# H7_130 - Math 415 Homework 7 10.2.21 f(x = x2/2 2 x 2 f(x 4...

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Math 415 Homework 7 10.2.21 f ( x ) = x 2 / 2 , - 2 x 2; f ( x + 4) = f ( x ) . (a) Sketch the graph of the given function for three periods. -6 -4 -2 0 2 4 6 1 1 2 3 4 5 (b) Find the Fourier series. a 0 = 1 2 Z 2 - 2 x 2 2 dx = 1 12 x 3 ± ± ± ± 2 - 2 = 4 3 . a m = 1 4 Z 2 - 2 x 2 cos mπx 2 = ² 1 2 x 2 sin mπx 2 ± ± ± ± 2 - 2 - 1 Z 2 - 2 x sin mπx 2 dx = ² 1 2 x 2 sin mπx 2 ± ± ± ± 2 - 2 + ² 2 m 2 π 2 x cos mπx 2 ± ± ± ± 2 - 2 - 2 m 2 π 2 Z 2 - 2 cos mπx 2 dx = ² 1 2 x 2 sin mπx 2 ± ± ± ± 2 - 2 + ² 2 m 2 π 2 x cos mπx 2 ± ± ± ± 2 - 2 - ² 4 m 3 π 3 sin mπx 2 ± ± ± ± 2 - 2 = 0 + 8 m 2 π 2 ( - 1) m + 0 . Finally, since f is an even function and sine is an odd function b m = R 2 - 2 x 2 sin( mπx/ 2) dx is the integral of an odd function over a symmetric domain, so is zero. Hence the Fourier series is 2 3 + X m =1 8( - 1) m m 2 π 2 cos mπx 2 . (c) Plot s m ( x ) versus x for m = 5, 10, and 20. See ﬁgure 1. (d) By the Fourier convergence theorem s m f as m → ∞ since f has no discontinuities. But looking at ﬁgure 1 we can see that it converges quite quickly everywhere but at x = 2 + 4 k for any integer k . 1

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01 2 1 2 Figure 1: The dotted line is a graph of x 2 that I included for reference. If you above x = 2, the graph closest to y = x 2 is s 20 , the graph furthest is s 5 . 10.3.17 Assuming that f ( x ) = a 0 2 + X n =1 ± a n cos nπx L + b n sin nπx L ² , (1) Show that 1 L Z L - L [ f ( x )]
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H7_130 - Math 415 Homework 7 10.2.21 f(x = x2/2 2 x 2 f(x 4...

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