57. Taylor Series
 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

f ( z )   an ( z  z0 ) n , (| z  z0 | R0 )
n 0
f ( n ) ( z0 )
an 
, (n  0,1, 2,...)
n!
That is, series converges to f (z) when z
lies in the stated open disk.
1
f ( z )dz
an 
2 i C ( z  z0 )n1 Refer to pp.167
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School of Software
57. Taylor Series
 Maclaurin Series
When z0=0 in the Taylor Series become the Maclauin Series

f ( z)  
n 0
y=ex
f ( n ) (0) n
z ,(| z  z0 | R0 )
n!
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
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School of Software
58. Proof the Taylor’s Theorem

f ( z)  
n 0
f ( n ) (0) n
z ,(| z  z0 | R0 )
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
1
f ( s)ds
f ( z) 
, z,| z | R0

2 i C0 s  z
1
1
1
1 1


, w  ( z / s),| w | 1
s  z s 1  ( z / s) s 1  w
Refer to pp.187
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School of Software
58. Proof the Taylor’s Theorem
N 1
1
1 n
1
N
  n 1 z  z
s  z n 0 s
(s  z )s N
f ( z) 
1
f ( s)ds
2 i C0 s  z
N 1
1
f ( s )ds n
1 N
f ( s )ds
f ( z)  
z 
z 
n 1
N

2

i
s
2

i
(
s

z
)
s
n 0
C0
C0
Refer to pp.167
N 1
f ( z)  
n 0
f ( n ) (0)
n!
ρN
f ( n ) (0) n z N
f (s)ds
z 
n!
2 i C0 ( s  z )s N
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School of Software
58. Proof the Taylor’s Theorem
zN
f ( s)ds
 N  lim
0
When Nlim

N  2 i  ( s  z ) s N
C0
N 1
f ( z )  lim(
N 
n 0


f ( n ) (0) n
f ( n ) (0) n
f ( n ) (0) n
z  N )  
z 0  
z
n!
n!
n!
n 0
n 0
zN
f (s)ds
| r |N
M
|  N ||
|

2 r0
N
N

2 i C0 (s  z )s
2 (r0  r )r0
Where M denotes the maximum value of |f(s)| on C0
|  N |
Mr0 r N
( )
r0  r r0
r
( ) 1
r0
lim  N  0
N 
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School of Software
59. Examples
 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
n

z
e z   , (| z | )
n 0 n !
Note that if z=x+i0, the above expansion becomes

n
x
e x   ,(  x  )
n 0 n !
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School of Software
60. Laurent Series
 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


bn
,( R1 | z  z0 | R2 )
n
n 1 ( z  z0 )
f ( z )   an ( z  z0 )  
n
n 0
7
an 
1
f ( z )dz
, (n  0,1, 2,...)
n 1

2 i C ( z  z0 )
bn 
1
f ( z )dz
, (n  1, 2,...)
 n 1

2 i C ( z  z0 )
School of Software
60. Laurent Series
 Theorem (Cont’)


bn
,( R1 | z  z0 | R2 )
n
n 1 ( z  z0 )
f ( z )   an ( z  z0 )  
n
n 0
an 
1
f ( z )dz
, (n  0,1, 2,...)
2 i C ( z  z0 )n1
bn 
1
f ( z )dz
, (n  1, 2,...)
 n 1

2 i C ( z  z0 )
1
1
b n
n

b
(
z

z
)


n
0
n
n  ( z  z0 )
n 
f ( z) 

n
c
(
z

z
)
, ( R1 | z  z0 | R2 )
 n
0
n 
b n , n  1
cn  
an , n  0
1
f ( z )dz
cn 
,(n  0, 1, 2,...)
n 1

2 i C ( z  z0 )
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School of Software
60. Laurent Series
 Laurent’s Theorem
If f is analytic throughout the disk |z-z0|<R2,


bn
,( R1 | z  z0 | R2 )
n
n 1 ( z  z0 )
f ( z )   an ( z  z0 )  
n
n 0
reduces to Taylor
Series about z0
bn 
1
f ( z )dz
1
n 1

(
z

z
)
f ( z )dz,(n  1, 2,...)
0
 n 1


2 i C ( z  z0 )
2 i C
Analytic in the region |z-z0|<R2
bn  0, (n  1, 2,...)

f ( z )   an ( z  z 0 ) n
n 0
f ( n ) ( z0 )
1
f ( z )dz
an 

, (n  0,1, 2,...)
n 1

2 i C ( z  z0 )
n!
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School of Software