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SalasSV_11_09_ex - 702 CHAPTER 11 INFINITE SERIES From...

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Unformatted text preview: 702 CHAPTER 11 INFINITE SERIES From Problem 2 you know that the binomial series converges on the open interval (−1, 1) and defines there an infinitely differentiable function. The next thing to show is that this function (the one defined by the series) is actually (1 + x)α . To do this, you first need some other results. Problem 3. Verify the identity (k + 1) Problem 4. α α α +k =α . k +1 k k Use the identity of Problem 3 to show that the sum of the binomial series ∞ φ (x ) = k =0 αk x k satisfies the differential equation (1 + x)φ (x) = αφ (x) together with the initial condition φ (0) = 1. You are now in a position to prove the main result. Problem 5. (11.9.2) for all x in (−1, 1), Show that (1 + x)α = ∞ k =0 αk x k for all x in (−1, 1). You can probably get a better feeling for the series by writing out the first few terms: (11.9.3) (1 + x)α = 1 + α x + α (α − 1) 2 α (α − 1)(α − 2) 3 x+ x + ··· . 2! 3! EXERCISES 11.9 In Exercises 1–10, expand f in powers of x up to x4 . √ √ 1. f (x) = 1 + x. 2. f (x) = 1 − x. √ √ 4. f (x) = 1 − x2 . 3. f (x) = 1 + x2 . 1 . 5. f (x) = √ 1+x √ 7. f (x) = 4 1 − x. 9. f (x) = (4 + x)3/2 . 1 6. f (x) = √ . 3 1+x 8. f (x) = √ 4 10. f (x) = √ 1 1+x . (b) Use the series for f in part (a) to find the Taylor series for F (x) = sinh−1 x and give the radius of convergence. 1 + x4 . c Estimate by using the first three terms of a binomial expansion, rounding off your answer to four decimal places. √ √ 1 98 = (100 − 2)1/2 = 10(1 − 50 )1/2 . 13. 98. HINT: √ √ 14. 5 36. 15. 3 9. √ 16. 4 620. 17. 17−1/4 . 18. 9−1/3 . c Approximate each integral to within 0.001. 11. (a) Use a binomial series to find the Taylor series of √ 2 in powers of x . 1/ 3 1/ 5 f (x) = 1/ 1 − x 1 + x3 dx. 20. 1 + x4 dx. 19. (b) Use the series for f in part (a) to find the Taylor series 0 0 −1 for F (x) = sin x and give the radius of convergence. 1/ 2 1/2 1 1 21. dx. 22. dx. √ √ 12. (a) Use a binomial series to find the Taylor series of √ 2 1+x 1 − x3 0 0 f (x) = 1/ 1 + x2 in powers of x. ...
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