4
$\begingroup$

Given a function $f(x)$ define the forward shift operator by $Ef(x)=f(x+1)$ and the discrete derivative $\delta f(x)=(E-1)f(x)=f(x+1)-f(x)$.

Given a partition $\lambda=(\lambda_1,\lambda_2,\dots,\lambda_k)$, where $\lambda_1\geq\lambda_2\geq\dots\geq\lambda_k\geq1$ and $k>0$, define the operator $$L_{\lambda}=\frac{(E^{\lambda_1}-1)\cdots(E^{\lambda_k}-1)}{\delta}.$$

Let $(x)_n=x(x-1)\cdots(x-n+1)$ be the falling factorial.

Question. If $\lambda\vdash n$ then is it true $\Phi_n(x)=L_{\lambda}(x)_n$ is either an even or an odd polynomial, with non-negative integer coefficients? It appears to be so.

Example. If $\lambda=(n)$ then $L_{\lambda}(x)_n=\frac{(x+n)_{n+1}-(x)_{n+1}}{n+1}$ indeed satisfies the claim (check!).

$\endgroup$
  • $\begingroup$ In the definition of $L_\lambda$, do you mean to divide by $\delta^{\lvert\lambda\rvert}$, or really just by $\delta$? $\endgroup$ – LSpice May 23 '17 at 0:31
  • 1
    $\begingroup$ Just $\delta$. It is not a mistake. Higher powers are not giving the same result. $\endgroup$ – T. Amdeberhan May 23 '17 at 0:32
  • $\begingroup$ Is the numerator of $L_{\lambda}$ a product or a composition? $\endgroup$ – Włodzimierz Holsztyński May 23 '17 at 5:25
  • $\begingroup$ It's a composition. Example: $(E^2-1)(E-1)f(x)=(E^3-E^2-E+1)f(x)=f(x+3)-f(x+2)-f(x+1)+f(x)$. $\endgroup$ – T. Amdeberhan May 23 '17 at 5:35
  • $\begingroup$ 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, ... $\endgroup$ – darij grinberg May 25 '17 at 13:17
2
$\begingroup$

I have found a proof. I hope someone else can give a more conceptual argument.

Let $\Psi_{\lambda}(x)=L_{\lambda}(x)_n$. We approach the expansion of $\Psi$ differently. Begin with \begin{align} \prod_{i=1}^k(E^{\lambda_i}-1) &=\sum_{T\subset\lambda}(-1)^{k-\#T}E^{\vert T\vert} \\ &=\frac12\left(\sum_{T\subset\lambda}(-1)^{k-\#T}E^{\vert T\vert}+ \sum_{T^c\subset\lambda}(-1)^{k-\#T^c}E^{\vert T^c\vert}\right)\\ &=\frac12\sum_{T\subset\lambda}(-1)^{\#T}\left( (-1)^kE^{\vert T\vert}+E^{n-\vert T\vert}\right);\end{align} where $\#T=$ the cardinality of $T$ (if empty then zero), $\vert T\vert=$ sum of elements of $T$ and $T^c$ is the complement of $T$ in the set $\lambda$.

The next step uses a couple of key facts, namely: $$(x+n-q)_{n+1}=(-1)^{n+1}(-x+q)_{n+1} \qquad \text{and} \qquad \frac1{\delta}(x)_n=\frac{(x)_{n+1}}{n+1}.$$ We thus compute \begin{align} \Psi_{\lambda}(x)&=\frac1{\delta}\prod_{i=1}^k(E^{\lambda_i}-1)(x)_n\\ &=\frac1{2(n+1)}\sum_{T\subset\lambda}(-1)^{\#T}\left((-1)^k (x+\vert T\vert)_{n+1}+(x+n-\vert T\vert)_{n+1}\right)\\ &=\frac1{2(n+1)}\sum_{T\subset\lambda}(-1)^{\#T}\left((-1)^k (x+\vert T\vert)_{n+1}+(-1)^{n+1}(-x+\vert T\vert)_{n+1}\right)\\ &=\frac{(-1)^{n+k+1}}{2(n+1)}\sum_{T\subset\lambda}(-1)^{\#T}\left((-1)^{n+1} (x+\vert T\vert)_{n+1}+(-1)^k(-x+\vert T\vert)_{n+1}\right)\\ &=(-1)^{n+k+1}\Psi_{\lambda}(-x).\end{align} The proof is complete. $\square$

$\endgroup$
  • $\begingroup$ Note that you seem to understand $\lambda$ as the multiset comprising the numbers $\lambda_1, \ldots, \lambda_k$ here when you write "$T \subset \lambda$". (Just pointing this out, since it confused me.) $\endgroup$ – darij grinberg May 25 '17 at 13:22

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.