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Unformatted text preview: Math 185. Sample Answers to Problem Set #14 Page 296 1. The function F ( s ) has simple poles at s = 2 and 2 i . Therefore, we may take = 2. For all s { 2 , 2 i } , we have Res s = s ( e st F ( s ) ) = 2 s 3 e s t 4 s 3 = e s t 2 . Therefore, f ( t ) = Res s = 2 e st F ( s ) + Res s = 2 e st F ( s ) + Res s = 2 i e st F ( s ) + Res s = 2 i e st F ( s ) = e 2 t + e 2 t + e 2 it + e 2 it 2 = cosh 2 t + cos 2 t , provided the italicized statement on page 289 is true. To check this, we have, for all R > 2 + 2,  F (2 + e i )  = 2(2 + Re i ) 3 (2 + Re i ) 4 4 2(2 + R ) 3 ( R 2) 4 4 , which goes to 0 as R . Thus the above expression for f ( t ) is true. 5. The function F ( s ) has poles at ai , so we may take = 1. To check the italicized statement on page 289, we have  F (1 + Re i )  = 8 a 3  1 + Re i  2  (1 + Re i ) 2 + a 2  3 8 a 3 (1 + R ) 2 (( R 1) 2 a 2 ) 3 for all R > 1 + a . This bound goes to zero when R , so the italicized statement is true....
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 Fall '07
 Lim
 Math

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