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!).

  • $\begingroup$ 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
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    $\begingroup$ 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

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. $\square$

  • $\begingroup$ 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

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