These polynomials are always either even or odd [duplicate]

Question

The falling factorials are $(x)_n=x(x-1)\cdots(x-n+1)$ with $(x)_0:=1$. Define the forward shift $E$ and the discrete derivative $\delta=E-1$, respectively, by $$Ef(x)=f(x+1) \qquad \text{and} \qquad \delta f(x)=f(x+1)-f(x).$$ Let $\lambda=(\lambda_1,\dots,\lambda_k)\vdash n$ be a partition of a positive integer $n$. Now, define the operator $\Psi_{\lambda}$ to act on polynomials according to $$\Psi_{\lambda}=(E-1)^{-1}\prod_{j=1}^{k}(E^{\lambda_j}-1)$$ (where $k $ is the length of $\lambda$) so in particular denote $\Psi_{\lambda}(x):=\Psi_{\lambda}((x)_n)$.

For example, if $n=9$ and $\lambda=(4,2,2,1)$ then $(E-1)^{-1}(E^4-1)(E^2-1)^2(E-1)=(E^4-1)(E^2-1)^2=E^8-2E^6+2E^2-1$ and hence $\Psi_{\lambda}(x)=(x+8)_9-2(x+6)_9+2(x+2)_9-(x)_9$, or $$\Psi_{\lambda}(x)=8064x^6+141120x^4+213696x^2.$$ Experiments suggest to ask

is it true that $\Psi_{\lambda}(x)$ is always an even or an odd polynomial?

Answer 1

We first note that $\Psi_\lambda$ is either symmetric or odd-symmetric in $E$. In other words, if we think of $\Psi_\lambda$ as a polynomial $P$ applied to the operator $E$ (of degree $n - 1$), then $P(t)=\pm t^{n-1} P(\frac{1}{t})$ (plus if the number of elements $|\lambda|$of the partition is odd, minus if even). This is easy to see from your product expression, as every multiplicand is odd-symmetric. Write $P(t)=\sum a_i t^i$; symmetry says $a_{n-1-i}=a_i$, while odd-symmetry says $a_{n-1-i}=-a_i$. Together, we have $a_{n-1-i}=-(-1)^{|\lambda|} a_i$.

Simple examination shows that $(x)_n = (-1)^n (n-1-x)_n$. Therefore:

$$ \begin{aligned}\Psi_\lambda(x) &= \sum a_i E^i (x)_n = -(-1)^{n+|\lambda|} \sum a_{n-1-i} E^i (n-1-x)_n \\ &= -(-1)^{n+|\lambda|} \sum a_{n-1-i} \left(E^{n-1-i} (y)_n\right)_{y=-x} = -(-1)^{n+|\lambda|} \Psi_\lambda(-x), \end{aligned}$$ where the third equality uses the fact that $E$ goes the "other way" for $-x$.

So your function is odd iff $n+|\lambda|$ is even, and vice versa.

Answer 2

Here is a combinatorial proof. Let $z_\lambda$ denote the number of permutations in the symmetric group $\mathfrak{S}_n$ that commute with a fixed permutation of cycle type $\lambda$. (There is a simple formula for $z_\lambda$ which is irrelevant here.) It is shown in Theorem 3.1 of http://math.mit.edu/~rstan/papers/cycles.pdf (stated in terms of the backwards shift operator) that
$$ \Psi_\lambda(x) = z_\lambda\sum_w x^{\kappa((1,2,\dots,n)\cdot w)}, $$ where $w$ ranges over all permutations in $\mathfrak{S}_n$ of cycle type $\lambda$, and where $\kappa((1,2,\dots,n)\cdot w)$ denotes the number of cycles of the product of the cycle $(1,2,\dots,n)$ and $w$. All products $(1,2,\dots,n)\cdot w$ have the same parity (that is, all are either even permutations or odd permutations), and the result follows since a permutation $v\in\mathfrak{S}_n$ is even if and only if $n-\kappa(v)$ is even.