3Objectives Understand the definition of a power series. Find the radius and interval of convergence of a power series.Determine the endpoint convergence of a power series.Differentiate and integrate a power series.
4We’ll strip you of any fears, plug you in,drill you on the essentials,make everything plane,and hopefully electrify you!
6We’ve seen power series before… 𝑎 𝑛 𝑥 𝑛We’ve seen power series before…Specifically when 𝑎 𝑛 = 𝑓 𝑛 (0) 𝑛!Does the name “Maclaurin” ring a bell?
7Power SeriesAn important function f(x) = ex can be represented exactly by an infinite series called a power series. For example, the Maclaurin series we found is a power series representation for exFor each real number x, it can be shown that the infinite series converges to the number ex.
9Example 1 – Power Series The following power series is centered at 0. c. The following power series is centered at 1.
10Power Series A power series can be viewed as a function of x. The domain of a power series is the set of all x for whichthe power series converges. Every power series convergesat its center becauseSo c always lies in the domain of f.
13Power Series The domain of a power series can be a single point, an interval centered at c, or the entire real line.
14Convergence of Power Series Theorem The series converges absolutely for all 𝑥The series converges only for 𝑥=0The series converges on some symmetric interval,such as −𝑏, 𝑏 , −𝑏, 𝑏 , −𝑏, 𝑏 or −𝑏,𝑏(𝑏 is called the 𝐫𝐚𝐝𝐢𝐮𝐬 of convergence.)Given a Power Series 𝑎 𝑛 𝑥 𝑛 , then eitherThe only tools we have to find said interval of convergence are the Ratio and Root Tests.Recall that both tests show convergence when lim 𝑛→∞ |𝑎 𝑛 𝑥 𝑛 |<1
15Example 2 – Finding the Radius of Convergence Find the radius of convergence ofSolution: (use the Ratio Test)Therefore, by the Ratio Test, the series diverges for|x| > 0 and converges only at its center, 0.So, the radius of convergence is R = 0.
16Why is the interval of convergence symmetric? Well…if a series is convergent then the ratio test yields a value 𝑎 𝑏 <1.Hence for the power function 𝑎 𝑛 𝑥 𝑛 we would get:lim 𝑛→∞ 𝑎 𝑛+1 𝑎 𝑛 ∙ |𝑥| 𝑛+1 |𝑥| 𝑛 = 𝑎 𝑏 𝑥 <1→ 𝑥 < 𝑏 𝑎→− 𝑏 𝑎 <𝑥< 𝑏 𝑎
17For example, in this Power Series: a n 𝑥 𝑛 if 𝑎 𝑛 =1, we getThis is a geometric series which converges for 𝑥 𝜖 −1, 1 .For example, in this Power Series: a n 𝑥 𝑛𝑥 𝑛 𝑛!if 𝑎 𝑛 = 1 𝑛! , we getwhich converges to 𝑓 𝑥 = 𝑒 𝑥 for all 𝑥.We have already studied a subsetof all Power Series.
19Endpoint ConvergenceA power series whose radius of convergence is a finite number R, says nothing about the convergence at the endpoints of the interval of convergence. Each endpoint must be tested separately for convergence or divergence.As a result, the interval of convergence of a power series can take any one of the six forms shown in Figure 9.18.Figure 9.18
20Example 3 – Finding the Interval of Convergence Find the interval of convergence ofSolution (using the Ratio Test):
21Example 3 – Solutioncont'dSo, by the Ratio Test, the radius of convergence is R = 1.Moreover, because the series is centered at 0, it converges in the interval (–1, 1).This interval, however, is not necessarily the interval of convergence.To determine this, you must test for convergence at each endpoint.When x = 1, you obtain the divergent harmonic series
22Example 3 – Solutioncont'dWhen x = –1, you obtain the convergent alternating harmonic seriesSo, the interval of convergence for the series is [–1, 1):
23The interval of convergence for this series is [−1,1) 𝑛=1 ∞ 𝑥 𝑛 𝑛Ex: #4 Find the interval of convergence for the series.Since we are using a test that is only valid for positive series, we must put absolute value around any quantity that might be negative.Convergence by the Root Test for 𝑥𝜖(−1, 1)lim 𝑛→∞ 𝑛 |𝑥| 𝑛 𝑛= lim 𝑛→∞ |𝑥| 𝑛 𝑛= lim 𝑛→∞ |𝑥|=|𝑥|<1Since the Root Test fails if the limit equals 1, we test the endpoints separately.lim 𝑛→∞ 𝑛 𝑛 =1If 𝑥=1; 𝑛=1 ∞ 1 𝑛 𝑛 ;divergent p−seriesThe interval of convergence for this series is [−1,1)If 𝑥=−1; 𝑛=1 ∞ −1 𝑛 𝑛 ;convergent by the A.S.T.
24This series is convergent on (−∞,∞) by the Root Test. 𝑛=1 ∞ (−1) 𝑛 𝑥 𝑛 𝑛 𝑛Ex: #5lim 𝑛→∞ 𝑛 |−1| 𝑛 |𝑥| 𝑛 𝑛 𝑛= lim 𝑛→∞ |𝑥| 𝑛=0<1 for all 𝑥This series is convergent on (−∞,∞) by the Root Test.If you get tired, it’s no wonder.We just did infinitely many problemstwice!
25Convergence for 𝑥 <10 𝑜𝑟 −10<𝑥<10 𝑛=1 ∞ 𝑛 10 𝑥 𝑛 1 0 𝑛Ex: #61lim 𝑛→∞ 𝑛 𝑛 10 |𝑥| 𝑛 1 0 𝑛= lim 𝑛→∞ 𝑛 𝑛 10 |𝑥| 10= lim 𝑛→∞ |𝑥| 10= |𝑥| 10 <1Convergence for 𝑥 <10 𝑜𝑟 −10<𝑥<10Now check the 2 endpoints.This series converges for 𝑥 𝜖 (−10, 10)If 𝑥=10; 𝑛=1 ∞ 𝑛 𝑛 1 0 𝑛; very divergent!If 𝑥=−10; 𝑛=1 ∞ 𝑛 10 −10 𝑛 1 0 𝑛 = 𝑛=1 ∞ 𝑛 10 −1 𝑛; also divergent by nth Term Test
26𝑛=1 ∞ (2𝑥−1) 𝑛 𝑛 4 +16In this case 𝑥 has been replaced with 2𝑥−1, so the center of the symmetric interval would be at 𝑥=1/2.Ex: #7lim 𝑛→∞ 𝑛 |2𝑥−1| 𝑛 𝑛 4 +16= lim 𝑛→∞ 𝑛 |2𝑥−1| 𝑛 𝑛 4= lim 𝑛→∞ |2𝑥−1| 𝑛 𝑛 4= 2𝑥−1 <1−1<2𝑥−1<1Check the endpoints.0<2𝑥<2If 𝑥=0; 𝑛=1 ∞ (−1) 𝑛 𝑛 ;converges by A.S.T.0<𝑥<1If 𝑥=1; 𝑛=1 ∞ (1) 𝑛 𝑛 ;compares to convergent p−seriesThe interval of convergence for this series converges is 0, 1 .The radius of convergence is 1/2.
27The interval of convergence is −2, 2 . 𝑛=0 ∞ 𝑥 𝑛Ex: #28lim 𝑛→∞ 𝑛 𝑥 𝑛= lim 𝑛→∞ 𝑥= 𝑥<1−1< 𝑥 <1−5<𝑥 2 +1<5Check the endpoints.−6<𝑥 2 <4→ 0<𝑥 2 <4→−2<𝑥<2𝑥=±2; 𝑛=0 ∞ 𝑛 = 𝑛=0 ∞ 1 𝑛 ;diverges by nth term test.The interval of convergence is −2, 2 .
28Ex: Determine the radius and interval of convergence: 𝑛=1 ∞ (−1) 𝑛 𝑛 4 𝑛 𝑥+3 𝑛Series convergesSeries diverges
29Ex: Determine the radius and interval of convergence: 𝑛=1 ∞ (−1) 𝑛 𝑛 4 𝑛 𝑥+3 𝑛Radius of Convergence = 4Test fails (where L=1) Need to test for convergence at endpoints of interval.Power series diverges at each endpoint. Interval of convergence is
30Ex: Determine the radius and interval of convergence: 𝑛=1 ∞ 2 𝑛 𝑛 4𝑥−8 𝑛
31Ex: Determine the radius and interval of convergence: 𝑛=1 ∞ (𝑥−6) 𝑛 𝑛 𝑛
32Ex: Determine the radius and interval of convergence: 𝑛=1 ∞ 2𝑛 (−3) 𝑛 2𝑛
33Ex: Determine the radius and interval of convergence:
35Rats! We have to use the Ratio Test when there is a factorial! 𝑛=1 ∞ 1∙3∙5∙⋯∙(2𝑛+1) 𝑛!Ex: Find the intervalof convergenceRats! We have to use the Ratio Test when there is a factorial!
36The interval of convergence is − 1 2 , 1 2 𝑛=1 ∞ 1∙3∙5∙⋯∙(2𝑛+1) 𝑛! 𝑥 𝑛Ex: #20 (read note in book)lim 𝑛→∞ 1∙3∙5∙⋯∙(2𝑛+1)(2 𝑛+1 +1) |𝑥| 𝑛+1 (𝑛+1)! 1∙3∙5∙⋯∙(2𝑛+1)|𝑥 | 𝑛 𝑛!= lim 𝑛→∞ 2𝑛+3 𝑛!|𝑥| 𝑛+1 𝑛!→ 𝑥 < 1 2=2|𝑥|<1Evidently, checking the endpoints is beyond the scope of this course as the text said: “don’t bother checking…the series diverges at the endpoints.”The interval of convergence is − 1 2 , 1 2
40Example 8 – Intervals of Convergence for f(x), f'(x), and ∫f(x)dx Consider the function given byFind the interval of convergence for each of the following.∫f(x)dxf(x)f'(x)
41Example 8 – Solution By Theorem 9.21, you have and cont'dBy Theorem 9.21, you haveandBy the Ratio Test, you can show that each series has aradius of convergence of R = 1.Considering the interval (–1, 1) you have the following.
42Example 8(a) – Solution For ∫f(x)dx, the series cont'dFor ∫f(x)dx, the seriesconverges for x = ±1, and its interval of convergence is[–1, 1 ]. See Figure 9.21(a).Figure 9.21(a)
43Example 8(b) – Solution For f(x), the series cont'dFor f(x), the seriesconverges for x = –1, and diverges for x = 1.So, its interval of convergence is [–1, 1).See Figure 9.21(b).Figure 9.21(b)
44Example 8(c) – Solution For f'(x), the series cont'dFor f'(x), the seriesdiverges for x = ±1, and its interval of convergence is (–1, 1). See Figure 9.21(c).Figure 9.21(c)
45EXAMPLE 15.2. Determine the radius and interval of convergence for the power series