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# sol4 - Problem 3.2.2(d The Fourier coecients are a0 = 1 2L...

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Problem 3.2.2(d) The Fourier coefficients are: a 0 = 1 2 L L 0 xdx = L 4 , a k = 1 L L 0 x cos π L kx dx = 1 πk x sin π L kx L 0 - L 0 sin π L kx dx = L π 2 k 2 (cos( πk ) - 1) = L π 2 k 2 (( - 1) k - 1) , and b k = 1 L L 0 x sin π L kx dx = - 1 πk x cos π L kx L 0 - L 0 cos π L kx dx = ( - 1) k +1 L πk + L π 2 k sin π L kx L 0 = ( - 1) k +1 L πk . Problem 3.3.2(b) The Fourier coefficients equal b k = 2 L L/ 6 0 sin π L kx dx + 3 L/ 2 L/ 6 sin π L kx dx = 2 πk 1 + 2 cos πk 6 - 3 cos πk 2 . Problem 2.3.5(c) The Fourier coefficients equal a 0 = 1 L L L/ 2 xdx = 3 L 8 , and a k = 2 L L L/ 2 x cos π L kx dx = 2 πk x sin π L kx L L/ 2 - L L/ 2 sin π L kx dx = - L πk sin π 2 k + 2 L π 2 k 2 ( - 1) k - cos π 2 k . Problem 3.3.15 The property of a function f ( x ) to be odd around x = L/ 2 means that (1) f L 2 - y = - f L 2 + y . 1

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2 Then, for even coefficients of the Fourier cosine series one gets a 2 k = 2 L L 0 f ( x ) cos 2 πk L x dx = 2 L L/ 2 0 f ( x ) cos
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