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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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This note was uploaded on 12/22/2011 for the course MAP 2302 taught by Professor Bell,d during the Fall '08 term at UNF.
 Fall '08
 BELL,D

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