pset_5sol

# pset_5sol - ACM 95/100b Problem Set 5 Solutions Faisal...

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ACM 95/100b Problem Set 5 Solutions Faisal Amlani February 17, 2009 Problem 1 Consider the function given by f ( x ) = (1 x 2 /L 2 ) 0 < x < L. (a) Develop the Fourier sine series over the interval 0 < x < L . (b) Develop the Fourier cosine series over the interval 0 < x < L . (c) Develop the fully periodic Fourier series over the interval 0 < x < L . ( N.B. note the interval ). (d) What is the value of each of the series at x = 0 ? Explain in each case why you get a given value. SOLUTION ( a ) : For this interval, we are looking for a series of the form f ( x ) = s n =1 A n sin p nπx L P . Multiplying both sides by sin ( mπx L ) and integrating over the interval gives i L 0 f ( x ) sin p mπx L P dx = s n =1 A n i L 0 sin p nπx L P sin p mπx L P dx. But by the orthogonality of the sine series, i L 0 f ( x ) sin p mπx L P = A m i L 0 sin 2 p mπx L P dx = A m L 2 . And so for all n = 1 , 2 , ... , A n = 2 L i L 0 f ( x ) sin p nπx L P dx = 2 L i L 0 (1 x 2 /L 2 ) sin p nπx L P dx = 2 L b 2 L cos( ) + Ln 2 π 2 + 2 L n 3 π 3 B = 2 n 3 π 3 [2 + n 2 π 2 2( 1) n ] 1

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so that f ( x ) = s n =1 2 n 3 π 3 [2 + n 2 π 2 2( 1) n ] sin p nπx L P . SOLUTION ( b ) : Similarly, we are looking for a series of the form f ( x ) = s n =0 B n cos p nπx L P . Multiplying both sides by cos ( mπx L ) and integrating over the interval gives i L 0 f ( x ) cos p mπx L P dx = s n =0 B n i L 0 cos p nπx L P cos p mπx L P dx. But by the orthogonality of the sine series, B 0 = 1 L i L 0 f ( x ) dx i L 0 f ( x ) cos p mπx L P dx = B m i L 0 cos p nπx L P cos p mπx L P dx = B m L 2 . And so for n = 0,1,2,. .. B 0 = 1 L i L 0 (1 x 2 /L 2 ) dx = 2 3 B n = 2 L i L 0 f ( x ) cos p nπx L P dx = 2 L i L 0 (1 x 2 /L 2 ) cos p nπx L P dx = 2 L b 2 Lnπ cos( ) n 3 π 3 B = 4( 1) n n 2 π 2 so that f ( x ) = 2 3 + s n =1 4( 1) n n 2 π 2 cos p nπx L P . SOLUTION ( c ) : 2
To develop the fully periodic series, we want these functions to be periodic on 0 < x < L . That is, we look for f ( x ) = s n =0 B n cos p 2 nπx L P + s n =1 A n sin p 2 nπx L P . Multiplying both sides by cos ( 2 mπx L ) and integrating over 0 < x < L and using orthogonality again yields B 0 = 1 L i L 0 (1 x 2 /L 2 ) dx = 2 3 , B n = 2 L i L 0 f ( x ) cos p 2 nπx L P dx = 2 L i L 0 (1 x 2 /L 2 ) cos p 2 nπx L P dx = 2 L L b 2 cos(2 ) 4 n 3 π 3 B = 1 n 2 π 2 . Similarly, multiplying both sides by sin ( 2 mπx L ) and integrating over 0 < x < L and using orthogonality again yields i L 0 f ( x ) sin p 2 mπx L P = A m i L 0 sin 2 p 2 mπx L P dx = A m L/ 2 . And so for all n = 1 , 2 , ... , A n = 2 L i L 0 f ( x ) sin p 2 nπx L P dx = 1 . Thus, the fully periodic series is given by f ( x ) = 2 3 + s n =1 ( 1) n 2 π 2 cos p 2 nπx L P + s n =1 1 sin p 2 nπx L P . SOLUTION ( d ) : Clearly by the above results, at x = 0 , f (0) = 0 for the sine series, 3

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The sine series approximates an odd (periodic) extension of (1 x 2 /L 2 ) to ( L, 0) .
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