Question

If the quadrature rule \[ \int_{-1}^1 f(x) \, dx \approx f(\alpha) + \gamma f(\beta), \] where \( \alpha, \beta \) and \( \gamma \) are real constants, is exact for all polynomials of degree \( \leq 3 \), then \( \gamma + 3 (\alpha^2 + \beta^2) + (\alpha^3 + \beta^3) \) is equal to ________.

Hint:

For Gaussian quadrature, the points \( \alpha \) and \( \beta \) are chosen so that the quadrature rule is exact for polynomials of degree \( \leq 3 \). Always check the symmetry of the weights and points.

Answer 1

Correct Answer: 3

The quadrature rule given is a linear approximation to the integral, where the integral of \( f(x) \) over the interval from \( -1 \) to \( 1 \) is approximated by values at \( \alpha \) and \( \beta \). The exactness condition tells us that this approximation works for polynomials of degree \( \leq 3 \), meaning it integrates linear, quadratic, and cubic polynomials exactly. The general form of a two-point Gaussian quadrature rule is: \[ \int_{-1}^1 f(x) \, dx = w_1 f(\alpha) + w_2 f(\beta) \] where \( \alpha \) and \( \beta \) are the quadrature points, and \( w_1 \) and \( w_2 \) are the weights. Since this rule is exact for polynomials up to degree 3, and considering the known properties of the Gaussian quadrature, we know that: \[ \alpha = -\frac{1}{\sqrt{3}}, \quad \beta = \frac{1}{\sqrt{3}}, \quad \gamma = 1. \] Now, we compute: \[ \gamma + 3 (\alpha^2 + \beta^2) + (\alpha^3 + \beta^3) \] Substitute the values of \( \alpha \) and \( \beta \): \[ \gamma + 3 \left( \left( -\frac{1}{\sqrt{3}} \right)^2 + \left( \frac{1}{\sqrt{3}} \right)^2 \right) + \left( \left( -\frac{1}{\sqrt{3}} \right)^3 + \left( \frac{1}{\sqrt{3}} \right)^3 \right) \] \[ = 1 + 3 \left( \frac{1}{3} + \frac{1}{3} \right) + \left( -\frac{1}{3\sqrt{3}} + \frac{1}{3\sqrt{3}} \right) \] \[ = 1 + 3 \times \frac{2}{3} + 0 = 1 + 2 = 3 \] Thus, the value of \( \gamma + 3 (\alpha^2 + \beta^2) + (\alpha^3 + \beta^3) \) is: \[ \boxed{3} \]