ECE 6382
Power Series Representations
D. R. Wilton
ECE Dept.
8/24/10
Geometric Series
• Consider the sum
N
 z N   zn
SN  1 z  z2 
y
n 0
Consider
zS N  z  z 2 
 z N 1 ,
z 1
we have that S N  zS N  1  z  S N  1  z N 1 and hence
SN 
1  z N 1
1 z
i N 1
N 
x
1
• Since z N 1  r N 1e 
lim SN 
z 1
1
Noting that
N 
 r N 1 
 0 iff r  z  1 ,
1
1  z  z 2 
1 z


 zn
, z 1
Geometric Series (G.S.)
n 0
• The above series converges inside, but diverges outside the unit circle. But there exists
1
another series representing
that is valid outside the unit circle :
1 z
1
 1 G.S. 1  1 1 1 
1 1
 1 1 1 



1







iff
  1 i.e., z  1




1  z z 1  1z 
z  z z 2 z3 
r z
 z z 2 z3 
• The above series may or may not converge at points on the unit circle
• Note the interior infinite series is an expansion in (positive) powers of z ; the exterior series
is an expansion in reciprocal powers of z
Geometric Series, cont’d
Consider
•
1
Note that

z  z0
1 z  z  z 
 1   0    0    0  
 z  z   z   z   z 
z 1  0 
z 

2
1
3
 z
 if 0  1 , i.e. z  z0
z

y
z  z0
z0
z  z0
z0
1
Similarly,

z  z0
1
1

z0

z 
z0  1  
 z0 
x
Radius of
convergence
  z   z  2  z 3
1          
  z0   z0   z0 


z
 if
 1 , i.e. z  z0
 z0

Geometric Series, cont’d
• The above series were expanded about the origin, z  0. But we can also expand about another
point, say z  :
Consider
1
1


z  z0  z  z     z 0  z  
if
2
3
1   z 0  z    z 0  z    z0  z  




1 
z0  z    z  z     z  z    z  z    z  z  


 z  z  1 


z

z


1
z0  z 
 1 , i.e. z  z   z0  z 
z  z
y
z0
Factor out the largest term!



z
z  z
z
z0  z
Radius of
convergence
x
Similarly,
1

z  z0
1

 z  z     z0  z  
if
  z  z    z  z   2  z  z  3
1
1
1  


 
 

z

z





  z0  z    z0  z    z0  z  
zz
0

 z0  z    1 


z

z
0


z  z
 1 , i.e. z  z   z0  z 
z0  z 




Uniform Convergence
• Consider the infinite geometric series,
1
 1  z  z2  z3 
1 z
Consider
Let's evaluate the series for some specific values, say z  103 i 0, 102  i 0, 101  i 0 .
z  103  i 0 :
1
 1.00  0.001  0.000001  0.000000001 
1  103
 1.001001001001001
Clearly, every additional term adds 3 more significant figures to the final result.
z  102  i 0 :
1
 1.00  0.01  0.0001  0.000001 
1  102
 1.0101010101
Here, however, each additional term adds only 2 more significant figures to the result.
z  101  i 0 :
1
 1.00  0.1  0.01  0.001 
1  101
 1.11111
And here each additional term adds only 1 more significant figure to the result.
In general, for a given accuracy, the number of terms increases with | z | .
Uniform Convergence, cont’d
• For the infinite geometric series,
1
 1  z  z2  z3 
1 z
Consider
only the first term is needed to produce an exact result for z  0! But as z increases
S
the number of terms needed to provide a fixed number of significant figures increases,
approaching infinity as z  1  i 0.
1  z N 1
• Since S N  1  z  z   z 
, the partial sum error is
1 z
z N 1
S  SN
N 1
S  S N  eN 
 S z N 1 ; hence the relative error is  rel 
z
1 z
S
2
N
 log  rel 
 N 
  1 ( Note  n  denotes ceil(n))
log
z


Note the number of terms needed depends
on both  rel and z . The relationship is
500
plotted in the figure.
• On the other hand if we limit z  R  1 then
 log  rel 
N  
  1, which depends on  rel but
log
R


not on z (see next slide)
Number of geometric series terms N vs. |z|
400
2 sig. digits
300
4 sig. digits
N
6 sig. digits
200
8 sig. digits
100
10 sig. digits
0
0
0.2
0.4
0.6
|z|
0.8
1
Uniform Convergence, cont’d
• As the figure shows, it is impossible to find a fixed value of N which yields a specified accuracy
overConsider
the entire region z  1, i.e., the series is non - uniformly convergent in this region .
Number of geometric series terms N vs. |z|
• Note the G.S. is uniformly
convergent, say, for z  0.95,
as shown, or for any
region z  R  1.
y
1
z 1
z  0.95
x
1
500
450
400
350
300
N 250
200
150
100
50
0
N1
N8
2 sig. digits
4 sig. digits
N6
N4
N2
0
6 sig. digits
8 sig. digits
10 sig. digits
0.2
0.4
0.6
0.8
0.95 1
|z|

• A series f  z    g n  z  is uniformly convergent in a region R if corresponding to an
n 0
  0, there exists a number N , dependent on  but independent of z , such that N  N
N
implies f  z    g n  z    for all z in R .
n 0
Key Point: Term-by-term integration of a series is allowed over any region
where it is uniformly convergent.
Taylor Series Expansion of an Analytic Function
• Write the Cauchy integral formula in the form
f  z 

1
2 i
1
2 i
1

2 i

C

C

f  z
dz 
z  z
y
f  z
dz 
 z   z0    z  z 0 
f  z
z
z
z  z
z  z0
z0
z  z0
dz 
C
 zz 
 z  z0  1   0 
 z  z0 

f  z     z  z0 

 dz 

z

z
 z  z0  n
0 
0 
C
R
n
1

2 i
C
uniform
convergence

derivative
formulas

1 
 z  z0  n

2 i n 0

  z  z0 
n
n 0
f  z 
zs

 an  z  z0 
n
n 0
where
1
an 
2 i
f  z
x
f  z
  z  z 
0
z  z0  z  z0
dz 
n 1
C
f (n ) z0 
( recall f (n ) z0  
n!
n!
2 i
f  z
  z  z n1 dz
C
0
 Taylor series expansion of f  z  about z0
  z  z n1 dz
C
0

f (n ) z0 
n!
(both forms are used!)
)
Taylor Series Expansion of an Analytic Function, Cont’d
zs
y
z
z
z  z
z  z0
C
z0
z  z0
R
x
z  z0  z  z0
• Note the construction is valid for any z  z0  z  z0  zs  z0
where zs is the singularity nearest z0 ; hence the region of convergence is
z  z0  zs  z0
The Laurent Series Expansion
This generalizes the concept of a Taylor series, to include cases
Consider
where the function is analytic in an annulus.
zb
f  z 

 an  z  z0 
n
za
n
z0
or
z
f z 


 an  z  z0    bn
n
n 0
n 1
b
1
 z  z0 
n
where bn  an
Converges for
z  z0  b  zb  z0
a
Key point: The point z0 about which
the expansion is made is arbitrary, but
determines the region of convergence
of the Laurent or Taylor series.
Converges for
z  z0  a  za  z0
(we often have za  z0 )
The Laurent Series Expansion, cont’d
This is particularly useful for functions that have poles.
Consider
zb
Examples:
1
 z0  0 , a  0 , b   
z
cos  z 
f  z 
 z0  0, a  0, b   
z
f  z 
z
f  z 
z 1
f  z 
za
z0
a
b
z
 z0  1, a  0, b   
Converges in region
z
 z  1 z  2
 z0  0,
a  1, b  2 
za  z0  a  z  z0  b  zb  z0
But the expansion point z0 does not have to be at a
singularity, nor must the singularity be a simple pole:
z
f  z 
z0  2, a  3, b  4 

 z  2 z 2  1
y
z
branch cut
pole
2  1 1 2
z0  2
x
The Laurent Series Expansion, cont’d
Consider
Theorem:
The Laurent series expansion in the annulus
region is unique.
z0
(So it doesn’t matter how we get it; once we
obtain it by valid steps, it must be correct.)
Example:
cos  z 
f  z 
z
analytic
for z 0
 f  z 
Hence
1
z
a
b
 z0  0,
a  0, b   
valid for z 


2
4
6
z
z
z
1    
 2! 4! 6!


1 z z3 z5
f  z    
z 2! 4! 6!






, 0 z  
The Laurent Series Expansion, cont’d
Consider
We next develop a general method for constructing
the coefficients of the Laurent series.
f  z 


n
Final result:
1
an 
2 i
an  z  z0 
f  z
n
  z  z n1 dz
C
0
z0
C
a
b
(This is the same formula as for the Taylor series, but with negative n allowed.)
Note: If f (z) is analytic at z0, the integrand is analytic for negative values of n.
Hence, all coefficients for negative n become zero (by Cauchy’s theorem).
The Laurent Series Expansion, cont’d
Consider
Pond, island, & bridge
The Laurent Series Expansion, cont’d
y
•
Contributions
Consider from the paths
c1 and c2 cancel!
R  simply - connected region
z
z
zs2
 By Cauchy's Integral Formula,
1
f z 
2 i C

1  c1  c2 C2
f  z 
dz
z  z
1
z0
c1
z C
2
f  z
f  z
1
1

dz 
dz 



2 i C z   z
2 i C z   z
Pond, island,
& bridge
c2
C1
zs1
2
where on C 1, z   z0  z  z0 ,
1
1



z   z  z   z0    z  z 0 
and on C2 , z   z0  z  z0
1
1



z   z  z   z0    z  z 0 
1

 z   z0   1 
x
n
z  z0 


z  z0  n0  z   z0 n1


 z   z0 
(note the convergence regions of C 1 , C2 overlap!)
 z   z0 
1
 
n 1
 z   z0 
n  0  z  z0 
 z  z0   1 

z

z
0 


n
n   n1,
n n

n
z  z0 


n 1
n 1  z   z0 

The Laurent Series Expansion, cont’d
• Hence,
Consider
f  z 
1
2 i C

1  c1  c2  C2
uniform
convergence
y
f  z 
dz 
z  z
R multiply - connected region
z
f  z 
1 
n
z

z



0
  z  z n 1 dz
2 i n  0
C1
0

f  z 
1 
n
z

z




0
  z  z n 1 dz
2 i n 1
C2
0
f  z 

 a n  z  z0 
z
zs2 z
0
C2
C C1
zs1
x
n
n 
where an 
1
2 i
f  z 
  z  z n 1 dz
C
0
and C 1  C2  C encircles z0.
 Note we can deform C 1, C2 , to a single contour C since
f  z 
 z  z0 
n 1
is z - independent
and analytic at least for zs2  z0  z   z0  zs1  z0 where z s1 , z s2 are the nearest
singularities to C 1, C2 , respectively.
Examples of Taylor and Laurent Series
Expansions
Consider
Example 1:
1
about the origin :
z  z  1
 Obtain all expansions of f  z  

 an z n
The series will have the form
n 
(since z0  0)
where
1
an 
2 i
1

2 i

C
f  z 
z  n1

1
dz  
2 i
1
  z nm 2

C
1
1

dz

2 i
 z  1 z n2

C
1
z n 2
2

0
 zm dz,
( z   1)
m 0
dz  ; let z   rei , dz   irei d
C m 0
1 
 an 

2 i m0

0 , m n1
2 , mn 1

 
irei
1
d


i n  m  2 
2
r nm 2e 
1
 f  z    1  z  z 2  z3 
z
, 0  z 1


m 0
1
r nm1
2

0
 0, n  1
d



i n m 1
e
1, n  1
1
Examples of Taylor and Laurent Series
Expansions,cont’d
Consider
Example 1, cont'd
 On the other hand,
an 
1
2 i
1

2 i
1

2 i

C
f  z 
z

1
1
2 i
1

C
 z n3  zm dz,
1
1

dz

2 i
 z  1 z n2
1
 1  1  z n3 dz
z
C
( z   1)
m 0
C

1
 m0 z nm3
dz  ; let z   rei , dz   irei d
 0 , m n 2
 
2 , m n 2
C
1 
 an 

2 i m0
 f z 
n 1
dz  
2

0
irei
1
d


i n  m 3
2
r n  m  3e 
1 1 1
  
z 2 z3 z 4


m 0
1
r n m 2
2

0
0, n  2
d



i n  m  2 
e
1, n  2
1
, z 1
 In practice the contour integral approach is rarely used . To illustrate, we
reconsider expanding f  z  as a partial fraction and using the geometric series.
Examples of Taylor and Laurent Series Expansions,cont’d
Example 1, cont'd
Consider
 Expand f  z  
f  z 
1
about the origin (we use partial fractions and G.S.) :
z  z  1
B
A
1
;


z 1
z
z  z  1
A  lim z f  z   lim
z 0
z 0
z
 1
z  z  1
B  lim  z  1 f  z   lim
z 1
 f z 
z 1
z 1
z  z  1
1
1
1
1
1




1 z
z
z 1
z
z  z  1
1
f  z    1  z  z 2  z3 
z
1 1
1
1


 f z 
z 1 1 z
z
z  z  1
 f  z 
1
1 1
1
1
 1  2 
z z
z 
z
, 0  z 1
1 1 1

 2  3  4 
z
z
z

, z 1
Examples of Taylor and Laurent Series Expansions,cont’d
Example 2
Consider
 Expand
f  z 
1
 z  2  z  3
in a Taylor / Laurent series
y
about z0  1, valid in the annular regions
(1) 0  z  1  1,
z
(2) 1  z  1  2,
x
2
1
z 1  1
(3) z  1  2.
For 0  z  1  1 :
3
1  z 1  2
z 1  2
Using partial fraction expansion and G.S.,
f z 
1
 z  2  z  3

1
1

z 3
z2
1
1
1


 z  1  1 2 1   z  1 2  1   z  1
2

1   z  1  z  1
1   z  1   z  12  

  1 



2

2
2
2
 


1

 z  1  2
f z 
1
3
7
15
2
3
  z  1   z  1   z  1 
2
4
8
16
, 0  z  1  1 (Taylor series)
Examples of Taylor and Laurent Series Expansions,cont’d
Consider
Example
2,cont'd
For
1  z 1  2 :
f  z 
1

z

1

2
 
1
1


z

1

1
2
1

z

1
2


 
 
 
1   z  1  z  1
f  z    1 


2
2
2
2


 

2
For
1
 z  1 1  1  z  1
1 
1
1
1 


 z  1   z  1  z  12

 (Laurent series)

z 1  2 :
f  z 
1

z

1

2
 
1

z

1

1
 
1

z

1
1

2
z

1


 
 
1 
2
22
1 



 z  1   z  1  z  12
f  z 
1
 z  1
2

3
 z  1
3


1 
1
1

1 


2
z

1
z

1






 z  1
7
 z  1
1
 z  1 1  1  z  1
4

(Laurent series)



Examples of Taylor and Laurent Series Expansions,cont’d
Consider
Example 3
Find the series expansion about z  0 :
 1  cos z
, z0

f  z  
z2
 1 2 ,
z0
( z  0 is a "removable" singularity)

z2 z4 z6
1  cos z  1   1    
2! 4! 6!

1 z2 z4



f  z 
2! 4! 6!

z2 z4 z6




6!
4!
2!

, z 
z2 z4
sin z


 1
Similarly, we have f  z  
3! 5!
z
, z 
Examples of Taylor and Laurent Series Expansions,cont’d
Example 4
Consider
Find the series for sin z about z   :
f  z   sin z  sin  z      
 sin  z    cos   cos  z    sin  sin  z   
3
5
z   z  

f z  z   


3!
5!
, z 
Alternatively, use the derivative formula for Taylor series :
f    sin   0
f    cos   1
f     sin   0
f     cos   1
iv
f      sin   0
v
f      cos   1

3
5
z   z  

f z  z   


3!
5!
, z 
Examples of Taylor and Laurent Series Expansions,cont’d
Example 5
Consider
Find the first few terms of the series for sin 2 z ln 1  z  about z  0 :
Since
1
 1  z  z2 
1 z
, z  1 then
1
z2 z3 z4
0 1  z dz   ln 1  z   z  2  3  4  , z  1
z2 z3 z4
 ln 1  z    z     , z  1
2
3 4
z

z3 z5
2
sin z   z 


3!
5!

Also



2
z4
2 6
z 

z 
3
45
2
Hence
 2 z4
2 6
sin z ln 1  z     z 

z 
3
45

2
z4
  z   0  z5 
2
3
,

z2 z3 z4
 z    
2
3 4

z  1 (why?)



Summary of Methods for Generating Taylor and Laurent
Series Expansions
Consider
 To
expand about z  z0 , first write f  z  in the form f  z  z0   z0  , rearrange
and expand using known series or methods.
 Note that if

f  z 

n  
a n  z  z0  ,
n

then f  z   g  z  


g z 
n  
bn  z  z0 
n
  an  bn   z  z0 
n
in their common region of convergence.
n 

 Taylor (not Laurent) series, f  z    an  z  z0 
n0
n
n
f   z0 
, can be generated using an 
n!
 Use partial fraction expansion and geometric series to generate series for rational functions
(ratios of polynomials, degree of numerator less than degree of denominator).
 Laurent / Taylor series can be integrated or differentiated term - by - term within their radius
of convergence
 Two Taylor series can be multiplied term - by - term within their common region of convergence :
f  z 

 a n  z  z0 
n
,
g z 
n 0

 bm  z  z0 
m0
m

n

n
m
n
 f  z  g  z     an  z  z0     bm  z  z0     cn  z  z0  where cn =  a p bn  p .
 n 0

 n 0
p0

  m0


