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Unformatted text preview: Chapter 7 Replacing in Problem 52 of this section n by n + 1 yields L 1 1 s n +(3 / 2) ( t ) = 2 n +1 t n +(1 / 2) 1 3 5 (2 n + 1) , so that L 1 s 3 / 2 e 1 /s = L 1 X n =0 ( 1) n n ! s n +3 / 2 = X n =0 ( 1) n n ! L 1 1 s n +(3 / 2) = X n =0 ( 1) n n ! 2 n +1 t n +(1 / 2) 1 3 5 (2 n + 1) . Multiplying the numerator and denominator of the n th term by 2 4 (2 n ) = 2 n n !, we obtain L 1 s 3 / 2 e 1 /s ( t ) = X n =0 ( 1) n 2 n +1 2 n t n +(1 / 2) (2 n + 1)! = X n =0 ( 1) n 2 2 n +1 (2 n + 1)! t (2 n +1) / 2 = 1 X n =0 ( 1) n (2 t ) 2 n +1 (2 n + 1)! = 1 sin 2 t . (See (7.17).) 58. (a) Since u ( t a ) = , t < a 1 , t > a, we have (i) for t < 0, u ( t ) u ( t a ) = 0 0 = 0; (ii) for 0 < t < a , u ( t ) u ( t a ) = 1 0 = 1; (iii) for t > a , u ( t ) u ( t a ) = 1 1 = 0 . Thus, u ( t ) u ( t a ) = G a ( t ). (b) We use now formula (4) from the text to get L{ G a } ( s ) = L{ u ( t ) u ( t a ) } ( s ) = L{ u ( t ) } ( s ) L{ u ( t a ) } ( s ) = 1 s e as s = 1 e as s . 236 Exercises 7.6 (c) Since G a ( t b ) = u ( t b ) u [( t b ) a ] = u ( t b ) u [ t ( a + b )] , similarly to part (b) we have L{ G a ( t b ) } ( s ) = L{ u ( t b ) u [ t ( a + b )] } ( s ) = L{ u ( t b ) } ( s ) L{ u [ t ( a + b )] } ( s ) = e bs s e ( a + b ) s s = e bs e ( a + b ) s s ....
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 Spring '08
 MAZMANI
 Equations

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