Chem Differential Eq HW Solutions Fall 2011 91

Chem Differential Eq HW Solutions Fall 2011 91 - Section...

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Section 5.6 Legendre Polynomials and Legendre Series Expansions 91 Solutions to Exercises 5.6 1. Bonnet’s relation says: For n =1 , 2 ,..., ( n +1) P n +1 ( x )+ nP n - 1 ( x )=(2 n xP n ( x ) . We have P 0 ( x ) = 1 and P 1 ( x )= x .Tak e n = 1, then 2 P 2 ( x P 0 ( x )=3 xP 1 ( x ) , 2 P 2 ( x x · x - 1 , P 2 ( x 1 2 (3 x 2 - 1) . Take n = 2 in Bonnet’s relation, then 3 P 3 ( x )+2 P 1 ( x )=5 xP 2 ( x ) , 3 P 3 ( x x ( 1 2 (3 x 2 - 1) ) - 2 x, P 3 ( x 5 2 x 3 - 3 2 x. n = 3 in Bonnet’s relation, then 4 P 4 ( x )+3 P 2 ( x )=7 xP 3 ( x ) , 4 P 4 ( x x ( 5 2 x 3 - 3 2 x ) - 3 2 ( x 2 - 1 ) , P 4 ( x 1 4 ± 35 2 x 4 - 15 x 2 + 3 2 ² . 5. By Bonnet’s relation with n =3, 7 xP 3 ( x )=4 P 4 ( x P 2 ( x ) , xP 3 ( x 4 7 P 4 ( x 3 7 P 2 ( x ) . So ³ 1 - 1 xP 2 ( x ) P 3 ( x ) dx = ³ 1 - 1 ´ 4 7 P 4 ( x 3 7 P 2 ( x ) µ P 2 ( x ) dx = 4 7 ³ 1 - 1 P 4 ( x ) P 2 ( x ) dx + 3 7 ³ 1 - 1 [ P 2 ( x )] 2 dx =0 + 3 7 2 5 = 6 35 , where we have used Theorem 1(i) and (ii) to evaluate the last two integrals.
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This note was uploaded on 12/22/2011 for the course MAP 3305 taught by Professor Stuartchalk during the Fall '11 term at UNF.

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