1 (1) power series (2) convergence of power series and the radius of convergence (3) the...
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1
(1) Power series
(2) Convergence of Power series and the Radius of Convergence
(3) The Cauchy-Hadamard formula
(4) Taylor’s series
Section 6
SECTION 6Power Series I - Taylor Series
2
Section 6Why Series ?
?/ C
zdzze
3
4
2
32/ 1
!4
1
!3
1
2 zzzzze z
We can expand the integrand….
dz
zdz
zdzdzzdzze
CCCCC
z2
32/ 1
!3
1
2
where C is 2z C
Analytic ( )by Formula for derivatives ( )
3
Section 6
Can we expand all functions in series ?
We can expand analytic functions in specialseries called “Power Series”
How do we find these series ?
(1) Using Taylor’s Theorem(2) Using other known series
(and other tricks)
4
Section 6Power Series
0
221 )()()(
nooo
non zzazzaazza
A series in powers of )( ozz
0
2)(2
1)(1)(
!
1
n
n jzjzjzn
e.g.
5
Section 6Power Series
0
221 )()()(
nooo
non zzazzaazza
centre
A series in powers of )( ozz
0
2)(2
1)(1)(
!
1
n
n jzjzjzn
e.g.
6
Section 6Power Series
0
221 )()()(
nooo
non zzazzaazza
coefficientscentre
A series in powers of )( ozz
0
2)(2
1)(1)(
!
1
n
n jzjzjzn
e.g.
7
Section 6Convergence of Power SeriesPower series often converge for some values of z butdiverge for other values.
For example the series
0
321n
n zzzz
converges for z 1 but diverges for z 1
diverges
converges
(geometric series)
8
Section 6Convergence of Power SeriesPower series often converge for some values of z butdiverge for other values.
For example the series
0
321n
n zzzz
converges for z 1 but diverges for z 1
Radius ofConvergence
R=1
diverges
converges
(geometric series)
9
Section 6
0
32 !321!n
n zzzzn
Example:
series diverges for all z (except z0)
0
32
!3!21
!
1
n
n zzzz
n
Example:
series converges for all z
Radius of Convergence“at infinity”; R
Zero Radius ofConvergence; R0
10
: converges
0
221 )()()(
nooo
non zzazzaazza
Section 6
(1) The power series always converges at zzo
oz
(2) There is a Radius of Convergence R for which:
Rzz o
: divergesRzz o
11
Section 6Is there a quick way to find the radius of convergence ?
Ra
a
n
n
n
1lim 1
… the Cauchy-Hadamard formula:
Example:
0
22
)3(6)3(21)3()!(
)!2(
n
n jzjzjzn
n
Rn
nn
n
n
n
n
a
ann
n
n
n
14
)1(
)12)(22(lim
)!2(
)!(
)!1(
!)1(2limlim
2
2
21
4
1R
12
Section 6
0
22
)3(6)3(21)3()!(
)!2(
n
n jzjzjzn
n
jzo 3
13
Section 6
0
22
)3(6)3(21)3()!(
)!2(
n
n jzjzjzn
n
: converges
jzo 3
4/13 jz
14
Section 6
0
22
)3(6)3(21)3()!(
)!2(
n
n jzjzjzn
n
: converges
jzo 3
4/13 jz
: diverges4/13 jz
15
Section 6Another Example:
0
2)4()4(1)4(n
n jzjzjz
Ra
an
n
n
n
11
1
1limlim 1
1R
: converges
jzo 4
14 jz
: diverges14 jz
16
Section 6Another Example:
0
22 )1(2)1()1(n
nn zezezne
Re
n
ne
ne
en
a
ann
n
nn
n
n
11lim
)1(limlim
11
eR /1
: convergesez /11
: divergesez /11 1oz
17
Section 6
0
2
25
)2(
5
)2(1
5
)2(
nn
n jzjzjz
n
n
n a
a 1lim
R
Question:
Where is the centre of the series?What is the radius of convergence?
18
Section 6Note:We have used the Cauchy-Hadamard formulato find the radius of convergence.
There are many other tests (which can be used when theabove test fails).
e.g. (1) Root test (2) if
(3) Comparison test
(4) The Ratio test
diverges1lim
nn
nz
diverges0lim0
n
nnn
zz
converges andif converges00
n
nnnn
n bbzz
diverges1lim 1
n
n
n z
z
19
Section 6Power Series and Analytic functions
Every analytic function f (z) can be represented by a powerseries with a radius of convergence R0. The function isanalytic at every point within the radius of convergence.
Example:
0
321n
n zzzz
series converges for z 1
Radius ofConvergence
zzf
1
1)(
20
Section 6
These power series which represent analytic functions f (z)are called Taylor’s series.
0
)( )(!
1,)()(
no
nn
non zf
nazzazf
(Cauchy, 1831)
How do we derive these Power Series?
They are given by the formula
21
Section 6
Example:
32
0
0
)(
0
1
!
)0(
1
1
zzz
z
zn
f
zaz
n
n
n
nn
n
nn
zzf
1
1)(
(1) centre z0 :
!3)0(
2)0(
1)0(
1)0(
f
f
f
f
432 1
!3)(,
1
2)(,
1
1)(,
1
1)(
zzf
zzf
zzf
zzf
1z0oz
centresingular point
1R
Derive the Taylor series for
22
Section 6
Example:
32
0
0
)(
0
1
!
)0(
1
1
zzz
z
zn
f
zaz
n
n
n
nn
n
nn
zzf
1
1)(
(1) centre z0 :
!3)0(
2)0(
1)0(
1)0(
f
f
f
f
432 1
!3)(,
1
2)(,
1
1)(,
1
1)(
zzf
zzf
zzf
zzf
1z0oz
centresingular point
1R
Derive the Taylor series for
23
Section 6
Example:
32
0
0
)(
0
1
!
)0(
1
1
zzz
z
zn
f
zaz
n
n
n
nn
n
nn
zzf
1
1)(
(1) centre z0 :
!3)0(
2)0(
1)0(
1)0(
f
f
f
f
432 1
!3)(,
1
2)(,
1
1)(,
1
1)(
zzf
zzf
zzf
zzf
1z0oz
centresingular point
1R
Derive the Taylor series for
24
Section 6
Example:
32
0
0
)(
0
1
!
)0(
1
1
zzz
z
zn
f
zaz
n
n
n
nn
n
nn
zzf
1
1)(
(1) centre z0 :
!3)0(
2)0(
1)0(
1)0(
f
f
f
f
432 1
!3)(,
1
2)(,
1
1)(,
1
1)(
zzf
zzf
zzf
zzf
1z0oz
centresingular point
1R
Derive the Taylor series for
25
Section 6
3
212
21
21
0211
0212
1)(
021
16842
2
!
)(
1
1
zzz
z
zn
f
zaz
n
nn
n
nn
n
nn
(2) centre z12 :
421
321
221
21
2!3)(
2.2)(
2)(
2)(
f
f
f
f
432 1
!3)(,
1
2)(,
1
1)(,
1
1)(
zzf
zzf
zzf
zzf
1z21oz
centresingular point
21R
26
Section 6An analytic function f (z) can be represented bydifferent power series with different centres zo
(although there will only be one unique series for each centre)
At least one singular point of f (z) will be on the circle of convergence
1z21oz
1z0oz
3211
1zzz
z
3
212
21
21 16842
1
1zzz
z
27
Section 6An analytic function f (z) can be represented bydifferent power series with different centres zo
(although there will only be one unique series for each centre)
At least one singular point will be on the circle of convergence
1z21oz
1z0oz
3211
1zzz
z
3
212
21
21 16842
1
1zzz
z
28
Section 6
Another Example:
!321
!
!
)0(
32
0
0
)(
0
zzz
n
z
zn
f
zae
n
n
n
nn
n
nn
z
zezf )( with centre z0
1)0(
1)0(
1)0(
1)0(
f
f
f
f
zzzz ezfezfezfezf )(,)(,)(,)(
0oz
centre
Rno singular points!
29
Section 6Deriving Taylor series directly from the formula
can be very tricky
0
)( )(!
1,)()(
no
nn
non zf
nazzazf
We usually use other methods:(1) Use the Geometric Series(2) Use the Binomial Series
(3) Use other series (exp., cos, etc.)
(4) Use other tricks
32
0
11
1zzzz
z n
n
32
0 !3
)2)(1(
!2
)1(1
)1(
1z
mmmz
mmmzz
n
m
z n
nm
!5!3)!12()1(sin
53
0
12 zzz
n
zz
n
nn
30
Section 6
Example: 21
1)(
zzf
Expand about z0
(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 R1
31
Section 6
Example: 21
1)(
zzf
centre
Expand about z0
(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 R1
32
Section 6
Example: 21
1)(
zzf
centre
Expand about z0
(use the geometric series)
First, draw the centre and singular points to see what’s going on:
singular points:
jjz ,
So it looks like the radius of convergence should be R1
33
Section 6
Example: 21
1)(
zzf
centre
Expand about z0
(use the geometric series)
First, draw the centre and singular points to see what’s going on:
singular points:
jjz ,
So it looks like the radius of convergence should be R1
34
Section 6
3211
1zzz
zWe know that
64222
1)(1
1
1
1zzz
zzTherefore
The geometric series converges for z1
Therefore our series converges for z21- which is the same as z1, as predicted
35
Section 6
3211
1zzz
zWe know that
64222
1)(1
1
1
1zzz
zzTherefore
The geometric series converges for z1
Therefore our series converges for z21- which is the same as z1, as predicted
36
Section 6
3211
1zzz
zWe know that
64222
1)(1
1
1
1zzz
zzTherefore
The geometric series converges for z1
Therefore our series converges for z21- which is the same as z1, as predicted
37
Section 6
3211
1zzz
zWe know that
64222
1)(1
1
1
1zzz
zzTherefore
The geometric series converges for z1
Therefore our series converges for z21- which is the same as z1, as predicted
38
Section 6
Example:z
zf23
1)(
centre
Expand about z1
(use the geometric series)
First, draw the centre and singular points to see what’s going on:
singular point:
2/3z
So it looks like the radius of convergence should be R1/2
39
Section 6
3211
1zzz
zWe know that
2)1(4)1(21)1(21
1
23
1zz
zzTherefore
The geometric series converges for z1
Therefore our series converges for 2(z1)1
- which is the same as z 1 1/2, as predicted
40
Section 6
Example: 2)1(
1)(
zzf
centre
Expand about z0
(use the binomial series)
First, draw the centre and singular points to see what’s going on:
singular point:1z
So it looks like the radius of convergence should be R1
41
Section 6
32
0 !3
)2)(1(
!2
)1(1
)1(
1z
mmmz
mmmzz
n
m
z n
nm
The binomial series is
3222
4321)](1[
1
)1(
1zzz
zzTherefore
The binomial series converges for z1
Therefore our series converges for z1- which is the same as z1, as predicted
as we could guess,since issingular at z1
mz )1(
42
Section 6
32
0 !3
)2)(1(
!2
)1(1
)1(
1z
mmmz
mmmzz
n
m
z n
nm
The binomial series is
3222
4321)](1[
1
)1(
1zzz
zzTherefore
The binomial series converges for z1
Therefore our series converges for z1- which is the same as z1, as predicted
as we could guess,since issingular at z1
mz )1(
43
Section 6
32
0 !3
)2)(1(
!2
)1(1
)1(
1z
mmmz
mmmzz
n
m
z n
nm
The binomial series is
3222
4321)](1[
1
)1(
1zzz
zzTherefore
The binomial series converges for z1
Therefore our series converges for z1- which is the same as z1, as predicted
as we could guess,since issingular at z1
mz )1(
44
Section 6Example:
)2()4(
34152)(
2
2
zz
zzzf
centre
Expand about z0
singular points:2,4 z
… the radius of convergence should be R2
2
1
)4(
134152)(
22
zzzzzf
need to expand in powers of z
45
Section 6
2)4()4()2()4(
34152)(
22
2
z
C
z
B
z
A
zz
zzzf
Use partial fractions:
22 )4()2)(4()2(34152 zCzzBzAzz
2
2
)4(
1)(
2
zzzf
32
2222
44
43
421
16
1
)]4/(1[4
1
)4(
1
)4(
1
zzz
zzz
Now
414/ zzconverges for
46
Section 6
z
zzzzzz
zzzf
32
15
16
17
2221
44
43
42
4
1
2
2
)4(
1)(
32
5
3
4
2
32
2
32
2221
)2/(1
1
2
2
2
2
zzz
zzzand
212/ zzconverges for
so
2zconverges for
47
Section 6
centre
5
3
4
2
32 44
43
42
4
1 zzz
convergesinside here
32
2221
zzz
convergesinside here whole thing
convergesinside overlap
48
Section 6
Other useful series:
132
)1(Ln
!5!3sinh
!3!21
!4!21cos
!5!3sin
32
53
32
42
53
zzz
zz
zzz
zz
zzz
ze
zzz
z
zzz
zz
z
49
Section 6Example:
22sin)( zzf Expand about z0no singular points … the radius of convergence should be R
!5!3
sin53 zz
zzUse the series
!5
)2(
!3
)2(2
2sin)(5232
2
2
zzz
zzf
(of using other series)
50
Section 6Summary
You should be able to find the power series of afunction and its radius of convergence using one ofthese 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 6Summary
You should be able to find the power series of afunction and its radius of convergence using one ofthese methods:
1. Taylor’s Theorem/Formula
2. Binomial Series
3. Geometric Series
4. Using other known series, e.g. sin, cos, exp, etc.
52
Section 6Summary
You should be able to find the power series of afunction and its radius of convergence using one ofthese methods:
1. Taylor’s Theorem/Formula
2. Geometric Series
3. Geometric Series
4. Using other known series, e.g. sin, cos, exp, etc.
53
Section 6
You should be able to find the power series of afunction and its radius of convergence using one ofthese methods:
1. Taylor’s Theorem/Formula
2. Geometric Series
3. Binomial Series
4. Using other known series, e.g. sin, cos, exp, etc.
Summary
54
Section 6Summary
You should be able to find the power series of afunction and its radius of convergence using one ofthese methods:
1. Taylor’s Theorem/Formula
2. Geometric Series
3. Binomial Series
4. Using other known series, e.g. sin, cos, exp, etc.
55
Section 6Topics not Covered
(1) Proof of Cauchy-Hadamard formula (follows from ratio test)
(2) Proof of Taylor’s Theorem - an analytic function can be writtenas a power series (use Cauchy’s Integral Formula)
(3) The concept of uniform convergence and proof that power seriesare uniformlyconvergent
(4) Some other practical methods of deriving power series:e.g. use of differentiation/integration of series, differentialequations, undetermined coefficients, …
(5) Analytic Continuation
56
Section 6(6) In some very exceptional cases, a singular point mayalso arise inside the circle of convergence.
0z
zLn singular (not analytic) along herejumps from to + as we cross this line
zLn
centre jzo 1
n
n
n
o
n
nn
jna
ja
jzaz
)1(
)1(
)1(Ln
)1(Ln
1
0
2
1
1
1
)1()1)(1(
)1()1(limlim
11
21
j
jn
jn
a
ann
nn
nn
n
n
2 R
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