Power Series is an infinite polynomial in x Is a power series centered at x = 0. Is a power series centered at x = a. and.

Power Series is an infinite polynomial in x

20 1 2

0

..... ...n nn n

n

c x c c x c x c x

20 1 2

0

( ) ( ) ( ) ..... ( ) ...n nn n

n

c x a c c x a c x a c x a

Is a power series centered at x = 0.

Is a power series centered at x = a.

and

Examples of Power Series 2 3

0

1 ...! 2 3!

n

n

x x xx

n

2

0

( 1) 1 1 1( 1) 1 ( 1) ( 1) ..... ( 1) ...

3 93 3

nn n

n nn

x x x x

Is a power series centered at x = 0.

Is a power series centered at x = -1.

and

Geometric Power Series2 3 4

0

1 ...n n

n

x x x x x x

1, 1

1 1

1

aS x

r x

a and r x

1

22

2 33

1

1

1

P x

P x x

P x x x

. The graph of f(x) = 1/(1-x) and four of its polynomial approximations

Convergence of a Power SeriesThere are three possibilities 1)There is a positive number R such that the

series diverges for |x-a|> R but converges for |x-a|< R. The series may or may not converge at the endpoints, x = a - R and x = a + R.

2)The series converges for every x. (R = .)

3)The series converges at x = a and diverges elsewhere. (R = 0)

Review of tests for convergence

1. Geometric Series a) converges for | r | <1b) diverges for | r | >1 and | r | = 1

2. Using the ratio test: a) Series converges for L < 1 b) Series diverges for L >1 c) Test is inconclusive if L = 1

1lim nn

n

uL

u

3. Using the root test: a) Series converges for L < 1 b) Series diverges for L >1 c) Test is inconclusive if L = 1

lim | |nn na L

What is the interval of convergence?

0

( 1)n

n

Since r = x, the series converges |x| <1, or -1 < x < 1.

Test endpoints of –1 and 1.

0

(1)n

n

Series diverges Series diverges

2 3 4

0

1 ...n n

n

x x x x x x

interval of convergence is (-1,1).-1 1

Geometric Power Series

1 3 3

1 3 ( 1) 41

11 ( 1)

3

1( 1)

3

a and r x

x

aS

r x x

2

0

( 1) 1 1 1( 1) 1 ( 1) ( 1) ..... ( 1) ...

3 93 3

nn n

n nn

x x x x

11

( 13

3 3

)

1

x

x

1.Find the function

2. Find the radius of convergence

4 2x R = 3

0

( 1)( 1)

3

nn

nn

x

For x = -2,

0 0 0

( 1) ( 1) ( 1) 1( 2 1)

3 3 3

n n nn

n n nn n n

Geometric series with r < 1, converges

0 0 0 0

( 1) ( 1) ( 3) 3( 4 1) 1

3 3 3

n n n nn

n n nn n n n

By nth term test, the series diverges.

For x = 4

2 4x Interval of convergence

3. Find the interval of convergence2 4x

Test endpoints

Find interval of convergence

Use the ratio test:2 1 1 2 3

1( 1) ( 1)

(2 1)! (2 3)!n n

n n n n

u and ux x

n n

2 21lim * 0* 0

(2 3)(2 2)n x xn n

2 1

0

( 1)

(2 1)!

n n

n

x

n

(-, )Interval of convergence

Series converges for all reals.

1 2 31

2 1

( 1) (2 1)!lim lim *

(2 3)! ( 1)

n nn

n n n nn

u x n

u n x

R=

2

lim(2 3)(2 2)n

x

n n

0 < 1 for all reals

Finding interval of convergence

1 for all realsnot true

Use the ratio test: 1

1! ( 1)!n nn nu x n and u x n

lim ( 1) *n n x x

0

!n

n

x n

[0, 0]Interval of convergence

Series converges only for center point

11 ( 1)!

lim lim!

nn

n n nn

u n x

u n x

R=0

Finding interval of convergence

1x

Use the ratio test:1

1 1

n n

n nx x

u and un n

1

1lim lim1

nn

n n nn

u x nx

u n x

-1< x <1 R=1

For x = 1For x = -1

0

n

n

x

n

(-1, 1)

0

1

n n

0

( 1)n

n n

Harmonic series diverges

Alternating Harmonic series converges

[-1, 1)Interval of convergence

Test endpoints

Differentiation and Integration of Power Series

20 1 2

0

( ) ( ) ( ) ..... ( ) ...n nn n

n

c x a c c x a c x a c x a

If the function is given by the series

Has a radius of convergence R > 0, on the interval (c-R, c+R) the function is continuous, differentiable and integrable where:

1

0

( ) ( )nn

n

f x nc x a

1

0

( )( )

1

n

nn

x af x dx C c

n

The radius of convergence is the same but the interval of convergence may differ at the endpoints.

and

Constructing Power SeriesIf a power series exists has a radius of convergence = RIt can be differentiated

20 1 2( ) ( ) ( ) ..... ( ) ...n

nf x c c x a c x a c x a

2 11 2 3( ) 2 ( ) 3 ( ) ..... ( ) ...n

nf x c c x a c x a nc x a

22 3( ) 2 2*3 ( ) 3* 4( ) ....f x c c x a x a

23 4( ) 1* 2*3 2*3* 4 ( ) 3* 4*5( ) ...f x c c x a x a

( )( ) ! ( )nnf x n c terms with factor of x a

So the nth derivative is

( )( ) ! ( )nnf x n c terms with factor of x a

All derivatives for f(x) must equal the seriesDerivatives at x = a.

1

2

3

( )

( ) 1* 2

( ) 1* 2*3

f a c

f a c

f a c

( )( ) !nnf a n c

( )( )

!

n

nf a

cn

Finding the coefficients for a Power Series

( )2 3

0

( )

( ) ( ) ( )( ) ( ) ( ) ( ) ...

! 2! 3!

( )...

!

k

k

nn

f a f a f af a f x a x a x a

k

f ax

n

If f has a series representation centered at x=a, the series must be

If f has a series representation centered at x=0, the series must be

( )2 3

0

( )

(0) (0) (0)( ) (0) ( ) ( ) ...

! 2! 3!

(0)...

!

k

k

nn

f f ff a f x a x a

k

fx

n

Form a Taylor Polynomial of order 3 for sin x at a =n f(n)(x) f(n)(a) f(n)(a)/n!

0 sin x

1 cos x

2 -sin x

3 -cos x

2

2

2

2

2

2

2

2

2

2

2

2

2

2* 2!

2

2*3!

4

0 1 2 32 3sin ( ) ( ) ( ) .

4 4 4x x xc c c xc

2 32 2 2 2sin ( ) ( ) ( ) .

2 2 4 2* 2! 4 2*3! 4x x x x

The graph of f(x) = ex and its Taylor polynomials

2 3 4

0

1 ...! 2 3! 4!

n

n

x x x xx

n

Find the derivative and the integral

2 1 3 5 7

0

1...

(2 1)! 3! 5! 7!

n n

n

x x x xx

n

Taylor polynomials for f(x) = cos (x)