1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

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250 CHAPTER 7 • TAYLOR AND LAURENT SERIES

00

If Sn (z) is the nth partial sum of the series I:; Ck (z - a/, Statement (7-1)

k=O
becomes

if n;::: N .... , then I~ ck (zo -a)k - f (zo)I < c:.
k=O

For a given value of e, the integer N•,zo needed to satisfy Statement (7-1)
often depends on our choice of zo. This is not t he case if the sequence {S.,}
converges uniformly. For a uniformly convergent sequence, it is possible to find
an integer N, (depending only one) that guarantees Statement (7-1) no matter
what value for zo E T we pick. In other words, if n is large enough, the function
Sn is uniformly close to the function f for all z E T. Formally, we have the
following definition.

I Definition 7 .1: Uniform convergence


The sequence { S,. ( z)} converges uniformly to f ( z) on t he set T if for every
c; > 0, there exists a positive integer N. (depending only on c:) such that

if n 2". N,, then IS,, (z) - f (z)I < c:, for all z ET. (7-2)
00
HS., (z) is the nth partial sum of the series L: Ck (z - a)k, we say that the series
k=O
00 k
I:; ck (z - a) converges uniformly to f (z) on the set T.
k=O


  • EXAMPLE 7.1 The sequence {S., (z)} = { e• + i } converges uniformly to
    the function f (z) = e• on the entire complex plane because for any c: > 0,
    Statement (7-2) is satisfied for all z for n 2". N., where N,, is any integer greater
    than ~-We leave the details of showing this result as an exercise.


A good example of a sequence of functions that does not converge uniformly
is the sequence of partial sums forming the geomet ri c series. Recall that the
n-1

geometric series has S,. (z) = I:; zk converging to f (z) = 1 :_, for z E D 1 (O).

k=O
Because the real numbers are a subset of the complex numbers, we can show
that Statement (7-2) is not satisfied by demonstrating that it does not bold
when we restrict our attention to the real numbers. In that context, D 1 (0)
becomes the open interval (-1, 1), and the inequality ISn (z) - f (z)I < c: be-
comes IS,.. (x) - f (x)I < c:, which for real variables is equivalent to the inequality
f (x) -c: <Sn (x) < f (x) +e. If Statement (7-2) were to be satisfied, then given

e > 0, Sn (x) would be within an e-bandwidth off (x) for all x in the interval

( -1, 1) provided n were large enough. Figure 7 .1 illust rates that there is an c:

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