SalasSV_11_07 - 680 CHAPTER 11 INFINITE SERIES EXERCISES...

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Unformatted text preview: 680 CHAPTER 11 INFINITE SERIES EXERCISES 11.6 Find the Taylor polynomial of the function f for the given values of a and n, and give the Lagrange form of the remainder. √ 1. f (x) = x; a = 4, n = 3. 2. f (x) = cos x; a = π/3, n = 4. 3. f (x) = sin x; a = π/4, n = 4. 4. f (x) = ln x; a = 1, n = 5. 5. f (x) = tan−1 x; a = 1, n = 3. 6. f (x) = cos π x; a = 1 , n = 4. 2 Expand g (x) as indicated and specify the values of x for which the expansion is valid. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. g (x) = 3x3 − 2x2 + 4x + 1 in powers of x − 1. g (x) = x4 − x3 + x2 − x + 1 in powers of x − 2. g (x) = 2x5 + x2 − 3x − 5 in powers of x + 1. g (x) = x−1 in powers of x − 1. g (x) = (1 + x)−1 in powers of x − 1. g (x) = (b + x)−1 in powers of x − a, a = −b. g (x) = (1 − 2x)−1 in powers of x + 2. g (x) = e−4x in powers of x + 1. g (x) = sin x in powers of x − π . g (x) = sin x in powers of x − 1 π . 2 g (x) = cos x in powers of x − π . g (x) = cos x in powers of x − 1 π . 2 g (x) = sin 1 π x in powers of x − 1. 2 g (x) = sin π x in powers of x − 1. g (x) = ln (1 + 2x) in powers of x − 1. g (x) = ln (2 + 3x) in powers of x − 4. 29. g (x) = cos2 x 31. g (x) = x 33. (a) (b) (c) 34. (a) (b) n in powers of x − π . in powers of x − 2. in powers of x. 30. g (x) = (1 + 2x)−4 32. g (x) = (x − 1)n x in powers of x − 1. Expand e in powers of x − a. Use the expansion to show that e x1 +x2 = e x1 e x2 . Expand e−x in powers of x − a. Expand sin x and cos x in powers of x − a. Show that both series absolutely convergent for all real x. (c) As noted earlier (Section 11.4), Riemann proved that the order of the terms of an absolutely convergent series may be changed without altering the sum of the series. Use Riemann’s discovery and the Taylor expansions of part (a) to derive the addition formulas sin (x1 + x2 ) = sin x1 cos x2 + cos x1 sin x2 , cos (x1 + x2 ) = cos x1 cos x2 − sin x1 sin x2 . Expand g (x) as indicated. 23. g (x) = x ln x in powers of x − 2. 24. g (x) = x2 + e3x in powers of x − 2. 25. g (x) = x sin x in powers of x. 26. g (x) = ln (x2 ) 27. g (x) = (1 − 2x) 28. g (x) = sin x 2 in powers of x − 1. −3 in powers of x + 2. in powers of x − 1 π . 2 c 35. (a) Find the Taylor polynomial Pn in powers of (x − π/6) of least degree that approximates sin 35◦ with four decimal place accuracy. (b) Evaluate the polynomial that you found in part (a) to obtain your approximation of sin 35◦ . c 36. (a) Find the Taylor polynomial Pn in powers of (x − π/3) of least degree that approximates cos 57◦ with four decimal place accuracy. (b) Evaluate the polynomial that you found in part (a) to obtain your approximation of cos 57◦ . √ c 37. Choose an appropriate Taylor polynomial of f (x) = x to √ approximate 38 with three decimal place accuracy. √ c 38. Choose an appropriate Taylor polynomial for f (x) = x to √ approximate 61 with three decimal place accuracy. c 39. Use a CAS to determine the Taylor polynomial P6 in powers of (x − 1) for f (x) = tan−1 x. c 40. Use a CAS to determine the Taylor polynomial P8 in powers of (x − 2) for f (x) = cosh 2x. 11.7 POWER SERIES You have become familiar with Taylor series ∞ k =0 f (k ) (0) k x k! ∞ and k =0 f (k ) (a) (x − a)k . k! Here we study series of the form ∞ ∞ ak x k k =0 and k =0 ak (x − a)k 11.7 POWER SERIES 681 without regard to how the coefficients ak have been generated. Such series are called power series: the first is a power series in x; the second is a power series in x − a. Since a simple translation converts ∞ ∞ ak (x − a)k k =0 into k =0 ak x k , we can focus our attention on power series of the form ∞ ak x k . k =0 When detailed indexing is unnecessary, we will omit it and write ak x k . DEFINITION 11.7.1 A power series (i) at c if ak x k is said to converge ak x k converges for each x ∈ S . ak ck converges; (ii) on the set S if The following result is fundamental. THEOREM 11.7.2 If ak x k converges at c = 0, then it converges absolutely for all x such that ak x k diverges at d , then it diverges for all x such that |x| > |d |. |x| < |c|. If If ak ck converges, then ak ck → 0 as k → ∞. In particular, for k sufficiently large, PROOF |ak ck | ≤ 1, x c k and thus For |x| < |c|, we have |ak x k | = |ak ck | ≤ x c k . x < 1. c The convergence of |ak x k | follows by comparison with the geometric series. This proves the first statement. 682 CHAPTER 11 INFINITE SERIES Suppose now that with |x| > |d | such that absolute convergence of ak d k diverges. By a similar argument, there cannot exist x ak x k converges. The existence of such an x would imply the ak d k . This proves the second statement. It follows from the theorem we just proved that there are exactly three possibilities for a power series: Case I. The series converges only at x = 0. This is what happens with k k xk . For x = 0, lim k k xk = 0, and so the series cannot converge (Theorem k →∞ 11.1.6). Case II. The series is absolutely convergent for all real numbers x. This is what happens with the exponential series xk . k! Case III. There exists a positive number r such that the series converges for |x| < r and diverges for |x| > r . This is what happens with the geometric series xk . For the geometric series, there is absolute convergence for |x| < 1 and divergence for |x| > 1. Associated with each case is a radius of convergence: In Case I, we say that the radius of convergence is 0. In Case II, we say that the radius of convergence is ∞. In Case III, we say that the radius of convergence is r . The three cases are pictured in Figure 11.7.1. convergence only at 0 0 case I: radius of convergence = 0 convergence everywhere 0 case II: radius of convergence = ∞ divergence convergence divergence –r 0 r case III: radius of convergence = r Figure 11.7.1 In general, the behavior of a power series at −r and at r is not predictable. For example, the series xk , ( − 1)k k x, k 1k x, k 1k x k2 all have radius of convergence 1, but the first series converges only on (−1, 1) the second series converges on (−1, 1], the third on [−1, 1), and the fourth on [−1, 1]. 11.7 POWER SERIES 683 The maximal interval on which a power series converges is called the interval of convergence. For a series with infinite radius of convergence, the interval of convergence is (−∞, ∞). For a series with radius of convergence r , the interval of convergence can be [−r , r ], (−r , r ], [−r , r ), or (−r , r ). For a series with radius of convergence 0, the interval of convergence reduces to a point, {0}. Example 1 Verify that the series (1) has interval of convergence ( −1, 1]. SOLUTION (−1)k k x k First we show that the radius of convergence is 1(that is, the series converges absolutely for |x| < 1 and diverges for |x| > 1). We do this by forming the series (2) (−1)k k x= k 1k |x| k and applying the ratio test. We set bk = and note that bk +1 k |x|k +1 |x|k +1 /(k + 1) k = |x| → |x| = = k /k k bk |x | k + 1 |x| k +1 as k → ∞. 1k |x| k By the ratio test, series (2) converges for |x| < 1 and diverges for |x| > 1. † It follows that series (1) converges absolutely for |x| < 1 and diverges for |x| > 1. The radius of convergence is therefore 1. Now we test the endpoints x = −1 and x = 1. At x = −1, (−1)k k x k becomes (−1)k (−1)k = k 1 . k This is the harmonic series, which, as you know, diverges. At x = 1, (−1)k k x k becomes (−1)k . k This is a (conditionally) convergent alternating series. We have shown that series (1) converges absolutely for |x| < 1, diverges at −1, and converges at 1. The interval of convergence is (−1, 1]. Remark The same arguments can be used to show that the series 1k x k converges on [−1, 1). † We could also have used the root test: (bk )1/k = 1 k 1/k |x| = 1 |x| → |x|. k 1/k 684 CHAPTER 11 INFINITE SERIES Example 2 Verify that the series (1) has interval of convergence [−1, 1]. SOLUTION 1k x k2 First we examine the series 1k x= k2 1k |x | . k2 (2) Here again we use the ratio test. We set bk = and note that b k +1 k 2 |x|k +1 = = bk (k + 1)2 |x|k k k +1 2 1k |x| k2 |x| → |x| as k → ∞. By the ratio test, (2) converges for |x| < 1 and diverges for |x| > 1. † This shows that (1) converges absolutely for |x| < 1 and diverges for |x| > 1. The radius of convergence is therefore 1. Now for the endpoints. At x = −1, 1k x k2 takes the form (−1)k = −1 + k2 1 4 − 1 9 + 1 16 − ··· . This is an absolutely convergent alternating series. At x = 1, 1k x k2 becomes 1 . k2 This is a convergent p-series. The interval of convergence is therefore the closed interval [−1, 1]. Example 3 Find the interval of convergence of the series kk x. 6k (1) SOLUTION We begin by examining the series kk x= 6k kk |x| . 6k kk |x| 6k (2) We set bk = † Once again we could have used the root test: (bk )1/k = 1 |x| → |x| k 2/k as k → ∞. 11.7 POWER SERIES 685 and apply the root test. (The ratio test will also work.) Since (bk )1/k = 1 k 1/k |x| → 1 |x| 6 6 you can see that (2) converges for and diverges for 1 |x| 6 1 |x| 6 as k → ∞, (recall k 1/k → 1) < 1, that is, for |x| < 6, > 1, that is, for |x| > 6. Testing the endpoints, we have: at at x = 6, x = −6, kk 6= k, 6k k (−6)k = 6k which is divergent; (−1)k k , which is also divergent. Thus, the interval of convergence is (−6, 6). Example 4 Find the interval of convergence of the series k! k x. (3k )! (1) SOLUTION Again, we begin by examining the series k! k x= (3k )! bk = k! |x|k . (3k )! (2) Set k! |x|k . (3k )! Since factorials are involved, we will use the ratio test. Note that (k + 1)! (3k )! |x|k +1 k +1 bk +1 = · · = |x | k bk [3(k + 1)]! k! |x| (3k + 3)(3k + 2)(3k + 1) 1 = |x |. 3(3k + 2)(3k + 1) Since 1 →0 3(3k + 2)(3k + 1) as k → ∞, the ratio bk +1 /bk tends to 0 no matter what x is. By the ratio test, (2) converges for all x and therefore (1) converges absolutely for all x. The radius of convergence is ∞ and the interval of convergence is (−∞, ∞). Example 5 SOLUTION Find the interval of convergence of the series kk k |x| 2k kk k x. 2k We set bk = 686 CHAPTER 11 INFINITE SERIES and apply the root test. Since (bk )1/k = 1 k |x| → ∞ as k → ∞ for every x = 0, the 2 series diverges for all x = 0; the series converges only at x = 0. Example 6 Find the interval of convergence of the series (−1)k (x + 2)k . k 2 3k SOLUTION We consider the series (−1)k (x + 2)k = k 2 3k 1 k 2 3k 1 k 2 3k |x + 2|k . Set bk = |x + 2|k and apply the ratio test (the root test will work equally well): k 2 3k |x + 2|k +1 k2 bk +1 = · = |x + 2| → 1 |x + 2| as k → ∞. 3 bk (k + 1)2 3k +1 |x + 2|k 3(k + 1)2 The series is absolutely convergent for −5 < x < 1: 1 |x 3 + 2| < 1 iff |x + 2| < 3 iff − 5 < x < 1. We now check the endpoints. At x = −5, (−1)k (−3)k = k 2 3k This is a convergent p-series. At x = 1, (−1)k k (3) = k 2 3k (−1)k . k2 1 . k2 This is a convergent alternating series. The interval of convergence is the closed interval [−5, 1]. EXERCISES 11.7 1. Suppose that the power series ∞ 0 ak x k converges at x = 3, k= that is, suppose that ∞ 0 ak 3k converges. What can you k= say about the convergence or divergence of the following? ∞ ∞ (c) k =0 ak 4k . ∞ (d) k =0 (−1)k ak 3k . (a) k =0 ∞ ak 2k . ak (−3)k . ∞ Find the interval of convergence. 3. 5. 7. 9. kx k . 1k x. (2k )! (−k )2k x2k . 1k x. k 2k 4. 6. 8. 10. 1k x. k 2k k x. k2 (−1)k k √ x. k 1 k 2 2k xk . (b) k =0 ∞ ak (−2)k . (c) k =0 (d) k =0 ak 4 k . 2. Suppose that the power series ∞ 0 ak x k converges at x = k= −3 and diverges at x = 5. What can you say about the convergence or divergence of the following? ∞ (a) k =0 ak 2k . ∞ (b) k =0 ak (−6)k . 11.7 POWER SERIES 687 11. 13. 15. 17. 19. 21. 23. 25. 27. 29. 31. 32. 33. 35. 36. k 100 k xk . 12. 14. 16. 18. 20. 22. 24. 26. 28. 30. k2 xk . 1 + k2 1k x. ln k kak xk . 3k 2 k x. ek 7k k x. k! k !xk . 2k k x. (2k )! (−e)k k x. k2 ln k (x + 1)k . k k (x − 4)k . ek 3 2k √ xk . k k −1 k x. k kk x. 10k xk . kk (−1)k (x − 2)k . kk 2k (−1)k k +1 xk . 3 (−1)k k! (x − 1)k . k3 (x + 2)k k . k −1 2k 2 43. Let ak x k be a power series with radius of convergence r > 0. (a) Show that if the power series is absolutely convergent at one endpoint of its interval of convergence, then it is absolutely convergent at the other endpoint. (b) Show that if the interval of convergence is (−r , r ], then the power series is conditionally convergent at r . 44. Let r > 0 be arbitrary. Give an example of a power series ak x k with radius of convergence r . ∞ 45. The power series k =0 ak x k has the property that ak +3 = ak for all k ≥ 0. (a) Show that the power series has radius of convergence r = 1. (b) Find an explicit formula in terms of a0 , a1 , a2 for the sum of the series. 46. Find the interval of convergence of the series sk xk where sk is the k th partial sum of the series ∞ n=1 k k (−1)k (x + 3)k . (k + 1)! 1kk x. k (−1)k ak (x − a)k . k2 ln k (x − 2)k . 2k 1+ (−1)k ( 2 )k (x + 1)k . 3 21/k π k (x − 2)k . k (k + 1)(k + 2) 1 . n 34. 1 (x − 1)k . ( ln k )k 47. Let ak x be a power series, and let r be its radius of convergence. (a) Given that |ak |1/k → ρ , show that, if ρ = 0, then r = 1/ρ and, if ρ = 0, then r = ∞. (b) Given that |ak +1 /ak | → λ, show that, if λ = 0, then r = 1/λ and, if λ = 0, then r = ∞. 48. Let ak x k be a power series and let r , 0 < r < ∞, be its radius of convergence. Prove that the power series ak x2k √ has radius of convergence r . x 2x 2 3x 3 4x4 + − + − ··· . 2 4 8 16 (x − 1) 4 9 16 38. + 4 (x − 1)2 + 6 (x − 1)3 + 8 (x − 1)4 + · · · . 52 5 5 5 2 4 6 8 9x 27x 81x 3x + + + + ··· . 39. 4 9 16 25 1 2 3 4 40. 16 (x + 1) − 25 (x + 1)2 + 36 (x + 1)3 − 49 (x + 1)4 + · · · . 37. 1 − 41. Suppose that the power series ∞ 0 ak (x − 1)k converges k= at x = 3. What can you say about the convergence or divergence of the following? ∞ c 49. Use a CAS to implement the ratio test to find the radius of convergence of the power series. ∞ 2k (a) xk . k =0 k ! (b) (c) k =0 2k k x. k =0 k ∞ ∞ (−1)k c 50. Consider the power series (−1)k +1 k x. k k =1 (a) Graph the partial sums s2 (x), s4 (x), s6 (x), s8 (x), s10 (x) together. ∞ x 2 k . (a) k =0 ∞ ak . (−1)k ak . (−1)k ak 2k . (b) On the same screen, graph f (x) = ln (1 + x). (c) On what interval does it appear that sn (x) may be converging to f (x)? (d) Experiment with sn (x) for larger values of n. ∞ 1 c 51. Repeat Exercise 50 with xk and f (x) = . 1−x k =0 (b) k =0 ∞ (c) k =0 42. Suppose that the power series ∞ 0 ak (x + 2)k converges at k= x = 4. At what other values of x must ∞ 0 ak (x + 2)k k= converge? Does the power series converge at x = −8? Explain. c 52. Repeat Exercise 50 with f (x)= tan−1 x. (−1)k +1 2k −1 and x k =1 2k − 1 ∞ ...
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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.

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