Answer to Question #265800 in Quantitative Methods for Terro

Question

Consider three-points Gaussian quadrature. The three points on the x-axis and the associated weights are:

x₁ =0, x₂ = -$\sqrt{3/5}$ , x₃ = $\sqrt{3/5}$ ; w₁ = 8/9, w₂ = w₃ = 5/9

Apply this 3-points Gaussian quadrature to the polynomial

P(x) = 2x⁵-x⁴+x²-1

Answer 1

$P(x)=2x^5-x^4+x^2-1$

$P(x_1)=P(0)=-1$

$P(x_2)=P\big(-\sqrt{\tfrac{3}{5}}\big)=-2\cdot \big(\tfrac{3}{5}\big)^2\cdot \sqrt{\tfrac{3}{5}}—\big(\tfrac{3}{5}\big)^2+\tfrac{3}{5}-1=-\tfrac{18}{25}\cdot \sqrt{\tfrac{3}{5}}-\tfrac{19}{25}$

$P(x_3)=P\big(\sqrt{\tfrac{3}{5}}\big)=2\cdot \big(\tfrac{3}{5}\big)^2\cdot \sqrt{\tfrac{3}{5}}—\big(\tfrac{3}{5}\big)^2+\tfrac{3}{5}-1=\tfrac{18}{25}\cdot \sqrt{\tfrac{3}{5}}-\tfrac{19}{25}$

$\int\limits_{-1}^1P(x)dx\approx\sum_{i=1}^3w_iP(x_i)=w_1P(x_1)+w_2P(x_2)+w_3P(x_3)=\tfrac{8}{9}\cdot (-1)+\tfrac{5}{9}(P(x_2)+P(x_3))=-\tfrac{8}{9}+\tfrac{5}{9}\cdot \big(-\tfrac{38}{25}\big)=-\tfrac{8}{9}-\tfrac{38}{45}=-\tfrac{40+38}{45}=-\tfrac{78}{45}=-\tfrac{26}{15}=-1\tfrac{11}{15}$

Answer: $\int\limits_{-1}^1P(x)dx\approx-1\tfrac{11}{15}$ .