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Unformatted text preview: 688 CHAPTER 11 INFINITE SERIES 11.8 DIFFERENTIATION AND INTEGRATION OF POWER SERIES Differentiation of Power Series We begin with a simple but important result. THEOREM 11.8.1 If k = a k x k = a + a 1 x + a 2 x 2 + + a n x n + converges on ( c , c ), then k = d dx ( a k x k ) = k = 1 ka k x k 1 = a 1 + 2 a 2 x + + na n x n 1 + also converges on ( c , c ). PROOF Assume that k = a k x k converges on ( c , c ). By Theorem 11.7.2, it converges absolutely on this interval. Now let x be some fixed number in ( c , c ) and choose > 0 such that  x  <  x  + < c . Since  x  + lies within the interval of convergence, k =  a k (  x  + ) k  converges. As you are asked to show in Exercise 48, for all k sufficiently large,  k x k 1  (  x  + ) k . It follows that for all such k ,  ka k x k 1   a k (  x  + ) k  . Since k =  a k (  x  + ) k  converges, we can conclude that k = d dx ( a k x k ) = k = 1  ka k x k 1  converges, 11.8 DIFFERENTIATION AND INTEGRATION OF POWER SERIES 689 and thus that k = d dx ( a k x k ) = k = 1 ka k x k 1 converges. Repeated application of the theorem shows that k = d 2 dx 2 ( a k x k ), k = d 3 dx 3 ( a k x k ), k = d 4 dx 4 ( a k x k ), and so on, all converge on ( c , c ). Example 1 Since the geometric series k = x k = 1 + x + x 2 + x 3 + x 4 + x 5 + x 6 + converges on ( 1, 1), the series k = d dx ( x k ) = k = 1 kx k 1 = 1 + 2 x + 3 x 2 + 4 x 3 + 5 x 4 + 6 x 5 + , k = d 2 dx 2 ( x k ) = k = 2 k ( k 1) x k 2 = 2 + 6 x + 12 x 2 + 20 x 3 + 30 x 4 + , k = d 3 dx 3 ( x k ) = k = 3 k ( k 1)( k 2) x k 3 = 6 + 24 x + 60 x 2 + 120 x 3 + , all converge on ( 1, 1). Remark Even though a k x k and its derivative ka k x k 1 have the same radius of convergence, their intervals of convergence may be different due to possible differences in behavior at the endpoints. For example, the interval of convergence of the series k = 1 1 k 2 x k is [ 1, 1], whereas the interval of convergence of its derivative k = 1 1 k x k 1 is [ 1, 1). Endpoints must always be checked separately. Suppose now that k = a k x k converges on ( c , c ). 690 CHAPTER 11 INFINITE SERIES Then, as we have seen, k = d dx ( a k x k ) also converges on ( c , c ). Using the first series, we can define a function f on ( c , c ) by setting f ( x ) = k = a k x k . Using the second series, we can define a function g on ( c , c ) by setting g ( x ) = k = d dx ( a k x k ). The crucial point is that f ( x ) = g ( x )....
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This note was uploaded on 10/12/2010 for the course MATH 12345 taught by Professor Smith during the Spring '10 term at University of Houston  Downtown.
 Spring '10
 SMITH
 Infinite Series, Power Series

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