Setting the Stage with Geometric Series

Setting the Stage with Geometric Series One of the difficult things about teaching infinite series at the end of an Calculus BC course is trying to make the students see the topic as something other than a four-week detour down a side track just when the train ought to be coming into the station. It is particularly difficult to disabuse students of this notion if the teacher secretly shares their concern. That is why it is easier for both teacher and students if the emphasis is on functions from the beginning, and if calculus becomes part of the picture shortly thereafter. There are good reasons to talk about series in an introductory calculus course, and it is helpful to keep them in mind when thinking about how to teach the topic. Infinite series are another important application of limits. Moreover, as limits they are easier to understand than either the derivative or the integral because they do not involve those mysterious differentials. The construction of Taylor series is not only a nice application of the derivative but also a nice review of such topics as linear approximation, slope, and concavity. If a function can be represented by a power series, it can be differentiated or integrated as easily as a polynomial. This provides another approach to evaluating x expressions like ∫ e dx that would be difficult or impossible otherwise. Series figured prominently in the early history of calculus, so they ought to play some role in an introductory course. Series figure prominently in higher-level analysis courses, so it is useful to lay the groundwork for our better students as soon as we can. Notice that most of these good reasons for teaching series in a calculus course involve series as functions, not series as infinite sums of numbers. In fact, series of numbers are only important when considering the various tests for convergence. One 2 Special Focus: Calculus school of thought is that students need to see all those tests before seeing their first power series, necessitating a detour away from functions and calculus that is difficult to motivate. A more intuitive (and historically faithful) approach is to play with power series from the beginning and see what can be done with them. The question of convergence eventually must be confronted, but in the meantime it is sufficient for students to know that convergence is an issue. Happily, there is a way to introduce students to power series right away and simultaneously make them aware of the question of convergence, all while building on their existing knowledge of a previous topic: geometric series. Geometric Series Basics Even if students have not studied geometric series by name, they have encountered them in various convergent forms. For example, a 1"-by-1" square can be cut into two halves. One half can then be cut into two quarters, one quarter can be cut into two eighths, and so on ad infinitum. This process of infinite subdivision, the basis of some of Zeno’s ancient paradoxes, leads to the inevitable conclusion that 1 1 1 1 n + + + L+   + L = 1 .  2 2 4 8 For another example, some well-known rational numbers have familiar decimal expansions that are actually convergent geometric series: 3 3 3 1 n 1  0. 3 = + + +L + 3 +L = .   10 100 1000 10 3 So infinite sums can converge to finite numbers, but obviously not all of them n+ 1 do. For example, 1 + 1 + 1 + … is infinite, and 1 − 1 + 1 − 1 + L + ( −1) + L is at least ambiguous. The latter series shows the necessity of defining the sum of an infinite series as the limit of its sequence of partial sums. There is a formula for the nth partial sum of a geometric series in which r ≠ 1: a − ar n a + ar + ar 2 + ar 3 + L + ar n−1 = . 1− r 10 Setting the Stage with Geometric Series (There are several ways to prove this, and most precalculus texts do.) The limit of the nth partial sum as n →∞ depends entirely on the fate of rn, which goes to zero if and only if r < 1 . If r ≥ 1 , the limit diverges. Thus, whether or not students have formally studied geometric series, they can be led very quickly to the realization that a series of the form a + ar + ar 2 + ar 3 + L + ar n−1 + L converges if and only if r < 1, in which case the sum a is . 1− r At this point, students are ready for x. A Geometric Series for f ( x ) = 1 1- x 1 is the sum of the geometric series 1 + x + x 2 + x 3 + L + x n−1 + L. 1− x The latter expression looks like a polynomial of infinite degree, but since there is If x < 1, then no such thing we must give it a new name. We call it a power series (a series of powers of x). It is an interesting example of a function with domain (–1, 1), since we can technically plug in values of x outside the domain, but we get expressions like 1 + 2 + 22 + 23 +L that are simply meaningless. If, on the other hand, we plug in a 1 value of x inside the domain, we get . The interval (–1, 1) is called the interval of 1− x convergence. The partial sums of a power series are polynomials. If we graph them, we get a dramatic visualization of why the interval of convergence matters, as in the graphs below of the fourth and fifth partial sums of 1 + x + x 2 + x 3 + L + x n−1 + L compared to 1 the graph of : 1− x 11 Special Focus: Calculus The quartic and quintic polynomial partial sums do a very good job of approximating 1 in the interval of convergence (–1, 1), but outside that interval they are not even 1− x close. The partial sums of higher degree approximate the curve progressively more closely on the interval of convergence, but with no better success outside that interval. Exploring the Implications • Find a power series to represent 1 and give its interval of convergence. 1+ x Solution: We use a geometric series with first term 1 and ratio –x. 1 − x + x 2 − x 3 + L + ( −1) n−1 x n−1 + L . It converges for − x < 1 , so the interval of convergence is (–1, 1). • Find a power series to represent 1 and give its interval of convergence 1+ x2 2 Solution: We use a geometric series with first term 1 and ratio − x . 1 − x 2 + x 4 − x 6 + L + (−1) n−1 x 2 n− 2 + L. 2 It converges for − x < 1 , so the interval of convergence is (–1, 1). • −1 Find a power series to represent tan ( x ) and give its interval of convergence. Solution: We just found a series to represent 1 − x 2 + x 4 − x 6 + L + (−1) n−1 x 2 n− 2 + L 12 . d 1 tan −1 ( x )) = : ( dx 1+ x2 Setting the Stage with Geometric Series The interval of convergence of this series is (–1, 1). • For x in the interval of convergence, we can write the equation: 1 = 1 − x 2 + x 4 − x 6 + L + (−1) n−1 x 2 n− 2 + L 2 1+ x The antiderivatives should then differ by a constant: tan −1 ( x ) + C = x − x3 x5 x7 x 2 n− 1 + − + L + ( −1) n−1 +L 3 5 7 2n − 1 Now we can find the constant by letting x = 0: tan −1 (0) + C = 0 − 0+C = 0 0 3 05 0 7 0 2 n− 1 + − + L + (−1) n−1 +L 3 5 7 2n − 1 C =0 So tan −1 ( x ) = x − x3 x5 x7 x 2 n− 1 + − + L + ( −1) n−1 + L. 3 5 7 2n − 1 We expect this equation at least to be valid on the interval (–1, 1), but in fact we have also picked up convergence at the endpoints. For example, plugging x = 1 into the series yields 1 1 1 1 1 − + − + L + (−1) n−1 +L . 3 5 7 2n − 1 This is an example of an alternating series of terms of diminishing magnitude that tend to a limit of 0. Such series always converge. (This is the alternating series test.) Students can easily be convinced of this fact by tracking the partial sums of this series on a number line: L 13 Special Focus: Calculus The partial sums bounce back and forth on the number line, but because each added or subtracted term is smaller than the previous one, they get closer to a limit in the process. Since the terms tend to zero, there is a limit L that becomes the sum. It seems logical (and it can be proved using a deeper understanding of limits) that the number to which 1 − 1 + 1 − 1 + L + ( −1) n −1 1 + L converges ought to be 3 5 7 2n − 1 π −1 tan (1) , which we know to be . This series can, in fact, be used to compute π, but 4 it is so close to the threshold of divergence that it converges too slowly to be of much practical value. (Students can convince themselves of this fact by trying the partial sums on their calculators. You know that some of them will.) Beginning with geometric series, you will have thus exposed students to the idea that functions can be represented by power series, that such series have intervals of convergence for which such representations are valid, and that series can be manipulated using calculus to yield new functions. You will have shown them power series that they will eventually recognize as the Maclaurin series for three different functions (along with their intervals of convergence), and you will even have exposed them to a valuable convergence test for series of constants. In all likelihood you will have been able to accomplish this during a single class period. Reaping the Benefits Once students have seen that a power series can represent a function on some interval of convergence, they can discover some significant results on their own. Here are just two examples. Challenge: Consider the function f defined by the infinite series f ( x) = 1 + x + x2 x3 xn + + L+ +L . 2! 3! n! (a) Find f(0). (b) Find f'(x). What is interesting about it? (c) What can you conclude about the function f? Students will easily guess that the function is f(x) - ex, but only the best of them will recognize that they have the information required to conclude that f(x) must be ex. In part (b) they discover the differential equation f’(x) = f(x), and in part (a) they discover the initial condition f(0) = 1. The unique solution to this initial value problem is f(x) = ex. 14 Setting the Stage with Geometric Series Challenge: Construct a fifth-degree polynomial P such that: P (0 ) = 2 P ′ (0 ) = 3 P ′′(0) = 5 P ′′′(0) = 7 P ( 4 ) (0) = 11 P (5 ) (0) = 13 Students will usually succeed at this, and in the process they will see that the P ( n ) (0 ) n coefficient of x must be . You are then ready to ask them to build a polynomial n! whose first n derivatives match the first n derivatives of a function f at 0 (the sine f ( n ) (0 ) n function, for example). The coefficient of each x will then be . Your students n! will have discovered Maclaurin series! ∞ t n  Let S be the series S = ∑   where t ≠ 0. n= 0 1 + t (a) Find the value to which S converges when t = 1. (b) Determine the values of t for which S converges. Justify your answer. (c) Find all values of t that make the sum of the series S greater than 10. Students are ready for this as soon as they understand geometric series. Curiously, as they learn more about series they become overqualified to solve something this simple. Challenge BC-3 Solution ∞ 1 n (a) When t = 1, S = ∑   . This is a geometric series with first term a = 1 and   n= 0 2 1 a 1 common ratio r = . Hence S = = = 2. 2 1 − r  1 − 1  2 t (b) The series will converge if and only if < 1 where t ≠ 0. This will be true for t 1+ t 1 > 0 and for − < t < 0 . 2 15 (c) Since the series is a geometric series with first term a = 1 and common ratio t a 1 1+ t r= , we will have S = = = = 1+ t . t+1 1 − r  1 − t  (1 + t − t )  1+ t Thus S > 10 when t > 9. Once your students have grasped the general idea of what series are and how they behave, they will be ready to tackle the rest of the topics for infinite series—eventually even those other tests for convergence!