Quadrature Rule Exactness Analysis

The given quadrature rule is $\int_{-1}^{1} f(x)dx \approx f(\alpha) + \gamma f(\beta)$. This rule must be exact for all polynomials of degree $\le 3$. We need to find the value of the expression $\gamma + 3 (\alpha^2 + \beta^2) + (\alpha^3 + \beta^3)$.

Step 1: Determine Constants $\alpha, \beta, \gamma$

Exactness for basis polynomials $1, x, x^2, x^3$ provides equations to find the constants.

• For $f(x) = 1$:
  • Integral value: $\int_{-1}^{1} 1 dx = [x]_{-1}^{1} = 1 - (-1) = 2$.
  • Rule approximation: $f(\alpha) + \gamma f(\beta) = 1 + \gamma(1) = 1 + \gamma$.
  • Equating values: $1 + \gamma = 2 \implies \gamma = 1$.
• For $f(x) = x$:
  • Integral value: $\int_{-1}^{1} x dx = [\frac{x^2}{2}]_{-1}^{1} = \frac{1}{2} - \frac{1}{2} = 0$.
  • Rule approximation: $f(\alpha) + \gamma f(\beta) = \alpha + \gamma \beta$.
  • Equating values: $\alpha + \gamma \beta = 0$. Since $\gamma = 1$, we get $\alpha + \beta = 0$, which implies $\beta = -\alpha$.
• For $f(x) = x^2$:
  • Integral value: $\int_{-1}^{1} x^2 dx = [\frac{x^3}{3}]_{-1}^{1} = \frac{1}{3} - (-\frac{1}{3}) = \frac{2}{3}$.
  • Rule approximation: $f(\alpha) + \gamma f(\beta) = \alpha^2 + \gamma \beta^2$.
  • Equating values: $\alpha^2 + \gamma \beta^2 = \frac{2}{3}$. With $\gamma = 1$, this is $\alpha^2 + \beta^2 = \frac{2}{3}$.
  • Substituting $\beta = -\alpha$: $\alpha^2 + (-\alpha)^2 = \frac{2}{3} \implies 2\alpha^2 = \frac{2}{3} \implies \alpha^2 = \frac{1}{3}$.
  • Consequently, $\beta^2 = (-\alpha)^2 = \alpha^2 = \frac{1}{3}$.
• For $f(x) = x^3$:
  • Integral value: $\int_{-1}^{1} x^3 dx = [\frac{x^4}{4}]_{-1}^{1} = \frac{1}{4} - \frac{1}{4} = 0$.
  • Rule approximation: $f(\alpha) + \gamma f(\beta) = \alpha^3 + \gamma \beta^3$.
  • Equating values: $\alpha^3 + \gamma \beta^3 = 0$. Using $\gamma = 1$ and $\beta = -\alpha$, we have $\alpha^3 + (1)(-\alpha)^3 = \alpha^3 - \alpha^3 = 0$. This condition is satisfied and confirms consistency.

Step 2: Evaluate the Expression

We need to calculate $\gamma + 3 (\alpha^2 + \beta^2) + (\alpha^3 + \beta^3)$ using the derived constants.

• Substitute the values:
• $\gamma = 1$.
• $\alpha^2 + \beta^2 = \frac{1}{3} + \frac{1}{3} = \frac{2}{3}$.
• $\alpha^3 + \beta^3 = \alpha^3 + (-\alpha)^3 = \alpha^3 - \alpha^3 = 0$.

The expression evaluates to:

$1 + 3 \left(\frac{2}{3}\right) + 0$

$1 + 2 + 0 = 3$

Final Result

The value of the expression $\gamma + 3 (\alpha^2 + \beta^2) + (\alpha^3 + \beta^3)$ is 3.