Chem Differential Eq HW Solutions Fall 2011 95

Chem Differential Eq HW Solutions Fall 2011 95 - Section...

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Section 6.2 Sturm-Liouville Theory 95 For the third integral, we have 1 - 1 g ( x ) h ( x ) w ( x ) dx = 1 - 1 2 x ( - 1 + 4 x 2 ) 1 - x 2 dx = 0 , because we are integrating an odd function over a symmetric interval. 9. In order for the functions 1 and a + bx + x 2 to be orthogonal, we must have 1 - 1 1 · ( a + bx + x 2 ) dx = 0 Evaluating the integral, we find ax + b 2 x 2 + 1 3 x 3 1 - 1 = 2 a + 2 3 = 0 a = - 1 3 . In order for the functions x and 1 3 + bx + x 2 to be orthogonal, we must have 1 - 1 1 · ( 1 3 + bx + x 2 ) xdx = 0 Evaluating the integral, we find 1 6 x 2 + b 3 x 3 + 1 4 x 4 1 - 1 = b 3 = 0 b = 0 . 13. Using Theorem 1, Section 5.6, we find the norm of P n ( x ) to be P n = 1 - 1 P n ( x ) 2 dx 1 2 = 2 2 n + 1 1 2 = 2 2 n + 1 . Thus the orthonormal set of functions obtained from the Legendre polynomials is 2 2 n
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