30.3 Taylor Series 951 22. (a) Expand f(x)=(a + x) 4 by multiplying out or by using Pascal’s triangle. (b) Rewrite f(x) as [a(1 + x a )] 4 = a 4 1 + x a 4 . Use the binomial series to expand 1 + x a 4 , multiply by a 4 , and demonstrate that the result is the same as in part (a). 23. Find the Maclaurin series for 1 1+x 2 . What is the radius of convergence? 24. Use the binomial series to find the Maclaurin series for 1 √ 1−x 2 . What is the radius of convergence? In Problems 25 through 34, use any method to find the Maclaurin series for f(x). (Strive for efficiency.) Determine the radius of convergence. 25. f(x)=xe −x 26. f(x)=sin 3x 27. f(x)=cos x 2 28. f(x)=3e 2x 29. f(x)=cos(x 2 ) 30. f(x)=3 x 31. f(x)=x 2 cos(−x) 32. f(x)=cos 2 x (Hint: use a trigonometric identity) 33. f(x)=(a + x) p , where “a” and “p” are constants and p is not a positive integer. 34. f(x)= 1 2x+3 35. Pathological Example: Let f(x)= e − 1 x 2 for x = 0, 0 for x = 0. (a) Graph f(x) on the following domains: [−20, 20], [−2, 2], and [−0.5, 0.5]. (A graphing instrument can be used.) (b) It can be shown that f is infinitely differentiable at x = 0 and that f (k) (0) = 0 for all k. Conclude that the Maclaurin series for f(x)converges for all x but only converges to f(x)at x = 0. 36. Find the Maclaurin series for 1 √ e x . What is its radius of convergence? 37. For x ∈ (−1, 1], ln(1 +x) =x − x 2 2 + x 3 3 − x 4 4 +···+(−1) k x k+1 k+1 +···. (a) Find the Maclaurin series for ln(1 + 2x). What is its interval of convergence? (b) Find the Maclaurin series for ln(e + ex). What is its interval of convergence? (c) Find the Maclaurin series for log 10 (1 + x). 952 CHAPTER 30 Series 38. Discover something wonderful. We know e u = 1 + u + u 2 2! + u 3 3! +···+ u k k! +···for all real u. Now define e raised to a complex number, a + bi where i = √ −1, to be e a · e bi where e bi = 1 + (bi) + (bi) 2 2! + (bi) 3 3! +···+ (bi) k k! +···. (a) Use the fact that i 2 =−1, i 3 =−i,and i 4 = 1 to simplify the expression for e ib . Gather together the real terms (the ones without i’s) and the terms with a factor of i. Express e ib as a sum of two familiar functions (one of them multiplied by i). (b) Use your answer to part (a) to evaluate e πi . 39. The hyperbolic functions, hyperbolic cosine, abbreviated cosh, and hyperbolic sine, abbreviated sinh, are defined as follows. cosh x = e x + e −x 2 sinh x = e x − e −x 2 (a) Graph cosh x and sinh x, each on its own set of axes. Do this without using a computer or graphing calculator, except possibly to check your work. (b) Find the Maclaurin series for cosh x. (c) Find the MacLaurin series for sinh x. Remark: From the graphs of cosh x and sinh x one might be surprised by the choice of names for these functions. After finding their Maclaurin series the choice should seem more natural. (d) Do some research and find out how these functions, known as hyperbolic functions, are used. The arch in St. Louis, the shape of many pottery kilns, and the shape of a hanging cable are all connected to hyperbolic trigonometric functions. 30.4 WORKING WITH SERIES AND POWER SERIES Absolute and Conditional Convergence There are many ways in which power series can be treated very much as we treat polyno- mials, but there are ways in which they can behave differently and must be treated with caution. This makes sense; there are ways in which series and finite sums behave very dif- ferently. In order to sort this out a bit, not only do we need to steer clear of divergent series and power series outside their interval of convergence, but we need to refine our notion of convergence to distinguish between absolute and conditional convergence. Definition A series ∞ n=1 a n is absolutely convergent if ∞ n=1 |a n | converges. Note that if the terms of a series are either all positive or all negative, then convergence implies absolute convergence. There is only an issue when some terms are positive and some terms are negative. 11 11 Actually, there is not an issue provided there exists a constant k such that a n is either positive for all n>k or negative for all n>k. 30.4 Working with Series and Power Series 953 Fact: If a series converges absolutely, it converges. This is proven in Appendix H. Definition A series ∞ n=1 a n is conditionally convergent if it is convergent but not absolutely convergent. Why is this distinction handy? Well, one might hope that the order of the terms in a sum could be rearranged without altering the sum, yet for infinite series this is true only if the series converges absolutely. In fact, it can be proven that if ∞ k=0 a n is conditionally convergent, then the order of the terms can be rearranged to produce any finite number. This unsettling fact is enough to make one wary of conditionally convergent series. It’s hard to be wary of something without a concrete example, so we will take this opportunity to look at alternating series. You will find that alternating series are fascinating in their own right, and that this excursion into the topic of alternating series will produce as a by-product an error estimate that will prove useful when dealing with many Taylor polynomials. Alternating Series Definition A series whose successive terms alternate in sign is called an alternating series. For any fixed x the Maclaurin series for sin x and cos x are alternating series. The Maclaurin series for e x will alternate when x is negative and the one for ln(1 + x) will alternate when x is positive. There is a simple convergence test, proved by Leibniz, that can be applied to alternating series. We know that for a general series, ∞ k=1 a k , the characteristic lim k→∞ a k = 0 is necessary but not sufficient for convergence. The divergence of the harmonic series 1 + 1 2 + 1 3 + ···+ 1 n + ···illustrates this fact. However, if a series is alternating, then if the magnitude of the terms decreases monotonically towards zero, this is enough to assure convergence. Alternating Series Test An alternating series, ∞ k=1 (−1) k a k or ∞ k=1 (−1) k+1 a k for a k > 0, converges if i. a k+1 <a k , the terms are decreasing in magnitude and ii. lim k→∞ a k = 0, the terms are approaching zero. The Basic Idea Behind the Alternating Series Test 12 Consider the series a 1 − a 2 + a 3 − a 4 + ···+(−1) k+1 a k +···for a k > 0. Suppose that conditions (i) and (ii) are satisfied. In Figure 30.8 we plot partial sums. 12 This is not a rigorous argument, but it can be made rigorous using the theorem that every bounded monotonic sequence is convergent. 954 CHAPTER 30 Series 0 S 2 S 6 S 5 S S 3 S 1 S 1 = a 1 S 4 a 6 a 5 a 4 a 2 a 1 a 3 Figure 30.8 S 1 = a 1 is to the right of zero. S 2 = a 1 − a 2 lies between 0 and S 1 because a 2 <a 1 . S 3 =a 1 −a 2 +a 3 lies between S 2 and S 1 because a 3 <a 2 . . . . Picture starting at the zero. Take a big step forward to S 1 , then a smaller step backward to S 2 , then an even smaller step forward to S 3 , and so on. The partial sums oscillate; S n is between S n−1 and S n−2 because a n <a n−1 .The distance between S n−1 and S n is a n and lim n→∞ a n = 0. Therefore the sequence of partial sums is approaching a finite limit L, with successive partial sums alternately overshooting then undershooting L. This argument can be made rigorous by considering the increasing but bounded se- quence of partial sums, S 2 , S 4 , S 6 , ,and the decreasing but bounded sequence of partial sums S 1 , S 3 , S 5 , ,and showing that both sequences converge to the same limit. Our analysis provides us with an easy-to-use error estimate. If an alternating series satisfies the two conditions of the Alternating Series Test and if we approximate the sum, L, using a partial sum S n , then the magnitude of the error will be less than a n+1 , the magnitude of the first unused term of the series. Furthermore, if the last term of the partial sum is positive, then the partial sum is larger than L; if its last term is negative, then the partial sum is smaller than L. We refer to this as the Alternating Series Error Estimate. ◆ EXAMPLE 30.20 Consider the alternating harmonic series 1 − 1 2 + 1 3 − 1 4 + 1 5 −···+(−1) k+1 1 k +···. (a) Show that this series converges conditionally. (b) It can be shown that ∞ k=1 (−1) k+1 1 k converges to ln 2. How many terms of the series must be used in order to approximate ln 2 with error less than 0.001? SOLUTION (a) The series ∞ k=1 (−1) k+1 1 k is alternating. It satisfies the conditions of the Alternating Series Test: i. The terms are decreasing in magnitude: 1 k+1 < 1 k . ii. The terms approach zero: lim k→∞ a k = lim k→∞ 1 k = 0. Therefore the series converges. But ∞ k=1 (−1) k+1 1 k = ∞ k=1 1 k is the harmonic se- ries, which diverges. Therefore the alternating harmonic series converges conditionally. (b) By the Alternating Series Error Estimate we know that the magnitude of the error is less than the magnitude of the first omitted term. Therefore we use the estimate 30.4 Working with Series and Power Series 955 ln 2 ≈ 999 k=1 (−1) k+1 1 k ; we need 999 terms. This series for ln 2 converges very slowly! ◆ ◆ EXAMPLE 30.21 Estimate 1 √ e with error less than 10 −3 . SOLUTION e x = 1 + x + x 2 2! + x 3 3! +···+ x k k! +···Thus 1 √ e = e − 1 2 = 1 − 1 2 + 1 2 2 · 2! − 1 2 3 · 3! +···+(−1) k 1 2 k · k! +···. This series is alternating, its terms are decreasing in magnitude, and its terms tend toward zero. Therefore, we can apply the Alternating Series Error Estimate. We must find k such that 1 2 k · k! < 1 1000 , or equivalently, 2 k · k! > 1000. We do this by trial and error. 2 4 · 4! = 384 but 2 5 · 5! = 3840 > 1000. 1 2 5 ·5! < 1 1000 ,sowedon’tneed to use this term. e − 1 2 ≈ 1 − 1 2 + 1 2 2 · 2! − 1 2 3 · 3! + 1 2 4 · 4! = 1 − 1 2 + 1 8 − 1 48 + 1 384 1 √ e ≈ .6068. Notice that the Alternating Series Error Estimate is simpler to apply than Taylor’s Remainder. ◆ Let’s return to the disturbing remark made before introducing alternating series. The assertion was that if a series converges conditionally, then rearranging the order of the terms of the series can change the sum. We’re now ready to demonstrate this. 1 − 1 2 + 1 3 − 1 4 + 1 5 − 1 6 + 1 7 − 1 8 + 1 9 − 1 10 + 1 11 −···=ln 2 Multiplying both sides by 2 gives 2 − 2 2 + 2 3 − 2 4 + 2 5 − 2 6 + 2 7 − 2 8 + 2 9 − 2 10 + 2 11 −···=2ln2=ln 4 (30.5) Rearrange the order of the terms in Equation (30.5) so that after each positive term there are two negative terms as follows. 2 − 1 − 2 4 + 2 3 − 2 6 − 2 8 + 2 5 − 2 10 − 2 12 + 2 7 − 2 14 − 2 16 + 2 19 −··· =(2−1)− 2 4 + 2 3 − 2 6 − 2 8 + 2 5 − 2 10 − 2 12 + 2 7 − 2 14 − 2 16 +··· =1− 1 2 + 1 3 − 1 4 + 1 5 − 1 6 + 1 7 − 1 8 +··· =ln 2. By rearranging the order of the terms we changed the sum from ln 4 to ln 2. Riemann proved that by rearranging the order of the terms we can actually get the sum to be any real number. 956 CHAPTER 30 Series On the other hand, it can be proven that if a series converges absolutely to a sum of S, then any rearrangement of the terms has a sum of S as well. This is one of the reasons we prefer to work with absolutely convergent series whenever possible. Manipulating Power Series Having defined absolute convergence, we can return to the theorem on the convergence of a power series and state a stronger form. (See Appendix H for justification.) Theorem on the Convergence of a Power Series For a given power series ∞ k=0 a n (x − b) k , one of the following is true: i. The series converges absolutely for all x. ii. The series converges only when x = b. iii. There is a number R, R>0,such that the series converges absolutely for all x such that |x − b| <Rand diverges for all x such that |x − b| >R. The points x = b + R and x = b − R must be studied separately. At these endpoints the series could converge conditionally, converge absolutely, or diverge. For the sake of simplicity we will generally restrict our attention to the interval (b − R, b + R) in which the power series converges absolutely. Differentiation and Integration of Power Series Differentiation and Integration of Power Series Let ∞ k=0 a k (x − b) k be a power series with radius of convergence R, where R>0, R possibly ∞. Then the function f(x)= ∞ k=0 a k (x − b) k can be differentiated term by term or integrated term by term on (b − R, b + R). That is, f (x) = ∞ k=0 d dx a k (x − b) k = ∞ k=1 ka k (x − b) k−1 with radius of convergence R and f(x)dx= ∞ k=0 a k (x − b) k dx = C + ∞ n=0 a k (x − b) k+1 k + 1 with radius of convergence R. This result, whose proof is omitted, says that the radius of convergence remains the same after integration or differentiation; it gives no information about convergence or divergence at x = b ± R. 13 13 The original series may diverge at an endpoint and yet converge once integrated, or vice versa. 30.4 Working with Series and Power Series 957 This Theorem gives us convenient ways of generating new Taylor series from familiar ones and provides a tool for integrating functions that don’t have elementary antiderivatives. ◆ EXAMPLE 30.22 Find the Maclaurin series for arctan x. What is the radius of convergence? SOLUTION This is unwieldy to compute by taking derivatives. Instead, we’ll use the fact that 1 1 + x 2 dx = arctan x + C. We know 1 1−u = 1 + u + u 2 + u 3 +···+u k +···for |u| < 1. Let u =−x 2 . 1 1 − (−x 2 ) = 1 + (−x 2 ) + (−x 2 ) 2 + (−x 2 ) 3 +···+(−x 2 ) k +··· for |−x 2 |<1 1 1 + x 2 = 1 − x 2 + x 4 − x 6 +···+(−1) k x 2k +··· for |x| < 1 1 1 + x 2 dx = 1 − x 2 + x 4 − x 6 +···+(−1) k x 2k +··· dx arctan x = C + x − x 3 3 + x 5 5 −···+(−1) k x 2k+1 2k + 1 +··· Todetermine C, evaluate both sides at x = 0. arctan 0 = C,soC=0. arctan x = x − x 3 3 + x 5 5 − x 7 7 +···+(−1) k x 2k+1 2k + 1 +··· The radius of convergence is 1, so the series converges absolutely for x ∈ (−1, 1) and diverges for |x| > 1. In fact, although we have only shown convergence for x ∈ (−1, 1), the series converges to arctan x for x =±1aswell. When evaluated at x = 1, the series is π 4 = 1 − 1 3 + 1 5 − 1 7 +···. 14 ◆ ◆ EXAMPLE 30.23 Find the Maclaurin series for ln(1 + x) by integrating the series for 1 1+x . What advantage does this approach have over computing the series by taking derivatives? SOLUTION We know 1 1−u = 1 + u + u 2 +···+u k +···for |u| < 1. Let u =−x. 1 1 + x = 1 1 − (−x) = 1 − x + x 2 − x 3 +···+(−1) k x k +··· for |−x|<1, i.e., |x| < 1 1 1 + x dx = C + x − x 2 2 + x 3 3 − x 4 4 +···+(−1) k x k+1 k + 1 +··· for |x| < 1 So ln(1 + x) = C + x − x 2 x + x 3 3 − x 4 4 +···(−1) k x k+1 k + 1 +···. 14 You will find this series is carved in stone at the entrance to Coimbra University’s department of mathematics building in Coimbra, Portugal. 958 CHAPTER 30 Series To determine C, evaluate at x = 0. ln(1 + 0) = C,soC=0. ln(1 + x) = x − x 2 2 + x 3 3 − x 4 4 +···+(−1) k x k+1 k + 1 +··· An advantage of this method of arriving at the series is that we know the radius of conver- gence is 1, and that the series converges to ln(1 + x) for |x| < 1. ◆ Once we know ln(1 + x) = ∞ k=0 (−1) k x k+1 k+1 for |x| < 1 we can set u = x + 1, (x = u − 1) and find ln(u) = ∞ k=0 (−1) k (u−1) k+1 k+1 = (u − 1) − (u−1) 2 2 + (u−1) 3 3 − ···+ (−1) k (u−1) k+1 k+1 + ···.When −1 <x<1, we know that 0 <x+1<2, so the series for ln u about u = 1 must converge on u ∈ (0, 2). In fact, it can be shown that both of these series converge at the right-hand endpoint of the respective interval of convergence. ln(1 + x) = x − x 2 2 + x 3 3 −···+(−1) k x k+1 k + 1 +··· for x ∈ (−1, 1] ln u = (u − 1) − (u − 1) 2 2 + (u − 1) 3 3 −···+(−1) k (u − 1) k+1 k + 1 for u ∈ (0, 2] REMARK We saw in Example 30.20 that the series 1 − 1 2 + 1 3 − 1 4 + ···converges very slowly. Similarly, observe that 1 − 1 3 + 1 5 − 1 7 + ···converges to π 4 very slowly. This series is aesthetically pleasing but computationally inefficient. For practical purposes the rate at which a series converges is important. For instance, it is more efficient to approximate ln 2 by looking at the following: − ln 2 = ln 1 2 = ln 1 − 1 2 =− 1 2 − 1 2 2 · 2 − 1 2 3 · 3 − 1 2 4 · 4 +··· =− 1 2 − 1 8 − 1 24 − 1 64 − 1 160 −···. x= 1 2 is closer to the center of the series than is x = 1, so the series converges more rapidly at 1 2 than at 1. For even more efficiency in approximating ln 2 we can find the Maclaurin series for ln 1+x 1−x and evaluate it at x = 1 3 . This is the topic of one of the problems at the end of this section. One reason that it is so useful to be able to represent a function as a power series is that a power series is simple to integrate. The use of power series expansions as an integration tool figured prominently in Newton’s work and continues to be important in the integration of otherwise intractable functions. Consider, for example, f(x)=e −x 2 ,afunction that has no elementary antiderivative. The graph of f is a bell-shaped curve which, with minor modifications, gives the standard normal distribution that plays such an important role in probability and statistics. It is crucial to know the area under the normal distribution, and 30.4 Working with Series and Power Series 959 for this we must compute a definite integral. The following example indicates how Taylor series can be used in such a computation. ◆ EXAMPLE 30.24 Approximate 0.2 0 e −x 2 dx with error less than 10 −8 . SOLUTION From Example 30.17 we know that e −x 2 = 1 − x 2 + x 4 2! − x 6 3! +···+(−1) k x 2k k! +··· for all x. 0.2 0 e −x 2 dx = 0.2 0 1 − x 2 + x 4 2! − x 6 3! +···+(−1) k x 2k k! +··· dx = x − x 3 3 + x 5 5 · 2! − x 7 7 · 3! +···+(−1) k x 2k+1 (2k + 1)k! +··· 0.2 0 = 0.2 − (0.2) 3 3 + (0.2) 5 5 · 2! − (0.2) 7 7 · 3! +···+(−1) k (0.2) 2k+1 (2k + 1)k! +··· Wecan apply the Alternating Series Error Estimate because the series above is alternating, its terms are decreasing in magnitude, and its terms tend toward zero. We look for a term whose magnitude is less than 10 −8 (0.2) 7 7 · 3! = 2 7 7 · 3! · 10 7 ≈ 3 × 10 −7 : not small enough (0.2) 9 9 · 4! = 2 9 9 · 4! · 10 9 ≈ 2.4 × 10 −9 < 10 −8 Therefore 0.2 0 e −x 2 dx ≈ 0.2 − (0.2) 3 3 + (0.2) 5 5 · 2 − (0.2) 7 7 · 6 with error less than 10 −8 . 0.2 0 e −x 2 dx ≈ 0.197365029 ◆ There are three main reasons for our interest in representing functions as power series. Such representations are useful in approximating functions by polynomials and approximating function values numeri- cally, integrating functions that don’t have elementary antiderivatives, and solving differential equations. Although we have illustrated the first two applications of power series, we have yet to give an example of the third. The Theorem on Differentiation of a Power Series plays the major role in this application. Power Series and Differential Equations The next example illustrates how power series can be used in solving differential equations. 960 CHAPTER 30 Series ◆ EXAMPLE 30.25 Use power series to solve the differential equation y =−y. SOLUTION Let f(x) be a solution to the differential equation. Assume that f(x) has a power series expansion. f(x)=a 0 +a 1 x +a 2 x 2 +a 3 x 3 +···+a k x k +··· f (x) = a 1 + 2a 2 x + 3a 3 x 2 + 4a 4 x 3 +···+ka k x k−1 +··· f (x) = 2a 2 + 3 · 2 · a 3 x + 4 · 3 · a 4 x 2 +···+k(k − 1)a k x k−2 +··· If f(x)is a solution to y =−y,then f (x) =−f(x). 2a 2 +3·2·a 3 x +4·3·a 4 x 2 +···+k(k − 1)a k x k−2 +···=−a 0 −a 1 x−a 2 x 2 − ···−a k x k ··· The key notion is that for these two polynomials to be equal the coefficients of corresponding powers of x must be equal. In other words, the constant terms must be equal, the coefficients of x must be equal, and so on. 2a 2 =−a 0 3·2·a 3 =−a 1 4·3·a 4 =−a 2 5·4·a 5 =−a 3 . . . k(k − 1) · a k =−a k−2 . . . Wecan solve for all the coefficients in terms of a 0 and a 1 . Let a 0 = C 0 , a 1 = C 1 .We’ll solve for a k , k = 2, 3, ···in terms of C 0 and C 1 . a 2 =− a 0 2 = −C 0 2! a 3 = −a 1 3 · 2 = −C 1 3! a 4 = −a 2 4 · 3 = C 0 4 · 3 · 2! = C 0 4! a 5 = −a 3 5 · 4 = C 1 5 · 4 · 3! = C 1 5! a 6 = −a 4 6 · 5 = −C 0 6 · 5 · 4! = −C 0 6! a 7 = −a 5 7 · 6 = −C 1 7 · 6 · 5! = −C 1 7! a 8 = −a 6 8 · 7 = C 0 8 · 7 · 6! = C 0 8! a 9 = −a 7 9 · 8 = C 1 9! . . . . . . a 2n = −a 2n−2 (2n)(2n − 1) = (−1) n C 0 (2n)! a 2n+1 = −a 2n−1 (2n + 1)(2n) = (−1) n C 1 (2n + 1)! . simple to integrate. The use of power series expansions as an integration tool figured prominently in Newton’s work and continues to be important in the integration of otherwise intractable functions. . these functions, known as hyperbolic functions, are used. The arch in St. Louis, the shape of many pottery kilns, and the shape of a hanging cable are all connected to hyperbolic trigonometric functions. 30.4. the other hand, it can be proven that if a series converges absolutely to a sum of S, then any rearrangement of the terms has a sum of S as well. This is one of the reasons we prefer to work with