Chapter 12 - 5E-12(pp 736-745 10:08 AM Page 736 CHAPTER 12 Bessel functions which are used to model the vibrations of drumheads and cymbals are

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Unformatted text preview: 5E-12(pp 736-745) 1/18/06 10:08 AM Page 736 CHAPTER 12 Bessel functions, which are used to model the vibrations of drumheads and cymbals, are defined as sums of infinite series in Section 12.8. Notice how closely the computergenerated models (which involve Bessel functions and cosine functions) match the photographs of a vibrating rubber membrane. I nfinite Sequences and Series 5E-12(pp 736-745) 1/18/06 10:08 AM Page 737 Infinite sequences and series were introduced briefly in A Preview of Calculus in connection with Zeno’s paradoxes and the decimal representation of numbers. Their importance in calculus stems from Newton’s idea of representing functions as sums of infinite series. For instance, in finding areas he often integrated a function by first expressing it as a series and then integrating each term of the series. We will pursue his idea in Section 12.10 in order to integrate such functions as e x2 . (Recall that we have previously been unable to do this.) Many of the functions that arise in mathematical physics and chemistry, such as Bessel functions, are defined as sums of series, so it is important to be familiar with the basic concepts of convergence of infinite sequences and series. Physicists also use series in another way, as we will see in Section 11.12. In studying fields as diverse as optics, special relativity, and electromagnetism, they analyze phenomena by replacing a function with the first few terms in the series that represents it. |||| 12.1 Sequences A sequence can be thought of as a list of numbers written in a definite order: a 1, a 2, a 3, a 4, . . . , a n, . . . The number a 1 is called the first term, a 2 is the second term, and in general a n is the nth term. We will deal exclusively with infinite sequences and so each term a n will have a successor a n 1 . Notice that for every positive integer n there is a corresponding number a n and so a sequence can be defined as a function whose domain is the set of positive integers. But we usually write a n instead of the function notation f n for the value of the function at the number n. NOTATION ■ The sequence {a 1 , a 2 , a 3 , . . .} is also denoted by an or an n1 EXAMPLE 1 Some sequences can be defined by giving a formula for the n th term. In the following examples we give three descriptions of the sequence: one by using the preceding notation, another by using the defining formula, and a third by writing out the terms of the sequence. Notice that n doesn’t have to start at 1. (a) (b) n n 1 1nn 3n an n1 1 an n n 1234 n , , , ,..., ,... 2345 n1 1 1nn 3n 1 23 ,, 39 45 , ,..., 27 81 1nn 3n 1 ,... 737 5E-12(pp 736-745) 738 ❙❙❙❙ 1/18/06 10:08 AM Page 738 CHAPTER 12 INFINITE SEQUENCES AND SERIES (c) (d) {sn cos 3 }n sn 3, n cos n ,n 6 {0, 1, s2, s3, . . . , sn 3 an n 6 an 3 0 n0 1, 3, . . .} n s3 1 , , 0, . . . , cos ,... 22 6 EXAMPLE 2 Find a formula for the general term a n of the sequence 3 , 5 4 5 , , 25 125 6 7 , ,... 625 3125 assuming that the pattern of the first few terms continues. SOLUTION We are given that a1 3 5 4 25 a2 a3 5 125 6 625 a4 a5 7 3125 Notice that the numerators of these fractions start with 3 and increase by 1 whenever we go to the next term. The second term has numerator 4, the third term has numerator 5; in general, the nth term will have numerator n 2. The denominators are the powers of 5, so a n has denominator 5 n. The signs of the terms are alternately positive and negative, so we need to multiply by a power of 1. In Example 1(b) the factor 1 n meant we started with a negative term. Here we want to start with a positive term and so we use 1 n 1 or 1 n 1. Therefore, an 1 n1 n 2 5n EXAMPLE 3 Here are some sequences that don’t have a simple defining equation. (a) The sequence pn , where pn is the population of the world as of January 1 in the year n. (b) If we let a n be the digit in the nth decimal place of the number e, then a n is a welldefined sequence whose first few terms are 7, 1, 8, 2, 8, 1, 8, 2, 8, 4, 5, . . . (c) The Fibonacci sequence fn is defined recursively by the conditions f1 1 f2 1 fn fn 1 fn 2 n 3 Each term is the sum of the two preceding terms. The first few terms are 1, 1, 2, 3, 5, 8, 13, 21, . . . This sequence arose when the 13th-century Italian mathematician known as Fibonacci solved a problem concerning the breeding of rabbits (see Exercise 65). a¡ 0 FIGURE 1 1 2 a™ a£ a¢ 1 A sequence such as the one in Example 1(a), a n n n 1 , can be pictured either by plotting its terms on a number line as in Figure 1 or by plotting its graph as in Figure 2. Note that, since a sequence is a function whose domain is the set of positive integers, its graph consists of isolated points with coordinates 1, a1 2, a2 3, a3 ... n, a n ... 5E-12(pp 736-745) 1/18/06 10:08 AM Page 739 S ECTION 12.1 SEQUENCES an From Figure 1 or 2 it appears that the terms of the sequence a n approaching 1 as n becomes large. In fact, the difference 1 1 7 a¶= 8 0 n 1234567 n n nn ❙❙❙❙ 739 1 are 1 1 n 1 can be made as small as we like by taking n sufficiently large. We indicate this by writing FIGURE 2 n lim n nl 1 1 In general, the notation lim a n nl L means that the terms of the sequence a n approach L as n becomes large. Notice that the following definition of the limit of a sequence is very similar to the definition of a limit of a function at infinity given in Section 4.4. 1 Definition A sequence a n has the limit L and we write lim a n nl L or a n l L as n l if we can make the terms a n as close to L as we like by taking n sufficiently large. If lim n l a n exists, we say the sequence converges (or is convergent). Otherwise, we say the sequence diverges (or is divergent). Figure 3 illustrates Definition 1 by showing the graphs of two sequences that have the limit L. an an L L FIGURE 3 Graphs of two sequences with lim an= L 0 0 n n n ` A more precise version of Definition 1 is as follows. 2 Definition A sequence an has the limit L and we write lim an nl L or a n l L as n l |||| Compare this definition with Definition 4.4.5. if for every 0 there is a corresponding integer N such that an L whenever n N 5E-12(pp 736-745) 740 ❙❙❙❙ 1/18/06 10:08 AM Page 740 CHAPTER 12 INFINITE SEQUENCES AND SERIES Definition 2 is illustrated by Figure 4, in which the terms a 1 , a 2 , a 3 , . . . are plotted on a number line. No matter how small an interval L ,L is chosen, there exists an N such that all terms of the sequence from a N 1 onward must lie in that interval. a¡ F IGURE 4 a£ a™ aˆ 0 aN+1 aN+2 L-∑ L a˜ aß a∞ a¢ a¶ L+∑ Another illustration of Definition 2 is given in Figure 5. The points on the graph of an must lie between the horizontal lines y L and y L if n N . This picture must be valid no matter how small is chosen, but usually a smaller requires a larger N. y y=L+∑ L y=L-∑ 0 FIGURE 5 1234 n N Comparison of Definition 2 and Definition 4.4.5 shows that the only difference between lim n l a n L and lim x l f x L is that n is required to be an integer. Thus, we have the following theorem, which is illustrated by Figure 6. 3 Theorem If lim x l f x lim n l an L and f n a n when n is an integer, then L. y y=ƒ L FIGURE 6 0 x 1234 In particular, since we know that limx l we have 1 lim 0 4 nl nr 1 xr if r 0 when r 0 (Theorem 4.4.4), 0 If a n becomes large as n becomes large, we use the notation lim n l a n lowing precise definition is similar to Definition 4.4.7. 5 Definition lim n l an integer N such that . The fol- means that for every positive number M there is an an M whenever n N If lim n l a n , then the sequence a n is divergent but in a special way. We say that a n diverges to . 5E-12(pp 736-745) 1/18/06 10:08 AM Page 741 S ECTION 12.1 SEQUENCES ❙❙❙❙ 741 The Limit Laws given in Section 2.3 also hold for the limits of sequences and their proofs are similar. If a n and bn are convergent sequences and c is a constant, then Limit Laws for Sequences lim a n bn lim a n bn lim ca n c lim a n nl nl nl lim a n lim a n lim bn nl lim c nl lim a n bn nl nl nl c nl lim a n nl lim lim bn nl lim bn nl nl lim a n an bn nl if lim bn lim bn 0 if p 0 and a n nl nl [ lim a ] lim a np nl p n nl 0 The Squeeze Theorem can also be adapted for sequences as follows (see Figure 7). Squeeze Theorem for Sequences If a n bn cn for n n 0 and lim a n lim cn nl L, then lim bn nl nl L. cn Another useful fact about limits of sequences is given by the following theorem, whose proof is left as Exercise 69. bn an 0 6 Theorem If lim a n 0, then lim a n nl 0. nl n FIGURE 7 The sequence bn is squeezed between the sequences a n and cn . EXAMPLE 4 Find lim nl n n 1 . SOLUTION The method is similar to the one we used in Section 4.4: Divide numerator and denominator by the highest power of n and then use the Limit Laws. lim nl n n 1 nl 1 1 |||| This shows that the guess we made earlier from Figures 1 and 2 was correct. 1 Here we used Equation 4 with r 1. lim 1 1 lim 0 nl 1 n 1 lim 1 nl lim nl 1 n 5E-12(pp 736-745) 742 ❙❙❙❙ 1/18/06 10:08 AM Page 742 CHAPTER 12 INFINITE SEQUENCES AND SERIES EXAMPLE 5 Calculate lim nl ln n . n SOLUTION Notice that both numerator and denominator approach infinity as n l . We can’t apply l’Hospital’s Rule directly because it applies not to sequences but to functions of a real variable. However, we can apply l’Hospital’s Rule to the related function fx ln x x and obtain lim xl ln x x lim xl 1x 1 0 Therefore, by Theorem 3 we have ln n n lim nl an 0 1 n is convergent or divergent. EXAMPLE 6 Determine whether the sequence a n 1 SOLUTION If we write out the terms of the sequence, we obtain 0 1 2 3 4 n 1, 1, 1, 1, 1, 1, 1, . . . _1 The graph of this sequence is shown in Figure 8. Since the terms oscillate between 1 and 1 infinitely often, a n does not approach any number. Thus, lim n l 1 n does not exist; n 1 is divergent. that is, the sequence FIGURE 8 |||| The graph of the sequence in Example 7 is shown in Figure 9 and supports the answer. EXAMPLE 7 Evaluate lim nl 1 n n if it exists. SOLUTION an 1 1 n lim nl n lim nl 1 n 0 Therefore, by Theorem 6, 0 1 n lim nl 1 n n 0 _1 n! n n, where EXAMPLE 8 Discuss the convergence of the sequence a n FIGURE 9 n! 123 n. SOLUTION Both numerator and denominator approach infinity as n l but here we have no corresponding function for use with l’Hospital’s Rule (x! is not defined when x is not an integer). Let’s write out a few terms to get a feeling for what happens to a n as n gets large: a1 7 a2 1 an 12 22 a3 123 nnn 123 333 n n It appears from these expressions and the graph in Figure 10 that the terms are decreas- 5E-12(pp 736-745) 1/18/06 10:08 AM Page 743 ❙❙❙❙ S ECTION 12.1 SEQUENCES |||| CREATING GRAPHS OF SEQUENCES Some computer algebra systems have special commands that enable us to create sequences and graph them directly. With most graphing calculators, however, sequences can be graphed by using parametric equations. For instance, the sequence in Example 8 can be graphed by entering the parametric equations x t y ing and perhaps approach 0. To confirm this, observe from Equation 7 that 1 n an 23 nn n n Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator. So 1 0 an n t! t t and graphing in dot mode starting with t 1, setting the t-step equal to 1. The result is shown in Figure 10. We know that 1 n l 0 as n l . Therefore, a n l 0 as n l Theorem. 1 by the Squeeze EXAMPLE 9 For what values of r is the sequence r n convergent? SOLUTION We know from Section 4.4 and the graphs of the exponential functions in Section 7.2 (or Section 7.4*) that lim x l a x for a 1 and lim x l a x 0 for 0 a 1. Therefore, putting a r and using Theorem 2, we have 0 10 lim r n if r if 0 0 nl FIGURE 10 1 r 1 It is obvious that lim 1n nl If 1 r 0, then 0 lim 0 n and 1 0 nl 1, so r lim r n lim r nl n 0 nl and therefore lim n l r n 0 by Theorem 6. If r 1, then r n diverges as in Example 6. Figure 11 shows the graphs for various values of r. (The case r 1 is shown in Figure 8.) an an r>1 1 _1<r<0 1 0 r=1 n 1 0 FIGURE 11 The sequence an=r 743 1 0<r<1 n r<_1 n The results of Example 9 are summarized for future use as follows. The sequence r n is convergent if values of r. 8 lim r n nl 0 1 1 r 1 and divergent for all other if 1 r if r 1 1 5E-12(pp 736-745) 744 ❙❙❙❙ 1/18/06 10:08 AM Page 744 CHAPTER 12 INFINITE SEQUENCES AND SERIES a n 1 for all n 1, that 9 Definition A sequence an is called increasing if a n is, a1 a2 a3 . It is called decreasing if a n a n 1 for all n 1. It is called monotonic if it is either increasing or decreasing. 3 EXAMPLE 10 The sequence n is decreasing because 5 3 n for all n 5 3 1 n 3 5 n 6 1. (The right side is smaller because it has a larger denominator.) n EXAMPLE 11 Show that the sequence a n SOLUTION 1 We must show that a n n2 is decreasing. 1 a n , that is, 1 n n 1 1 n 2 n2 1 1 This inequality is equivalent to the one we get by cross-multiplication: n 1 n 1 n 1 n3 1 Since n 1, we know that the inequality n 2 so a n is decreasing. x2 x Thus, f is decreasing on 1, x 2x 2 2 n n2 n 1 1 2 nn 1 1 n3 1 2n 2 2n n 1 is true. Therefore, a n 1 a n and : 1 x2 x2 12 0 and so f n fn 1 . Therefore, a n is decreasing. 1 2 n2 x SOLUTION 2 Consider the function f x fx 1 n2 n &? 1 2 &? &? n 2 1 2 whenever x 2 1 1 0 Definition A sequence a n is bounded above if there is a number M such that an M for all n 1 It is bounded below if there is a number m such that m an for all n 1 If it is bounded above and below, then a n is a bounded sequence. 5E-12(pp 736-745) 1/18/06 10:08 AM Page 745 S ECTION 12.1 SEQUENCES an M L 0 1 23 FIGURE 12 n ❙❙❙❙ 745 For instance, the sequence a n n is bounded below a n 0 but not above. The sequence a n n n 1 is bounded because 0 a n 1 for all n. We know that not every bounded sequence is convergent [for instance, the sequence an 1 n satisfies 1 a n 1 but is divergent from Example 6] and not every monotonic sequence is convergent a n n l . But if a sequence is both bounded and monotonic, then it must be convergent. This fact is proved as Theorem 11, but intuitively you can understand why it is true by looking at Figure 12. If a n is increasing and a n M for all n, then the terms are forced to crowd together and approach some number L. The proof of Theorem 11 is based on the Completeness Axiom for the set of real numbers, which says that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S ), then S has a least upper bound b. (This means that b is an upper bound for S, but if M is any other upper bound, then b M .) The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line. 1 1 Monotonic Sequence Theorem Every bounded, monotonic sequence is convergent. Proof Suppose a n is an increasing sequence. Since a n is bounded, the set S a n n 1 has an upper bound. By the Completeness Axiom it has a least upper bound L . Given is not an upper bound for S (since L is the least upper 0, L bound). Therefore aN L for some integer N But the sequence is increasing so a n a N for every n an since a n L N we have L 0 so N . Thus, if n an L . Thus L an whenever n N so lim n l a n L . A similar proof (using the greatest lower bound) works if a n is decreasing. The proof of Theorem 11 shows that a sequence that is increasing and bounded above is convergent. (Likewise, a decreasing sequence that is bounded below is convergent.) This fact is used many times in dealing with infinite series. EXAMPLE 12 Investigate the sequence an defined by the recurrence relation a1 2 an 1 1 2 an 6 for n 1, 2, 3, . . . SOLUTION We begin by computing the first several terms: a2 a1 2 a4 1 2 a7 5.9375 5 6 5.5 1 2 a3 1 2 a5 5.75 a6 5.875 a8 5.96875 a9 5.984375 2 6 4 4 6 5 5E-12(pp 746-755) 746 ❙❙❙❙ 1/18/06 10:52 AM Page 746 CHAPTER 12 INFINITE SEQUENCES AND SERIES |||| Mathematical induction is often used in dealing with recursive sequences. See page 59 for a discussion of the Principle of Mathematical Induction. These initial terms suggest that the sequence is increasing and the terms are approaching 6. To confirm that the sequence is increasing we use mathematical induction to show that a n 1 a n for all n 1. This is true for n 1 because a 2 4 a 1. If we assume that it is true for n k, then we have ak so ak 1 2 and 1 6 1 ak ak ak 1 2 6 1 Thus ak 6 ak ak 2 6 1 We have deduced that a n 1 a n is true for n k 1. Therefore, the inequality is true for all n by induction. Next we verify that a n is bounded by showing that a n 6 for all n. (Since the sequence is increasing, we already know that it has a lower bound: a n a 1 2 for all n.) We know that a 1 6, so the assertion is true for n 1. Suppose it is true for n k. Then ak so 6 ak 1 2 ak 6 12 1 2 6 Thus ak 12 6 6 1 This shows, by mathematical induction, that a n 6 for all n. Since the sequence an is increasing and bounded, Theorem 11 guarantees that it has a limit. The theorem doesn’t tell us what the value of the limit is. But now that we know L lim n l an exists, we can use the recurrence relation to write 1 2 an |||| A proof of this fact is requested in Exercise 52. Since a n l L , it follows that a n 1 l L , too (as n l , n 1 nl 1 2 6 1 2 L Solving this equation for L, we get L |||| 12.1 ( lim a lim lim a n nl L nl n ) 6 1 2 L 6 1 l , too). So we have 6 6, as predicted. Exercises 1. (a) What is a sequence? (b) What does it mean to say that lim n l a n (c) What does it mean to say that lim n l a n 3–8 8? ? 2. (a) What is a convergent sequence? Give two examples. (b) What is a divergent sequence? Give two examples. |||| 3. a n 5. a n List the first five terms of the sequence. 1 3 0.2 1 n! n 4. a n n 3n 1 1 n 6. 2 4 6 2n 5E-12(pp 746-755) 1/18/06 10:52 AM Page 747 S ECTION 12.1 SEQUENCES 7. a 1 ■ 3, ■ 9–14 an 2a n 1 ■ ■ 8. a 1 1 ■ ■ ■ 4, ■ an an 1 ■ an ■ ■ 47. a n |||| 1 {1 , 1 , 1 , 16 , . . .} 248 10. 11. 2, 7, 12, 17, . . . 13. {1, ■ 24 39 ■ , . . .} 8 27 ,, ■ 12. ■ 48. a n {1 , 1 , 1 , 1 , . . . } 2468 34 { 1 , 2 , 16 , 25 , . . .} 49 ■ ■ ■ ■ ■ ■ ■ |||| Determine whether the sequence converges or diverges. If it converges, find the limit. nn 16. a n 1 5n 2 n2 17. a n 3 n 19. a n 2n 3n 1 20. a n n 1 n2 23. a n 22. a n 24. a n cos n 2 2n 2n 25. 1! 1! 29. n 2e ■ 1 2 an 3a n 1 ■ ■ ■ ■ ■ ■ if a n is an even number if a n is an odd number 1 lim a n cos 2 n 25. Make a conjecture about 1 lim a n nl (b) A sequence a n is defined by a 1 1 and a n 1 1 1 a n for n 1. Assuming that a n is convergent, find its limit. 53. Suppose you know that a n is a decreasing sequence and 28. n all its terms lie between the numbers 5 and 8. Explain why the sequence has a limit. What can you say about the value of the limit? ln n ln 2 n 30. n cos n 32. a n ln n 1 54–60 |||| Determine whether the sequence is increasing, decreasing, or not monotonic. Is the sequence bounded? ln n 33. a n n sin 1 n 34. a n sn sn 2 1 36. a n sin 2 n 1 sn 1n 2 n 1 38. 37. 0, 1, 0, 0, 1, 0, 0, 0, 1, . . . n! 2n ■ {, ■ ■ ■ arctan 1 n n 1 n 2n 2n 1 3 n! ■ ■ ■ 42. a n 44. 2n 3n 58. a n ne 60. a n n 55. a n 3 4 57. a n 1 2n n ■ ■ 2 sin n sn 2 n ■ ■ ■ n 59. a n 3 cos n , , , , , , , . . .} |||| Use a graph of the sequence to decide whether the sequence is convergent or divergent. If the sequence is convergent, guess the value of the limit from the graph and then prove your guess. (See the margin note on page 743 for advice on graphing sequences.) 41. a n 1 5n 11111111 13243546 40. a n ■ 54. a n 56. a n ; 41–48 43. ■ nl cos n 2n ■ ■ 52. (a) If a n is convergent, show that 31. a n 39. a n ■ 1 n n3 2n2 1 2 35. a n n! ■ 51. For what values of r is the sequence nr n convergent? 26. arctan 2 n en e n e 2n 1 27. 1 and a1 11. Do the same if a1 this type of sequence. sn n3 1 2n 135 an n 1 2n n 2n 50. Find the first 40 terms of the sequence defined by sn sn 1 135 1 1 n1 1 21. a n 18. a n n 3n 5n after n years the investment is worth a n 1000 1.06 n dollars. (a) Find the first five terms of the sequence a n . (b) Is the sequence convergent or divergent? Explain. ■ 15–40 15. a n n s3 n 49. If $1000 is invested at 6% interest, compounded annually, then 14. 5, 1, 5, 1, 5, 1, . . . ■ 46. a n 747 ■ Find a formula for the general term a n of the sequence, assuming that the pattern of the first few terms continues. 9. n3 n! 45. a n 1 ❙❙❙❙ 2 n n2 1 1 n ■ ■ ■ ■ ■ ■ ■ ■ ■ 61. Find the limit of the sequence {s2, s2s2, s2s2s2, . . .} 62. A sequence a n is given by a 1 s2, a n 1 s2 a n . (a) By induction or otherwise, show that a n is increasing and bounded above by 3. Apply Theorem 11 to show that lim n l a n exists. (b) Find lim n l a n. ■ 5E-12(pp 746-755) 748 ❙❙❙❙ 1/18/06 10:52 AM Page 748 CHAPTER 12 INFINITE SEQUENCES AND SERIES 63. Show that the sequence defined by a 1 1, a n 1 3 1 a n is 3 for all n. Deduce that a n is convergent increasing and a n and find its limit. ( f) Use Theorem 11 to show that lim n l (The limit is e. See Equation 7.4.9.) a1 b. Let a 1 be their arithmetic mean and b1 their geometric mean: an 2 1 1 3 a a1 an satisfies 0 a n 2 and is decreasing. Deduce that the sequence is convergent and find its limit. 65. (a) Fibonacci posed the following problem: Suppose that rabbits live forever and that every month each pair produces a new pair which becomes productive at age 2 months. If we start with one newborn pair, how many pairs of rabbits will we have in the nth month? Show that the answer is fn , where fn is the Fibonacci sequence defined in Example 3(c). (b) Let a n fn 1 fn and show that a n 1 1 1 a n 2. Assuming that a n is convergent, find its limit. 66. (a) Let a 1 a, a 2 f a , a 3 f a 2 f f a ,..., a n 1 f a n , where f is a continuous function. If lim n l a n L, show that f L L. (b) Illustrate part (a) by taking f x cos x, a 1, and estimating the value of L to five decimal places. lim nl sab b1 Repeat this process so that, in general, an bn an 1 bn 2 sa n bn 1 (a) Use mathematical induction to show that an an bn 1 bn 1 (b) Deduce that both a n and bn are convergent. (c) Show that lim n l a n lim n l bn . Gauss called the common value of these limits the arithmetic-geometric mean of the numbers a and b. 72. (a) Show that if lim n l a 2 n L and lim n l a2n a n is convergent and lim n l a n L. (b) If a 1 1 and an 5 b 2 ; 67. (a) Use a graph to guess the value of the limit 1 1 L, then 1 1 1 an find the first eight terms of the sequence a n . Then use part (a) to show that lim n l a n s2. This gives the continued fraction expansion n n! (b) Use a graph of the sequence in part (a) to find the smallest values of N that correspond to 0.1 and 0.001 in Definition 2. 68. Use Definition 2 directly to prove that lim n l r n 1 1 s2 2 0 1 2 73. The size of an undisturbed fish population has been modeled 1. by the formula 69. Prove Theorem 6. [Hint: Use either Definition 2 or the Squeeze Theorem.] 1 n (a) Show that if 0 70. Let a n 1 n n exists. 71. Let a and b be positive numbers with a 64. Show that the sequence defined by when r 1 1 bn n . a 1 b b, then an a 1 n 1 bn (b) Deduce that b n n 1 a n b a n 1. (c) Use a 1 1 n 1 and b 1 1 n in part (b) to show that a n is increasing. (d) Use a 1 and b 1 1 2 n in part (b) to show that a 2 n 4. (e) Use parts (c) and (d) to show that a n 4 for all n. pn 1 a b pn pn where pn is the fish population after n years and a and b are positive constants that depend on the species and its environment. Suppose that the population in year 0 is p 0 0. (a) Show that if pn is convergent, then the only possible values for its limit are 0 and b a. (b) Show that pn 1 b a pn. (c) Use part (b) to show that if a b, then lim n l pn 0; in other words, the population dies out. (d) Now assume that a b. Show that if p 0 b a, then pn is increasing and 0 pn b a. Show also that if p 0 b a, then pn is decreasing and pn b a. Deduce that if a b, then lim n l pn b a. 5E-12(pp 746-755) 1/18/06 10:52 AM Page 749 S ECTION 12.2 SERIES ❙❙❙❙ 749 LABORATORY PROJECT C AS L ogistic Sequences A sequence that arises in ecology as a model for population growth is defined by the logistic difference equation pn 1 k pn 1 pn where pn measures the size of the population of the nth generation of a single species. To keep the numbers manageable, pn is a fraction of the maximal size of the population, so 0 pn 1. Notice that the form of this equation is similar to the logistic differential equation in Section 10.5. The discrete model—with sequences instead of continuous functions—is preferable for modeling insect populations, where mating and death occur in a periodic fashion. An ecologist is interested in predicting the size of the population as time goes on, and asks these questions: Will it stabilize at a limiting value? Will it change in a cyclical fashion? Or will it exhibit random behavior? Write a program to compute the first n terms of this sequence starting with an initial population p0 , where 0 p0 1. Use this program to do the following. 1. Calculate 20 or 30 terms of the sequence for p0 1 2 and for two values of k such that 1 k 3. Graph the sequences. Do they appear to converge? Repeat for a different value of p0 between 0 and 1. Does the limit depend on the choice of p0? Does it depend on the choice of k ? 2. Calculate terms of the sequence for a value of k between 3 and 3.4 and plot them. What do you notice about the behavior of the terms? 3. Experiment with values of k between 3.4 and 3.5. What happens to the terms? 4. For values of k between 3.6 and 4, compute and plot at least 100 terms and comment on the behavior of the sequence. What happens if you change p0 by 0.001? This type of behavior is called chaotic and is exhibited by insect populations under certain conditions. |||| 12.2 Series If we try to add the terms of an infinite sequence a n a1 1 a2 a3 n1 we get an expression of the form an which is called an infinite series (or just a series) and is denoted, for short, by the symbol an or an n1 But does it make sense to talk about the sum of infinitely many terms? It would be impossible to find a finite sum for the series 1 2 3 4 5 n because if we start adding the terms we get the cumulative sums 1, 3, 6, 10, 15, 21, . . . and, after the n th term, we get n n 1 2, which becomes very large as n increases. However, if we start to add the terms of the series 1 2 1 4 1 8 1 16 1 32 1 64 1 2n 5E-12(pp 746-755) ❙❙❙❙ 750 1/18/06 10:52 AM Page 750 CHAPTER 12 INFINITE SEQUENCES AND SERIES n Sum of first n terms 1 2 3 4 5 6 7 10 15 20 25 0.50000000 0.75000000 0.87500000 0.93750000 0.96875000 0.98437500 0.99218750 0.99902344 0.99996948 0.99999905 0.99999997 we get 1 , 3 , 7 , 15 , 31 , 63 , . . . , 1 1 2 n, . . . . The table shows that as we add more and more 2 4 8 16 32 64 terms, these partial sums become closer and closer to 1. (See also Figure 11 in A Preview of Calculus, page 7.) In fact, by adding sufficiently many terms of the series we can make the partial sums as close as we like to 1. So it seems reasonable to say that the sum of this infinite series is 1 and to write n1 1 2n 1 2 1 4 1 8 1 16 1 2n 1 We use a similar idea to determine whether or not a general series (1) has a sum. We consider the partial sums s1 a1 s2 a1 a2 s3 a1 a2 a3 s4 a1 a2 a3 a4 and, in general, n sn a1 a2 a3 an ai i1 These partial sums form a new sequence sn , which may or may not have a limit. If lim n l sn s exists (as a finite number), then, as in the preceding example, we call it the sum of the infinite series a n . 2 Definition Given a series n1 an a1 a2 , let sn denote its nth a3 partial sum: n sn ai a1 a2 an i1 If the sequence sn is convergent and lim n l sn s exists as a real number, then the series a n is called convergent and we write a1 a2 an s or an s n1 The number s is called the sum of the series. Otherwise, the series is called divergent. Thus, when we write n 1 a n s we mean that by adding sufficiently many terms of the series we can get as close as we like to the number s. Notice that |||| Compare with the improper integral n y 1 f x dx lim tl y t 1 an f x dx To find this integral we integrate from 1 to t and then let t l . For a series, we sum from 1 to n and then let n l . n1 lim nl ai i1 EXAMPLE 1 An important example of an infinite series is the geometric series a ar ar 2 ar 3 ar n 1 ar n n1 1 a 0 5E-12(pp 746-755) 1/18/06 10:52 AM Page 751 S ECTION 12.2 SERIES |||| Figure 1 provides a geometric demonstration of the result in Example 1. If the triangles are constructed as shown and s is the sum of the series, then, by similar triangles, s a a a ar so s a 1 r ar a rsn ar 2 ar n 1 ar sn ar 2 ar n 1 and ar@ ar n Subtracting these equations, we get ar@ sn ar ar a 751 Each term is obtained from the preceding one by multiplying it by the common ratio r. (We have already considered the special case where a 1 and r 1 on page 749.) 2 2 If r 1, then sn a a a na l . Since lim n l sn doesn’t exist, the geometric series diverges in this case. If r 1, we have ar# a-ar ❙❙❙❙ s a If 1 ar n a rn r a1 1 sn 3 1, we know from (12.1.8) that r n l 0 as n l , so r lim sn nl a rsn lim nl rn r a1 1 a 1 a r 1 r lim r n nl a 1 r Thus, when r 1 the geometric series is convergent and its sum is a 1 r . If r 1 or r 1, the sequence r n is divergent by (12.1.8) and so, by Equation 3, lim n l sn does not exist. Therefore, the geometric series diverges in those cases. FIGURE 1 We summarize the results of Example 1 as follows. 4 The geometric series ar n 1 a ar 2 ar n1 |||| In words: The sum of a convergent geometric series is 1 is convergent if r 1 and its sum is first term common ratio ar n a 1 1 n1 If r r r 1 1, the geometric series is divergent. EXAMPLE 2 Find the sum of the geometric series 5 10 3 20 9 40 27 SOLUTION The first term is a 5 and the common ratio is r series is convergent by (4) and its sum is 5 10 3 20 9 40 27 5 1 ( 2) 3 2 3 . Since r 5 5 3 3 2 3 1, the 5E-12(pp 746-755) 752 ❙❙❙❙ 1/18/06 10:52 AM Page 752 CHAPTER 12 INFINITE SEQUENCES AND SERIES |||| What do we really mean when we say that the sum of the series in Example 2 is 3? Of course, we can’t literally add an infinite number of terms, one by one. But, according to Definition 2, the total sum is the limit of the sequence of partial sums. So, by taking the sum of sufficiently many terms, we can get as close as we like to the number 3. The table shows the first ten partial sums sn and the graph in Figure 2 shows how the sequence of partial sums approaches 3. sn n sn 1 2 3 4 5 6 7 8 9 10 5.000000 1.666667 3.888889 2.407407 3.395062 2.736626 3.175583 2.882945 3.078037 2.947975 3 0 20 n FIGURE 2 22n31 EXAMPLE 3 Is the series n convergent or divergent? n1 SOLUTION Let’s rewrite the nth term of the series in the form ar n 1: |||| Another way to identify a and r is to write out the first few terms: 4 16 3 22n31 64 9 4n 3n 1 n n1 n1 n1 4( 4 ) 3 n1 We recognize this series as a geometric series with a series diverges by (4). EXAMPLE 4 Write the number 2.317 4 and r 4 3 . Since r 1, the 2.3171717. . . as a ratio of integers. SOLUTION 2.3171717. . . 17 10 3 2.3 17 10 5 17 10 7 After the first term we have a geometric series with a 17 10 3 and r Therefore 17 17 3 10 1000 2.317 2.3 2.3 1 99 1 2 10 100 23 17 1147 10 990 495 x n, where x EXAMPLE 5 Find the sum of the series 1 10 2. 1. n0 0 and so the first term is x 0 1 even when x 0.) Thus SOLUTION Notice that this series starts with n Module 12.2 explores a series that depends on an angle in a triangle and enables you to see how rapidly the series converges when varies. series, we adopt the convention that x xn 1 0 x2 x x3 1. (With x4 n0 This is a geometric series with a (4) gives 1 and r xn 5 n0 x. Since r 1 1 x x 1, it converges and 5E-12(pp 746-755) 1/18/06 10:53 AM Page 753 S ECTION 12.2 SERIES 1 EXAMPLE 6 Show that the series nn n1 ❙❙❙❙ 753 is convergent, and find its sum. 1 SOLUTION This is not a geometric series, so we go back to the definition of a convergent series and compute the partial sums. n 1 sn ii i1 1 1 12 1 1 1 23 34 nn 1 We can simplify this expression if we use the partial fraction decomposition 1 ii 1 i 1 1 i 1 (see Section 8.4). Thus, we have n i1 ii 1 1 1 2 n i 1 1 3 1 3 1 4 1 n 1 n 1 lim sn lim nl 1 1 nl n 1 1 0 ) 1 1 Therefore, the given series is convergent and 1 nn n1 1 1 EXAMPLE 7 Show that the harmonic series 1 n1 sn 1 1 and so |||| Figure 3 illustrates Example 6 by showing the graphs of the sequence of terms a n 1 [n n 1 ] and the sequence sn of partial sums. Notice that a n l 0 and sn l 1. See Exercises 54 and 55 for two geometric interpretations of Example 6. 1 i i1 1 2 1 |||| Notice that the terms cancel in pairs. This is an example of a telescoping sum: Because of all the cancellations, the sum collapses (like an old-fashioned collapsing telescope) into just two terms. n 1 sn 1 n 1 2 1 1 3 1 4 is divergent. SOLUTION s1 s2 an 0 FIGURE 3 n 1 1 1 2 s4 1 1 2 (1 3 1 4 s8 1 1 2 1 4 1 1 2 (1 3 (1 4 1 1 2 1 2 1 1 2 1 1 2 (1 3 (1 4 1 1 2 1 2 s16 ) 1 ) 1 4) (1 5 (1 8 1 2 1 6 1 7 1 8 1 8 1 8 1 8 (1 5 (1 8 1 4 1 2 1 4 1 2 2 2 ) ) 3 2 1 ) 1 4) (1 4 1 2 ) 1 8) 1 8 1 4 2 (1 9 1 ( 16 1 16 ) ) 1 16 1 5E-12(pp 746-755) 754 ❙❙❙❙ 1/18/06 10:53 AM Page 754 CHAPTER 12 INFINITE SEQUENCES AND SERIES Similarly, s32 1 5 2 , s64 6 2 1 , and in general s2 n |||| The method used in Example 7 for showing that the harmonic series diverges is due to the French scholar Nicole Oresme (1323–1382). This shows that s2 n l series diverges. 6 as n l Theorem If the series n 2 1 and so sn is divergent. Therefore, the harmonic a n is convergent, then lim an nl n1 0. Proof Let sn a1 a2 a n . Then a n sn sn 1. Since a n is convergent, the sequence sn is convergent. Let lim n l sn s. Since n 1 l as n l , we also have lim n l sn 1 s. Therefore lim a n lim sn nl s NOTE 1 s lim sn 1 nl lim sn nl 1 0 ■ NOTE 2 sn nl ■ With any series a n we associate two sequences: the sequence sn of its partial sums and the sequence a n of its terms. If a n is convergent, then the limit of the sequence sn is s (the sum of the series) and, as Theorem 6 asserts, the limit of the sequence a n is 0. | The converse of Theorem 6 is not true in general. If lim n l a n 0, we cannot conclude that a n is convergent. Observe that for the harmonic series 1 n we have a n 1 n l 0 as n l , but we showed in Example 7 that 1 n is divergent. 7 The Test for Divergence If lim a n does not exist or if lim a n nl series 0, then the nl a n is divergent. n1 The Test for Divergence follows from Theorem 6 because, if the series is not divergent, then it is convergent, and so lim n l a n 0. n2 EXAMPLE 8 Show that the series n1 5n 2 4 diverges. SOLUTION lim a n nl lim nl n2 5n 2 4 lim nl 5 1 4 n2 1 5 0 So the series diverges by the Test for Divergence. NOTE 3 If we find that lim n l a n 0, we know that a n is divergent. If we find that lim n l a n 0, we know nothing about the convergence or divergence of a n. Remember the warning in Note 2: If lim n l a n 0, the series a n might converge or it might diverge. ■ 5E-12(pp 746-755) 1/18/06 10:53 AM Page 755 S ECTION 12.2 SERIES 8 Theorem If a n and (where c is a constant), (i) ca n c an n1 (iii) bn are convergent series, then so are the series a n bn , and a n bn , and (ii) an n1 an bn an n1 bn an n1 ❙❙❙❙ 755 ca n bn n1 n1 bn n1 n1 These properties of convergent series follow from the corresponding Limit Laws for Sequences in Section 11.1. For instance, here is how part (ii) of Theorem 8 is proved: Let n sn n ai s an i1 tn bi n1 The nth partial sum for the series t bn i1 an n1 bn is n un ai bi i1 and, using Equation 5.2.9, we have n lim u n nl n lim nl ai bi lim nl i1 n ai lim sn an lim bi i1 bi nl i1 i1 lim tn nl Therefore, i1 n lim nl n ai s nl t bn is convergent and its sum is an bn s t an n1 n1 3 EXAMPLE 9 Find the sum of the series n1 SOLUTION The series bn n1 nn 1 . 2n 1 1 2 n 1 2 is a geometric series with a n1 1 2 1 2n 1 2 , so 1 1 2 1 and r In Example 6 we found that 1 n1 nn 1 1 So, by Theorem 8, the given series is convergent and 3 n1 nn 1 1 2n 1 3 n1 31 nn 1 1 4 n1 1 2n 5E-12(pp 756-765) 756 ❙❙❙❙ 1/18/06 10:09 AM Page 756 CHAPTER 12 INFINITE SEQUENCES AND SERIES NOTE 4 A finite number of terms doesn’t affect the convergence or divergence of a series. For instance, suppose that we were able to show that the series ■ n n3 1 2 9 n4 3 28 is convergent. Since n n1 n 3 1 2 1 n n4 n 3 1 it follows that the entire series n 1 n n 3 1 is convergent. Similarly, if it is known that the series n N 1 a n converges, then the full series N an an n1 n1 an nN1 is also convergent. |||| 12.2 Exercises 14. 1 1. (a) What is the difference between a sequence and a series? 0.4 0.16 0.064 (b) What is a convergent series? What is a divergent series? 2. Explain what it means to say that n 5( 2 ) 3 15. 5. 1 an |||| Find at least 10 partial sums of the series. Graph both the sequence of terms and the sequence of partial sums on the same screen. Does it appear that the series is convergent or divergent? If it is convergent, find the sum. If it is divergent, explain why. n1 3n 4n 17. n1 n1 5. 4. n n1 6. tan n n1 1 1 n1 1 n 1.5 ■ 1 ■ n 1.5 1 ■ ■ ■ nn ■ 1 ■ ■ ■ 2n . 3n 1 (a) Determine whether a n is convergent. (b) Determine whether n 1 a n is convergent. ■ 13. 4 3 2 5 2 8 9 25 8 n 0.3 n ln 32. arctan n 2n cos 1 5 k k1 3 nn ■ ■ ■ n1 n1 30. n1 aj 0.8 3 5 4n 3 ■ ■ 34. n1 ■ ■ 3 5n ■ 2 n ■ ■ ■ ■ i1 12. 125 32 2 4n n1 33. n Determine whether the series is convergent or divergent. If it is convergent, find its sum. 2 n2 28. n n1 |||| 11. 3 12 2 n1 2n 6 31. aj and n nn 26. n s2 35–40 11–34 n1 1 3n 29. (b) Explain the difference between ai 3 n 24. 1 k n1 j1 i1 n1 2 k2 k2 27. n n en 3n 1 22. 5 n1 and i1 n1 n2 25. 10. (a) Explain the difference between n (s2 ) n 20. 2 n2 9. Let a n ai n0 n 23. 1 n2 ■ 1 18. n1 n 1 n1 0.6 8. 3 21. n1 7. ■ 2n2 n2 n0 n1 12 5 1 n 19. n1 6 5n 16. n1 ; 3–8 3. n1 1 8 1 4 1 2 1 |||| 35. 0.2 Express the number as a ratio of integers. 36. 0.73 0.2222 . . . 37. 3.417 38. 6.254 3.417417417 . . . 39. 0.123456 ■ ■ ■ 0.73737373 . . . 40. 5.6021 ■ ■ ■ ■ ■ ■ ■ ■ ■ 5E-12(pp 756-765) 1/18/06 10:09 AM Page 757 ❙❙❙❙ S ECTION 12.2 SERIES 41–45 |||| Find the values of x for which the series converges. Find the sum of the series for those values of x. n1 43. xn 3n 42. 4nx n 41. 44. 4 (a) Assuming that the ball continues to bounce indefinitely, find the total distance that it travels. (Use the fact that the ball falls 1 t t 2 meters in t seconds.) 2 (b) Calculate the total time that the ball travels. (c) Suppose that each time the ball strikes the surface with velocity v it rebounds with velocity k v, where 0 k 1. How long will it take for the ball to come to rest? n 3 x n n1 n0 x 2 n0 n n 45. n0 ■ cos x 2n ■ 53. What is the value of c if ■ ■ ■ ■ ■ ■ ■ ■ ■ 1 n 1 n1 47–48 |||| Use the partial fraction command on your CAS to find a convenient expression for the partial sum, and then use this expression to find the sum of the series. Check your answer by using the CAS to sum the series directly. 1 47. n1 ■ ■ 4n 1 4n ■ 3 ■ n1 ■ ■ ■ 49. If the nth partial sum of a series sn find a n and n1 n n n1 3n n2 ■ n1 1 n ■ 2 ■ ■ ■ nn 1 1 55. The figure shows two circles C and D of radius 1 that touch at P. T is a common tangent line; C1 is the circle that touches C, D, and T ; C2 is the circle that touches C, D, and C1; C3 is the circle that touches C, D, and C2. This procedure can be continued indefinitely and produces an infinite sequence of circles Cn . Find an expression for the diameter of Cn and thus provide another geometric demonstration of Example 6. an is 1 1 an . 50. If the nth partial sum of a series find a n and n2 48. 2? x n, 0 x 1, for n 0, 1, 2, 3, 4, . . . on a common screen. By finding the areas between successive curves, give a geometric demonstration of the fact, shown in Example 6, that is another series with this property. C AS n c ; 54. Graph the curves y whose terms approach 0. Show that ln 1 1 n2 ■ 46. We have seen that the harmonic series is a divergent series n1 757 n1 an is sn 3 P n 2 n, an . 51. When money is spent on goods and services, those that receive the money also spend some of it. The people receiving some of the twice-spent money will spend some of that, and so on. Economists call this chain reaction the multiplier effect. In a hypothetical isolated community, the local government begins the process by spending D dollars. Suppose that each recipient of spent money spends 100c% and saves 100s% of the money that he or she receives. The values c and s are called the marginal propensity to consume and the marginal propensity to save and, of course, c s 1. (a) Let Sn be the total spending that has been generated after n transactions. Find an equation for Sn. (b) Show that lim n l Sn kD, where k 1 s. The number k is called the multiplier. What is the multiplier if the marginal propensity to consume is 80%? Note: The federal government uses this principle to justify deficit spending. Banks use this principle to justify lending a large percentage of the money that they receive in deposits. C£ C™ 1 C 1 D C¡ T 56. A right triangle ABC is given with A and AC b. CD is drawn perpendicular to AB, DE is drawn perpendicular to BC, EF AB, and this process is continued indefinitely as shown in the figure. Find the total length of all the perpendiculars CD DE EF FG in terms of b and . A D ¨ F H b 52. A certain ball has the property that each time it falls from a height h onto a hard, level surface, it rebounds to a height rh, where 0 r 1. Suppose that the ball is dropped from an initial height of H meters. B G E C 5E-12(pp 756-765) 758 ❙❙❙❙ 1/18/06 10:10 AM Page 758 CHAPTER 12 INFINITE SEQUENCES AND SERIES nitely many numbers. Give examples of some numbers in the Cantor set. (b) The Sierpinski carpet is a two-dimensional counterpart of the Cantor set. It is constructed by removing the center oneninth of a square of side 1, then removing the centers of the eight smaller remaining squares, and so on. (The figure shows the first three steps of the construction.) Show that the sum of the areas of the removed squares is 1. This implies that the Sierpinski carpet has area 0. 57. What is wrong with the following calculation? 0 0 0 1 0 1 1 1 1 1 1 1 1 0 1 1 1 1 1 0 1 1 1 1 0 1 1 1 (Guido Ubaldus thought that this proved the existence of God because “something has been created out of nothing.”) 58. Suppose that n1 series. Prove that a n a n 0 is known to be a convergent n 1 1 a n is a divergent series. 59. Prove part (i) of Theorem 8. 60. If 61. If a n is divergent and c 0, show that ca n is divergent. a n is convergent and bn is divergent, show that the series a n bn is divergent. [Hint: Argue by contradiction.] 62. If a n and divergent? bn are both divergent, is an bn necessarily 63. Suppose that a series a n has positive terms and its partial sums sn satisfy the inequality sn 1000 for all n. Explain why a n must be convergent. 64. The Fibonacci sequence was defined in Section 12.1 by the equations f1 1, f2 1, fn fn 1 fn 2 n 3 Show that each of the following statements is true. 1 1 1 (a) fn 1 fn 1 fn 1 fn fn fn 1 n2 1 fn 1 fn n2 fn (b) (c) fn f 1 1 2 1n1 65. The Cantor set, named after the German mathematician Georg Cantor (1845–1918), is constructed as follows. We start with the closed interval [0, 1] and remove the open interval ( 1 , 2 ). 33 That leaves the two intervals [0, 1 ] and [ 2, 1] and we remove 3 3 the open middle third of each. Four intervals remain and again we remove the open middle third of each of them. We continue this procedure indefinitely, at each step removing the open middle third of every interval that remains from the preceding step. The Cantor set consists of the numbers that remain in [0, 1] after all those intervals have been removed. (a) Show that the total length of all the intervals that are removed is 1. Despite that, the Cantor set contains infi- 66. (a) A sequence a n is defined recursively by the equation a n 1 a n 1 a n 2 for n 3, where a 1 and a 2 can be any 2 real numbers. Experiment with various values of a 1 and a 2 and use your calculator to guess the limit of the sequence. (b) Find lim n l a n in terms of a 1 and a 2 by expressing a n 1 a n in terms of a 2 a 1 and summing a series. 67. Consider the series n n1 n 1! (a) Find the partial sums s1, s2, s3, and s4. Do you recognize the denominators? Use the pattern to guess a formula for sn . (b) Use mathematical induction to prove your guess. (c) Show that the given infinite series is convergent, and find its sum. 68. In the figure there are infinitely many circles approaching the vertices of an equilateral triangle, each circle touching other circles and sides of the triangle. If the triangle has sides of length 1, find the total area occupied by the circles. 5E-12(pp 756-765) 1/18/06 10:10 AM Page 759 S ECTION 12.3 THE INTEGRAL TEST AND ESTIMATES OF SUMS |||| 12.3 ❙❙❙❙ 759 The Integral Test and Estimates of Sums In general, it is difficult to find the exact sum of a series. We were able to accomplish this for geometric series and the series 1 n n 1 because in each of those cases we could find a simple formula for the nth partial sum sn . But usually it isn’t easy to compute lim n l sn. Therefore, in the next few sections we develop several tests that enable us to determine whether a series is convergent or divergent without explicitly finding its sum. (In some cases, however, our methods will enable us to find good estimates of the sum.) Our first test involves improper integrals. We begin by investigating the series whose terms are the reciprocals of the squares of the positive integers: n n sn i1 5 10 50 100 500 1000 5000 1 i2 1.4636 1.5498 1.6251 1.6350 1.6429 1.6439 1.6447 n1 1 n2 1 12 1 22 1 32 1 42 1 52 There’s no simple formula for the sum sn of the first n terms, but the computer-generated table of values given in the margin suggests that the partial sums are approaching a number near 1.64 as n l and so it looks as if the series is convergent. We can confirm this impression with a geometric argument. Figure 1 shows the curve y 1 x 2 and rectangles that lie below the curve. The base of each rectangle is an interval of length 1; the height is equal to the value of the function y 1 x 2 at the right endpoint of the interval. So the sum of the areas of the rectangles is 1 12 1 22 y 1 32 1 42 1 52 n1 1 n2 y= 1 ≈ area= 1 1@ 0 1 2 area= 1 2@ FIGURE 1 3 area= 1 3@ 4 area= 1 4@ 5 x area= 1 5@ If we exclude the first rectangle, the total area of the remaining rectangles is smaller than the area under the curve y 1 x 2 for x 1, which is the value of the integral x1 1 x 2 dx. In Section 8.8 we discovered that this improper integral is convergent and has value 1. So the picture shows that all the partial sums are less than 1 12 y 1 1 dx x2 2 Thus, the partial sums are bounded. We also know that the partial sums are increasing (because all the terms are positive). Therefore, the partial sums converge (by the Monotonic Sequence Theorem) and so the series is convergent. The sum of the series (the limit of the partial sums) is also less than 2: n1 1 n2 1 12 1 22 1 32 1 42 2 5E-12(pp 756-765) 760 ❙❙❙❙ 1/18/06 10:10 AM Page 760 CHAPTER 12 INFINITE SEQUENCES AND SERIES n n sn i1 5 10 50 100 500 1000 5000 1 si 3.2317 5.0210 12.7524 18.5896 43.2834 61.8010 139.9681 [The exact sum of this series was found by the Swiss mathematician Leonhard Euler (1707–1783) to be 2 6, but the proof of this fact is quite difficult. (See Problem 6 in the Problems Plus following Chapter 16.)] Now let’s look at the series n1 1 sn 1 s1 1 s2 1 s3 1 s4 1 s5 The table of values of sn suggests that the partial sums aren’t approaching a finite number, so we suspect that the given series may be divergent. Again we use a picture for confirmation. Figure 2 shows the curve y 1 sx, but this time we use rectangles whose tops lie above the curve. y y= 1 x œ„ 0 1 2 area= 1 1 œ„ FIGURE 2 3 area= 1 œ2 „ 4 area= 1 œ3 „ 5 x area= 1 œ4 „ The base of each rectangle is an interval of length 1. The height is equal to the value of the function y 1 sx at the left endpoint of the interval. So the sum of the areas of all the rectangles is 1 1 1 1 1 1 s1 s2 s3 s4 s5 n 1 sn This total area is greater than the area under the curve y 1 sx for x 1, which is equal to the integral x1 (1 sx ) dx. But we know from Section 8.8 that this improper integral is divergent. In other words, the area under the curve is infinite. So the sum of the series must be infinite; that is, the series is divergent. The same sort of geometric reasoning that we used for these two series can be used to prove the following test. (The proof is given at the end of this section.) The Integral Test Suppose f is a continuous, positive, decreasing function on 1, and let a n f n . Then the series n 1 a n is convergent if and only if the improper integral x1 f x d x is convergent. In other words: (i) If y f x d x is convergent, then 1 a n is convergent. n1 (ii) If y f x d x is divergent, then 1 NOTE ■ gral at n a n is divergent. n1 When we use the Integral Test it is not necessary to start the series or the inte1. For instance, in testing the series 1 n4 n 3 2 we use y 4 1 x 3 2 dx 5E-12(pp 756-765) 1/18/06 10:10 AM Page 761 S ECTION 12.3 THE INTEGRAL TEST AND ESTIMATES OF SUMS ❙❙❙❙ 761 Also, it is not necessary that f be always decreasing. What is important is that f be ultimately decreasing, that is, decreasing for x larger than some number N. Then n N a n is convergent, so n 1 a n is convergent by Note 4 of Section 12.2. 1 EXAMPLE 1 Test the series n1 n2 1 1 x2 so we use the Integral Test: SOLUTION The function f x 1, 1 y x2 1 1 dx for convergence or divergence. 1 is continuous, positive, and decreasing on lim y tl 1 t 1 x2 1 lim tan 1t tl 4 tl lim tan 1x dx 2 t 1 4 4 Thus, x1 1 x 2 1 d x is a convergent integral and so, by the Integral Test, the series 1 n 2 1 is convergent. 1 convergent? np EXAMPLE 2 For what values of p is the series n1 p 0, then lim n l 1 n . If p 0, then lim n l 1 n p 1. In either p case lim n l 1 n 0, so the given series diverges by the Test for Divergence (12.2.7). If p 0, then the function f x 1 x p is clearly continuous, positive, and decreasing on 1, . We found in Chapter 8 [see (8.8.2)] that SOLUTION If p y 1 1 dx converges if p xp 1 and diverges if p 1 It follows from the Integral Test that the series 1 n p converges if p 1 and diverges if 0 p 1. (For p 1, this series is the harmonic series discussed in Example 7 in Section 12.2.) The series in Example 2 is called the p-series. It is important in the rest of this chapter, so we summarize the results of Example 2 for future reference as follows. 1 The p-series n1 1 is convergent if p np 1 and divergent if p EXAMPLE 3 (a) The series n1 1 n3 1 13 1 23 1 33 is convergent because it is a p-series with p (b) The series n1 1 n1 3 n1 1 3 sn 1 is divergent because it is a p-series with p 3 1. 1 3 s2 1 3 1 43 1 3 s3 1. 1 3 s4 1. 5E-12(pp 756-765) 762 ❙❙❙❙ 1/18/06 10:10 AM Page 762 CHAPTER 12 INFINITE SEQUENCES AND SERIES NOTE We should not infer from the Integral Test that the sum of the series is equal to the value of the integral. In fact, ■ n1 2 1 n2 6 1 dx x2 y whereas 1 1 Therefore, in general, y an 1 n1 f x dx ln n converges or diverges. n EXAMPLE 4 Determine whether the series n1 SOLUTION The function f x ln x x is positive and continuous for x 1 because the logarithm function is continuous. But it is not obvious whether or not f is decreasing, so we compute its derivative: 1 x x ln x x2 fx Thus, f x 0 when ln x 1, that is, x and so we can apply the Integral Test: y 1 ln x dx x x ln x dx x t 1 ln t lim tl 2 tl 0 x y=ƒ Rn Rn FIGURE 4 ln n n is also divergent by the s sn an 1 an an 2 3 The remainder Rn is the error made when sn, the sum of the first n terms, is used as an approximation to the total sum. We use the same notation and ideas as in the Integral Test, assuming that f is decreasing on n, . Comparing the areas of the rectangles with the area under y f x for x n in Figure 3, we see that y n+1 1 2 ... FIGURE 3 0 t Suppose we have been able to use the Integral Test to show that a series a n is convergent and we now want to find an approximation to the sum s of the series. Of course, any partial sum sn is an approximation to s because lim n l sn s. But how good is such an approximation? To find out, we need to estimate the size of the remainder n an+1 an+2 2 e Estimating the Sum of a Series y=ƒ an+1 an+2 ln x 2 lim Since this improper integral is divergent, the series Integral Test. y ln x 2 e. It follows that f is decreasing when x lim y tl 1 ... an 1 an 2 y f x dx y f x dx n Similarly, we see from Figure 4 that x Rn an 1 an 2 n1 5E-12(pp 756-765) 1/18/06 10:10 AM Page 763 SECTION 12.3 THE INTEGRAL TEST AND ESTIMATES OF SUMS ❙❙❙❙ 763 So we have proved the following error estimate. 2 Remainder Estimate for the Integral Test Suppose f k ous, positive, decreasing function for x then y f x dx n1 a k, where f is a continua n is convergent. If Rn s sn , n and y Rn f x dx n EXAMPLE 5 (a) Approximate the sum of the series 1 n 3 by using the sum of the first 10 terms. Estimate the error involved in this approximation. (b) How many terms are required to ensure that the sum is accurate to within 0.0005? SOLUTION In both parts (a) and (b) we need to know xn f x d x. With f x 1 x 3, we have y n 1 dx x3 t 1 2x 2 lim tl 1 2t 2 lim tl n 1 2n 2 1 2n 2 (a) n1 1 n3 1 13 s10 1 23 1 33 1 10 3 1.1975 According to the remainder estimate in (2), we have y R10 10 1 dx x3 1 2 10 1 200 2 So the size of the error is at most 0.005. (b) Accuracy to within 0.0005 means that we have to find a value of n such that Rn 0.0005. Since Rn y n 1 2n 2 we want 1 dx x3 1 2n 2 0.0005 Solving this inequality, we get n2 1 0.001 1000 or n s1000 We need 32 terms to ensure accuracy to within 0.0005. If we add sn to each side of the inequalities in (2), we get 3 sn y n1 f x dx s sn y n f x dx 31.6 5E-12(pp 756-765) 764 ❙❙❙❙ 1/18/06 10:10 AM Page 764 CHAPTER 12 INFINITE SEQUENCES AND SERIES because sn Rn s. The inequalities in (3) give a lower bound and an upper bound for s. They provide a more accurate approximation to the sum of the series than the partial sum sn does. EXAMPLE 6 Use (3) with n 10 to estimate the sum of the series n1 1 . n3 SOLUTION The inequalities in (3) become y s10 11 1 dx x3 s y s10 10 1 dx x3 From Example 5 we know that y n so 1 2 11 s10 1 dx x3 1 2n 2 1 2 10 s s10 1.201664 Using s10 2 s 1.202532 2 1.197532, we get If we approximate s by the midpoint of this interval, then the error is at most half the length of the interval. So n1 y 1 n3 1.2021 with error 0.0005 If we compare Example 6 with Example 5, we see that the improved estimate in (3) can be much better than the estimate s sn . To make the error smaller than 0.0005 we had to use 32 terms in Example 5 but only 10 terms in Example 6. y=ƒ Proof of the Integral Test a™ a£ a¢ a∞ 0 1 2 3 4 an 5 ... nx FIGURE 5 y We have already seen the basic idea behind the proof of the Integral Test in Figures 1 and 2 for the series 1 n 2 and 1 sn. For the general series a n look at Figures 5 and 6. The area of the first shaded rectangle in Figure 5 is the value of f at the right endpoint of 1, 2 , that is, f 2 a 2 . So, comparing the areas of the shaded rectangles with the area under y f x from 1 to n, we see that y=ƒ 4 an-1 a¡ a™ a£ a¢ 0 1 FIGURE 6 2 3 4 5 ... a2 a3 an y n 1 f x dx (Notice that this inequality depends on the fact that f is decreasing.) Likewise, Figure 6 shows that nx 5 y n 1 f x dx a1 a2 an 1 5E-12(pp 756-765) 1/18/06 10:10 AM Page 765 S ECTION 12.3 THE INTEGRAL TEST AND ESTIMATES OF SUMS ❙❙❙❙ 765 (i) If y f x d x is convergent, then (4) gives 1 n y ai y f x dx 1 i2 since f x n 1 f x dx 0. Therefore n sn a1 ai y a1 Since sn f x dx 1 i2 M, say M for all n, the sequence sn is bounded above. Also sn sn 1 an 1 sn since a n 1 f n 1 0. Thus, sn is an increasing bounded sequence and so it is convergent by the Monotonic Sequence Theorem (12.1.11). This means that a n is convergent. (ii) If x1 f x d x is divergent, then x1n f x d x l as n l because f x 0. But (5) gives y n 1 and so sn |||| 12.3 1 n 1.3 1 1 8 12. 1 1 dx x 1.3 y 1 2 s2 6 1 5 n1 6 f x dx ai ai i1 n1 1 n4 4. n1 n 7. n1 |||| 9. n1 n1 n ■ ■ ■ ■ 3n n1 ne n n 8. ■ ■ ■ n2 n 10. n n1 1.4 3n 1 4 s4 1 5 s5 n3 n1 ■ 25. n2 2 1 1 4n 2 n1 ln n n2 n1 n4 n3 1 n ln n ln ln n 24. n ■ ■ ■ ■ 5 n 22. ■ |||| n 20. 1 25–28 3n nn 18. 2 n1 n3 ■ n 16. n2 n1 ■ 5 14. 1 n ln n 23. 1.2 1 3 s3 1 n2 ■ 1 125 4 2 ne 21. 2 1 ■ 1 64 n1 1 Determine whether the series is convergent or divergent. 2 n 0.85 19. 1 5. n1 ■ 9–24 1 4 sn 1 27 n 17. i2 Use the Integral Test to determine whether the series is convergent or divergent. ■ a n diverges. 1 15. |||| e 1 2 sn n3 n1 x 1 and an f n . By drawing a picture, rank the following three quantities in increasing order: y 5 13. 2. Suppose f is a continuous positive decreasing function for 6. and so 11. 1 What can you conclude about the series? n1 sn Exercises n2 3. ai i1 l . This implies that sn l 1 1. Draw a picture to show that 3–8 n1 f x dx ■ 1 ■ ■ ■ Find the values of p for which the series is convergent. 1 n ln n p 26. n3 1 n ln n ln ln n p ■ 5E-12(pp 766-775) 766 ❙❙❙❙ 27. 1/18/06 10:54 AM Page 766 CHAPTER 12 INFINITE SEQUENCES AND SERIES n2 n1 p 28. n1 ■ ■ n1 ■ ■ ■ ■ 29. The Riemann zeta-function ■ ln n np ■ CAS ■ ■ ■ ■ is defined by x n1 1 nx and is used in number theory to study the distribution of prime numbers. What is the domain of ? 1 n 4. Estimate the error in using s10 as an approximation to the sum of the series. (b) Use (3) with n 10 to give an improved estimate of the sum. (c) Find a value of n so that sn is within 0.00001 of the sum. 30. (a) Find the partial sum s10 of the series n1 ln n 2 n 2 is convergent. (b) Find an upper bound for the error in the approximation s sn . (c) What is the smallest value of n such that this upper bound is less than 0.05? (d) Find sn for this value of n. 36. (a) Show that the series 37. (a) Use (4) to show that if sn is the nth partial sum of the har- monic series, then sn n1 1 n 5 correct to three decimal 1 2 1 tn n1 n 32 to within 0.01. 34. How many terms of the series n 2 1 n ln n to add to find its sum to within 0.01? 2 would you need 35. Show that if we want to approximate the sum of the series n 1.001 so that the error is less than 5 in the ninth decimal place, then we need to add more than 10 11,301 terms! n1 |||| 12.4 1 3 1 n ln n has a limit. (The value of the limit is denoted by and is called Euler’s constant.) (a) Draw a picture like Figure 6 with f x 1 x and interpret tn as an area [or use (5)] to show that tn 0 for all n. (b) Interpret places. 33. Estimate ln n 38. Use the following steps to show that the sequence 31. (a) Use the sum of the first 10 terms to estimate the sum of the 32. Find the sum of the series 1 (b) The harmonic series diverges, but very slowly. Use part (a) to show that the sum of the first million terms is less than 15 and the sum of the first billion terms is less than 22. tn series n 1 1 n 2. How good is this estimate? (b) Improve this estimate using (3) with n 10. (c) Find a value of n that will ensure that the error in the approximation s sn is less than 0.001. n1 tn ln n 1 1 ln n 1 n 1 as a difference of areas to show that tn tn 1 0. Therefore, tn is a decreasing sequence. (c) Use the Monotonic Sequence Theorem to show that tn is convergent. 39. Find all positive values of b for which the series n1 b ln n converges. The Comparison Tests In the comparison tests the idea is to compare a given series with a series that is known to be convergent or divergent. For instance, the series 1 1 n1 2 n 1 reminds us of the series n 1 1 2 n, which is a geometric series with a 1 and r 1 and 2 2 is therefore convergent. Because the series (1) is so similar to a convergent series, we have the feeling that it too must be convergent. Indeed, it is. The inequality 1 2n 1 1 2n shows that our given series (1) has smaller terms than those of the geometric series and therefore all its partial sums are also smaller than 1 (the sum of the geometric series). This means that its partial sums form a bounded increasing sequence, which is convergent. It 5E-12(pp 766-775) 1/18/06 10:54 AM Page 767 S ECTION 12.4 THE COMPARISON TESTS ❙❙❙❙ 767 also follows that the sum of the series is less than the sum of the geometric series: 1 2 n1 n 1 1 Similar reasoning can be used to prove the following test, which applies only to series whose terms are positive. The first part says that if we have a series whose terms are smaller than those of a known convergent series, then our series is also convergent. The second part says that if we start with a series whose terms are larger than those of a known divergent series, then it too is divergent. The Comparison Test Suppose that (i) If (ii) If |||| It is important to keep in mind the distinction between a sequence and a series. A sequence is a list of numbers, whereas a series is a sum. With every series a n there are associated two sequences: the sequence a n of terms and the sequence sn of partial sums. bn is convergent and a n bn is divergent and a n a n and bn are series with positive terms. bn for all n, then a n is also convergent. bn for all n, then a n is also divergent. Proof (i) Let n n sn ai tn i1 bi t bn i1 n1 Since both series have positive terms, the sequences sn and tn are increasing sn 1 sn a n 1 sn . Also tn l t, so tn t for all n. Since a i bi , we have sn tn . Thus, sn t for all n. This means that sn is increasing and bounded above and therefore converges by the Monotonic Sequence Theorem. Thus, a n converges. (ii) If bn is divergent, then tn l (since tn is increasing). But a i bi so sn tn . Thus, sn l . Therefore, a n diverges. Standard Series for Use with the Comparison Test In using the Comparison Test we must, of course, have some known series bn for the purpose of comparison. Most of the time we use either a p-series [ 1 n p converges if p 1 and diverges if p 1; see (12.3.1)] or a geometric series [ ar n 1 converges if r 1 and diverges if r 1; see (12.2.4)]. EXAMPLE 1 Determine whether the series n1 5 4n 2n 2 3 converges or diverges. SOLUTION For large n the dominant term in the denominator is 2 n 2 so we compare the given series with the series 5 2 n 2 . Observe that 2n 2 5 4n 5 2n 2 3 because the left side has a bigger denominator. (In the notation of the Comparison Test, a n is the left side and bn is the right side.) We know that n1 5 2n 2 5 2 n1 1 n2 is convergent because it’s a constant times a p-series with p n1 2n 2 5 4n is convergent by part (i) of the Comparison Test. 3 2 1. Therefore 5E-12(pp 766-775) 768 ❙❙❙❙ 1/18/06 10:54 AM Page 768 CHAPTER 12 INFINITE SEQUENCES AND SERIES bn or a n bn in the Comparison Test is given for N OTE 1 Although the condition a n all n, we need verify only that it holds for n N, where N is some fixed integer, because the convergence of a series is not affected by a finite number of terms. This is illustrated in the next example. ■ EXAMPLE 2 Test the series n1 ln n for convergence or divergence. n SOLUTION This series was tested (using the Integral Test) in Example 4 in Section 12.3, but it is also possible to test it by comparing it with the harmonic series. Observe that ln n 1 for n 3 and so ln n n 1 n n 3 We know that 1 n is divergent ( p-series with p gent by the Comparison Test. 1). Thus, the given series is diver- NOTE 2 The terms of the series being tested must be smaller than those of a convergent series or larger than those of a divergent series. If the terms are larger than the terms of a convergent series or smaller than those of a divergent series, then the Comparison Test doesn’t apply. Consider, for instance, the series ■ 1 n1 2 n 1 The inequality 1 2 n 1 1 2n n ( 1 ) is convergent is useless as far as the Comparison Test is concerned because bn 2 n 1 ought to be convergent and a n bn. Nonetheless, we have the feeling that 1 2 1n because it is very similar to the convergent geometric series ( 2 ) . In such cases the following test can be used. The Limit Comparison Test Suppose that |||| Exercises 40 and 41 deal with the cases c 0 and c . a n and lim nl where c is a finite number and c diverge. an bn bn are series with positive terms. If c 0, then either both series converge or both Proof Let m and M be positive numbers such that m c for large n, there is an integer N such that c M . Because a n bn is close to m and so an bn M when n N m bn an Mbn when n N 5E-12(pp 766-775) 1/18/06 10:54 AM Page 769 S ECTION 12.4 THE COMPARISON TESTS ❙❙❙❙ 769 If bn converges, so does Mbn . Thus, a n converges by part (i) of the Comparison Test. If bn diverges, so does mbn and part (ii) of the Comparison Test shows that a n diverges. 1 EXAMPLE 3 Test the series n1 2n for convergence or divergence. 1 SOLUTION We use the Limit Comparison Test with 1 an 2n 1 2n bn 1 and obtain lim nl an bn lim nl 1 2n 1 1 2n lim nl 2n 2 n lim 1 nl 1 1 1 2n 1 0 Since this limit exists and 1 2 n is a convergent geometric series, the given series converges by the Limit Comparison Test. 2n 2 s5 EXAMPLE 4 Determine whether the series n1 3n converges or diverges. n5 SOLUTION The dominant part of the numerator is 2 n 2 and the dominant part of the denom- inator is sn5 n 5 2. This suggests taking 2n 2 s5 an bn nl an bn lim nl 2n 2 s5 n1 2 2 3n n5 2 lim 2n 2 n5 2 lim 3n n5 3 n lim nl 2 5 n5 nl 2 n1 2 2n 5 2 3n 3 2 2 s5 n 5 20 2 s0 1 1 Since bn 2 1 n 1 2 is divergent ( p-series with p by the Limit Comparison Test. 1 2 1 1), the given series diverges Notice that in testing many series we find a suitable comparison series only the highest powers in the numerator and denominator. bn by keeping Estimating Sums If we have used the Comparison Test to show that a series a n converges by comparison with a series bn, then we may be able to estimate the sum a n by comparing remainders. As in Section 12.3, we consider the remainder Rn For the comparison series s sn an 1 an 2 bn we consider the corresponding remainder Tn t tn bn 1 bn 2 5E-12(pp 766-775) ❙❙❙❙ 770 1/18/06 10:54 AM Page 770 CHAPTER 12 INFINITE SEQUENCES AND SERIES Since a n bn for all n, we have Rn Tn . If bn is a p-series, we can estimate its remainder Tn as in Section 12.3. If bn is a geometric series, then Tn is the sum of a geometric series and we can sum it exactly (see Exercises 35 and 36). In either case we know that Rn is smaller than Tn . EXAMPLE 5 Use the sum of the first 100 terms to approximate the sum of the series 1 n3 1 . Estimate the error involved in this approximation. SOLUTION Since 1 n 3 1 n3 1 the given series is convergent by the Comparison Test. The remainder Tn for the comparison series 1 n 3 was estimated in Example 5 in Section 12.3 using the Remainder Estimate for the Integral Test. There we found that y Tn n 1 dx x3 1 2n 2 Therefore, the remainder Rn for the given series satisfies Rn With n 1 2n 2 Tn 100 we have 1 2 100 R100 2 0.00005 Using a programmable calculator or a computer, we find that 100 1 n1 n 3 1 n1 1 n3 1 0.6864538 with error less than 0.00005. |||| 12.4 Exercises 1. Suppose a n and bn are series with positive terms and bn is known to be convergent. (a) If a n bn for all n, what can you say about a n? Why? (b) If a n bn for all n, what can you say about a n? Why? n 7. 1 n n1 a n and bn are series with positive terms and bn is known to be divergent. (a) If a n bn for all n, what can you say about a n ? Why? (b) If a n bn for all n, what can you say about a n ? Why? cos n n2 1 10. 12. n2 n2 n3 n1 n1 n4n 11. 13. 3–32 |||| 3. Determine whether the series converges or diverges. n1 n n1 2 1 n 2 5 5. 3n 1 2 4. n1 n n2 n 3 4 1 6. 2 15. sn n1 17. n1 n1 1 1 1 1 n 3n 4 1 sin n 10 n n0 sn 14. n2 1 n sn 1 sn 2 n 2 n1 2. Suppose 2 n1 2 9. 3n 4 8. 2 1 n n1 1 sn 3 n1 2n n 16. 1 1 1 18. 3 5E-12(pp 766-775) 1/18/06 10:54 AM Page 771 S ECTION 12.5 ALTERNATING SERIES 2n 19. n1 1 3 20. n n1 1 21. n1 1 sn n1 5 1 n1 1 s1 23. 25. 27. n n2 n2 n6 31. 26. 38. For what values of p does the series 39. Prove that if a n 5n n1 n n 3 sn 7 2 n 2 7n 3 n 2 5n 1 n! nn 28. 1 n! 30. 1 n n1 ■ 32. n1 ■ ■ ■ ■ ■ n lim n n1 e 1 nl 5 n2 lim 1 n1 1 n ■ ■ |||| Use the sum of the first 10 terms to approximate the sum of the series. Estimate the error. 33. n1 34. n2 n1 1 35. n1 ■ n4 ■ 1 2 n1 ■ ■ ■ 42. Give an example of a pair of series a n and bn with positive terms where lim n l a n bn 0 and bn diverges, but a n converges. [Compare with Exercise 40.] cos n n5 43. Show that if a n n 36. n ■ 1 ■ ■ n 13 ■ an bn then an is also divergent. (b) Use part (a) to show that the series diverges. 1 ln n (ii) (i) n n 2 ln n n1 33–36 1 an and bn are series with positive terms bn is divergent. Prove that if nl ■ 0 41. (a) Suppose that and ■ an bn then an is also convergent. (b) Use part (a) to show that the series converges. ln n ln n (ii) (i) n3 sn e n n1 n1 n ■ 2 a n also a n converges, then an and bn are series with positive terms bn is convergent. Prove that if and 3 1 n p ln n converge? 40. (a) Suppose that n1 2 0 and n2 771 converges. 2 13 n1 1 n sin ■ n n n2 24. n1 n1 n3 2n n2 2 1 29. 22. 2n 3n 1 1 ❙❙❙❙ 0 and lim n l n a n 0, then a n is divergent. n ■ ■ 44. Show that if a n ■ 0 and a n is convergent, then ln 1 an is convergent. 37. The meaning of the decimal representation of a number 45. If 0.d1 d2 d3 . . . (where the digit d i is one of the numbers 0, 1, 2, . . . , 9) is that d2 d1 d3 d4 0.d1 d2 d3 d4 . . . 2 3 10 10 10 10 4 a n is a convergent series with positive terms, is it true that sin a n is also convergent? 46. If a n and bn are both convergent series with positive terms, is it true that a n bn is also convergent? Show that this series always converges. |||| 12.5 Alternating Series The convergence tests that we have looked at so far apply only to series with positive terms. In this section and the next we learn how to deal with series whose terms are not necessarily positive. Of particular importance are alternating series, whose terms alternate in sign. An alternating series is a series whose terms are alternately positive and negative. Here are two examples: 1 1 2 1 2 1 3 1 4 1 5 1 6 n1 2 3 3 4 4 5 5 6 6 7 n1 1n n 1 1 n n n 1 5E-12(pp 766-775) 772 ❙❙❙❙ 1/18/06 10:54 AM Page 772 CHAPTER 12 INFINITE SEQUENCES AND SERIES We see from these examples that the n th term of an alternating series is of the form an 1 n1 bn or 1 nbn an where bn is a positive number. (In fact, bn a n .) The following test says that if the terms of an alternating series decrease toward 0 in absolute value, then the series converges. The Alternating Series Test If the alternating series 1 n1 bn b1 b2 b3 b4 b5 b6 bn 0 n1 satisfies (i) bn 1 bn (ii) lim bn 0 for all n nl then the series is convergent. Before giving the proof let’s look at Figure 1, which gives a picture of the idea behind the proof. We first plot s1 b1 on a number line. To find s2 we subtract b2 , so s2 is to the left of s1 . Then to find s3 we add b3 , so s3 is to the right of s2 . But, since b3 b2 , s3 is to the left of s1 . Continuing in this manner, we see that the partial sums oscillate back and forth. Since bn l 0, the successive steps are becoming smaller and smaller. The even partial sums s2 , s4 , s6 , . . . are increasing and the odd partial sums s1 , s3 , s5 , . . . are decreasing. Thus, it seems plausible that both are converging to some number s, which is the sum of the series. Therefore, in the following proof we consider the even and odd partial sums separately. b¡ -b™ +b£ -b¢ +b∞ -bß FIGURE 1 s™ 0 s¢ sß s∞ s s£ s¡ Proof of the Alternating Series Test We first consider the even partial sums: s2 s2 n b2 0 s4 In general b1 s2 b3 b4 s2 1 b2 n s2 n s2 n 2 0 Thus b2 n s2 s4 since b2 s6 since b4 2 b1 b3 b2 n since b2 n 1 s2 n But we can also write s2 n b1 b2 b3 b4 b5 b2 n 2 b2 n 1 b2 n 5E-12(pp 766-775) 1/18/06 10:55 AM Page 773 S ECTION 12.5 ALTERNATING SERIES ❙❙❙❙ 773 Every term in brackets is positive, so s2 n b1 for all n. Therefore, the sequence s2 n of even partial sums is increasing and bounded above. It is therefore convergent by the Monotonic Sequence Theorem. Let’s call its limit s, that is, lim s2 n s nl Now we compute the limit of the odd partial sums: lim s2 n nl lim s2 n 1 b2 n nl lim s2 n lim b2 n nl s 1 nl 1 0 [by condition (ii)] s Since both the even and odd partial sums converge to s, we have lim n l sn Exercise 72 in Section 12.1) and so the series is convergent. |||| Figure 2 illustrates Example 1 by showing 1 n 1 n and the graphs of the terms a n the partial sums sn . Notice how the values of sn zigzag across the limiting value, which appears to be about 0.7. In fact, the exact sum of the series is ln 2 0.693 (see Exercise 36). 1 2 1 an 1 1 4 bn 1n n n1 satisfies EXAMPLE 2 The series 1 because n1 1 1 n 1 n 3n is alternating but 4n 1 n lim bn nl FIGURE 2 1 3 n1 1 (ii) lim bn lim 0 nl nl n so the series is convergent by the Alternating Series Test. sn 0 EXAMPLE 1 The alternating harmonic series (i) bn 1 s (see lim nl 3n 4n 1 3 lim nl 4 1 n 3 4 so condition (ii) is not satisfied. Instead, we look at the limit of the n th term of the series: lim a n lim nl nl 1 n 3n 4n 1 This limit does not exist, so the series diverges by the Test for Divergence. EXAMPLE 3 Test the series 1 n1 n2 n1 n3 1 for convergence or divergence. SOLUTION The given series is alternating so we try to verify conditions (i) and (ii) of the Alternating Series Test. Unlike the situation in Example 1, it is not obvious that the sequence given by bn n 2 n 3 1 is decreasing. However, if we consider the related function 5E-12(pp 766-775) 774 ❙❙❙❙ 1/18/06 10:55 AM Page 774 CHAPTER 12 INFINITE SEQUENCES AND SERIES x2 x3 fx 1 , we find that |||| Instead of verifying condition (i) of the Alternating Series Test by computing a derivative, we could verify that bn 1 bn directly by using the technique of Solution 1 of Example 11 in Section 12.1. x3 12 x2 x3 fx Since we are considering only positive x, we see that f x 0 if 2 x 3 0, that is, 3 3 x s2. Thus, f is decreasing on the interval (s2, ). This means that f n 1 fn and therefore bn 1 bn when n 2. (The inequality b2 b1 can be verified directly but all that really matters is that the sequence bn is eventually decreasing.) Condition (ii) is readily verified: lim bn n lim nl n3 nl 1 n 2 lim 1 nl 1 n3 1 0 Thus, the given series is convergent by the Alternating Series Test. Estimating Sums A partial sum sn of any convergent series can be used as an approximation to the total sum s, but this is not of much use unless we can estimate the accuracy of the approximation. The error involved in using s sn is the remainder Rn s sn . The next theorem says that for series that satisfy the conditions of the Alternating Series Test, the size of the error is smaller than bn 1 , which is the absolute value of the first neglected term. Alternating Series Estimation Theorem If s |||| You can see geometrically why the Alternating Series Estimation Theorem is true by looking at Figure 1 (on page 772). Notice that b6, and so on. Notice s s4 b 5, s s5 also that s lies between any two consecutive partial sums. 1 n1 bn is the sum of an alternating series that satisfies (i) 0 bn then and bn 1 Rn s (ii) lim bn nl sn bn 0 1 Proof We know from the proof of the Alternating Series Test that s lies between any two consecutive partial sums sn and sn 1 . It follows that s sn sn EXAMPLE 4 Find the sum of the series (By definition, 0! n0 1.) sn 1 1 n! bn 1 n correct to three decimal places. SOLUTION We first observe that the series is convergent by the Alternating Series Test because (i) (ii) 1 1 n 0 1! 1 n! n! n 1 1 n! 1 1 l 0 so l 0 as n l n n! 5E-12(pp 766-775) 1/18/06 10:55 AM Page 775 S ECTION 12.5 ALTERNATING SERIES ❙❙❙❙ 775 To get a feel for how many terms we need to use in our approximation, let’s write out the first few terms of the series: s 1 0! 1 1 1! 1 2! 1 2 1 1 3! 1 6 1 24 s6 and 1 1 2 1 1 120 1 5040 b7 Notice that 1 4! 1 5! 1 720 1 5000 1 6 1 24 1 6! 1 7! 1 5040 0.0002 1 120 1 720 0.368056 By the Alternating Series Estimation Theorem we know that s s6 b7 0.0002 This error of less than 0.0002 does not affect the third decimal place, so we have |||| In Section 12.10 we will prove that n ex n 0 x n! for all x, so what we have obtained in Example 4 is actually an approximation to the number e 1. s 0.368 correct to three decimal places. | |||| 12.5 NOTE The rule that the error (in using sn to approximate s) is smaller than the first neglected term is, in general, valid only for alternating series that satisfy the conditions of the Alternating Series Estimation Theorem. The rule does not apply to other types of series. ■ Exercises 1. (a) What is an alternating series? (b) Under what conditions does an alternating series converge? (c) If these conditions are satisfied, what can you say about the remainder after n terms? |||| 2. 1 3 4. n1 cos n n3 4 16. n1 1 n sin 17. n1 2–20 3. 15. 4 7 Test the series for convergence or divergence. 2 4 3 5 4 8 4 9 1 ln 2 5. n1 7. 1n sn n 1 n1 11. 1 n2 1 ln 4 ■ 1 ln 5 1 ln 6 1 6. n1 1 1 8. 3n 1 1 n2 n 4n 2 n n n 1 n ln n 4 12. n1 1 n1 14. 1 n1 21. 2n 1 n1 3 ■ nn n! ■ 20. n1 ■ ■ ■ ■ ■ n 5 n ■ ■ ■ ■ |||| Calculate the first 10 partial sums of the series and graph both the sequence of terms and the sequence of partial sums on the same screen. Estimate the error in using the 10th partial sum to approximate the total sum. n1 n1 10. n1 n n n1 ; 21–22 1 3n 2n 1 n1 4 11 1 n1 13. 4 10 n1 1 4n 2 1 19. 5 7 1 ln 3 n1 9. 4 6 1 n cos 18. n sin n 2 n! n1 1 sn 2 sn n1 ■ ■ 1 n3 n1 22. 2 n1 ■ ■ ■ ■ 23–26 ■ ■ 1n n3 ■ 1 ■ ■ ■ |||| How many terms of the series do we need to add in order to find the sum to the indicated accuracy? e1 n n 23. ln n n 24. 1 n1 1n n2 1 n1 1n n4 ( error 0.01) ( error 0.001) 5E-12(pp 776-785) 776 ❙❙❙❙ 1/18/06 10:58 AM Page 776 CHAPTER 12 INFINITE SEQUENCES AND SERIES 2 n! 25. n1 n ( 0.01) error 34. 1 n2 ■ 1 nn 4n 26. n1 ■ ■ 27–30 ( ■ |||| ■ error ■ 0.002) ■ n1 ■ ■ ■ 1 ■ ■ ■ n1 1 n 10 n 29. n1 ■ ■ ■ 2 30. n1 ■ ■ ■ ■ ■ ■ ■ ■ ■ hn 1 n 1 n an overestimate or an underestimate of the total sum? Explain. 32. n1 1 |||| 12.6 ■ 1 33. n1 n ■ ■ ■ ■ ■ n1 1n n 1 ln 2 as n l ln n l and therefore h2 n For what values of p is each series convergent? 1n np ■ Let hn and sn be the partial sums of the harmonic and alternating harmonic series. (a) Show that s2 n h2 n hn. (b) From Exercise 38 in Section 12.3 we have 1n n 3 n! n1 |||| ■ n1 31. Is the 50th partial sum s50 of the alternating series 32–34 ■ 36. Use the following steps to show that 1 nn 8n 28. n1 ■ p 1 bn , where bn 1 n if n is odd and bn 1 n 2 if n is even, is divergent. Why does the Alternating Series Test not apply? Approximate the sum of the series correct to four decimal 1n n5 ■ ln n n 35. Show that the series places. 27. n1 as n l ln 2 n l Use these facts together with part (a) to show that s2 n l ln 2 as n l . n p Absolute Convergence and the Ratio and Root Tests Given any series a n , we can consider the corresponding series an a1 a2 a3 n1 whose terms are the absolute values of the terms of the original series. |||| We have convergence tests for series with positive terms and for alternating series. But what if the signs of the terms switch back and forth irregularly? We will see in Example 3 that the idea of absolute convergence sometimes helps in such cases. 1 Definition A series a n is called absolutely convergent if the series of absolute values a n is convergent. Notice that if a n is a series with positive terms, then a n vergence is the same as convergence in this case. a n and so absolute con- EXAMPLE 1 The series 1n n2 n1 1 1 1 22 1 32 1 42 1 n2 1 1 22 1 32 is absolutely convergent because n1 1n n2 is a convergent p-series ( p 1 n1 2). 1 42 5E-12(pp 776-785) 1/18/06 10:58 AM Page 777 S ECTION 12.6 ABSOLUTE CONVERGENCE AND THE RATIO AND ROOT TESTS ❙❙❙❙ 777 EXAMPLE 2 We know that the alternating harmonic series 1n n n1 1 1 1 2 1 3 1 4 is convergent (see Example 1 in Section 12.5), but it is not absolutely convergent because the corresponding series of absolute values is 1n n n1 1 1 n n1 1 2 1 which is the harmonic series ( p-series with p 1 3 1 4 1) and is therefore divergent. 2 Definition A series a n is called conditionally convergent if it is convergent but not absolutely convergent. Example 2 shows that the alternating harmonic series is conditionally convergent. Thus, it is possible for a series to be convergent but not absolutely convergent. However, the next theorem shows that absolute convergence implies convergence. 3 Theorem If a series a n is absolutely convergent, then it is convergent. Proof Observe that the inequality 0 an an 2 an is true because a n is either a n or a n . If a n is absolutely convergent, then a n is convergent, so 2 a n is convergent. Therefore, by the Comparison Test, (a n an ) is convergent. Then (a n an an ) an is the difference of two convergent series and is therefore convergent. |||| Figure 1 shows the graphs of the terms a n and partial sums sn of the series in Example 3. Notice that the series is not alternating but has positive and negative terms. EXAMPLE 3 Determine whether the series n1 cos n n2 cos 1 12 cos 2 22 cos 3 32 is convergent or divergent. 0.5 sn an 0 FIGURE 1 n SOLUTION This series has both positive and negative terms, but it is not alternating. (The first term is positive, the next three are negative, and the following three are positive. The signs change irregularly.) We can apply the Comparison Test to the series of absolute values cos n cos n n2 n2 n1 n1 Since cos n 1 for all n, we have cos n n2 1 n2 5E-12(pp 776-785) 778 ❙❙❙❙ 1/18/06 10:58 AM Page 778 CHAPTER 12 INFINITE SEQUENCES AND SERIES We know that 1 n 2 is convergent ( p-series with p 2) and therefore cos n n 2 is 2 convergent by the Comparison Test. Thus, the given series cos n n is absolutely convergent and therefore convergent by Theorem 3. The following test is very useful in determining whether a given series is absolutely convergent. The Ratio Test an 1 L 1, then the series nl an n (and therefore convergent). (i) If lim an 1 an is divergent. (ii) If lim an 1 an 1 or lim L nl nl a n is absolutely convergent 1 , then the series an n1 an 1 1, the Ratio Test is inconclusive; that is, no conclusion can nl an be drawn about the convergence or divergence of a n. (iii) If lim Proof L (i) The idea is to compare the given series with a convergent geometric series. Since 1, we can choose a number r such that L r 1. Since lim nl the ratio a n such that 1 an 1 an L and L r a n will eventually be less than r ; that is, there exists an integer N an 1 an r whenever n N or, equivalently, an 4 an r 1 Putting n successively equal to N , N whenever n 1, N N 2, . . . in (4), we obtain aN 1 aN r aN 2 aN 1 r aN r 2 aN 3 aN 2 r aN r 3 and, in general, aN 5 k aN r k for all k 1 Now the series aN r k k1 aN r aN r 2 aN r 3 5E-12(pp 776-785) 1/18/06 10:58 AM Page 779 S ECTION 12.6 ABSOLUTE CONVERGENCE AND THE RATIO AND ROOT TESTS is convergent because it is a geometric series with 0 r together with the Comparison Test, shows that the series an aN nN1 aN k aN 1 ❙❙❙❙ 779 1. So the inequality (5), aN 2 3 k1 is also convergent. It follows that the series n 1 a n is convergent. (Recall that a finite number of terms doesn’t affect convergence.) Therefore, a n is absolutely convergent. (ii) If a n 1 a n l L 1 or a n 1 a n l , then the ratio a n 1 a n will eventually be greater than 1; that is, there exists an integer N such that an 1 an This means that a n 1 whenever n a n whenever n 1 N and so lim a n 0 nl Therefore, NOTE N a n diverges by the Test for Divergence. Part (iii) of the Ratio Test says that if lim n l a n 1 a n information. For instance, for the convergent series 1 n 2 we have ■ 1, the test gives no 1 an 1 an n 2 1 n2 1 n2 n whereas for the divergent series 1 1 2 1 l1 2 1 n as n l 1 n we have 1 an 1 an n 1 n 1 n n 1 1 l1 1 n 1 as n l 1, the series a n might converge or it might diverge. In Therefore, if lim n l a n 1 a n this case the Ratio Test fails and we must use some other test. EXAMPLE 4 Test the series 1 n n1 |||| ESTIMATING SUMS In the last three sections we used various methods for estimating the sum of a series—the method depended on which test was used to prove convergence. What about series for which the Ratio Test works? There are two possibilities: If the series happens to be an alternating series, as in Example 4, then it is best to use the methods of Section 12.5. If the terms are all positive, then use the special methods explained in Exercise 34. n3 for absolute convergence. 3n 1 n n 3 3 n: SOLUTION We use the Ratio Test with a n | 1 3 an 1 an 1 n1 n n 1 3 3n 1 1 nn 3 3n 1 n 3 1 3 | n 1 3 1 3 3n n3 n1 1 n 3 l 1 3 1 Thus, by the Ratio Test, the given series is absolutely convergent and therefore convergent. 5E-12(pp 776-785) 780 ❙❙❙❙ 1/18/06 10:58 AM Page 780 CHAPTER 12 INFINITE SEQUENCES AND SERIES E XAMPLE 5 Test the convergence of the series n1 n SOLUTION Since the terms a n an 1 an n n! are positive, we don’t need the absolute value signs. 1n 1 1! n n n 1 n! nn n 1 (See Equation 7.4.9 or 7.4*.9.) Since e Ratio Test. n 1n 1 n 1 n! 1 n n n NOTE nn . n! n le n! nn as n l 1, the given series is divergent by the Although the Ratio Test works in Example 5, an easier method is to use the Test for Divergence. Since ■ nn n! an nnn 123 n n n it follows that a n does not approach 0 as n l . Therefore, the given series is divergent by the Test for Divergence. The following test is convenient to apply when n th powers occur. Its proof is similar to the proof of the Ratio Test and is left as Exercise 38. The Root Test n (i) If lim s a n nl L 1, then the series a n is absolutely convergent n1 (and therefore convergent). n (ii) If lim s a n nl n (iii) If lim s a n nl L n 1 or lim s a n a n is divergent. , then the series nl n1 1, the Root Test is inconclusive. n 1, then part (iii) of the Root Test says that the test gives no inforIf lim n l s a n mation. The series a n could converge or diverge. (If L 1 in the Ratio Test, don’t try the Root Test because L will again be 1.) EXAMPLE 6 Test the convergence of the series n1 2n 3n 3 2 n . SOLUTION an n s an 2n 3n 2n 3n 3 2 3 2 n 2 3 Thus, the given series converges by the Root Test. 3 n 2 l 3 2 n 1 5E-12(pp 776-785) 1/18/06 10:59 AM Page 781 S ECTION 12.6 ABSOLUTE CONVERGENCE AND THE RATIO AND ROOT TESTS ❙❙❙❙ 781 Rearrangements The question of whether a given convergent series is absolutely convergent or conditionally convergent has a bearing on the question of whether infinite sums behave like finite sums. If we rearrange the order of the terms in a finite sum, then of course the value of the sum remains unchanged. But this is not always the case for an infinite series. By a rearrangement of an infinite series a n we mean a series obtained by simply changing the order of the terms. For instance, a rearrangement of a n could start as follows: a1 a2 a5 a3 a4 a 15 a6 a7 a 20 It turns out that if a n is an absolutely convergent series with sum s, then any rearrangement of a n has the same sum s. However, any conditionally convergent series can be rearranged to give a different sum. To illustrate this fact let’s consider the alternating harmonic series 1 2 1 6 1 3 1 4 1 5 1 6 1 7 1 8 ln 2 1 2 (See Exercise 36 in Section 12.5.) If we multiply this series by , we get 1 2 1 4 1 6 1 8 1 2 ln 2 Inserting zeros between the terms of this series, we have |||| Adding these zeros does not affect the sum of the series; each term in the sequence of partial sums is repeated, but the limit is the same. 0 7 1 2 0 1 4 1 6 0 0 1 8 1 2 ln 2 Now we add the series in Equations 6 and 7 using Theorem 12.2.8: 1 8 1 3 1 2 1 5 1 7 1 4 3 2 ln 2 Notice that the series in (8) contains the same terms as in (6), but rearranged so that one negative term occurs after each pair of positive terms. The sums of these series, however, are different. In fact, Riemann proved that if a n is a conditionally convergent series and r is any real number whatsoever, then there is a rearrangement of a n that has a sum equal to r. A proof of this fact is outlined in Exercise 40. |||| 12.6 Exercises 1. What can you say about the series a n in each of the following cases? (a) lim nl n0 an 1 an an 1 (c) lim nl an 2–28 8 (b) lim nl an 1 an 2. n1 n2 2n 10 n! n 4. 1 0.8 1 5. n1 7. 1 sn n1 11. n1 n1 n n n1 9. 6. 4 1 n1 n1 n1 Determine whether the series is absolutely convergent, conditionally convergent, or divergent. |||| 3. 5 1 2n ! 1 ne 1 n n3 n 8. n 1 n4 1 e n n! n1 12. n1 n n1 n1 10. 2n n4 sin 4 n 4n n2 1 5E-12(pp 776-785) 782 ❙❙❙❙ n1 15. 3 4 14. n2 n1 n1 cos n 3 n! n1 23. n1 31 2 5 n1 n an 1 1 ■ ■ the series ■ 2 ■ 37. Prove that if ■ ■ ■ ■ a n is absolutely convergent, then ■ an 29. The terms of a series are defined recursively by the equations a1 an 2 Determine whether 30. A series 5n 4n 1 1 an 3 a n converges or diverges. a n is defined by the equations a1 an 1 Determine whether 2 1 cos n an sn a n converges or diverges. n1 (that is, it fails to give a definite answer)? 1 n (a) (b) 3 n n1n n12 n1 3n sn 1 (d) n1 1 sn n2 32. For which positive integers k is the following series convergent? n1 33. (a) Show that n n! 2 kn ! n 0 x n! converges for all x. (b) Deduce that lim n l x n n! 0 for all x. an n1 38. Prove the Root Test. [Hint for part (i): Take any number r 1 and use the fact that there is an integer N r whenever n N .] such that L r n such that s a n 39. Given any series a n we define a series a n whose terms are all the positive terms of a n and a series a n whose terms are all the negative terms of a n. To be specific, we let an 31. For which of the following series is the Ratio Test inconclusive (c) n 2n Use Exercise 34 to estimate the error. 3n ■ n1 36. Use the sum of the first 10 terms to approximate the sum of 2 n n! ■ 1 1 n2 n. Use Exercise 34 to estimate the error in using s5 as an approximation to the sum of the series. (b) Find a value of n so that sn is within 0.00005 of the sum. Use this value of n to approximate the sum of the series. 2 6 10 14 5 8 11 14 5 8 11 n1 1 rn 35. (a) Find the partial sum s5 of the series n1 n 1, show, by 1 an 1 1L Rn n n! 1 3 (b) If rn is an increasing sequence, show that 2n n1 an 2 an Rn 135 1357 5! 7! 135 2n 1 n1 1 2n 1 ! 2 6 10 5 8 11 an (a) If rn is a decreasing sequence and rn summing a geometric series, that n 1n arctan n 24. 246 28. Rn 1n n ln n n 1 1 26 58 27. ■ n2 n2 13 3! 25. 1 1 ln n 22. n2 2n 2 a n be a series with positive terms and let rn a n 1 a n. Suppose that lim n l rn L 1, so a n converges by the Ratio Test. As usual, we let Rn be the remainder after n terms, that is, n! nn 20. 3n 34. Let 3 cos n n2 3 2 18. nn 21. n 22n n! n1 16. 1 1n ln n 19. n1 1 n1 10 n 1 42n n 17. Page 782 n n1 n1 26. 10:59 AM CHAPTER 12 INFINITE SEQUENCES AND SERIES n 13. 1/18/06 an an 2 an an an 2 Notice that if a n 0, then a n a n and a n 0, whereas if a n 0, then a n a n and a n 0. (a) If a n is absolutely convergent, show that both of the series a n and a n are convergent. (b) If a n is conditionally convergent, show that both of the series a n and a n are divergent. 40. Prove that if a n is a conditionally convergent series and r is any real number, then there is a rearrangement of a n whose sum is r . [Hints: Use the notation of Exercise 39. Take just enough positive terms a n so that their sum is greater than r . Then add just enough negative terms a n so that the cumulative sum is less than r . Continue in this manner and use Theorem 12.2.6.] 5E-12(pp 776-785) 1/18/06 10:59 AM Page 783 S ECTION 12.7 STRATEGY FOR TESTING SERIES |||| 12.7 ❙❙❙❙ 783 Strategy for Testing Series We now have several ways of testing a series for convergence or divergence; the problem is to decide which test to use on which series. In this respect testing series is similar to integrating functions. Again there are no hard and fast rules about which test to apply to a given series, but you may find the following advice of some use. It is not wise to apply a list of the tests in a specific order until one finally works. That would be a waste of time and effort. Instead, as with integration, the main strategy is to classify the series according to its form. 1 n p, it is a p-series, which we know to be convergent if p 1 and divergent if p 1. 2. If the series has the form ar n 1 or ar n, it is a geometric series, which converges if r 1 and diverges if r 1. Some preliminary algebraic manipulation may be required to bring the series into this form. 1. If the series is of the form 3. If the series has a form that is similar to a p-series or a geometric series, then 4. 5. 6. 7. 8. one of the comparison tests should be considered. In particular, if a n is a rational function or algebraic function of n (involving roots of polynomials), then the series should be compared with a p-series. Notice that most of the series in Exercises 12.4 have this form. (The value of p should be chosen as in Section 12.4 by keeping only the highest powers of n in the numerator and denominator.) The comparison tests apply only to series with positive terms, but if a n has some negative terms, then we can apply the Comparison Test to a n and test for absolute convergence. If you can see at a glance that lim n l a n 0, then the Test for Divergence should be used. If the series is of the form 1 n 1bn or 1 nbn , then the Alternating Series Test is an obvious possibility. Series that involve factorials or other products (including a constant raised to the n th power) are often conveniently tested using the Ratio Test. Bear in mind that a n 1 a n l 1 as n l for all p-series and therefore all rational or algebraic functions of n. Thus, the Ratio Test should not be used for such series. If a n is of the form bn n, then the Root Test may be useful. If a n f n , where x1 f x d x is easily evaluated, then the Integral Test is effective (assuming the hypotheses of this test are satisfied). In the following examples we don’t work out all the details but simply indicate which tests should be used. EXAMPLE 1 n1 Since a n l 1 2 EXAMPLE 2 n1 n 2n 1 1 0 as n l , we should use the Test for Divergence. sn 3 1 3n 3 4n 2 2 Since a n is an algebraic function of n, we compare the given series with a p-series. 5E-12(pp 776-785) ❙❙❙❙ 784 1/18/06 11:00 AM Page 784 CHAPTER 12 INFINITE SEQUENCES AND SERIES The comparison series for the Limit Comparison Test is bn EXAMPLE 3 ne sn 3 3n 3 n3 2 3n 3 bn , where 1 3n 3 2 n2 n1 Since the integral x1 xe also works. EXAMPLE 4 1 n1 x2 dx is easily evaluated, we use the Integral Test. The Ratio Test n3 n n4 1 Since the series is alternating, we use the Alternating Series Test. EXAMPLE 5 k1 2k k! Since the series involves k!, we use the Ratio Test. 1 EXAMPLE 6 n1 2 3n Since the series is closely related to the geometric series Test. |||| 12.7 1–38 |||| 1. n1 Exercises Test the series for convergence or divergence. n2 n2 1 n 2. n1 1 3. n1 n 2 n1 7. n2 n n2 4. n 3n 2 3n 5. n1 1 1 6. n1 8. k1 19. 1 n n1 1 n s ln n 1 k n n2 1 n tan 1 n k1 25. n1 2 k k! k 2! 27. n3 29. n1 n1 11. n2 1n1 n ln n n 13. n1 3n n! 12. n0 258 1 n21 n 17. n1 n2 25 14. 3n 2 n1 n2 1n 1 sn 1 18. 1 1 n1 n (s2 37. n n1 e1 n n2 1 32. 4k n1 34. n1 n2 36. 1 n2 n 1) j n 1 ln n n n1 5 2n n n 2n (s2 38. sj j j1 n n1 1 5n 30. sin 1 n sn 35. n2 28. tan 1n n sn 3k 33. n2 n3 16. k ln k k 13 n1 sin n 3 n1 k1 n1 cos n 2 n 2 4n 26. 5k 31. sn 2 1 n 2n 2 5 n1 n! 2 en n n 2 n! 15. 1 n1 24. n1 n1 n 5 5k k1 22. n1 23. k 20. 2 2n nn 21. 3n 1 8n n 2e 10. ln n sn n n1 k1 k 2e 9. 1 3 n, we use the Comparison 1 n cos2 n ln n 1) 5E-12(pp 776-785) 1/18/06 11:00 AM Page 785 SECTION 12.8 POWER SERIES |||| 12.8 ❙❙❙❙ 785 Power Series A power series is a series of the form cn x n 1 c0 c2 x 2 c1 x c3 x 3 n0 where x is a variable and the cn’s are constants called the coefficients of the series. For each fixed x, the series (1) is a series of constants that we can test for convergence or divergence. A power series may converge for some values of x and diverge for other values of x. The sum of the series is a function |||| TRIGONOMETRIC SERIES A power series is a series in which each term is a power function. A trigonometric series a n cos n x bn sin n x fx c0 c2 x 2 c1 x cn x n whose domain is the set of all x for which the series converges. Notice that f resembles a polynomial. The only difference is that f has infinitely many terms. For instance, if we take cn 1 for all n, the power series becomes the geometric series n0 is a series whose terms are trigonometric functions. This type of series is discussed on the web site www.stewartcalculus.com Click on Additional Topics and then on Fourier Series. xn 1 x x2 xn n0 which converges when 1 x 1 and diverges when x More generally, a series of the form cn x 2 a n c0 c1 x a 1 (see Equation 12.2.5). c2 x a 2 n0 is called a power series in x a or a power series centered at a or a power series about a. Notice that in writing out the term corresponding to n 0 in Equations 1 and 2 we have adopted the convention that x a 0 1 even when x a. Notice also that when x a all of the terms are 0 for n 1 and so the power series (2) always converges when x a. n! x n convergent? EXAMPLE 1 For what values of x is the series n0 SOLUTION We use the Ratio Test. If we let a n , as usual, denote the n th term of the series, then a n n! x n. If x 0, we have an 1 an lim nl n lim nl 1 !xn n! x n By the Ratio Test, the series diverges when x when x 0. 1 lim n nl 0. Thus, the given series converges only x EXAMPLE 2 For what values of x does the series n1 SOLUTION Let a n x 3 n 1x 3 n n converge? n. Then an 1 an x 3 n 1 1 1 n1 1 n x n x 3 n 3lx 3 as n l 5E-12(pp 786-795) 786 ❙❙❙❙ 1/18/06 10:36 AM Page 786 CHAPTER 12 INFINITE SEQUENCES AND SERIES By the Ratio Test, the given series is absolutely convergent, and therefore convergent, when x 3 1 and divergent when x 3 1. Now x 3 1 1 &? x 3 1 2 &? x 4 so the series converges when 2 x 4 and diverges when x 2 or x 4. The Ratio Test gives no information when x 3 1 so we must consider x 2 and x 4 separately. If we put x 4 in the series, it becomes 1 n, the harmonic series, which is divergent. If x 2, the series is 1 n n, which converges by the Alternating Series Test. Thus, the given power series converges for 2 x 4. We will see that the main use of a power series is that it provides a way to represent some of the most important functions that arise in mathematics, physics, and chemistry. In particular, the sum of the power series in the next example is called a Bessel function, after the German astronomer Friedrich Bessel (1784–1846), and the function given in Exercise 33 is another example of a Bessel function. In fact, these functions first arose when Bessel solved Kepler’s equation for describing planetary motion. Since that time, these functions have been applied in many different physical situations, including the temperature distribution in a circular plate and the shape of a vibrating drumhead (see the photographs on page 736). EXAMPLE 3 Find the domain of the Bessel function of order 0 defined by 1 nx 2n 2 n! 2 J0 x 2n n0 1 n x 2 n 2 2 n n! 2 . Then SOLUTION Let a n an 1 an 1 2 2 n 1 2n 1 2n 2 n 1! x 2n 2 n 1 2 n! x2 4n 2 2 n n! 2 1 nx 2 n x 2n 1 1 2 2 2 2 n n! x 2n 2 l0 1 2 for all x Thus, by the Ratio Test, the given series converges for all values of x. In other words, the domain of the Bessel function J0 is , . Recall that the sum of a series is equal to the limit of the sequence of partial sums. So when we define the Bessel function in Example 3 as the sum of a series we mean that, for every real number x, n J0 x lim sn x where nl 1 ix 2i 2 i! 2 sn x 2i i0 The first few partial sums are s0 x s3 x 1 x2 4 1 s1 x x4 64 x6 2304 1 x2 4 s4 x s2 x 1 x2 4 1 x4 64 x2 4 x4 64 x6 2304 x8 147,456 5E-12(pp 786-795) 1/18/06 10:36 AM Page 787 S ECTION 12.8 POWER SERIES y s™ s¸ 1 s¢ 0 x 1 s¡ s£ Figure 1 shows the graphs of these partial sums, which are polynomials. They are all approximations to the function J0 , but notice that the approximations become better when more terms are included. Figure 2 shows a more complete graph of the Bessel function. For the power series that we have looked at so far, the set of values of x for which the series is convergent has always turned out to be an interval [a finite interval for the geometric series and the series in Example 2, the infinite interval , in Example 3, and a collapsed interval 0, 0 0 in Example 1]. The following theorem, proved in Appendix F, says that this is true in general. 3 Partial sums of the Bessel function J¸ y 1 y=J¸(x) 10 0 787 J¸ FIGURE 1 _10 ❙❙❙❙ x FIGURE 2 Theorem For a given power series cn x a n there are only three n0 possibilities: (i) The series converges only when x a. (ii) The series converges for all x. (iii) There is a positive number R such that the series converges if x and diverges if x a R. a R The number R in case (iii) is called the radius of convergence of the power series. By convention, the radius of convergence is R 0 in case (i) and R in case (ii). The interval of convergence of a power series is the interval that consists of all values of x for which the series converges. In case (i) the interval consists of just a single point a. In case (ii) the interval is , . In case (iii) note that the inequality x a R can be rewritten as a R x a R. When x is an endpoint of the interval, that is, x a R, anything can happen—the series might converge at one or both endpoints or it might diverge at both endpoints. Thus, in case (iii) there are four possibilities for the interval of convergence: a R, a R a R, a R a R, a R a R, a R The situation is illustrated in Figure 3. convergence for | x-a |<R a-R FIGURE 3 a+R a divergence for | x-a |>R We summarize here the radius and interval of convergence for each of the examples already considered in this section. Series Radius of convergence Interval of convergence xn R 1 1, 1 n! x n Geometric series R 0 0 R 1 2, 4 n0 Example 1 n0 x Example 2 3 n n1 n n0 1 nx 2n 2 2 n n! 2 Example 3 R , 5E-12(pp 786-795) 788 ❙❙❙❙ 1/18/06 10:37 AM Page 788 CHAPTER 12 INFINITE SEQUENCES AND SERIES In general, the Ratio Test (or sometimes the Root Test) should be used to determine the radius of convergence R. The Ratio and Root Tests always fail when x is an endpoint of the interval of convergence, so the endpoints must be checked with some other test. EXAMPLE 4 Find the radius of convergence and interval of convergence of the series 3 nx n sn 1 n0 3 n x n sn SOLUTION Let a n 1. Then 3 n 1x n sn 2 an 1 an 1 1 3 1 sn 1 3 nx n 1n 2n 3x x l3x n n 1 2 as n l By the Ratio Test, the given series converges if 3 x 1 and diverges if 3 x 1. 1 1 Thus, it converges if x 3 and diverges if x 3 . This means that the radius of convergence is R 1 . 3 We know the series converges in the interval ( 1 , 1 ), but we must now test for 33 1 convergence at the endpoints of this interval. If x 3 , the series becomes n n0 3 n( 1 ) 3 sn 1 1 n0 sn 1 s1 1 1 s2 1 s3 1 s4 which diverges. (Use the Integral Test or simply observe that it is a p-series with p 1 1.) If x 1 , the series is 2 3 n0 3 sn n1n 3 () 1 n 1 n0 sn 1 which converges by the Alternating Series Test. Therefore, the given power series converges when 1 x 1 , so the interval of convergence is ( 1 , 1 ]. 3 3 33 EXAMPLE 5 Find the radius of convergence and interval of convergence of the series n0 SOLUTION If a n nx an 1 an 2 n nx 2 3n 1 n 3 n 1, then n 1 1x 2 3n 2 1 n x n1 2 3 3n 1 nx 2 l x n 2 3 as n l Using the Ratio Test, we see that the series converges if x 2 3 1 and it diverges if x 2 3 1. So it converges if x 2 3 and diverges if x 2 3. Thus, the radius of convergence is R 3. 5E-12(pp 786-795) 1/18/06 10:38 AM Page 789 ❙❙❙❙ S ECTION 12.8 POWER SERIES 3 can be written as 5 The inequality x 2 5, the series is the endpoints 5 and 1. When x n n0 3 3n 1, so we test the series at n 1 3 1 1 nn n0 1 nn doesn’t converge to 0]. When x which diverges by the Test for Divergence [ the series is n3n 3n 1 n0 x 789 1 3 1, n n0 which also diverges by the Test for Divergence. Thus, the series converges only when 5 x 1, so the interval of convergence is 5, 1 . |||| 12.8 Exercises 1. What is a power series? 2. (a) What is the radius of convergence of a power series? 27. n2 28. Find the radius of convergence and interval of convergence of the series. 3. n1 n1 7. n0 4. n0 1 n 1xn n3 5. 1 n 246 135 ■ 2n xn 2n 1 ■ ■ ■ ■ ■ ■ ■ ■ ■ n n 0 cn 4 is convergent, does it follow that the following series are convergent? cn (a) 2 n cn (b) n0 4 n n0 n 4 and diverges n 0 cn x converges when x when x 6. What can be said about the convergence or divergence of the following series? xn n3n (a) (c) n1 xn 5nn5 10. 12. xn 4 ln n 14. n n2 sn x 1 1 n n3 x 16. 17. 1 x n n1 x 2 n 2 n2 n 18. n1 n 20. n n1 n x bn a, n! 2 x 1 n b 0 1 ncn 9 n (d) n0 n0 n! k n x kn ! ; 32. Graph the first several partial sums sn x of the series 2 sn x 3x 2 n3n 3 n n x n, 33. The function J1 defined by 4n 1 n1 n0 together with the sum function f x 1 1 x , on a common screen. On what interval do these partial sums appear to be converging to f x ? n n2xn 246 24. n n 5 nx n3 22. 3 the series n1 n cn 31. If k is a positive integer, find the radius of convergence of x 2n 2n ! n0 n n0 n0 n cn 8 n (b) cn n0 n0 n0 n1 n n 29. If 1x 1 n1 2 nx n sn 13. 23. n2 2x 3 n ln n 30. Suppose that 4 n1 n1 n 1 n1 11. 21. 26. n nx n 8. n1 n1 n x ln n ■ n1 1 nn 4 n x n 19. ■ sn x n 6. xn n! 9. 15. n1 nn x sn n 2 n |||| n 1 n n1 How do you find it? (b) What is the interval of convergence of a power series? How do you find it? 3–28 4x 25. J1 x n0 2n ; 1 nx 2n 1 n! n 1 ! 2 2 n 1 is called the Bessel function of order 1. (a) Find its domain. (b) Graph the first several partial sums on a common screen. 5E-12(pp 786-795) ❙❙❙❙ 790 CAS 1/18/06 10:39 AM Page 790 CHAPTER 12 INFINITE SEQUENCES AND SERIES that is, its coefficients are c2 n 1 and c2 n 1 2 for all n Find the interval of convergence of the series and find an explicit formula for f x . (c) If your CAS has built-in Bessel functions, graph J1 on the same screen as the partial sums in part (b) and observe how the partial sums approximate J1. n cn for all n 0, find the n 0 cn x , where cn 4 interval of convergence of the series and a formula for f x . 36. If f x 34. The function A defined by Ax ; CAS x3 23 1 x6 2356 x9 235689 n 37. Show that if lim n l s cn c, where c 0, then the radius of convergence of the power series cn x n is R 1 c. is called the Airy function after the English mathematician and astronomer Sir George Airy (1801–1892). (a) Find the domain of the Airy function. (b) Graph the first several partial sums sn x on a common screen. (c) If your CAS has built-in Airy functions, graph A on the same screen as the partial sums in part (b) and observe how the partial sums approximate A. cn x a n satisfies c n 0 for all n. Show that if lim n l cn cn 1 exists, then it is equal to the radius of convergence of the power series. 38. Suppose that the power series cn x n has radius of convergence 2 and the series dn x n has radius of convergence 3. What is the radius of convergence of the series cn dn x n ? 39. Suppose the series 40. Suppose that the radius of convergence of the power series 35. A function f is defined by fx |||| 12.9 1 2x 0. x 2 2x 3 x cn x n is R. What is the radius of convergence of the power series cn x 2 n ? 4 Representations of Functions as Power Series In this section we learn how to represent certain types of functions as sums of power series by manipulating geometric series or by differentiating or integrating such a series. You might wonder why we would ever want to express a known function as a sum of infinitely many terms. We will see later that this strategy is useful for integrating functions that don’t have elementary antiderivatives, for solving differential equations, and for approximating functions by polynomials. (Scientists do this to simplify the expressions they deal with; computer scientists do this to represent functions on calculators and computers.) We start with an equation that we have seen before: |||| A geometric illustration of Equation 1 is shown in Figure 1. Because the sum of a series is the limit of the sequence of partial sums, we have 1 lim sn x nl 1x where sn x 1 x x2 1 1 1 x 1 x x2 x3 xn x We first encountered this equation in Example 5 in Section 12.2, where we obtained it by observing that it is a geometric series with a 1 and r x. But here our point of view is different. We now regard Equation 1 as expressing the function f x 1 1 x as a sum of a power series. xn s¡¡ y is the n th partial sum. Notice that as n increases, sn x becomes a better approximation to f x for 1 x 1. sˆ s∞ f s™ FIGURE 1 1 ƒ= and some partial sums 1-x 1 n0 _1 0 1 x 5E-12(pp 786-795) 1/18/06 10:39 AM Page 791 S ECTION 12.9 REPRESENTATIONS OF FUNCTIONS AS POWER SERIES ❙❙❙❙ 791 x 2 as the sum of a power series and find the interval of EXAMPLE 1 Express 1 1 convergence. x 2 in Equation 1, we have SOLUTION Replacing x by 1 1 x x2 x2 1 n n0 1 nx 2n 1 2 1 x2 x4 x6 x8 n0 Because this is a geometric series, it converges when 1, that is, x 2 1, or x2 1. Therefore, the interval of convergence is 1, 1 . (Of course, we could have x determined the radius of convergence by applying the Ratio Test, but that much work is unnecessary here.) EXAMPLE 2 Find a power series representation for 1 x 2. SOLUTION In order to put this function in the form of the left side of Equation 1 we first factor a 2 from the denominator: 1 1 2 x This series converges when is 2, 2 . x 2 21 1 2 1 x 2 n0 x 2 21 n n0 2 1, that is, x x2 n 1 n1 xn 2. So the interval of convergence EXAMPLE 3 Find a power series representation of x 3 x 2. 3 SOLUTION Since this function is just x times the function in Example 2, all we have to do is to multiply that series by x 3: |||| It’s legitimate to move x 3 across the sigma sign because it doesn’t depend on n. [Use Theorem 12.2.8(i) with c x 3.] x3 x 2 x3 1 2 x3 1 x 2 1 4 2n n0 1 8 x4 n 1 x3 x5 1 16 1 n 1 xn n0 2n 1 xn 3 x6 Another way of writing this series is as follows: x3 x 2 n3 As in Example 2, the interval of convergence is 1 2n n1 2 xn 2, 2 . Differentiation and Integration of Power Series The sum of a power series is a function f x a n whose domain is the intern 0 cn x val of convergence of the series. We would like to be able to differentiate and integrate such functions, and the following theorem (which we won’t prove) says that we can do so by differentiating or integrating each individual term in the series, just as we would for a polynomial. This is called term-by-term differentiation and integration. 5E-12(pp 786-795) 792 ❙❙❙❙ 1/18/06 10:40 AM Page 792 CHAPTER 12 INFINITE SEQUENCES AND SERIES 2 Theorem If the power series a n has radius of convergence R cn x 0, then the function f defined by fx c0 c1 x a c2 x a 2 cn x a n n0 is differentiable (and therefore continuous) on the interval a (i) f x c1 2c2 x a 3c3 x 2 a R, a ncn x a R and n1 n1 |||| In part (ii), x c0 dx c0 x C1 is written C, where C C1 ac0, so as c0 x a all the terms of the series have the same form. (ii) yf x dx C c0 x C a cn x a n n0 x c1 a 2 c2 2 x a 3 3 n1 1 The radii of convergence of the power series in Equations (i) and (ii) are both R. NOTE 1 (iii) (iv) ■ d dx Equations (i) and (ii) in Theorem 2 can be rewritten in the form cn x a n n0 y cn x n0 a n n0 d cn x dx yc dx n a n a n dx x n0 We know that, for finite sums, the derivative of a sum is the sum of the derivatives and the integral of a sum is the sum of the integrals. Equations (iii) and (iv) assert that the same is true for infinite sums, provided we are dealing with power series. (For other types of series of functions the situation is not as simple; see Exercise 36.) NOTE 2 Although Theorem 2 says that the radius of convergence remains the same when a power series is differentiated or integrated, this does not mean that the interval of convergence remains the same. It may happen that the original series converges at an endpoint, whereas the differentiated series diverges there. (See Exercise 37.) ■ NOTE 3 The idea of differentiating a power series term by term is the basis for a powerful method for solving differential equations. We will discuss this method in Chapter 18. ■ EXAMPLE 4 In Example 3 in Section 12.8 we saw that the Bessel function J0 x n0 1 nx 2n 2 2 n n! 2 is defined for all x. Thus, by Theorem 2, J0 is differentiable for all x and its derivative is found by term-by-term differentiation as follows: J0 x n0 EXAMPLE 5 Express 1 1 the radius of convergence? d 1 nx 2 n 2n dx 2 n! 2 n1 1 n 2nx 2 n 2 2 n n! 2 1 x 2 as a power series by differentiating Equation 1. What is 5E-12(pp 786-795) 1/18/06 10:41 AM Page 793 S ECTION 12.9 REPRESENTATIONS OF FUNCTIONS AS POWER SERIES ❙❙❙❙ 793 SOLUTION Differentiating each side of the equation 1 1 1 x 1 1 2 x x3 xn n0 1 we get x2 x 3x 2 2x nx n 1 n1 If we wish, we can replace n by n 1 and write the answer as 1 1 1 xn n 2 x n0 According to Theorem 2, the radius of convergence of the differentiated series is the same as the radius of convergence of the original series, namely, R 1. EXAMPLE 6 Find a power series representation for ln 1 convergence. x and its radius of SOLUTION We notice that, except for a factor of 11 1, the derivative of this function is x . So we integrate both sides of Equation 1: ln 1 y1 x 1 x x2 2 x n1 x xn n x2 x dx C n0 C x xn 1 n1 1 x3 3 n1 xn n 1 2 1 2 ln 2 1 8 1 24 1 64 n1 tan 1x y C C. 1. ln 2, we 1 n2n tan 1x. 1 1 x 2 and find the required series by integrating found in Example 1. SOLUTION We observe that f x x 0 1 in the result of Example 6. Since ln 1 2 EXAMPLE 7 Find a power series representation for f x the power series for 1 1 ln 1 x The radius of convergence is the same as for the original series: R Notice what happens if we put x see that C 0 in this equation and obtain x2 2 x 1 x3 3 To determine the value of C we put x Thus, C 0 and ln 1 y dx 2 1 x2 1 x dx y x3 3 x5 5 1 x2 x7 7 x4 x6 dx 5E-12(pp 786-795) 794 ❙❙❙❙ 1/18/06 10:42 AM Page 794 CHAPTER 12 INFINITE SEQUENCES AND SERIES |||| The power series for tan 1x obtained in Example 7 is called Gregory’s series after the Scottish mathematician James Gregory (1638–1675), who had anticipated some of Newton’s discoveries. We have shown that Gregory’s series is valid when 1 x 1, but it turns out (although it isn’t easy to prove) that it is also valid when x 1. Notice that when x 1 the series becomes 4 1 1 3 1 5 1 7 This beautiful result is known as the Leibniz formula for . To find C we put x tan 1 0 0 and obtain C tan 1x x3 3 x x5 5 0. Therefore x7 7 1 n n0 x 2n 1 2n 1 x 2 is 1, the radius of conver- Since the radius of convergence of the series for 1 1 gence of this series for tan 1x is also 1. EXAMPLE 8 (a) Evaluate x 1 1 x 7 d x as a power series. (b) Use part (a) to approximate x00.5 1 1 x 7 d x correct to within 10 7. SOLUTION (a) The first step is to express the integrand, 1 1 x 7 , as the sum of a power series. As in Example 1, we start with Equation 1 and replace x by x 7: 1 1 x x7 x7 1 n n0 1 n x 7n 1 7 1 x7 x 14 n0 |||| This example demonstrates one way in which power series representations are useful. Integrating 1 1 x 7 by hand is incredibly difficult. Different computer algebra systems return different forms of the answer, but they are all extremely complicated. (If you have a CAS, try it yourself.) The infinite series answer that we obtain in Example 8(a) is actually much easier to deal with than the finite answer provided by a CAS. Now we integrate term by term: y1 1 x7 1 n x 7n d x y dx C 1 n0 C n0 x8 8 x x 15 15 n x 7n 1 7n 1 x 22 22 This series converges for x7 1, that is, for x 1. (b) In applying the Fundamental Theorem of Calculus it doesn’t matter which antiderivative we use, so let’s use the antiderivative from part (a) with C 0: y 1 0.5 0 1 x7 dx x 1 2 x8 8 1 8 28 x 15 15 12 x 22 22 0 1 15 2 1 15 22 2 22 7n 1n 1 2 7n 1 This infinite series is the exact value of the definite integral, but since it is an alternating series, we can approximate the sum using the Alternating Series Estimation Theorem. If we stop adding after the term with n 3, the error is smaller than the term with n 4: 1 29 2 29 6.4 10 11 So we have y 1 0.5 0 1 x7 dx 1 2 1 8 28 1 15 2 15 1 22 2 22 0.49951374 5E-12(pp 786-795) 1/18/06 10:43 AM Page 795 SECTION 12.9 REPRESENTATIONS OF FUNCTIONS AS POWER SERIES |||| 12.9 n0 cn x n 15–18 |||| Find a power series representation for the function and determine the radius of convergence. is 10, what is the radius of convergence of the series n1 ? Why? n 1 ncn x 2. Suppose you know that the series 15. f x n bn x converges for 2. What can you say about the following series? Why? n0 n n0 x 1 5. f x 7. f x 9. f x ■ ■ 11–12 1 1 1 x x 9 ■ ■ ■ ■ ■ ■ 19. f x 21. f x ln x ■ 1 x a3 ■ x3 ■ ■ Express the function as the sum of a power series by first using partial fractions. Find the interval of convergence. |||| 11. f x 12. f x ■ ■ x2 2 7x ■ ■ ■ ■ ■ ■ 1 1 x x ■ 1 20. f x x x2 22. f x ■ ■ 1 tan ■ ■ 25 ■ ■ 2x ■ ■ ■ 23. y1 25. y t t8 x tan x3 ■ 1 ■ ln 1 24. x dx ■ ■ ■ y 26. dt y tan t dt t ■ ■ 1 x 2 dx ■ ■ ■ ■ |||| Use a power series to approximate the definite integral to six decimal places. 2x ■ ■ ■ 27–30 1 3x 2 ■ ■ 3 x arctan x 3 23–26 |||| Evaluate the indefinite integral as a power series. What is the radius of convergence? 2 ■ 18. f x ■ ln 3 4x 2 ■ 9x 2 1 10. f x x2 ■ 1 8. f x 5 ■ 2 x4 1 6. f x x3 1 x 2 |||| Find a power series representation for f , and graph f and several partial sums sn x on the same screen. What happens as n increases? 3 4. f x x x2 1 2x ; 19–22 3–10 |||| Find a power series representation for the function and determine the interval of convergence. 1 16. f x x3 n1 ■ x ln 5 17. f x bn 3. f x 795 Exercises 1. If the radius of convergence of the power series x ❙❙❙❙ 1 ■ ■ ■ ■ ■ ■ ■ ■ 13. (a) Use differentiation to find a power series representation for fx 1 1 x 29. y 1 0.2 13 0 ■ x 2 tan ■ 28. 1 x 4 dx ■ ■ ■ y 30. dx x5 1 0 y ■ x x6 1 ■ ■ ■ 32. Show that the function fx 3 n0 3 14. (a) Find a power series representation for f x What is the radius of convergence? (b) Use part (a) to find a power series for f x (c) Use part (a) to find a power series for f x 1 nx 2n 2n ! is a solution of the differential equation fx x dx 0.5 0 decimal places. x2 1 x 4 dx ln 1 ■ 31. Use the result of Example 6 to compute ln 1.1 correct to five 1 1 0.4 0 2 (c) Use part (b) to find a power series for fx y ■ What is the radius of convergence? (b) Use part (a) to find a power series for fx 27. fx 0 33. (a) Show that J0 (the Bessel function of order 0 given in ln 1 x. x ln 1 ln x 2 x. 1. Example 4) satisfies the differential equation x 2J 0 x x J0 x x 2J0 x 0 (b) Evaluate x01 J0 x d x correct to three decimal places. ■ 5E-12(pp 796-805) 796 ❙❙❙❙ 1/18/06 10:29 AM Page 796 CHAPTER 12 INFINITE SEQUENCES AND SERIES 38. (a) Starting with the geometric series 34. The Bessel function of order 1 is defined by n0 x n, find the sum of the series J1 x n0 1 nx 2n 1 n! n 1 !2 2 n nx n 1 x J1 x (b) Show that J0 x x2 1 J1 x 0 (i) 35. (a) Show that the function 1 x n, nn x 1 n2 fx n0 xn n! n2 (ii) n 2 n2 (iii) n n1 39. Use the power series for tan is a solution of the differential equation n2 2n 1 x to prove the following expresas the sum of an infinite series: sion for (b) Show that f x 1 (c) Find the sum of each of the following series. J1 x . fx x (b) Find the sum of each of the following series. n (i) (ii) n x n, x 1 2n n1 n1 (a) Show that J1 satisfies the differential equation x 2J1 x 1 n1 fx 2s3 x e. n0 1n 2n 1 3 n 40. (a) By completing the square, show that sin n x n 2. Show that the series fn x converges for all values of x but the series of derivatives f n x diverges when x 2 n , n an integer. For what values of x does the series f n x converge? 36. Let fn x y dx x 12 x2 0 1 3s3 3 (b) By factoring x 1 as a sum of cubes, rewrite the integral in part (a). Then express 1 x 3 1 as the sum of a power series and use it to prove the following formula for : 37. Let fx n1 xn n2 3 s3 4 Find the intervals of convergence for f , f , and f . |||| 12.10 1 8n n0 n 2 3n 1 1 3n 2 Taylor and Maclaurin Series In the preceding section we were able to find power series representations for a certain restricted class of functions. Here we investigate more general problems: Which functions have power series representations? How can we find such representations? We start by supposing that f is any function that can be represented by a power series 1 fx c0 c1 x a c2 x a 2 c3 x a 3 c4 x a 4 x a R Let’s try to determine what the coefficients cn must be in terms of f . To begin, notice that if we put x a in Equation 1, then all terms after the first one are 0 and we get fa c0 By Theorem 12.9.2, we can differentiate the series in Equation 1 term by term: 2 fx c1 2c2 x and substitution of x a 3c3 x a 2 4c4 x a in Equation 2 gives fa c1 a 3 x a R 5E-12(pp 796-805) 1/18/06 10:30 AM Page 797 S ECTION 12.10 TAYLOR AND MACLAURIN SERIES ❙❙❙❙ 797 Now we differentiate both sides of Equation 2 and obtain 3 fx 2c2 2 3c3 x Again we put x a 3 4c4 x a 2 x a R a in Equation 3. The result is fa 2c2 Let’s apply the procedure one more time. Differentiation of the series in Equation 3 gives 4 f x 2 3c3 2 3 4c4 x and substitution of x a 3 4 5c5 x a 2 x a R a in Equation 4 gives fa 2 3c3 3! c3 By now you can see the pattern. If we continue to differentiate and substitute x obtain n f a 234 n cn a, we n! cn Solving this equation for the nth coefficient cn , we get f cn n a n! This formula remains valid even for n 0 if we adopt the conventions that 0! f0 f . Thus, we have proved the following theorem. 5 1 and Theorem If f has a power series representation (expansion) at a, that is, if fx cn x a n x a R n0 then its coefficients are given by the formula cn f n a n! Substituting this formula for cn back into the series, we see that if f has a power series expansion at a, then it must be of the following form. 6 f fx n0 fa n n! a x fa 1! a x n a fa 2! x a 2 fa 3! x a 3 5E-12(pp 796-805) 798 ❙❙❙❙ 1/18/06 10:30 AM Page 798 CHAPTER 12 INFINITE SEQUENCES AND SERIES |||| The Taylor series is named after the English mathematician Brook Taylor (1685–1731) and the Maclaurin series is named in honor of the Scottish mathematician Colin Maclaurin (1698–1746) despite the fact that the Maclaurin series is really just a special case of the Taylor series. But the idea of representing particular functions as sums of power series goes back to Newton, and the general Taylor series was known to the Scottish mathematician James Gregory in 1668 and to the Swiss mathematician John Bernoulli in the 1690s. Taylor was apparently unaware of the work of Gregory and Bernoulli when he published his discoveries on series in 1715 in his book Methodus incrementorum directa et inversa. Maclaurin series are named after Colin Maclaurin because he popularized them in his calculus textbook Treatise of Fluxions published in 1742. The series in Equation 6 is called the Taylor series of the function f at a (or about a or centered at a). For the special case a 0 the Taylor series becomes f fx 7 n 0 xn n! n0 f0 x 1! f0 f0 2 x 2! This case arises frequently enough that it is given the special name Maclaurin series. NOTE We have shown that if f can be represented as a power series about a, then f is equal to the sum of its Taylor series. But there exist functions that are not equal to the sum of their Taylor series. An example of such a function is given in Exercise 62. ■ EXAMPLE 1 Find the Maclaurin series of the function f x e x and its radius of convergence. e x, then f n x e x, so f n 0 e0 Taylor series for f at 0 (that is, the Maclaurin series) is SOLUTION If f x f n 0 n! n0 xn n0 xn n! xn n 1 1! x3 3! x n n!. Then To find the radius of convergence we let a n an 1 an x2 2! x 1! 1 1 for all n. Therefore, the n! xn x n 1 l0 1 so, by the Ratio Test, the series converges for all x and the radius of convergence is R . The conclusion we can draw from Theorem 5 and Example 1 is that if e x has a power series expansion at 0, then ex n0 xn n! So how can we determine whether e x does have a power series representation? Let’s investigate the more general question: Under what circumstances is a function equal to the sum of its Taylor series? In other words, if f has derivatives of all orders, when is it true that fn a fx x an n! n0 As with any convergent series, this means that f x is the limit of the sequence of partial sums. In the case of the Taylor series, the partial sums are n Tn x i0 fa f i i! a x fa 1! a x i a fa 2! x a 2 f n n! a x a n 5E-12(pp 796-805) 1/18/06 10:31 AM Page 799 S ECTION 12.10 TAYLOR AND MACLAURIN SERIES y y=T£(x) T1 x y=T™(x) (0, 1) y=T¡(x) 0 799 Notice that Tn is a polynomial of degree n called the nth-degree Taylor polynomial of f at a. For instance, for the exponential function f x e x, the result of Example 1 shows that the Taylor polynomials at 0 (or Maclaurin polynomials) with n 1, 2, and 3 are y=´ y=T™(x) ❙❙❙❙ x 1 x T2 x 1 x2 2! x T3 x 1 x2 2! x x3 3! The graphs of the exponential function and these three Taylor polynomials are drawn in Figure 1. In general, f x is the sum of its Taylor series if y=T£(x) fx FIGURE 1 |||| As n increases, Tn x appears to approach e x in Figure 1. This suggests that e x is equal to the sum of its Taylor series. lim Tn x nl If we let Rn x fx Tn x so that fx Tn x Rn x then Rn x is called the remainder of the Taylor series. If we can somehow show that lim n l Rn x 0, then it follows that lim Tn x nl lim f x Rn x nl fx lim Rn x nl fx We have therefore proved the following. 8 Theorem If f x mial of f at a and Tn x Rn x , where Tn is the nth-degree Taylor polynolim Rn x 0 nl for x a R, then f is equal to the sum of its Taylor series on the interval xa R. In trying to show that lim n l Rn x following fact. Taylor’s Inequality If f 0 for a specific function f , we usually use the n1 x M for x of the Taylor series satisfies the inequality 9 Rn x M n 1! x a a n1 d, then the remainder Rn x for x To see why this is true for n 1, we assume that f x fx M , so for a x a d we have y x a f t dt y x a a d M . In particular, we have M dt An antiderivative of f is f , so by Part 2 of the Fundamental Theorem of Calculus, we have fx fa Mx a or fx fa Mx a 5E-12(pp 796-805) 800 ❙❙❙❙ 1/18/06 10:31 AM Page 800 CHAPTER 12 INFINITE SEQUENCES AND SERIES |||| As alternatives to Taylor’s Inequality, we have the following formulas for the remainder term. If f n 1 is continuous on an interval I and x I , then 1x Rn x y x t n f n 1 t dt n! a This is called the integral form of the remainder term. Another formula, called Lagrange’s form of the remainder term, states that there is a number z between x and a such that fn1z Rn x x an 1 n 1! This version is an extension of the Mean Value Theorem (which is the case n 0). Proofs of these formulas, together with discussions of how to use them to solve the examples of Sections 12.10 and 12.12, are given on the web site y Thus x a fx fx But R1 x fa fx fx Mt a M x 2 a M x 2 a dt x M a 2 2 2 a . So fax R1 x A similar argument, using f x fa fax a fa x a fa fax T1 x y f t dt a 2 M , shows that M x 2 R1 x a 2 www.stewartcalculus.com Click on Additional Topics and then on Formulas for the Remainder Term in Taylor series. So M x 2 R1 x a 2 Although we have assumed that x a, similar calculations show that this inequality is also true for x a. This proves Taylor’s Inequality for the case where n 1. The result for any n is proved in a similar way by integrating n 1 times. (See Exercise 61 for the case n 2.) NOTE In Section 12.12 we will explore the use of Taylor’s Inequality in approximating functions. Our immediate use of it is in conjunction with Theorem 8. In applying Theorems 8 and 9 it is often helpful to make use of the following fact. ■ lim 10 nl xn n! for every real number x 0 This is true because we know from Example 1 that the series and so its nth term approaches 0. x n n! converges for all x EXAMPLE 2 Prove that e x is equal to the sum of its Maclaurin series. SOLUTION If f x x d, then f says that e x, then f n x ex 1 n1 x e x for all n. If d is any positive number and d e . So Taylor’s Inequality, with a 0 and M e d, ed Rn x n nl ed n x n1 for x d e d works for every value of n. But, from Equa- Notice that the same constant M tion 10, we have lim 1! 1! x n1 e d lim nl x n n1 1! 0 5E-12(pp 796-805) 1/18/06 10:32 AM Page 801 ❙❙❙❙ S ECTION 12.10 TAYLOR AND MACLAURIN SERIES 801 0 and therefore It follows from the Squeeze Theorem that lim n l Rn x lim n l Rn x 0 for all values of x. By Theorem 8, e x is equal to the sum of its Maclaurin series, that is, ex 11 n0 |||| In 1748 Leonard Euler used Equation 12 to find the value of e correct to 23 digits. In 2000 Xavier Gourdon and S. Kondo, again using the series in (12), computed e to more than twelve billion decimal places. The special techniques they employed to speed up the computation are explained on the web page xn n! for all x In particular, if we put x 1 in Equation 11, we obtain the following expression for the number e as a sum of an infinite series: 1 n! e 12 n0 1 1! 1 1 2! 1 3! www.numbers.computation.free.fr e x at a EXAMPLE 3 Find the Taylor series for f x SOLUTION We have f (6), we get n 2 2 e and so, putting a f n 2 x n! n0 2 2. 2 in the definition of a Taylor series n n0 e2 x n! 2 n Again it can be verified, as in Example 1, that the radius of convergence is R 0, so Example 2 we can verify that lim n l Rn x ex 13 n0 e2 x n! 2 n . As in for all x We have two power series expansions for e x, the Maclaurin series in Equation 11 and the Taylor series in Equation 13. The first is better if we are interested in values of x near 0 and the second is better if x is near 2. EXAMPLE 4 Find the Maclaurin series for sin x and prove that it represents sin x for all x. SOLUTION We arrange our computation in two columns as follows: fx sin x f0 0 fx cos x f0 1 0 fx sin x f0 f cos x f0 f x 4 x sin x f 4 0 1 0 Since the derivatives repeat in a cycle of four, we can write the Maclaurin series as follows: f0 f0 2 f03 f0 x x x 1! 2! 3! x x3 3! x5 5! x7 7! 1 n0 n x 2n 1 2n 1 ! 5E-12(pp 796-805) 802 ❙❙❙❙ 1/18/06 10:33 AM Page 802 CHAPTER 12 INFINITE SEQUENCES AND SERIES Since f take M |||| Figure 2 shows the graph of sin x together with its Taylor (or Maclaurin) polynomials T1 x x x3 3! x x3 3! n1 x is sin x or cos x, we know that f 1 in Taylor’s Inequality: x 1 for all x. So we can x T3 x n1 T5 x x5 5! M Rn x 14 n 1! xn x 1 n1 n 1! By Equation 10 the right side of this inequality approaches 0 as n l , so Rn x l 0 by the Squeeze Theorem. It follows that Rn x l 0 as n l , so sin x is equal to the sum of its Maclaurin series by Theorem 8. Notice that, as n increases, Tn x becomes a better approximation to sin x. y We state the result of Example 4 for future reference. T¡ 1 T∞ y=sin x sin x 15 0 x3 3! x x 1 n0 T£ x7 7! x 2n 1 2n 1 ! n 1 FIGURE 2 x5 5! for all x EXAMPLE 5 Find the Maclaurin series for cos x. SOLUTION We could proceed directly as in Example 4 but it’s easier to differentiate the Maclaurin series for sin x given by Equation 15: cos x d sin x dx 1 |||| The Maclaurin series for e x, sin x, and cos x that we found in Examples 2, 4, and 5 were discovered, using different methods, by Newton. (There is evidence that the series for sin x and cos x were known to Indian astronomers more than a century before Newton, but this knowledge didn’t spread to the rest of the world.) These equations are remarkable because they say we know everything about each of these functions if we know all its derivatives at the single number 0. 3x 2 3! d dx x3 3! x 5x 4 5! x5 5! 7x 6 7! x7 7! x2 2! 1 x4 4! x6 6! Since the Maclaurin series for sin x converges for all x, Theorem 2 in Section 12.9 tells us that the differentiated series for cos x also converges for all x. Thus cos x 16 1 x2 2! 1 n0 x4 4! n x 2n 2n ! x6 6! for all x EXAMPLE 6 Find the Maclaurin series for the function f x x cos x. SOLUTION Instead of computing derivatives and substituting in Equation 7, it’s easier to multiply the series for cos x (Equation 16) by x: x cos x x 1 n0 EXAMPLE 7 Represent f x n x 2n 2n ! 1 n0 n x 2n 1 2n ! sin x as the sum of its Taylor series centered at 3. 5E-12(pp 796-805) 1/18/06 10:34 AM Page 803 S ECTION 12.10 TAYLOR AND MACLAURIN SERIES ❙❙❙❙ 803 SOLUTION Arranging our work in columns, we have fx fx |||| We have obtained two different series representations for sin x, the Maclaurin series in Example 4 and the Taylor series in Example 7. It is best to use the Maclaurin series for values of x near 0 and the Taylor series for x near 3. Notice that the third Taylor polynomial T3 in Figure 3 is a good approximation to sin x near 3 but not as good near 0. Compare it with the third Maclaurin polynomial T3 in Figure 2, where the opposite is true. y cos x 3 f s3 2 3 f x 1 2 1 2 and this pattern repeats indefinitely. Therefore, the Taylor series at f f 3 1! 3 s3 2 x sin x s3 2 3 f f y=sin x π 3 cos x 3 f fx f 0 sin x x 1 3 x 2 1! x x x 3 3 1 3 3 3 3! 3 2 s3 2 2! 3 f 2 3 2! 3 is x 2 3! 3 The proof that this series represents sin x for all x is very similar to that in Example 4. [Just replace x by x 3 in (14).] We can write the series in sigma notation if we separate the terms that contain s3 : T£ sin x FIGURE 3 n0 1 ns3 x 2 2n ! 2n 1 3 n0 2 2n n 1! 2n 1 x 3 The power series that we obtained by indirect methods in Examples 5 and 6 and in Section 12.9 are indeed the Taylor or Maclaurin series of the given functions because cn x a n Theorem 5 asserts that, no matter how a power series representation f x n is obtained, it is always true that cn f a n!. In other words, the coefficients are uniquely determined. We collect in the following table, for future reference, some important Maclaurin series that we have derived in this section and the preceding one. Important Maclaurin series and their intervals of convergence 1 1 x ex n0 xn 1 x2 x x3 1, 1 x3 3! , n0 xn n! x 1! 1 x2 2! 1 n x 2n 1 2n 1 ! 1 sin x n x 2n 2n ! n0 cos x n0 tan 1x 1 n0 n x 2n 1 2n 1 x x3 3! x2 2! 1 x x5 5! x4 4! x3 3 x7 7! x6 6! x5 5 , , x7 7 1, 1 5E-12(pp 796-805) 804 ❙❙❙❙ 1/18/06 10:34 AM Page 804 CHAPTER 12 INFINITE SEQUENCES AND SERIES Module 12.10/12.12 enables you to see how successive Taylor polynomials approach the original function. One reason that Taylor series are important is that they enable us to integrate functions that we couldn’t previously handle. In fact, in the introduction to this chapter we mentioned that Newton often integrated functions by first expressing them as power series and 2 then integrating the series term by term. The function f x e x can’t be integrated by techniques discussed so far because its antiderivative is not an elementary function (see Section 8.5). In the following example we use Newton’s idea to integrate this function. EXAMPLE 8 2 (a) Evaluate x e x dx as an infinite series. 2 (b) Evaluate x01 e x dx correct to within an error of 0.001. SOLUTION 2 (a) First we find the Maclaurin series for f x e x . Although it’s possible to use the direct method, let’s find it simply by replacing x with x 2 in the series for e x given in the table of Maclaurin series. Thus, for all values of x, x2 n! x2 e n0 n 1 n0 n x 2n n! x2 1! 1 x4 2! x6 3! Now we integrate term by term: ye x2 y dx x2 1! 1 C x x4 2! x3 3 1! x6 3! 1 x5 5 2! n x 2n n! dx x7 7 3! 1 This series converges for all x because the original series for e (b) The Evaluation Theorem gives |||| We can take C in part (a). 0 in the antiderivative y 1 0 e x2 dx x3 3 1! x x5 5 2! 1 1 3 1 10 1 42 1 3 1 10 1 42 1 216 x 2n 1 2n 1 n! converges for all x. x9 9 4! 1 0 1 216 1 x7 7 3! x2 n 0.7475 The Alternating Series Estimation Theorem shows that the error involved in this approximation is less than 1 11 5! 1 1320 0.001 Another use of Taylor series is illustrated in the next example. The limit could be found with l’Hospital’s Rule, but instead we use a series. EXAMPLE 9 Evaluate lim xl0 ex 1 x2 x . 5E-12(pp 796-805) 1/18/06 10:35 AM Page 805 S ECTION 12.10 TAYLOR AND MACLAURIN SERIES ❙❙❙❙ 805 SOLUTION Using the Maclaurin series for e x, we have |||| Some computer algebra systems compute limits in this way. lim xl0 ex 1 x2 1 x x 1! x2 2! x2 2! x3 3! x4 4! lim x3 3! x2 xl0 lim xl0 x x2 xl0 lim 1 1 2 x2 4! x 3! x3 5! 1 2 because power series are continuous functions. Multiplication and Division of Power Series If power series are added or subtracted, they behave like polynomials (Theorem 12.2.8 shows this). In fact, as the following example illustrates, they can also be multiplied and divided like polynomials. We find only the first few terms because the calculations for the later terms become tedious and the initial terms are the most important ones. EXAMPLE 10 Find the first three nonzero terms in the Maclaurin series for (a) e x sin x and (b) tan x. SOLUTION (a) Using the Maclaurin series for e x and sin x in the table on page 803, we have e x sin x x2 2! x 1! 1 x3 3! x x3 3! We multiply these expressions, collecting like terms just as for polynomials: 1 6 1 6 x3 x3 x2 1 2 1 6 x3 x3 x Thus x2 x 1 x2 1 3 x3 x x 1 2 e x sin x x2 x 1 6 1 6 1 3 x4 x4 x3 (b) Using the Maclaurin series in the table, we have tan x sin x cos x x 1 x3 3! x2 2! x5 5! x4 4! 5E-12(pp 806-815) ❙❙❙❙ 806 1/18/06 10:18 AM Page 806 CHAPTER 12 INFINITE SEQUENCES AND SERIES We use a procedure like long division: x x2 Thus 1 24 x x x4 tan x 1 3 x x5 x3 x3 2 15 1 120 1 24 x3 x3 1 30 1 6 x5 x5 2 15 1 2 x3 1 6 1 2 1 3 1 3 1 1 3 x5 2 15 x3 x5 x5 x5 Although we have not attempted to justify the formal manipulations used in Example 10, they are legitimate. There is a theorem which states that if both f x cn x n and n tx bn x converge for x R and the series are multiplied as if they were polynomials, then the resulting series also converges for x R and represents f x t x . For division we require b0 0; the resulting series converges for sufficiently small x . |||| 12.10 1. If f x Exercises n0 5 n for all x, write a formula for b 8. bn x 11–18 |||| Find the Taylor series for f x centered at the given value of a. [Assume that f has a power series expansion. Do not show that Rn x l 0.] 2. (a) The graph of f is shown. Explain why the series 1.6 0.8 x 1 0.4 x 1 2 0.1 x 1 3 11. f x 1 x 2, a x 12. f x 0 cos x, sin x, a 1 sx, a 18. f x 2 ■ (b) Explain why the series 2.8 0.5 x 2 1.5 x 2 0.1 x 2 a ■ ■ 2 9 1 ■ ■ ■ ■ ■ ■ 5. f x 1 7. f x 21. Prove that the series obtained in Exercise 9 represents sinh x for all x. 22. Prove that the series obtained in Exercise 10 represents cosh x for all x. sin 2 x 6. f x ln 1 8. f x xe x 23. f x cos x 24. f x e 9. f x 10. f x cosh x 25. f x x tan 1x 26. f x sin x 4 ■ ■ 27. f x x 2e 28. f x x cos 2 x 3 e5x sinh x x ■ ■ ■ ■ for all x. 4. f x cos x ■ x, 2 20. Prove that the series obtained in Exercise 16 represents sin x |||| Find the Maclaurin series for f x using the definition of a Maclaurin series. [Assume that f has a power series expansion. Do not show that Rn x l 0.] Also find the associated radius of convergence. ■ a all x. 3 3–10 ■ ■ 1 3 ■ 19. Prove that the series obtained in Exercise 3 represents cos x for 2 is not the Taylor series of f centered at 2. 3. f x 2 a 17. f x x 1 ln x, 16. f x 1 a 15. f x f a e x, 14. f x y x, 13. f x is not the Taylor series of f centered at 1. 3 ■ 23–32 |||| Use a Maclaurin series derived in this section to obtain the Maclaurin series for the given function. x ■ ■ ■ x x2 5E-12(pp 806-815) 1/18/06 10:18 AM Page 807 S ECTION 12.10 TAYLOR AND MACLAURIN SERIES sin 2x 29. f x 1 cos 2 x .] 49. lim ■ sin x x 1 31. f x x 32. f x if x 0 if x if x ■ 0 ■ ■ ■ ■ ■ 34. f x 2 ■ ■ x2 e 36. f x ■ ■ 2 ■ ■ e 53. y 0.2 ■ ■ correct to five 55–60 ■ ■ 3 40. dx 1 42. 1 dx ■ ■ ■ ■ ■ y ■ ■ ■ 57. ■ |||| Use series to approximate the definite integral to within the indicated accuracy. y 44. y 45. 46. x cos x 3 d x 0.2 ln 2 ■ 0.5 0 2 xe ■ x2 ■ |||| 47. lim xl0 x f sin x 3 d x ■ error 10 ■ ■ ■ n0 ■ n ■ ■ 2n 2n ! n 58. 1! n0 3 5n n! 81 4! ln 2 2! ■ 6 2n 2 3 ln 2 3! ■ ■ ■ ■ ■ ■ error M for x 2, that is, prove that if d, then a M x 6 0.001) ■ ■ R2 x ) ■ a 3 for x a d 62. (a) Show that the function defined by ■ ■ ■ ■ fx ■ Use series to evaluate the limit. tan 1x x3 x 8 ( dx (five decimal places) ( x3 s1 0 47–49 ■ 1 2n 1 27 3! 9 2! ■ (three decimal places) x3 dx 0.1 ■ 1 tan 0 y ■ x 61. Prove Taylor’s Inequality for n 1 0 y 4 1 2n 60. 1 ■ 43–46 43. ■ 56. 2n 1 59. 3 ■ ■ x 4n n! n n0 sin x dx x ex 1 dx x y ■ e x ln 1 Find the sum of the series. 55. Evaluate the indefinite integral as an infinite series. y sx ■ sec x 54. y ■ |||| n0 41. ■ x 52. y cos x n y x cos x ■ ■ five decimal places. 39. tan x x3 x sin x x ■ x2 51. y 38. Use the Maclaurin series for sin x to compute sin 3 correct to 3 ■ |||| Use multiplication or division of power series to find the first three nonzero terms in the Maclaurin series for each function. decimal places. |||| ■ 51–54 cos x ■ ■ 37. Use the Maclaurin series for e x to calculate e 39–42 ■ ■ ■ ■ ■ We found this limit in Example 4 in Section 7.7 using l’Hospital’s Rule three times. Which method do you prefer? x cos x ■ ■ 0 |||| Find the Maclaurin series of f (by any method) and its radius of convergence. Graph f and its first few Taylor polynomials on the same screen. What do you notice about the relationship between these polynomials and f ? ■ ■ xl0 ; 33–36 35. f x ■ 50. Use the series in Example 10(b) to evaluate if x s1 x3 5 lim sin x x3 ■ 33. f x x ■ 1 6 x 807 0 1 6 ■ sin x xl0 cos 2x 30. f x ■ 1 2 [Hint: Use sin 2x ❙❙❙❙ 1 48. lim xl0 1 cos x x ex ; e 0 1 x2 if x if x 0 0 is not equal to its Maclaurin series. (b) Graph the function in part (a) and comment on its behavior near the origin. 5E-12(pp 806-815) 808 ❙❙❙❙ 1/18/06 10:18 AM Page 808 CHAPTER 12 INFINITE SEQUENCES AND SERIES LABORATORY PROJECT C AS An Elusive Limit This project deals with the function fx sin tan x arcsin arctan x tan sin x arctan arcsin x 1. Use your computer algebra system to evaluate f x for x 1, 0.1, 0.01, 0.001, and 0.0001. Does it appear that f has a limit as x l 0? 2. Use the CAS to graph f near x 0. Does it appear that f has a limit as x l 0? 3. Try to evaluate lim x l 0 f x with l’Hospital’s Rule, using the CAS to find derivatives of the numerator and denominator. What do you discover? How many applications of l’Hospital’s Rule are required? 4. Evaluate lim x l 0 f x by using the CAS to find sufficiently many terms in the Taylor series of the numerator and denominator. (Use the command taylor in Maple or Series in Mathematica.) 5. Use the limit command on your CAS to find lim x l 0 f x directly. (Most computer algebra systems use the method of Problem 4 to compute limits.) 6. In view of the answers to Problems 4 and 5, how do you explain the results of Problems 1 and 2? |||| 12.11 The Binomial Series You may be acquainted with the Binomial Theorem, which states that if a and b are any real numbers and k is a positive integer, then a b k ak ka k 1b kk 1 a k 2b 2 2! kk 1k kk 2 k 1k 3! n 2 1 a k nb n n! kab k 1 a k 3b 3 bk The traditional notation for the binomial coefficients is k 0 1 k n kk 1k 2 k n n! 1 n which enables us to write the Binomial Theorem in the abbreviated form k a b k n0 k knn ab n 1, 2, . . . , k 5E-12(pp 806-815) 1/18/06 10:19 AM Page 809 SECTION 12.11 THE BINOMIAL SERIES In particular, if we put a 1 and b 809 x, we get k 1 1 ❙❙❙❙ kn x n k x n0 One of Newton’s accomplishments was to extend the Binomial Theorem (Equation 1) to the case in which k is no longer a positive integer. (See the Writing Project on page 812.) In this case the expression for 1 x k is no longer a finite sum; it becomes an infinite series. To find this series we compute the Maclaurin series of 1 x k in the usual way: k fx 1 x f0 fx k1 x fx kk 11 x f x kk 1k 21 f0 x kk 1 k n k f0 kk 1 f0 . . . n f0 kk 1k 2 kk 1 k k1 k2 k3 x . . . f 1 n 11 x Therefore, the Maclaurin series of f x f n 0 n! n0 kn 1 x k is 1 kk xn n 1 k n 1 n! n0 xn This series is called the binomial series. If its nth term is a n , then an 1 an kk k n 1 n 1 k n n xn n 1k 1! k n 1 n 1 x 1 1 kk xlx n! k 1 n 1 xn as n l Thus, by the Ratio Test, the binomial series converges if x 1 and diverges if x 1 . The following theorem states that 1 x k is equal to the sum of its Maclaurin series. It is possible to prove this by showing that the remainder term Rn x approaches 0, but that turns out to be quite difficult. The proof outlined in Exercise 19 is much easier. 2 The Binomial Series If k is any real number and x 1 x k 1 k n kk 1 2! x2 kk 1k 3! 2 x3 kn x n n0 where kk kx 1, then 1 k n! n 1 n 1 and k 0 1 5E-12(pp 806-815) 810 ❙❙❙❙ 1/18/06 10:19 AM Page 810 CHAPTER 12 INFINITE SEQUENCES AND SERIES 1, the question of whether Although the binomial series always converges when x or not it converges at the endpoints, 1, depends on the value of k. It turns out that the series converges at 1 if 1 k 0 and at both endpoints if k 0. Notice that if k is a k k positive integer and n k, then the expression for ( n ) contains a factor k k , so ( n ) 0 for n k. This means that the series terminates and reduces to the ordinary Binomial Theorem (Equation 1) when k is a positive integer. As we have seen, the binomial series is just a special case of the Maclaurin series; it occurs so frequently that it is worth remembering. 1 EXAMPLE 1 Expand 1 x 2 as a power series. SOLUTION We use the binomial series with k 2 n 2 3 2. The binomial coefficient is 4 2 n 1 n! 1 n2 3 4 nn 1 1nn n! and so, when x 1 1, 1 1 x 1 2 2n x n 2 x n0 1nn 1 xn 1 3x 2 2x 4x3 n0 1 EXAMPLE 2 Find the Maclaurin series for the function f x s4 convergence. SOLUTION As given, f x is not quite of the form 1 1 s4 1 x x 4 1 2 Using the binomial series with k 1 s4 x 1 2 1 1 2 x 4 1 12 1 2 1 2 1 2 n0 x 4 2 1 2 n! 1 2 1 1 x 8 132 x 2!8 2 x 4 2! ( x 4 1353 x 3!8 3 12 x 4 1 x 4, we have n ( 1 )( 3 ) 2 2 ( 1)( 3)( 5) 2 2 2 1 2 and with x replaced by x 4 n 1 and its radius of x k so we rewrite it as follows: 1 41 x n 2 ( 1)( 3)( 5) 2 2 2 x 4 3! 1) x 4 3 n 135 2n n!8 n 1 xn 5E-12(pp 806-815) 1/18/06 10:19 AM Page 811 SECTION 12.11 THE BINOMIAL SERIES x4 We know from (2) that this series converges when radius of convergence is R 4. y y= 1 1, that is, x 1 œ4-x œ„„„„ ❙❙❙❙ 811 4, so the T£ T™ T¡ |||| A binomial series is a special case of a Taylor series. Figure 1 shows the graphs of the first three Taylor polynomials computed from the answer to Example 2. 0 _4 x 4 FIGURE 1 |||| 12.11 1–8 Exercises x 1 x 2 as a power series. (b) Use part (a) to find the sum of the series 15. (a) Expand f x Use the binomial series to expand the function as a power series. State the radius of convergence. |||| 1. s1 1 3. 2 x 4 5. s1 7. 2. x ■ ; 9–10 x n1 x x 2 1 x 3 as a power series. (b) Use part (a) to find the sum of the series 16. (a) Expand f x 23 1 5 s32 x x2 8. s2 x x2 ■ x 4 6. 8x x s4 1 4. 1 3 ■ ■ ■ ■ ■ ■ ■ n1 ■ ■ ■ ■ 34 ■ 3 10. s1 ■ ■ ■ ■ ■ fx 1 s1 x 3. (b) Use part (a) to evaluate f ■ (a) Let t x ■ ■ 11. (a) Use the binomial series to expand 1 s1 x 2. (b) Use part (a) to find the Maclaurin series for sin 1x. 12. (a) Use the binomial series to expand 1 s1 x 2. (b) Use part (a) to find the Maclaurin series for sinh 1x. 13. (a) Expand s1 n0 4 14. (a) Expand 1 s1 x as a power series. 4 (b) Use part (a) to estimate 1 s1.1 correct to three decimal places. 0. 9 0. k ( n )x n. Differentiate this series to show that kt x 1x tx ■ (b) Let h x 1x (c) Deduce that t x 1 x 1 k t x and show that h x 1 x k. 0. 20. In Exercise 53 in Section 11.2 it was shown that the length of the ellipse x a sin , y 3 x as a power series. 3 (b) Use part (a) to estimate s1.01 correct to four decimal places. 10 19. Use the following steps to prove (2). 4x ■ fx s1 x 2. (b) Use part (a) to evaluate f 18. (a) Use the binomial series to find the Maclaurin series of Use the binomial series to expand the function as a Maclaurin series and to find the first three Taylor polynomials T1, T2, and T3. Graph the function and these Taylor polynomials in the interval of convergence. 2x n2 2n 17. (a) Use the binomial series to find the Maclaurin series of |||| 9. 1 n 2n 1 L 4a y 0 2 a cos , where a s1 b 0, is e 2 sin2 d where e sa 2 b 2 a is the eccentricity of the ellipse. Expand the integrand as a binomial series and use the result of Exercise 44 in Section 8.1 to express L as a series in powers of the eccentricity up to the term in e 6. 5E-12(pp 806-815) 812 ❙❙❙❙ 1/18/06 10:19 AM Page 812 CHAPTER 12 INFINITE SEQUENCES AND SERIES WRITING PROJECT H ow Newton Discovered the Binomial Series The Binomial Theorem, which gives the expansion of a b k, was known to Chinese mathematicians many centuries before the time of Newton for the case where the exponent k is a positive integer. In 1665, when he was 22, Newton was the first to discover the infinite series expansion of a b k when k is a fractional exponent (positive or negative). He didn’t publish his discovery, but he stated it and gave examples of how to use it in a letter (now called the epistola prior) dated June 13, 1676, that he sent to Henry Oldenburg, secretary of the Royal Society of London, to transmit to Leibniz. When Leibniz replied, he asked how Newton had discovered the binomial series. Newton wrote a second letter, the epistola posterior of October 24, 1676, in which he explained in great detail how he arrived at his discovery by a very indirect route. He was investigating the areas under the curves y 1 x 2 n 2 from 0 to x for n 0, 1, 2, 3, 4, . . . . These are easy to calculate if n is even. By observing patterns and interpolating, Newton was able to guess the answers for odd values of n. Then he realized he could get the same answers by expressing 1 x 2 n 2 as an infinite series. Write a report on Newton’s discovery of the binomial series. Start by giving the statement of the binomial series in Newton’s notation (see the epistola prior on page 285 of [4] or page 402 of [2]). Explain why Newton’s version is equivalent to Theorem 2 on page 809. Then read Newton’s epistola posterior (page 287 in [4] or page 404 in [2]) and explain the patterns that Newton discovered in the areas under the curves y 1 x 2 n 2. Show how he was able to guess the areas under the remaining curves and how he verified his answers. Finally, explain how these discoveries led to the binomial series. The books by Edwards [1] and Katz [3] contain commentaries on Newton’s letters. 1. C. H. Edwards, The Historical Development of the Calculus (New York: Springer-Verlag, 1979), pp. 178–187. 2. John Fauvel and Jeremy Gray, eds., The History of Mathematics: A Reader (London: MacMillan Press, 1987). 3. Victor Katz, A History of Mathematics: An Introduction (New York: HarperCollins, 1993), pp. 463–466. 4. D. J. Struik, ed., A Sourcebook in Mathematics, 1200–1800 (Princeton, N.J.: Princeton University Press, 1969). |||| 12.12 Applications of Taylor Polynomials In this section we explore two types of applications of Taylor polynomials. First we look at how they are used to approximate functions––computer scientists like them because polynomials are the simplest of functions. Then we investigate how physicists and engineers use them in such fields as relativity, optics, blackbody radiation, electric dipoles, the velocity of water waves, and building highways across a desert. Approximating Functions by Polynomials Suppose that f x is equal to the sum of its Taylor series at a: f fx n0 n n! a x a n 5E-12(pp 806-815) 1/18/06 10:20 AM Page 813 ❙❙❙❙ S ECTION 12.12 APPLICATIONS OF TAYLOR POLYNOMIALS 813 In Section 12.10 we introduced the notation Tn x for the nth partial sum of this series and called it the n th-degree Taylor polynomial of f at a. Thus n Tn x i0 fa f i i! a x fa 1! a i x a fa 2! x a f 2 n a n! Since f is the sum of its Taylor series, we know that Tn x l f x as n l be used as an approximation to f : f x Tn x . Notice that the first-degree Taylor polynomial x a n and so Tn can y T1 x y=´ y=T£(x) y=T™(x) y=T™(x) (0, 1) y=T¡(x) 0 x y=T£(x) FIGURE 1 x 0.2 x 3.0 T2 x T4 x T6 x T8 x T10 x 1.220000 1.221400 1.221403 1.221403 1.221403 8.500000 16.375000 19.412500 20.009152 20.079665 ex 1.221403 20.085537 fa fax a is the same as the linearization of f at a that we discussed in Section 3.10. Notice also that T1 and its derivative have the same values at a that f and f have. In general, it can be shown that the derivatives of Tn at a agree with those of f up to and including derivatives of order n (see Exercise 36). To illustrate these ideas let’s take another look at the graphs of y e x and its first few Taylor polynomials, as shown in Figure 1. The graph of T1 is the tangent line to y e x at 0, 1 ; this tangent line is the best linear approximation to e x near 0, 1 . The graph of T2 is the parabola y 1 x x 2 2, and the graph of T3 is the cubic curve y 1 x x 2 2 x 3 6, which is a closer fit to the exponential curve y e x than T2. The next Taylor polynomial T4 would be an even better approximation, and so on. The values in the table give a numerical demonstration of the convergence of the Taylor polynomials Tn x to the function y e x. We see that when x 0.2 the convergence is very rapid, but when x 3 it is somewhat slower. In fact, the farther x is from 0, the more slowly Tn x converges to e x. When using a Taylor polynomial Tn to approximate a function f , we have to ask the questions: How good an approximation is it? How large should we take n to be in order to achieve a desired accuracy? To answer these questions we need to look at the absolute value of the remainder: Rn x fx Tn x There are three possible methods for estimating the size of the error: 1. If a graphing device is available, we can use it to graph Rn x and thereby estimate the error. 2. If the series happens to be an alternating series, we can use the Alternating Series Estimation Theorem. 3. In all cases we can use Taylor’s Inequality (Theorem 12.10.9), which says that if fn1 x M , then Rn x M n 1! x a n1 EXAMPLE 1 3 (a) Approximate the function f x sx by a Taylor polynomial of degree 2 at a (b) How accurate is this approximation when 7 x 9? 8. 5E-12(pp 806-815) 814 ❙❙❙❙ 1/18/06 10:20 AM Page 814 CHAPTER 12 INFINITE SEQUENCES AND SERIES S OLUTION (a) 1 3 fx 2 9 10 27 x f8 x 53 2 f8 23 x fx f x1 3 3 sx fx 1 12 1 144 f8 83 x Thus, the second-degree Taylor polynomial is T2 x f8 1! f8 2 1 12 x x 1 288 8 f8 2! 8 x 8 x 8 2 2 The desired approximation is 3 sx T2 x 1 12 2 x 1 288 8 x 8 2 (b) The Taylor series is not alternating when x 8, so we can’t use the Alternating Series Estimation Theorem in this example. But we can use Taylor’s Inequality with n 2 and a 8: M x 3! R2 x where f x M . Because x f x 7, we have x 8 3 10 27 1 x 10 27 83 Therefore, we can take M 0.0021. Also 7 x8 1. Then Taylor’s Inequality gives 2.5 T™ R2 x 8 0.0021 3! x 3 7 8 3 and so 1 78 3 0.0021 9, so 1 0.0021 6 13 x 8 1 and 0.0004 #„ y= œx Thus, if 7 x 9, the approximation in part (a) is accurate to within 0.0004. 15 0 Let’s use a graphing device to check the calculation in Example 1. Figure 2 shows that 3 the graphs of y sx and y T2 x are very close to each other when x is near 8. Figure 3 shows the graph of R2 x computed from the expression FIGURE 2 0.0003 R2 x y=|R™(x)| T2 x We see from the graph that R2 x 7 3 sx 0.0003 9 0 FIGURE 3 when 7 x 9. Thus, the error estimate from graphical methods is slightly better than the error estimate from Taylor’s Inequality in this case. 5E-12(pp 806-815) 1/18/06 10:20 AM Page 815 S ECTION 12.12 APPLICATIONS OF TAYLOR POLYNOMIALS ❙❙❙❙ 815 EXAMPLE 2 (a) What is the maximum error possible in using the approximation sin x x3 3! x x5 5! when 0.3 x 0.3? Use this approximation to find sin 12 correct to six decimal places. (b) For what values of x is this approximation accurate to within 0.00005? SOLUTION (a) Notice that the Maclaurin series sin x x3 3! x x5 5! x7 7! is alternating for all nonzero values of x, and the successive terms decrease in size because x 1, so we can use the Alternating Series Estimation Theorem. The error in approximating sin x by the first three terms of its Maclaurin series is at most x7 7! If 0.3 x 0.3, then x x7 5040 0.3, so the error is smaller than 0.3 7 5040 4.3 10 8 To find sin 12 we first convert to radian measure. sin 12 sin 12 180 sin 3 15 15 15 5 1 3! 15 1 5! 0.20791169 Thus, correct to six decimal places, sin 12 0.207912. (b) The error will be smaller than 0.00005 if x7 5040 0.00005 Solving this inequality for x, we get x 7 0.252 or x 0.252 17 0.821 So the given approximation is accurate to within 0.00005 when x Module 12.10/12.12 graphically shows the remainders in Taylor polynomial approximations. What if we use Taylor’s Inequality to solve Example 2? Since f have f 7 x 1 and so 1 R6 x x7 7! 0.82. 7 x cos x, we So we get the same estimates as with the Alternating Series Estimation Theorem. 5E-12(pp 816-824) 816 ❙❙❙❙ 1/18/06 10:22 AM Page 816 CHAPTER 12 INFINITE SEQUENCES AND SERIES 4.3 What about graphical methods? Figure 4 shows the graph of 10–* R6 x (x sin x y=| Rß(x)| _0.3 0.3 0 FIGURE 4 0.00006 y=0.00005 1 6 x3 T1 x 1 0 FIGURE 5 x5) and we see from it that R6 x 4.3 10 8 when x 0.3. This is the same estimate that we obtained in Example 2. For part (b) we want R6 x 0.00005, so we graph both y R6 x and y 0.00005 in Figure 5. By placing the cursor on the right intersection point we find that the inequality is satisfied when x 0.82. Again this is the same estimate that we obtained in the solution to Example 2. If we had been asked to approximate sin 72 instead of sin 12 in Example 2, it would have been wise to use the Taylor polynomials at a 3 (instead of a 0) because they are better approximations to sin x for values of x close to 3. Notice that 72 is close to 60 (or 3 radians) and the derivatives of sin x are easy to compute at 3. Figure 6 shows the graphs of the Maclaurin polynomial approximations y=| Rß(x)| _1 1 120 T5 x x T3 x x3 3! x5 5! x x3 3! T7 x x x x3 3! x5 5! x7 7! to the sine curve. You can see that as n increases, Tn x is a good approximation to sin x on a larger and larger interval. y T¡ T∞ x 0 y=sin x T£ FIGURE 6 T¶ One use of the type of calculation done in Examples 1 and 2 occurs in calculators and computers. For instance, when you press the sin or e x key on your calculator, or when a computer programmer uses a subroutine for a trigonometric or exponential or Bessel function, in many machines a polynomial approximation is calculated. The polynomial is often a Taylor polynomial that has been modified so that the error is spread more evenly throughout an interval. Applications to Physics Taylor polynomials are also used frequently in physics. In order to gain insight into an equation, a physicist often simplifies a function by considering only the first two or three terms in its Taylor series. In other words, the physicist uses a Taylor polynomial as an approximation to the function. Taylor’s Inequality can then be used to gauge the accuracy of the approximation. The following example shows one way in which this idea is used in special relativity. EXAMPLE 3 In Einstein’s theory of special relativity the mass of an object moving with velocity v is m0 m s1 v 2 c 2 where m0 is the mass of the object when at rest and c is the speed of light. The kinetic 5E-12(pp 816-824) 1/18/06 10:22 AM Page 817 S ECTION 12.12 APPLICATIONS OF TAYLOR POLYNOMIALS ❙❙❙❙ 817 energy of the object is the difference between its total energy and its energy at rest: mc 2 K m0 c 2 (a) Show that when v is very small compared with c, this expression for K agrees with classical Newtonian physics: K 1 m0v 2. 2 (b) Use Taylor’s Inequality to estimate the difference in these expressions for K when v 100 m s. SOLUTION (a) Using the expressions given for K and m, we get mc 2 |||| The upper curve in Figure 7 is the graph of the expression for the kinetic energy K of an object with velocity v in special relativity. The lower curve shows the function used for K in classical Newtonian physics. When v is much smaller than the speed of light, the curves are practically identical. 1 12 x 1 1 2 1 1 and 1 c2 1 2 ( 1 )( 3 ) x 2 2 2 x 3 8 1 ( 1 )( 3 )( 5) x 3 2 2 2 2! x 5 16 2 1 v2 2 c2 m0 c 2 K x 1 2 is most easily computed as a 1 because v c.) Therefore, we have 1 v2 2 c2 x 3! 3 3 v4 8 c4 3 v4 8 c4 5 v6 16 c 6 1 5 v6 16 c 6 If v is much smaller than c, then all terms after the first are very small when compared with the first term. If we omit them, we get 1 K =2 m¸√ @ c 12 x m0 c 2 0 v2 m0 c 2 v2 c2 s1 With x v 2 c 2, the Maclaurin series for 1 1 binomial series with k 2 . (Notice that x K K=mc@-m¸c@ m0 c 2 m0 c 2 m0 c 2 K √ m0 c 2 K FIGURE 7 (b) If x fx 1 v2 2 c2 1 2 m0 v 2 v 2 c 2, f x m0 c 2 1 x 1 2 1 , and M is a number such that M , then we can use Taylor’s Inequality to write M2 x 2! R1 x 3 4 We have f x m0 c 2 1 x and we are given that v 3m0 c 2 4 1 v2 c2 fx 3 1 2 52 41 3m0 c 2 100 2 c 2 100 m s, so M 52 10 8 m s, R1 x Thus, with c 52 41 3m0 c 2 100 2 c 2 52 100 4 c4 4.17 10 10 m0 So when v 100 m s, the magnitude of the error in using the Newtonian expression for kinetic energy is at most 4.2 10 10 m0. 5E-12(pp 816-824) 818 ❙❙❙❙ 1/18/06 10:22 AM Page 818 CHAPTER 12 INFINITE SEQUENCES AND SERIES Another application to physics occurs in optics. Figure 8 is adapted from Optics, 4th ed., by Eugene Hecht (Reading, MA: Addison-Wesley, 2002), page 153. It depicts a wave from the point source S meeting a spherical interface of radius R centered at C. The ray SA is refracted toward P. ¨r A ¨i Lo h R V ˙ ¨t Li S C so P si n¡ FIGURE 8 n™ Refraction at a spherical interface Using Fermat’s principle that light travels so as to minimize the time taken, Hecht derives the equation n1 n2 o 1 i 1 R n2si n1so i o where n1 and n2 are indexes of refraction and o , i , so , and si are the distances indicated in Figure 8. By the Law of Cosines, applied to triangles ACS and ACP, we have cos o cos sR 2 so R 2 2R so R cos i 2 |||| Here we use the identity sR 2 si R 2 2R si R cos Because Equation 1 is cumbersome to work with, Gauss, in 1841, simplified it by using 1 for small values of . (This amounts to using the the linear approximation cos Taylor polynomial of degree 1.) Then Equation 1 becomes the following simpler equation [as you are asked to show in Exercise 32(a)]: n1 so 3 n2 si n2 n1 R The resulting optical theory is known as Gaussian optics, or first-order optics, and has become the basic theoretical tool used to design lenses. A more accurate theory is obtained by approximating cos by its Taylor polynomial of degree 3 (which is the same as the Taylor polynomial of degree 2). This takes into account rays for which is not so small, that is, rays that strike the surface at greater distances h above the axis. In Exercise 32(b) you are asked to use this approximation to derive the more accurate equation 4 n1 so n2 si n2 n1 R h2 n1 2so 1 so 1 R 2 n2 2si 1 R 1 si 2 The resulting optical theory is known as third-order optics. Other applications of Taylor polynomials to physics and engineering are explored in Exercises 30, 31, 33, 34, and 35 and in the Applied Project on page 821. 5E-12(pp 816-824) 1/18/06 10:23 AM Page 819 SECTION 12.12 APPLICATIONS OF TAYLOR POLYNOMIALS |||| 12.12 20. f x a ln x, 4. f x x e, 5. f x sin x, a 6. f x cos x, a 7. f x arcsin x, 8. f x ln x , x 10. f x ■ a 1, a 2x xe n 2, 2 a 1, a , 0, s3 x, a ■ ■ n ■ 27. sin x 12. f x tan x, n ■ ■ n n ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 4, n 2, 4 a 1, n 2, 0.9 x ■ a 1, n 3, 0.8 x 15. f x x 23 16. f x cos x, a 17. f x tan x, a 18. f x ln 1 2x , 19. f x x2 e, , a 3, n 0, 0, n a n ■ ■ ■ 0.01) error x4 24 ( ■ error ■ x t 1.2 x, x2 2 ( 0.005) ■ ■ ■ ■ ■ ■ of a conducting wire is the reciprocal of the conductivity and is measured in units of ohm-meters ( -m). The resistivity of a given metal depends on the temperature according to the equation ■ 1.1 14. f x x do you 30. The resistivity ■ a 2 ■ at a given instant. Using a second-degree Taylor polynomial, estimate how far the car moves in the next second. Would it be reasonable to use this polynomial to estimate the distance traveled during the next minute? |||| sx, ■ 29. A car is moving with speed 20 m s and acceleration 2 m s2 (a) Approximate f by a Taylor polynomial with degree n at the number a. (b) Use Taylor’s Inequality to estimate the accuracy of the approximation f x Tn x when x lies in the given interval. ; (c) Check your result in part (b) by graphing Rn x . 13. f x 1 ■ 1, 3, 5, 7, 9 ■ x3 6 28. cos x 2 2, 4, 6, 8 ■ x 3 |||| Use a computer algebra system to find the Taylor polynomials Tn at a 0 for the given values of n. Then graph these polynomials and f on the same screen. sec x, ■ |||| Use the Alternating Series Estimation Theorem or Taylor’s Inequality to estimate the range of values of x for which the given approximation is accurate to within the stated error. Check your answer graphically. 11–12 11. f x 1 ■ ; 27–28 4 3 1, x ■ 1 the Maclaurin series for e x that should be used to estimate e 0.1 to within 0.00001. 3 n 2 1 ■ x 25. Use Taylor’s Inequality to determine the number of terms of 3 n n 5, ■ 1 26. How many terms of the Maclaurin series for ln 1 n 0, n 0, ■ 4, 1.5 need to use to estimate ln 1.4 to within 0.001? 3, a n x correct to five decimal places. 3 6, ■ 0, 0.5 24. Use the information from Exercise 16 to estimate cos 69 4 n ■ 3, n correct to five decimal places. |||| 3. f x 13–22 sinh 2 x, ■ 1, a 23. Use the information from Exercise 5 to estimate sin 35 Find the Taylor polynomial Tn x for the function f at the number a. Graph f and Tn on the same screen. ■ a ■ 1x centered at a 1. Graph f and these polynomials on a common screen. (b) Evaluate f and these polynomials at x 0.9 and 1.3. (c) Comment on how the Taylor polynomials converge to f x . CAS x sin x, 22. f x ; 2. (a) Find the Taylor polynomials up to degree 3 for f x ■ x ln x, 21. f x fx cos x centered at a 0. Graph f and these polynomials on a common screen. (b) Evaluate f and these polynomials at x 4, 2, and . (c) Comment on how the Taylor polynomials converge to f x . 9. f x 819 Exercises ; 1. (a) Find the Taylor polynomials up to degree 6 for ; 3–10 ❙❙❙❙ 4, 3, 1, n 3, 0 4.2 0 0 x x 3, x 2 3 6 0.5 0.1 x ; 1.5 20 e t 20 where t is the temperature in C. There are tables that list the values of (called the temperature coefficient) and 20 (the resistivity at 20 C) for various metals. Except at very low temperatures, the resistivity varies almost linearly with temperature and so it is common to approximate the expression for t by its first- or second-degree Taylor polynomial at t 20. (a) Find expressions for these linear and quadratic approximations. 0.0039 C and (b) For copper, the tables give 1.7 10 8 -m. Graph the resistivity of copper 20 and the linear and quadratic approximations for 250 C t 1000 C. 5E-12(pp 816-824) ❙❙❙❙ 820 ; 1/18/06 10:24 AM Page 820 CHAPTER 12 INFINITE SEQUENCES AND SERIES (c) For what values of t does the linear approximation agree with the exponential expression to within one percent? 31. An electric dipole consists of two electric charges of equal magnitude and opposite signs. If the charges are q and q and are located at a distance d from each other, then the electric field E at the point P in the figure is E q q D2 D d 2 By expanding this expression for E as a series in powers of d D, show that E is approximately proportional to 1 D 3 when P is far away from the dipole. q _q P d D 32. (a) Derive Equation 3 for Gaussian optics from Equation 1 by approximating cos in Equation 2 by its first-degree Taylor polynomial. (b) Show that if cos is replaced by its third-degree Taylor polynomial in Equation 2, then Equation 1 becomes Equation 4 for third-order optics. [Hint: Use the first two terms in the binomial series for o 1 and i 1. Also, use sin .] 33. If a water wave with length L moves with velocity v across a body of water with depth d, as in the figure, then tL 2d tanh 2 L v2 (a) If the water is deep, show that v stL 2 . (b) If the water is shallow, use the Maclaurin series for tanh to show that v std. (Thus, in shallow water the velocity of a wave tends to be independent of the length of the wave.) (c) Use the Alternating Series Estimation Theorem to show that if L 10d, then the estimate v 2 td is accurate to within 0.014tL. 0 L (1 t 1 4 k2) T L4 t4 2 3k 2 4k 2 (c) Use the inequalities in part (b) to estimate the period of a pendulum with L 1 meter and 0 10 . How does it compare with the estimate T 2 sL t ? What if 0 42 ? 35. If a surveyor measures differences in elevation when making plans for a highway across a desert, corrections must be made for the curvature of the Earth. (a) If R is the radius of the Earth and L is the length of the highway, show that the correction is C R sec L R R (b) Use a Taylor polynomial to show that L2 2R C 5L 4 24R 3 (c) Compare the corrections given by the formulas in parts (a) and (b) for a highway that is 100 km long. (Take the radius of Earth to be 6370 km.) C L R R mating a root r of the equation f x 0, and from an initial approximation x 1 we obtained successive approximations x 2 , x 3 , . . . , where with the vertical is T L t 4 y xn 2 0 s1 dx k 2 sin 2x where k sin( 1 0 ) and t is the acceleration due to gravity. (In 2 Exercise 40 in Section 8.7 we approximated this integral using Simpson’s Rule.) (a) Expand the integrand as a binomial series and use the result of Exercise 44 in Section 8.1 to show that 2 2 37. In Section 4.9 we considered Newton’s method for approxi- 34. The period of a pendulum with length L that makes a maxi- T (b) Notice that all the terms in the series after the first one have coefficients that are at most 1. Use this fact to compare this 4 series with a geometric series and show that 36. Show that Tn and f have the same derivatives at a up to order n. L d mum angle If 0 is not too large, the approximation T 2 sL t, obtained by using only the first term in the series, is often used. A better approximation is obtained by using two terms: L T2 (1 1 k 2 ) 4 t L t 1 12 2 k 22 12 3 2 4 k 2 242 12 3 25 2 6 k 2 2426 2 xn 1 f xn f xn Use Taylor’s Inequality with n 1, a x n, and x r to show that if f x exists on an interval I containing r, x n , and x n 1, and f x M, f x K for all x I , then xn 1 r M xn 2K r 2 [This means that if x n is accurate to d decimal places, then x n 1 is accurate to about 2d decimal places. More precisely, if the error at stage n is at most 10 m, then the error at stage n 1 is at most M 2K 10 2 m.] 5E-12(pp 816-824) 1/18/06 10:24 AM Page 821 A PPLIED PROJECT RADIATION FROM THE STARS ❙❙❙❙ 821 A PPLIED PROJECT Radiation from the Stars Any object emits radiation when heated. A blackbody is a system that absorbs all the radiation that falls on it. For instance, a matte black surface or a large cavity with a small hole in its wall (like a blastfurnace) is a blackbody and emits blackbody radiation. Even the radiation from the Sun is close to being blackbody radiation. Proposed in the late 19th century, the Rayleigh-Jeans Law expresses the energy density of blackbody radiation of wavelength as 8 kT f 4 where is measured in meters, T is the temperature in kelvins (K), and k is Boltzmann’s constant. The Rayleigh-Jeans Law agrees with experimental measurements for long wavelengths but disagrees drastically for short wavelengths. [The law predicts that f l as l 0 but experiments have shown that f l 0.] This fact is known as the ultraviolet catastrophe. In 1900 Max Planck found a better model (known now as Planck’s Law) for blackbody radiation: 8 hc f where e hc 5 kT 1 is measured in meters, T is the temperature (in kelvins), and h Planck’s constant 6.6262 c speed of light k Boltzmann’s constant 10 34 Js 8 2.997925 10 m s 1.3807 10 23 JK 1. Use l’Hospital’s Rule to show that lim f l0 0 and lim f l 0 for Planck’s Law. So this law models blackbody radiation better than the Rayleigh-Jeans Law for short wavelengths. 2. Use a Taylor polynomial to show that, for large wavelengths, Planck’s Law gives approxi- mately the same values as the Rayleigh-Jeans Law. ; 3. Graph f as given by both laws on the same screen and comment on the similarities and differences. Use T 5700 K (the temperature of the Sun). (You may want to change from meters to the more convenient unit of micrometers: 1 m 10 6 m.) 4. Use your graph in Problem 3 to estimate the value of for which f is a maximum under Planck’s Law. ; 5. Investigate how the graph of f changes as T varies. (Use Planck’s Law.) In particular, graph f for the stars Betelgeuse (T 3400 K), Procyon (T 6400 K), and Sirius (T 9200 K) as well as the Sun. How does the total radiation emitted (the area under the curve) vary with T ? Use the graph to comment on why Sirius is known as a blue star and Betelgeuse as a red star. 5E-12(pp 816-824) ❙❙❙❙ 822 |||| 1/18/06 10:24 AM Page 822 CHAPTER 12 INFINITE SEQUENCES AND SERIES 12 Review ■ CONCEPT CHECK 1. (a) What is a convergent sequence? (c) If a series is convergent by the Alternating Series Test, how do you estimate its sum? (b) What is a convergent series? (c) What does lim n l an 3 mean? (d) What does n 1 an 3 mean? 8. (a) Write the general form of a power series. (b) What is the radius of convergence of a power series? (c) What is the interval of convergence of a power series? 2. (a) What is a bounded sequence? (b) What is a monotonic sequence? (c) What can you say about a bounded monotonic sequence? 3. (a) What is a geometric series? Under what circumstances is it 9. Suppose f x is the sum of a power series with radius of con- vergence R. (a) How do you differentiate f ? What is the radius of convergence of the series for f ? (b) How do you integrate f ? What is the radius of convergence of the series for x f x d x? convergent? What is its sum? (b) What is a p-series? Under what circumstances is it convergent? 4. Suppose an 3 and sn is the nth partial sum of the series. What is lim n l an? What is lim n l sn? 10. (a) Write an expression for the nth-degree Taylor polynomial 5. State the following. (a) (b) (c) (d) (e) (f) (g) of f centered at a. (b) Write an expression for the Taylor series of f centered at a. (c) Write an expression for the Maclaurin series of f . (d) How do you show that f x is equal to the sum of its Taylor series? (e) State Taylor’s Inequality. The Test for Divergence The Integral Test The Comparison Test The Limit Comparison Test The Alternating Series Test The Ratio Test The Root Test 11. Write the Maclaurin series and the interval of convergence for 6. (a) What is an absolutely convergent series? each of the following functions. (a) 1 1 x (b) e x (c) sin x (d) cos x (e) tan 1x (b) What can you say about such a series? (c) What is a conditionally convergent series? 7. (a) If a series is convergent by the Integral Test, how do you estimate its sum? (b) If a series is convergent by the Comparison Test, how do you estimate its sum? ■ 2. The series 3. If lim n l a n 0, then n1 n sin 1 a n is convergent. 1 4. If cn6 n is convergent, then cn 5. If cn6 n is convergent, then cn 6 n is convergent. 6. If cn x n diverges when x 6, then it diverges when x 7. The Ratio Test can be used to determine whether 1 n3 converges. 8. The Ratio Test can be used to determine whether n0 n 1 e 1 a n is divergent, then f0 2x 2. x2 bn diverges, then 1 3 0. a n is divergent. x3 converges for all x, then 14. If a n and bn are divergent, then a n 10. bn is divergent. 15. If a n and bn are divergent, then a n bn is divergent. 16. If a n is decreasing and a n 0 for all n, then a n is 1 n! 17. If a n bn and n 1, then lim n l convergent. converges. an 1 n! 10. 13. If f x 2 n is convergent. 9. If 0 ■ 12. If L. x k. What is the radius of convergence of this series? 11. If is convergent. L, then lim n l a 2 n 12. Write the binomial series expansion of 1 TRUE-FALSE QUIZ Determine whether the statement is true or false. If it is true, explain why. If it is false, explain why or give an example that disproves the statement. 1. If lim n l a n ■ a n diverges. 0 and 18. If a n 0 and lim n l a n converges, then an 1 an 1 n a n converges. 1, then lim n l a n 0. 5E-12(pp 816-824) 1/18/06 10:25 AM Page 823 CHAPTER 12 REVIEW ■ EXERCISES 1–8 |||| Determine whether the sequence is convergent or divergent. If it is convergent, find its limit. n3 2n 3 2 1 1. a n 2. a n 4. a n 1 n 5. a n 7. 3n ■ ■ cos n 6. a n 2 n sin n n2 1 1 27–31 2 ln n sn n1 4n 8. ■ ■ ■ ■ ■ n 10 ■ 2 1 n0 n! 31. 1 ■ ■ 9. A sequence is defined recursively by the equations a 1 a n 1 1 a n 4 . Show that a n is increasing and a n 3 all n. Deduce that a n is convergent and find its limit. 1, 2 for n1 n 3 12. 1 n1 n2 n3 14. n1 n 5n 16. n2 1 n s ln n 18. n1 cos 3n 1 1.2 13. 15. 17. n1 2n n1 sn 22. ■ n2n 1 2n2 ■ 1 n1 ■ ■ n ■ 1 2 1 ■ 13 n1 5n 39. Prove that if the series sn n1 1 ■ ■ 24. 1 3n n ■ n1 ■ ■ ■ ■ ■ ■ ■ ■ n1 ln x n converge? 1 correct to four n1 1 . Estimate the error involved in this nn is convergent. 2n ! 0. n1 an is absolutely convergent, then the series 1 ■ ■ 1 n n 40–43 |||| Find the radius of convergence and interval of convergence of the series. 41. 2n x 2 n n 2! 43. 40. 1 nsn ln n ■ xn n 25 n n1 3 1 42. ■ ■ ■ an is also absolutely convergent. ■ n1 n1 1 26. 1 ■ 1n n5 n ■ ■ x 2 for all x. n1 n1 1nn 22n ■ n1 ■ n1 ■ 1 n1 ■ ■ nn (b) Deduce that lim nl 2n ! |||| Determine whether the series is conditionally convergent, absolutely convergent, or divergent. 25. ■ 38. (a) Show that the series n 23–26 23. e4 4! n n1 ■ ■ the series n 1 2 approximation. 1 n1 sn e3 3! 37. Use the sum of the first eight terms to approximate the sum of n 3n 21. 1 e2 2! e 1 5 nn! 5 2n n 9n 2n ! 35. Find the sum of the series 1 2 n 1 n 6 and estimate the error in using it as an approximation to the sum of the series. (b) Find the sum of this series correct to five decimal places. n1 n n1 20. ln n 34. For what values of x does the series n sn 3 2n 36. (a) Find the partial sum s5 of the series n1 135 19. 1 3 tan 1n 1 decimal places. 1 1 3 n 33. Show that cosh x n n 11. 1 nn 32. Express the repeating decimal 4.17326326326 . . . as a fraction. Determine whether the series is convergent or divergent. |||| n1 ■ 0 and use a graph to find the smallest value of N that corresponds to 0.1 in the precise definition of a limit. 11–22 1 28. 5n tan ■ 4 ; 10. Show that lim n l n e 2n 1 n1 30. ■ Find the sum of the series. |||| 29. n3 3. a n 823 ■ 27. 9n 1 10 n ❙❙❙❙ ■ ■ n ■ x n1 n0 ■ ■ ■ ■ ■ 2 n4 n n 2n x sn 3n 3 ■ ■ ■ ■ 5E-12(pp 816-824) 824 ❙❙❙❙ 1/18/06 10:26 AM Page 824 CHAPTER 12 INFINITE SEQUENCES AND SERIES 44. Find the radius of convergence of the series n1 57. f x 58. f x 2n ! n x n! 2 s x, sec x, ■ 45. Find the Taylor series of f x sin x at a 6. 46. Find the Taylor series of f x cos x at a ■ a ■ 1, a n 0, n ■ ■ xl0 47–54 |||| Find the Maclaurin series for f and its radius of convergence. You may use either the direct method (definition of a Maclaurin series) or known series such as geometric series, binomial series, or the Maclaurin series for e x, sin x, and tan 1x. 1 48. f x 50. f x 49. f x ln 1 51. f x 53. f x 4 1 s16 ■ ■ ■ 55. Evaluate y ■ x ■ ■ 1 ■ ■ ■ 3x ; 5 ■ ■ ■ x 4 d x correct to two decimal places. 57–58 6 ■ ■ ■ ■ ■ sin x x x3 mtR2 Rh F x2 ex d x as an infinite series. x 1 56. Use series to approximate x0 s1 x ■ above the surface of the Earth is 10 x 54. f x sin x 4 x 1 0 1.1 60. The force due to gravity on an object with mass m at a height h xe 2 x 52. f x x tan 2, x 59. Use series to evaluate the following limit. lim x2 0.9 ■ 3. 47. f x 3, where R is the radius of the Earth and t is the acceleration due to gravity. (a) Express F as a series in powers of h R. (b) Observe that if we approximate F by the first term in the series, we get the expression F m t that is usually used when h is much smaller than R. Use the Alternating Series Estimation Theorem to estimate the range of values of h for which the approximation F m t is accurate to within 1%. (Use R 6400 km.) cn x n for all x. (a) If f is an odd function, show that 61. Suppose that f x n0 c0 |||| (a) Approximate f by a Taylor polynomial with degree n at the number a. ; (b) Graph f and Tn on a common screen. (c) Use Taylor’s Inequality to estimate the accuracy of the approximation f x Tn x when x lies in the given interval. ; (d) Check your result in part (c) by graphing Rn x . 2 c2 c4 0 (b) If f is an even function, show that c1 62. If f x 2 c3 e x , show that f c5 2n 0 0 2n ! . n! 5E-12(pp 825-827) 1/18/06 10:27 AM Page 825 PROBLEMS PLUS sin x 3 , find f 1. If f x 15 0. 2. A function f is defined by fx lim nl x 2n x 2n 1 1 Where is f continuous? 1 cot 1 x 2 3. (a) Show that tan 2 x 2 cot x. (b) Find the sum of the series n1 P¢ 4 4. Let Pn be a sequence of points determined as in the figure. Thus AP1 P£ 2 8 P∞ FIGURE FOR PROBLEM 4 1 x tan n 2n 2 P™ 1 A 1 P¡ Pn Pn 1 2 n 1, and angle APn Pn 1 is a right angle. Find lim n l 1, Pn APn 1 . 5. To construct the snowflake curve, start with an equilateral triangle with sides of length 1. Step 1 in the construction is to divide each side into three equal parts, construct an equilateral triangle on the middle part, and then delete the middle part (see the figure). Step 2 is to repeat Step 1 for each side of the resulting polygon. This process is repeated at each succeeding step. The snowflake curve is the curve that results from repeating this process indefinitely. (a) Let sn , ln , and pn represent the number of sides, the length of a side, and the total length of the n th approximating curve (the curve obtained after Step n of the construction), respectively. Find formulas for sn , ln , and pn . (b) Show that pn l as n l . (c) Sum an infinite series to find the area enclosed by the snowflake curve. Parts (b) and (c) show that the snowflake curve is infinitely long but encloses only a finite area. 1 2 3 6. Find the sum of the series 1 1 2 1 3 1 4 1 6 1 8 1 9 1 12 where the terms are the reciprocals of the positive integers whose only prime factors are 2s and 3s. 7. (a) Show that for xy 1, arctan x if the left side lies between (b) Show that arctan y 2 and 120 arctan 119 arctan x 1 y xy 2. 1 arctan 239 4 (c) Deduce the following formula of John Machin (1680–1751): 4 arctan 1 5 1 arctan 239 4 825 5E-12(pp 825-827) 1/18/06 10:27 AM Page 826 (d) Use the Maclaurin series for arctan to show that 1 0.197395560 0.004184075 0.197395562 arctan 5 arctan 239 (e) Show that 1 0.004184077 (f) Deduce that, correct to seven decimal places, 3.1415927 Machin used this method in 1706 to find correct to 100 decimal places. Recently, with the aid of computers, the value of has been computed to increasingly greater accuracy. In 1999, Takahashi and Kanada, using methods of Borwein and Brent Salamin, calculated the value of to 206,158,430,000 decimal places! 8. (a) Prove a formula similar to the one in Problem 7(a) but involving arccot instead of arctan. (b) Find the sum of the series arccot n 2 n 1 n0 9. Find the interval of convergence of 10. If a 0 a1 0, show that a2 ak lim (a0 sn a1 sn nl n 3x n and find its sum. n1 a2 sn 1 ak sn 2 k) 0 If you don’t see how to prove this, try the problem-solving strategy of using analogy (see page 58). Try the special cases k 1 and k 2 first. If you can see how to prove the assertion for these cases, then you will probably see how to prove it in general. 1 . n2 ln 1 11. Find the sum of the series n2 12. Suppose you have a large supply of books, all the same size, and you stack them at the edge 1 16 8 FIGURE FOR PROBLEM 12 1 4 1 2 of a table, with each book extending farther beyond the edge of the table than the one beneath it. Show that it is possible to do this so that the top book extends entirely beyond the table. In fact, show that the top book can extend any distance at all beyond the edge of the table if the stack is high enough. Use the following method of stacking: The top book extends half its length beyond the second book. The second book extends a quarter of its length beyond the third. The third extends one-sixth of its length beyond the fourth, and so on. (Try it yourself with a deck of cards.) Consider centers of mass. u 14. If p v3 w3 x3 3! x6 6! x9 9! x x4 4! x7 7! x 10 10! w Show that u 3 1 v 13. Let x2 2! 3u vw x8 8! 1. 1, evaluate the expression 1 1 826 x5 5! 1 2p 1 2p 1 3p 1 3p 1 4p 1 4p 5E-12(pp 825-827) 1/18/06 10:28 AM Page 827 15. Suppose that circles of equal diameter are packed tightly in n rows inside an equilateral tri- angle. (The figure illustrates the case n 4.) If A is the area of the triangle and An is the total area occupied by the n rows of circles, show that lim nl An A 2 s3 16. A sequence a n is defined recursively by the equations a0 a1 1 Find the sum of the series FIGURE FOR PROBLEM 15 nn n0 1 an n 1n 2 an n 1 3 an 2 an. x 17. Taking the value of x at 0 to be 1 and integrating a series term-by-term, show that y 1 0 P¡ P∞ P™ P˜ Pˆ Pß P¡¸ 1n nn x x dx n1 1 18. Starting with the vertices P1 0, 1 , P2 1, 1 , P3 1, 0 , P4 0, 0 of a square, we construct further points as shown in the figure: P5 is the midpoint of P1 P2, P6 is the midpoint of P2 P3, P7 is the midpoint of P3 P4, and so on. The polygonal spiral path P1 P2 P3 P4 P5 P6 P7 . . . approaches a point P inside the square. (a) If the coordinates of Pn are x n, yn , show that 1 x n x n 1 x n 2 x n 3 2 and find a 2 similar equation for the y-coordinates. (b) Find the coordinates of P. 19. If f x m0 cm x m has positive radius of convergence and e f x n0 dn x n, show that n P¢ P¶ P£ i ci dn ndn n i 1 i1 FIGURE FOR PROBLEM 18 20. Right-angled triangles are constructed as in the figure. Each triangle has height 1 and its base is the hypotenuse of the preceding triangle. Show that this sequence of triangles makes indefinitely many turns around P by showing that n is a divergent series. 1 1 1 1 ¨£ ¨™ ¨¡ 1 P 21. Consider the series whose terms are the reciprocals of the positive integers that can be written in base 10 notation without using the digit 0. Show that this series is convergent and the sum is less than 90. 22. (a) Show that the Maclaurin series of the function fx 1 x x x2 fn x n is n1 where fn is the nth Fibonacci number, that is, f1 1, f2 1, and fn fn 1 fn 2 for n 3. [Hint: Write x 1 x x 2 c0 c1 x c2 x 2 . . . and multiply both sides of this equation by 1 x x 2.] (b) By writing f x as a sum of partial fractions and thereby obtaining the Maclaurin series in a different way, find an explicit formula for the nth Fibonacci number. 827 ...
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This note was uploaded on 02/04/2010 for the course M 56435 taught by Professor Hamrick during the Fall '09 term at University of Texas at Austin.

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