Convergence of Taylor and Maclaurin Series

Convergence of Taylor and Maclaurin Series Instructional Unit Overview Focus: How to determine the radius and interval of convergence of a Taylor or Maclaurin series In this three-day unit students are introduced to the idea of the interval and radius of convergence of a Maclaurin or Taylor series. The first lesson has students determine this interval visually and by checking suspected endpoints of the interval. Students 1 start by considering the series for sin(x) and also the series for since this 1− x expression can be viewed as the sum of a geometric series. Next they investigate a series that is not geometric but appears to have a limited interval of convergence. In the process of exploring this series, the harmonic series is introduced along with the integral test. In a homework problem, students will encounter an alternating harmonic series and need to reason out its convergence. An investigation is also provided that leads students through much of this material independently or in a small-group setting rather than in a large class format. The second lesson introduces the ratio test and formalizes the alternating series test. Time is spent practicing these tests on portions of previous free-response problems. An investigation is provided that asks students to explore the ratio of the terms of a series in the limit and generalize the ratio test from their results. The third lesson addresses some interesting questions about series, such as “Does a Taylor series always converge to the function from which it was constructed?” “Can a series converge to more than one function?”, and “Is every series that converges Special Focus: Calculus to a function the Taylor series for that function?” Reflecting on questions like these and extending beyond the normal range of material helps students gain perspective on the larger picture. Assumed skills and knowledge: Students should be able to use a calculator or computer to graph a function and evaluate it at a point. They should understand the meaning of convergence and divergence for an infinite series. Students should know the formula for the sum of a geometric series. Students should be able to construct a Taylor or Maclaurin series for a given function. They should know the Lagrange form of the error and be able to use it to estimate the accuracy of an approximation made with a Taylor polynomial. Background information on Taylor series: When finding the Taylor series for a function, it is important to determine for what values of x the series approximates the generating function. Finding this interval of convergence is sometimes a very straightforward procedure. Often the greatest challenges come in determining convergence at the endpoints of the interval. The Taylor series for a function about x = a is constructed according to the formula f ′′ ( a) f ′′′ ( a) f n ( a) ( x − a )2 + ( x − a)3 + L + ( x − a ) n + L. f ( x ) = f ( a) + f ′ ( a) ( x − a) + 2! 3! n! Students should recognize the first two terms of this series as the linear approximation for a function, or the equation of the line tangent to f(x) at x = a. I like to check the students’ understanding of the information presented in the first few terms of the series by using exercises like the following: 1. If the Taylor series is constructed for the function shown at right, centered at the point (1, 3), what can be determined about the coefficients of the first three terms of the series? Solution: The first term is 2, the value of f(1). The coefficient of the second term is positive because the first derivative or slope of the tangent is positive. The coefficient of the third term is also positive because the function is concave up near the point (1, 2). 18 Convergence of Taylor and Maclaurin Series 2. The third-degree Taylor polynomial T about x = 1 for a function, f (x), is T ( x ) = 3 + 2 ( x − 1)2 − 4 ( x − 1)3 . What information do you know about f(x)? Solution: From the first term you know that f(1)=3. There is no linear term so the first derivative at the point (1, 3) is 0, and this is a critical point. The coefficient of the second-degree term is positive, so the function is concave up near the point (1, 3), making this critical point a minimum. The second derivative has a value of 4 and the third derivative has a value of -24. Other problems that test students’ understanding of the construction of Taylor series give an expression for the nth derivative of the function. In student work one common error is to neglect to include the factorials in the denominators. A nice example of this sort of problem comes from the Calculus BC Exam, question 6. BC#6 Let f be a function with derivatives of all orders and for which f (2) = 7. When n is odd, the nth derivative of f at x = 2 is 0. When n is even and n ≥ 2, the nth derivative of f at ( n − 1)! ( n) x = 2 is given by f ( 2 ) = . 3n a. Write the sixth-degree Taylor polynomial for f about x = 2. Solution: P6 ( x ) = 7 + 1! 1 3! 1 5! 1 · ( x − 2 )2 + 4 · ( x − 2 )4 + 6 · ( x − 2 )6 2 3 2! 3 4! 3 6! Students must pay close attention to the fact that the odd derivatives are 0. They must also realize that the sixth-degree Taylor polynomial includes terms up to and including (x – 2)6, and not necessarily six terms. b. For the Taylor series for f about x = 2, what is the coefficient of (x – 2)2n for n ≥ 1? Solution: ( 2 n − 1)! 1 1 · = 2n 2n ( 2 n)! 3 ( 2 n) 3 Students can either substitute 2n for n in the general term of the series, or observe the pattern in the terms of the series with powers of 2 ∙ 0, 2 ∙ 1, 2 ∙ 2, and 2 ∙ 3, then create a general term based on that pattern. Once again they must be careful to remember to include the factorial in the denominator and not just state the value of the derivative. c. Find the interval of convergence of the Taylor series for f about x = 2. Show the work that leads to your answer. 19 Special Focus: Calculus The first two parts of this question deal with constructing the Taylor polynomial and series, and can be done before studying the interval of convergence. The problem of determining the interval of convergence is the focus of the next three lessons. Day 1: Introducing the Interval of Convergence To introduce the idea of the interval of convergence I have students look at the Taylor series for f (x) = sin(x) and graph partial sums on their calculators. We then discuss how it appears that as one adds more terms of the series, the partial sums “fit” the function better and better. I ask them if they continued this forever, would the series fit the entire function? Or is there some x value beyond which the series just won’t fit, no matter how many terms are added on? How do they know? This usually prompts a good discussion and provides a motivation for looking for these x values where the series “fits” or converges to the function. Next I like to introduce a second example that has a limited interval of convergence, 1 for instance, the function g ( x ) = , and look at several of its partial sums. In 1− x this case it becomes readily apparent that this series is a good fit only in a small interval. Students easily see that it is not good for any approximations when x is greater than 1, and they may also guess the lower bound of the interval to be –1. I ask them why this happens, and since we have discussed infinite geometric series previously, we can usually come to the realization that this function can be thought of as the sum of a geometric series with first term 1 and ratio x. They then make the connection that it converges only when x < 1 . 20 Convergence of Taylor and Maclaurin Series The next issue we consider is what happens if the series doesn’t appear to converge for all values of x but is not geometric. How can we determine the interval of convergence? We are now ready to return to part (c) of BC, #6. If we take our answer 1 ·( x − 2 )2 n . from part (b) and write the general term of the series we get 2 n 3 ( 2 n) 1  x − 2 2n · Or by rearranging a bit we can write it as 2 n  3  . In this form it looks like some relative of a geometric series. We look at the graphs of some of the partial sums and try to make an estimate as to where the series will converge. Based on the graphs students usually agree that the series converges for x = 3 and x = 4, but might wonder about x = 5 and anything greater. So we try out one or more of these x values to see what the series looks like if we substitute them in place of x. 21 Special Focus: Calculus 1  1 2n · For instance, if x = 3 the general term becomes . The second factor here 2n  3 is the nth term of a convergent geometric series. Students can then reason that the first factor is less than 1 and thus makes the terms of the actual series smaller than those of the convergent geometric series. So the series converges when x = 3. I have students investigate other values of x. They quickly realize that for x > 5 the geometric part of this series diverges and the first factor probably doesn’t compensate for this divergence. A good discussion usually surrounds the case of x = 5. In this case the first factor essentially becomes one-half the nth term of the harmonic series. I like to encourage this discussion as students wrestle with this important series. Eventually we are led to looking at the series graphically and thinking about the integral 1 associated with the function f ( x ) = . x ∞ The areas of the rectangles represent the terms of the harmonic series 1 ∑ n . The n= 1 sum of their areas is clearly larger than the area under the curve. We have studied ∞ 1 this function and the improper integral ∫ dx and know that it is divergent. So it is x 1 logical that the harmonic series is also divergent. We often have to go back and redo this integral, but since this is such an important result, I think the time spent is worth it. Students are now ready to state the upper bound of the interval of convergence as 5 (with divergence at x = 5). By similar reasoning they realize that the series will diverge if x < –1. The next question is whether it converges at x = –1. Because of the 2n in the exponent, the series is identical to that when x = 5 so it diverges here as well. 22 Convergence of Taylor and Maclaurin Series At this point I like students to try these ideas on their own, and I ask them to ∞ x n−1 investigate the Taylor series for cos(x) along with another series such as ∑ n for n = 1 n·3 their homework. They should be able to use the graphs of the partial sums and their reasoning about geometric series, along with the harmonic series to find the intervals ∞ x n−1 of convergence for these series. In ∑ n , one endpoint of the interval creates an n = 1 n·3 alternating series. I ask students to think about this and try to reason out on their own whether it will converge. They usually decide that if the partial sums are oscillating and those oscillations are getting smaller, the series will converge. This informal understanding of the ideas behind the alternating series test helps them when we study it formally later. They are also motivated to learn the ratio test the next day in order to make this search for the interval of convergence a bit easier. Day 2: A Procedure for Determining the Interval of Convergence During the next class we look at the ratio test and practice its use. I do not formally prove the test, but rather present it with the sort of informal reasoning outlined below. The Ratio Test an+1 L either exists or is infinite. =L n →∞ a n Suppose the limit lim Then a. If L < 1, the series ∞ ∑a n converges. n diverges. n= 1 ∞ b. If L > 1, the series ∑a n= 1 c. If L = 1, the test is inconclusive. If L < 1, this means that successive terms are getting larger and it is logical that the series diverges. If L < 1, this means that successive terms are getting smaller and it is possible that the series converges. Look a bit more closely at the terms of a power series ∞ ∑c n= 0 n x n , where an = cnxn. The power on x increases regularly. This is like a geometric 23 Special Focus: Calculus series. Of course, the coefficients don’t always change by the same ratio, but we can still think of this as a sort of cousin to a geometric series. The ratio formed in the limit is less than some number r for a geometric series. Since the limit is less than 1, you an +1 can pick a number for r that is also less than 1 and above the ratio . So our series an is less than this convergent geometric series and it makes sense that if L < 1, the series will converge. When L = 1 the result is inconclusive. In the harmonic series, L = 1 and the series ∞ 1 diverges. However, the series ∑ 2 has L = 1 again, but this series converges. So if L n= 1 n = 1, we can’t say whether the series converges. We go back and apply the ratio test to part (c) from Calculus BC, #6, andget the following: 1 1 · 2( n + 1) ( x − 2 )2( n + 1) 2n 32 n n ( x − 2 )2 ( x − 2 )2 2 ( n + 1) 3 lim = lim · 2 n 2 ·( x − 2 )2 = lim · = . 1 1 n→ ∞ n + 1 n→∞ n→∞ 2 ( n + 1) 3 3 9 9 · 2 n ( x − 2 )2 n 2n 3 The series converges when this limit is less than 1: ( x − 2 )2 9 < 1 or ( x − 2 )2 < 9 or x − 2 < 3 . Written this way we can see the set of solutions to this inequality as an interval centered at 2 with a radius of 3. Here ideas of function transformations can help a student to understand the relationship of an inequality like this to the parent inequality x < 1 . The radius of this interval is called the radius of convergence. In this case, −1 < x < 5 is a subset of the interval of convergence. In order to complete the analysis of this interval of convergence, the behavior at the endpoints must be investigated. After the hard work of graphing partial sums and checking values, students find using the ratio test to be much simpler and more direct. Checking the endpoints then becomes our challenge, and we review the work we did the previous day. ∞ ( −3 )2 n 1 ∞ 1 For x = –1, the series becomes 7 + ∑ = 7 + ∑ . This series is the 2n 2 n=1 n n = 1 2 n·3 harmonic series and it diverges. 24 Convergence of Taylor and Maclaurin Series ∞ 32 n 1 ∞ 1 = 7 + ∑ . This series diverges 2n 2 n=1 n n = 1 2 n·3 because of its relationship to the harmonic series. Thus the interval of convergence is –1 < x < 5. Since quite often when evaluating the endpoints of an interval, the series can be compared to a harmonic or alternating harmonic series, it is now appropriate to formalize the alternating series test. For x = 5, the series becomes 7 + ∑ Alternating Series Test ∞ If a series is of the form ∑ ( −1) n+1 n=1 an , where all an > 0 and 0 < an+1 < an for all n greater than some integer N, and lim an = 0 , then the series converges. n→∞ When the hypotheses are not satisfied, other convergence tests must be used. See the article “Overview of Tests for Convergence of Infinite Series” for more details on this topic. I next like to work through several problems where students have the opportunity to put these tests into practice. 2000 BC, #3 The Taylor series about x = 5 for a certain function f converges to f(x) for all values of x in the interval of convergence. The nth derivative of f at x = 5 is given by ( −1) n n! 1 f (5 ) = n , and f (5 ) = . 2 ( n + 2) 2 ( n) a. Write the third-degree Taylor polynomial for f about x = 5. ( −1) n f ( n) (5 ) Solution: The coefficients of the polynomial are given by n! = 2 n ( n + 2 ) . using n = 1, 2, and 3 creates P3 ( x ) = 1 1 1 1 − ( x − 5 ) + ( x − 5 )2 − ( x − 5 )3 . 2 6 16 40 b. Find the radius of convergence of the Taylor series for f about x = 5. Solution: Using the ratio test 25 Special Focus: Calculus ( −1) n + 1 ( x − 5 ) n + 1 −1( n + 2 ) ( x − 5 ) 1 ( n + 2) x −5 2 n+1 ( n + 3) lim = lim = lim · x −5 = n n n →∞ n →∞ n →∞ ( −1) ( x − 5 ) 2 ( n + 3) 2 ( n + 3) 2 n 2 ( n + 2) . x -5 < 1 or x - 5 < 2 . If this limit is less than 1, the series will converge: 2 From here we can see the interval of convergence is centered at 5 with a radius of 2. The problem did not ask for the actual interval, but it’s good practice to go ahead and work this out. So far we have the interval 3 < x < 7. Now check the endpoints. (−1) (−2) = (2) = 1 If x = 3, the series becomes ∑ . It diverges ∑ ∑ 2 (n + 2 ) 2 (n + 2 ) (n + 2) n ∞ n n n=0 n ∞ n=0 n ∞ n=0 because of its relationship to the harmonic series. (−1) (2) = (−1) , which is an alternating If x = 7, the series becomes ∑ ∑ (n + 2) 2 (n + 2 ) n ∞ n n n=0 n ∞ n=0 harmonic series and thus converges. So the interval of convergence is 3 < x ≤ 7. c. Show that the sixth-degree Taylor polynomial for f about x = 5 approximates f (6) with error less than 1 . 1000 Solution: Since x = 6 is in the interval of convergence, we can use the series to approximate the value of f(6). But how good is our approximation? If students substitute x = 6 in the sixth-degree polynomial, it will look like this: f(6) ≈ 1 2 – 1 6 + 1 16 – 1 40 + 1 96 – 1 224 + 1 512 = 6787 ≈ 0.3787. 17920 Ask students if this is more or less than the sum of the entire series. By looking at the signs they should realize that this is more than the actual sum. Ask students how much the next term will change this total. Since the next term is subtracted, it will 1 . Then ask them to think about how close they were to the decrease the sum by 1152 26 Convergence of Taylor and Maclaurin Series 1 causes the total to go from too big 1152 to too small, you were less than this far from the actual sum. Since this is less than 1 the requested accuracy, the error in this approximation is within of the actual 1000 value. This sort of reasoning should lead students to the realization that when a series satisfies the conditions of the alternating series test, the error in any partial sum is less than the next term. actual sum. Since decreasing the sum by 2002 BC, #6 The Maclaurin series for the function f is given by ∞ f (x ) = ∑ n=0 (2x ) (2x ) + … 4x 2 8x 3 16x 4 = 2x + + + +…+ n+1 2 3 4 n+1 n+1 n+1 on its interval of convergence. a. Find the interval of convergence of the Maclaurin series for f. Justify your answer. Parts (b) and (c) deal with manipulation of Taylor and Maclaurin series, which is covered in the “Manipulation of Power Series” section of this Special Focus. Solution: (2x ) n+ 2 2 = lim (n + 1) 2x = 2x . Apply the ratio test lim n + n+1 n→∞ n→∞ (n + 2 ) (2x ) n+1 This indicates the series converges if − 1 If x = − , the series becomes 2 converges. If x = 1 , the series becomes 2 ∞ ∑ n=0 ∞ (−1) 1 1 < x < . But what about the endpoints? 2 2 n+1 n+1 , an alternating harmonic series, so it 1 ∑ n + 1 , the harmonic series, so it diverges. n=0 1 1 Thus the interval of convergence is - ≤ x < . 2 2 27 Special Focus: Calculus After discussing these problems I present students with a problem like BC,Form B, #3. This is a nice problem that ties together understanding the meanings ofthe first several terms of the Taylor series, constructing the series from a definitionof the derivatives and finding the radius of convergence. Doing this as a homework assignment helps students to discover just where they have questions on this process. In class we also work to find the actual interval of convergence to make the problem a complete summary of the work done so far. Evaluating convergence at the endpointsis not as easy here as in the problems presented thus far, and it will take some work together as a group to come to the right conclusions and justifications. BC, #3 The Taylor series about x = 0 for a certain function f converges to f ( x) for all values of x in the interval of convergence. The nth derivative of f at x = 0 is given by f (n) (−1) (n + 1)! for n ≥ 2. The graph of f has a horizontal tangent line at x = 0 (0) = 5 (n − 1) n+1 2 n and f (0) = 6. a.D etermine whether f has a relative maximum, a relative minimum, or neither at x = 0. Justify your answer. Solution: By the second derivative test, f has a relative maximum at x = 0 because f ′(0) = 0 (due to the horizontal tangent) and f ′′(0) < 0. b. Write the third-degree Taylor polynomial for f about x = 0. Solution: Using the given information f (0 ) = 6, f ′ (0 ) = 0, f ′′ (0 ) = − P (x ) = 6 − 3 2 1 3 x + x 25 125 3! 6 4! 6 = − , f ′′′ (0 ) = 3 2 = 2 2 51 25 52 125 c.F ind the radius of convergence of the Taylor series for f about x = 0. Show the work that leads to your answer. Solution: Using the ratio test: (−1) (n + 2)x n+ 2 5n+1 n 2 lim n→∞ (−1)n+1 (n + 1) 28 5n (n − 1) 2 n+1 x = lim n n→∞ −1 (n + 2 )(n − 1) x 2 5 (n + 1)(n ) 2 2 x  n + 2  n − 1 x = lim  = n→∞  n + 1    n  5 5 Convergence of Taylor and Maclaurin Series The series converges if |x| < 5 or –5 < x < 5 . The radius of convergence is 5. But what about those pesky endpoints? If x = –5 , the series becomes (−1) (n + 1)(−5) 6+∑ 5 (n − 1) n+1 ∞ n 2 n n= 2 ∞ = 6+∑ n= 2 (−1) (n + 1) (n − 1) 2 ∞ = 6−∑ n= 2 (n + 1) . (n − 1) 2 Because the numerator of each term of this series is one degree less than the degree of the denominator, this series is reminiscent of the harmonic series. Note that n +1 2 (n − 1) > n −1 2 (n − 1) = 1 , the nth term of a shifted harmonic series, which is n −1 ∞ divergent. So the series ∞ series 6 − ∑ n= 2 (n + 1) . (n − 1) (n + 1) ∑ (n − 1) n=2 2 diverges by comparison, and thus so does the 2 (−1) (n + 1)(5) If x = 5, the series becomes 6 + ∑ 5 (n − 1) ∞ n= 2 n+1 n n 2 (−1) (n + 1) = 6+∑ (n − 1) . ∞ n= 2 n+1 2 In this case, the series alternates and satisfies the conditions of the alternating series test, so it converges. The interval of convergence is then –5 < x ≤ 5. Day 3: Some Interesting Questions About Convergence Students are now familiar with constructing Taylor series and identifying the radius of convergence. The verification of endpoint convergence is something we will continue to work on as we develop the convergence tests more fully. However, it is helpful to take some time to discuss questions about convergence that may not have otherwise occurred to students. We have seen that the series for sin(x)converges for all values of x, but how do we know it converges to the function sin(x) and not something else? This question may seem silly at first to students, but it is worth examining. It is possible for a series to converge and yet not converge to the function used to construct it. In order to show that a series does indeed converge to the original function, we must examine what happens to the error. 29 Special Focus: Calculus If f and all its derivatives are continuous and Pn(x) is the nth degree Taylor approximation to f (x) about x = a, then the error, En (x ) = f (x )− Pn (x ) ≤ M n+1 x − a , where M is the maximum value of f (n+1) on (n + 1)! the interval between x and a. Applying this formula to f (x)= sin(x) and its series expansion ( −1) x 2n +1 ≤ 1 x n +1 , where x 3 x 5 x7 En ( x ) = sin x − x − + − +L+ 3! 5! 7 ! ( 2n + 1)! (n + 1)! we use M = 1 because all derivatives of f (x)= sin(x) have a maximum value of 1, and a = 0 because the series is centered at 0. If this error has a limit of 0, then the series converges to the function f (x)= sin(x) for all values of x in the interval of n+1 x convergence. This means we must show that lim = 0 . Using a calculator n→∞ (n + 1)! and experimenting with various values of x and large values of n provides evidence that this limit statement is true. To consider the limit more analytically, select a x 1 positive integer N such that N > 2|x|. Note that for n ≥ N, < , the expression € n 2 n x n+ 1 x x L . For n ≥ N, each · · ⋅ … ( n + 1)! ( n + 2 ) ( n + 3 ) € 1 x factor of in the expression will be less than . So for n ≥ N we can state 2 (n + k ) in the limit contains factors such as x n+1 xN x x xN 1 ≤ · ·L· ≤ · . Thus the limit will go to 0 as n→∞. Therefore ( n + 1)! N ! N + 1 n N ! 2 n- N € €the Taylor series for sin(x) converges to this function for all values of x. Students can work through similar arguments to show that the series for cos(x) converges to cos(x) for all x, and that the series for ex converges to ex for all values of x. It may be instructive to look at a series that does not converge to the function e −1/x 2 x ≠ 0 that is used to construct it. For instance, the function f (x ) =  has x=0  0 derivatives at x = 0 that are all equal to 0. (We need to use the limit definition of the derivative to show this.) The Taylor series for the function is simply a series of zeroes. This series certainly converges regardless of the value of x, and it converges to 0 for all values of x. So it does not converge to f ( x) for any values of x but 0. Students do not 30 Convergence of Taylor and Maclaurin Series generally encounter functions like this in an introductory course, but should be aware that they exist and not take for granted that every Taylor series converges to the function used to construct it. A second interesting question is, “Can a Taylor series converge to different functions on different parts of its interval of convergence?” Again this may seem (−1) x counterintuitive, but it can happen in some sense. Consider the series ∑ (2n)! ∞ n n n=0 . Graphing a few of the partial sums shows different behavior on each side of 0. Using the ratio test we find that the series converges for all values of x. This happens to be the series you get if you replace x in the series for cos(x) with x . The function cos ( x is not defined for x < 0 if we stay in the real number system. ) So what does it converge to when x < 0? Students may not have learned about the ( ) hyperbolic functions, but this series actually converges to cos x when x < 0. For some students it will be sufficient to see that this function does “match” the series when x is negative. An activity sheet is provided to help students work through the demonstration of the relationship. A third question that can be explored is, “Is every series that converges to some function a Taylor series, for that function?” Since students have been focusing on Taylor series, it is natural for them to think that any convergent series is a Taylor series. ∞ x2 2 n . However, consider the series ∑ n=0 1 + x ( ) This is a geometric series. Using the principles for finding where a geometric series converges, you 1 find the following: < 1 so 1 < 1 + x 2 or x 2 > 0 . 2 1+ x If we use x = 0, the series is convergent to 0 because all terms are 0. So this series converges for all values of x. If you apply the formula for the sum of a convergent geometric series, you get: x2 1− 1 1 + x2 ( = ) x2 x2 = = 1 + x 2 for x ≠ 0. So here we have a series that 1 + x2 − 1 x2 1 + x2 1 + x2 converges everywhere, but to different functions for different values of x. 31 Special Focus: Calculus However, it is clearly not a Taylor series since the Taylor series for a polynomial is simply that polynomial. While students are not apt to encounter series like this in this course, it is nevertheless valuable for them to realize that such things can happen and that not everything is as simple as applying the ratio test and recognizing harmonic series. The materials that follow are investigations that can be given to students. Investigation 1: Convergence of a Taylor Series 1. Write the Taylor series for f (x)=sin(x). 2. Graph the following Taylor polynomials for f (x)=sin(x) along with f (x)=sin(x):T5 (x), T11(x), T17 (x). Record a sketch of your graphs here. 3. Based on your graph above, for what values of x do you think you could use the series to approximate the values of f (x)=sin(x)? Explain. 32 Convergence of Taylor and Maclaurin Series 1 4. Write the Maclaurin series for f (x ) = 1 − x . (Hint: Think of this function as the sum of an infinite geometric series.) 1 1 along with f (x ) = : 1− x 1− x T4 (x ),T7 (x ),T12 (x ). Record a sketch of your graphs here. 5. Graph the following Taylor polynomials for f (x ) = 6. Based on your graph above, for what values of x do you think you could use the 1 series to approximate the values of f (x ) = ? Justify your answer using ideas 1− x of convergent geometric series. 7. When you know the function that was used to create a Taylor series, you can often look at the graphs of some of the polynomials and estimate the interval of convergence from those graphs. However, when you are simply given the series 33 Special Focus: Calculus for an unknown function, it may be a bit harder to determine its interval of ∞ x n−1 convergence from the graph. Consider the series ∑ n . Write the first four terms n=1 n·3 of the series. 8. Graph several of the partial sums of this series and estimate the interval of convergence based on your graphs. 9. To check your interval of convergence, you can look at the series you obtain if you substitute a value for x. For instance, if you evaluate the series for x = 2, you have n ∞ 2n−1  2 1 ∑ n·3n or ∑  3  2n . This new series should remind you of a geometric series n=1 n=1 ∞ with its nth term multiplied by another factor. Study the following argument for the convergence of this series: ∞ (i) ∑ n=1 (ii) 1 < 1 for all values of n ≥ 1. 2n ∞ € (iii)  2 n 1 converges because its terms (all of which are positive) are less   ⋅  3  2n ∑ n=1 €  2 n   converges because it is a geometric series with |r| <1.  3 than those of the geometric series. So this sum will be less than the sum of the geometric series. € Think about putting other values in place of x. How large can x be if the series is still to converge? In other words, what is the upper bound of the interval of convergence? 34 Convergence of Taylor and Maclaurin Series 10. If you substitute a negative value for x, the terms of the series will alternate in sign. You will learn an official test for alternating series convergence later, but for now think about what happens when you add up a string of numbers that alternate in sign. If the terms you are summing are getting closer and closer to zero then the partial sums behave something like the picture below, where the upward arrows represent positive terms and the downward arrows represent negative terms. If more terms were added you can imagine that the series would converge to some point between the high and low, something like a spring bouncing and finally coming to rest at an equilibrium position. x n−1 For the series ∑ n , what negative values of x would result in the convergence n=1 n·3 ∞ of the series? In other words, what is the lower bound of the interval of convergence? Investigation 2: The Ratio Test When a series is geometric, it is easy to tell when it converges. However, if the series is not geometric and not closely related to a geometric series, it can be more difficult to find this interval of convergence. In this investigation you will look for relationships based on the ideas of geometric series that will be useful in determining the convergence of other series. an+1 is always the an same, no matter which two consecutive terms you use. The series converges if this ratio has an absolute value less than one. In nongeometric series this ratio is not constant, it can change with each pair of terms. However, since the series is infinite it is reasonable to look at the limit of this ratio to discover the behavior of the terms in the long run. 1. The terms of a geometric series have a constant ratio. That is, 35 Special Focus: Calculus an+1 . n→∞ a n Below are two lists of series. For each one determine lim Converging Series Diverging Series 2 ∑ n2 + 1 n=0 1  3 ∑ n  2  n=1 ∞ n ∑n+1 n=1 2 ∑ n! n=0 ∞ (n!)2 n ∑ (2n)! ∞ n! ∑n n=1 2 2n ∑ n2 n=1 ∞ ∞ n n ∞  6 ∑  7  n=0 ∞ ∞ n=0 2. If the limit of this ratio is less than 1, what seems to be true about the series? Why is this reasonable? 3. If the limit of the ratio is more than 1, what seems to be true about the series? Why is this reasonable? 4. If the limit of the ratio is 1, what seems to be true about the series? Why is this reasonable? 5. Using what you have discovered, for what interval of values do you think the series ∞ ∑ n=0 36 (x − 2) n 10n will converge? Convergence of Taylor and Maclaurin Series 6. The results you discovered are summarized as the ratio test for convergence. It will be one of the most useful tests you learn for finding the interval of convergence of a Taylor series. The Ratio Test Suppose the limit lim n→∞ an+1 = LL either exists or is infinite. an Then ∞ ∑a If L > 1, the series ∑ a a. If L < 1, the series n converges. n diverges. n=1 ∞ b. n=1 c. If L = 1, the € test is inconclusive. € Investigation 3: A Series That Converges to Two Different Functions (in Form) 1. Write the first four nonzero terms and general term of the Maclaurin series for f ( x) =cos( x). 2. You know that you can create a series for cos( g ( x)) by replacing each x in the series with g ( x). You have done this with such functions as g ( x) = x2. Now use this idea to write a series for f (x ) = cos x . ( ) 3. Find the interval of€convergence for your series. 37 Special Focus: Calculus 4. Verify by graphing that the series seems to converge to the original function for x ≥ 0. Sketch a graph of f (x ) = cos x and two different series approximations to show this convergence. ( ) € 5. What happens when x < 0? The interval of convergence indicates that the series will still converge, but f (x ) = cos x is not defined for x < 0. What will the series converge to then? We can find an answer to this question by examining the e x + e− x function (called the hyperbolic cosine function) cosh ( x ) = . Write the 2 € series for ex here. ( ) 6. Write the series for e-x here. 7. Use the two series from steps 5 and 6 to get a series for cosh( x). ) 8. Use the series from step 7 to get a series for cos ( x . ) 9. For x < 0, use the series from step 8 to get a series for cosh ( − x . 10. Let h( x) be the function represented by the series in step 2. Express h( x) in terms of the cos ( x and cosh ( − x . Verify by graphing cosh ( − x and two different series approximations to illustrate convergence for x < 0. ) 38 ) ) Convergence of Taylor and Maclaurin Series Teacher Notes on the Investigations Investigation 1: Convergence of a Taylor Series 1. Write the Taylor series for f ( x) = sin( x). f (x) = x − x3 x5 x7 x 2 n+1 + ... + − + ... + ( −1) n ( 2 n + 1)! 3! 5 ! 7 ! 2. Graph the following Taylor polynomials for f ( x) = sin( x) along with f ( x) = sin( x): T5 ( x), T11 ( x), T17 ( x). Record a sketch of your graphs here. 3. Based on your graph above, for what values of x do you think you could use the series to approximate the values of f ( x) = sin( x)? Explain. It appears that as you add more terms to the polynomial, it fits a larger portion of the curve. Since the series is infinitely long, it is reasonable that the series would fit the entire function. So you could use it to approximate values of f ( x) = sin( x) for all values of x. However, if the x value is far from zero, it would take many terms to have a good approximation. 1 4. Write the Maclaurin series for f (x ) = . (Hint: Think of this function as the sum 1− x of an infinite geometric series.) f ( x ) = 1 + x + x 2 + x 3 + ... + x n + ... 39 Special Focus: Calculus 1 1 along with f (x ) = : 1− x 1− x T4 ( x), T7 ( x), T12 ( x). Record a sketch of your graphs here. 5. Graph the following Taylor polynomials for f (x ) = 6. Based on your graph above, for what values of x do you think you could use the 1 series to approximate the values of f (x ) = ? Justify your answer using ideas 1− x of convergent geometric series. The series appears to “match” the function well in the interval –1 < x < 1. This is reasonable because it is a geometric series with ratio x, and thus converges when |x| < 1. 7. When you know the function that was used to create the Taylor series, you can often look at the graphs of some of the polynomials and estimate the interval of convergence from those graphs. However, when you are simply given the series for an unknown function, it may be a bit harder to determine its interval of ∞ x n−1 convergence from the graph. Consider the series ∑ n . Write the first four terms n=1 n·3 of the series. 1 x x2 x3 + + + 3 2·3 2 3·33 4·3 4 8. Graph several of the partial sums of this series and estimate the interval of convergence based on your graphs. 40 Convergence of Taylor and Maclaurin Series Based on the graphs of partial sums students may estimate the interval of convergence to be –4 < x < 3. This is not the correct interval, but because the series actually converges for x = –3, students may think the interval extends beyond this value. 9. To check your interval of convergence, you can look at the series you obtain if you substitute a value for x. For instance, if you evaluate the series for x = 2, you have n ∞ ∞ 2n−1  2 1 ∑ n·3n or ∑  3  2n . This new series should remind you of a geometric series n=1 n=1 with its nth term multiplied by another factor. Study the following argument for the convergence of this series: ∞ (i) ∑ n=1 (ii) € € € 1 < 1 for all n ≥ 1. 2n ∞ (iii)  2 n   converges since it is a geometric series with |r| < 1.  3  2 n 1 converges because its terms (all of which are positive) are less   ⋅  3  2n ∑ n=1 than those of the geometric series. So this sum will be less than the sum of the geometric series. Think about putting other values in place of x. How large can x be if the series is still to converge? In other words, what is the upper bound of the interval of convergence? 41 Special Focus: Calculus If x = 3, this series becomes the harmonic series and will diverge. For any positive value less than 3, however, the series will converge. 10. If you substitute a negative value for x, the terms of the series will alternate in sign. You will learn an official test for alternating series convergence later, but for now think about what happens when you add up a string of numbers that alternate in sign. If the terms you are summing are getting closer and closer to zero, then the partial sums behave something like the picture below, where the upward arrows represent positive terms and the downward arrows represent negative terms. If more terms were added you can imagine that the series would converge to some point between the high and low, something like a spring bouncing and finally coming to rest at an equilibrium position. x n−1 , what negative values of x would result in the n n=1 n·3 convergence of the series? In other words, what is the lower bound of the interval of convergence? ∞ For the series ∑ When trying to find the lower bound by substituting values of x, students will find that if they use a number less than –3, the “geometric” factor in the series is divergent. Using x = –3, they create an alternating harmonic series. Students should be able to reason that this will converge because the terms being added alternate in sign and get smaller in absolute value. When discussing this question, you may wish to formally state the alternating series test and show how it applies here. Investigation 2: The Ratio Test When a series is geometric, it is easy to tell when it converges. However, if the series is not geometric and not closely related to a geometric series, it can be more difficult to find this interval of convergence. In this investigation you will look for relationships 42 Convergence of Taylor and Maclaurin Series based on the ideas of geometric series that will be useful in determining the convergence of other series. an+1 is always the an same, no matter which two consecutive terms you use. The series converges if this ratio has an absolute value less than 1. In nongeometric series this ratio is not constant; it can change with each pair of terms. However, since the series is infinite it is reasonable to look at the limit of this ratio to discover the behavior of the terms in the long run. a Below are two lists of series. For each one determine lim n +1 . n→∞ a n 1. The terms of a geometric series have a constant ratio. That is, Converging Series Diverging Series an+1 2 ∑ n2 + 1 lim n→∞ a n=0 n n ∞ an+1  6 ∑  7  lim n→∞ a n=0 n n ∞ an+1 2 lim ∑ n! n→∞ a n=1 n 2 ∞ (n!) lim an+1 ∑ (2n)! n→∞ a n=0 n 1  3 ∑ n  2  n=1 ∞ n ∑n+1 n=0 ∞ n! ∑ n2 n=1 ∞ 2n ∑ n2 n=1 6 7 =0 = 1 4 = =1 ∞ ∞ n an+1 n→∞ a n an+1 lim n→∞ a n an+1 lim n→∞ a n an+1 lim n→∞ a n lim = 3 2 =1 =∞ =2 2. If the limit of this ratio is less than 1, what seems to be true about the series? Why is this reasonable? The series converges. When the limit of the ratio is less than 1, it means the terms of the series are approaching 0 and are doing so in a way such that each pair of successive terms has a smaller ratio than the pair before. This is just like a geometric series with a ratio less than 1, and thus the series converges. 3. If the limit of the ratio is more than 1, what seems to be true about the series? Why is this reasonable? The series diverges. When the limit of the ratio is greater than 1, it means the terms of the series are growing in size. With each additional term the total changes by more than it did when the previous term was added. Thus the sum cannot be approaching a single value and the series diverges. 4. If the limit of the ratio is 1, what seems to be true about the series? Why is this reasonable? 43 Special Focus: Calculus The series might converge or it might diverge. Both outcomes are possible if the limit of the ratio is one. When the limit of the ratio is one this means that the terms are becoming more and more alike. If they are also becoming smaller at a fast enough rate, the series will converge. However, if they are either not approaching a limit of 0 or not doing so quickly enough, as in the harmonic series, the series will diverge. 5. Using what you have discovered, for what interval of values do you think the series ∞ (x − 2)n will converge? ∑ 10n n=0 x−2 This is a geometric series with ratio so the series will converge when 10 x−2 < 1 or x − 2 < 10 . This describes an interval of radius 10 centered at 2. 10 So the interval of convergence is –8 < x < 12. There is no convergence at the endpoints since this is a geometric series. 6. The results you discovered are summarized as the ratio test for convergence. It will be one of the most useful tests you learn for finding the interval of convergence of a Taylor series. The Ratio Test Suppose the limit lim n→∞ Then an+1 = LL either exists or is infinite. an ∞ a. If L < 1, the series ∑a n converges. n=1 ∞ b. If L > 1, the series ∑a n diverges. n=1 € c. If L = 1, the test is inconclusive. € Series That Converges to Two Different Functions Investigation 3: A (in Form) 1. Write the first four nonzero terms and general term of the Maclaurin series for f ( x) = cos( x). 2n x2 x4 x6 n x 1− + − + ... + ( −1) + ... ( 2 n)! 2! 4! 6! 44 Convergence of Taylor and Maclaurin Series 2. You know that you can create a series for cos( g ( x))by replacing each x in the series with g ( x). You have done this with such functions as g ( x) = x 2 . Now use this idea to write the series for f (x ) = cos x . ( ) x x2 x3 xn + − + ... + ( −1) n + ... ( 2 n)! 2! 4! 6! € 3. Find the interval of convergence for your series. Using the ratio test, the series converges for all values of x. 1− 4. Verify by graphing that the series seems to converge to the original function for x > 0. Sketch a graph of f (x ) = cos x and two different series approximations to show this convergence. ( ) € 5. What happens when x = 0? The interval of convergence indicates that the series will still converge, but f (x ) = cos x is not defined for x < 0. What will the series converge to then? We can find an answer to this question by examining the e x + e−x function (called the hyperbolic cosine function) cosh(x ) = . Write the series 2 € for ex here. ( ) ee−xx = 1− 1+ x + x 2 x 33 x 44 x5 −+ + −+ +L L +L 2! 3! 4! 5! € 6. Write the series for e–x here. €€ €€ x 2x 2 x 3x 3 x 4x 4 x 5x 5 e−xe−x= =1− 1−x x+ + − − + + − −+ + + + LL L 2!2! 3!3! 4!4! 5!5! 45 Special Focus: Calculus 7. Use the two series from steps 5 and 6 to get a series for cosh( x). Adding the two series from steps 5 and 6 we get e x + e−x = 1+ x + x2 x3 x4 x2 x3 x4 x2 x4 + + +L +L L+ 1− x + − + L= 2 + 2 + 2 + L L. 2! 3! 4! 2! 3! 4! 2! 4! Thus € 1 x2 x4 cosh(x ) = (e x + e−x )= 1+ + +L = 2 2! 4! ∞ ∑ n= 0 x 2n . (2n )! ) 8. Use the series from step 7 to get a series for cosh ( x . € 2 4 2n ( x) + ( x) +L = ( x) = x cosh( x )= 1+ ∑ (2n)! ∑ (2n)! 2! 4! ∞ ∞ n= 0 n n= 0 ) 9. For x < 0, use the series from step 8 to get a series for cosh ( − x . € € ∞ ( )∑ cosh −x = 2n ( −x ) n= 0 ∞ n ∞ n −x ) −1) x n ( ( = = 2n )! ∑ (2n )! (2n)! ∑ ( n= 0 n= 0 10. Let h (x ) be the function represented by the series in step 2. Express h( x) in terms of the cos ( x and cosh ( − x . Verify by graphing cosh ( − x and € two different series approximations to illustrate convergence for x < 0.  €  cos x if x ≥ 0 if x ≥ 0  cos x h (x ) =  =  cosh −x if x < 0 cosh x if x < 0 ) ( ) ( ) A graph is given by: € 46 ) ( ) ( ) ) Convergence of Taylor and Maclaurin Series Students are likely to be unfamiliar with the hyperbolic functions. The connections between the hyperbolic functions and the trigonometric functions are interesting and somewhat unexpected. One can also use complex numbers to explore the relationships between these two types of functions. For example, since i2n equals 1 when n is even and –1 when n is odd, it is easy to use the Maclaurin series for cos( x) to check that cos( i x) = cosh( x). Alternatively, this identity follows from the identity eiθ = cos(θ) + isin (θ), which plays a very important role in complex analysis. Many students find complex numbers fascinating, and when they realize that there is a meaning to the sine or cosine of an imaginary number they are intrigued. Students interested in this topic could investigate what happens with the series for sin( x) when x is replaced by x . This will lead to the identity sin( ix) = isinh( x). 47