tch-prob1 chap 5. series series representations of analytic functions 43. convergence of sequences...

tch-prob 1

Chap 5. Series

Series representations of analytic functions

43. Convergence of Sequences and Series

An infinite sequence 數列

1 2 nz , z , ...........,z ,..........

of complex numbers has a limit z if, for each positive , there exists a positive integer n0 such that

0. whenever

nz z n n

0

For sufficieutly large , the points lie in any given neighborhood of ., in general, depends on .

nz

nz

n

z1

z3

z2

zn

z

tch-prob 2

The limit z is unique if it exists. (Exercise 6). When the limit exists, the sequence is said to converge to z.

limn

z zn

Otherwise, it diverges.

Thm 1.

Suppose

lim lim

then iff and lim .

n n n

n n nn n

z x iy , z x iy z z x x, y yn

tch-prob 3

1 21

1 21

(6)

of complex numbers converges to the sum if the sequence

( 1 2 )

of partial sums con

n nn

N

nN Nn

z

z

z z ......... z ...............

S

S z z .......... z N , , ......

1

1 1 1

verges to .We then write

Thm 2. Suppose ,

Then iff and

nn

n n n

n n nn n n

z

iy

z x x y

S

S

z x S x iy

S y

An infinite series

tch-prob 4

A necessary condition for the convergence of series (6) is that lim 0 znn

The terms of a convergent series of complex numbers are, therefore, bounded, .,

nz Mi.e.

Absolute convergence:2 2

1 1 convergesn n n

n nz x y

Since

,

2 2 2 2,

and also converges1 1

and converge1 1

From Thm. 2 converge1

x x y x

x y

x y

z

y yn n n n n n

n nn n

, n nn n

nn

Absolute convergence of a series of complex numbers implies convergence of that series.

tch-prob 5

44. Taylor Series

Thm. Suppose that a function f is analytic throughout an open disk

0 0.z z R Then at each point z in that disk, f(z) has the series

representation

0

( )0

( )

( )

( )0

where

0 1 2!

n z

nf z a z- znn

f a n , , , .....n n

That is, the power series here converges to f(z)

0 0.when z- z R

z

z0

R0

tch-prob 6

This is the expansion of f(z) into a Taylor series about the point z0

Any function that is known to be analytic at a point z0 must have a Taylor series about that point.

(For, if f is analytic at z0, it is analytic in some neighborhood

0 0of (sec 20 55)z z ε z . , p . may serve as R0 is the

statement of Taylor’s Theorem)

0( )

0

0 0

( )

. (a) when 0

0) prove ( ) (!0

Let : Circle

nz

z

pf z

f nf z z Rn n

C r

Positively oriented within

and z is interior to it.

0z R

~ Maclaurin series.z0=0的 case

0

C0

R0r r0

sz

tch-prob 7

The Cauchy integral formula applies:

0

1

0

1

0

1

10

( )( )

( )1( )21 1 1now

1

1and1 1

1 11

( )

N Nn

n

NN n

n

nN N

n Nn

s z

zs

z

zsz

s zs

z

s

f s dsf z cπi

s z s -

z - z z

( )s z s

z s z s

( 1), 1zs

zs

6). sec. 13, (Exeicise( 1)z

tch-prob 8

1

100 00( )1

00( )1

0 0

( )

(0)2

!

(0)( )

!

( ) ( )( )

( ) ( )

( )1( )2

wher

Nn N

c n Nn

nNn N

Nn

nNn

Nn

f s ds

s

fi

n

fz z

n

f s ds f s ds z zc cs z s z s

f s ds z z c s z s

f s dsf z ρc s zπi

0( ) ( )e

2 ( )N

N Nz f s dsz ρ cπi s z s

0

0

Suppose that

then if is a point on

z r

s Cs z s z r r

tch-prob 9

0 0

0 00 0

0( )

( ) )

lim ( )

(0)

If ( )

2(

2 ( )

But 1 0

( )!0

NN

N N

NN

nn

z

z

f s M

M π r M rr r ρ r r rπ r r rr , ρ r

f f z znn

1zs

主要

原因

(b) For arbitrary z0

Suppose f is analytic when0 0

z z R and note that the

composite function 0( )f z z

must be analytic when0 0 0

( )z z z R

tch-prob 10

0).Let ( ) ( g z f z z

The analyticity of g(z) in the disk 0z R ensures

the existence of a Maclaurin series representation:

( )

0

( )0

0

0( )

00

( )

( ))

( ))

0) ( ) (!0

or (!0

Replace by -

( ) (!0

nn

nn

nn

z

z

g g z z z Rnn

f f z z z

nnz z z ,

f f z z z

nn

tch-prob 11

45 ExamplesEx1. Since is entire ( ) zf z e

It has a Maclaurin series representation which is valid for all z.

( ) ( )

2 3

2 3 2

2

( ) (0) 1

( )!0

( ) is also entire.

3!03 ( )

( 2)!2

n n

n

z

nz n

nn

z zf e , f ,zz e znn

f z z e

z e znn

z zn-n

tch-prob 12

Ex2. Find Maclaurin series representation of

(2 ) (2 1)

2 1

(0) (0) ( 1)

( )0 0 1 2

( 1) ( )(2 1)!0

n n n

nn

f z sin zf f , n , , , ....

zsin z - znn

sin ( )sinh z -i iz

Ex3.

(2 ) (2 1)

2

( (0)

( ) cos

0) ( 1) 0

cos ( 1) ( )(2 )!0

cosh cos ( )

nn n

nn

f z z

f fz z - znn

z iz

tch-prob 13

Ex4.

( )1

( )

( )

(0)

1 ( 1)1 0

!since (1 )!

substitute - for1 ( 1) 1

1 0substitute 1 for

n

nn

n

n n

z

z zz n

n fz

f nz z

z ( z )z n

- z z

1 ( 1) ( 1) ( 1 1)0

(1 ) ( 1) ( 1)0 0

n n

n n n

z z -z n

- z - z-n n

tch-prob 14

Ex5.)

)

22 2(1 11 2 1( )3 5 3 211 1(23 21

can not find a Maclaurin series for ( )since it is not analytic at 0But

1 2 4 61 121

Hence, when

zzf z z z z z

-z z

f z z .

z z z ............( z )z

0 11 2 4 6( ) (2 13

1 1 3 53

z ,

f z - z z z ..............)z

z z z ...........zz

為 Laurent series 預告

tch-prob 15

46. Laurent Series

If a function f fails to be analytic at a point z0, we can not apply Taylor’s theorem at that point.

However, we can find a series representation for f(z) involving both positive and negative powers of (z-z0).

Thm. Suppose that a function f is analytic in a domain

1 0 2 ,R z z R and let C denote any positively oriented

simple closed contour around z0 and lying in that domain. Then at each z in the domain

00

))

( ) ( (1)(0 1

bn nf z a z- z n nz- zn n

tch-prob 16

where

0

0

0

)

)

)

( )1 ( 0 1 2 )2 1(

( )1 ( 1 2 )2 1(

or

( ) (

f z dz a n , , , ......n cπ i nz- zf z dz b n , , .......n cπ i nz- z

n f z C z- z nn -

1 0 2

0

)

)

( (4)

( )1where ( 0 1 2 ) (5)2 1(

R z z R

f z dz C n , , , ..... n cπ i nz- z

Pf: see textbook.

2 1

( ) ( ) ( ) 0

2 ( )

f s f s ds f s ds ds-c cs z s z s z

πi f z

rZ

R1Z0

R2

CC1

C2

(1) (4) are Laureut Series

tch-prob 17

47. Examples

The coefficients in a Laurent series are generally found by means other than by appealing directly to their integral representation.

1

2 3

1

)!

(0

1 replace by

1 1 1 11 (0 )! 2! 3!0

no positive powers of appear.

since

n

zn

z zn

z e n

z z

e ......... zzn z z zn z

b

1

1

1 12

2

z

z

e dzcπi

e dz πic

Ex1.

Alterative way to calculate

tch-prob 18

Ex2.

0

2

1( ) is already in the form of a laurent series, where .2( )

( ) ( ) (0 - )

where 1, all other coefficients are zero.( )1 1Since

2 1( )

f z z iz i

nf z C z i z innC

f z dzC n cπ i nz-i

1

2 3( )0 when n -2

3 2 when n=-2( )

11[ 2 0]

2 2 !( ) ( )

dzci nz i

dz c n iz-i

dz idzdz dz π i c c nz i z-i

tch-prob 19

０D1

D2D3

21

Ex3.1 1 1( )

1 2( 1)( 2)f z

z zz z

has two singular points z=1 and z=2, and is analytic in the domains

Recall that

1 ( 1)1 0

nz zz n

(a) f(z) in D1

1

1 1 1( ) 1- 2 1- 2

Since 1 and 1 in D2

nn -n-1( ) (2 1)n 120 0 0

f z zz

zz

z nf z z zn n n

( 1)z

1

2

3

:

: 1 2

: 2

1

z

D z

D z

D

tch-prob 20

(b) f(z) in D2

) )

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

1Since 1 , 1 , when 1 221 1 1( ) ( (

2 20 01 (1 2)n 1 120 0

1 1n nn 120 1

f z zzz

z z ,z

zn nf z z zn nnz z

nzn n

zzn n

(1 2)z

tch-prob 21

(c) f(z) in D3

1 1 1 1( ) . 1 21- 1-

1 2since 1 , 1 when 2

1 2( ) n 1 10 0

n1-2 (2 )n 1n 0

11 2 (2 )n 1

f z z zz z

zz zn

f znz zn n

zz

nznz

tch-prob 22

48. Absolute and uniform convergence of power series

Thm1.

0

1 1 0

0 1 1 1 0.

If a power series ( - ) convergesn 0

when (z ), then it is absolutelyconvergent at each point in the open disk

- , where

nna z z

z z zz

z z R R z z

R1

z1

z z01

0 1.

Converges at Converges at all points

inside the circle -

z

z z R

(1)

tch-prob 23

• The greatest circle centered at z0 such that series (1) converges at each point inside is called the circle of convergence of series (1).

• The series CANNOT converge at any point z2 outside that circle, according to the theorem; otherwise circle of convergence is bigger.

0

0

0

0

Suppose that (z-z ) (1)n 0

has circle of convergence -

Let

( ) ( - )0

N-1 ( ) ( - )

n 0

N

nan

z z R

nS z a z znn

nS z a z zn

0z z R

tch-prob 24

0write ( ) ( ) ( ) ( - )

since (1) converges, ( ) 0 as or for each , there is a swch that

( ) whenever

N N

N

N

z S z S z z z R

z NN

z N N

When the choice of depends only on the value of and is independent of the point z taken in a specified region within the circle of convergence, the convergence is said to be uniform in that region.

N

tch-prob 25

z0

R

R1z

z1

Corollary.

0

0

A power series )( represents a continuous function ( )0

at each point inside its circle of convergence .

n a z- z S znn

z- z R

1 0 0Thm 2. If is in c. of. c. - of ( - )

n 0nz z z R a z zn

then that series is uniformly convergent in the closed disk

0 1 1 1 0., where z z R R z z

tch-prob 26

49. Integration and Differentiation of power seriesHave just seen that a power series

0( ) (1)( )

0z nS z a zn

n

represents continuous function at each point interior to its circle of convergence.

We state in this section that the sum S(z) is actually analytic within the circle.

Thm1. Let C denote any contour interior to the circle of convergence of the power series (1), and let g(z) be any function that is continuous on C. The series formed by multiplying each term of the power series by g(z) can be integrated term by term over C; that is,

0)( ) ( ) ( ) (

0

ng z S z dz a g z z- z dznc cn

tch-prob 27

Corollary. The sum S(z) of power series (1) is analytic at each point z interior to the circle of convergence of that series.

Ex1.sin when 0

want to show ( )1 when 0

2 1since sin ( 1) represent sin for every ,

(2 1)!02 1

( 1)2 2 4(2 1)!0 ( 1) 1

3! 5!(2 1)!0

z z z f z z

nzn z - z znn

nzn - nn z z znn z nn

(4)

converges to ( ) when 0

............

f z z

is entire

But series (4) clearly converges to f(0) when z=0. Hence f(z) is an entire function.

tch-prob 28

Thm2. The power series (1) can be differentiated term by term. That is, at each point z interior to the circle of convergence of that series,

0) 1( ) (

0nS' z n a z- zn

n

Ex2. (1 ( 1) 1) ( 1 1)0

zn n - z -z n

Diff.1 1( 1) ( 1) 1 12 1

1 ( 1) ( 1)( 1) 1 12 0

n n- - n z- z -z n

n n - n z z -z n

tch-prob 29

50. Uniqueness of series representation

Thm 1. If a series

0 ( ) converges to ( )

0na z z f zn

n

at all points interior to some circle , then it is the Taylor series expansion for f in powers of .

0z z R

0z z

Thm 2. If a series

0)( n C z- znn

converges to f(z) at all points in some annular domain about z0, then it is the Laurent series expansion for f in powers of for that domain.0

z z

tch-prob 30

51. Multiplication and Division of Power SeriesSuppose

0

0

0

( )

( )

( )0

and ( )0

converges within ,

z

z

z

n f z a znn

ng z b znn

z R

then f(z) and g(z) are analytic functions in 0z z R

has a Taylor series expansion Their product

0 0( )( ) ( )

0

Cauchy product0

z

c

nf z g z c z z z Rnn

n a b n k n-kk

tch-prob 31

Ex1.1z The Maclaurin series for is valid in disk

11 12 3 2 3(1 )(1 )2 6

1 12 31 12 3

z

z

zez

z z ..... - z z z ........

z ..... z

Ex2.Zero of the entire function sinh z

)

are the mumbers ( 0 1 2 )1 1 1So

2 3 2 4sinh 1 3! 5!has laurent series in the punctured disk 0

1 1 1 7 (0 )2 3 360sinh

(si2

16

nh

z nπi n , , , ...

z z z z z( .......)

z π

. z ..... z πzz z zz ze ez

1zez