Question

If \( P(x) \) is a polynomial of degree 5 and \[ \alpha = \sum_{i=0}^{6} P(x_i) \left( \prod_{\substack{j=0
j\neq i}}^{6} (x_i - x_j)^{-1} \right), \] where \( x_0, x_1, \ldots, x_6 \) are distinct points in the interval \([2,3]\), then the value of \( \alpha^2 - \alpha + 1 \) is __________.

Hint:

In Lagrange interpolation, the sum of the Lagrange basis polynomials equals 1 when summed over all distinct interpolation points.

Answer 1

Correct Answer: 1

The given expression for \( \alpha \) is a form of the Lagrange interpolation formula. This formula is used to express a polynomial passing through a set of points. Given that \( P(x) \) is a polynomial of degree 5, the expression for \( \alpha \) will essentially sum the values of the polynomial at each of the distinct points \( x_0, x_1, \ldots, x_6 \), weighted by their corresponding Lagrange basis polynomials.
The key observation is that this interpolation formula sums the values of the polynomial \( P(x) \) evaluated at distinct points, and since the polynomial \( P(x) \) has degree 5, we know that \( P(x) \) is uniquely determined by these points. The structure of the formula for \( \alpha \) suggests that \( \alpha = 1 \), based on the properties of the Lagrange interpolation and the fact that the sum of Lagrange basis polynomials for a complete set of distinct points equals 1.
Thus, \( \alpha = 1 \). Now, we compute \( \alpha^2 - \alpha + 1 \): \[ \alpha^2 - \alpha + 1 = 1^2 - 1 + 1 = 1. \] Thus, the value of \( \alpha^2 - \alpha + 1 \) is \(\boxed{1}\).