Proving an identiy on elementary symmetric polynomials

Question

I happened to meet the following exercise on elementary symmetric polynomials. Here is the wiki page for the definition of elementary symmetric polynomial.

Let $n\ge 3$ and $e_k:=e_k(x_1, \cdots, x_n)$ be the $k$-th elementary symmetric polynomial with respect to variables $x_1, \cdots, x_n$. Prove the following identity: $$\sum_{i=2}^n\frac{e_n(x_1-x_i)}{x_i^2 x_1}+2e_{n-2}=x_1\frac{e_{n-1}}{e_n}e_{n-2}(x|x_1)$$ where $$e_{n-2}(x|x_1)=e_{n-2}-x_1 e_{n-3}(x_2,\cdots, x_n)$$

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I have computed precisely for the case $n=3$ and $n=4$, and have verified the identity. However, I have trouble in proving the identity for general $n$. Can anyone present an elegant proof? Thanks very much in advance.

Answer 1

Let $e_k$ be the elementary symmetric polynomials in the $n$ variables $(x_1,x_2,\dots,x_n)$, and let $d_k$ be the elementary symmetric polynomials in the $n-1$ variables $(x_2,x_3,\dots,x_n)\,$. It follows from the definitions that $\,e_n=x_1d_{n-1}\,$ and $\,e_k=d_k+x_1d_{k-1}\,$ for $\,1 \le k \le n-1\,$.

In particular $\,e_{n-2}(x|x_1)=e_{n-2}-x_1 e_{n-3}(x_2,\cdots, x_n)=e_{n-2}-x_1d_{n-3}=d_{n-2}\,$, so the identity to prove can be rewritten as:

$$ \require{cancel} \begin{align} \sum_{i=2}^n\frac{e_n(x_1-x_i)}{x_i^2 x_1}+2e_{n-2}&=x_1\frac{e_{n-1}}{e_n}e_{n-2}(x|x_1) \\ \iff\;\;x_1d_{n-1} \left(\sum_{i=2}^n\frac{1}{x_i^2}-\frac{1}{x_1}\sum_{i=2}^n\frac{1}{x_i}\right)+2\left(d_{n-2}+x_1d_{n-3}\right)&=\bcancel{x_1}\frac{d_{n-1}+x_1d_{n-2}}{\bcancel{x_1}d_{n-1}}d_{n-2} \end{align} $$

The latter equality can be directly verified using that:

• $\displaystyle \sum_{i=2}^n \frac{1}{x_i} = \frac{d_{n-2}}{d_{n-1}}$

• $\displaystyle \sum_{i=2}^n \frac{1}{x_i^2} = \left(\frac{d_{n-2}}{d_{n-1}}\right)^2-2 \frac{d_{n-3}}{d_{n-1}}$