Need help with this question. Thank you :) Consider the quadrature ruleQ[f; -1, 1] = wof(−1) + w₁f(x₁) + w₂f(1)with weights wo,W₁,W₂ER and nodes x。=-1, x₁€(-1,1) and x2=1. Is it possible to specify w₁, W₁, W₂ and ₁ in such a way that the degree of exactness of Q is m=5?If so, what is the product X₁W₁?O a. X1W₁=-1O b. X₁W₁=0C. X₁W₁=1O d. W₁=√2e. X₁W₁=√3f. x₁W₁=3√2g. X₁W₁=3√3Oh. X₁W₁=3√5Oi. X₁W₁1=5√3O j. It is impossible to choose wo, W₁, W₂ and X₁ in such a way that the degree of exactness of Q is m=5.

suppose that the degree of exactness is 5 . So we must have Q[f;-1,1]= ∫-11 f(x)dx =w0f(-1)+ w1…

Answered: Consider the quadrature rule Q[f; -1,…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Need help with this question. Thank you :)

Consider the quadrature rule
Q[f; -1, 1] = wof(−1) + w₁f(x₁) + w₂f(1)
with weights wo,W₁,W₂ER and nodes x。=-1, x₁€(-1,1) and x2=1. Is it possible to specify w₁, W₁, W₂ and ₁ in such a way that the degree of exactness of Q is m=5?
If so, what is the product X₁W₁?
O a. X1W₁=-1
O b. X₁W₁=0
C. X₁W₁=1
O d. W₁=√2
e. X₁W₁=√3
f. x₁W₁=3√2
g. X₁W₁=3√3
Oh. X₁W₁=3√5
Oi. X₁W₁1=5√3
O j. It is impossible to choose wo, W₁, W₂ and X₁ in such a way that the degree of exactness of Q is m=5.

Expert Solution

Step 1

suppose that the degree of exactness is 5 .

So we must have $Q [f; - 1, 1] = \int_{- 1}^{1} f (x) d x = w_{0} f (- 1) + w_{1} f (x_{1}) + w_{2} f (1)$

is exact for all polynomials with degree at-most 5.

Putting f(x) = 1,x,x²,x³,x⁴, x⁵ respectively on the equation we should have the following system of equations-

$\int_{- 1}^{1} 1 d x = 2 = w_{0} + w_{1} + w_{2} - - - - - (1) \int_{- 1}^{1} x d x = 0 = - w_{0} + w_{1} x_{1} + w_{2} - - - - - - (2) (\sin c e x i s o d d f u n c t i o n) \int_{- 1}^{1} x^{2} d x = \frac{2}{3} = w_{0} + w_{1} {x_{1}}^{2} + w_{2} - - - - - (3) \int_{- 1}^{1} x^{3} d x = 0 = - w_{0} + w_{1} {x_{1}}^{3} + w_{2} - - - - - (4) (\sin c e x^{3} i s o d d f u n c t i o n) \int_{- 1}^{1} x^{4} d x = \frac{2}{5} = w_{0} + w_{1} {x_{1}}^{4} + w_{2} - - - - - - (5) \int_{- 1}^{1} x^{5} d x = 0 = - w_{0} + w_{1} {x_{1}}^{5} + w_{2} - - - - - (6) (\sin c e x^{5} i s o d d f u n c t i o n)$