Explanation The Fourier-Legendre series of a function f defined on the interval ( − 1 , 1 ) is given as follows: f ( x ) = ∑ n = 0 ∞ c n P n ( x ) c n = 2 n + 1 2 ∫ − 1 1 f ( x ) P n ( x ) d x Case 1: If P n ( x ) is an even function that is P 2 n ( x ) and f ( x ) is an even function, then f ( x ) P 2 n ( x ) is even because product of two even functions is even. Then, by the property of finite integral it can be written as follows: ∫ − 1 1 f ( x ) P 2 n ( x ) d x = 2 ∫ 0 1 f ( x ) P 2 n ( x ) d x And c 2 n is obtained as follows: c 2 n = 2 ( 2 n ) + 1 2 ∫ − 1 1 f ( x ) P 2 n ( x ) d x = 4 n + 1 2 2 ∫ 0 1 f ( x ) P 2 n ( x ) d x = ( 4 n + 1 ) ∫ 0 1 f ( x ) P 2 n ( x ) d x

Differential Equations with Boundary-Value Problems

9th Edition

ISBN: 9781337632515

Author: Dennis G. Zill

Publisher: Cengage Learning US

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Chapter 11.5, Problem 19E

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To show: If f is an even function on interval (−1,1), then f(x)=∑n=0∞c2nP2n(x) and c2n=(4n+1)∫01f(x)P2n(x) dx.

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Chapter 11.5, Problem 18E

Chapter 11.5, Problem 20E

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Differential Equations with Boundary-Value Problems

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If f is an even function on interval ( − 1 , 1 ) , then f ( x ) = ∑ n = 0 ∞ c 2 n P 2 n ( x ) and c 2 n = ( 4 n + 1 ) ∫ 0 1 f ( x ) P 2 n ( x ) d x .

Solution Summary: The author explains the Fourier-Legendre series of a function f defined on the interval (-1,1).

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To show: If f is an even function on interval (−1,1), then f(x)=∑n=0∞c2nP2n(x) and c2n=(4n+1)∫01f(x)P2n(x) dx.

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Chapter 11 Solutions