example ex. for what values of x is the power series convergent? sol. by ratio test, the power...

Example Ex. For what values of x is the power series

convergent? Sol. By ratio test,

the power series absolutely converges when

and diverges when When we easily

see that it diverges when x=4 and converges when x=1. Thus

the power series converges for

1

( 3)n

n

x

n

1 ( 3)lim lim | 3 |

1n

n nn

a n xx

a n

| 3 | 1x | 3 | 1.x | 3 | 1,x

2 4.x

Example Ex. Find the domain of the Bessel function defined by

Sol. By ratio test,

the power series absolutely converges for all x. In other words,

The domain of the Bessel function is

2

2 20

( 1)( )

2 ( !)

n n

nn

xJ x

n

21

2lim lim 0

4( 1)n

n nn

a x

a n

( , ).

Characteristic of convergence Theorem For a given power series

there are only three possibilities:

(i) The series converges only when x=a.

(ii) The series converges for all x.

(iii) There is a positive number R such that the series

converges if and diverges if The number R is called the radius of convergence of the

power series. By convention, in case (i) the radius of

convergence is R=0, and in case (ii)

0

( )nnn

c x a

.R

| |x a R | | .x a R

Characteristic of convergence The interval of convergence of a power series is the

interval that consists of all x for which the series converges. To find the interval of convergence, we need to determine

whether the series converges or diverges at endpoints |x-a|=R. Ex. Find the radius of convergence and interval of

convergence of the series

Sol

radius of convergence is 1/3.

At two endpoints: diverge at 5/3, converge at 7/3. (5/3,7/3]

0

( 3) ( 2)

1

n n

n

x

n

1lim | | 3 | 2 |n

nn

ax

a

3 | 2 | 1 | 2 | 1/ 3x x R

Radius and interval of convergence From the above example, we found that the ratio test or the

root test can be used to determine the radius of convergence. Generally, by ratio test, if then

By root test, if then Ex. Find the interval of convergence of the series

Sol. when |x|=1/e, the general does not have limit zero, so diverge. (-1/e,1/e)

2

1

1( )n n

n

nx

n

1lim | | ,n

nn

c

c

1/ .R

lim | | ,nn

nc

1/ .R

lim | | 1/nn

nc e R e

Representations of functions as power series

We know that the power series converges to

when –1<x<1. In other words, we can represent the function

as a power series

Ex. Express as the sum of a power series and find the interval of convergence.

Sol. Replacing x by in the last equation, we have

0

n

n

x

1

1 x

2 3

0

11 (| | 1)

1n

n

x x x x xx

21/(1 )x

2x

2 2 4 6 82 2

0

1 1( ) 1

1 1 ( )n

n

x x x x xx x

Example Ex. Find a power series representation for Sol.

The series converges when |-x/2|<1, that is |x|<2. So the

interval of convergence is (-2,2).

Question: find a power series representation for

Sol.

1 1 1 1 1

2 2 1 / 2 2 1 ( / 2)x x x

10 0

1 1 ( 1)( )

2 2 2 2

nn n

nn n

xx

x

1/( 2).x

3 /( 2).x x 3

3 3 31 1

0 0

1 ( 1) ( 1)

2 2 2 2

n nn n

n nn n

xx x x x

x x

Differentiation and integration Theorem If the power series has radius of

convergence R>0, then the sum function

is differentiable on the interval and

(i)

(ii)

The above two series have same radius of convergence R.

( )nnc x a

20 1 2

0

( ) ( ) ( ) ( )nn

n

f x c x a c c x a c x a

( , )a R a R 1

1 21

( ) 2 ( ) ( )nnn

f x c c x a nc x a

2 1

0 10

( ) ( )( ) ( )

2 1

n

nn

x a x af x dx C c x a c C c

n

Example The above formula are called term-by-term

differentiation and integration. Ex. Express as a power series and find the radius

of convergence. Sol. Differentiating gives

By the theorem, the radius of convergence is same as the

original series, namely, R=1.

21/(1 )x

0

1

1n

n

xx

1 22

1

11 2 3

(1 )n

n

nx x xx

Example Ex. Find a power series representation of Sol.

( ) arctan .f x x

2 2 2 4 62

0 0

1( ) ( 1) 1

1n n n

n n

x x x x xx

2 1

20

1arctan ( 1)

1 2 1

nn

n

xx dx C

x n

2 1

0

arctan 0 0 0 arctan ( 1)2 1

nn

n

xC x

n

Example Ex. Find a power series representation for

and its radius of convergence.

Sol.

( ) ln(1 )f x x

2 31ln(1 ) (1 )

1x dx x x x dx

x

2

12

n

n

x xC x C

n

1

ln(1 0) 0 0 ln(1 ) .n

n

xC x

n

1

1ln 2.

2nn n

1.R

Taylor series Theorem If f has a power series representation

(expansion) at a, that is, if

Then its coefficients are given by the formula

This is called the Taylor series of f at a (or about a)

( ) ( )

!

n

n

f ac

n

0

( ) ( ) | |nn

n

f x c x a x a R

( )2

0

( ) ( ) ( )( ) ( ) ( ) ( ) ( )

! 1! 2!

nn

n

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

n

Maclaurin series The Taylor series of f at a=0 is called Maclaurin series

Ex. Find the Maclaurin series of the function and its radius of convergence.

( ) xf x e

2 3

0

1! 1! 2! 3!

nx

n

x x x xe x

n

( )2

0

(0) (0) (0)( ) (0)

! 1! 2!

nn

n

f f ff x x f x x

n

Maclaurin series Ex. Find the Maclaurin series for sinx. Sol.

So the Maclaurin series is

3 5 2 1

0

( 1)

3! 5! (2 1)!

n n

n

x x xx

n

( ) ( )( ) sin( ) (0) sin2 2

n nn nf x x f

Important Maclaurin series Important Maclaurin series and their convergence interval

2 3

0

1 ( , )! 1! 2! 3!

nx

n

x x x xe

n

2 1 3 5 7

0

( 1)sin ( , )

(2 1)! 3! 5! 7!

n n

n

x x x xx x

n

2 3

0

11 ( 1,1)

1n

n

x x x xx

2 2 4 6

0

( 1)cos 1 ( , )

(2 )! 2! 4! 6!

n n

n

x x x xx

n

Example Ex. Find the Maclaurin series of Sol.

( ) 1 .f x x

1/ 2 1/ 21( ) ( 1) ( ) ( 1) ,

2f x x f x x

3/ 2 5/ 21 3( ) ( 1) , ( ) ( 1) ,

4 8f x x f x x

( ) 1 (2 1) / 23 5 (2 3)( ) ( 1) ( 1)

2n n n

n

nf x x

( ) 1 1(2 3)!! (2 3)!!(0) ( 1) , ( 1)

2 2 !n n n

nn n

n nf a

n

1

2

(2 3)!!1 1 ( 1) .

2 2 !n n

nn

x nx x

n

Multiplication of power series Ex. Find the first 3 terms in the Maclaurin series for

Sol I. Find and the Maclaurin series is found.

Sol II. Multiplying the Maclaurin series of and sinx

collecting terms:

sinxe x

(0), (0), (0)f f f

xe

2 3 3

sin (1 )( )2 6 6

x x x xe x x x

32

3

xx x

Division of power series Ex. Find the first 3 terms in the Maclaurin series for

Sol I. Find and the Maclaurin series is found.

Sol II. Use long division

tan x

(0), (0), (0)f f f

3 5

3 5

2 4

26 120tan3 15

12 24

x xx x x

x xx x

Application of power series Ex. Find

Sol. Let then s=S(-1/2).

To find S(x), we rewrite it as

2

0

1( 1) .

2n

nn

n ns

2

0

( ) ( 1) ,n

n

S x n n x

1

1 0

( ) ( 1) ( ) 1/(1 ),n n

n n

S x x n n x x xT x x

1 2

0 01

( ) /(1 )x x n

n

T x dx dx x x x

2

3

2( )

1 (1 )

xT x

x x

3

1 2 2 10.

2 (3 / 2) 3 27s

Exercise Ex. Find

Sol.

2

0

1.

3nn

n ns

2

0

( ) ( 1) ,n

n

S x n n x

2 2 2

2 0

( ) ( 1) ( ) 1/(1 ),n n

n n

S x x n n x x x T x x

2

0 02

( ) /(1 )x x n

n

T x dx dx x x x

9

.4

s

Application of power series Ex. Find by the Maclaurin series expansion.

Sol.

where in the last limit, we have used the fact that power

series are continuous functions.

20

1lim

x

x

e x

x

2 3

2 20 0

(1 ) 11 2! 3!lim limx

x x

x xx xe x

x x

2 3

2

20 0

1 12! 3!lim lim ,2 3! 4! 2x x

x xx x

x

Homework 25 Section 11.8: 10, 17, 24, 35

Section 11.9: 12, 18, 25, 38, 39

Section 11.10: 41, 47, 48