Lect6PowerSeries
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Transcript Lect6PowerSeries
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
zz
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 103 i 0, 102 i 0, 101 i 0 .
z 103 i 0 :
1
1.00 0.001 0.000001 0.000000001
1 103
1.001001001001001
Clearly, every additional term adds 3 more significant figures to the final result.
z 102 i 0 :
1
1.00 0.01 0.0001 0.000001
1 102
1.0101010101
Here, however, each additional term adds only 2 more significant figures to the result.
z 101 i 0 :
1
1.00 0.1 0.01 0.001
1 101
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
zz
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 n1 dz
C
0
Taylor series expansion of f z about z0
z z n1 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 an
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 n1 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 n0 z z0 n1
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 n1,
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 n1
1
dz
2 i
1
z nm 2
C
1
1
dz
2 i
z 1 z n2
C
1
z n 2
2
0
zm dz,
( z 1)
m 0
dz ; let z rei , dz irei d
C m 0
1
an
2 i m0
0 , m n1
2 , mn 1
irei
1
d
i n m 2
2
r nm 2e
1
f z 1 z z 2 z3
z
, 0 z 1
m 0
1
r nm1
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 n3 zm dz,
1
1
dz
2 i
z 1 z n2
1
1 1 z n3 dz
z
C
( z 1)
m 0
C
1
m0 z nm3
dz ; let z rei , dz irei d
0 , m n 2
2 , m n 2
C
1
an
2 i m0
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
z2
1
1
1
z 1 1 2 1 z 1 2 1 z 1
2
1 z 1 z 1
1 z 1 z 12
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 12
(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 12
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
, z0
f z
z2
1 2 ,
z0
( 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
n0
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
m0
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
p0
m0