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ОБЛАСТИ СХОДИМОСТИ РЯДА ТЕЙЛОРА
Х. Г. Хаитова –
Бухарский государственный университет, преподаватель кафедры
«Математический анализ»
Email:
Аннотация. В этой статье представлены некоторые концепции
разложения функций в ряд Тейлора при некотором значении. Кроме того,
представлены сходимости рядами Тейлора и анализ области сходимости.
Здесь мы опишем некоторые из простейших примеров областей в
пространстве
𝐶
𝑛
.
Как обычно, область - это открытое связное
множество, где открытость означает, что вместе с любой его точкой
множество также содержит окрестность этой точки, а связность
открытого множества
𝐷
означает, что для любых точек
𝑧
′
, 𝑧
»
∈ 𝐶
𝑛
существует непрерывное множество
𝛾: [0,1] → 𝑑
для которого
𝛾(0) = 𝑧ʹ
и
𝛾(1) = 𝑧ʹʹ.
Ключевые слова. Степенный ряды, комплексные числа, несколько
переменных, точка, функция.
THE DOMAINS OF CONVERGENCE OF THE TAYLOR SERIES
Kh.G.Khayitova –
teacher of the «Mathematical Analysis» department, Bukhara State University,
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Annotation. This article presents some concepts of Taylor series expansion
of functions for some value. In addition, Taylor series convergence and
convergence domain analysis are presented. Here we shall describe some of the
simplest examples of domains in the space
𝐶
𝑛
. As usual, a domain is an open
connected set, where openness means that along with any point of it the set also
contains a neighborhood of that point, and connectedness of an open set
𝐷
means
that, for any points
𝑧ʹ, 𝑧ʹʹ𝜖 𝐷
there exists a continuous are
𝛾: [0,1] → 𝑑
for which
𝛾(0) = 𝑧ʹ
and
𝛾(1) = 𝑧ʹʹ.
Key words: degree series, several variables, complex numbers, point,
function.
One of the main theorems of the theory of a complex variable is
Theorem:
Let
𝑓 𝜖 𝑂(𝐷)
and let
𝑧0 𝜖 𝐷
be an arbitrary point in
𝐷
. Then the
function
𝑓
may be represented as a sum of a convergent power series
𝑓(𝑧) = ∑ 𝑐
𝑛
(𝑧 − 𝑧
0
)
𝑛
∞
𝑛=0
inside any disk
𝑈 = {ǀ𝑧 − 𝑧
0
ǀ < 𝑅} ⊂ 𝐷
.
Proof
. Let
𝑧𝜖𝑈
be an arbitrary point. Choose
𝑟 > 0
so that
|𝑧 − 𝑧_0 | < 𝑟 <
𝑅
and denote by
{𝛾
𝑟
= ϛ ∶ |ϛ − 𝑧| = 𝑟}
. The integral Cauchy formula implies that
𝑓(𝑧) =
1
2𝜋𝑖
∫
𝑓(ϛ)
ϛ − 𝑧
𝐷
𝑑ϛ.
in order to represent f as a power series let us represent the kernel of this integral
as the sum of a geometric series:
1
ϛ − 𝑧
= [(ϛ − 𝑧
0
) (1 −
𝑧 − 𝑧
0
ϛ − 𝑧
0
)]
−1
= ∑
(𝑧 − 𝑧
0
)
𝑛
(ϛ − 𝑧
0
)
𝑛+1
∞
𝑛=0
.
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We multiply both sides by
1
2𝜋𝑖
𝑓(ϛ)
and integrate the series term-wise along
𝛾
𝑟
.
The series converges uniformly on
𝛾
𝑟
since
|
𝑧 − 𝑧
0
ϛ − 𝑧
0
| =
|𝑧 − 𝑧
0
|
𝑟
= 𝑞 < 1
for all
ϛ𝜖𝛾
𝑟
. Uniform convergence is preserved under multiplication by a
continuous and hence bounded function
1
2𝜋𝑖
𝑓(ϛ)
. Therefore our term-wise
integration is legitimate and we obtain [1-7]
𝑓(𝑧) =
1
2𝜋𝑖
∫
𝑓(ϛ)𝑑ϛ
(ϛ − 𝑧
0
)
𝑛+1
(𝑧 − 𝑧
0
)
𝑛
𝐷
= ∑ 𝑐
𝑛
∞
𝑛=0
(𝑧 − 𝑧
0
)
𝑛
where
𝑐
𝑛
=
1
2𝜋𝑖
∫
𝑓(ϛ)𝑑ϛ
(ϛ − 𝑧
0
)
𝑛+1
𝐷
𝑛 = 0,1, …
Definition
:
The power series with coefficient given by is the Taylor series of
the function
𝑓
at the point
𝑧
0
(or centered at
𝑧
0
).
The Cauchy theorem implies that the coefficients
𝑐
𝑛
of the Taylor series
defined by do not depend on the radius
𝑟
of the circle
𝛾
𝑟
, 0 < 𝑟 < R.
Exercise 1.
Find the radius of the largest disk where the function
𝑧 ⁄ 𝑠𝑖𝑛 𝑧
may be represented by a Taylor series centered at
𝑧
0
= 0.
Exercice 2.
Let
f
be holomorphic in
𝐶.
Show that
(𝑎)
𝑓
is even if and only
if its Taylor series at
𝑧 = 0
contains only even powers;
(𝑏)
𝑓
is real on the real axis
if and only if
𝑓(𝑧̅) = 𝑓(𝑧)
̅̅̅̅̅̅
for all
𝑧 ∈ 𝐶
. We present some simple corollaries of
Theorem [7-13].
The Cauchy inequalities
. Let the function
𝑓
be holomorphic on a closed
disk
U
̅ = {|𝑧 − 𝑧
0
| ≤ 𝑟}
and let its absolute value on the circle
𝛾
𝑟
= 𝜕U
be
bounded by a constant
𝑀.
then the coefficients of the Taylor series of
𝑓
at
𝑧
0
satisfy
the inequalities
|𝑐| ≤ M 𝑟
𝑛
⁄
(𝑛 = 0,1, … )
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Proof
. We deduce from exercise 1 using the fact that
|𝑓(ϛ)| ≤ M
for all
ϛ𝜖𝛾
𝑟
:
|𝑐
𝑛
| ≤
1
2𝜋
M
𝑟
𝑛+1
2𝜋𝑟 =
M
𝑟
𝑛
.
Exercise 3.
Let
𝑃(𝑧)
be a polynomial in
𝑧
of degree
𝑛.
Show that if
|𝑝(𝑧)| ≤ M
for
|𝑧| = 1
then
|𝑝(𝑧)| ≤ m|𝑧|
n
for all
|𝑧| ≥ 1.
The Cauchy inequalities simply the interesting.
Theorem
.
(Liouville) If the function
𝑓
is holomorphic in the whole complex
plane and bounded then it is equal identically to a constant.
Proof.
According to theorem the function
𝑓
may be represented by a Taylor
series
𝑓(𝑧) = ∑ 𝑐
𝑛
𝑧
𝑛
∞
𝑛=0
in any closed disk
U
̅ = {|𝑧| ≤ 𝑟}, 𝑟 < ∞
with the coefficients that do not depend
on
𝑅.
Since f is bounded in
𝐶,
say
|𝑓(𝑧)| ≤ 𝑀
then the Cauchy inequalities imply
that for any
𝑛 = 0,1, ….
we have
|𝑐
𝑛
| ≤ M R
𝑛
.
⁄
We may take
𝑅
to be arbitrary
large and hence the right side tends to zero as
𝑅 → +∞
while the left side is
independent of
𝑅.
Therefore the two properties of a function-to be holomorphic and
bounded are realized simultaneously only for the trivial functions that are equal
identically to a constant [10-25].
Theorem
:
If a function
𝑓
is holomorphic in the closed complex plane
𝐶
then
it is equal identically to a constant
.
Proof.
If the function
𝑓
is holomorphic at infinity the limit
lim
𝑧 → ∞
𝑓(𝑧)
exists
and is finite. Therefore
𝑓
is bounded in a neighborhood
𝑈 = {|𝑧| > 𝑟}
of this point.
However,
𝑓
is also bounded in the complement
U
𝑐
= {|𝑧| ≤ 𝑟}
since it is
continuous there and the set
U
𝑐
is compact. Therefore
𝑓
is holomorphic and
bounded in
𝐶
and thus Theorem implies that is equal to a constant.
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Theorem
:
Claims that any function holomorphic in a disk may be
represented as a sum of a convergent power series inside this disk. We would like
to show now that, conversely, the sum of a convergent power series is a
holomorphic function. Let us first recall some properties of power series that are
familiar from the real analysis.
Lemma
:
If the terms of a power series
∑ 𝑐
𝑛
(𝑧 − 𝑎)
𝑛
∞
𝑛=0
are bounded at some point
𝑧
0
𝜖𝑐
, that is
|𝑐
𝑛
(𝑧
0
− 𝑎)
𝑛
| ≤ 𝑀, (𝑛 = 0,1,2 … )
Then the series converges in the disk
𝑈 = {𝑧 ∶ |𝑧 − 𝑎| < |𝑧
0
− 𝑎|}
. Moreover, it
converges absolutely and uniformly on any set
𝐾
that is properly contained in
𝑈.
Proof.
We may assume that
𝑧
0
≠ 𝑎
, so that
|𝑧
0
− 𝑎| = 𝜌 > 0
, otherwise the
set
𝑈
is empty. Let
𝐾
be properly contained in
𝑈,
then there exists
𝑞 < 1
so that
|𝑧 − 𝑎| 𝜌 ≤ 𝑞 < 1
⁄
for all
𝑧 ∈ 𝑈.
Therefore for any
𝑧 ∈ 𝐾
and any
𝑛 ∈ 𝑁
we
have
|𝑐
𝑛
(𝑧
0
− 𝑎)| ≤ |𝑐|𝜌
𝑛
𝑝
𝑛
. However, assumption implies that
|𝑐|𝜌
𝑛
≤ 𝑀
so that the
series is majorized by a convergent series
𝑀 ∑
𝑞
𝑛
∞
𝑛=0
for all
𝑧 ∈ 𝐾.
Therefore the
series converges uniformly and absolutely on
𝐾.
This proves the second statement
of this lemma. The first one follows from the second since any point
𝑧 ∈ 𝑈
belongs
to a disk
{|𝑧 − 𝑎| < 𝜌ʹ}
with
𝜌ʹ < 𝜌
, that is properly contained in
𝑈.
Theorem: (Abel) Let the power converge at a point
𝑧
0
∈ 𝐶.
Then this series
converges in the disk
𝑈 = {𝑧 ∶ |𝑧 − 𝑎| < |𝑧
0
− 𝑎|}
and, moreover, it converges
uniformly and absolutely on every compact subset of
𝑈
.
Proof
. Since the series converges at a point
𝑧
0
the terms
𝑐
𝑛
(𝑧
0
−
𝑎)
𝑛
converge to zero as
𝑛 → ∞
. However, every converging sequence is bounded,
and hence the assumptions of the previous lemma are satisfied both claims of the
present theorem follow from this lemma [12-25].
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The Cauchy – Hadamard formula
.
Let the coefficients of the power series
satisty
lim
𝑛→∞
𝑠𝑢𝑝|𝑐
𝑛
|
1 𝑛
⁄
=
1
R
,
with
0 ≤ 𝑟 ≤ ∞.
Then the series
∑ 𝑐
𝑛
(𝑧 − 𝑎)
𝑛
∞
𝑛=0
converges at all
𝑧
such that
|𝑧 − 𝑎| < 𝑅
and diverges at all
𝑧
such that
|𝑧 − 𝑎| > 𝑅.
Proof
. Recall that
𝐴 = 𝑙𝑖𝑚
𝑛→∞
𝑠𝑢𝑝𝑎
𝑛
if there exists a subsequence
𝛼
𝑛
𝑘
→ A
as
𝑘 → ∞,
and for any
𝜀 > 0
there exists
𝑛 ∈ 𝑁
so that
𝛼
𝑛
< 𝑎 + 𝜀
for all
𝑛 ≥
N.
This includes the cases
𝐴 = ±∞
. However, if
𝐴 = +∞
then condition is not
necessary, and if
𝐴 = −∞
then the number
𝐴 + ԑ
in condition is replaced by an
arbitrary number. It is shown in real analysis that
lim
𝑛→∞
𝑠𝑢𝑝𝛼
𝑛
exists for any
sequence
𝛼
𝑛
∈ 𝑟.
Let
0 < 𝑟 < ∞
, then for any
𝜀 > 0
we may find
𝑁
such that for all
𝑛 ≥
𝑁
we have
|𝑐
𝑛
|
1 𝑛
⁄
≤
1
R
+ 𝜀
. Therefore, we have [11-25]
|𝑐
𝑛
(𝑧 − 𝑎)
𝑛
| < {(
1
R
+ 𝜀) |𝑧 − 𝑎|}
𝑛
.
Furthermore, given
𝑧 ∈ 𝐶
such that
|𝑧 − 𝑎| < 𝑅
we may choose
ԑ
so small that we
have
(
1
𝑅
+ ԑ) |𝑧 − 𝑎| = 𝑞 < 1
. Then shows that the terms of the series are
majorized by a convergent geometric series
𝑞
𝑛
for
𝑛 ≥ 𝑁,
and hence the series
converges when
|𝑧 − 𝑎| < 𝑅.
Condition in the definition of
lim
𝑛→∞
𝑠𝑢𝑝𝛼
𝑛
implies that for any
𝜀 > 0
one
may find a subsequence
𝑐
𝑛
𝑘
so that
|𝑐
𝑛
𝑘
|
1 𝑛
𝑘
⁄
>
1
R
− 𝜀
and hence
|𝑐
𝑛
𝑘
(𝑧 − 𝑎)
𝑛
𝑘
| > {(
1
R
− 𝜀) |𝑧 − 𝑎|}
𝑛
𝑘
.
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Then, given
𝑧 ∈ 𝐶
such that
|𝑧 − 𝑎| > 𝑅
we may choose
ԑ
so small that we have
(
1
R
− 𝜀) |𝑧 − 𝑎| > 1
. Then implies that
|𝑐
𝑛
𝑘
(𝑧 − 𝑎)
𝑛
𝑘
| > 1
for all
𝑘
and hence the
𝑛 −
th term of the power series does not vanish as
𝑛 → ∞
so that the series diverges
|𝑧 − 𝑎| > R
.
We leave the proof in the special case
𝑅 = 0
and
𝑅 = ∞
as an exercise for
the reader.
Definition. The domain of convergence of a power series is the interior of
the set
𝐸
of the points
𝑧 ∈ 𝐶
where the series converges.
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