527solns9 - 642:527 SOLUTIONS ASSIGNMENT 9 FALL 2009...

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642:527 SOLUTIONS: ASSIGNMENT 9 FALL 2009 Section 17.3: 4 (g) f ( x ) = | sin x | has period π (since sin( x + π ) = - sin x ) so its Fourier series will have the form FS f = a 0 + summationdisplay n =1 [ a n cos 2 nx + b n sin 2 nx ] . Since f ( x ) is even ( | sin( - x ) | = | - sin x | = | sin x | ) the coefficients b n will all vanish. Moreover, using sin θ = ( e - e ) / 2 i and cos θ = ( e + e ) / 2, we have a 0 = 1 π integraldisplay π 0 f ( x ) dx = 1 π integraldisplay π 0 sin( x ) dx = 2 π , a n = 2 π integraldisplay π 0 f ( x ) cos(2 nx ) dx = 2 π integraldisplay π 0 sin( x ) cos(2 nx ) dx = 1 2 integraldisplay π 0 bracketleftBig e (2 n +1) ix - e (2 n 1) ix + e ( 2 n +1) ix - e ( 2 n 1) ix bracketrightBig dx = - 1 2 π bracketleftbigg e (2 n +1) ix 2 n + 1 - e (2 n 1) ix 2 n - 1 - e (2 n 1) ix 2 n - 1 + e (2 n +1) ix 2 n + 1 bracketrightbigg π 0 = - 4 π (4 n 2 - 1) . That is, FS f = 2 π - 4 π summationdisplay n =1 cos 2 nx 4 n 2 - 1 . 16 (b) The complex Fourier series of f ( x ) is summationdisplay n = −∞ c n e iπnx with c n = 1 2 integraldisplay 2 0 f ( x ) e iπnx dx = 1 2 integraldisplay 2 0 e (1 iπn ) x dx = e 2 - 1 2(1 - iπn ) . Section 17.4: 1 (b) To derive the QRC formulas (5) we start with the function f ( x ) defined on [0 , L ] and let f 3 ( x ) be the extension which is symmetric around x = 0 and antisymmetric around x = L (see Figure 3(c)); specifically we define f 3 ( x ) = f ( x ) for 0 x L , then f 3 ( x ) = - f 3 (2 L - x ) for L < x 2 L , and then f 3 ( x ) = f 3 ( - x ) for - 2 L < x < 0; f 3 ( x ) is now defined for - 2 L < x 2 L and we extend it to a function periodic of period 4 L defined for all x . The QRC for f is just the Fourier series of f 3 : FS f 3 = a 0 + summationdisplay n =1 bracketleftBig a n cos nπx 2 L + b n sin nπx 2 L bracketrightBig . (12.1) Here (repeatedly using integraltext 2 L L g ( x ) dx = integraltext L 0 g (2 L - x ) dx from the change of variables x 2 L - x ) a 0 = 1 4 L integraldisplay 2 L 2 L f 3 ( x ) dx = 1 2 L integraldisplay 2 L 0 f 3 ( x ) dx 1
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642:527 SOLUTIONS: ASSIGNMENT 9 FALL 2009 = 1 2 L parenleftBigg integraldisplay L 0 f ( x ) dx - integraldisplay 2 L L f (2 L - x ) dx parenrightBigg = 0 , a n = 1 2 L integraldisplay 2 L 2 L f 3 ( x ) cos nπx 2 L dx = 1 L integraldisplay 2 L 0 f 3 ( x ) cos nπx 2 L dx = 1 L parenleftBigg integraldisplay L 0 f ( x ) cos nπx 2 L dx - integraldisplay 2 L L f (2 L - x ) cos nπx 2 L dx parenrightBigg = 1 L parenleftBigg integraldisplay L 0 f ( x ) cos nπx 2 L dx - integraldisplay L 0 f ( x ) cos (2 L - x ) 2 L dx parenrightBigg (12.2) = 1 - ( - 1) n L integraldisplay L 0 f ( x ) cos nπx 2 L dx = 2 L integraldisplay L 0 f ( x ) cos nπx 2 L dx, n odd, 0 , n even b n = 1 2 L integraldisplay 2 L 2 L f 3 ( x ) sin nπx 2 L dx = 0 . In finding a n we have used cos( - θ ) = cos cos θ + sin sin θ = ( - 1) n cos θ , and in finding b n = 0 we have used the fact that f 3 ( x ) sin( nπx/ 2 L ) is odd. Equations (12.1) and (12.2) correspond to (5) of the text.
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