Power Series. Part 2 Differentiation & Integration; Multiplication of Power Series. J. Gonzalez-Zugasti, University of Massachusetts - Lowell

Transcription

1 Power Series Part 2 Differentiation & Integration; Multiplication of Power Series 1

2 Theorem 1 If a n x n converges absolutely for x < R, then a n f x n converges absolutely for any continuous function f on f x < R. 2

3 Example 1 Since 1 1 x = xk, for x < 1 Theorem 1 tells us that 1 k = 4x2 1 4x2, for 4x 2 < 1 3

4 Example 2 Find the interval of convergence of e x 4 k and, within this interval, the sum of the series as a function of x. 4

5 Example 2 (continued) Solution: e x 4 k ; Using the Ratio Test for Absolute Convergence: u k+1 e x 4 k+1 ρ = lim = lim k u k k e x 4 k = lim e x 4 k = e x 4 lim 1 k = e x 4 Therefore the series converges absolutely when ρ = e x 4 < 1. 5

6 Example 2 (continued) ρ = e x 4 < 1 1 < e x 4 < 1 3 < e x < 5 ln 3 < x < ln 5 Let s check what happens to the series at the endpoint of this interval. 6

7 Example 2 (continued) At x = ln 3, the series becomes e ln 3 4 k which diverges. At x = ln 5, the series becomes e ln 5 4 k which diverges. = 3 4 k = 1 k = 5 4 k = 1 7

8 Example 2 (continued) The series e x 4 k is a convergent geometric series (a = 1, r = e x 4) when ln 3 < x < ln 5 and the sum is a 1 r = 1 1 e x 4 = 1 5 e x. 8

9 Power Series as Functions If c k x a k converges for x a < R (that is, a R < x < a + R), then define f x = c k x a k, a R < x < a + R We can find f x and f x dx as follows: 9

10 Term by Term Differentiation and Integration (a) f x = d dd = kc k x a k 1 (b) f x dx = = c k k+1 k+1 x a c k x a k c k x a k dx + C Both have radius of convergence R and interval of convergence x a < R. 10

11 Series Multiplication If a n x n and b n x n converge absolutely for x < R and n then c n = a k b n k a n x n b n x n = c n x n which also converges for x < R. 11

12 The series Example 3 e x = 1 + x + x2 2! + x3 3! + x4 4! + x5 5! + converges to e x for all x. (a) Find the series for d dd ex. (b) Find the series for e x dx. (c) Find the series for e x. (d) Multiply the series for e x and e x to find e x e x. 12

13 Example 3 (continued) Solution (a): e x = 1 + x + x2 d dd ex = d dd x k + x3 + x4 + x5 2! 3! 4! = k xk 1 + = x k 5! = xk 1 k 1! k=1 = xk = e x 13

14 Example 3 (continued) Solution (b): e x = e x dx = xk dx x k = xk+1 + C k + 1 = xk+1 + C k + 1! = x + x2 2! + x3 3! + x4 4! + x5 5! + + C = x + x2 2! + x3 3! + x4 4! + x5 5! + + C = 1 + xk + C = xk + C = e x + C 14

15 Example 3 (continued) Solution (c): e x = x k e x = x k = 1 k xk 15

16 Example 3 (continued) Solution (d): e x =, e x = 1 x k k xk e x e x = xk 1 k xk = c n x n n=0 where a k = 1 and b k = c n = n a k b n k. 1 k and 16

17 Example 3 (continued) n c n = a k b n k n = 1 1 n k n n k! = 1 n k n k! c 0 = c 1 = ! 0 0! = ! 1 0! c 2 = + 0! 2 0! 1! 2 1! etc. c n = 0, n ! 1 1! = = ! 2 2! = = 0 17

18 Example 3 (continued) e x e x = xk 1 k xk = c n x n n=0 = 1 x x x x 3 + = 1 18

19 Advice Read the Power Series section in your textbook (including its exercises) it will provide you with some excellent examples of how to identify a power series as a function by looking at either the derivative, the antiderivative of the series, or the product to two known series. 19

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