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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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 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.)

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 

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. 

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. 

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 

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: 

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 

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)

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? 

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 

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

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

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

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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