Explanation Definition used: Fourier series with period 2 π : An infinite sum of sine and cosine waves with period 2 π is, f ( x ) = { a 0 + a 1 cos x + a 2 cos 2 x + ⋅ ⋅ ⋅ + a n cos n x + ⋅ ⋅ ⋅ + b 1 sin x + b 2 sin 2 x + ⋅ ⋅ ⋅ + b n sin n x + ⋅ ⋅ ⋅ } . Here, the values of a 0 = 1 2 π ∫ − π π f ( x ) d x , a n = 1 π ∫ − π π f ( x ) cos n x d x and b n = 1 π ∫ − π π f ( x ) sin n x d x . Calculation: It is given that the function is f ( x ) = { − x − π ≤ x < 0 x 0 ≤ x < π . Obtain the value of a 0 as follows. a 0 = 1 2 π ∫ − π 0 ( − x ) d x + 1 2 π ∫ 0 π ( x ) d x = − 1 2 π [ x 2 2 ] − π 0 + 1 2 π [ x 2 2 ] 0 π = − 1 4 π [ ( 0 ) 2 − ( − π ) 2 ] + 1 4 π [ ( π ) 2 − ( 0 ) 2 ] = 1 4 π [ π 2 ] + 1 4 π [ π 2 ] That is, a 0 = 1 4 π [ 2 π 2 ] = π 2 . Obtain the value of a n as follows. a n = 1 π ∫ − π 0 ( − x ) cos n x d x + 1 π ∫ 0 π ( x ) cos n x d x = − 1 π [ x sin n x n + cos n x n 2 ] − π 0 + 1 π [ x sin n x n + cos n x n 2 ] 0 π = − 1 π [ 0 sin n 0 n + cos n 0 n 2 − ( − π ) ⋅ sin n ( − π ) n − cos n ( − π ) n 2 ] + 1 π [ π sin n π n + cos n π n 2 − 0 ⋅ sin 0 n − cos 0 n 2 ] = − 1 π [ 0 + 1 n 2 + 0 − ( − 1 ) n n 2 ] + 1 π [ 0 + ( − 1 ) n n 2 − 0 − 1 n 2 ] That is, a n = 2 π [ ( − 1 ) n n 2 − 1 n 2 ] = 2 n 2 π [ ( − 1 ) n − 1 ] . When n = 1 , the value of a 1 becomes a 1 = 2 ( 1 ) 2 π ( ( − 1 ) − 1 ) = − 4 π

11th Edition

ISBN: 9780134437705

Author: Washington

Publisher: PEARSON

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Chapter 30.6, Problem 11E

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To find: The Fourier series of the function.

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Chapter 30 Solutions

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Ch. 30.1 - Prob. 1PE Ch. 30.1 - Prob. 2PE Ch. 30.1 - Prob. 1E Ch. 30.1 - Prob. 2E Ch. 30.1 - Prob. 3E Ch. 30.1 - Prob. 4E Ch. 30.1 - Prob. 5E Ch. 30.1 - Prob. 6E Ch. 30.1 - Prob. 7E Ch. 30.1 - Prob. 8E

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