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Complex power series

在文檔中 Graduate Texts in Mathematics (頁 34-42)

Power Series

3.1. Complex power series

Let A be a subset of C, and let {fn} = {fn}n=0 be a sequence of functions defined on A (in the previous chapter, sequences were indexed by Z>0; for convenience, in this chapter, they will be indexed by Z0). We form the new sequence (known as a series) {SN}, where SN(z) =

N n=0

fn(z), and the formal infinite series n=0

fn(z) .

The sequence of complex numbers {SN(z)}n=0 is also known as the sequence of partial sums associated with the infinite series

n=0

fn(z) at the point z A. When the indices of summation are clear from the context, we often omit them. For example,

fn(ζ) usually means the infinite sum

n=0

fn(ζ). Other similar abbreviations are used.

Definition 3.1. i) We say that the infinite series n=0

fn(z) con- verges at a pointζ ∈Aif{SN(ζ)}converges. In this case, we write

n=0

fn(ζ) = lim

n→∞SN(ζ).

ii) We say that the infinite series n=0

fn(z)converges pointwise inA if {SN(ζ)} converges for everyζ ∈A.

iii) We say that the infinite series n=0

fn(z) converges absolutely at a point ζ ∈A if the infinite series

n=0

|fn(ζ)|converges.

iv) We say that the infinite series n=0

fn(z) converges uniformly in A if the sequence of partial sums {SN(z)} converges uniformly in A.

3.1. COMPLEX POWER SERIES 25

v) We say that the infinite series n=0

fn(z)converges normally on a set B A if there exists a sequence of positive constants {Mn} such that

(1) |fn(z)| ≤Mn for allz ∈B and all n, and

(2)

Mn<∞.

vi) We say that the infinite series n=0

fn(z) diverges at a point of A if it does not converge at that point.

We speak of the pointwise, uniform, absolute, or normal conver- gence of a series as well as of thedivergence of a series.

Remark 3.2. The fact that the sequence {SN(ζ)}converges if and only if {SN(ζ)} is Cauchy allows us to rephrase conditions (i–iv) as if and only if statements when useful: (i)

fn(z) converges at a point ζ ∈Aif and only if{SN(ζ)}is Cauchy; (ii)

fn(z)converges pointwise in A if and only if {SN(ζ)} is Cauchy for every ζ A; (iii)

fn(z) converges absolutely at a point ζ A if and only if the infinite series |fn(ζ)| converges; and (iv)

fn(z) converges uniformly inA if and only if the sequence of partial sums {SN(z)}converges uniformly inA.

Remark3.3. Many questions on convergence of complex sequences are reduced to the real case by the trivial but important observation that absolute convergence at a point implies convergence at that point.

Remark 3.4. Two other such observations, both trivial but impor- tant, are that if an infinite series

n=0

fn(z) converges at a point ζ, then

nlim→∞fn(ζ) = 0 and lim

n→∞

k=n

fk(ζ) = 0.

Some relationships between some different types of convergence of a series are given in the following result.

Weierstrass M-test. Normal convergence implies uniform and absolute convergence.

Proof. With notation as in the last definition, if N1 < N are positive integers, then

|SN(z)−SN1(z)| ≤ N n=N1+1

Mn for all z ∈A (needed for the uniform convergence argument), and

| |SN(z)| − |SN1(z)| | ≤ N n=N1+1

Mn for allz ∈A (needed for the absolute convergence argument).

Given any > 0, we can find a positive integer N0 such that N >

N1 > N0 implies N n=N1+1

Mn< .

We shall be mostly interested in series of the form

n=0anzn with anC(these are known aspower series) and theassociated real-valued series

n=0|an|rn where r =|z|. We define, for each N Z0, SN (r) =

N n=0

|an|rn for r∈R0, and we observe that

SN+1(r) SN(r) for all N Z0 and for all r R0 . We refer to SN(r) as thereal partial sum at r.

An elementary but most important example is provided by the geo- metric series with r R0:

1 +r+r2+. . . .

Note thatSN(1) =N + 1, and SN(r) = 1−rN+1

1−r for 0≤r <1, as well as for r >1.

Thus n=0

rn = 1

1−r if 0≤r <1 and

rn diverges if r≥1.

We now introduce two special cases of divergence of a sequence of real numbers. It will be useful to regard these sequences as convergent sequences with infinite limits.

3.1. COMPLEX POWER SERIES 27

Definition3.5. A sequence of real numbers{bn}converges to+ if and only if for all M >0 there exists an N Z>0 such thatbn > M for all n > N. In this case we shall write limn→∞bn = +. A similar definition applies to real sequences converging to −∞.

With this notation either:

(a) lim

N→∞SN(r) exists and is finite: That is,

n=0|an|rnconverges.

In this case we write

n=0|an|rn<+; or

(b) lim

N→∞SN(r) = +: That is,

n=0|an|rn diverges (in the pre- vious sense). In this case we write

n=0|an|rn= +. For real sequences, we have thecomparison test. Let 0≤an ≤bn.

(a) If

an= +, then

bn = +. (b) If

bn<+, then

an <+.

Abel’s lemma. Let 0< r < r0. Assume there exists an M R>0

such that

|anrn0| ≤M for all n∈Z>0 . Then the series

anznconverges normally for all zwith |z| ≤r. In particular, it converges absolutely and uniformly for all z with |z| ≤r.

Proof. For|z| ≤r we have

|anzn|=|an| |z|n ≤ |an| |r|n=|an| r

r0

n

rn0 ≤M r

r0

n

. Comparison with the geometric series [or an application of the Weierstrass M-test with Mn = M(rr

0)n] shows the normal conver-

gence.

IfSis any nonempty set of real numbers, then the least upper bound or supremum of S is denoted by supS and the greatest lower bound or infimum is denoted by infS. The possibilities that supS = + or infS=−∞ are allowed.

Definition 3.6. The radius of convergence ρ of the power series anzn is given by

ρ= sup{r 0;

|an|rn <+∞} .

Note that 0 ρ +. As a result of the next theorem it makes sense to define{z C; |z|< ρ}as thedisk of convergenceof the power series

anzn.

Theorem 3.7. Let

anznbe a power series with radius of conver- gence ρ.

(a) If 0< r < ρ, then

anzn converges normally, absolutely, and uniformly for |z| ≤r.

(b) The series

anzn diverges for |z|> ρ.

Proof. (a) Choose r0 with r < r0 < ρsuch that

|an|r0n<+. Thus there exists an M > 0 with |an|rn0 M for all n in Z>0. Now apply Abel’s lemma.

(b) We claim that for |z| > ρ, the sequence {|an| |z|n} is not even bounded. Otherwise Abel’s lemma (with r0 = |z|) would guarantee the existence of an r with ρ < r < |z| and

|an|rn < +. This

contradicts the definition of ρ.

Corollary 3.8. Let

anzn be a power series with radius of con- vergence ρ. Then the function defined by S(z) =

anzn is continuous for |z|< ρ.

Proof. It follows immediately from Theorems 2.23 and 3.7.

We now turn to the obvious and important question: How do we compute ρ?

To answer this question, we introduce the concepts of lim sup and lim inf.

Definition 3.9. Let {un} be a real sequence. We use to indi- cate equivalent names [we also use this same notation with a different meaning in other places, such as f 0 or f g, to emphasize that these functions are (identically) equal] and define

limn unlim sup

n

un upper limit of {un}

limit superior of {un}= lim

p→∞sup

np{un}= inf

p sup

np{un}, and

lim

n

unlim inf

n un lower limit of {un}

limit inferior of {un}= lim

p→∞inf

np{un}= sup

p

ninfp{un}. Note that every real sequence has a limit superior as well as a limit inferior, which are either real numbers or +or −∞.

Properties of limits superior and inferior. Let{un}and {vn} be real sequences. Then:

3.1. COMPLEX POWER SERIES 29

(a) lim inf

n unlim sup

n

un. (b) lim inf

n (−un) =lim sup

n

un. (c) If r >0, then

lim inf

n (run) =rlim inf

n un, and lim sup

n

(run) =rlim sup

n

un. (d) If un≤vn for all n, then

lim inf

n unlim inf

n vn, and lim sup

n

unlim sup

n

vn. (e) lim

n un =L if and only if lim inf

n un=L= lim sup

n

un (L is allowed to be±∞ in this context).

(f) lim inf

n (un+vn)lim inf

n un+ lim inf

n vn, and lim sup

n

(un+vn)lim sup

n

un+ lim sup

n

vn.

Remark and exercise. Iff is a real-valued function of a complex variable, defined on a set S in C and ifζ is a limit point of S, then it is possible to define lim

zζf(z) and lim

zζ

f(z).

Example 3.10. Let un = sin

2

, n = 0,1,2, . . . . This is the sequence {0,1,0,−1, . . .}. Hence for all p∈Z0,

ap = sup

np

un= 1 and thus lim sup

n

un= lim

p→∞ap = 1, and bp = inf

np un=1 and thus lim inf

n un= lim

p→∞bp =1.

The ratio test will be well known to most readers (see Exercise 3.4). Suppose that vn>0 for all non-negative integers n.

(a) If lim

n→∞

vn+1

vn =L <1, then

vn converges.

(b) If lim

n→∞

vn+1 vn

=L >1, then

vn diverges.

Perhaps less familiar is the root test. Suppose that vn0 all n.

(a) If lim

n (vn)n1 =L <1, then

vn converges.

(b) If lim

n (vn)n1 =L >1, then

vn diverges.

Proof. (a) Choose > 0 such that 0< L+ <1. There exists a P Z>0 such that

sup

np

vn1n

< L+ for all p≥P . Thus

vn<(L+)n for all n≥P,

and comparison with the geometric series yields convergence.

(b) Suppose that

vn<+. Then lim

n vn = 0. Thus there exists a P in Z>0 such that n ≥P implies that vn <1 for all n ≥P. Hence (vn)n1 <1 for all n≥P and therefore lim

n (vn)n1 1.

Remark 3.11. The root test applies (with the same value of L) whenever the ratio test applies. However, the converse is not true.

To see this take a sequence where the ratios are alternately 12 and 18. Then the root test will apply with L = 14, but the sequence of ratios, obviously, will not converge.

Example 3.12. Consider the sequence ns with s a positive real number. Both the ratio and the root test end up with L = 1. The series diverges (converges to +) for 0 < s 1 and converges for s >1.

We now return to the study of the complex power series

n=0anzn and to the problem of computing the radius of convergence.

Theorem3.13. Let

anzn be a power series. Suppose thatan = 0 for all n and that the limit lim

n

an+1 an

=L exists, with 0≤L≤+∞. Then the radius of convergence ρ of the power series

n=0anzn is

1

L; in other words,

1

ρ = lim

n→∞

an+1

an , where 1

+ is to be understood as being = 0.

The hypotheses required for this result to hold are strong. As pointed out in Remark 3.11, the ratio test is stronger than the root test. The next result provides a way of computing the radius of con- vergence for any power series.

3.1. COMPLEX POWER SERIES 31

Theorem 3.14 (Hadamard). The radius of convergence ρ of the power series

n=0anzn is given by 1 ρ = lim

n |an|n1 . Proof. Let L = lim

n |an|1n. Thus lim

n |anrn|1n = rL for all r 0 and we conclude by the root test that the associated series

|an|rn converges for 0 ≤r < L1 and diverges forr > L1. Thus ρ= L1.

Lemma 3.15. Let

un and

vn be two absolutely convergent se- ries. Define

wn= n

p=0

upvnp.

Then

wn is absolutely convergent and

wn= (

un) ( vn).

Proof. Letαp =

np

|un| and βp =

np

|vn|. Then limp αp = 0 = lim

p βp . Also

N n=0

|wn| ≤

n=0

|un|

n=0

|vn|=α0β0 <+∞,

and therefore, n=0

|wn| < +. Thus we have proven the absolute convergence of the new series.

To show the required equality, choose m and n with m 2n and consider

m k=0

wk n

k=0

uk

n k=0

vk

=L.

We have to show that L 0 as n → ∞ (we already know each of the above series converges). We rewrite

L=

m k=0

k i=0

uivki n

j=0

n i=0

uivj

.

5

4

3

2

1

-1

k

2 4 i

(a) The first sum

5

4

3

2

1

-1 k

2 4 i

(b) The second sum

Figure 3.1. The (i, k) plane

By looking at the diagrams in the (i, k) plane shown in Figure 3.1, we see that

m k=0

k i=0

uivki= m

i=0

m k=i

uivki

= m

i=0

ui m

k=i

vki = m

i=0

ui

mi

j=0

vj = m

i=0 mi

j=0

uivj . Thus we can estimate

L=

n i=0

mi

j=0

uivj n

j=0

uivj

+ m i=n+1

mi

j=0

uivj

n i=0

mi

j=n+1

uivj +

m i=n+1

mi

j=0

uivj

i=0

j=n+1

|ui| |vj|+ i=n+1

j=0

|ui| |vj|= (α0βn+1+β0αn+1), and the last expression approaches 0 as n goes to.

在文檔中 Graduate Texts in Mathematics (頁 34-42)