Sol4 - Problem 3.2.2(d The Fourier coefficients are a0 = 1 2L L xdx = 0 L 4 ak = 1 kx dx = x sin kx L k L 0 L L = 2 2(cos(k 1 = 2 2-1)k 1 k k 1 L x

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Unformatted text preview: Problem 3.2.2(d) The Fourier coefficients are: a0 = 1 2L L xdx = 0 L , 4 ak = 1 kx dx = x sin kx L k L 0 L L = 2 2 (cos(k) - 1) = 2 2 ((-1)k - 1), k k 1 L x cos L L L - 0 0 sin kx dx L and bk = 1 L L x sin 0 1 kx dx = - x cos kx L k L L 0 L L - 0 0 cos kx dx L L L + 2 sin kx = (-1)k+1 k k L L = (-1)k+1 . k Problem 3.3.2(b) The Fourier coefficients equal bk = = 2 L L/6 sin 0 kx dx + 3 L - 3 cos L/2 sin L/6 kx dx L 2 k 1 + 2 cos k 6 k 2 . Problem 2.3.5(c) The Fourier coefficients equal a0 = and ak = 2 L L 1 L L xdx = L/2 3L , 8 x cos L/2 2 kx dx = x sin kx L k L L L - L/2 L/2 sin kx dx L =- L 2L sin k + 2 2 (-1)k - cos k k 2 k 2 Problem 3.3.15 . The property of a function f (x) to be odd around x = L/2 means that (1) f L -y 2 = -f 1 L +y . 2 2 Then, for even coefficients of the Fourier cosine series one gets a2k = 2 L L L f (x) cos 0 2k 2 x dx = L L L/2 f (x) cos 0 2k x dx L 2k + f (x) cos x dx . L L/2 In the first integral, we make a substitution x = (L/2) - y, and in the second integral we make a substitution x = (L/2) + y. Notice that cos and cos so a2k = (-1)k because of (1). 2 L 0 2k L 2k L L -y 2 L +y 2 L/2 = cos k - 2k y L 2k y L = (-1)k cos 2k y L 2k y , L = cos k + = (-1)k cos f L L -y +f +y 2 2 cos 2k y dy = 0 L ...
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This note was uploaded on 09/02/2008 for the course MAE 105 taught by Professor Neiman-nassat during the Summer '07 term at UCSD.

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Sol4 - Problem 3.2.2(d The Fourier coefficients are a0 = 1 2L L xdx = 0 L 4 ak = 1 kx dx = x sin kx L k L 0 L L = 2 2(cos(k 1 = 2 2-1)k 1 k k 1 L x

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