FourierSeries-1

Y 17 a show that if x2 1 52 19 use the result of

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Unformatted text preview: −π 1 2π 0 −π dx − f (x) cos nx dx = 1 π 1 f (x) sin nx dx = π 0 2 = − [1 − cos(−nπ)] = − 4 nπ nπ bn = (b) f (x) = ∞ k=0 π −π − π 0 dx = 0. 0 −π cos nx dx − 1 π π 0 cos nx dx = 0 [since cos nx is even]. 0 −π sin nx dx − 1 π π 0 sin nx dx = 2 π 0 −π sin nx dx [since sin nx is odd] if n even if n odd 4 sin(2k + 1)x when −π < x < 0 and 0 < x < π . (2k + 1)π (c) y 1 0.5 0 -2.5 -1.25 0 1.25 2.5 x -0.5 -1 3. (a) a0 = 1 2π π −π f (x) dx = 1 2π π −π x dx = 0. an = 1 π π −π f (x) cos nx dx = 1 π π −π x cos nx dx = 0 [because x cos nx is odd] bn = 1 π π −π f (x) sin nx dx = 1 π π −π x sin nx dx = =− (b) f (x) = 2 cos nπ n ∞ n=1 [using integration by parts] = (−1)n+1 2 sin nx n 2 π π 0 x sin nx dx [since x sin nx is odd] −(2/n) if n even (2/n) if n odd y (c) 2.5 when −π < x < π. 1.25 0 -2.5 -1.25 0 1.25 2.5 x -1.25 -2.5 12 ■ FOURIER SERIES 5. (a) a0 = an = bn = = 1 2π π −π 1 π 1 π (b) f (x) = 1 2π f (x) dx = π 0 cos x dx = 0 π −π f (x) cos nx dx = 1 π π 0 π −π f (x) sin nx dx = 1 π π 0 1 2 if n = 1 0 cos x cos nx dx = if n = 1 cos x sin nx dx 2n π (n2 − 1) if n even 0 if n odd 1 2 cos x + [by symmetry about x = π 2 using an integral table, and simplified using the addition formula for cos(a + b) 4k sin(2k) when −π < x < 0, 0 < x < π. π (4k2 − 1) k=1 ∞ y (c) 1 0.5 0 -2.5 -1.25 0 1.25 2.5 x -0.5 -1 0 7. Use f (x) = 1 0 a0 = 1 2L L −L 1 an = L L −L 1 L L −L bn = if − 2 ≤ x ≤ −1 if − 1 < x < 1, L = 2. if 1 ≤ x ≤ 2 f (x) dx = 1 4 1 −1 dx = nπx dx = f (x) cos L 1 2 nπx dx = L 1 2 f (x) sin 1 2 + 1 2 Fourier Series: + 1 2 1 −1 1 −1 0 2 π nπx 2/nπ dx = sin n = cos 2 nπ 2 −2/nπ sin k=1 if n = 4n + 1 if n = 4n + 3 nπx dx = 0 2 2 πx 2 3πx cos − cos π 2 3π 2 ∞ if n even + 2 5πx cos 5π 2 −··· 2 π 2 π sin (4k + 1) − sin (4k + 3) (4k + 1) π 2 (4k + 3) π 2 y 1 0.75 0.5 0.25 0 -5 -2.5 0 2.5 5 x F OURIER SERIES ■ 13 −x a0 = 1 2L an = 1 L L −L L −L if − 4 ≤ x < 0 0 9....
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