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18.112 Functions of a Complex Variable
Fall 2008
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Solution for 18.112 ps 6
1(Prob 1 on P193).
Solution: The partial product
n �
�
�
1
Pn =
1− 2
k
k=2
��
�
n �
�
1
1
=
1−
1+
k
k
thus
=
k=2
n
�
=
n+1
,
2n
k + 1 k − 1
k
k
k=2
�
∞ �
�
1
1 − 2 = lim Pn
n→∞
n
n=2
n+1
n→∞ 2n
1
= .
2
= lim
2(Prob 2 on P193).
Method 1. Note that
n−1
(1 − z)Pn = (1 − z)(1 + z) · · · (1 + z 2
2n
=1−z .
Taking limit, we get
(1 − z)P = lim (1 − z)Pn
n→∞
n
= lim (1 − z 2 )
n→∞
= 1,
1
)
thus
P =
1
.
1−z
Method 2. You can use induction to prove
Pn =
n −1
2�
k=0
n
1 − z2
z =
,
1−z
thus
P =
k
1
.
1−z
Method 3. You can also use the uniqueness of 2-adic expansion to get
Pn =
n −1
2�
zk .
k=0
3(Prob 3 on P193).
Method 1. By Theorem 5, it is enough to prove
∞
�
�
z −z �
log (1 + )e n
n
n=1
converges absolutely and uniformly on every compact set. Take integer M big
enough such that the given compact set is bounded by M/2. Then
1+
|z| |z|2
+ 2 +··· < 2
n
n
for n ≥ M. So
∞ �
∞
�
z
z ��
z − z �� � ��
�
n
�log[(1 + )e ]� =
�[log(1 + ) − ]�
n
n
n
n=M
n=M
�
�
∞
�
� 1 z 2 1 z 3 1 z 4
�
�[− ( ) + ( ) − ( ) + · · · ]�
=
� 2 n
�
3 n
4 n
n=M
�
∞ � � �
�
|z| |z|2
� z �2
≤
+ 2 +···
� � 1+
n
n
n
n=M
∞
1 2� 1
≤ M
.
2
n2
n=M
2
This proves
∞
�
�
�
z
z
log (1 + )e− n
n
n=1
converges absolutely and uniformly on any compact set.
Method 2. By Theorem 6, it is enough to prove
∞ �
�
�
z z
� − nz
�
�e + e− n − 1�
n
n=1
converges uniformly on |z| < R for any R. This is true, since
�
�
���
∞
� z
�
�
i
�
(−z) 1
1
� − n z − nz
� �
�
−
�e + e − 1� = �
�
i
�
n
n
i! (i − 1)! �
i=2
�i−2
∞ �
|z|2 � |z|
1
≤ 2
n i=2 n
(i − 2)!
|z|2 |z|
en
n2
R2 R
≤ 2 en.
n
≤
3