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Part 3

Manual

Kyiv

«NAU-druk Publishing»

2009

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UDC 51(075.8)

ББК В11я7

Н 65

Authors:

V.P. Denisiuk, L. I. Grishina, O. V. Karupu, T. A. Oleshko, V. V. Pakhnenko, V. K. Repeta

Reviewers:

O. P. Besklinska — candidate of science (physics and mathematics), associate professor (Kyiv National Linguistic University)

P.T. Kachanov — candidate of science (engineering), associate professor (State University of Information and Communication Technologies)

English language adviser

O. Yu. Kravchuk, associate professor

Approved by the Methodical

and Editorial Board of the National Aviation University (Minutes № 8/09 of 17.12.2009)

The manual fully corresponds to the educational program of higher mathematics for engineering universities. It contains basic theoretical information, examples of problems solution, class, home and self-test assignments.

The manual is intended for the second year students of all specialties

ISBN 978–966–598–612–6	© Denisiuk V. P., Grishina L. I.,
ISBN 978–966–598–612–6	Karupu O. V. and fo on, 2009
ISBN 978–966–598–613–3 (Part 3)	© NAU, 2009

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INTRODUCTION

This book offers the modular technology of studying higher mathematics course by the students of engineering specialties in the third term.

Each module represents logically compiled sections of the subject. The material of the third term is divided into two modules:

1.Series.

2.Multiple integrals. Line and surface integrals. Field theory. These modules contain several micromodules.

Every module begins with the general statements, which represent the

themes of the sections, basic concepts, and main tasks. Each micromodule covers:

1)basic theoretical information,

2)examples of problem solutions,

3)class and home assignments,

4)self-test assignments.

The theoretical part presents all necessary material for mastering the theme in question (lecture summary). References to the literature are applied to all the themes which enable students to learn theoretical material.

The practical part contains examples of problem solutions illustrating the theoretical material and problems with answers for class and home work.

Each micromodule is provided with self-test assignments containing a great number of problems to be done by the students in a written form.

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Моdule

SERIES

MODULE STRUCTURE

Micromodule 1. Number series. Principal concepts and definitions. Convergence. Properties of number series. The necessary condition for convergence. Tests for convergence of positive terms series (comparison test, D’Alembert’s test, Cauchy’s test). Alternating series. Leibniz test. Absolute and conditional convergence.

Micromodule 2. Functional series. General definitions. Uniform convergence. Weierstrass’ test. Properties of uniformly convergent series. Power series. Abel’s theorem. Interval and radius of convergence of a power series. Taylor’s and Maclaurin’s series. Expansion of a function in a series. Applications of the series.

Micromodule 3. Trigonometric Fourier series. Fourier coefficients. Dirichlet’s theorem. Fourier series for odd and even functions. Fourier series for 2π-and 2l-periodic functions. Fourier series for functions defined on a segment [0; l] or on arbitrary segment [a; b]. Complex form of Fourier series.

Micromodule 4. Fourier integral. Fourier transformation. Fourier integral for odd and even functions. Complex form of Fourier integral. Fourier transformation.

Micromodule 1

BASIC THEORETICAL INFORMATION. NUMBER SERIES

Principal concepts and definitions. Convergence. Properties of number series. The necessary condition for convergence. Tests for Convergence of positive terms series (comparison test, D’Alembert’s test, Cauchy’s test). Alternating series. Leibniz’s test. Absolute and conditional convergence.

Key words: number (numerical) series — числовий ряд; partial sum — частинна сума; to converge — збігатися; to diverge — розбігатися; remainder — залишок; harmonic series — гармонічний ряд; geometric progression —

геометрична прогресія; alternating series — альтерновний ряд, absolute convergence — абсолютназбіжність, conditional convergence — умовназбіжність.

Literature: [3, chapter 5, section 5.1 — 5.3], [9, chapter 9, § 1], [14, chapter 3, § 2], [15, chapter 13, section 13.1], [16, chapter 16, § 1—8], [17, chapter 4, § 13—15].

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1.1. Definitions. Sum of a Series

Consider a number sequence a1, a2 , a3 , ..., an , ... .

Definition. An expression of the form

∞
a1 + a2 + a3 + ...+ an + ... = ∑ an	(1.1)
n=1

is called a number series or simply a series. The numbers a1, a2 , ..., an are

called the terms (or members) of the series, and the number an is called the n-th

or general term of the series.

Definition. The sum of the finite number n of the terms of the series is called the nth partial sum of the series:

			n
	Sn = a1 + a2 + a3 + ... + an = ∑ ai .					(1.2)
			i=1
Consider the sequence {Sn } of partial sums of the series (1)
	S1 = a1 ,
	S2	= a1 + a2 ,
	.............
	Sn	= a1 + a2 + a3 + ... + an ,... .
Definition. If the sequence		{Sn }	has a finite limit,	lim Sn = S,	i. e.,	{Sn }
				∞
converges to S, then the limit is called the sum of the series ∑ an					and the se-
		∞		n=1
		∞	= S. If there is no	lim Sn , i.e., {Sn }
ries is said to be convergent: ∑ an			= S. If there is no	lim Sn , i.e., {Sn }		does
		n=1
		∞
not converge, then the series		∑ an	is said to be divergent and does not have
any sum.		n=1
any sum.

Example. Examine convergence of the series
			∞			(1.3)
	1+1+1+...+1+... = ∑1n					(1.3)
			n=1
We consider the nth partial sum of the series
	Sn =1+1+1+...+1= n.
Then lim Sn = lim n = ∞.
n→∞	n→∞

The given series diverges.

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Example. Examine convergence of the series

1−1+1−1+...+

∞

(1.4)

(−1)n−1 +... = ∑(−1)n−1

n=1

We consider the sequence of partial sums {Sn}

of the series (1.4)

S1 = 1,

S2 = 0,

S3 = 1, S4 = 0, ..., S2n−1 = 1, S2n = 0

and understand that lim Sn

does not exist.

n→∞

Then the given series diverges.

Consider a series known as a geometric series with a ratio q

∞

−1, (a ≠ 0).

a + aq + aq2 + aq3 + ... + aqn−1 + ... = ∑ aqn

n=1

The sum of n terms is

= a + aq + aq2 + aq3 + ... + aqn−1 =

a − aqn

−

aqn

(q ≠ 1).

1− q

− q

1− q

< 1 , then lim qn

= 0 , and so

n→∞

aqn

lim Sn = lim

−

− q

1− q

− q

n→∞

i.e., the series converges and its sum is

, or

− q

∑ aqn−1

a .

∞

n=1

1− q

then lim qn

> 1,

= ∞ and hence lim Sn

= ∞,

i. e., the series diverges.

n→∞

q = −1 we obtain the divergent series a − a + a − a + ... (a ≠ 0).

Its par-

a for odd n,

tial sum is Sn =

q =1

0 for even n.

a + a + a + a + ... ,

for which Sn = an, and

we will have the series

hence lim Sn = lim an = ∞,

i. e., the series diverges.

n→∞

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The geometric series

∞

−1,

(a ≠ 0)

∑ aqn

n=1

converges for

< 1 and diverges for

≥ 1 .

Definition. Consider the number series

(1.5)

1+ 1 + 1 + ... +

1 + ... = ∑ 1,

∞

n=1

known as the harmonic series.

We will prove that the series is divergent. We remember that

1 n

> e , for any n N.

Passing to the natural logarithm we obtain

n ln 1+

< 1

n + 1

> ln(n

+ 1) − ln(n).

and let’s write

Hence

n = 1 then

1 > ln 2 − ln1,

n = 2 then

> ln 3 − ln 2,

n = 3 then

> ln 4 − ln 3,

. . . . . . . . . . . . . . . . . . . . . . . . . . .

then

> ln(n + 1) − ln(n).

Let’s find the sum of the right and left hand parts

+ ... +

> (ln 2 − ln1) + (ln 3 − ln 2) + (ln 4 − ln 3) + ...

... + (ln(n + 1) − ln(n)) = ln(n + 1).

So we have Sn

> ln(n + 1) .

Passing to the limit we obtain lim Sn

= ∞, then the harmonic series is diver-

gent.

n→∞

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1.2. Operations on Series

Operations on number series may be deduced from the following theorems:

∞

Theorem 1 If the series a1 + a2 + ... + an + ... = ∑ an converges, then so does

n=1

the series obtained from it by discarding any finite number of terms in the beginning. Conversely, if a series obtained from the given series by discarding a finite number of terms in the beginning converges, then the given series converges.

Proof. In the partial sum

Sn = a1 + a2 + ... + an ,

of the series we denote by σk the sum of the first k(k < n) discarded terms. We get

Sn = a1 + a2 + ... + ak + ak +1 + ... + an = σk + Sn− k .

It follows that if there exists lim Sn , then there is lim Sn− k (k = const) and,
n→∞	n→∞
conversely, if there is lim Sn− k , then there exists lim Sn .
n→∞	n→∞
Remark. The resultant series a	ak +1 + ak + 2 + ... has the sum S = S − σk , S
being the sum of the original series.

∞

Theorem 2 Let the series a1 + a2 + ... + an + ... = ∑ an be convergent and

n=1

∞

λ ≠ 0 be a number. Then the series λa1 + λa2 + ... + λan + ... = ∑ λan converges and

n=1

			∞	∞
			∑ λan = λ ∑ an .			(1.6)
			n=1	n=1
				∞	∞
Proof. We write partial sums for the series ∑ an and					∑ λan :
				n=1	n=1
Sn	= a1 + a2 + ... + an and			σn = λa1 + λa2 + ... + λan .
It is obvious that σn		= λSn .
As stated, the series		∞				Sn , then from the
As stated, the series		∑ an converges, i. e., there exists lim				Sn , then from the
last equality there is		n=1			n→∞
last equality there is	lim σn , so that
	n→∞
		lim σn	= lim λSn	= λ lim Sn , i. e.,
		n→∞	n→∞	n→∞
			∞	∞
			∑ λan = λ ∑ an .
			n=1	n=1
8

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			∞		∞
Theorem 3	If the series ∑ an				and ∑ bn	converge, then their sum and dif-
			n=1		n=1
	∞		∞
ference, i. e., ∑ (an + bn )			and ∑ (an − bn ), converge, and
n=1			n=1
			∞		∞	∞
			∑ (an ± bn ) = ∑ an ± ∑ bn .
			n=1		n=1	n=1
Proof. Let Sn		= a1 + a2 + ... + an ,
			σn = b1 + b2 + ...+ bn
and		Cn = (a1 + b1 ) + (a2 + b2 ) + ... + (an + bn ),
			∞		∞	∞
be partial sums of the series ∑ an ,					∑ bn and	∑ (an + bn ),		respectively.
Clearly, σn + Sn = Cn .			n=1		n=1	n=1
Clearly, σn + Sn = Cn .
						∞		∞
As by the statement of the theorem the series ∑ an and ∑ bn , converge, i.e.,
there exist lim Sn						n=1		n=1
there exist lim Sn		and lim σn , it follows from the last equality, which holds for
n→∞		n→∞
all n, that there exists lim Cn , and
		n→∞		(Sn + σn ) = lim Sn + lim σn ,
		lim Cn	= lim	(Sn + σn ) = lim Sn + lim σn ,
		n→∞	n→∞			n→∞	n→∞
which is equivalent to
			∞		∞	∞
			∑ (an	+ bn ) = ∑ an + ∑ bn .
			n=1		n=1	n=1
								∞
In the same manner, we can prove the convergence of ∑(an − bn ) .
								n=1
Let introduce the concept of the remainder of a series. If we discard the first
					a1 + a2 + ... + an + an+1 +			∞
n terms in the convergent series					a1 + a2 + ... + an + an+1 +			+ an+ 2 + ... = ∑ an we
								n=1
								∞
will obtain the convergent series					an+1 + an+ 2 + ... ... + an+ k + ... = ∑ an+ k which is
								k =1
							Rn	∞
called the n-th remainder of the series and denoted by							Rn	= ∑ an+ k for a fixed n.
								k =1

∞

The original series can then be written as ∑ an = Sn + Rn .

n=1

Theorem 4 If S is the sum of the series, then the remainder will be

Rn = S − Sn for any n = 1, 2, ...

and lim Rn = 0.

n→∞

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1.3. Tests for Convergence of Series

Cauchy’s criterion for convergence of a sequence allows formulating the general criterion for convergence of a number series.

Theorem 5 (Cauchy’s criterion). A necessary and sufficient condition for a

∞

number series ∑ an to converge is that for any number ε > 0 there exists a

n=1

number N = N (ε) such that for any n > N there holds

an + an+1 +...+ an+ p

< ε

(1.7)

for all p = 0, 1, 2, ... .

∞

In terms of the partial sums Sn+ p and Sn−1 of ∑ an , we can write (1.4) as

n=1

Sn+ p − Sn−1 < ε.

From the Cauchy’s criterion follows the necessary test for convergence of a number series.

Theorem 6 (The necessary condition for convergence of a series). If series

∞

∑ an converges, then its general term tends to zero, i.e.

n=1

lim an = 0.

n→∞

∞

Proof. Let series ∑ an

n=1

n−1

where Sn−1 = ∑ ak , but an

k =1

So that lim an = 0.

n→∞

converges. Then exists lim Sn = S	and lim Sn−1 = S,
n→∞	n→∞
= Sn − Sn−1.

	Corollary. If lim an	∞
	Corollary. If lim an	is not equal to zero or does not exist, then ∑ an di-
	n→∞	n=1
	verges.	n=1
	verges.	∞
		∞
	Proof. Suppose that	∑ an converges. Then by Theorem 6, there must be
		n=1

lim an that is equal to zero. Our assumption has led us to a contradiction, hence

n→∞

it is wrong. Therefore, the series is divergent.

Remark. Theorem 6 gives a necessary condition for a series to converge,

which is not, however, sufficient, i.e., the condition lim an = 0 may be satisfied

n→∞

∞

for a divergent series ∑ an as well.

n=1

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