9.1Power Series

This is an example of an infinite series.

1

1

Start with a square one unit by one unit:

1

21

21

4

1

4

1

81

8

1

161

16

1

32 1

64

1

32

1

64 1

This series converges (approaches a limiting value.)

Many series do not converge:1 1 1 1 1

1 2 3 4 5

In an infinite series: 1 2 31

n kk

a a a a a

a1, a2,… are terms of the series. an is the nth term.

Partial sums: 1 1S a

2 1 2S a a

3 1 2 3S a a a

1

n

n kk

S a

nth partial sum

If Sn has a limit as , then the series converges,

otherwise it diverges.

n

Geometric Series:

In a geometric series, each term is found by multiplying the preceding term by the same number, r.

2 3 1 1

1

n n

n

a ar ar ar ar ar

This converges to if , and diverges if .1

a

r1r 1r

1 1r is the interval of convergence.

Example 1:

3 3 3 3

10 100 1000 10000

.3 .03 .003 .0003 .333...

1

3

310

11

10

a

r

3109

10

3

9

1

3

1 1 11

2 4 8

11

12

11

12

13

2

2

3

a

r

Example 2:

The partial sum of a geometric series is: 1

1

n

n

a rS

r

If then1r 1

lim1

n

n

a r

r

1

a

r

0

If and we let , then:1x r x

2 31 x x x 1

1 x

The more terms we use, the better our approximation (over the interval of convergence.)

A power series is in this form:

c x c c x c x c x c xnn

nn

n

0 1 22

33

0

or

c x a c c x a c x a c x a c x ann

nn

n

( ) ( ) ( ) ( ) ( )

0 1 22

33

0

The coefficients c0, c1, c2… 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.)

Once we have a series that we know, we can find a new series by doing the same thing to the left and right hand sides of the equation.

This is a geometric series where r=-x.

1

x

xTo find a series for multiply both sides by x.

2 311

1x x x

x

2 3 4

1

xx x x x

x

1

1 xExample 3:

Example 4:

Given: 2 311

1x x x

x

find:

2

1

1 x

1

1

d

dx x 11

dx

dx 2

1 1x

2

1

1 x

So:

2 32

11

1

dx x x

dxx

2 31 2 3 4x x x

We differentiated term by term.

Example 5:

Given: 2 311

1x x x

x

find: ln 1 x

1ln 1

1dx x c

x

2 311

1t t t

t

hmm?

Example 5: 2 311

1t t t

t

2 3

0 0

11

1

x xdt t t t dt

t

2 3 4

00

1 1 1ln 1

2 3 4

xx

t t t t t

2 3 41 1 1ln 1 ln 1 0

2 3 4x x x x x

2 3 41 1 1ln 1

2 3 4x x x x x 1 1x

2 3 41 1 1ln 1

2 3 4x x x x x 1 1x

The previous examples of infinite series approximated simple functions such as or .1

3

1

1 x

This series would allow us to calculate a transcendental function to as much accuracy as we like using only pencil and paper!

p