Unformatted text preview: smooth derivative. We have f ± ( x ) = -c d if 0 < x < d, i f d < | x | < p, c d if-d < x < . The Fourier series of f ± is obtained by diﬀerentiating term by term the Fourier series of f (by Exercise 26). Now the function in this exercise is obtained by multiplying f ± ( x ) by-d c . So the desired Fourier series is-d c f ± ( x ) =-d c 2 cp dπ 2 ∞ ² n =1 1-cos dnπ p n 2 »-nπ p ¼ sin nπ p x = 2 π ∞ ² n =1 1-cos dnπ p n sin nπ p x....
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- Fall '11