Книжная полка Сохранить
Размер шрифта:
А
А
А
|  Шрифт:
Arial
Times
|  Интервал:
Стандартный
Средний
Большой
|  Цвет сайта:
Ц
Ц
Ц
Ц
Ц

Continuous mathematics: theory and practice. Limit of a sequence and limit of a function, continuous and differentiable functions

Покупка
Новинка
Основная коллекция
Доступ онлайн
239 ₽
В корзину
The textbook gives a brief description of theoretical material on the studied sections of the course. There are given and analyzed numerous examples illustrating various types of tasks and methods for solving them. At the end of each chapter, there are given tasks for independent solution. All these tasks are provided with answers. The tutorial contains a lot of illustrations. The given textbook is intended to help the students of the training program 02.03.02 "Computer Science Informatics and Information Technologies” in studying the practical part of the course "Continuous Mathematics” in the first semester.
Абрамян, А. В. Abramyan, A. V. Continuous mathematics: theory and practice. Limit of a sequence and limit of a function, continuous and differentiable functions : textbook / A. V. Abramyan ; Southern Federal University - Rostov-on-Don ; Taganrog : Southern Federal University Press, 2020. - 183 p. - Текст : электронный. - URL: https://znanium.com/catalog/product/1894465 (дата обращения: 09.12.2022). – Режим доступа: по подписке.
Фрагмент текстового слоя документа размещен для индексирующих роботов. Для полноценной работы с документом, пожалуйста, перейдите в ридер.
UDC
BBC
     А16

Published by decision of the educational-methodical commission of the I. I. Vorovich Institute of Mathematics, Mechanics, and Computer Science of the Southern Federal University (minutes No. 11 dated November 10, 2020)

Reviewers:
Doctor of physical and mathematical sciences, professor of the Department of Applied Mathematics of the Platov South-Russian State Polytechnic University (NPI), Professor A.E. Pasenchuk;

Candidate of physical and mathematical sciences, associate professor of the Department of Algebra and Discrete Mathematics of the Southern Federal University,
Associate professor A.V. Kozak

     Abramyan, A. V.
А16       Continuous mathematics: theory and practice. Limit of a sequence and
     limit of a function, continuous and differentiable functions : textbook / A. V. Abramyan ; Southern Federal University - Rostov-on-Don ; Taganrog : Southern Federal University Press, 2020. - 183 p.
          ISBN
          The textbook gives a brief description of theoretical material on the studied sections of the course. There are given and analyzed numerous examples illustrating various types of tasks and methods for solving them. At the end of each chapter, there are given tasks for independent solution. All these tasks are provided with answers. The tutorial contains a lot of illustrations.
          The given textbook is intended to help the students of the training program 02.03.02 “Computer Science Informatics and Information Technologies” in studying the practical part of the course “Continuous Mathematics” in the first semester.

                                                UDC
                                                BBC
ISBN
                                     © Southern Federal University, 2020
                                     © Abramyan A.V., 2020
Contents
Preface ................................................................. 6
Chapter 1. Introduction.................................................. 7
  1.1. Mathematical induction............................................ 7
  1.2. Problems ........................................................ 16
Chapter 2. The limit of a sequence ..................................... 19
  2.1. Definition of convergent sequence................................ 19
    2.1.1. Sequences with the finite limit.............................. 19
    2.1.2 Infinitely large sequences.................................... 25
    2.1.3. Sequences tending to +ж.......................................29
    2.1.4. Sequences tending to —ж.......................................32
  2.2. Calculation of the limit of a sequence........................... 34
    Properties of limits of sequences................................... 34
  2.3. Problems ........................................................ 43
Chapter 3. Limits of functions.......................................... 45
  3.1. Definition of the limit of a function ........................... 45
    3.1.1. Limit of a function in the case x ^ a, a E R..................45
    3.1.2. Limit of a function in the case x ^ +ж........................50
    3.1.3. Limit of a function in the case x ^ —ж........................53
    3.1.4. Limit of a function in the case x ^ ж.........................56
    3.1.5. Infinitely large functions as x ^ a...........................57
    3.1.6. Functions tending to +ж as    x ^ a...........................58
    3.1.7. Functions tending to —ж as    x ^ a...........................59
    3.1.8. One-sided limits ............................................ 60
    3.1.9. Infinitely large functions as x ^ ж...........................61
  3.2. Finding the limits .............................................. 62
    3.2.1. Limit of quotient of polynomials .............................. 63
    3.2.2. Limits of irrational functions ................................ 67
    3.2.3. Usage ofequivalences......................................... 70
    3.2.3. Limit of u.(x)v⁽x')............................................80
  3.3. Continuity and discontinuity points ............................. 85
  3.4. Problems ........................................................ 88
Chapter 4. Derivatives of functions of one variable..................... 92
  4.1. Derivatives of explicit functions ............................... 92
  4.2. Higher derivatives ............................................. 102
  4.3. Derivatives of inverse functions................................ 106
  4.4. Derivatives of functions defined parametrically ................ 108
  4.5. Derivatives of implicit functions............................... 111
  4.6. Derivative of a function defined in polar coordinates .......... 113
  4.7. Problems ....................................................... 114
Chapter 5. Applications of derivatives...................................117
  5.1. L’Hospital’s Rule.................................................117
  5.2. Taylor’s formula ................................................ 124
    5.2.1. Little о notation.............................................124
    5.2.2. Expansion of a function by Taylor’s formula ................. 126
    5.2.3. Approximate calculations .................................... 133
  5.3. Taylor's formula for evaluating indeterminate forms.............. 138
  5.4. Problems ........................................................ 142
Chapter 6. Plotting functions ...........................................145
  6.1. Preliminary information.......................................... 145
    6.1.1. A criterion for monotonicity of a function .................. 145
    6.1.2. Extremum of a function ...................................... 145
    6.1.3. Convex functions ............................................ 146
    6.1.4. Tangent to the graph of a function .......................... 146
    6.1.5. Inflection points ........................................... 146
    6.1.6. Asymptotes .................................................. 147
  6.2. Graphing functions .............................................. 147
  6.3. Problems ........................................................ 164
Chapter 7. Reference information ....................................... 165
  7.1. Formulas for exponentiation...................................... 165
  7.2. Abbreviated multiplication formulas.............................. 166
  7.3. Arithmetic progression .......................................... 166
  7.4. Geometric progression............................................ 166
  7.5. Trigonometric formulas .......................................... 167
    7.5.1. Values and relations with inverse functions.................. 167
    7.5.2. Reduction formulas........................................... 168
    7.5.3. Period of sine and cosine ................................... 168
    7.5.4. Period of tan and cot ....................................... 168
    7.5.5. Sum and difference of trigonometric functions................ 168
    7.5.6. Functions of sum and difference of arguments ................ 169
    7.5.7. Double argument functions ................................... 169
    7.5.8. Degree reduction formulas ................................... 169
  7.6. Equivalent functions ............................................ 169
  7.7. Table of derivatives ............................................ 169
  7.8. Taylor’s formula ................................................ 170
Answers to the problems ................................................ 171
References...............................................................184
Preface




     The textbook gives a brief description of theoretical material on the studied sections of the course. There are given and analyzed numerous examples illustrating various types of tasks and methods for solving them. At the end of each chapter, there are given tasks for independent work. All these tasks are provided with answers. The tutorial contains a lot of illustrations.
     An extensive set of examples and problems included in the textbook allows the reader after reading the examples on the selected topic to consolidate the studied techniques solving problems on the same topic. For a number of examples, several solutions are given. Some formulations of examples and problems were taken from the classical problem book [3].
     The textbook consists of seven chapters. Chapter 1 is an introduction. It provides a detailed description of various versions of the method of mathematical induction, which is widely used in proving mathematical statements and solving practical problems. Chapter 2 is devoted to the study of the concept of a numerical sequence and its limit, as well as to various methods for calculating limits of numerical sequences. Chapter 3 introduces the concept of the limit of a function and various ways of calculating it; in addition, it discusses the concept of a continuous function and gives a classification of discontinuity points. Chapter 4 introduces the concept of derivative of a function and proposes various methods for its calculation for functions defined in explicit form, parametrically, in polar coordinates and implicitly. Chapters 5 and 6 are devoted to applications of the concept of derivative. Chapter 5 discusses the use of derivatives for evaluating indeterminate forms in the limits and expansion of functions using Taylor’s formula. Chapter 6 is devoted to geometrical applications of derivatives. Chapter 7 contains the background information needed to study the sections in the course.
     The given textbook is intended to help the students of the training program 02.03.02 “Theoretical Computer Science and Information Technologies” in studying the practical part of the course “Continuous Mathematics” in the first semester.
Chapter 1. Introduction




            1.1. Mathematical induction


    Mathematical induction is the method to prove that some property P(n) holds for every natural number n. The proof consists of two steps:
   1. prove that the statement P(1) is true (the base case);
   2. prove that if the statement P(k) holds for some arbitrary natural number k, then the property P(k + 1) holds (inductive step).
The hypothesis that the statement P(k) holds in the inductive step is called the induction hypothesis or inductive hypothesis.


     Example 1.1.      Using the method of mathematical induction prove that for an arbitrary natural n the following equality holds
1³ + 2³ + 3³ + - + n³ = П2⁽П + ¹⁾².
                                                4
     Solution. We denote

Sₙ = 1³ + 2³ + 3³ + - + n³.
Then the assertion P (n) states that
                                   n²(n + 1)²
Sⁿ =      4     .
     1) The base case: we prove validity of the statement P (1):
51 = 1³ = 1,
1²(1 + 1)² _
                                   4      = .
Conclusion: the statement P (1) is true.
     2) Inductive step: suppose that for some к > 1 the property P(k) holds: k²(k + 1)²
fc =      4     .
We are going to prove that the statement P(k + 1) holds:
                              _(k + 1)²(k + 2)²
5fc+1 =         4        .
Let us compare the sums Sₖ and Sₖ₊₁:
Sₖ = 1³ + 2³ + 3³ + - + k³, Sₖ₊₁ = l³ + 2³ + 3³ + ••• + к³ + (k + 1)³.

=sₖ

We see that there is valid the equality
Sₖ+! = sₖ+ (fc + 1)³.

Using the induction hypothesis that Sₖ = k ⁽k⁺¹⁾ , we obtain: -                                  4      4

5fc+i =

k²(k + 1)² 4

+ (к + 1)³
(к + 1)2 •

, к² + 4к + 4
(к + 1)2--------------

⁽Т ⁺⁽fc ⁺ ¹⁾)

The statement Р(к + 1) is proved.

(к + 1)2 (к + 2)2 4

     Example 1.2.      Using the method of mathematical induction prove that for an arbitrary natural n the following equality holds
1 • 2 • 3 + 2 • 3 • 4 + - + n • (n + 1) • (n + 2) n(n + 1)(n + 2)(n + з)


4

     Solution. Let’s denote

Sₙ = 1 • 2 • 3 + 2 • 3 • 4 + - + n • (n +1) • (n +2).
Then the assertion P(ri) states that
                             n(n + 1)(n + 2)(n + 3)
Sn =------------₄------------.
     1)      The base case (we prove validity of the statement P(1)). The left-hand side of the equality under consideration in the case n =1 equals

51 = 1 • 2 • 3 = 6,

the right-hand side of the proved equality in the case n = 1 equals
                              1 • 2 • 3 • 4
-----=----= 6.
4
Conclusion: the statement P (1) is true.
     2)     Inductive step: suppose that for some к > 1 the property P(k) holds: k(k + 1)(k + 2)(k + 3)
sₖ =------------₄-----------.
We will prove that the statement P(k + 1) is valid:
                          (k + 1)(k + 2)(k + 3)(k + 4)
s.₊₁ =---------------₄--------------.
Comparing the sums Sₖ and Sₖ₊₁:
Sₖ = 1 • 2 • 3 + 2 • 3 • 4 + - + k- (k + 1) • (k + 2), Sₖ₊₁ = 1 • 2 • 3 + 2 • 3 • 4 + - + k- (k + 1) • (k + 2)

=sₖ
+ (k + 1) • (k + 2) • (k + 3),

we see that there holds the equality
$k+i = $k ⁺ ⁽k ⁺ 1) • ⁽k ⁺ ²⁾ • ⁽k ⁺ ³⁾.
Substituting in the right-hand side of the last equality the relation
                              k(k + 1)(k + 2)(k + 3)
SK =-------------₄------------,
which is valid according the induction hypothesis, we obtain:
к(к + 1)(к + 2)(к + 3)
Sₖ₊i =  -------   +    -----+ + (к + 1)(к + 2)(к + 3)
4
= ⁽к + 1) • ⁽к + ²⁾ • ⁽к + ³⁾ • (^ + 1)
(к + 1)(к + 2)(к + 3)(к + 4)
=              4               ■
The statement Р(к + 1) is proved.


     Example 1.3. Using the method of mathematical induction prove that for an arbitrary natural n the following equality holds
12     22              n²            n(n + 1)
FT ⁺ FT ⁺ ”’ ⁺ (2n - 1) • (2n + 1) = 2(2n + 1)■
     Solution. We denote
               ₑ _ 12      22 ₍      ₍        n²
Sⁿ = 1~3 ⁺ FT ⁺ ”’⁺ (2n - 1) • (2n + 1)■
Then the assertion P(n) states that the following equality is valid:
n(n + 1)
Sⁿ = 2(2n + 1) ■
     1)      The base case (we prove validity of the statement P(1)). The left-hand side in the case n = 1 takes the value

_ 12 _ 1
                                ⁵¹ = 1"T = 3 ' the right-hand side in the case n = 1 takes the value 1(1 + 1)                                    1

2(2 + 1)     3 ■
Conclusion: the statement P(1) is true.
     2)      Induction hypothesis: suppose that for some к > 1 the statement P (k) is valid:

_ _ k⁽k + 1) k = 2(2k + 1).

Inductive step: we prove that then the statement P (k + 1) is valid:



(k + 1)(k + 2) 2(2k + 3)

We compare the sums Sₖ and Sₖ₊₁:
              ᵣ _ 12     22 ₍     ₍        k²
sk = 173 ⁺  3-5 ⁺ ^ ⁺ (2ₖ — 1) • (2ₖ + 1) ,
              e _ 12      ²2   .  .        k2
sk+1 = 173 ⁺ 375 ⁺ - ⁺ (2ₖ — 1) • (2ₖ + 1)

=sₖ
.      (fc + 1)²
⁺ (2k + 1) • (2k + 3) '
and get that there holds the equality
e _ e ,           (k +1)²
Sk⁺¹ - Sₖ ⁺ (2ₖ + i) ^ (2ₖ + 3) .

Substituting in the right-hand side of the last equality the relation Sₖ — which is true by the induction hypothesis, we get:
                   ᵣ ₌ к(к + 1)             (к + 1)²
⁵fc⁺¹ 2(2k + 1) ⁺ (2k + 1) • (2k + 3)
             _k + 1 zfc к + 1\ _k + 1 2k² + 5k + 2
             ⁻   2k + 1 ^ \2 ⁺ 2k + 3/ ⁻ 2k + 1 ’ 2(2k + 3)
к + 1 (2k + 1) • (k + 2)     (k + 1)(k + 2)
⁻  2k + 1      2(2k + 3)     ⁻    2(2k + 3)    ’
The statement P(k + 1) is proved.


fc(fc+i) 2(2fc+1),

     Example 1.4.       Using the method of mathematical induction prove that for an arbitrary natural n the following equality holds
1³ — 2³ + 3³ — 4³ + —+ (2n — 1)³ — (2n)³ - — n²(4n + 3).
     Solution. We denote
             Sₙ - 1³ — 2³ + 3³ — 4³ + - + (2n — 1)³ — (2n)³.
Then the statement P(n) says that the following equality holds:
Sₙ - — n² (4n + 3).
     1)      We check the base case (validity of the statement P(1)). The left-hand side of the equality being proved in the case n - 1 takes the value
                             S1 - 1³ — 2³ - —7, the right-hand side for n - 1 equals
(—1)² • (4 • 1 + 3) - —7.
Conclusion: the statement P(1) is true.
     2)      Induction hypothesis. Suppose that for к > 1 there is valid the statement Р(кУ.
Sₖ - —k²(4k + 3).
      Inductive step. We must prove that under this assumption there is valid the assertion P (k + 1):
$k+1 ⁻ —⁽k ⁺ ¹⁾²⁽⁴^ ⁺ ⁷⁾.
Let us compare the consecutive sums Sₖ and Sₖ₊₁:
             Sₖ - 1³ — 2³ + 3³ — 4³ + - + (2k — 1)³ — (2k)³,
Sₖ₊₁ - 1³ — 2³ + 3³ — 4³ + - + (2k — 1)³ — (2k)³
=Sk
+ (2k + 1)³ — (2k + 2)³.
We see that there holds the equality
Sk+1 = Sk + (2k + 1)³ - (2k + 2)³.
Substituting in the right-hand side of the last equality the relation ^fc = — k²⁽4k + 3), which is true by the induction hypothesis, we find that
$k+i = —k²(4k + 3) + (2k + 1)³ — (2k + 2)³ = —4k³ — 15k² — 18k — 7 = —(к + 1)² (4k + 7).
The statement P(k + 1) is proved.


    Example 1.5.    Using the method of mathematical induction prove that for an arbitrary natural n the following equality holds
(1—1).(1—1)........(1_____l_ ) = ⁿ⁺².
⁽¹  4) ⁽¹   9)     ⁽¹  (n + 1)²⁾ 2(n + 1).
    Solution. We denote
Aⁿ=⁽¹—4H¹—9)...........⁽¹—c^W
Then the statement P(ri) says that the following equality is valid:
n + 2
Aⁿ = 2(n + 1).
    1)     We check the base case (validity of the statement P (1)). The left-hand side of the equality under consideration in the case n =1 takes the following value:

Л1

= 1

                       13 — = —.
                       4  4'

the right-hand side of the equality in the same case equals
1+2                                   3

2(1 + 1)   4.
Conclusion: the statement P(1) is true.
     2)     Induction hypothesis. Suppose that for some к > 1 the statement P(k) holds:
л _ k + 2
Ak = 2(k + 1).
Inductive step. We must prove validity of the statement P(k + 1):
к + 3
Лк⁺¹ = 2(k + 2).
Comparing the following expressions
Ak = ⁽¹ —     ⁽¹ — 9......⁽11— (Г+1)²⁾'    !
Ak⁺¹ =⁽¹ — 4H¹ — 9)' - ^⁽¹ — a+ 1)2) •⁽¹ — (fc+W
                             =^k
we see that there holds the equality
k+2        , J .
—----, we obtain
2(k+1)

Ak⁺¹ Ak ^ ⁽¹ (к + 2)²)’
Taking into account the induction hypothesis saying that Aₖ that
                           _ k+2       (        1    \
Ak⁺¹ = 2(k + 1) ^ ⁽¹ - (k + 2)²⁾
_ к + 2 к² + 4k + 3
= 2(k + 1) ^ (k + 2)2
(k + 2) (k + 1)(k + 3)
= 2(k + 1) ^   (k + 2)2
к + 3
                                = 2(k + 2)' The statement P(k + 1) is proved.

     Example 1.6. Prove that the sum of the cubes of three consecutive positive integers is divisible by 9.
     Solution. We denote
Sₙ = n³ + (n + 1)³ + (n + 2)³.
Let P(n) be the statement “Sₙ is divisible by 9”.
     1) We check the base case (validity of the statement P(1)):
S1 = 1³ + 2³ + 3³ = 1 + 8 + 27 = 36 = 9 • 4.
Conclusion: the statement P(1) is valid (S1 is divisible by 9).
     2)      Induction hypothesis. Suppose that for some к > 1 there is valid the assertion P(k): “Sₖ is divisible by 9”.
     Inductive step. We must prove that under this assumption there is valid the assertion P(k + 1): “Sₖ₊₁ is divisible by 9”.
Let’s find a connection between Sₖ and Sₖ₊₁. We take into account that Sₖ+i = (к + 1)³ + (к + 2)³ + (к + 3)³
= (к³ + (к + 1)³ + (n + 2)³) + (к + 3)³ - k³
                             =sₖ
= $k ⁺ ⁽fc ⁺ ³⁾³ — ^³,
Removing the parentheses in the expression (к + 3)³, using the formula for the cube of the sum, we obtain:
(k + 3)³-k³
= k³ + 9k² + 27k + 27 - k³
= 9(k² + 3k + 3).
So, we get that Sₖ₊₁ = Sₖ + 9(k² + 3k + 3). In the right-hand side of the last equality the both terms are divisible by 9. Hence Sₖ₊₁ is divisible by 9. The statement P(k + 1) is proved.
Example 1.7.       Using the method of mathematical induction prove that for an arbitrary natural n the value of the expression 4ⁿ + 15n — 1 is divisible by 9.
     Solution. We denote aₙ = 4ⁿ + 15n — 1, let P(n) be the following assertion: “aₙ is divisible by 9”.
     1) Let’s check the base case (validity of the statement P(1)):
a1 = 4 + 15 — 1 = 18.
Conclusion: the statement P(1) is valid (a1 is divisible by 9).
     2)      Induction hypothesis. Suppose that for some к > 1 the statement P (k) is valid: “aₖ is divisible by 9”.
     Inductive step. We are going to prove that then there is valid the statement P(k + 1): “aₖ₊₁ is divisible by 9”.
     Let’s find a connection between aₖ and aₖ₊₁:
aₖ+i = 4k⁺¹ + 15(k + 1) — 1
= 4 • 4k + 15k + 14.
From the equality aₖ = 4k + 15k — 1 we get 4k = aₖ — 15k + 1. Then
aₖ+i = 4 • (aₖ — 15k + 1) + 15k + 14.
Removing the brackets and collecting similar terms, we get:
aₖ+i = 4aₖ — 45k + 18
= 4aₖ — 9(5k — 2).
From the inductive hypothesis P(k) (“aₖ is divisible by 9”) and the last equality we obtain validity of the statement P(k + 1) (“aₖ₊₁ is divisible by 9”).

     Example 1.8.       Using method of mathematical induction prove that for an arbitrary natural n the value of the expression 5ⁿ⁺³ + 11³ⁿ⁺¹ is divisible by 17.
     Solution. We denote aₙ = 5ⁿ⁺³ + 11³ⁿ⁺¹, let P(n) be the following assertion “aₙ is divisible by 17”.
     1) Let’s check the base case (validity of the statement P(1)):
a1 = 5¹⁺³ + 11³⁺¹ = 15266 = 17 • 898.
Conclusion: the statement P(1) is true.
     2)      Induction hypothesis. Suppose that for some к > 1 the assertion P(k): “aₖ is divisible by 17” is valid.
     Inductive step. We must prove that the statement P(k + 1): “aₖ₊₁ is divisible by 17” is valid.
     Let’s find a connection between aₖ and aₖ₊₁:
aₖ₊₁ = 5k⁺⁴ + 11³k⁺⁴
= 5k⁺⁴ + 11³^ 11³k⁺¹.
From the equality aₖ = 5k⁺³ + 11³k⁺¹ we get that 11³k⁺¹ = aₖ — 5k⁺³, then aₖ+1 = 5fc⁺⁴ + 11³ • (aₖ — 5k⁺³).
Removing the brackets and collecting similar terms, we get:
aₖ₊₁ = 11³ • aₖ + 5k⁺⁴ — 11³ • 5k⁺³
= 11³ • aₖ + 5k⁺³ • (5 — 11³)
  • document_id: 415246
  • product_id: 1894465
  • ins_time: 2022-06-09 00:56:50
  • upd_time: 2022-06-09 00:56:50
  • upp_upd_date: 2022-06-08
  • Full PDF: WARN Путь не доступен (не определен) /mnt/znanium_fullpdf/booksfull/done/1894/1894465.pdf
  • PDF pages: WARN Количество страниц документа (183) не соответствует физическому наличию (184). Путь /mnt/resources/resources/1894/1894465/pdf
  • XML pages: WARN Количество страниц документа (183) не соответствует физическому наличию (184). Путь: /mnt/resources/resources/1894/1894465/xml
  • text *.idx: OK
  • Full text: OK /mnt/resources/resources/1894/1894465/txt/1894465.txt
  • Оглавления: OK Путь /mnt/resources/resources/1894/1894465/txt/1894465.toc.txt
Доступ онлайн
239 ₽
В корзину