398_pdfsam_math 54 differential equation solutions odd

398_pdfsam_math 54 differential equation solutions odd -...

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Chapter 7 (h) The function 3 + cos 4 t is bounded because | 3+cos4 t |≤ 3+ | cos 4 t |≤ 4 . Therefore, by the triangle inequality, | f ( t ) |≥ ± ± ± e t 2 ± ± ± −| 3+cos4 t |≥ e t 2 4 , and, therefore, for any Fxed α , f ( t ) grows faster than e αt (because e t 2 does, and the other term is bounded). So, f ( t )is not of exponential order. (i) Clearly, for any t> 0, t 2 t +1 = t t +1 t< (1) t = t. Therefore, e t 2 / ( t +1) <e t , and DeFnition 3 holds with M =1 , α =1,and T =0. (j) Since, for any x , 1 sin x 1, the given function is bounded. Indeed, ± ± ± sin ² e t 2 ³ + e sin t ± ± ± ± ± ± sin ² e t 2 ³ ± ± ± + e sin t 1+ e Thus it is of exponential order 0. 31. (a) L ´ e ( a + ib ) t µ ( s ): = Z 0 e st e ( a + ib ) t dt = Z 0 e ( a + ib s ) t dt = lim N →∞ N Z 0 e ( a + ib s ) t dt = lim N →∞ e ( a +
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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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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at Berkeley.

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