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P. 88
Unit 8: Series
Notes
< ,
3 3 3
Now suppose, we have a sequence (f ) of continuous functions which converges uniformly on a
n
n
n
contour C to the function f. Then the sequence f (z)dz converges to f (z)dz. This is easy to
C C
see. Let > 0. Now let N be so that |f (z) f(z)| < A for n > N, where A is the length of C. Then,
n
n
f (z)dz f(z)dz = (f (z) f(z))dz
n
C C C
< A
A
whenever n > N.
Now suppose (f ) is a sequence of functions each analytic on some region D, and suppose the
n
sequence converges uniformly on D to the function f. Then f is analytic. This result is in marked
contrast to what happens with real functionsexamples of uniformly convergent sequences of
differentiable functions with a non-differentiable limit abound in the real case. To see that this
uniform limit is analytic, let z D, and let S = {z : |z z | < r} D. Now consider any simple
0
0
0
closed curve C S. Each f is analytic, and so f (z)dz for every n. From the uniform
n
n
C
n
convergence of (f ), we know that f(z)dz is the limit of the sequence f (z)dz , and so
n
C C
f(z)dz 0. Moreras theorem now tells us that f is analytic on S, and hence at z . Truly a miracle.
0
C
8.2 Series
A series is simply a sequence (s ) in which s = a + a + ... + a . In other words, there is sequence
2
1
n
n
n
(a ) so that s = s n 1 + a . The s are usually called the partial sums. Recall from Mrs. Turners class
n
n
n
n
n
that if the series a j has a limit, then it must be true that lim(a ) 0.
n n
j 1
n
Consider a series fj(z) of functions. Chances are this series will converge for some values
j 1
of z and not converge for others. A useful result is the celebrated Weierstrass M-test: Suppose
(M) is a sequence of real numbers such that M 0 for all j > J, where J is some number., and
j
j
n
suppose also that the series M j converges. If for all z D, we have |fj(z)| M for all j > J,
j
j 1
n
then the series f (z) converges uniformly on D.
j
j 1
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