7 Power series

7.1 Basic theory

Def. A power series P

f_k (z) is shown as follows:

c₀ + c₁(z z₀) + c₂(z z₀)² + ::::: + c_k(z z₀)^k + ::::;

where c₀; c₁; .... are complex constants and z, z₀ 2 C: ^In particular, z₀ = 0; the series becomes

c₀ + c₁z + c₂z² + ::::: + c_kz^k + :::::

Ex. 1 + z + 2²z² + 3³z³ + :::: + nⁿzⁿ + ::: Please nd the convergence set of it.

sol. 8z⁰ 6= 0; jnz⁰j = n jz⁰j > 2 ^if n > ²

jz⁰j: ) jnⁿz₀ⁿj = jnz⁰jⁿ > 2ⁿ; and lim

n!1 jnz⁰jⁿ = 1:

Therefore lim

n!1 nⁿz₀ⁿ = 1; P

nⁿz₀ⁿ diverges. Thus we nd that the series converges only at z = 0:

Ex. Show that the series 1 + z + ^z₂²2 + :::: + _n^znn + :::: converges everywhere.

pf. 8z⁰ 6= 0; ^jz_n^0j < ¹₂ for suf ciently large n:

) ^jz_n^0njn < ₂¹_n for suf ciently large n:

Therefore, by the convergence of P ₁

2ⁿ and comparison test, we nd that P _zⁿ

nⁿ converges absolutely.

Ex. The series 1 +z +z²+ ::: +zⁿ+ ::: converges (absolutely)

Complex Analysis 94 on jzj < 1 and diverges on jzj 1:

Theorem 39 P

c_nzⁿ converges at z₀ 6= 0 )P

c_nzⁿ is absolutely convergent for all z such that jzj < jz⁰j :

pf. 8z ^with jzj < jz⁰j : ^Since P

c_nz₀ⁿ converges, we have

nlim!1c_nz₀ⁿ = 0: i. e.

9M > 0 ^{such that} jcⁿz₀ⁿj M; 8n:

) jcⁿzⁿj = jcⁿz₀ⁿj jzj jz⁰j

n

M rⁿ;

where r = jz=z⁰j < 1: Use comparison test, we obtain that Pc_nzⁿ is absolutely convergent.

Theorem 40 P

c_nzⁿ is convergent for some but not all non-zero values of z:

) 9R > 0 ^{such that} P

c_nzⁿ is absolutely convergent for all jzj < R ^and P

c_nzⁿ is divergent for all jzj > R:

pf. Draw any line on the complex plane from the origin O: Then contains a nonzero point at which the series P

c_nzⁿ converges and a point at which it diverges. (The validity of this assumption can be drawn from the previous theorem.) Then consider the set

= n

OP P 2 ; X

c_nzⁿ converges at P:o

Complex Analysis 95 We may nd R = sup exists and R > 0: Let's consider the following conditions:

i). 8z ^with jzj < R :

9P 2 ^{such that} jzj < OP R: Then obviously P

c_nzⁿ converges at P: By the previous theorem, P

c_nzⁿ is absolutely convergent.

ii). 8z ^with jzj > R :

9P 2 ^{such that} jzj > OP > R: ^{We nd that} P

c_nzⁿ cannot converge at z —otherwise, P

c_nzⁿ becomes convergent at P; which contradicts to the de nition of R:

Def. i). The number R in the above theorem is called the radius of convergence of P

c_nzⁿ:

ii). C : jzj = R is called the circle of convergence.

Note: i). If P

c_nzⁿ converges only at 0; R = 0:

ii). If P

c_nzⁿ converges everywhere on C; R = 1:

Theorem 41 P

c_nzⁿ is a series with radius of convergence R:

Then it converges uniformly on every closed disk jzj r < R:

pf. 8z ^{such that} jzj r; we nd that jcⁿzⁿj < jcⁿj rⁿ: Since Pc_nrⁿ is absolutely convergent for r < R; we have P

c_nzⁿ converges uniformly on jzj r:

Theorem 42 The sum S (z) of P

c_nzⁿ with radius of conver-gence R is a continuous function on the open disk jzj < R:

Complex Analysis 96 pf. For all z₀ with jz⁰j < R; let's take an r such that

jz⁰j < r < R: ^Then P

c_nzⁿ is uniformly convergent on jzj r;

and hence S (z) is continuous on jzj r:

In particular, S (z) is continuous at z₀; while z₀ is an arbitrary point on jzj < R: ^Therefore, S (z) is continuous on jzj < R:

Theorem 43 S (z) = P

c_nzⁿ is a power series with radius of convergence R:

) S (z) is analytic on the disk jzj < R: ^Moreover, S⁰(z) =

X1 n=1

nc_nz^{n 1}; ( )

and the differentiated series above has the same radius of conver-gence R:

pf. Since P

c_nzⁿ is uniformly convergent on every closed disk jzj r < R; we may nd that P

c_nzⁿ is also uniformly convergent on every closed and bounded domain D on jzj < R:

By Weierstrass theorem, S (z) is analytic and S (z) = X

c_nzⁿ for all z on the disk jzj < R:

Next, let's set R⁰ being the radius of convergence of

differentiated series ( ) ; and assume that R⁰ > R: Then choose a

Complex Analysis 97 point z₀ such that R⁰ > jz⁰j > R; and we see that

Z

C

X1 n=1

nc_{n 1}z^{n 1}dz =

X1 n=1

c_nz₀ⁿ;

where C is any contour lying on jzj < R^{0 joining} 0 and z₀; and the result is by the uniform convergence (on C) of series ( ) : Hence

c₀ +

X1 n=1

c_nz₀ⁿ also exists.

i. e. P

c_nzⁿ is convergent at z₀ while jz⁰j > R; which contradicts to the fact that R is the radius of convergence. Therefore, we must have R⁰ R; or simply R⁰ = R:

Remark 13 By induction and result of the above theorem, we nd that

S^[k] (z) =

X1 n=k

c_n n

k k!z^{n k} with the same radius of convergence as S (z) :

Complex Analysis 98 7.2 Determination of radius of convergence

Def. Given a sequence faⁿg¹_n=1 of non-negative real numbers.

Then we say that

i). faⁿg is ”unbounded” if

nlim!1a_n = lim

n!1sup a_n = 1:

ii). faⁿg is ”bounded” if

nlim!1a_n = l;

where l is the largest limit point of faⁿg and is a nite number.

Remark 14 lim

n!1a_n = l means

i). 8" > 0; 9N > 0 : 8n > N ) aⁿ < l + ":

ii). 8" > 0; 8N > 0 : 9n > N ) aⁿ > l ":

Theorem 44 (Cauchy-Hadamard) Given a complex seriesP

c_nzⁿ; and set l = lim

n!1

pn

jcⁿj: ^Then the radius of convergence R = ¹_l; with the understanding that R = 1; ^if l = 0;

0; if l = 1:

pf. To prove the theorem, just consider the following 3 cases:

Case 1: l = 1:

Complex Analysis 99 l = 1 ^{means that}

n pn

jcⁿj

o is unbounded. Suppose that there is some z₀ 6= 0 ^{such that} P

c_nz₀ⁿ converges. Then obviously

nlim!1c_nz₀ⁿ = 0; which also means that

This contradicts to the fact that l = 1: So we may conclude that for all z₀ 6= 0; P test, then we see that P

c_nz₀ⁿ also converges. Since z₀ is arbitrary,

R = 1: ]

Case 3: 0 < l < 1 ⁽l is nite.) To show that P

c_nzⁿ is absolutely convergent at every point z₁ such that jz¹j < ¹_l and is divergent at every point z₂ such that jz²j > ¹_l: First, consider the following:

Complex Analysis 100 By using comparison test, we nd that P

c_nz₁ⁿ is absolutely

holds for in nitely many terms of n and note that l " > ¹

jz²j:

Ex. Find the radius of convergence for the series P₁

k=0 z^k2:

Complex Analysis 101

So 0 and 1 are the limit points of the sequence n pn

Ex. Find the radius of convergence for the series P₁

k=0 z^k k!: Sol. i). In this problem, Cauchy-Hadamard theorem is rather dif cult for us to use, so we may utilize ratio test instead. From ratio test, we have

R = lim

ii). If we want to use Cauchy-Hadamard theorem, consider the following:

Complex Analysis 102

Ex. Find the radius of convergence for the series P₁

k=0

Ex. The following series 1 + z + z² all have radius of convergence R = 1:

Ex. Find the convergence set for the series P₁

k=0 k!z^k:

n!: Also from the previous 3rd example, pⁿ

n! p

n; we have

nlim!1

pn

jcⁿj = 1:

Complex Analysis 103 So the convergence set is f0g ^only.

Ex. Find the convergence set for the following series:

i). 1 + z + z² + :::: + z^k + ::::

ii). 1 + z + z²

2 + :::: + z^k

k + ::::

iii). 1 + z + z²

2² + :::: + z^k

k² + ::::

Sol. i). c_n = 1; lim

n!1

pn

jcⁿj = 1: But the series diverges everywhere on the circle of convergence jzj = 1:

ii). c_n = _n¹; lim

n!1

pn

jcⁿj = 1: It is also convergent at some points (such as z = 1) of the circle C : jzj = 1 and divergent at some other points of C^.

iii). c_n = _n¹₂; lim

n!1

pn

jcⁿj = 1: The series is convergent absolutely everywhere on the circle C : jzj = 1 (known by using comparison test.)

Complex Analysis 104

在文檔中 Complex Analysis Lecturenotes Nai-Sher Yeh (頁 93-104)