Do Now: #18 and 20 on p.466Find the interval of convergence and the function of xrepresented by the given geometric series.

0

1 1n n

n

x

0

1n

n

x

1a 1r x

This series will only converge when : 1 1x

1 1 1x 1 1 1x 0 2x

Interval of convergence: 2,0

Do Now: #18 and 20 on p.466Find the interval of convergence and the function of xrepresented by the given geometric series.

0

1 1n n

n

x

0

1n

n

x

1a 1r x

1ar

Sum of the series: 1

1 1x

12x

So, this series represents the function

1 ,2

f xx

2 0x Graphical support?

Do Now: #18 and 20 on p.466Find the interval of convergence and the function of xrepresented by the given geometric series.

0

132

n

n

x

3a 12xr

This series will only converge when :1 12x

Interval of convergence: 1,3

11 12x

2 1 2x 1 3x

Do Now: #18 and 20 on p.466Find the interval of convergence and the function of xrepresented by the given geometric series.

0

132

n

n

x

3a 12xr

1ar

Sum of the series: 3

1 1 2x

So, this series represents the function

6 ,3

f xx

1 3x Graphical support?

33 2x

63x

Section 9.1bPower Series

Similar Problems: #42 and 44 on p.468Find a power series to represent the given function andidentify its interval of convergence.

11 3x

Compare to:1ar

1a 3r x

Series: 21 3 9 3 nx x x

The series converges when :3 1x 1 3 1x 1 13 3

x

Interval of convergence:1 1,3 3

Similar Problems: #42 and 44 on p.468Find a power series to represent the given function andidentify its interval of convergence.

3

31 x

Compare to:1ar

3a 3r x

Series: 3 6 33 3 3 3 nx x x

The series converges when :3 1x 31 1x

Interval of convergence: 1,1

1 1x

Finding a Power Series by DifferentiationGiven that 1/(1 – x) is represented by the power series

21 ,nx x x find a power series to represent

1 1,x

21 1 x

Notice that this is the derivative of this !!!

21 11

nd d x x xdx x dx

2 3 1

2

1 1 2 3 41

nx x x nxx

With the same interval of convergence: 1,1

Graphical support?

Theorem: Term-by-Term Differentiation

If 0 10

nn

n

f x c x a c c x a

2

2n

nc x a c x a

converges for , then the seriesx a R

11 2

1

2nn

n

nc x a c c x a

2 1

33n

nc x a nc x a

obtained by differentiating the series for term by term,converges on the same interval and represents onthat interval.

f f x

Finding a Power Series by IntegrationGiven that 2 31 1 ,

1nx x x x

x

1 1,x find the power series to represent ln 1 x

2 3

0 0

1 1 11x x n ndt t t t t dt

t

2 3 4 1

00

ln 1 12 3 4 1

xnx nt t t tt t

n

2 3 4 1

ln 1 12 3 4 1

nnx x x xx xn

With the same interval of convergence: 1,1

Theorem: Term-by-Term IntegrationIf 0 1

0

nn

n

f x c x a c c x a

2

2n

nc x a c x a converges for , then the seriesx a R

1 2

0 10 1 2

n

nn

x a x ac c x a c

n

3 1

2 3 1

n

n

x a x ac c

n

obtained by integrating the series for term by term,converges on the same interval and represents on that interval.

f

x

af t dt

Exploration 3 on p.465

2 3 4

12! 3! 4! !

nx x x xf x xn

1. 2 3

12! 3! !

nx x xf x xn

2. 0 1 0 0 1f

Since this function is its own derivative and takes on thevalue 1 at x = 0…. perhaps it is the exponential function?

3.

4. ,y f x ,dy ydx

0 1f

Exploration 3 on p.465

2 3 4

12! 3! 4! !

nx x x xf x xn

5. ,y f x ,dy ydx

0 1f

dy dxy

ln y x C

xy Ke01 1Ke K

xy e6.

7.

It appears that the first three partial sums may convergeon (–1, 1)…

The actual interval of convergence is all real numbers!!!