Mathematical analysis- 9786010416933

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Al-FARABI KAZAKH NATIONAL UNIVERSITY

G. Abduakhitova U. Kusherbayeva G. Rzaeva

MATHEMATICAL ANALYSIS-1 Educational manual

Almaty Qazaq university 2016

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UDC 517 (075.8) LBC 22.161 я 73 A 22 Recommended for publication by the Academic Council of the Faculty of Mechanics and Mathematics, Editorial-Publishing Council, and Educational and methodical association of the Republic educational-methodical council of higher and postgraduate education of Ministry of education and science of the Republic of Kazakhstan on the basis of Al-Farabi of Kazakh National University (Protocol №1 from 07.10.2015) Rewievers: Doctor of mathematical and physical sciences, professor A.B. Tungatarov Doctor of mathematical and physical sciences, professor M.N. Kalymoldaev Doctor of mathematical and physical sciences, professor B.D. Koshanov

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Abduakhitova G. Mathematical analysis-1: educational manual / G. Abduakhitova, U. Kusherbayeva, G. Rzaeva. – Almaty: Qazaq university, 2016. – 134 p. ISBN 978-601-04-1693-3 Educational manual is designed for the students of all specialties of faculty of mechanics and mathematics. The purpose of this educational manual is to familiarize students with the fundamental concepts of mathematical analysis. The following educational manual contains theoretical topics from syllabus and examples. Graphs and illustrations accompany examples and exercises. Учебное пособие предназначено для студентов всех специальностей механико-математического факультета. Цель этого учебного пособия ознакомить студентов с основными понятиями математического анализа. Темы, изложенные в учебном пособии, соответствуют основному учебному плану дисциплины «Математический анализ-1». Каждый параграф содержит справочный материал, набор типовых примеров с решениями и задачи для самостоятельной работы. Примеры и упражнения сопровождаются графиками и иллюстрациями.

UDC 517 (075.8) LBC 22.161 я 73 ISBN 978-601-04-1693-3

© Abduakhitova G., Kusherbayeva U., Rzaeva G., 2016 © Al-Farabi KazNU, 2016

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INTRODUCTION

This textbook is designed to be completely accessible to the beginning calculus. The importance of visualization cannot be overemphasized in developing students’ understanding of mathematical concepts. For that reason, some graphs, illustrations accompany the examples and exercises. In this textbook an emphasisis made on problem solving which offers students the opportunity to investigate a variety of special topics that supplement a textmaterial. «Mathematical analysis-1» is devoted to the study of single calculus. While applications from the sciences, engineering, and economics are often used to motivate or illustrate mathematical ideas, the emphasis is on the three basic concepts of calculus: limit, derivative and integral. The discipline «Mathematical analysis-1» considers theory of sequences of real numbers, limit of function of one variable, continuous of function, derivative of function and analysis of function using the plotting of the graph of function’s derivative. Textbook consists of two chapters: 1. Introduction to mathematical analysis and sequences of real numbers; 2. Theory of functions of one variable. Since Mathematical analysis is a basis of higher mathematics, knowledge obtained during this discipline serves as a fundament for the students while studying other subjects such as Mathematical analysis-2, Mathematical analysis-3, linear algebra, differential geometry, theory of functions of complex variables, functional analysis, theory of probability and mathematical statistics, discrete mathematics, differential equations and mathematical physics. Moreover, there are various areas of application of the knowledge obtained during this course such as mechanics, economics and other spheres of science. 3

Chapter 1

THE REAL NUMBERS

Theoretical questions. 1. Real numbers. The need to extend the set of rational numbers. The set of real numbers. Classification of subsets of the set of real numbers. Exact bounds of set of numbers. 2. Principle of Mathematical induction. Absolute value of real numbers, and its properties. 3. The theory of sequences. The numerical sequences. Classification of sequences of numbers. 4. Limit of a sequence. Convergent sequences of numbers and their properties. 5. Infinitely small and large sequence, and its properties. Supremum of a set and infimum of a set. 6. The structure of an arbitrary numerical sequence: subsequence, partial limit, upper and lower limits. 7. Monotonic sequences. A number е. 8. Cauchy criterion of the existence of the limit.

The real number system The real number system (which we will often call simply the reals) is first of all a set on which the operations of addition and multiplication are defined so that every pair of real numbers has a unique sum and product, both real numbers, with the following properties. 1. and (commutative laws). 2. and (associative laws). 3. (distributive law). 4. There are distinct real numbers 0 and 1 such that and for all .

4

5. For each there is a real number – such that and if , there is a real number such that . A set on which two operations are defined so as to have properties (1)–(5) is called a field. The real number system is by no means the only field. The rational numbers (which are the real numbers that can be written as , where and are integers and also form a field under addition and multiplication. The simplest possible field consists of two elements, which we denote by 0 and 1, with addition defined by 0+0=1+(-1)=0, 1+0=0+1=1

(1)

and multiplication defined by (2)

The Order Relation The real number system is ordered by the relation 0 and xn 

,

then the sequence

 xn n1 

there is

such that

is called infinitely large

quantity. Example. Prove that xn  (1)n n is infinitely large quantity. Proof:  > 0. We should find 13

such that xn 

.

xn    (1)n n    n   .

Thus,

n     1 .

Definition. If for every positive number

xn   ,

such that

and

there is

 xn n1 is 

then the sequence

called

infinitely small quantity (or infinitesimal). 1 Example. Prove that xn  is an infinitesimal. n! Proof:  > 0 . We should find such that xn   . xn   

1 1  .   n! n!

1 as n!

We can write

1 1 1 1    n 1   . n! 1  2  3  ...  n 1  2  2  ...  2 2 The inequality

2n 1 

1   can be written 2n 1

1



 n  1  log 2

 n  log 2   1 .    1

1

Definition. If for every positive number such that

and

xn  a   , then the sequence

there is

 xn n1 

is called

convergent sequence and a number а is called the limit of sequence and we write a  lim xn . n 

Example. Prove that lim xn  1 , if xn  n 

14

n , n 1

 n  1,2,...

Proof:  > 0 . We should find xn  1   

such that xn  1   .

1 n 1   1      n 1 n 1 n 1 1 1  n 1   n  1 





n ,  n  1,2,... has limit 1. n 1 Now, we should prove some useful limits. №1. Prove that nk lim n  0 , a  1 , n  a where k is an integer. Therefore, the sequence xn 

We should prove that the sequence xn 

Proof: Let

an infinitesimal, i.e. we should find

such that

(4) nk is an

xn   

k

n  . an 1) If k  0 then (4) is true.  2) If k  0  we get indeterminate form. So 

0

nk  an

k

 n  n k     k n  . k n a  a 

nk

  k

an

k

Therefore, we can find a number lity. Since a  1 

k

which satisfies last inequa-

a 1 k a 1

15



k



a 1 ,

we can use Newton’s binomial: k

a  1   n



k





n k

a

n





n

a 1   1 n 



k



k



2

n





a 1 

n  n  1

n

a 1 

n  n  1 2

k



a 1

2



k

n  n  1 2!





k



2

a  1  ...

2

a 1 

2

 n  1 

k



a 1

2

k.

By solving last inequality we obtain n 1  k





2 k



a 1

2

n k





2 k



a 1

2

 1.

  nk 2 x  This gives n   . Consequently, is an 2 n 2 an  k  k a  1  infinitesimal. Remark. Any exponential function increases faster than power function. №2. Prove that lim n  q n  0 , q  1 . (5)





n 

Proof: q  1  1  q  1 . 1) For q  0 the result of (5) is obvious.

n  q n  lim 2) 0  q  1 : lim n  n 

k  1,

n 1   q

a

n

0

1 1 . q 16

3) 1  q  0 : q   q  2

lim n  q n  lim  1 n  q   1 lim n  q  0. n

n

n 

n

n

n 

n 

n Therefore, xn  nq is an infinitesimal.

2n  0. n  n ! Proof: For   0 we should find a number 2n 2n 0   . We can write as n! n!

№3. Prove that lim

such that

2n 2  2  2  ...  2 2 2 2 2   2    ...   2   n! 1  2  3  ...  n 3 4 n 3

n2

 .

The inequality can be written 2   3

n2



 2

 . 3 2

 n  2  log 2

(Note, the inequality has been reversed.) We have found the number   n  log 2   3 .  3 2

2n is an infinitesimal. n! Remark. It is possible to find a number expression:

Consequently, xn 

from the following n 3

2n 2  2  2  ...  2 2 2 2 22   2    ...   2      , n! 1  2  3  ...  n 3 4 n 3 4 and so on. It is important that the base of exponent must be less than 1. 17

№4. Prove that an 0 n  n !

(6)

lim

Proof: 1) For a  1 , the result of (6) is obvious. 2) We consider case a  1. a) Let a  1 . For   0 we should find a number such that n a 0    . For this purpose integer part of we denote by : n!

a  k  k  a  k  1 . a n a  a  ...  a a a a a a a a ak  a       ...     ...    n! 1  2  ...  n 1 2 3 k k 1 k  2 n k!  k 1

nk

 .

From last inequality we obtain  a     k 1

nk



k ! k ! k !  n  k  log a k  n  log a k  k . k a k 1 a k 1 a

(Note the base of logarithmic is less than 1). Consequently,  k !  n  log a k   k  1 .  k 1 a  б) Let a  1  a   a :

a a a an n n lim  lim  1    1 lim 0 . n  n ! n  n  n ! n! n

n

Remark. Factorial increases faster than exponential function.

18

№5. Prove that lim n n  1 , n  1 . n 

Proof: We show that the sequence xn  n n is convergent. It is certainly true that n  1  n n  1 . By definition of convergent sequence, for every positive number we should find a number such that 0  n n  1   . For this purpose number is expressed as

n

 n n

 1  

n



n



n 1  

n

then use the Newton’s binomial:

n  1  



n







n

n 1   1  n 



n

n  n  1 2





n

n

n 1 



n





n 1 

n  n  1 2

2

n 1  n 



n  n  1

 n

2 n



n 1





n



2

n  1  ...

2

2

n 1 

2 . n 1

This gives n

n 1 

2  n 1

(7)

2 2 2   2  n  1  2  n  2  1. n 1   Consequently, 2 n   2   2 .  

Therefore, the sequence xn  n n ,

19

 n  1,2,...

converges to 1.

№6. Prove that lim n a  1 , a  0 . n 

(8)

Proof: We should show that the sequence xn  n а converges. 1) If a  1 then the expression (8) is true. 2) Let a  1 . This gives 1 n

0  a 1    a    1, a    1 . 1 1  log a    1  n  . n log a    1 n

n

Consequently,

  1 n    1.  log a    1  1 2 1  1. n  1 1 n a

3) 0  a  1 : lim n a  lim n 

Thus, the sequence xn  n а ,  n  1,2,... converges to 1. №7. Prove that log n lim a  0 , a  1 . n  n

(9)

log a n іs n infinitesimal. For this purpose we should find for every a log a n number that satisfies 0    . This inequality can be n 1 written as log a n   or n Proof: We are going to show that the sequence xn 

20

log a n n   . Since a  1 , then n n  a . We can use formula (7) and write n

n 1 

2 / n 1



n

2  a  1 . n 1

n 1 

By solving the inequality we obtain 2 2 2   a  1  n  1  2 n 1  a  1



Consequently, n  

2

  a  12 

 n

a

2 

 1

2

1.

  2 .  

Remark. Power function increases faster than logarithmic function. Limit of sequences We know that the sequence n

 1 xn  1    e.  n  n  1 Now we are going to show the sequence yn  1    n decreasing and converges to a number е . Proof: Since

n 1

 is

n 

n 1

 1 1   yn 1 n 1 1 n  1 n3  n 2  n  1 , n        3 2 1 n n n yn 1  n n n n n 1   1  1 2 1  1      2 n 1  n 1  n 1

we have yn  yn 1 .This confirms that the sequence is decreasing. 21

n 1

n

n

 1  1  1  1 n  N , yn  1    1   1   > 1    xn ,  n  n  n  n  1  1  lim yn  lim  xn 1     lim xn  lim 1    e 1  e . n  n  n  n   n   n

From this expressions n  N , x n < е < y n we conclude that

 1 1  1   < е < 1  n  n    n

n 1

.

We can therefore write  1  1 n ln 1   < 1 <  n  1 ln 1   .  n  n

This gives: n  N ,

1  1 1 < ln 1   < . n 1  n n

Definition. The sequence is said to be increasing if , nondecreasing if , decreasing if , nonincreasing if . A sequence that satisfies any of these conditions is called monotonic. If the sequence  xn n 1 has only positive numbers, i.e., n  N 

xn > 0 then monotonicity of sequence can be shown by conditions: x the sequence is increasing if n 1 > 1, the sequence is nondecreasing xn x x if n 1  1, the sequence is decreasing if n 1 < 1, and the sequence xn xn is nonincreasing if xn 1  1. xn 22

Example. By theorem about monotonic and bounded sequence 10 11 n  9 prove that the sequence xn  . ... converges. 1 3 2n  1 Solution. 1 We show that the sequence is monotonic: 10 11 n  9 n  10 . ... . xn 1 1 3 2n  1 2n  1 n  10    1, 10 11 n  9 xn 2n  1 . ... 1 3 2n  1

if n  10  2n  1, n  9. For n  9 the sequence decreases, that is xn1  xn .

2  The sequence is bounded:

0  xn  max x1 , x10  By theorem about monotonic and bounded sequence we conclude 10 11 n  9 that the sequence xn  . ... converges. 1 3 2n  1 Cauchy’s Convergence Criterion Theorem (Cauchy’s Convergence Criterion). The sequence x n n1 converges  if it is fundamental   >0 n  N :

n  n , p  N  xn  p  xn <  . Example. Use Cauchy’s Convergence Criterion to prove that the cos1! cos 2! cos n! sequence xn  converges.     1 2 23 n  n  1 Solution. To verify that this statement conforms to the Cauchy’s Convergence Criterion , we must show that for each there exists a positive integer such that if then 23

xn  p  xn <  . We have   0,   N . So we introduce the

difference xn  p  xn as

xn    xn  

cos  n  1!

  

cos  n   !

 n  1 n  2   n    n    1 cos  n  1! cos  n   !      n  1 n  2   n    n    1



x   

1

 n  1 n  2 

  

1   n    n    1

 1  1   1 1  1  1             n 1 n  2   n  2 n  3   n   n   1 

1 1 1 1      n 1  , n 1 n   1 n 1 

1 This gives n    . We have found a number . Consequently,   a given sequence is fundamental , therefore  lim xn   . n 

Example. Find inf xn , sup xn , limxn , lim xn for the sequence

xn  1  2  1

n 1

 3   1

n n 1 2

 n  1,2,... .

Solution. By Bolzano-Weierstrass theorem from any bounded sequence it is possible to choose convergent subsequence. We can choose subsequences:

24

1 n  4k  3,

x4k 3  1  2  3  6   6, k  2  n  4k  2,

x4k 2  1  2  3  4  4, k  3 n  4k  1,

x4k 1  1  2  3  0   0, k  4  n  4k ,

x4k  1  2  3  2   2, k 

We construct the set of partial limits:

Limxn  4,0,2,6. The set of partial limits consists of four elements. Therefore, inf xn  4, sup xn  6, limxn  4, lim xn  6.

Example. Find limxn , lim xn for the sequence

xn 

2n  1  n sin , nN. n 1 2

Solution. We consider the sequence as the product of two factors: 2n  1 first factor is the sequence of real numbers, it converges to 2. n 1 n A function sin takes values: 2

n sin

n 2

1

2

3

4

5

6

7

8…

1

0

-1

0

1

0

-1

0…

25

From this sequence we can choose four subsequences:

n  4k  3 : x4 k 3  n  4k  2; : x4 k  2 

n  4k  1 : x4 k 1 

2  4k  3   1 4k  3  1

k 

2  (4k  2)  1  0  0, k  4k  2  1

2  4k  1  1

n  4k ; x4 k 

 1  2,

4k  1  1

  1  -2, k 

2  4k  1  0  0. k  4k  1

Therefore, the set of partial limits consists of four elements:

Limxn  2;0;2;0 , This gives, limxn  sup Limxn  2 , n 

limxn  inf Limxn  2 .

Examination questions: 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12.

What are the properties of the real number system? What is a field of the real number system? What is an ordered field? How triangle inequality looks like? Explain. What is the principle of mathematical induction? What are the operations over sets (union, intersection and symmetric difference of sets)? Explain. What is infimum of sets? What is supremum of sets? What are the infinitely small and large sequences? What is the limit of sequences? Explain Cauchy’s Convergence Criterion. Explain Bolzano-Weierstrass theorem.

26

 Independent work 1. Use mathematical induction to prove following assertions: 1.1. 1  2  2  3     n  1  1.2. 13  23  33    n3  1.3. 14  24  34    n4 

 n  1 n  n  1 , 3

n2  n  1 4

2

,

n  n  1 2n  1  3n 2  3n  1 30

,

1.4. 1  4  2  7  3  10      n  3n  1  n  n  1 , 2

1.5.

1 1 1 n ,      1 5 5  9  4n  3 4n  1 4n  1

1.6. 1  22  2  32     n  1 n2 

n  n2  1  3n  2  12 n  n  1

,

1.7.

12 22 n2 ,      1 3 3  5  2n  1 2n  1 2  2n  1

1.8.

1 1 1 1 n       , 45 56 67  n  3 n  4  4  n  4 

1.9. 1  2  3  2  3  4    n  n  1 n  2  

n  n  1 n  2  n  3

1  n2  1  1    1.10. 1  1      1  , 2   4  9    n  1  2n  2

1.11. n3  5n , n  N , divisible by 6. 1.12. 2n3  3n2  7n , n  N , divisible by 6. 1.13. n  n , n  N , divisible by 30. 5





2 4 1.14. n n  1 , n  N divisible by 60.

1.15. 10  18n  28 , n  N , divisible by 27. n

1.16. 1.17. 1.18. 1.19. 1.20.

5n 3  113n 1 , n  N , divisible by 17. 62n  3n  2  3n , n  N , divisible by 11. 9n 1  18n  91 , n  N , divisible by 18. 72 n  1 , n  N , divisible by 48. 52n 1  3n  22n 1 , n  N , divisible by 19.

27

4

,

2. Find

 3n 2  2.1. A   2   2n  1nN n2  2.2. A    4  n  1 nN n 1  2.3. A     3n  1nN

 2n 2  2.4. A   2   n  1nN n3  2.5. A     5n  1nN  n2  2.6. A   2   2n  1nN n 1  2.7. A    2  n  1 nN  4n 2  2.8. A   2   3n  1nN  n2  2.9. A   2   5n  1 nN n2 2.10. A     3n  1nN 3. Solve inequalities by both analytical and graphical methods. 3.1. x  1  x  1  1 3.2. x  3  x  3  3 3.3. x  1  x  1  1 3.4. x  1  x  1  3 3.5. x  2  x  1 3.6. x  3  x 3.7. x  2  x  2  6 3.8. x  1  x  1  1 3.9. 2 x  1  x  1 3.10. x  x  2

28

4. Prove: 4.1. A / ( A / B)  A

B

4.2. ( A / B) ( B / A)  ( A B) / ( A B) 4.3. ( A / B) / C  A / ( B C ) 4.4. ( A / B) C  ( A C) / ( B C) 4.5. A ( B / C)  ( A B) / C 4.6. ( A C) / B  ( A / B) C 4.7. A ( B C)  ( A B) C 4.8. A ( B C)  ( A B) C 4.9. ( A B) C  ( A C) ( B C) 4.10. ( A B) C  ( A C) ( B C) 5. Prove by definition:

3n n  10 1  0 ; b) lim  n  n ! n  5n  1 5 5n  0 ; b) lim  3n  53   5.2. а) lim n  n ! n  2 n 1  3n  1 5.3. a) lim n  0 ; b) lim n  4 n  1  3n n3 n 1 1  5.4. a) lim n  0 ; b) lim n  3 n  7 n  3 7 2 n log 8 n 5.5. a) lim n  0 ; b) lim 0 n  6 n  n n3 2 5.6. a) lim n  0 ; b) lim 5n  5    n  5 n  7n n  10 1 0  ; b) lim 5.7. a) lim n  n ! n  2n  1 2 n3 n 1 lim 0 lim  5.8. a) ; b) n  4 n n  2n  1 2 5.1. a) lim

5.9. а) lim

n 

2n n  1 ; b) lim 25  1 n 2n

5.10. а) lim

n 

log8 n 3n 3 0  ; b) lim n  n 2n  1 2 3

n n2  1  1 ; b) lim n  0 n  9 n  n 2

5.11. а) lim

29

6. Prove by Cauchy’s Convergence Criterion (6.1-6.10) 6.1. xn 

sin1 sin 2 sin n      1 2 2  3 n  n  1

6.2. xn 

cos1 cos 2 cos n      1 3 2  4 n  n  2

cos1! cos 2! cos n!  2    n  2 2 2 sin1 sin 2 sin n  2    n  6.4. xn  3 3 3 1 1 1      6.5. xn  1 2 2  3 n  n  1 6.3. xn 

1 1 1     n  4 42 4 cos 2 cos 4 cos 2n  2     6.7. xn  2 2 2n 1 1 1 6.8. xn  1    ...  2! 3! n! cos cos 2 cosn       6.9. xn  5 52 5n 6.6. xn 

 1  1 1     1 2 2  3 n  n  1 n 1

6.10. xn 

7. Applying theorem about monotonic and bounded sequence, prove that following sequence converges. n! 7.1. xn   2n  1!! 7.2. x1  1, xn 1  12  xn

1 1 1  3    n  2 2 3 n  2n !!

7.3. xn  1  7.4. xn 

 2n  1!!

7.5. x1  0, xn 1  6  xn 7.6. x1  8, x2 

8 11 8 11 3n  5  ,..., xn    ...  1 7 1 7 6n  5

30

7.7. xn 

2n n!

 2  2   2  7.8. xn  1   1   ...1   2  3  3  4    n  1 n  2  

8. Find inf xn , sup xn , lim xn , lim xn n 

n 

2  3   n  3 n 1  8.2. xn   1  4   n   1 n 8.3. xn   1  2   n  8.1. xn   1

n 1

4 n 8.4. xn   1  5   n  3 n 1  8.5. xn   1 1    n 8.6. xn   1

n 1

8.7. xn   1

3 2

n

8.8. xn   cos

3   2   n 

2n  1 n 2 n   3 

n



31

TEST

Select three correct answers from eight V2 0 0 0 0 0 0 0 0

V2 0 0 0 0 0 0 0

№1 For following sets sup A  2 is true: A   1,0

A  0,2 A   ,2 A   1,

A   1,1 A   1,0 A   1,2

A  1,3

№2 For following sets inf A  2 is true: A   2,0

A  0,1

A   ,1

A   2,

A  1,2

A   1,0 A   2,1

№3 V2

n

For sequence xn  1  1  is true, that: 

0 0

n

n

 1 lim 1    e n   n xn - is unbounded above

32

0 0

xn - is increasing sequence xn - is unbounded sequence

0

x1  2

0

xn - is decreasing sequence xn - diverges

0 0

n

 1 lim 1    e n 0  n

№4 V2

For sequence xn  1  1  

n 1

n

n 1

0

 1 lim 1   n 0  n

0

xn - is unbounded above xn - is increasing sequence xn - is unbounded sequence

0 0 0 0 0 0

is true, that:

e

x1  4

xn - is decreasing sequence xn - diverges  1 lim 1   n   n

n 1

e

№5 V2

 2 For sequence xn   1n 1   is true, that:  n

0

limxn  lim xn  1

0

limxn  limxn

0

limxn  1

0

limxn  1

33

0

limxn  1

0

limxn  2

0

lim x n 

0

V2 0 0 0 0 0 0 0 0

V2 0 0 0 0 0

n

2

There is no limit №6 Convergence sequence can be: Is infinitesimal Is infinitely large Is decreasing Is increasing unbounded Is fundamental sequence bounded There is a unique limit №7 The infinitely large sequence are: n 3  n  2n 2 xn  n2  1 n3 xn  n n 1 xn  2 n n n 1 xn   1 n xn  4 n

0

2 xn    3

0

 1  xn     5n  2  3n xn  2n

0

n

n

34

V2 0 0 0 0 0

№8 Convergence sequence are: n 3  n  2n 2 xn  n2  1 n3 xn  n n 1 xn  2 n n n 2  5n  1 xn  n2 n xn  4

0

4 xn    3

0

 5n  2  xn     5n 

0

xn 

n

n

2  3n 3 2n

35

Chapter 2

FUNCTIONS OF ONE VARIABLE

Theoretical questions. 1. Functions of one variable. The limit of the function. Functions that have limit at the point and their properties. Cauchy’s Criterion. 2. One sided limits. First and second remarkable limits. 3. Limit of composition functions. Limits of infinitely small and large functions. Comparison of infinitesimals. 4. Continuous of function. Continuous functions and their local and global properties. Points of discontinuity. Classification of points of discontinuity. The comparison of functions, "o", "O", "~" symbols. 5. Uniform Continuity. Cantor’s Theorem. 6. The derivative of function. One-sided derivatives. Differentiation of function: differentiable of a function of one variable at the point; the derivative at the point; differential and its geometric meaning; mechanical meaning of the derivative; Differential. 7. Geometric significance of derivative and differential. Differentiation of composite and inverse functions. Differentiation of functions given parametrically. 8. Invariance of the form of first differential. Basic theorems of differentiation of functions. Main properties of differentiation of functions. 9. Higher order derivatives and differential. Leibniz formula. 10. Fermat's, Rolle's, Lagrange's, Cauchy's theorems. 11. Taylor’s formula. The remainder term of the Taylor's formula. Maclaurin’s formula and main expansions of elementary functions. L’Hospital’s Rule. 12. Research of function by using a derivative. Extremum of the function. Concavity and convexity of the graph. The inflection points of the graph of function. Asymptotes of the function; The scheme of the plotting.

Functions Definition of a function. If to every value of a variable , which belongs to some collection (set) , there corresponds one and only one finite value of the quantity then is said to be a function

36

(single-valued) of or a dependent variable defined on the set , is the argument or independent variable. The fact that is a function of is expressed in brief form by the notation If to every value of , belonging to some set E there corresponds one or several values of the variable then is called a multiplevalued function of defined on From now on we shall use the word “function” only in the meaning of a single-valued function, if not otherwise stated. The domain of a function. The collection of values of for which the given function is defined is called the domain of this function. In the simplest cases, the domain of a function is either a closed interval which is the set of real numbers that satisfy the inequalities or an open interval which is the set of real numbers that satisfy the inequalities Also possible is a more complex structure of the domain of function. Example. Determine the domain of the function

y

1 x2  1

.

Solution. The function is defined if

that is, if Thus, the domain of the function is a set of two intervals: and Inverse function. If the equation may be solved uniquely for the variable that is, if there is a function x such that

then the function x or, in standard notation, is the inverse of Obviously, that is the function is the inverse of (and vise versa).

37

In the general case, the equation defines a multiplevalued inverse function such that for all that are values of the function Example. Determine the inverse of the function (1) Solution. Solving equation (1) for

we have

and

x

log 2 (1  y ) . log 2 2

(2)

Obviously, the domain of the function (2) is Composite and implicit functions. A function of defined by a series of equalities where u etc., is called a composite function. A function defined by an equation not solved for the dependent variable is called an implicit function. For example, defines as an implicit function of The graph of a function. A set of points in the -plane, whose coordinates are connected by the equation is called the graph of the given function. Graphs of functions are mainly constructed by marking a sufficiently dense net of points where and by connecting the points with a line that takes account of intermediate points.

The hyperbolic functions Certain combinations of the exponentials and occur so frequently in mathematical applications that they are given special names. 38

The hyperbolic functions: X , Y  R ; y  f ( x) : X  Y , where X is a domain of the function and Y a is range of the function. The function f ( x) establishes one-to-one correspondence between the elements of sets X ,Y . x  f 1 ( y) : Y  X , where f 1 ( y) is inverse function. We define the hyperbolic functions: e x  e x 1. y  chx  : R  1;   is hyperbolic cosine. 2 y  chx Properties of hyperbolic cosine: 1) is continuous on the real line ( . 2) 3) is an even function. The graph is therefore symmetric about the y-axis. 4) The function decreases on (−∞, 0] and increases on [0,∞). e x  e x 2. y  shx  : R  R is hyperbolic sine. 2

y  shx The function establishes one-toone correspondence between the domain and range of function,

 x  Arshy  ln( y  1  y 2 ) : R  R. Properties of hyperbolic sine: 1) is continuous on the real line (s . 2) 3) is an odd function and thus the graph is symmetric about the origin.

39

4)

, the hyperbolic sine increases everywhere. shx e x  e x 3. y  thx   : R  (1;1) is hyperbolic tangent. chx e x  e x y  thx, x  Arthy The function establishes one-toone correspondence between the domain and range of function, 1 1 y  x  Arthy  ln : (1;1)  R . 2 1 y Properties of hyperbolic tangent: is continuous on the real line (t .

1) 2) 3) is an odd function and thus the graph is symmetric about theorigin. 4) , the hyperbolic sine increases everywhere. chx e x  e x 4. y  cthx   : R \ 0  R \  1;1 - is hyperbolic shx e x  e x cotangent. y  cthx

Properties of hyperbolic cotangent: 1) is a point of discontinuity ( 2) is an odd function and thus the graph is symmetric about the origin. 3) The function decreases everywhere.

Limit of a function

y  f ( x) : X  R; a, b  R , where X is a domain of the function and a is limit point of X .

40

Definition 1. We say that lim f ( x)  b , if   0 there exists a

 ( )  0 such that if

x a

then

(Figure 1). f(x)

b+ε b b-ε

a - δ(ε)

a

a + δ(ε)

Figure 1

Definition 2. (The right-hand limit of ). We write lim f ( x)  b to indicate that as approaches x a  0

the right, approaches , i.e., if conventionally The number

, then we write f (a  0)  lim f ( x) is x a  0

called the limit from the right of the function. Definition 3. (The left-hand limit of ). We write lim f ( x)  b to indicate that as x a  0

the left, approaches conventionally

i.e. if The number

from

approaches

from

, then we write f (a  0)  lim f ( x) is

called the limit from the left of the function. Definition 2. (According to Cauchy, i.e., “ limit from the right).

x a  0

definition of

  0    ( )  0: x  X , a  x  a    f ( x)  b   . Definition 3. (According to Cauchy, i.e., “ limit from the left).

definition of

  0    ( )  0: x  X , a    x  a  f ( x)  b   . 41

Definition . ( “neighborhood” definition of limit from the right).

V (b) V (a) : x  X  V (a)  f ( x) V (b). Definition . ( “neighborhood” definition of limit from the left).

V (b) V (a) : x  X  V (a)  f ( x) V (b). Definition. The right-hand limit and left-hand limit are called one-sided limits. For a full limit to exist, both one-sided limits have to exist and they have to be equal. Example. Show that lim(2 x  1)  3. x 2

Solution. Finding a    ( ). (Figure 2). Let   0 . We seek a number

such that if

then

.

What we have to do first is establish a connection between and

.

The connection is evident: (1) To make less than we need to make which we can accomplish by making .

 2

This suggests that we choose   . Showing that the  ( ) “works”. If and, by (1)

42

, then

Figure 2

 Remark. In Example1 we chose   , but we could have chosen 2 any positive number δ less than the one we have chosen which is 1  . In general, if a certain “works” for a given , then any δ less 2 than will also work. Example. Show that lim x 2  9. x 3

Solution. Finding a    ( ). (Figure 3). Let   0 . We seek a number

such that if then

.

What we have to do first is establish a connection between and . The connection between and can be found by factoring: and thus (2) 43

At this point, we need to get an estimate for the size of for close to 3. For convenience, we will take within one unit of 3. If

, then 2