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Chapter 9.8 - Power Series . A power series is in this form: or The coefficients c 0, c 1, c 2 … are constants. The center “a” is also a constant. (The.

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Presentation on theme: "Chapter 9.8 - Power Series . A power series is in this form: or The coefficients c 0, c 1, c 2 … are constants. The center “a” is also a constant. (The."— Presentation transcript:

1 Chapter Power Series 

2 A power series is in this form: or The coefficients c 0, c 1, c 2 … are constants. The center “a” is also a constant. (The first series would be centered at the origin if you graphed it. The second series would be shifted left or right. “a” is the new center.)

3 1) The series converges absolutely for all x. the radius of convergence R is ∞ 2) The series converges absolutely at only. The radius of convergence R is 0. 3) The series converges absolutely on a finite interval called the interval of convergence. The interval is centered at The series might or might not converge at the endpoints 

4 Example : #1 Find the interval of convergence for the series: Using the ratio test with Since ρ = 0 no matter what x equals, the series converges absolutely for all x. 

5 Example : #2 Find the interval of convergence for the series: Using the ratio test with Since ρ = ∞ except when x = 0, the series converges when x = 0 but diverges for all other x values. 

6 Find the interval of absolute convergence for the series: In absolute value, use ratio test. This makes the current interval of convergence x = (-1, 1). If ρ = 1, the ratio test is inconclusive so we must check the endpoints separately Example : #3 

7 Let x = 1 and the series becomes: If x = –1, then the series becomes: Which also diverges as the harmonic series. The interval of absolute convergence is thus (-1, 1) with a radius of convergence R = 1. Which diverges as the harmonic series 1 Starting with the original power series in absolute value: 

8 Find the interval of conditional convergence for the series: The first part is the same as before. Since ρ = x, the series converges on the interval (-1,1) but we need to check the endpoints to see if they converge conditionally without absolute value. Example : #3 

9 Let x = 1 and the series becomes: If x = –1, then the series becomes: Which diverges as the harmonic series. The interval of conditional convergence is thus (-1, 1] with a radius of convergence R = 1. Which converges as the alternating harmonic series 1 Starting with the original power series:  (2n is always even so (-1) 2n is always positive)

10 Do you recognize this series? A geometric series with a = 1 and r = x What is the sum of this series? What is its interval of convergence? 

11 Watch what happens to the graphs of the partial sums on the interval of convergence as we add more terms to the polynomial power function 

12

13 On the TI-83, set the Y= and WINDOW to view the graph 

14 New series from old series Begin with the geometric series substitute ( x) for x or Substitution: 

15 New series from old series Begin with the alternating geometric series differentiate with respect to x. Differentiation: 

16 New series from old series Begin with the alternating geometric series integrate with respect to x. Integration: Remember the alternating harmonic series? If you let x = 1 that is what you get. 


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