e^s$ to $\left(n-k\right)! e^s$ ? | |
| Aug 21, 2018 at 22:38 | history | answered | MTyson | CC BY-SA 4.0 |