Explanation Given: p ( x ) = 2 x 3 + x 2 + 5 . Approach: The used theorem is as follows, Let f and f ′ be continuous on the interval ( − 1 , 1 ) . Then, for − 1 < x < 1 , f ( x ) = a 0 P 0 ( x ) + a 1 P 1 ( x ) + ... + a n P n ( x ) + ... = ∑ n = 0 ∞ a n P n ( x ) . Here, a n = 2 n + 1 2 ∫ − 1 1 f ( x ) P n ( x ) d x . The first five Legendre polynomials are given in table below, N Legendre Polynomial of Degree N 0 P 0 ( x ) = 1 1 P 1 ( x ) = x 2 P 2 ( x ) = 1 2 ( 3 x 2 − 1 ) 3 P 3 ( x ) = 1 2 x ( 5 x 2 − 3 ) 4 P 4 ( x ) = 1 8 ( 35 x 4 − 30 x 2 + 3 ) Table ( 1 ) Calculation: Write the equation to find the value of a n . a n = 2 n + 1 2 ∫ − 1 1 f ( x ) P n ( x ) d x a n = 2 n + 1 2 ∫ − 1 1 ( 2 x 3 + x 2 + 5 ) P n ( x ) d x ... ( 1 ) The above function has 3 degree so it can be represent as, f ( x ) = a 0 P 0 ( x ) + a 1 P 1 ( x ) + a 3 P 3 ( x ) 2 x 3 + x 2 + 5 = a 0 P 0 ( x ) + a 1 P 1 ( x ) + a 2 P 2 ( x ) + a 3 P 3 ( x ) ... ( 2 ) Substitute 0 for n and 1 for P 0 in equation to find the value of a 0 . a 0 = 2 ( 0 ) + 1 2 ∫ − 1 1 ( 2 x 3 + x 2 + 5 ) P 0 ( x ) d x = 0 + 1 2 ∫ − 1 1 ( 2 x 3 + x 2 + 5 ) ( 1 ) d x = 16 3 Substitute 1 for n and x for P 1 in equation to find the value of a 1

Differential Equations and Linear Algebra (4th Edition)

4th Edition

ISBN: 9780321964670

Author: Stephen W. Goode, Scott A. Annin

Publisher: PEARSON

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Chapter 11.3, Problem 5P

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The linear combination of Legendre Polynomial.

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Differential Equations and Linear Algebra (4th Edition)

Ch. 11.1 - True-False Review For Questions a-j, decide if the...Ch. 11.1 - Prob. 2TFR Ch. 11.1 - Prob. 3TFR Ch. 11.1 - Prob. 4TFR Ch. 11.1 - Prob. 5TFR Ch. 11.1 - True-False Review For Questions a-j, decide if the...Ch. 11.1 - Prob. 7TFR Ch. 11.1 - Prob. 8TFR Ch. 11.1 - Prob. 9TFR Ch. 11.1 - Prob. 10TFR

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