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Unformatted text preview: s ) t a + ib s N = 1 a + ib s lim N ( e ( a s + ib ) N 1 ) . (7.1) Since e ( a s + ib ) x = e ( a s ) x e ibx , where the Frst factor vanishes at if a s < 0 while the second factor is a bounded ( e ibx 1) and periodic function, the limit in (7.1) exists if and only if a s < 0. Assuming that s > a , we get 1 a + ib s lim N ( e ( a s + ib ) N 1 ) = 1 a + ib s (0 1) = 1 s ( a + ib ) . 394...
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 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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