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11_1_a

# 11_1_a - f x = a 2 ∞ X n =1 a n cos nx(19 = 1(20 18 1...

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11/1 hw 10.4 4, 5, 9, 17, 18(graded) November 2, 2010 10.4.4 Determine if the function is even, odd, or neither. g ( x ) = sec x (1) = 1 cos x (2) evaluating g ( - x ), g ( - x ) = 1 cos - x (3) = 1 cos x (4) = g ( x ) (5) so the function is even. 5 g ( x ) = | x | 3 (6) evaluating g ( - x ) g ( - x ) = | - x | 3 (7) = | x | 3 (8) = g ( x ) (9) so the function is even. 9 A function f is given on an interval of length L . Sketch the even and odd extensions of f of period 2 L . f ( x ) = 2 - x, 0 < x < 2 (10) Figure 1: even extension of f Figure 2: odd extension of f 17 & 18 a) Find the Fourier series for the given function b) Sketch the graph of the function to which the series converges over 3 periods 17 f ( x ) = 1 , 0 x π ; cosine series period 2 π (11) To obtain the cosine series we take the even extension of f ( x ) and hence all the sine coefficients , b n , will be zero. Computing a 0 and a n , a 0 = 2 L L 0 f ( x ) dx (12) = 2 L L 0 dx (13) = 2 (14) a n = 2 L L 0 f ( x ) cos nπx L dx (15) = 2 π π 0 cos nxdx (16) = 2 π sin nx n π 0 (17) = 0 (18) The Fourier cosine series is,

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Unformatted text preview: f ( x ) = a 2 + ∞ X n =1 a n cos nx (19) = 1 (20) 18 1 Figure 3: graph of the function to which the series con-verges f ( x ) = 1 , < x < π ; sine series period 2 π (21) To obtain the sine series we take the odd extension of f ( x ) and hence all the cosine coeﬃcients , a n , will be zero. Computing b n , b n = 2 L Z π f ( x ) sin nπx L dx (22) = 2 π Z π sin nxdx (23) = 2 nπ-cos nx ± ± ± π (24) = 2 nπ ² 2 , n = odd , n = even (25) the corresponding Fourier sine series is, f ( x ) = ∞ X n =1 b n sin nx (26) = 4 π ∞ X n = odd sin nx n (27) = 4 π ∞ X n =1 sin(2 n-1) x 2 n-1 (28) Figure 4: graph of the function to which the series con-verges 2...
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