2.3 Complex Power Series
Power series in real analysis can be generalized into complex series. Particularly, concepts such as uniform convergence are analogous to those in real analysis:
Definition 2.3.1 (Uniform Convergence).
For a set , a function sequence uniformly converges to a function on iff , such that , , .
Remark.
The definition above is equivalent to the following definition.
For a set , a function sequence uniformly converges to iff
Informally, we will use the notation to represent uniform convergence.
Theorem 2.3.1 (Cauchy Criterion).
For a set , a function sequence uniformly converges to a function iff , such that , , .
Proof.
If uniformly converges to , then for , such that and ,
Then,
For the converse, refer to the analogous proof in Theorem 1.2.12.
Function series are defined to be a sequence formed by the partial sums of function sequences. There are many ways to verify the uniform convergence of a function series. Perhaps the most widely known is the Weierstrass -Test.
Theorem 2.3.2 (Weierstrass -Test).
Let be a region and be a function sequence on .
If such that , , and the series converges, then the series converges uniformly and absolutely on .
Proof.
By the convergence of , , such that ,
Since bounds , it follows that
and the result follows from Theorem 2.3.1.
The concept of uniform convergence is generalized to improper integrals with parameters, and the same theorems from series have a corresponding counterpart.
In both complex and real analysis, the concept of power series, a unique type of function series, is of trivial importance. Similar to real power series, complex series have the form
where are constants.
Let denote the open disk centered at with radius . For simplicity, from now on we will have denote the unit open disk, or . We will now observe the convergence of power series.
Theorem 2.3.3 (Abel's Theorem).
For a power series , there exists a constant , known as the radius of convergence such that:
- absolutely converges on , and , uniformly converges on .
- diverges when .
- is holomorphic over and can be obtained by termwise differentiation, or , which also has a convergence radius of .
The disk is known as the disk of convergence, a direct generalization of the interval of convergence for real series. There are many ways to determine the radius of convergence:
Theorem 2.3.4 (Cauchy–Hadamard).
The radius of convergence of the power series in the form can be determined by
Of course, a convergence radius of implies that the series is divergent everywhere except for possibly at , and a convergence radius of means that the series absolutely converges everywhere.
Proof.
We will prove that the value in (2.3.1) satisfies the criteria in Theorem 2.3.3.
Assume . Then , and consequently, . By Definition 1.2.3 and (2.3.1), such that , and . It follows that for all . Then converges.
Let . Similarly, such that , , and . Then . By the Weierstrass -Test (Theorem 2.3.2), is uniformly bounded for by the convergent series , and thus uniformly converges on . This proves part 1.
Assume that . For all , . By Definition 1.2.3, , such that . It follows that . Since ,
by the Cauchy Criterion (Theorem 1.2.12), is divergent. Thus, part 2 is satisfied.
To prove part 3, first observe that and have the same convergence radius since . For , let , where
Let
Let be positive and . Then we aim to show that
By analyzing the difference,
Since as , it follows that , such that , . Since
for , with ,
Since is absolutely convergent, is the remainder term of a convergent series. Then, such that , .
Finally, for a fixed , such that ,
From (2.3.2), we get
confirming part 3.
Obviously, a substitution of where translates the disk of convergence to . The subsequent results on uniform convergence hold for complex functions:
Theorem 2.3.5 (Uniform Limit).
Let be a sequence of continuous functions on and uniformly convergent to . Then is continuous on .
Proof.
Fix and . By uniform convergence, such that , and . By continuity, such that , .
By the triangle inequality,
for all . Then the continuity of is satisfied.
Lastly, the sufficient criteria to pass a limit through an integral:
Theorem 2.3.6.
Let be a rectifiable curve on which the function sequence is continuous. If uniformly converges to , then
Proof.
Since uniformly converges to on , , there exists such that for all ,
Since each is continuous and is rectifiable, each integral is convergent and well-defined.
Then ,
Therefore,
Remark.
For a uniformly convergent series , the commutation between the limit and the integral becomes a summation-integral switch: