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Notes For 110.311 - Complex Analysis: 1 Sequences and Uniform Convergence

1) The lecture discusses convergence of series and Taylor series of complex functions. 2) It defines uniform convergence of sequences and series, and gives an example where a series converges pointwise but not uniformly on a domain. 3) Taylor's theorem for complex functions is presented, stating that if a function is analytic in a disk, its Taylor series converges to the function uniformly in smaller disks contained within the original disk.

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75 views3 pages

Notes For 110.311 - Complex Analysis: 1 Sequences and Uniform Convergence

1) The lecture discusses convergence of series and Taylor series of complex functions. 2) It defines uniform convergence of sequences and series, and gives an example where a series converges pointwise but not uniformly on a domain. 3) Taylor's theorem for complex functions is presented, stating that if a function is analytic in a disk, its Taylor series converges to the function uniformly in smaller disks contained within the original disk.

Uploaded by

Joe Bloe
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as PDF, TXT or read online on Scribd
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Notes for 110.

311 Complex Analysis


16th Lecture
29 October 2015
Abstract
This lecture discusses convergence of series as well as the Taylor Series of a complex function.

Sequences and Uniform Convergence

A Cauchy
n0 . Likewise, the
P sequence an is such that for all  > 0, n0 such that |am an | <  if m, n P

series j=0 cj converges if  > 0n0 such that |cn + + cm | <  if m n n0 . If j=0 cj converges

P

m
then limj cj = 0. We call a sequence uniformly Cauchy if for  > 0n0 such that j=n fj (z) <  if
m n n0 and z D for some domain D.
Example 1: Uniform Convergence
Fix a C; then
P
(a) For which z does the series j=0
The series

zj
j=0 aj

j=0


z j
a

zj
aj

converge?


converging az < 1 |z| < |a|.

(b) What is the limit function?


For |z| < |a|, the series is geometric, so the limit function is

1
1z/a

a
az .

(c) Does this series converge uniformly?


No. The sequence for this
is not uniformly Cauchy, since for  = 12 , for any n 0 , n n0 and z
z series
n
with |z| < |a| such that a > 12 . In fact, one could take any n n0 and z with az sufficiently close
to 1, depending on n.
P j
(d) For r < |a| show that j=0 az j converges uniformly on |z| < r
P j
P
j
We know that j=0 ar j converges; for any , there is a n0 such that for all n n0 , j=n+1 ar j < , then




P
zj
P
zj
j=n+1 aj <  z with |z| r. This shows that
j=n+1 aj < , or, equivalently, there is a limit
P

j


function F (z) such that j=n+1 az j F (z) <  n n0 and |z| r (see Figure 1).
|a|
r

Figure 1: Diagram used to explain Example 1.

Taylor Series

If f (z) has derivatives of all orders at z = a, then the Taylor series around z = a is

X
f j (a)
j=0

j!

(z a)j = f (a) + f 0 (a)(z a) +

f 00 (a)
(z a)2 + . . .
2

The Taylor series around z = 0 is called the Maclaurin series. In the calculus of real variables, there are
badly behaved functions such that all of their derivatives are defined at the point x0 but the series does never
converges to f (x) except for x = x0 . Interestingly enough, this does not happen for complex functions.
Theorem 1. If f is an analytic function on |z a| < R, then its Taylor series around the point z = a
converges to f (z) z |z a| < R. Furthermore, for any r < R, the series converges uniformly on
|z a| < r.
We need to prove everything after the furthermore part; look to Figure 3 for context.

z
a

Figure 2: Diagram used to explain Theorem 1.


Proof. We employ the following procedure:
1. Draw the circle c = {|z a| = r}, r < r0 < R.
2. From the Cauchy Integral Formula, f (j) (a) =
1
2i

f (w)
c wz

w
c (wa)j+1

dw.

dw.

3. f (z) =

j!
2i

r0

a
r

Figure 3: Diagram used to explain the proof for Theorem 1.


4. Write

1
wz

as a Taylor series around

1
a:

1
1
1
1
=
=

za
wz
(w a) (z a)
w a 1 wa
j X


X
1
za
1
=

=
(z a)j
j+1
w a j=0 w a
(w

a)
j=0
For which z is this true?

5.
1
2i

1
(z a)j dw
j+1
(w

a)
c
j=0

Z

X
1
f (w)
=
dw (z a)j
j+1
2i
(w

a)
c
j=0

f (a) =

f (w)

The integrand is f j (a), thus completing the proof.

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