1

(1) Power series

(2) Convergence of Power series and the Radius of Convergence

(3) The Cauchy-Hadamard formula

(4) Taylor’s series

Section 6

SECTION 6Power Series I - Taylor Series

2

Section 6Why Series ?

?/ C

zdzze

3

4

2

32/ 1

!4

1

!3

1

2 zzzzze z

We can expand the integrand….

dz

zdz

zdzdzzdzze

CCCCC

z2

32/ 1

!3

1

2

where C is 2z C

Analytic ( )by Formula for derivatives ( )

3

Section 6

Can we expand all functions in series ?

We can expand analytic functions in specialseries called “Power Series”

How do we find these series ?

(1) Using Taylor’s Theorem(2) Using other known series

(and other tricks)

4

Section 6Power Series

0

221 )()()(

nooo

non zzazzaazza

A series in powers of )( ozz

0

2)(2

1)(1)(

!

1

n

n jzjzjzn

e.g.

5

Section 6Power Series

0

221 )()()(

nooo

non zzazzaazza

centre

A series in powers of )( ozz

0

2)(2

1)(1)(

!

1

n

n jzjzjzn

e.g.

6

Section 6Power Series

0

221 )()()(

nooo

non zzazzaazza

coefficientscentre

A series in powers of )( ozz

0

2)(2

1)(1)(

!

1

n

n jzjzjzn

e.g.

7

Section 6Convergence of Power SeriesPower series often converge for some values of z butdiverge for other values.

For example the series

0

321n

n zzzz

converges for z 1 but diverges for z 1

diverges

converges

(geometric series)

8

Section 6Convergence of Power SeriesPower series often converge for some values of z butdiverge for other values.

For example the series

0

321n

n zzzz

converges for z 1 but diverges for z 1

Radius ofConvergence

R=1

diverges

converges

(geometric series)

9

Section 6

0

32 !321!n

n zzzzn

Example:

series diverges for all z (except z0)

0

32

!3!21

!

1

n

n zzzz

n

Example:

series converges for all z

Radius of Convergence“at infinity”; R

Zero Radius ofConvergence; R0

10

: converges

0

221 )()()(

nooo

non zzazzaazza

Section 6

(1) The power series always converges at zzo

oz

(2) There is a Radius of Convergence R for which:

Rzz o

: divergesRzz o

11

Section 6Is there a quick way to find the radius of convergence ?

Ra

a

n

n

n

1lim 1

… the Cauchy-Hadamard formula:

Example:

0

22

)3(6)3(21)3()!(

)!2(

n

n jzjzjzn

n

Rn

nn

n

n

n

n

a

ann

n

n

n

14

)1(

)12)(22(lim

)!2(

)!(

)!1(

!)1(2limlim

2

2

21

4

1R

12

Section 6

0

22

)3(6)3(21)3()!(

)!2(

n

n jzjzjzn

n

jzo 3

13

Section 6

0

22

)3(6)3(21)3()!(

)!2(

n

n jzjzjzn

n

: converges

jzo 3

4/13 jz

14

Section 6

0

22

)3(6)3(21)3()!(

)!2(

n

n jzjzjzn

n

: converges

jzo 3

4/13 jz

: diverges4/13 jz

15

Section 6Another Example:

0

2)4()4(1)4(n

n jzjzjz

Ra

an

n

n

n

11

1

1limlim 1

1R

: converges

jzo 4

14 jz

: diverges14 jz

16

Section 6Another Example:

0

22 )1(2)1()1(n

nn zezezne

Re

n

ne

ne

en

a

ann

n

nn

n

n

11lim

)1(limlim

11

eR /1

: convergesez /11

: divergesez /11 1oz

17

Section 6

0

2

25

)2(

5

)2(1

5

)2(

nn

n jzjzjz

n

n

n a

a 1lim

R

Question:

Where is the centre of the series?What is the radius of convergence?

18

Section 6Note:We have used the Cauchy-Hadamard formulato find the radius of convergence.

There are many other tests (which can be used when theabove test fails).

e.g. (1) Root test (2) if

(3) Comparison test

(4) The Ratio test

diverges1lim

nn

nz

diverges0lim0

n

nnn

zz

converges andif converges00

n

nnnn

n bbzz

diverges1lim 1

n

n

n z

z

19

Section 6Power Series and Analytic functions

Every analytic function f (z) can be represented by a powerseries with a radius of convergence R0. The function isanalytic at every point within the radius of convergence.

Example:

0

321n

n zzzz

series converges for z 1

Radius ofConvergence

zzf

1

1)(

20

Section 6

These power series which represent analytic functions f (z)are called Taylor’s series.

0

)( )(!

1,)()(

no

nn

non zf

nazzazf

(Cauchy, 1831)

How do we derive these Power Series?

They are given by the formula

21

Section 6

Example:

32

0

0

)(

0

1

!

)0(

1

1

zzz

z

zn

f

zaz

n

n

n

nn

n

nn

zzf

1

1)(

(1) centre z0 :

!3)0(

2)0(

1)0(

1)0(

f

f

f

f

432 1

!3)(,

1

2)(,

1

1)(,

1

1)(

zzf

zzf

zzf

zzf

1z0oz

centresingular point

1R

Derive the Taylor series for

22

Section 6

Example:

32

0

0

)(

0

1

!

)0(

1

1

zzz

z

zn

f

zaz

n

n

n

nn

n

nn

zzf

1

1)(

(1) centre z0 :

!3)0(

2)0(

1)0(

1)0(

f

f

f

f

432 1

!3)(,

1

2)(,

1

1)(,

1

1)(

zzf

zzf

zzf

zzf

1z0oz

centresingular point

1R

Derive the Taylor series for

23

Section 6

Example:

32

0

0

)(

0

1

!

)0(

1

1

zzz

z

zn

f

zaz

n

n

n

nn

n

nn

zzf

1

1)(

(1) centre z0 :

!3)0(

2)0(

1)0(

1)0(

f

f

f

f

432 1

!3)(,

1

2)(,

1

1)(,

1

1)(

zzf

zzf

zzf

zzf

1z0oz

centresingular point

1R

Derive the Taylor series for

24

Section 6

Example:

32

0

0

)(

0

1

!

)0(

1

1

zzz

z

zn

f

zaz

n

n

n

nn

n

nn

zzf

1

1)(

(1) centre z0 :

!3)0(

2)0(

1)0(

1)0(

f

f

f

f

432 1

!3)(,

1

2)(,

1

1)(,

1

1)(

zzf

zzf

zzf

zzf

1z0oz

centresingular point

1R

Derive the Taylor series for

25

Section 6

3

212

21

21

0211

0212

1)(

021

16842

2

!

)(

1

1

zzz

z

zn

f

zaz

n

nn

n

nn

n

nn

(2) centre z12 :

421

321

221

21

2!3)(

2.2)(

2)(

2)(

f

f

f

f

432 1

!3)(,

1

2)(,

1

1)(,

1

1)(

zzf

zzf

zzf

zzf

1z21oz

centresingular point

21R

26

Section 6An analytic function f (z) can be represented bydifferent power series with different centres zo

(although there will only be one unique series for each centre)

At least one singular point of f (z) will be on the circle of convergence

1z21oz

1z0oz

3211

1zzz

z

3

212

21

21 16842

1

1zzz

z

27

Section 6An analytic function f (z) can be represented bydifferent power series with different centres zo

(although there will only be one unique series for each centre)

At least one singular point will be on the circle of convergence

1z21oz

1z0oz

3211

1zzz

z

3

212

21

21 16842

1

1zzz

z

28

Section 6

Another Example:

!321

!

!

)0(

32

0

0

)(

0

zzz

n

z

zn

f

zae

n

n

n

nn

n

nn

z

zezf )( with centre z0

1)0(

1)0(

1)0(

1)0(

f

f

f

f

zzzz ezfezfezfezf )(,)(,)(,)(

0oz

centre

Rno singular points!

29

Section 6Deriving Taylor series directly from the formula

can be very tricky

0

)( )(!

1,)()(

no

nn

non zf

nazzazf

We usually use other methods:(1) Use the Geometric Series(2) Use the Binomial Series

(3) Use other series (exp., cos, etc.)

(4) Use other tricks

32

0

11

1zzzz

z n

n

32

0 !3

)2)(1(

!2

)1(1

)1(

1z

mmmz

mmmzz

n

m

z n

nm

!5!3)!12()1(sin

53

0

12 zzz

n

zz

n

nn

30

Section 6

Example: 21

1)(

zzf

Expand about z0

(use the geometric series)

First, draw the centre and singular points to see what’s going on:

singular points:

So it looks like the radius of convergence should be R1

31

Section 6

Example: 21

1)(

zzf

centre

Expand about z0

(use the geometric series)

First, draw the centre and singular points to see what’s going on:

singular points:

So it looks like the radius of convergence should be R1

32

Section 6

Example: 21

1)(

zzf

centre

Expand about z0

(use the geometric series)

First, draw the centre and singular points to see what’s going on:

singular points:

jjz ,

So it looks like the radius of convergence should be R1

33

Section 6

Example: 21

1)(

zzf

centre

Expand about z0

(use the geometric series)

First, draw the centre and singular points to see what’s going on:

singular points:

jjz ,

So it looks like the radius of convergence should be R1

34

Section 6

3211

1zzz

zWe know that

64222

1)(1

1

1

1zzz

zzTherefore

The geometric series converges for z1

Therefore our series converges for z21- which is the same as z1, as predicted

35

Section 6

3211

1zzz

zWe know that

64222

1)(1

1

1

1zzz

zzTherefore

The geometric series converges for z1

Therefore our series converges for z21- which is the same as z1, as predicted

36

Section 6

3211

1zzz

zWe know that

64222

1)(1

1

1

1zzz

zzTherefore

The geometric series converges for z1

Therefore our series converges for z21- which is the same as z1, as predicted

37

Section 6

3211

1zzz

zWe know that

64222

1)(1

1

1

1zzz

zzTherefore

The geometric series converges for z1

Therefore our series converges for z21- which is the same as z1, as predicted

38

Section 6

Example:z

zf23

1)(

centre

Expand about z1

(use the geometric series)

First, draw the centre and singular points to see what’s going on:

singular point:

2/3z

So it looks like the radius of convergence should be R1/2

39

Section 6

3211

1zzz

zWe know that

2)1(4)1(21)1(21

1

23

1zz

zzTherefore

The geometric series converges for z1

Therefore our series converges for 2(z1)1

- which is the same as z 1 1/2, as predicted

40

Section 6

Example: 2)1(

1)(

zzf

centre

Expand about z0

(use the binomial series)

First, draw the centre and singular points to see what’s going on:

singular point:1z

So it looks like the radius of convergence should be R1

41

Section 6

32

0 !3

)2)(1(

!2

)1(1

)1(

1z

mmmz

mmmzz

n

m

z n

nm

The binomial series is

3222

4321)](1[

1

)1(

1zzz

zzTherefore

The binomial series converges for z1

Therefore our series converges for z1- which is the same as z1, as predicted

as we could guess,since issingular at z1

mz )1(

42

Section 6

32

0 !3

)2)(1(

!2

)1(1

)1(

1z

mmmz

mmmzz

n

m

z n

nm

The binomial series is

3222

4321)](1[

1

)1(

1zzz

zzTherefore

The binomial series converges for z1

Therefore our series converges for z1- which is the same as z1, as predicted

as we could guess,since issingular at z1

mz )1(

43

Section 6

32

0 !3

)2)(1(

!2

)1(1

)1(

1z

mmmz

mmmzz

n

m

z n

nm

The binomial series is

3222

4321)](1[

1

)1(

1zzz

zzTherefore

The binomial series converges for z1

Therefore our series converges for z1- which is the same as z1, as predicted

as we could guess,since issingular at z1

mz )1(

44

Section 6Example:

)2()4(

34152)(

2

2

zz

zzzf

centre

Expand about z0

singular points:2,4 z

… the radius of convergence should be R2

2

1

)4(

134152)(

22

zzzzzf

need to expand in powers of z

45

Section 6

2)4()4()2()4(

34152)(

22

2

z

C

z

B

z

A

zz

zzzf

Use partial fractions:

22 )4()2)(4()2(34152 zCzzBzAzz

2

2

)4(

1)(

2

zzzf

32

2222

44

43

421

16

1

)]4/(1[4

1

)4(

1

)4(

1

zzz

zzz

Now

414/ zzconverges for

46

Section 6

z

zzzzzz

zzzf

32

15

16

17

2221

44

43

42

4

1

2

2

)4(

1)(

32

5

3

4

2

32

2

32

2221

)2/(1

1

2

2

2

2

zzz

zzzand

212/ zzconverges for

so

2zconverges for

47

Section 6

centre

5

3

4

2

32 44

43

42

4

1 zzz

convergesinside here

32

2221

zzz

convergesinside here whole thing

convergesinside overlap

48

Section 6

Other useful series:

132

)1(Ln

!5!3sinh

!3!21

!4!21cos

!5!3sin

32

53

32

42

53

zzz

zz

zzz

zz

zzz

ze

zzz

z

zzz

zz

z

49

Section 6Example:

22sin)( zzf Expand about z0no singular points … the radius of convergence should be R

!5!3

sin53 zz

zzUse the series

!5

)2(

!3

)2(2

2sin)(5232

2

2

zzz

zzf

(of using other series)

50

Section 6Summary

You should be able to find the power series of afunction and its radius of convergence using one ofthese methods:

1. Taylor’s Theorem/Formula

2. Binomial Series

3. Geometric Series

4. Using other known series, e.g. sin, cos, exp, etc.

51

Section 6Summary

You should be able to find the power series of afunction and its radius of convergence using one ofthese methods:

1. Taylor’s Theorem/Formula

2. Binomial Series

3. Geometric Series

4. Using other known series, e.g. sin, cos, exp, etc.

52

Section 6Summary

You should be able to find the power series of afunction and its radius of convergence using one ofthese methods:

1. Taylor’s Theorem/Formula

2. Geometric Series

3. Geometric Series

4. Using other known series, e.g. sin, cos, exp, etc.

53

Section 6

You should be able to find the power series of afunction and its radius of convergence using one ofthese methods:

1. Taylor’s Theorem/Formula

2. Geometric Series

3. Binomial Series

4. Using other known series, e.g. sin, cos, exp, etc.

Summary

54

Section 6Summary

You should be able to find the power series of afunction and its radius of convergence using one ofthese methods:

1. Taylor’s Theorem/Formula

2. Geometric Series

3. Binomial Series

4. Using other known series, e.g. sin, cos, exp, etc.

55

Section 6Topics not Covered

(1) Proof of Cauchy-Hadamard formula (follows from ratio test)

(2) Proof of Taylor’s Theorem - an analytic function can be writtenas a power series (use Cauchy’s Integral Formula)

(3) The concept of uniform convergence and proof that power seriesare uniformlyconvergent

(4) Some other practical methods of deriving power series:e.g. use of differentiation/integration of series, differentialequations, undetermined coefficients, …

(5) Analytic Continuation

56

Section 6(6) In some very exceptional cases, a singular point mayalso arise inside the circle of convergence.

0z

zLn singular (not analytic) along herejumps from to + as we cross this line

zLn

centre jzo 1

n

n

n

o

n

nn

jna

ja

jzaz

)1(

)1(

)1(Ln

)1(Ln

1

0

2

1

1

1

)1()1)(1(

)1()1(limlim

11

21

j

jn

jn

a

ann

nn

nn

n

n

2 R