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Trying to understand a result in a representation theoretical paper, I realized that it implies the following elementary identity for symmetric functions. My question is whether this identity is true, known, and whether it has a direct proof. It might also happen that this identity is simpler than I think.

Let $d$ be a natural number. For $0\leq k\leq d$ let $e_k(z_1\dots,z_d)$ denote the $k$th elementary symmetric polynomial in $d$ variables, namely $$e_k(z_1\dots,z_d)=\sum_{1\leq i_1<\dots <i_k\leq d}z_{i_1}\dots z_{i_k}.$$

Let $x$ be a variable. The relevant identity is \begin{eqnarray*} \sum_{k=0}^d (-1)^k(x-d)^{2k}\cdot e_{d-k}((x-d+1)^2,\dots,(x-1)^2,x^2)=\\ d!\cdot (2x-2d+1)\dots (2x-d-1) (2x-d).\end{eqnarray*}

Notice that the right hand side is a polynomial in $x$ of degree $d$, while the left hand side has a priori degree at most $2d$ only.

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1 Answer 1

up vote 11 down vote accepted

Viete formula gives $\sum_{k=0}^{d} t^k e_{d-k}\left(\alpha_0,\ldots,\alpha_{d-1}\right)=(t+\alpha_1)\ldots(t+\alpha_d).$

Therefore (substituing $t=-(x-d)^2$ and $\alpha_j=(x-j)^2$), LHS is equal to $\prod_{j=0}^{d-1} \left(-(x-d)^2+(x-j)^2\right)= \prod_{j=0}^{d-1} (2x-j-d)(d-j)=d!\prod_{k=d}^{2d-1} (2x-k)$

which is exactly the RHS.

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