{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

SalasSV_11_08

# SalasSV_11_08 - 688 CHAPTER 11 INFINITE SERIES 11.8...

This preview shows pages 1–4. Sign up to view the full content.

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 = 0 a k x k = a 0 + a 1 x + a 2 x 2 + · · · + a n x n + · · · converges on ( c , c ), then k = 0 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 = 0 a k x k converges on ( c , c ). By Theorem 11.7.2, it converges absolutely on this interval. Now let x be some fi xed number in ( c , c ) and choose > 0 such that | x | < | x | + < c . Since | x | + lies within the interval of convergence, k = 0 | a k ( | x | + ) k | converges. As you are asked to show in Exercise 48, for all k suf fi ciently 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 = 0 | a k ( | x | + ) k | converges, we can conclude that k = 0 d dx ( a k x k ) = k = 1 | ka k x k 1 | converges,

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
11.8 DIFFERENTIATION AND INTEGRATION OF POWER SERIES 689 and thus that k = 0 d dx ( a k x k ) = k = 1 ka k x k 1 converges. Repeated application of the theorem shows that k = 0 d 2 dx 2 ( a k x k ), k = 0 d 3 dx 3 ( a k x k ), k = 0 d 4 dx 4 ( a k x k ), and so on, all converge on ( c , c ). Example 1 Since the geometric series k = 0 x k = 1 + x + x 2 + x 3 + x 4 + x 5 + x 6 + · · · converges on ( 1, 1), the series k = 0 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 = 0 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 = 0 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 = 0 a k x k converges on ( c , c ).
690 CHAPTER 11 INFINITE SERIES Then, as we have seen, k = 0 d dx ( a k x k ) also converges on ( c , c ). Using the fi rst series, we can de fi ne a function f on ( c , c ) by setting f ( x ) = k = 0 a k x k . Using the second series, we can de fi ne a function g on ( c , c ) by setting g ( x ) = k = 0 d dx ( a k x k ). The crucial point is that f ( x ) = g ( x ). THEOREM 11.8.2 THE DIFFERENTIABILITY THEOREM If f ( x ) = k = 0 a k x k for all x in ( c , c ), then f is differentiable on ( c , c ) and f ( x ) = k = 0 d dx ( a k x k ) for all x in ( c , c ). By applying this theorem to f , you can see that f is itself differentiable. This in turn implies that f is differentiable, and so on. In short, f has derivatives of all orders. The discussion up to this point can be summarized as follows: In the interior of its interval of convergence a power series de fi nes an in fi nitely differentiable function, the derivatives of which can be obtained by differentiating term by term: d n dx n k = 0 a k x k = k = 0 d n dx n ( a k x k ) for all n .

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}