Transcript Chap. 5 - Sun Yat

Chapter 5. Series
Weiqi Luo (骆伟祺)
School of Software
Sun Yat-Sen University
Email：[email protected] Office：# A313
Chapter 5: Series








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
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55. Convergence of Sequences
 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
| zn  z | 
Denoted as
lim zn  z
n 
Note that the limit must be unique if it exists;
Otherwise it diverges
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55. Convergence of Sequences
 Theorem
Suppose that zn = xn + iyn (n = 1, 2, . . .) and z = x + iy.
Then
lim zn  z
n 
If and only if
lim xn  x & lim yn  y
n 
n 
Proof:
xn  x & lim yn  y
If lim
n 
n 
then, for each positive number ε, there exists n1 and n2,
such that


| xn  x |
2
, n  n1
| yn  y |
4
2
, n  n2
School of Software
55. Convergence of Sequences
Let n0=max(n1,n2), then when n>n0
| xn  x |

2
& | yn  y |

2
| zn  z || ( xn  iyn )  ( x  iy) || ( xn  x)  i( yn  y) |
| xn  x |  | yn  y |

2


2

Conversely, if lim zn  z we have that for each positive ε, there
n 
exists a positive integer n0 such that, when n>n0
| zn  z || ( xn  iyn )  ( x  iy) || ( xn  x)  i( yn  y) | 
| xn  x || ( xn  x)  i( yn  y) | 
| yn  y || ( xn  x)  i( yn  y) | 
lim xn  x & lim yn  y
n 
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n 
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55. Convergence of Sequences
 Example 1
The sequence
1
zn  3  i, (n  1, 2,...)
n
converges to i since
1
1
lim( 3  i )  lim 3  i lim1  0  i1  i
n  n
n  n
n 
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55. Convergence of Sequences
 Example 2
When
(1)n
zn  2  i 2 , (n  1, 2,...)
n
(1)n
The theorem tells us that lim zn  lim(2)  i lim( 2 )  2  i0  2
n 
n 
n 
n
If using polar coordinates, we write rn | zn | &n  Argzn ,(n  1, 2,...)
  n  
1
lim rn  lim 4  4  2
n 
n 
n
lim Argz2 n  
n 
lim Argz2 n 1  
Why?
Evidently, the limit of Θn does not exist as n
tends to infinity.
n 
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56. Convergence of Series
 Convergence of Series
An infinite series

z
n 1
n
 z1  z2  ...  zn  ...
Series
of complex number converges to the sum S if the
N
sequence
S N   zn  z1  z2  ...  zN , N  (1, 2,...)
n 1
of partial sums converges to S; we then write

z
n 1
n
S
The series has at most one limit,
otherwise it diverges
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Sequence
56. Convergence of Series
 Theorem
Suppose that zn = xn + iyn (n = 1, 2, . . .) and S = X + iY.

Then
z

n 1
n
S

If and only if  xn  X &  yn  Y
n 1

z
n 1
n
S
n 1
N
N
n 1
n 1
S N   xn  i  yn  X N  iYN
lim S N  S
N 
lim X N  X & lim YN  Y
N 

x
n 1
n
N 

 X &  yn  Y
n 1
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56. Convergence of Series
 Corollary 1
If a series of complex numbers converges, the nth term
converges to zero as n tends to infinity.

Assuming that  zn  z1  z2  ...  zn  ... converges, based on the
n 1
theorem, both the two following real series converse.


x & y
n 1
n
n 1
n
Then we get that xn and yn converge to zero as n tends to infinity
(why?), and thus
lim zn  lim xn  i lim yn  0  i 0  0
n 
n 
n 
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56. Convergence of Series
 Absolutely convergent
If the series


n 1
n 1
2
2
|
z
|

x

y
 n  n n ,( zn  xn  iyn )
of real number
xn2  y n2
converges,
then the series is said to be absolutely convergent.
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56. Convergence of Series
 Corollary 2
The absolute convergence of a series of complex
numbers implies the convergence of that series.
| xn | x n  y n
2
2
| yn | x n2  y n2

z
n 1
n

| x
n 1
n


|   xn  y n

2
| x
2
n 1
n 1

2
2
|
y
|

x

y
 n  n n
n 1
n
Converge

| y
n 1
n 1

n

x y
 z1  z2  ...  zn  ...
n 1
Converge
n
n 1
n
Converge
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56. Convergence of Series
 The remainder ρN after N terms

S   zn  z1  z2  ...  zN  zN 1  z N 2  ...
n 1
ρN
SN
N  S  SN
| N  0 || S  SN |
Therefore, a series converges to a number S if and only if the sequence of
remainders tends to zero.
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56. Convergence of Series
 Example
With the aid of remainders, it is easy to verify that when |z| <1,

1
z 

1 z
n 0
n
Note that
n 1
1

z
1  z  z 2  ...  z n 
, z 1
1 z
N
1

z
The partial sums S N ( z )   z n  1  z  z 2  ...  z N 1 
,z 1
1 z
n 0
N
1
z
, z  1 then  N ( z )  S ( z )  S N ( z ) 
If S ( z ) 
, z 1
1 z
1 z
N 1
| z |N
|  N ( z ) |
|1  z |
When |z|<1 ρN tends to zero, but not when |z|>1
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56. Homework
pp.188-189
Ex. 2, Ex. 3, Ex. 5, Ex. 9
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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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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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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
f ( z) 
1
f (s)ds
, 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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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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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
r
( ) ( ) 1
r0  r r0
r0
lim  N  0
N 
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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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59. Examples
 Example 1 (Cont’)
The entire function z2e3z also has a Maclaurin series
expansion,

n
Replace z by 3z
z
e   ,(| z | )
n 0 n !

n
3
z 2e3 z   z n 2 , (| z | )
n 0 n !
z
If replace n by n-2, we have
n2
3
z 2 e3 z  
z n , (| z | )
n  2 (n  2)!

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59. Example2
 Example 2
Trigonometric Functions
2 n 1

eiz  eiz
z
sin z 
  (1)n
,(| Z | )
2i
(2n  1)!
n 0
2n

eiz  eiz
n z
cos z 
  (1)
,(| Z | )
2
(2n)!
n 0
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59. Examples
 Example 4
Another Maclaurin series representation is

1
  z n , (| z | 1)
1  z n 0
since the derivative of the function f(z)=1/(1-z), which
fails to be analytic at z=1, are
f
In particular,
(n)
n!
( z) 
, (n  0,1, 2,...)
n 1
(1  z )
f ( n) (0)  n!,(n  0,1, 2,...)
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59. Examples
 Example 4 (Cont’)
substitute –z for z

1
  (1)n z n , (| z | 1)
1  z n 0

1
  z n , (| z | 1)
1  z n 0
replace z by 1-z
1 
  (1)n ( z  1)n , (| z  1| 1)
z n 0
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59. Examples
 Example 5
1  2 z 2 1 2(1  z 2 )  1 1
1
f ( z)  3 5  3

(2

)
2
3
2
z z
z
1 z
z
1 z
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
1
2
4
6
8

1

z

z

z

z
 ...(| z | 1)
2
1 z
Hence, when 0<|z|<1
Negative powers
1
1 1
2
4
6
8
f ( z )  3 (2  1  z  z  z  z  ...)  3   z  z 3  z 5  ...
z
z
z
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59. Homework
pp. 195-197
Ex. 2, Ex. 3, Ex. 7, Ex. 11
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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
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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 )
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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
)
 n
0 , ( R1 | z  z0 | R2 )
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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60. Laurent Series
 Laurent’s Theorem
If f is analytic throughout the disk |z-z0|<R2,


bn
,( R1 | z  z0 | R2 )
n
(
z

z
)
n 1
0
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  z0 ) 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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62. Examples
 Example 1
Replacing z by 1/z in the Maclaurin series expansion

zn
z z 2 z3
e    1     ...(| z | )
1! 2! 3!
n 0 n !
z
We have the Laurent series representation

1/ z
e
1
1
1
1

1



 ...(0 | z | )
n
2
3
1! z 2! z 3! z
n 0 n ! z

There is no positive powers of z, and all coefficients of the positive powers are zeros.
bn 
1
f ( z )dz
where c is any positively oriented simple closed
,(
n

1,
2,...)
contours around the origin
2 i C ( z  0) n1
1
e1/ z dz
1
1/ z
1/ z
e
dz  2 i
1  b1 

e
dz

11


2 i C ( z  0)
2 i C
C
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62. Examples
 Example 2
The function f(z)=1/(z-i)2 is already in the form of a
Laurent series, where z0=i,. That is

1
n

c
(
z

i
)
, (0 | z  i | )

n
2
( z  i)
n 
where c-2=1 and all of the other coefficients are zero.
1
dz
cn 
,(n  0, 1, 2,...)
n 3

2 i C ( z  z0 )
0, n  2
dz
C ( z  i)n3  2 i, n  2
where c is any positively oriented simple contour
around the point z0=i
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62. Examples
Consider the following function
1
1
1
f ( z) 


( z  1)( z  2) z  1 z  2
which has the two singular points z=1 and z=2, is analytic in the domains
D1 :| z | 1
D2 :1 | z | 2
D3 : 2 | z | 
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62. Examples
 Example 3
The representation in D1 is Maclaurin series.
f ( z) 
1
1
1
1
1


 
z 1 z  2
1  z 2 1  ( z / 2)
Refer to pp. 194 Example 4
where |z|<1 and |z/2|<1


n

z
f ( z )   z n   n1   (2 n1  1) z n ,(| z | 1)
n 0
n 0 2
n 0
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62. Examples
 Example 4
Because 1<|z|<2 when z is a point in D2, we know
1
1
1
1
1
1
f ( z) 

 
 
z  1 z  2 z 1  (1/ z ) 2 1  ( z / 2)
Refer to pp. 194 Example 4
where |1/z|<1 and |z/2|<1


zn
1  zn
f ( z )   n1   n1   n   n1 ,(1 | z | 2)
n 0 z
n 0 2
n 1 z
n 0 2
1

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School of Software
62. Examples
 Example 5
Because 2<|z|<∞ when z is a point in D3, we know
1
1
1
1
1
1
f ( z) 

 
 
z  1 z  2 z 1  (1/ z ) z 1  (2 / z )
Refer to pp. 194 Example 4
where |1/z|<1 and |2/z|<1

2n
1  2n  1  2n1
f ( z )   n1   n1  n1  
,(2 | z | )
n
z
n 0 z
n 0 z
n 0 z
n 1

1

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School of Software
62. Homework
pp. 205-208
Ex. 3, Ex. 4, Ex. 6, Ex. 7
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School of Software
63~66 Some Useful Theorems
 Theorem 1 (pp.208)
If a power series

n
a
(
z

z
)
 n
0
n 0
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|
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School of Software
63~66 Some Useful Theorems
 Theorem 2 (pp.210)
If z1 is a point inside the circle of convergence |z − z0| =
R of a power series 
n
a
(
z

z
)
 n
0
n 0
then that series must be uniformly convergent in the
closed disk |z − z0| ≤ R1, where R1 = |z1 − z0|
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School of Software
63~66 Some Useful Theorems
 Theorem (pp.211)
A power series

 a (z  z )
n 0
n
n
0
represents a continuous function S(z) at each point
inside its circle of convergence |z − z0| = R.
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School of Software
63~66 Some Useful Theorems
 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,

n
g
(
z
)
S
(
z
)
dz

a
g
(
z
)(
z

z
)
dz
 n
0

C
n 0
C

S ( z )   an ( z  z0 )n
n 0
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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School of Software
63~66 Some Useful Theorems
 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,


S '( z )   (an ( z  z0 ) ) '  nan ( z  z0 )
n
n 0
n 1
n 0
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School of Software
63~66 Some Useful Theorems
The uniqueness of Taylor/Laurent series representations
 Theorem 1 (pp.217)

If a series
 a (z  z )
n 0
n
n
0
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) 

bn
n
n 1 ( z  z0 )
 cn ( z  z0 )n   an ( z  z0 )n  
If a series
n 
n 0
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.
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School of Software