Methods of Solving Calculus Problems for UNM-Gallup Students

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The Educational Publisher Columbus, 2017 METHODS OF SOLVING CALCULUS PROBLEMS FOR UNM-GALLUP STUDENTS โ–  Constantin Dumitrescu โ–  Florentin Smarandache โ– 

Transcript of Methods of Solving Calculus Problems for UNM-Gallup Students

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The Educational

Publisher Columbus, 2017

METHODS OF SOLVING CALCULUS PROBLEMS

FOR UNM-GALLUP STUDENTS

โ–  Constantin Dumitrescu โ–  Florentin Smarandache โ– 

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โ–  C. Dumitrescu โ– F. Smarandache โ– 

Methods of Solving Calculus Problems for UNM-Gallup Students

โ–  Theory and Exercises โ– 

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Methods of Solving Calculus ProblemsFor UNM-Gallup Students

(AdSumus Cultural and Scientific Society)

The Educational Publisher

1313 Chesapeake Ave.

Columbus, Ohio 43212, USA

Email:

[email protected]

Editura Ferestre

13 Cantemir str.

410473 Oradea, Romania

ISBN 9781599735237

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โ–  Constantin Dumitrescu โ–  Florentin Smarandache โ– 

Methods of Solving

Calculus Problems

For UNM-Gallup StudentsTheory and Exercises

The Educational

Publisher Columbus, 2017

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Motto:

Love is the purpose, the sacred command of existence.

Driven by it, eagles search each other in spaces,

Female dolphins brace the sea looking for their grooms

Even the stars in the skies cluster in constellations.

(Vasile Voiculescu)

ยฉ Educational Publisher, Columbus,2017.

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Contents Foreword ..................................................................................................... 9

I. Sets Theory ...................................................................................... 11

Operations with sets ........................................................................... 11

1. The Intersection ......................................................................... 12

2. The Union ................................................................................... 12

3. The Difference............................................................................ 13

4. The Complement ....................................................................... 13

5. The Symmetrical Difference .................................................... 14

6. The Cartesian Product ............................................................... 14

Methods for Proving the Equality of Two Sets ............................. 15

Properties of the Characteristic Function ....................................... 16

Exercises ............................................................................................... 17

II. Functions ......................................................................................... 31

Function definition ............................................................................. 31

Examples .......................................................................................... 34

The inverse of a function ................................................................... 35

Examples .......................................................................................... 37

Observation ..................................................................................... 38

The Graphic of the Inverse Function ............................................. 38

Methods to show that a function is bijective ................................. 39

1. Using the definition ................................................................... 39

2. The graphical method ................................................................ 39

3. Using the theorem ...................................................................... 40

4. Using the proposition 1 ............................................................ 41

5. Using the proposition 2 ............................................................ 41

Exercises ............................................................................................... 42

Monotony and boundaries for sequences and functions ............. 48

Example ........................................................................................... 50

Methods for the study of monotony and bounding ..................... 53

Exercises .......................................................................................... 54

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III. The limits of sequences and functions .................................. 61

The limits of functions ....................................................................... 61

Exercises .......................................................................................... 66

The limits of sequences ...................................................................... 71

Examples .......................................................................................... 72

Calculation methods for the limits of functions and sequences . 73

Joint methods for sequences and functions ............................... 73

Specific methods for sequences ................................................. 108

IV. Continuity and derivability ..................................................... 146

Continuity ........................................................................................... 146

(A) Methods for the study of continuity .................................. 146

(B) Types of discontinuity points .............................................. 149

(C) The extension through continuity ....................................... 149

(D) The continuity of composite functions ............................. 150

Derivability ......................................................................................... 154

(A) Definition. Geometric interpretation. Consequences ..... 154

(B) The derivation of composite functions .............................. 155

(C) Derivatives of order ๐‘›........................................................... 158

(D) The study of derivability ...................................................... 158

(E) Applications of the derivative in economics ..................... 162

V. Fermat, Rolle, Lagrange, Cauchy Theorems ............................ 169

(A) Fermatโ€™s theorem ....................................................................... 169

1. Enunciation ............................................................................... 169

2. Geometric and algebraic interpretation ................................ 169

(B) Rolleโ€™s theorem ........................................................................... 172

1. Enunciation ............................................................................... 172

2. Geometric and algebraic interpretation ................................ 172

3. Consequences ........................................................................... 173

Methods for the study of an equationโ€™s roots.............................. 177

1. Using Rollesโ€™s theorem (its algebraic interpretation) ......... 177

2. Using Darbouxโ€™s properties ................................................... 177

3. Using Rolleโ€™s sequence ............................................................ 177

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4. The graphical method. ............................................................. 178

5. Vietteโ€™s relations ....................................................................... 178

6. Using the theorem of the average ......................................... 179

(C) Lagrangeโ€™s Theorem .................................................................. 181

1. Enunciation ............................................................................... 181

2. Geometric and algebraic interpretation ................................ 181

3. The corollary of Lagrangeโ€™s Theorem .................................. 181

(D) Cauchyโ€™s Theorem ..................................................................... 186

1.Enunciation ................................................................................ 186

2. Geometric and algebraic interpretation ................................ 186

VI. Equalities and Inequalities ..................................................... 189

Equalities ............................................................................................ 189

Inequalities ......................................................................................... 190

Method 1. Using Lagrangeโ€™s or Cauchyโ€™s theorem. ................ 190

Method 2: The method of the minimum ................................. 192

Method 3: (Inequalities for integrals, without the calculation of

the integrals) .................................................................................. 193

Method 4: Using the inequality from the definition of convex

(concave) functions ...................................................................... 195

VII. Primitives .................................................................................. 204

Connections with other notions specific to functions ................ 204

(A) Methods to show that a function has primitives .................. 209

(B) Methods to show that a function doesnโ€™t have primitives .. 210

Logical Scheme for Solving Problems ...................................... 221

VIII. Integration ................................................................................ 222

Connections between integration and other notions specific to

functions ............................................................................................. 227

Methods to show that a function is integrable ............................. 229

Methods to show that a function is not integrable .................... 229

Examples ........................................................................................ 229

Exercises ........................................................................................ 235

References ............................................................................................... 240

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Foreword

We rejoice expressing our gratitude towards all those who,

better or less known, along the years, have helped us to get to

this particular book. And they are many, many indeed whom

have helped us! Some provided suggestions, others, ideas; at

times, we have struggled together to decipher a certain elusive

detail; other times we learned from the perceptive or sometime

naรฏve questions of our interlocutors.

The coming into being of this book represents a very

good example of altruism, kindness and selflessly giving, of the

journey, made together, on the road to discovering lifeโ€™s beauty.

He who thinks not only of his mother or father, of

himself or his family, will discover an even bigger family, will

discover life everywhere, which he will start to love more and

more in all its manifestations.

Mathematics can help us in our journey, stimulating and

enriching not just our mental capacity, our reasoning, logic and

the decision algorithms, but also contributing in many ways to

our spiritual betterment. It helps shed light in places we

considered, at first, fit only for an empirical evolution โ€“ our own

selves.

We offer this succession of methods encountered in the

study of UNM-Gallup calculus to students and instrcutors, to

higher education entry examination candidates, to all those

interested, in order to allow them to reduce as many diverse

problems as possible to already known work schemes.

This textbook is for Calculus I, II and III students at the University of New Mexico, Gallup Campus.

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We tried to present in a methodical manner, advantageous

for the reader, the most frequent calculation methods

encountered in the study of the calculus at this level.

In this book, one can find:

methods for proving the equality of sets,

methods for proving the bijectivity of functions,

methods for the study of the monotonic sequences

and functions,

mutual methods for the calculation of the limits of

sequences and functions,

specific methods for the calculation of the limits of

sequences,

methods for the study of continuity and

differentiation,

methods to determine the existence of an equationโ€™s

root,

applications of Fermatโ€™s, Rolleโ€™s, Lagrangeโ€™s and

Cauchyโ€™s theorems,

methods for proving equalities and inequalities,

methods to show that a function has primitives,

methods to show that a function does not have

primitives,

methods to show that a function is integrable,

methods to show that a function is not integrable.

We welcome your suggestions and observations for the

improvement of this presentation.

C. Dumitrescu, F. Smarandache

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I. Sets Theory

A set is determined with the help of one or more properties

that we demand of its elements to fulfill.

Using this definition might trick us into considering that any

totality of objects constitutes a set. Nevertheless it is not so.

If we imagine, against all reason, that any totality of objects is

a set, then the totality of sets would form, in its own turn, a set that

we can, for example, note with ๐‘€ . Then the family ๐œŒ(๐‘€) of its

parts would form a set. We would thus have ๐œŒ(๐‘€) โˆˆ ๐‘€.

Noting with card๐‘€ the number of elements belonging to ๐‘€,

we will have:

๐‘๐‘Ž๐‘Ÿ๐‘‘ ๐œŒ(๐‘€) โ‰ค ๐‘๐‘Ž๐‘Ÿ๐‘‘๐‘€.

However, a theorem owed to Cantor shows that we always

have

๐‘๐‘Ž๐‘Ÿ๐‘‘ ๐‘€ < ๐‘๐‘Ž๐‘Ÿ๐‘‘ ๐œŒ(๐‘€).

Therefore, surprisingly maybe, not any totality of objects

can be considered a set.

Operations with sets DEFINITION: The set of mathematical objects that we

work with at some point is called a total set, notated with ๐‘‡.

For example,

drawing sets on a sheet of paper in the notebook, the

total set is the sheet of paper;

drawing sets on the blackboard, the total set is the set

of all the points on the blackboard.

It follows that the total set is not unique, it depends on the

type of mathematical objects that we work with at some given point

in time.

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In the following diagrams we will represent the total set

using a rectangle, and the subsets of ๐‘‡ by the inner surfaces of this

rectangle. This sort of diagram is called an Euler-Venn diagram.

We take into consideration the following operations with

sets:

1. The Intersection ๐ด โˆฉ ๐ต = {๐‘ฅ โˆˆ ๐‘‡ โ”‚ ๐‘ฅ โˆˆ ๐ด ๐‘Ž๐‘›๐‘‘ ๐‘ฅ โˆˆ ๐ต}

Because at a given point we work only with elements

belonging to ๐‘‡, the condition ๐‘ฅ โˆˆ ๐‘‡ is already implied, so we can

write:

๐ด โˆฉ ๐ต = {๐‘ฅ โ”‚ ๐‘ฅ โˆˆ ๐ด ๐‘Ž๐‘›๐‘‘ ๐‘ฅ โˆˆ ๐ต}

๐‘จ โˆฉ ๐‘ฉ

More generally,

Letโ€™s observe that:

๐‘ฅ โˆ‰ ๐ด โˆฉ ๐ต <===> ๐‘ฅ โˆ‰ ๐ด ๐‘œ๐‘Ÿ ๐‘ฅ โˆ‰ ๐ต

2. The Union ๐ด โˆช ๐ต = {๐‘ฅ โ”‚ ๐‘ฅ โˆˆ ๐ด ๐‘œ๐‘Ÿ๐‘ฅ โˆˆ ๐ต}

๐‘จ โˆช ๐‘ฉ

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As in the case of the intersection, we can consider:

We see that:

๐‘ฅ โˆ‰ ๐ด โˆช ๐ต <===> ๐‘ฅ โˆ‰ ๐ด ๐‘Ž๐‘›๐‘‘ ๐‘ฅ โˆ‰ ๐ต

3. The Difference๐ด โˆ’ ๐ต = {๐‘ฅ โ”‚ ๐‘ฅ โˆˆ ๐ด ๐‘Ž๐‘›๐‘‘ ๐‘ฅ โˆ‰ ๐ต}

We retain that:

๐‘ฅ โˆ‰ ๐ด โˆ’ ๐ต <===> ๐‘ฅ โˆ‰ ๐ด ๐‘œ๐‘Ÿ ๐‘ฅ โˆˆ ๐ต

4. The ComplementThe complement of a set ๐ด is the difference between the

total set and ๐ด.

๐ถ๐‘‡๐ด = { ๐‘ฅ | ๐‘ฅ โˆˆ ๐‘‡ ๐‘Ž๐‘›๐‘‘ ๐‘ฅ โˆ‰ ๐ด} = { ๐‘ฅ | ๐‘ฅ โˆ‰ ๐ด}

The complement of a set is noted with ๐ถ๐ด or with ๏ฟฝฬ…๏ฟฝ.

๐‘ช๐‘จ

Letโ€™s observe that

๐‘ฅ โˆ‰ ๐ถ๐ด <===> ๐‘ฅ โˆˆ ๐ด

More generally, we can talk about the complement of a set to

another, random set. Thus,

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๐ถ๐ด๐ต = { ๐‘ฅ | ๐‘ฅ โˆˆ ๐ต ๐‘Ž๐‘›๐‘‘ ๐‘ฅ โˆ‰ ๐ด}

is the complement of set ๐ต to set ๐ด.

5. The Symmetrical Difference๐ดโˆ†๐ต = (๐ด โˆ’ ๐ต) โˆช (๐ต โˆ’ ๐ด)

๐‘จโˆ†๐‘ฉ

We have ๐‘ฅ โˆ‰ ๐ดโˆ†๐ต <===> ๐‘ฅ โˆ‰ ๐ด โˆ’ ๐ต and ๐‘ฅ โˆ‰ ๐ต โˆ’ ๐ด .

6. The Cartesian Product๐ด ร— ๐ต = {(๐‘ฅ, ๐‘ฆ)โ”‚๐‘ฅ โˆˆ ๐ด ๐‘Ž๐‘›๐‘‘ ๐‘ฆ โˆˆ ๐ต}

The Cartesian product of two sets is a set of an ordered

pairs of elements, the first element belonging to the first set and the

second element belonging to the second set.

For example, โ„ ร— โ„ = {(๐‘ฅ, ๐‘ฆ)|๐‘ฅ โˆˆ โ„ , ๐‘ฆ โˆˆ โ„} , and an

intuitive representation of this set is provided in Figure 1.1.

By way of analogy, the Cartesian product of three sets is a

set of triplets:

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๐ด ร— ๐ต ร— ๐ถ = {(๐‘ฅ, ๐‘ฆ, ๐‘ง)|๐‘ฅ โˆˆ ๐ด , ๐‘ฆ โˆˆ ๐ต, ๐‘ง โˆˆ ๐ถ}.

An intuitive image of โ„3 = โ„ ร— โ„ ร— โ„ is given in the figure

below.

More generally, ๐‘›๐‘‹

๐‘– = 1๐ด๐‘– = {(๐‘‹1, ๐‘‹2, โ€ฆ , ๐‘‹๐‘›)| ๐‘‹๐‘– โˆˆ ๐ด๐‘– ๐‘“๐‘œ๐‘Ÿ ๐‘Ž๐‘›๐‘ฆ ๐‘– โˆˆ 1, ๐‘›ฬ…ฬ… ฬ…ฬ…ฬ…}.

Methods for Proving the Equality of Two Sets

The difficulties and surprises met on the process of trying to

rigorously define the notion of sets are not the only ones we

encounter in the theory of sets.

Another surprise is that the well-known (and the obvious, by

intuition) method to prove the equality of two sets using the

double inclusion is โ€“ at the level of a rigorous definition of sets โ€“

just an axiom.

๐ด = ๐ต <===> ๐ด โŠ† ๐ต ๐‘Ž๐‘›๐‘‘ ๐ต โŠ† ๐ด (1.1)

By accepting this axiom, we will further exemplify two

methods to prove the equality of sets:

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(A) THE DOUBLE INCLUSION [expressed through the

equivalence (1.1)];

(B) UTILIZING THE CHARACTERISTIC FUNCTION

OF A SET.

We will first provide a few details of this second method that

is much faster in practice, hence more convenient to use than the

first method.

DEFINITION. We call a characteristic function of the set ๐ด

the function ๐œ‘๐ด: ๐‘‡ โŸถ {0,1} defined through:

๐œ‘๐ด(๐‘ฅ) = {1 ๐‘–๐‘“ ๐‘ฅ โˆˆ ๐ด0 ๐‘–๐‘“ ๐‘ฅ โˆ‰ ๐ด

As we can observe, the numbers 0 and 1 are used to divide

the elements of the total set ๐‘‡ in two categories:

(1) one category contains those elements ๐‘ฅ in which the

value of ๐œ‘๐ด is 1 (the elements that belong to ๐ด);

(2) all the elements in which the value of ๐œ‘๐ด is 0 (elements

that do not belong to ๐ด) belong to the second category.

Method (B) of proving the equality of two sets is based on

the fact that any set is uniquely determined by its characteristic

function, in the sense that: there exists a bijection from set๐‹(๐‘ป)

of the subsets of ๐‘ป to the set ๐œธ(๐‘ป) of the characteristic

functions defined on ๐‘ป (see Exercise VII).

Hence,

๐ด = ๐ต <===> ๐œ‘๐ด = ๐œ‘๐ต.

(1.2)

Properties of the Characteristic Function

๐œ‘1) ๐œ‘๐ดโˆฉ๐ต(๐‘ฅ) = ๐œ‘๐ด(๐‘ฅ). ๐œ‘๐ต(๐‘ฅ)

๐œ‘2) ๐œ‘๐ดโˆช๐ต(๐‘ฅ) = ๐œ‘๐ด(๐‘ฅ) + ๐œ‘๐ต(๐‘ฅ) โˆ’ ๐œ‘๐ดโˆฉ๐ต(๐‘ฅ)

๐œ‘3) ๐œ‘๐ดโˆ’๐ต(๐‘ฅ) = ๐œ‘๐ด(๐‘ฅ) โˆ’ ๐œ‘๐ดโˆฉ๐ต(๐‘ฅ)

๐œ‘4) ๐œ‘๐ถ๐ด(๐‘ฅ) = 1 โˆ’ ๐œ‘๐ด(๐‘ฅ)

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๐œ‘5) ๐œ‘๐ดโˆ†๐ต(๐‘ฅ) = ๐œ‘๐ด(๐‘ฅ) + ๐œ‘๐ต(๐‘ฅ) โˆ’ 2. ๐œ‘๐ดโˆฉ๐ต(๐‘ฅ)

๐œ‘6) ๐œ‘๐ดร—๐ต(๐‘ฅ, ๐‘ฆ) = ๐œ‘๐ด(๐‘ฅ) ร— ๐œ‘๐ต(๐‘ฆ)

๐œ‘7) ๐œ‘๐ด2(๐‘ฅ) = ๐œ‘๐ด(๐‘ฅ)

๐œ‘8) ๐œ‘โˆ…(๐‘ฅ) โ‰ก 0, ๐œ‘๐‘‡(๐‘ฅ) โ‰ก 1

๐œ‘9) ๐ด โŠ† ๐ต <===> ๐œ‘๐ด(๐‘ฅ) โ‰ค ๐œ‘๐ต(๐‘ฅ) for any ๐‘ฅ โˆˆ ๐‘‡.

Letโ€™s prove, for example ๐œ‘1). For this, letโ€™s observe that the

total set ๐‘‡ is divided by sets ๐ด and ๐ต in four regions, at most:

1) for the points ๐‘ฅ that do belong to neither ๐ด, nor ๐ต, we

have ๐œ‘๐ด(๐‘ฅ) = ๐œ‘๐ต(๐‘ฅ) = 0 and ๐œ‘๐ดโˆฉ๐ต(๐‘ฅ) = 0.

2) for the points ๐‘ฅ that are in ๐ด but are not in ๐ต, we have

๐œ‘๐ด(๐‘ฅ) = 1, ๐œ‘๐ต(๐‘ฅ) = 0 ๐‘Ž๐‘›๐‘‘ ๐œ‘๐ดโˆฉ๐ต(๐‘ฅ) = 0 .

3) if ๐‘ฅ โˆ‰ ๐ด ๐‘Ž๐‘›๐‘‘ ๐‘ฅ โˆˆ ๐ต,

the equalities follow analogously.

4) if ๐‘ฅ โˆˆ ๐ต ๐‘Ž๐‘›๐‘‘ ๐‘ฅ โˆ‰ ๐ด

Exercises I. Using methods (A) and (B) show that:

SOLUTIONS:

(A) (the double inclusion)

Letโ€™s notice that solving a mathematical problem requires we

successively answer two groups of questions:

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(q) WHAT DO WE HAVE TO PROVE?

(Q) HOW DO WE PROVE?

Answering question (Q) leads us to a new (q) question. Thus,

for exercise 1, the answer to the question (q) is:

(r1): We have to prove an equality of sets,

and the answer to the question (Q) is:

(R1): We can prove this equality using two methods: using

the double inclusions or using the properties of the characteristic

function.

We choose the first method and end up again at (q). This

time the answer is:

(r2): We have to prove two inclusions,

and the answer to the corresponding question (Q) is:

(R2): We prove each inclusion, one at a time.

Then,

(r3): We have to prove one inclusion;

(R3): We prove that an arbitrary element belonging to the

โ€œincluded setโ€ belongs to the โ€œset that includesโ€.

It is obvious that this entire reasoning is mental, the solution

starts with:

a) Let ๐‘ฅ โˆˆ ๐ด โˆ’ (๐ต โˆ’ ๐ถ), irrespective [to continue we will read the

final operation (the difference)], <===> ๐‘ฅ โˆˆ ๐ด and ๐‘ฅ โˆ‰ ๐ต โˆฉ ๐ถ [we will

read the operation that became the last (the union)] <===>

๐‘ฅ โˆˆ ๐ด and {๐‘ฅ โˆ‰ ๐ต

๐‘œ๐‘Ÿ๐‘ฅ โˆ‰ ๐ถ

<===> {๐‘ฅ โˆˆ ๐ด ๐‘Ž๐‘›๐‘‘ ๐‘ฅ โˆ‰ ๐ต

๐‘œ๐‘Ÿ๐‘ฅ โˆˆ ๐ด ๐‘Ž๐‘›๐‘‘ ๐‘ฅ โˆ‰ ๐ถ

(1.3)

From this point on we have no more operations to explain

and we can regroup the enunciation we have reached in several

ways (many implications flow from (1.3), but we are only interested

in one - the conclusion of the exercise), that is why we have to

update the conclusion:

๐‘ฅ โˆˆ (๐ด โˆ’ ๐ต) โˆช (๐ด โˆ’ ๐ถ) <===> ๐‘ฅ โˆˆ ๐ด โˆ’ ๐ต ๐‘œ๐‘Ÿ ๐‘ฅ โˆˆ ๐ด โˆ’ ๐ถ.

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We observe from (1.3) that this conclusion follows

immediately, so the inclusion is proven.

For the second inclusion: Let ๐‘ฅ โˆˆ (๐ด โˆ’ ๐ต) โˆช (๐ด โˆ’ ๐ถ), [we

will read the final operation (the reunion)] <===>

{๐‘ฅ โˆˆ ๐ด โˆ’ ๐ต

๐‘œ๐‘Ÿ๐‘ฅ โˆˆ ๐ด โˆ’ ๐ถ

<===> [we will read the operation that became the last]

<===>

{๐‘ฅ โˆˆ ๐ด ๐‘Ž๐‘›๐‘‘ ๐‘ฅ โˆ‰ ๐ต

๐‘œ๐‘Ÿ๐‘ฅ โˆˆ ๐ด ๐‘Ž๐‘›๐‘‘ ๐‘ฅ โˆ‰ ๐ถ

<===> ๐‘ฅ โˆˆ ๐ด and {๐‘ฅ โˆ‰ ๐ต

๐‘œ๐‘Ÿ๐‘ฅ โˆ‰ ๐ถ

(1.4)

We have no more operations to explain, so we will describe

in detail the conclusion we want to reach: ๐‘ฅ โˆˆ ๐ด โˆ’ (๐ต โˆฉ ๐ถ), i.e. ๐‘ฅ โˆˆ

๐ด and ๐‘ฅ โˆ‰ ๐ต โˆฉ ๐ถ, an affirmation that immediately issues from (1.4).

5. (r1): We have to prove an equality of sets;

(R1): We prove two inclusions;

(r2): We have to prove an inclusion;

(R2): We prove that each element belonging to the included

set is in the set that includes.

Let ๐‘ฅ โˆˆ ๐ด โˆ’ (๐ด โˆ’ ๐ต) , irrespective [we will read the final

operation (the difference)], <===> ๐‘ฅ โˆˆ ๐ด and ๐‘ฅ โˆ‰ ๐ด โˆ’ ๐ต [we will

read the operation that became the last] <===>

๐‘ฅ โˆˆ ๐ด and {๐‘ฅ โˆ‰ ๐ด

๐‘œ๐‘Ÿ๐‘ฅ โˆˆ ๐ต.

(1.5)

We have no more operations to explain, so we update the

conclusion:

๐‘ฅ โˆˆ ๐ด โˆฉ ๐ต ๐‘–. ๐‘’. ๐‘ฅ โˆˆ ๐ด ๐‘Ž๐‘›๐‘‘ ๐‘ฅ โˆˆ ๐ต.

We observe that (1.5) is equivalent to:

{๐‘ฅ โˆˆ ๐ด ๐‘Ž๐‘›๐‘‘ ๐‘ฅ โˆ‰ ๐ด

๐‘œ๐‘Ÿ๐‘ฅ โˆˆ ๐ด ๐‘Ž๐‘›๐‘‘ ๐‘ฅ โˆˆ ๐ต.

The first affirmation is false, and the second one proves

that๐‘ฅ โˆˆ ๐ด โˆฉ ๐ต.

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We will solve the same exercises using the properties of the

characteristic function and the equivalence (1.2).

1. We have to prove that ๐œ‘๐ดโˆ’(๐ตโˆฉ๐ถ) = ๐œ‘(๐ดโˆ’๐ต)โˆช(๐ดโˆ’๐ถ);

(r1): We have to prove an equality of functions;

(R1): As the domain and codomain of the two functions

coincide, we have to further prove that their values coincide, i.e.

(r2): ๐œ‘๐ดโˆ’(๐ตโˆฉ๐ถ)(๐‘ฅ) = ๐œ‘(๐ดโˆ’๐ต)โˆช(๐ดโˆ’๐ถ)(๐‘ฅ) for any ๐‘ฅ โˆˆ ๐‘‡.

The answer to the corresponding (Q) question is

(R2): We explain both members of the previous equality.

๐œ‘๐ดโˆ’(๐ตโˆฉ๐ถ)(๐‘ฅ) = [we will read the final operation (the

difference) and we will apply property ๐œ‘3 ] = ๐œ‘๐ด(๐‘ฅ) โˆ’

๐œ‘๐ดโˆฉ(๐ตโˆฉ๐ถ)(๐‘ฅ) = [we will read the operation that became the last and

we will apply property๐œ‘1] =

๐œ‘๐ด(๐‘ฅ) โˆ’ ๐œ‘๐ด(๐‘ฅ). ๐œ‘๐ต(๐‘ฅ). ๐œ‘๐ถ(๐‘ฅ) (1.6)

The second member of the equality becomes successive:

5. In order to prove the equality ๐œ‘๐ดโˆ’(๐ดโˆ’๐ต)(๐‘ฅ) โˆ’ ๐œ‘๐ดโˆฉ๐ต(๐‘ฅ)

we notice that:

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II. Prove that:

SOLUTIONS.

Method (A) (the double inclusion)

1. Let ๐‘ฅ โˆˆ (๐ด โˆฉ ๐ต) ร— (๐ถ โˆฉ ๐ท) , irrespective [we explain the

final operation (the scalar product)] <===> ๐‘ฅ = (๐›ผ, ๐›ฝ), with ๐›ผ โˆˆ ๐ด โˆฉ

๐ต and ๐›ฝ โˆˆ ๐ถ โˆฉ ๐ท <===>[we explain the operations that became the last

ones] <===> ๐‘ฅ = (๐›ผ, ๐›ฝ) with ๐›ผ โˆˆ ๐ด , ๐›ผ โˆˆ ๐ต and ๐›ฝ โˆˆ ๐ถ , ๐›ฝ โˆˆ ๐ท

[we have no more operations to explain so we update the conclusion: ๐‘ฅ โˆˆ (๐ด ร—

๐ถ) โˆฉ (๐ต ร— ๐ท), so ๐‘ฅ belongs to an intersection (the last operation)] <===

> ๐‘ฅ = (๐›ผ, ๐›ฝ) with ๐›ผ โˆˆ ๐ด , ๐›ฝ โˆˆ ๐ถ and ๐›ผ โˆˆ ๐ต , ๐›ฝ โˆˆ ๐ท <===> ๐‘ฅ โˆˆ

๐ด ร— ๐ถ and ๐‘ฅ โˆˆ ๐ต ร— ๐ท.

3. Let ๐‘ฅ โˆˆ (๐ด โˆ’ ๐ต) ร— ๐ถ , irrespective <===> ๐‘ฅ โˆˆ (๐›ผ, ๐›ฝ)

with ๐›ผ โˆˆ ๐ด โˆ’ ๐ต and ๐›ฝ โˆˆ ๐ถ <===> ๐‘ฅ โˆˆ (๐›ผ, ๐›ฝ) with ๐›ผ โˆˆ ๐ด and

๐›ผ โˆ‰ ๐ต and ๐›ฝ โˆˆ ๐ถ [by detailing the conclusion we deduce the following steps]

<===> ๐‘ฅ = (๐›ผ, ๐›ฝ), ๐›ผ โˆˆ ๐ด, ๐›ฝ โˆˆ ๐ถ and ๐‘ฅ = (๐›ผ, ๐›ฝ), ๐›ผ โˆ‰ ๐ต , ๐›ฝ โˆˆ ๐ถ

===> ๐‘ฅ โˆˆ ๐ด ร— ๐ถ and ๐‘ฅ โˆ‰ ๐ต ร— ๐ถ.

Reciprocally, let ๐‘ฅ โˆˆ (๐ด ร— ๐ถ) โˆ’ (๐ต ร— ๐ถ) , irrespective

<===> ๐‘ฅ = (๐›ผ, ๐›ฝ) with (๐›ผ, ๐›ฝ) โˆˆ ๐ด ร— ๐ถ and ร— ๐ถ <===> ๐‘ฅ =

(๐›ผ, ๐›ฝ), ๐›ผ โˆˆ ๐ด and ๐›ฝ โˆˆ ๐ถ and

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{๐›ผ โˆ‰ ๐ต

๐‘œ๐‘Ÿ๐›ฝ โˆ‰ ๐ถ

<===> {๐›ผ โˆˆ ๐ด ๐‘Ž๐‘›๐‘‘ ๐›ฝ โˆˆ ๐ถ ๐‘Ž๐‘›๐‘‘ ๐›ผ โˆ‰ ๐ต

๐‘œ๐‘Ÿ๐›ผ โˆˆ ๐ด ๐‘Ž๐‘›๐‘‘ ๐›ฝ โˆˆ ๐ถ ๐‘Ž๐‘›๐‘‘ ๐›ฝ โˆ‰ ๐ถ (๐‘“๐‘Ž๐‘™๐‘ )

===> (๐›ผ, ๐›ฝ) โˆˆ (๐ด โˆ’ ๐ต) ร— ๐ถ.

Method (B) (using the characteristic function)

Explaining ๐œ‘(๐ดโˆ†๐ต)โˆ†๐ถ(๐‘ฅ), in an analogue manner, we obtain

the desired equality. If we observe that the explanation

of ๐œ‘(๐ดโˆ†๐ต)โˆ†๐ถ(๐‘ฅ) from above is symmetrical, by utilizing the

commutativity of addition and multiplication, the required equality

follows easily.

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Equalities 4)-5) were easy to prove using the characteristic

function; they are, however, more difficult to prove using the first

method. The characteristic function is usually preferred, due to the

ease of use and the rapidity of reaching the result.

III. Prove the following equivalences:

SOLUTIONS:

1. r1: We have to prove an equivalence;

R1: We prove two implications;

r2: ๐ด โˆช ๐ต โŠ‚ ๐ถ => ๐ด โŠ‚ ๐ถ ๐‘Ž๐‘›๐‘‘ ๐ต โŠ‚ ๐ถ

R2: We prove that ๐ด โŠ‚ ๐ถ ๐‘Ž๐‘›๐‘‘ ๐ต โŠ‚ ๐ถ (two inclusions).

Let ๐‘ฅ โˆˆ ๐ด irrespective [we have no more operations to

explain, therefore we will update the conclusion: ๐‘ฅ โˆˆ ๐ถ; but this is

not evident from ๐‘ฅ โˆˆ ๐ด, but only with the help of the hypothesis]

๐‘ฅ โˆˆ ๐ด => ๐‘ฅ โˆˆ ๐ด โˆช ๐ต =

> [๐‘กโ„Ž๐‘Ÿ๐‘œ๐‘ข๐‘”โ„Ž ๐‘กโ„Ž๐‘’ โ„Ž๐‘ฆ๐‘๐‘œ๐‘กโ„Ž๐‘’๐‘กโ„Ž๐‘–๐‘  ๐ด โˆช ๐ต โŠ‚ ๐ถ] =

> ๐‘ฅ โˆˆ ๐ถ

Let ๐‘ฅ โˆˆ ๐ต irrespective => ๐‘ฅ โˆˆ ๐ด โˆช ๐ต โŠ‚ ๐ถ => ๐‘ฅ โˆˆ ๐ถ.

Reciprocally (the second implication):

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r1: We have to prove that ๐ด โˆช ๐ต โŠ‚ ๐ถ (an inclusion)

R1: Let ๐‘ฅ โˆˆ ๐ด โˆช ๐ต irrespective (we explain the last

operation):

=> {๐‘ฅ โˆˆ ๐ด => ๐‘ฅ โˆˆ ๐ถ (๐‘‘๐‘ข๐‘’ ๐‘ก๐‘œ ๐‘กโ„Ž๐‘’ โ„Ž๐‘ฆ๐‘๐‘œ๐‘กโ„Ž๐‘’๐‘ ๐‘–๐‘ )

๐‘œ๐‘Ÿ๐‘ฅ โˆˆ ๐ต => ๐‘ฅ โˆˆ ๐ถ(๐‘‘๐‘ข๐‘’ ๐‘ก๐‘œ ๐‘กโ„Ž๐‘’ โ„Ž๐‘ฆ๐‘๐‘œ๐‘กโ„Ž๐‘’๐‘ ๐‘–๐‘ )

5. r1: We have to prove an equivalence;

R1: We prove two implications;

r2: (๐ด โˆ’ ๐ต) โˆช ๐ต = ๐ด => ๐ต โŠ‚ ๐ด;

R2: We prove the conclusion (an inclusion), using the

hypothesis.

Let ๐‘ฅ โˆˆ ๐ต irrespective => ๐‘ฅ โˆˆ (๐ด โˆ’ ๐ต) โˆช ๐ต => ๐‘ฅ โˆˆ ๐ด

r3: We have to prove the implication: ๐ต โŠ‚ ๐ด => (๐ด โˆ’ ๐ต) โˆช

๐ต = ๐ด;

R3: We prove an equality of sets (two inclusions).

For the first inclusion, let ๐‘ฅ โˆˆ (๐ด โˆ’ ๐ต) โˆช ๐ต irrespective [we

explain the last operation] =>

=> {๐‘ฅ โˆˆ ๐ด โˆ’ ๐ต

๐‘œ๐‘Ÿ๐‘ฅ โˆˆ ๐ต

=> {๐‘ฅ โˆˆ ๐ด ๐‘Ž๐‘›๐‘‘ ๐‘ฅ โˆ‰ ๐ต (๐‘Ž)

๐‘œ๐‘Ÿ๐‘ฅ โˆˆ ๐ต (๐‘)

[We have no more operation to explain, so we update the

conclusion]. We thus observe that from (a) it follows that ๐‘ฅ โˆˆ ๐ด,

and from (b), with the help of the hypothesis, ๐ต โŠ‚ ๐ด , we also

obtain ๐‘ฅ โˆˆ ๐ด.

Method 2:

We have to prove two inequalities among functions.

Through the hypothesis, ๐œ‘๐ดโˆช๐ต < ๐œ‘๐ถ and, moreover ๐œ‘๐ด < ๐œ‘๐ดโˆช๐ต

(indeed, ๐œ‘๐ด < ๐œ‘๐ดโˆช๐ต <=> ๐œ‘๐ด < ๐œ‘๐ด + ๐œ‘๐ต โˆ’ ๐œ‘๐ด๐œ‘๐ต <=> ๐œ‘๐ต(1 โˆ’

๐œ‘๐ด) > 0 โ€“ true, because the characteristic functions take two

values: 0 and 1).

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It follows that ๐œ‘๐ด < ๐œ‘๐ดโˆช๐ต < ๐œ‘๐ถ and, analogously ๐œ‘๐ต <

๐œ‘๐ดโˆช๐ต < ๐œ‘๐ถ .

Reciprocally, for the inverse implication, we have to prove

that: ๐œ‘๐ดโˆช๐ต < ๐œ‘๐ถ , i.e.

๐œ‘๐ด(๐‘ฅ) + ๐œ‘๐ต(๐‘ฅ) โˆ’ ๐œ‘๐ดโˆฉ๐ต(๐‘ฅ) < ๐œ‘๐ถ(x) (1.7)

in the hypothesis:

๐œ‘๐ด < ๐œ‘๐ถ and ๐œ‘๐ต < ๐œ‘๐ถ (1.8)

But ๐œ‘๐ด, ๐œ‘๐ต, ๐œ‘๐ถ can take only two values: 0 and 1. From

the eight possible cases in which (1.7) must be checked, due to (1.8),

there remain only four possibilities:

In each of these cases (1.7) checks out.

5. We have to prove that

But

For the direct implication, through the hypothesis, we

have:

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Consequently:

{๐œ‘๐ต(๐‘ฅ) = 0 => ๐œ‘๐ต(๐‘ฅ) < ๐œ‘๐ด(๐‘ฅ) (๐‘‘๐‘ข๐‘’ ๐‘ก๐‘œ ๐‘กโ„Ž๐‘’ ๐‘ฃ๐‘Ž๐‘™๐‘ข๐‘’๐‘  0 ๐‘Ž๐‘›๐‘‘ 1 ๐‘กโ„Ž๐‘Ž๐‘ก ๐‘กโ„Ž๐‘’ ๐‘“๐‘ข๐‘›๐‘๐‘ก๐‘–๐‘œ๐‘› ๐œ‘ ๐‘ก๐‘Ž๐‘˜๐‘’๐‘ )

๐‘œ๐‘Ÿ1 โˆ’ ๐œ‘๐ด(๐‘ฅ) = 0 => ๐œ‘๐ด(๐‘ฅ) = 1 ๐‘ ๐‘œ ๐œ‘๐ต(๐‘ฅ) < ๐œ‘๐ด(๐‘ฅ)

Reciprocally, if ๐œ‘๐ต(๐‘ฅ) < ๐œ‘๐ด(๐‘ฅ), we cannot have ๐œ‘๐ต(๐‘ฅ) =

1 and ๐œ‘๐ด(๐‘ฅ) = 0, and from (1.9) we deduce:

- equality that is true for the three remaining cases:

IV. Using the properties of the characteristic function, solve

the following equations and systems of equations with sets:

SOLUTION:

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We have to analyze the following cases:

a) if ๐‘ฅ โˆ‰ ๐ด and ๐‘ฅ โˆ‰ ๐ต, we have:

๐œ‘๐ถ(๐ดโˆฉ๐ต)(๐‘ฅ) = 1 and ๐œ‘๐ดโˆฉ๐ถ๐ต(๐‘ฅ) = 1 , so for (1.10) to be

accomplished we must have ๐œ‘๐‘ฅ(๐‘ฅ) = 0. So, any point that doesnโ€™t

belong to ๐ด or ๐ต, doesnโ€™t belong to ๐‘‹ either.

b) if ๐‘ฅ โˆˆ ๐ด and ๐‘ฅ โˆ‰ ๐ต, we have:

๐œ‘๐ถ(๐ดโˆฉ๐ต)(๐‘ฅ) = 1 and ๐œ‘๐ดโˆฉ๐ถ๐ต(๐‘ฅ) = 1 , so for (1.10) to be

accomplished we must have ๐œ‘๐‘ฅ(๐‘ฅ) = 1. Therefore, any point from

๐ด โˆ’ ๐ต is in ๐‘‹.

c) if ๐‘ฅ โˆ‰ ๐ด and ๐‘ฅ โˆˆ ๐ต and for (1.10) to be accomplished, we

must have ๐œ‘๐‘ฅ(๐‘ฅ) = 0, consequently no point from ๐ต โˆ’ ๐ด is not in

๐‘‹.

d) if ๐‘ฅ โˆˆ ๐ด and ๐‘ฅ โˆˆ ๐ต we can have ๐œ‘๐‘ฅ(๐‘ฅ) = 0 or ๐œ‘๐‘ฅ(๐‘ฅ) =

1.

So any point from ๐ด โˆฉ ๐ต can or cannot be in ๐‘‹. Therefore

๐‘‹ = (๐ด\๐ต) โˆช ๐ท, where ๐ท โŠ‚ ๐ด โˆฉ ๐ต, arbitrarily.

We have to analyze four cases, as can be seen from the

following table:

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So ๐‘‹ = (๐ด โˆ’ ๐ต) โˆช ๐ถ.

V. Determine the following sets:

7. Let ๐‘ท โˆˆ โ„ค(๐’™) a ๐’ degree polynomial and ๐’’ โˆˆ โ„ค .

Knowing that ๐‘ท(๐’’) = ๐Ÿ๐Ÿ“, determine the set:

INDICATIONS:

1. An integer maximal part is highlighted. This is obtained

making the divisions:

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10 is dividable by 2๐‘› + 1 <=> ๐‘› โˆˆ {0, 1, โˆ’2}

so ๐ธ โˆˆ โ„ค <=> 27 is dividable by 2๐‘ฅ + 1 and 4๐‘ฅ2 โˆ’ 2๐‘ฅ โˆ’ 11 +27

2๐‘ฅ+1 is a multiple of 8 <=> ๐ด โˆˆ {โˆ’14, โˆ’5, โˆ’2, โˆ’1, 0,1, 4, 13}.

so ๐‘ฅ and ๐‘ฆ are integer solutions of systems the shape of:

{3๐‘ฆ โˆ’ ๐‘ฅ โˆ’ 1 = ๐‘ข3๐‘ฆ + ๐‘ฅ + 1 = ๐‘ฃ

with ๐‘ข and ๐‘ฃ as divisors of 32.

So ๐‘ƒ(๐‘ฅ)

๐‘ฅโˆ’๐‘žโˆˆ โ„ค <=> 15 is dividable by ๐‘ฅ โˆ’ ๐‘ž.

8. Method 1: For ๐‘Ž โ‰  โˆ’1 we have

so, we must have ๐‘ฅ2 + 6๐‘ฅ โˆ’ 3 > 0 . As ๐‘ฅ1,2 = โˆ’3 ยฑ 2โˆš3 , we

deduce:

Method 2: We consider the function ๐‘“(๐‘Ž) =๐‘Ž2โˆ’๐‘Ž+1

๐‘Ž+1 and ๐ด =

๐‘“(๐ท), with ๐ท = โ„{โˆ’1} the domain of ๐‘“. ๐ด is obtained from the

variation table of ๐‘“.

VI. Determine the following sets, when ๐‘Ž, ๐‘ โˆˆ โ„•:

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where [๐‘ก] is the integer part of ๐‘ก.

SOLUTIONS: We use the inequalities: [๐‘ก] < ๐‘ก < [๐‘ก] + 1, so:

VII. The application ๐‘“: ๐’ซ(๐‘‡) โŸถ โ„ฑ๐‘‡ defined through

๐‘“(๐ด) = ๐œ‘๐ด is a bijection from the family of the parts belonging to

๐‘‡ to the family of the characteristic functions defined on ๐‘‡.

SOLUTION: To demonstrate the injectivity, let ๐ด, ๐ต โˆˆ ๐’ซ(๐‘‡)

with ๐ด โ‰  ๐ต . From ๐ด โ‰  ๐ต we deduce that there exists ๐‘ฅ0 โˆˆ ๐‘‡ so

that:

a) ๐‘ฅ0 โˆˆ ๐ด and ๐‘ฅ0 โˆ‰ ๐ต or b) ๐‘ฅ0 โˆˆ ๐ต and ๐‘ฅ0 โˆ‰ ๐ด

In the first case ๐œ‘๐ด(๐‘ฅ0) = 1 , case ๐œ‘๐ต(๐‘ฅ0) = 0 , so ๐œ‘๐ด โ‰ 

๐œ‘๐ต. In the second case, also, ๐œ‘๐ด โ‰  ๐œ‘๐ต.

The surjectivity comes back to:

Let then ๐œ‘ โˆˆ โ„ฑ๐‘‡ irrespective. Then for ๐ด = {๐‘ฅ|๐œ‘(๐‘ฅ) = 1}

we have ๐œ‘ = ๐œ‘๐ด.

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II. Functions

Function definition A function is determined by three elements ๐ท, ๐ธ and ๐‘“, with

the following significations: ๐ท and ๐ธ are sets, named the domain

and the codomain, respectively, of the function, and ๐‘“ is a

correspondence law from ๐ท to ๐ธ that causes:

each element ๐’™ โˆˆ ๐‘ซ to have one,

and only one corresponding element ๐’š โˆˆ ๐‘ฌ. (F)

So we can say a function is a triplet (๐ท, ๐ธ, ๐‘“), the elements of

this triplet having the signification stated above.

We usually note this triplet with ๐‘“: ๐ท โ†’ ๐ธ. Two functions

are equal if they are equal also as triplets, i.e. when their three

constitutive elements are respectively equal (not just the

correspondence laws).

To highlight the importance the domain and codomain have

in the definition of functions, we will give some examples below, in

which, keeping the correspondence law ๐‘“ unchanged and modifying

just the domain or (and) the codomain, we can encounter all

possible situations, ranging from those where the triplet is a

function to those where it is a bijective function.

In order to do this, we have to first write in detail condition

(๐น) that characterizes a function.

We can consider this condition as being made up of two sub

conditions, namely:

(๐‘“1) each element ๐‘ฅ from the domain has a corresponding element, in the

sense of โ€œat least one elementโ€, in the codomain.

Using a diagram as the one drawn below, the proposition can

be stated like this:

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โ€œ(At least) one arrow can be drawn from each point of

the domain.โ€

More rigorously, this condition is expressed by:

The second sub condition regarding ๐‘“ in the definition of a

function is:

(๐‘“2) the element from the codomain that corresponds to ๐‘ฅ is unique.

For the type of diagram in fig. 2.1, this means that the arrow

that is drawn from one point is unique.

This condition has the equivalent formulation:

โ€œIf two arrows have the same starting point then they

also have the same arrival point.โ€

Namely,

.

Conditions (๐‘“1) and (๐‘“2) are necessary and sufficient for any

correspondence law ๐‘“ to be a function. These conditions are easy to

use in practice.

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In Figure 2.1, the correspondence law from a) satisfies (๐‘“1)

and doesnโ€™t satisfy (๐‘“2); in b) the correspondence law doesnโ€™t satisfy

(๐‘“1) , but satisfies (๐‘“2) . In diagram c), neither(๐‘“1) , nor (๐‘“2) are

satisfied, and in d) the correspondence law satisfies both (๐‘“1) and

(๐‘“2), so it is the (only) function.

A function is therefore a triplet ( ๐ท, ๐ธ, ๐‘“ ), in which the

correspondence law ๐‘“ satisfies (๐‘“1) and (๐‘“2).

Letโ€™s observe that these two condition only refer to the

domain of the function:

(๐‘“1) โ€“ from each point of the domain stems at least one arrow;

(๐‘“2) โ€“ the arrow that stems from one point of the domain is

unique.

It is known that the graphic of a function is made up of the

set of pairs of points (๐‘ฅ, ๐‘“(๐‘ฅ)), when ๐‘ฅ covers the domain of the

function.

HOW CAN WE RECOGNIZE ON A GRAPHIC IF A

CORRESPONDANCE LAW IS A FUNCTION [if it satisfies (๐‘“1)

and (๐‘“2)] ?

In order to answer this question we will firstly remember (for

the case ๐ท, ๐ธ โŠ‚ ๐‘…) the answer to two other questions:

1. Given ๐‘ฅ, how do we obtain โ€“ using the graphic โ€“ ๐‘“(๐‘ฅ)

[namely, the image of ๐‘ฅ (or images, if more than one, and in this case

the correspondence law ๐‘“ is, of course, not a function)] ?

Answer: We trace a parallel from ๐‘ฅ to 0๐‘ฆ to the point it

touches the graphic, and from the intersection point (points) we

then trace a parallel (parallels) to 0๐‘ฅ . The intersection points of

these parallels with 0๐‘ฆ are the images ๐‘“(๐‘ฅ) of ๐‘ฅ.

2. Reciprocally, given ๐‘ฆ, to obtain the point (points) ๐‘ฅ having

the property ๐‘“(๐‘ฅ) = ๐‘ฆ, we trace a parallel to 0๐‘ฅ through ๐‘ฆ , and

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from the point (points) of intersection with the graphic, we then

trace a parallel (parallels) to 0๐‘ฆ.

Examples 1. A circle with the center at the origin and with the radius ๐‘Ÿ

is not the graphic of a function ๐‘“: โ„ โŸถ โ„, because it does not

satisfy condition (๐‘“1) (there are points in the domain that do not have any

image, namely, all the points through which the parallel 0๐‘ฆ doesnโ€™t

intersect the graphic) or (๐‘“2), because there are points in the domain that

have more than one image (all the points ๐‘ฅ โˆˆ (โˆ’๐‘Ÿ, ๐‘Ÿ) have two images).

2. A circle with the center at the origin and with the radius ๐‘Ÿ

is not the graphic of a function ๐‘“: [โˆ’๐‘Ÿ, ๐‘Ÿ] โŸถ โ„, because it does not

satisfy condition (๐‘“2)(this time there are no more points in the

domain that do not have an image, but there are points that have two

images).

3. A circle with the center at the origin and with the radius ๐‘Ÿ

is not the graphic of a function ๐‘“: โ„ โŸถ [0, โˆž), because it does not

satisfy condition (๐‘“1). With the codomain [0, โˆž), the points have

one image at most. There are points, however that donโ€™t have an image.

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4. A circle with the center at the origin and with the radius ๐‘Ÿ

is the graphic of a function ๐‘“: [โˆ’๐‘Ÿ, ๐‘Ÿ] โŸถ [0,1], because all the points

in the domain have one, and only one image.

5. A circle with the center at the origin and with the radius ๐‘Ÿ

is the graphic of a function ๐‘“: [โˆ’๐‘Ÿ, ๐‘Ÿ] โŸถ [0,1].

In all these examples, the correspondence law has remained

unchanged (a circle with the center at the origin and with the radius

๐‘Ÿ, having therefore the equation ๐‘ฅ2 + ๐‘ฆ2 = ๐‘Ÿ2 , which yields ๐‘ฆ =

ยฑ(๐‘Ÿ2 โˆ’ ๐‘ฅ21/2).

By modifying just the domain and (or) the codomain, we

highlighted all possible situations, starting from the situation where

none of the required conditions that define a function were fulfilled,

to the situation where both conditions were fulfilled.

WITH THE HELP OF THE PARALLELS TO THE

COORDINATES AXES, WE RECOGNIZE THE

FULFILLMENT OF THE CONDITIONS (๐‘“1) and (๐‘“2), THUS:

a) A graphic satisfies the condition (๐‘“1) if and only if any parallel to

0๐‘ฆ traced through the points of the domain touches the graphic in at least one

point.

b) A graphic satisfies the condition (๐‘“2) if and only if any parallel to

0๐‘ฆ traced through the points of the domain touches the graphic in one point at

most.

The inverse of a function We donโ€™t always obtain a function by inverting the

correspondence law (inverting the arrow direction) for a randomly

given function ๐‘“: ๐ท โŸถ ๐ธ. Hence, in the Figure 2.3, ๐‘“ is a function,

but ๐‘“โˆ’1 (obtained by inverting the correspondence law ๐‘“) is not a

function, because it does not satisfy (๐‘“2) (there are points that have

more than an image).

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With the help of this diagram, we observe that the inverse

doesnโ€™t satisfy (๐‘“2) every time there are points in the codomain of ๐‘“

that are the image of at least two points from the domain of ๐‘“.

In other words, ๐‘“โˆ’1 doesnโ€™t satisfy (๐‘“2) every time there are

different points that have the same image through ๐‘“.

Therefore, as by inverting the correspondence law, the

condition ๐’‡ still be satisfied, it is necessary and sufficient for

the different points through the direct function have different

images, namely

A second situation where ๐‘“โˆ’1 is not a function is when it

does not satisfy condition (๐‘“1) :

We observe this happens every time there are points in the

codomain through the direct function that are not the images of any

point in the domain.

So as by inverting the correspondence law, the condition

(๐‘“1) still be satisfied, it is necessary and sufficient for all the points in

the codomain be consummated through the direct function, namely

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In conclusion,

๐‘“โˆ’1 satisfies (๐‘“2) if and only if ๐‘“ satisfies (๐‘“3)

๐‘“โˆ’1 satisfies (๐‘“1) if and only if ๐‘“ satisfies (๐‘“4)

๐‘“โˆ’1 is a function if and only if ๐‘“ satisfies (๐‘“3) + (๐‘“4).

As it is known, a function that satisfies (๐‘“3 ) is called an

injective function, a function that satisfies (๐‘“4 ) is called a

surjective function, and a function that satisfies (๐‘“3) + (๐‘“4 ) is

called a bijective function.

We see then that the affirmation: โ€œA function has an inverse

if and only if it is bijectiveโ€, has the meaning that the inverse ๐‘“โˆ’1

(that always exists, as a correspondence law, even if ๐‘“ is not

bijective) is a function if and only if ๐‘“ is bijective.

With the help of the parallels to the axes of coordinates, we

recognize if a graphic is the graphic of an injective or surjective

function; thus:

c) a graphic is the graphic of an injective function if and only if any

parallel to 0๐‘ฅ traced through the points of the codomain touches the graphic in

one point, at most [namely ๐‘“โˆ’1 satisfies (๐‘“2)].

d) a graphic is the graphic of a surjective function if and only if any

parallel to 0๐‘ฅ traced through the points of the codomain touches the graphic in

at least one point [namely ๐‘“โˆ’1 satisfies (๐‘“1)].

Examples 1. A circle with the center at the origin and with the radius ๐‘Ÿ

is the graphic of a function ๐‘“: [0, ๐‘Ÿ] โŸถ [0, โˆž) that is injective but

not surjective.

2. A circle with the center at the origin and with the radius ๐‘Ÿ

is the graphic of a function ๐‘“: [โˆ’๐‘Ÿ, ๐‘Ÿ] โŸถ [0, ๐‘Ÿ) that is surjective

but not injective.

3. A circle with the center at the origin and with the radius ๐‘Ÿ

is the graphic of a bijective function ๐‘“: [0, ๐‘Ÿ] โŸถ [0, ๐‘Ÿ).

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So, by only modifying the domain and the codomain, using a

circle with the center at the origin and with the radius ๐‘Ÿ , all

situations can be obtained, starting from the situation where none

of the two required conditions that define a function were fulfilled,

to a bijective function.

Observation ๐‘“โˆ’1 is obtained by inverting the correspondence law ๐น, namely

in other words

In the case of the exponential function, for example, the

equivalence (2.1) becomes:

because the inverse of the exponential function is noted by

๐‘“โˆ’1(๐‘ฆ) = ๐‘™๐‘œ๐‘”๐›ผ๐‘ฆ. The relation (2.2) defines the logarithm:

The logarithm of a number ๐‘ฆ in a given base, ๐‘Ž, is the exponent ๐‘ฅ to

which the base has to be raised to obtain ๐‘ฆ.

The Graphic of the Inverse Function If ๐ท and ๐ธ are subsets of ๐‘… and the domain ๐ท of ๐‘“ (so the

codomain of ๐‘“โˆ’1) is represented on the axis 0๐‘ฅ, and on the axis

0๐‘ฆ, the codomain ๐ธ (so the domain of ๐‘“โˆ’1), then the graphic of ๐‘“

and ๐‘“โˆ’1 coincides, because ๐‘“โˆ’1 only inverses the correspondence

law (it inverses the direction of the arrows).

But if we represent for ๐‘“โˆ’1 the domain, horizontally, and the

codomain vertically, so we represent ๐ธ on 0๐‘ฅ and ๐ท on 0๐‘ฆ, then

any random point on the initial graphic ๐บ๐‘“ (that, without the

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aforementioned convention is the graphic for both ๐‘“ and ๐‘“โˆ’1 ),

such a point (๐‘ฅ, ๐‘“(๐‘ฅ)) becomes (๐‘“(๐‘ฅ), ๐‘ฅ).

The points (๐‘ฅ, ๐‘“(๐‘ฅ)) and (๐‘“(๐‘ฅ), ๐‘ฅ) are symmetrical in

relation to the first bisector, so we obtain another graphic ๐บ๐‘“๐‘Ž

besides ๐บ๐‘“ if the domain of ๐’‡โˆ’๐Ÿ is on ๐ŸŽ๐’™.

Agreeing to represent the domains of all the functions on

0๐‘ฅ it follows that ๐บ๐‘“๐‘Ž is a graphic of ๐‘“โˆ’1.

With this convention, the graphics of ๐‘“ and ๐‘“โˆ’1 are

symmetrical in relation to the first bisector.

Methods to show that a function is bijective

1. Using the definition- to study the injectivity, we verify if the function

fulfills the condition (๐‘“3);

- to study the surjectivity, we verify if the condition

(๐‘“4) is fulfilled.

2. The graphical method- to study the injectivity we use proposition c);

- to study the surjectivity we use proposition d).

Important observation If we use the graphical method, it is essential, for functions

defined on branches, to trace as accurately as possible the graphic

around the point (points) of connection among branches.

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Examples

1. The function ๐‘“: โ„ โŸถ โ„, ๐‘“(๐‘ฅ) = {2๐‘ฅ + 1 ๐‘–๐‘“ ๐‘ฅ โ‰ค 1๐‘ฅ + 3 ๐‘–๐‘“ ๐‘ฅ > 1

is

injective, but is not surjective. The graphic is represented in Figure

2.5 a).

2. The function ๐‘“: โ„ โŸถ โ„ , ๐‘“(๐‘ฅ) = {3๐‘ฅ โˆ’ 1 ๐‘–๐‘“ ๐‘ฅ โ‰ค 22๐‘ฅ โˆ’ 1 ๐‘–๐‘“ ๐‘ฅ > 2

is

surjective, but is not injective. The graphic is in Figure 2.5 b).

3. The function ๐‘“: โ„ โŸถ โ„, ๐‘“(๐‘ฅ) = {2๐‘ฅ + 1 ๐‘–๐‘“ ๐‘ฅ โ‰ค 1๐‘ฅ + 2 ๐‘–๐‘“ ๐‘ฅ > 1

is

bijective. The graphic is represented in Figure 2.5 c).

3. Using the theorem A strictly monotonic function is injective.

To study the surjectivity, we verify if the function is

continuous and, in case it is, we calculate the limits at the

extremities of the domain.

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Example Letโ€™s show that ๐‘“(๐‘ฅ) = ๐‘ก๐‘” ๐‘ฅ, ๐‘“: (โˆ’

๐œ‹

2,

๐œ‹

2) โŸถ โ„ is bijective.

(i) injectivity: because ๐‘“(๐‘ฅ) =1

๐‘๐‘œ๐‘ 2๐‘ฅ is strictly positive, we

deduce that the function is strictly ascending, so, it is injective.

(ii) surjectivity: the function is continuous on all the definition

domain, so it has Darbouxโ€™s property and

from where it follows that it is surjective.

4. Using the proposition 1The function , ๐‘“: ๐ท โŸถ ๐ธ is bijective if and only if

โฉ ๐‘ฆ โˆˆ ๐ธ the equation ๐‘“(๐‘ฅ) = ๐‘ฆ has an unique solution

(see the Algebra workbook grade XII)

Example Let ๐ท = โ„š ร— โ„š and the matrix ๐ด = (3

412). Then the function

is bijective (Algebra, grade XII).

Letโ€™s observe that in the proposition used at this point, the

affirmation โ€œthe equation ๐‘“(๐‘ฅ) = ๐‘ฆ has a solutionโ€ ensures the

surjectivity of the function and the affirmation โ€œthe solution is

uniqueโ€ ensures the injectivity.

5. Using the proposition 2If ๐‘“, ๐‘”: ๐ท โŸถ ๐ท and ๐‘”. ๐‘“ = 1๐ท then ๐‘“ is injective and ๐‘” is

surjective (Algebra, grade XII).

Example Let ๐ท = โ„ค ร— โ„ค and ๐ด = ( 2

โˆ’13

โˆ’2) . Then the function

๐‘“๐ด: ๐ท โŸถ ๐ท defined through

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satisfies ๐‘“๐ด โˆ˜ ๐‘“๐ด = 1๐ท, so it is bijective ( algebra, grade XII).

Letโ€™s observe that the used proposition can be generalized, at

this point, thus:

โ€œLet ๐‘“: ๐ท โŸถ ๐ธ, ๐‘”: ๐ธ โŸถ ๐น, so that ๐‘” โˆ˜ ๐‘“ is bijective. Then ๐‘“

is injective and ๐‘” is surjective.โ€

Exercises I. Trace the graphics of the following functions and specify,

in each case, if the respective function is injective, surjective or

bijective.

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SOLUTIONS.

It is known that:

In order to more easily explain the conditions from the

inequalities over ๐‘ข, ๐‘ฃ ๐‘Ž๐‘›๐‘‘ ๐‘ค we will proceed as follows:

1. we draw the table with the signs of the functions ๐‘ข โˆ’

๐‘ฃ, ๐‘ข โˆ’ ๐‘ค and ๐‘ฃ โˆ’ ๐‘ค

2. using the table we can easily explain the inequalities,

because for example ๐‘ข(๐‘ฅ) โ‰ค ๐‘ฃ(๐‘ฅ) <===> ๐‘ข(๐‘ฅ) โ‰ค 0.

1. For ๐‘ข(๐‘ฅ) = ๐‘ฅ + 1 , ๐‘ฃ(๐‘ฅ) = ๐‘ฅ2 + 2 and ๐‘ค(๐‘ฅ) = 3๐‘ฅ we

have the following table:

So:

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Therefore,

5. To explain ๐‘“ we proceed as follows:

(1) we draw the variation table of the function ๐‘ฆ(๐‘ก) = ๐‘ก2

(2) considering ๐‘ฅ in the first monotonic interval (deduced

from the table), at the right of โˆ’1 (because we are only interested in

the values ๐‘ก โ‰ฅ โˆ’1 ), we calculate the minimum of ๐‘ฆ(๐‘ก) on the

interval ๐‘ก โˆˆ [โˆ’1, ๐‘ฅ] etc.

a) for the first monotonic interval, ๐‘ฅ โˆˆ (โˆ’1,0), the function

๐‘ฆ(๐‘ก) = ๐‘ก2 has a single minimum point on the interval [โˆ’1, ๐‘ฅ] ,

situated in ๐‘ก = ๐‘ฅ; its values is ๐‘ฆ(๐‘ฅ) = ๐‘ฅ2. Being a single minimum

point it is also the global minimum (absolute) of ๐‘ฆ(๐‘ก) on the

interval [โˆ’1, ๐‘ฅ], so, the value of ๐‘“ is ๐‘“(๐‘ฅ) = ๐‘ฅ2 for ๐‘ฅ โˆˆ (โˆ’1,0].

b) for the second monotonic interval, ๐‘ฅ โˆˆ (0, โˆž) , the

function ๐‘ฆ(๐‘ก) = ๐‘ก2 has a single minimum point on the interval

[โˆ’1, ๐‘ฅ], in ๐‘ก = 0; its value is ๐‘ฆ(0) = 0. Being a single minimum

point, we have ๐‘“(๐‘ฅ) = 0 for ๐‘ฅ โˆˆ (0, โˆž), so

Letโ€™s observe that using the same variation table we can

explain the function ๐‘”(๐‘ฅ) = ๐‘š๐‘Ž๐‘ฅ1โ‰ค1โ‰ค๐‘ฅ

๐‘ก2. Hence:

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a) for ๐‘ฅ โˆˆ (โˆ’1,0] , the function ๐‘ฆ(๐‘ก) = ๐‘ก2 , has only one

maximum, namely [โˆ’1, ๐‘ฅ] in ๐‘ก = โˆ’1; its value is ๐‘€ = ๐‘ฆ(1) = 1.

So ๐‘”(๐‘ฅ) = 1, for ๐‘ฅ โˆˆ (1,0].

b) for ๐‘ฅ โˆˆ (0, +โˆž), ๐‘ฆ(๐‘ก) = ๐‘ก2 has two maximum points, in

๐‘ก1 = โˆ’1 and ๐‘ก2 = ๐‘ฅ ; their values are ๐‘€1 = ๐‘ฆ(1) = 1 and ๐‘€2 =

๐‘ฆ(๐‘ฅ) = ๐‘ฅ2. The values of ๐‘“ is the global maximum, so:

๐‘”(๐‘ฅ) = max (1, ๐‘ฅ) for ๐‘ฅ โˆˆ (0, +โˆž).

It follows that ๐‘”(๐‘ฅ) = {1 ๐‘–๐‘“ ๐‘ฅ โˆˆ [โˆ’1,0]

max(1, ๐‘ฅ2) ๐‘–๐‘“ ๐‘ฅ > 0

9. From the variation table of ๐‘ฆ(๐‘ก) =๐‘ก4

(๐‘ก+7)3

1. for ๐‘ฅ โˆˆ (โˆ’2, ๐‘œ] , ๐‘ฆ(๐‘ก) has a single minimum on the

interval [โˆ’2, ๐‘ฅ], in ๐‘ก = ๐‘ฅ; its value is ๐‘š = ๐‘ฆ(๐‘ฅ) =๐‘ฅ4

(๐‘ฅ+7)3

2. for ๐‘ฅ > 0, ๐‘ฆ(๐‘ก) has a single minimum, namely [โˆ’2, ๐‘ฅ], in

๐‘ก = 0; its value is ๐‘š = ๐‘ฆ(0) = 0. So

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Analogously, for

10. From the variation table of ๐‘ฆ(๐‘ก) = ๐‘ก2 in ๐‘ก

we deduce:

(a) if ๐‘ฅ โˆˆ (0,1

โˆš๐‘’] , the function ๐‘ฆ(๐‘ก) = ๐‘ก2 in ๐‘ก has, for ๐‘ก โˆˆ

(0, ๐‘ฅ], a single supremum ๐‘† =๐‘™๐‘–๐‘š

๐‘ก โ†’ ๐‘œ ๐‘ก > 0

๐‘ฆ(๐‘ก) = 0 .

(b) ๐‘ฅ โˆˆ (โˆ’1

โˆš๐‘’, โˆž], the function ๐‘ฆ(๐‘ก) has two supreme values:

๐‘†1 = 0 and ๐‘€1 = ๐‘ฆ(๐‘ฅ) = ๐‘ฅ21๐‘› ๐‘ฅ.

So:

II. Study the bijectivity of:

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2. ๐‘“(๐‘ฅ) = ๐‘ƒ(๐‘ฅ), ๐‘ƒ being an uneven degree polynomial.

III. Study the irreversibility of the hyperbolic functions:

and show their inverses.

IV. 1. Show that the functions ๐‘“(๐‘ฅ) = ๐‘ฅ2 โˆ’ ๐‘ฅ + 1 ,

๐‘“: [1

2, โˆž) โŸถ โ„ and

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are inverse one to the other.

2. Show that ๐‘“(๐‘ฅ) =1โˆ’๐‘ฅ

1+๐‘ฅ coincides with its inverse.

3. Determine the parameters ๐‘Ž, ๐‘, ๐‘, ๐‘‘ so that ๐‘“(๐‘ฅ) =๐‘Ž๐‘ฅ+๐‘

๐‘๐‘ฅ+๐‘‘

coincides with its inverse.

4. Show that the function ๐‘“: (0,1] โŸถ โ„ defined

through๐‘“(๐‘ฅ) =4

3๐‘› โˆ’ ๐‘ฅ, if ๐‘ฅ โˆˆ [1

3๐‘› ,1

3๐‘›โˆ’1] is bijective.

5. On what subinterval is the function ๐‘“: [โˆ’1

2, โˆž) โŸถ โ„

๐‘“(๐‘ฅ) = โˆš๐‘ฅ โˆ’ โˆš2๐‘ฅ โˆ’ 1 bijective?

INDICATIONS.

4. To simplify the reasoning, sketch the graphic of function

๐‘“.

5. Use the superposed radical decomposition formula:

where ๐ถ = โˆš๐ด2 โˆ’ ๐ต.

Monotony and boundaries for sequences and functions

A sequence is a function defined on the natural numbersโ€™ set.

The domain of such a function is, therefore, the set of natural

numbers โ„• . The codomain and the correspondence law can

randomly vary.

Hereinafter, we will consider just natural numbersโ€™

sequences, i.e. sequences with the codomain represented by the set

of real numbers, โ„.

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๐‘“: โ„• โŸถ โ„ REAL NUMBERS SEQUENCE

For this kind of function, with the same domain and

codomain each time, only the correspondence law ๐‘“ should be

added, which comes down to specifying the set of its values:

๐‘“(0), ๐‘“(1), ๐‘“(2), โ€ฆ , ๐‘“(๐‘›), โ€ฆ

For the ease of notation, we write, for example:

So, to determine a sequence, it is necessary and sufficient to know

the set of its values: ๐‘Ž0, ๐‘Ž1, โ€ฆ , ๐‘Ž๐‘›, โ€ฆ. This set is abbreviated by

(๐‘Ž๐‘›)๐‘›โˆˆโ„•.

Consequently, a sequence is a particular case of function.

The transition from a function to a sequence is made in this

manner:

(๐‘ 1) by replacing the domain ๐ทwith โ„•

(๐‘ 2) by replacing the codomain ๐ธ with โ„

(๐‘ 3) by replacing the variable ๐‘ฅ with ๐‘› (or ๐‘š, or ๐‘–, etc)

(๐‘ 4) by replacing ๐‘“(๐‘ฅ) with ๐‘Ž๐‘›

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DOMAIN CODOMAIN VARIABLE CORRESP. LAW

Function D E x ๐‘“(๐‘ฅ) Sequence โ„• โ„ n ๐‘Ž๐‘›

Hereinafter we will use this transition method, from a

function to a sequence, in order to obtain the notations for

monotony, bounding and limit of a sequence as particular cases

of the same notations for functions.

We can frequently attach a function to a sequence (๐‘Ž๐‘›)๐‘›โˆˆโ„• ,

obtained by replacing ๐‘› with ๐‘ฅ in the expression of ๐‘Ž๐‘› ( ๐‘Ž๐‘›

becomes ๐‘“(๐‘ฅ)).

Example We can attach the function ๐‘“(๐‘ฅ) =

๐‘ฅ+3

2๐‘ฅ+1 to the sequence

๐‘Ž๐‘› =๐‘›+3

2๐‘›+1.

However, there are sequences that we cannot attach a

function to by using this method. For example:

It is easier to solve problems of monotony, bounding and

convergence of sequences by using the function attached to a

sequence, as we shall discover shortly.

Using functions to study the monotony, the bounding and

the limits of a sequence offers the advantage of using the derivative

and the table of variation.

On the other hand, it is useful to observe that the monotony,

as well as the bounding and the limit of a sequence are particular

cases of the same notion defined for functions. This

particularization is obtained in the four mentioned steps ( ๐‘ 1) โˆ’

(๐‘ 4) .

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Here is how we can obtain the particularization, first for the

monotony and bounding, then for the limit.

MONOTONIC FUNCTIONS MONOTONIC SEQUENCES

a) the function ๐‘“: ๐ท โŸถ โ„, ๐ท โŠ†โ„ is monotonically increasing if:

โฉ ๐‘ฅ1. ๐‘ฅ2 โˆˆ ๐ท, ๐‘ฅ1 โ‰ค ๐‘ฅ2 โ†’ ๐‘“(๐‘ฅ1)โ‰ค ๐‘“(๐‘ฅ2)

a) the sequence ๐‘“: โ„• โŸถ โ„ ismonotonically increasing if:

โฉ ๐‘›1, ๐‘›2 โˆˆ โ„•, ๐‘›1 โ‰ค ๐‘›2 โ†’๐‘Ž๐‘›1

โ‰ค ๐‘Ž๐‘›2 (1)

If condition (1) is fulfilled,

taking, particularly ๐‘›1 = ๐‘› and

๐‘›2 = ๐‘› + 1, we deduce

๐‘Ž๐‘› โ‰ค ๐‘Ž๐‘›+1 for any ๐‘› (2)

In conclusion (1) โŸถ (2) The reciprocal implication also stands true. Its demonstration can be deduced from the following example:

If (2) is fulfilled and we take

๐‘›1 = 7, ๐‘›2 = 11 , we have

๐‘›1 โ‰ค ๐‘›2 and step by step

and ๐‘Ž7 โ‰ค ๐‘Ž11.

Consequently, (1) โŸท (2).

Because conditions (1) and (2) are equivalent, and (2) is more convenient, we will use this condition to define the monotonically increasing sequence. But we must loose from sight the fact that it is equivalent to that particular condition that is obtained from the definition of monotonic functions, through the

particularizations ( ๐‘ 1) โˆ’ (๐‘ 4) , that define the transition from function to sequence.

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b) the function ๐‘“: ๐ท โŸถ โ„, ๐ท โŠ†โ„ is monotonically decreasing if:

โฉ ๐‘ฅ1, ๐‘ฅ2 โˆˆ ๐ท, ๐‘ฅ1 โ‰ค ๐‘ฅ2 โ†’ ๐‘“(๐‘ฅ1)โ‰ฅ ๐‘“(๐‘ฅ2)

b) the sequence ๐‘“: โ„• โŸถ โ„ is monotonically decreasing if:

โฉ ๐‘›1, ๐‘›2 โˆˆ โ„•, ๐‘›1 โ‰ค ๐‘›2 โ†’ ๐‘Ž๐‘›1

โ‰ฅ ๐‘Ž๐‘›2

It can be shown that this condition is equivalent to:

๐‘Ž๐‘› โ‰ฅ ๐‘Ž๐‘›+1 for any ๐‘› โˆˆ โ„• (4)

LIMITED FUNCTIONS LIMITED SEQUENCES

a) ๐‘“: ๐ท โŸถ โ„ , ๐ท โŠ† โ„ is of inferiorly bounded if it doesnโ€™t

have values towards โˆ’โˆž , i.e.

โˆƒ๐‘Ž โˆˆ โ„,โฉ ๐‘ฅ โˆˆ ๐ท, ๐‘“(๐‘ฅ) โ‰ฅ ๐‘Ž

b) ๐‘“: ๐ท โŸถ โ„ , ๐ท โŠ† โ„ is of superiorly bounded if it doesnโ€™t

have values towards +โˆž

i.e. โˆƒ๐‘ โˆˆ โ„,โฉ ๐‘ฅ โˆˆ ๐ท, ๐‘“(๐‘ฅ) โ‰ค ๐‘

c) ๐‘“: ๐ท โŸถ โ„ , ๐ท โŠ† โ„ is bounded, if it is inferiorly and superiorly bounded, i.e.

โˆƒ๐‘Ž, ๐‘ โˆˆ โ„,โฉ ๐‘ฅ โˆˆ โ„• ๐‘Ž โ‰ค ๐‘“(๐‘ฅ) โ‰ค๐‘. In other words, there exists an

interval [๐‘Ž, ๐‘] that contains all the values of the function. This interval is not, generally, symmetrical relative to the origin, but we can consider it so, by enlarging one of the extremities

wide enough. In this case [๐‘Ž, ๐‘] becomes [โˆ’๐‘€, ๐‘€] and the limiting condition is:

โˆƒ ๐‘€ > 0, ,โฉ ๐‘ฅ โˆˆ ๐ท โˆ’๐‘€ โ‰ค๐‘“(๐‘ฅ) โ‰ค ๐‘€ i.e.

โˆƒ ๐‘€ > 0, ,โฉ ๐‘ฅ โˆˆ ๐ท |๐‘“(๐‘ฅ)| โ‰ค ๐‘€

a) the sequence ๐‘“: โ„• โŸถ โ„ is of inferiorly bounded if it doesnโ€™t have values towards

โˆ’โˆž , i.e.

โˆƒ๐‘Ž โˆˆ โ„,โฉ ๐‘› โˆˆ โ„•, ๐‘Ž๐‘› โ‰ฅ ๐‘Ž

b) the sequence ๐‘“: โ„• โŸถ โ„ is of superiorly bounded if it doesnโ€™t have values towards

+โˆž , i.e.

โˆƒ๐‘ โˆˆ โ„,โฉ ๐‘› โˆˆ โ„•, ๐‘Ž๐‘› โ‰ฅ ๐‘

c) the sequence ๐‘“: โ„• โŸถ โ„ is bounded if it is inferiorly and superiorly bounded, i.e

โˆƒ๐‘Ž, ๐‘ โˆˆ โ„,โฉ ๐‘› โˆˆ โ„• ๐‘Ž โ‰ค ๐‘Ž๐‘› โ‰ค๐‘ OR

โˆƒ ๐‘€ > 0,โฉ ๐‘› โˆˆ โ„• |๐‘Ž๐‘›| โ‰ค ๐‘€

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Methods for the study of monotony

and bounding

METHODS FOR FUNCTIONS

METHODS FOR SEQUENCES

The study of monotony

1. Using the definition:

We consider ๐‘ฅ1 โ‰ค ๐‘ฅ2 and we

compare the difference ๐‘“(๐‘ฅ1) =๐‘“(๐‘ฅ2) to zero. This can be done through successive minorings and majorings or by applying Lagrangeโ€™s theorem to function

๐‘“ on the interval [๐‘ฅ1, ๐‘ฅ2].

2. Using the variation table (inthe case of differentiable functions) As it is known, the variation table of a differentiable function offers precise information on monotonic and bounding functions.

3. Using Lagrangeโ€™s theoremIt allows the replacement of the

difference ๐‘“(๐‘ฅ2) โˆ’ ๐‘“(๐‘ฅ1) with

๐‘“(๐‘) that is then compared to zero.

1. Using the definition:We compare the difference

๐‘Ž๐‘›+1 โˆ’ ๐‘Ž๐‘› to zero, and for sequences with positive terms we can compare the quotient

๐‘Ž๐‘›+1/๐‘Ž๐‘› to one. We can make successive minorings and majorings or by applying Lagrangeโ€™s theorem to the attached considered sequence. 2. Using the variation tablefor the attached function We study the monotony of the given sequence and, using the sequences criterion, we deduce that the monotony of the sequence is given by the monotony of this function, on

the interval [0, โˆž), (see Method 10 point c) 3. Using Lagrangeโ€™s theoremfor the attached function.

The study of bounding

1. Using the definition2. Using the variation table

1.. Using the definition 2. Using the variation tablefor the attached function 3. If the sequence

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decomposes in a finite number of bounded subsequences, it is bounded. 4. Using the monotony.If a sequence is monotonic, at least half of the bounding problem is solved, namely: a) if the sequence ismonotonically increasing, it is inferiorly bounded by the first term and only the superior bound has to be found. b) if the sequence ismonotonically decreasing, it is superiorly bounded by the first term, and only the inferior bound must be found. c) IF THE SEQUENCE ISMONOTONICALLY DECREASING AND HAS POSITIVE TERMS, IT IS BOUNDED.

Exercises I. Study the monotony and the bounding of the functions:

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ANSWERS:

3. We can write ๐‘“(๐‘ฅ) = ๐‘’๐‘™๐‘›ร—ln (ln ๐‘ฅ) and we have the

variation table:

so the function is decreasing on the interval (1, ๐‘’1/๐‘’) and increasing

on (๐‘’1/๐‘’, โˆž) . It is inferiorly bounded, its minimum being ๐‘š =

๐‘’โˆ’1/๐‘’ and it is not superiorly bounded.

5. We use the superposed radicalsโ€™ decomposition formula:

6. Let ๐‘ฅ1 < ๐‘ฅ2. In order to obtain the sign of the difference

๐‘“(๐‘ฅ1) โˆ’ ๐‘“(๐‘ฅ2) we can apply Lagrangeโ€™s theorem to the given

function, on the interval [๐‘ฅ1, ๐‘ฅ2] (the conditions of the theorem are

fulfilled): ๐‘ โˆˆ โ„ exists, so that:

so the function is increasing.

II. Study the monotony and the bounding of the sequences:

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also determine the smallest term of the sequence.

also determine the smallest term of the sequence.

๐›พ๐‘› being the intermediary value that is obtained by applying

Lagrangeโ€™s theorem to the function ๐‘“(๐‘ฅ) = ln ๐‘ฅ on the intervals

[๐‘›, ๐‘› + 1].

ANSWERS:

2. The function attached to the sequence is ๐‘“(๐‘ฅ) = ๐‘ฅ1/๐‘ฅ .

From the variation table:

we deduce, according to the sequence criterion (Heineโ€™s criterion,

see Method 10, point c), the type of monotony of the studied

sequence. The sequence has the same monotony as the attached

function, so it is decreasing for ๐‘› โ‰ฅ 3 . Because the sequence is

decreasing if ๐‘› โ‰ฅ 3, in order to discover the biggest term, we have

to compare ๐‘Ž2 and ๐‘Ž3 . We deduce that the biggest term of the

sequence is:

3. The function attached to the sequence is ๐‘“(๐‘ฅ) = ๐‘ฅ๐‘Ž๐‘ฅ .

From the variation table for ๐‘Ž โˆˆ (0,1):

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using point c) from Method 10, we deduce that the given sequence

is increasing for ๐‘›, smaller or equal to the integer part of (โˆ’ ln ๐‘Ž)โˆ’1

and is decreasing for ๐‘› โ‰ฅ [(โˆ’ ln ๐‘Ž)โˆ’1] + 1.

Because

we deduce that the interval [1, [(โˆ’ ln ๐‘Ž)โˆ’1]], in which the sequence

is increasing, can be no matter how big or small.

6. The sequence of the general term ๐‘Ž๐‘› is increasing, the

sequence of the general term ๐‘๐‘› is decreasing, and the sequence ๐‘๐‘›

is increasing if ๐‘Ž โˆˆ [0, ๐‘› โˆ’ ๐›พ๐‘›] and is decreasing for ๐‘Ž โˆˆ [, ๐‘› โˆ’

๐›พ๐‘›, 1].

III. Using Lagrangeโ€™s theorem.

1. Prove the inequality | sin ๐‘ฅ| โ‰ค |๐‘ฅ| for any real ๐‘ฅ , then

show that the sequence ๐‘Ž1 = sin ๐‘ฅ, ๐‘Ž2 = sin sin ๐‘ฅ, โ€ฆ , ๐‘Ž๐‘› =

sin sin โ€ฆ sin ๐‘ฅ, is monotonic and bounded, irrespective of ๐‘ฅ, and

its limit is zero.

2. Study the monotony and the bounding of the sequences:

ANSWERS:

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From Lagrangeโ€™s theorem applied to the function ๐‘“(๐‘›) = ๐‘ ๐‘–๐‘› ๐‘ก on

the interval [๐‘ฅ, 0] or [0, ๐‘ฅ], depending on ๐‘ฅ being either negative or

positive, we obtain |sin ๐‘ฅ

๐‘ฅ| = | cos ๐‘| โ‰ค 1. The sequence satisfies the

recurrence relation: ๐‘Ž๐‘›+1 = sin ๐‘Ž๐‘›, so if sin ๐‘ฅ > 0, we deduce that

the sequence is decreasing, and if sin ๐‘ฅ < 0 the sequence is

increasing. It is evidently bounded by โˆ’1 and 1. Noting with ๐ฟ its

limit and passing to the limit in the recurrence relation, we obtain

๐ฟ = sin ๐ฟ, so ๐ฟ = 0.

where ๐‘“(๐‘ฅ) = (1 +1

๐‘ฅ)๐‘ฅ defines the function attached to the

sequence.

b) Applying Lagrangeโ€™s theorem to the function attached to

the sequence, on the interval [๐‘›, ๐‘› + 1], we deduce๐‘Ž๐‘›+1 โˆ’ ๐‘Ž๐‘› =

๐‘“(๐‘๐‘›). We have to further determine the sign of the derivation:

namely, the sign of

From the table below we deduce ๐‘”(๐‘ฅ) > 0 for ๐‘ฅ > 0, so ๐‘Ž๐‘›+1 โˆ’

๐‘Ž๐‘› > 0.

IV. 1. Let I be a random interval and ๐‘“: ๐ผ โ†’ ๐ผ a function.

Show that the sequence defined through the recurrence relation

๐‘Ž๐‘›+1 = ๐‘“(๐‘Ž๐‘›), with ๐‘Ž๐‘œ being given, is:

a) increasing, if ๐‘“(๐‘ฅ) โˆ’ ๐‘ฅ > 0 on ๐ผ

b) decreasing if ๐‘“(๐‘ฅ) โˆ’ ๐‘ฅ < 0 on ๐ผ.

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2. If the sequence (๐‘Ž๐‘›)๐‘›โˆˆโ„• is increasing, and the sequence

(๐‘๐‘›)๐‘›โˆˆโ„• is decreasing and ๐‘Ž๐‘› โ‰ค ๐‘๐‘› for any ๐‘›, then:

a) the two sequences are bounded, and so convergent.

b) if ๐‘™๐‘–๐‘š๐‘›โ†’โˆž

(๐‘๐‘› โˆ’ ๐‘Ž๐‘›) = 0 then they have the same limit.

c) apply these results to the sequences given by the

recurrence formulas:

with ๐‘Ž๐‘œ and ๐‘๐‘œ being given.

SOLUTIONS:

1. ๐‘Ž๐‘›+1 โˆ’ ๐‘Ž๐‘› = ๐‘“(๐‘Ž๐‘›) โˆ’ ๐‘Ž๐‘› > 0 in the first case.

2. the two sequences are bounded between ๐‘Ž1 and ๐‘1, so the

sequences are convergent, because they are also monotonic. Let ๐‘™1

and ๐‘™2 represent their limits. From the hypothesis from (b) it

follows that ๐‘™1 = ๐‘™2.

(c) the sequences have positive terms and:

V. Study the bounding of the sequences:

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ANSWER: Each sequence is decomposed in two bounded

subsequences (convergent even, having the same limit), so they are

bounded (convergent).

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III. The limits of sequencesand functions

The definition of monotony and bounding of sequences has

been deduced from the corresponding definitions of functions,

using the method shown, in which ๐‘ฅ is replaced with ๐‘› and ๐‘“(๐‘ฅ) is

replaced with ๐‘Ž๐‘› . We will use the same method to obtain the

definition of the limit of a sequence from the definition of the limit

of a function. This method of particularization of a definition of

functions with the purpose of obtaining the analogous definition for

sequences highlights the connection between sequences and

functions. Considering that sequences are particular cases of

functions, it is natural that the definitions of monotony, bounding

and the limit of a sequence are particular cases of the corresponding

definitions of functions.

The limits of functions The definition of a functionโ€™s limit is based on the notion of

vicinity of a point.

Intuitively, the set ๐‘‰ โŠ‚ โ„ is in the vicinity of point ๐‘ฅ๐‘œ โˆˆ โ„,

if (1) ๐‘ฅ โˆˆ ๐‘‰, and, moreover, (2) ๐‘‰ also contains points that are in

the vicinity of ๐‘ฅ๐‘œ (at its left and right).

Obviously, if there is any open interval (๐‘Ž, ๐‘), so that ๐‘ฅ๐‘œ โˆˆ

(๐‘Ž, ๐‘) โŠ‚ ๐‘‰, then the two conditions are met.

Particularly, any open interval (๐’‚, ๐’ƒ) that contains ๐’™๐’

satisfies both conditions, so it represents a vicinity for ๐’™๐’.

For finite points, we will consider as vicinities open intervals

with the center in those points, of the type:

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and sets that contain such intervals.

For (+โˆž), ๐‘‰โˆž = (โˆž โˆ’ ํœ€, โˆž + ํœ€) doesnโ€™t make sense, so we

have to find a different form for the vicinities of +โˆž. In order to

accomplish this, we observe that to the right of +โˆž we cannot

consider any points, so โˆž + ํœ€ has to be replaced with +โˆž .

Furthermore, because โˆž โˆ’ ํœ€ = โˆž, we replace โˆž โˆ’ ํœ€ with ํœ€. So, we

will consider the vicinities of +โˆž with the form:

๐‘‰โˆž = (ํœ€, โˆž)

and sets that contain such intervals.

Analogously, the vicinities of โˆ’โˆž have the form:

๐‘‰โˆ’โˆž = (โˆ’ โˆž, ํœ€)

The definition of a functionโ€™s limit in a point expresses the

condition according to which when ๐‘ฅ approaches ๐‘ฅ๐‘œ , ๐‘“(๐‘ฅ) will

approach the value ๐‘™ of the limit.

With the help of vicinities, this condition is described by:

As we have shown, for the vicinities ๐‘‰๐‘™ and ๐‘‰๐‘ฅ๐‘œ

from the

definition (โˆ—โˆ—โˆ—), there exist three essential forms, corresponding to

the finite points and to + โˆž and โˆ’ โˆž respectively.

CONVENTION:

We use the letter ํœ€ to write the vicinities of ๐‘™, and the letter ๐›ฟ

designates the vicinities of ๐‘ฅ๐‘œ.

We thus have the following cases:

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and condition ๐‘ฅ โˆˆ and ๐‘‰๐‘ฅ๐‘œ becomes ๐‘ฅ > ๐›ฟ

and condition ๐‘ฅ๐‘œ โˆˆ and ๐‘‰๐‘ฅ๐‘œ becomes ๐‘ฅ < ๐›ฟ

All these cases are illustrated in Table 3.1.

Considering each corresponding situation to ๐‘ฅ๐‘œ and ๐‘™, in the

definition (โˆ—โˆ—โˆ—) , we have 9 total forms for the definition of a

function. we have the following 9 situations:

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Exercises I. Using the definition, prove that:

ANSWERS:

In all cases, we have ๐‘ฅ๐‘œ โˆ’ ๐‘“๐‘–๐‘›๐‘–๐‘ก๐‘’.

2. We go through the following stages:

a) we particularize the definition of limit

(taking into account the form of the vicinity).

b) we consider ํœ€ > 0, irrespective; we therefore search for

๐›ฟ .

c) we make calculation in the expression |๐‘“(๐‘ฅ) โˆ’ ๐‘™| ,

highlighting the module |๐‘ฅ โˆ’ ๐‘ฅ๐‘œ|.

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d) we increase (decrease) the obtained expression, keeping in

mind we are ony interested in those values of ๐‘ฅ for which |๐‘ฅ โˆ’

๐‘ฅ๐‘œ| < ๐›ฟ .

So:

e) if the expression that depends on ๐‘ฅ is bounded

(continuous) in a vicinity of ๐‘ฅ๐‘œ, we can further increase, to eliminate

๐‘ฅ. We have:

because in the interval [0,2], for example, that is a vicinity of ๐‘ฅ๐‘œ =

1 , the function ๐‘”(๐‘ฅ) =1

|3๐‘ฅ+2| is continuous, and therefore

bounded: ๐‘”(๐‘ฅ) โˆˆ [1

5,

1

2], if ๐‘ฅ โˆˆ [0,2].

f) we determine ๐›ฟ , setting the condition that the expression

we have reached (and that doesnโ€™t depend on ๐‘ฅ , just on ๐›ฟ ) be

smaller than ํœ€:

Any ๐›ฟ that fulfills this condition is satisfactory. We can, for

example, choose ๐›ฟ3

2ํœ€ or ๐›ฟ = ํœ€ etc.

In order to use the increase made to function ๐‘”, we must

also have ๐›ฟ โ‰ค 1, so, actually ๐›ฟ = min (1,3

2ํœ€ ) (for example).

It follows that:

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b) let ํœ€ > 0; we search for ๐›ฟ

because in the interval [1,3], for example, the function:

is bounded by:

f) we determine ๐›ฟ from the condition: ๐›ฟ๐œ€

3< ํœ€ . We obtain

๐›ฟ < 3ํœ€, so we can, for example, take ๐›ฟ = ํœ€. But in order for the

increase we have made to function ๐‘” to be valid, we must also have

๐›ฟ โ‰ค 1, so, actually ๐›ฟ = min(1, ํœ€).

It follows that:

so the condition from the definition of the limit is in this case, also,

fulfilled.

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b) let ํœ€ any (not necessarily positive, because ๐‘™ = +โˆž , so

๐‘‰๐‘™ = (ํœ€, +โˆž)only makes sense for any ํœ€, not just for ํœ€ positive).

so we can take ๐›ฟ =1

2โˆš

Then:

II. Using the definition, prove that:

ANSWERS:

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b) let ํœ€ > 0 any; we search for ๐›ฟ .

c) because ๐‘ฅ๐‘œ isnโ€™t finite anymore, we cannot highlight |๐‘ฅ โˆ’

๐‘ฅ๐‘œ |. We will proceed differently: we consider the inequality

|๐‘“(๐‘ฅ) โˆ’ ๐‘™| < ํœ€ as an inequation in the unknown ๐‘ฅ (using also the

fact that ๐‘ฅ โ†’ โˆž , so, we can consider |2๐‘ฅ + 1| = 2๐‘ฅ + 1 ). We

have:

b) let ํœ€ > 0 any; we determine ๐›ฟ .

c) we express ๐‘ฅ from the inequality ๐‘“(๐‘ฅ) > ํœ€ :

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If ๐‘ฅ1 and ๐‘ฅ2 are the roots of the attached second degree equation

and we assume ๐‘ฅ1 < ๐‘ฅ2, we have ๐‘“(๐‘ฅ) > ํœ€ โŸบ ๐‘ฅ > max (100, ๐‘ฅ2).

So we can take ๐›ฟ = max (100, ๐‘ฅ2).

The limits of sequences The definition of a sequenceโ€™s limit is deduced from the

definition of a functionโ€™s limit, making the mentioned

particularizations:

(1) ๐‘ฅ is replaced with ๐‘›,

(2) ๐‘“(๐‘ฅ) is replaced with ๐‘Ž๐‘›,

(3) ๐‘ฅ๐‘œ is replaced with ๐‘›๐‘œ.

Moreover, we have to observe that ๐‘›๐‘œ = โˆ’โˆž and finite ๐‘›๐‘œ

donโ€™t make sense (for example ๐‘› โ†’ 3 doesnโ€™t make sense, because

๐‘›, being a natural number, cannot come no matter how close to 3).

So, from the nine forms of a functionโ€™s limit, only 3 are

particularized: the ones corresponding to ๐‘ฅ๐‘œ = +โˆž.

We therefore have:

(๐‘™10) ๐‘™ โˆ’ ๐‘“๐‘–๐‘›๐‘–๐‘ก๐‘’ (we transpose the definition (๐‘™2))

Because in the inequality ๐‘› > ๐›ฟ , ๐‘› is a natural number, we

can consider ๐›ฟ as a natural number also. To highlight this, we will

write ๐‘› instead of ๐›ฟ . We thus have:

(๐‘™11) ๐‘™ = +โˆž (we transpose the definition (๐‘™5))

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(๐‘™12) ๐‘™ = โˆ’โˆž (we transpose the definition (๐‘™8))

Examples I. Using the definition, prove that:

ANSWERS:

1. We adapt the corresponding stages for functionsโ€™ limits, in

the case of ๐‘ฅ๐‘œ = +โˆž.

b) let ํœ€ > 0 any; we determine ๐‘› โˆˆ โ„•.

c) we consider the inequality: |๐‘Ž๐‘› โˆ’ ๐‘™| < ํœ€ as an inequation

with the unknown ๐‘›:

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so, ๐‘› = [2โˆ’

3] + 1

b) let ํœ€ > 0 any; we determine ๐‘› โˆˆ โ„• as follows:

Let ๐‘›1and ๐‘›2 represent the solutions to this second degree

equation (๐‘›1 < ๐‘›2). Then, for ๐‘›1 > ๐‘›2, the required inequality is

satisfied. So, ๐‘› = [๐‘›2] + 1.

Calculation methods for the limits of functions and sequences

Letโ€™s now look at the most common (for the high school

workbook level) calculation methods of the limits of functions and

sequences.

Joint methods for sequences and functions

1. DefinitionIn the first chapter, we have shown how to use the definition

to demonstrate the limits of both functions and sequences. We will

add to what has already been said, the following set of exercises:

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I. Using the definition, find out if there limits to the

following sequences and functions:

2. Giving the forced common factorThe method is frequently used to eliminate the

indeterminations of type: โˆž

โˆž, โˆž โˆ’ โˆž ,

0

0. By removing the forced

common factor we aim at obtaining as many expressions as possible

that tend towards zero. For this, we commonly employ the

following three limits:

In order to have expressions that tend towards zero we aim

therefore to:

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obtain as many terms as possible with the form ๐‘ฅ๐›ผ, with

๐›ผ < 0, if ๐‘ฅ โ†’ โˆž.

obtain as many terms as possible with the form ๐‘ฅ๐›ผ, with

๐›ผ > 0, if ๐‘ฅ โ†’ 0.

Examples I. Calculate:

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ANSWERS:

2. Given the common factor 9๐‘› we obtain terms that have

the form ๐‘ฅ๐‘›, with sub unitary ๐‘ฅ.

expression that tends to 1

5 when ๐‘ฅ tends to infinity.

II. Trace the graphics of the functions:

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8. For any rational function, non-null, ๐‘… with real

coefficients, we have:

3. Amplification with the conjugateIt is used to eliminate the indeterminations that contain

radicals. If the indetermination comes from an expression with the

form:

than we amplify with:

with the purpose of eliminating the radicals from the initial

expression. The sum from (3.1) is called a conjugate of order ๐‘.

If the indetermination comes from an expression with the

form:

with ๐‘ โˆ’ uneven number, the conjugate is:

An example of application for this formula is exercise 5,

below.

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Exercises

ANSWERS:

7. We replace, for example ๐‘Ž0 = โˆ’๐‘Ž1 โˆ’ ๐‘Ž2 โˆ’ โ‹ฏ โˆ’ ๐‘Ž๐‘˜, in the

given sequence and we calculate ๐‘˜ limits of the form:

4. Using fundamental limits Hereinafter, we will name fundamental limits the following

limits:

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OBSERVATION:

From (a) we deduce that:

From (b) we deduce that:

From (c) we deduce that:

Exercises I. Calculate:

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ANSWERS:

12. Noting with ๐›ผ = arcsin ๐‘ฅ โˆ’ ๐‘Ž๐‘Ÿ๐‘๐‘ก๐‘” ๐‘ฅ, we observe that ๐›ผ

tends to zero when ๐‘ฅ tends to zero, so we aim to obtain ๐›ผ

sin ๐›ผ. To

accomplish this we amplify with:

Noting with:

it follows:

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II. Calculate:

ANSWERS:

6. In the limits where there are indeterminations with

logarithms, we aim to permute the limit with the logarithm:

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= (we permute again the limit with the logarithm) =

III. Calculate:

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ANSWERS:

and noting ๐‘ฅ โˆ’ ๐‘Ž = ๐›ผ, we obtain:

4. We consider the function:

obtained from ๐‘Ž๐‘›, by replacing ๐‘› with ๐‘ฅ. We calculate:

According to the criterion with sequences (see Method 10,

point c), we also have: ๐‘™๐‘–๐‘š

๐‘› โ†’ โˆž๐‘Ž๐‘› = ln 2.

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5. Something that is bounded multiplied with something that tends to zero, tends to zero

(A) If (๐‘Ž๐‘›)๐‘›โˆˆโ„• is bounded and ๐‘™๐‘–๐‘š

๐‘› โ†’ โˆž๐‘๐‘› = 0, then:

๐‘™๐‘–๐‘š๐‘› โ†’ โˆž

๐‘Ž๐‘›. ๐‘๐‘› = 0

(B) If ๐‘“ is bounded in a vicinity of ๐‘ฅ0 and ๐‘™๐‘–๐‘š

๐‘ฅ โ†’ ๐‘ฅ0๐‘”(๐‘ฅ) = 0,

then ๐‘™๐‘–๐‘š

๐‘ฅ โ†’ ๐‘ฅ0๐‘“(๐‘ฅ). ๐‘”(๐‘ฅ) = 0.

Examples

because (โˆ’1)๐‘› is bounded by โˆ’1 and 1 and 1

๐‘› tends to zero;

because 1 โˆ’ cos ๐‘ฅ is bounded by 0 and 2 and 1

๐‘ฅ2 tends to zero;

where ๐‘Ž๐‘› is the nth decimal fraction of the number ๐œ‹;

where ๐›ฝ๐‘› is the approximation with ๐‘› exact decimal fractions of the

number ๐‘’.

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INDICATIONS:

4. (๐‘Ž๐‘›)๐‘›โˆˆโ„• is bounded by 0 and 9 , being composed of

numbers;

5. (๐›ฝ๐‘›)๐‘›โˆˆโ„• is bounded by 2 and 3;

6. The majoring and minoring method(A) If functions ๐‘” and โ„Ž exist, so that in a vicinity of ๐‘ฅ๐‘œ we

have:

then:

Schematically:

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(B) If the sequences (๐‘)๐‘›โˆˆโ„• and (๐‘๐‘›)๐‘›โˆˆโ„• exist so that: ๐‘๐‘› โ‰ค

๐‘Ž๐‘› โ‰ค ๐‘๐‘›, starting from the rank ๐‘›0 and ๐‘™๐‘–๐‘š

๐‘› โ†’ โˆž๐‘๐‘› =

๐‘™๐‘–๐‘š๐‘› โ†’ โˆž

๐‘๐‘› = ๐‘™,

then: ๐‘™๐‘–๐‘š

๐‘› โ†’ โˆž๐‘Ž๐‘› = ๐‘™.

Schematically:

OBSERVATION:

If ๐‘™ = +โˆž , we can eliminate โ„Ž ( (๐‘๐‘›)๐‘›โˆˆโ„• ๐‘Ÿ๐‘’๐‘ ๐‘๐‘’๐‘๐‘ก๐‘–๐‘ฃ๐‘’๐‘™๐‘ฆ) ,

and if ๐‘™ = โˆ’โˆž, we can eliminate ๐‘” ((๐‘๐‘›)๐‘›โˆˆโ„• ๐‘Ÿ๐‘’๐‘ ๐‘๐‘’๐‘๐‘ก๐‘–๐‘ฃ๐‘’๐‘™๐‘ฆ).

This method is harder to apply, due to the majorings and

minorings it presupposes. These have to lead to expressions, as least

different as the initial form as possible, not to modify the limit.

We mention a few methods, among the most frequently

used, to obtain the sequences (๐‘๐‘›)๐‘›โˆˆโ„• and (๐‘๐‘›)๐‘›โˆˆโ„•.

Examples 1. We put the smallest (biggest) term of the sequence

(๐‘Ž๐‘›)๐‘›โˆˆโ„•, instead of all the terms.

ANSWERS:

a) In the sum that expresses the sequence (๐‘Ž๐‘›)๐‘›โˆˆโ„•, there are:

the smallest term 1

โˆš๐‘›2+๐‘› and the biggest term

1

โˆš๐‘›2+1. By applying

Method 1, we have:

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2. We minorate (majorate) the terms that make up the

sequence ๐‘Ž๐‘› with the same quantity.

ANSWERS:

a) This time there isnโ€™t a smallest (biggest) term in the sum

that makes up the sequence ๐‘Ž๐‘› . This is due to the fact that the

function ๐‘“(๐‘ฅ) = sin ๐‘ฅ isnโ€™t monotonic. Keeping into consideration

that we have: โˆ’1 โ‰ค sin ๐‘ฅ โ‰ค 1 , we can majorate, replacing sin ๐‘ฅ

with 1, all over. We obtain:

Analogously, we can minorate, replacing sin ๐‘ฅ with โˆ’1 ,

then, by applying the first method it follows that:

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3. We minorate (majorate), eliminating the last terms that

make up the sequence ๐‘Ž๐‘›.

ANSWERS:

Exercises I. Calculate the limits of the sequences:

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ANSWERS:

We majorate using the smallest and the biggest denominator.

Thus, we obtain common denominators for the sum.

and we make majorings and minorings using the biggest and the

smallest denominator.

7. Exercises that feature the integer partThe most common methods to solve them is:

(A) The minoring and majoring method using the double inequality:

that helps to encase the function (the sequence) whose limit we

have to calculate, between two functions (sequences) that do not

contain the integer part.

Using this method when dealing with functions, most of the

times, we donโ€™t obtain the limit directly, instead we use the lateral

limits.

Example: For the calculation of the limit:

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we proceed as follows:

(a) We use the inequalities (3.2) to obtain expressions that

do not contain the integer part:

(b) To obtain the function whose limit is required, we

amplify the previous inequalities with ๐‘ฅ2+2๐‘ฅ

7, keeping into

consideration that:

- on the left of โˆ’2 we have: ๐‘ฅ2+2๐‘ฅ

7> 0, so:

- on the right of โˆ’2 we have: ๐‘ฅ2+2๐‘ฅ

7< 0, so:

(c) Moving on to the limit in these inequalities(the majoring

and minoring method), we obtain the limit to the left:๐‘™๐‘ (โˆ’2) =2

7

and the limit to the right: ๐‘™๐‘Ÿ(โˆ’2) =2

7, so ๐‘™ =

2

7.

(B) If the double inequality cannot be used (3.2), for

๐‘™๐‘–๐‘š๐‘ฅ โ†’ ๐‘›

๐‘“(๐‘ฅ) , with ๐‘› โˆˆ โ„ค , we calculate the lateral limits directly,

keeping into consideration that:

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Example: For ๐‘™๐‘–๐‘š

๐‘› โ†’ โˆž(โˆ’1[๐‘ฅ])/(๐‘ฅ โˆ’ 1) we cannot use (3.2)

because the expressions obtained for the calculation of the lateral

limits donโ€™t have the same limit. That is why we use the fact that an

the left of 1 we have [๐‘ฅ] = 0, and on the right of 1 we have [๐‘ฅ] =

1, so:

It follows that ๐‘™ = โˆ’โˆž

(C) If ๐‘ฅ โ†’ โˆž, we can replace [๐‘ฅ], keeping into consideration

that for ๐‘ฅ โˆˆ [๐‘š, ๐‘š + 1) , we have: [๐‘ฅ] = ๐‘š (obviously, ๐‘ฅ โ†’ โˆž

implies ๐‘š โ†’ โˆž).

Example:

The variable ๐‘ฅ that tends to infinity is certainly between two

consecutive numbers: ๐‘š โ‰ค ๐‘ฅ โ‰ค ๐‘š + 1, so:

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Exercises

8. Using the definition of the derivativeAs it is known, the derivative of a function is:

if this limit exists and is finite. It can be used to (A) eliminate the

indeterminations 0

0 for limits of functions that can be written in the

form:

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๐‘ฅ๐‘œ being a point in which ๐‘” is differentiable.

Example:

We notice that by noting ๐‘”(๐‘ฅ) = ๐‘Ž๐‘ฅ , we have ๐‘”(๐‘Ž) = ๐‘Ž๐‘Ž

and the limit becomes:

This limit has the value ๐‘”,(๐‘Ž) because ๐‘” is a function

differentiable in ๐‘Ž.

We have thus reduced the calculation of the given limit to

the calculation of the value of ๐‘”,(๐‘Ž), calculation that can be made

by deriving ๐‘”:

Observation: this method can be used like this: โ€œlet ๐‘”(๐‘ฅ) =

๐‘Ž๐‘ฅ . We have ๐‘”,(๐‘ฅ) = ๐‘Ž๐‘ฅ ln ๐‘Ž . ๐‘”,(๐‘Ž) = ๐‘Ž๐‘Ž ln ๐‘Ž = ๐‘™ . We donโ€™t

recommend this synthetic method for the elaboration of the

solution at exams because it can misguide the examiners. That is

why the elaboration: โ€œWe observe that by noting ๐‘”(๐‘ฅ) = โ‹ฏ, we

have ๐‘”(๐‘Ž) = โ‹ฏ and the limit becomes โ€ฆ = ๐‘”,(๐‘Ž) because ๐‘” is

differentiable in ๐‘Ž โ€ฆโ€is much more advisable.

(B) The calculation of the limits of sequences, using attached

functions. The function attached to the sequence is not obtained,

this time, by replacing ๐‘› with ๐‘ฅ, in the expression of ๐‘Ž๐‘›, because for

๐‘ฅ โ†’ โˆž we canโ€™t calculate the derivative, but by replacing ๐‘› with 1

๐‘ฅ

(which leads to the calculation of the derivative in zero).

Example: For ๐‘™๐‘–๐‘š

๐‘› โ†’ โˆž๐‘›( โˆš2

๐‘›โˆ’ 1) we consider the function

๐‘“(๐‘ฅ) =2๐‘ฅโˆ’1

๐‘ฅobtained by replacing ๐‘› with

1

๐‘ฅ in the expression of

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๐‘Ž๐‘› . We calculate ๐‘™๐‘–๐‘š

๐‘› โ†’ 0๐‘“(๐‘ฅ). For this we observe that by noting

๐‘”(๐‘ฅ) = ๐‘ฅ2 , we have ๐‘”(๐‘œ) = 1 and the limit

becomes: ๐‘™๐‘–๐‘š

๐‘› โ†’ 0๐‘”(๐‘ฅ)โˆ’๐‘”(๐‘œ)

๐‘ฅโˆ’0= ๐‘”,(0) , because ๐‘” is a differentiable

function in zero. We have: ๐‘”,(๐‘ฅ) = 2๐‘ฅ ln 2, so ๐‘”,(0) = ln 2. Then ,

according to the criterion with sequences, we obtain ๐‘™๐‘–๐‘š

๐‘› โ†’ โˆž๐‘Ž๐‘› =

ln 2.

Exercises

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II. For ๐‘Ž > 0 calculate:

ANSWER:

4. We divide the numerator and the denominator with ๐‘ฅ โˆ’ ๐‘Ž.

III. Show that we cannot use the definition of the derivative

for the calculation of the limits:

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ANSWERS:

2. We consider the function ๐‘“(๐‘ฅ) =2๐‘ฅโˆ’1

3๐‘ฅโˆ’1, obtained by

replacing the sequence ๐‘Ž๐‘› on ๐‘› with 1

๐‘ฅ. We calculate

๐‘™๐‘–๐‘š๐‘ฅ โ†’ 0

๐‘“(๐‘ฅ) .

To use the definition of the derivative, we divide the numerator and

the denominator with ๐‘ฅ, so:

We observe that by noting ๐‘”(๐‘ฅ) = 2๐‘ฅ , we have ๐‘”(0) = 1

and the limit from the numerator becomes: ๐‘™๐‘–๐‘š

๐‘ฅ โ†’ 0๐‘”(๐‘ฅ)โˆ’๐‘”(0)

๐‘ฅ=

๐‘”,(๐‘ฅ) because ๐‘” is differentiable in zero.

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๐‘”,(๐‘ฅ) = 2๐‘ฅ ln 2 , so, according to the criterion with

sequences ๐‘™1 = ๐‘”,(0) = ln 2 . Analogously, by noting โ„Ž(๐‘ฅ) = 3๐‘ฅ ,

the limit from the numerator becomes: ๐‘™๐‘–๐‘š

๐‘ฅ โ†’ 0โ„Ž(๐‘ฅ)โˆ’โ„Ž(0)

๐‘ฅ= โ„Ž,(0) ,

because โ„Ž is differentiable in zero.

โ„Ž(๐‘ฅ) = 3๐‘ฅ ln 3 , so, according to the criterion with

sequences๐‘™2 = โ„Ž,(0) = ln 3, and ๐‘™ =ln 2

ln 3.

5. The limit contains an indetermination with the form 1โˆž,

so we will first apply method 4.

We now consider the function ๐‘“(๐‘ฅ) =1

๐‘ฅ(๐‘Ž1

๐‘ฅ + ๐‘Ž2๐‘ฅ โˆ’ 2)

obtained by replacing ๐‘› with 1

๐‘ฅ in the expression above. We

calculate ๐‘™๐‘–๐‘š

๐‘ฅ โ†’ 0๐‘“(๐‘ฅ). For this, we observe that by noting ๐‘”(๐‘ฅ) =

๐‘Ž1๐‘ฅ + ๐‘Ž2

๐‘ฅ we have ๐‘”(0) = 2 and the limit can be written:

๐‘™๐‘–๐‘š๐‘ฅ โ†’ 0

๐‘”(๐‘ฅ)โˆ’๐‘”(0)

๐‘ฅ= ๐‘”,(0) because ๐‘” is differentiable in zero.

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Now ๐‘”,(๐‘ฅ) = ๐‘Ž1๐‘ฅ ln ๐‘Ž1 + ๐‘Ž2

๐‘ฅ ln ๐‘Ž2, so ๐‘”,(0) = ln ๐‘Ž1 + ln ๐‘Ž2

According to the criterion with sequences:

and the initial limit is โˆš๐‘Ž1. ๐‘Ž2.

9. Using Lโ€™Hospitalโ€™s theorem The theorem. [lโ€™Hospital (1661-1704)] If the functions ๐‘“ and ๐‘”

fulfill the conditions:

1. are continuous on [๐‘Ž, ๐‘] and differentiable on (๐‘Ž, ๐‘){๐‘ฅ๐‘œ}

2. ๐‘“(๐‘ฅ๐‘œ) = ๐‘”(๐‘ฅ๐‘œ) = 0

3. ๐‘”, is not annulled in a vicinity of ๐‘ฅ๐‘œ

4. there exists ๐‘™๐‘–๐‘š

๐‘ฅ โ†’ ๐‘ฅ๐‘œ

๐‘“,(๐‘ฅ)

๐‘”,(๐‘ฅ)= ๐›ผ , finite or infinite,

then there exists ๐‘™๐‘–๐‘š

๐‘ฅ โ†’ ๐‘ฅ๐‘œ

๐‘“(๐‘ฅ)

๐‘”(๐‘ฅ)= ๐›ผ.

This theorem can be used for:

(A) The calculation of limits such as ๐‘™๐‘–๐‘š

๐‘ฅ โ†’ ๐‘ฅ๐‘œ

๐‘“(๐‘ฅ)

๐‘”(๐‘ฅ) when these

contain an indetermination of the form 0

0 or

โˆž

โˆž.

(B) The calculation of limits such as ๐‘™๐‘–๐‘š

๐‘ฅ โ†’ ๐‘ฅ๐‘œ๐‘“(๐‘ฅ). ๐‘”(๐‘ฅ)

when these contain an indetermination of the form โˆž. 0.

This indetermination can be brought to form (A) like this:

(indetermination with the form โˆž

โˆž)

(indetermination with the form 0

0)

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Observation: sometimes it is essential if we write the

indetermination 0

0 or

โˆž

โˆž.

Example:

If we write:

(indetermination 0

0) and we derive, we obtain:

(indetermination with the form 0

0 with a higher numerator degree).

If we highlight the indetermination โˆž

โˆž:

and we derive, we obtain ๐‘™ = 0.

(C) The calculation of limits such as ๐‘™๐‘–๐‘š

๐‘ฅ โ†’ ๐‘ฅ๐‘œ(๐‘“(๐‘ฅ) โˆ’ ๐‘”(๐‘ฅ)),

with indetermination โˆž โˆ’ โˆž.

This indetermination can be brought to form (B) giving ๐‘“(๐‘ฅ)

or ๐‘”(๐‘ฅ) as forced common factor.

We have, for example:

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and because

we distinguish two cases:

(namely, ๐‘”(๐‘ฅ)

๐‘“(๐‘ฅ) doesnโ€™t tend to 1), we have:

the indetermination is of the form (B).

(D) The calculation of limits such as ๐‘™๐‘–๐‘š

๐‘ฅ โ†’ ๐‘ฅ๐‘œ๐‘“(๐‘ฅ)๐‘”(๐‘ฅ), with

indeterminations 1โˆž, 00, โˆž0.

These indeterminations can be brought to the form (C),

using the formula:

that for ๐‘Ž = ๐‘’, for example, becomes ๐ด = ๐‘’ln ๐ด, so:

The last limit contains an indetermination of the form (C).

For sequences, this method is applied only through one of

the two functions attached to the sequence (obtained by replacing ๐‘›

with ๐‘ฅ or by replacing ๐‘› with 1

๐‘ฅ in the expression of ๐‘Ž๐‘›). According

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to the criterion with sequences it follows that the limit of the

sequence is equal to the limit of the attached function.

Attention: by replacing ๐‘› with ๐‘ฅ we calculate the limit in โˆž,

and by replacing ๐‘› with 1

๐‘ฅ, we calculate the limit in 0.

Exercises I. Calculate:

(a polynomial increases faster to infinity than a logarithm)

(a polynomial increases slower to infinity than an exponential)

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II. Show that lโ€™Hospitalโ€™s rule cannot be applied for:

ANSWER:

1. For ๐‘“(๐‘ฅ) = ๐‘ฅ2๐‘ ๐‘–๐‘›1

๐‘ฅ and ๐‘”(๐‘ฅ) = sin ๐‘ฅ we have:

- doesnโ€™t exist, so the condition 4 from the theorem isnโ€™t

met. Still, the limit can be calculated observing that:

10. Using the criterion with sequences (the Heine Criterion)

The criterion enunciation:

with the properties:

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Observation: If ๐‘ฅ0 = โˆž , condition ๐‘) doesnโ€™t make sense

anymore.

This criterion cannot be used to eliminate indeterminations

because the sequence (๐‘ฅ๐‘›)๐‘›โˆˆโ„• from the enunciation is random, so,

by modifying the sequences (an infinite number of sequences), we

cannot eliminate the indetermination. Still, the criterion can be used

at:

(A) The calculation of limits that donโ€™t contain indeterminations.

Example: Using the criterion with sequences, show that:

Solution: (a) we have to show that:

โฉ (๐‘ฅ๐‘›)๐‘›โˆˆโ„• , with the properties:

(b) let (๐‘ฅ๐‘›)๐‘›โˆˆโ„• be any sequence with the

properties ๐‘Ž), ๐‘), ๐‘)

(c) in ๐‘™๐‘–๐‘š

๐‘› โ†’ โˆž๐‘“(๐‘ฅ๐‘›) we use the properties of the

operations of limits of sequences to highlight ๐‘™๐‘–๐‘š

๐‘› โ†’ โˆž๐‘ฅ๐‘› (it is

possible because there are no indeterminations), then we use the

fact that ๐‘™๐‘–๐‘š

๐‘› โ†’ โˆž๐‘ฅ๐‘› = ๐‘ฅ0:

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(B) To show that a function doesnโ€™t have a limit in a point.

To accomplish this we can proceed as follows:

1๐ต . we find two sequences: (๐‘ฅ๐‘›)๐‘›โˆˆโ„• and (๐‘ฆ๐‘›)๐‘›โˆˆโ„• with the

properties ๐‘Ž), ๐‘), ๐‘) so that:

or:

2๐ต. we find a single sequence with the properties ๐‘Ž), ๐‘), ๐‘)

so that ๐‘™๐‘–๐‘š

๐‘› โ†’ โˆž๐‘“(๐‘ฅ๐‘›) doesnโ€™t exist.

Example: ๐‘™๐‘–๐‘š

๐‘ฅ โ†’ โˆžsin ๐‘ฅ doesnโ€™t exist

1๐ต. let ๐‘ฅ๐‘› = 2๐‘›๐œ‹ and ๐‘ฆ๐‘› = 2๐‘›๐œ‹ +๐œ‹

2. We have:

So the limit doesnโ€™t exist.

2๐ต. let๐‘ฅ๐‘› =๐‘›๐œ‹

2. We have:

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(C) The criterion with sequences can be used in the calculation of the

limits of sequences as follows:

(a) let ๐‘“(๐‘ฅ) be the function attached to the sequence, by the

replacement of, for example, ๐‘› with ๐‘ฅ in the expression of ๐‘Ž๐‘› ( or

of ๐‘› with 1

๐‘ฅ, but then we calculate

๐‘™๐‘–๐‘š๐‘ฅ โ†’ 0

๐‘“(๐‘ฅ))

(b) we calculate ๐‘™๐‘–๐‘š

๐‘ฅ โ†’ โˆž๐‘“(๐‘ฅ) = ๐‘™ (

๐‘™๐‘–๐‘š๐‘› โ†’ 0

๐‘“(๐‘ฅ), respectively).

(c) according to the criterion with sequences lim ๐‘“(๐‘ฅ) = ๐‘™

means: for any sequence (๐‘ฅ๐‘›)๐‘›โˆˆโ„• with the properties:

we have ๐‘™๐‘–๐‘š

๐‘› โ†’ โˆž๐‘“(๐‘ฅ๐‘›) = ๐‘™

We observe that the sequence ๐‘ฅ๐‘› = ๐‘› fulfills the conditions

๐‘Ž) and ๐‘) and, moreover, for this sequence we have:

Exercises I. Using the criterion with sequences, calculate:

II. Show that the following donโ€™t exist:

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7. If ๐‘“has a limit to its left (right) in ๐‘ฅ0, then:

8. If๐‘™๐‘–๐‘š

๐‘› โ†’ โˆž๐‘“(๐‘ฅ) = ๐‘™ , then

๐‘™๐‘–๐‘š๐‘› โ†’ โˆž

๐‘“(๐‘›) = ๐‘™ , but not the

other way around.

9. ๐‘™๐‘–๐‘š

๐‘ฅ โ†’ โˆž๐‘“(๐‘ฅ) doesnโ€™t exist if ๐‘“: ๐ท โ†’ โ„ is a periodic, non-

constant function. Consequences: the following do not exist:

Answer: 9. This exercise is the generalization of exercises 1.-6.

For the solution, we explain the hypothesis:

We now use method 1๐ต . The following sequences fulfill

conditions ๐‘Ž) and ๐‘):

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and:

III. Determine the limit points of the functions: ๐‘“, ๐‘”, ๐‘“ โˆ˜

๐‘”, ๐‘” โˆ˜ ๐‘“, for:

Answer: let ๐‘ฅ0 โˆˆ ๐ท = โ„ randonm. We check if ๐‘“ has a limit

in ๐‘ฅ0. For this we observe it is essential if (๐‘ฅ๐‘›)๐‘›โˆˆโ„• is rational or

irrational. We firstly consider the situation where the sequence

(๐‘ฅ๐‘›)๐‘›โˆˆโ„• is made up only of rational points (or only of irrational

points).

Let (๐‘ฅ๐‘›)๐‘›โˆˆโ„• be a sequence of rational points with the

properties ๐‘Ž), ๐‘), ๐‘). We have:

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So the function doesnโ€™t have a limit in any point ๐‘ฅ0 for

which ๐‘ฅ02 โ‰  2, i.e. ๐‘ฅ0 and ยฑ2. For ๐‘ฅ0 = โˆš2 we consider a random

sequence (๐‘ฅ๐‘›)๐‘›โˆˆโ„• with the properties ๐‘Ž), ๐‘), ๐‘). ๐ธ1decomposes in

two subsequences: (๐‘ฅ๐‘›, )๐‘›โˆˆโ„• made up of solely rational terms and

(๐‘ฅ๐‘›, )๐‘›โˆˆโ„• made up of irrational terms. Because:

we deduce that:

The same goes for ๐‘ฅ = โˆ’โˆš2.

Specific methods for sequences

11. Any bounded and monotonic sequence is convergent

Applying this method comes back to the study of monotony

and bounding, and for the determination of the limit we pass over

to the limit in the recurrence relation of the given sequence. If such

a relation is not initially provided, it can be obtained at the study of

the monotony.

Example: ๐‘Ž๐‘› =๐‘›!

๐‘›๐‘›

(a) For the study of monotony and bounding we cannot

apply the method of the attached function (๐‘“(๐‘ฅ) =๐‘ฅ!

๐‘ฅ๐‘ฅ doesnโ€™t make

sense). We employ the method of the report (we begin by studying

the monotony because (1) if the sequence is monotonic, at least

half of the bounding problem is solved, as it has been already

proven; (2) if the sequence isnโ€™t monotonic, we canโ€™t apply the

method so we wonโ€™t study the bounding.

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We therefore have:

So the sequence is decreasing.

(b) The bounding. The sequence is bounded between 0 and

๐‘Ž1, being decreasing and with positive terms. Therefore ๐‘™๐‘–๐‘š

๐‘› โ†’ โˆž๐‘Ž๐‘›

exists.

(c) For the calculation of ๐‘™ we observe that studying the

monotony we have obtained the recurrence relation:

Moving on to the limit in this equality, we obtain:

EXERCISES:

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where (๐‘๐‘›)๐‘›โˆˆโ„• fulfills conditions:

ANSWERS:

3. We observe the recurrence relation:

a) The monotony:

In order to compare this difference to zero, we make a

choice, for example ๐‘Ž๐‘›+1 โ‰ฅ ๐‘Ž๐‘› , that we then transform through

equivalences:

So ๐‘Ž๐‘›+1 โ‰ฅ ๐‘Ž๐‘›๐‘Ž๐‘› โˆˆ (โˆ’1,2).

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We have thus reached the conclusion that the sequence is

increasing if and only if it is bounded between โˆ’๐Ÿ and ๐Ÿ. We

obviously have ๐‘Ž๐‘› > 0 > โˆ’1 so we still have to prove that ๐‘Ž๐‘› < 2.

This inequality can be proven by induction:

Verification:

So we have proven the monotony and the bounding.

Therefore ๐‘™ =๐‘™๐‘–๐‘š

๐‘› โ†’ โˆž๐‘Ž๐‘› exists. To determine ๐‘™ we move on to the

limit in the recurrence relation (this time this type of relation is

given initially):

As ๐‘™ = โˆ’1 is impossible, because ๐‘Ž๐‘› > 0, we deduce ๐‘™ = 2.

6. We decompose the fraction2๐‘˜+3

5๐‘˜ in a difference of

consecutive terms:

To determine the four parameters:

a) we use the identification method and

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b) we state the condition that the terms from the right

member in (3.4) be consecutive.

c) the denominators are consecutive, so we state the

condition that the numerators be in the same relation as

consecutives; the denominator of the first fraction is obtained from

the denominator of the second one, by replacing ๐‘˜ with ๐‘˜ โˆ’ 1, so

we must have the same relation between the numerators:

We have thus obtained the system:

with the solution:

It follows that:

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consequently:

7. We have:

so:

8. We apply the conclusion from 7.

II. Study the convergence of the sequences defined through:

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Indications:

4. sequence with positive terms (induction) and decreasing.

Let ๐‘™1 and ๐‘™2 be the limits of the subsequences of even and

uneven value. Show that ๐‘™1 = ๐‘™2.

12. Using the Cesaro-Stolz and Rizzoli Lemmas Lemma: (Cesaro-Stolz) Let (๐›ผ๐‘›)๐‘›โˆˆโ„• be a random sequence and

(๐›ฝ๐‘›)๐‘›โˆˆโ„• a strictly increasing sequence, having the limit

infinite. If ๐‘™๐‘–๐‘š

๐‘› โ†’ โˆž๐›ผ๐‘›+1โˆ’๐›ผ๐‘›

๐›ฝ๐‘›+1โˆ’๐›ฝ๐‘›= ๐‘™ exists โ€“ finite or infinite,

then:

๐‘™๐‘–๐‘š๐‘› โ†’ โˆž

๐›ผ๐‘›

๐›ฝ๐‘›= ๐‘™.

This lemma can be used to calculate the limits of sequences

that can be expressed as a fraction:

and of the limits of the form:

presupposing that this limit exists and (๐›ฝ๐‘›)๐‘›โˆˆโ„• is strictly

monotonic.

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CONSEQUENCES:

Indeed,

Example:

because for ๐‘Ž๐‘› = ๐‘›, we have:

(B) The limits of the arithmetic, geometric and harmonic

averages of the first ๐‘› terms of a sequence having the limit ๐‘™, have

the value ๐‘™ also:

Observation: a variant for the Cesaro-Stolz lemma for the case

when ๐‘Ž๐‘›, ๐›ฝ๐‘› โ†’ 0 has been proven by I.Rizzoli [Mathematic Gazette,

no 10-11-12, 1992, p. 281-284]:

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Lemma: (I.Rizzoli) If (๐›ผ๐‘›)๐‘›โˆˆโ„•, (๐›ฝ๐‘›)๐‘›โˆˆโ„• are two sequences

of natural numbers that fulfill the conditions:

(ii) the sequence (๐›ฝ๐‘›)๐‘›โˆˆโ„• is strictly monotonic

(increasing or decreasing)

(iii) there exists:

Then: ๐‘™๐‘–๐‘š

๐‘› โ†’ โˆž๐›ผ๐‘›

๐›ฝ๐‘›= ๐‘™.

EXERCISES:

8. ๐‘Ž๐‘› =1โˆ™2โˆ™ โ€ฆ.โˆ™๐‘˜+2โˆ™3โˆ™โ€ฆโˆ™(๐‘˜+1)+โ‹ฏ+(๐‘›โˆ’๐‘˜+1)โˆ™(๐‘›โˆ’๐‘˜+2)โˆ™โ€ฆ โˆ™๐‘›

๐‘›๐‘˜

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(Indications: amplification with the conjugate, the

minoring/majoring method, the Cesaro-Stolz lemma).

5. Let ๐‘ฅ๐‘›+1 = ln(1 + ๐‘ฅ๐‘›), ๐‘ฅ0 > 0. Using the Cesaro-Stolz

lemma show that ๐‘™๐‘–๐‘š

๐‘› โ†’ โˆž๐‘›. ๐‘ฅ๐‘› = 2.

(Indications: ๐‘›๐‘ฅ๐‘› =๐‘›1

๐‘ฅ๐‘›

)

III. Calculate the limits of the following sequences,

presupposing they exist:

knowing that:

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where ๐‘ > 0 and ๐‘ is Eulerโ€™s constant.

Answers:

1. We apply the Cesaro-Stolz lemma for:

We have:

and through the hypothesis it follows that:

6. We consider:

We have:

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and keeping into account that:

(we can use the attached function), we obtain:

13. Utilization of Lagrangeโ€™s theoremTheorem: (Lagrange, (1736-1813)). If ๐‘“: [๐‘Ž, ๐‘] โ†’ โ„ is

continuous on [๐‘Ž, ๐‘] and differentiable on

(๐‘Ž, ๐‘),

then: โˆƒโŠ‚โˆˆ (๐‘Ž, ๐‘)๐‘Ž. ๐‘–.๐‘“(๐‘)โˆ’๐‘“(๐‘Ž)

๐‘โˆ’๐‘Ž= ๐‘“ ,(๐‘).

This theorem can be used for:

(A) The calculation of the limits of sequences in which we

can highlight an expression with the form:

for the attached function.

Example:

Solution:

a) we highlight an expression of the form (3.6), considering:

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b) we apply Lagrangeโ€™s theorem on function ๐‘“ on the

interval [๐‘›, ๐‘› + 1]:

c) we replace in ๐‘Ž๐‘› the expression ๐‘“(๐‘)โˆ’๐‘“(๐‘Ž)

๐‘โˆ’๐‘Ž with ๐‘“ ,(๐‘๐‘›). We

obtain:

d) we use the inequalities: ๐‘› < ๐‘๐‘› < ๐‘› + 1 to obtain

convenient majorings (minorings). In our case, we have:

It follows that:

(B) The calculation of the limits of sequences that can be put

under one of the formulas:

๐‘“ being a function that is subject to Lagrangeโ€™s theorem on the

intervals: [๐‘›, ๐‘› + 1], ๐‘› โˆˆ โ„•.

Observation: the sequences of the form ๐Ÿ๐‘ฉ are always

monotonic and bounded (so convergent) if ๐’‡ is a function

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differentiable on โ„ and so that ๐’‡ and ๐’‡, have different

monotonies.

Demonstration: to make a choice, letโ€™s assume ๐‘“ is increasing

and ๐‘“ , is decreasing:

1. the monotony:

2. we apply Lagrangeโ€™s theorem to function ๐‘“ on the interval

[๐‘›, ๐‘› + 1]:

so:

(๐‘“ ,- decreasing), wherefrom we deduce the sequence is decreasing.

3. the bounding: being decreasing, the sequence is superiorily

bounded by ๐‘Ž1. We have only to find the inferior bound. For this,

we write (3.8) starting with ๐‘› = 1 and we thus obtain the

possibility of minoring (majoring) the sequence ๐‘Ž๐‘›.

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We obtain a convenient replacement for the sum:

We write the relation (3.7) starting ๐‘› = 1 and we add the

obtained equalities:

it follows that:

consequently, the sequence is inferiorly bounded. It is thus

convergent. Its limit is a number between โˆ’๐‘“(1) and ๐‘Ž1.

Exercises I. Using Lagrangeโ€™s theorem, calculate the limits of the

sequences:

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II. Study the convergence of the sequences:

(the limit of this sequence is called Eulerโ€™s constant ๐‘ โˆˆ

(0,1))

(the partial sums of the harmonic sequence โˆ‘1

๐‘–

โˆž๐‘–=1 )

(the partial sums of the generalized harmonic sequence)

Indication: 4. In order to have one of the forms 1๐ต or 2๐ต, we

determine ๐‘“, keeping into account that:

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We apply for this function the steps from the demonstration

at the beginning of this paragraph.

III. Show that:

Indication: 1. we must show that:

for:

The harmonic sequence

The sequence 1 +1

2+

1

3+ โ‹ฏ +

1

๐‘› is called a harmonic

sequence because it has the property: any three consecutive terms

are in a harmonic progression. Indeed, if we note ๐‘๐‘› =1

๐‘›, we have:

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For a long time, it was believed that this sequence has a

defined sum ๐‘ . In antiquity they sought to obtain the approximate

value of ๐‘ , by calculating the sum of as many terms of the sequence

as possible. Today it is a fact that ๐‘  = โˆž (the harmonic sequence is

divergent). We mention a few of the methods that help prove this,

methods that use exercises from high school manuals, or exercises

of high school level:

1. Lagrangeโ€™s theorem (exercise I.2)

2. The inequality:

that is proven by induction in the X grade. Indeed, if ๐‘  had a finite

value:

then, by noting:

we would have:

and because ๐‘  =๐‘™๐‘–๐‘š

๐‘› โ†’ โˆž๐‘Ž๐‘›, we deduce

๐‘™๐‘–๐‘š๐‘› โ†’ โˆž

๐‘…๐‘› = 0.

But:

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so ๐‘…๐‘› doesnโ€™t tend to zero. Therefore, ๐‘  isnโ€™t finite either.

3. Using the limit studied in the XII grade:

4. Using the inequality:

5. Using the definition of the integral (Riemann sums) (see

method 16, exercise II).

14. Sequences given by recurrence relations

(A) The linear Recurrence

1. The first degree linear recurrence Is of the form:

The expression of the general term follows from the

observation that:

so:

Example:

We have:

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2. Second degree linear recurrenceIs of the form:

In order to find the expression of the general term, in this

case, we will use the expression of the general term from the first

order recurrence. We saw that for this recurrence we have: ๐‘Ž๐‘› =

๐‘Ž0๐‘ž๐‘›. For the second order recurrence, we search for the general

term with the same form:

๐‘Ž๐‘› = ๐‘ โˆ™ ๐‘ž๐‘›, ๐‘ being an undetermined constant.

By replacing in the recurrence relation, we obtain:

from where, dividing by ๐‘ž๐‘›โˆ’1 , we obtain the characteristic

equation:

We consider the following cases:

a) โˆ†> 0 (the characteristic equation has real and distinct

roots ๐‘ž1 and ๐‘ž2)

In this case, having no reason to neglect one of the roots, we

modify the expression ๐‘Ž๐‘›, considering it of the form:

and we determine the constants ๐‘1 and ๐‘2 so that the first two

terms of the sequence have the initially given values.

Example:

(Fibonacciโ€™s sequence). The characteristic equation is:

with the roots:

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Considering ๐‘Ž๐‘› of the form:

from the system:

we obtain:

so:

Observation: ๐‘ž2 =1+โˆš5

2 is known from antiquity as the golden

ratio. It is the limit of the sequence:

(๐‘› radical) (method 11 applies). This number is often found in

nature (the arrangement of branches on trees, the proportion of the

human body etc. (For details, see, for example Matila Ghika:

Esthetics and art theory, Encyclopedic and Scientific Press, Bucharest,

1981).

b) โˆ†= 0 (the characteristic equation has equal roots ๐‘ž1 =

๐‘ž2). In this case we consider ๐‘Ž๐‘› of the form:

(namely ๐‘Ž๐‘› = ๐‘ž1๐‘› โˆ™ ๐‘ƒ1(๐‘›), with ๐‘ƒ1- first degree polynomial) and we

determine ๐‘1and ๐‘2, stating the same condition, that the first two

terms have the same initially given values.

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Example:

Taking ๐‘Ž๐‘› = ๐‘ โˆ™ ๐‘ž๐‘›, we obtain the characteristic equation:

Considering:

from the system:

we obtain: ๐‘1 = โˆ’4, ๐‘2 = 6, so:

c) ) โˆ†< 0 (the characteristic equation has complex roots)

Let these be:

We have:

but, in order to use only natural numbers, we show that we can

replace ๐‘ž1๐‘› and ๐‘ž๐‘ง

๐‘› respectively with:

Then we can write:

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3. Linear recurrence of degree h (bigger than 2)Is of the form:

with the first โ„Ž terms given.

Proceeding as with the second order recurrence, we combine

๐‘Ž๐‘› = ๐‘ โˆ™ ๐‘ž and we state the condition that the recurrence relation

be met. We obtain the characteristic equation:

We distinguish the following cases:

a) if the characteristic equation has all the roots real and

distinct: ๐‘ž1, ๐‘ž2, โ€ฆ , ๐‘žโ„Ž, we will consider ๐‘Ž๐‘› of the form:

and we determine the constants ๐‘๐‘– , ๐‘– โˆˆ 1, โ„Žฬ…ฬ… ฬ…ฬ…ฬ…, stating the condition

that the first ๐‘˜ terms of the sequence have the initially given values.

b) if a root, for example, ๐‘ž1 is multiple of order ๐‘  , (๐‘ž1 =

๐‘ž2 = โ‹ฏ = ๐‘ž๐‘ ), we replace the sum:

from the expression of ๐‘Ž๐‘› with:

๐‘ƒ๐‘ โˆ’1 being a polynomial of degree ๐‘  โˆ’ 2 , whose coefficient is

revealed by stating the condition that the first ๐‘  terms have the

initially given values.

c) if a root, for example, ๐‘ž1, is complex, then its conjugate is

also a root of the characteristic equation (let for example, ๐‘ž2 = ๐‘ž1ฬ…ฬ…ฬ…).

In this case, we replace in the expression of ๐‘Ž๐‘› the sum:

with:

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and if the root ๐‘ž1 is a multiple of order ๐‘  (๐‘ž1 = ๐‘ž2 = โ‹ฏ ๐‘ž๐‘  and

we replace in the expression of ๐‘Ž๐‘› the terms that contain the

complex root with:

EXERCISES:

I. Determine the expression of the general term and calculate

the limit for:

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II. 1. Determine ๐›ผ so that the sequence given by the

recurrence relation:

has a limit and calculate that limit.

2. Write the recurrence relations and the general terms of the

sequences for which: ๐‘Ž1 = 1 , ๐‘Ž2 = 2 , and the roots of the

characteristic equation are:

3. Let : ๐‘Ž๐‘› = ๐ด โˆ™ ๐›ผ๐‘› + ๐ต โˆ™ ๐›ฝ๐‘› , with ๐‘Ž, ๐ต, ๐›ผ, ๐›ฝ โˆˆ โ„ and

๐ด, ๐ต โ‰  0, |๐›ผ| โ‰  | ๐›ฝ|. Determine ๐›ผ ๐‘Ž๐‘›๐‘‘ ๐›ฝso that:

a) the sequence (๐‘ฅ๐‘›) is convergent;

b) ๐‘™๐‘–๐‘š

๐‘› โ†’ โˆž๐‘ฅ๐‘› = โˆž

c) ๐‘™๐‘–๐‘š

๐‘› โ†’ โˆž๐‘ฅ๐‘› = โˆ’โˆž

4. Let ๐‘Ž๐‘› = ๐ด โˆ™ ๐›ผ๐‘› + ๐ต โˆ™ ๐‘› โˆ™ ๐›ฝ๐‘› , ๐‘› โ‰ฅ 1, with ๐‘Ž, ๐ต, ๐›ผ, ๐›ฝ โˆˆ โ„

and ๐ด, ๐ต โ‰  0 . Determine ๐›ผ ๐‘Ž๐‘›๐‘‘ ๐›ฝ so that the sequence is

convergent.

(B) Nonlinear recurrence

1. Recurrence of the form ๐‘Ž๐‘›+1 = ๐›ผ โˆ™ ๐‘Ž๐‘› + ๐›ฝ,

with ๐‘Ž0 given The expression of the general term is obtained by observing

that if ๐‘™ is a root for the equation ๐‘™ = ฮฑ โˆ™ l + ฮฒ (obtained by

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replacing an+1and an with ๐‘™ in the recurrence relation), then the

sequence (an โˆ’ ๐‘™)๐‘›โˆˆโ„•

is a geometric progression.

Example:

From the equation ๐‘™ =๐‘™+1

2 we deduce ๐‘™ = ๐‘™.

Then the sequence (an โˆ’ ๐‘™)๐‘›โˆˆโ„•

is a geometric progression.

Let ๐‘๐‘› = ๐‘Ž๐‘› โˆ’ 1. We have:

The ratio of the geometric progression is, consequently: ๐‘ž =1

2. We obtain:

2. Recurrence of the form ๐‘Ž๐‘›+1 = ๐›ผ โˆ™ ๐‘Ž๐‘› + ๐‘“(๐‘›) (๐‘“ beinga random function)

The general term is found based on the observation that, if

(๐‘๐‘›)๐‘›โˆˆโ„• is a sequence that verifies the same recurrence relation,

then any sequence (๐›ผ๐‘›)๐‘›โˆˆโ„• verifies the given relation of the form:

In practice we chose ๐‘๐‘› of the form:

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the coefficients of ๐‘“ being undetermined. We determine these

coefficients by stating the condition that (๐‘๐‘›)๐‘›โˆˆโ„• verifies the

recurrence relation.

Particular case: a recurrence with the form:

with ๐‘ƒ a polynomial of degree ๐‘ .

Example:

We search for the sequence ๐‘๐‘› with the form:

We state the condition that (๐‘๐‘›)๐‘›โˆˆโ„• verifies the given

recurrence relation:

We obtain:

By identification, it follows that:

so:

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3. Recurrence of the form ๐‘Ž๐‘›+1 = ๐‘“(๐‘Ž๐‘›) (with ๐‘“: [๐‘Ž, ๐‘] โ†’โ„ continuous function)

Theorem: 1. If the function ๐‘“: [๐‘Ž, ๐‘] โ†’ โ„ is

continuous and increasing on [๐‘Ž, ๐‘], then:

a) if ๐‘Ž1 > ๐‘Ž0, the sequence (๐›ผ๐‘›)๐‘›โˆˆโ„• is increasing,

b) if ๐‘Ž1 < ๐‘Ž0, the sequence (๐›ผ๐‘›)๐‘›โˆˆโ„• is decreasing.

The limit of the sequence is a fixed point of ๐‘“, i.e. a

characteristic equation ๐‘“(๐‘ฅ) = ๐‘ฅ.

2. If ๐‘“ is decreasing on [๐‘Ž, ๐‘], then the subsequences of

even and uneven value, respectively, of (๐›ผ๐‘›)๐‘›โˆˆโ„• are

monotonic and of different monotonies. If these two

subsequences have limits, then they are equal.

Example:

We have:

from the variation table we deduce that ๐‘“ is decreasing on (0, โˆš2)

and increasing on (โˆš2, โˆž). By way of induction, we prove that

๐›ผ๐‘› > โˆš2 , and, because ๐‘Ž2 = 5,1 < ๐‘Ž1 , then the sequence is

decreasing.

The bounding can be deduced from the property of

continuous functions defined on closed intervals and bounded of

being bounded.

The characteristic equation ๐‘“(๐‘ฅ) = ๐‘ฅ has the roots ๐‘ฅ1,2 =

ยฑโˆš2, so ๐‘™ = โˆš2.

Particular case: the homographic recurrence:

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We have:

called a homographic function and:

so the monotony of ๐‘“ depends on the sign of the expression ๐›ผ โˆ™

๐›ฟ โˆ’ ๐›ฝ โˆ™ ๐›พ.

If ๐‘ฅ1 โ‰  ๐‘ฅ2 are the roots of the characteristic equation

๐‘“(๐‘ฅ) = ๐‘ฅ, by noting:

it can be shown that the general term ๐›ผ๐‘› of the sequence is given by

the relation:

Example:

Prove that:

From here it follows that the sequence is convergent (it

decomposes in two convergent sequences). It is deduced that the

limit of the sequence is 1.

Exercises I. Determine the general term and study the convergence of

the sequences:

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Determine coefficients ๐‘Ž and ๐‘ so that the sequence has a

finite limit. Calculate this limit.

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15. Any Cauchy sequence of natural numbers is convergent

One of the recapitulative exercises from the XI grade

analysis manual requires to be shown that if a sequence (๐›ผ๐‘›)๐‘›โˆˆโ„• of

real numbers is convergent, then:

A sequence that satisfies condition (3.8) is called a Cauchy

sequence or a fundamental sequence. It is proven that a natural

numbersโ€™ sequence is convergent if and only if it is a Cauchy

sequence. This propositions enables the demonstration of the

convergence of sequences, showing that they are Cauchy sequences.

In this type of exercises the following condition is used:

which is equivalent to (3.9) but easier to use.

Example:

It has to be proven that:

(b) let ํœ€ > 0. We check to see if ๐‘› exists so that property

(๐‘Ž) takes place.

(c) we have:

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(d) we state the condition: 2

๐‘›< ํœ€ and we obtain ๐‘› > :

2

๐‘›, so

we can take: ๐‘› = [ 2

๐‘›] + 1.

Exercises I. Show that the following sequences are fundamental:

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Show that |๐‘Ž๐‘›+๐‘ โˆ’ ๐‘Ž๐‘›| < ๐‘๐‘›+1 for any ๐‘›, ๐‘ โˆˆ โ„• . Deduce

from this that the sequence is a Cauchy sequence.

INDICATIONS:

and we decompose in simple fractions.

If ๐‘ is uneven, we deduce:

and if it is even:

So:

From the condition: 1

๐‘›+1< ํœ€, we obtain: ๐‘› = [

1โˆ’ 1] + 1.

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9. We proceed as we did in the previous exercise, where we

had:

16. Using the definition of the integralIt is known that the Riemann sum attached to the function

๐‘“: [๐‘Ž, ๐‘] โ†’ โ„, corresponding to a division:

and to the intermediate points:

is:

If the points ๐‘ฅ๐‘– are equidistant, then:

and we have:

A function is integrable if:

exists and is finite. The value of the limit is called the integral of

function ๐‘“ on the interval [๐‘Ž, ๐‘]:

To use the definition of the integral in the calculation of the

limits of sequences, we observe that for divisions โˆ† formed with

equidistant points, we have:

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So we can proceed as follows:

a) we show that the general term ๐‘Ž๐‘› can be written in the

form:

with ๐‘“ as a continuous function on [๐‘Ž, ๐‘], and ๐œ‰๐‘– being the points

of an equidistant division.

b) ๐‘“ , being continuous, it is also integrable, and:

Example:

a) we write ๐‘Ž๐‘› in the form:

highlighting the common factor 1

๐‘›. We have:

b) in:

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we make the equidistant points: (๐‘โˆ’๐‘Ž)๐‘–

๐‘› appear:

and we deduce the function ๐‘“. We have:

c) we arrange the equidistant points on a straight line:

and we deduce the division:

and the interval [๐‘Ž, ๐‘].

d) we observe that ๐‘Ž๐‘› is the Riemann sum corresponding to

function ๐‘“ (continuous, thus integrable) and to the deduced division

โˆ†, on interval [๐‘Ž, ๐‘]. As the points of โˆ† are equidistant, we have:

Exercises: I. Using the definition of the integral, calculate the limits of

the following sequences:

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Indications:

6. By making logarithms, we obtain:

II. a) Write the Riemann sum

corresponding to the function:

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the division:

and the intermediate points:

b) Calculate:

c) Calculate:

and deduce that:

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IV. Continuity and derivability

Continuity DEFINITION: the function ๐‘“: ๐ท โ†’ โ„ is continuous in ๐‘ฅ0 โˆˆ

๐ท if:

1. it has a limit in ๐‘ฅ0 โˆˆ,

2. the limit is equal to ๐‘“(๐‘ฅ0),

i.e.

THE CONSEQUENCE: Any continuous function in a

point has a limit in that point.

(A) Methods for the study of continuity

1. Using the definitionExample:

is continuous in ๐‘ฅ = 1.

Indeed, letโ€™s show that:

Let there be any ํœ€ > 0. We must determine ๐›ฟ . We have:

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because on the interval [0,2], for example, [2๐‘ฅ + 3] is bounded

between 3 and 7. We determine ๐›ฟ from the condition:

We can take, for example ๐›ฟ = 2ํœ€. Then:

therefore ๐‘“ is continuous in ๐‘ฅ = 1.

2. Using the criterion with the lateral limits๐’‡ is continuous in:

where

are the left and right limit, respectively in ๐‘ฅ0.

EXAMPLE: Letโ€™s study the continuity of the function:

for ๐›ผ โˆˆ [0, 2๐œ‹].

SOLUTION: Because we are required to study the

continuity, with a certain point being specified, we must make the

study on the whole definition domain. The points of the domain

appear to belong to two categories:

1. Connection points between branches, in which continuity

can be studied using the lateral limits.

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2. The other points, in which the function is continuous,

being expressed by continuous functions (elementary) (but this

must be specified each time).

In our case:

(a) in any point ๐‘ฅ0 โ‰  1, the function is continuous being

expressed through continuous functions.

(b) we study the continuity in ๐‘ฅ0 โ‰  1. We have:

The value of the function in the point is ๐‘“(1) =1

2, so ๐‘“ is

continuous in:

To explain the module we use the table with the sign of the

function sin ๐›ผ โˆ’1

2:

1. if ๐›ผ โˆˆ [0,๐œ‹

6] โˆช [

5๐œ‹

6, 2๐œ‹], the equation becomes:

2. if ๐›ผ โˆˆ (๐œ‹

6,

5๐œ‹

6), we obtain sin ๐›ผ = 1, so ๐›ผ =

๐œ‹

2.

For this four values of ๐›ผ the function is continuous also in

point ๐‘ฅ = 1, so it is continuous on โ„.

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3. Using the criterion with sequencesThis criterion is obtained from the criterion with sequences

for the limits of functions, by replacing ๐‘™ with ๐‘“(๐‘ฅ0) and

eliminating the condition ๐‘ฅ๐‘› โ‰  ๐‘ฅ0 (which is essential in the case of

limits, because we can study the limit of a function in a point where

๐‘“(๐‘ฅ) doesnโ€™t exist).

EXAMPLE:

Proceeding as we did with Method 10, with the two

adaptations mentioned just above, it follows that ๐‘“ is continuous in

the points ๐‘ฅ0 for which:

(B) Types of discontinuity points The point ๐‘ฅ0 โˆˆ ๐ท in which ๐‘“ is not continuous is said to be

a first order discontinuity point if the lateral limits in ๐‘ฅ0 exist and

are finite.

Any other discontinuity point is said to be of a second order.

(C) The extension through continuity To extend the function ๐‘“: ๐ท โ†’ โ„ means to add new points

to the domain ๐ท, where we define the correspondence law willingly.

If ๐‘€ is the set of added points and โ„Ž is the correspondence

law on ๐‘€, the extension will be:

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For example:

can be extended to the entire set โ„ by putting:

There are many extensions to a function. Nevertheless, if ๐‘“

is continuous on ๐ท and has a finite limit in a point ๐‘ฅ0 โˆˆ ๐ท, there is

only one extension:

that is continuous, named the extension through continuity of ๐‘“ in

point ๐‘ฅ0.

So, for our example, because ๐‘™๐‘–๐‘š

๐‘ฅ โ†’ 0sin ๐‘ฅ

๐‘ฅ= 1, the function:

is the only extension of sin ๐‘ฅ/๐‘ฅ that is continuous on โ„.

(D) The continuity of composite functions If ๐‘“ is continuous in ๐‘ฅ0 and ๐‘” is continuous in ๐‘“(๐‘ฅ0), then

๐‘” โˆ˜ ๐‘“ is continuous in ๐‘ฅ0.

So the composition of two continuous functions is a

continuous function. Reciprocally it isnโ€™t true: it is possible that ๐‘“

and ๐‘” are not continuous and ๐‘” โˆ˜ ๐‘“ is.

For example:

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isnโ€™t continuous in any point, but the function ๐‘”(๐‘ฅ) = (๐‘“ โˆ˜ ๐‘“)(๐‘ฅ)

is the identically equal function to 1, so it is continuous in any point

from โ„.

Exercises I. Study, on the maximum definition domain, the continuity

of the functions defined by:

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II. Specify the type of discontinuity points for the functions:

Determine the extensions using the continuity, where

possible.

III. Study the continuity of the functions๐‘“, ๐‘”, ๐‘“ โˆ˜ ๐‘”, ๐‘” โˆ˜ ๐‘“ ,

for:

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IV. 1. Let ๐‘“, ๐‘”: โ„ โ†’ โ„ be a continuous function so that

๐‘“(๐‘ฅ) = ๐‘”(๐‘ฅ) for any ๐‘ฅ โˆˆ โ„š. Show that ๐‘“ = ๐‘”. (Manual)

2. Let ๐‘“: ๐ผ โ†’ โ„ , ๐ผ โŠ†โ†’ โ„ be a function with Darbouxโ€™s

property. If ๐‘“ has lateral limits in any point ๐ผ, then ๐‘“ is continuous

on ๐ผ. (Manual)

3. ๐‘“: โ„ โ†’ โ„ so that:

for any ๐‘ฅ, ๐‘ฆ โˆˆ โ„. Prove that ๐‘Ž > 0 exists so that for any ๐‘ฅ with

|๐‘ฅ| โ‰ค ๐‘Ž we have |๐‘“(๐‘ฅ)| < ๐‘Ž. Deduce the existence of a fixed point

for ๐‘“. (๐‘ฅ0 is a fixed point for ๐‘“ if ๐‘“(๐‘ฅ0) = ๐‘ฅ0). (Manual)

4. Let ๐‘“: [๐‘Ž, ๐‘] โ†’ โ„, continuous. Then:

(Manual)

A function with the property (4.1) is called an uniform

continuously function on [๐’‚, ๐’ƒ].

The uniform continuity is therefore defined on an interval,

while continuity can be defined in a point.

Taking ๐‘ฆ = ๐‘ฅ0 in (4.1) it is deduced that any uniform

continuous function on an interval is continuous on that interval.

Exercise 4 from above affirms the reciprocal of this

proposition, which is true if the interval is closed and bounded.

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Derivability

(A) Definition. Geometric interpretation. Consequences

Definition The function ๐‘“: ๐ท โ†’ โ„ is differentiable in ๐‘ฅ0 โˆˆ ๐ท if the

following limit exists and is finite:

The value of this limit is noted by ๐‘“ ,(๐‘ฅ0) and it is called the

derivative of ๐‘“ in ๐‘ฅ0.

Geometric interpretation The derivative of a function in a point ๐‘ฅ0 is the slope of the

tangent to the graphic of ๐’‡in the abscissa point ๐’™๐ŸŽ.

When ๐‘ฅ tends to ๐‘ฅ0 , the chord ๐ด๐ต tends towards the

tangent in ๐ด to the graphic, so the slope of the chord

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tends to the slope of the tangent. Therefore, ๐‘“ ,(๐‘ฅ0) is the slope of

the tangent in the abscissa point ๐‘ฅ0 to the graphic.

The geometric interpretation of the derivative thus follows

from the next sequence of implications:

Consequences 1. Any function differentiable in a point is continuous in that

point.

2. The equation of the tangent to the graphic of a

differentiable function.

It is known that equation of a straight line that passes

through the point (๐‘ฅ0, ๐‘“(๐‘ฅ0)) is:

3. The equation of the normal (the perpendicular on the

tangent) is deduced from the condition of perpendicularity of two

straight lines (๐‘š1 โˆ™ ๐‘š2 = โˆ’1) and it is:

(B) The derivation of composite functions

From formula (4.2) it is deduced that:

So, in order to derive a composite function we will proceed

as follows:

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(a) we derive the last function that is composed (in our case

๐‘“3) and we replace in this function the variable ๐‘ฅ with the

expression issuing from the composition of the other function (in

our case ๐‘“2 โˆ˜ ๐‘“1).

(b) by neglecting the derived function in the previous step (in

our case ๐‘“3), we derive the function that became last (for us ๐‘“2) and

we also replace the variable ๐‘ฅ with the expression that resulted from

the composition of the functions that have no yet been derived (in

this step we only have ๐‘“1(๐‘ฅ).

(c) we continue this procedure until all the functions are

derived.

EXAMPLE:

1. By applying formula (4.2) we have:

We can obtain the same results much faster, using the

generalization of formula (4.2) presented above.

Therefore, the composite functions are:

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CONSEQUENCES:

1. The derivative of the inverse function.

From ๐‘“โˆ’1(๐‘“(๐‘ฅ)) = ๐‘ฅ, by applying formula (4.2), it follows

that:

or by noting ๐‘ฆ = ๐‘“(๐‘ฅ):

EXAMPLE: For ๐‘“(๐‘ฅ) = ln ๐‘ฅ, ๐‘“: (0, โˆž) โ†’ โ„ we have:

and as ๐‘ฆ = ln ๐‘ฅ => ๐‘ฅ = ๐‘’๐‘ฆ we obtain:

2. High order derivatives of composite functions.

From (๐‘“(๐‘ข(๐‘ฅ)) = ๐‘“ ,(๐‘ข(๐‘ฅ)). ๐‘ข,(๐‘ฅ) , deriving again in

relation to ๐‘ฅ it follows that:

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We have therefore obtained the second order derivative of

the composite function and the procedure can go on.

EXAMPLE: Letโ€™s calculate ๐‘“ ,,(๐‘ฅ) if ๐‘“(๐‘ฅ) = ๐‘”(๐‘”โˆ’๐‘ฅ) , ๐‘”

being a function two times differentiable on โ„.

We have:

and

(C) Derivatives of order ๐‘› For the calculation of the derivative ๐‘“(๐‘›)(๐‘ฅ) we can use one

of the following methods:

1. We calculate a few derivatives (๐‘“ ,, ๐‘“ ,,, ๐‘“ ,,,, โ€ฆ ) in order to

deduce the expression of ๐‘“(๐‘›), which we then demonstrate by way

of induction.

2. We use Leibniz formula for the derivation of the product

of two functions:

The formula can be applied to a random quotient, because:

(D) The study of derivability In order to study the derivability of a function:

(a) we distinguish two categories of points that the domain

has: points in which we know that the function is differentiable

(being expressed by differentiable functions) and points in which we

pursue the study of derivability (generally connection points

between branches).

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(b) for the study of derivability in this second category of

points, we use one of the following methods:

(1) we calculate the lateral derivatives using the

definition

(2) we calculate the lateral derivatives using the

Corollary of Lagrangeโ€™s theorem: If ๐‘“ is

differentiable in a vicinity of ๐‘ฅ0 and if it

continues in ๐‘ฅ0 and if there exists:

then, ๐‘“ has a derivative on the left (on the right,

respectively) in ๐‘ฅ0 and:

The continuity condition in ๐’™๐ŸŽ that is required in the

hypothesis of the corollary is essential for its application, but it

doesnโ€™t constitute a restriction in the study of derivability,

because if ๐‘“ isnโ€™t continuous in ๐‘ฅ0, it isnโ€™t differentiable either.

EXAMPLE: Letโ€™s determine the parameters ๐›ผ, ๐›ฝ โˆˆ โ„ so

that the function:

is differentiable. What is the geometric interpretation of the result?

Answer: Method 1. (using the lateral derivatives)

a) in any point ๐‘ฅ โ‰  ๐‘’ , the function is differentiable, being

expressed through differentiable functions.

b) we study the derivability in ๐‘ฅ = ๐‘’.

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Because of the proposition: ๐’‡ non continuous ==> ๐’‡ non

differentiable, we study the continuity first.

so: ๐‘“ continuous in ๐‘’ <=> ๐›ผ๐‘’ + ๐›ฝ = 1. (4.3)

If the continuity condition in ๐‘ฅ = ๐‘’ isnโ€™t met, the function is

also not differentiable in this point, so we study the derivability

assuming the continuity condition is fulfilled.

For the calculation of this limit we can:

(1) apply the definition of the derivative,

(2) permute the limit with the logarithm,

(3) use lโ€™Hospitalโ€™s rule.

(1) We observe that by noting ๐‘”(๐‘ฅ) = ๐‘™๐‘›2๐‘ฅ , we have

๐‘”(๐‘’) = 1 and the limit becomes:

(because ๐‘” is differentiable in ๐‘’)

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Therefore, ๐‘“ is differentiable in:

and from the continuity condition we obtain ๐›ฝ = โˆ’1 . We thus

have:

The geometric interpretation of the result: the straight

line ๐‘ฆ =2

๐‘’๐‘ฅ โˆ’ 1 is tangent to the curve ๐‘ฆ = ln ๐‘ฅ2.

If the branch of a function is expressed through a

straight line (๐’š = ๐œถ๐’™ + ๐œท) , the function is

differentiable in the point ๐’™๐ŸŽ of connection

between the branches, if and only if the

respective straight line is tangent in the point of

abscissa ๐’™๐ŸŽ to the graphic of the other branch.

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Method 2: (Using the corollary of Lagrangeโ€™s theorem)

(a) In any point ๐‘ฅ โ‰  ๐‘’, the function is differentiable, being

expressed by differentiable function and we have:

(b) We study the derivability in ๐‘ฅ = ๐‘’ . We first state the

continuity condition:

In order to use the corollary of Lagrangeโ€™s theorem we

calculate the lateral limits of the derivative:

From the corollary we deduce ๐‘“ ,(๐‘’) = ๐›ผ. Analogously:

If ๐‘“ is differentiable in ๐‘ฅ = ๐‘’ if and only if๐›ผ =2

๐‘’. From the

continuity condition we deduce ๐›ฝ = โˆ’1.

(E) Applications of the derivative in economics

(See the Calculus manual, IXth grade)

1. Let ๐›ฝ(๐‘ฅ) be the benefit obtained for an expenditure of ๐‘ฅ

lei. For any additional expenditure of โ„Ž lei, the supplementary

benefit on any spent leu is:

If โ„Ž is sufficiently small, this relation gives an indication of

the variation of the benefit corresponding to the sum of ๐‘ฅ lei.

If this limit exists:

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it is called the bound benefit corresponding to the sum of ๐’™ lei.

2. Let ๐›พ(๐‘) be the total cost for the production of ๐‘ units of

a particular product. Then, the cost per each supplementary unit of

product is:

and the limit of this relation, when โ„Ž tends to zero, if it exists, is

called a bound cost of production for ๐‘ units of the considered

product.

EXAMPLE: The benefit obtained for an expenditure of ๐‘ฅ

lei is:

For any additional expenditure of โ„Ž lei, calculate the

supplementary benefit per spent leu and the bound benefit

corresponding to the sum of 1000 lei.

Answer: The supplementary benefit per spent leu is:

For ๐‘ฅ = 1000 this is equal to 1997 + โ„Ž and

Exercises I. The following are required:

1. The equation of the tangent to the graphic ๐‘“(๐‘ฅ) =

ln โˆš1 + ๐‘ฅ2 in the abscissa point ๐‘ฅ0 = 1.

2. The equation of the tangent to the curve ๐‘“(๐‘ฅ) =

โˆš๐‘ฅ2 โˆ’ ๐‘˜2, that is parallel to 0๐‘ฅ.

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3. The equation of the tangent to the curve ๐‘“(๐‘ฅ) = ๐‘ฅ3 ,

parallel to the first bisector.

4. The equation of the tangent to the curve ๐‘ฆ =3๐‘ฅ+2

2๐‘ฅ+6, that is

parallel to the chord that unites the abscissae points ๐‘ฅ = 1 and ๐‘ฅ =

3.

5. Determine ๐›ผ and ๐›ฝ so that ๐‘ฆ = ๐›ผ๐‘ฅ + ๐›ฝ and ๐‘ฆ =๐‘ฅโˆ’1

๐‘ฅ are

tangent in ๐‘ฅ = 1. Write down their common tangent.

6. Show that the straight line ๐‘ฆ = 7๐‘ฅ โˆ’ 2 is tangent to the

curve ๐‘ฆ = ๐‘ฅ2 + 4๐‘ฅ. (Manual)

II. 1. Calculate the derivative of the function:

2. Let ๐‘“: (โˆ’โˆž, 0) โ†’ โ„ , ๐‘“(๐‘ฅ) = ๐‘ฅ2 โˆ’ 3๐‘ฅ . Determine a

subinterval ๐ฝ โŠ† โ„, so that ๐‘“: (โˆ’โˆž, 0) โ†’ ๐ฝ is bijective. Let ๐‘” be its

inverse. Calculate ๐‘”,(โˆ’1) and ๐‘”,,(โˆ’1). (Manual)

3. If:

with ๐›ผ > 0 for any ๐‘ฅ, ๐‘ฆ โˆˆ ๐ผ, the function ๐‘“ is constant on ๐ผ.

4. If ๐‘“ has a limit in point ๐‘Ž, then the function:

is differentiable in ๐‘Ž.

5. If ๐‘“ is bounded in a vicinity of ๐‘ฅ0 , then ๐‘”(๐‘ฅ) =

(๐‘ฅ โˆ’ ๐‘ฅ0)2๐‘“(๐‘ฅ) is differentiable in ๐‘ฅ0. Particular case:

Indications:

1. Let ๐‘“ be a primitive of the function:

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We have:

3. For ๐‘ฅ โ‰  ๐‘ฆ the inequality from the enunciation is

equivalent to:

from where, for ๐‘ฆ โŸถ ๐‘ฅ we obtain ๐‘“(๐‘ฅ) = 0, so ๐‘“ is constant on ๐ผ.

III. Calculate the derivatives of order ๐‘› for:

10. ๐‘“(๐‘ฅ) = ๐‘’๐‘Ž๐‘ฅ + ๐‘’๐‘๐‘ฅ from the expression of the derivative

of order ๐‘›, deduce Newton's binomial formula.

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IV. 1. Show that:

2. Applying Leibnizโ€™s formula for:

show that:

Find similar formulas, using the functions:

3. Let ๐ผ = (0,1) and the functions ๐‘ข, ๐‘ฃ: ๐ผ โŸถ โ„ , ๐‘ข(๐‘ฅ) =

๐‘ข(๐‘ฅ) =๐‘–๐‘›๐‘“

๐‘ฆ โˆˆ ๐ผ(๐‘ฅ โˆ’ ๐‘ฆ)2 , ๐‘ฃ(๐‘ฅ) ==

๐‘ ๐‘ข๐‘๐‘ฆ โˆˆ ๐ผ(๐‘ฅ โˆ’ ๐‘ฆ)2 . Study the

derivability of the functions ๐‘ข and ๐‘ฃ and calculate ๐‘ ๐‘ข๐‘

๐‘ฅ โˆˆ ๐ผ๐‘ข(๐‘ฅ) ,

๐‘–๐‘›๐‘“๐‘ฅ โˆˆ ๐ผ

๐‘ฃ(๐‘ฅ). (Manual)

4. If ๐‘“๐‘›(๐‘ฅ) is a sequence of differentiable function, having a

limit in any point ๐‘ฅ, then:

V. Study the derivability of the functions:

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is indefinitely differentiable on โ„ and:

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๐‘Ž โˆˆ โ„.

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V. Fermat, Rolle, Lagrange, Cauchy Theorems

(A) Fermatโ€™s theorem

1. EnunciationIf ๐‘“: [๐‘Ž, ๐‘] โ†’ โ„ is continuous on [๐‘Ž, ๐‘] and differentiable

on (๐‘Ž, ๐‘), then in any point of extremum from (๐‘Ž, ๐‘) (from

the interior of the interval), the derivative is annulled.

(Fermat, 1601-1665)

Observation: ๐‘ฅ0 โˆˆ (๐‘Ž, ๐‘) point of extremum ==> ๐‘“ ,(๐‘ฅ0) =

0. If ๐‘ฅ0 is a point of extremum at the ends of the interval, it is

possible that the derivative isnโ€™t annulled in ๐‘ฅ0.

Example:

has two points of extremum: in ๐‘ฅ1 = โˆ’1 and in ๐‘ฅ2 = 2, and:

2. Geometric and algebraic interpretationa) The geometric interpretation results from the

geometrical interpretation of the derivative: if the conditions of

Fermatโ€™s theorem are fulfilled in any point from the interior of the

interval, the tangent to the functionโ€™s graphic is parallel to the axis

0๐‘ฅ.

b) The algebraic interpretation: if the conditions of

Fermatโ€™s theorem are fulfilled on [๐‘Ž, ๐‘] , any point of extremum

from (๐‘Ž, ๐‘) is a root to the equation ๐‘“ ,(๐‘ฅ) = 0.

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Exercises 1. If ๐‘Ž1, ๐‘Ž2, โ€ฆ ๐‘Ž๐‘› are positive numbers, so that:

for any ๐‘ฅ โˆˆ โ„, then we have:

2. Let ๐‘Ž1, ๐‘Ž2, โ€ฆ ๐‘Ž๐‘›, ๐‘1, ๐‘2, โ€ฆ ๐‘๐‘› be positive real numbers, so

that:

for any ๐‘ฅ โˆˆ โ„. Then :

(generalization of the previous exercise)

3. If ๐‘“ is continuous on [๐‘Ž, ๐‘] and differentiable on (๐‘Ž, ๐‘),

and ๐‘“(๐‘Ž) = ๐‘“(๐‘) = 0, then:

(i) if ๐‘“ , is increasing, it follows that ๐‘“(๐‘ฅ) โ‰ค 0 on [๐‘Ž, ๐‘];

(ii) ) if ๐‘“ ,, is decreasing, it follows that ๐‘“(๐‘ฅ) โ‰ฅ 0 on [๐‘Ž, ๐‘].

4. If ๐‘Ž๐‘ฅ โ‰ฅ ๐‘ฅ๐‘Ž for any ๐‘ฅ > 0, then ๐‘Ž = ๐‘’.

5. ๐‘ก๐‘” ๐‘Ž

๐‘ก๐‘” ๐‘<

๐‘Ž

๐‘ if 0 < ๐‘Ž < ๐‘ <

๐œ‹

2.

SOLUTIONS:

1. The exercise is a particular case of exercise 2.

2. We highlight a function that has a point of extremum

(global) on โ„, observing that the inequality from the hypothesis can

be written:

Because this inequality is true for any real ๐‘ฅ, it follows that ๐‘ฅ = 0 is

a global point of minimum for ๐‘“. According to Fermatโ€™s theorem,

in this point the derivative is annulled, so ๐‘“ ,(0) = 0. We have:

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3. (i) Letโ€™s assume there exists ๐‘ โˆˆ (๐‘Ž, ๐‘) so that ๐‘“(๐‘) > 0.

We can even assume that ๐‘ is a point of maximum, because such a

point exists, according to Rolleโ€™s theorem, so we have ๐‘“ ,(๐‘) = 0.

Between ๐‘ and ๐‘ at least one point ๐‘‘ exists in which

๐‘“ ,(๐‘‘) < 0 , because, if, letโ€™s say ๐‘“ ,(๐‘ฅ) โ‰ฅ 0 for any ๐‘ฅ โˆˆ (๐‘, ๐‘) , it

follows that ๐‘“ is increasing on (๐‘, ๐‘) and so ๐‘“(๐‘) < ๐‘“(๐‘) = 0

implies ๐‘“(๐‘) = 0. Then, from ๐‘“ ,(๐‘‘) =โ‰ค 0 = ๐‘“ ,(๐‘) we deduce the

contradiction:

4. ๐‘Ž๐‘ฅ โ‰ฅ ๐‘ฅ๐‘Ž <=> ๐‘ฅ. ln ๐‘Ž โ‰ฅ ๐‘Ž. ln ๐‘ฅ <=>ln ๐‘Ž

๐‘Žโ‰ฅ

ln ๐‘ฅ

๐‘ฅ (for any

๐‘ฅ > 0). So ๐‘Ž is the abscissa of the maximum of the function:

We have:

It is sufficient to prove that ๐‘“(๐‘ฅ) =๐‘ก๐‘” ๐‘ฅ

๐‘ฅ is increasing, i.e.

๐‘“ ,(๐‘ฅ) > 0.

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(B) Rolleโ€™s theorem

1. Enunciation If ๐‘“: [๐‘Ž, ๐‘] โ†’ โ„ is continuous on [๐‘Ž, ๐‘], differentiable

on (๐‘Ž, ๐‘) and ๐‘“(๐‘Ž) = ๐‘“(๐‘), then there exists ๐‘ โˆˆ

(๐‘Ž, ๐‘) so that ๐‘“ ,(๐‘) = 0. (Rolle, 1652-1719)

Fermatโ€™s theorem states that in a point of extremum from

the interior of an interval the derivative is annulled, but it doesnโ€™t

mention when such a point exists.

Rolleโ€™s theorem provides a sufficient condition for the

existence of at least one such point, adding to the hypothesis from

Fermatโ€™s theorem the condition: ๐‘“(๐‘Ž) = ๐‘“(๐‘).

2. Geometric and algebraic interpretation a) The geometric interpretation: if the conditions of

Rolleโ€™s theorem are met, there exists at least one point in the

interval (๐‘Ž, ๐‘)in which the tangent to the graphic is parallel to 0๐‘ฅ.

b) The algebraic interpretation: if the conditions of

Rolleโ€™s theorem are met, the equation ๐’‡,(๐’™) = ๐ŸŽ has at least one

root in the interval (๐‘Ž, ๐‘).

The algebraic interpretation of the theorem highlights a

method used to prove that the equation ๐‘“(๐‘ฅ) = 0 has at least one

root in the interval (๐‘Ž, ๐‘). For this, it is sufficient to consider a

primitive ๐น of ๐‘“, for which the conditions of Rolleโ€™s theorem are

fulfilled on [๐‘Ž, ๐‘]. It results that the equation ๐น,(๐‘ฅ) = 0 has at least

one root in the interval (๐‘Ž, ๐‘), i.e. ๐‘“(๐‘ฅ) has a root in the interval

(๐‘Ž, ๐‘).

A second method to show that the equation ๐‘“(๐‘ฅ) = 0 has

at least one root in the interval (๐‘Ž, ๐‘) is to show just that ๐‘“ is

continuous and ๐‘“(๐‘Ž). ๐‘“(๐‘) < 0.

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3. Consequences1. Between two roots of a differentiable function on an

interval, there exists at least one root of the derivative.

2. Between two roots of a functionโ€™s derivative on an

interval, there exists, at most a root of the function.

Consequence 2. allows us to determine the number of roots of

a function on an interval with the help of its derivativeโ€™s roots

(Rolleโ€™s sequence):

Let ๐‘ฅ1, ๐‘ฅ2, โ€ฆ , ๐‘ฅ๐‘› be the derivativeโ€™s roots.

Then ๐‘“ has as many real, simple roots as there are variations

of sign in the sequence:

(if we have ๐‘“(๐‘ฅ๐‘–) = 0, then ๐‘ฅ๐‘– is a multiple root)

Exercises I. Study the applicability of Rolleโ€™s Theorem for the

functions:

What is the geometric interpretation of the result?

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Indication: 1. We obtain ๐‘ โˆˆ [4,5] . In any point from the

interval [4,5], the graphic coincides with the tangent to the graphic.

II. Given:

Show that the equation:

has at least one solution in the interval (0, 2๐œ‹). (Manual)

2. If:

then the equation:

has at least one root in the interval (0,1).

3. (i) If:

then the equation:

has at least one root in the interval (0,1).

(ii) in what conditions the same equation has at least one

root in the interval (โˆ’1,0)?

(iii) The same question for the equations:

on the interval (โˆ’1,1).

4. Let ๐‘“: โ„ โ†’ โ„ be differentiable and ๐‘Ž1 < ๐‘Ž2 < โ‹ฏ < ๐‘Ž๐‘› ,

roots of ๐‘“. Show that ๐‘“ ,has at least ๐‘› โˆ’ 1 roots.

(Manual)

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Consequences:

a) A polynomial function of degree ๐‘› has, at most, ๐‘› real,

distinct zeros.

b) If all the roots of a polynomial are real and distinct, its

derivative has the same property.

c) If all the roots of a polynomial are real, then its derivative

has the same property.

5. Given:

Show that the equation ๐‘“๐‘›(๐‘ฅ) = 0 has ๐‘› distinct roots in

the interval (โˆ’1,1).

6. If ๐‘“ is ๐‘› times differentiable on ๐ผ and it has ๐‘› + 1 distinct

roots on ๐ผ, then ๐‘“๐‘›(๐‘ฅ) has at least one root on ๐ผ.

7. Let ๐‘“, ๐‘”: [๐‘Ž, ๐‘] โ†’ โ„, continuous on [๐‘Ž, ๐‘], differentiable

on (๐‘Ž, ๐‘) with ๐‘”(๐‘ฅ) โ‰  0 and ๐‘”,(๐‘ฅ) โ‰  0 on [a,b], and ๐‘“(๐‘Ž)

๐‘”(๐‘Ž)=

๐‘“(๐‘)

๐‘”(๐‘).

Show that ๐‘ โˆˆ (๐‘Ž, ๐‘) exists, so that:

8. Let๐‘“: [๐‘Ž, ๐‘] โ†’ โ„, continuous on (๐‘Ž, ๐‘), differentiable on

(๐‘Ž, ๐‘). Then between two roots of ๐‘“ there exists at least a root of

๐›ผ. ๐‘“ + ๐‘“ ,.

9. If the differentiable functions ๐‘“ and ๐‘” have the property:

on an interval, then between two of its roots there is a root of ๐‘” and

vice versa.

Consequence: If ๐‘“ ,(๐‘ฅ). cos ๐‘ฅ + ๐‘“(๐‘ฅ). sin ๐‘ฅ โ‰  0 , in any

length interval bigger than ๐œ‹ there is at least one root of ๐‘“.

10. If ๐‘“, ๐‘”: [๐‘Ž, ๐‘] โ†’ โ„, continuous on [๐‘Ž, ๐‘], differentiable

on (๐‘Ž, ๐‘) , ๐‘”(๐‘ฅ) โ‰  0, ๐‘”,(๐‘ฅ) โ‰  0 and ๐‘“(๐‘Ž) = ๐‘“(๐‘) = 0, then ๐‘๐‘› โˆˆ

(๐‘Ž, ๐‘) so that:

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SOLUTIONS:

7. The required equality can also been written:

This equality appears in points ๐‘ that are roots of the

derivative of โ„Ž(๐‘ฅ) =๐‘“(๐‘ฅ)

(๐‘ฅ)๐‘”, so it is sufficient to demonstrate that the

function โ„Ž fulfills the conditions of Rolleโ€™s theorem on [๐‘Ž, ๐‘].

8. The equation ๐›ผ. ๐‘“(๐‘ฅ) + ๐‘“ ,(๐‘ฅ) = 0 comes from the

equalizing with zero the derivative of the function ๐น(๐‘ฅ) =

๐‘’๐›ผ๐‘ฅ๐‘“(๐‘ฅ), so it is sufficient to show that ๐น satisfies the conditions of

Rolleโ€™s theorem.

9. Let ๐‘ฅ1, ๐‘ฅ2 be roots of ๐‘“. By hypothesis, we have๐‘”(๐‘ฅ1) โ‰ 

0 and ๐‘”(๐‘ฅ2) โ‰  0. If, for example, there can be found a root of ๐‘”

between ๐‘ฅ1, ๐‘ฅ2 , this means that the function โ„Ž(๐‘ฅ) =๐‘“(๐‘ฅ)

(๐‘ฅ)๐‘” satisfies

the conditions of Rolleโ€™s theorem and so โ„Ž,(๐‘ฅ) is annulled in at least

one point between ๐‘ฅ1 and ๐‘ฅ2, which contradicts the hypothesis.

has at least one root in (๐‘Ž, ๐‘) <=> โ„Ž,(๐‘ฅ) = 0 has at least one root

in (๐‘Ž, ๐‘), where โ„Ž(๐‘ฅ) =๐‘“(๐‘ฅ)

๐‘”๐‘›(๐‘ฅ).

We will present next, two methods for the study of an

equationโ€™s roots. The first of these uses Rolleโ€™s Theorem and the

other is a consequence of the fact that any continuous function has

Darbouxโ€™s property.

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Methods for the study of an equationโ€™s roots

1. Using Rollesโ€™s theorem (its algebraicinterpretation) 2. Using Darbouxโ€™s properties (if ๐‘“ has Darbouxโ€™s

property (particularly if it is continuous) on [๐‘Ž, ๐‘] and:

๐‘“(๐‘Ž). ๐‘“(๐‘) โ‰ค 0, then ๐‘“ has a root in [๐‘Ž, ๐‘]).

Examples:

1) If ๐‘“: [๐‘Ž, ๐‘] โ†’ [๐‘Ž, ๐‘] is a continuous function, then ๐‘ข, ๐‘ฃ โˆˆ

[๐‘Ž, ๐‘] exists so that ๐‘“(๐‘ข) = ๐‘ข and ๐‘“(๐‘ฃ) = ๐‘ฃ. (Manual)

2) Let ๐‘“: [0, 2๐œ‹] โ†’ โ„ be a continuous function so that

๐‘“(0) = ๐‘“(2๐œ‹). Show that ๐‘ โˆˆ [0, ๐œ‹] exists so that ๐‘“(๐‘) = ๐‘“(๐‘ +

๐‘›). (Manual)

Consequences: If a traveler leaves in the morning from spot

๐ดand arrives, in the evening in spot ๐ต, and the next day he leaves

and reaches again spot ๐ด, show that there is a point between ๐ด and

๐ต where the traveler has been, at the same hour, in both days.

Indication: If ๐‘†(๐‘ก) is the space covered by the traveler, we

have: ๐‘†(0) = ๐‘†(24) and so ๐‘ก0 โˆˆ [0,12] exists, so that: ๐‘†(๐‘ก0) =

๐‘†(๐‘ก0 + 12).

By formulating the problem like this, the result might seem

surprising, but it is equal to saying that two travelers that leave, one

from spot ๐ด heading to spot ๐ต, and the other from ๐ต to ๐ด, meet on

the way.

3. Using Rolleโ€™s sequenceExample: Letโ€™s study the nature of the equationโ€™s roots:

๐‘Ž being a real parameter.

Answer: The derivativeโ€™s roots are:

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and we have:

The results are illustrated in the following table:

4. The graphical method. The equation ๐น(๐‘ฅ, ๐‘š) = 0 is

reduced to the form ๐‘“(๐‘ฅ) = ๐‘Ž. The number of roots is equal to the

number of intersection points between the graphics: ๐‘ฆ = ๐‘“(๐‘ฅ) and

๐‘ฆ = ๐‘š.

Example:

The number of roots of the given equation is equal to the

number of intersection points between the graphics:

5. Vietteโ€™s relationsExample: Show that the equation:

with ๐‘Ž, ๐‘, ๐‘ โˆˆ โ„ has, at most, two real roots.

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Answer: we have:

so the equation has at least two complex roots.

6. Using the theorem of the averageTheorem: If ๐‘“: [๐‘Ž, ๐‘] โ†’ โ„ is a Riemann integral (so it is

bounded), ๐œ‡ โˆˆ [๐‘š, ๐‘€] exists, so that:

Example: Let ๐‘“: [0,1] โ†’ โ„ be continuous, so that:

Show that the equation ๐‘“(๐‘ฅ) = ๐‘ฅ has a root ๐‘ฅ0 โˆˆ (0,1).

Answer: ๐‘“(๐‘ฅ) = ๐‘ฅ <=> ๐‘“(๐‘ฅ) โˆ’ ๐‘ฅ = 0 , so by noting:

โ„Ž(๐‘ฅ) = ๐‘“(๐‘ฅ) โˆ’ ๐‘ฅ we have to show that the equation โ„Ž(๐‘ฅ) = 0 has

a root in the interval (0,1). We have:

But: ๐‘“02 โ„Ž(๐‘ฅ) = 0, according to the theorem of the average,

with ๐œ‡ โˆˆ (๐‘š, ๐‘€). The function ๐‘“, being continuous, has Darbouxโ€™s

property, so, for ๐œ‡ there exists ๐‘ฅ0 โˆˆ (0,1)so that ๐œ‡ = โ„Ž(๐‘ฅ0).

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Exercises 1. The equations:

canโ€™t all have real roots if ๐›ผ, ๐›ฝ, ๐›พ โˆˆ โ„.

Indication:

2. If ๐‘“: [0,1] โ†’ โ„ is continuous and with the property:

then, the equation ๐‘“(๐‘ฅ) โˆ’ sin ๐œ‹๐‘ฅ = 0 has one root in (0,1).

3. If ๐‘“: [0,1] โ†’ โ„ is continuous and with the property:

then, the equation ๐‘“(๐‘ฅ) โˆ’ ๐‘Ž๐‘ฅ2 โˆ’ ๐‘๐‘ฅ โˆ’ ๐‘ = 0 has one root in

(0,1).

4. If ๐‘“: [0,1] โ†’ โ„ is continuous and๐‘› > 1 exists for which:

the equation: (1 โˆ’ ๐‘ฅ)๐‘“(๐‘ฅ) = 1 โˆ’ ๐‘ฅ๐‘› has one root in (0,1).

Indication: We apply the theorem of the average to the

function:

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(C) Lagrangeโ€™s Theorem 1. Enunciation:

Let ๐‘“: [๐‘Ž, ๐‘] โ†’ โ„ be a continuous function

on [๐‘Ž, ๐‘] and differentiable on (๐‘Ž, ๐‘) . Then ๐‘ โˆˆ

(๐‘Ž, ๐‘) exists, so that:

(Lagrange, 1736 โ€“ 1813)

2. Geometric and algebraic interpretation

a) Geometric interpretation:If the conditions of the theorem are met, there exists a point

๐‘ โˆˆ (๐‘Ž, ๐‘) in which the tangent to the graphic is parallel to the

chord that unites the graphicโ€™s extremities.

b) Algebraic interpretationIf the conditions of the theorem are met, the equation:

has at least one root in the interval (๐‘Ž, ๐‘).

3. The corollary of Lagrangeโ€™s TheoremIf ๐‘“ is continuous on an interval ๐ผ and differentiable on

๐ผ\{๐‘ฅ0} and in ๐‘ฅ0 there exists the derivativeโ€™s limit:

then ๐‘“ is differentiable in ๐‘ฅ0 and:

This corollary facilitates the study of a functionโ€™s derivability

in a point in an easier manner that by using the lateral derivatives.

Examples:

1. Study the derivability of the function:

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Solution: In any point ๐‘ฅ โ‰  1, the function is differentiable,

being expressed by differentiable functions. We study the

derivability in ๐‘ฅ = 1 using the corollary of Lagrangeโ€™s theorem.

- the continuity in ๐‘ฅ = 1:

๐‘“(1) = ln 1 = 0 , so ๐‘“ is continuous on โ„ (and

differentiable on โ„{1}).

- derivability. For the study of derivability in ๐‘ฅ = 1, we have:

(we know that ๐‘“ , exists just on โ„ โˆ’ (1)).

Exercises I. Study the applicability of Lagrangeโ€™s theorem and

determine the intermediary points ๐‘ for:

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Indication:

2. The continuity condition in ๐‘ฅ =๐œ‹

6 is:

and the derivability condition (the corollary to Lagrangeโ€™s theorem)

is:

Point ๐‘ is the solution of the equation:

II. 1. Let ๐‘“(๐‘ฅ) = ๐‘Ž๐‘ฅ2 + ๐‘๐‘ฅ + ๐‘. Apply the theorem of the

finite increases on the interval [๐‘ฅ1, ๐‘ฅ2] , finding the intermediary

point ๐‘ . Deduce from this a way to build the tangent to the

parabola, in one of its given points.

2. We have ๐‘š โ‰ค ๐‘“ ,(๐‘ฅ) โ‰ค ๐‘€ for any ๐‘ฅ โˆˆ ๐ผ, if and only if:

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3. (The generalization of Rolleโ€™s theorem). If ๐‘“ is continuous

on [๐‘Ž, ๐‘]and differentiable on (๐‘Ž, ๐‘), then ๐‘ โˆˆ (๐‘Ž, ๐‘) exists so that:

4. Supposing ๐‘“ is twice differentiable in a vicinity ๐‘‰ of point

๐‘Ž, show that for โ„Ž, small enough, there exists ๐‘, ๐‘ž โˆˆ ๐‘‰ so that:

(Manual)

5. Let ๐‘“: [0, โˆž) โ†’ โ„ be a differentiable function so that:

and let ๐‘Ž > 0 be fixed. Applying the Lagrange theorem on each

interval:

show there exists a sequence (๐‘ฅ๐‘›)๐‘›โˆˆโ„•, having the limit infinite and

so that:

6. The theorem of finite increases can also be written:

Apply this formula to the functions:

and study the values corresponding to the real point ๐‘.

SOLUTIONS:

1. ๐‘ =๐‘ฅ1+๐‘ฅ2

2, so, the abscissa point ๐‘ being given, in order to

build the tangent in point (๐‘, ๐‘“(๐‘)) of the graphic, we consider two

points, symmetrical to ๐‘:

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The tangent passes through (๐‘, ๐‘“(๐‘)) and is parallel to the

chord determined by the points:

2. The necessity: For ๐‘ฅ = ๐‘ฆ (๐ฟ1) is verified, and for ๐‘ฅ โ‰  ๐‘ฆ,

we have:

inequalities which are true due to the hypothesis.

Reciprocally, let ๐‘ฅ โˆˆ ๐ผ, randomly. Making ๐‘ฆ tend to ๐‘ฅ, for:

we have:

3. We distinguish three cases:

In this case we must prove that ๐‘ โˆˆ (๐‘Ž, ๐‘)exists, so that:

๐‘“ ,(๐‘) > 0. If we had, letโ€™s say, ๐‘“ ,(๐‘) โ‰ค 0 on (๐‘Ž, ๐‘), it would follow

that ๐‘“ is decreasing, so ๐‘“(๐‘Ž) > ๐‘“(๐‘).

We are situated within the conditions of Rolleโ€™s theorem.

(3) The case ๐‘“(๐‘Ž) > ๐‘“(๐‘) is analogous to (1).

4. a) We apply Lagrangeโ€™s theorem to ๐‘“ on [๐‘Ž โˆ’ โ„Ž, ๐‘Ž + โ„Ž].

b) We apply Lagrangeโ€™s theorem twice to the function

๐‘”(๐‘ฅ) = ๐‘“(๐‘Ž + ๐‘ฅ) โˆ’ ๐‘“(๐‘ฅ) on the interval [๐‘Ž โˆ’ โ„Ž, ๐‘Ž].

6. a) ๐‘ = 1/2 , c) is obtained ๐‘šโ„Ž = ๐‘šโ„Ž , so ๐‘ cannot be

determined.

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(D) Cauchyโ€™s Theorem 1.Enunciation

Let If ๐‘“, ๐‘”: [๐‘Ž, ๐‘] โ†’ โ„ be continuous on[๐‘Ž, ๐‘] and

differentiable on (๐‘Ž, ๐‘) so that:

Then ๐‘ โˆˆ (๐‘Ž, ๐‘) exists, so that:

(Cauchy, 1789-1857)

2. Geometric and algebraic interpretationa) The geometric interpretation: in the conditions of the

theorem, there exists a point in which the relation of the tangentsโ€™

slopes to the two graphics is equal to the relation of the chordsโ€™

slopes that unite the extremities of the graphics.

b) The algebraic interpretation: in the conditions of the

theorem, the equation:

has at least one root in (๐‘Ž, ๐‘).

Exercises I. Study the applicability of Cauchyโ€™s theorem and determine

the intermediary point ๐‘ for:

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II. If ๐‘“ has derivatives of any order on [๐‘Ž, ๐‘], by applying

Cauchyโ€™s formula to the functions:

a) ๐‘”(๐‘ฅ) = ๐‘“(๐‘ฅ), โ„Ž(๐‘ฅ) = ๐‘ โˆ’ ๐‘ฅ, show that ๐‘ โˆˆ (๐‘Ž, ๐‘) exists,

so that:

show that ๐‘ โˆˆ (๐‘Ž, ๐‘) exists, so that:

show that ๐‘ โˆˆ (๐‘Ž, ๐‘) exists, so that:

show that ๐‘ โˆˆ (๐‘Ž, ๐‘) exists, so that:

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2. Applying the conclusion from point ๐‘) show that there

exists no functions ๐‘“: โ„ โ†’ โ„ twice differentiable, so that:

๐‘“(๐‘ฅ) โ‰ฅ 0 and ๐‘“ ,,(๐‘ฅ) < 0

(there are no non-negative and strictly concave functions on the

entire real axis).

SOLUTION:

2. From ๐‘“ ,,(๐‘ฅ) < 0 for any ๐‘ฅ โˆˆ โ„ , we deduce that ๐‘“ , is

strictly increasing and so we cannot have ๐‘“ ,(๐‘ฅ) = 0 for any ๐‘ฅ โˆˆ โ„.

Therefore, for any ๐‘ฅ0 โˆˆ โ„ exists, so that ๐‘“ ,(๐‘ฅ) โ‰  0.

But then, from the relation:

it follows that:

We have two situations:

(1) if ๐‘“ ,(๐‘ฅ0) > 0, making ๐‘ฅ tend to โˆ’โˆž in (๐ถ1), we have:

consequently:

2. If ๐‘“ ,(๐‘ฅ0) < 0, we obtain the same contradiction if we

make ๐‘ฅ tend to +โˆž.

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VI. Equalities and Inequalities

Equalities It is known that if the derivative of a function is equal to

zero on an interval, the respective function is constant on that

interval. This observation allows us to prove certain equalities of

the form:

๐‘“(๐‘ฅ) = ๐‘”(๐‘ฅ) + ๐ถ.

or, in particular,

๐‘“(๐‘ฅ) = ๐‘”(๐‘ฅ) and ๐‘“(๐‘ฅ) = ๐ถ = ๐‘๐‘œ๐‘›๐‘ ๐‘ก๐‘Ž๐‘›๐‘ก.

Indeed:

and in order to prove this equality it is sufficient to prove that:

(๐‘“(๐‘ฅ) โˆ’ ๐‘”(๐‘ฅ)) = 0.

Observation: If the derivative is equal to zero on a reunion of

disjunctive intervals, the constant may vary from one interval to the

next.

To determine the constant ๐ถ, we can use two methods:

1. We calculate the expression ๐‘“(๐‘ฅ) โˆ’ ๐‘”(๐‘ฅ) in a

conveniently chosen point from the considered interval.

2. It the previous method cannot be applied, we can

calculate:

๐‘ฅ0 also being a conveniently chosen point (one end of the interval).

Example: Show that we have:

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Solution: For:

we have โ„Ž,(๐‘ฅ) = 0 for any ๐‘ฅ โˆˆ โ„(โˆ’1), domain that is a reunion of

disjunctive intervals. We calculate the value of the constant on each

interval.

(1) For ๐‘ฅ > โˆ’1, we choose the point ๐‘ฅ = 0 where we can

easily make the calculations:

(2) For ๐‘ฅ < โˆ’1 , we canโ€™t find a point in which to easily

calculate the value of โ„Ž, but we observe that we can calculate:

Inequalities Method 1. Using Lagrangeโ€™s or Cauchyโ€™s

theorem.

In some inequalities, we can highlight an expression of the

form:

๐‘“ and the points ๐‘Ž, ๐‘being conveniently chosen. In this case, we can

use Lagrangeโ€™s theorem to prove the respective inequality, by

replacing the expression:

from the inequality, with ๐‘“ ,(๐‘). The new form of the inequality can

be proven taking into account the fact that ๐‘Ž < ๐‘ < ๐‘ and the

monotony of ๐‘“ ,.

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A similar method can be used if in the inequality an

expression of the following form is highlighted:

using Cauchyโ€™s theorem.

EXAMPLES:

1. Using Lagrangeโ€™s theorem, show that:

if 0 < ๐‘Ž < ๐‘.

Solution: We go through these steps:

(a) We highlight an expression of the form:

observing that:

and dividing by ๐‘ โˆ’ ๐‘Ž. The inequality becomes:

(b) we apply Lagrangeโ€™s theorem to the function ๐‘“(๐‘ฅ) = ln ๐‘ฅ

on the interval [๐‘Ž, ๐‘]. ๐‘ โˆˆ (๐‘Ž, ๐‘) exists so that:

(c) the inequality becomes:

This new form of the inequality is proven considering the

fact that ๐‘Ž < ๐‘ < ๐‘ and that ๐‘“ ,(๐‘ฅ) =1

๐‘ฅ is decreasing.

2. Using Lagrangeโ€™s theorem, show that ๐‘’๐‘ฅ > 1 + ๐‘ฅ for any

๐‘ฅ โ‰  0.

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Solution:

To use Lagrangeโ€™s theorem, we search for an interval [๐‘Ž, ๐‘]

and a function so that we can highlight the relation ๐‘“(๐‘)โˆ’๐‘“(๐‘Ž)

๐‘โˆ’๐‘Ž in our

inequality. For this we observe that, if ๐‘ฅ โ‰  0 , we have two

situations:

(1) ๐‘ฅ > 0. In this case we can consider the interval [๐‘œ, ๐‘ฅ].

a) We highlight, in the given inequality, an expression of the

form: ๐‘“(๐‘ฅ)โˆ’๐‘“(๐‘Ž)

๐‘ฅโˆ’0, observing that, if ๐‘ฅ > 0, we have:

(b) We apply Lagrangeโ€™s theorem to the function:

on the interval [0, ๐‘ฅ]. ๐‘ โˆˆ (0, ๐‘ฅ) exists so that:

(c) The inequality becomes:

This form of the inequality is proven considering the fact

that 0 < ๐‘ < ๐‘ฅ and the derivative ๐‘“ ,(๐‘ฅ) = ๐‘’๐‘ฅis increasing (so ๐‘’0 <

๐‘’๐‘).

(2) If ๐‘ฅ < 0 the demonstration is done analogously.

Method 2: The method of the minimum This method is based on the observation that if ๐‘ฅ0 is a global

point of minimum (the smallest minimum) for a function โ„Ž on a

domain ๐ท and if โ„Ž(๐‘ฅ0) โ‰ฅ 0 (the smallest value of โ„Ž is non-

negative), then โ„Ž(๐‘ฅ) โ‰ฅ 0 for any ๐‘ฅ โˆˆ ๐ท (all the values of โ„Ž are

non-negative). We can demonstrate therefore inequalities of the

form ๐‘“(๐‘ฅ) โ‰ฅ ๐‘”(๐‘ฅ), i.e. ๐‘“(๐‘ฅ) โˆ’ ๐‘”(๐‘ฅ) โ‰ฅ 0, in other words, of the

form: โ„Ž(๐‘ฅ) โ‰ฅ 0.

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Example: Using the method of the minimum, show that ๐‘’๐‘ฅ >

1 + ๐‘ฅ for any ๐‘ฅ โ‰  0.

Solution: In order to use this method, we proceed as follows:

(a) We write the inequality under the form: โ„Ž(๐‘ฅ) โ‰ฅ 0:

(b) Let โ„Ž(๐‘ฅ) = ๐‘’๐‘ฅ โˆ’ 1 โˆ’ ๐‘ฅ . Using the variation table, we

calculate the global minimum (the smallest minimum) of โ„Ž:

From the table it can be observed that โ„Ž has a single

minimum, in ๐‘ฅ = 0, so this is also the global minimum. Its value is

โ„Ž(0) = 0. So, for ๐‘ฅ โ‰  0, we have โ„Ž(๐‘ฅ) > 0.

Method 3: (Inequalities for integrals, without the calculation of the integrals)

To prove an inequality of the form:

without calculating the integral, we use the observation that if ๐‘š, or

๐‘€ respectively, are values of the global minimum and maximum of

๐‘“ on the interval [๐‘Ž, ๐‘], we have:

and so, by integrating, we obtain:

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so, in order to prove the required inequalities, the following

inequalities (numerical) still have to be proven:

Example: Show that:

Solution: (a) The inequality can be written under the form:

(b) Using the variation table, we calculate the

global minimum and maximum of the function ๐‘“(๐‘ฅ) = ๐‘’๐‘ฅ2+

๐‘’1โˆ’๐‘ฅ2on the interval [0,1]:

We thus have ๐‘š = 2โˆš๐‘’ and ๐‘€ = 1 + ๐‘’

(c) We integrate the inequalities ๐‘š โ‰ค ๐‘“(๐‘ฅ) โ‰ค ๐‘€

on the interval [๐‘Ž, ๐‘] and we obtain:

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Method 4: Using the inequality from the definition of convex (concave) functions

Let ๐ผ be an interval on a real axis.

Definition: The function ๐‘“: ๐ผ โ†’ โ„ is convex on ๐ผ if between any

two points ๐‘ฅ1, ๐‘ฅ2 โˆˆ ๐ผ , the graphic of ๐‘“ is situated

underneath the chord that unites the abscissae points

๐‘ฅ1 and ๐‘ฅ2.

With the notations from the adjacent graphic, this condition

is expressed through the inequality:๐‘“(๐‘ฅ) โ‰ค ๐‘ฆ.

In order to explain ๐‘ฅ and ๐‘ฆ, we observe that:

1. ๐‘ฅ is in the interval [๐‘ฅ1, ๐‘ฅ2] if and only if it can be written

in the form:

Indeed, if ๐‘ฅ โˆˆ [๐‘ฅ1, ๐‘ฅ2], it is sufficient that we take ๐›ผ = (๐‘ฅ โˆ’

๐‘ฅ2)/๐‘ฅ1 โˆ’ ๐‘ฅ2) to meet the required equality.

Reciprocally, if ๐‘ฅ has the expression from (๐ธ. ๐ผ. 1), in order

to show that ๐‘ฅ โˆˆ [๐‘ฅ1, ๐‘ฅ2], the inequality system must be checked:

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The first inequality is equivalent to ๐›ผ โ‰ค 1, and the second

inequality is equivalent to ๐›ผ โ‰ค 0.

Observation: The condition ๐›ผ โˆˆ [0,1] from (๐ธ. ๐ผ. 1) is

therefore essential.

The inequality from the convexity becomes:

2. To express ๐‘ฆ, we observe that:

so:

from which it follows:

By replacing ๐‘ฅ and ๐‘ฆ, we obtain:

๐‘“: ๐ผ โ†’ โ„ is convex <=>

๐‘“: ๐ผ โ†’ โ„ is said to be concave if, for any ๐‘ฅ1, ๐‘ฅ2 โˆˆ ๐ผ, between

the abscissae points ๐‘ฅ1 and ๐‘ฅ2 , the graphic of the function ๐‘“ is

above the chord that unites these points, namely:

It is known that we can verify the convexity and concavity of

a function twice differentiable on an interval with the use of the

second derivative:

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Using the two ways of expressing convexity, we can

formulate and prove certain inequalities.

Example: Show that for any ๐‘ฅ1 and ๐‘ฅ2 from โ„, we have:

Solution: The function ๐‘“(๐‘ฅ) = ๐‘’๐‘ฅ is convex on โ„(๐‘“ ,,(๐‘ฅ) >

0), so:

from where, for ๐›ผ =1

2 we obtain the inequality from the

enunciation.

Exercises I. Show that:

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5. For the intervals ๐ผ1, ๐ผ2 โŠ‚ โ„ letโ€™s consider the continuous

and injective functions:

and the constant ๐‘ in the interior of ๐‘“(๐ผ1) . From the

condition: ๐‘“(๐‘ข(๐‘ฅ)) = ๐‘ + ๐‘”(๐‘ฅ) it follows that ๐‘ข(๐‘ฅ) = ๐‘“โˆ’1(๐‘ +

๐‘”(๐‘ฅ) for any ๐‘ฅ that fulfills the condition ๐‘ + ๐‘”(๐‘ฅ) โˆˆ ๐‘“(๐ผ1). Using

this observation, construct equalities of the form โ„Ž(๐‘ฅ) = ๐‘ for:

Answer: (a) We have:

We determine function ๐‘ข from the condition ๐‘“(๐‘ข(๐‘ฅ)) =

๐‘ + ๐‘”(๐‘ฅ) that becomes:

For ๐‘ =๐œ‹

4, for example, we have:

and from the condition:

we deduce:

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and so we have:

II. Using Lagrangeโ€™s theorem, prove the inequalities:

6. Using Cauchyโ€™s theorem applied to the functions:

show that:

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III. Using the method of the minimum, demonstrate the

inequalities:

IV. Without calculating the integrals, show that:

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V. Without calculating the integral, show which one of the

following integrals has the highest value:

Indication: Using the method of the minimum, we prove

inequalities of the form ๐‘“1(๐‘ฅ) < ๐‘“2(๐‘ฅ) or ๐‘“1(๐‘ฅ) > ๐‘“2(๐‘ฅ) on the

considered intervals. We then integrate the obtained inequality.

VI. 1. If ๐‘“: ๐ผ โ†’ โ„ is convex, then for any ๐‘ฅ1, ๐‘ฅ2, โ€ฆ , ๐‘ฅ๐‘› โˆˆ ๐ผ,

we have the inequality (Jensenโ€™s inequality):

(Manual)

2. Using Jensenโ€™s inequality applied to the convex function

๐‘“(๐‘ฅ) = ๐‘’๐‘ฅ, prove the inequalities of the averages:

3. Show that for any ๐‘ฅ1, ๐‘ฅ2, โ€ฆ , ๐‘ฅ๐‘› โˆˆ [๐‘œ,๐œ‹

2], we have:

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What are the maximum domains that contain zero and in

which these inequalities take place? Give examples of other

domains, that donโ€™t contain the origin and that host inequalities like

those illustrated above. In what domains do inverse inequalities take

place?

4. Using the concavity of the logarithmic function, prove that

for any ๐‘ฅ1, ๐‘ฅ2, โ€ฆ , ๐‘ฅ๐‘› > 0 we have:

5. Show that the function ๐‘“(๐‘ฅ) = ๐‘ฅ๐›ผ , ๐›ผ > 1 , ๐‘ฅ > 0 is

convex. Using this property show that for any non-negative

numbers ๐‘ฅ1, ๐‘ฅ2, โ€ฆ , ๐‘ฅ๐‘›, the following inequality takes place:

6. Applying Jensenโ€™s inequality to the function ๐‘“(๐‘ฅ) = ๐‘ฅ2 ,

show that for any ๐‘ฅ1, ๐‘ฅ2, โ€ฆ , ๐‘ฅ๐‘› โˆˆ โ„, we have:

What analogous inequality can be deduced from the convex

function ๐‘“(๐‘ฅ) = ๐‘ฅ3 , for ๐‘ฅ > 0? And from the concave function

๐‘“(๐‘ฅ) = ๐‘ฅ3, for ๐‘ฅ โ‰ค 0?

7. Applying Jensenโ€™s inequality for the convex function

๐‘“(๐‘ฅ) =1

๐‘ฅ, for ๐‘ฅ > 0, show that, if:

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is a polynomial with all real and positive roots, we have:

What can be said if the polynomial has all the roots real and

negative?

SOLUTIONS:

By comparing the right member of the required inequality

with the right member of Jensenโ€™s inequality, we deduce that the

points ๐‘ฅ1, ๐‘ฅ2, โ€ฆ , ๐‘ฅ๐‘› are deduced from the conditions:

We have:

With these points, Jensenโ€™s inequality, for ๐‘“(๐‘ฅ) = ๐‘’๐‘ฅ, becomes:

Making the calculations in the left member we obtain the inequality

of the averages.

3. On the interval [0,๐œ‹

2], the functions sin ๐‘ฅ and cos ๐‘ฅ are

concave. The biggest interval that contains the origin and in which

sin ๐‘ฅ is concave, is the interval [0, ๐œ‹]. On [๐œ‹, 2๐œ‹], for example, the

same function is convex, so the inequality from the enunciation is

changing.

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VII. Primitives

Connections with other notions specific to functions

Definition : The function ๐‘“: ๐ผ โ†’ โ„ has primitives on the interval

๐ผ if there exists a function ๐น: ๐ผ โ†’ โ„, differentiable

and ๐น,(๐‘ฅ) = ๐‘“(๐‘ฅ) on ๐ผ.

It is known that two primitives differ through a constant:

The set of all primitives is called an undefined integral of the

function ๐‘“ and is noted by:

In order to enunciate the method employed to show that

a function has primitives and the method to show that a

function doesnโ€™t have primitives, we recall some of the

connections that exist between the main notions of calculus studied

in high school, relative to functions:

1. THE LIMIT

2. THE CONTINUITY

3. THE DERIVABILITY

4. DARBOUXโ€™S PROPERTY

5. THE PRIMTIVE

6. THE INTEGRAL

The connections between these notions are given by the

following properties:

๐‘ƒ1 ) Any continuous function in a point

๐‘ฅ0 has a limit in this point:

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๐‘ƒ2 ) Any differentiable function in a point

๐‘ฅ0 is continuous in this point:

๐‘ƒ3 ) Any continuous function on [๐‘Ž, ๐‘] has Darbouxโ€™s

property on [๐‘Ž, ๐‘]:

๐‘ƒ4 ) Any continuous function on [๐‘Ž, ๐‘] has primitives on

[๐‘Ž, ๐‘]:

๐‘ƒ5 ) Any continuous function on [๐‘Ž, ๐‘] is integrable on

[๐‘Ž, ๐‘]:

๐‘ƒ6) Any function that has primitives on [๐‘Ž, ๐‘] has Darbouxโ€™s

property on [๐‘Ž, ๐‘]:

๐‘ƒ7) If a function has Darbouxโ€™s property on [๐‘Ž, ๐‘], then in

any point ๐‘ฅ0 in which the limit (the lateral limit) exists, it is equal to

๐‘“(๐‘ฅ0).

Consequences: 1. A function with Darbouxโ€™s property (so a

function that admits primitives) canโ€™t have an infinite limit (lateral

limit) in a point ๐‘ฅ0.

The functions for which at least one lateral limit is

infinite or different from ๐’‡(๐’™๐ŸŽ) doesnโ€™t have primitives.

2. If ๐‘“ has Darbouxโ€™s property on [๐‘Ž, ๐‘] and ๐‘ฅ0 is a

discontinuity point, AT LEAST ONE LATERAL LIMIT

DOESNโ€™T EXIST.

Due to the implications stated above, we can form the Table:

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OBSERVATION: If ๐‘ฅ0 is a discontinuity point for ๐‘“, one of

the following situations is possible:

(1) the lateral limits exist and they are finite

(2) a lateral limit is finite, the other is infinite

(3) both lateral limits are infinite

(4) one lateral limit is infinite, the other doesnโ€™t exist

(5) one lateral limit is finite, the other doesnโ€™t exist

(6) both lateral limits donโ€™t exist

FROM ALL THE FUNCTIONS DISCONTINUED IN

AT LEAST ONE POINT, THE ONLY ONES THAT CAN

HAVE PRIMITIVES ARE THOSE FOR WHICH AT LEAST

ONE LATERAL LIMIT DOESNโ€™T EXIST (AND THE OTHER,

IF IT IS EXISTS, IS FINITE).

Also, letโ€™s observe that, although in case (1) the function

doesnโ€™t have primitives on [๐‘Ž, ๐‘] (doesnโ€™t have Darbouxโ€™s

property), still, if, moreover, the lateral limits are equal, then ๐‘“ has

primitives on [๐‘Ž, ๐‘]\{๐‘ฅ0} through the function ๐‘“๐‘ โ€“ the extension

through continuity of ๐‘“โ€ฒ๐‘  restriction to the domain [๐‘Ž, ๐‘]\{๐‘ฅ0}.

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Example:

has Darbouxโ€™s property for any ๐‘Ž โˆˆ [โˆ’1,1] , but doesnโ€™t have

primitives if ๐‘Ž โ‰  0.

Indeed, assuming ad absurdum that ๐‘“ has primitives, let ๐น be

one of its primitives. Then ๐น must be of the form:

In order to calculate โˆซ sin1

๐‘ฅ๐‘‘๐‘ฅ, we observe that sin

1

๐‘ฅ comes

from the derivation of the function ๐‘ฅ2 cos1

๐‘ฅ. Indeed,

so:

The function ๐‘”(๐‘ฅ) = ๐‘ฅ cos1

๐‘ฅ has primitives on (0, โˆž)

because it is continuous on this interval, but doesnโ€™t have primitives

on [0, โˆž) (we aim for the interval to be closed at zero in order to

study the continuity and derivability od ๐น).

We observe that, because ๐‘™๐‘–๐‘š

๐‘ฅ โ†’ 0๐‘ฅ > 0

๐‘”(๐‘ฅ) = 0, we can consider

its extension through continuity:

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This function, being continuous on [0, โˆž), has primitives on

this interval. Let ๐บ be one of these primitives. Then:

We state the condition that ๐น be continuous in ๐‘ฅ0 = 0. We

have:

so we must have ๐ถ1 = ๐ถ2 + 2. ๐บ(0) and so, for the primitive, we

obtain the following formula:

We state the condition that ๐น be differentiable in zero:

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because ๐บ is differentiable in zero. We have ๐บ ,(0) = ๐‘”๐‘ƒ(0) = 0 (๐บ

is a primitive of ๐‘”๐‘ƒ ), so ๐น๐‘‘, (0) = 0 . Consequently, ๐น is not

differentiable in zero if ๐‘Ž โ‰  0.

Observation: For the study of derivability we have mentioned

two methods: using the definition and using the corollary of

Lagrangeโ€™s theorem. Letโ€™s observe that in the example from above,

we canโ€™t use the corollary of Lagrangeโ€™s theorem because

๐‘™๐‘–๐‘š๐‘ฅ โ†’ 0๐‘ฅ > 0

๐น,(๐‘ฅ) doesnโ€™t exist.

The table at page 193 allows us to formulate methods to

show that a function has primitives and methods to show that a

function doesnโ€™t have primitives.

We will enunciate and exemplify these methods, firstly with

exercises from the calculus manual, XII grade, in order to highlight

the necessity of familiarizing oneself with the different notions

frequently used in high school manuals.

(A) Methods to show that a function has primitives

The first three methods are deduced from the table, using

implications of the form ๐‘Ž โ†’ ๐‘:

(๐ธ๐‘ƒ1) We show that the function is differentiable.

(๐ธ๐‘ƒ2) We show that the function is continuous.

(๐ธ๐‘ƒ3) We construct the primitive.

(๐ธ๐‘ƒ4) We show that the function is a sum of functions that

admit primitives.

Keep in mind that method ๐ธ๐‘ƒ2 is preferred over ๐ธ๐‘ƒ1

because derivability is easier to prove that continuity, and the

method ๐ธ๐‘ƒ3 is used only if the function doesnโ€™t have a limit in a

discontinuity point (this is the only situation where it can still have

primitives).

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(B) Methods to show that a function doesnโ€™t have primitives

The first three methods are deduced from the table, using

the equivalence: ๐‘Ž โ†’ ๐‘ โŸท ๐‘›๐‘œ๐‘› ๐‘ โŸถ ๐‘›๐‘œ๐‘› ๐‘Ž.

We therefore have:

This way, we deduce the following methods to show that a

function doesnโ€™t have primitives:

(๐‘๐‘ƒ1) We show that in a discontinuity point, both lateral

limits exist and are finite or at least one lateral limit is infinite..

(๐‘๐‘ƒ2) We show that the function doesnโ€™t have Darbouxโ€™s

property..

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(๐‘๐‘ƒ3) We assume ad absurdum that ๐‘“ has primitives and we

obtain a contradiction.

(๐‘๐‘ƒ4 ) If ๐‘“ is the sum of two function, one of them

admitting primitives, the other not admitting primitives, it follows

that ๐‘“ doesnโ€™t have primitives.

Examples:

has primitives because ๐‘“ is continuous.

Solution: We observe that 2๐‘ฅ. ๐‘ ๐‘–๐‘›1

๐‘ฅโˆ’ ๐‘๐‘œ๐‘ 

1

๐‘ฅ comes from the

derivation of the function ๐‘ฅ2๐‘ ๐‘–๐‘›1

๐‘ฅ, so, a primitive of ๐‘“ must have

the form:

We state the condition that ๐น be continuous in zero:

so ๐›ผ = ๐ถ and we have:

We state the condition that ๐น be differentiable in zero:

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(something bounded multiplied with something that tends to zero,

tends to zero), so ๐น is differentiable in zero. But we must also have

๐น,(0) = ๐‘“(0), equality which is also met, so ๐‘“ has primitives.

We have:

where ๐‘”(๐‘ฅ) =1

2 and:

and the functions ๐‘” and โ„Ž are primitives.

(๐‘๐‘ƒ1 ) a) ๐‘“(๐‘ฅ) = [๐‘ฅ] โˆ’ ๐‘ฅ doesnโ€™t have primitives on โ„

because in points ๐‘ฅ0 = ๐‘› is discontinuous and both lateral limits are

finite.

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doesnโ€™t have primitives because ๐‘™๐‘–๐‘š

๐‘ฅ โ†’ 0๐‘ฅ > 0

๐‘“(๐‘ฅ) = โˆ’โˆž.

(๐‘๐‘ƒ2) The functions in which this method is used in the

manual are of the form:

with ๐‘” and โ„Ž, continuous functions, ๐‘” โ‰  โ„Ž. We will prove on this

general case that ๐‘“ doesnโ€™t have Darbouxโ€™s property. The

demonstration can be adapted to any particular case.

We must prove there exists an interval whose image through

๐‘“ isnโ€™t an interval. We use the two conditions from the hypothesis.

Solving the exercise: We must prove there exists an interval

whose image through ๐‘“ isnโ€™t an interval, using the two conditions

from the hypothesis.

From the condition ๐‘” โ‰  โ„Ž we deduce there exists at least

one point ๐‘ฅ0 in which ๐‘”(๐‘ฅ0) โ‰  โ„Ž(๐‘ฅ0). Then, for ํœ€ small enough,

the intervals (๐‘”(๐‘ฅ0) โˆ’ ํœ€, ๐‘”(๐‘ฅ0) + ํœ€) and (โ„Ž(๐‘ฅ0) โˆ’ ํœ€, โ„Ž(๐‘ฅ0) + ํœ€)

are disjunctive (see the Figure below).

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From the conditions ๐‘” and โ„Ž continuous in ๐‘ฅ0 we deduce:

Considering, as we said ํœ€ small enough for the intervals

(๐‘”(๐‘ฅ0) โˆ’ ํœ€, ๐‘”(๐‘ฅ0) + ํœ€) and (โ„Ž(๐‘ฅ0) โˆ’ ํœ€, โ„Ž(๐‘ฅ0) + ํœ€) to be

disjunctive, and ๐›ฟ = min {๐›ฟ1, ๐›ฟ2}, it follows that the image through

the function ๐‘“ of the interval (๐‘ฅ0 โˆ’ ๐›ฟ, ๐‘ฅ0 + ๐›ฟ) isnโ€™t an interval,

because:

(๐‘๐‘ƒ3) This method presupposes the same steps as in

method ๐ธ๐‘ƒ3.

(a) we look for primitives for ๐‘“โ€ฒ๐‘  branches and we obtain a

first expression of ๐น;

(b) we state the condition that ๐น be continuous in the

connection point (points) between the branches;

(c) we state the condition that ๐น be differentiable in these

points:

- if ๐น is differentiable and ๐น,(๐‘ฅ0) = ๐‘“(๐‘ฅ0),

we have covered method ๐ธ๐‘ƒ3.

- if ๐น isnโ€™t differentiable in ๐‘ฅ0 or ๐น,(๐‘ฅ0) โ‰ 

๐‘“(๐‘ฅ0), we have covered method (๐‘๐‘ƒ3).

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doesnโ€™t have primitives on โ„.

Indeed, we have ๐‘“(๐‘ฅ) = ๐‘”(๐‘ฅ) + โ„Ž(๐‘ฅ) with:

and ๐‘” is continuous so it has primitives and โ„Ž has lateral limits,

finite in zero, so it doesnโ€™t have primitives. If, letโ€™s say by absurd, ๐‘“

had primitives, it would follow that โ„Ž(๐‘ฅ) = ๐‘“(๐‘ฅ) โˆ’ ๐‘”(๐‘ฅ) has

primitives.

Exercises I. Using the method (๐ธ๐‘ƒ2) , show that the following

functions have primitives in โ„:

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II. Using methods (๐ธ๐‘ƒ3) โˆ’ (๐‘๐‘ƒ3) , study if the following

functions have primitives:

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if ๐‘”: โ„ โ†’ โ„ and if it satisfies: ๐‘”(๐‘ฅ) = 0 <=> ๐‘ฅ = 0.

10. The product from a function โ„Ž that admits primitives

and a function ๐‘” differentiable with the continuous derivative is a

function that admits primitives.

Indications: If ๐ป is the primitive of โ„Ž, then ๐ป. ๐‘”, is continuous,

so it admits primitives. Let ๐บ be one of its primitives. We have

(๐ป. ๐‘” โˆ’ ๐บ), = โ„Ž. ๐‘”.

๐‘” being a differentiable function with a continuous derivative on โ„.

III. Using method (๐‘๐‘ƒ1), show that the following functions

donโ€™t have primitives:

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IV. Using method (๐‘๐‘ƒ2), show that the following functions

donโ€™t have primitives:

Answer:

1. The functions ๐‘”(๐‘ฅ) = 3๐‘ฅ and โ„Ž(๐‘ฅ) = 2๐‘ฅ2 + 1 are

continuous on โ„ and ๐‘” โ‰  โ„Ž . We will show that ๐‘“ doesnโ€™t have

Darbouxโ€™s property. Let ๐‘ฅ0 be a point in which we have (๐‘ฅ0) =

โ„Ž(๐‘ฅ0 ), for example ๐‘ฅ0 = 2. We have ๐‘”(2) = 6, โ„Ž(2) = 9.

โ„Ž โˆ’ ๐‘๐‘œ๐‘›๐‘ก๐‘–๐‘›๐‘ข๐‘œ๐‘ข๐‘  ๐‘–๐‘› ๐‘ฅ0 = 2 <=>

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๐‘” โˆ’ ๐‘๐‘œ๐‘›๐‘ก๐‘–๐‘›๐‘ข๐‘œ๐‘ข๐‘  ๐‘–๐‘› ๐‘ฅ0 = 2 <=>

Let there be ํœ€ so that the intervals:

are disjunctive. For example ํœ€ = 1 and let there be ๐›ฟ =

min (๐›ฟ1, ๐›ฟ2). Then the image of the interval (2 โˆ’ ๐›ฟ, 2 + ๐›ฟ) through

๐‘“ isnโ€™t an interval because for:

we have ๐‘“(๐‘ฅ) = 3๐‘ฅ โˆˆ (5,7) and for:

we have ๐‘“(๐‘ฅ) = 2๐‘ฅ2 + 1 โˆˆ (8,10).

V. Using methods (๐ธ๐‘ƒ4) โˆ’ (๐‘๐‘ƒ4) , study if the following

functions admit primitives:

Indication:

so ๐‘“(๐‘ฅ) = ๐‘”(๐‘ฅ) + โ„Ž(๐‘ฅ), with:

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Indication:

We obtain ๐‘Ž = 0.

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Logical Scheme for Solving Problems Solving problems that require studying if a function has

primitives can be done according to the following logical scheme:

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VIII. Integration

An integral is defined with the help of a limit. It is much

easier to understand the theoretical aspects regarding the integral if

one has understood why it is necessary to renounce the two broad

categories of limits considered up to this point:

1. the limit when ๐‘ฅ tends to ๐‘ฅ0,

2. the limit when ๐‘› tends to infinity,

and a new category of limits is introduced:

3. the limit when the divisionโ€™s norm tends to zero.

The specifications that follow are aimed exactly at clarifying

this aspect.

The theory of the integral has appeared out of a practical

necessity, in order to calculate the area that lays between the graphic

of a function and the axis 0๐‘ฅ.

Given a bounded function and (for now) non-negative

๐‘“: [๐‘Ž, ๐‘] โ†’ โ„ we can approximate the desired area with the help of

certain rectangles having the base on 0๐‘ฅ.

For this, we consider the points:

The set of these points form a division of the interval [๐‘Ž, ๐‘]

and they are noted with โˆ†:

The vertical strips made using the points ๐‘ฅ๐‘– have an area as

hard to calculate as the initial area, because of the superior outline.

We obtain rectangles if we replace the superior outlines with

horizontal segments.

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In order that the obtained approximation be reasonable, it is

only natural that these horizontal segments meet the graphic of the

function. We observe that such a horizontal is uniquely determined

by a point ๐œ‰๐‘– situated between ๐‘ฅ๐‘–โˆ’1 and ๐‘ฅ๐‘– . In this way, we have

obtained the rectangles with which we aimed to approximate the

desired area.

The areas of the rectangles are:

The sum of these areas is called the Riemann sum attached

to the function ๐‘“, to the division โˆ† and to the intermediary points

๐œ‰๐‘– . This sum:

approximates the desired area.

The necessity of introducing the third category of limits is

due to the necessity that the approximation be as accurate as

possible.

And now, a question: Is the approximation better when:

(a) the rectangles increase in number, or when

(b) the rectangles are narrower and narrower?

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We observe that the rectangles โ€œincrease in numberโ€ if and

only if ๐‘› tends to infinity, but making ๐‘› tend to infinity we do not

obtain a โ€œbetter approximationโ€. Indeed, if we leave the first

rectangle unchanged, for example, and raising the number of points

from its location to its right as much as we want (making ๐‘› tend to

infinity), the approximation remains coarse. The approximation becomes

much โ€œbetterโ€ if the rectangles are โ€œthinner and thinnerโ€.

In order to concretize this intuitive intuition, we observe that

the rectangles are thinner and thinner if the โ€œthickest of themโ€

becomes thinner and thinner.

The biggest thickness of the rectangles determined by a

division โˆ† is called the norm of the division โˆ†.

Therefore, โ€œbetter approximationsโ€ <=> โ€œthinner and

thinner rectanglesโ€ <=> โ€œthe biggest thickness tends to zeroโ€ <=>

โ€œthe norm of โˆ† tends to zeroโ€ <=> ||โˆ†|| โ†’ 0.

The area of the stretch situated between the graphic of the

function ๐‘“ and the axis 0๐‘ฅ is, therefore:

This limit is noted with โˆซ ๐‘“(๐‘ฅ)๐‘

๐‘Ž๐‘‘๐‘ฅ and it is called the

integral of function ๐‘“ on the interval [๐‘Ž, ๐‘].

For functions that are not necessarily non-negative on [๐‘Ž, ๐‘],

the integral is represented by the difference between the area

situated above the axis 0๐‘ฅ and the area situated below the axis 0๐‘ฅ.

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Consequently, in order to obtain the area of the stretch

situated between the graphic and the axis 0๐‘ฅ, in such a case, we

have to take into consideration the module of ๐‘“:

Coming back to the affirmations (๐‘Ž) and (๐‘) from above,

we observe that: โ€œthe rectangles are thinner and thinnerโ€ => โ€œthe

rectangles increase in numberโ€, namely (๐‘) => (๐‘Ž).

In other words, ||โˆ†|| โ†’ 0 = ๐‘› โ†’ โˆž.

Reciprocally, this statement isnโ€™t always true. That is why we

cannot renounce ||โˆ†|| โ†’ ๐ŸŽ in favor of ๐’ โ†’ โˆž . Nevertheless,

there is a case where the reciprocal implication is true: when the

points of the division are equidistant.

๐‘› โ†’ โˆž => ||โˆ†|| โ†’ 0 for equidistant points.

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In this case, the rectangles have the same dimension of the

base, namely ๐‘ฅ๐‘– โˆ’ ๐‘ฅ๐‘–โˆ’1 =๐‘โˆ’๐‘Ž

๐‘› for any ๐‘– = 1,2, โ€ฆ , ๐‘› , and the

Riemann sum becomes:

For divisions such as these, the following equivalence takes

place:

||โˆ†|| โ†’ 0 <=> ๐‘› โ†’ โˆž

that allows, as it was revealed by Method 16 for the calculation of

limits, for the utilization of the definition of the integral for the

calculation of sequencesโ€™ limits.

Before we enumerate and exemplify the methods for the

study of integration, letโ€™s observe that, if the function is

continuous, among the Riemann sums ๐œŽโˆ†(๐‘“, ๐œ‰) that can be

obtained by modifying just the heights of the rectangles (modifying

just the intermediary points ๐œ‰๐‘–), there exists a biggest and smallest

Riemann sum, namely the Riemann sum having the highest value is

obtained by choosing the points ๐œ‰๐‘– โˆˆ [๐‘ฅ๐‘–โˆ’1, ๐‘ฅ๐‘–] for which:

and the Riemann sum having the smallest value is obtained by

choosing the points ๐œ‰๐‘– for which:

By noting with ๐‘†โˆ†(๐‘“) and ๐‘ โˆ†(๐‘“), respectively, these sums (called

the superior Darboux sum, the inferior Darboux sum, respectively),

we have:

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The points ๐œ‰๐‘– that realizes the supreme, respectively, the

infimum of ๐‘“ on the intervals [๐‘ฅ๐‘–โˆ’1, ๐‘ฅ๐‘–] exists because it is known

that a continuous function, on a closed and bounded interval is

bounded and it touches its bounds. It is known (see Theorem 37

from the Calculus manual, grade XI) that a function is integrable if and

only if:

Connections between integration and other notions specific to functions

In order to formulate such connections, we fill the list of the

propositions ๐‘ƒ1 โˆ’ ๐‘ƒ7 enunciated in the previous chapter, with the

following:

๐‘ƒ8) any function integrable on [๐‘Ž, ๐‘] is bounded,

๐‘ƒ9) any function monotonous on [๐‘Ž, ๐‘] is integrable,

๐‘ƒ10) any function continuous on [๐‘Ž, ๐‘] is integrable,

๐‘ƒ11) any function that has on [๐‘Ž, ๐‘] a finite number of

discontinuity points of first order is integrable.

The demonstration of ๐‘ƒ10 follows from the fact that if we

modify the values of a function integrable on [๐‘Ž, ๐‘], in a finite

number of points, we obtain also an integrable function and from

the property of additivity in relation to the interval of the integral:

Using the proposition stated above, we can make the

following table:

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and the table containing implications between all the notions that

we have dealt with is the following ( ๐‘™1 and ๐‘™2 being one of the

lateral limits):

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Methods to show that a function is

integrable (Riemann) on [๐‘Ž, ๐‘] (๐ผ1) using the definition;

(๐ผ2) we show that it is continuous;

(๐ผ3) we show that it has a finite number of discontinuity

points of first order;

(๐ผ4) we show it is monotonic;

(๐ผ5) we show that the function is a sum of integrable

functions.

Methods to show that a function is not

integrable (Riemann) on [๐‘Ž, ๐‘] (๐‘1) using the definition (we show that ๐œŽโˆ†(๐‘“, ๐œ‰) doesnโ€™t

have a finite limit for ||โˆ†|| โ†’ 0);

(๐‘2) we show that the function is not bounded;

(๐‘3) we show that the function is the sum of an integrable

function and a non-integrable function.

Examples (๐ผ1) (a) Using the definition of the integral, show that any

differentiable function with the derivative bounded on [๐‘Ž, ๐‘] is

integrable on [๐‘Ž, ๐‘].

Solution: Let ๐‘“: [๐‘Ž, ๐‘] โ†’ โ„ be differentiable. We have to

show that ๐œŽโˆ†(๐‘“, ๐œ‰) has a finite limit when ||โˆ†|| โ†’ 0 . The given

function being differentiable, it is continuous, so it has primitives.

Let ๐น be one of its primitives. We can apply Lagrangeโ€™s theorem to

function ๐น , on any interval [๐‘ฅ๐‘–โˆ’1, ๐‘ฅ๐‘–]. There exists therefore ๐‘๐‘– โˆˆ

[๐‘ฅ๐‘–โˆ’1, ๐‘ฅ๐‘–] so that:

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It follows that:

We calculate the two sums separately:

We show that the second sum tends to zero when ||โˆ†|| โ†’ 0.

For this, we take into consideration that ๐‘“ , is bounded, in other

words, ๐‘€ > 0 exists, so that |๐‘“ ,(๐‘ฅ) < ๐‘€| for any ๐‘ฅ from [๐‘Ž, ๐‘] and

we apply the theorem of Lagrange to the function ๐‘“, but on the

intervals of extremities ๐œ‰๐‘– and ๐‘๐‘– . There exists ๐œƒ๐‘– between ๐œ‰๐‘– and ๐‘๐‘–

so that:

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It follows therefore that:

For the second sum, we therefore have:

The condition from this exerciseโ€™s hypothesis is very strong.

As it is known, it can be shown that a function that fulfills only the

continuity condition is also integrable. Still, the method used in this

exercise can be easily adapted to most exercises that require

showing that a function is integrable.

(๐ผ1) (b) Show that ๐‘“) ๐‘ฅ=sin ๐‘ฅ is integrable on any interval

[๐‘Ž, ๐‘]. (Manual).

Solution: We will adapt the previous demonstration. The

given function is differentiable and has the derivative ๐‘“ ,(๐‘ฅ) = cos ๐‘ฅ

bounded. Being differentiable, it is continuous and it has primitives.

Let ๐น be one of its primitives, ๐น(๐‘ฅ) = โˆ’ cos ๐‘ฅ and let:

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be a division of [๐‘Ž, ๐‘] . We apply Langrangeโ€™s theorem to the

function ๐น on each interval [๐‘ฅ๐‘–โˆ’1, ๐‘ฅ๐‘–]. There exists therefore ๐‘๐‘– โˆˆ

[๐‘ฅ๐‘–โˆ’1, ๐‘ฅ๐‘–] so that:

The function ๐น has a bounded derivative: | cos ๐‘ฅ| < 1 for

any ๐‘ฅ from โ„ , so for any ๐‘ฅ from [๐‘Ž, ๐‘] . We have to show that

๐œŽโˆ†(๐‘“, ๐œ‰) has a finite limit when ||โˆ†|| โ†’ 0. We will show that this

limit is (as it is known) ๐น(๐‘) โˆ’ ๐น(๐‘Ž) = โˆ’ cos ๐‘ + cos ๐‘Ž .

We have:

Applying the theorem of Lagrange to the function ๐‘“(๐‘ฅ) =

sin ๐‘ฅ, on the interval determined by the points ๐œ‰๐‘– and ๐‘๐‘– , we deduce

there exists ๐œƒ๐‘– between ๐œ‰๐‘– and ๐‘๐‘– so that:

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(๐ผ2) The function ๐‘“: [0,3] โ†’ โ„ defined by:

is integrable. Indeed, it is continuous in any point ๐‘ฅ โ‰  2 from the

definition domain, being expressed through continuous functions.

We study the continuity in ๐‘ฅ = 2:

The function is continuous on the interval [0,3], so it is integrable.

(๐ผ3) ๐‘“(๐‘ฅ) = ๐‘ฅ โˆ’ [๐‘ฅ], ๐‘“: [0, 2โˆš3] โ†’ โ„ can be discontinuous

just in the points where [๐‘ฅ] is discontinuous. These are of the form

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๐‘ฅ = ๐‘›, ๐‘› โˆˆ โ„ค. From these, only points 0,1,2,3, are in the functionโ€™s

domain. We study the limit in these points. We have:

so the function has a finite number of discontinuity points of first

order and is therefore integrable.

(๐ผ4) For the function ๐‘“(๐‘ฅ) = [๐‘ฅ] we can say it is integrable

either by using the previous criterium, either using the fact that it is

monotonic on any interval [๐‘Ž, ๐‘].

(๐ผ5) The function ๐‘“: [โˆ’1,1] โ†’ โ„ through ๐‘“(๐‘ฅ) = ๐‘ ๐‘”๐‘› ๐‘ฅ +

|๐‘ฅ| + [๐‘ฅ] is integrable because it is the sum of three integrable

functions: ๐‘“1(๐‘ฅ) = ๐‘ ๐‘”๐‘› ๐‘ฅ is integrable because it has a finite

number of discontinuity points (a single point, ๐‘ฅ = 0 ) and the

discontinuity is of the first order; the function ๐‘“2(๐‘ฅ) = |๐‘ฅ| is

integrable because it is continuous, and the function ๐‘“3(๐‘ฅ) = [๐‘ฅ] is

integrable because it is monotonic (or because it has a single f

discontinuity point, of the first order).

isnโ€™t integrable because, by choosing in the Riemann sum:

the intermediary points ๐œ‰๐‘– , rational, we have: ๐‘“(๐œ‰๐‘–) = 0, so in this

case:

and if ๐œ‰๐‘– are irrational, we have:

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isnโ€™t integrable (Riemann) on [0,1] because it is not bounded.

is the sum of the functions:

If ๐‘“ would be integrable, from ๐‘“ = ๐‘“1 + ๐‘“2 it would follow

that ๐‘“2 = ๐‘“ โˆ’ ๐‘“1 so ๐‘“2 would be integrable on [0,1], as the sum of

two integrable functions.

Exercises I. Using the definition, show that the following functions are

integrable:

II. Study the integration of the functions:

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8. The characteristic function of the set [0,1] โŠ‚ โ„.

III. For which values of the parameters ๐‘Ž and ๐‘ are the

functions below integrable? For which values of the parameters do

they have primitives?

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INDICATIONS:

1. ๐‘™๐‘‘(0) = โˆ’๐œ‹

2, so for any value of the parameter, the

function has only one discontinuity, of first order on [0,1].

2. For any value of ๐‘Ž โˆˆ (0,1) the function has a limited

number of discontinuity points of first order.

4. The function ๐‘’๐‘ฅ๐‘๐‘œ๐‘ 1

๐‘ฅ comes from the derivation of

๐‘ฅ2๐‘’๐‘ฅ sin1

๐‘ฅ.

IV. 1. Show that any continuous function on an interval

[๐‘Ž, ๐‘] is integrable.

2. Adapt the demonstration from the previous point to show

that the following functions are integrable on the interval [0,1]:

ANSWERS:

1. We will show that:

We have:

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where:

We now use two properties of continuous functions on a

closed and bounded interval:

(1) a continuous function on a closed, bounded interval is

bounded and it touches its bounds; so, there exists ๐‘ข๐‘– , ๐‘ฃ๐‘– โˆˆ [๐‘Ž, ๐‘] so

that ๐‘€๐‘– = ๐‘“(๐‘ข๐‘–), ๐‘š๐‘– = ๐‘“(๐‘ฃ๐‘–).

Therefore:

In order to evaluate the difference ๐‘“(๐‘ข๐‘–) โˆ’ ๐‘“(๐‘ฃ๐‘–), we use

the property:

(2) a continuous function on a closed and bounded interval is

uniformly continuous (see Chapter ๐ผ๐‘‰), namely:

In order to replace ๐‘ฅ1 with ๐‘ข๐‘– and ๐‘ฅ2 with ๐‘ฃ๐‘– , we have to

consider a division โˆ† having a smaller norm than ๐›ฟ (which is

possible because we are making the limit for ||โˆ†|| โ†’ 0). Therefore,

assuming ||โˆ†|| < ๐›ฟ we have:

i.e. ๐‘€๐‘– โˆ’ ๐‘š๐‘– < ํœ€ and so:

Because ํœ€ is arbitrary (small), we deduce that:

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so the function is integrable.

2. We show that ๐‘†โˆ†(๐‘“) โˆ’ ๐‘ โˆ†(๐‘“) tends to zero when ||โˆ†||

tends to zero. We have:

where:

because on the interval [0,1] the function ๐‘“(๐‘ฅ) = ๐‘ฅ2is increasing.

So:

The function ๐‘“(๐‘ฅ) = ๐‘ฅ2 is continuous on [0,1] so it is

uniformly continuous, namely:

Considering the division โˆ† so that ||โˆ†|| < ๐›ฟ (possibly

because ||โˆ†|| โ†’ 0), we have ๐‘ฅ๐‘–2 โˆ’ ๐‘ฅ๐‘–โˆ’1

2 < ํœ€, so:

Because ํœ€ is arbitrarily small, we deduce that:

i.e. ๐‘“(๐‘ฅ) = ๐‘ฅ2 is integrable on [0,1].

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References

T. ANDREESCU & collective: Mathematical problems from competitions and

exams sustained in 1983, printed at the Polygraphic Enterprise โ€œBanatโ€,

Timisoara, 1984.

L. ARAMA, T. MOROZAN: Problems of differential and integral calculus,

Ed. Tehnica, Bucharest, 1978.

C. AVADANEI, N. AVADANEI, C. BORS, C. CHIREA: From

elementary to superior mathematics, Ed. Academiei R.S.R, Bucharest, 1977.

D. M. BATINETU: Mathematical problems for the secondary high school, Ed.

Albatros, Bucharest, 1979.

D. M. BATINETU-GIURGIU, V. GHIORGHITA, I. V. MAFTEI, I.

TOMESCU, F. VORNICESCU: Problems given at mathematical Olympiads

for high schools, Ed. Stiintifica, Bucharest, 1992.

D. BUSNEAG, I. MAHTEI: Homework for circles and mathematical contests

for students, Ed. Scrisul Romanesc, Craiova, 1983.

A. CORDUNEANU, G. RADU, L. POP, V. GRAMADA:

Mathematical problems workbook, Ed. Junimea, Iasi, 1972.

GH. CALUGARITA, V. MANGU: Mathematical problems for the

elementary and secondary level in high school, Ed. Albatros, Bucharest 1977.

M. COCUZ: Mathematical problems workbook, Ed. Academiei R.S.R,

Bucharest, 1984.

N. DONCIU, D. FLONDOR: Algebra and calculus, workbook with

problems, E.D.P., Bucharest, 1978.

M. DANGA: Homework and mathematical problems, Ed. Tehnica,

Bucharest, 1991.

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B. R. GELBAUM, J. M. H. OLMSTED: Counterexamples in the analysis,

Ed. Stiintifica, Bucharest, 1973.

I. GIURGIU, F. TURTOIU: Mathematical problems workbook, E.D.P,

Bucharest, 1980.

I. ILIESCU, B. IONESCU, D. RADU: Mathematical problems for the

admission in the higher education, E.D.P., Bucharest, 1976.

C. IONESCU-TIU, L. PIRSAN: Differential and integral calculus for the

admission in college, Ed. Albatros, Bucharest, 1975.

AL. LEONTE, C.P.NICULESCU: Workbook of algebraic and calculus

problems, Ed. Scrisul Romanesc, Craiova, 1984.

I. A. MARON: Problems in calculus of one variable, Mir Publishers,

Moscow, 1973.

A. S. MURESAN, V. MURESAN: Algebraic and calculus problem, admission

contests 1981-1990, Ed. Tehnica, Bucharest, 1991.

C. NASTASESCU, C. NITA, M. BRANDIBURU, D.JOITA: Algebraic

exercises and problems, E. D. P., Bucharest, 1981.

C. P. NICOLESCU: Calculus, Ed. Albatros, Bucharest, 1987.

C. P. NICOLESCU: Mathematical synthesis, Ed. Albatros, Bucharest,

1990.

G. RIZESCU, E. RIZESCU: Homework for high school mathematical circles,

Ed. Tehnica, Bucharest, 1977.

V. SCHNEIDER, GH. A. SCHNEIDER: Calculus problems workbook for

grades XI-XII, Ed. Hyperion-Apollo, Craiova, 1993.

GH. SIRETCHI: Calculus problems, Typography of the Bucharest

University, 1975.

N. TEODORESCU, A. CONSTANTINESCU, M. MIHAI, L.

PIRSAN, E. PERJERIU, A. POPESCU-ZORICA, P. RADOVICI-

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C. Dumitrescu โ–  F. Smarandache

MARCULESCU, M. TENA : Problems from the mathematical gazette,

methodological and selective edition, Ed. Tehnica, Bucharest, 1984.

R. TRANDAFIR, AL. LEONTE: Mathematical problems and exercises

workbook, Ed. Junimea, Iasi, 1975.

*** Exercises workbook for the admission in the higher education, Mathematics,

Physiscs and Chemistry, Ed. St. Encicl., Bucharest, editions 1984 and 1989.

*** High school manuals, different editions.

*** The Mathematical Gazette collection.

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In this book, we discuss a succession of methods encountered in the study of Calculus I, II and III to students and instructors at the University of New Mexico - Gallup, to higher education entry examination candidates, to all those interested, in order to allow them to reduce as many diverse problems as possible to already known work schemes. We tried to present in a methodical manner, advantageous for the reader, the most frequent calculation methods encountered in the study of the calculus at this level.

ISBN 9781599735237