30.3 Taylor Series 941 30.3 TAYLOR SERIES Defining Taylor Series In many examples in this chapter we’ve observed that the higher the degree of the Taylor polynomial generated by f at x = b, the better it approximates f(x) for x near b.For functions such as sin x and cos x, the higher the degree of the Taylor polynomial the longer the interval over which the polynomial follows the undulations of the function’s graph. Letting the degree of the polynomial increase without bound gives us the Taylor series for f . Definition If a function f has derivatives of all orders at x = b, then the Taylor series of f at (or about) x = b is defined to be f(b)+f (b)(x − b) + f (b) 2! (x − b) 2 +···+ f (n) (b) n! (x − b) n +···, that is, ∞ k=0 f (k) (b) k! (x − b) k . We refer to this series as the Taylor expansion of f about x = b or centered at x = b. In the special case where b = 0, the series ∞ k=0 f (k) (0) k! x k can be called the Maclaurin series for f . From the work we’ve done with Taylor polynomials, we can easily find the Maclaurin series for e x , sin x, and cos x. ◆ EXAMPLE 30.11 Find the Maclaurin series for f(x)=e x . SOLUTION All derivatives of e x are e x . When evaluated at x = 0, e x is 1. Maclaurin series for e x : 1 + x + x 2 2! +···+ x k k! +···= ∞ k=0 x k k! ◆ ◆ EXAMPLE 30.12 Find the Maclaurin series for f(x)=sin x. SOLUTION Even order derivatives Odd order derivatives f(x)=sin xf(0)=0 f (x) = cos xf (0)=1 f (x) =−sin xf (0) = 0 f (x) =−cos xf (0) =−1 f (4) (x) = sin xf (4) (0)=0 f (5) (x) = cos xf (5) (0)=1 . . . . . . . . . . . . f (2k) (x) = (−1) k sin xf (2k) (0) = 0 f (2k+1) (x) = (−1) k cos xf (2k+1) (0)=(−1) k Maclaurin series for sin x: x − x 3 3! + x 5 5! −···+(−1) k x 2k+1 (2k + 1)! +···= ∞ k=0 (−1) k x 2k+1 (2k + 1)! ◆ 942 CHAPTER 30 Series ◆ EXAMPLE 30.13 Find the Maclaurin series for f(x)= 1 1−x . SOLUTION f(x)=(1−x) −1 f(0)=1 f (x) = (1 − x) −2 f (0) = 1 f (x) = 2(1 − x) −3 f (0) = 2 f (x) = 3 · 2(1 − x) −4 f (0) = 3! . . . . . . f (k) (x) = k!(1 − x) −(k+1) f (k) (0) = k! Maclaurin series for 1 1−x : 1 + x + 2x 2 2! + 3!x 3 3! + 4!x 4 4! +···+ k!x k k! +··· =1+x+x 2 +x 3 +···+x k +···= ∞ k=0 x k The Maclaurin series for 1 1−x should look familiar. ∞ k=0 x k is a geometric series with a = 1 and r = x. Therefore we know that it converges to 1 1−x for |x| < 1 and diverges for |x|≥1. ◆ This observation at the end of Example 30.13 highlights the important question “What is the significance of the Taylor series for f(x)?”Forinstance, for what x does the Maclaurin series for sin x converge? When it converges, to what does it converge? In particular, does sin 0.1 = 0.1 − (0.1) 3 3! + (0.1) 5 5! −···+(−1) k (0.1) 2k+1 (2k+1)! +···? Or, more generally, for which values of x is it true that sin x = x − x 3 3! + x 5 5! − x 7 7! +···+(−1) k x 2k+1 (2k + 1)! +···? These latter questions can be answered using Taylor’s Theorem. f(x)=P n (x) + R n (x) Taking the limit as n increases without bound gives f(x)= lim n→∞ P n (x) + lim n→∞ R n (x). Therefore, f(x)is the sum of its Taylor series if and only if lim n→∞ R n (x) = 0. We state this more precisely below. Theorem on Convergence of Taylor Series If f is infinitely differentiable on an interval I centered around x = b, then the Taylor series for f at x = b converges to f(x)for all x ∈ I if and only if lim n→∞ R n (x) = 0 for all x ∈ I , where R n (x) is the Taylor remainder. 30.3 Taylor Series 943 In applying this theorem we frequently use the fact that lim n→∞ |x| n n! = 0 for every x. Think about this; it should make sense that eventually n! will be much larger than x n for fixed x. We prove this below. Fact: lim n→∞ |x| n n! = 0 for every real number x. Proof: 0 ≤ |x| n n! = |x| 1 · |x| 2 · |x| 3 ··· |x| n Let p be a positive constant integer such that 0 ≤ |x| p < 1. Then 0 ≤ |x| 1 · |x| 2 · |x| 3 ··· |x| p p positive terms, each less than or equal to |x| · |x| p + 1 ··· |x| n n−p positive terms, each less than |x| p ≤|x| p · |x| p n−p So 0 ≤ |x| n n! < |x| p · |x| p n−p (30.4) If 0 ≤ r<1, then lim n→∞ r n = 0. Therefore lim n→∞ r n−p = 0 for 0 ≤ r<1and p constant. But 0 ≤ |x| p < 1, so lim n→∞ |x| p n−p = 0. Return to (30.4) and let n increase without bound. lim n→∞ 0 ≤ lim n→∞ |x| n n! ≤ lim n→∞ |x| p · |x| p n−p 0 ≤ lim n→∞ |x| n n! ≤|x| p ·0=0 Therefore lim n→∞ |x| n n! = 0, by the Sandwich Theorem. We are now ready to show that sin x and e x are equal to their respective Taylor series. ◆ EXAMPLE 30.14 Show that sin x = ∞ k=0 (−1) k x 2k+1 (2k+1)! for all x. SOLUTION For each x there exists a c between 0 and x such that 0 ≤|R n (x)|= |f (n+1) (c)| (n + 1)! |x| n+1 . Therefore |R n (x)|≤ 1·|x| n+1 (n+1)! . The latter inequality holds because |f (n+1) (c)|=|sin c| or | cos c| and both are bounded by 1. lim n→∞ 0 ≤ lim n→∞ |R n (x)|≤ lim n→∞ |x| n+1 (n + 1)! 0 ≤ lim n→∞ |R n (x)|≤0 From the Sandwich Theorem we conclude that lim n→∞ |R n (x)|=0 and therefore lim n→∞ R n (x) = 0 for all x. Thus, sin x is equal to its Taylor expansion about zero for all x. ◆ 944 CHAPTER 30 Series ◆ EXAMPLE 30.15 Show that e x = ∞ k=0 x k k! for all x. SOLUTION For each x there exists a c between 0 and x such that 0 ≤|R n (x)|= |f (n+1) (c)| (n + 1)! |x| n+1 = e c |x| n+1 (n + 1)! e x is an increasing function, so e c ≤ e |x| . 0 ≤|R n (x)|≤e |x| |x| n+1 (n + 1)! lim n→∞ 0 ≤ lim n→∞ |R n (x)|≤ lim n→∞ e |x| |x| n+1 (n + 1)! But lim n→∞ e |x| |x| n+1 (n + 1)! = e |x| lim n→∞ |x| n+1 (n + 1)! = e |x| · 0 = 0. 0 ≤ lim n→∞ |R n (x)|≤0 So lim n→∞ |R n (x)|=0bythe Sandwich Theorem. Therefore, lim n→∞ R n (x) = 0. We conclude that e x = 1 + x + x 2 2! + x 3 3! +···for all x. ◆ EXERCISE 30.4 Show that cos x is equal to its Maclaurin series for all x. Take a moment to reflect upon the rather remarkable results we have accumulated. Not only can we express e x , sin x, and cos x as infinite “polynomials” (called power series), but we determined the coefficients using information about derivatives evaluated only at x = 0. We think of a derivative as giving local information, yet somehow information generating the entire function is encoded in the set of infinitely many derivatives. This is philosophically intriguing. Let’s take inventory on convergence issues. A Taylor series might converge to its generating function for all x. For example, consider the Maclaurin series for e x , sin x, and cos x. A Taylor series might converge to its generating function only over a certain interval. For example, 1 1−x = ∞ k=0 x k only for x ∈ (−1, 1). At minimum a Taylor series will be equal to the value of its generating function at its center. 6 Power Series We’ll put Taylor series in a broader context by discussing power series. 6 It is possible for a Taylor series to converge, but not to its generating function, except at its center. This pathology is illustrated in Problem 35 at the end of this section. 30.3 Taylor Series 945 Definition A power series in x is an infinite series of the form ∞ k=0 a k x k .Apower series in (x − b), or a power series centered at b, is a series of the form ∞ k=0 a k (x − b) k . Uniqueness Theorem for Power Series Expansions If f has a power series expansion (or representation) at b, that is, if f(x)= ∞ k=0 a k (x − v) k for |x − b| <R,then that power series is the Taylor series for f at x = b. The Uniqueness Theorem can be verified by repeatedly differentiating the power series expansion term by term and evaluating each successive derivative at x = b. The Uniqueness Theorem carries with it computational power. For example, we could have avoided computing derivatives in Example 30.13 by using the fact that 1 1 − x = 1 + x + x 2 +··· for x ∈ (−1, 1). This is a power series expansion of 1 1−x , and therefore it must be the Taylor series for 1 1−x at x = 0. Convergence of a Power Series 7 Theorem on the Convergence of a Power Series 8 For a given power series ∞ k=0 a k (x − b) k , one of the following is true: i. The series converges for all x. ii. The series converges only when x = b. iii. There is a number R, R>0such that the series converges for all x such that |x − b| <R (x is within R of the center) and diverges for all x such that |x − b| >R. Ris called the radius of convergence. If the series converges for all x, we say R =∞; if the series converges only at its center, we say R = 0. The set of all x for which a power series converges is called the interval of converge of the series. From the theorem stated above we see that a power series in (x − b) will have an interval of convergence centered around x = b. At the endpoints of the interval the series could either converge or diverge; further investigation is necessary. In other words, if the radius of convergence is R, the interval of convergence will be one of the following: 7 The student or instructor who prefers a thorough discussion of convergence before a discussion of the convergence of a power series can turn to page 964 (Section 30.5), and, after completing that section, return to this point. 8 Justification is given in Appendix H. 946 CHAPTER 30 Series (b – R, b + R] b R [b – R, b + R) b R [b – R, b + R] b R (b – R, b + R) b R The behavior of a power series at the points b + R and b − R can be tricky, but for |x − b| <Rwe will find the behavior reassuringly like that of polynomials in many respects. We will use substitution, integration, and differentiation of power series on |x − b| <Rto obtain new Taylor series from familiar ones. Before moving in this direction we must add one more very important Taylor series to our list of “familiar” ones. The Binomial Series ◆ EXAMPLE 30.16 THE BINOMIAL SERIES Find the Maclaurin series generated by f(x)=(1+x) p , where p is constant. This series is called the binomial series. SOLUTION The Maclaurin series is given by ∞ k=0 f (k) (0) k! x k . f(x)=(1+x) p f(0)=1 f (x) = p(1 + x) p−1 f (0) = p f (x) = p(p − 1)(1 + x) p−2 f (0) = p(p − 1) f (x) = p(p − 1)(p − 2)(1 + x) p−3 f (0) = p(p − 1)(p − 2) . . . . . . f (k) (x) = p(p − 1)(p − 2) ···(p − k + 1)(1 + x) p−k f (k) (0) = p(p − 1) ···(p − k + 1) . . . . . . Therefore the Maclaurin series is 1 + px + p(p − 1) 2! x 2 + p(p − 1)(p − 2) 3! x 3 +···+ p(p − 1)(p − 2) ···(p − k + 1) k! x k +··· 9 ◆ Fact: The Maclaurin series for (1 + x) p converges to (1 + x) p for x ∈ (−1, 1) and diverges for |x| > 1. (1 + x) p = 1 + px + p(p − 1) 2! x 2 + p(p − 1)(p − 2) 3! x 3 +···+ p(p − 1) ···(p − k + 1) k! x k +··· for x ∈ (−1, 1) Proving this fact by showing that lim n→∞ R n (x) = 0 is difficult, but possible. We omit the proof. 10 REMARKS CONCERNING THE BINOMIAL SERIES 1. In the case that p is a positive integer the series terminates with the x p term; subsequent coefficients all contain a factor (p − p).Weare left with an expansion of the polynomial 9 The coefficients match those given by Pascal’s Triangle. 10 By the end of Section 30.5 you will be able to show that the radius of convergence of the binomial series is 1. 30.3 Taylor Series 947 (1 + x) p . As an exercise, show that if p = 4 the binomial series becomes (1 + x) 4 = 1 + 4x + 6x 2 + 4x 3 + x 4 . 2. The notation p k is often used as an abbreviation for the binomial coefficients where p k = p(p−1)(p−2)···(p−k+1) k! for k ≥ 1 and p 0 = 1. Using this notation we can write (1 + x) p = ∞ k=0 p k x k for x ∈ (−1, 1). 3. The binomial expansion is valuable to know, as applications of it abound. Examples 30. 9, 30.10, and 30.13 all involve binomial expansions. Often one uses the first and second order approximations, (1 + x) p ≈ 1 + px or (1 + x) p ≈ 1 + px + p(p − 1) 2! x 2 for |x| small, in computations in applied science. EXERCISE 30.5 By letting p =−1, use the binomial series to find the Maclaurin series for 1 1+u . Then let u =−xto arrive at the Maclaurin series for 1 1−x . Below we list some commonly used Taylor expansions together with their intervals of convergence. e x = 1 + x + x 2 2! + x 3 3! +···+ x k k! +··· for all x sin x = x − x 3 3! + x 5 5! −···+(−1) k x 2k+1 (2k + 1)! +··· for all x cos x = 1 − x 2 2! + x 4 4! −···+(−1) k x 2k (2k)! +··· for all x (1 + x) p = 1 + px + p(p − 1) 2! x 2 +···+ p(p − 1) ···(p − k + 1) k! x k +··· for |x| < 1 1 1 − x = 1 + x + x 2 + x 3 +···+x k +··· for |x| < 1 You will find it useful to know these series off the top of your head because other series can be derived directly from these. Obtaining New Taylor Series From Familiar Ones: Substitution ◆ EXAMPLE 30.17 Find the Taylor expansion for e −x 2 about x = 0. SOLUTION Calculating this series by computing derivatives very quickly becomes unwieldy. Instead, we’ll use substitution. 948 CHAPTER 30 Series e u = 1 +u + u 2 2! + u 3 3! +···+ u k k! +··· for all u. Let u =−x 2 . e −x 2 =1+(−x 2 )+ (−x 2 ) 2 2! + (−x 2 ) 3 3! +···+ (−x 2 ) k k! +··· e −x 2 =1=x 2 + x 4 2! − x 6 3 +···+(−1) k x 2k k! +··· for all x. By the Uniqueness Theorem, this is the Taylor series for e −x 2 about x = 0. ◆ ◆ EXAMPLE 30.18 Find the Maclaurin series for f(x)=2xsin x cos x. SOLUTION f(x)=x ·2sin x cos x =x · sin(2x) sin u = ∞ k=0 (−1) k u 2k+1 (2k + 1)! for all u. Let u = 2x. sin(2x) = ∞ k=0 (−1) k (2x) 2k+1 (2k + 1)! = ∞ k=0 (−1) k 2 2k+1 x 2k+1 (2k + 1)! for all x x · sin(2x) =x ∞ k=0 (−1) k 2 2k+1 x 2k+1 (2k + 1)! = ∞ k=0 (−1) k 2 2k+1 x 2k+2 (2k + 1)! We can write this out as 2x 2 − 2 3 x 4 3! + 2 5 x 6 5! −···+(−1) k 2 2k+1 x 2k+2 (2k + 1)! +···. By the Uniqueness Theorem, this is the Maclaurin series for 2x sin x cos x. Note the difference between substituting 2x for u in the first step and multiplying the whole series by x in the second step. ◆ ◆ EXAMPLE 30.19 Find the fourth degree Taylor polynomial for f(x)= √ 9 − x 2 about x = 0. For what x- values does the Taylor series converge to f ? SOLUTION Let’s transform this function so that we can use the binomial series. 9 − x 2 = 9 1 − x 2 9 = 3 1 − x 2 9 = 3 1 − x 2 9 1 2 From the binomial series we know (1 + u) p = 1 +pu + p(p − 1) 2! u 2 +··· for |u| < 1. so (1 + u) 1 2 = 1 + 1 2 u + 1 2 − 1 2 2! u 2 +··· (1+u) 1 2 = 1 + 1 2 u − 1 8 u 2 +··· for |u| < 1. Let u =− x 2 9 . 30.3 Taylor Series 949 1 + − x 2 9 1 2 = 1 + 1 2 − x 2 9 − 1 8 − x 2 9 2 +··· for − x 2 9 < 1 1 − x 2 9 = 1 − 1 18 x 2 − 1 648 x 4 +··· for |x 2 | < 9 3 1 − x 2 9 = 3 − 1 6 x 2 − 1 216 x 4 +··· for x ∈ (−3, 3) Thus, the fourth degree Taylor polynomial is 3 − 1 6 x 2 − 1 216 x 4 . The Taylor series for f(x) converges to f(x)on (−3, 3). ◆ Note that in Examples 30.17 and 30.18 the “old” series being used converge for all real numbers. In Example 30.19 this was not the case; the new interval of convergence was obtained by substitution. PROBLEMS FOR SECTION 30.3 1. Find the Maclaurin series for cos x and show that it is equal to cos x for all x. 2. (a) Find the Maclaurin series for ln(1 + x). (b) On the same set of axes, graph ln(1 + x) and P 6 (x). Observe that the polynomial approximation to ln(1 + x) is good for |x| < 1. (c) Graph R 6 (x) =ln(1 +x) −P 6 (x). Observe that R 6 (x) is close to zero on |x| < 1. In the next section we will show that the radius of convergence of the Maclaurin series for ln(1 + x) is 1. 3. The interval of convergence of the Maclaurin series for ln(1 + u) is u ∈ (−1, 1]. On this interval the series converges to ln(1 +u). (a) Find the Maclaurin series for ln(1 + u). (b) By setting u = x − 1 in part (a), find the Taylor series for ln x centered at x = 1. (c) Find the Taylor series for ln x at x = 1 by taking derivatives. Make sure your answers to parts (b) and (c) agree. (d) What is the interval of convergence for the Taylor series for ln x centered at x = 1? (e) Graph ln x and several of its Taylor polynomials at x = 1 to be sure your answer to part (d) is reasonable. In Problems 4 through 9, find the Taylor series for f(x)centered at the indicated value of b. 4. f(x)=sin x, b =π 5. f(x)=2cos x, b = π 2 6. f(x)=10 x , b =0 7. f(x)= 1 √ x , b =1 950 CHAPTER 30 Series 8. f(x)=(3+2x) 3 , b =0 9. f(x)=(1+x) 5 , b =0 10. A power series centered at b = 0 has a radius of convergence of 5. For each value of x given below, determine whether the series converges, diverges, or there is not enough information available to determine. (a) x = 0 (b) x = 3 (c) x = 5 (d) x = 7 (e) x =−1.8 (f) x =− √ 5 (g) x =−5 (h) x =−6 11. A power series of the form ∞ k=0 a k (x − 2) k has a radius of convergence of 3. (a) For what values of x can you say with confidence that the series converges? (b) For what values of x can you say with confidence that the series diverges? (c) For what values of x are you given inadequate information to determine conver- gence? 12. The interval of convergence of a power series is (−2, 5]. (a) What is the radius of convergence? (b) What is the center of the series? 13. A power series is of the form ∞ k=0 a k (x + 3) k . Which of the intervals given below could conceivably be the interval of convergence of the series? For each option ruled out, explain the rationale. (a) (0, ∞) (b) (2, 4) (c) [−10, 4) (d) [−3, 3] (e) (−4, 2) (f) (−5, −1] (g) (−∞, ∞) In Problems 14 through 21, use your knowledge of the binomial series to find the nth degree Taylor polynomial for f(x)about x = 0. Give the radius of convergence of the corresponding Maclaurin series. One of these “series” converges for all x. 14. f(x)= √ 1 + 3x, n = 3 15. f(x)= 1 √ 1+x , n = 2 16. f(x)=(1−x) 2 3 , n = 3 17. f(x)= 3 √ 1 + x 2 , n = 5 18. f(x)=(1+3x) 5 , n =6 19. f(x)= 1 (1+x) 2 , n =5 20. f(x)=2(9−x) 1 2 , n = 3 21. f(x)= x √ 4+x , n = 3 . instructor who prefers a thorough discussion of convergence before a discussion of the convergence of a power series can turn to page 964 (Section 30.5), and, after completing that section, return to. left with an expansion of the polynomial 9 The coefficients match those given by Pascal’s Triangle. 10 By the end of Section 30.5 you will be able to show that the radius of convergence of the binomial. What is the interval of convergence for the Taylor series for ln x centered at x = 1? (e) Graph ln x and several of its Taylor polynomials at x = 1 to be sure your answer to part (d) is reasonable. In