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An empty product would not be divisible by $\delta$
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darij grinberg
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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!).

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 $\lambda_1\geq\lambda_2\geq\dots\geq\lambda_k\geq1$, 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!).

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!).

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T. Amdeberhan
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Operational action A discrete operator begets even/odd polynomials

added 126 characters in body
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T. Amdeberhan
  • 43.2k
  • 5
  • 57
  • 217

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 $\lambda_1\geq\lambda_2\geq\dots\geq\lambda_k\geq1$, 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!).

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 $\lambda_1\geq\lambda_2\geq\dots\geq\lambda_k\geq1$, 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.

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 $\lambda_1\geq\lambda_2\geq\dots\geq\lambda_k\geq1$, 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!).

added 29 characters in body
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T. Amdeberhan
  • 43.2k
  • 5
  • 57
  • 217
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Source Link
T. Amdeberhan
  • 43.2k
  • 5
  • 57
  • 217
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