Chap
.
15
Power Series
,
Taylor Series
These are analogs of power series and Taylor series (which are power series, too) known to you from calculus.
Perhaps you take a look at your calculus book to refresh your memory before you begin to study this chapter.
These series are important by themselves, in fact much more important than real power series in
calculus. Moreover, this chapter is also needed for the understanding of the next chapter on Laurent series
(which have no useful analog in calculus) and a complex integration method based on these series.
Sec
.
15
.
1
Sequences
,
Series
,
Convergence Tests
This is similar to sequences and series in calculus. Most important from a practical point of view is the ratio
test in Theorem 7.
Problem Set 15
.
1
.
Page 672
1
.
Sequence
.
The terms
z
n
u003d
ue0a2
u2212
1
ue0a3
n
u002b
i
/2
n
,
n
u003d
0,1,
u22ef
, are
i
u002b
i
,
u2212
1
u002b
1
2
i
,
1
u002b
1
4
i
,
u22ef
.
For the absolute value you obtain

z
n

2
u003d
1
u002b
1
2
2
n
u2264
1
u002b
1
u003d
2,
which shows boundedness. Since Im
z
u003d
1/2
n
u2192
0 as
n
u2192
u221e
and Re
z
u003d u00b1
1, you see that
ue0a6
z
n
ue0a7
has the
two limit points 1 and
u2212
1. Conclude from this that
ue0a6
z
n
ue0a7
diverges.
9
.
Sequence
.
Calculate

z
n

u003d
0.9
u002b
0.1
i

2
n
u003d
ue0a2
0.9
u002b
0.1
i

2
ue0a3
n
u003d
ue0a2
0.81
u002b
0.01
ue0a3
n
u003d
0.82
n
u2192
0
as
n
u2192
0.
Conclude that the sequence converges absolutely to 0.
15
.
Boundedness
.
Let
ue0a6
z
n
ue0a7
be bounded, say, 
z
n

u003c
K
for some
K
and all
n
. Set
z
n
u003d
x
n
u002b
iy
n
as in the text.
Then boundedness of the sequences
ue0a6
x
n
ue0a7
and
ue0a6
y
n
ue0a7
can be seen from

x
n

u2264

z
n

u003c
K
,

y
n

u2264

z
n

u003c
K
.
Here it was used that for
z
u003d
x
u002b
iy
you always have
x
2
u2264
x
2
u002b
y
2
u003d

z

2
hence

x

u2264

z

and similarly for the imaginary part
y
, namely, 
y

u2264

z
.
Conversely, let
ue0a6
x
n
ue0a7
and
ue0a6
y
n
ue0a7
be bounded, say,

x
n

u003c
K
,

y
n

u003c
K
.
Then
x
n
2
u003c
K
2
,
y
n
2
u003c
K
2
, so that

z
n

2
u003d
x
n
2
u002b
y
n
2
u003c
2
K
2
By taking square roots this gives

z
n

u003c
k
ue0a2
k
u003d
K
2
ue0a3
.
Hence
ue0a6
z
n
ue0a7
is bounded.
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144
Complex Analysis
Part D
17
.
Ratio test
.
For the ratio test in Theorem 7 you need
z
n
u003d
ue0a2
u2212
1
ue0a3
n
ue0a2
1
u002b
2
i
ue0a3
2
n
u002b
1
ue0a2
2
n
u002b
1
ue0a3
!
.
Calculate the test ratio
z
n
u002b
1
z
n
u003d
ue0a2
u2212
1
ue0a3
n
u002b
1
ue0a2
1
u002b
2
i
ue0a3
2
ue0a2
n
u002b
1
ue0a3
u002b
1
/
ue0a2
2
ue0a2
n
u002b
1
ue0a3
u002b
1
ue0a3
!
ue0a2
u2212
1
ue0a3
n
ue0a2
1
u002b
2
i
ue0a3
2
n
u002b
1
/
ue0a2
2
n
u002b
1
ue0a3
!
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 Spring '12
 M.lee
 Calculus, Maclaurin Series, Taylor Series, Mathematical Series

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