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Objectives 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.

We’ll strip you of any fears,
plug you in, drill you on the essentials, make everything plane, and hopefully electrify you!

Power Series

We’ve seen power series before…
𝑎 𝑛 𝑥 𝑛 We’ve seen power series before… Specifically when 𝑎 𝑛 = 𝑓 𝑛 (0) 𝑛! Does the name “Maclaurin” ring a bell?

Power Series An 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 ex For each real number x, it can be shown that the infinite series converges to the number ex.

Power Series

Example 1 – Power Series The following power series is centered at 0.
c. The following power series is centered at 1.

Power 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 which the power series converges. Every power series converges at its center because So c always lies in the domain of f.

Power Series The domain of a power series can be a single point,
an interval centered at c, or the entire real line.

Convergence of Power Series Theorem
The series converges absolutely for all 𝑥 The series converges only for 𝑥=0 The 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 either The 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

Example 2 – Finding the Radius of Convergence
Find the radius of convergence of Solution: (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.

Why 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

For example, in this Power Series: a n 𝑥 𝑛
if 𝑎 𝑛 =1, we get This is a geometric series which converges for 𝑥 𝜖 −1, 1 . For example, in this Power Series: a n 𝑥 𝑛 𝑥 𝑛 𝑛! if 𝑎 𝑛 = 1 𝑛! , we get which converges to 𝑓 𝑥 = 𝑒 𝑥 for all 𝑥. We have already studied a subset of all Power Series.

Endpoint Convergence

Endpoint Convergence A 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

Example 3 – Finding the Interval of Convergence
Find the interval of convergence of Solution (using the Ratio Test):

Example 3 – Solution cont'd So, 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

Example 3 – Solution cont'd When x = –1, you obtain the convergent alternating harmonic series So, the interval of convergence for the series is [–1, 1):

The interval of convergence for this series is [−1,1)
𝑛=1 ∞ 𝑥 𝑛 𝑛 Ex: #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|<1 Since the Root Test fails if the limit equals 1, we test the endpoints separately. lim 𝑛→∞ 𝑛 𝑛 =1 If 𝑥=1; 𝑛=1 ∞ 1 𝑛 𝑛 ;divergent p−series The interval of convergence for this series is [−1,1) If 𝑥=−1; 𝑛=1 ∞ −1 𝑛 𝑛 ;convergent by the A.S.T.

This series is convergent on (−∞,∞) by the Root Test.
𝑛=1 ∞ (−1) 𝑛 𝑥 𝑛 𝑛 𝑛 Ex: #5 lim 𝑛→∞ 𝑛 |−1| 𝑛 |𝑥| 𝑛 𝑛 𝑛 = lim 𝑛→∞ |𝑥| 𝑛 =0<1 for all 𝑥 This series is convergent on (−∞,∞) by the Root Test. If you get tired, it’s no wonder. We just did infinitely many problems twice!

Convergence for 𝑥 <10 𝑜𝑟 −10<𝑥<10
𝑛=1 ∞ 𝑛 10 𝑥 𝑛 1 0 𝑛 Ex: #6 1 lim 𝑛→∞ 𝑛 𝑛 10 |𝑥| 𝑛 1 0 𝑛 = lim 𝑛→∞ 𝑛 𝑛 10 |𝑥| 10 = lim 𝑛→∞ |𝑥| 10 = |𝑥| 10 <1 Convergence for 𝑥 <10 𝑜𝑟 −10<𝑥<10 Now 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

𝑛=1 ∞ (2𝑥−1) 𝑛 𝑛 4 +16 In this case 𝑥 has been replaced with 2𝑥−1, so the center of the symmetric interval would be at 𝑥=1/2. Ex: #7 lim 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<1 Check the endpoints. 0<2𝑥<2 If 𝑥=0; 𝑛=1 ∞ (−1) 𝑛 𝑛 ;converges by A.S.T. 0<𝑥<1 If 𝑥=1; 𝑛=1 ∞ (1) 𝑛 𝑛 ;compares to convergent p−series The interval of convergence for this series converges is 0, 1 . The radius of convergence is 1/2.

The interval of convergence is −2, 2 .
𝑛=0 ∞ 𝑥 𝑛 Ex: #28 lim 𝑛→∞ 𝑛 𝑥 𝑛 = lim 𝑛→∞ 𝑥 = 𝑥 <1 −1< 𝑥 <1 −5<𝑥 2 +1<5 Check the endpoints. −6<𝑥 2 <4 → 0<𝑥 2 <4 →−2<𝑥<2 𝑥=±2; 𝑛=0 ∞ 𝑛 = 𝑛=0 ∞ 1 𝑛 ;diverges by nth term test. The interval of convergence is −2, 2 .

Ex: Determine the radius and interval of convergence:
𝑛=1 ∞ (−1) 𝑛 𝑛 4 𝑛 𝑥+3 𝑛 Series converges Series diverges

Ex: Determine the radius and interval of convergence:
𝑛=1 ∞ (−1) 𝑛 𝑛 4 𝑛 𝑥+3 𝑛 Radius of Convergence = 4 Test fails (where L=1) Need to test for convergence at endpoints of interval. Power series diverges at each endpoint. Interval of convergence is

Ex: Determine the radius and interval of convergence:
𝑛=1 ∞ 2 𝑛 𝑛 4𝑥−8 𝑛

Ex: Determine the radius and interval of convergence:
𝑛=1 ∞ (𝑥−6) 𝑛 𝑛 𝑛

Ex: Determine the radius and interval of convergence:
𝑛=1 ∞ 2𝑛 (−3) 𝑛 2𝑛

Ex: Determine the radius and interval of convergence:

Just 1 more example!

Rats! We have to use the Ratio Test when there is a factorial!
𝑛=1 ∞ 1∙3∙5∙⋯∙(2𝑛+1) 𝑛! Ex: Find the interval of convergence Rats! We have to use the Ratio Test when there is a factorial!

The interval of convergence is − 1 2 , 1 2
𝑛=1 ∞ 1∙3∙5∙⋯∙(2𝑛+1) 𝑛! 𝑥 𝑛 Ex: #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|<1 Evidently, 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

Homework Day 1: Page 666: odd Day 2: MMM

Differentiation and Integration of Power Series

Differentiation and Integration of Power Series

Example 8 – Intervals of Convergence for f(x), f'(x), and ∫f(x)dx
Consider the function given by Find the interval of convergence for each of the following. ∫f(x)dx f(x) f'(x)

Example 8 – Solution By Theorem 9.21, you have and
cont'd By Theorem 9.21, you have and By the Ratio Test, you can show that each series has a radius of convergence of R = 1. Considering the interval (–1, 1) you have the following.

Example 8(a) – Solution For ∫f(x)dx, the series
cont'd For ∫f(x)dx, the series converges for x = ±1, and its interval of convergence is [–1, 1 ]. See Figure 9.21(a). Figure 9.21(a)

Example 8(b) – Solution For f(x), the series
cont'd For f(x), the series converges 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)

Example 8(c) – Solution For f'(x), the series
cont'd For f'(x), the series diverges for x = ±1, and its interval of convergence is (–1, 1). See Figure 9.21(c). Figure 9.21(c)

EXAMPLE 15.2. Determine the radius and interval of convergence for the power series