Question #265800

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

x1 =0, x2 = -3/5\sqrt{3/5} , x3 = 3/5\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



1
Expert's answer
2021-11-15T16:45:35-0500

P(x)=2x5x4+x21P(x)=2x^5-x^4+x^2-1

P(x1)=P(0)=1P(x_1)=P(0)=-1

P(x2)=P(35)=2(35)235(35)2+351=1825351925P(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(x3)=P(35)=2(35)235(35)2+351=1825351925P(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}


11P(x)dxi=13wiP(xi)=w1P(x1)+w2P(x2)+w3P(x3)=89(1)+59(P(x2)+P(x3))=89+59(3825)=893845=40+3845=7845=2615=11115\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: 11P(x)dx11115\int\limits_{-1}^1P(x)dx\approx-1\tfrac{11}{15} .


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