Unformatted text preview: Exercises 7.2
for all α > 0. Thus, ﬁxed α > 0, for some T = T (α) > 0, we have t3  < eαt for all t > T , and so t3 sin t ≤ t3 < eαt , t > T. Therefore, t3 sin t is of exponential order α, for any α > 0. (b) Clearly, for any t, f (t) = 100e49t , and so Deﬁnition 3 is satisﬁed with M = 100, α = 49, and any T . Hence, f (t) is of exponential order 49. (c) Since f (t) 3 2 = lim et −αt = lim e(t −α)t = ∞, αt t→∞ e t→∞ t→∞ αt we see that f (t) grows faster than e for any α. Thus f (t) is not of exponential order. lim (d) Similarly to (a), for any α > 0, we get t ln t t ln t ln t + 1 1/t = lim αt = lim = lim 2 αt = 0 , αt αt t→∞ e t→∞ e t→∞ αe t→∞ α e lim and so f (t) is of exponential order α for any positive α. (e) Since,
2 12 et + e−t > et f (t) = cosh t = 2 2 t2 αt and e grows faster than e for any ﬁxed α (see page 357 in the text), we conclude that cosh (t2 ) is not of exponential order. (f) This function is bounded: f (t) = t2 1 1 = 1, = f r 1t2 + 1 ≤ +1 0+1 2 2 and so Deﬁnition 3 is satisﬁed with M = 1 and α = 0. Hence, f (t) is of exponential order 0. (g) The function sin (t2 ) is bounded, namely, sin (t2 ) ≤ 1. For any ﬁxed β > 0, the limit of t4 /eβt , as t → ∞, is 0, which implies that t4 ≤ eβt for all t > T = T (β ). Thus, sin t2 + t4 e6t ≤ 1 + eβt e6t = 2eβ +6t . This means that f (t) is of exponential order α for any α > 6. 393 ...
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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.
 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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