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

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Unformatted text preview: 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 + ∞ 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 iθ- e − iθ ) / 2 i and cos θ = ( e iθ + e − iθ ) / 2, we have a = 1 π integraldisplay π f ( x ) dx = 1 π integraldisplay π sin( x ) dx = 2 π , a n = 2 π integraldisplay π f ( x ) cos(2 nx ) dx = 2 π integraldisplay π sin( x ) cos(2 nx ) dx = 1 2 iπ integraldisplay π 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 =- 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 f ( x ) e − iπnx dx = 1 2 integraldisplay 2 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 + ∞ 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 g (2 L- x ) dx from the change of variables x → 2 L- x ) a = 1 4 L integraldisplay 2 L − 2 L f 3 ( x ) dx = 1 2 L integraldisplay 2 L f 3 ( x ) dx 1 642:527 SOLUTIONS: ASSIGNMENT 9 FALL 2009 = 1 2 L parenleftBigg integraldisplay L 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 f 3 ( x ) cos nπx 2 L dx = 1 L parenleftBigg integraldisplay L 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 f ( x ) cos nπx 2 L dx- integraldisplay L f ( x ) cos nπ (2 L- x ) 2 L dx parenrightBigg (12.2) = 1- (- 1) n L integraldisplay L f ( x ) cos nπx 2 L dx = 2 L integraldisplay L f ( x ) cos nπx 2 L dx, n odd, , n even b n = 1 2 L integraldisplay 2 L − 2 L f 3 ( x ) sin nπx 2 L dx = 0 ....
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This note was uploaded on 12/15/2009 for the course MATH 527 taught by Professor Staff during the Fall '08 term at Rutgers.

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

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