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Course: MATH 1005, Fall 2009
School: ECCD
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Word Count: 518

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Let 1. f (x) = series of f . MATH 1005B - Solutions 6 0; 1; 2 x &lt; 0 0x2 , and f (x + 2) = f(x) for all x. Find the Fourier The Fourier series of f , with L = 2, is given by 1 nx nx i a0 X h + an cos + bn sin ; 2 2 2 n=1 with bn dx 2 2 0 nx 2 n 1 = 1 [1 cos(n)] = 1 (1) ; n 1; = cos n 2 0 n n Z 2 1 = dx = 1; 2 0 f (x) sin sin 2 1 = 2 Z 2 nx 1 dx = 2 Z 2 nx a0 and for n 1, 1 an...

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Let 1. f (x) = series of f . MATH 1005B - Solutions 6 0; 1; 2 x < 0 0x2 , and f (x + 2) = f(x) for all x. Find the Fourier The Fourier series of f , with L = 2, is given by 1 nx nx i a0 X h + an cos + bn sin ; 2 2 2 n=1 with bn dx 2 2 0 nx 2 n 1 = 1 [1 cos(n)] = 1 (1) ; n 1; = cos n 2 0 n n Z 2 1 = dx = 1; 2 0 f (x) sin sin 2 1 = 2 Z 2 nx 1 dx = 2 Z 2 nx a0 and for n 1, 1 an = 2 Z 2 cos 0 Thus, the Fourier series of f is nx 2 1 nx 1 X 1 (1)n + sin : 2 n=1 n 2 2. Let f(x) = x on [1; 1]. Find the Fourier series of f . nx 2 1 = 0: dx = sin n 2 0 Since f is odd on [1; 1] with L = 1, its Fourier series is the sine series 1 X n=1 bn sin(nx); with bn = 2 Z 1 0 2 2(1)n cos(n) = ; n 1: n n Thus, the Fourier series of f is given by = 1 X 2(1)n1 n=1 # 1 Z 1 1 1 x sin(nx) dx = 2 x cos(nx) + cos(nx) dx n n 0 0 " n sin(nx): 2 3. Let f(x) = jxj on [1; 1]. Find the Fourier series of f . Since f is even on [1; 1] with L = 1, its Fourier series is the cosine series a0 X + an cos(nx); 2 n=1 with a0 = 2 an Z Z 1 1 x dx = 1; 0 1 Since (1)n 1 = 0 if n = 2k is even, and (1)n 1 = 2 if n = 2k 1 is odd, a2k = 0 4 for k 1, and a2k1 = for k 1. Thus, the Fourier series of f is given by (2k 1)2 2 1 X 4 + cos [(2k 1)x] : 2 k=1 (2k 1)2 2 4. Let f(x) = x on [0; 1]. Find the Fourier sine series of f . In order to obtain a sine series, f must be extended [1; to 1] as an odd function, i.e., f (x) = x for 1 x 1. The sine series of f is then given by the series in Problem 2. 5. Let f(x) = x on [0; 1]. Find the Fourier cosine series of f . In order to obtain a cosine series, f must be extended to [1; 1] as an even function, i.e., f(x) = jxj for 1 x 1. The cosine series of f is then given by the series in Problem 3. 6. Let f(x) = sin(x) on [0; 3]. Find the Fourier cosine series of f . In order to obtain a cosine series, f must be extended to [3; 3] as an even function, sin(x); 0 x 3 i.e., f(x) = . Then, with L = 3, the Fourier series of f sin(x); 3 x 0 reduces to the cosine series nx a0 X + an cos ; 2 3 n=1 1 1 # 1 Z 1 1 1 = 2 x cos(nx) dx = 2 sin(nx) dx x sin(nx) n n 0 0 0 1 2 2 [(1)n 1] 2 cos(nx) = 2 2 [cos(n) 1] = ; n 1: = n2 2 n n2 2 0 " 3 2 with an = 3...

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