Explanation Given: The function is f ( x ) = e x , − 1 < x < 1 . Calculation: The Fourier-Legendre series of a function f ( x ) defined on the interval ( − 1 , 1 ) is given as follows: f ( x ) = ∑ n = 0 ∞ c n P n ( x ) ...... ( 1 ) Here, c n is defined as follows: c n = 2 n + 1 2 ∫ − 1 1 f ( x ) P n ( x ) d x ...... ( 2 ) Substitute 0 for n in equation (2) and solve. c 0 = 2 ( 0 ) + 1 2 ∫ − 1 1 f ( x ) P 0 ( x ) d x = 1 2 ∫ − 1 1 e x ⋅ 1 d x = 1 2 ∫ − 1 1 e x d x = 1 2 ( e − e − 1 ) Substitute 1 for n in equation (2) and solve. c 1 = 2 ( 1 ) + 1 2 ∫ − 1 1 f ( x ) P 1 ( x ) d x = 3 2 ∫ − 1 1 e x ⋅ x d x = 3 2 [ x e x − e x ] − 1 1 = 3 e − 1 Substitute 2 for n in equation (2) and solve. c 2 = 2 ( 2 ) + 1 2 ∫ − 1 1 f ( x ) P 2 ( x ) d x = 5 2 ∫ − 1 1 e x ⋅ 1 2 ( 3 x 2 − 1 ) d x = 5 4 [ ( 3 x 2 − 1 ) e x − ∫ 6 x e x d x ] − 1 1 = 5 4 [ ( 3 x 2 − 1 ) e x − 6 ( x e x − e x ) ] − 1 1 Simplify further as follows. c 2 = 5 4 [ ( 3 ( 1 ) 2 − 1 ) e 1 − 6 ( 1 e 1 − e 1 ) − ( 3 ( − 1 ) 2 − 1 ) e − 1 + 6 ( ( − 1 ) e − 1 − e − 1 ) ] = 5 4 [ 2 e − 2 e − 1 − 12 e − 1 ] = 5 4 [ 2 e − 14 e − 1 ] = 5 2 [ e − 7 e − 1 ] Substitute 3 for n in equation (2) and solve. c 3 = 2 ( 3 ) + 1 2 ∫ − 1 1 f ( x ) P 3 ( x ) d x = 7 2 ∫ − 1 1 e x ⋅ 1 2 ( 5 x 3 − 3 x ) d x = 7 4 [ ( 5 x 3 − 3 x ) e x − ∫ ( 15 x 2 − 3 ) e x d x ] − 1 1 = 7 4 [ ( 5 x 3 − 3 x ) e x − ( 15 x 2 − 3 ) e x + ∫ 30 x e x d x ] − 1 1 Simplify further as follows

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 16E

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The first five nonzero terms in the Fourier-Legendre expansion for the function f(x)=ex, −1<x<1.

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Chapter 11.5, Problem 15E

Chapter 11.5, Problem 17E

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Chapter 11 Solutions

Differential Equations with Boundary-Value Problems

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The first five nonzero terms in the Fourier-Legendre expansion for the function f ( x ) = e x , − 1 < x < 1 .

Solution Summary: The author calculates the Fourier-Legendre series of a function f(x) defined on the interval

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The first five nonzero terms in the Fourier-Legendre expansion for the function f(x)=ex, −1<x<1.

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Chapter 11 Solutions

The first five nonzero terms in the Fourier-Legendre expansion for the function f ( x ) = e x , − 1 &lt; x &lt; 1 .

Solution Summary: The author calculates the Fourier-Legendre series of a function f(x) defined on the interval

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The first five nonzero terms in the Fourier-Legendre expansion for the function f(x)=ex, −1<x<1.

Want to see the full answer?

Chapter 11 Solutions

The first five nonzero terms in the Fourier-Legendre expansion for the function f ( x ) = e x , − 1 < x < 1 .