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Unformatted text preview: 18.03 Problem Set 7: Solutions 26. [Part I problems: 4] (a) [10] G ( s ) = Z f ( at ) e st dt . To make this look more like F ( s ) = Z f ( t ) e st dt , make the substitution u = at . Then du = adt and G ( s ) = Z f ( u ) e su/a du a = 1 a Z f ( u ) e ( s/a ) u du = 1 a F ( s/a ) . For example, take f ( t ) = sin t , so F ( s ) = 1 s 2 + 1 , g ( t ) = sin( at ), G ( s ) = a s 2 + a 2 . Now compute 1 a F s a = 1 a 1 ( s/a ) 2 + 1 = a s 2 + a 2 = G ( s ). (b) [10] Compute F ( s ) G ( s ) = Z Z f ( x ) e sx g ( y ) e sy dxdy = ZZ R f ( x ) g ( y ) e s ( x + y ) dxdy , where R is the first quadrant. The suggested substitution is x = t , y = . To convert to these coordinates, note that the Jacobian is det x t x y t y = det 1 1 1 = 1 For fixed , t ranges over numbers greater than , and ranges over positive numbers; or, what is the same, t ranges over positive numbers and ranges over numbers between 0 and t . So, continuing, and noticing that x + y = t , F ( s ) G ( s ) = Z Z t f ( t ) g ( ) e st ddt = Z Z t f ( t ) g ( ) d e st dt = Z ( f ( t ) * g ( t )) e st...
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This note was uploaded on 02/06/2009 for the course 18 18.03 taught by Professor Unknown during the Spring '09 term at MIT.
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