Chem Differential Eq HW Solutions Fall 2011 108

# Chem Differential Eq HW Solutions Fall 2011 108 - 108...

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108 Chapter 7 The Fourier Transform and its Applications Multiplying both sides by ± π 2 and using the linearity of the Fourier transform, it follows that F ( ² π 2 f ( x ))( w )= g ( w ) . So F g = FF ³² π 2 f ( x ) ´ = ² π 2 f ( x ) , by the reciprocity relation. Using the symbol w as a variable, we get F ³ sin ax x ´ = ² π 2 f ( w µ ± π 2 if | w | <a , 0 otherwise . 17. (a) Consider ±rst the case a> 0. Using the de±nition of the Fourier transform and a change of variables F ( f ( ax ))( w 1 2 π -∞ f ( ax ) e - iωx dx = 1 a 1 2 π -∞ f ( x ) e - i ω a x dx ( ax = X, dx = 1 a dX ) = 1 a F ( f )( w a ) . If a< 0, then F ( f ( ax ))( w 1 2 π -∞ f ( ax ) e - iωx dx = 1 a 1 2 π -∞ f ( x ) e - i ω a x dx = - 1 a F ( f )( w a ) . Hence for all a ± = 0, we can write = 1 | a | F ( f ) · ω a ¸ . (b) We have F ( e -| x | )( w ² 2 π 1 1+ w 2 . By (a), F ( e - 2 | x | )( w 1 2 ² 2 π 1 1+( w/ 2) 2 = ² 2 π 2 4+ w 2 . (c) Let
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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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