methods of solving calculus problems for unm-gallup students
TRANSCRIPT
Methods of Solving Calculus Problems for UNM-G Students
1
The Educational
Publisher Columbus, 2017
METHODS OF SOLVING CALCULUS PROBLEMS
FOR UNM-GALLUP STUDENTS
β Constantin Dumitrescu β Florentin Smarandache β
C. Dumitrescu β F. Smarandache
2
β C. Dumitrescu β F. Smarandache β
Methods of Solving Calculus Problems for UNM-Gallup Students
β Theory and Exercises β
Methods of Solving Calculus Problems
3
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 β F. Smarandache
4
β Constantin Dumitrescu β Florentin Smarandache β
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)
Β© Educational Publisher, Columbus,2017.
C. Dumitrescu β 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 π........................................................... 158
(D) The study of derivability ...................................................... 158
(E) Applications of the derivative in economics ..................... 162
V. Fermat, Rolle, Lagrange, Cauchy Theorems ............................ 169
(A) Fermatβs theorem ....................................................................... 169
1. Enunciation ............................................................................... 169
2. Geometric and algebraic interpretation ................................ 169
(B) Rolleβs theorem ........................................................................... 172
1. Enunciation ............................................................................... 172
2. Geometric and algebraic interpretation ................................ 172
3. Consequences ........................................................................... 173
Methods for the study of an equationβs roots.............................. 177
1. Using Rollesβs theorem (its algebraic interpretation) ......... 177
2. Using Darbouxβs properties ................................................... 177
3. Using Rolleβs sequence ............................................................ 177
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4. The graphical method. ............................................................. 178
5. Vietteβs relations ....................................................................... 178
6. Using the theorem of the average ......................................... 179
(C) Lagrangeβs Theorem .................................................................. 181
1. Enunciation ............................................................................... 181
2. Geometric and algebraic interpretation ................................ 181
3. The corollary of Lagrangeβs Theorem .................................. 181
(D) Cauchyβs Theorem ..................................................................... 186
1.Enunciation ................................................................................ 186
2. Geometric and algebraic interpretation ................................ 186
VI. Equalities and Inequalities ..................................................... 189
Equalities ............................................................................................ 189
Inequalities ......................................................................................... 190
Method 1. Using Lagrangeβs or Cauchyβs 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 doesnβt 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
naΓ―ve 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 lifeβs 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 β 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 β 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 equationβs
root,
applications of Fermatβs, Rolleβs, Lagrangeβs and
Cauchyβs 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 π . Then the family π(π) of its
parts would form a set. We would thus have π(π) β π.
Noting with cardπ the number of elements belonging to π,
we will have:
ππππ π(π) β€ πππππ.
However, a theorem owed to Cantor shows that we always
have
ππππ π < ππππ π(π).
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 π.
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 π 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 π΄ β© π΅ = {π₯ β π β π₯ β π΄ πππ π₯ β π΅}
Because at a given point we work only with elements
belonging to π, the condition π₯ β π is already implied, so we can
write:
π΄ β© π΅ = {π₯ β π₯ β π΄ πππ π₯ β π΅}
π¨ β© π©
More generally,
Letβs observe that:
π₯ β π΄ β© π΅ <===> π₯ β π΄ ππ π₯ β π΅
2. The Union π΄ βͺ π΅ = {π₯ β π₯ β π΄ πππ₯ β π΅}
π¨ βͺ π©
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As in the case of the intersection, we can consider:
We see that:
π₯ β π΄ βͺ π΅ <===> π₯ β π΄ πππ π₯ β π΅
3. The Differenceπ΄ β π΅ = {π₯ β π₯ β π΄ πππ π₯ β π΅}
We retain that:
π₯ β π΄ β π΅ <===> π₯ β π΄ ππ π₯ β π΅
4. The ComplementThe complement of a set π΄ is the difference between the
total set and π΄.
πΆππ΄ = { π₯ | π₯ β π πππ π₯ β π΄} = { π₯ | π₯ β π΄}
The complement of a set is noted with πΆπ΄ or with οΏ½Μ οΏ½.
πͺπ¨
Letβs observe that
π₯ β πΆπ΄ <===> π₯ β π΄
More generally, we can talk about the complement of a set to
another, random set. Thus,
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πΆπ΄π΅ = { π₯ | π₯ β π΅ πππ π₯ β π΄}
is the complement of set π΅ to set π΄.
5. The Symmetrical Differenceπ΄βπ΅ = (π΄ β π΅) βͺ (π΅ β π΄)
π¨βπ©
We have π₯ β π΄βπ΅ <===> π₯ β π΄ β π΅ and π₯ β π΅ β π΄ .
6. The Cartesian Productπ΄ Γ π΅ = {(π₯, π¦)βπ₯ β π΄ πππ π¦ β π΅}
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, β Γ β = {(π₯, π¦)|π₯ β β , π¦ β β} , 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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π΄ Γ π΅ Γ πΆ = {(π₯, π¦, π§)|π₯ β π΄ , π¦ β π΅, π§ β πΆ}.
An intuitive image of β3 = β Γ β Γ β is given in the figure
below.
More generally, ππ
π = 1π΄π = {(π1, π2, β¦ , ππ)| ππ β π΄π πππ πππ¦ π β 1, πΜ Μ Μ Μ Μ }.
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 β at the level of a rigorous definition of sets β
just an axiom.
π΄ = π΅ <===> π΄ β π΅ πππ π΅ β π΄ (1.1)
By accepting this axiom, we will further exemplify two
methods to prove the equality of sets:
C. Dumitrescu β F. Smarandache
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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 π΄
the function ππ΄: π βΆ {0,1} defined through:
ππ΄(π₯) = {1 ππ π₯ β π΄0 ππ π₯ β π΄
As we can observe, the numbers 0 and 1 are used to divide
the elements of the total set π in two categories:
(1) one category contains those elements π₯ in which the
value of ππ΄ is 1 (the elements that belong to π΄);
(2) all the elements in which the value of ππ΄ is 0 (elements
that do not belong to π΄) 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 setπ(π»)
of the subsets of π» to the set πΈ(π») of the characteristic
functions defined on π» (see Exercise VII).
Hence,
π΄ = π΅ <===> ππ΄ = ππ΅.
(1.2)
Properties of the Characteristic Function
π1) ππ΄β©π΅(π₯) = ππ΄(π₯). ππ΅(π₯)
π2) ππ΄βͺπ΅(π₯) = ππ΄(π₯) + ππ΅(π₯) β ππ΄β©π΅(π₯)
π3) ππ΄βπ΅(π₯) = ππ΄(π₯) β ππ΄β©π΅(π₯)
π4) ππΆπ΄(π₯) = 1 β ππ΄(π₯)
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π5) ππ΄βπ΅(π₯) = ππ΄(π₯) + ππ΅(π₯) β 2. ππ΄β©π΅(π₯)
π6) ππ΄Γπ΅(π₯, π¦) = ππ΄(π₯) Γ ππ΅(π¦)
π7) ππ΄2(π₯) = ππ΄(π₯)
π8) πβ (π₯) β‘ 0, ππ(π₯) β‘ 1
π9) π΄ β π΅ <===> ππ΄(π₯) β€ ππ΅(π₯) for any π₯ β π.
Letβs prove, for example π1). For this, letβs observe that the
total set π is divided by sets π΄ and π΅ in four regions, at most:
1) for the points π₯ that do belong to neither π΄, nor π΅, we
have ππ΄(π₯) = ππ΅(π₯) = 0 and ππ΄β©π΅(π₯) = 0.
2) for the points π₯ that are in π΄ but are not in π΅, we have
ππ΄(π₯) = 1, ππ΅(π₯) = 0 πππ ππ΄β©π΅(π₯) = 0 .
3) if π₯ β π΄ πππ π₯ β π΅,
the equalities follow analogously.
4) if π₯ β π΅ πππ π₯ β π΄
Exercises I. Using methods (A) and (B) show that:
SOLUTIONS:
(A) (the double inclusion)
Letβs notice that solving a mathematical problem requires we
successively answer two groups of questions:
C. Dumitrescu β F. Smarandache
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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
βincluded setβ belongs to the βset that includesβ.
It is obvious that this entire reasoning is mental, the solution
starts with:
a) Let π₯ β π΄ β (π΅ β πΆ), irrespective [to continue we will read the
final operation (the difference)], <===> π₯ β π΄ and π₯ β π΅ β© πΆ [we will
read the operation that became the last (the union)] <===>
π₯ β π΄ and {π₯ β π΅
πππ₯ β πΆ
<===> {π₯ β π΄ πππ π₯ β π΅
πππ₯ β π΄ πππ π₯ β πΆ
(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:
π₯ β (π΄ β π΅) βͺ (π΄ β πΆ) <===> π₯ β π΄ β π΅ ππ π₯ β π΄ β πΆ.
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We observe from (1.3) that this conclusion follows
immediately, so the inclusion is proven.
For the second inclusion: Let π₯ β (π΄ β π΅) βͺ (π΄ β πΆ), [we
will read the final operation (the reunion)] <===>
{π₯ β π΄ β π΅
πππ₯ β π΄ β πΆ
<===> [we will read the operation that became the last]
<===>
{π₯ β π΄ πππ π₯ β π΅
πππ₯ β π΄ πππ π₯ β πΆ
<===> π₯ β π΄ and {π₯ β π΅
πππ₯ β πΆ
(1.4)
We have no more operations to explain, so we will describe
in detail the conclusion we want to reach: π₯ β π΄ β (π΅ β© πΆ), i.e. π₯ β
π΄ and π₯ β π΅ β© πΆ, 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 π₯ β π΄ β (π΄ β π΅) , irrespective [we will read the final
operation (the difference)], <===> π₯ β π΄ and π₯ β π΄ β π΅ [we will
read the operation that became the last] <===>
π₯ β π΄ and {π₯ β π΄
πππ₯ β π΅.
(1.5)
We have no more operations to explain, so we update the
conclusion:
π₯ β π΄ β© π΅ π. π. π₯ β π΄ πππ π₯ β π΅.
We observe that (1.5) is equivalent to:
{π₯ β π΄ πππ π₯ β π΄
πππ₯ β π΄ πππ π₯ β π΅.
The first affirmation is false, and the second one proves
thatπ₯ β π΄ β© π΅.
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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 ππ΄β(π΅β©πΆ) = π(π΄βπ΅)βͺ(π΄βπΆ);
(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): ππ΄β(π΅β©πΆ)(π₯) = π(π΄βπ΅)βͺ(π΄βπΆ)(π₯) for any π₯ β π.
The answer to the corresponding (Q) question is
(R2): We explain both members of the previous equality.
ππ΄β(π΅β©πΆ)(π₯) = [we will read the final operation (the
difference) and we will apply property π3 ] = ππ΄(π₯) β
ππ΄β©(π΅β©πΆ)(π₯) = [we will read the operation that became the last and
we will apply propertyπ1] =
ππ΄(π₯) β ππ΄(π₯). ππ΅(π₯). ππΆ(π₯) (1.6)
The second member of the equality becomes successive:
5. In order to prove the equality ππ΄β(π΄βπ΅)(π₯) β ππ΄β©π΅(π₯)
we notice that:
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II. Prove that:
SOLUTIONS.
Method (A) (the double inclusion)
1. Let π₯ β (π΄ β© π΅) Γ (πΆ β© π·) , irrespective [we explain the
final operation (the scalar product)] <===> π₯ = (πΌ, π½), with πΌ β π΄ β©
π΅ and π½ β πΆ β© π· <===>[we explain the operations that became the last
ones] <===> π₯ = (πΌ, π½) with πΌ β π΄ , πΌ β π΅ and π½ β πΆ , π½ β π·
[we have no more operations to explain so we update the conclusion: π₯ β (π΄ Γ
πΆ) β© (π΅ Γ π·), so π₯ belongs to an intersection (the last operation)] <===
> π₯ = (πΌ, π½) with πΌ β π΄ , π½ β πΆ and πΌ β π΅ , π½ β π· <===> π₯ β
π΄ Γ πΆ and π₯ β π΅ Γ π·.
3. Let π₯ β (π΄ β π΅) Γ πΆ , irrespective <===> π₯ β (πΌ, π½)
with πΌ β π΄ β π΅ and π½ β πΆ <===> π₯ β (πΌ, π½) with πΌ β π΄ and
πΌ β π΅ and π½ β πΆ [by detailing the conclusion we deduce the following steps]
<===> π₯ = (πΌ, π½), πΌ β π΄, π½ β πΆ and π₯ = (πΌ, π½), πΌ β π΅ , π½ β πΆ
===> π₯ β π΄ Γ πΆ and π₯ β π΅ Γ πΆ.
Reciprocally, let π₯ β (π΄ Γ πΆ) β (π΅ Γ πΆ) , irrespective
<===> π₯ = (πΌ, π½) with (πΌ, π½) β π΄ Γ πΆ and Γ πΆ <===> π₯ =
(πΌ, π½), πΌ β π΄ and π½ β πΆ and
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{πΌ β π΅
πππ½ β πΆ
<===> {πΌ β π΄ πππ π½ β πΆ πππ πΌ β π΅
πππΌ β π΄ πππ π½ β πΆ πππ π½ β πΆ (ππππ )
===> (πΌ, π½) β (π΄ β π΅) Γ πΆ.
Method (B) (using the characteristic function)
Explaining π(π΄βπ΅)βπΆ(π₯), in an analogue manner, we obtain
the desired equality. If we observe that the explanation
of π(π΄βπ΅)βπΆ(π₯) from above is symmetrical, by utilizing the
commutativity of addition and multiplication, the required equality
follows easily.
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Equalities 4)-5) were easy to prove using the characteristic
function; they are, however, more difficult to prove using the first
method. The characteristic function is usually preferred, due to the
ease of use and the rapidity of reaching the result.
III. Prove the following equivalences:
SOLUTIONS:
1. r1: We have to prove an equivalence;
R1: We prove two implications;
r2: π΄ βͺ π΅ β πΆ => π΄ β πΆ πππ π΅ β πΆ
R2: We prove that π΄ β πΆ πππ π΅ β πΆ (two inclusions).
Let π₯ β π΄ irrespective [we have no more operations to
explain, therefore we will update the conclusion: π₯ β πΆ; but this is
not evident from π₯ β π΄, but only with the help of the hypothesis]
π₯ β π΄ => π₯ β π΄ βͺ π΅ =
> [π‘βπππ’πβ π‘βπ βπ¦πππ‘βππ‘βππ π΄ βͺ π΅ β πΆ] =
> π₯ β πΆ
Let π₯ β π΅ irrespective => π₯ β π΄ βͺ π΅ β πΆ => π₯ β πΆ.
Reciprocally (the second implication):
C. Dumitrescu β F. Smarandache
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r1: We have to prove that π΄ βͺ π΅ β πΆ (an inclusion)
R1: Let π₯ β π΄ βͺ π΅ irrespective (we explain the last
operation):
=> {π₯ β π΄ => π₯ β πΆ (ππ’π π‘π π‘βπ βπ¦πππ‘βππ ππ )
πππ₯ β π΅ => π₯ β πΆ(ππ’π π‘π π‘βπ βπ¦πππ‘βππ ππ )
5. r1: We have to prove an equivalence;
R1: We prove two implications;
r2: (π΄ β π΅) βͺ π΅ = π΄ => π΅ β π΄;
R2: We prove the conclusion (an inclusion), using the
hypothesis.
Let π₯ β π΅ irrespective => π₯ β (π΄ β π΅) βͺ π΅ => π₯ β π΄
r3: We have to prove the implication: π΅ β π΄ => (π΄ β π΅) βͺ
π΅ = π΄;
R3: We prove an equality of sets (two inclusions).
For the first inclusion, let π₯ β (π΄ β π΅) βͺ π΅ irrespective [we
explain the last operation] =>
=> {π₯ β π΄ β π΅
πππ₯ β π΅
=> {π₯ β π΄ πππ π₯ β π΅ (π)
πππ₯ β π΅ (π)
[We have no more operation to explain, so we update the
conclusion]. We thus observe that from (a) it follows that π₯ β π΄,
and from (b), with the help of the hypothesis, π΅ β π΄ , we also
obtain π₯ β π΄.
Method 2:
We have to prove two inequalities among functions.
Through the hypothesis, ππ΄βͺπ΅ < ππΆ and, moreover ππ΄ < ππ΄βͺπ΅
(indeed, ππ΄ < ππ΄βͺπ΅ <=> ππ΄ < ππ΄ + ππ΅ β ππ΄ππ΅ <=> ππ΅(1 β
ππ΄) > 0 β true, because the characteristic functions take two
values: 0 and 1).
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It follows that ππ΄ < ππ΄βͺπ΅ < ππΆ and, analogously ππ΅ <
ππ΄βͺπ΅ < ππΆ .
Reciprocally, for the inverse implication, we have to prove
that: ππ΄βͺπ΅ < ππΆ , i.e.
ππ΄(π₯) + ππ΅(π₯) β ππ΄β©π΅(π₯) < ππΆ(x) (1.7)
in the hypothesis:
ππ΄ < ππΆ and ππ΅ < ππΆ (1.8)
But ππ΄, ππ΅, ππΆ 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 β F. Smarandache
26
Consequently:
{ππ΅(π₯) = 0 => ππ΅(π₯) < ππ΄(π₯) (ππ’π π‘π π‘βπ π£πππ’ππ 0 πππ 1 π‘βππ‘ π‘βπ ππ’πππ‘πππ π π‘ππππ )
ππ1 β ππ΄(π₯) = 0 => ππ΄(π₯) = 1 π π ππ΅(π₯) < ππ΄(π₯)
Reciprocally, if ππ΅(π₯) < ππ΄(π₯), we cannot have ππ΅(π₯) =
1 and ππ΄(π₯) = 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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27
We have to analyze the following cases:
a) if π₯ β π΄ and π₯ β π΅, we have:
ππΆ(π΄β©π΅)(π₯) = 1 and ππ΄β©πΆπ΅(π₯) = 1 , so for (1.10) to be
accomplished we must have ππ₯(π₯) = 0. So, any point that doesnβt
belong to π΄ or π΅, doesnβt belong to π either.
b) if π₯ β π΄ and π₯ β π΅, we have:
ππΆ(π΄β©π΅)(π₯) = 1 and ππ΄β©πΆπ΅(π₯) = 1 , so for (1.10) to be
accomplished we must have ππ₯(π₯) = 1. Therefore, any point from
π΄ β π΅ is in π.
c) if π₯ β π΄ and π₯ β π΅ and for (1.10) to be accomplished, we
must have ππ₯(π₯) = 0, consequently no point from π΅ β π΄ is not in
π.
d) if π₯ β π΄ and π₯ β π΅ we can have ππ₯(π₯) = 0 or ππ₯(π₯) =
1.
So any point from π΄ β© π΅ can or cannot be in π. Therefore
π = (π΄\π΅) βͺ π·, where π· β π΄ β© π΅, arbitrarily.
We have to analyze four cases, as can be seen from the
following table:
C. Dumitrescu β F. Smarandache
28
So π = (π΄ β π΅) βͺ πΆ.
V. Determine the following sets:
7. Let π· β β€(π) a π degree polynomial and π β β€ .
Knowing that π·(π) = ππ, 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 2π + 1 <=> π β {0, 1, β2}
so πΈ β β€ <=> 27 is dividable by 2π₯ + 1 and 4π₯2 β 2π₯ β 11 +27
2π₯+1 is a multiple of 8 <=> π΄ β {β14, β5, β2, β1, 0,1, 4, 13}.
so π₯ and π¦ are integer solutions of systems the shape of:
{3π¦ β π₯ β 1 = π’3π¦ + π₯ + 1 = π£
with π’ and π£ as divisors of 32.
So π(π₯)
π₯βπβ β€ <=> 15 is dividable by π₯ β π.
8. Method 1: For π β β1 we have
so, we must have π₯2 + 6π₯ β 3 > 0 . As π₯1,2 = β3 Β± 2β3 , we
deduce:
Method 2: We consider the function π(π) =π2βπ+1
π+1 and π΄ =
π(π·), with π· = β{β1} the domain of π. π΄ is obtained from the
variation table of π.
VI. Determine the following sets, when π, π β β:
C. Dumitrescu β F. Smarandache
30
where [π‘] is the integer part of π‘.
SOLUTIONS: We use the inequalities: [π‘] < π‘ < [π‘] + 1, so:
VII. The application π: π«(π) βΆ β±π defined through
π(π΄) = ππ΄ is a bijection from the family of the parts belonging to
π to the family of the characteristic functions defined on π.
SOLUTION: To demonstrate the injectivity, let π΄, π΅ β π«(π)
with π΄ β π΅ . From π΄ β π΅ we deduce that there exists π₯0 β π so
that:
a) π₯0 β π΄ and π₯0 β π΅ or b) π₯0 β π΅ and π₯0 β π΄
In the first case ππ΄(π₯0) = 1 , case ππ΅(π₯0) = 0 , so ππ΄ β
ππ΅. In the second case, also, ππ΄ β ππ΅.
The surjectivity comes back to:
Let then π β β±π irrespective. Then for π΄ = {π₯|π(π₯) = 1}
we have π = ππ΄.
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II. Functions
Function definition A function is determined by three elements π·, πΈ and π, with
the following significations: π· and πΈ are sets, named the domain
and the codomain, respectively, of the function, and π is a
correspondence law from π· to πΈ that causes:
each element π β π« to have one,
and only one corresponding element π β π¬. (F)
So we can say a function is a triplet (π·, πΈ, π), the elements of
this triplet having the signification stated above.
We usually note this triplet with π: π· β πΈ. 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 π 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
(πΉ) that characterizes a function.
We can consider this condition as being made up of two sub
conditions, namely:
(π1) each element π₯ from the domain has a corresponding element, in the
sense of βat least one elementβ, in the codomain.
Using a diagram as the one drawn below, the proposition can
be stated like this:
C. Dumitrescu β F. Smarandache
32
β(At least) one arrow can be drawn from each point of
the domain.β
More rigorously, this condition is expressed by:
The second sub condition regarding π in the definition of a
function is:
(π2) the element from the codomain that corresponds to π₯ 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:
βIf two arrows have the same starting point then they
also have the same arrival point.β
Namely,
.
Conditions (π1) and (π2) are necessary and sufficient for any
correspondence law π to be a function. These conditions are easy to
use in practice.
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In Figure 2.1, the correspondence law from a) satisfies (π1)
and doesnβt satisfy (π2); in b) the correspondence law doesnβt satisfy
(π1) , but satisfies (π2) . In diagram c), neither(π1) , nor (π2) are
satisfied, and in d) the correspondence law satisfies both (π1) and
(π2), so it is the (only) function.
A function is therefore a triplet ( π·, πΈ, π ), in which the
correspondence law π satisfies (π1) and (π2).
Letβs observe that these two condition only refer to the
domain of the function:
(π1) β from each point of the domain stems at least one arrow;
(π2) β 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 (π₯, π(π₯)), when π₯ covers the domain of the
function.
HOW CAN WE RECOGNIZE ON A GRAPHIC IF A
CORRESPONDANCE LAW IS A FUNCTION [if it satisfies (π1)
and (π2)] ?
In order to answer this question we will firstly remember (for
the case π·, πΈ β π ) the answer to two other questions:
1. Given π₯, how do we obtain β using the graphic β π(π₯)
[namely, the image of π₯ (or images, if more than one, and in this case
the correspondence law π is, of course, not a function)] ?
Answer: We trace a parallel from π₯ to 0π¦ to the point it
touches the graphic, and from the intersection point (points) we
then trace a parallel (parallels) to 0π₯ . The intersection points of
these parallels with 0π¦ are the images π(π₯) of π₯.
2. Reciprocally, given π¦, to obtain the point (points) π₯ having
the property π(π₯) = π¦, we trace a parallel to 0π₯ through π¦ , and
C. Dumitrescu β F. Smarandache
34
from the point (points) of intersection with the graphic, we then
trace a parallel (parallels) to 0π¦.
Examples 1. A circle with the center at the origin and with the radius π
is not the graphic of a function π: β βΆ β, because it does not
satisfy condition (π1) (there are points in the domain that do not have any
image, namely, all the points through which the parallel 0π¦ doesnβt
intersect the graphic) or (π2), because there are points in the domain that
have more than one image (all the points π₯ β (βπ, π) have two images).
2. A circle with the center at the origin and with the radius π
is not the graphic of a function π: [βπ, π] βΆ β, because it does not
satisfy condition (π2)(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 π
is not the graphic of a function π: β βΆ [0, β), because it does not
satisfy condition (π1). With the codomain [0, β), the points have
one image at most. There are points, however that donβt have an image.
Methods of Solving Calculus Problems for UNM-Gallup Students
35
4. A circle with the center at the origin and with the radius π
is the graphic of a function π: [βπ, π] βΆ [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 π
is the graphic of a function π: [βπ, π] βΆ [0,1].
In all these examples, the correspondence law has remained
unchanged (a circle with the center at the origin and with the radius
π, having therefore the equation π₯2 + π¦2 = π2 , which yields π¦ =
Β±(π2 β π₯21/2).
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 (π1) and (π2), THUS:
a) A graphic satisfies the condition (π1) if and only if any parallel to
0π¦ traced through the points of the domain touches the graphic in at least one
point.
b) A graphic satisfies the condition (π2) if and only if any parallel to
0π¦ traced through the points of the domain touches the graphic in one point at
most.
The inverse of a function We donβt always obtain a function by inverting the
correspondence law (inverting the arrow direction) for a randomly
given function π: π· βΆ πΈ. Hence, in the Figure 2.3, π is a function,
but πβ1 (obtained by inverting the correspondence law π) is not a
function, because it does not satisfy (π2) (there are points that have
more than an image).
C. Dumitrescu β F. Smarandache
36
With the help of this diagram, we observe that the inverse
doesnβt satisfy (π2) every time there are points in the codomain of π
that are the image of at least two points from the domain of π.
In other words, πβ1 doesnβt satisfy (π2) every time there are
different points that have the same image through π.
Therefore, as by inverting the correspondence law, the
condition π still be satisfied, it is necessary and sufficient for
the different points through the direct function have different
images, namely
A second situation where πβ1 is not a function is when it
does not satisfy condition (π1) :
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
(π1) still be satisfied, it is necessary and sufficient for all the points in
the codomain be consummated through the direct function, namely
Methods of Solving Calculus Problems for UNM-Gallup Students
37
In conclusion,
πβ1 satisfies (π2) if and only if π satisfies (π3)
πβ1 satisfies (π1) if and only if π satisfies (π4)
πβ1 is a function if and only if π satisfies (π3) + (π4).
As it is known, a function that satisfies (π3 ) is called an
injective function, a function that satisfies (π4 ) is called a
surjective function, and a function that satisfies (π3) + (π4 ) is
called a bijective function.
We see then that the affirmation: βA function has an inverse
if and only if it is bijectiveβ, has the meaning that the inverse πβ1
(that always exists, as a correspondence law, even if π is not
bijective) is a function if and only if π 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 0π₯ traced through the points of the codomain touches the graphic in
one point, at most [namely πβ1 satisfies (π2)].
d) a graphic is the graphic of a surjective function if and only if any
parallel to 0π₯ traced through the points of the codomain touches the graphic in
at least one point [namely πβ1 satisfies (π1)].
Examples 1. A circle with the center at the origin and with the radius π
is the graphic of a function π: [0, π] βΆ [0, β) that is injective but
not surjective.
2. A circle with the center at the origin and with the radius π
is the graphic of a function π: [βπ, π] βΆ [0, π) that is surjective
but not injective.
3. A circle with the center at the origin and with the radius π
is the graphic of a bijective function π: [0, π] βΆ [0, π).
C. Dumitrescu β 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 π , 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 πβ1 is obtained by inverting the correspondence law πΉ, 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
πβ1(π¦) = ππππΌπ¦. The relation (2.2) defines the logarithm:
The logarithm of a number π¦ in a given base, π, is the exponent π₯ to
which the base has to be raised to obtain π¦.
The Graphic of the Inverse Function If π· and πΈ are subsets of π and the domain π· of π (so the
codomain of πβ1) is represented on the axis 0π₯, and on the axis
0π¦, the codomain πΈ (so the domain of πβ1), then the graphic of π
and πβ1 coincides, because πβ1 only inverses the correspondence
law (it inverses the direction of the arrows).
But if we represent for πβ1 the domain, horizontally, and the
codomain vertically, so we represent πΈ on 0π₯ and π· on 0π¦, then
any random point on the initial graphic πΊπ (that, without the
Methods of Solving Calculus Problems for UNM-Gallup Students
39
aforementioned convention is the graphic for both π and πβ1 ),
such a point (π₯, π(π₯)) becomes (π(π₯), π₯).
The points (π₯, π(π₯)) and (π(π₯), π₯) are symmetrical in
relation to the first bisector, so we obtain another graphic πΊππ
besides πΊπ if the domain of πβπ is on ππ.
Agreeing to represent the domains of all the functions on
0π₯ it follows that πΊππ is a graphic of πβ1.
With this convention, the graphics of π and πβ1 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 (π3);
- to study the surjectivity, we verify if the condition
(π4) 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 β F. Smarandache
40
Examples
1. The function π: β βΆ β, π(π₯) = {2π₯ + 1 ππ π₯ β€ 1π₯ + 3 ππ π₯ > 1
is
injective, but is not surjective. The graphic is represented in Figure
2.5 a).
2. The function π: β βΆ β , π(π₯) = {3π₯ β 1 ππ π₯ β€ 22π₯ β 1 ππ π₯ > 2
is
surjective, but is not injective. The graphic is in Figure 2.5 b).
3. The function π: β βΆ β, π(π₯) = {2π₯ + 1 ππ π₯ β€ 1π₯ + 2 ππ π₯ > 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 Letβs show that π(π₯) = π‘π π₯, π: (β
π
2,
π
2) βΆ β is bijective.
(i) injectivity: because π(π₯) =1
πππ 2π₯ 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 Darbouxβs property and
from where it follows that it is surjective.
4. Using the proposition 1The function , π: π· βΆ πΈ is bijective if and only if
β© π¦ β πΈ the equation π(π₯) = π¦ has an unique solution
(see the Algebra workbook grade XII)
Example Let π· = β Γ β and the matrix π΄ = (3
412). Then the function
is bijective (Algebra, grade XII).
Letβs observe that in the proposition used at this point, the
affirmation βthe equation π(π₯) = π¦ has a solutionβ ensures the
surjectivity of the function and the affirmation βthe solution is
uniqueβ ensures the injectivity.
5. Using the proposition 2If π, π: π· βΆ π· and π. π = 1π· then π is injective and π is
surjective (Algebra, grade XII).
Example Let π· = β€ Γ β€ and π΄ = ( 2
β13
β2) . Then the function
ππ΄: π· βΆ π· defined through
C. Dumitrescu β F. Smarandache
42
satisfies ππ΄ β ππ΄ = 1π·, so it is bijective ( algebra, grade XII).
Letβs observe that the used proposition can be generalized, at
this point, thus:
βLet π: π· βΆ πΈ, π: πΈ βΆ πΉ, so that π β π is bijective. Then π
is injective and π is surjective.β
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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43
SOLUTIONS.
It is known that:
In order to more easily explain the conditions from the
inequalities over π’, π£ πππ π€ we will proceed as follows:
1. we draw the table with the signs of the functions π’ β
π£, π’ β π€ and π£ β π€
2. using the table we can easily explain the inequalities,
because for example π’(π₯) β€ π£(π₯) <===> π’(π₯) β€ 0.
1. For π’(π₯) = π₯ + 1 , π£(π₯) = π₯2 + 2 and π€(π₯) = 3π₯ we
have the following table:
So:
C. Dumitrescu β F. Smarandache
44
Therefore,
5. To explain π we proceed as follows:
(1) we draw the variation table of the function π¦(π‘) = π‘2
(2) considering π₯ in the first monotonic interval (deduced
from the table), at the right of β1 (because we are only interested in
the values π‘ β₯ β1 ), we calculate the minimum of π¦(π‘) on the
interval π‘ β [β1, π₯] etc.
a) for the first monotonic interval, π₯ β (β1,0), the function
π¦(π‘) = π‘2 has a single minimum point on the interval [β1, π₯] ,
situated in π‘ = π₯; its values is π¦(π₯) = π₯2. Being a single minimum
point it is also the global minimum (absolute) of π¦(π‘) on the
interval [β1, π₯], so, the value of π is π(π₯) = π₯2 for π₯ β (β1,0].
b) for the second monotonic interval, π₯ β (0, β) , the
function π¦(π‘) = π‘2 has a single minimum point on the interval
[β1, π₯], in π‘ = 0; its value is π¦(0) = 0. Being a single minimum
point, we have π(π₯) = 0 for π₯ β (0, β), so
Letβs observe that using the same variation table we can
explain the function π(π₯) = πππ₯1β€1β€π₯
π‘2. Hence:
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45
a) for π₯ β (β1,0] , the function π¦(π‘) = π‘2 , has only one
maximum, namely [β1, π₯] in π‘ = β1; its value is π = π¦(1) = 1.
So π(π₯) = 1, for π₯ β (1,0].
b) for π₯ β (0, +β), π¦(π‘) = π‘2 has two maximum points, in
π‘1 = β1 and π‘2 = π₯ ; their values are π1 = π¦(1) = 1 and π2 =
π¦(π₯) = π₯2. The values of π is the global maximum, so:
π(π₯) = max (1, π₯) for π₯ β (0, +β).
It follows that π(π₯) = {1 ππ π₯ β [β1,0]
max(1, π₯2) ππ π₯ > 0
9. From the variation table of π¦(π‘) =π‘4
(π‘+7)3
1. for π₯ β (β2, π] , π¦(π‘) has a single minimum on the
interval [β2, π₯], in π‘ = π₯; its value is π = π¦(π₯) =π₯4
(π₯+7)3
2. for π₯ > 0, π¦(π‘) has a single minimum, namely [β2, π₯], in
π‘ = 0; its value is π = π¦(0) = 0. So
C. Dumitrescu β F. Smarandache
46
Analogously, for
10. From the variation table of π¦(π‘) = π‘2 in π‘
we deduce:
(a) if π₯ β (0,1
βπ] , the function π¦(π‘) = π‘2 in π‘ has, for π‘ β
(0, π₯], a single supremum π =πππ
π‘ β π π‘ > 0
π¦(π‘) = 0 .
(b) π₯ β (β1
βπ, β], the function π¦(π‘) has two supreme values:
π1 = 0 and π1 = π¦(π₯) = π₯21π π₯.
So:
II. Study the bijectivity of:
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47
2. π(π₯) = π(π₯), π being an uneven degree polynomial.
III. Study the irreversibility of the hyperbolic functions:
and show their inverses.
IV. 1. Show that the functions π(π₯) = π₯2 β π₯ + 1 ,
π: [1
2, β) βΆ β and
C. Dumitrescu β F. Smarandache
48
are inverse one to the other.
2. Show that π(π₯) =1βπ₯
1+π₯ coincides with its inverse.
3. Determine the parameters π, π, π, π so that π(π₯) =ππ₯+π
ππ₯+π
coincides with its inverse.
4. Show that the function π: (0,1] βΆ β defined
throughπ(π₯) =4
3π β π₯, if π₯ β [1
3π ,1
3πβ1] is bijective.
5. On what subinterval is the function π: [β1
2, β) βΆ β
π(π₯) = βπ₯ β β2π₯ β 1 bijective?
INDICATIONS.
4. To simplify the reasoning, sketch the graphic of function
π.
5. Use the superposed radical decomposition formula:
where πΆ = βπ΄2 β π΅.
Monotony and boundaries for sequences and functions
A sequence is a function defined on the natural numbersβ set.
The domain of such a function is, therefore, the set of natural
numbers β . The codomain and the correspondence law can
randomly vary.
Hereinafter, we will consider just natural numbersβ
sequences, i.e. sequences with the codomain represented by the set
of real numbers, β.
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π: β βΆ β REAL NUMBERS SEQUENCE
For this kind of function, with the same domain and
codomain each time, only the correspondence law π should be
added, which comes down to specifying the set of its values:
π(0), π(1), π(2), β¦ , π(π), β¦
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: π0, π1, β¦ , ππ, β¦. This set is abbreviated by
(ππ)πββ.
Consequently, a sequence is a particular case of function.
The transition from a function to a sequence is made in this
manner:
(π 1) by replacing the domain π·with β
(π 2) by replacing the codomain πΈ with β
(π 3) by replacing the variable π₯ with π (or π, or π, etc)
(π 4) by replacing π(π₯) with ππ
C. Dumitrescu β F. Smarandache
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DOMAIN CODOMAIN VARIABLE CORRESP. LAW
Function D E x π(π₯) Sequence β β n ππ
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 (ππ)πββ ,
obtained by replacing π with π₯ in the expression of ππ ( ππ
becomes π(π₯)).
Example We can attach the function π(π₯) =
π₯+3
2π₯+1 to the sequence
ππ =π+3
2π+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 ( π 1) β
(π 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 π: π· βΆ β, π· ββ is monotonically increasing if:
β© π₯1. π₯2 β π·, π₯1 β€ π₯2 β π(π₯1)β€ π(π₯2)
a) the sequence π: β βΆ β ismonotonically increasing if:
β© π1, π2 β β, π1 β€ π2 βππ1
β€ ππ2 (1)
If condition (1) is fulfilled,
taking, particularly π1 = π and
π2 = π + 1, we deduce
ππ β€ ππ+1 for any π (2)
In conclusion (1) βΆ (2) The reciprocal implication also stands true. Its demonstration can be deduced from the following example:
If (2) is fulfilled and we take
π1 = 7, π2 = 11 , we have
π1 β€ π2 and step by step
and π7 β€ π11.
Consequently, (1) β· (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 ( π 1) β (π 4) , that define the transition from function to sequence.
C. Dumitrescu β F. Smarandache
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b) the function π: π· βΆ β, π· ββ is monotonically decreasing if:
β© π₯1, π₯2 β π·, π₯1 β€ π₯2 β π(π₯1)β₯ π(π₯2)
b) the sequence π: β βΆ β is monotonically decreasing if:
β© π1, π2 β β, π1 β€ π2 β ππ1
β₯ ππ2
It can be shown that this condition is equivalent to:
ππ β₯ ππ+1 for any π β β (4)
LIMITED FUNCTIONS LIMITED SEQUENCES
a) π: π· βΆ β , π· β β is of inferiorly bounded if it doesnβt
have values towards ββ , i.e.
βπ β β,β© π₯ β π·, π(π₯) β₯ π
b) π: π· βΆ β , π· β β is of superiorly bounded if it doesnβt
have values towards +β
i.e. βπ β β,β© π₯ β π·, π(π₯) β€ π
c) π: π· βΆ β , π· β β is bounded, if it is inferiorly and superiorly bounded, i.e.
βπ, π β β,β© π₯ β β π β€ π(π₯) β€π. In other words, there exists an
interval [π, π] 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 [π, π] becomes [βπ, π] and the limiting condition is:
β π > 0, ,β© π₯ β π· βπ β€π(π₯) β€ π i.e.
β π > 0, ,β© π₯ β π· |π(π₯)| β€ π
a) the sequence π: β βΆ β is of inferiorly bounded if it doesnβt have values towards
ββ , i.e.
βπ β β,β© π β β, ππ β₯ π
b) the sequence π: β βΆ β is of superiorly bounded if it doesnβt have values towards
+β , i.e.
βπ β β,β© π β β, ππ β₯ π
c) the sequence π: β βΆ β is bounded if it is inferiorly and superiorly bounded, i.e
βπ, π β β,β© π β β π β€ ππ β€π OR
β π > 0,β© π β β |ππ| β€ π
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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 π₯1 β€ π₯2 and we
compare the difference π(π₯1) =π(π₯2) to zero. This can be done through successive minorings and majorings or by applying Lagrangeβs theorem to function
π on the interval [π₯1, π₯2].
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 Lagrangeβs theoremIt allows the replacement of the
difference π(π₯2) β π(π₯1) with
π(π) that is then compared to zero.
1. Using the definition:We compare the difference
ππ+1 β ππ to zero, and for sequences with positive terms we can compare the quotient
ππ+1/ππ to one. We can make successive minorings and majorings or by applying Lagrangeβs 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, β), (see Method 10 point c) 3. Using Lagrangeβs 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 β 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 π(π₯) = πππΓln (ln π₯) and we have the
variation table:
so the function is decreasing on the interval (1, π1/π) and increasing
on (π1/π, β) . It is inferiorly bounded, its minimum being π =
πβ1/π and it is not superiorly bounded.
5. We use the superposed radicalsβ decomposition formula:
6. Let π₯1 < π₯2. In order to obtain the sign of the difference
π(π₯1) β π(π₯2) we can apply Lagrangeβs theorem to the given
function, on the interval [π₯1, π₯2] (the conditions of the theorem are
fulfilled): π β β exists, so that:
so the function is increasing.
II. Study the monotony and the bounding of the sequences:
C. Dumitrescu β F. Smarandache
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also determine the smallest term of the sequence.
also determine the smallest term of the sequence.
πΎπ being the intermediary value that is obtained by applying
Lagrangeβs theorem to the function π(π₯) = ln π₯ on the intervals
[π, π + 1].
ANSWERS:
2. The function attached to the sequence is π(π₯) = π₯1/π₯ .
From the variation table:
we deduce, according to the sequence criterion (Heineβs 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 π β₯ 3 . Because the sequence is
decreasing if π β₯ 3, in order to discover the biggest term, we have
to compare π2 and π3 . We deduce that the biggest term of the
sequence is:
3. The function attached to the sequence is π(π₯) = π₯ππ₯ .
From the variation table for π β (0,1):
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using point c) from Method 10, we deduce that the given sequence
is increasing for π, smaller or equal to the integer part of (β ln π)β1
and is decreasing for π β₯ [(β ln π)β1] + 1.
Because
we deduce that the interval [1, [(β ln π)β1]], in which the sequence
is increasing, can be no matter how big or small.
6. The sequence of the general term ππ is increasing, the
sequence of the general term ππ is decreasing, and the sequence ππ
is increasing if π β [0, π β πΎπ] and is decreasing for π β [, π β
πΎπ, 1].
III. Using Lagrangeβs theorem.
1. Prove the inequality | sin π₯| β€ |π₯| for any real π₯ , then
show that the sequence π1 = sin π₯, π2 = sin sin π₯, β¦ , ππ =
sin sin β¦ sin π₯, is monotonic and bounded, irrespective of π₯, and
its limit is zero.
2. Study the monotony and the bounding of the sequences:
ANSWERS:
C. Dumitrescu β F. Smarandache
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From Lagrangeβs theorem applied to the function π(π) = π ππ π‘ on
the interval [π₯, 0] or [0, π₯], depending on π₯ being either negative or
positive, we obtain |sin π₯
π₯| = | cos π| β€ 1. The sequence satisfies the
recurrence relation: ππ+1 = sin ππ, so if sin π₯ > 0, we deduce that
the sequence is decreasing, and if sin π₯ < 0 the sequence is
increasing. It is evidently bounded by β1 and 1. Noting with πΏ its
limit and passing to the limit in the recurrence relation, we obtain
πΏ = sin πΏ, so πΏ = 0.
where π(π₯) = (1 +1
π₯)π₯ defines the function attached to the
sequence.
b) Applying Lagrangeβs theorem to the function attached to
the sequence, on the interval [π, π + 1], we deduceππ+1 β ππ =
π(ππ). We have to further determine the sign of the derivation:
namely, the sign of
From the table below we deduce π(π₯) > 0 for π₯ > 0, so ππ+1 β
ππ > 0.
IV. 1. Let I be a random interval and π: πΌ β πΌ a function.
Show that the sequence defined through the recurrence relation
ππ+1 = π(ππ), with ππ being given, is:
a) increasing, if π(π₯) β π₯ > 0 on πΌ
b) decreasing if π(π₯) β π₯ < 0 on πΌ.
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2. If the sequence (ππ)πββ is increasing, and the sequence
(ππ)πββ is decreasing and ππ β€ ππ for any π, then:
a) the two sequences are bounded, and so convergent.
b) if ππππββ
(ππ β ππ) = 0 then they have the same limit.
c) apply these results to the sequences given by the
recurrence formulas:
with ππ and ππ being given.
SOLUTIONS:
1. ππ+1 β ππ = π(ππ) β ππ > 0 in the first case.
2. the two sequences are bounded between π1 and π1, so the
sequences are convergent, because they are also monotonic. Let π1
and π2 represent their limits. From the hypothesis from (b) it
follows that π1 = π2.
(c) the sequences have positive terms and:
V. Study the bounding of the sequences:
C. Dumitrescu β F. Smarandache
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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 π₯ is replaced with π and π(π₯) is
replaced with ππ . 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 functionβs limit is based on the notion of
vicinity of a point.
Intuitively, the set π β β is in the vicinity of point π₯π β β,
if (1) π₯ β π, and, moreover, (2) π also contains points that are in
the vicinity of π₯π (at its left and right).
Obviously, if there is any open interval (π, π), so that π₯π β
(π, π) β π, then the two conditions are met.
Particularly, any open interval (π, π) that contains ππ
satisfies both conditions, so it represents a vicinity for ππ.
For finite points, we will consider as vicinities open intervals
with the center in those points, of the type:
C. Dumitrescu β F. Smarandache
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and sets that contain such intervals.
For (+β), πβ = (β β ν, β + ν) doesnβt make sense, so we
have to find a different form for the vicinities of +β. In order to
accomplish this, we observe that to the right of +β we cannot
consider any points, so β + ν has to be replaced with +β .
Furthermore, because β β ν = β, we replace β β ν with ν. So, we
will consider the vicinities of +β with the form:
πβ = (ν, β)
and sets that contain such intervals.
Analogously, the vicinities of ββ have the form:
πββ = (β β, ν)
The definition of a functionβs limit in a point expresses the
condition according to which when π₯ approaches π₯π , π(π₯) will
approach the value π of the limit.
With the help of vicinities, this condition is described by:
As we have shown, for the vicinities ππ and ππ₯π
from the
definition (βββ), there exist three essential forms, corresponding to
the finite points and to + β and β β respectively.
CONVENTION:
We use the letter ν to write the vicinities of π, and the letter πΏ
designates the vicinities of π₯π.
We thus have the following cases:
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and condition π₯ β and ππ₯π becomes π₯ > πΏ
and condition π₯π β and ππ₯π becomes π₯ < πΏ
All these cases are illustrated in Table 3.1.
Considering each corresponding situation to π₯π and π, in the
definition (βββ) , we have 9 total forms for the definition of a
function. we have the following 9 situations:
C. Dumitrescu β F. Smarandache
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C. Dumitrescu β F. Smarandache
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Exercises I. Using the definition, prove that:
ANSWERS:
In all cases, we have π₯π β πππππ‘π.
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 ν > 0, irrespective; we therefore search for
πΏ .
c) we make calculation in the expression |π(π₯) β π| ,
highlighting the module |π₯ β π₯π|.
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d) we increase (decrease) the obtained expression, keeping in
mind we are ony interested in those values of π₯ for which |π₯ β
π₯π| < πΏ .
So:
e) if the expression that depends on π₯ is bounded
(continuous) in a vicinity of π₯π, we can further increase, to eliminate
π₯. We have:
because in the interval [0,2], for example, that is a vicinity of π₯π =
1 , the function π(π₯) =1
|3π₯+2| is continuous, and therefore
bounded: π(π₯) β [1
5,
1
2], if π₯ β [0,2].
f) we determine πΏ , setting the condition that the expression
we have reached (and that doesnβt depend on π₯ , just on πΏ ) be
smaller than ν:
Any πΏ that fulfills this condition is satisfactory. We can, for
example, choose πΏ3
2ν or πΏ = ν etc.
In order to use the increase made to function π, we must
also have πΏ β€ 1, so, actually πΏ = min (1,3
2ν ) (for example).
It follows that:
C. Dumitrescu β F. Smarandache
68
b) let ν > 0; we search for πΏ
because in the interval [1,3], for example, the function:
is bounded by:
f) we determine πΏ from the condition: πΏπ
3< ν . We obtain
πΏ < 3ν, so we can, for example, take πΏ = ν. But in order for the
increase we have made to function π to be valid, we must also have
πΏ β€ 1, so, actually πΏ = min(1, ν).
It follows that:
so the condition from the definition of the limit is in this case, also,
fulfilled.
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b) let ν any (not necessarily positive, because π = +β , so
ππ = (ν, +β)only makes sense for any ν, not just for ν positive).
so we can take πΏ =1
2β
Then:
II. Using the definition, prove that:
ANSWERS:
C. Dumitrescu β F. Smarandache
70
b) let ν > 0 any; we search for πΏ .
c) because π₯π isnβt finite anymore, we cannot highlight |π₯ β
π₯π |. We will proceed differently: we consider the inequality
|π(π₯) β π| < ν as an inequation in the unknown π₯ (using also the
fact that π₯ β β , so, we can consider |2π₯ + 1| = 2π₯ + 1 ). We
have:
b) let ν > 0 any; we determine πΏ .
c) we express π₯ from the inequality π(π₯) > ν :
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If π₯1 and π₯2 are the roots of the attached second degree equation
and we assume π₯1 < π₯2, we have π(π₯) > ν βΊ π₯ > max (100, π₯2).
So we can take πΏ = max (100, π₯2).
The limits of sequences The definition of a sequenceβs limit is deduced from the
definition of a functionβs limit, making the mentioned
particularizations:
(1) π₯ is replaced with π,
(2) π(π₯) is replaced with ππ,
(3) π₯π is replaced with ππ.
Moreover, we have to observe that ππ = ββ and finite ππ
donβt make sense (for example π β 3 doesnβt make sense, because
π, being a natural number, cannot come no matter how close to 3).
So, from the nine forms of a functionβs limit, only 3 are
particularized: the ones corresponding to π₯π = +β.
We therefore have:
(π10) π β πππππ‘π (we transpose the definition (π2))
Because in the inequality π > πΏ , π is a natural number, we
can consider πΏ as a natural number also. To highlight this, we will
write π instead of πΏ . We thus have:
(π11) π = +β (we transpose the definition (π5))
C. Dumitrescu β F. Smarandache
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(π12) π = ββ (we transpose the definition (π8))
Examples I. Using the definition, prove that:
ANSWERS:
1. We adapt the corresponding stages for functionsβ limits, in
the case of π₯π = +β.
b) let ν > 0 any; we determine π β β.
c) we consider the inequality: |ππ β π| < ν as an inequation
with the unknown π:
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so, π = [2β
3] + 1
b) let ν > 0 any; we determine π β β as follows:
Let π1and π2 represent the solutions to this second degree
equation (π1 < π2). Then, for π1 > π2, the required inequality is
satisfied. So, π = [π2] + 1.
Calculation methods for the limits of functions and sequences
Letβs 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:
C. Dumitrescu β F. Smarandache
74
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: β
β, β β β ,
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 π₯πΌ, with
πΌ < 0, if π₯ β β.
obtain as many terms as possible with the form π₯πΌ, with
πΌ > 0, if π₯ β 0.
Examples I. Calculate:
C. Dumitrescu β F. Smarandache
76
ANSWERS:
2. Given the common factor 9π we obtain terms that have
the form π₯π, with sub unitary π₯.
expression that tends to 1
5 when π₯ tends to infinity.
II. Trace the graphics of the functions:
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8. For any rational function, non-null, π 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 π.
If the indetermination comes from an expression with the
form:
with π β uneven number, the conjugate is:
An example of application for this formula is exercise 5,
below.
C. Dumitrescu β F. Smarandache
78
Exercises
ANSWERS:
7. We replace, for example π0 = βπ1 β π2 β β― β ππ, in the
given sequence and we calculate π 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:
C. Dumitrescu β F. Smarandache
80
ANSWERS:
12. Noting with πΌ = arcsin π₯ β ππππ‘π π₯, we observe that πΌ
tends to zero when π₯ tends to zero, so we aim to obtain πΌ
sin πΌ. 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:
C. Dumitrescu β F. Smarandache
82
= (we permute again the limit with the logarithm) =
III. Calculate:
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ANSWERS:
and noting π₯ β π = πΌ, we obtain:
4. We consider the function:
obtained from ππ, by replacing π with π₯. We calculate:
According to the criterion with sequences (see Method 10,
point c), we also have: πππ
π β βππ = ln 2.
C. Dumitrescu β F. Smarandache
84
5. Something that is bounded multiplied with something that tends to zero, tends to zero
(A) If (ππ)πββ is bounded and πππ
π β βππ = 0, then:
ππππ β β
ππ. ππ = 0
(B) If π is bounded in a vicinity of π₯0 and πππ
π₯ β π₯0π(π₯) = 0,
then πππ
π₯ β π₯0π(π₯). π(π₯) = 0.
Examples
because (β1)π is bounded by β1 and 1 and 1
π tends to zero;
because 1 β cos π₯ is bounded by 0 and 2 and 1
π₯2 tends to zero;
where ππ is the nth decimal fraction of the number π;
where π½π is the approximation with π exact decimal fractions of the
number π.
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INDICATIONS:
4. (ππ)πββ is bounded by 0 and 9 , being composed of
numbers;
5. (π½π)πββ is bounded by 2 and 3;
6. The majoring and minoring method(A) If functions π and β exist, so that in a vicinity of π₯π we
have:
then:
Schematically:
C. Dumitrescu β F. Smarandache
86
(B) If the sequences (π)πββ and (ππ)πββ exist so that: ππ β€
ππ β€ ππ, starting from the rank π0 and πππ
π β βππ =
ππππ β β
ππ = π,
then: πππ
π β βππ = π.
Schematically:
OBSERVATION:
If π = +β , we can eliminate β ( (ππ)πββ πππ ππππ‘ππ£πππ¦) ,
and if π = ββ, we can eliminate π ((ππ)πββ πππ ππππ‘ππ£πππ¦).
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 (ππ)πββ and (ππ)πββ.
Examples 1. We put the smallest (biggest) term of the sequence
(ππ)πββ, instead of all the terms.
ANSWERS:
a) In the sum that expresses the sequence (ππ)πββ, there are:
the smallest term 1
βπ2+π and the biggest term
1
βπ2+1. By applying
Method 1, we have:
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2. We minorate (majorate) the terms that make up the
sequence ππ with the same quantity.
ANSWERS:
a) This time there isnβt a smallest (biggest) term in the sum
that makes up the sequence ππ . This is due to the fact that the
function π(π₯) = sin π₯ isnβt monotonic. Keeping into consideration
that we have: β1 β€ sin π₯ β€ 1 , we can majorate, replacing sin π₯
with 1, all over. We obtain:
Analogously, we can minorate, replacing sin π₯ with β1 ,
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 ππ.
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 donβt obtain the limit directly, instead we use the lateral
limits.
Example: For the calculation of the limit:
C. Dumitrescu β F. Smarandache
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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 π₯2+2π₯
7, keeping into
consideration that:
- on the left of β2 we have: π₯2+2π₯
7> 0, so:
- on the right of β2 we have: π₯2+2π₯
7< 0, so:
(c) Moving on to the limit in these inequalities(the majoring
and minoring method), we obtain the limit to the left:ππ (β2) =2
7
and the limit to the right: ππ(β2) =2
7, so π =
2
7.
(B) If the double inequality cannot be used (3.2), for
ππππ₯ β π
π(π₯) , with π β β€ , we calculate the lateral limits directly,
keeping into consideration that:
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Example: For πππ
π β β(β1[π₯])/(π₯ β 1) we cannot use (3.2)
because the expressions obtained for the calculation of the lateral
limits donβt have the same limit. That is why we use the fact that an
the left of 1 we have [π₯] = 0, and on the right of 1 we have [π₯] =
1, so:
It follows that π = ββ
(C) If π₯ β β, we can replace [π₯], keeping into consideration
that for π₯ β [π, π + 1) , we have: [π₯] = π (obviously, π₯ β β
implies π β β).
Example:
The variable π₯ that tends to infinity is certainly between two
consecutive numbers: π β€ π₯ β€ π + 1, so:
C. Dumitrescu β F. Smarandache
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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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π₯π being a point in which π is differentiable.
Example:
We notice that by noting π(π₯) = ππ₯ , we have π(π) = ππ
and the limit becomes:
This limit has the value π,(π) because π is a function
differentiable in π.
We have thus reduced the calculation of the given limit to
the calculation of the value of π,(π), calculation that can be made
by deriving π:
Observation: this method can be used like this: βlet π(π₯) =
ππ₯ . We have π,(π₯) = ππ₯ ln π . π,(π) = ππ ln π = π . We donβt
recommend this synthetic method for the elaboration of the
solution at exams because it can misguide the examiners. That is
why the elaboration: βWe observe that by noting π(π₯) = β―, we
have π(π) = β― and the limit becomes β¦ = π,(π) because π is
differentiable in π β¦βis 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 π with π₯, in the expression of ππ, because for
π₯ β β we canβt calculate the derivative, but by replacing π with 1
π₯
(which leads to the calculation of the derivative in zero).
Example: For πππ
π β βπ( β2
πβ 1) we consider the function
π(π₯) =2π₯β1
π₯obtained by replacing π with
1
π₯ in the expression of
C. Dumitrescu β F. Smarandache
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ππ . We calculate πππ
π β 0π(π₯). For this we observe that by noting
π(π₯) = π₯2 , we have π(π) = 1 and the limit
becomes: πππ
π β 0π(π₯)βπ(π)
π₯β0= π,(0) , because π is a differentiable
function in zero. We have: π,(π₯) = 2π₯ ln 2, so π,(0) = ln 2. Then ,
according to the criterion with sequences, we obtain πππ
π β βππ =
ln 2.
Exercises
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II. For π > 0 calculate:
ANSWER:
4. We divide the numerator and the denominator with π₯ β π.
III. Show that we cannot use the definition of the derivative
for the calculation of the limits:
C. Dumitrescu β F. Smarandache
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ANSWERS:
2. We consider the function π(π₯) =2π₯β1
3π₯β1, obtained by
replacing the sequence ππ on π with 1
π₯. We calculate
ππππ₯ β 0
π(π₯) .
To use the definition of the derivative, we divide the numerator and
the denominator with π₯, so:
We observe that by noting π(π₯) = 2π₯ , we have π(0) = 1
and the limit from the numerator becomes: πππ
π₯ β 0π(π₯)βπ(0)
π₯=
π,(π₯) because π is differentiable in zero.
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π,(π₯) = 2π₯ ln 2 , so, according to the criterion with
sequences π1 = π,(0) = ln 2 . Analogously, by noting β(π₯) = 3π₯ ,
the limit from the numerator becomes: πππ
π₯ β 0β(π₯)ββ(0)
π₯= β,(0) ,
because β is differentiable in zero.
β(π₯) = 3π₯ ln 3 , so, according to the criterion with
sequencesπ2 = β,(0) = ln 3, and π =ln 2
ln 3.
5. The limit contains an indetermination with the form 1β,
so we will first apply method 4.
We now consider the function π(π₯) =1
π₯(π1
π₯ + π2π₯ β 2)
obtained by replacing π with 1
π₯ in the expression above. We
calculate πππ
π₯ β 0π(π₯). For this, we observe that by noting π(π₯) =
π1π₯ + π2
π₯ we have π(0) = 2 and the limit can be written:
ππππ₯ β 0
π(π₯)βπ(0)
π₯= π,(0) because π is differentiable in zero.
C. Dumitrescu β F. Smarandache
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Now π,(π₯) = π1π₯ ln π1 + π2
π₯ ln π2, so π,(0) = ln π1 + ln π2
According to the criterion with sequences:
and the initial limit is βπ1. π2.
9. Using LβHospitalβs theorem The theorem. [lβHospital (1661-1704)] If the functions π and π
fulfill the conditions:
1. are continuous on [π, π] and differentiable on (π, π){π₯π}
2. π(π₯π) = π(π₯π) = 0
3. π, is not annulled in a vicinity of π₯π
4. there exists πππ
π₯ β π₯π
π,(π₯)
π,(π₯)= πΌ , finite or infinite,
then there exists πππ
π₯ β π₯π
π(π₯)
π(π₯)= πΌ.
This theorem can be used for:
(A) The calculation of limits such as πππ
π₯ β π₯π
π(π₯)
π(π₯) when these
contain an indetermination of the form 0
0 or
β
β.
(B) The calculation of limits such as πππ
π₯ β π₯ππ(π₯). π(π₯)
when these contain an indetermination of the form β. 0.
This indetermination can be brought to form (A) like this:
(indetermination with the form β
β)
(indetermination with the form 0
0)
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Observation: sometimes it is essential if we write the
indetermination 0
0 or
β
β.
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 β
β:
and we derive, we obtain π = 0.
(C) The calculation of limits such as πππ
π₯ β π₯π(π(π₯) β π(π₯)),
with indetermination β β β.
This indetermination can be brought to form (B) giving π(π₯)
or π(π₯) as forced common factor.
We have, for example:
C. Dumitrescu β F. Smarandache
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and because
we distinguish two cases:
(namely, π(π₯)
π(π₯) doesnβt tend to 1), we have:
the indetermination is of the form (B).
(D) The calculation of limits such as πππ
π₯ β π₯ππ(π₯)π(π₯), with
indeterminations 1β, 00, β0.
These indeterminations can be brought to the form (C),
using the formula:
that for π = π, for example, becomes π΄ = πln π΄, 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 π
with π₯ or by replacing π with 1
π₯ in the expression of ππ). 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 π with π₯ we calculate the limit in β,
and by replacing π with 1
π₯, we calculate the limit in 0.
Exercises I. Calculate:
(a polynomial increases faster to infinity than a logarithm)
(a polynomial increases slower to infinity than an exponential)
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II. Show that lβHospitalβs rule cannot be applied for:
ANSWER:
1. For π(π₯) = π₯2π ππ1
π₯ and π(π₯) = sin π₯ we have:
- doesnβt exist, so the condition 4 from the theorem isnβt
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 π₯0 = β , condition π) doesnβt make sense
anymore.
This criterion cannot be used to eliminate indeterminations
because the sequence (π₯π)πββ 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 donβt contain indeterminations.
Example: Using the criterion with sequences, show that:
Solution: (a) we have to show that:
β© (π₯π)πββ , with the properties:
(b) let (π₯π)πββ be any sequence with the
properties π), π), π)
(c) in πππ
π β βπ(π₯π) we use the properties of the
operations of limits of sequences to highlight πππ
π β βπ₯π (it is
possible because there are no indeterminations), then we use the
fact that πππ
π β βπ₯π = π₯0:
C. Dumitrescu β F. Smarandache
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(B) To show that a function doesnβt have a limit in a point.
To accomplish this we can proceed as follows:
1π΅ . we find two sequences: (π₯π)πββ and (π¦π)πββ with the
properties π), π), π) so that:
or:
2π΅. we find a single sequence with the properties π), π), π)
so that πππ
π β βπ(π₯π) doesnβt exist.
Example: πππ
π₯ β βsin π₯ doesnβt exist
1π΅. let π₯π = 2ππ and π¦π = 2ππ +π
2. We have:
So the limit doesnβt exist.
2π΅. letπ₯π =ππ
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 π(π₯) be the function attached to the sequence, by the
replacement of, for example, π with π₯ in the expression of ππ ( or
of π with 1
π₯, but then we calculate
ππππ₯ β 0
π(π₯))
(b) we calculate πππ
π₯ β βπ(π₯) = π (
ππππ β 0
π(π₯), respectively).
(c) according to the criterion with sequences lim π(π₯) = π
means: for any sequence (π₯π)πββ with the properties:
we have πππ
π β βπ(π₯π) = π
We observe that the sequence π₯π = π fulfills the conditions
π) and π) and, moreover, for this sequence we have:
Exercises I. Using the criterion with sequences, calculate:
II. Show that the following donβt exist:
C. Dumitrescu β F. Smarandache
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7. If πhas a limit to its left (right) in π₯0, then:
8. Ifπππ
π β βπ(π₯) = π , then
ππππ β β
π(π) = π , but not the
other way around.
9. πππ
π₯ β βπ(π₯) doesnβt exist if π: π· β β 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 1π΅ . The following sequences fulfill
conditions π) and π):
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and:
III. Determine the limit points of the functions: π, π, π β
π, π β π, for:
Answer: let π₯0 β π· = β randonm. We check if π has a limit
in π₯0. For this we observe it is essential if (π₯π)πββ is rational or
irrational. We firstly consider the situation where the sequence
(π₯π)πββ is made up only of rational points (or only of irrational
points).
Let (π₯π)πββ be a sequence of rational points with the
properties π), π), π). We have:
C. Dumitrescu β F. Smarandache
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So the function doesnβt have a limit in any point π₯0 for
which π₯02 β 2, i.e. π₯0 and Β±2. For π₯0 = β2 we consider a random
sequence (π₯π)πββ with the properties π), π), π). πΈ1decomposes in
two subsequences: (π₯π, )πββ made up of solely rational terms and
(π₯π, )πββ made up of irrational terms. Because:
we deduce that:
The same goes for π₯ = ββ2.
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: ππ =π!
ππ
(a) For the study of monotony and bounding we cannot
apply the method of the attached function (π(π₯) =π₯!
π₯π₯ doesnβt 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 isnβt monotonic, we canβt apply the
method so we wonβt 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
π1, being decreasing and with positive terms. Therefore πππ
π β βππ
exists.
(c) For the calculation of π 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 β F. Smarandache
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where (ππ)πββ 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 ππ+1 β₯ ππ , that we then transform through
equivalences:
So ππ+1 β₯ ππππ β (β1,2).
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We have thus reached the conclusion that the sequence is
increasing if and only if it is bounded between βπ and π. We
obviously have ππ > 0 > β1 so we still have to prove that ππ < 2.
This inequality can be proven by induction:
Verification:
So we have proven the monotony and the bounding.
Therefore π =πππ
π β βππ exists. To determine π we move on to the
limit in the recurrence relation (this time this type of relation is
given initially):
As π = β1 is impossible, because ππ > 0, we deduce π = 2.
6. We decompose the fraction2π+3
5π in a difference of
consecutive terms:
To determine the four parameters:
a) we use the identification method and
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b) we state the condition that the terms from the right
member in (3.4) be consecutive.
c) the denominators are consecutive, so we state the
condition that the numerators be in the same relation as
consecutives; the denominator of the first fraction is obtained from
the denominator of the second one, by replacing π with π β 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 β F. Smarandache
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Indications:
4. sequence with positive terms (induction) and decreasing.
Let π1 and π2 be the limits of the subsequences of even and
uneven value. Show that π1 = π2.
12. Using the Cesaro-Stolz and Rizzoli Lemmas Lemma: (Cesaro-Stolz) Let (πΌπ)πββ be a random sequence and
(π½π)πββ a strictly increasing sequence, having the limit
infinite. If πππ
π β βπΌπ+1βπΌπ
π½π+1βπ½π= π exists β finite or infinite,
then:
ππππ β β
πΌπ
π½π= π.
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 (π½π)πββ is strictly
monotonic.
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CONSEQUENCES:
Indeed,
Example:
because for ππ = π, we have:
(B) The limits of the arithmetic, geometric and harmonic
averages of the first π terms of a sequence having the limit π, have
the value π also:
Observation: a variant for the Cesaro-Stolz lemma for the case
when ππ, π½π β 0 has been proven by I.Rizzoli [Mathematic Gazette,
no 10-11-12, 1992, p. 281-284]:
C. Dumitrescu β F. Smarandache
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Lemma: (I.Rizzoli) If (πΌπ)πββ, (π½π)πββ are two sequences
of natural numbers that fulfill the conditions:
(ii) the sequence (π½π)πββ is strictly monotonic
(increasing or decreasing)
(iii) there exists:
Then: πππ
π β βπΌπ
π½π= π.
EXERCISES:
8. ππ =1β2β β¦.βπ+2β3ββ¦β(π+1)+β―+(πβπ+1)β(πβπ+2)ββ¦ βπ
ππ
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(Indications: amplification with the conjugate, the
minoring/majoring method, the Cesaro-Stolz lemma).
5. Let π₯π+1 = ln(1 + π₯π), π₯0 > 0. Using the Cesaro-Stolz
lemma show that πππ
π β βπ. π₯π = 2.
(Indications: ππ₯π =π1
π₯π
)
III. Calculate the limits of the following sequences,
presupposing they exist:
knowing that:
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where π > 0 and π is Eulerβs 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 Lagrangeβs theoremTheorem: (Lagrange, (1736-1813)). If π: [π, π] β β is
continuous on [π, π] and differentiable on
(π, π),
then: βββ (π, π)π. π.π(π)βπ(π)
πβπ= π ,(π).
This theorem can be used for:
(A) The calculation of the limits of sequences in which we
can highlight an expression with the form:
for the attached function.
Example:
Solution:
a) we highlight an expression of the form (3.6), considering:
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b) we apply Lagrangeβs theorem on function π on the
interval [π, π + 1]:
c) we replace in ππ the expression π(π)βπ(π)
πβπ with π ,(ππ). We
obtain:
d) we use the inequalities: π < ππ < π + 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:
π being a function that is subject to Lagrangeβs theorem on the
intervals: [π, π + 1], π β β.
Observation: the sequences of the form ππ© are always
monotonic and bounded (so convergent) if π is a function
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differentiable on β and so that π and π, have different
monotonies.
Demonstration: to make a choice, letβs assume π is increasing
and π , is decreasing:
1. the monotony:
2. we apply Lagrangeβs theorem to function π on the interval
[π, π + 1]:
so:
(π ,- decreasing), wherefrom we deduce the sequence is decreasing.
3. the bounding: being decreasing, the sequence is superiorily
bounded by π1. We have only to find the inferior bound. For this,
we write (3.8) starting with π = 1 and we thus obtain the
possibility of minoring (majoring) the sequence ππ.
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We obtain a convenient replacement for the sum:
We write the relation (3.7) starting π = 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 βπ(1) and π1.
Exercises I. Using Lagrangeβs 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 Eulerβs constant π β
(0,1))
(the partial sums of the harmonic sequence β1
π
βπ=1 )
(the partial sums of the generalized harmonic sequence)
Indication: 4. In order to have one of the forms 1π΅ or 2π΅, we
determine π, keeping into account that:
C. Dumitrescu β 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+ β― +
1
π is called a harmonic
sequence because it has the property: any three consecutive terms
are in a harmonic progression. Indeed, if we note ππ =1
π, we have:
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For a long time, it was believed that this sequence has a
defined sum π . In antiquity they sought to obtain the approximate
value of π , by calculating the sum of as many terms of the sequence
as possible. Today it is a fact that π = β (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. Lagrangeβs theorem (exercise I.2)
2. The inequality:
that is proven by induction in the X grade. Indeed, if π had a finite
value:
then, by noting:
we would have:
and because π =πππ
π β βππ, we deduce
ππππ β β
π π = 0.
But:
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so π π doesnβt tend to zero. Therefore, π isnβt 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: ππ =
π0ππ. For the second order recurrence, we search for the general
term with the same form:
ππ = π β ππ, π being an undetermined constant.
By replacing in the recurrence relation, we obtain:
from where, dividing by ππβ1 , we obtain the characteristic
equation:
We consider the following cases:
a) β> 0 (the characteristic equation has real and distinct
roots π1 and π2)
In this case, having no reason to neglect one of the roots, we
modify the expression ππ, considering it of the form:
and we determine the constants π1 and π2 so that the first two
terms of the sequence have the initially given values.
Example:
(Fibonacciβs sequence). The characteristic equation is:
with the roots:
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Considering ππ of the form:
from the system:
we obtain:
so:
Observation: π2 =1+β5
2 is known from antiquity as the golden
ratio. It is the limit of the sequence:
(π 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) β= 0 (the characteristic equation has equal roots π1 =
π2). In this case we consider ππ of the form:
(namely ππ = π1π β π1(π), with π1- first degree polynomial) and we
determine π1and π2, stating the same condition, that the first two
terms have the same initially given values.
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Example:
Taking ππ = π β ππ, we obtain the characteristic equation:
Considering:
from the system:
we obtain: π1 = β4, π2 = 6, so:
c) ) β< 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 π1π and ππ§
π respectively with:
Then we can write:
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3. Linear recurrence of degree h (bigger than 2)Is of the form:
with the first β terms given.
Proceeding as with the second order recurrence, we combine
ππ = π β π 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: π1, π2, β¦ , πβ, we will consider ππ of the form:
and we determine the constants ππ , π β 1, βΜ Μ Μ Μ Μ , stating the condition
that the first π terms of the sequence have the initially given values.
b) if a root, for example, π1 is multiple of order π , (π1 =
π2 = β― = ππ ), we replace the sum:
from the expression of ππ with:
ππ β1 being a polynomial of degree π β 2 , whose coefficient is
revealed by stating the condition that the first π terms have the
initially given values.
c) if a root, for example, π1, is complex, then its conjugate is
also a root of the characteristic equation (let for example, π2 = π1Μ Μ Μ ).
In this case, we replace in the expression of ππ the sum:
with:
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and if the root π1 is a multiple of order π (π1 = π2 = β― ππ and
we replace in the expression of ππ 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 πΌ 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: π1 = 1 , π2 = 2 , and the roots of the
characteristic equation are:
3. Let : ππ = π΄ β πΌπ + π΅ β π½π , with π, π΅, πΌ, π½ β β and
π΄, π΅ β 0, |πΌ| β | π½|. Determine πΌ πππ π½so that:
a) the sequence (π₯π) is convergent;
b) πππ
π β βπ₯π = β
c) πππ
π β βπ₯π = ββ
4. Let ππ = π΄ β πΌπ + π΅ β π β π½π , π β₯ 1, with π, π΅, πΌ, π½ β β
and π΄, π΅ β 0 . Determine πΌ πππ π½ so that the sequence is
convergent.
(B) Nonlinear recurrence
1. Recurrence of the form ππ+1 = πΌ β ππ + π½,
with π0 given The expression of the general term is obtained by observing
that if π is a root for the equation π = Ξ± β l + Ξ² (obtained by
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replacing an+1and an with π in the recurrence relation), then the
sequence (an β π)πββ
is a geometric progression.
Example:
From the equation π =π+1
2 we deduce π = π.
Then the sequence (an β π)πββ
is a geometric progression.
Let ππ = ππ β 1. We have:
The ratio of the geometric progression is, consequently: π =1
2. We obtain:
2. Recurrence of the form ππ+1 = πΌ β ππ + π(π) (π beinga random function)
The general term is found based on the observation that, if
(ππ)πββ is a sequence that verifies the same recurrence relation,
then any sequence (πΌπ)πββ verifies the given relation of the form:
In practice we chose ππ of the form:
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the coefficients of π being undetermined. We determine these
coefficients by stating the condition that (ππ)πββ verifies the
recurrence relation.
Particular case: a recurrence with the form:
with π a polynomial of degree π .
Example:
We search for the sequence ππ with the form:
We state the condition that (ππ)πββ verifies the given
recurrence relation:
We obtain:
By identification, it follows that:
so:
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3. Recurrence of the form ππ+1 = π(ππ) (with π: [π, π] ββ continuous function)
Theorem: 1. If the function π: [π, π] β β is
continuous and increasing on [π, π], then:
a) if π1 > π0, the sequence (πΌπ)πββ is increasing,
b) if π1 < π0, the sequence (πΌπ)πββ is decreasing.
The limit of the sequence is a fixed point of π, i.e. a
characteristic equation π(π₯) = π₯.
2. If π is decreasing on [π, π], then the subsequences of
even and uneven value, respectively, of (πΌπ)πββ 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 π is decreasing on (0, β2)
and increasing on (β2, β). By way of induction, we prove that
πΌπ > β2 , and, because π2 = 5,1 < π1 , 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 π(π₯) = π₯ has the roots π₯1,2 =
Β±β2, so π = β2.
Particular case: the homographic recurrence:
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We have:
called a homographic function and:
so the monotony of π depends on the sign of the expression πΌ β
πΏ β π½ β πΎ.
If π₯1 β π₯2 are the roots of the characteristic equation
π(π₯) = π₯, by noting:
it can be shown that the general term πΌπ 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 π and π 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 (πΌπ)πββ 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
numbersβ 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 ν > 0. We check to see if π exists so that property
(π) takes place.
(c) we have:
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(d) we state the condition: 2
π< ν and we obtain π > :
2
π, so
we can take: π = [ 2
π] + 1.
Exercises I. Show that the following sequences are fundamental:
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Show that |ππ+π β ππ| < ππ+1 for any π, π β β . Deduce
from this that the sequence is a Cauchy sequence.
INDICATIONS:
and we decompose in simple fractions.
If π is uneven, we deduce:
and if it is even:
So:
From the condition: 1
π+1< ν, we obtain: π = [
1β 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
π: [π, π] β β, corresponding to a division:
and to the intermediate points:
is:
If the points π₯π 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 π on the interval [π, π]:
To use the definition of the integral in the calculation of the
limits of sequences, we observe that for divisions β formed with
equidistant points, we have:
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So we can proceed as follows:
a) we show that the general term ππ can be written in the
form:
with π as a continuous function on [π, π], and ππ being the points
of an equidistant division.
b) π , being continuous, it is also integrable, and:
Example:
a) we write ππ in the form:
highlighting the common factor 1
π. We have:
b) in:
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we make the equidistant points: (πβπ)π
π appear:
and we deduce the function π. We have:
c) we arrange the equidistant points on a straight line:
and we deduce the division:
and the interval [π, π].
d) we observe that ππ is the Riemann sum corresponding to
function π (continuous, thus integrable) and to the deduced division
β, on interval [π, π]. As the points of β 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 π: π· β β is continuous in π₯0 β
π· if:
1. it has a limit in π₯0 β,
2. the limit is equal to π(π₯0),
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 π₯ = 1.
Indeed, letβs show that:
Let there be any ν > 0. We must determine πΏ . We have:
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because on the interval [0,2], for example, [2π₯ + 3] is bounded
between 3 and 7. We determine πΏ from the condition:
We can take, for example πΏ = 2ν. Then:
therefore π is continuous in π₯ = 1.
2. Using the criterion with the lateral limitsπ is continuous in:
where
are the left and right limit, respectively in π₯0.
EXAMPLE: Letβs study the continuity of the function:
for πΌ β [0, 2π].
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 π₯0 β 1, the function is continuous being
expressed through continuous functions.
(b) we study the continuity in π₯0 β 1. We have:
The value of the function in the point is π(1) =1
2, so π is
continuous in:
To explain the module we use the table with the sign of the
function sin πΌ β1
2:
1. if πΌ β [0,π
6] βͺ [
5π
6, 2π], the equation becomes:
2. if πΌ β (π
6,
5π
6), we obtain sin πΌ = 1, so πΌ =
π
2.
For this four values of πΌ the function is continuous also in
point π₯ = 1, so it is continuous on β.
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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 π with π(π₯0) and
eliminating the condition π₯π β π₯0 (which is essential in the case of
limits, because we can study the limit of a function in a point where
π(π₯) doesnβt exist).
EXAMPLE:
Proceeding as we did with Method 10, with the two
adaptations mentioned just above, it follows that π is continuous in
the points π₯0 for which:
(B) Types of discontinuity points The point π₯0 β π· in which π is not continuous is said to be
a first order discontinuity point if the lateral limits in π₯0 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 π: π· β β means to add new points
to the domain π·, where we define the correspondence law willingly.
If π is the set of added points and β is the correspondence
law on π, the extension will be:
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For example:
can be extended to the entire set β by putting:
There are many extensions to a function. Nevertheless, if π
is continuous on π· and has a finite limit in a point π₯0 β π·, there is
only one extension:
that is continuous, named the extension through continuity of π in
point π₯0.
So, for our example, because πππ
π₯ β 0sin π₯
π₯= 1, the function:
is the only extension of sin π₯/π₯ that is continuous on β.
(D) The continuity of composite functions If π is continuous in π₯0 and π is continuous in π(π₯0), then
π β π is continuous in π₯0.
So the composition of two continuous functions is a
continuous function. Reciprocally it isnβt true: it is possible that π
and π are not continuous and π β π is.
For example:
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isnβt continuous in any point, but the function π(π₯) = (π β π)(π₯)
is the identically equal function to 1, so it is continuous in any point
from β.
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 functionsπ, π, π β π, π β π ,
for:
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IV. 1. Let π, π: β β β be a continuous function so that
π(π₯) = π(π₯) for any π₯ β β. Show that π = π. (Manual)
2. Let π: πΌ β β , πΌ ββ β be a function with Darbouxβs
property. If π has lateral limits in any point πΌ, then π is continuous
on πΌ. (Manual)
3. π: β β β so that:
for any π₯, π¦ β β. Prove that π > 0 exists so that for any π₯ with
|π₯| β€ π we have |π(π₯)| < π. Deduce the existence of a fixed point
for π. (π₯0 is a fixed point for π if π(π₯0) = π₯0). (Manual)
4. Let π: [π, π] β β, continuous. Then:
(Manual)
A function with the property (4.1) is called an uniform
continuously function on [π, π].
The uniform continuity is therefore defined on an interval,
while continuity can be defined in a point.
Taking π¦ = π₯0 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 π: π· β β is differentiable in π₯0 β π· if the
following limit exists and is finite:
The value of this limit is noted by π ,(π₯0) and it is called the
derivative of π in π₯0.
Geometric interpretation The derivative of a function in a point π₯0 is the slope of the
tangent to the graphic of πin the abscissa point ππ.
When π₯ tends to π₯0 , the chord π΄π΅ tends towards the
tangent in π΄ to the graphic, so the slope of the chord
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tends to the slope of the tangent. Therefore, π ,(π₯0) is the slope of
the tangent in the abscissa point π₯0 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 (π₯0, π(π₯0)) is:
3. The equation of the normal (the perpendicular on the
tangent) is deduced from the condition of perpendicularity of two
straight lines (π1 β π2 = β1) 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
π3) and we replace in this function the variable π₯ with the
expression issuing from the composition of the other function (in
our case π2 β π1).
(b) by neglecting the derived function in the previous step (in
our case π3), we derive the function that became last (for us π2) and
we also replace the variable π₯ with the expression that resulted from
the composition of the functions that have no yet been derived (in
this step we only have π1(π₯).
(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 πβ1(π(π₯)) = π₯, by applying formula (4.2), it follows
that:
or by noting π¦ = π(π₯):
EXAMPLE: For π(π₯) = ln π₯, π: (0, β) β β we have:
and as π¦ = ln π₯ => π₯ = ππ¦ we obtain:
2. High order derivatives of composite functions.
From (π(π’(π₯)) = π ,(π’(π₯)). π’,(π₯) , deriving again in
relation to π₯ 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: Letβs calculate π ,,(π₯) if π(π₯) = π(πβπ₯) , π
being a function two times differentiable on β.
We have:
and
(C) Derivatives of order π For the calculation of the derivative π(π)(π₯) we can use one
of the following methods:
1. We calculate a few derivatives (π ,, π ,,, π ,,,, β¦ ) in order to
deduce the expression of π(π), 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 Lagrangeβs theorem: If π is
differentiable in a vicinity of π₯0 and if it
continues in π₯0 and if there exists:
then, π has a derivative on the left (on the right,
respectively) in π₯0 and:
The continuity condition in ππ that is required in the
hypothesis of the corollary is essential for its application, but it
doesnβt constitute a restriction in the study of derivability,
because if π isnβt continuous in π₯0, it isnβt differentiable either.
EXAMPLE: Letβs determine the parameters πΌ, π½ β β 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 π₯ β π , the function is differentiable, being
expressed through differentiable functions.
b) we study the derivability in π₯ = π.
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Because of the proposition: π non continuous ==> π non
differentiable, we study the continuity first.
so: π continuous in π <=> πΌπ + π½ = 1. (4.3)
If the continuity condition in π₯ = π isnβt 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 lβHospitalβs rule.
(1) We observe that by noting π(π₯) = ππ2π₯ , we have
π(π) = 1 and the limit becomes:
(because π is differentiable in π)
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Therefore, π is differentiable in:
and from the continuity condition we obtain π½ = β1 . We thus
have:
The geometric interpretation of the result: the straight
line π¦ =2
ππ₯ β 1 is tangent to the curve π¦ = ln π₯2.
If the branch of a function is expressed through a
straight line (π = πΆπ + π·) , the function is
differentiable in the point ππ of connection
between the branches, if and only if the
respective straight line is tangent in the point of
abscissa ππ to the graphic of the other branch.
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Method 2: (Using the corollary of Lagrangeβs theorem)
(a) In any point π₯ β π, the function is differentiable, being
expressed by differentiable function and we have:
(b) We study the derivability in π₯ = π . We first state the
continuity condition:
In order to use the corollary of Lagrangeβs theorem we
calculate the lateral limits of the derivative:
From the corollary we deduce π ,(π) = πΌ. Analogously:
If π is differentiable in π₯ = π if and only ifπΌ =2
π. From the
continuity condition we deduce π½ = β1.
(E) Applications of the derivative in economics
(See the Calculus manual, IXth grade)
1. Let π½(π₯) be the benefit obtained for an expenditure of π₯
lei. For any additional expenditure of β lei, the supplementary
benefit on any spent leu is:
If β is sufficiently small, this relation gives an indication of
the variation of the benefit corresponding to the sum of π₯ lei.
If this limit exists:
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it is called the bound benefit corresponding to the sum of π lei.
2. Let πΎ(π) be the total cost for the production of π units of
a particular product. Then, the cost per each supplementary unit of
product is:
and the limit of this relation, when β tends to zero, if it exists, is
called a bound cost of production for π units of the considered
product.
EXAMPLE: The benefit obtained for an expenditure of π₯
lei is:
For any additional expenditure of β 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 π₯ = 1000 this is equal to 1997 + β and
Exercises I. The following are required:
1. The equation of the tangent to the graphic π(π₯) =
ln β1 + π₯2 in the abscissa point π₯0 = 1.
2. The equation of the tangent to the curve π(π₯) =
βπ₯2 β π2, that is parallel to 0π₯.
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3. The equation of the tangent to the curve π(π₯) = π₯3 ,
parallel to the first bisector.
4. The equation of the tangent to the curve π¦ =3π₯+2
2π₯+6, that is
parallel to the chord that unites the abscissae points π₯ = 1 and π₯ =
3.
5. Determine πΌ and π½ so that π¦ = πΌπ₯ + π½ and π¦ =π₯β1
π₯ are
tangent in π₯ = 1. Write down their common tangent.
6. Show that the straight line π¦ = 7π₯ β 2 is tangent to the
curve π¦ = π₯2 + 4π₯. (Manual)
II. 1. Calculate the derivative of the function:
2. Let π: (ββ, 0) β β , π(π₯) = π₯2 β 3π₯ . Determine a
subinterval π½ β β, so that π: (ββ, 0) β π½ is bijective. Let π be its
inverse. Calculate π,(β1) and π,,(β1). (Manual)
3. If:
with πΌ > 0 for any π₯, π¦ β πΌ, the function π is constant on πΌ.
4. If π has a limit in point π, then the function:
is differentiable in π.
5. If π is bounded in a vicinity of π₯0 , then π(π₯) =
(π₯ β π₯0)2π(π₯) is differentiable in π₯0. Particular case:
Indications:
1. Let π be a primitive of the function:
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We have:
3. For π₯ β π¦ the inequality from the enunciation is
equivalent to:
from where, for π¦ βΆ π₯ we obtain π(π₯) = 0, so π is constant on πΌ.
III. Calculate the derivatives of order π for:
10. π(π₯) = πππ₯ + πππ₯ from the expression of the derivative
of order π, deduce Newton's binomial formula.
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IV. 1. Show that:
2. Applying Leibnizβs formula for:
show that:
Find similar formulas, using the functions:
3. Let πΌ = (0,1) and the functions π’, π£: πΌ βΆ β , π’(π₯) =
π’(π₯) =πππ
π¦ β πΌ(π₯ β π¦)2 , π£(π₯) ==
π π’ππ¦ β πΌ(π₯ β π¦)2 . Study the
derivability of the functions π’ and π£ and calculate π π’π
π₯ β πΌπ’(π₯) ,
ππππ₯ β πΌ
π£(π₯). (Manual)
4. If ππ(π₯) is a sequence of differentiable function, having a
limit in any point π₯, then:
V. Study the derivability of the functions:
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is indefinitely differentiable on β and:
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π β β.
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V. Fermat, Rolle, Lagrange, Cauchy Theorems
(A) Fermatβs theorem
1. EnunciationIf π: [π, π] β β is continuous on [π, π] and differentiable
on (π, π), then in any point of extremum from (π, π) (from
the interior of the interval), the derivative is annulled.
(Fermat, 1601-1665)
Observation: π₯0 β (π, π) point of extremum ==> π ,(π₯0) =
0. If π₯0 is a point of extremum at the ends of the interval, it is
possible that the derivative isnβt annulled in π₯0.
Example:
has two points of extremum: in π₯1 = β1 and in π₯2 = 2, and:
2. Geometric and algebraic interpretationa) The geometric interpretation results from the
geometrical interpretation of the derivative: if the conditions of
Fermatβs theorem are fulfilled in any point from the interior of the
interval, the tangent to the functionβs graphic is parallel to the axis
0π₯.
b) The algebraic interpretation: if the conditions of
Fermatβs theorem are fulfilled on [π, π] , any point of extremum
from (π, π) is a root to the equation π ,(π₯) = 0.
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Exercises 1. If π1, π2, β¦ ππ are positive numbers, so that:
for any π₯ β β, then we have:
2. Let π1, π2, β¦ ππ, π1, π2, β¦ ππ be positive real numbers, so
that:
for any π₯ β β. Then :
(generalization of the previous exercise)
3. If π is continuous on [π, π] and differentiable on (π, π),
and π(π) = π(π) = 0, then:
(i) if π , is increasing, it follows that π(π₯) β€ 0 on [π, π];
(ii) ) if π ,, is decreasing, it follows that π(π₯) β₯ 0 on [π, π].
4. If ππ₯ β₯ π₯π for any π₯ > 0, then π = π.
5. π‘π π
π‘π π<
π
π if 0 < π < π <
π
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 β, observing that the inequality from the hypothesis can
be written:
Because this inequality is true for any real π₯, it follows that π₯ = 0 is
a global point of minimum for π. According to Fermatβs theorem,
in this point the derivative is annulled, so π ,(0) = 0. We have:
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3. (i) Letβs assume there exists π β (π, π) so that π(π) > 0.
We can even assume that π is a point of maximum, because such a
point exists, according to Rolleβs theorem, so we have π ,(π) = 0.
Between π and π at least one point π exists in which
π ,(π) < 0 , because, if, letβs say π ,(π₯) β₯ 0 for any π₯ β (π, π) , it
follows that π is increasing on (π, π) and so π(π) < π(π) = 0
implies π(π) = 0. Then, from π ,(π) =β€ 0 = π ,(π) we deduce the
contradiction:
4. ππ₯ β₯ π₯π <=> π₯. ln π β₯ π. ln π₯ <=>ln π
πβ₯
ln π₯
π₯ (for any
π₯ > 0). So π is the abscissa of the maximum of the function:
We have:
It is sufficient to prove that π(π₯) =π‘π π₯
π₯ is increasing, i.e.
π ,(π₯) > 0.
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(B) Rolleβs theorem
1. Enunciation If π: [π, π] β β is continuous on [π, π], differentiable
on (π, π) and π(π) = π(π), then there exists π β
(π, π) so that π ,(π) = 0. (Rolle, 1652-1719)
Fermatβs theorem states that in a point of extremum from
the interior of an interval the derivative is annulled, but it doesnβt
mention when such a point exists.
Rolleβs theorem provides a sufficient condition for the
existence of at least one such point, adding to the hypothesis from
Fermatβs theorem the condition: π(π) = π(π).
2. Geometric and algebraic interpretation a) The geometric interpretation: if the conditions of
Rolleβs theorem are met, there exists at least one point in the
interval (π, π)in which the tangent to the graphic is parallel to 0π₯.
b) The algebraic interpretation: if the conditions of
Rolleβs theorem are met, the equation π,(π) = π has at least one
root in the interval (π, π).
The algebraic interpretation of the theorem highlights a
method used to prove that the equation π(π₯) = 0 has at least one
root in the interval (π, π). For this, it is sufficient to consider a
primitive πΉ of π, for which the conditions of Rolleβs theorem are
fulfilled on [π, π]. It results that the equation πΉ,(π₯) = 0 has at least
one root in the interval (π, π), i.e. π(π₯) has a root in the interval
(π, π).
A second method to show that the equation π(π₯) = 0 has
at least one root in the interval (π, π) is to show just that π is
continuous and π(π). π(π) < 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 functionβs 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 derivativeβs roots
(Rolleβs sequence):
Let π₯1, π₯2, β¦ , π₯π be the derivativeβs roots.
Then π has as many real, simple roots as there are variations
of sign in the sequence:
(if we have π(π₯π) = 0, then π₯π is a multiple root)
Exercises I. Study the applicability of Rolleβs Theorem for the
functions:
What is the geometric interpretation of the result?
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Indication: 1. We obtain π β [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, 2π). (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 (β1,0)?
(iii) The same question for the equations:
on the interval (β1,1).
4. Let π: β β β be differentiable and π1 < π2 < β― < ππ ,
roots of π. Show that π ,has at least π β 1 roots.
(Manual)
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Consequences:
a) A polynomial function of degree π has, at most, π 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 ππ(π₯) = 0 has π distinct roots in
the interval (β1,1).
6. If π is π times differentiable on πΌ and it has π + 1 distinct
roots on πΌ, then ππ(π₯) has at least one root on πΌ.
7. Let π, π: [π, π] β β, continuous on [π, π], differentiable
on (π, π) with π(π₯) β 0 and π,(π₯) β 0 on [a,b], and π(π)
π(π)=
π(π)
π(π).
Show that π β (π, π) exists, so that:
8. Letπ: [π, π] β β, continuous on (π, π), differentiable on
(π, π). Then between two roots of π there exists at least a root of
πΌ. π + π ,.
9. If the differentiable functions π and π have the property:
on an interval, then between two of its roots there is a root of π and
vice versa.
Consequence: If π ,(π₯). cos π₯ + π(π₯). sin π₯ β 0 , in any
length interval bigger than π there is at least one root of π.
10. If π, π: [π, π] β β, continuous on [π, π], differentiable
on (π, π) , π(π₯) β 0, π,(π₯) β 0 and π(π) = π(π) = 0, then ππ β
(π, π) so that:
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SOLUTIONS:
7. The required equality can also been written:
This equality appears in points π that are roots of the
derivative of β(π₯) =π(π₯)
(π₯)π, so it is sufficient to demonstrate that the
function β fulfills the conditions of Rolleβs theorem on [π, π].
8. The equation πΌ. π(π₯) + π ,(π₯) = 0 comes from the
equalizing with zero the derivative of the function πΉ(π₯) =
ππΌπ₯π(π₯), so it is sufficient to show that πΉ satisfies the conditions of
Rolleβs theorem.
9. Let π₯1, π₯2 be roots of π. By hypothesis, we haveπ(π₯1) β
0 and π(π₯2) β 0. If, for example, there can be found a root of π
between π₯1, π₯2 , this means that the function β(π₯) =π(π₯)
(π₯)π satisfies
the conditions of Rolleβs theorem and so β,(π₯) is annulled in at least
one point between π₯1 and π₯2, which contradicts the hypothesis.
has at least one root in (π, π) <=> β,(π₯) = 0 has at least one root
in (π, π), where β(π₯) =π(π₯)
ππ(π₯).
We will present next, two methods for the study of an
equationβs roots. The first of these uses Rolleβs Theorem and the
other is a consequence of the fact that any continuous function has
Darbouxβs property.
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Methods for the study of an equationβs roots
1. Using Rollesβs theorem (its algebraicinterpretation) 2. Using Darbouxβs properties (if π has Darbouxβs
property (particularly if it is continuous) on [π, π] and:
π(π). π(π) β€ 0, then π has a root in [π, π]).
Examples:
1) If π: [π, π] β [π, π] is a continuous function, then π’, π£ β
[π, π] exists so that π(π’) = π’ and π(π£) = π£. (Manual)
2) Let π: [0, 2π] β β be a continuous function so that
π(0) = π(2π). Show that π β [0, π] exists so that π(π) = π(π +
π). (Manual)
Consequences: If a traveler leaves in the morning from spot
π΄and arrives, in the evening in spot π΅, and the next day he leaves
and reaches again spot π΄, show that there is a point between π΄ and
π΅ where the traveler has been, at the same hour, in both days.
Indication: If π(π‘) is the space covered by the traveler, we
have: π(0) = π(24) and so π‘0 β [0,12] exists, so that: π(π‘0) =
π(π‘0 + 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 π΄ heading to spot π΅, and the other from π΅ to π΄, meet on
the way.
3. Using Rolleβs sequenceExample: Letβs study the nature of the equationβs roots:
π being a real parameter.
Answer: The derivativeβs roots are:
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and we have:
The results are illustrated in the following table:
4. The graphical method. The equation πΉ(π₯, π) = 0 is
reduced to the form π(π₯) = π. The number of roots is equal to the
number of intersection points between the graphics: π¦ = π(π₯) and
π¦ = π.
Example:
The number of roots of the given equation is equal to the
number of intersection points between the graphics:
5. Vietteβs relationsExample: Show that the equation:
with π, π, π β β 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 π: [π, π] β β is a Riemann integral (so it is
bounded), π β [π, π] exists, so that:
Example: Let π: [0,1] β β be continuous, so that:
Show that the equation π(π₯) = π₯ has a root π₯0 β (0,1).
Answer: π(π₯) = π₯ <=> π(π₯) β π₯ = 0 , so by noting:
β(π₯) = π(π₯) β π₯ we have to show that the equation β(π₯) = 0 has
a root in the interval (0,1). We have:
But: π02 β(π₯) = 0, according to the theorem of the average,
with π β (π, π). The function π, being continuous, has Darbouxβs
property, so, for π there exists π₯0 β (0,1)so that π = β(π₯0).
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Exercises 1. The equations:
canβt all have real roots if πΌ, π½, πΎ β β.
Indication:
2. If π: [0,1] β β is continuous and with the property:
then, the equation π(π₯) β sin ππ₯ = 0 has one root in (0,1).
3. If π: [0,1] β β is continuous and with the property:
then, the equation π(π₯) β ππ₯2 β ππ₯ β π = 0 has one root in
(0,1).
4. If π: [0,1] β β is continuous andπ > 1 exists for which:
the equation: (1 β π₯)π(π₯) = 1 β π₯π has one root in (0,1).
Indication: We apply the theorem of the average to the
function:
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(C) Lagrangeβs Theorem 1. Enunciation:
Let π: [π, π] β β be a continuous function
on [π, π] and differentiable on (π, π) . Then π β
(π, π) exists, so that:
(Lagrange, 1736 β 1813)
2. Geometric and algebraic interpretation
a) Geometric interpretation:If the conditions of the theorem are met, there exists a point
π β (π, π) in which the tangent to the graphic is parallel to the
chord that unites the graphicβs extremities.
b) Algebraic interpretationIf the conditions of the theorem are met, the equation:
has at least one root in the interval (π, π).
3. The corollary of Lagrangeβs TheoremIf π is continuous on an interval πΌ and differentiable on
πΌ\{π₯0} and in π₯0 there exists the derivativeβs limit:
then π is differentiable in π₯0 and:
This corollary facilitates the study of a functionβs 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 π₯ β 1, the function is differentiable,
being expressed by differentiable functions. We study the
derivability in π₯ = 1 using the corollary of Lagrangeβs theorem.
- the continuity in π₯ = 1:
π(1) = ln 1 = 0 , so π is continuous on β (and
differentiable on β{1}).
- derivability. For the study of derivability in π₯ = 1, we have:
(we know that π , exists just on β β (1)).
Exercises I. Study the applicability of Lagrangeβs theorem and
determine the intermediary points π for:
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Indication:
2. The continuity condition in π₯ =π
6 is:
and the derivability condition (the corollary to Lagrangeβs theorem)
is:
Point π is the solution of the equation:
II. 1. Let π(π₯) = ππ₯2 + ππ₯ + π. Apply the theorem of the
finite increases on the interval [π₯1, π₯2] , finding the intermediary
point π . Deduce from this a way to build the tangent to the
parabola, in one of its given points.
2. We have π β€ π ,(π₯) β€ π for any π₯ β πΌ, if and only if:
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3. (The generalization of Rolleβs theorem). If π is continuous
on [π, π]and differentiable on (π, π), then π β (π, π) exists so that:
4. Supposing π is twice differentiable in a vicinity π of point
π, show that for β, small enough, there exists π, π β π so that:
(Manual)
5. Let π: [0, β) β β be a differentiable function so that:
and let π > 0 be fixed. Applying the Lagrange theorem on each
interval:
show there exists a sequence (π₯π)πββ, 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 π.
SOLUTIONS:
1. π =π₯1+π₯2
2, so, the abscissa point π being given, in order to
build the tangent in point (π, π(π)) of the graphic, we consider two
points, symmetrical to π:
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The tangent passes through (π, π(π)) and is parallel to the
chord determined by the points:
2. The necessity: For π₯ = π¦ (πΏ1) is verified, and for π₯ β π¦,
we have:
inequalities which are true due to the hypothesis.
Reciprocally, let π₯ β πΌ, randomly. Making π¦ tend to π₯, for:
we have:
3. We distinguish three cases:
In this case we must prove that π β (π, π)exists, so that:
π ,(π) > 0. If we had, letβs say, π ,(π) β€ 0 on (π, π), it would follow
that π is decreasing, so π(π) > π(π).
We are situated within the conditions of Rolleβs theorem.
(3) The case π(π) > π(π) is analogous to (1).
4. a) We apply Lagrangeβs theorem to π on [π β β, π + β].
b) We apply Lagrangeβs theorem twice to the function
π(π₯) = π(π + π₯) β π(π₯) on the interval [π β β, π].
6. a) π = 1/2 , c) is obtained πβ = πβ , so π cannot be
determined.
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(D) Cauchyβs Theorem 1.Enunciation
Let If π, π: [π, π] β β be continuous on[π, π] and
differentiable on (π, π) so that:
Then π β (π, π) 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 tangentsβ
slopes to the two graphics is equal to the relation of the chordsβ
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 (π, π).
Exercises I. Study the applicability of Cauchyβs theorem and determine
the intermediary point π for:
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II. If π has derivatives of any order on [π, π], by applying
Cauchyβs formula to the functions:
a) π(π₯) = π(π₯), β(π₯) = π β π₯, show that π β (π, π) exists,
so that:
show that π β (π, π) exists, so that:
show that π β (π, π) exists, so that:
show that π β (π, π) exists, so that:
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2. Applying the conclusion from point π) show that there
exists no functions π: β β β twice differentiable, so that:
π(π₯) β₯ 0 and π ,,(π₯) < 0
(there are no non-negative and strictly concave functions on the
entire real axis).
SOLUTION:
2. From π ,,(π₯) < 0 for any π₯ β β , we deduce that π , is
strictly increasing and so we cannot have π ,(π₯) = 0 for any π₯ β β.
Therefore, for any π₯0 β β exists, so that π ,(π₯) β 0.
But then, from the relation:
it follows that:
We have two situations:
(1) if π ,(π₯0) > 0, making π₯ tend to ββ in (πΆ1), we have:
consequently:
2. If π ,(π₯0) < 0, we obtain the same contradiction if we
make π₯ tend to +β.
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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:
π(π₯) = π(π₯) + πΆ.
or, in particular,
π(π₯) = π(π₯) and π(π₯) = πΆ = ππππ π‘πππ‘.
Indeed:
and in order to prove this equality it is sufficient to prove that:
(π(π₯) β π(π₯)) = 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 πΆ, we can use two methods:
1. We calculate the expression π(π₯) β π(π₯) in a
conveniently chosen point from the considered interval.
2. It the previous method cannot be applied, we can
calculate:
π₯0 also being a conveniently chosen point (one end of the interval).
Example: Show that we have:
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Solution: For:
we have β,(π₯) = 0 for any π₯ β β(β1), domain that is a reunion of
disjunctive intervals. We calculate the value of the constant on each
interval.
(1) For π₯ > β1, we choose the point π₯ = 0 where we can
easily make the calculations:
(2) For π₯ < β1 , we canβt find a point in which to easily
calculate the value of β, but we observe that we can calculate:
Inequalities Method 1. Using Lagrangeβs or Cauchyβs
theorem.
In some inequalities, we can highlight an expression of the
form:
π and the points π, πbeing conveniently chosen. In this case, we can
use Lagrangeβs theorem to prove the respective inequality, by
replacing the expression:
from the inequality, with π ,(π). The new form of the inequality can
be proven taking into account the fact that π < π < π and the
monotony of π ,.
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A similar method can be used if in the inequality an
expression of the following form is highlighted:
using Cauchyβs theorem.
EXAMPLES:
1. Using Lagrangeβs theorem, show that:
if 0 < π < π.
Solution: We go through these steps:
(a) We highlight an expression of the form:
observing that:
and dividing by π β π. The inequality becomes:
(b) we apply Lagrangeβs theorem to the function π(π₯) = ln π₯
on the interval [π, π]. π β (π, π) exists so that:
(c) the inequality becomes:
This new form of the inequality is proven considering the
fact that π < π < π and that π ,(π₯) =1
π₯ is decreasing.
2. Using Lagrangeβs theorem, show that ππ₯ > 1 + π₯ for any
π₯ β 0.
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Solution:
To use Lagrangeβs theorem, we search for an interval [π, π]
and a function so that we can highlight the relation π(π)βπ(π)
πβπ in our
inequality. For this we observe that, if π₯ β 0 , we have two
situations:
(1) π₯ > 0. In this case we can consider the interval [π, π₯].
a) We highlight, in the given inequality, an expression of the
form: π(π₯)βπ(π)
π₯β0, observing that, if π₯ > 0, we have:
(b) We apply Lagrangeβs theorem to the function:
on the interval [0, π₯]. π β (0, π₯) exists so that:
(c) The inequality becomes:
This form of the inequality is proven considering the fact
that 0 < π < π₯ and the derivative π ,(π₯) = ππ₯is increasing (so π0 <
ππ).
(2) If π₯ < 0 the demonstration is done analogously.
Method 2: The method of the minimum This method is based on the observation that if π₯0 is a global
point of minimum (the smallest minimum) for a function β on a
domain π· and if β(π₯0) β₯ 0 (the smallest value of β is non-
negative), then β(π₯) β₯ 0 for any π₯ β π· (all the values of β are
non-negative). We can demonstrate therefore inequalities of the
form π(π₯) β₯ π(π₯), i.e. π(π₯) β π(π₯) β₯ 0, in other words, of the
form: β(π₯) β₯ 0.
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Example: Using the method of the minimum, show that ππ₯ >
1 + π₯ for any π₯ β 0.
Solution: In order to use this method, we proceed as follows:
(a) We write the inequality under the form: β(π₯) β₯ 0:
(b) Let β(π₯) = ππ₯ β 1 β π₯ . Using the variation table, we
calculate the global minimum (the smallest minimum) of β:
From the table it can be observed that β has a single
minimum, in π₯ = 0, so this is also the global minimum. Its value is
β(0) = 0. So, for π₯ β 0, we have β(π₯) > 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 π, or
π respectively, are values of the global minimum and maximum of
π on the interval [π, π], 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 π(π₯) = ππ₯2+
π1βπ₯2on the interval [0,1]:
We thus have π = 2βπ and π = 1 + π
(c) We integrate the inequalities π β€ π(π₯) β€ π
on the interval [π, π] and we obtain:
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Method 4: Using the inequality from the definition of convex (concave) functions
Let πΌ be an interval on a real axis.
Definition: The function π: πΌ β β is convex on πΌ if between any
two points π₯1, π₯2 β πΌ , the graphic of π is situated
underneath the chord that unites the abscissae points
π₯1 and π₯2.
With the notations from the adjacent graphic, this condition
is expressed through the inequality:π(π₯) β€ π¦.
In order to explain π₯ and π¦, we observe that:
1. π₯ is in the interval [π₯1, π₯2] if and only if it can be written
in the form:
Indeed, if π₯ β [π₯1, π₯2], it is sufficient that we take πΌ = (π₯ β
π₯2)/π₯1 β π₯2) to meet the required equality.
Reciprocally, if π₯ has the expression from (πΈ. πΌ. 1), in order
to show that π₯ β [π₯1, π₯2], the inequality system must be checked:
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The first inequality is equivalent to πΌ β€ 1, and the second
inequality is equivalent to πΌ β€ 0.
Observation: The condition πΌ β [0,1] from (πΈ. πΌ. 1) is
therefore essential.
The inequality from the convexity becomes:
2. To express π¦, we observe that:
so:
from which it follows:
By replacing π₯ and π¦, we obtain:
π: πΌ β β is convex <=>
π: πΌ β β is said to be concave if, for any π₯1, π₯2 β πΌ, between
the abscissae points π₯1 and π₯2 , the graphic of the function π 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 π₯1 and π₯2 from β, we have:
Solution: The function π(π₯) = ππ₯ is convex on β(π ,,(π₯) >
0), so:
from where, for πΌ =1
2 we obtain the inequality from the
enunciation.
Exercises I. Show that:
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5. For the intervals πΌ1, πΌ2 β β letβs consider the continuous
and injective functions:
and the constant π in the interior of π(πΌ1) . From the
condition: π(π’(π₯)) = π + π(π₯) it follows that π’(π₯) = πβ1(π +
π(π₯) for any π₯ that fulfills the condition π + π(π₯) β π(πΌ1). Using
this observation, construct equalities of the form β(π₯) = π for:
Answer: (a) We have:
We determine function π’ from the condition π(π’(π₯)) =
π + π(π₯) that becomes:
For π =π
4, for example, we have:
and from the condition:
we deduce:
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and so we have:
II. Using Lagrangeβs theorem, prove the inequalities:
6. Using Cauchyβs 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 π1(π₯) < π2(π₯) or π1(π₯) > π2(π₯) on the
considered intervals. We then integrate the obtained inequality.
VI. 1. If π: πΌ β β is convex, then for any π₯1, π₯2, β¦ , π₯π β πΌ,
we have the inequality (Jensenβs inequality):
(Manual)
2. Using Jensenβs inequality applied to the convex function
π(π₯) = ππ₯, prove the inequalities of the averages:
3. Show that for any π₯1, π₯2, β¦ , π₯π β [π,π
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 donβt 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 π₯1, π₯2, β¦ , π₯π > 0 we have:
5. Show that the function π(π₯) = π₯πΌ , πΌ > 1 , π₯ > 0 is
convex. Using this property show that for any non-negative
numbers π₯1, π₯2, β¦ , π₯π, the following inequality takes place:
6. Applying Jensenβs inequality to the function π(π₯) = π₯2 ,
show that for any π₯1, π₯2, β¦ , π₯π β β, we have:
What analogous inequality can be deduced from the convex
function π(π₯) = π₯3 , for π₯ > 0? And from the concave function
π(π₯) = π₯3, for π₯ β€ 0?
7. Applying Jensenβs inequality for the convex function
π(π₯) =1
π₯, for π₯ > 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 Jensenβs inequality, we deduce that the
points π₯1, π₯2, β¦ , π₯π are deduced from the conditions:
We have:
With these points, Jensenβs inequality, for π(π₯) = ππ₯, becomes:
Making the calculations in the left member we obtain the inequality
of the averages.
3. On the interval [0,π
2], the functions sin π₯ and cos π₯ are
concave. The biggest interval that contains the origin and in which
sin π₯ is concave, is the interval [0, π]. On [π, 2π], 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 π: πΌ β β has primitives on the interval
πΌ if there exists a function πΉ: πΌ β β, differentiable
and πΉ,(π₯) = π(π₯) on πΌ.
It is known that two primitives differ through a constant:
The set of all primitives is called an undefined integral of the
function π 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 doesnβt 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. DARBOUXβS PROPERTY
5. THE PRIMTIVE
6. THE INTEGRAL
The connections between these notions are given by the
following properties:
π1 ) Any continuous function in a point
π₯0 has a limit in this point:
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π2 ) Any differentiable function in a point
π₯0 is continuous in this point:
π3 ) Any continuous function on [π, π] has Darbouxβs
property on [π, π]:
π4 ) Any continuous function on [π, π] has primitives on
[π, π]:
π5 ) Any continuous function on [π, π] is integrable on
[π, π]:
π6) Any function that has primitives on [π, π] has Darbouxβs
property on [π, π]:
π7) If a function has Darbouxβs property on [π, π], then in
any point π₯0 in which the limit (the lateral limit) exists, it is equal to
π(π₯0).
Consequences: 1. A function with Darbouxβs property (so a
function that admits primitives) canβt have an infinite limit (lateral
limit) in a point π₯0.
The functions for which at least one lateral limit is
infinite or different from π(ππ) doesnβt have primitives.
2. If π has Darbouxβs property on [π, π] and π₯0 is a
discontinuity point, AT LEAST ONE LATERAL LIMIT
DOESNβT EXIST.
Due to the implications stated above, we can form the Table:
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OBSERVATION: If π₯0 is a discontinuity point for π, 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 doesnβt exist
(5) one lateral limit is finite, the other doesnβt exist
(6) both lateral limits donβt 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 DOESNβT EXIST (AND THE OTHER,
IF IT IS EXISTS, IS FINITE).
Also, letβs observe that, although in case (1) the function
doesnβt have primitives on [π, π] (doesnβt have Darbouxβs
property), still, if, moreover, the lateral limits are equal, then π has
primitives on [π, π]\{π₯0} through the function ππ β the extension
through continuity of πβ²π restriction to the domain [π, π]\{π₯0}.
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Example:
has Darbouxβs property for any π β [β1,1] , but doesnβt have
primitives if π β 0.
Indeed, assuming ad absurdum that π has primitives, let πΉ be
one of its primitives. Then πΉ must be of the form:
In order to calculate β« sin1
π₯ππ₯, we observe that sin
1
π₯ comes
from the derivation of the function π₯2 cos1
π₯. Indeed,
so:
The function π(π₯) = π₯ cos1
π₯ has primitives on (0, β)
because it is continuous on this interval, but doesnβt have primitives
on [0, β) (we aim for the interval to be closed at zero in order to
study the continuity and derivability od πΉ).
We observe that, because πππ
π₯ β 0π₯ > 0
π(π₯) = 0, we can consider
its extension through continuity:
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This function, being continuous on [0, β), has primitives on
this interval. Let πΊ be one of these primitives. Then:
We state the condition that πΉ be continuous in π₯0 = 0. We
have:
so we must have πΆ1 = πΆ2 + 2. πΊ(0) and so, for the primitive, we
obtain the following formula:
We state the condition that πΉ be differentiable in zero:
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because πΊ is differentiable in zero. We have πΊ ,(0) = ππ(0) = 0 (πΊ
is a primitive of ππ ), so πΉπ, (0) = 0 . Consequently, πΉ is not
differentiable in zero if π β 0.
Observation: For the study of derivability we have mentioned
two methods: using the definition and using the corollary of
Lagrangeβs theorem. Letβs observe that in the example from above,
we canβt use the corollary of Lagrangeβs theorem because
ππππ₯ β 0π₯ > 0
πΉ,(π₯) doesnβt 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 doesnβt 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 π β π:
(πΈπ1) We show that the function is differentiable.
(πΈπ2) We show that the function is continuous.
(πΈπ3) We construct the primitive.
(πΈπ4) We show that the function is a sum of functions that
admit primitives.
Keep in mind that method πΈπ2 is preferred over πΈπ1
because derivability is easier to prove that continuity, and the
method πΈπ3 is used only if the function doesnβt 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 doesnβt have primitives
The first three methods are deduced from the table, using
the equivalence: π β π β· πππ π βΆ πππ π.
We therefore have:
This way, we deduce the following methods to show that a
function doesnβt have primitives:
(ππ1) We show that in a discontinuity point, both lateral
limits exist and are finite or at least one lateral limit is infinite..
(ππ2) We show that the function doesnβt have Darbouxβs
property..
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(ππ3) We assume ad absurdum that π has primitives and we
obtain a contradiction.
(ππ4 ) If π is the sum of two function, one of them
admitting primitives, the other not admitting primitives, it follows
that π doesnβt have primitives.
Examples:
has primitives because π is continuous.
Solution: We observe that 2π₯. π ππ1
π₯β πππ
1
π₯ comes from the
derivation of the function π₯2π ππ1
π₯, so, a primitive of π must have
the form:
We state the condition that πΉ be continuous in zero:
so πΌ = πΆ and we have:
We state the condition that πΉ be differentiable in zero:
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(something bounded multiplied with something that tends to zero,
tends to zero), so πΉ is differentiable in zero. But we must also have
πΉ,(0) = π(0), equality which is also met, so π has primitives.
We have:
where π(π₯) =1
2 and:
and the functions π and β are primitives.
(ππ1 ) a) π(π₯) = [π₯] β π₯ doesnβt have primitives on β
because in points π₯0 = π is discontinuous and both lateral limits are
finite.
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doesnβt have primitives because πππ
π₯ β 0π₯ > 0
π(π₯) = ββ.
(ππ2) The functions in which this method is used in the
manual are of the form:
with π and β, continuous functions, π β β. We will prove on this
general case that π doesnβt have Darbouxβs property. The
demonstration can be adapted to any particular case.
We must prove there exists an interval whose image through
π isnβt an interval. We use the two conditions from the hypothesis.
Solving the exercise: We must prove there exists an interval
whose image through π isnβt an interval, using the two conditions
from the hypothesis.
From the condition π β β we deduce there exists at least
one point π₯0 in which π(π₯0) β β(π₯0). Then, for ν small enough,
the intervals (π(π₯0) β ν, π(π₯0) + ν) and (β(π₯0) β ν, β(π₯0) + ν)
are disjunctive (see the Figure below).
C. Dumitrescu β F. Smarandache
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From the conditions π and β continuous in π₯0 we deduce:
Considering, as we said ν small enough for the intervals
(π(π₯0) β ν, π(π₯0) + ν) and (β(π₯0) β ν, β(π₯0) + ν) to be
disjunctive, and πΏ = min {πΏ1, πΏ2}, it follows that the image through
the function π of the interval (π₯0 β πΏ, π₯0 + πΏ) isnβt an interval,
because:
(ππ3) This method presupposes the same steps as in
method πΈπ3.
(a) we look for primitives for πβ²π branches and we obtain a
first expression of πΉ;
(b) we state the condition that πΉ be continuous in the
connection point (points) between the branches;
(c) we state the condition that πΉ be differentiable in these
points:
- if πΉ is differentiable and πΉ,(π₯0) = π(π₯0),
we have covered method πΈπ3.
- if πΉ isnβt differentiable in π₯0 or πΉ,(π₯0) β
π(π₯0), we have covered method (ππ3).
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doesnβt have primitives on β.
Indeed, we have π(π₯) = π(π₯) + β(π₯) with:
and π is continuous so it has primitives and β has lateral limits,
finite in zero, so it doesnβt have primitives. If, letβs say by absurd, π
had primitives, it would follow that β(π₯) = π(π₯) β π(π₯) has
primitives.
Exercises I. Using the method (πΈπ2) , show that the following
functions have primitives in β:
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II. Using methods (πΈπ3) β (ππ3) , study if the following
functions have primitives:
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if π: β β β and if it satisfies: π(π₯) = 0 <=> π₯ = 0.
10. The product from a function β that admits primitives
and a function π differentiable with the continuous derivative is a
function that admits primitives.
Indications: If π» is the primitive of β, then π». π, is continuous,
so it admits primitives. Let πΊ be one of its primitives. We have
(π». π β πΊ), = β. π.
π being a differentiable function with a continuous derivative on β.
III. Using method (ππ1), show that the following functions
donβt have primitives:
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IV. Using method (ππ2), show that the following functions
donβt have primitives:
Answer:
1. The functions π(π₯) = 3π₯ and β(π₯) = 2π₯2 + 1 are
continuous on β and π β β . We will show that π doesnβt have
Darbouxβs property. Let π₯0 be a point in which we have (π₯0) =
β(π₯0 ), for example π₯0 = 2. We have π(2) = 6, β(2) = 9.
β β ππππ‘πππ’ππ’π ππ π₯0 = 2 <=>
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π β ππππ‘πππ’ππ’π ππ π₯0 = 2 <=>
Let there be ν so that the intervals:
are disjunctive. For example ν = 1 and let there be πΏ =
min (πΏ1, πΏ2). Then the image of the interval (2 β πΏ, 2 + πΏ) through
π isnβt an interval because for:
we have π(π₯) = 3π₯ β (5,7) and for:
we have π(π₯) = 2π₯2 + 1 β (8,10).
V. Using methods (πΈπ4) β (ππ4) , study if the following
functions admit primitives:
Indication:
so π(π₯) = π(π₯) + β(π₯), with:
C. Dumitrescu β F. Smarandache
220
Indication:
We obtain π = 0.
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Logical Scheme for Solving Problems Solving problems that require studying if a function has
primitives can be done according to the following logical scheme:
C. Dumitrescu β 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 π₯ tends to π₯0,
2. the limit when π tends to infinity,
and a new category of limits is introduced:
3. the limit when the divisionβs 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 0π₯.
Given a bounded function and (for now) non-negative
π: [π, π] β β we can approximate the desired area with the help of
certain rectangles having the base on 0π₯.
For this, we consider the points:
The set of these points form a division of the interval [π, π]
and they are noted with β:
The vertical strips made using the points π₯π have an area as
hard to calculate as the initial area, because of the superior outline.
We obtain rectangles if we replace the superior outlines with
horizontal segments.
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In order that the obtained approximation be reasonable, it is
only natural that these horizontal segments meet the graphic of the
function. We observe that such a horizontal is uniquely determined
by a point ππ situated between π₯πβ1 and π₯π . 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 π, to the division β and to the intermediary points
ππ . 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 β F. Smarandache
224
We observe that the rectangles βincrease in numberβ if and
only if π tends to infinity, but making π tend to infinity we do not
obtain a βbetter approximationβ. 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 π tend to
infinity), the approximation remains coarse. The approximation becomes
much βbetterβ if the rectangles are βthinner and thinnerβ.
In order to concretize this intuitive intuition, we observe that
the rectangles are thinner and thinner if the βthickest of themβ
becomes thinner and thinner.
The biggest thickness of the rectangles determined by a
division β is called the norm of the division β.
Therefore, βbetter approximationsβ <=> βthinner and
thinner rectanglesβ <=> βthe biggest thickness tends to zeroβ <=>
βthe norm of β tends to zeroβ <=> ||β|| β 0.
The area of the stretch situated between the graphic of the
function π and the axis 0π₯ is, therefore:
This limit is noted with β« π(π₯)π
πππ₯ and it is called the
integral of function π on the interval [π, π].
For functions that are not necessarily non-negative on [π, π],
the integral is represented by the difference between the area
situated above the axis 0π₯ and the area situated below the axis 0π₯.
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Consequently, in order to obtain the area of the stretch
situated between the graphic and the axis 0π₯, in such a case, we
have to take into consideration the module of π:
Coming back to the affirmations (π) and (π) from above,
we observe that: βthe rectangles are thinner and thinnerβ => βthe
rectangles increase in numberβ, namely (π) => (π).
In other words, ||β|| β 0 = π β β.
Reciprocally, this statement isnβt always true. That is why we
cannot renounce ||β|| β π in favor of π β β . Nevertheless,
there is a case where the reciprocal implication is true: when the
points of the division are equidistant.
π β β => ||β|| β 0 for equidistant points.
C. Dumitrescu β F. Smarandache
226
In this case, the rectangles have the same dimension of the
base, namely π₯π β π₯πβ1 =πβπ
π for any π = 1,2, β¦ , π , and the
Riemann sum becomes:
For divisions such as these, the following equivalence takes
place:
||β|| β 0 <=> π β β
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 sequencesβ limits.
Before we enumerate and exemplify the methods for the
study of integration, letβs observe that, if the function is
continuous, among the Riemann sums πβ(π, π) that can be
obtained by modifying just the heights of the rectangles (modifying
just the intermediary points ππ), there exists a biggest and smallest
Riemann sum, namely the Riemann sum having the highest value is
obtained by choosing the points ππ β [π₯πβ1, π₯π] for which:
and the Riemann sum having the smallest value is obtained by
choosing the points ππ for which:
By noting with πβ(π) and π β(π), respectively, these sums (called
the superior Darboux sum, the inferior Darboux sum, respectively),
we have:
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The points ππ that realizes the supreme, respectively, the
infimum of π on the intervals [π₯πβ1, π₯π] 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 π1 β π7 enunciated in the previous chapter, with the
following:
π8) any function integrable on [π, π] is bounded,
π9) any function monotonous on [π, π] is integrable,
π10) any function continuous on [π, π] is integrable,
π11) any function that has on [π, π] a finite number of
discontinuity points of first order is integrable.
The demonstration of π10 follows from the fact that if we
modify the values of a function integrable on [π, π], 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 β F. Smarandache
228
and the table containing implications between all the notions that
we have dealt with is the following ( π1 and π2 being one of the
lateral limits):
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Methods to show that a function is
integrable (Riemann) on [π, π] (πΌ1) using the definition;
(πΌ2) we show that it is continuous;
(πΌ3) we show that it has a finite number of discontinuity
points of first order;
(πΌ4) we show it is monotonic;
(πΌ5) we show that the function is a sum of integrable
functions.
Methods to show that a function is not
integrable (Riemann) on [π, π] (π1) using the definition (we show that πβ(π, π) doesnβt
have a finite limit for ||β|| β 0);
(π2) we show that the function is not bounded;
(π3) we show that the function is the sum of an integrable
function and a non-integrable function.
Examples (πΌ1) (a) Using the definition of the integral, show that any
differentiable function with the derivative bounded on [π, π] is
integrable on [π, π].
Solution: Let π: [π, π] β β be differentiable. We have to
show that πβ(π, π) has a finite limit when ||β|| β 0 . The given
function being differentiable, it is continuous, so it has primitives.
Let πΉ be one of its primitives. We can apply Lagrangeβs theorem to
function πΉ , on any interval [π₯πβ1, π₯π]. There exists therefore ππ β
[π₯πβ1, π₯π] so that:
C. Dumitrescu β F. Smarandache
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It follows that:
We calculate the two sums separately:
We show that the second sum tends to zero when ||β|| β 0.
For this, we take into consideration that π , is bounded, in other
words, π > 0 exists, so that |π ,(π₯) < π| for any π₯ from [π, π] and
we apply the theorem of Lagrange to the function π, but on the
intervals of extremities ππ and ππ . There exists ππ between ππ and ππ
so that:
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231
It follows therefore that:
For the second sum, we therefore have:
The condition from this exerciseβs 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.
(πΌ1) (b) Show that π) π₯=sin π₯ is integrable on any interval
[π, π]. (Manual).
Solution: We will adapt the previous demonstration. The
given function is differentiable and has the derivative π ,(π₯) = cos π₯
bounded. Being differentiable, it is continuous and it has primitives.
Let πΉ be one of its primitives, πΉ(π₯) = β cos π₯ and let:
C. Dumitrescu β F. Smarandache
232
be a division of [π, π] . We apply Langrangeβs theorem to the
function πΉ on each interval [π₯πβ1, π₯π]. There exists therefore ππ β
[π₯πβ1, π₯π] so that:
The function πΉ has a bounded derivative: | cos π₯| < 1 for
any π₯ from β , so for any π₯ from [π, π] . We have to show that
πβ(π, π) has a finite limit when ||β|| β 0. We will show that this
limit is (as it is known) πΉ(π) β πΉ(π) = β cos π + cos π .
We have:
Applying the theorem of Lagrange to the function π(π₯) =
sin π₯, on the interval determined by the points ππ and ππ , we deduce
there exists ππ between ππ and ππ so that:
Methods of Solving Calculus Problems for UNM-Gallup Students
233
(πΌ2) The function π: [0,3] β β defined by:
is integrable. Indeed, it is continuous in any point π₯ β 2 from the
definition domain, being expressed through continuous functions.
We study the continuity in π₯ = 2:
The function is continuous on the interval [0,3], so it is integrable.
(πΌ3) π(π₯) = π₯ β [π₯], π: [0, 2β3] β β can be discontinuous
just in the points where [π₯] is discontinuous. These are of the form
C. Dumitrescu β F. Smarandache
234
π₯ = π, π β β€. From these, only points 0,1,2,3, are in the functionβs
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.
(πΌ4) For the function π(π₯) = [π₯] we can say it is integrable
either by using the previous criterium, either using the fact that it is
monotonic on any interval [π, π].
(πΌ5) The function π: [β1,1] β β through π(π₯) = π ππ π₯ +
|π₯| + [π₯] is integrable because it is the sum of three integrable
functions: π1(π₯) = π ππ π₯ is integrable because it has a finite
number of discontinuity points (a single point, π₯ = 0 ) and the
discontinuity is of the first order; the function π2(π₯) = |π₯| is
integrable because it is continuous, and the function π3(π₯) = [π₯] is
integrable because it is monotonic (or because it has a single f
discontinuity point, of the first order).
isnβt integrable because, by choosing in the Riemann sum:
the intermediary points ππ , rational, we have: π(ππ) = 0, so in this
case:
and if ππ are irrational, we have:
Methods of Solving Calculus Problems for UNM-Gallup Students
235
isnβt integrable (Riemann) on [0,1] because it is not bounded.
is the sum of the functions:
If π would be integrable, from π = π1 + π2 it would follow
that π2 = π β π1 so π2 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 β F. Smarandache
236
8. The characteristic function of the set [0,1] β β.
III. For which values of the parameters π and π 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. ππ(0) = βπ
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 π β (0,1) the function has a limited
number of discontinuity points of first order.
4. The function ππ₯πππ 1
π₯ comes from the derivation of
π₯2ππ₯ sin1
π₯.
IV. 1. Show that any continuous function on an interval
[π, π] 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 β 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 π’π , π£π β [π, π] so
that ππ = π(π’π), ππ = π(π£π).
Therefore:
In order to evaluate the difference π(π’π) β π(π£π), we use
the property:
(2) a continuous function on a closed and bounded interval is
uniformly continuous (see Chapter πΌπ), namely:
In order to replace π₯1 with π’π and π₯2 with π£π , we have to
consider a division β having a smaller norm than πΏ (which is
possible because we are making the limit for ||β|| β 0). Therefore,
assuming ||β|| < πΏ we have:
i.e. ππ β ππ < ν and so:
Because ν 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 πβ(π) β π β(π) tends to zero when ||β||
tends to zero. We have:
where:
because on the interval [0,1] the function π(π₯) = π₯2is increasing.
So:
The function π(π₯) = π₯2 is continuous on [0,1] so it is
uniformly continuous, namely:
Considering the division β so that ||β|| < πΏ (possibly
because ||β|| β 0), we have π₯π2 β π₯πβ1
2 < ν, so:
Because ν is arbitrarily small, we deduce that:
i.e. π(π₯) = π₯2 is integrable on [0,1].
C. Dumitrescu β F. Smarandache
240
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In this book, we discuss a succession of methods encountered in the study of Calculus I, II and III to students and instructors at the University of New Mexico - Gallup, to higher education entry examination candidates, to all those interested, in order to allow them to reduce as many diverse problems as possible to already known work schemes. We tried to present in a methodical manner, advantageous for the reader, the most frequent calculation methods encountered in the study of the calculus at this level.
ISBN 9781599735237