2011_Solution_5 - MA 1505 Mathematics I Tutorial 5...

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MA 1505 Mathematics I Tutorial 5 Solutions 1. Rewrite the function: f ( x ) = 1 2 ( x + | x | ) = ( 0 - π < x < 0 x 0 < x < π The Fourier series of f ( x ) is given by a 0 + X n =1 ( a n cos nx + b n sin nx ) . a 0 = 1 2 π Z π 0 x dx = π 4 . a n = 1 π Z π 0 x cos nx dx = 1 π x sin nx n + cos nx n 2 π 0 = ( - 1) n - 1 πn 2 . b n = 1 π Z π 0 x sin nx dx = 1 π - x cos nx n + sin nx n 2 π 0 = ( - 1) n +1 n . So the Fourier series is f ( x ) = π 4 + X n =1 ( - 1) n - 1 πn 2 cos nx + ( - 1) n +1 n sin nx ± . More explicitly, we have f ( x ) = π 4 - 2 π ² cos x + cos3 x 3 2 + cos5 x 5 2 + ··· + ² sin x - sin2 x 2 + sin3 x 3 - sin4 x 4 + ··· 2. From the graph, the function is given by : f ( x ) = ( 2 - π < x < 0 1 0 < x < π The Fourier series of f ( x ) is given by a 0 + X n =1 ( a n cos nx + b n sin nx ) . a 0 = 1 2 π Z 0 - π 2 dx + 1 2 π Z π 0 1 dx = 3 2 . a n = 1 π Z 0 - π 2cos nx dx + 1 π Z π 0 cos nx dx = 0.
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MA1505 Tutorial 5 Solutions b n = 1 π Z 0 - π 2sin nx dx + 1 π Z
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This note was uploaded on 01/05/2012 for the course MATHEMATIC MA1505 taught by Professor Freudleong during the Spring '10 term at National University of Singapore.

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2011_Solution_5 - MA 1505 Mathematics I Tutorial 5...

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