Methods of Solving Calculus Problems for UNM-Gallup Students
Transcript of Methods of Solving Calculus Problems for UNM-Gallup Students
Methods of Solving Calculus Problems for UNM-G Students
1
The Educational
Publisher Columbus, 2017
METHODS OF SOLVING CALCULUS PROBLEMS
FOR UNM-GALLUP STUDENTS
โ Constantin Dumitrescu โ Florentin Smarandache โ
C. Dumitrescu โ F. Smarandache
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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
Methods of Solving Calculus Problems
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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:
Editura Ferestre
13 Cantemir str.
410473 Oradea, Romania
ISBN 9781599735237
C. Dumitrescu โ F. Smarandache
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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
Methods of Solving Calculus Problems
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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.
C. Dumitrescu โ F. Smarandache
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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) Fermatโs theorem ....................................................................... 169
1. Enunciation ............................................................................... 169
2. Geometric and algebraic interpretation ................................ 169
(B) Rolleโs theorem ........................................................................... 172
1. Enunciation ............................................................................... 172
2. Geometric and algebraic interpretation ................................ 172
3. Consequences ........................................................................... 173
Methods for the study of an equationโs roots.............................. 177
1. Using Rollesโs theorem (its algebraic interpretation) ......... 177
2. Using Darbouxโs properties ................................................... 177
3. Using Rolleโs sequence ............................................................ 177
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4. The graphical method. ............................................................. 178
5. Vietteโs relations ....................................................................... 178
6. Using the theorem of the average ......................................... 179
(C) Lagrangeโs Theorem .................................................................. 181
1. Enunciation ............................................................................... 181
2. Geometric and algebraic interpretation ................................ 181
3. The corollary of Lagrangeโs Theorem .................................. 181
(D) Cauchyโs Theorem ..................................................................... 186
1.Enunciation ................................................................................ 186
2. Geometric and algebraic interpretation ................................ 186
VI. Equalities and Inequalities ..................................................... 189
Equalities ............................................................................................ 189
Inequalities ......................................................................................... 190
Method 1. Using Lagrangeโs or Cauchyโs 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 doesnโt 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
Methods of Solving Calculus Problems for UNM-Gallup Students
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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 lifeโs 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.
C. Dumitrescu โ F. Smarandache
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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 equationโs
root,
applications of Fermatโs, Rolleโs, Lagrangeโs and
Cauchyโs 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(๐) of its
parts would form a set. We would thus have ๐(๐) โ ๐.
Noting with card๐ the number of elements belonging to ๐,
we will have:
๐๐๐๐ ๐(๐) โค ๐๐๐๐๐.
However, a theorem owed to Cantor shows that we always
have
๐๐๐๐ ๐ < ๐๐๐๐ ๐(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 ๐ด โฉ ๐ต = {๐ฅ โ ๐ โ ๐ฅ โ ๐ด ๐๐๐ ๐ฅ โ ๐ต}
Because at a given point we work only with elements
belonging to ๐, the condition ๐ฅ โ ๐ is already implied, so we can
write:
๐ด โฉ ๐ต = {๐ฅ โ ๐ฅ โ ๐ด ๐๐๐ ๐ฅ โ ๐ต}
๐จ โฉ ๐ฉ
More generally,
Letโs observe that:
๐ฅ โ ๐ด โฉ ๐ต <===> ๐ฅ โ ๐ด ๐๐ ๐ฅ โ ๐ต
2. The Union ๐ด โช ๐ต = {๐ฅ โ ๐ฅ โ ๐ด ๐๐๐ฅ โ ๐ต}
๐จ โช ๐ฉ
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As in the case of the intersection, we can consider:
We see that:
๐ฅ โ ๐ด โช ๐ต <===> ๐ฅ โ ๐ด ๐๐๐ ๐ฅ โ ๐ต
3. The Difference๐ด โ ๐ต = {๐ฅ โ ๐ฅ โ ๐ด ๐๐๐ ๐ฅ โ ๐ต}
We retain that:
๐ฅ โ ๐ด โ ๐ต <===> ๐ฅ โ ๐ด ๐๐ ๐ฅ โ ๐ต
4. The ComplementThe complement of a set ๐ด is the difference between the
total set and ๐ด.
๐ถ๐๐ด = { ๐ฅ | ๐ฅ โ ๐ ๐๐๐ ๐ฅ โ ๐ด} = { ๐ฅ | ๐ฅ โ ๐ด}
The complement of a set is noted with ๐ถ๐ด or with ๏ฟฝฬ ๏ฟฝ.
๐ช๐จ
Letโs observe that
๐ฅ โ ๐ถ๐ด <===> ๐ฅ โ ๐ด
More generally, we can talk about the complement of a set to
another, random set. Thus,
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๐ถ๐ด๐ต = { ๐ฅ | ๐ฅ โ ๐ต ๐๐๐ ๐ฅ โ ๐ด}
is the complement of set ๐ต to set ๐ด.
5. The Symmetrical Difference๐ดโ๐ต = (๐ด โ ๐ต) โช (๐ต โ ๐ด)
๐จโ๐ฉ
We have ๐ฅ โ ๐ดโ๐ต <===> ๐ฅ โ ๐ด โ ๐ต and ๐ฅ โ ๐ต โ ๐ด .
6. The Cartesian Product๐ด ร ๐ต = {(๐ฅ, ๐ฆ)โ๐ฅ โ ๐ด ๐๐๐ ๐ฆ โ ๐ต}
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 = {(๐ฅ, ๐ฆ)|๐ฅ โ โ , ๐ฆ โ 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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๐ด ร ๐ต ร ๐ถ = {(๐ฅ, ๐ฆ, ๐ง)|๐ฅ โ ๐ด , ๐ฆ โ ๐ต, ๐ง โ ๐ถ}.
An intuitive image of โ3 = โ ร โ ร โ is given in the figure
below.
More generally, ๐๐
๐ = 1๐ด๐ = {(๐1, ๐2, โฆ , ๐๐)| ๐๐ โ ๐ด๐ ๐๐๐ ๐๐๐ฆ ๐ โ 1, ๐ฬ 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.
๐ด = ๐ต <===> ๐ด โ ๐ต ๐๐๐ ๐ต โ ๐ด (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 ๐ด
the function ๐๐ด: ๐ โถ {0,1} defined through:
๐๐ด(๐ฅ) = {1 ๐๐ ๐ฅ โ ๐ด0 ๐๐ ๐ฅ โ ๐ด
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 ๐ฅ in which the
value of ๐๐ด is 1 (the elements that belong to ๐ด);
(2) all the elements in which the value of ๐๐ด is 0 (elements
that do not belong to ๐ด) 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(๐ป)
of the subsets of ๐ป to the set ๐ธ(๐ป) of the characteristic
functions defined on ๐ป (see Exercise VII).
Hence,
๐ด = ๐ต <===> ๐๐ด = ๐๐ต.
(1.2)
Properties of the Characteristic Function
๐1) ๐๐ดโฉ๐ต(๐ฅ) = ๐๐ด(๐ฅ). ๐๐ต(๐ฅ)
๐2) ๐๐ดโช๐ต(๐ฅ) = ๐๐ด(๐ฅ) + ๐๐ต(๐ฅ) โ ๐๐ดโฉ๐ต(๐ฅ)
๐3) ๐๐ดโ๐ต(๐ฅ) = ๐๐ด(๐ฅ) โ ๐๐ดโฉ๐ต(๐ฅ)
๐4) ๐๐ถ๐ด(๐ฅ) = 1 โ ๐๐ด(๐ฅ)
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๐5) ๐๐ดโ๐ต(๐ฅ) = ๐๐ด(๐ฅ) + ๐๐ต(๐ฅ) โ 2. ๐๐ดโฉ๐ต(๐ฅ)
๐6) ๐๐ดร๐ต(๐ฅ, ๐ฆ) = ๐๐ด(๐ฅ) ร ๐๐ต(๐ฆ)
๐7) ๐๐ด2(๐ฅ) = ๐๐ด(๐ฅ)
๐8) ๐โ (๐ฅ) โก 0, ๐๐(๐ฅ) โก 1
๐9) ๐ด โ ๐ต <===> ๐๐ด(๐ฅ) โค ๐๐ต(๐ฅ) for any ๐ฅ โ ๐.
Letโs prove, for example ๐1). For this, letโs observe that the
total set ๐ is divided by sets ๐ด and ๐ต in four regions, at most:
1) for the points ๐ฅ that do belong to neither ๐ด, nor ๐ต, we
have ๐๐ด(๐ฅ) = ๐๐ต(๐ฅ) = 0 and ๐๐ดโฉ๐ต(๐ฅ) = 0.
2) for the points ๐ฅ that are in ๐ด but are not in ๐ต, we have
๐๐ด(๐ฅ) = 1, ๐๐ต(๐ฅ) = 0 ๐๐๐ ๐๐ดโฉ๐ต(๐ฅ) = 0 .
3) if ๐ฅ โ ๐ด ๐๐๐ ๐ฅ โ ๐ต,
the equalities follow analogously.
4) if ๐ฅ โ ๐ต ๐๐๐ ๐ฅ โ ๐ด
Exercises I. Using methods (A) and (B) show that:
SOLUTIONS:
(A) (the double inclusion)
Letโs 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
โincluded setโ belongs to the โset that includesโ.
It is obvious that this entire reasoning is mental, the solution
starts with:
a) Let ๐ฅ โ ๐ด โ (๐ต โ ๐ถ), irrespective [to continue we will read the
final operation (the difference)], <===> ๐ฅ โ ๐ด and ๐ฅ โ ๐ต โฉ ๐ถ [we will
read the operation that became the last (the union)] <===>
๐ฅ โ ๐ด and {๐ฅ โ ๐ต
๐๐๐ฅ โ ๐ถ
<===> {๐ฅ โ ๐ด ๐๐๐ ๐ฅ โ ๐ต
๐๐๐ฅ โ ๐ด ๐๐๐ ๐ฅ โ ๐ถ
(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:
๐ฅ โ (๐ด โ ๐ต) โช (๐ด โ ๐ถ) <===> ๐ฅ โ ๐ด โ ๐ต ๐๐ ๐ฅ โ ๐ด โ ๐ถ.
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We observe from (1.3) that this conclusion follows
immediately, so the inclusion is proven.
For the second inclusion: Let ๐ฅ โ (๐ด โ ๐ต) โช (๐ด โ ๐ถ), [we
will read the final operation (the reunion)] <===>
{๐ฅ โ ๐ด โ ๐ต
๐๐๐ฅ โ ๐ด โ ๐ถ
<===> [we will read the operation that became the last]
<===>
{๐ฅ โ ๐ด ๐๐๐ ๐ฅ โ ๐ต
๐๐๐ฅ โ ๐ด ๐๐๐ ๐ฅ โ ๐ถ
<===> ๐ฅ โ ๐ด and {๐ฅ โ ๐ต
๐๐๐ฅ โ ๐ถ
(1.4)
We have no more operations to explain, so we will describe
in detail the conclusion we want to reach: ๐ฅ โ ๐ด โ (๐ต โฉ ๐ถ), i.e. ๐ฅ โ
๐ด and ๐ฅ โ ๐ต โฉ ๐ถ, 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 ๐ฅ โ ๐ด โ (๐ด โ ๐ต) , irrespective [we will read the final
operation (the difference)], <===> ๐ฅ โ ๐ด and ๐ฅ โ ๐ด โ ๐ต [we will
read the operation that became the last] <===>
๐ฅ โ ๐ด and {๐ฅ โ ๐ด
๐๐๐ฅ โ ๐ต.
(1.5)
We have no more operations to explain, so we update the
conclusion:
๐ฅ โ ๐ด โฉ ๐ต ๐. ๐. ๐ฅ โ ๐ด ๐๐๐ ๐ฅ โ ๐ต.
We observe that (1.5) is equivalent to:
{๐ฅ โ ๐ด ๐๐๐ ๐ฅ โ ๐ด
๐๐๐ฅ โ ๐ด ๐๐๐ ๐ฅ โ ๐ต.
The first affirmation is false, and the second one proves
that๐ฅ โ ๐ด โฉ ๐ต.
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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 ๐๐ดโ(๐ตโฉ๐ถ) = ๐(๐ดโ๐ต)โช(๐ดโ๐ถ);
(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): ๐๐ดโ(๐ตโฉ๐ถ)(๐ฅ) = ๐(๐ดโ๐ต)โช(๐ดโ๐ถ)(๐ฅ) for any ๐ฅ โ ๐.
The answer to the corresponding (Q) question is
(R2): We explain both members of the previous equality.
๐๐ดโ(๐ตโฉ๐ถ)(๐ฅ) = [we will read the final operation (the
difference) and we will apply property ๐3 ] = ๐๐ด(๐ฅ) โ
๐๐ดโฉ(๐ตโฉ๐ถ)(๐ฅ) = [we will read the operation that became the last and
we will apply property๐1] =
๐๐ด(๐ฅ) โ ๐๐ด(๐ฅ). ๐๐ต(๐ฅ). ๐๐ถ(๐ฅ) (1.6)
The second member of the equality becomes successive:
5. In order to prove the equality ๐๐ดโ(๐ดโ๐ต)(๐ฅ) โ ๐๐ดโฉ๐ต(๐ฅ)
we notice that:
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II. Prove that:
SOLUTIONS.
Method (A) (the double inclusion)
1. Let ๐ฅ โ (๐ด โฉ ๐ต) ร (๐ถ โฉ ๐ท) , irrespective [we explain the
final operation (the scalar product)] <===> ๐ฅ = (๐ผ, ๐ฝ), with ๐ผ โ ๐ด โฉ
๐ต and ๐ฝ โ ๐ถ โฉ ๐ท <===>[we explain the operations that became the last
ones] <===> ๐ฅ = (๐ผ, ๐ฝ) with ๐ผ โ ๐ด , ๐ผ โ ๐ต and ๐ฝ โ ๐ถ , ๐ฝ โ ๐ท
[we have no more operations to explain so we update the conclusion: ๐ฅ โ (๐ด ร
๐ถ) โฉ (๐ต ร ๐ท), so ๐ฅ belongs to an intersection (the last operation)] <===
> ๐ฅ = (๐ผ, ๐ฝ) with ๐ผ โ ๐ด , ๐ฝ โ ๐ถ and ๐ผ โ ๐ต , ๐ฝ โ ๐ท <===> ๐ฅ โ
๐ด ร ๐ถ and ๐ฅ โ ๐ต ร ๐ท.
3. Let ๐ฅ โ (๐ด โ ๐ต) ร ๐ถ , irrespective <===> ๐ฅ โ (๐ผ, ๐ฝ)
with ๐ผ โ ๐ด โ ๐ต and ๐ฝ โ ๐ถ <===> ๐ฅ โ (๐ผ, ๐ฝ) with ๐ผ โ ๐ด and
๐ผ โ ๐ต and ๐ฝ โ ๐ถ [by detailing the conclusion we deduce the following steps]
<===> ๐ฅ = (๐ผ, ๐ฝ), ๐ผ โ ๐ด, ๐ฝ โ ๐ถ and ๐ฅ = (๐ผ, ๐ฝ), ๐ผ โ ๐ต , ๐ฝ โ ๐ถ
===> ๐ฅ โ ๐ด ร ๐ถ and ๐ฅ โ ๐ต ร ๐ถ.
Reciprocally, let ๐ฅ โ (๐ด ร ๐ถ) โ (๐ต ร ๐ถ) , irrespective
<===> ๐ฅ = (๐ผ, ๐ฝ) with (๐ผ, ๐ฝ) โ ๐ด ร ๐ถ and ร ๐ถ <===> ๐ฅ =
(๐ผ, ๐ฝ), ๐ผ โ ๐ด and ๐ฝ โ ๐ถ and
C. Dumitrescu โ F. Smarandache
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{๐ผ โ ๐ต
๐๐๐ฝ โ ๐ถ
<===> {๐ผ โ ๐ด ๐๐๐ ๐ฝ โ ๐ถ ๐๐๐ ๐ผ โ ๐ต
๐๐๐ผ โ ๐ด ๐๐๐ ๐ฝ โ ๐ถ ๐๐๐ ๐ฝ โ ๐ถ (๐๐๐๐ )
===> (๐ผ, ๐ฝ) โ (๐ด โ ๐ต) ร ๐ถ.
Method (B) (using the characteristic function)
Explaining ๐(๐ดโ๐ต)โ๐ถ(๐ฅ), in an analogue manner, we obtain
the desired equality. If we observe that the explanation
of ๐(๐ดโ๐ต)โ๐ถ(๐ฅ) from above is symmetrical, by utilizing the
commutativity of addition and multiplication, the required equality
follows easily.
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23
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: ๐ด โช ๐ต โ ๐ถ => ๐ด โ ๐ถ ๐๐๐ ๐ต โ ๐ถ
R2: We prove that ๐ด โ ๐ถ ๐๐๐ ๐ต โ ๐ถ (two inclusions).
Let ๐ฅ โ ๐ด irrespective [we have no more operations to
explain, therefore we will update the conclusion: ๐ฅ โ ๐ถ; but this is
not evident from ๐ฅ โ ๐ด, but only with the help of the hypothesis]
๐ฅ โ ๐ด => ๐ฅ โ ๐ด โช ๐ต =
> [๐กโ๐๐๐ข๐โ ๐กโ๐ โ๐ฆ๐๐๐กโ๐๐กโ๐๐ ๐ด โช ๐ต โ ๐ถ] =
> ๐ฅ โ ๐ถ
Let ๐ฅ โ ๐ต irrespective => ๐ฅ โ ๐ด โช ๐ต โ ๐ถ => ๐ฅ โ ๐ถ.
Reciprocally (the second implication):
C. Dumitrescu โ F. Smarandache
24
r1: We have to prove that ๐ด โช ๐ต โ ๐ถ (an inclusion)
R1: Let ๐ฅ โ ๐ด โช ๐ต irrespective (we explain the last
operation):
=> {๐ฅ โ ๐ด => ๐ฅ โ ๐ถ (๐๐ข๐ ๐ก๐ ๐กโ๐ โ๐ฆ๐๐๐กโ๐๐ ๐๐ )
๐๐๐ฅ โ ๐ต => ๐ฅ โ ๐ถ(๐๐ข๐ ๐ก๐ ๐กโ๐ โ๐ฆ๐๐๐กโ๐๐ ๐๐ )
5. r1: We have to prove an equivalence;
R1: We prove two implications;
r2: (๐ด โ ๐ต) โช ๐ต = ๐ด => ๐ต โ ๐ด;
R2: We prove the conclusion (an inclusion), using the
hypothesis.
Let ๐ฅ โ ๐ต irrespective => ๐ฅ โ (๐ด โ ๐ต) โช ๐ต => ๐ฅ โ ๐ด
r3: We have to prove the implication: ๐ต โ ๐ด => (๐ด โ ๐ต) โช
๐ต = ๐ด;
R3: We prove an equality of sets (two inclusions).
For the first inclusion, let ๐ฅ โ (๐ด โ ๐ต) โช ๐ต irrespective [we
explain the last operation] =>
=> {๐ฅ โ ๐ด โ ๐ต
๐๐๐ฅ โ ๐ต
=> {๐ฅ โ ๐ด ๐๐๐ ๐ฅ โ ๐ต (๐)
๐๐๐ฅ โ ๐ต (๐)
[We have no more operation to explain, so we update the
conclusion]. We thus observe that from (a) it follows that ๐ฅ โ ๐ด,
and from (b), with the help of the hypothesis, ๐ต โ ๐ด , we also
obtain ๐ฅ โ ๐ด.
Method 2:
We have to prove two inequalities among functions.
Through the hypothesis, ๐๐ดโช๐ต < ๐๐ถ and, moreover ๐๐ด < ๐๐ดโช๐ต
(indeed, ๐๐ด < ๐๐ดโช๐ต <=> ๐๐ด < ๐๐ด + ๐๐ต โ ๐๐ด๐๐ต <=> ๐๐ต(1 โ
๐๐ด) > 0 โ true, because the characteristic functions take two
values: 0 and 1).
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It follows that ๐๐ด < ๐๐ดโช๐ต < ๐๐ถ and, analogously ๐๐ต <
๐๐ดโช๐ต < ๐๐ถ .
Reciprocally, for the inverse implication, we have to prove
that: ๐๐ดโช๐ต < ๐๐ถ , i.e.
๐๐ด(๐ฅ) + ๐๐ต(๐ฅ) โ ๐๐ดโฉ๐ต(๐ฅ) < ๐๐ถ(x) (1.7)
in the hypothesis:
๐๐ด < ๐๐ถ and ๐๐ต < ๐๐ถ (1.8)
But ๐๐ด, ๐๐ต, ๐๐ถ 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:
C. Dumitrescu โ F. Smarandache
26
Consequently:
{๐๐ต(๐ฅ) = 0 => ๐๐ต(๐ฅ) < ๐๐ด(๐ฅ) (๐๐ข๐ ๐ก๐ ๐กโ๐ ๐ฃ๐๐๐ข๐๐ 0 ๐๐๐ 1 ๐กโ๐๐ก ๐กโ๐ ๐๐ข๐๐๐ก๐๐๐ ๐ ๐ก๐๐๐๐ )
๐๐1 โ ๐๐ด(๐ฅ) = 0 => ๐๐ด(๐ฅ) = 1 ๐ ๐ ๐๐ต(๐ฅ) < ๐๐ด(๐ฅ)
Reciprocally, if ๐๐ต(๐ฅ) < ๐๐ด(๐ฅ), we cannot have ๐๐ต(๐ฅ) =
1 and ๐๐ด(๐ฅ) = 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 ๐ฅ โ ๐ด and ๐ฅ โ ๐ต, we have:
๐๐ถ(๐ดโฉ๐ต)(๐ฅ) = 1 and ๐๐ดโฉ๐ถ๐ต(๐ฅ) = 1 , so for (1.10) to be
accomplished we must have ๐๐ฅ(๐ฅ) = 0. So, any point that doesnโt
belong to ๐ด or ๐ต, doesnโt belong to ๐ either.
b) if ๐ฅ โ ๐ด and ๐ฅ โ ๐ต, we have:
๐๐ถ(๐ดโฉ๐ต)(๐ฅ) = 1 and ๐๐ดโฉ๐ถ๐ต(๐ฅ) = 1 , so for (1.10) to be
accomplished we must have ๐๐ฅ(๐ฅ) = 1. Therefore, any point from
๐ด โ ๐ต is in ๐.
c) if ๐ฅ โ ๐ด and ๐ฅ โ ๐ต and for (1.10) to be accomplished, we
must have ๐๐ฅ(๐ฅ) = 0, consequently no point from ๐ต โ ๐ด is not in
๐.
d) if ๐ฅ โ ๐ด and ๐ฅ โ ๐ต we can have ๐๐ฅ(๐ฅ) = 0 or ๐๐ฅ(๐ฅ) =
1.
So any point from ๐ด โฉ ๐ต can or cannot be in ๐. Therefore
๐ = (๐ด\๐ต) โช ๐ท, where ๐ท โ ๐ด โฉ ๐ต, arbitrarily.
We have to analyze four cases, as can be seen from the
following table:
C. Dumitrescu โ F. Smarandache
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So ๐ = (๐ด โ ๐ต) โช ๐ถ.
V. Determine the following sets:
7. Let ๐ท โ โค(๐) a ๐ degree polynomial and ๐ โ โค .
Knowing that ๐ท(๐) = ๐๐, 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 โ {0, 1, โ2}
so ๐ธ โ โค <=> 27 is dividable by 2๐ฅ + 1 and 4๐ฅ2 โ 2๐ฅ โ 11 +27
2๐ฅ+1 is a multiple of 8 <=> ๐ด โ {โ14, โ5, โ2, โ1, 0,1, 4, 13}.
so ๐ฅ and ๐ฆ are integer solutions of systems the shape of:
{3๐ฆ โ ๐ฅ โ 1 = ๐ข3๐ฆ + ๐ฅ + 1 = ๐ฃ
with ๐ข and ๐ฃ as divisors of 32.
So ๐(๐ฅ)
๐ฅโ๐โ โค <=> 15 is dividable by ๐ฅ โ ๐.
8. Method 1: For ๐ โ โ1 we have
so, we must have ๐ฅ2 + 6๐ฅ โ 3 > 0 . As ๐ฅ1,2 = โ3 ยฑ 2โ3 , we
deduce:
Method 2: We consider the function ๐(๐) =๐2โ๐+1
๐+1 and ๐ด =
๐(๐ท), with ๐ท = โ{โ1} the domain of ๐. ๐ด is obtained from the
variation table of ๐.
VI. Determine the following sets, when ๐, ๐ โ โ:
C. Dumitrescu โ F. Smarandache
30
where [๐ก] is the integer part of ๐ก.
SOLUTIONS: We use the inequalities: [๐ก] < ๐ก < [๐ก] + 1, so:
VII. The application ๐: ๐ซ(๐) โถ โฑ๐ defined through
๐(๐ด) = ๐๐ด 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 ๐ด, ๐ต โ ๐ซ(๐)
with ๐ด โ ๐ต . From ๐ด โ ๐ต we deduce that there exists ๐ฅ0 โ ๐ so
that:
a) ๐ฅ0 โ ๐ด and ๐ฅ0 โ ๐ต or b) ๐ฅ0 โ ๐ต and ๐ฅ0 โ ๐ด
In the first case ๐๐ด(๐ฅ0) = 1 , case ๐๐ต(๐ฅ0) = 0 , so ๐๐ด โ
๐๐ต. In the second case, also, ๐๐ด โ ๐๐ต.
The surjectivity comes back to:
Let then ๐ โ โฑ๐ irrespective. Then for ๐ด = {๐ฅ|๐(๐ฅ) = 1}
we have ๐ = ๐๐ด.
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II. Functions
Function definition A function is determined by three elements ๐ท, ๐ธ and ๐, with
the following significations: ๐ท and ๐ธ are sets, named the domain
and the codomain, respectively, of the function, and ๐ is a
correspondence law from ๐ท to ๐ธ that causes:
each element ๐ โ ๐ซ to have one,
and only one corresponding element ๐ โ ๐ฌ. (F)
So we can say a function is a triplet (๐ท, ๐ธ, ๐), the elements of
this triplet having the signification stated above.
We usually note this triplet with ๐: ๐ท โ ๐ธ. 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
(๐น) that characterizes a function.
We can consider this condition as being made up of two sub
conditions, namely:
(๐1) each element ๐ฅ from the domain has a corresponding element, in the
sense of โat least one elementโ, in the codomain.
Using a diagram as the one drawn below, the proposition can
be stated like this:
C. Dumitrescu โ F. Smarandache
32
โ(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:
(๐2) the element from the codomain that corresponds to ๐ฅ 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:
โIf two arrows have the same starting point then they
also have the same arrival point.โ
Namely,
.
Conditions (๐1) and (๐2) 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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33
In Figure 2.1, the correspondence law from a) satisfies (๐1)
and doesnโt satisfy (๐2); in b) the correspondence law doesnโt satisfy
(๐1) , but satisfies (๐2) . In diagram c), neither(๐1) , nor (๐2) are
satisfied, and in d) the correspondence law satisfies both (๐1) and
(๐2), so it is the (only) function.
A function is therefore a triplet ( ๐ท, ๐ธ, ๐ ), in which the
correspondence law ๐ satisfies (๐1) and (๐2).
Letโs observe that these two condition only refer to the
domain of the function:
(๐1) โ from each point of the domain stems at least one arrow;
(๐2) 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 (๐ฅ, ๐(๐ฅ)), when ๐ฅ covers the domain of the
function.
HOW CAN WE RECOGNIZE ON A GRAPHIC IF A
CORRESPONDANCE LAW IS A FUNCTION [if it satisfies (๐1)
and (๐2)] ?
In order to answer this question we will firstly remember (for
the case ๐ท, ๐ธ โ ๐ ) the answer to two other questions:
1. Given ๐ฅ, how do we obtain โ using the graphic โ ๐(๐ฅ)
[namely, the image of ๐ฅ (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 ๐ฅ to 0๐ฆ to the point it
touches the graphic, and from the intersection point (points) we
then trace a parallel (parallels) to 0๐ฅ . The intersection points of
these parallels with 0๐ฆ are the images ๐(๐ฅ) of ๐ฅ.
2. Reciprocally, given ๐ฆ, to obtain the point (points) ๐ฅ having
the property ๐(๐ฅ) = ๐ฆ, we trace a parallel to 0๐ฅ through ๐ฆ , and
C. Dumitrescu โ F. Smarandache
34
from the point (points) of intersection with the graphic, we then
trace a parallel (parallels) to 0๐ฆ.
Examples 1. A circle with the center at the origin and with the radius ๐
is not the graphic of a function ๐: โ โถ โ, because it does not
satisfy condition (๐1) (there are points in the domain that do not have any
image, namely, all the points through which the parallel 0๐ฆ doesnโt
intersect the graphic) or (๐2), because there are points in the domain that
have more than one image (all the points ๐ฅ โ (โ๐, ๐) have two images).
2. A circle with the center at the origin and with the radius ๐
is not the graphic of a function ๐: [โ๐, ๐] โถ โ, because it does not
satisfy condition (๐2)(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
is not the graphic of a function ๐: โ โถ [0, โ), because it does not
satisfy condition (๐1). With the codomain [0, โ), the points have
one image at most. There are points, however that donโt have an image.
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4. A circle with the center at the origin and with the radius ๐
is the graphic of a function ๐: [โ๐, ๐] โถ [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
is the graphic of a function ๐: [โ๐, ๐] โถ [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, having therefore the equation ๐ฅ2 + ๐ฆ2 = ๐2 , which yields ๐ฆ =
ยฑ(๐2 โ ๐ฅ21/2).
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 (๐1) and (๐2), THUS:
a) A graphic satisfies the condition (๐1) if and only if any parallel to
0๐ฆ traced through the points of the domain touches the graphic in at least one
point.
b) A graphic satisfies the condition (๐2) if and only if any parallel to
0๐ฆ traced through the points of the domain touches the graphic in one point at
most.
The inverse of a function We donโt always obtain a function by inverting the
correspondence law (inverting the arrow direction) for a randomly
given function ๐: ๐ท โถ ๐ธ. Hence, in the Figure 2.3, ๐ is a function,
but ๐โ1 (obtained by inverting the correspondence law ๐) is not a
function, because it does not satisfy (๐2) (there are points that have
more than an image).
C. Dumitrescu โ F. Smarandache
36
With the help of this diagram, we observe that the inverse
doesnโt satisfy (๐2) 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, ๐โ1 doesnโt satisfy (๐2) 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 ๐โ1 is not a function is when it
does not satisfy condition (๐1) :
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
(๐1) 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,
๐โ1 satisfies (๐2) if and only if ๐ satisfies (๐3)
๐โ1 satisfies (๐1) if and only if ๐ satisfies (๐4)
๐โ1 is a function if and only if ๐ satisfies (๐3) + (๐4).
As it is known, a function that satisfies (๐3 ) is called an
injective function, a function that satisfies (๐4 ) is called a
surjective function, and a function that satisfies (๐3) + (๐4 ) is
called a bijective function.
We see then that the affirmation: โA function has an inverse
if and only if it is bijectiveโ, has the meaning that the inverse ๐โ1
(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 0๐ฅ traced through the points of the codomain touches the graphic in
one point, at most [namely ๐โ1 satisfies (๐2)].
d) a graphic is the graphic of a surjective function if and only if any
parallel to 0๐ฅ traced through the points of the codomain touches the graphic in
at least one point [namely ๐โ1 satisfies (๐1)].
Examples 1. A circle with the center at the origin and with the radius ๐
is the graphic of a function ๐: [0, ๐] โถ [0, โ) that is injective but
not surjective.
2. A circle with the center at the origin and with the radius ๐
is the graphic of a function ๐: [โ๐, ๐] โถ [0, ๐) that is surjective
but not injective.
3. A circle with the center at the origin and with the radius ๐
is the graphic of a bijective function ๐: [0, ๐] โถ [0, ๐).
C. Dumitrescu โ F. Smarandache
38
So, by only modifying the domain and the codomain, using a
circle with the center at the origin and with the radius ๐ , 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 ๐โ1 is obtained by inverting the correspondence law ๐น, 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
๐โ1(๐ฆ) = ๐๐๐๐ผ๐ฆ. The relation (2.2) defines the logarithm:
The logarithm of a number ๐ฆ in a given base, ๐, is the exponent ๐ฅ to
which the base has to be raised to obtain ๐ฆ.
The Graphic of the Inverse Function If ๐ท and ๐ธ are subsets of ๐ and the domain ๐ท of ๐ (so the
codomain of ๐โ1) is represented on the axis 0๐ฅ, and on the axis
0๐ฆ, the codomain ๐ธ (so the domain of ๐โ1), then the graphic of ๐
and ๐โ1 coincides, because ๐โ1 only inverses the correspondence
law (it inverses the direction of the arrows).
But if we represent for ๐โ1 the domain, horizontally, and the
codomain vertically, so we represent ๐ธ on 0๐ฅ and ๐ท on 0๐ฆ, then
any random point on the initial graphic ๐บ๐ (that, without the
Methods of Solving Calculus Problems for UNM-Gallup Students
39
aforementioned convention is the graphic for both ๐ and ๐โ1 ),
such a point (๐ฅ, ๐(๐ฅ)) becomes (๐(๐ฅ), ๐ฅ).
The points (๐ฅ, ๐(๐ฅ)) and (๐(๐ฅ), ๐ฅ) are symmetrical in
relation to the first bisector, so we obtain another graphic ๐บ๐๐
besides ๐บ๐ if the domain of ๐โ๐ is on ๐๐.
Agreeing to represent the domains of all the functions on
0๐ฅ it follows that ๐บ๐๐ is a graphic of ๐โ1.
With this convention, the graphics of ๐ and ๐โ1 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 (๐3);
- to study the surjectivity, we verify if the condition
(๐4) 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.
C. Dumitrescu โ F. Smarandache
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Examples
1. The function ๐: โ โถ โ, ๐(๐ฅ) = {2๐ฅ + 1 ๐๐ ๐ฅ โค 1๐ฅ + 3 ๐๐ ๐ฅ > 1
is
injective, but is not surjective. The graphic is represented in Figure
2.5 a).
2. The function ๐: โ โถ โ , ๐(๐ฅ) = {3๐ฅ โ 1 ๐๐ ๐ฅ โค 22๐ฅ โ 1 ๐๐ ๐ฅ > 2
is
surjective, but is not injective. The graphic is in Figure 2.5 b).
3. The function ๐: โ โถ โ, ๐(๐ฅ) = {2๐ฅ + 1 ๐๐ ๐ฅ โค 1๐ฅ + 2 ๐๐ ๐ฅ > 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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41
Example Letโs show that ๐(๐ฅ) = ๐ก๐ ๐ฅ, ๐: (โ
๐
2,
๐
2) โถ โ is bijective.
(i) injectivity: because ๐(๐ฅ) =1
๐๐๐ 2๐ฅ 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 Darbouxโs property and
from where it follows that it is surjective.
4. Using the proposition 1The function , ๐: ๐ท โถ ๐ธ is bijective if and only if
โฉ ๐ฆ โ ๐ธ the equation ๐(๐ฅ) = ๐ฆ has an unique solution
(see the Algebra workbook grade XII)
Example Let ๐ท = โ ร โ and the matrix ๐ด = (3
412). Then the function
is bijective (Algebra, grade XII).
Letโs observe that in the proposition used at this point, the
affirmation โthe equation ๐(๐ฅ) = ๐ฆ has a solutionโ ensures the
surjectivity of the function and the affirmation โthe solution is
uniqueโ ensures the injectivity.
5. Using the proposition 2If ๐, ๐: ๐ท โถ ๐ท and ๐. ๐ = 1๐ท then ๐ is injective and ๐ is
surjective (Algebra, grade XII).
Example Let ๐ท = โค ร โค and ๐ด = ( 2
โ13
โ2) . Then the function
๐๐ด: ๐ท โถ ๐ท defined through
C. Dumitrescu โ F. Smarandache
42
satisfies ๐๐ด โ ๐๐ด = 1๐ท, so it is bijective ( algebra, grade XII).
Letโs observe that the used proposition can be generalized, at
this point, thus:
โLet ๐: ๐ท โถ ๐ธ, ๐: ๐ธ โถ ๐น, so that ๐ โ 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 ๐ข, ๐ฃ ๐๐๐ ๐ค we will proceed as follows:
1. we draw the table with the signs of the functions ๐ข โ
๐ฃ, ๐ข โ ๐ค and ๐ฃ โ ๐ค
2. using the table we can easily explain the inequalities,
because for example ๐ข(๐ฅ) โค ๐ฃ(๐ฅ) <===> ๐ข(๐ฅ) โค 0.
1. For ๐ข(๐ฅ) = ๐ฅ + 1 , ๐ฃ(๐ฅ) = ๐ฅ2 + 2 and ๐ค(๐ฅ) = 3๐ฅ we
have the following table:
So:
C. Dumitrescu โ F. Smarandache
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Therefore,
5. To explain ๐ we proceed as follows:
(1) we draw the variation table of the function ๐ฆ(๐ก) = ๐ก2
(2) considering ๐ฅ in the first monotonic interval (deduced
from the table), at the right of โ1 (because we are only interested in
the values ๐ก โฅ โ1 ), we calculate the minimum of ๐ฆ(๐ก) on the
interval ๐ก โ [โ1, ๐ฅ] etc.
a) for the first monotonic interval, ๐ฅ โ (โ1,0), the function
๐ฆ(๐ก) = ๐ก2 has a single minimum point on the interval [โ1, ๐ฅ] ,
situated in ๐ก = ๐ฅ; its values is ๐ฆ(๐ฅ) = ๐ฅ2. Being a single minimum
point it is also the global minimum (absolute) of ๐ฆ(๐ก) on the
interval [โ1, ๐ฅ], so, the value of ๐ is ๐(๐ฅ) = ๐ฅ2 for ๐ฅ โ (โ1,0].
b) for the second monotonic interval, ๐ฅ โ (0, โ) , the
function ๐ฆ(๐ก) = ๐ก2 has a single minimum point on the interval
[โ1, ๐ฅ], in ๐ก = 0; its value is ๐ฆ(0) = 0. Being a single minimum
point, we have ๐(๐ฅ) = 0 for ๐ฅ โ (0, โ), so
Letโs observe that using the same variation table we can
explain the function ๐(๐ฅ) = ๐๐๐ฅ1โค1โค๐ฅ
๐ก2. Hence:
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a) for ๐ฅ โ (โ1,0] , the function ๐ฆ(๐ก) = ๐ก2 , has only one
maximum, namely [โ1, ๐ฅ] in ๐ก = โ1; its value is ๐ = ๐ฆ(1) = 1.
So ๐(๐ฅ) = 1, for ๐ฅ โ (1,0].
b) for ๐ฅ โ (0, +โ), ๐ฆ(๐ก) = ๐ก2 has two maximum points, in
๐ก1 = โ1 and ๐ก2 = ๐ฅ ; their values are ๐1 = ๐ฆ(1) = 1 and ๐2 =
๐ฆ(๐ฅ) = ๐ฅ2. The values of ๐ is the global maximum, so:
๐(๐ฅ) = max (1, ๐ฅ) for ๐ฅ โ (0, +โ).
It follows that ๐(๐ฅ) = {1 ๐๐ ๐ฅ โ [โ1,0]
max(1, ๐ฅ2) ๐๐ ๐ฅ > 0
9. From the variation table of ๐ฆ(๐ก) =๐ก4
(๐ก+7)3
1. for ๐ฅ โ (โ2, ๐] , ๐ฆ(๐ก) has a single minimum on the
interval [โ2, ๐ฅ], in ๐ก = ๐ฅ; its value is ๐ = ๐ฆ(๐ฅ) =๐ฅ4
(๐ฅ+7)3
2. for ๐ฅ > 0, ๐ฆ(๐ก) has a single minimum, namely [โ2, ๐ฅ], in
๐ก = 0; its value is ๐ = ๐ฆ(0) = 0. So
C. Dumitrescu โ F. Smarandache
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Analogously, for
10. From the variation table of ๐ฆ(๐ก) = ๐ก2 in ๐ก
we deduce:
(a) if ๐ฅ โ (0,1
โ๐] , the function ๐ฆ(๐ก) = ๐ก2 in ๐ก has, for ๐ก โ
(0, ๐ฅ], a single supremum ๐ =๐๐๐
๐ก โ ๐ ๐ก > 0
๐ฆ(๐ก) = 0 .
(b) ๐ฅ โ (โ1
โ๐, โ], the function ๐ฆ(๐ก) has two supreme values:
๐1 = 0 and ๐1 = ๐ฆ(๐ฅ) = ๐ฅ21๐ ๐ฅ.
So:
II. Study the bijectivity of:
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2. ๐(๐ฅ) = ๐(๐ฅ), ๐ being an uneven degree polynomial.
III. Study the irreversibility of the hyperbolic functions:
and show their inverses.
IV. 1. Show that the functions ๐(๐ฅ) = ๐ฅ2 โ ๐ฅ + 1 ,
๐: [1
2, โ) โถ โ and
C. Dumitrescu โ F. Smarandache
48
are inverse one to the other.
2. Show that ๐(๐ฅ) =1โ๐ฅ
1+๐ฅ coincides with its inverse.
3. Determine the parameters ๐, ๐, ๐, ๐ so that ๐(๐ฅ) =๐๐ฅ+๐
๐๐ฅ+๐
coincides with its inverse.
4. Show that the function ๐: (0,1] โถ โ defined
through๐(๐ฅ) =4
3๐ โ ๐ฅ, if ๐ฅ โ [1
3๐ ,1
3๐โ1] is bijective.
5. On what subinterval is the function ๐: [โ1
2, โ) โถ โ
๐(๐ฅ) = โ๐ฅ โ โ2๐ฅ โ 1 bijective?
INDICATIONS.
4. To simplify the reasoning, sketch the graphic of function
๐.
5. Use the superposed radical decomposition formula:
where ๐ถ = โ๐ด2 โ ๐ต.
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 โถ 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), โฆ , ๐(๐), โฆ
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: ๐0, ๐1, โฆ , ๐๐, โฆ. This set is abbreviated by
(๐๐)๐โโ.
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 ๐ทwith โ
(๐ 2) by replacing the codomain ๐ธ with โ
(๐ 3) by replacing the variable ๐ฅ with ๐ (or ๐, or ๐, etc)
(๐ 4) by replacing ๐(๐ฅ) with ๐๐
C. Dumitrescu โ F. Smarandache
50
DOMAIN CODOMAIN VARIABLE CORRESP. LAW
Function D E x ๐(๐ฅ) Sequence โ โ n ๐๐
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 (๐๐)๐โโ ,
obtained by replacing ๐ with ๐ฅ in the expression of ๐๐ ( ๐๐
becomes ๐(๐ฅ)).
Example We can attach the function ๐(๐ฅ) =
๐ฅ+3
2๐ฅ+1 to the sequence
๐๐ =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) โ
(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: ๐ท โถ โ, ๐ท โโ is monotonically increasing if:
โฉ ๐ฅ1. ๐ฅ2 โ ๐ท, ๐ฅ1 โค ๐ฅ2 โ ๐(๐ฅ1)โค ๐(๐ฅ2)
a) the sequence ๐: โ โถ โ ismonotonically increasing if:
โฉ ๐1, ๐2 โ โ, ๐1 โค ๐2 โ๐๐1
โค ๐๐2 (1)
If condition (1) is fulfilled,
taking, particularly ๐1 = ๐ and
๐2 = ๐ + 1, we deduce
๐๐ โค ๐๐+1 for any ๐ (2)
In conclusion (1) โถ (2) The reciprocal implication also stands true. Its demonstration can be deduced from the following example:
If (2) is fulfilled and we take
๐1 = 7, ๐2 = 11 , we have
๐1 โค ๐2 and step by step
and ๐7 โค ๐11.
Consequently, (1) โท (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) โ (๐ 4) , that define the transition from function to sequence.
C. Dumitrescu โ F. Smarandache
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b) the function ๐: ๐ท โถ โ, ๐ท โโ is monotonically decreasing if:
โฉ ๐ฅ1, ๐ฅ2 โ ๐ท, ๐ฅ1 โค ๐ฅ2 โ ๐(๐ฅ1)โฅ ๐(๐ฅ2)
b) the sequence ๐: โ โถ โ is monotonically decreasing if:
โฉ ๐1, ๐2 โ โ, ๐1 โค ๐2 โ ๐๐1
โฅ ๐๐2
It can be shown that this condition is equivalent to:
๐๐ โฅ ๐๐+1 for any ๐ โ โ (4)
LIMITED FUNCTIONS LIMITED SEQUENCES
a) ๐: ๐ท โถ โ , ๐ท โ โ is of inferiorly bounded if it doesnโt
have values towards โโ , i.e.
โ๐ โ โ,โฉ ๐ฅ โ ๐ท, ๐(๐ฅ) โฅ ๐
b) ๐: ๐ท โถ โ , ๐ท โ โ is of superiorly bounded if it doesnโt
have values towards +โ
i.e. โ๐ โ โ,โฉ ๐ฅ โ ๐ท, ๐(๐ฅ) โค ๐
c) ๐: ๐ท โถ โ , ๐ท โ โ is bounded, if it is inferiorly and superiorly bounded, i.e.
โ๐, ๐ โ โ,โฉ ๐ฅ โ โ ๐ โค ๐(๐ฅ) โค๐. In other words, there exists an
interval [๐, 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, ๐] becomes [โ๐, ๐] and the limiting condition is:
โ ๐ > 0, ,โฉ ๐ฅ โ ๐ท โ๐ โค๐(๐ฅ) โค ๐ i.e.
โ ๐ > 0, ,โฉ ๐ฅ โ ๐ท |๐(๐ฅ)| โค ๐
a) the sequence ๐: โ โถ โ is of inferiorly bounded if it doesnโt have values towards
โโ , i.e.
โ๐ โ โ,โฉ ๐ โ โ, ๐๐ โฅ ๐
b) the sequence ๐: โ โถ โ is of superiorly bounded if it doesnโt have values towards
+โ , i.e.
โ๐ โ โ,โฉ ๐ โ โ, ๐๐ โฅ ๐
c) the sequence ๐: โ โถ โ is bounded if it is inferiorly and superiorly bounded, i.e
โ๐, ๐ โ โ,โฉ ๐ โ โ ๐ โค ๐๐ โค๐ OR
โ ๐ > 0,โฉ ๐ โ โ |๐๐| โค 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 ๐ฅ1 โค ๐ฅ2 and we
compare the difference ๐(๐ฅ1) =๐(๐ฅ2) to zero. This can be done through successive minorings and majorings or by applying Lagrangeโs theorem to function
๐ on the interval [๐ฅ1, ๐ฅ2].
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 Lagrangeโs theoremIt allows the replacement of the
difference ๐(๐ฅ2) โ ๐(๐ฅ1) with
๐(๐) that is then compared to zero.
1. Using the definition:We compare the difference
๐๐+1 โ ๐๐ to zero, and for sequences with positive terms we can compare the quotient
๐๐+1/๐๐ to one. We can make successive minorings and majorings or by applying Lagrangeโs 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, โ), (see Method 10 point c) 3. Using Lagrangeโs 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
C. Dumitrescu โ F. Smarandache
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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(๐ฅ) = ๐๐๐รln (ln ๐ฅ) and we have the
variation table:
so the function is decreasing on the interval (1, ๐1/๐) and increasing
on (๐1/๐, โ) . It is inferiorly bounded, its minimum being ๐ =
๐โ1/๐ and it is not superiorly bounded.
5. We use the superposed radicalsโ decomposition formula:
6. Let ๐ฅ1 < ๐ฅ2. In order to obtain the sign of the difference
๐(๐ฅ1) โ ๐(๐ฅ2) we can apply Lagrangeโs theorem to the given
function, on the interval [๐ฅ1, ๐ฅ2] (the conditions of the theorem are
fulfilled): ๐ โ โ exists, so that:
so the function is increasing.
II. Study the monotony and the bounding of the sequences:
C. Dumitrescu โ F. Smarandache
56
also determine the smallest term of the sequence.
also determine the smallest term of the sequence.
๐พ๐ being the intermediary value that is obtained by applying
Lagrangeโs theorem to the function ๐(๐ฅ) = ln ๐ฅ on the intervals
[๐, ๐ + 1].
ANSWERS:
2. The function attached to the sequence is ๐(๐ฅ) = ๐ฅ1/๐ฅ .
From the variation table:
we deduce, according to the sequence criterion (Heineโs 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 โฅ 3 . Because the sequence is
decreasing if ๐ โฅ 3, in order to discover the biggest term, we have
to compare ๐2 and ๐3 . We deduce that the biggest term of the
sequence is:
3. The function attached to the sequence is ๐(๐ฅ) = ๐ฅ๐๐ฅ .
From the variation table for ๐ โ (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 (โ ln ๐)โ1
and is decreasing for ๐ โฅ [(โ ln ๐)โ1] + 1.
Because
we deduce that the interval [1, [(โ ln ๐)โ1]], in which the sequence
is increasing, can be no matter how big or small.
6. The sequence of the general term ๐๐ is increasing, the
sequence of the general term ๐๐ is decreasing, and the sequence ๐๐
is increasing if ๐ โ [0, ๐ โ ๐พ๐] and is decreasing for ๐ โ [, ๐ โ
๐พ๐, 1].
III. Using Lagrangeโs theorem.
1. Prove the inequality | sin ๐ฅ| โค |๐ฅ| for any real ๐ฅ , then
show that the sequence ๐1 = sin ๐ฅ, ๐2 = sin sin ๐ฅ, โฆ , ๐๐ =
sin sin โฆ sin ๐ฅ, is monotonic and bounded, irrespective of ๐ฅ, and
its limit is zero.
2. Study the monotony and the bounding of the sequences:
ANSWERS:
C. Dumitrescu โ F. Smarandache
58
From Lagrangeโs theorem applied to the function ๐(๐) = ๐ ๐๐ ๐ก on
the interval [๐ฅ, 0] or [0, ๐ฅ], depending on ๐ฅ being either negative or
positive, we obtain |sin ๐ฅ
๐ฅ| = | cos ๐| โค 1. The sequence satisfies the
recurrence relation: ๐๐+1 = sin ๐๐, so if sin ๐ฅ > 0, we deduce that
the sequence is decreasing, and if sin ๐ฅ < 0 the sequence is
increasing. It is evidently bounded by โ1 and 1. Noting with ๐ฟ its
limit and passing to the limit in the recurrence relation, we obtain
๐ฟ = sin ๐ฟ, so ๐ฟ = 0.
where ๐(๐ฅ) = (1 +1
๐ฅ)๐ฅ defines the function attached to the
sequence.
b) Applying Lagrangeโs theorem to the function attached to
the sequence, on the interval [๐, ๐ + 1], we deduce๐๐+1 โ ๐๐ =
๐(๐๐). We have to further determine the sign of the derivation:
namely, the sign of
From the table below we deduce ๐(๐ฅ) > 0 for ๐ฅ > 0, so ๐๐+1 โ
๐๐ > 0.
IV. 1. Let I be a random interval and ๐: ๐ผ โ ๐ผ a function.
Show that the sequence defined through the recurrence relation
๐๐+1 = ๐(๐๐), with ๐๐ being given, is:
a) increasing, if ๐(๐ฅ) โ ๐ฅ > 0 on ๐ผ
b) decreasing if ๐(๐ฅ) โ ๐ฅ < 0 on ๐ผ.
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2. If the sequence (๐๐)๐โโ is increasing, and the sequence
(๐๐)๐โโ is decreasing and ๐๐ โค ๐๐ for any ๐, then:
a) the two sequences are bounded, and so convergent.
b) if ๐๐๐๐โโ
(๐๐ โ ๐๐) = 0 then they have the same limit.
c) apply these results to the sequences given by the
recurrence formulas:
with ๐๐ and ๐๐ being given.
SOLUTIONS:
1. ๐๐+1 โ ๐๐ = ๐(๐๐) โ ๐๐ > 0 in the first case.
2. the two sequences are bounded between ๐1 and ๐1, so the
sequences are convergent, because they are also monotonic. Let ๐1
and ๐2 represent their limits. From the hypothesis from (b) it
follows that ๐1 = ๐2.
(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 ๐ฅ is replaced with ๐ and ๐(๐ฅ) is
replaced with ๐๐ . 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 functionโs limit is based on the notion of
vicinity of a point.
Intuitively, the set ๐ โ โ is in the vicinity of point ๐ฅ๐ โ โ,
if (1) ๐ฅ โ ๐, and, moreover, (2) ๐ also contains points that are in
the vicinity of ๐ฅ๐ (at its left and right).
Obviously, if there is any open interval (๐, ๐), so that ๐ฅ๐ โ
(๐, ๐) โ ๐, then the two conditions are met.
Particularly, any open interval (๐, ๐) that contains ๐๐
satisfies both conditions, so it represents a vicinity for ๐๐.
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 (+โ), ๐โ = (โ โ ํ, โ + ํ) doesnโt make sense, so we
have to find a different form for the vicinities of +โ. In order to
accomplish this, we observe that to the right of +โ we cannot
consider any points, so โ + ํ has to be replaced with +โ .
Furthermore, because โ โ ํ = โ, we replace โ โ ํ with ํ. So, we
will consider the vicinities of +โ with the form:
๐โ = (ํ, โ)
and sets that contain such intervals.
Analogously, the vicinities of โโ have the form:
๐โโ = (โ โ, ํ)
The definition of a functionโs limit in a point expresses the
condition according to which when ๐ฅ approaches ๐ฅ๐ , ๐(๐ฅ) 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 ๐๐ and ๐๐ฅ๐
from the
definition (โโโ), there exist three essential forms, corresponding to
the finite points and to + โ and โ โ respectively.
CONVENTION:
We use the letter ํ to write the vicinities of ๐, and the letter ๐ฟ
designates the vicinities of ๐ฅ๐.
We thus have the following cases:
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and condition ๐ฅ โ and ๐๐ฅ๐ becomes ๐ฅ > ๐ฟ
and condition ๐ฅ๐ โ and ๐๐ฅ๐ becomes ๐ฅ < ๐ฟ
All these cases are illustrated in Table 3.1.
Considering each corresponding situation to ๐ฅ๐ and ๐, in the
definition (โโโ) , 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 ๐ฅ๐ โ ๐๐๐๐๐ก๐.
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 ํ > 0, irrespective; we therefore search for
๐ฟ .
c) we make calculation in the expression |๐(๐ฅ) โ ๐| ,
highlighting the module |๐ฅ โ ๐ฅ๐|.
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d) we increase (decrease) the obtained expression, keeping in
mind we are ony interested in those values of ๐ฅ for which |๐ฅ โ
๐ฅ๐| < ๐ฟ .
So:
e) if the expression that depends on ๐ฅ is bounded
(continuous) in a vicinity of ๐ฅ๐, we can further increase, to eliminate
๐ฅ. We have:
because in the interval [0,2], for example, that is a vicinity of ๐ฅ๐ =
1 , the function ๐(๐ฅ) =1
|3๐ฅ+2| is continuous, and therefore
bounded: ๐(๐ฅ) โ [1
5,
1
2], if ๐ฅ โ [0,2].
f) we determine ๐ฟ , setting the condition that the expression
we have reached (and that doesnโt depend on ๐ฅ , just on ๐ฟ ) be
smaller than ํ:
Any ๐ฟ that fulfills this condition is satisfactory. We can, for
example, choose ๐ฟ3
2ํ or ๐ฟ = ํ etc.
In order to use the increase made to function ๐, we must
also have ๐ฟ โค 1, so, actually ๐ฟ = min (1,3
2ํ ) (for example).
It follows that:
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b) let ํ > 0; we search for ๐ฟ
because in the interval [1,3], for example, the function:
is bounded by:
f) we determine ๐ฟ from the condition: ๐ฟ๐
3< ํ . We obtain
๐ฟ < 3ํ, so we can, for example, take ๐ฟ = ํ. But in order for the
increase we have made to function ๐ to be valid, we must also have
๐ฟ โค 1, so, actually ๐ฟ = min(1, ํ).
It follows that:
so the condition from the definition of the limit is in this case, also,
fulfilled.
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b) let ํ any (not necessarily positive, because ๐ = +โ , so
๐๐ = (ํ, +โ)only makes sense for any ํ, not just for ํ positive).
so we can take ๐ฟ =1
2โ
Then:
II. Using the definition, prove that:
ANSWERS:
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b) let ํ > 0 any; we search for ๐ฟ .
c) because ๐ฅ๐ isnโt finite anymore, we cannot highlight |๐ฅ โ
๐ฅ๐ |. We will proceed differently: we consider the inequality
|๐(๐ฅ) โ ๐| < ํ as an inequation in the unknown ๐ฅ (using also the
fact that ๐ฅ โ โ , so, we can consider |2๐ฅ + 1| = 2๐ฅ + 1 ). We
have:
b) let ํ > 0 any; we determine ๐ฟ .
c) we express ๐ฅ from the inequality ๐(๐ฅ) > ํ :
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If ๐ฅ1 and ๐ฅ2 are the roots of the attached second degree equation
and we assume ๐ฅ1 < ๐ฅ2, we have ๐(๐ฅ) > ํ โบ ๐ฅ > max (100, ๐ฅ2).
So we can take ๐ฟ = max (100, ๐ฅ2).
The limits of sequences The definition of a sequenceโs limit is deduced from the
definition of a functionโs limit, making the mentioned
particularizations:
(1) ๐ฅ is replaced with ๐,
(2) ๐(๐ฅ) is replaced with ๐๐,
(3) ๐ฅ๐ is replaced with ๐๐.
Moreover, we have to observe that ๐๐ = โโ and finite ๐๐
donโt make sense (for example ๐ โ 3 doesnโt make sense, because
๐, being a natural number, cannot come no matter how close to 3).
So, from the nine forms of a functionโs limit, only 3 are
particularized: the ones corresponding to ๐ฅ๐ = +โ.
We therefore have:
(๐10) ๐ โ ๐๐๐๐๐ก๐ (we transpose the definition (๐2))
Because in the inequality ๐ > ๐ฟ , ๐ is a natural number, we
can consider ๐ฟ as a natural number also. To highlight this, we will
write ๐ instead of ๐ฟ . We thus have:
(๐11) ๐ = +โ (we transpose the definition (๐5))
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(๐12) ๐ = โโ (we transpose the definition (๐8))
Examples I. Using the definition, prove that:
ANSWERS:
1. We adapt the corresponding stages for functionsโ limits, in
the case of ๐ฅ๐ = +โ.
b) let ํ > 0 any; we determine ๐ โ โ.
c) we consider the inequality: |๐๐ โ ๐| < ํ as an inequation
with the unknown ๐:
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so, ๐ = [2โ
3] + 1
b) let ํ > 0 any; we determine ๐ โ โ as follows:
Let ๐1and ๐2 represent the solutions to this second degree
equation (๐1 < ๐2). Then, for ๐1 > ๐2, the required inequality is
satisfied. So, ๐ = [๐2] + 1.
Calculation methods for the limits of functions and sequences
Letโs 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: โ
โ, โ โ โ ,
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 ๐ฅ๐ผ, with
๐ผ < 0, if ๐ฅ โ โ.
obtain as many terms as possible with the form ๐ฅ๐ผ, with
๐ผ > 0, if ๐ฅ โ 0.
Examples I. Calculate:
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ANSWERS:
2. Given the common factor 9๐ we obtain terms that have
the form ๐ฅ๐, with sub unitary ๐ฅ.
expression that tends to 1
5 when ๐ฅ 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 โ uneven number, the conjugate is:
An example of application for this formula is exercise 5,
below.
C. Dumitrescu โ F. Smarandache
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Exercises
ANSWERS:
7. We replace, for example ๐0 = โ๐1 โ ๐2 โ โฏ โ ๐๐, 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 ๐ผ = arcsin ๐ฅ โ ๐๐๐๐ก๐ ๐ฅ, we observe that ๐ผ
tends to zero when ๐ฅ tends to zero, so we aim to obtain ๐ผ
sin ๐ผ. 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 ๐ฅ โ ๐ = ๐ผ, we obtain:
4. We consider the function:
obtained from ๐๐, by replacing ๐ with ๐ฅ. We calculate:
According to the criterion with sequences (see Method 10,
point c), we also have: ๐๐๐
๐ โ โ๐๐ = ln 2.
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5. Something that is bounded multiplied with something that tends to zero, tends to zero
(A) If (๐๐)๐โโ is bounded and ๐๐๐
๐ โ โ๐๐ = 0, then:
๐๐๐๐ โ โ
๐๐. ๐๐ = 0
(B) If ๐ is bounded in a vicinity of ๐ฅ0 and ๐๐๐
๐ฅ โ ๐ฅ0๐(๐ฅ) = 0,
then ๐๐๐
๐ฅ โ ๐ฅ0๐(๐ฅ). ๐(๐ฅ) = 0.
Examples
because (โ1)๐ is bounded by โ1 and 1 and 1
๐ tends to zero;
because 1 โ cos ๐ฅ is bounded by 0 and 2 and 1
๐ฅ2 tends to zero;
where ๐๐ is the nth decimal fraction of the number ๐;
where ๐ฝ๐ is the approximation with ๐ exact decimal fractions of the
number ๐.
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INDICATIONS:
4. (๐๐)๐โโ is bounded by 0 and 9 , being composed of
numbers;
5. (๐ฝ๐)๐โโ is bounded by 2 and 3;
6. The majoring and minoring method(A) If functions ๐ and โ exist, so that in a vicinity of ๐ฅ๐ we
have:
then:
Schematically:
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(B) If the sequences (๐)๐โโ and (๐๐)๐โโ exist so that: ๐๐ โค
๐๐ โค ๐๐, starting from the rank ๐0 and ๐๐๐
๐ โ โ๐๐ =
๐๐๐๐ โ โ
๐๐ = ๐,
then: ๐๐๐
๐ โ โ๐๐ = ๐.
Schematically:
OBSERVATION:
If ๐ = +โ , we can eliminate โ ( (๐๐)๐โโ ๐๐๐ ๐๐๐๐ก๐๐ฃ๐๐๐ฆ) ,
and if ๐ = โโ, we can eliminate ๐ ((๐๐)๐โโ ๐๐๐ ๐๐๐๐ก๐๐ฃ๐๐๐ฆ).
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 (๐๐)๐โโ and (๐๐)๐โโ.
Examples 1. We put the smallest (biggest) term of the sequence
(๐๐)๐โโ, instead of all the terms.
ANSWERS:
a) In the sum that expresses the sequence (๐๐)๐โโ, there are:
the smallest term 1
โ๐2+๐ and the biggest term
1
โ๐2+1. By applying
Method 1, we have:
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2. We minorate (majorate) the terms that make up the
sequence ๐๐ with the same quantity.
ANSWERS:
a) This time there isnโt a smallest (biggest) term in the sum
that makes up the sequence ๐๐ . This is due to the fact that the
function ๐(๐ฅ) = sin ๐ฅ isnโt monotonic. Keeping into consideration
that we have: โ1 โค sin ๐ฅ โค 1 , we can majorate, replacing sin ๐ฅ
with 1, all over. We obtain:
Analogously, we can minorate, replacing sin ๐ฅ with โ1 ,
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 ๐๐.
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 donโt 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 ๐ฅ2+2๐ฅ
7, keeping into
consideration that:
- on the left of โ2 we have: ๐ฅ2+2๐ฅ
7> 0, so:
- on the right of โ2 we have: ๐ฅ2+2๐ฅ
7< 0, so:
(c) Moving on to the limit in these inequalities(the majoring
and minoring method), we obtain the limit to the left:๐๐ (โ2) =2
7
and the limit to the right: ๐๐(โ2) =2
7, so ๐ =
2
7.
(B) If the double inequality cannot be used (3.2), for
๐๐๐๐ฅ โ ๐
๐(๐ฅ) , with ๐ โ โค , we calculate the lateral limits directly,
keeping into consideration that:
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Example: For ๐๐๐
๐ โ โ(โ1[๐ฅ])/(๐ฅ โ 1) we cannot use (3.2)
because the expressions obtained for the calculation of the lateral
limits donโt have the same limit. That is why we use the fact that an
the left of 1 we have [๐ฅ] = 0, and on the right of 1 we have [๐ฅ] =
1, so:
It follows that ๐ = โโ
(C) If ๐ฅ โ โ, we can replace [๐ฅ], keeping into consideration
that for ๐ฅ โ [๐, ๐ + 1) , we have: [๐ฅ] = ๐ (obviously, ๐ฅ โ โ
implies ๐ โ โ).
Example:
The variable ๐ฅ that tends to infinity is certainly between two
consecutive numbers: ๐ โค ๐ฅ โค ๐ + 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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๐ฅ๐ being a point in which ๐ is differentiable.
Example:
We notice that by noting ๐(๐ฅ) = ๐๐ฅ , we have ๐(๐) = ๐๐
and the limit becomes:
This limit has the value ๐,(๐) because ๐ is a function
differentiable in ๐.
We have thus reduced the calculation of the given limit to
the calculation of the value of ๐,(๐), calculation that can be made
by deriving ๐:
Observation: this method can be used like this: โlet ๐(๐ฅ) =
๐๐ฅ . We have ๐,(๐ฅ) = ๐๐ฅ ln ๐ . ๐,(๐) = ๐๐ ln ๐ = ๐ . We donโt
recommend this synthetic method for the elaboration of the
solution at exams because it can misguide the examiners. That is
why the elaboration: โWe observe that by noting ๐(๐ฅ) = โฏ, we
have ๐(๐) = โฏ and the limit becomes โฆ = ๐,(๐) because ๐ is
differentiable in ๐ โฆโis 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 ๐ฅ, in the expression of ๐๐, because for
๐ฅ โ โ we canโt calculate the derivative, but by replacing ๐ with 1
๐ฅ
(which leads to the calculation of the derivative in zero).
Example: For ๐๐๐
๐ โ โ๐( โ2
๐โ 1) we consider the function
๐(๐ฅ) =2๐ฅโ1
๐ฅobtained by replacing ๐ with
1
๐ฅ in the expression of
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๐๐ . We calculate ๐๐๐
๐ โ 0๐(๐ฅ). For this we observe that by noting
๐(๐ฅ) = ๐ฅ2 , we have ๐(๐) = 1 and the limit
becomes: ๐๐๐
๐ โ 0๐(๐ฅ)โ๐(๐)
๐ฅโ0= ๐,(0) , because ๐ is a differentiable
function in zero. We have: ๐,(๐ฅ) = 2๐ฅ ln 2, so ๐,(0) = ln 2. Then ,
according to the criterion with sequences, we obtain ๐๐๐
๐ โ โ๐๐ =
ln 2.
Exercises
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II. For ๐ > 0 calculate:
ANSWER:
4. We divide the numerator and the denominator with ๐ฅ โ ๐.
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(๐ฅ) =2๐ฅโ1
3๐ฅโ1, obtained by
replacing the sequence ๐๐ on ๐ with 1
๐ฅ. We calculate
๐๐๐๐ฅ โ 0
๐(๐ฅ) .
To use the definition of the derivative, we divide the numerator and
the denominator with ๐ฅ, so:
We observe that by noting ๐(๐ฅ) = 2๐ฅ , we have ๐(0) = 1
and the limit from the numerator becomes: ๐๐๐
๐ฅ โ 0๐(๐ฅ)โ๐(0)
๐ฅ=
๐,(๐ฅ) because ๐ is differentiable in zero.
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๐,(๐ฅ) = 2๐ฅ ln 2 , so, according to the criterion with
sequences ๐1 = ๐,(0) = ln 2 . Analogously, by noting โ(๐ฅ) = 3๐ฅ ,
the limit from the numerator becomes: ๐๐๐
๐ฅ โ 0โ(๐ฅ)โโ(0)
๐ฅ= โ,(0) ,
because โ is differentiable in zero.
โ(๐ฅ) = 3๐ฅ ln 3 , so, according to the criterion with
sequences๐2 = โ,(0) = ln 3, and ๐ =ln 2
ln 3.
5. The limit contains an indetermination with the form 1โ,
so we will first apply method 4.
We now consider the function ๐(๐ฅ) =1
๐ฅ(๐1
๐ฅ + ๐2๐ฅ โ 2)
obtained by replacing ๐ with 1
๐ฅ in the expression above. We
calculate ๐๐๐
๐ฅ โ 0๐(๐ฅ). For this, we observe that by noting ๐(๐ฅ) =
๐1๐ฅ + ๐2
๐ฅ we have ๐(0) = 2 and the limit can be written:
๐๐๐๐ฅ โ 0
๐(๐ฅ)โ๐(0)
๐ฅ= ๐,(0) because ๐ is differentiable in zero.
C. Dumitrescu โ F. Smarandache
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Now ๐,(๐ฅ) = ๐1๐ฅ ln ๐1 + ๐2
๐ฅ ln ๐2, so ๐,(0) = ln ๐1 + ln ๐2
According to the criterion with sequences:
and the initial limit is โ๐1. ๐2.
9. Using LโHospitalโs theorem The theorem. [lโHospital (1661-1704)] If the functions ๐ and ๐
fulfill the conditions:
1. are continuous on [๐, ๐] and differentiable on (๐, ๐){๐ฅ๐}
2. ๐(๐ฅ๐) = ๐(๐ฅ๐) = 0
3. ๐, is not annulled in a vicinity of ๐ฅ๐
4. there exists ๐๐๐
๐ฅ โ ๐ฅ๐
๐,(๐ฅ)
๐,(๐ฅ)= ๐ผ , finite or infinite,
then there exists ๐๐๐
๐ฅ โ ๐ฅ๐
๐(๐ฅ)
๐(๐ฅ)= ๐ผ.
This theorem can be used for:
(A) The calculation of limits such as ๐๐๐
๐ฅ โ ๐ฅ๐
๐(๐ฅ)
๐(๐ฅ) when these
contain an indetermination of the form 0
0 or
โ
โ.
(B) The calculation of limits such as ๐๐๐
๐ฅ โ ๐ฅ๐๐(๐ฅ). ๐(๐ฅ)
when these contain an indetermination of the form โ. 0.
This indetermination can be brought to form (A) like this:
(indetermination with the form โ
โ)
(indetermination with the form 0
0)
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Observation: sometimes it is essential if we write the
indetermination 0
0 or
โ
โ.
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 โ
โ:
and we derive, we obtain ๐ = 0.
(C) The calculation of limits such as ๐๐๐
๐ฅ โ ๐ฅ๐(๐(๐ฅ) โ ๐(๐ฅ)),
with indetermination โ โ โ.
This indetermination can be brought to form (B) giving ๐(๐ฅ)
or ๐(๐ฅ) as forced common factor.
We have, for example:
C. Dumitrescu โ F. Smarandache
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and because
we distinguish two cases:
(namely, ๐(๐ฅ)
๐(๐ฅ) doesnโt tend to 1), we have:
the indetermination is of the form (B).
(D) The calculation of limits such as ๐๐๐
๐ฅ โ ๐ฅ๐๐(๐ฅ)๐(๐ฅ), with
indeterminations 1โ, 00, โ0.
These indeterminations can be brought to the form (C),
using the formula:
that for ๐ = ๐, for example, becomes ๐ด = ๐ln ๐ด, 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 ๐ฅ or by replacing ๐ with 1
๐ฅ in the expression of ๐๐). 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 ๐ฅ we calculate the limit in โ,
and by replacing ๐ with 1
๐ฅ, 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)
C. Dumitrescu โ F. Smarandache
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II. Show that lโHospitalโs rule cannot be applied for:
ANSWER:
1. For ๐(๐ฅ) = ๐ฅ2๐ ๐๐1
๐ฅ and ๐(๐ฅ) = sin ๐ฅ we have:
- doesnโt exist, so the condition 4 from the theorem isnโt
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 ๐ฅ0 = โ , condition ๐) doesnโt make sense
anymore.
This criterion cannot be used to eliminate indeterminations
because the sequence (๐ฅ๐)๐โโ 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 donโt contain indeterminations.
Example: Using the criterion with sequences, show that:
Solution: (a) we have to show that:
โฉ (๐ฅ๐)๐โโ , with the properties:
(b) let (๐ฅ๐)๐โโ be any sequence with the
properties ๐), ๐), ๐)
(c) in ๐๐๐
๐ โ โ๐(๐ฅ๐) we use the properties of the
operations of limits of sequences to highlight ๐๐๐
๐ โ โ๐ฅ๐ (it is
possible because there are no indeterminations), then we use the
fact that ๐๐๐
๐ โ โ๐ฅ๐ = ๐ฅ0:
C. Dumitrescu โ F. Smarandache
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(B) To show that a function doesnโt have a limit in a point.
To accomplish this we can proceed as follows:
1๐ต . we find two sequences: (๐ฅ๐)๐โโ and (๐ฆ๐)๐โโ with the
properties ๐), ๐), ๐) so that:
or:
2๐ต. we find a single sequence with the properties ๐), ๐), ๐)
so that ๐๐๐
๐ โ โ๐(๐ฅ๐) doesnโt exist.
Example: ๐๐๐
๐ฅ โ โsin ๐ฅ doesnโt exist
1๐ต. let ๐ฅ๐ = 2๐๐ and ๐ฆ๐ = 2๐๐ +๐
2. We have:
So the limit doesnโt exist.
2๐ต. let๐ฅ๐ =๐๐
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(๐ฅ) be the function attached to the sequence, by the
replacement of, for example, ๐ with ๐ฅ in the expression of ๐๐ ( or
of ๐ with 1
๐ฅ, but then we calculate
๐๐๐๐ฅ โ 0
๐(๐ฅ))
(b) we calculate ๐๐๐
๐ฅ โ โ๐(๐ฅ) = ๐ (
๐๐๐๐ โ 0
๐(๐ฅ), respectively).
(c) according to the criterion with sequences lim ๐(๐ฅ) = ๐
means: for any sequence (๐ฅ๐)๐โโ with the properties:
we have ๐๐๐
๐ โ โ๐(๐ฅ๐) = ๐
We observe that the sequence ๐ฅ๐ = ๐ fulfills the conditions
๐) and ๐) and, moreover, for this sequence we have:
Exercises I. Using the criterion with sequences, calculate:
II. Show that the following donโt exist:
C. Dumitrescu โ F. Smarandache
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7. If ๐has a limit to its left (right) in ๐ฅ0, then:
8. If๐๐๐
๐ โ โ๐(๐ฅ) = ๐ , then
๐๐๐๐ โ โ
๐(๐) = ๐ , but not the
other way around.
9. ๐๐๐
๐ฅ โ โ๐(๐ฅ) doesnโt exist if ๐: ๐ท 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 1๐ต . The following sequences fulfill
conditions ๐) and ๐):
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and:
III. Determine the limit points of the functions: ๐, ๐, ๐ โ
๐, ๐ โ ๐, for:
Answer: let ๐ฅ0 โ ๐ท = โ randonm. We check if ๐ has a limit
in ๐ฅ0. For this we observe it is essential if (๐ฅ๐)๐โโ is rational or
irrational. We firstly consider the situation where the sequence
(๐ฅ๐)๐โโ is made up only of rational points (or only of irrational
points).
Let (๐ฅ๐)๐โโ be a sequence of rational points with the
properties ๐), ๐), ๐). We have:
C. Dumitrescu โ F. Smarandache
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So the function doesnโt have a limit in any point ๐ฅ0 for
which ๐ฅ02 โ 2, i.e. ๐ฅ0 and ยฑ2. For ๐ฅ0 = โ2 we consider a random
sequence (๐ฅ๐)๐โโ with the properties ๐), ๐), ๐). ๐ธ1decomposes in
two subsequences: (๐ฅ๐, )๐โโ made up of solely rational terms and
(๐ฅ๐, )๐โโ made up of irrational terms. Because:
we deduce that:
The same goes for ๐ฅ = โโ2.
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: ๐๐ =๐!
๐๐
(a) For the study of monotony and bounding we cannot
apply the method of the attached function (๐(๐ฅ) =๐ฅ!
๐ฅ๐ฅ doesnโt 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 isnโt monotonic, we canโt apply the
method so we wonโt 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
๐1, being decreasing and with positive terms. Therefore ๐๐๐
๐ โ โ๐๐
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:
C. Dumitrescu โ F. Smarandache
110
where (๐๐)๐โโ 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 ๐๐+1 โฅ ๐๐ , that we then transform through
equivalences:
So ๐๐+1 โฅ ๐๐๐๐ โ (โ1,2).
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We have thus reached the conclusion that the sequence is
increasing if and only if it is bounded between โ๐ and ๐. We
obviously have ๐๐ > 0 > โ1 so we still have to prove that ๐๐ < 2.
This inequality can be proven by induction:
Verification:
So we have proven the monotony and the bounding.
Therefore ๐ =๐๐๐
๐ โ โ๐๐ 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 = โ1 is impossible, because ๐๐ > 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
C. Dumitrescu โ F. Smarandache
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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 โ 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:
C. Dumitrescu โ F. Smarandache
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Indications:
4. sequence with positive terms (induction) and decreasing.
Let ๐1 and ๐2 be the limits of the subsequences of even and
uneven value. Show that ๐1 = ๐2.
12. Using the Cesaro-Stolz and Rizzoli Lemmas Lemma: (Cesaro-Stolz) Let (๐ผ๐)๐โโ be a random sequence and
(๐ฝ๐)๐โโ a strictly increasing sequence, having the limit
infinite. If ๐๐๐
๐ โ โ๐ผ๐+1โ๐ผ๐
๐ฝ๐+1โ๐ฝ๐= ๐ exists โ finite or infinite,
then:
๐๐๐๐ โ โ
๐ผ๐
๐ฝ๐= 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 (๐ฝ๐)๐โโ is strictly
monotonic.
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CONSEQUENCES:
Indeed,
Example:
because for ๐๐ = 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 ๐๐, ๐ฝ๐ โ 0 has been proven by I.Rizzoli [Mathematic Gazette,
no 10-11-12, 1992, p. 281-284]:
C. Dumitrescu โ F. Smarandache
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Lemma: (I.Rizzoli) If (๐ผ๐)๐โโ, (๐ฝ๐)๐โโ are two sequences
of natural numbers that fulfill the conditions:
(ii) the sequence (๐ฝ๐)๐โโ is strictly monotonic
(increasing or decreasing)
(iii) there exists:
Then: ๐๐๐
๐ โ โ๐ผ๐
๐ฝ๐= ๐.
EXERCISES:
8. ๐๐ =1โ2โ โฆ.โ๐+2โ3โโฆโ(๐+1)+โฏ+(๐โ๐+1)โ(๐โ๐+2)โโฆ โ๐
๐๐
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(Indications: amplification with the conjugate, the
minoring/majoring method, the Cesaro-Stolz lemma).
5. Let ๐ฅ๐+1 = ln(1 + ๐ฅ๐), ๐ฅ0 > 0. Using the Cesaro-Stolz
lemma show that ๐๐๐
๐ โ โ๐. ๐ฅ๐ = 2.
(Indications: ๐๐ฅ๐ =๐1
๐ฅ๐
)
III. Calculate the limits of the following sequences,
presupposing they exist:
knowing that:
C. Dumitrescu โ F. Smarandache
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where ๐ > 0 and ๐ is Eulerโs 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 Lagrangeโs theoremTheorem: (Lagrange, (1736-1813)). If ๐: [๐, ๐] โ โ is
continuous on [๐, ๐] and differentiable on
(๐, ๐),
then: โโโ (๐, ๐)๐. ๐.๐(๐)โ๐(๐)
๐โ๐= 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:
C. Dumitrescu โ F. Smarandache
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b) we apply Lagrangeโs theorem on function ๐ on the
interval [๐, ๐ + 1]:
c) we replace in ๐๐ the expression ๐(๐)โ๐(๐)
๐โ๐ with ๐ ,(๐๐). We
obtain:
d) we use the inequalities: ๐ < ๐๐ < 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 Lagrangeโs theorem on the
intervals: [๐, ๐ + 1], ๐ โ โ.
Observation: the sequences of the form ๐๐ฉ 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, letโs assume ๐ is increasing
and ๐ , is decreasing:
1. the monotony:
2. we apply Lagrangeโs 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 ๐1. 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 ๐๐.
C. Dumitrescu โ F. Smarandache
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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 โ๐(1) and ๐1.
Exercises I. Using Lagrangeโs 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 Eulerโs constant ๐ โ
(0,1))
(the partial sums of the harmonic sequence โ1
๐
โ๐=1 )
(the partial sums of the generalized harmonic sequence)
Indication: 4. In order to have one of the forms 1๐ต or 2๐ต, we
determine ๐, keeping into account that:
C. Dumitrescu โ F. Smarandache
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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+ โฏ +
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 ๐๐ =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 = โ (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. Lagrangeโs 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 =๐๐๐
๐ โ โ๐๐, we deduce
๐๐๐๐ โ โ
๐ ๐ = 0.
But:
C. Dumitrescu โ F. Smarandache
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so ๐ ๐ doesnโt tend to zero. Therefore, ๐ isnโt 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: ๐๐ =
๐0๐๐. For the second order recurrence, we search for the general
term with the same form:
๐๐ = ๐ โ ๐๐, ๐ being an undetermined constant.
By replacing in the recurrence relation, we obtain:
from where, dividing by ๐๐โ1 , we obtain the characteristic
equation:
We consider the following cases:
a) โ> 0 (the characteristic equation has real and distinct
roots ๐1 and ๐2)
In this case, having no reason to neglect one of the roots, we
modify the expression ๐๐, considering it of the form:
and we determine the constants ๐1 and ๐2 so that the first two
terms of the sequence have the initially given values.
Example:
(Fibonacciโs sequence). The characteristic equation is:
with the roots:
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Considering ๐๐ of the form:
from the system:
we obtain:
so:
Observation: ๐2 =1+โ5
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) โ= 0 (the characteristic equation has equal roots ๐1 =
๐2). In this case we consider ๐๐ of the form:
(namely ๐๐ = ๐1๐ โ ๐1(๐), with ๐1- first degree polynomial) and we
determine ๐1and ๐2, stating the same condition, that the first two
terms have the same initially given values.
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Example:
Taking ๐๐ = ๐ โ ๐๐, we obtain the characteristic equation:
Considering:
from the system:
we obtain: ๐1 = โ4, ๐2 = 6, so:
c) ) โ< 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 ๐1๐ and ๐๐ง
๐ respectively with:
Then we can write:
C. Dumitrescu โ F. Smarandache
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3. Linear recurrence of degree h (bigger than 2)Is of the form:
with the first โ terms given.
Proceeding as with the second order recurrence, we combine
๐๐ = ๐ โ ๐ 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: ๐1, ๐2, โฆ , ๐โ, we will consider ๐๐ of the form:
and we determine the constants ๐๐ , ๐ โ 1, โฬ 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, ๐1 is multiple of order ๐ , (๐1 =
๐2 = โฏ = ๐๐ ), we replace the sum:
from the expression of ๐๐ with:
๐๐ โ1 being a polynomial of degree ๐ โ 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, ๐1, is complex, then its conjugate is
also a root of the characteristic equation (let for example, ๐2 = ๐1ฬ ฬ ฬ ).
In this case, we replace in the expression of ๐๐ the sum:
with:
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and if the root ๐1 is a multiple of order ๐ (๐1 = ๐2 = โฏ ๐๐ and
we replace in the expression of ๐๐ 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 ๐ผ 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: ๐1 = 1 , ๐2 = 2 , and the roots of the
characteristic equation are:
3. Let : ๐๐ = ๐ด โ ๐ผ๐ + ๐ต โ ๐ฝ๐ , with ๐, ๐ต, ๐ผ, ๐ฝ โ โ and
๐ด, ๐ต โ 0, |๐ผ| โ | ๐ฝ|. Determine ๐ผ ๐๐๐ ๐ฝso that:
a) the sequence (๐ฅ๐) is convergent;
b) ๐๐๐
๐ โ โ๐ฅ๐ = โ
c) ๐๐๐
๐ โ โ๐ฅ๐ = โโ
4. Let ๐๐ = ๐ด โ ๐ผ๐ + ๐ต โ ๐ โ ๐ฝ๐ , ๐ โฅ 1, with ๐, ๐ต, ๐ผ, ๐ฝ โ โ
and ๐ด, ๐ต โ 0 . Determine ๐ผ ๐๐๐ ๐ฝ so that the sequence is
convergent.
(B) Nonlinear recurrence
1. Recurrence of the form ๐๐+1 = ๐ผ โ ๐๐ + ๐ฝ,
with ๐0 given The expression of the general term is obtained by observing
that if ๐ is a root for the equation ๐ = ฮฑ โ l + ฮฒ (obtained by
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replacing an+1and an with ๐ in the recurrence relation), then the
sequence (an โ ๐)๐โโ
is a geometric progression.
Example:
From the equation ๐ =๐+1
2 we deduce ๐ = ๐.
Then the sequence (an โ ๐)๐โโ
is a geometric progression.
Let ๐๐ = ๐๐ โ 1. We have:
The ratio of the geometric progression is, consequently: ๐ =1
2. We obtain:
2. Recurrence of the form ๐๐+1 = ๐ผ โ ๐๐ + ๐(๐) (๐ beinga random function)
The general term is found based on the observation that, if
(๐๐)๐โโ is a sequence that verifies the same recurrence relation,
then any sequence (๐ผ๐)๐โโ verifies the given relation of the form:
In practice we chose ๐๐ of the form:
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the coefficients of ๐ being undetermined. We determine these
coefficients by stating the condition that (๐๐)๐โโ verifies the
recurrence relation.
Particular case: a recurrence with the form:
with ๐ a polynomial of degree ๐ .
Example:
We search for the sequence ๐๐ with the form:
We state the condition that (๐๐)๐โโ verifies the given
recurrence relation:
We obtain:
By identification, it follows that:
so:
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3. Recurrence of the form ๐๐+1 = ๐(๐๐) (with ๐: [๐, ๐] โโ continuous function)
Theorem: 1. If the function ๐: [๐, ๐] โ โ is
continuous and increasing on [๐, ๐], then:
a) if ๐1 > ๐0, the sequence (๐ผ๐)๐โโ is increasing,
b) if ๐1 < ๐0, the sequence (๐ผ๐)๐โโ is decreasing.
The limit of the sequence is a fixed point of ๐, i.e. a
characteristic equation ๐(๐ฅ) = ๐ฅ.
2. If ๐ is decreasing on [๐, ๐], then the subsequences of
even and uneven value, respectively, of (๐ผ๐)๐โโ 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, โ2)
and increasing on (โ2, โ). By way of induction, we prove that
๐ผ๐ > โ2 , and, because ๐2 = 5,1 < ๐1 , 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(๐ฅ) = ๐ฅ has the roots ๐ฅ1,2 =
ยฑโ2, so ๐ = โ2.
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 ๐ผ โ
๐ฟ โ ๐ฝ โ ๐พ.
If ๐ฅ1 โ ๐ฅ2 are the roots of the characteristic equation
๐(๐ฅ) = ๐ฅ, by noting:
it can be shown that the general term ๐ผ๐ 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 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 (๐ผ๐)๐โโ 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 ํ > 0. We check to see if ๐ exists so that property
(๐) takes place.
(c) we have:
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(d) we state the condition: 2
๐< ํ 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 |๐๐+๐ โ ๐๐| < ๐๐+1 for any ๐, ๐ โ 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< ํ, we obtain: ๐ = [
1โ 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, ๐] โ โ, corresponding to a division:
and to the intermediate points:
is:
If the points ๐ฅ๐ 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, ๐]:
To use the definition of the integral in the calculation of the
limits of sequences, we observe that for divisions โ formed with
equidistant points, we have:
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So we can proceed as follows:
a) we show that the general term ๐๐ can be written in the
form:
with ๐ as a continuous function on [๐, ๐], and ๐๐ being the points
of an equidistant division.
b) ๐ , being continuous, it is also integrable, and:
Example:
a) we write ๐๐ in the form:
highlighting the common factor 1
๐. We have:
b) in:
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we make the equidistant points: (๐โ๐)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, ๐].
d) we observe that ๐๐ is the Riemann sum corresponding to
function ๐ (continuous, thus integrable) and to the deduced division
โ, on interval [๐, ๐]. As the points of โ 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: ๐ท โ โ is continuous in ๐ฅ0 โ
๐ท if:
1. it has a limit in ๐ฅ0 โ,
2. the limit is equal to ๐(๐ฅ0),
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 ๐ฅ = 1.
Indeed, letโs show that:
Let there be any ํ > 0. We must determine ๐ฟ . We have:
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because on the interval [0,2], for example, [2๐ฅ + 3] is bounded
between 3 and 7. We determine ๐ฟ from the condition:
We can take, for example ๐ฟ = 2ํ. Then:
therefore ๐ is continuous in ๐ฅ = 1.
2. Using the criterion with the lateral limits๐ is continuous in:
where
are the left and right limit, respectively in ๐ฅ0.
EXAMPLE: Letโs study the continuity of the function:
for ๐ผ โ [0, 2๐].
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 ๐ฅ0 โ 1, the function is continuous being
expressed through continuous functions.
(b) we study the continuity in ๐ฅ0 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 ๐ผ โ1
2:
1. if ๐ผ โ [0,๐
6] โช [
5๐
6, 2๐], the equation becomes:
2. if ๐ผ โ (๐
6,
5๐
6), we obtain sin ๐ผ = 1, so ๐ผ =
๐
2.
For this four values of ๐ผ the function is continuous also in
point ๐ฅ = 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(๐ฅ0) and
eliminating the condition ๐ฅ๐ โ ๐ฅ0 (which is essential in the case of
limits, because we can study the limit of a function in a point where
๐(๐ฅ) doesnโt 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 ๐ฅ0 for which:
(B) Types of discontinuity points The point ๐ฅ0 โ ๐ท in which ๐ is not continuous is said to be
a first order discontinuity point if the lateral limits in ๐ฅ0 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: ๐ท โ โ means to add new points
to the domain ๐ท, where we define the correspondence law willingly.
If ๐ is the set of added points and โ 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 ๐ท and has a finite limit in a point ๐ฅ0 โ ๐ท, there is
only one extension:
that is continuous, named the extension through continuity of ๐ in
point ๐ฅ0.
So, for our example, because ๐๐๐
๐ฅ โ 0sin ๐ฅ
๐ฅ= 1, the function:
is the only extension of sin ๐ฅ/๐ฅ that is continuous on โ.
(D) The continuity of composite functions If ๐ is continuous in ๐ฅ0 and ๐ is continuous in ๐(๐ฅ0), then
๐ โ ๐ is continuous in ๐ฅ0.
So the composition of two continuous functions is a
continuous function. Reciprocally it isnโt true: it is possible that ๐
and ๐ are not continuous and ๐ โ ๐ is.
For example:
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isnโt continuous in any point, but the function ๐(๐ฅ) = (๐ โ ๐)(๐ฅ)
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 โ ๐, ๐ โ ๐ ,
for:
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IV. 1. Let ๐, ๐: โ โ โ be a continuous function so that
๐(๐ฅ) = ๐(๐ฅ) for any ๐ฅ โ โ. Show that ๐ = ๐. (Manual)
2. Let ๐: ๐ผ โ โ , ๐ผ โโ โ be a function with Darbouxโs
property. If ๐ has lateral limits in any point ๐ผ, then ๐ is continuous
on ๐ผ. (Manual)
3. ๐: โ โ โ so that:
for any ๐ฅ, ๐ฆ โ โ. Prove that ๐ > 0 exists so that for any ๐ฅ with
|๐ฅ| โค ๐ we have |๐(๐ฅ)| < ๐. Deduce the existence of a fixed point
for ๐. (๐ฅ0 is a fixed point for ๐ if ๐(๐ฅ0) = ๐ฅ0). (Manual)
4. Let ๐: [๐, ๐] โ โ, continuous. Then:
(Manual)
A function with the property (4.1) is called an uniform
continuously function on [๐, ๐].
The uniform continuity is therefore defined on an interval,
while continuity can be defined in a point.
Taking ๐ฆ = ๐ฅ0 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: ๐ท โ โ is differentiable in ๐ฅ0 โ ๐ท if the
following limit exists and is finite:
The value of this limit is noted by ๐ ,(๐ฅ0) and it is called the
derivative of ๐ in ๐ฅ0.
Geometric interpretation The derivative of a function in a point ๐ฅ0 is the slope of the
tangent to the graphic of ๐in the abscissa point ๐๐.
When ๐ฅ tends to ๐ฅ0 , the chord ๐ด๐ต tends towards the
tangent in ๐ด to the graphic, so the slope of the chord
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tends to the slope of the tangent. Therefore, ๐ ,(๐ฅ0) is the slope of
the tangent in the abscissa point ๐ฅ0 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 (๐ฅ0, ๐(๐ฅ0)) is:
3. The equation of the normal (the perpendicular on the
tangent) is deduced from the condition of perpendicularity of two
straight lines (๐1 โ ๐2 = โ1) 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
๐3) and we replace in this function the variable ๐ฅ with the
expression issuing from the composition of the other function (in
our case ๐2 โ ๐1).
(b) by neglecting the derived function in the previous step (in
our case ๐3), we derive the function that became last (for us ๐2) and
we also replace the variable ๐ฅ with the expression that resulted from
the composition of the functions that have no yet been derived (in
this step we only have ๐1(๐ฅ).
(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 ๐โ1(๐(๐ฅ)) = ๐ฅ, by applying formula (4.2), it follows
that:
or by noting ๐ฆ = ๐(๐ฅ):
EXAMPLE: For ๐(๐ฅ) = ln ๐ฅ, ๐: (0, โ) โ โ we have:
and as ๐ฆ = ln ๐ฅ => ๐ฅ = ๐๐ฆ we obtain:
2. High order derivatives of composite functions.
From (๐(๐ข(๐ฅ)) = ๐ ,(๐ข(๐ฅ)). ๐ข,(๐ฅ) , deriving again in
relation to ๐ฅ 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: Letโs calculate ๐ ,,(๐ฅ) if ๐(๐ฅ) = ๐(๐โ๐ฅ) , 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)(๐ฅ) we can use one
of the following methods:
1. We calculate a few derivatives (๐ ,, ๐ ,,, ๐ ,,,, โฆ ) 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 Lagrangeโs theorem: If ๐ is
differentiable in a vicinity of ๐ฅ0 and if it
continues in ๐ฅ0 and if there exists:
then, ๐ has a derivative on the left (on the right,
respectively) in ๐ฅ0 and:
The continuity condition in ๐๐ that is required in the
hypothesis of the corollary is essential for its application, but it
doesnโt constitute a restriction in the study of derivability,
because if ๐ isnโt continuous in ๐ฅ0, it isnโt differentiable either.
EXAMPLE: Letโs determine the parameters ๐ผ, ๐ฝ โ 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 ๐ฅ โ ๐ , the function is differentiable, being
expressed through differentiable functions.
b) we study the derivability in ๐ฅ = 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 <=> ๐ผ๐ + ๐ฝ = 1. (4.3)
If the continuity condition in ๐ฅ = ๐ isnโt 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 lโHospitalโs rule.
(1) We observe that by noting ๐(๐ฅ) = ๐๐2๐ฅ , 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 ๐ฝ = โ1 . We thus
have:
The geometric interpretation of the result: the straight
line ๐ฆ =2
๐๐ฅ โ 1 is tangent to the curve ๐ฆ = ln ๐ฅ2.
If the branch of a function is expressed through a
straight line (๐ = ๐ถ๐ + ๐ท) , the function is
differentiable in the point ๐๐ of connection
between the branches, if and only if the
respective straight line is tangent in the point of
abscissa ๐๐ to the graphic of the other branch.
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Method 2: (Using the corollary of Lagrangeโs theorem)
(a) In any point ๐ฅ โ ๐, the function is differentiable, being
expressed by differentiable function and we have:
(b) We study the derivability in ๐ฅ = ๐ . We first state the
continuity condition:
In order to use the corollary of Lagrangeโs theorem we
calculate the lateral limits of the derivative:
From the corollary we deduce ๐ ,(๐) = ๐ผ. Analogously:
If ๐ is differentiable in ๐ฅ = ๐ if and only if๐ผ =2
๐. From the
continuity condition we deduce ๐ฝ = โ1.
(E) Applications of the derivative in economics
(See the Calculus manual, IXth grade)
1. Let ๐ฝ(๐ฅ) be the benefit obtained for an expenditure of ๐ฅ
lei. For any additional expenditure of โ lei, the supplementary
benefit on any spent leu is:
If โ is sufficiently small, this relation gives an indication of
the variation of the benefit corresponding to the sum of ๐ฅ lei.
If this limit exists:
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it is called the bound benefit corresponding to the sum of ๐ lei.
2. Let ๐พ(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 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 ๐ฅ
lei is:
For any additional expenditure of โ 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 ๐ฅ = 1000 this is equal to 1997 + โ and
Exercises I. The following are required:
1. The equation of the tangent to the graphic ๐(๐ฅ) =
ln โ1 + ๐ฅ2 in the abscissa point ๐ฅ0 = 1.
2. The equation of the tangent to the curve ๐(๐ฅ) =
โ๐ฅ2 โ ๐2, that is parallel to 0๐ฅ.
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3. The equation of the tangent to the curve ๐(๐ฅ) = ๐ฅ3 ,
parallel to the first bisector.
4. The equation of the tangent to the curve ๐ฆ =3๐ฅ+2
2๐ฅ+6, that is
parallel to the chord that unites the abscissae points ๐ฅ = 1 and ๐ฅ =
3.
5. Determine ๐ผ and ๐ฝ so that ๐ฆ = ๐ผ๐ฅ + ๐ฝ and ๐ฆ =๐ฅโ1
๐ฅ are
tangent in ๐ฅ = 1. Write down their common tangent.
6. Show that the straight line ๐ฆ = 7๐ฅ โ 2 is tangent to the
curve ๐ฆ = ๐ฅ2 + 4๐ฅ. (Manual)
II. 1. Calculate the derivative of the function:
2. Let ๐: (โโ, 0) โ โ , ๐(๐ฅ) = ๐ฅ2 โ 3๐ฅ . Determine a
subinterval ๐ฝ โ โ, so that ๐: (โโ, 0) โ ๐ฝ is bijective. Let ๐ be its
inverse. Calculate ๐,(โ1) and ๐,,(โ1). (Manual)
3. If:
with ๐ผ > 0 for any ๐ฅ, ๐ฆ โ ๐ผ, the function ๐ is constant on ๐ผ.
4. If ๐ has a limit in point ๐, then the function:
is differentiable in ๐.
5. If ๐ is bounded in a vicinity of ๐ฅ0 , then ๐(๐ฅ) =
(๐ฅ โ ๐ฅ0)2๐(๐ฅ) is differentiable in ๐ฅ0. Particular case:
Indications:
1. Let ๐ be a primitive of the function:
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We have:
3. For ๐ฅ โ ๐ฆ the inequality from the enunciation is
equivalent to:
from where, for ๐ฆ โถ ๐ฅ we obtain ๐(๐ฅ) = 0, so ๐ is constant on ๐ผ.
III. Calculate the derivatives of order ๐ for:
10. ๐(๐ฅ) = ๐๐๐ฅ + ๐๐๐ฅ from the expression of the derivative
of order ๐, deduce Newton's binomial formula.
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IV. 1. Show that:
2. Applying Leibnizโs formula for:
show that:
Find similar formulas, using the functions:
3. Let ๐ผ = (0,1) and the functions ๐ข, ๐ฃ: ๐ผ โถ โ , ๐ข(๐ฅ) =
๐ข(๐ฅ) =๐๐๐
๐ฆ โ ๐ผ(๐ฅ โ ๐ฆ)2 , ๐ฃ(๐ฅ) ==
๐ ๐ข๐๐ฆ โ ๐ผ(๐ฅ โ ๐ฆ)2 . Study the
derivability of the functions ๐ข and ๐ฃ and calculate ๐ ๐ข๐
๐ฅ โ ๐ผ๐ข(๐ฅ) ,
๐๐๐๐ฅ โ ๐ผ
๐ฃ(๐ฅ). (Manual)
4. If ๐๐(๐ฅ) is a sequence of differentiable function, having a
limit in any point ๐ฅ, then:
V. Study the derivability of the functions:
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is indefinitely differentiable on โ and:
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๐ โ โ.
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V. Fermat, Rolle, Lagrange, Cauchy Theorems
(A) Fermatโs theorem
1. EnunciationIf ๐: [๐, ๐] โ โ is continuous on [๐, ๐] and differentiable
on (๐, ๐), then in any point of extremum from (๐, ๐) (from
the interior of the interval), the derivative is annulled.
(Fermat, 1601-1665)
Observation: ๐ฅ0 โ (๐, ๐) point of extremum ==> ๐ ,(๐ฅ0) =
0. If ๐ฅ0 is a point of extremum at the ends of the interval, it is
possible that the derivative isnโt annulled in ๐ฅ0.
Example:
has two points of extremum: in ๐ฅ1 = โ1 and in ๐ฅ2 = 2, and:
2. Geometric and algebraic interpretationa) The geometric interpretation results from the
geometrical interpretation of the derivative: if the conditions of
Fermatโs theorem are fulfilled in any point from the interior of the
interval, the tangent to the functionโs graphic is parallel to the axis
0๐ฅ.
b) The algebraic interpretation: if the conditions of
Fermatโs theorem are fulfilled on [๐, ๐] , any point of extremum
from (๐, ๐) is a root to the equation ๐ ,(๐ฅ) = 0.
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Exercises 1. If ๐1, ๐2, โฆ ๐๐ are positive numbers, so that:
for any ๐ฅ โ โ, then we have:
2. Let ๐1, ๐2, โฆ ๐๐, ๐1, ๐2, โฆ ๐๐ be positive real numbers, so
that:
for any ๐ฅ โ โ. Then :
(generalization of the previous exercise)
3. If ๐ is continuous on [๐, ๐] and differentiable on (๐, ๐),
and ๐(๐) = ๐(๐) = 0, then:
(i) if ๐ , is increasing, it follows that ๐(๐ฅ) โค 0 on [๐, ๐];
(ii) ) if ๐ ,, is decreasing, it follows that ๐(๐ฅ) โฅ 0 on [๐, ๐].
4. If ๐๐ฅ โฅ ๐ฅ๐ for any ๐ฅ > 0, then ๐ = ๐.
5. ๐ก๐ ๐
๐ก๐ ๐<
๐
๐ if 0 < ๐ < ๐ <
๐
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 ๐ฅ, it follows that ๐ฅ = 0 is
a global point of minimum for ๐. According to Fermatโs theorem,
in this point the derivative is annulled, so ๐ ,(0) = 0. We have:
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3. (i) Letโs assume there exists ๐ โ (๐, ๐) so that ๐(๐) > 0.
We can even assume that ๐ is a point of maximum, because such a
point exists, according to Rolleโs theorem, so we have ๐ ,(๐) = 0.
Between ๐ and ๐ at least one point ๐ exists in which
๐ ,(๐) < 0 , because, if, letโs say ๐ ,(๐ฅ) โฅ 0 for any ๐ฅ โ (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) =โค 0 = ๐ ,(๐) we deduce the
contradiction:
4. ๐๐ฅ โฅ ๐ฅ๐ <=> ๐ฅ. ln ๐ โฅ ๐. ln ๐ฅ <=>ln ๐
๐โฅ
ln ๐ฅ
๐ฅ (for any
๐ฅ > 0). So ๐ is the abscissa of the maximum of the function:
We have:
It is sufficient to prove that ๐(๐ฅ) =๐ก๐ ๐ฅ
๐ฅ is increasing, i.e.
๐ ,(๐ฅ) > 0.
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(B) Rolleโs theorem
1. Enunciation If ๐: [๐, ๐] โ โ is continuous on [๐, ๐], differentiable
on (๐, ๐) and ๐(๐) = ๐(๐), then there exists ๐ โ
(๐, ๐) so that ๐ ,(๐) = 0. (Rolle, 1652-1719)
Fermatโs theorem states that in a point of extremum from
the interior of an interval the derivative is annulled, but it doesnโt
mention when such a point exists.
Rolleโs theorem provides a sufficient condition for the
existence of at least one such point, adding to the hypothesis from
Fermatโs theorem the condition: ๐(๐) = ๐(๐).
2. Geometric and algebraic interpretation a) The geometric interpretation: if the conditions of
Rolleโs theorem are met, there exists at least one point in the
interval (๐, ๐)in which the tangent to the graphic is parallel to 0๐ฅ.
b) The algebraic interpretation: if the conditions of
Rolleโs theorem are met, the equation ๐,(๐) = ๐ has at least one
root in the interval (๐, ๐).
The algebraic interpretation of the theorem highlights a
method used to prove that the equation ๐(๐ฅ) = 0 has at least one
root in the interval (๐, ๐). For this, it is sufficient to consider a
primitive ๐น of ๐, for which the conditions of Rolleโs theorem are
fulfilled on [๐, ๐]. It results that the equation ๐น,(๐ฅ) = 0 has at least
one root in the interval (๐, ๐), i.e. ๐(๐ฅ) has a root in the interval
(๐, ๐).
A second method to show that the equation ๐(๐ฅ) = 0 has
at least one root in the interval (๐, ๐) is to show just that ๐ is
continuous and ๐(๐). 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 functionโs 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 derivativeโs roots
(Rolleโs sequence):
Let ๐ฅ1, ๐ฅ2, โฆ , ๐ฅ๐ be the derivativeโs roots.
Then ๐ has as many real, simple roots as there are variations
of sign in the sequence:
(if we have ๐(๐ฅ๐) = 0, then ๐ฅ๐ is a multiple root)
Exercises I. Study the applicability of Rolleโs Theorem for the
functions:
What is the geometric interpretation of the result?
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Indication: 1. We obtain ๐ โ [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). (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 (โ1,0)?
(iii) The same question for the equations:
on the interval (โ1,1).
4. Let ๐: โ โ โ be differentiable and ๐1 < ๐2 < โฏ < ๐๐ ,
roots of ๐. Show that ๐ ,has at least ๐ โ 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 ๐๐(๐ฅ) = 0 has ๐ distinct roots in
the interval (โ1,1).
6. If ๐ is ๐ times differentiable on ๐ผ and it has ๐ + 1 distinct
roots on ๐ผ, then ๐๐(๐ฅ) has at least one root on ๐ผ.
7. Let ๐, ๐: [๐, ๐] โ โ, continuous on [๐, ๐], differentiable
on (๐, ๐) with ๐(๐ฅ) โ 0 and ๐,(๐ฅ) โ 0 on [a,b], and ๐(๐)
๐(๐)=
๐(๐)
๐(๐).
Show that ๐ โ (๐, ๐) exists, so that:
8. Let๐: [๐, ๐] โ โ, continuous on (๐, ๐), differentiable on
(๐, ๐). Then between two roots of ๐ there exists at least a root of
๐ผ. 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 ,(๐ฅ). cos ๐ฅ + ๐(๐ฅ). sin ๐ฅ โ 0 , in any
length interval bigger than ๐ there is at least one root of ๐.
10. If ๐, ๐: [๐, ๐] โ โ, continuous on [๐, ๐], differentiable
on (๐, ๐) , ๐(๐ฅ) โ 0, ๐,(๐ฅ) โ 0 and ๐(๐) = ๐(๐) = 0, then ๐๐ โ
(๐, 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(๐ฅ) =๐(๐ฅ)
(๐ฅ)๐, so it is sufficient to demonstrate that the
function โ fulfills the conditions of Rolleโs theorem on [๐, ๐].
8. The equation ๐ผ. ๐(๐ฅ) + ๐ ,(๐ฅ) = 0 comes from the
equalizing with zero the derivative of the function ๐น(๐ฅ) =
๐๐ผ๐ฅ๐(๐ฅ), so it is sufficient to show that ๐น satisfies the conditions of
Rolleโs theorem.
9. Let ๐ฅ1, ๐ฅ2 be roots of ๐. By hypothesis, we have๐(๐ฅ1) โ
0 and ๐(๐ฅ2) โ 0. If, for example, there can be found a root of ๐
between ๐ฅ1, ๐ฅ2 , this means that the function โ(๐ฅ) =๐(๐ฅ)
(๐ฅ)๐ satisfies
the conditions of Rolleโs theorem and so โ,(๐ฅ) is annulled in at least
one point between ๐ฅ1 and ๐ฅ2, which contradicts the hypothesis.
has at least one root in (๐, ๐) <=> โ,(๐ฅ) = 0 has at least one root
in (๐, ๐), where โ(๐ฅ) =๐(๐ฅ)
๐๐(๐ฅ).
We will present next, two methods for the study of an
equationโs roots. The first of these uses Rolleโs Theorem and the
other is a consequence of the fact that any continuous function has
Darbouxโs property.
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Methods for the study of an equationโs roots
1. Using Rollesโs theorem (its algebraicinterpretation) 2. Using Darbouxโs properties (if ๐ has Darbouxโs
property (particularly if it is continuous) on [๐, ๐] and:
๐(๐). ๐(๐) โค 0, then ๐ has a root in [๐, ๐]).
Examples:
1) If ๐: [๐, ๐] โ [๐, ๐] is a continuous function, then ๐ข, ๐ฃ โ
[๐, ๐] exists so that ๐(๐ข) = ๐ข and ๐(๐ฃ) = ๐ฃ. (Manual)
2) Let ๐: [0, 2๐] โ โ be a continuous function so that
๐(0) = ๐(2๐). Show that ๐ โ [0, ๐] exists so that ๐(๐) = ๐(๐ +
๐). (Manual)
Consequences: If a traveler leaves in the morning from spot
๐ดand arrives, in the evening in spot ๐ต, and the next day he leaves
and reaches again spot ๐ด, show that there is a point between ๐ด and
๐ต where the traveler has been, at the same hour, in both days.
Indication: If ๐(๐ก) is the space covered by the traveler, we
have: ๐(0) = ๐(24) and so ๐ก0 โ [0,12] exists, so that: ๐(๐ก0) =
๐(๐ก0 + 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 ๐ด heading to spot ๐ต, and the other from ๐ต to ๐ด, meet on
the way.
3. Using Rolleโs sequenceExample: Letโs study the nature of the equationโs roots:
๐ being a real parameter.
Answer: The derivativeโs roots are:
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and we have:
The results are illustrated in the following table:
4. The graphical method. The equation ๐น(๐ฅ, ๐) = 0 is
reduced to the form ๐(๐ฅ) = ๐. The number of roots is equal to the
number of intersection points between the graphics: ๐ฆ = ๐(๐ฅ) and
๐ฆ = ๐.
Example:
The number of roots of the given equation is equal to the
number of intersection points between the graphics:
5. Vietteโs relationsExample: Show that the equation:
with ๐, ๐, ๐ โ 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, ๐] โ โ is a Riemann integral (so it is
bounded), ๐ โ [๐, ๐] exists, so that:
Example: Let ๐: [0,1] โ โ be continuous, so that:
Show that the equation ๐(๐ฅ) = ๐ฅ has a root ๐ฅ0 โ (0,1).
Answer: ๐(๐ฅ) = ๐ฅ <=> ๐(๐ฅ) โ ๐ฅ = 0 , so by noting:
โ(๐ฅ) = ๐(๐ฅ) โ ๐ฅ we have to show that the equation โ(๐ฅ) = 0 has
a root in the interval (0,1). We have:
But: ๐02 โ(๐ฅ) = 0, according to the theorem of the average,
with ๐ โ (๐, ๐). The function ๐, being continuous, has Darbouxโs
property, so, for ๐ there exists ๐ฅ0 โ (0,1)so that ๐ = โ(๐ฅ0).
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Exercises 1. The equations:
canโt all have real roots if ๐ผ, ๐ฝ, ๐พ โ โ.
Indication:
2. If ๐: [0,1] โ โ is continuous and with the property:
then, the equation ๐(๐ฅ) โ sin ๐๐ฅ = 0 has one root in (0,1).
3. If ๐: [0,1] โ โ is continuous and with the property:
then, the equation ๐(๐ฅ) โ ๐๐ฅ2 โ ๐๐ฅ โ 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 โ ๐ฅ)๐(๐ฅ) = 1 โ ๐ฅ๐ has one root in (0,1).
Indication: We apply the theorem of the average to the
function:
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(C) Lagrangeโs Theorem 1. Enunciation:
Let ๐: [๐, ๐] โ โ be a continuous function
on [๐, ๐] and differentiable on (๐, ๐) . Then ๐ โ
(๐, 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 โ (๐, ๐) in which the tangent to the graphic is parallel to the
chord that unites the graphicโs extremities.
b) Algebraic interpretationIf the conditions of the theorem are met, the equation:
has at least one root in the interval (๐, ๐).
3. The corollary of Lagrangeโs TheoremIf ๐ is continuous on an interval ๐ผ and differentiable on
๐ผ\{๐ฅ0} and in ๐ฅ0 there exists the derivativeโs limit:
then ๐ is differentiable in ๐ฅ0 and:
This corollary facilitates the study of a functionโs 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 ๐ฅ โ 1, the function is differentiable,
being expressed by differentiable functions. We study the
derivability in ๐ฅ = 1 using the corollary of Lagrangeโs theorem.
- the continuity in ๐ฅ = 1:
๐(1) = ln 1 = 0 , so ๐ is continuous on โ (and
differentiable on โ{1}).
- derivability. For the study of derivability in ๐ฅ = 1, we have:
(we know that ๐ , exists just on โ โ (1)).
Exercises I. Study the applicability of Lagrangeโs theorem and
determine the intermediary points ๐ for:
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Indication:
2. The continuity condition in ๐ฅ =๐
6 is:
and the derivability condition (the corollary to Lagrangeโs theorem)
is:
Point ๐ is the solution of the equation:
II. 1. Let ๐(๐ฅ) = ๐๐ฅ2 + ๐๐ฅ + ๐. Apply the theorem of the
finite increases on the interval [๐ฅ1, ๐ฅ2] , 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 โค ๐ ,(๐ฅ) โค ๐ for any ๐ฅ โ ๐ผ, if and only if:
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3. (The generalization of Rolleโs theorem). If ๐ is continuous
on [๐, ๐]and differentiable on (๐, ๐), then ๐ โ (๐, ๐) exists so that:
4. Supposing ๐ is twice differentiable in a vicinity ๐ of point
๐, show that for โ, small enough, there exists ๐, ๐ โ ๐ so that:
(Manual)
5. Let ๐: [0, โ) โ โ be a differentiable function so that:
and let ๐ > 0 be fixed. Applying the Lagrange theorem on each
interval:
show there exists a sequence (๐ฅ๐)๐โโ, 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 =๐ฅ1+๐ฅ2
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 ๐ฅ = ๐ฆ (๐ฟ1) is verified, and for ๐ฅ โ ๐ฆ,
we have:
inequalities which are true due to the hypothesis.
Reciprocally, let ๐ฅ โ ๐ผ, randomly. Making ๐ฆ tend to ๐ฅ, for:
we have:
3. We distinguish three cases:
In this case we must prove that ๐ โ (๐, ๐)exists, so that:
๐ ,(๐) > 0. If we had, letโs say, ๐ ,(๐) โค 0 on (๐, ๐), it would follow
that ๐ is decreasing, so ๐(๐) > ๐(๐).
We are situated within the conditions of Rolleโs theorem.
(3) The case ๐(๐) > ๐(๐) is analogous to (1).
4. a) We apply Lagrangeโs theorem to ๐ on [๐ โ โ, ๐ + โ].
b) We apply Lagrangeโs theorem twice to the function
๐(๐ฅ) = ๐(๐ + ๐ฅ) โ ๐(๐ฅ) on the interval [๐ โ โ, ๐].
6. a) ๐ = 1/2 , c) is obtained ๐โ = ๐โ , so ๐ cannot be
determined.
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(D) Cauchyโs Theorem 1.Enunciation
Let If ๐, ๐: [๐, ๐] โ โ be continuous on[๐, ๐] and
differentiable on (๐, ๐) so that:
Then ๐ โ (๐, 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, ๐).
Exercises I. Study the applicability of Cauchyโs theorem and determine
the intermediary point ๐ for:
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II. If ๐ has derivatives of any order on [๐, ๐], by applying
Cauchyโs formula to the functions:
a) ๐(๐ฅ) = ๐(๐ฅ), โ(๐ฅ) = ๐ โ ๐ฅ, show that ๐ โ (๐, ๐) exists,
so that:
show that ๐ โ (๐, ๐) exists, so that:
show that ๐ โ (๐, ๐) exists, so that:
show that ๐ โ (๐, 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(๐ฅ) โฅ 0 and ๐ ,,(๐ฅ) < 0
(there are no non-negative and strictly concave functions on the
entire real axis).
SOLUTION:
2. From ๐ ,,(๐ฅ) < 0 for any ๐ฅ โ โ , we deduce that ๐ , is
strictly increasing and so we cannot have ๐ ,(๐ฅ) = 0 for any ๐ฅ โ โ.
Therefore, for any ๐ฅ0 โ โ exists, so that ๐ ,(๐ฅ) โ 0.
But then, from the relation:
it follows that:
We have two situations:
(1) if ๐ ,(๐ฅ0) > 0, making ๐ฅ tend to โโ in (๐ถ1), we have:
consequently:
2. If ๐ ,(๐ฅ0) < 0, we obtain the same contradiction if we
make ๐ฅ tend to +โ.
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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(๐ฅ) = ๐(๐ฅ) + ๐ถ.
or, in particular,
๐(๐ฅ) = ๐(๐ฅ) and ๐(๐ฅ) = ๐ถ = ๐๐๐๐ ๐ก๐๐๐ก.
Indeed:
and in order to prove this equality it is sufficient to prove that:
(๐(๐ฅ) โ ๐(๐ฅ)) = 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 ๐ถ, we can use two methods:
1. We calculate the expression ๐(๐ฅ) โ ๐(๐ฅ) in a
conveniently chosen point from the considered interval.
2. It the previous method cannot be applied, we can
calculate:
๐ฅ0 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,(๐ฅ) = 0 for any ๐ฅ โ โ(โ1), domain that is a reunion of
disjunctive intervals. We calculate the value of the constant on each
interval.
(1) For ๐ฅ > โ1, we choose the point ๐ฅ = 0 where we can
easily make the calculations:
(2) For ๐ฅ < โ1 , we canโt find a point in which to easily
calculate the value of โ, but we observe that we can calculate:
Inequalities Method 1. Using Lagrangeโs or Cauchyโs
theorem.
In some inequalities, we can highlight an expression of the
form:
๐ and the points ๐, ๐being conveniently chosen. In this case, we can
use Lagrangeโs 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 < 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 Cauchyโs theorem.
EXAMPLES:
1. Using Lagrangeโs theorem, show that:
if 0 < ๐ < ๐.
Solution: We go through these steps:
(a) We highlight an expression of the form:
observing that:
and dividing by ๐ โ ๐. The inequality becomes:
(b) we apply Lagrangeโs theorem to the function ๐(๐ฅ) = ln ๐ฅ
on the interval [๐, ๐]. ๐ โ (๐, ๐) exists so that:
(c) the inequality becomes:
This new form of the inequality is proven considering the
fact that ๐ < ๐ < ๐ and that ๐ ,(๐ฅ) =1
๐ฅ is decreasing.
2. Using Lagrangeโs theorem, show that ๐๐ฅ > 1 + ๐ฅ for any
๐ฅ โ 0.
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Solution:
To use Lagrangeโs theorem, we search for an interval [๐, ๐]
and a function so that we can highlight the relation ๐(๐)โ๐(๐)
๐โ๐ in our
inequality. For this we observe that, if ๐ฅ โ 0 , we have two
situations:
(1) ๐ฅ > 0. In this case we can consider the interval [๐, ๐ฅ].
a) We highlight, in the given inequality, an expression of the
form: ๐(๐ฅ)โ๐(๐)
๐ฅโ0, observing that, if ๐ฅ > 0, we have:
(b) We apply Lagrangeโs theorem to the function:
on the interval [0, ๐ฅ]. ๐ โ (0, ๐ฅ) exists so that:
(c) The inequality becomes:
This form of the inequality is proven considering the fact
that 0 < ๐ < ๐ฅ and the derivative ๐ ,(๐ฅ) = ๐๐ฅis increasing (so ๐0 <
๐๐).
(2) If ๐ฅ < 0 the demonstration is done analogously.
Method 2: The method of the minimum This method is based on the observation that if ๐ฅ0 is a global
point of minimum (the smallest minimum) for a function โ on a
domain ๐ท and if โ(๐ฅ0) โฅ 0 (the smallest value of โ is non-
negative), then โ(๐ฅ) โฅ 0 for any ๐ฅ โ ๐ท (all the values of โ are
non-negative). We can demonstrate therefore inequalities of the
form ๐(๐ฅ) โฅ ๐(๐ฅ), i.e. ๐(๐ฅ) โ ๐(๐ฅ) โฅ 0, in other words, of the
form: โ(๐ฅ) โฅ 0.
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Example: Using the method of the minimum, show that ๐๐ฅ >
1 + ๐ฅ for any ๐ฅ โ 0.
Solution: In order to use this method, we proceed as follows:
(a) We write the inequality under the form: โ(๐ฅ) โฅ 0:
(b) Let โ(๐ฅ) = ๐๐ฅ โ 1 โ ๐ฅ . Using the variation table, we
calculate the global minimum (the smallest minimum) of โ:
From the table it can be observed that โ has a single
minimum, in ๐ฅ = 0, so this is also the global minimum. Its value is
โ(0) = 0. So, for ๐ฅ โ 0, we have โ(๐ฅ) > 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, or
๐ respectively, are values of the global minimum and maximum of
๐ on the interval [๐, 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(๐ฅ) = ๐๐ฅ2+
๐1โ๐ฅ2on the interval [0,1]:
We thus have ๐ = 2โ๐ and ๐ = 1 + ๐
(c) We integrate the inequalities ๐ โค ๐(๐ฅ) โค ๐
on the interval [๐, ๐] and we obtain:
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Method 4: Using the inequality from the definition of convex (concave) functions
Let ๐ผ be an interval on a real axis.
Definition: The function ๐: ๐ผ โ โ is convex on ๐ผ if between any
two points ๐ฅ1, ๐ฅ2 โ ๐ผ , the graphic of ๐ is situated
underneath the chord that unites the abscissae points
๐ฅ1 and ๐ฅ2.
With the notations from the adjacent graphic, this condition
is expressed through the inequality:๐(๐ฅ) โค ๐ฆ.
In order to explain ๐ฅ and ๐ฆ, we observe that:
1. ๐ฅ is in the interval [๐ฅ1, ๐ฅ2] if and only if it can be written
in the form:
Indeed, if ๐ฅ โ [๐ฅ1, ๐ฅ2], it is sufficient that we take ๐ผ = (๐ฅ โ
๐ฅ2)/๐ฅ1 โ ๐ฅ2) to meet the required equality.
Reciprocally, if ๐ฅ has the expression from (๐ธ. ๐ผ. 1), in order
to show that ๐ฅ โ [๐ฅ1, ๐ฅ2], the inequality system must be checked:
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The first inequality is equivalent to ๐ผ โค 1, and the second
inequality is equivalent to ๐ผ โค 0.
Observation: The condition ๐ผ โ [0,1] from (๐ธ. ๐ผ. 1) is
therefore essential.
The inequality from the convexity becomes:
2. To express ๐ฆ, we observe that:
so:
from which it follows:
By replacing ๐ฅ and ๐ฆ, we obtain:
๐: ๐ผ โ โ is convex <=>
๐: ๐ผ โ โ is said to be concave if, for any ๐ฅ1, ๐ฅ2 โ ๐ผ, between
the abscissae points ๐ฅ1 and ๐ฅ2 , 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 ๐ฅ1 and ๐ฅ2 from โ, we have:
Solution: The function ๐(๐ฅ) = ๐๐ฅ is convex on โ(๐ ,,(๐ฅ) >
0), so:
from where, for ๐ผ =1
2 we obtain the inequality from the
enunciation.
Exercises I. Show that:
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5. For the intervals ๐ผ1, ๐ผ2 โ โ letโs consider the continuous
and injective functions:
and the constant ๐ in the interior of ๐(๐ผ1) . From the
condition: ๐(๐ข(๐ฅ)) = ๐ + ๐(๐ฅ) it follows that ๐ข(๐ฅ) = ๐โ1(๐ +
๐(๐ฅ) for any ๐ฅ that fulfills the condition ๐ + ๐(๐ฅ) โ ๐(๐ผ1). Using
this observation, construct equalities of the form โ(๐ฅ) = ๐ for:
Answer: (a) We have:
We determine function ๐ข from the condition ๐(๐ข(๐ฅ)) =
๐ + ๐(๐ฅ) that becomes:
For ๐ =๐
4, for example, we have:
and from the condition:
we deduce:
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and so we have:
II. Using Lagrangeโs theorem, prove the inequalities:
6. Using Cauchyโs 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 ๐1(๐ฅ) < ๐2(๐ฅ) or ๐1(๐ฅ) > ๐2(๐ฅ) on the
considered intervals. We then integrate the obtained inequality.
VI. 1. If ๐: ๐ผ โ โ is convex, then for any ๐ฅ1, ๐ฅ2, โฆ , ๐ฅ๐ โ ๐ผ,
we have the inequality (Jensenโs inequality):
(Manual)
2. Using Jensenโs inequality applied to the convex function
๐(๐ฅ) = ๐๐ฅ, prove the inequalities of the averages:
3. Show that for any ๐ฅ1, ๐ฅ2, โฆ , ๐ฅ๐ โ [๐,๐
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 donโt 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 ๐ฅ1, ๐ฅ2, โฆ , ๐ฅ๐ > 0 we have:
5. Show that the function ๐(๐ฅ) = ๐ฅ๐ผ , ๐ผ > 1 , ๐ฅ > 0 is
convex. Using this property show that for any non-negative
numbers ๐ฅ1, ๐ฅ2, โฆ , ๐ฅ๐, the following inequality takes place:
6. Applying Jensenโs inequality to the function ๐(๐ฅ) = ๐ฅ2 ,
show that for any ๐ฅ1, ๐ฅ2, โฆ , ๐ฅ๐ โ โ, we have:
What analogous inequality can be deduced from the convex
function ๐(๐ฅ) = ๐ฅ3 , for ๐ฅ > 0? And from the concave function
๐(๐ฅ) = ๐ฅ3, for ๐ฅ โค 0?
7. Applying Jensenโs inequality for the convex function
๐(๐ฅ) =1
๐ฅ, for ๐ฅ > 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 Jensenโs inequality, we deduce that the
points ๐ฅ1, ๐ฅ2, โฆ , ๐ฅ๐ are deduced from the conditions:
We have:
With these points, Jensenโs inequality, for ๐(๐ฅ) = ๐๐ฅ, becomes:
Making the calculations in the left member we obtain the inequality
of the averages.
3. On the interval [0,๐
2], the functions sin ๐ฅ and cos ๐ฅ are
concave. The biggest interval that contains the origin and in which
sin ๐ฅ is concave, is the interval [0, ๐]. On [๐, 2๐], 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: ๐ผ โ โ has primitives on the interval
๐ผ if there exists a function ๐น: ๐ผ โ โ, differentiable
and ๐น,(๐ฅ) = ๐(๐ฅ) on ๐ผ.
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 doesnโt 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. DARBOUXโS PROPERTY
5. THE PRIMTIVE
6. THE INTEGRAL
The connections between these notions are given by the
following properties:
๐1 ) Any continuous function in a point
๐ฅ0 has a limit in this point:
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๐2 ) Any differentiable function in a point
๐ฅ0 is continuous in this point:
๐3 ) Any continuous function on [๐, ๐] has Darbouxโs
property on [๐, ๐]:
๐4 ) Any continuous function on [๐, ๐] has primitives on
[๐, ๐]:
๐5 ) Any continuous function on [๐, ๐] is integrable on
[๐, ๐]:
๐6) Any function that has primitives on [๐, ๐] has Darbouxโs
property on [๐, ๐]:
๐7) If a function has Darbouxโs property on [๐, ๐], then in
any point ๐ฅ0 in which the limit (the lateral limit) exists, it is equal to
๐(๐ฅ0).
Consequences: 1. A function with Darbouxโs property (so a
function that admits primitives) canโt have an infinite limit (lateral
limit) in a point ๐ฅ0.
The functions for which at least one lateral limit is
infinite or different from ๐(๐๐) doesnโt have primitives.
2. If ๐ has Darbouxโs property on [๐, ๐] and ๐ฅ0 is a
discontinuity point, AT LEAST ONE LATERAL LIMIT
DOESNโT EXIST.
Due to the implications stated above, we can form the Table:
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OBSERVATION: If ๐ฅ0 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 doesnโt exist
(5) one lateral limit is finite, the other doesnโt exist
(6) both lateral limits donโt 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 DOESNโT EXIST (AND THE OTHER,
IF IT IS EXISTS, IS FINITE).
Also, letโs observe that, although in case (1) the function
doesnโt have primitives on [๐, ๐] (doesnโt have Darbouxโs
property), still, if, moreover, the lateral limits are equal, then ๐ has
primitives on [๐, ๐]\{๐ฅ0} through the function ๐๐ โ the extension
through continuity of ๐โฒ๐ restriction to the domain [๐, ๐]\{๐ฅ0}.
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Example:
has Darbouxโs property for any ๐ โ [โ1,1] , but doesnโt have
primitives if ๐ โ 0.
Indeed, assuming ad absurdum that ๐ has primitives, let ๐น be
one of its primitives. Then ๐น must be of the form:
In order to calculate โซ sin1
๐ฅ๐๐ฅ, we observe that sin
1
๐ฅ comes
from the derivation of the function ๐ฅ2 cos1
๐ฅ. Indeed,
so:
The function ๐(๐ฅ) = ๐ฅ cos1
๐ฅ has primitives on (0, โ)
because it is continuous on this interval, but doesnโt have primitives
on [0, โ) (we aim for the interval to be closed at zero in order to
study the continuity and derivability od ๐น).
We observe that, because ๐๐๐
๐ฅ โ 0๐ฅ > 0
๐(๐ฅ) = 0, we can consider
its extension through continuity:
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This function, being continuous on [0, โ), has primitives on
this interval. Let ๐บ be one of these primitives. Then:
We state the condition that ๐น be continuous in ๐ฅ0 = 0. We
have:
so we must have ๐ถ1 = ๐ถ2 + 2. ๐บ(0) and so, for the primitive, we
obtain the following formula:
We state the condition that ๐น be differentiable in zero:
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because ๐บ is differentiable in zero. We have ๐บ ,(0) = ๐๐(0) = 0 (๐บ
is a primitive of ๐๐ ), so ๐น๐, (0) = 0 . Consequently, ๐น is not
differentiable in zero if ๐ โ 0.
Observation: For the study of derivability we have mentioned
two methods: using the definition and using the corollary of
Lagrangeโs theorem. Letโs observe that in the example from above,
we canโt use the corollary of Lagrangeโs theorem because
๐๐๐๐ฅ โ 0๐ฅ > 0
๐น,(๐ฅ) doesnโt 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 doesnโt 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 โ ๐:
(๐ธ๐1) We show that the function is differentiable.
(๐ธ๐2) We show that the function is continuous.
(๐ธ๐3) We construct the primitive.
(๐ธ๐4) We show that the function is a sum of functions that
admit primitives.
Keep in mind that method ๐ธ๐2 is preferred over ๐ธ๐1
because derivability is easier to prove that continuity, and the
method ๐ธ๐3 is used only if the function doesnโt 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 doesnโt have primitives
The first three methods are deduced from the table, using
the equivalence: ๐ โ ๐ โท ๐๐๐ ๐ โถ ๐๐๐ ๐.
We therefore have:
This way, we deduce the following methods to show that a
function doesnโt have primitives:
(๐๐1) We show that in a discontinuity point, both lateral
limits exist and are finite or at least one lateral limit is infinite..
(๐๐2) We show that the function doesnโt have Darbouxโs
property..
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(๐๐3) We assume ad absurdum that ๐ has primitives and we
obtain a contradiction.
(๐๐4 ) If ๐ is the sum of two function, one of them
admitting primitives, the other not admitting primitives, it follows
that ๐ doesnโt have primitives.
Examples:
has primitives because ๐ is continuous.
Solution: We observe that 2๐ฅ. ๐ ๐๐1
๐ฅโ ๐๐๐
1
๐ฅ comes from the
derivation of the function ๐ฅ2๐ ๐๐1
๐ฅ, so, a primitive of ๐ must have
the form:
We state the condition that ๐น be continuous in zero:
so ๐ผ = ๐ถ and we have:
We state the condition that ๐น be differentiable in zero:
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(something bounded multiplied with something that tends to zero,
tends to zero), so ๐น is differentiable in zero. But we must also have
๐น,(0) = ๐(0), equality which is also met, so ๐ has primitives.
We have:
where ๐(๐ฅ) =1
2 and:
and the functions ๐ and โ are primitives.
(๐๐1 ) a) ๐(๐ฅ) = [๐ฅ] โ ๐ฅ doesnโt have primitives on โ
because in points ๐ฅ0 = ๐ is discontinuous and both lateral limits are
finite.
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doesnโt have primitives because ๐๐๐
๐ฅ โ 0๐ฅ > 0
๐(๐ฅ) = โโ.
(๐๐2) The functions in which this method is used in the
manual are of the form:
with ๐ and โ, continuous functions, ๐ โ โ. We will prove on this
general case that ๐ doesnโt have Darbouxโs property. The
demonstration can be adapted to any particular case.
We must prove there exists an interval whose image through
๐ isnโt an interval. We use the two conditions from the hypothesis.
Solving the exercise: We must prove there exists an interval
whose image through ๐ isnโt an interval, using the two conditions
from the hypothesis.
From the condition ๐ โ โ we deduce there exists at least
one point ๐ฅ0 in which ๐(๐ฅ0) โ โ(๐ฅ0). Then, for ํ small enough,
the intervals (๐(๐ฅ0) โ ํ, ๐(๐ฅ0) + ํ) and (โ(๐ฅ0) โ ํ, โ(๐ฅ0) + ํ)
are disjunctive (see the Figure below).
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From the conditions ๐ and โ continuous in ๐ฅ0 we deduce:
Considering, as we said ํ small enough for the intervals
(๐(๐ฅ0) โ ํ, ๐(๐ฅ0) + ํ) and (โ(๐ฅ0) โ ํ, โ(๐ฅ0) + ํ) to be
disjunctive, and ๐ฟ = min {๐ฟ1, ๐ฟ2}, it follows that the image through
the function ๐ of the interval (๐ฅ0 โ ๐ฟ, ๐ฅ0 + ๐ฟ) isnโt an interval,
because:
(๐๐3) This method presupposes the same steps as in
method ๐ธ๐3.
(a) we look for primitives for ๐โฒ๐ branches and we obtain a
first expression of ๐น;
(b) we state the condition that ๐น be continuous in the
connection point (points) between the branches;
(c) we state the condition that ๐น be differentiable in these
points:
- if ๐น is differentiable and ๐น,(๐ฅ0) = ๐(๐ฅ0),
we have covered method ๐ธ๐3.
- if ๐น isnโt differentiable in ๐ฅ0 or ๐น,(๐ฅ0) โ
๐(๐ฅ0), we have covered method (๐๐3).
Methods of Solving Calculus Problems for UNM-Galllup Students
215
doesnโt have primitives on โ.
Indeed, we have ๐(๐ฅ) = ๐(๐ฅ) + โ(๐ฅ) with:
and ๐ is continuous so it has primitives and โ has lateral limits,
finite in zero, so it doesnโt have primitives. If, letโs say by absurd, ๐
had primitives, it would follow that โ(๐ฅ) = ๐(๐ฅ) โ ๐(๐ฅ) has
primitives.
Exercises I. Using the method (๐ธ๐2) , show that the following
functions have primitives in โ:
C. Dumitrescu โ F. Smarandache
216
II. Using methods (๐ธ๐3) โ (๐๐3) , study if the following
functions have primitives:
Methods of Solving Calculus Problems for UNM-Gallup Students
217
if ๐: โ โ โ and if it satisfies: ๐(๐ฅ) = 0 <=> ๐ฅ = 0.
10. The product from a function โ that admits primitives
and a function ๐ differentiable with the continuous derivative is a
function that admits primitives.
Indications: If ๐ป is the primitive of โ, then ๐ป. ๐, is continuous,
so it admits primitives. Let ๐บ be one of its primitives. We have
(๐ป. ๐ โ ๐บ), = โ. ๐.
๐ being a differentiable function with a continuous derivative on โ.
III. Using method (๐๐1), show that the following functions
donโt have primitives:
C. Dumitrescu โ F. Smarandache
218
IV. Using method (๐๐2), show that the following functions
donโt have primitives:
Answer:
1. The functions ๐(๐ฅ) = 3๐ฅ and โ(๐ฅ) = 2๐ฅ2 + 1 are
continuous on โ and ๐ โ โ . We will show that ๐ doesnโt have
Darbouxโs property. Let ๐ฅ0 be a point in which we have (๐ฅ0) =
โ(๐ฅ0 ), for example ๐ฅ0 = 2. We have ๐(2) = 6, โ(2) = 9.
โ โ ๐๐๐๐ก๐๐๐ข๐๐ข๐ ๐๐ ๐ฅ0 = 2 <=>
Methods of Solving Calculus Problems for UNM-Gallup Students
219
๐ โ ๐๐๐๐ก๐๐๐ข๐๐ข๐ ๐๐ ๐ฅ0 = 2 <=>
Let there be ํ so that the intervals:
are disjunctive. For example ํ = 1 and let there be ๐ฟ =
min (๐ฟ1, ๐ฟ2). Then the image of the interval (2 โ ๐ฟ, 2 + ๐ฟ) through
๐ isnโt an interval because for:
we have ๐(๐ฅ) = 3๐ฅ โ (5,7) and for:
we have ๐(๐ฅ) = 2๐ฅ2 + 1 โ (8,10).
V. Using methods (๐ธ๐4) โ (๐๐4) , study if the following
functions admit primitives:
Indication:
so ๐(๐ฅ) = ๐(๐ฅ) + โ(๐ฅ), with:
C. Dumitrescu โ F. Smarandache
220
Indication:
We obtain ๐ = 0.
Methods of Solving Calculus Problems for UNM-Gallup Students
221
Logical Scheme for Solving Problems Solving problems that require studying if a function has
primitives can be done according to the following logical scheme:
C. Dumitrescu โ F. Smarandache
222
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 ๐ฅ tends to ๐ฅ0,
2. the limit when ๐ tends to infinity,
and a new category of limits is introduced:
3. the limit when the divisionโs 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 0๐ฅ.
Given a bounded function and (for now) non-negative
๐: [๐, ๐] โ โ we can approximate the desired area with the help of
certain rectangles having the base on 0๐ฅ.
For this, we consider the points:
The set of these points form a division of the interval [๐, ๐]
and they are noted with โ:
The vertical strips made using the points ๐ฅ๐ 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.
Methods of Solving Calculus Problems for UNM-Gallup Students
223
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 ๐๐ situated between ๐ฅ๐โ1 and ๐ฅ๐ . 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 โ and to the intermediary points
๐๐ . 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?
C. Dumitrescu โ F. Smarandache
224
We observe that the rectangles โincrease in numberโ if and
only if ๐ tends to infinity, but making ๐ tend to infinity we do not
obtain a โbetter 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 โbetterโ if the rectangles are โthinner and thinnerโ.
In order to concretize this intuitive intuition, we observe that
the rectangles are thinner and thinner if the โthickest of themโ
becomes thinner and thinner.
The biggest thickness of the rectangles determined by a
division โ is called the norm of the division โ.
Therefore, โbetter approximationsโ <=> โthinner and
thinner rectanglesโ <=> โthe biggest thickness tends to zeroโ <=>
โthe norm of โ tends to zeroโ <=> ||โ|| โ 0.
The area of the stretch situated between the graphic of the
function ๐ and the axis 0๐ฅ is, therefore:
This limit is noted with โซ ๐(๐ฅ)๐
๐๐๐ฅ and it is called the
integral of function ๐ on the interval [๐, ๐].
For functions that are not necessarily non-negative on [๐, ๐],
the integral is represented by the difference between the area
situated above the axis 0๐ฅ and the area situated below the axis 0๐ฅ.
Methods of Solving Calculus Problems for UNM-Gallup Students
225
Consequently, in order to obtain the area of the stretch
situated between the graphic and the axis 0๐ฅ, in such a case, we
have to take into consideration the module of ๐:
Coming back to the affirmations (๐) and (๐) from above,
we observe that: โthe rectangles are thinner and thinnerโ => โthe
rectangles increase in numberโ, namely (๐) => (๐).
In other words, ||โ|| โ 0 = ๐ โ โ.
Reciprocally, this statement isnโt always true. That is why we
cannot renounce ||โ|| โ ๐ in favor of ๐ โ โ . Nevertheless,
there is a case where the reciprocal implication is true: when the
points of the division are equidistant.
๐ โ โ => ||โ|| โ 0 for equidistant points.
C. Dumitrescu โ F. Smarandache
226
In this case, the rectangles have the same dimension of the
base, namely ๐ฅ๐ โ ๐ฅ๐โ1 =๐โ๐
๐ for any ๐ = 1,2, โฆ , ๐ , and the
Riemann sum becomes:
For divisions such as these, the following equivalence takes
place:
||โ|| โ 0 <=> ๐ โ โ
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, letโs observe that, if the function is
continuous, among the Riemann sums ๐โ(๐, ๐) that can be
obtained by modifying just the heights of the rectangles (modifying
just the intermediary points ๐๐), there exists a biggest and smallest
Riemann sum, namely the Riemann sum having the highest value is
obtained by choosing the points ๐๐ โ [๐ฅ๐โ1, ๐ฅ๐] for which:
and the Riemann sum having the smallest value is obtained by
choosing the points ๐๐ for which:
By noting with ๐โ(๐) and ๐ โ(๐), respectively, these sums (called
the superior Darboux sum, the inferior Darboux sum, respectively),
we have:
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227
The points ๐๐ that realizes the supreme, respectively, the
infimum of ๐ on the intervals [๐ฅ๐โ1, ๐ฅ๐] 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 ๐1 โ ๐7 enunciated in the previous chapter, with the
following:
๐8) any function integrable on [๐, ๐] is bounded,
๐9) any function monotonous on [๐, ๐] is integrable,
๐10) any function continuous on [๐, ๐] is integrable,
๐11) any function that has on [๐, ๐] a finite number of
discontinuity points of first order is integrable.
The demonstration of ๐10 follows from the fact that if we
modify the values of a function integrable on [๐, 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:
C. Dumitrescu โ F. Smarandache
228
and the table containing implications between all the notions that
we have dealt with is the following ( ๐1 and ๐2 being one of the
lateral limits):
Methods of Solving Calculus Problems for UNM- Gallup Students
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Methods to show that a function is
integrable (Riemann) on [๐, ๐] (๐ผ1) using the definition;
(๐ผ2) we show that it is continuous;
(๐ผ3) we show that it has a finite number of discontinuity
points of first order;
(๐ผ4) we show it is monotonic;
(๐ผ5) we show that the function is a sum of integrable
functions.
Methods to show that a function is not
integrable (Riemann) on [๐, ๐] (๐1) using the definition (we show that ๐โ(๐, ๐) doesnโt
have a finite limit for ||โ|| โ 0);
(๐2) we show that the function is not bounded;
(๐3) we show that the function is the sum of an integrable
function and a non-integrable function.
Examples (๐ผ1) (a) Using the definition of the integral, show that any
differentiable function with the derivative bounded on [๐, ๐] is
integrable on [๐, ๐].
Solution: Let ๐: [๐, ๐] โ โ be differentiable. We have to
show that ๐โ(๐, ๐) has a finite limit when ||โ|| โ 0 . The given
function being differentiable, it is continuous, so it has primitives.
Let ๐น be one of its primitives. We can apply Lagrangeโs theorem to
function ๐น , on any interval [๐ฅ๐โ1, ๐ฅ๐]. There exists therefore ๐๐ โ
[๐ฅ๐โ1, ๐ฅ๐] so that:
C. Dumitrescu โ F. Smarandache
230
It follows that:
We calculate the two sums separately:
We show that the second sum tends to zero when ||โ|| โ 0.
For this, we take into consideration that ๐ , is bounded, in other
words, ๐ > 0 exists, so that |๐ ,(๐ฅ) < ๐| for any ๐ฅ from [๐, ๐] and
we apply the theorem of Lagrange to the function ๐, but on the
intervals of extremities ๐๐ and ๐๐ . There exists ๐๐ between ๐๐ and ๐๐
so that:
Methods of Solving Calculus Problems for UNM-Gallup Students
231
It follows therefore that:
For the second sum, we therefore have:
The condition from this exerciseโs 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.
(๐ผ1) (b) Show that ๐) ๐ฅ=sin ๐ฅ is integrable on any interval
[๐, ๐]. (Manual).
Solution: We will adapt the previous demonstration. The
given function is differentiable and has the derivative ๐ ,(๐ฅ) = cos ๐ฅ
bounded. Being differentiable, it is continuous and it has primitives.
Let ๐น be one of its primitives, ๐น(๐ฅ) = โ cos ๐ฅ and let:
C. Dumitrescu โ F. Smarandache
232
be a division of [๐, ๐] . We apply Langrangeโs theorem to the
function ๐น on each interval [๐ฅ๐โ1, ๐ฅ๐]. There exists therefore ๐๐ โ
[๐ฅ๐โ1, ๐ฅ๐] so that:
The function ๐น has a bounded derivative: | cos ๐ฅ| < 1 for
any ๐ฅ from โ , so for any ๐ฅ from [๐, ๐] . We have to show that
๐โ(๐, ๐) has a finite limit when ||โ|| โ 0. We will show that this
limit is (as it is known) ๐น(๐) โ ๐น(๐) = โ cos ๐ + cos ๐ .
We have:
Applying the theorem of Lagrange to the function ๐(๐ฅ) =
sin ๐ฅ, on the interval determined by the points ๐๐ and ๐๐ , we deduce
there exists ๐๐ between ๐๐ and ๐๐ so that:
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233
(๐ผ2) The function ๐: [0,3] โ โ defined by:
is integrable. Indeed, it is continuous in any point ๐ฅ โ 2 from the
definition domain, being expressed through continuous functions.
We study the continuity in ๐ฅ = 2:
The function is continuous on the interval [0,3], so it is integrable.
(๐ผ3) ๐(๐ฅ) = ๐ฅ โ [๐ฅ], ๐: [0, 2โ3] โ โ can be discontinuous
just in the points where [๐ฅ] is discontinuous. These are of the form
C. Dumitrescu โ F. Smarandache
234
๐ฅ = ๐, ๐ โ โค. From these, only points 0,1,2,3, are in the functionโs
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.
(๐ผ4) For the function ๐(๐ฅ) = [๐ฅ] we can say it is integrable
either by using the previous criterium, either using the fact that it is
monotonic on any interval [๐, ๐].
(๐ผ5) The function ๐: [โ1,1] โ โ through ๐(๐ฅ) = ๐ ๐๐ ๐ฅ +
|๐ฅ| + [๐ฅ] is integrable because it is the sum of three integrable
functions: ๐1(๐ฅ) = ๐ ๐๐ ๐ฅ is integrable because it has a finite
number of discontinuity points (a single point, ๐ฅ = 0 ) and the
discontinuity is of the first order; the function ๐2(๐ฅ) = |๐ฅ| is
integrable because it is continuous, and the function ๐3(๐ฅ) = [๐ฅ] is
integrable because it is monotonic (or because it has a single f
discontinuity point, of the first order).
isnโt integrable because, by choosing in the Riemann sum:
the intermediary points ๐๐ , rational, we have: ๐(๐๐) = 0, so in this
case:
and if ๐๐ are irrational, we have:
Methods of Solving Calculus Problems for UNM-Gallup Students
235
isnโt integrable (Riemann) on [0,1] because it is not bounded.
is the sum of the functions:
If ๐ would be integrable, from ๐ = ๐1 + ๐2 it would follow
that ๐2 = ๐ โ ๐1 so ๐2 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:
C. Dumitrescu โ F. Smarandache
236
8. The characteristic function of the set [0,1] โ โ.
III. For which values of the parameters ๐ and ๐ are the
functions below integrable? For which values of the parameters do
they have primitives?
Methods of Solving Calculus Problems for UNM-Gallup Students
237
INDICATIONS:
1. ๐๐(0) = โ๐
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 โ (0,1) the function has a limited
number of discontinuity points of first order.
4. The function ๐๐ฅ๐๐๐ 1
๐ฅ comes from the derivation of
๐ฅ2๐๐ฅ sin1
๐ฅ.
IV. 1. Show that any continuous function on an interval
[๐, 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:
C. Dumitrescu โ F. Smarandache
238
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 ๐ข๐ , ๐ฃ๐ โ [๐, ๐] so
that ๐๐ = ๐(๐ข๐), ๐๐ = ๐(๐ฃ๐).
Therefore:
In order to evaluate the difference ๐(๐ข๐) โ ๐(๐ฃ๐), we use
the property:
(2) a continuous function on a closed and bounded interval is
uniformly continuous (see Chapter ๐ผ๐), namely:
In order to replace ๐ฅ1 with ๐ข๐ and ๐ฅ2 with ๐ฃ๐ , we have to
consider a division โ having a smaller norm than ๐ฟ (which is
possible because we are making the limit for ||โ|| โ 0). Therefore,
assuming ||โ|| < ๐ฟ we have:
i.e. ๐๐ โ ๐๐ < ํ and so:
Because ํ is arbitrary (small), we deduce that:
Methods of Solving Calculus Problems for UNM-Gallup Students
239
so the function is integrable.
2. We show that ๐โ(๐) โ ๐ โ(๐) tends to zero when ||โ||
tends to zero. We have:
where:
because on the interval [0,1] the function ๐(๐ฅ) = ๐ฅ2is increasing.
So:
The function ๐(๐ฅ) = ๐ฅ2 is continuous on [0,1] so it is
uniformly continuous, namely:
Considering the division โ so that ||โ|| < ๐ฟ (possibly
because ||โ|| โ 0), we have ๐ฅ๐2 โ ๐ฅ๐โ1
2 < ํ, so:
Because ํ is arbitrarily small, we deduce that:
i.e. ๐(๐ฅ) = ๐ฅ2 is integrable on [0,1].
C. Dumitrescu โ F. Smarandache
240
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243
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