Show that $\prod_{i=1}^n\frac{z-tx_i}{z-x_i}=1+\sum_{i=1}^n\frac{(1-t)x_i}{z-x_i}\prod_{\substack{j\ne i,\\j\in [n]}}\frac{x_i-tx_j}{x_i-x_j}.$

Question

Let $x_1,...,x_n,t,z$ are independent indeterminants over the ring of integer $\mathbb Z$, how to prove that:

$$\prod_{i=1}^n\frac{z-tx_i}{z-x_i}=1+\sum_{i=1}^n\frac{(1-t)x_i}{z-x_i}\prod_{\substack{j\ne i,\\j\in [n]}}\frac{x_i-tx_j}{x_i-x_j}.$$

This identity is from page 210 of Macdonald's book: Symmetric Functions and Hall Polynomials. I have tried to use induction but no luck after hours of computation. Any hint?

Answer 1

Hint: Use partial fractions. Suppose there are $\alpha _i(z)$ such that $$LHS=\sum _{i=1}^n\frac{\alpha _i(z)}{z-x_i},$$ notice that the numerator is $$\sum _{i=1}^n\alpha _i(z)\prod _{j\neq i}(z-x_i).$$ Equate numerators and evaluate at $z=x_1,x_2,\cdots ,x_n.$