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homework 1 solutions

homework 1 solutions - Homework 1 Solutions and remarks on...

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Homework 1: Solutions and remarks on selected problems Greenberg § 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 . (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. However, for a given c , you can deduce that constants K and T exist making (1) true, if you can show lim t →∞ | f ( t ) | e - ct < , (2) and so by using (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 , (3) then f is not of exponential order as t → ∞ . Thus for problem 5(e), observe that lim t →∞ e - ct sinh( t 2 ) = lim t →∞ (1 / 2) e t 2 - ct - e - t 2 - ct = for every c > 0 because lim t →∞ t 2 - ct = for every c . Thus sinh( t 2 ) is not of exponential order as t → ∞ . For problem 5(h), repeated application of L’Hˆ opital’s rule shows that for any c > 0, lim t →∞ t 100 e - ct = 0, and hence t 100 is of exponential order as t → ∞ . Pick a c > 0. By using the first derivative test, one finds that t 100 e - ct , as a function on the domain 0 t < , achieves its maximum at ˜ t = 100 /c . Thus, for all t 0, t 100 e - ct (100 /c ) 100 e - 100 . Hence t 100 (100 /c ) 100 e 100 e ct for all t 0, which establishes (1) with explicit K and T corresponding to c . Greenberg, § 5.2, #5,6. Suppose R b a f ( t ) dt is an improper integral. This happens if one or both of the limits of integration are infinite and/or if the function f has a vertical asymptote at a finite limit of integration. The comparison test for improper integrals implies that if
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