tch-prob1 chap 5. series series representations of analytic functions 43. convergence of sequences...
TRANSCRIPT
tch-prob 1
Chap 5. Series
Series representations of analytic functions
43. Convergence of Sequences and Series
An infinite sequence 數列
1 2 nz , z , ...........,z ,..........
of complex numbers has a limit z if, for each positive , there exists a positive integer n0 such that
0. whenever
nz z n n
0
For sufficieutly large , the points lie in any given neighborhood of ., in general, depends on .
nz
nz
n
z1
z3
z2
zn
z
tch-prob 2
The limit z is unique if it exists. (Exercise 6). When the limit exists, the sequence is said to converge to z.
limn
z zn
Otherwise, it diverges.
Thm 1.
Suppose
lim lim
then iff and lim .
n n n
n n nn n
z x iy , z x iy z z x x, y yn
tch-prob 3
1 21
1 21
(6)
of complex numbers converges to the sum if the sequence
( 1 2 )
of partial sums con
n nn
N
nN Nn
z
z
z z ......... z ...............
S
S z z .......... z N , , ......
1
1 1 1
verges to .We then write
Thm 2. Suppose ,
Then iff and
nn
n n n
n n nn n n
z
iy
z x x y
S
S
z x S x iy
S y
An infinite series
tch-prob 4
A necessary condition for the convergence of series (6) is that lim 0 znn
The terms of a convergent series of complex numbers are, therefore, bounded, .,
nz Mi.e.
Absolute convergence:2 2
1 1 convergesn n n
n nz x y
Since
,
2 2 2 2,
and also converges1 1
and converge1 1
From Thm. 2 converge1
x x y x
x y
x y
z
y yn n n n n n
n nn n
, n nn n
nn
Absolute convergence of a series of complex numbers implies convergence of that series.
tch-prob 5
44. Taylor Series
Thm. Suppose that a function f is analytic throughout an open disk
0 0.z z R Then at each point z in that disk, f(z) has the series
representation
0
( )0
( )
( )
( )0
where
0 1 2!
n z
nf z a z- znn
f a n , , , .....n n
That is, the power series here converges to f(z)
0 0.when z- z R
z
z0
R0
tch-prob 6
This is the expansion of f(z) into a Taylor series about the point z0
Any function that is known to be analytic at a point z0 must have a Taylor series about that point.
(For, if f is analytic at z0, it is analytic in some neighborhood
0 0of (sec 20 55)z z ε z . , p . may serve as R0 is the
statement of Taylor’s Theorem)
0( )
0
0 0
( )
. (a) when 0
0) prove ( ) (!0
Let : Circle
nz
z
pf z
f nf z z Rn n
C r
Positively oriented within
and z is interior to it.
0z R
~ Maclaurin series.z0=0的 case
0
C0
R0r r0
sz
tch-prob 7
The Cauchy integral formula applies:
0
1
0
1
0
1
10
( )( )
( )1( )21 1 1now
1
1and1 1
1 11
( )
N Nn
n
NN n
n
nN N
n Nn
s z
zs
z
zsz
s zs
z
s
f s dsf z cπi
s z s -
z - z z
( )s z s
z s z s
( 1), 1zs
zs
6). sec. 13, (Exeicise( 1)z
tch-prob 8
1
100 00( )1
00( )1
0 0
( )
(0)2
!
(0)( )
!
( ) ( )( )
( ) ( )
( )1( )2
wher
Nn N
c n Nn
nNn N
Nn
nNn
Nn
f s ds
s
fi
n
fz z
n
f s ds f s ds z zc cs z s z s
f s ds z z c s z s
f s dsf z ρc s zπi
0( ) ( )e
2 ( )N
N Nz f s dsz ρ cπi s z s
0
0
Suppose that
then if is a point on
z r
s Cs z s z r r
tch-prob 9
0 0
0 00 0
0( )
( ) )
lim ( )
(0)
If ( )
2(
2 ( )
But 1 0
( )!0
NN
N N
NN
nn
z
z
f s M
M π r M rr r ρ r r rπ r r rr , ρ r
f f z znn
1zs
主要
原因
(b) For arbitrary z0
Suppose f is analytic when0 0
z z R and note that the
composite function 0( )f z z
must be analytic when0 0 0
( )z z z R
tch-prob 10
0).Let ( ) ( g z f z z
The analyticity of g(z) in the disk 0z R ensures
the existence of a Maclaurin series representation:
( )
0
( )0
0
0( )
00
( )
( ))
( ))
0) ( ) (!0
or (!0
Replace by -
( ) (!0
nn
nn
nn
z
z
g g z z z Rnn
f f z z z
nnz z z ,
f f z z z
nn
tch-prob 11
45 ExamplesEx1. Since is entire ( ) zf z e
It has a Maclaurin series representation which is valid for all z.
( ) ( )
2 3
2 3 2
2
( ) (0) 1
( )!0
( ) is also entire.
3!03 ( )
( 2)!2
n n
n
z
nz n
nn
z zf e , f ,zz e znn
f z z e
z e znn
z zn-n
tch-prob 12
Ex2. Find Maclaurin series representation of
(2 ) (2 1)
2 1
(0) (0) ( 1)
( )0 0 1 2
( 1) ( )(2 1)!0
n n n
nn
f z sin zf f , n , , , ....
zsin z - znn
sin ( )sinh z -i iz
Ex3.
(2 ) (2 1)
2
( (0)
( ) cos
0) ( 1) 0
cos ( 1) ( )(2 )!0
cosh cos ( )
nn n
nn
f z z
f fz z - znn
z iz
tch-prob 13
Ex4.
( )1
( )
( )
(0)
1 ( 1)1 0
!since (1 )!
substitute - for1 ( 1) 1
1 0substitute 1 for
n
nn
n
n n
z
z zz n
n fz
f nz z
z ( z )z n
- z z
1 ( 1) ( 1) ( 1 1)0
(1 ) ( 1) ( 1)0 0
n n
n n n
z z -z n
- z - z-n n
tch-prob 14
Ex5.)
)
22 2(1 11 2 1( )3 5 3 211 1(23 21
can not find a Maclaurin series for ( )since it is not analytic at 0But
1 2 4 61 121
Hence, when
zzf z z z z z
-z z
f z z .
z z z ............( z )z
0 11 2 4 6( ) (2 13
1 1 3 53
z ,
f z - z z z ..............)z
z z z ...........zz
為 Laurent series 預告
tch-prob 15
46. Laurent Series
If a function f fails to be analytic at a point z0, we can not apply Taylor’s theorem at that point.
However, we can find a series representation for f(z) involving both positive and negative powers of (z-z0).
Thm. Suppose that a function f is analytic in a domain
1 0 2 ,R z z R and let C denote any positively oriented
simple closed contour around z0 and lying in that domain. Then at each z in the domain
00
))
( ) ( (1)(0 1
bn nf z a z- z n nz- zn n
tch-prob 16
where
0
0
0
)
)
)
( )1 ( 0 1 2 )2 1(
( )1 ( 1 2 )2 1(
or
( ) (
f z dz a n , , , ......n cπ i nz- zf z dz b n , , .......n cπ i nz- z
n f z C z- z nn -
1 0 2
0
)
)
( (4)
( )1where ( 0 1 2 ) (5)2 1(
R z z R
f z dz C n , , , ..... n cπ i nz- z
Pf: see textbook.
2 1
( ) ( ) ( ) 0
2 ( )
f s f s ds f s ds ds-c cs z s z s z
πi f z
rZ
R1Z0
R2
CC1
C2
(1) (4) are Laureut Series
tch-prob 17
47. Examples
The coefficients in a Laurent series are generally found by means other than by appealing directly to their integral representation.
1
2 3
1
)!
(0
1 replace by
1 1 1 11 (0 )! 2! 3!0
no positive powers of appear.
since
n
zn
z zn
z e n
z z
e ......... zzn z z zn z
b
1
1
1 12
2
z
z
e dzcπi
e dz πic
Ex1.
Alterative way to calculate
tch-prob 18
Ex2.
0
2
1( ) is already in the form of a laurent series, where .2( )
( ) ( ) (0 - )
where 1, all other coefficients are zero.( )1 1Since
2 1( )
f z z iz i
nf z C z i z innC
f z dzC n cπ i nz-i
1
2 3( )0 when n -2
3 2 when n=-2( )
11[ 2 0]
2 2 !( ) ( )
dzci nz i
dz c n iz-i
dz idzdz dz π i c c nz i z-i
tch-prob 19
0D1
D2D3
21
Ex3.1 1 1( )
1 2( 1)( 2)f z
z zz z
has two singular points z=1 and z=2, and is analytic in the domains
Recall that
1 ( 1)1 0
nz zz n
(a) f(z) in D1
1
1 1 1( ) 1- 2 1- 2
Since 1 and 1 in D2
nn -n-1( ) (2 1)n 120 0 0
f z zz
zz
z nf z z zn n n
( 1)z
1
2
3
:
: 1 2
: 2
1
z
D z
D z
D
tch-prob 20
(b) f(z) in D2
) )
1 1 1 1( ) 1 2 1-1-( ) 2
1Since 1 , 1 , when 1 221 1 1( ) ( (
2 20 01 (1 2)n 1 120 0
1 1n nn 120 1
f z zzz
z z ,z
zn nf z z zn nnz z
nzn n
zzn n
(1 2)z
tch-prob 21
(c) f(z) in D3
1 1 1 1( ) . 1 21- 1-
1 2since 1 , 1 when 2
1 2( ) n 1 10 0
n1-2 (2 )n 1n 0
11 2 (2 )n 1
f z z zz z
zz zn
f znz zn n
zz
nznz
tch-prob 22
48. Absolute and uniform convergence of power series
Thm1.
0
1 1 0
0 1 1 1 0.
If a power series ( - ) convergesn 0
when (z ), then it is absolutelyconvergent at each point in the open disk
- , where
nna z z
z z zz
z z R R z z
R1
z1
z z01
0 1.
Converges at Converges at all points
inside the circle -
z
z z R
(1)
tch-prob 23
• The greatest circle centered at z0 such that series (1) converges at each point inside is called the circle of convergence of series (1).
• The series CANNOT converge at any point z2 outside that circle, according to the theorem; otherwise circle of convergence is bigger.
0
0
0
0
Suppose that (z-z ) (1)n 0
has circle of convergence -
Let
( ) ( - )0
N-1 ( ) ( - )
n 0
N
nan
z z R
nS z a z znn
nS z a z zn
0z z R
tch-prob 24
0write ( ) ( ) ( ) ( - )
since (1) converges, ( ) 0 as or for each , there is a swch that
( ) whenever
N N
N
N
z S z S z z z R
z NN
z N N
When the choice of depends only on the value of and is independent of the point z taken in a specified region within the circle of convergence, the convergence is said to be uniform in that region.
N
tch-prob 25
z0
R
R1z
z1
Corollary.
0
0
A power series )( represents a continuous function ( )0
at each point inside its circle of convergence .
n a z- z S znn
z- z R
1 0 0Thm 2. If is in c. of. c. - of ( - )
n 0nz z z R a z zn
then that series is uniformly convergent in the closed disk
0 1 1 1 0., where z z R R z z
tch-prob 26
49. Integration and Differentiation of power seriesHave just seen that a power series
0( ) (1)( )
0z nS z a zn
n
represents continuous function at each point interior to its circle of convergence.
We state in this section that the sum S(z) is actually analytic within the circle.
Thm1. Let C denote any contour interior to the circle of convergence of the power series (1), 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,
0)( ) ( ) ( ) (
0
ng z S z dz a g z z- z dznc cn
tch-prob 27
Corollary. The sum S(z) of power series (1) is analytic at each point z interior to the circle of convergence of that series.
Ex1.sin when 0
want to show ( )1 when 0
2 1since sin ( 1) represent sin for every ,
(2 1)!02 1
( 1)2 2 4(2 1)!0 ( 1) 1
3! 5!(2 1)!0
z z z f z z
nzn z - z znn
nzn - nn z z znn z nn
(4)
converges to ( ) when 0
............
f z z
is entire
But series (4) clearly converges to f(0) when z=0. Hence f(z) is an entire function.
tch-prob 28
Thm2. The power series (1) can be differentiated term by term. That is, at each point z interior to the circle of convergence of that series,
0) 1( ) (
0nS' z n a z- zn
n
Ex2. (1 ( 1) 1) ( 1 1)0
zn n - z -z n
Diff.1 1( 1) ( 1) 1 12 1
1 ( 1) ( 1)( 1) 1 12 0
n n- - n z- z -z n
n n - n z z -z n
tch-prob 29
50. Uniqueness of series representation
Thm 1. If a series
0 ( ) converges to ( )
0na z z f zn
n
at all points interior to some circle , then it is the Taylor series expansion for f in powers of .
0z z R
0z z
Thm 2. If a series
0)( n C z- znn
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 for that domain.0
z z
tch-prob 30
51. Multiplication and Division of Power SeriesSuppose
0
0
0
( )
( )
( )0
and ( )0
converges within ,
z
z
z
n f z a znn
ng z b znn
z R
then f(z) and g(z) are analytic functions in 0z z R
has a Taylor series expansion Their product
0 0( )( ) ( )
0
Cauchy product0
z
c
nf z g z c z z z Rnn
n a b n k n-kk
tch-prob 31
Ex1.1z The Maclaurin series for is valid in disk
11 12 3 2 3(1 )(1 )2 6
1 12 31 12 3
z
z
zez
z z ..... - z z z ........
z ..... z
Ex2.Zero of the entire function sinh z
)
are the mumbers ( 0 1 2 )1 1 1So
2 3 2 4sinh 1 3! 5!has laurent series in the punctured disk 0
1 1 1 7 (0 )2 3 360sinh
(si2
16
nh
z nπi n , , , ...
z z z z z( .......)
z π
. z ..... z πzz z zz ze ez
1zez