MA360Handout 3b

# MA360Handout 3b - 1 MA360 Handout 3b(week of Jan 23 Sp2006...

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Unformatted text preview: 1 MA360 Handout 3b (week of Jan 23) Sp2006 Kallfelz Assignment: P 41 Sheng: 1,2,4,5( a), b), c)),6( b), c)), 7a) Second session of Class: Remaining Material in Ch I, Sheng Lemma 3b.1: lim s →∞ L [ f ( t )] = lim s →∞ F ( s ) = 0, for any continuous 1 f . Proof: () [ ] ( ) () dt e t f dt e t f t f L st st- ∞ ∞- ∫ ∫ ≤ = Moreover, for any continuous f ( t ), there exist M , α , t such that: () t Me t f α < for all t > t (i.e., any continuous function, no matter how fast-growing, is bounded above by an exponential function for sufficiently large values of t .) Hence, choose M , α such when t 0 = 0: () t Me t f α < for all t > 0 So: () [ ] ( ) () ( ) ( ) ( ) ( ) { } ( ) ( ) ( ) { } ( ) { } ( ) α α α α α α α α α α---- ∞ →- ∞ →-- ∞ → ∞--- ∞- ∞ ∞- =- =- = = = < ≤ =---- ∫ ∫ ∫ ∫ ∫ s M s s e s e d d s e d d t s d t s st t st st M M M dt e M dt e M dt e Me dt e t f dt e t f t f L d s t s 1 lim lim lim Hence: ( ) ( ) ( ) ( ) lim lim lim lim = ∴ = < ≤ ∞ →- ∞ → ∞ → ∞ → s F s F s F s s M s s s α The above Lemma gives us a valuable insight into the unique aspects of s – space (the domain space where the LTs live, i.e. the domain of the functions F ( s ), given that t – space is the domain space where f ( t ) live, i.e. the arguments of the LTs). Its ‘asymptotic’ behavior is such that it always vanishes, i.e. F is non-zero on a finite interval. We can use Lemma 3b.1 to prove, for example, Thm 10: () ( ) () [ ] ) : ( ), ( lim lim t f L s F where s sF t f s t = = ∞ → → + 1 This Lemma can be generalized also to piecewise continuous functions. 2 Note1: we’re taking a right-hand limit on the left hand side (in t-space) since the domain of f is nonnegative. Proof: According to Thm2, for any differentiable (therefore continous) f : () [ ] () [ ] ( ) f t f sL t f L dt d- = Now, obviously, (since f is continuous 2 ): ( ) ( ) t f f t + → = lim Hence: () [ ] () [ ] ( ) () [ ] () t f t f sL f t f sL t f L t dt d + →- =- = lim Taking the s →∞ limit of both sides: () [ ] ( ) [ ] () t f t f sL t f L t s dt d s + → ∞ → ∞ →- = lim lim lim Note 2: The last term on the right hand side is a constant with respect to s . Therefore its limiting value is the same throughout. 3 But according to Lemma3b.1, the left hand side vanishes, if we assume the derivative of f is continuous. Hence: () [ ] ( ) [ ] ( ) () () [ ] t f sL t f t f t f sL t f L s t t s dt d s ∞ → → → ∞ → ∞ → = ∴- = = + + lim lim lim lim lim Thm 11 is the analogous version, this time when: s →∞ , and t → + Both Theorems 10 & 11 should be viewed as tracking the limiting behavior of...
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## This note was uploaded on 11/18/2010 for the course MA 360 taught by Professor Williamm.kallfelz during the Spring '06 term at Capitol College.

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MA360Handout 3b - 1 MA360 Handout 3b(week of Jan 23 Sp2006...

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