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Solution_4_09S2

Solution_4_09S2 - MA 1505 Mathematics I Tutorial 4...

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Unformatted text preview: MA 1505 Mathematics I Tutorial 4 Solutions (AY2009/2010 Semester 2) 1. Rewrite the function: 1 f (x) = (x + |x|) = 2 0 −π < x < 0 x 0<x<π The Fourier series of f (x) is given by ∞ a0 + (an cos nx + bn sin nx). n=1 a0 = π 1 2π 1 an = π 1 bn = π x dx = 0 π π . 4 1 x sin nx cos nx + π n n2 x cos nx dx = 0 π 0 1 x cos nx sin nx x sin nx dx = − + π n n2 π = 0 (−1)n − 1 . πn2 π = 0 (−1)n+1 . n So the Fourier series is π f (x) = + 4 (−1)n − 1 (−1)n+1 cos nx + sin nx . πn2 n ∞ n=1 More explicitly, we have f (x) = π 2 − 4π cos x + cos 3x cos 5x + + ··· 32 52 + sin x − sin 2x sin 3x sin 4x + − + ··· 2 3 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 ∞ a0 + (an cos nx + bn sin nx). n=1 a0 = an = 1 2π 1 π 0 2 dx + −π 0 1 2π π 1 dx = 0 2 cos nx dx + −π 1 π 3 . 2 π cos nx dx = 0. 0 MA1505 Tutorial 4 Solutions bn = = = = = = 1 π 0 2 sin nx dx + −π 0 2 cos nx 1 − π n 1 π 1 π 1 π + −π −2 + 2 cos nπ n cos nπ − 1 n (−1)n − 1 n 0 −2 π (2m−1) π sin nx dx 0 1 cos nx − π n + 1 π π 0 − cos nπ + 1 n if n = 2m is even if n = 2m − 1 is odd. So the Fourier series is f (x) = 1 π 3 2 − 2π ∞ n=1 sin(2n − 1)x . 2n − 1 More explicitly, we have f (x) = 32 − 2π sin x sin 3x sin 5x + + + ··· 1 3 5 . 3. The graph of f is given as follow: Since the graph is symmetrical about y -axis, f (x) is an even function. So bn = 0 for all n. ∞ The Fourier series of f (x) is given by a0 + (an cos nx). n=1 a0 = 2 1 2π π /2 1 dx 0 1 =. 2 2 MA1505 Tutorial 4 Solutions π /2 1 π an = 2 cos nx dx 0 2 sin nx π/2 π n 0 nπ 2 sin πn 2 0 2 m+1 π (2m−1) (−1) = = = if n = 2m is even if n = 2m − 1 is odd. So the Fourier series is f (x) = 4. The period 2L = 2π w ⇒L= 12 + 2π ∞ (−1)n+1 n=1 cos(2n − 1)x . 2n − 1 π w. The Fourier series of u(t) is given by ∞ a0 + (an cos nwt + bn sin nwt). n=1 π /w w 2π sin wt dt = 1 . π an = a0 = w π 0 = = = = π /w sin wt cos nwt dt 0 π /w w 2π [sin(1 − n)wt + sin(1 + n)wt] dt 0 w cos(1 − n)wt cos(1 + n)wt π/w − − (∗) 2π (1 − n)w (1 + n)w 0 1 − cos(1 − n)π + 1 − cos(1 + n)π + 1 + 2π 1−n 1+n 0 if n is odd, n = 1 if n is even. −2 (n−1)(n+1)π Note that the ﬁrst term in (∗) is not deﬁned at n = 1. Separately, a1 is computed below: a1 = w 2π π /w (0 + sin 2wt)dt 0 cos 2wt w − 2π 2w = 0. π /w = 3 0 MA1505 Tutorial 4 Solutions bn = = π /w w π sin wt sin nwt dt 0 w 2π π /w [cos(1 − n)wt − cos(1 + n)wt] dt 0 w sin(1 − n)wt sin(1 + n)wt − 2π (1 − n)w (1 + n)w 1 sin(1 − n)π sin(1 + n)π − = 2π 1−n 1+n = 0 if n ≥ 2. π /w = (∗∗) 0 Note that the ﬁrst term in (∗∗) is not deﬁned at n = 1. Separately, b1 is computed below: b1 = = = w 2π π /w (1 − cos 2wt) dt 0 w sin 2wt t− 2π 2w 1 . 2 π /w 0 So the Fourier series is u(t) = 11 2 + sin wt − π2 π 1 1 cos 2wt + cos 4wt + · · · 1·3 3·5 This can also be expressed as: 11 2 u(t) = + sin wt − π2 π ∞ 1 cos 2nwt. (2n − 1)(2n + 1) n=1 4 . MA1505 Tutorial 4 Solutions 5. Note that this function is an odd function with period 2L = 4: an = 0 for all n. bn = = = = = 2 L L nπx dx L 0 2 nπx dx (2 − x) sin 2 0 f (x) sin (2 − x) −2 nπ −4 − 0− nπ cos nπx 2 2 nπ 2 2 2 − (−1) 0 0 nπx sin 2 −2 nπ 2 0 4 . nπ So the Fourier series is 4 f (x) = π ∞ nπx 1 sin . n 2 n=1 6. Fourier sine half range expansion: 2π 2 x cos nx sin nx x sin nx dx = − + π0 π n n2 So the Fourier sine half range expansion is π bn = = 0 ∞ f (x) = (−1)n+1 n=1 5 (−1)n+1 2 . n 2 sin nx. n cos nπx dx 2 MA1505 Tutorial 4 Solutions Fourier cosine half range expansion: a0 = 1 π π x dx = 0 π . 2 2π x cos nx dx π0 2 x sin nx cos nx π + = π n n2 0 (−1)n − 1 =2 πn2 0 if n = 2m is even = −4 if n = 2m 1 is odd. π (2m−1)2 an = So the Fourier cosine half range expansion is ∞ f (x) = π 4 cos(2n − 1)x − . 2 π n=1 (2n − 1)2 6 ...
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