527solns4

527solns4 - 642:527 SOLUTIONS: ASSIGNMENT 4 FALL 2007 Some...

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Unformatted text preview: 642:527 SOLUTIONS: ASSIGNMENT 4 FALL 2007 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 > lim t →∞ e − ct cosh( t 4 ) = lim t →∞ parenleftBigg e t 4 − ct + e − t 4 − ct 2 parenrightBigg = ∞ , because lim t →∞ t 4- ct = ∞ for every c , so cosh( t 2 ) 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...
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This note was uploaded on 09/29/2009 for the course 642 527 taught by Professor Speer during the Fall '07 term at Rutgers.

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527solns4 - 642:527 SOLUTIONS: ASSIGNMENT 4 FALL 2007 Some...

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