Section 6
SECTION 6
Power Series I - Taylor Series
(1) Power series
(2) Convergence of Power series and the
Radius of Convergence
(3) The Cauchy-Hadamard formula
(4) Taylor’s series
1
Why Series ?
Section 6
 /z
ze
 dz  ? where C is z  2
C
C
We can expand the integrand….
ze / z  z   
2 1 3 1
2 z

3! z
2

4 1
4! z
3

2
3



1

 /z
ze
C dz  C zdz   C dz   2 C z dz   3!


Analytic ( )
1
C z 2 dz  
by Formula for
derivatives ( ) 2
Section 6
Can we expand all functions in series ?
We can expand analytic functions in special
series called “Power Series”
How do we find these series ?
(1) Using Taylor’s Theorem
(2) Using other known series
(and other tricks)
3
Section 6
Power Series
A series in powers of ( z  zo )

 a (z  z )
n 0
n
o
n
 ao  a1 ( z  zo )  a2 ( z  zo )  
2

e.g.
1
1
n
2
(
z

j
)

1

(
z

j
)

(
z

j
)


2
n  0 n!
4
Section 6
Power Series
A series in powers of ( z  zo )

 a (z  z )
n 0
n
n
o
 ao  a1 ( z  zo )  a2 ( z  zo )  
2
centre

e.g.
1
1
n
2
(
z

j
)

1

(
z

j
)

(
z

j
)


2
n  0 n!
5
Section 6
Power Series
A series in powers of ( z  zo )

 a (z  z )
n 0
n
n
o
centre
 ao  a1 ( z  zo )  a2 ( z  zo )  
2
coefficients

e.g.
1
1
n
2
(
z

j
)

1

(
z

j
)

(
z

j
)


2
n  0 n!
6
Convergence of Power Series
Section 6
Power series often converge for some values of z but
diverge for other values.

For example the series
z
n
 1 z  z  z 
2
3
n 0
(geometric
series)
converges for z 1 but diverges for z 1
diverges
converges
7
Convergence of Power Series
Section 6
Power series often converge for some values of z but
diverge for other values.

For example the series
z
n
 1 z  z  z 
2
3
n 0
(geometric
series)
converges for z 1 but diverges for z 1
diverges
Radius of
Convergence
R=1
converges
8
Radius of Convergence
“at infinity”; R
Example:
Section 6

1 n
z2 z3
z  1 z   

2! 3!
n 0 n!
series converges for all z
Example:

 n! z
n
Zero Radius of
Convergence; R0
 1  z  2 z  3! z  
2
3
n 0
series diverges for all z (except z0)
9
Section 6

 a (z  z )
n 0
n
o
n
 ao  a1 ( z  zo )  a2 ( z  zo )  
2
(1) The power series always converges at zzo
(2) There is a Radius of Convergence R for which:
zo
z  zo  R : diverges
z  zo  R : converges
10
Is there a quick way to find the radius of convergence ?
Section 6
… the Cauchy-Hadamard formula:
an 1 1
lim

n  a
R
n

(2n)!
n
2
(
z

3
j
)

1

2
(
z

3
j
)

6
(
z

3
j
)

Example: 
2
n  0 ( n!)
an1

2(n  1)! (n!) 2
(2n  2)( 2n  1)
1
lim
 lim
 lim
4
2
2
n a
n (n  1)! (2n)!
n 
(n  1)
R
n
1
R
4
11
Section 6

(2n)!
n
2
(
z

3
j
)

1

2
(
z

3
j
)

6
(
z

3
j
)


2
n  0 ( n!)
zo  3 j
12
Section 6

(2n)!
n
2
(
z

3
j
)

1

2
(
z

3
j
)

6
(
z

3
j
)


2
n  0 ( n!)
z  3 j  1 / 4 : converges
zo  3 j
13
Section 6

(2n)!
n
2
(
z

3
j
)

1

2
(
z

3
j
)

6
(
z

3
j
)


2
n  0 ( n!)
z  3 j  1 / 4 : converges
zo  3 j
z  3 j  1 / 4 : diverges
14
Another Example:
Section 6

n
2
(
z

4
j
)

1

(
z

4
j
)

(
z

4
j
)


n 0
an1
1
1
lim
 lim  1 
n  a
n  1
R
n
R 1
zo  4 j
z  4 j  1 : diverges
z  4 j  1 : converges
15
Another Example:
Section 6

n
n
2
2
ne
(
z

1
)

e
(
z

1
)

2
e
(
z

1
)


n 0
an1
(n  1)e n1
n 1
1
lim
 lim
 lim e
e
n
n a
n
n 
ne
n
R
n
R  1/ e
zo  1
z 1  1 / e : diverges
z 1  1 / e : converges
16
Section 6
Question:
( z  2 j)
( z  2 j) ( z  2 j)
 1



n
5
5
25
n 0

n
2
Where is the centre of the series?
What is the radius of convergence?
an 1
lim

n  a
n
R
17
Section 6
Note:
We have used the Cauchy-Hadamard formula
to find the radius of convergence.
There are many other tests (which can be used when the
above test fails).
e.g. (1) Root test lim n z n  1 diverges
n 

(2) if lim zn  0   zn diverges
n 
n 0
(3) Comparison test

z
n 0
n
converges if z n  bn and

b
n 0
zn1
 1 diverges
(4) The Ratio test lim
n  z
n
n
converges
18
Section 6
Power Series and Analytic functions
Every analytic function f (z) can be represented by a power
series with a radius of convergence R0. The function is
analytic at every point within the radius of convergence.
Example:
1
f ( z) 
1 z

  z  1 z  z  z 
n
2
3
n 0
series converges for z 1
Radius of
Convergence
19
Section 6
How do we derive these Power Series?
These power series which represent analytic functions f (z)
are called Taylor’s series.
They are given by the formula

f ( z )   an ( z  z o ) ,
n
n 0
1 (n)
an  f ( z o )
n!
(Cauchy, 1831)
20
Example:
Derive the Taylor series for
1
f ( z) 
1 z
Section 6
1
1
2
3!
f ( z) 
, f ( z ) 
, f ( z ) 
, f ( z ) 
2
3
1 z
1  z 
1  z 
1  z 4
(1) centre z0 :

1
  an z n
1  z n 0


n 0
f ( n ) ( 0) n
z
n!
zo  0
z 1

  zn
singular point
n 0
 1 z  z2  z3  
f ( 0)  1
f (0)  1
f (0)  2
f (0)  3!
centre
R 1
21
Example:
Derive the Taylor series for
1
f ( z) 
1 z
Section 6
1
1
2
3!
f ( z) 
, f ( z ) 
, f ( z ) 
, f ( z ) 
2
3
1 z
1  z 
1  z 
1  z 4
(1) centre z0 :

1
  an z n
1  z n 0


n 0
f ( n ) ( 0) n
z
n!
zo  0
z 1

  zn
singular point
n 0
 1 z  z2  z3  
f ( 0)  1
f (0)  1
f (0)  2
f (0)  3!
centre
R 1
22
Example:
Derive the Taylor series for
1
f ( z) 
1 z
Section 6
1
1
2
3!
f ( z) 
, f ( z ) 
, f ( z ) 
, f ( z ) 
2
3
1 z
1  z 
1  z 
1  z 4
(1) centre z0 :

1
  an z n
1  z n 0


n 0
f ( n ) ( 0) n
z
n!
zo  0
z 1

  zn
singular point
n 0
 1 z  z2  z3  
f ( 0)  1
f (0)  1
f (0)  2
f (0)  3!
centre
R 1
23
Example:
Derive the Taylor series for
1
f ( z) 
1 z
Section 6
1
1
2
3!
f ( z) 
, f ( z ) 
, f ( z ) 
, f ( z ) 
2
3
1 z
1  z 
1  z 
1  z 4
(1) centre z0 :

1
  an z n
1  z n 0


n 0
f ( n ) ( 0) n
z
n!
zo  0
z 1

  zn
singular point
n 0
 1 z  z2  z3  
f ( 0)  1
f (0)  1
f (0)  2
f (0)  3!
centre
R 1
24
1
1
2
3!
f ( z) 
, f ( z ) 
, f ( z ) 
, f ( z ) 
2
3
1 z
1  z 
1  z 
1  z 4
(2) centre z12 :
zo 

1
1 n
  an  z  2 
1  z n 0


n 0

1
2
z 1
f ( n ) ( 12 )
1 n
z  2 
n!
 2
n 1
z  
n 0
 2  4z 
centre
1 n
2
1
2
  8z  
1 2
2
 16z 

1 3
2

Section 6
f ( 12 )  2
f ( 12 )  2 2
f ( 12 )  2.23
f ( 12 )  3!2 4
singular point
R
1
2
25
An analytic function f (z) can be represented by Section 6
different power series with different centres zo
(although there will only be one unique series for each centre)
1
 1 z  z2  z3  
1 z
zo  0
z 1
1
2
3
 2  4z  12   8z  12   16z  12   
1 z
zo 
1
2
z 1
At least one singular point of f (z) will be on the
circle of convergence
26
An analytic function f (z) can be represented by Section 6
different power series with different centres zo
(although there will only be one unique series for each centre)
1
 1 z  z2  z3  
1 z
zo  0
z 1
1
2
3
 2  4z  12   8z  12   16z  12   
1 z
zo 
1
2
z 1
At least one singular point will be on the
circle of convergence
27
Another Example:
f ( z )  e with centre z0
Section 6
z
f ( z )  e z , f ( z )  e z , f ( z )  e z , f ( z )  e z

e   an z
z
n
n 0


n 0
f ( n ) (0) n
z
n!
zo  0
f ( 0)  1
f (0)  1
f (0)  1
f (0)  1

zn

n  0 n!
z2 z3
 1 z   
2 3!
R
centre
no singular points!
28
Deriving Taylor series directly from the formula
Section 6

1 (n)
f ( z )   an ( z  z o ) , an  f ( z o )
n!
n 0
n
can be very tricky
We usually use other methods:
(1) Use the Geometric Series
(2) Use the Binomial Series

1
  zn  1 z  z2  z3 
1  z n 0

  m n
1
m(m  1) 2 m(m  1)( m  2) 3



z

1

mz

z 
z 

m


(1  z )
2!
3!
n 0  n 
(3) Use other series (exp., cos, etc.)
z 2 n1
z3 z5
sin z   (1)
 z   
(2n  1)!
3! 5!
n 0

n
(4) Use other tricks
29
Example:
1
Expand f ( z ) 
2 about z0
1 z
Section 6
(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 R1
30
Example:
1
Expand f ( z ) 
2 about z0
1 z
Section 6
(use the geometric series)
First, draw the centre and singular points to see what’s going on:
singular points:
centre
So it looks like the radius of convergence should be R1
31
Example:
1
Expand f ( z ) 
2 about z0
1 z
Section 6
(use the geometric series)
First, draw the centre and singular points to see what’s going on:
singular points:
z   j,  j
centre
So it looks like the radius of convergence should be R1
32
Example:
1
Expand f ( z ) 
2 about z0
1 z
Section 6
(use the geometric series)
First, draw the centre and singular points to see what’s going on:
singular points:
z   j,  j
centre
So it looks like the radius of convergence should be R1
33
Section 6
We know that
Therefore
1
2
3
 1 z  z  z 
1 z
1
1
2
4
6


1

z

z

z

2
2
1 z
1  ( z )
The geometric series converges for z1
Therefore our series converges for z21
- which is the same as z1, as predicted
34
Section 6
We know that
Therefore
1
2
3
 1 z  z  z 
1 z
1
1
2
4
6


1

z

z

z

2
2
1 z
1  ( z )
The geometric series converges for z1
Therefore our series converges for z21
- which is the same as z1, as predicted
35
Section 6
We know that
Therefore
1
2
3
 1 z  z  z 
1 z
1
1
2
4
6


1

z

z

z

2
2
1 z
1  ( z )
The geometric series converges for z1
Therefore our series converges for z21
- which is the same as z1, as predicted
36
Section 6
We know that
Therefore
1
2
3
 1 z  z  z 
1 z
1
1
2
4
6


1

z

z

z

2
2
1 z
1  ( z )
The geometric series converges for z1
Therefore our series converges for z21
- which is the same as z1, as predicted
37
Example:
1
Expand f ( z ) 
about z1
3  2z
Section 6
(use the geometric series)
First, draw the centre and singular points to see what’s going on:
centre
singular point:
z  3/ 2
So it looks like the radius of convergence should be R1/2 38
Section 6
We know that
1
2
3
 1 z  z  z 
1 z
1
1

 1  2( z  1)  4( z  1) 2  
Therefore
3  2 z 1  2( z  1)
The geometric series converges for z1
Therefore our series converges for 2(z1)1
- which is the same as z 1 1/2, as predicted
39
Example:
1
Expand f ( z ) 
2 about z0
(1  z )
Section 6
(use the binomial series)
First, draw the centre and singular points to see what’s going on:
singular point:
centre
z 1
So it looks like the radius of convergence should be R1
40
Section 6
The binomial series is

  m n
1
m(m  1) 2 m(m  1)( m  2) 3
 z  1  mz 
  
z 
z 
m
(1  z )
2!
3!
n 0  n 
1
1
2
3


1

2
z

3
z

4
z

Therefore
2
2
(1  z )
[1  ( z )]
as we could guess,
since (1  z )  m is
singular at z1
The binomial series converges for z1
Therefore our series converges for z1
- which is the same as z1, as predicted
41
Section 6
The binomial series is

  m n
1
m(m  1) 2 m(m  1)( m  2) 3
 z  1  mz 
  
z 
z 
m
(1  z )
2!
3!
n 0  n 
1
1
2
3


1

2
z

3
z

4
z

Therefore
2
2
(1  z )
[1  ( z )]
as we could guess,
since (1  z )  m is
singular at z1
The binomial series converges for z1
Therefore our series converges for z1
- which is the same as z1, as predicted
42
Section 6
The binomial series is

  m n
1
m(m  1) 2 m(m  1)( m  2) 3
 z  1  mz 
  
z 
z 
m
(1  z )
2!
3!
n 0  n 
1
1
2
3


1

2
z

3
z

4
z

Therefore
2
2
(1  z )
[1  ( z )]
as we could guess,
since (1  z )  m is
singular at z1
The binomial series converges for z1
Therefore our series converges for z1
- which is the same as z1, as predicted
43
Example:
Section 6
2 z 2  15 z  34
Expand f ( z ) 
about z0
2
( z  4) ( z  2)
singular points:
z  4,  2
centre
… the radius of convergence
should be R2
 1  1 
f ( z )  2 z  15 z  34 

2 
 ( z  4)  z  2 

2

need to expand in powers of z
44
Section 6
Use partial fractions:
2 z 2  15 z  34
A
B
C
f ( z) 



2
2
( z  4) ( z  2) ( z  4) ( z  4) z  2
 2 z 2  15z  34  A( z  2)  B( z  4)( z  2)  C ( z  4) 2
1
2
 f ( z)  

2
( z  4)
z2
Now
1
1
1


 2
2
2
( z  4)
(4  z )
4 [1  ( z / 4)]2
2
3


1 
z z
z
  1  2  3   4   
16 
4  4

 4
converges for z / 4  1  z  4
45
and
2
2
1


z2
2 z
1  ( z / 2)
Section 6
 z  z  2  z 3

  1         
 2  2   2 

converges for z / 2  1  z  2
so f ( z )  
1
2

( z  4) 2 z  2
2
3


 1
  z z z
z
z
z
  2  2 3  3 4  4 5    1         
4
4
4

 4
  2  2   2 
17 15
   z 
16 32
2
3
converges for z  2
46
1
z
z2
z3
 2  2 3  3 4  4 5 
4
4
4
4
centre
converges
inside here
converges
inside here
2
1 
Section 6
3
z z z
      
2  2  2
whole thing
converges
inside overlap
47
Section 6
Other useful series:
3
5
z
z
sin z  z    
3! 5!
z2 z4
cos z  1    
2! 4!
2
3
z
z
ez  1 z   
2! 3!
3
5
z
z
sinh z  z    
3! 5!
z2 z3
Ln (1  z )  z    
2 3
z 
z 
z 
z 
z 1
48
Example: (of using other series)
Expand f ( z )  sin 2 z
2
Section 6
about z0
no singular points … the radius of convergence should be R
Use the series
z3 z5
sin z  z    
3! 5!
f ( z )  sin 2 z 2
2 3
2 5
(
2
z
)
(
2
z
)
2
 2z 


3!
5!
49
Section 6
Summary
You should be able to find the power series of a
function and its radius of convergence using one of
these methods:
1. Taylor’s Theorem/Formula
2. Binomial Series
3. Geometric Series
4. Using other known series, e.g. sin, cos, exp, etc.
50
Section 6
Summary
You should be able to find the power series of a
function and its radius of convergence using one of
these 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 6
Summary
You should be able to find the power series of a
function and its radius of convergence using one of
these methods:
1. Taylor’s Theorem/Formula
2. Geometric Series
3. Geometric Series
4. Using other known series, e.g. sin, cos, exp, etc.
52
Section 6
Summary
You should be able to find the power series of a
function and its radius of convergence using one of
these methods:
1. Taylor’s Theorem/Formula
2. Geometric Series
3. Binomial Series
4. Using other known series, e.g. sin, cos, exp, etc.
53
Section 6
Summary
You should be able to find the power series of a
function and its radius of convergence using one of
these methods:
1. Taylor’s Theorem/Formula
2. Geometric Series
3. Binomial Series
4. Using other known series, e.g. sin, cos, exp, etc.
54
Topics not Covered
Section 6
(1) Proof of Cauchy-Hadamard formula (follows from ratio test)
(2) Proof of Taylor’s Theorem - an analytic function can be written
as a power series (use Cauchy’s Integral Formula)
(3) The concept of uniform convergence and proof that power series
are uniformlyconvergent
(4) Some other practical methods of deriving power series:
e.g. use of differentiation/integration of series, differential
equations, undetermined coefficients, …
(5) Analytic Continuation
55
(6) In some very exceptional cases, a singular point may
also arise inside the circle of convergence.
Section 6

Ln z   an ( z  1  j ) n
centre zo  1  j
n 0
ao  Ln (1  j )
(1) n 1
an 
n(1  j ) n
an 1
(1) n  2 n(1  j ) n
lim
 lim
n  a
n  ( n  1)( 1  j ) n 1 ( 1) n 1
n
z 0
Lnz  
Lnz singular (not analytic) along here
jumps from  to + as we cross this line

1
1

1 j
2
R 2
56