Explanation The given function is f ( x ) = π 2 − x 2 , − π < x < π . Consider that if the given function is odd or even then use the sine or cosine series accordingly. If the function is odd then the sine series is evaluated and if it is even then the cosine series is evaluated. f ( x ) = π 2 − x 2 , − π < x < π . If the value of x is such that − π < x < π then the value of f ( x ) is f ( − x ) = π 2 − x 2 = π 2 − ( − x ) 2 = π 2 − x 2 It can be seen that f ( − x ) = f ( x ) Thus, the function is even in this interval. Hence, the cosine series has to be evaluated. Now, the cosine series of an even function is given by f ( x ) = a 0 2 + ∑ n = 1 ∞ a n cos n π x L . (1) Where a n is the coefficient given by a n = 2 L ∫ 0 L f ( x ) cos n π x L d x , n ∈ N First calculate the coefficient a n , a 0 = 2 π ∫ 0 π ( π 2 − x 2 ) d x = 2 π [ π 2 x − x 3 3 ] | 0 π = 2 π [ π 3 − π 3 3 ] = 4 π 2 3 Now, evaluate the next coefficients a n using the formula a n = 2 L ∫ 0 L f ( x ) cos n π x L d x , n ∈ N Put the value of f ( x ) in above equation

WEB ASSIGN FOR ZILL'S DIFFERENTIAL EQUAT

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ISBN: 9780357539545

Author: ZILL

Publisher: CENGAGE L

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Chapter 11.3, Problem 17E

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To expand the function f(x)=π2−x2, −π<x<π in sine or cosine series accordingly.

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Chapter 11.3, Problem 16E

Chapter 11.3, Problem 18E

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WEB ASSIGN FOR ZILL'S DIFFERENTIAL EQUAT

Ch. 11.1 - In problem 16 show that the given functions are...Ch. 11.1 - In problem 16 show that the given functions are...Ch. 11.1 - In problem 16 show that the given functions are...Ch. 11.1 - In problem 16 show that the given functions are...Ch. 11.1 - In problem 16 show that the given functions are...Ch. 11.1 - Prob. 6E Ch. 11.1 - Prob. 7E Ch. 11.1 - In Problems 712 show that the given set of...Ch. 11.1 - In Problems 712 show that the given set of...Ch. 11.1 - In Problems 712 show that the given set of...

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To expand the function f ( x ) = π 2 − x 2 , − π < x < π in sine or cosine series accordingly.

Solution Summary: The author explains how the function f(x) can be expanded in the cosine series.

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To expand the function f(x)=π2−x2, −π<x<π in sine or cosine series accordingly.

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Chapter 11 Solutions

To expand the function f ( x ) = π 2 − x 2 , − π &lt; x &lt; π in sine or cosine series accordingly.

Solution Summary: The author explains how the function f(x) can be expanded in the cosine series.

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To expand the function f(x)=π2−x2, −π<x<π in sine or cosine series accordingly.

Want to see the full answer?

Chapter 11 Solutions

To expand the function f ( x ) = π 2 − x 2 , − π < x < π in sine or cosine series accordingly.