MATH 338 Assignment 10 Solutions - MATH 338 Assignment 10 Solutions Sec 6.1 6 We can prove the orthogonality of 1 1 x(2 4x x2\/2 with respect to ex on[0

# MATH 338 Assignment 10 Solutions - MATH 338 Assignment 10...

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MATH 338 - Assignment 10 Solutions Sec 6.1 : 6* We can prove the orthogonality of 1, 1 - x , (2 - 4 x + x 2 ) / 2 with respect to e - x on [0 , ) by integrating directly, using techniques such as integration by parts. But we can observe something more general, concerning integrals of the form R 0 p ( x )e - x d x , where p ( x ) = a n x n + a n - 1 x n - 1 + · · · + a 1 x + a 0 is a polynomial. First, set I n = R 0 x n e - x d x ; we know that I 0 = Z 0 e - x d x = - e - x | 0 = 1 Next, we observe that I n = Z 0 x n e - x d x = Z 0 nx n - 1 e - x d x = nI n - 1 where we used integration by parts with u = x n , d v = e - x d x , and the fact that lim x →∞ x n e - x = 0 (which can be proved rigorously by repeated applications of l’Hopital’s rule). From these two facts we deduce I n = n ! and so Z 0 p ( x )e - x d x = n ! a n + ( n - 1)! a n - 1 + · · · + 2! a 2 + a 1 + a 0 Since 1 · (1 - x ) = - x + 1, we have a 1 = - 1 and a 0 = 1, so a 1 + a 0 = 0. Thus 1, 1 - x are orthogonal with respect to e - x on [0 , ). Likewise, 1, (2 - 4 x + x 2 ) / 2 are orthogonal with respect to e - x on [0 , ) since 1 · (2 - 4 x + x 2 ) / 2 = 1 2 x 2 - 2 x +1, we have a 2 = 1 2 , a 1 = - 2 and a 0 = 1, so 2! a 2 + a 1 + a 0 = 1 - 2 + 1 = 0. Finally, 1 - x , (2 - 4 x + x 2 ) / 2 are orthogonal with respect to e - x on [0 , ) since (1 - x ) · (2 - 4 x + x 2 ) / 2 = - 1 2 x 3 + 5 2 x 2 - 3 x + 1, so a 3 = - 1 2 , a 2 = 5 2 , a 1 = - 3 and a 0 = 1, so 3! a 3 + 2! a 2 + a 1 + a 0 = - 3 + 5 - 3 + 1 = 0.