Calculus II (part 4): Power Series and Taylor Series (by Evan Dummit, 2015, v. 2.01) Contents 8 Power Series and Taylor Series 1 8.1 Power Series . 1 8.1.1 Convergence of Power Series . 2 8.1.2 Power Series as Functions . 5 8.2 Taylor Series . 8 8.3 Taylor Polynomials and Convergence of Taylor Series . 12 8.3.1 Taylor's Theorem . 13 8.3.2 Convergence of Common Taylor Series . 13 8.3.3 Table of Common Taylor Series . 15 8.4 Applications of Taylor Series . 15 8.4.1 Summing Series Using Taylor Expansions . 16 8.4.2 Numerical Approximations Via Taylor Series . 17 8.4.3 Approximating Functions by Polynomials . 18 8.4.4 Computing Integrals Using Series Expansions . 19 8.4.5 Computing Limits Using Series Expansions . 21 8.4.6 Euler's Formula . 22 8 Power Series and Taylor Series In this chapter, we continue our discussion of innite series from the previous chapter. However, we will narrow 1 X n our focus to a particular kind of innite series, called a power series, which has the general form an(x − c) n=0 for real numbers an and c, and a parameter x. We will discuss the basic theory of power series and methods for representing functions as power series. We will then turn our attention to Taylor series, which are a special type of power series that arise in trying to nd good polynomial approximations to arbitrary functions, and conclude by outlining some of the more important applications of Taylor series. 8.1 Power Series 1 X n • Denition: A power series centered at x = c is a series of the form an(x − c) , where the center c and the n=0 coecients ai are constants. 1 X n ◦ We will usually be interested in power series centered at x = 0, which have the simpler form anx . n=0 1 X n • We will generally think of a given power series an(x − c) as a function of x. n=0 ◦ Our initial goal is to study for which x this power series converges. 1 ◦ We will then turn our attention to describing the resulting function of x (dened where the series converges). 1 X • Example: The geometric series xn is a power series centered at x = 0 all of whose coecients are equal n=0 to 1. 1 ◦ From our earlier analysis of geometric series, we know this series will converge to the limit whenever 1 − x −1 < x < 1, and that it will diverge for other x. ◦ By the denition of convergent series, this says that the sequence of polynomials 1, 1 + x, 1 + x + x2, 1 1 + x + x2 + x3, ... converges to the value whenever −1 < x < 1. 1 − x ◦ We can see this convergence explicitly from the graphs (the functions are 1+x, 1+x+x2, 1+x+x2 +x3, 1 1 + x + x2 + x3 + x4, and from bottom to top): 1 − x 5 4 3 2 1 -1.0 -0.5 0.0 0.5 1.0 8.1.1 Convergence of Power Series • In general, we can typically determine where a power series converges by using the Ratio or Root Tests. A useful technique is to combine the Ratio/Root Tests with the Absolute Convergence Theorem to obtain versions which apply to general series (possibly with negative terms): bn+1 ◦ Strengthened Ratio Test: If fbng has the property that lim exists and equals some constant ρ, n!1 bn 1 X then the sum bn converges if ρ < 1, and diverges if ρ > 1. n=1 pn ◦ Strengthened Root Test: If fbng has the property that lim jbnj exists and equals some constant ρ, n!1 1 X then the sum bn converges if ρ < 1, and diverges if ρ > 1. n=1 ◦ For both tests, if ρ = 1 then the test is inconclusive, while if ρ = 1 then the series diverges. 1 X 1 • Example: Determine the values of x for which the power series xn converges. n2 n=1 n n+1 2 2 We use the Ratio Test: with x , we have bn+1 x =(n + 1) n + 1 . ◦ bn = 2 = n 2 = jxj · n bn x =n n bn+1 ◦ Then lim = jxj. n!1 bn ◦ Thus, by the (strengthened) Ratio Test, we see that the series converges whenever jxj < 1 and diverges whenever jxj > 1. ◦ There are two cases where the test is inconclusive: x = 1 and x = −1. 1 X 1 ◦ When x = 1, the series is , which is a convergent p-series. n2 n=1 2 1 X (−1)n ◦ When x = −1, the series is , which converges by the Alternating Series Test (or by the fact n2 n=1 that its absolute value series is the one we just saw above). ◦ Therefore, the power series converges for − 1 ≤ x ≤ 1 and diverges for other x. 1 X 1 n • Example: Determine the values of x for which the power series (x − 2) converges. 3n n=1 n (x − 2) pn x − 2 ◦ We use the Root Test: with bn = , we have jbnj = . 3n 3 pn x − 2 ◦ Then lim jbnj = . n!1 3 x − 2 ◦ Thus, by the (strengthened) Ratio Test, we see that the series converges whenever < 1 and 3 x − 2 x − 2 diverges whenever > 1, while the test is inconclusive when = 1. 3 3 x − 2 x − 2 ◦ Notice that < 1 is equivalent to −1 < < 1, which is the same as −1 < x < 5. 3 3 1 1 X (−3)n X ◦ When x = −1, the series is = (−1)n, which diverges. 3n n=1 n=1 1 1 X 3n X ◦ When x = 5, the series is = 1, which also diverges. 3n n=1 n=1 ◦ Therefore, the power series converges for − 1 < x < 5 and diverges for other x. ◦ Notice, in particular, that the region of convergence is the interval (−1; 5), and its midpoint is the center of the power series. 1 X 1 • Example: Determine the values of x for which the power series xn converges. n! n=0 1 b xn+1=(n + 1)! n! jxj We use the Ratio Test: with n, we have n+1 . ◦ bn = x = n = x · = n! bn x =n! (n + 1)! n + 1 bn+1 jxj ◦ Then, for any xed value of x, we see that lim = lim = 0. n!1 bn n!1 n + 1 ◦ Thus, by the (strengthened) Ratio Test, we see that the series converges for all x . 1 X • Example: Determine the values of x for which the power series n! · xn converges. n=0 b (n + 1)! · xn+1 (n + 1)! We use the Ratio Test: with n, we have n+1 . ◦ bn = n!·x = n = x · = jxj·(n+1) bn n! · x n! bn+1 ◦ Then, for any nonzero value of x, we see that lim = lim (n + 1) jxj = +1. If x = 0, then the n!1 bn n!1 limit is clearly zero. ◦ Thus, by the (strengthened) Ratio Test, we see that the series converges only for x = 0 . 1 X n • In each of the examples above, notice that the set of x for which the power series an(x − c) converged n=0 was an interval whose midpoint was the center x = c of the power series. 3 ◦ Note that we have included the case of the degenerate interval [0; 0] consisting of a single point, as well as the innite interval (−∞; 1). • In fact, the region of convergence is always an interval. To show this, we rst need a preliminary result: 1 X n • Proposition: If the power series anx converges for x = d, then it converges absolutely for all x with n=0 jxj < jdj. 1 X n n ◦ Proof: Since the series and converges, by the Test for Divergence we know that lim and = 0. n!1 n=0 N ◦ By the denition of limit, in particular this implies that for large enough N, aN d ≤ 1: therefore, −N jaN j ≤ d . 1 1 X X x n x ◦ But then ja xnj ≤ , and this last series is a convergent geometric series when < 1. This n d d n=N n=N x implies the original series converges absolutely for < 1 i.e., whenever jxj < d. d • From this, we can conclude that the set of convergence of a power series must have a very particular form: 1 X n • Theorem (Power Series Convergence): For any power series an(x − c) , precisely one of the following three n=0 things holds: 1. The series converges absolutely for every x. 2. The series converges at x = c and diverges for other x. 3. There exists a positive real number R such that the series converges absolutely for x with jx − cj < R and diverges for jx − cj > R. The series may or may not converge at the two endpoints x = c ± R. ◦ Remark: The value of R is called the radius of convergence for the power series. It is conventional to say that R = 1 in the rst case and to say that R = 0 in the second case. (Thus, for example, the radius 1 1 X X 1 of convergence of the power series xn is 1, while the radius of convergence of xn is 1.) n! n=0 n=0 1 X n ◦ Proof: Let u = x − c: then the power series has the form anu . n=0 ∗ Consider the set of values of juj such that this series converges: if there is no upper bound, then (by the previous proposition applied to an increasing sequence of values of u) we conclude that the series converges absolutely for every value of u.