Exercise 64102 solve find the impulse response x 0 ax

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Exercise 6.4.102 : Solve (find the impulse response) x 0 + ax = δ ( t ) , x (0) = 0 , x 0 (0) = 0 . Exercise 6.4.103 : Suppose that Lx = δ ( t ) , x (0) = 0 , x 0 (0) = 0 , has the solution x ( t ) = cos ( t ) for t > 0 . Find (in closed form) the solution to Lx = sin( t ) , x (0) = 0 , x 0 (0) = 0 for t > 0 . Exercise 6.4.104 : Compute L - 1 n s 2 s 2 + 1 o . Exercise 6.4.105 : Compute L - 1 n 3 s 2 e - s + 2 s 2 o .
294 CHAPTER 6. THE LAPLACE TRANSFORM
Chapter 7 Power series methods 7.1 Power series Note: 1 or 1.5 lecture, §8.1 in [EP], §5.1 in [BD] Many functions can be written in terms of a power series X k = 0 a k ( x - x 0 ) k . If we assume that a solution of a di ff erential equation is written as a power series, then perhaps we can use a method reminiscent of undetermined coe ffi cients. That is, we will try to solve for the numbers a k . Before we can carry out this process, let us review some results and concepts about power series. 7.1.1 Definition As we said, a power series is an expression such as X k = 0 a k ( x - x 0 ) k = a 0 + a 1 ( x - x 0 ) + a 2 ( x - x 0 ) 2 + a 3 ( x - x 0 ) 3 + · · · , (7.1) where a 0 , a 1 , a 2 , . . . , a k , . . . and x 0 are constants. Let S n ( x ) = n X k = 0 a k ( x - x 0 ) k = a 0 + a 1 ( x - x 0 ) + a 2 ( x - x 0 ) 2 + a 3 ( x - x 0 ) 3 + · · · + a n ( x - x 0 ) n , denote the so-called partial sum . If for some x , the limit lim n →∞ S n ( x ) = lim n →∞ n X k = 0 a k ( x - x 0 ) k 295
296 CHAPTER 7. POWER SERIES METHODS exists, then we say that the series (7.1) converges at x . Note that for x = x 0 , the series always converges to a 0 . When (7.1) converges at any other point x , x 0 , we say that (7.1) is a convergent power series . In this case we write X k = 0 a k ( x - x 0 ) k = lim n →∞ n X k = 0 a k ( x - x 0 ) k . If the series does not converge for any point x , x 0 , we say that the series is divergent . Example 7.1.1: The series X k = 0 1 k ! x k = 1 + x + x 2 2 + x 3 6 + · · · is convergent for any x . Recall that k ! = 1 · 2 · 3 · · · k is the factorial. By convention we define 0! = 1. In fact, you may recall that this series converges to e x . We say that (7.1) converges absolutely at x whenever the limit lim n →∞ n X k = 0 | a k | | x - x 0 | k exists. That is, the series k = 0 | a k | | x - x 0 | k is convergent. If (7.1) converges absolutely at x , then it converges at x . However, the opposite implication is not true. Example 7.1.2: The series X k = 1 1 k x k converges absolutely for all x in the interval ( - 1 , 1). It converges at x = - 1, as k = 1 ( - 1) k k converges (conditionally) by the alternating series test. But the power series does not converge absolutely at x = - 1, because k = 1 1 k does not converge. The series diverges at x = 1. 7.1.2 Radius of convergence If a power series converges absolutely at some x 1 , then for all x such that | x - x 0 | ≤ | x 1 - x 0 | (that is, x is closer than x 1 to x 0 ) we have a k ( x - x 0 ) k a k ( x 1 - x 0 ) k for all k . As the numbers a k ( x 1 - x 0 ) k sum to some finite limit, summing smaller positive numbers a k ( x - x 0 ) k must also have a finite limit. Therefore, the series must converge absolutely at x . We have the following result. Theorem 7.1.1. For a power series (7.1) , there exists a number ρ (we allow ρ = ) called the radius of convergence such that the series converges absolutely on the interval ( x 0 - ρ, x 0 + ρ ) and diverges for x < x 0 - ρ and x > x 0 + ρ . We write ρ = if the series converges for all x.
7.1. POWER SERIES 297 x 0 x 0 + ρ x 0 - ρ diverges

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