Chem Differential Eq HW Solutions Fall 2011 109

# Chem Differential Eq HW Solutions Fall 2011 109 - Section...

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Section 7.2 The Fourier Transform 109 25 Let g ( x )= ± 1i f | x | < 1, 0 otherwise, and note that f ( x ) = cos( x ) g ( x ). Now F ( g ( x )) = ² 2 π sin w w . Using Exercise 20, we have F ( f ( x )) = F (cos xg ( x )) = 1 2 ³ 2 π ´ sin( w - 1) w - 1 + sin( w +1) w +1 µ . 29. Take a = 0 and relabel b = a in Exercise 27, you will get the function f ( x h if 0 <x<a . Its Fourier transform is f ( w )== ³ 2 π he - i a 2 w sin ( aw 2 ) w . Let g ( x ) denote the function in the ±gure. Then g ( x 1 a xf ( x ) and so, by Theo- rem 3, g ( w 1 a i d dw f ( w ) = i ³ 2 π h a d dw · e - i a 2 w sin ( 2 ) w ¸ = i ³ 2 π h a e - i a 2 w w · a cos ( 2 ) - i sin ( 2 ) 2 - sin ( 2 ) w ¸ = i ³ 2 π h a e - i a 2 w · - 2 sin ( 2 ) + awe - i a 2 w 2 w 2 ¸ . 33. Let g ( x ) denote the function in this exercise. By the reciprocity relation, since the function is even, we have F ( F ( g )) = g ( - x g ( x ). Taking inverse Fourier transforms, we obtain F - 1 ( g F ( g ). Hence it is enough to compute the Fourier transform. We use the notation and the result of Exercise 34. We have
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## This note was uploaded on 12/22/2011 for the course MAP 3305 taught by Professor Stuartchalk during the Fall '11 term at UNF.

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