91Ó°ÊÓ

Before proving the given result, let's get familiar with the definitions of the difference product and Vandermonde determinant. The difference product is defined as \(g=g\left(x_{i}\right)\), where \[g(x_{1}, \ldots, x_{n}) = \prod_{1\leq i<j\leq n}(x_j - x_i)\] The Vandermonde determinant, denoted as \(V_{n-1}(x)\), is a determinant of a matrix constructed from elements having the form \(x_{i}^{j}\) and is given by: \[V_{n-1}(x) = \left|\begin{array}{cccc} 1 & x_{1} & x_{1}^2 & \ldots & x_{1}^{n-1} \\ 1 & x_{2} & x_{2}^2 & \ldots & x_{2}^{n-1} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & x_{n} & x_{n}^2 & \ldots & x_{n}^{n-1} \end{array}\right|\]

To establish the relation between the difference product and Vandermonde determinant, first expand the Vandermonde determinant (\(V_{n-1}(x)\)) using row operations and looking for a common pattern. Taking the first row as pivot, add \(R_{i} = R_{i} - R_{1}\) for all \(i > 1\). The new matrix becomes: \[\begin{array}{ccccc} 1 & x_{1} & x_{1}^2 & \ldots & x_{1}^{n-1} \\ 0 & x_{2} - x_{1} & x_{2}^2 - x_{1}^2 & \ldots & x_{2}^{n-1} - x_{1}^{n-1} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & x_{n} - x_{1} & x_{n}^2 - x_{1}^2 & \ldots & x_{n}^{n-1} - x_{1}^{n-1} \end{array}\] Now, perform the following operations on the matrix: \\ \(R_2 = \frac{1}{x_2 - x_1} R_2\) \\ \(R_3 = \frac{1}{x_3 - x_1} R_3\) \\ \(\dots\) \\ \(R_n = \frac{1}{x_n - x_1} R_n\) The new matrix becomes: \[\begin{array}{ccccc} 1 & x_{1} & x_{1}^2 & \ldots & x_{1}^{n-1} \\ 0 & 1 & (x_2 - x_1) & \ldots & \frac{x_2^{n-1} - x_1^{n-1}}{x_2 - x_1} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 1 & (x_n - x_1) & \ldots & \frac{x_n^{n-1} - x_1^{n-1}}{x_n - x_1} \end{array}\] Now, set \(x = x_n\), we have: \[\begin{array}{ccccc} 1 & x_{1} & x_{1}^2 & \ldots & x_{1}^{n-1} \\ 0 & 1 & (x_2 - x_1) & \ldots & \frac{x_2^{n-1} - x_1^{n-1}}{x_2 - x_1} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 1 & (x - x_1) & \ldots & \frac{x^{n-1} - x_1^{n-1}}{x - x_1} \end{array}\] Let \(V_{n-1}(x)\) be the determinant of the above matrix.

We will compute the determinant of the matrix obtained in Step 2. To do this, we can expand the determinant along the first row. \[V_{n-1}(x) = x^{n-1}_{1} \left|\begin{array}{cccc} 1 & (x_{2} - x_{1}) & \ldots & \frac{x_{2}^{n-1} - x_{1}^{n-1}}{x_{2} - x_{1}} \\ \vdots & \vdots & \ddots & \vdots \\ 1 & (x - x_{1}) & \ldots & \frac{x^{n-1} - x_{1}^{n-1}}{x - x_{1}} \end{array}\right|\] We observe that the new matrix is a Vandermonde determinant of order \(n-2\), so the expansion pattern observed can be applied repeatedly. Thus, \[V_{n-1}(x) = x^{n-1}_{1} x^{n-2}_{2} \ldots x_{n-1} \cdot V_{0}(x)\] Since \(V_{0}(x)\) is a determinant of an empty matrix, which is considered zero, we can rewrite this as: \[V_{n-1}(x) = x^{n-1}_{1} x^{n-2}_{2} \ldots x_{n-1} \cdot 1\] Now, observe that the term \(x^{n-1}_{1} x^{n-2}_{2} \ldots x_{n-1}\) is the telescoping product, and it is equal to \(\frac{g\left(x_{1}, \ldots, x_{n}\right)}{(-1)^{n}} \) when \(x = x_{n}\). Therefore, we obtain the desired result: \[g\left(x_{1}, \ldots, x_{n}\right) = (-1)^{n} V_{n-1}(x)\] This proves the relation between the difference product and the Vandermonde determinant.