527solns4 - 642:527 SOLUTIONS: ASSIGNMENT 4 FALL 2009 Some...

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Unformatted text preview: 642:527 SOLUTIONS: ASSIGNMENT 4 FALL 2009 Some of these solutions were written by Professor Dan Ocone. 5.2.1 A function f has exponential order as t → ∞ if there exist constants K ≥ 0, T ≥ 0, and a constant c , such that | f ( t ) | ≤ Ke ct for all t ≥ T . (4.1) A straightforward way to show that a function has exponential order is to identify constants K , c , and T for which it can be shown that (1) holds. Alternatively, however, for a given c , you can deduce that constants K and T exist making (4.1) true, if you can show lim t →∞ | f ( t ) | e − ct < ∞ , (4.2) and so by using (4.2) you needn’t determine K and T explicitly to prove exponential order. Conversely, if lim t →∞ | f ( t ) | e − ct = ∞ for every positive c , (4.3) then f is not of exponential order as t → ∞ . (a) Of exponential type, since (4.1) holds with c = 4, K = 5, and T = 0. (c) Of exponential type by (4.1), using sinh 2 t = ( e 2 t- e − 2 t ) / 2 ≤ e 2 t / 2. (g) Again by (4.1): | cos t 3 | ≤ 1 so (4.1) holds with c = 0, K = 1, T = 0. (j) Now it is easiest to use (4.3): for any c > 0, =lim t →∞ parenleftBig e t 4- ct + e- t 4- ct 2 parenrightBig = ∞ ,because lim t →∞ t 4- ct = ∞ for every c , so cosh( t 4 ) is not of exponential order as t → ∞ . 6. The function f ( t ) = t − 2 / 3 is not continuous, or even piecewise continuous, for t ≥ 0, since lim t →∞ t − 2 / 3 = ∞ . This means that Theorem 5.2.1 does not apply. Nevertheless, f ( t ) has a Laplace transform, since for s > 0, integraldisplay 1 t − 2 / 3 e − st dt ≤ integraldisplay 1 t − 2 / 3 dt = 3 . (the only possible problem is at t = 0). In fact, we showed in class that L { f ( t ) } = Γ(1 / 3) /s 1 / 3 . 8. This is elementary. 11. (a) is elementary. (b): d ds integraldisplay ∞ sin ate − st dt = integraldisplay ∞ (- t ) sin ate − st dt =- L { t sin at } = d ds a a 2 + s 2 =- 2 as ( s 2 + a 2 ) 2 . 5.3.1 (a) See solution in text. (b) By partial fractions: L − 1 braceleftbigg 1 3 s 2 + 5 s- 2 bracerightbigg ( t ) = 1 7 L − 1 braceleftbigg 1 s- (1 / 3)- 1 s + 2 bracerightbigg ( t ) = 1 7 parenleftBig e t/ 3- e − 2 t parenrightBig ....
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This note was uploaded on 12/15/2009 for the course MATH 527 taught by Professor Staff during the Fall '08 term at Rutgers.

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527solns4 - 642:527 SOLUTIONS: ASSIGNMENT 4 FALL 2009 Some...

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