Chapter 5. Series

Chapter 5. Series

Weiqi Luo (骆伟祺 )School of Software

Sun Yat-Sen UniversityEmail ： [email protected] Office ： # A313

mailto:[email protected]

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Convergence of Sequences; Convergence of Series Taylor Series; Proof of Taylor's Theorem; Examples; Laurent Series; Proof of Laurent's Theorem; Examples Absolute and Uniform Covergence of Power Series Continuity of Sums of Power Series Integration and Differentiation of Power Series Uniqueness of Series Representations Multiplication and Division of Power Series

2

Chapter 5: Series

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The limit of Sequences An infinite sequence

z1, z2, …, zn, … of complex number has a limit z if, for each positive number ε, there exists a positive integer n0 such that when n>n0

55. Convergence of Sequences

3

| |nz z

Note that the limit must be unique if it exists;Otherwise it diverges

lim nn

z z

Denoted as

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Theorem Suppose that zn = xn + iyn (n = 1, 2, . . .) and z = x + iy.

Then

If and only if

Proof: If

then, for each positive number ε, there exists n1 and n2, such that


4

lim nn

z z

lim & limn nn n

x x y y

lim & limn nn n

x x y y

1| | ,2nx x n n

2| | ,2ny y n n

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5

Let n0=max(n1,n2), then when n>n0

| | & | |2 2n nx x y y

| | | ( ) ( ) | | ( ) ( ) |n n n n nz z x iy x iy x x i y y

| | | |2 2n nx x y y

Conversely, if we have that for each positive ε, there exists a positive integer n0 such that, when n>n0

lim nn

z z

| | | ( ) ( ) | | ( ) ( ) |n n n n nz z x iy x iy x x i y y

| | | ( ) ( ) |n n nx x x x i y y | | | ( ) ( ) |n n ny y x x i y y

lim & limn nn n

x x y y

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Example 1 The sequence

converges to i since


6

3

1, ( 1,2,...)nz i n

n

3 3

1 1lim( ) lim lim1 0 1n n n

i i i in n

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Example 2 When

The theorem tells us that

If using polar coordinates, we write


7

2

( 1)2 , ( 1,2,...)

n

nz i nn

2

( 1)lim lim( 2) lim( ) 2 0 2

n

nn n n

z i in

| | & , ( 1,2,...)n n n nr z Argz n

4

1lim lim 4 2nn n

rn

2lim nn

Argz

2 1lim nn

Argz

Why? Evidently, the limit of Θn does not exist as ntends to infinity.

n

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Convergence of Series An infinite series

of complex number converges to the sum S if the sequence

of partial sums converges to S; we then write

56. Convergence of Series

8

1 21

... ...n nn

z z z z

1 21

... , (1, 2,...)N

N n Nn

S z z z z N

1n

n

z S

The series has at most one limit, otherwise it diverges

Series

Sequence

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Theorem Suppose that zn = xn + iyn (n = 1, 2, . . .) and S = X + iY.

Then

If and only if


9

1n

n

z S

1 1

&n nn n

x X y Y

1n

n

z S

1 1

N N

N n n N Nn n

S x i y X iY

lim NN

S S

lim & limN NN N

X X Y Y

1 1

&n nn n

x X y Y

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Corollary 1

If a series of complex numbers converges, the nth term converges to zero as n tends to infinity.

Assuming that converges, based on the theorem, both the two following real series converse.

Then we get that xn and yn converge to zero as n tends to infinity (why?), and thus


10

1 21

... ...n nn

z z z z

1 1

&n nn n

x y

lim lim lim 0 0 0n n nn n nz x i y i

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Absolutely convergent If the series

of real number converges,

then the series is said to be absolutely convergent.


11

2 2

1 1

| | , ( )n nn n n n

n n

z x y z x iy

2 2

n nx y

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Corollary 2

The absolute convergence of a series of complex numbers implies the convergence of that series.


12

2 2| |n nnx x y

2 2| |n nny x y

2 2

1 1

| |n nn

n n

x x y

2 2

1 1

| |n nn

n n

y x y

1

| |nn

x

1

| |nn

y

Converge

1n

n

x

1n

n

y

Converge

1 21

... ...n nn

z z z z

Converge

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The remainder ρN after N terms


13

1 2 1 21

... ...n N N Nn

S z z z z z z

SN

N NS S | 0 | | |N NS S

Therefore, a series converges to a number S if and only if the sequence of remainders tends to zero.

ρN

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Example With the aid of remainders, it is easy to verify that when |z| <1,

Note that

The partial sums

If then


14

0

1

1n

n

zz

12 1

1 ... , 11

nn z

z z z zz

1

2 1

0

1( ) 1 ... , 1

1

NNn N

Nn

zS z z z z z z

z

1

( ) , 11

S z zz

( ) ( ) ( ) , 1

1

N

N N

zz S z S z z

z

| || ( ) |

|1 |

N

N

zz

z

When |z|<1 ρN tends to zero, but not when |z|>1

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pp.188-189

Ex. 2, Ex. 3, Ex. 5, Ex. 9

56. Homework

15

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Theorem Suppose that a function f is analytic throughout a disk

|z − z0| < R0, centered at z0 and with radius R0. Then f (z) has the power series representation

57. Taylor Series

16

0 0 00

( ) ( ) , (| | )nn

n

f z a z z z z R

( )

0( ), ( 0,1,2,...)

!

n

n

f za n

n

That is, series converges to f (z) when z lies in the stated open disk.

10

1 ( )

2 ( )n nC

f z dza

i z z Refer to pp.167

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Maclaurin Series

57. Taylor Series

17

When z0=0 in the Taylor Series become the Maclauin Series

In the following Section, we first prove the Maclaurin Series, in which case f is assumed to be assumed to be analytic throughout a disk |z|<R0

( )

0 00

(0)( ) ,(| | )

!

nn

n

ff z z z z R

n

y=ex

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58. Proof the Taylor’s Theorem

18

( )

0 00

(0)( ) ,(| | )

!

nn

n

ff z z z z R

n

Proof:

Let C0 denote and positively oriented circle |z|=r0, where r<r0<R0

Since f is analytic inside and on the circle C0 and since the point z is interior to C0, the Cauchy integral formula holds

0

0

1 ( )( ) , ,| |

2 C

f s dsf z z z R

i s z

1 1 1 1 1

, ( / ),| | 11 ( / ) 1

w z s ws z s z s s w

Refer to pp.187

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19

1

10

1 1 1

( )

Nn N

n Nn

z zs z s s z s

0

1 ( )( )

2 C

f s dsf z

i s z

0 0

1

10

1 ( ) 1 ( )( )

2 2 ( )

Nn N

n Nn C C

f s ds f s dsf z z z

i s i s z s

( ) (0)

!

nf

n

Refer to pp.167

0

( )1

0

(0) ( )( )

! 2 ( )

n NNn

Nn C

f z f s dsf z z

n i s z s

ρN

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20

0

( )lim lim 0

2 ( )

N

N NN NC

z f s ds

i s z s

( ) ( ) ( )1

0 0 0

(0) (0) (0)( ) lim( ) 0

! ! !

n n nNn n n

NNn n n

f f ff z z z z

n n n

When

0

00 0

( ) | || | | | 2

2 ( ) 2 ( )

N N

N N NC

z f s ds r Mr

i s z s r r r

Where M denotes the maximum value of |f(s)| on C0

0

0 0

| | ( )NN

Mr r

r r r

lim 0NN

0

( ) 1r

r

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Example 1 Since the function f (z) = ez is entire, it has a Maclaurin

series representation which is valid for all z. Here f(n)(z) = ez (n = 0, 1, 2, . . .) ; and because f(n)(0) = 1 (n = 0, 1, 2, . . .) , it follows that

Note that if z=x+i0, the above expansion becomes

59. Examples

21

0

, (| | )!

nz

n

ze z

n

0

, ( )!

nx

n

xe x

n

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Example 1 (Cont’) The entire function z2e3z also has a Maclaurin series

expansion,

If replace n by n-2, we have

59. Examples

22

2 3 2

0

3, (| | )

!

nz n

n

z e z zn

0

, (| | )!

nz

n

ze z

n

Replace z by 3z

22 3

2

3, (| | )

( 2)!

nz n

n

z e z zn

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Example 2 Trigonometric Functions

59. Example2

23

2 1

0

sin ( 1) , (| | )2 (2 1)!

iz iz nn

n

e e zz Z

i n

2

0

cos ( 1) , (| | )2 (2 )!

iz iz nn

n

e e zz Z

n

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Example 4

Another Maclaurin series representation is

since the derivative of the function f(z)=1/(1-z), which fails to be analytic at z=1, are

In particular,

59. Examples

24

0

1, (| | 1)

1n

n

z zz

( )1

!( ) , ( 0,1,2,...)

(1 )n

n

nf z n

z

( ) (0) !, ( 0,1,2,...)nf n n

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Example 4 (Cont’)

59. Examples

25

0

1( 1) , (| | 1)

1n n

n

z zz

0

1( 1) ( 1) , (| 1| 1)n n

n

z zz

0

1, (| | 1)

1n

n

z zz

substitute –z for z

replace z by 1-z

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Example 5

expand f(z) into a series involving powers of z.

We can not find a Maclaurin series for f(z) since it is not analytic at z=0. But we do know that expansion

Hence, when 0<|z|<1

59. Examples

26

2 2

3 5 3 2 3 2

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

1 1

z zf z

z z z z z z

2 4 6 82

11 ...(| | 1)

1z z z z z

z

2 4 6 8 3 53 3

1 1 1( ) (2 1 ...) ...f z z z z z z z z

z z z

Negative powers

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pp. 195-197

Ex. 2, Ex. 3, Ex. 7, Ex. 11

59. Homework

27

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Theorem Suppose that a function f is analytic throughout an annular domain

R1< |z − z0| < R2, centered at z0 , and let C denote any positively oriented simple closed contour around z0 and lying in that domain. Then, at each point in the domain, f (z) has the series representation

60. Laurent Series

28

0 1 0 20 1 0

( ) ( ) , ( | | )( )

n nn n

n n

bf z a z z R z z R

z z

10

1 ( ), ( 0,1,2,...)

2 ( )n nC

f z dza n

i z z

10

1 ( ), ( 1,2,...)

2 ( )n nC

f z dzb n

i z z

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Theorem (Cont’)

60. Laurent Series

29

0 1 0 2( ) ( ) , ( | | )nn

n

f z c z z R z z R

0 1 0 20 1 0

( ) ( ) , ( | | )( )

n nn n

n n

bf z a z z R z z R

z z

10

1 ( ), ( 0,1, 2,...)

2 ( )n nC

f z dza n

i z z

10

1 ( ), ( 1, 2,...)

2 ( )n nC

f z dzb n

i z z

10

1 ( ), ( 0, 1, 2,...)

2 ( )n nC

f z dzc n

i z z

1 1

00

( )( )

nnnn

n n

bb z z

z z

, 1

, 0n

nn

b nc

a n

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Laurent’s Theorem If f is analytic throughout the disk |z-z0|<R2,

60. Laurent Series

30

00

( ) ( )nn

n

f z a z z

101

0

1 ( ) 1( ) ( ) , ( 1,2,...)

2 ( ) 2n

n nC C

f z dzb z z f z dz n

i z z i

Analytic in the region |z-z0|<R2

0, ( 1,2,...)nb n

( )0

10

( )1 ( ), ( 0,1,2,...)

2 ( ) !

n

n nC

f zf z dza n

i z z n

reduces to Taylor Series about z0

0 1 0 20 1 0

( ) ( ) , ( | | )( )

n nn n

n n

bf z a z z R z z R

z z

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Example 1 Replacing z by 1/z in the Maclaurin series expansion

We have the Laurent series representation

62. Examples

31

2 3

0

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

nz

n

z z z ze z

n

1/2 3

0

1 1 1 11 ...(0 | | )

! 1! 2! 3!z

nn

e zn z z z z

There is no positive powers of z, and all coefficients of the positive powers are zeros.

1

1 ( ), ( 1,2,...)

2 ( 0)n nC

f z dzb n

i z

1/1/

1 1 1

1 11

2 ( 0) 2

zz

C C

e dzb e dz

i z i 1/ 2z

C

e dz i

where c is any positively oriented simple closedcontours around the origin

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Example 2

The function f(z)=1/(z-i)2 is already in the form of a Laurent series, where z0=i,. That is

where c-2=1 and all of the other coefficients are zero.

62. Examples

32

2

1( ) , (0 | | )

( )n

nn

c z i z iz i

30

1, ( 0, 1, 2,...)

2 ( )n nC

dzc n

i z z

3

0, 2

2 , 2( )nC

ndz

i nz i

where c is any positively oriented simple contouraround the point z0=i

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62. Examples

33

Consider the following function

1 1 1( )

( 1)( 2) 1 2f z

z z z z

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

1 :| | 1D z

3 : 2 | |D z

2 :1 | | 2D z

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Example 3 The representation in D1 is Maclaurin series.

62. Examples

34

1 1 1 1 1( )

1 2 1 2 1 ( / 2)f z

z z z z

Refer to pp. 194 Example 4

11

0 0 0

( ) (2 1) , (| | 1)2

nn n n

nn n n

zf z z z z

where |z|<1 and |z/2|<1

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Example 4 Because 1<|z|<2 when z is a point in D2, we know

62. Examples

35

1 1 1 1 1 1( )

1 2 1 (1/ ) 2 1 ( / 2)f z

z z z z z

where |1/z|<1 and |z/2|<1

1 1 10 0 1 0

1 1( ) , (1 | | 2)

2 2

n n

n n n nn n n n

z zf z z

z z


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Example 5 Because 2<|z|<∞ when z is a point in D3, we know

62. Examples

36

1 1 1 1 1 1( )

1 2 1 (1/ ) 1 (2 / )f z

z z z z z z

where |1/z|<1 and |2/z|<1

1

1 1 10 0 0 1

1 2 1 2 1 2( ) , (2 | | )

n n n

n n n nn n n n

f z zz z z z


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pp. 205-208

Ex. 3, Ex. 4, Ex. 6, Ex. 7

62. Homework

37

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Theorem 1 (pp.208) If a power series

converges when z = z1 (z1 ≠ z0), then it is absolutely convergent at each point z in the open disk |z − z0| < R1 where R1 = |z1 − z0|

63~66 Some Useful Theorems

38

00

( )nn

n

a z z

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Theorem 2 (pp.210) If z1 is a point inside the circle of convergence |z − z0| =

R of a power series

then that series must be uniformly convergent in the closed disk |z − z0| ≤ R1, where R1 = |z1 − z0|


39

00

( )nn

n

a z z

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Theorem (pp.211)

A power series

represents a continuous function S(z) at each point inside its circle of convergence |z − z0| = R.


40

00

( )nn

n

a z z

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Theorem 1 (pp.214) Let C denote any contour interior to the circle of convergence of

the power series S(z), 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,


41

00 C

( ) ( ) ( )( )nn

nC

g z S z dz a g z z z dz

00

( ) ( )nn

n

S z a z z

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

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Theorem 2 (pp.216) The power series S(z) can be differentiated term by

term. That is, at each point z interior to the circle of convergence of that series,


42

10 0

0 0

'( ) ( ( ) ) ' ( )n nn n

n n

S z a z z na z z

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The uniqueness of Taylor/Laurent series representations Theorem 1 (pp.217) If a series

converges to f (z) at all points interior to some circle |z − z0| = R, then it is the Taylor series expansion for f in powers of z − z0.

Theorem 2 (pp.218) If a series

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 z − z0 for that domain.


43

00

( )nn

n

a z z

0 00 1 0

( ) ( )( )

n n nn n n

n n n

bc z z a z z

z z

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Chapter 5. Series