Differential Equations Lecture Work Solutions 32

Differential Equations Lecture Work Solutions 32 - 3 2 1 0...

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Unformatted text preview: 3 2 1 0 −1 −2 −3 −10 −8 −6 −4 −2 0 2 4 6 8 10 Figure 12: Graph of f (x) = x 3 2 1 0 −1 −2 −3 −10 −8 −6 −4 −2 0 2 4 6 8 10 Figure 13: Graph of its periodic extension 2. a. f (x) = x Since the periodic extension of f (x) is discontinuous, the Fourier series is identical to (the periodic extension of) f (x) everywhere except at the points of discontinuities. At those point x = ±L (and similar points in each period), we have L + (−L) =0 2 Now we evaluate the coeﬃcients. a0 = 1 L L −L f (x) dx = 1 L L −L x dx = 0 Since we have integrated an odd function on a symmetric interval. Similarly for all an . bn = 1 L L −L f (x) sin −x cos nπ x L 1 nπ x dx = L L nπ L |L L + − L cos nπ x L nπ L −L This was a result of integration by parts. = −L cos nπ 1 L nπ L + −L cos(−nπ ) nπ L 32 + sin nπ x L L 2 | −L nπ L dx ...
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