Mathematical Analysis Course - ucv.ro · CHAPTER 1 Notions of set theory 1.1. Sets The notion of...

Mathematical Analysis Course

Contents

Chapter 1. Notions of set theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1. Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2. Binary relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.3. Real and extended real numbers sets. Intervals . . . . . . . . . . . . . . . . 51.4. Neighbourhoods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.5. Interior point of a set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81.6. Cluster point of a set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81.7. Isolated point of a set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.8. Adherent point of a set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.9. Boundary point of a set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.10. Real functions of a real variable . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

Chapter 2. Real sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132.1. Definitions and notations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132.2. Monotone sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142.3. Bounded sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152.4. Subsequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162.5. Convergent sequences. Limit point of a sequence . . . . . . . . . . . . . . 162.6. Convergence tests . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192.7. Existence results for limit of a sequence . . . . . . . . . . . . . . . . . . . . . . . 202.8. Operations with sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222.9. Study of undeterminate forms of limits . . . . . . . . . . . . . . . . . . . . . . . . 262.10. Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

Chapter 3. Real series. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 393.1. Definitions and notations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 393.2. Convergence tests for series with positive terms . . . . . . . . . . . . . . . 413.3. Convergence tests for series with real terms . . . . . . . . . . . . . . . . . . . 443.4. Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

Chapter 4. Taylor series. Taylor expansions . . . . . . . . . . . . . . . . . . 514.1. Taylor series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 514.2. Taylor’s formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 544.3. Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

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4 CONTENTS

Chapter 5. Continuous Functions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 635.1. The Euclidean structure of Rp . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 635.2. Vector sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 665.3. Continuity of real functions of vector variable . . . . . . . . . . . . . . . . . 685.4. Limits of real functions of vector variable . . . . . . . . . . . . . . . . . . . . . 695.5. Continuity of vector functions of vector variable . . . . . . . . . . . . . . . 715.6. Limits of vector functions of vector variable . . . . . . . . . . . . . . . . . . . 725.7. Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73

Chapter 6. Differentiable functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 776.1. Differentiability of real functions of real variable . . . . . . . . . . . . . . 77

6.1.1. The derivatives of certain elementary functions 806.1.2. Geometrical meaning of the derivative . . . . . . . . . . . . . . . . 816.1.3. Semitangents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 826.1.4. Angular point. Sharp point . . . . . . . . . . . . . . . . . . . . . . . . . . . 836.1.5. Operations with differentiable functions . . . . . . . . . . . . . . . 83

6.2. Derivatives of higher order . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 866.2.1. Local Extrema . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 866.2.2. L’Hospital Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90

6.3. Differentiability of vector functions of a real variable . . . . . . . . . . 936.4. Differentiability of real functions of vector variable . . . . . . . . . . . . 93

6.4.1. Local Extrema . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1026.5. Differentiability of vector functions of vector variables . . . . . . . . . 1056.6. Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1076.7. Representation of the grapph of a function . . . . . . . . . . . . . . . . . . . . 1176.8. Demonstrarea unor inegalitati . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1186.9. Studiul ecuatiilor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119

6.9.1. Ecuatii de tipul f (x) = g (x) . . . . . . . . . . . . . . . . . . . . . . . . . . 1196.9.2. Sirul lui Rolle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120

Chapter 7. Integrals. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1217.1. Definite Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121

7.1.1. Integration by parts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1227.1.2. Changing variable first method in undefinite integral . . 1237.1.3. Changing variable second method in undefinite integral1247.1.4. Integration of rational functions. . . . . . . . . . . . . . . . . . . . . . . 1257.1.5. Integration of irrational functions . . . . . . . . . . . . . . . . . . . . . 1277.1.6. Integration of trigonometric functions . . . . . . . . . . . . . . . . . 129

7.2. Definite Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1307.2.1. Integration by parts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1327.2.2. Changing variable first method in definite integral . . . . 1327.2.3. Changing method second variable in definite integral . . 133

7.3. Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133

CONTENTS 5

Chapter 8. Improper Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1418.1. Integrals on unbounded intervals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1418.2. Integrals of unbounded functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1438.3. Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144

Chapter 9. Function Sequences and Function Series . . . . . . . . . . 1499.1. Function sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1499.2. Function Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1509.3. Power Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1519.4. Analytical functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1529.5. Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153

Chapter 10. Partial derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15710.1. Linear spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15710.2. Linear and continuous operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15910.3. Vector-valued functions of a real variable . . . . . . . . . . . . . . . . . . . . 16010.4. Paths and curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16110.5. Real-valued functions of vector variable . . . . . . . . . . . . . . . . . . . . . . 16310.6. Partial derivatives. The differential of a function . . . . . . . . . . . . . 16410.7. Extrema . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16610.8. Vector-value functions of vector variables . . . . . . . . . . . . . . . . . . . . 16910.9. Implicit Functions Theorem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17110.10. Conditioned Extrema . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17310.11. Changing of variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17410.12. Surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18110.13. Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184

Chapter 11. Integrals with parameter . . . . . . . . . . . . . . . . . . . . . . . . 19111.1. Theorem of differentiability under the integral . . . . . . . . . . . . . . . 19111.2. Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193

Chapter 12. Double integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19712.1. Definitions and calculus methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19712.2. Properties of the double integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19812.3. Changing of variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19912.4. Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201

Chapter 13. Triple Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20713.1. Definitions and calculus methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20713.2. Properties of the triple integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20813.3. Changing of variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20913.4. Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211

Chapter 14. Curvilinear Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215

6 CONTENTS

14.1. Curvilinear Integrals of First Kind . . . . . . . . . . . . . . . . . . . . . . . . . . . 21514.1.1. Definitions and calculus methods . . . . . . . . . . . . . . . . . . . . 21514.1.2. Properties of the curvilinear integral of first kind. . . . . 216

14.2. Curvilinea integral of second kind . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21814.2.1. Definitions and calculus methods . . . . . . . . . . . . . . . . . . . . 21814.2.2. Properties of the curvilinear integral of second kind . . 21814.2.3. Independence of the curvilinear integral of second

order with respect to the path. . . . . . . . . . . . . . . . . . . . . . . . 21914.3. Green-Riemann Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22114.4. Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 221

Chapter 15. Surface Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22715.1. Surface Integrals of First Kind . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 227

15.1.1. Definitions and calculus methods . . . . . . . . . . . . . . . . . . . . 22715.1.2. Properties of the curvilinear integral of first kind. . . . . 228

15.2. Surface integral of second kind . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23115.2.1. Definitions and calculus methods . . . . . . . . . . . . . . . . . . . . 23115.2.2. Properties of the surface integral of second kind. . . . . . 233

15.3. Integral Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23315.3.1. Gauss-Ostrogradski Theorem . . . . . . . . . . . . . . . . . . . . . . . . 23415.3.2. Stokes’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 235

15.4. Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 236

Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 241

CHAPTER 1

Notions of set theory

1.1. Sets

The notion of the set, fundamental in mathematics, represents a collectionof different objects. These objects are called elements of the set.

For example, one can consider the set of all students of a group. Thenthe students are the elements of this set. If one considers the set of all primenumbers, then 2, 3, 5, and 11 are elements of this set.

The sets are conventionally denoted with capital letters: A, B, C, M, Xetc., while the elements of a set are usually denoted by small letters: a, b, c,m, x etc.

Definition 1.1. The set containing no elements is called the empty set andit is denoted by ∅.

By convention, particular symbols are reserved for the most important setsof numbers: N is the set of the natural numbers, Z is the set of the integers,Q is the set of the rational numbers, R is the set of the real numbers, C is theset of the complex numbers.

A set can be defined by two methods.1. By enumerating its elements. When we define a set by enumerating

its elements, we will enclose the list of the elements in curly brackets; e.g.{10, 30, 50, 70, 90} is the set whose elements are 10, 30, 50, 70, and 90.

2. By specifying the properties of each element of it. For example, if Mis the set of the natural numbers having the remainder 2 at the division by 3,then one defines M through

M = {x ∈ N, x mod 3 = 2}.Anyway, one may enumerate only a few elements of M , by writting M ={2, 5, 8, 11, 14, ...} .

Definition 1.2. Two sets A and B are called equal (respectively different)iff they contain (respectively do not contain) exactly the same elements. If thesets A and B are equal, then we write A = B, and if A and B are different,then we write A = B.

1

2 1. NOTIONS OF SET THEORY

For example, if A = {1, 3, 9, 27, 81} and B = {x ∈ N, x < 100, x is a powerof 3}, then A = B.

Definition 1.3. We say that an element x belongs (respectively does notbelong) to the set A, and we write x ∈ A (respectively x /∈ A) iff the elementx is (respectively is not) in the set A.

For example, if A is the set of prime numbers then 11 ∈ A and 12 /∈ A.

Definition 1.4. Let A and B be two sets. The union of the sets A and B,denoted A ∪ B, is the set of all elements belonging to A or B, i.e. A ∪ B ={x, |x ∈ A or x ∈ B} .

For example, if A = {1, 2, 4, 6, 8} and B = {4, 8, 12, 16, 20} , then A ∪ B ={1, 2, 4, 6, 8, 12, 16, 20} .

Definition 1.5. Let A and B be two sets. The intersection of the sets Aand B, denoted A ∩ B, is the set of all elements common to A and B, i.e.A ∩B = {x, x ∈ A and x ∈ B} .

For example, if A = {1, 2, 4, 6, 8} and B = {4, 8, 12, 16, 20} , then A ∩ B ={4, 8} .

Definition 1.6. Two sets having empty intersection are called disjunctive.

For example, the sets A = {x ∈ N, x even} and B = {x ∈ N, x odd} aredisjunctive sets.

Definition 1.7. Let A and B be two sets. We say that A is a subset of B(or the set A is enclosed in B, or the set B includes the set A) and we writeA ⊆ B or B ⊇ A, iff each element of A belongs to B, i.e. ∀x ∈ A, it followsx ∈ B.

In order to prove that two sets A and B are equal, it is sufficient to provethat A ⊆ B and B ⊆ A.

Definition 1.8. The set difference of set B from set A, denoted by A\B, isthe set of all elements of A which are not in B, i.e.

A\B = {x, x ∈ A and x /∈ B} .

For example, if A = {1, 2, 4, 6, 8} and B = {4, 8, 12, 16, 20} , then A\B ={1, 2, 6} and B\A = {12, 16, 20} .

1.2. BINARY RELATIONS 3

Definition 1.9. The symmetric difference of sets A and B, denoted bydenoted by A∆B is the set of elements which are in either of the sets and notin their intersection, i.e.

A∆B = (A\B) ∪ (B\A) .

For exampe, if A = {1, 2, 4, 6, 8} and B = {4, 8, 12, 16, 20} , then A∆B ={1, 2, 6, 12, 16, 20}.

Definition 1.10. If A is a subset of E, then the complement of a A, denotedCEA, is the set of all elements in E that are not in A.

i.e.

CEA = E\A = {x, x ∈ E and x /∈ A} .

Fo example if A = {4, 8} and E = {1, 2, 4, 6, 8} one has CEA = {1, 2, 6} .In the case of an universal set E, without the risk of confusion, we will

denote by CA the complement of A.

1.2. Binary relations

Definition 1.11. The cartesian product of two nonempty sets A and B is theset

A×B = {(x, y) , x ∈ A and y ∈ B} .

For example, if A = {1, 2, 3} and B = {1, 2}, thenA×B = {(1, 1) , (2, 1) , (3, 1) , (1, 2) , (2, 2) , (3, 2)} .

Definition 1.12. Every subset r of the cartesian product A×B is called binaryrelation from A to B; if (x, y) ∈ r, we will denote this xry, by reading “x isin relation r with y”.

For example, if A is a nonempty set and P (A) is the set of all subsets ofit, one can consider the binary relation r ⊂ P (A)× P (A) , defined by

M, N ∈ P (A) , MrN iff (∀x ∈M) =⇒ (x ∈ N) .

Let A be an arbitrary nonempty set.

Definition 1.13. A binary relation r ⊂ A×A having the properties:1) ∀x ∈ A, xrx (reflexivity);2) ∀x, y ∈ A, if xry and yrx,then x = y (antisymmetry);3) ∀x, y, z ∈ A, if xry and yrz then xrz (transitivity)

is called order relation.


In this case, A is called ordered set with respect to the relation r and wedenote this by (A, r) .

For example, the previous relation is an order relation, the set (A, r)being ordered.

Definition 1.14. A binary relation r ⊂ A×A having the properties:1) ∀x ∈ A, xrx (reflexivity);2) ∀x, y ∈ A, if xry then yrx (symmetry);3) ∀x, y, z ∈ A, if xry and yrz, then xrz (transitivity)

is called equivalence relation.

For example, if A este is a nonempty set and P (A) is the set of all subsetsof it, one can consider the binary relation r ⊂ P (A)×P (A) , defined by MrNiff M = N . Then r is an equivalence relation.

Definition 1.15. An order relation is called totally orderd if it fulfills thecondition

∀x, y ∈ A, xry or yrx.

For example, if A = {1, 2, 3, 4}, then the binary relation r ⊂ A×A definedby xry iff x divides y, is an order relation, but it is not total, since 3 does notdivide 4 and 4 does not divide 3. The same relation defined on A = {1, 2, 4} istotally ordered.

Let (A,≤) be an oderered set and B ⊂ A arbitary.

Definition 1.16. 1) Upper bound of the set B is any element x∗ ∈ A havingthe property

z ≤ x∗, ∀z ∈ B.

If x∗ ∈ B, then it is called the greatest element of B (if there exists one,then it is unique) and we denote it by x∗ = maxB.

If the set of all upper bounds of B has a smallest element, then it is calledsupremum of B and it is denoted by supB.

2) Lower bound of the set B is any element x∗ ∈ A having the property

x∗ ≤ z, ∀z ∈ B.

If x∗ ∈ B, then it is called the smallest element of B (if there existsone, then it is unique) and we denote it by x∗ = minB.

If the set of all lower bounds of B has a greatest element, then it is calledinfimum of B and it is denoted by inf B.

1.3. REAL AND EXTENDED REAL NUMBERS SETS. INTERVALS 5

3) If the set B has upper bounds (respectively lower bounds), then B iscalled bounded above (respectively bounded below). If the set B is boundedabove and below, then it is called bounded.

3) If supB and inf B do exist for each bounded set B, we say that (A,≤)is a complete (in order) set.

Remark 1.1. If the supremum or the infimum of a set exists, then it is unique.

Remark 1.2. The supremum or the infimum of a set may not belong to theset.

For example, let us consider B = {1, 2, 3, 4, 5, 6, 7, 8, 9} ⊂ N and the binaryrelation ≤ ⊂ B ×B, defined by: x ≤ y iff x divides y.

Then, ≤ is an order relation on B, but still not on Z∗ (since x divides −xand −x divides x does not impliy x = −x).

With respect to ≤ the set B ⊂ N admits upper bounds, since the leastcommon multiple x∗ := l.c.m. {1, 2, 3, 4, 5, 6, 7, 8, 9} /∈ B, with z ≤ x, ∀z ∈ B.This number is also supB and B is bounded above. The set B does not admita greatest element, since x∗ ∈ B.

With respect to ≤ the set B ⊂ N admits lower bounds, since the greatestcommon divisor x∗ := g.c.d. {1, 2, 3, 4, 5, 6, 7, 8, 9} = 1 ∈ B. This number isalso inf B, minB, and the set B is bounded below.

Axiom 1.1. (Completeness Axiom) Every nonempty and bounded above subsetof R admits supremum.

1.3. Real and extended real numbers sets. Intervals

Definition 1.17. The modulus (absolute value) of a real number x is thenumber

|x| :={x, if x ≥ 0,−x, if x < 0.

The modulus has the following properties:1) |x| ≥ 0, ∀x ∈ R; |x| = 0 if and only if x = 0;2) |x · y| = |x| · |y| , ∀x, y ∈ R;3) |x+ y| ≤ |x|+ |y| , ∀x, y ∈ R; (triangle inequality)4) − |x| ≤ x ≤ |x| , ∀x ∈ R;5) |x| ≤ a if and only if −a ≤ x ≤ a;6) |x| ≥ a > 0 if and only if x ≤ −a or x ≥ a;7) |∑n

i=1 xi| ≤∑n

i=1 |xi| , ∀x1, x2, ..., xn ∈ R;


8) ||x| − |y|| ≤ |x− y| , ∀x, y ∈ R;9) |x · y| = |x| · |y| , ∀x, y ∈ R;10)

∣∣∣xy ∣∣∣ = |x||y| , ∀x ∈ R, ∀y ∈ R\ {0};

11) |x|2k = x2k, ∀x ∈ R, ∀k ∈ N.Due to properties 1)-3), the mapping |·| : R → R is called Euclidean

norm and the function d : R× R → R,d (x, y) = |x− y|

fulfills the following properties:D1) d (x, y) ≥ 0, ∀x, y ∈ R, and d (x, y) = 0 if and only if x = y;D2) d(x, y) = d (y, x) , ∀x, y ∈ R;D3) d (x, y) ≤ d (x, z) + d (z, y) , ∀x, y, z ∈ R.

Definition 1.18. The mapping d is called Euclidean metric, d (x, y) is calledthe distance between x and y, and we say that (R, d) is a metric space.

We define to symbols, −∞ (minus infinity) and ∞ (infinity), by extendingthe set R by

−∞ < x, ∀x ∈ R,x < ∞, ∀x ∈ R.

Then the extended real numbers set is defined as R := R∪{−∞,∞} .The main property of R is that every nonempty subset of R admits supre-

mum and infimum in R. If A is a bounded above subset of R, then the exis-tence of the supremum follows by Axiom 1.1, while those subsets A that arenot bounded above necessarily has supA = ∞. Hence the fact that a subset Aof R is not bounded above in R, can be pointed out through supA = ∞. Thealgebraic structure of R can be extended, by defining the following operationswith infinite elements.

x+ (−∞) = (−∞) + x = −∞, ∀x ∈ R,(−∞) + (−∞) = −∞

x+∞ = ∞+ x = ∞, ∀x ∈ R,∞+∞ = ∞,

x · (−∞) = (−∞) · x =

{∞, if x ∈ R, x < 0,−∞, if x ∈ R, x > 0,

x · ∞ = ∞ · x =

{−∞, if x ∈ R, x < 0,∞, if x ∈ R, x > 0.

The following operations are not defined: ∞−∞, (−∞)+∞, (−∞)·0,0 · (−∞) , 0 · ∞, ∞ · 0, 0

0 ,∞∞ , 1

∞, ∞0, 00.

1.4. NEIGHBOURHOODS 7

Let a, b ∈ R with a < b. One can define the following subsets of R, calledintervals:

(a, b) : = {x ∈ R, a < x < b} , [a, b] := {x ∈ R, a ≤ x ≤ b}[a, b) : = {x ∈ R, a ≤ x < b} , (a, b] := {x ∈ R, a < x ≤ b}

representing segments (with or without boundaries),

(a,∞) : = {x ∈ R, a < x} , [a,∞) = {x ∈ R, a ≤ x}(−∞, a) : = {x ∈ R, x < a} , (−∞, a] := {x ∈ R, x ≤ a} ,

representing semi-lines (with or without origin) and

(−∞,∞) := R, [−∞,∞] := R,

representing the real line and the extended real line.

Definition 1.19. The set A ⊂ R is bounded iff there exists M ≥ 0 such that|x| ≤ M , ∀x ∈ A or, equivalently, there exists an interval (a, b) , such thatA ⊂ (a, b) .

For example, the set

A ={

n√2, n ∈ N\ {0, 1}

}is bounded, since A ⊂ (1, 2) .

1.4. Neighbourhoods

Definition 1.20. Let x ∈ R. Each open interval containing x is called neigh-bourhood of x. Each neighbourhood of x ∈ R of type (x− ε, x+ ε), whereε > 0 is called centered neighbourhood.

For example, (−1, 4) is a neighbourhood of 0.Clearly, every x ∈ R admits an infinity of neighbourhoods.

Theorem 1.1. Every neighbourhood of x contains a centered neighbourhood ofx and conversely.

Definition 1.21. Each interval of type (a,∞) , unde a ∈ R is called neigh-bourhood of ∞.


Denote by V (x) the family of all neighbourhoods of x, i.e.

V (x) = {V, V neighbourhood of x} .

The neighbourhoods of real numbers fulfill the following properties.1) Each x ∈ R belongs to each neighbourhood of x.2) Intersection of every two neighbourhoods of a point is still a neighbour-

hood of that point.3) For every two real numbers x, y, with x = y, there exist V1 ∈ V (x) and

V2 ∈ V (y) , such that V1 ∩ V2 = ∅.4) ∀x ∈ R, V ∈ V (x) , and ∀y ∈ V , there exists W ∈ V (y), with W ⊂ V.

1.5. Interior point of a set

Definition 1.22. The point a ∈ R is called interior point of the set A ⊆ Riff there exists V ∈ V (a) , such that V ⊂ A.

The set of all interior points of a set A is called the interior of set A andwe denote it by Int (A) .

For example, for a bounded interval A, having boundaries a and b, Int(A) = (a, b) . But every finite set A = {a1, a2, ..., an} has Int (A) = ∅.

Definition 1.23. The set A ⊆ R is called open set iff A = Int (A) .

For example, every open interval, bounded or unbounded, (a, b), (a,+∞) ,or (−∞, a), with a ∈ R is an open set. ∅, R are open sets. Set (a, b] is notopen, since Int ((a, b]) = (a, b) .

1.6. Cluster point of a set

Definition 1.24. a ∈ R is called cluster point of the nonempty set A ⊆ Riff in each neighbourhood of a one finds at least an point of A\{a}.

The set of all cluster points of a set A is called the derivative set of Aand it is denoted by A′.

That is, a ∈ A′ if and only if ∀V ∈ V (a), (V \ {a}) ∩A = ∅.For example, if A = [0, 1] , then A′ = [0, 1]; if A = (0, 1], then A′ = [0, 1] ;

N′ = {∞} ; if A = {0, 1}, then A′ = ∅; if A ={1, 12 , ...,

1n , ...

}, then A′ = ∅.

Remark that a cluster point of a set may belong or not to that set. Forexample, if A = (0, 1) , then 0 ∈ A′\A and 1 ∈ A′\A.

1.9. BOUNDARY POINT OF A SET 9

1.7. Isolated point of a set

Definition 1.25. a ∈ R is called isolated point of the nonempty set A ⊆ Riff there exists a neighbourhood of a, that does not contain any point of A\{a}.

That is, a is isolated point of the set A if and only if ∃V ∈ V (a), such that(V \ {a}) ∩A = ∅.

For example, for A = (0, 1)∪{2}, 2 is isolated, while 0 is not isolated point,it is cluster point.

Remark that for a nonempty set A ⊆ R, every a ∈ A may be cluster pointof isolated point of the set A.

1.8. Adherent point of a set

Definition 1.26. Let A ⊂ R be a nonempty set. a ∈ R is called adherentpoint of set A iff in each neighbnourhood of a se there exists at least an elementof A.

The set of all adherent points of set A is called the closure (adherence)of A and it is denoted by A.

That is, a ∈ A if and only if ∀V ∈ V (a), V ∩A = ∅.For example, if A = [0, 1] , then A = [0, 1]; if A = (0, 1], then A = [0, 1] ;

Q = R; if A = {0, 1}, then A = {0, 1} .Remark that an adherent point of a set may belong or not to that set.

For example, if A = (0, 1) , it follows that 0 ∈ A\A and 1 ∈ A\A; if A ={1, 12 , ...,

1n , ...

}, then A = A ∪ {0} .

Definition 1.27. Set A ⊆ R is called closed set iff CA is open.

Proposition 1.1. A set is closed if and only if is equal to its closure.

For example, [a, b] este is a closed set, while (a, b] is not closed, its closurebeing [a, b] . R and ∅ are closed sets.

1.9. Boundary point of a set

Definition 1.28. a ∈ R is called boundary point for the nonempty set A ⊂ Riff it is an adherent point of A or CA .

The set of all boundary points of a set A is called the boundary of A andit is denoted by Fr (A).

That iS, Fr (A) = A ∩ CA.


For example, Fr ([a, b]) = Fr ((a, b)) = Fr ((a, b]) = Fr ([a, b)) = {a, b}; ifA =

{1, 12 , ...,

1n , ...

}, then Fr (A) = A ∪ {0} .

1.10. Real functions of a real variable

Definition 1.29. A function f : A → B is called real function of a realvariable, iff A ⊆ R and B ⊆ R.

Definition 1.30. Function f : A→ B is called bounded iff f (A) a boundedset.

For example, the function f : [0, 1] → R, f (x) = x2 is bounded, becausef ([0, 1]) = [0, 1]; the function f : R → R, f (x) = x2 is not bounded, since theset f (R) = [0,∞) is unbounded.

Definition 1.31. Function f : A → B is called increasing, strictly in-creasing, decreasing or strictly decreasing on the nonempty set D ⊆ A, ifone has respectively

1. ∀x1, x2 ∈ D, x1 < x2, f (x1) ≤ f (x2) ;

2. ∀x1, x2 ∈ D, x1 < x2, f (x1) < f (x2) ;

3. ∀x1, x2 ∈ D, x1 < x2, f (x1) ≥ f (x2) ;

4. ∀x1, x2 ∈ D, x1 < x2, f (x1) > f (x2) .

In cases 1. and 3. we say that f is monotone, while in cases 2. and 4.we say that f is strictly monotone.

For example, function f : R → R, f (x) =

{x+ 1, if x ≤ 0x, if x > 0

is strictly

increasing on each of sets (−∞, 0], (0,∞) , and it is not strictly increasing norincreasing on R.

Theorem 1.2. Every strictly monotone set is injective.

The converse of this theorem doesn’t hold. For example, function f :

(−∞, 2] → [−2,∞), f (x) =

{−x, if x < 0x− 2, if x ∈ [0, 2]

is injective, but it is not

strictly monotone.

Theorem 1.3. If the function f : A → B is strictly monotone and invertible,then the function functia f−1 is strictly monotone.

1.10. REAL FUNCTIONS OF A REAL VARIABLE 11

Definition 1.32. The line x = x0 is axis of symmetry for the graph of afunction iff the symmetry of every point from the graph, with respect to the linex = x0, is still on the graph.

Fpr example, axis Oy is axis of symmetry for the graph of the functionf : R → R, f (x) = x4 + 1.

Theorem 1.4. The line x = x0 is symmetry axis for the graph of the functionf : A→ B, if and only if

f (x0 − x) = f (x0 + x) ,

for all x ∈ R with x0 − x, x0 + x ∈ A.

Definition 1.33. The function f : A→ B is called even iff:

1. ∀x ∈ A it follows −x ∈ A;2. f (−x) = f (x), ∀x ∈ A.Due to Theorem 1.4, it follows that Oy is axis of symmetry of each even

function.For example, the function f : R → R, f (x) = |x| cosx, ∀x ∈ R is an even

function and admits Oy as axis of symmetry.

Definition 1.34. The point M0 (x0, y0) is point of symmetry for the graphof a function, iff the symmetric of every point from the graph, with respect toM0, is still on the graph.

For example, the origin O is point of symmetry for the graph of the functionf : R → R, f (x) = x3 |sinx| .

Theorem 1.5. The point M0 (x0, y0) is point of symmetry for the graph of thefunction f : A→ B if and only if

f (x0 − x) + f (x0 + x) = 2y0,

for all x ∈ R with x0 − x, x0 + x ∈ A.

Definition 1.35. The function f : A→ B is called odd iff1. ∀x ∈ A it follows −x ∈ A;2. f (−x) = −f (x), ∀x ∈ A.

Due to Theorem 1.5, we get that the origin O is point of symmetry for thegraph of every even function.

For example, the function f : R → R, f (x) = x3, ∀x ∈ R is an evenfunction and admits O as point of symmetry.


Definition 1.36. The function f : A ⊂ R → R is called periodic iff thereexists T = 0 such that f (x+ T ) = f (x), ∀x ∈ A with x+ T ∈ A. The numberT is called period of function f. The smallest positive period, if it exists, iscalled least (principal) period.

If T is period of a function f : A→ R and for all x ∈ A one has x−T ∈ A,then −T is period of f, too.

For example, funcion f : R → R, f (x) = sin 2x3 , ∀x ∈ R is periodical having

least period 3π. There also exists periodic functions without least period. For

example, for the Dirichlet’s function, f : R → R, f (x) ={

1, if x ∈ Q−1, if x ∈ R\Q ,

every rational number is period and there is not least period.

Proposition 1.2. If T0 is the least period of a periodic function, then everyother period T is of form T = mT0, with m ∈ Z.

CHAPTER 2

Real sequences

2.1. Definitions and notations

Definition 2.1. A function x : N → R is called real sequence.The real number xn := f (n) , n ∈ N is called the general term (of rank

n) of the sequence x.

Usually a sequence is also denoted by (an)n≥0 .

Remark 2.1. A sequence may not be defined on the whole N. For example,since yn = 1

n exists for n = 0 and zn = 1n2−2n

exists for n = 0 and n = 2, the

sequences will be defined on subsets of N: (yn)n≥1 and (zn)n≥2 .

The sequences can be defined through different ways.1. Through a table. For example,

n 1 2 3 4 5 ...an 3 2 5 7 12 ...

2. Through a calculus formula. For example,

an =5n− 1

(−1)n (n+ 1), n ≥ 0

or

a2 = 1, an = a2

(1

2

)n−1

, n ≥ 3.

3. Through more calculus formulas. For example,

an =

{2n+ 3, if n is even,n4, if n is odd,

or

an =

n, if n = 4k,1n , if n = 4k + 1,1n3 , if n = 4k + 2,1n4 , if n = 4k + 3,

13

14 2. REAL SEQUENCES

for n ∈ N.4. Through a recurrence relation. For example,

a0 =1

2, an+1 =

√1− an, n ∈ N

or

a0 = 0, a1 = 1, an+2 = an+1 + an, n ∈ N(Fibonacci’s sequence).

2.2. Monotone sequences

Definition 2.2. The sequence an, n ∈ N is called increasing (decreasing)iff an ≤ an+1, ∀n ∈ N, (an ≥ an+1, ∀n ∈ N).

The sequence an, n ∈ N is called strictly increasing (strictly decreas-ing) iff an < an+1, ∀n ∈ N, (an > an+1, ∀n ∈ N).

In practice the monotony of a sequence can by established through differentways.

1. By using the definition. For example, the sequence an =√n

n+1 ,n ∈ N∗, is strictly decreasing for n ≥ 1, since an > an+1, ∀n ∈ N∗.

2. By estimating the difference an+1−an or an−an+1 and comparingit to 0. For example, if an = 1

12+ 1

22+ ...+ 1

n2 , n ∈ N∗, one has an+1 − an =1

(n+1)2> 0, ∀n ∈ N∗, hence (an)n∈N∗ is strictly increasing.

3. By estimating the ratio an+1

anor an

an+1and comparing it to 1.

For example, if an = (2n)!!(2n+1)!! , n ∈ N, then an+1

an= 2n+2

2n+3 < 1, ∀n ∈ N, hence(an)n∈N∗ is strictly decreasing.

4. By using mathematical induction. For example, if an =√6 + an−1,

∀n ≥ 1 and a0 = 1, one remarks that a0 < a1. If ak < ak+1 holds, thenak+1 =

√6 + ak <

√6 + ak+1 = ak+2. Therefore, the sequence is strictly

increasing.

5. By considering the sequence as restriction to N of a function

whose monotony is known. For example, the sequence an = exp(

1n2+1

),

n ∈ N, represents the restriction to N of the function f : R → R, f (x) =

exp(

1x2+1

), which is strictly decreasing on (0,+∞) .

6. Other methods. For example, if an =∑n

k=1 k · k!, n ∈ N∗, we remarkthat

an =n∑k=1

(k + 1) · k!−n∑k=1

1 · k! =n∑k=1

(k + 1)!−n∑k=1

k! = (n+ 1)!.

2.3. BOUNDED SEQUENCES 15

So, an < an+1, ∀n ∈ N∗, and the sequence is strictly increasing.

2.3. Bounded sequences

Definition 2.3. (an)n∈N is called bounded from below, iff there exists m ∈R, such that m ≤ an, ∀n ∈ N.

(an)n∈N is called bounded from above, iff there exists M ∈ R, such thatan ≤M , ∀n ∈ N.

(an)n∈N is called bounded, iff it is bounded from below and from above or,equivalently, there exists M > 0 such that |an| ≤M, ∀n ∈ N.

If the sequence is not bounded, we call it unbounded.

Obviuously, if the sequence ca an, n ∈ N is increasing, then it will bebounded from below by its first term, a0; if it is decreasing, then it will bebounded from above by its first term, a0.

For example, the sequence an = 2n2n+1 , n ∈ N is bounded, since 0 ≤ an < 1,

∀n ∈ N; the sequence bn = n2, n ∈ N (which is strictly increasing) is boundedfrom below by b0 = 0 and unbounded from above; the sequence cn = −n2 + 1,n ∈ N∗ (which is strictly decreasing) is bounded from above by c1 = 0 and itis unbounded from below; the sequence dn = (−1)n n2, n ∈ N is unbounded,being unbounded from below (and from above, too).

In practice the boundedness of a sequence can by established through dif-ferent ways.

1. By using estimates. For example, if an = 11! +

12! + ... + 1

n! , n ∈ N∗,then, obviously, 0 < an, n ∈ N∗ and

an <1

20+

1

21+

1

22+ ...+

1

2n−1=

(12

)n − 112 − 1

< 2, n ∈ N∗.

So, 0 < an < 2, ∀n ∈ N∗.If bn = − 1√

n2+1− 1√

n2+2− ...− 1√

n2+n, n ∈ N∗, then

−1 < − 1√n2 + 1

− 1√n2 + 1

− ...− 1√n2 + 1

< bn, n ∈ N∗

therefore (bn)n∈N∗ is bounded from below by −1.

2. By using the monotony. For example, for the strictly increasingsequence defined by an =

√6 + an−1, ∀n ≥ 1 and a0 = 1, one has 0 < an <√

6 + an, n ∈ N, hence a2n − an − 6 < 0, n ∈ N or an ∈ (−2, 3) , n ∈ N. Due tothe positivity, an ∈ (0, 3) , n ∈ N.

3. By using the mathematical induction. For example, if a1 =√2,

an+1 =(√

2)an

, n ∈ N∗, then a1 < 2 and, supposing an < 2, an+1 <(√

2)2

=2. Therefore, an ∈ (0, 2) , ∀n ∈ N∗.


4. By considering the sequence as restriction to N of a functionwhose boundedness is known. For example, the sequence an = cos 2nπ

n4+4,

n ∈ N is the restriction to N of the vounded function f : R → [−1, 1] , f (x) =cos 2xπ

x4+4.

5. Other methods. For example, if an = 2n2n+1 , n ∈ N, then an =

1 − 12n+1 < 1 and since an ≥ 0, n ∈ N, it follows that an ∈ (0, 1), n ∈ N; if

bn =∑n

k=11

(k+1)(k+2) , n ∈ N∗, then 0 ≤ bn =∑n

k=1

(1

k+1 − 1k+2

)= 1

2 −1

n+2 <12 , ∀n ∈ N∗ and so (bn)n∈N∗ is bounded.

2.4. Subsequences

Definition 2.4. Let (an)n∈N be a real the sequence. If (kn)n∈N is a strictlyincreasing sequence of naturals, (akn)n∈N is called subsequence of (an)n∈N.

For example, if an = 2n

n! , n ∈ N, then a subsequence of (an)n∈N is a3n+1 =23n+1

(3n+1)! , n ∈ N.By Definition 2.4, it follows that kn ≥ n, n ∈ N.

2.5. Convergent sequences. Limit point of a sequence

Definition 2.5. The sequence an, n ∈ N is called convergent iff there existsl ∈ R such that in each neighborhood V of l there are all the terms of thesequence, starting from a certain rank. The point l is called limit point of(an)n∈N. Otherwise, (an)n∈N is called divergent.

If l is limit point of (an)n∈N we also say that (an)n∈N converges (is con-vergent) to l and we denote this by an → l, as n→ ∞ or lim

n→∞an = l.

From this definition we can easily deduce that limn→∞

an = l if and only if

∀ε > 0, ∃n0 = n0 (ε) ∈ N such that ∀n ∈ N, n ≥ n0, one has

|an − l| < ε.

Exemples1. lim

n→∞1n = 0, because for every V = (−a, b) ∈ V (′), with a, b > 0, the

term 1n ∈ V if and only if n > 1

b . Therefore, if n0 =[1b

]+ 1, Definition 2.5 is

fulfilled.2. lim

n→∞13n = 0, because for every V = (−a, b) ∈ V (′), with a, b > 0, the

term 13n ∈ V if and only if 3n > 1

b . Therefore, if1b < 1, we can consider n0 = 0

and if 1b ≥ 1, n0 =

[log3

1b

]+ 1 and so Definition 2.5 is fulfilled.

2.5. CONVERGENT SEQUENCES. LIMIT POINT OF A SEQUENCE 17

3. limn→∞

n = +∞, because for every V = (a,+∞) ∈ V (+∞) , the term

n ∈ V if and only if n > a. If a < 0, we can consider n0 = 0 and if a ≥ 0,n0 = [a] + 1.

4. The sequence an = (−1)n , n ∈ N, is divergent, it only has the terms−1 and 1, and none of them could be limit point of the sequence.

5. The sequences bn = n, cn = (−1)n si dn = (−1)n · n are divergent.

Theorem 2.1. If a sequence is convergent to l, then l is unique.

Definition 2.6. Let an, n ∈ N be a real sequence. l ∈ R is called clusterpoint of (an)n∈N iff ∀ε > 0, ∀n ∈ N, ∃kn ≥ n, such that |xkn − l| < ε.

Obviously, every limit point of a sequence is also a cluster point of it. TheCesaro’s Lemma asserts that l ∈ R is a cluster point of a sequence if and onlyif a subsequence converges to l. The converse of this fact is not true, e.g., thesequence an = (−1)n , ∀n ∈ N has −1, 1 cluster points, but has none limitpoint.

Remark 2.2. A convergent sequence has a single cluster point, i.e. the limitof the sequence.

Theorem 2.2. If a sequence has limit, then each subsequence of it has thesame limit.

Let an, n ∈ N be a real sequence. We denote by γ (an) ⊂ R the set of allcluster points of (an)n∈N . Due to the Cesaro’s Lemma,

γ (an) ={l ∈ R, ∃akn subsequence of an, l = lim

n→∞akn

}.

previous One can prove that if (an)n∈N is bounded, then γ (an) is nonempty.

Since γ (an) admits supremum and infimum in R, we may denote

liman : = lim inf an := inf γ (an) ,

liman : = lim sup an := sup γ (an) .

An important result ensures us that a bounded sequence is convergent ifand only if lim inf an = lim sup an.

By Theorem 2.1 it follows that for a sequence an having limit, the set γ (an)contains a single point, the linit of the sequence.

Theorem 2.2 it follows that in order to prove the divergence of a sequence,it is enough either to find a subsequence without limit or two subsequenceswith different limits.


For example, the sequence an = sin nπ2 , n ∈ N does not admit limit, since

a2n = sinnπ = 0 → 0, a2n+1 = sin (2n+1)π2 = (−1)n has no limit. But γ (an) =

{−1, 0, 1} .

Theorem 2.3. limn→∞

an = a if and only if limn→∞

(an − a) = 0.

Let us present some examples.1. Prove that lim

n→∞2n

2n+1 = 1.

Let ε > 0 be arbitrary. Inequality∣∣∣ 2n2n+1 − 1

∣∣∣ < ε leads us to n > 12

(1ε − 1

).

Consider

n0 =

{0, if 1

ε − 1 < 0,[12

(1ε − 1

)]+ 1, if ε ≥ 0.

Then ∀ε > 0, ∃n0 = n0 (ε) ∈ N, such that∣∣∣ 2n2n+1 − 1

∣∣∣ < ε, ∀n ≥ n0. 2

2. Prove that limn→∞

(√2)n

= ∞.

Let ε > 0 be arbitrary. Inequality(√

2)n

> ε leads us to n > log√2 ε.Consider

n0 =

{0, if ε < 1,[log√2 ε

]+ 1, if ε ≥ 1.

Then ∀ε > 0, ∃n0 = n0 (ε) ∈ N, such that(√

2)n> ε, ∀n ≥ n0. 2

Teorema 2.5.5.

limn→∞

an =

+∞, if a > 11, if a = 10, if a ∈ (−1, 1)does not exist, if a < −1.

.

Definition 2.7. The sequence an, n ≥ 0, is called Cauchy (fundamental)sequence iff for all ε > 0, there exists n0 = n0 (ε) ∈ N, such that for all n,p ∈ N, n ≥ n0, one has

|an+1 − an| < ε.

For example, the sequence an =∑n

k=012k, n ∈ N is Cauchy sequence.

Indeed, let ε > 0 be arbitrary. Then∣∣∣∣∣n+p∑k=0

1

2k−

n∑k=0

1

2k

∣∣∣∣∣ =

∣∣∣∣∣n+p∑k=n

1

2k

∣∣∣∣∣ =n+p∑k=n

1

2k=

1

2n1−

(12

)p+1

1− 12

<

<1

2n−1

2.6. CONVERGENCE TESTS 19

and, imposing condition 12n−1 < ε, one finds

n0 =

{0, if ε > 2,2 +

[log2

(1ε

)], if 0 < ε ≤ 2.

Therefore, we found n0 = n0 (ε) ∈ N, such that ∀n, p ∈ N, n ≥ n0,|an+p − an| < ε and the sequence is fundamental.

We are considering now another example, bn = 1 + 12 + ... + 1

n , ∀n ≥ 1. Ifwe estimate

b2n − bn =1

n+ 1+

1

n+ 2+ ...+

1

n+ n

≥ 1

n+ n+

1

n+ n+ ...+

1

n+ n︸︷︷︸n times

=1

2> 0,

so the sequence is not Cauchy.

Remark 2.3. Let (an)n∈N si (bn)n∈N be two sequences having the property thatfrom a rank n0 ∈ N, an = bn, ∀n ≥ n0. Then an → a if and only is bn → a. Inother words this property shows that one can eliminate or add a finite numberof terms to a sequence, without loss of the convergence.

Theorem 2.4. (Bolzano-Weierstrass). Each bounded and infinite set of realnumbers has at least one cluster point.

2.6. Convergence tests

Theorem 2.5. Every convergent sequence is bounded.

The converse does not hold true. The classical example is the sequencean = (−1)n , n ∈ N, which is bounded (an ∈ {−1, 1}), without any limit.

Theorem 2.6. Every Cauchy sequence is bounded.

Theorem 2.7. Every Cauchy sequence which contains a convergent subse-quence is also convergent, having the same limit.

Theorem 2.8. Every convergent sequence is Cauchy.


2.7. Existence results for limit of a sequence

Theorem 2.9. (Weierstrass). 1.Every increasing sequence and bounded abovein R is convergent to the least upper bound of the set of its terms.

2. Every decreasing sequence and bounded below in R is convergent to thegreatest lower bound of the set of its terms.

3. Every increasing (decreasing) sequence and unbounded has the limit ∞(respectively −∞).

Weierstrass’s Theorem is only in fact a sufficient result to deduce the exis-tence of the limit of a sequence, without specifying any algorithm to determinethe limit. For example, the sequence an = n

2n , n ∈ N is bounded (since

0 ≤ an ≤ 12 , ∀n ∈ N) and it is decreasing, so it will be convergent; the se-

quence bn = n2 is unbounded and increasing, hence it will have the limit ∞;the sequence cn = −n2 is unbounded and decreasing, so it will have the limit−∞.

Theorem 2.10. Every monotone sequence admits limit (finite or infinite).

Theorem 2.11. (The number e). The sequence an =(1 + 1

n

)n, n ∈ N∗ is

convergent. Its limit is denoted by e.

The number e = 2, 71821828459045... ∈ (2, 3) is irrational.

Theorem 2.12. (Cauchy’s convergence general test). A sequence of real num-bers is convergent if and only if is fundamental.

Due to this Theorem, remark that the sequence an =∑n

k=012k, ∀n ∈ N is

convergent and the sequence bn = 1 + 12 + ...+ 1

n , ∀n ≥ 1 is divergent.

Notice that the Cauchy’s convergence general test allows us to say that Rendowed with the Euclidean distance is a complete metric space.

Let us remark that, in order to prove that a sequence is Cauchy, it is enoughto find a sequence (yn)n∈N, with limn→∞ yn = 0, such that

|an+p − an| ≤ yn, ∀n, p ∈ N.

In order to find the limit of the sequence, denoted by a := limn→∞ an,remark that, since

limn→∞

|a− an| = 0,

2.7. EXISTENCE RESULTS FOR LIMIT OF A SEQUENCE 21

we can approximate a with a term an of the sequence, the approximation beingas better as n is bigger. To do this, we pass to the limit in this inequality asp→ ∞.We get

|a− an| ≤ yn, ∀n ∈ N.

If we try to approximate the value of the limit with a given error, ε > 0,we search for the smallest rank n0 of n, for which yn < ε.

Then, the limit a will be approximated with the term an0 , the error in thisapproximation being smaller than ε.

For example, let us consider the sequence an =∑n

k=1

sin(k2+1)k(k+1) , ∀n ∈ N∗.

If is very hard even to intuit the limit of the sequence, due to the lack ofthe monotony. We try to see if the sequence is fundamental. To do this, weestimate

|an+p − an| =

∣∣∣∣∣n+p∑k=1

sin(k2 + 1

)k (k + 1)

−n∑k=1

sin(k2 + 1

)k (k + 1)

∣∣∣∣∣=

∣∣∣∣∣n+p∑

k=n+1

sin(k2 + 1

)k (k + 1)

∣∣∣∣∣ ≤n+p∑

k=n+1

∣∣sin (k2 + 1)∣∣

k (k + 1)

≤n+p∑

k=n+1

1

k (k + 1)=

n+p∑k=n+1

(1

k− 1

k + 1

)=

1

n+ 1− 1

n+ p+ 1<

1

n+ 1, ∀n, p ≥ 1.

So, by choosing yn := 1n+1 , ∀n ≥ 1, we have limn→∞ yn = 0 and the sequence

(an)n∈N∗ will be Cauchy and so convergent. Let a := limn→∞ an. By passingto limit as p→ ∞ in the inequality

|an+p − an| <1

n+ 1, ∀n, p ≥ 1,

we get that

|a− an| ≤1

n+ 1, ∀n ≥ 1.

If we want to approximate the value of a with two exact decimals, then weimpose condition 1

n+1 < 10−2. Solving this inequation, and choosing n0 := 100,

the smallest solution of it, we find that a ≃ a101 =∑101

k=1

sin(k2+1)k(k+1) with two

exact decimals.

Theorem 2.13. (The sequence of moduli) 1. If (an)n∈N is convergent to a,then (|an|)n∈N is convergent to |a| .


The converse is not true. See, e.g., the sequence an = (−1)n , n ∈ N. But,if a = 0 we have the following result.

Theorem 2.14. |an| → 0 if and only if an → 0.

Theorem 2.15. (Squeeze test). If an, bn, cn, n ∈ N are three sequences of realnumbers, fulfilling the conditions:

1. an ≤ bn ≤ cn, ∀n ≥ n0 ∈ N;2. lim

n→∞an = lim

n→∞cn = b ∈ R,

then the sequence bn has limit and, in addtion, limn→∞

bn = b.

For example, let us consider the sequence an = 1n3+1

+ 1n3+2

+ ... + 1n3+n

,n ≥ 1. Then 0 < an <

nn3+1

, ∀n ∈ N∗, and since limn→∞

0 = limn→∞

nn3+1

= 0, we

get that limn→∞

an = 0.

Theorem 2.16. (Comparison test) 1. Let (an)n∈N be a sequence of real num-bers, (bn)n∈N a sequence of positive numbers with the limit 0, and a ∈ R suchthat |an − a| ≤ bn, ∀n ∈ N. Then lim

n→∞an = a.

2. Let (an)n∈N be a sequence of real numbers, (bn)n∈N a sequence with thelimit ∞, such that bn ≤ an, ∀n ∈ N. Then lim

n→∞an = ∞.

3. Let (an)n∈N be a sequence of real numbers, (bn)n∈N a sequence with thelimit −∞, such that an ≤ bn, ∀n ∈ N. Then, lim

n→∞an = −∞.

Theorem 2.17. (Passing to limit in inequalities).If lim

n→∞an = a ∈ R, lim

n→∞bn = b ∈ R, and an < bn, ∀n ≥ n0 ∈ N, then a ≤ b.

The same result hold if an ≤ bn, ∀n ∈ N.

2.8. Operations with sequences

Theorem 2.18. If two sequences are convergent, then their sum (difference)is convergent to the sum (difference) of the limits of the sequences:(

limn→∞

an = a ∈ R, limn→∞

bn = b ∈ R)=⇒

(limn→∞

(an ± bn) = a± b).

Theorem 2.19. The product of a sequence convergent to 0 and a bounded setis a sequence convergent to 0.

For example, limn→∞

(1√n2+1

· cos πn3

2n2+3

)= 0, since lim

n→∞1√n2+1

= 0 and∣∣∣cos πn3

2n2+3

∣∣∣ ≤ 1.

2.8. OPERATIONS WITH SEQUENCES 23

Theorem 2.20. The product of two convergent sequences is convergent to theproduct of the limits of the sequences:(

limn→∞

an = a ∈ R, limn→∞

bn = b ∈ R)=⇒

(limn→∞

(an · bn) = a · b).

Theorem 2.21. 1. If an → a ∈ R and c ∈ R, then limn→∞

(c · an) = c · a;2. If an → a ∈ R, then lim

n→∞(−an) = −a;

3. If an → a ∈ R and k ∈ N∗, then limn→∞

akn = ak;

4. If an → a ∈ R and bn → b ∈ R∗, then limn→∞

anbn

= ab .

5. If an > 0, ∀n ≥ n0, limn→∞

an = a ∈ (0,+∞), limn→∞

bn = b ∈ R, then

limn→∞

abnn = ab.

6. If limn→∞

an = 0 and an > 0 (an < 0), ∀n ≥ n0, then limn→∞

1an

= ∞ (−∞) .

7. If limn→∞

an = a ∈ R, then limn→∞

k√an = k

√a, if k

√an and k

√a are well

defined.8. If lim

n→∞an = a ∈ R, then lim

n→∞sin an = sin a and lim

n→∞cos an = cos a.

9. If limn→∞

an = a ∈ R and an, a ∈ R\{π2 + kπ| k ∈ Z

}, ∀n ≥ n0, then

limn→∞

tg an = tg a.

10. If limn→∞

an = a and an, a > 0, ∀n ≥ n0, then limn→∞

logb an = logb a,

where b ∈ (0,+∞) \ {1} .11. If lim

n→∞an = 0 and an = 0, ∀n ≥ n0, then lim

n→∞sin anan

= 1.

12. If limn→∞

an = 0 and an = 0, ∀n ≥ n0, then limn→∞

tg anan

= 1.

Theorem 2.22. If an > 0, ∀n ≥ 0 and there exists

limn→∞

an+1

an= l ∈ [0, 1),

then

limn→∞

an = 0.

For example, the sequence an = nqn, with q ∈ (0, 1) , is convergent to 0.Indeed, estimating

limn→∞

an+1

an= lim

n→∞q

(1 +

1

n

)= q ∈ [0, 1)

it follows, by Theorem 2.22, that limn→∞

nqn = 0. A similar result one can also

establish if q ∈ (−1, 1), by estimating limn→∞

an+1

an.


Theorem 2.23. If an, n ∈ N is an increasing and unbounded sequence, withan = 0, ∀n ≥ n0, then lim

n→∞1an

= 0.

Theorem 2.24. (Cesaro-Stolz Lemma). If an, bn, n ∈ N are two sequencessuch that (bn)n∈N is increasing and unbounded, with bn = 0, ∀n ≥ n0, andthere exists

limn→∞

an+1 − anbn+1 − bn

= l ∈ R,

then

limn→∞

anbn

= l.

Theorem 2.25. If an, n ∈ N∗ is a sequence with positive terms and limn→∞

an+1

an=

a > 0, then limn→∞

n√an = a.

Theorem 2.26. 1. If limn→∞

an = ∞ and limn→∞

bn = b ∈ R then limn→∞

(an + bn) =∞.

2. If limn→∞

an = −∞ and limn→∞

bn = b ∈ R, then limn→∞

(an + bn) = −∞.

3. If limn→∞


bn = ∞, then limn→∞

(an + bn) = ∞.

4. If limn→∞


bn = − ∞, then limn→∞

(an + bn) = −∞.

Hence, one can consider the following conventions:

∞+ b = b+∞ = ∞, ∀b ∈ R,−∞+ b = b−∞ = −∞, ∀b ∈ R,∞+∞ = ∞,

−∞−∞ = −∞.

The operation ∞−∞ is undeterminate.5. If lim

n→∞an = ∞ and lim

n→∞bn = b = 0, then

limn→∞

an · bn =

{∞, if b > 0,−∞, if b < 0.

6. If limn→∞



an · bn = ∞.

7. If limn→∞


bn = −∞, then limn→∞

an · bn = ∞.

2.8. OPERATIONS WITH SEQUENCES 25

8. If limn→∞



an · bn = −∞. Hence, one

can consider the following conventions:

∞ · b =

{∞, if b > 0−∞, if b < 0

,

−∞ · b =

{−∞, if b > 0∞, if b < 0

,

∞ ·∞ = ∞,

(−∞) · (−∞) = ∞.

The operations 0 · ∞, ∞ · 0, 0 · (−∞) , (−∞) · 0, are undeterminate.9. If the sequences an, bn, n ∈ N have limt and the product of theirs limits

is defined, then (an · bn)n∈N has the limit equal to the product of the two limits.

10. If limn→∞

an = ±∞ and limn→∞

bn = b ∈ R then limn→∞

bnan

= 0.

11. If limn→∞

an = a ∈ R∗ and limn→∞

bn = ∞ then

limn→∞

bnan

=

{∞, if a > 0,−∞, if a < 0.


b

±∞= 0, ∀b ∈ R,

∞a

=

{∞, if a > 0,−∞, if a < 0.

The operations ±∞±∞ ,

00 are undeterminate.

12. If the sequences an, bn, n ∈ N have limits and the ratio of theirs limits

is determinate, then(anbn

)n∈N

has the limit equal to the ratio of the two limits.

13. If limn→∞

an = a > 1 and limn→∞


abnn = ∞.

14. If limn→∞

an = a ∈ (0, 1) and limn→∞


abnn = 0.

15. If limn→∞

an = a > 1 and limn→∞


abnn = 0.

16. If limn→∞

an = a ∈ (0, 1) and limn→∞


abnn = ∞.

17. If limn→∞


bn = b > 0, then limn→∞

abnn = ∞.

18. If limn→∞


bn = b < 0, then limn→∞

abnn = 0.


19. If limn→∞

an = 0, an > 0, n ∈ N and limn→∞


abnn = 0.


a∞ =

{∞, daca a > 1,

0, daca a ∈ (0, 1) .

a−∞ =

{0, daca a > 1,

∞, daca a ∈ (0, 1) .

∞b =

{∞, daca b > 0,0, daca b < 0.

0∞ = 0.

The operations 00, 1∞, 1−∞, ∞0 are undeterminate.20. If the sequences an, bn, n ∈ N has the limis a, respectively b ∈ R and if

abnn and ab are determinate, then(abnn)n∈N has the limit equal to ab.

2.9. Study of undeterminate forms of limits

Theorem 2.27. 1. Let xn = a0nk + a1n

k−1 + ...+ an−kn+ ak, ai ∈ R, a0 = 0,∀n ∈ N and let k ∈ N∗ be fixed. Then

limn→∞

xn = ∞ · a0.

2. Let xn =a0nk+a1nk−1+...+an−kn+akb0np+b1np−1+...+bn−pn+bp

, ai, bi ∈ R, a0, b0 = 0, ∀n ∈ N and k,

p ∈ N∗ be fixed. Then

limn→∞

xn =

∞a0

b0, if n > p,

a0b0, if n = p,

0, if n < p.

3. If limn→∞

xn = ∞, then limn→∞

(1 + 1

xn

)xn= e.

4. If limn→∞

xn = −∞, then limn→∞

(1 + 1

xn

)xn= e.

5. If limn→∞

xn = 0, then limn→∞

(1 + xn)1xn = e.

6. If limn→∞

xn = 1, limn→∞

yn = +∞, and limn→∞

(xn − 1) yn = l then limn→∞

xynn =

el.7. If a > 1 and k ∈ N are fixed, then lim

n→∞an

nk = ∞.

8. If a > 1, then limn→∞

an

nn = 0.

9. If an > 0, n ∈ N and limn→∞

an = ∞, then limn→∞

ln anan

= 0.

10. If a > 1, an > 0, n ∈ N and limn→∞

an = ∞, then limn→∞

loga anan

= 0.

11. If a ∈ (0,+∞) \ {1} and limn→∞

an = 0, then limn→∞

aan−1an

= ln a.

2.10. EXERCISES 27

12. If limn→∞

an = 0, then limn→∞

(1+an)a−1

an= a, a ∈ R.

2.10. Exercises

(1) Study the monotony and the boundedness of the following sequences:

a) an = n+2n+3 , n ∈ N; b) bn = −n+1

3n , n ∈ N∗; c) cn = 2√n, n ∈ N; d)

dn = 2n

n , n ∈ N∗; e) en = (2n−1)!!(2n)!! , n ∈ N∗; f) fn = 1

12+ 1

22+ 1

32+...+ 1

n2 ,

n ∈ N∗; g) gn = 112

+ 112+22

+ ...+ 112+22+...+n2 , n ∈ N.

A: a) Remark that an = 1− 1n+3 , so (an)n∈N is strictly increasing;

in addition, an ∈ (0, 1) , ∀n ∈ N, so (an)n∈N is bounded.

b) bn = −13 − 1

3n , so (bn)n∈N∗ is strictly increasing; moreover,

bn ∈(−2

3 , 0), ∀n ∈ N∗, so (bn)n∈N∗ is bounded.

c) cn < cn+1, ∀n ∈ N, so (cn)n∈N is strictly increasing; in addition,cn > 0, ∀n ∈ N, the sequence being bounded below and unboundedabove.

d) One has dn+1

dn=

2n+1

n+12n

n

= 2nn+1 ≥ 1, ∀n ∈ N∗, so (dn)n∈N∗ is

strictly increasing; moreover, dn > 0, ∀n ∈ N∗ and is unboundedabove.

e) One has en+1

en=

(2n+1)!!(2n+2)!!(2n−1)!!(2n)!!

= 2n+12n+2 < 1, ∀n ∈ N∗, so the sequence is

strictly decreasing; en ∈ (0, 1) , ∀n ∈ N∗, so the sequence is bounded.f) Obviously, fn+1 − fn = 1

(n+1)2> 0, ∀n ∈ N, so the sequence is

strictly increasing; in order to check the boundedness of this sequence,we have fn > 0, ∀n ∈ N∗. Taking into account that

1

k2<

1

k (k − 1)=

1

k − 1− 1

k, ∀k ≥ 2;

we get

1

22<

1

1− 1

21

32<

1

2− 1

3...........

1

n2<

1

n− 1− 1

n

and, by adding these inequalities, we find

fn =1

12+

1

22+

1

32+ ...+

1

n2<

1

12+

(1− 1

n

)= 2− 1

n< 2, ∀n ∈ N∗,


so an upper bound of the sequence is 2. Actually, one can provethat

limn→∞

(1

12+

1

22+

1

32+ ...+

1

n2

)=π2

6.

g)

1

12+

1

12 + 22+ ...+

1

12 + 22 + ...+ n2< fn

so gn ∈ (0, 2) , ∀n ∈ N. Since gn+1− gn = 112+22+...+(n+1)2

> 0, ∀n ∈ Nthe sequence is strictly increasing.

(2) If an → ∞, what can we say about limn→∞

sin an ?

A: Generally, the limit does not exist. For example, if an = nπ2 ,n ∈ N, lim

n→∞an = ∞, and

limn→∞

sin an = limn→∞

sinnπ

2=

=

limk→∞

sin 2kπ = 0, if n = 4k

limk→∞

sin(2kπ + π

2

)= 1, if n = 4k + 1

limk→∞

sin (2kπ + π) = 0, if n = 4k + 2

limk→∞

sin(2kπ + 3π

2

)= −1, if n = 4k + 3

and so limn→∞

sin an does not exist.

(3) Determine the general term of the sequence (an)n∈N∗ , defined through

a1 = 1, an+1 = an +(12

)n, n ∈ N∗.

A: One has

an = an−1 +

(1

2

)n−1

= an−2 +

(1

2

)n−2

+

(1

2

)n−1

= ... =

= a1 +

(1

2

)1

+

(1

2

)2

+ ...+

(1

2

)n−1

=

= a1 +1

2·1−

(12

)n−1

1− 12

= a1 +

(1−

(1

2

)n−1)

=

= 2−(1

2

)n−1

, ∀n ∈ N∗.

(4) Prove that the sequence defined by x0 = 0, x0 = 1, xn+1 = 1+xn1−xn ,

∀n ∈ N is periodical.

2.10. EXERCISES 29

A: We have successively,

xn+2 =1+xn+1

1−xn+1=

1+ 1+xn1−xn

1− 1+xn1−xn

= − 1xn,

xn+3 =1− 1

xn

1+ 1xn

= xn−1xn+1 ,

xn+4 =1+xn−1

xn+1

1−xn−1xn+1

= 2xn2 = xn, ∀n ∈ N,

and so the sequence is periodical of period 4.(5) Prove that thw sequence defined by x1, x2 ∈ R and xn+1+xn+xn−1 =

0, ∀n ≥ 2 is bounded.R: Due to the hypothesis

xn+1 + xn + xn−1 = 0, ∀n ≥ 2

and

xn+2 + xn+1 + xn = 0, ∀n ≥ 2

so we get

xn+2 = xn−1, ∀n ≥ 2.

This fact allows us to conclude that the sequence is periodical of period2, so it will be bounded, since xn ∈ {x1, x2} , ∀n ∈ N∗.

(6) Consider the neighbourhood V =(− 1

10 ,110

)∈ V (′). How many terms

of the following sequences do not enter V ?

a) an = (−1)n

n , n ∈ N∗; b) bn = nn2+1

, n ∈ N.

A: a) By condition an = (−1)n

n =

{1n , n even− 1n , n odd

/∈ V =(− 1

10 ,110

),

we deduce that there are 10 terms outside of V ;b) By condition bn = n

n2+1/∈(− 1

10 ,110

), we get that there are 9

terms outside of V.(7) Study the existence of the limit to the following sequences:

a) an = sin nπ2 , n ∈ N; b) bn = sinπn, n ∈ N; c) cn =

{1n , if n is even,0, if n is odd;

d) dn =

{1n , if n is even,1, if n is odd.

A: a) Since

an =

0, if n = 4k,1, if n = 4k + 1,0, if n = 4k + 2,−1, if n = 4k + 3,

an does not admit limit.b) bn = sinnπ = 0, ∀n ∈ N, so lim

n→∞bn = 0.


c) limn→∞

cn = 0.

d) We have limn→∞

d2n = limn→∞

12n = 0 and lim

n→∞d2n+1 = lim

n→∞1 = 1,

so (dn)n∈N has not limit.(8) Find the extreme limits of the following sequences:

a) an = 1+(−1)n

2 + (−1)n · n2n+1 , n ∈ N;

b) bn = n2·(−1)n

n , n ∈ N∗;

c) cn = 1n · n(−1)n + sin nπ

2 , n ∈ N∗.A: a) We have

an =

{1 + 2k

4k+1 , if n = 2k,

−2k+14k+3 , if n = 2k + 1.

Since 1+ 2k4k+1 → 3

2 si −2k+14k+3 → −1

2 , it follows that γ (an) ={−1

2 ,32

}.

Therefore, lim sup an = 32 and lim inf an = −1

2 .b) We have

an =

{2k, if n = 2k,

1(2k+1)3

, if n = 2k + 1.

It follows that b2k → ∞ si b2k+1 → 0; so γ (bn) = {0,∞} , lim sup an =∞ and lim inf an = 0.

c) We have

cn =

1, if n = 4k

1(4k+1)2

+ 1, if n = 4k + 1

1, if n = 4k + 21

(4k+3)2− 1, if n = 4k + 3

.

Since 1 → 1, 1(4k+1)2

+ 1 → 1, 1(4k+3)2

− 1 → −1, it follows that

γ (cn) = {−1, 1} , so lim sup an = 1 and lim inf an = −1.(9) Prove that the sequence

an =n2 + 1

2n2 − 1, n ∈ N

converges to 12 .

A: Fie ε > 0 be arbitrary. Inequality∣∣∣∣ n2 + 1

2n2 − 1− 1

2

∣∣∣∣ < ε

gives us√

3ε +

12 < n. So, by choosing n0 =

[√3ε +

12

]+1, we deduce,

via Definition 2.5, that limn→∞ an = 12 .

2.10. EXERCISES 31

(10) By applying the squeeze test, prove that

a) limn→∞

(1

n2+1+ 2

n2+2+ ...+ n

n2+n

)= 1

2 ; b) limn→∞

(sin 1n2+1

+ sin 2n2+1

+ ...+ sinnn2+1

)=

0; c) limn→∞

(1√n2+1

+ 1√n2+2

+ ...+ 1√n2+n

)= 1; d) lim

n→∞n√n = 1; e)

limn→∞

n√1p + 2p + ...+ np = 1, p > 0; f) lim

n→∞n

√1 +

√2 +

√3 + ...+

√n =

1.(11) A: a) We have

1 + 2 + ...+ n

n2 + n<

1

n2 + 1+

2

n2 + 2+ ...+

n

n2 + n<

1 + 2 + ...+ n

n2 + 1

and, since 1+2+...+nn2+n

=n(n+1)

2n2+n

→ 12 ,

1+2+...+nn2+1

=n(n+1)

2n2+1

→ 12 , it follows

that

limn→∞

(1

n2 + 1+

2

n2 + 2+ ...+

n

n2 + n

)=

1

2.

b) We have

−nn2 + n

<sin 1

n2 + 1+

sin 2

n2 + 1+ ...+

sinn

n2 + 1<

n

n2 + 1

and, since −nn2+n

→ 0, nn2+n

→ 0, it follows that

limn→∞

(sin 1

n2 + 1+

sin 2

n2 + 1+ ...+

sinn

n2 + 1

)= 0.

c) We have

n√n2 + n

<1√

n2 + 1+

1√n2 + 2

+ ...+1√

n2 + n<

n√n2 + 1

and, since n√n2+n

→ 1, n√n2+1

→ 1, it follows that

limn→∞

(1√

n2 + 1+

1√n2 + 2

+ ...+1√

n2 + n

)= 1.

d) Since

1 ≤ n√n = n

√1 · 1 · ... · 1︸︷︷︸n−2 times

·√n ·

√n ≤

≤

1 + 1 + ...+ 1︸︷︷︸n−2 times

+√n+

√n

n=n− 2 + 2

√n

n

and, since 1 → 1, n−2+2√n

n → 1, we deduce that

limn→∞

n√n = 1.


e) We have the inequalities

1 < n√1p + 2p + ...+ np < n

√n · np

and, since 1 → 1,n√np+1 = ( n

√n)p+1 → 1, we deduce that

limn→∞

n√1p + 2p + ...+ np = 1, p > 0.

f) We have the inequalities

1 <n

√1 +

√2 +

√3 + ...+

√n <

n

√n ·

√n

and, since 1 → 1, n√n ·

√n = ( n

√n)

32 → 1, it follows that

limn→∞

n

√1 +

√2 +

√3 + ...+

√n = 1.

(12) Establish if the following sequences are Cauchy:a) xn = cos 1!

1·2 + cos 2!2·3 + ... + cosn!

n·(n+1) , n ∈ N∗; b) xn = 1 + 1√2+

1√3+ ... + 1√

n, n ∈ N∗; c) xn = 1 + 1

22+ 1

32+ ... + 1

n2 , n ∈ N∗;

d) xn = 1 + 123

+ 133

+ ... + 1n3 , n ∈ N∗; e) xn = 3n+2

4n+3 , n ∈ N∗; f)

xn =∑n

k=1 sin2 πk , n ∈ N∗; g) xn =

∑nk=1

2k

k!(k+2) , n ∈ N∗.

A: a) We have

|xn+p − xn| =

∣∣∣∣∣n+p∑

k=n+1

cos k!

k (k + 1)

∣∣∣∣∣ ≤n+p∑

k=n+1

|cos k!|k (k + 1)

≤n+p∑

k=n+1

1

k (k + 1)=

n+p∑k=n+1

(1

k− 1

k + 1

)=

1

n+ 1− 1

n+ p+ 1<

1

n+ 1, ∀n, p ∈ N∗,

Therefore, since yn := 1n+1 → 0, (an)n≥1 is a Cauchy sequence, so it

will be convergent.b) The sequence is not Cauchy.c) The sequence is Cauchy.d) The sequence is Cauchy.e) The sequence is Cauchy.f) The sequence is Cauchy.g) The sequence is Cauchy.

(13) Study the convergence of the following sequences:a) an = 1

2n2+n+1

, n ∈ N;b) an = (−1)n

n , n ∈ N∗;

2.10. EXERCISES 33

c) an = sinnn , n ∈ N∗;

d) an = 1 + 12 + ...+ 1

n , n ≥ 1;e) an = 0, 25n + 0, 5n + 0, 75, n ∈ N;f) an = 2n+3n

4n , n ∈ N;g) an = cosn π

4 , n ∈ N.A: a) We have 0 < 1

2n2+n+1

< 1n , ∀n ∈ N∗, so lim

n→∞1

2n2+n+1

= 0.

b) We have∣∣∣ (−1)n

n

∣∣∣ = 1n → 0, so lim

n→∞(−1)n

n = 0.

c) We have∣∣ sinnn

∣∣ ≤ 1n → 0, so lim

n→∞sinnn = 0.

d) Due to the inequalities(1 +

1

n

)n< e <

(1 +

1

n

)n+1

, ∀n ∈ N∗.

we deduce that

ln (n+ 1)− lnn <1

n, ∀n ∈ N∗

and so

ln (n+ 1) < 1 +1

2+ ...+

1

n, ∀n ∈ N∗.

Since limn→∞

ln (n+ 1) = ∞, it follows that limn→∞

(1 + 1

2 + ...+ 1n

)= ∞.

e) limn→∞

an = 0, 75.

f)

limn→∞

an = limn→∞

2n + 3n

4n= lim

n→∞

3n[(

23

)n+ 1]

4n=

= limn→∞

(3

4

)n [(2

3

)n+ 1

]= 0.

g) limn→∞

an = 0.

(14) Study the convergence of the following sequences:a) an = n

2n+3 , n ∈ N;b) an = 2n2

n+1 , n ∈ N;c) an = n2 sin nπ

2 , n ∈ N;d) an = (1 + cosnπ) n2

n2+1, n ∈ N;

e) an = (2n−1)!!(2n)!! , n ∈ N∗.

A: a) The sequence is convergent, with the limit 12 .

b) The sequence is divergent, with the limit ∞.c) The sequence is divergent.d) The sequence is divergent.e) The sequence is divergent.


(15) Find the limits of the following sequences:a) an =

∑nk=1

14k2−1

, n ∈ N;b) an =

∑nk=1

k(k+1)! , n ∈ N;

c) an = n+12n∑n

k=12k

k , n ∈ N∗;

d) an = 1lnn

∑nk=1

1k , n ≥ 2;

e) an = 1n

n√(n+ 1) (n+ 2) ... (2n), n ≥ 2.

A: a) We have

an =1

2

n∑k=1

(1

2k − 1− 1

2k + 1

)=

=1

2

[(1

1− 1

3

)+

(1

3− 1

5

)+ ...+

(1

2n− 1− 1

2n+ 1

)]=

1

2

(1− 1

2n+ 1

)→ 1

2.

b) We have

an =n∑k=1

(k + 1)− 1

(k + 1)!=

n∑k=1

[1

k!− 1

(k + 1)!

]=

=

n∑k=1

[(1

1!− 1

2!

)+

(1

2!− 1

3!

)+ ...+

(1

n!− 1

(n+ 1)!

)]=

= 1− 1

(n+ 1)!→ 0.

c) limn→∞ an = 2.d) limn→∞ an = 1.e) limn→∞ an = 4

e .

(16) Calculate limn→∞

1p+2p+...+np

np+1 , p ∈ N.A: Let an = 1p + 2p + ...+ np and bn = np+1, n ∈ N∗. Estimate

l : = limn→∞

an+1 − anbn+1 − bn

= limn→∞

(n+ 1)p

(n+ 1)p+1 − np+1

= limn→∞

np + C1pn

p−1 + ...+ 1

C1p+1n

p + C2p+1n

p−1 + ...+ 1=

1

p+ 1.

Then, by applying Cesaro-Stolz Lemma, we deduce that

limn→∞

anbn

=1

p+ 1.

(17) Determine limn→∞

1+a+...+an

1+b+...+bn , if a, b ∈ (−1, 1) .

2.10. EXERCISES 35

A: We have

limn→∞

1 + a+ ...+ an

1 + b+ ...+ bn= lim

n→∞

1−an+1

1−a1−bn+1

1−b=

1− b

1− a.

(18) Let x1 ∈ [1, 2] and xn+1 = x2n − 2xn + 2, ∀n ∈ N∗. Prove that thesequence xn is convergent and find lim

n→∞xn.

A: limn→∞ xn = 1.(19) Let an, bn, n ∈ N bve two sequences fulfilling the conditions:

1) 0 < a0 < b0;

2) an =√an−1 · bn−1 si bn = an−1+bn−1

2 , ∀n ≥ 1,(20) Prove that the sequences an and bn are convergent, with the same

limit.A: By mathematical induction, one proves that ∀n ∈ N,a0 < a1 < ... < an−1 < an < bn < bn−1 < ... < b1 < b0,

So, (an)n∈N , (bn)n∈N are convergent. Let a := limn→∞

an and b :=

limn→∞

bn. By passing to the limit in relations from hypothesis 2), we

deduce

a =√a · b si b = a+ b

2.

Hence, a = b.(21) Study the convergence of the sequence xn, n ∈ N defined through

x0 = 1, x1 = 2, xn+2 =√xn+1 · xn, ∀n ∈ N.

A: By mathematical indiction, it follows that xn > 0, ∀n ∈ N. Byapplying ln to the reccurence relation and by denotting yn := lnxn,

yn+2 =1

2yn+1 +

1

2yn, n ∈ N.

Therefore,

yn+2 =1

2yn+1 +

1

2yn,

yn+1 =1

2yn +

1

2yn−1,

yn =1

2yn−1 +

1

2yn−2,

...........

y4 =1

2y3 +

1

2y2,

y3 =1

2y2 +

1

2y1,

y2 =1

2y1 +

1

2y0,


and

yn+2 = −1

2yn+1 + ln 2, n ∈ N

Setting n := n, n− 1, ..., 0, we get

yn+2 = −1

2yn+1 + ln 2,

yn+1 = −1

2yn + ln 2,

yn = −1

2yn−1 + ln 2,

..............

y4 = −1

2y3 + ln 2,

y3 = −1

2y2 + ln 2,

y2 = −1

2y1 + ln 2.

Hence,

yn+2 = ln 2

[1 +

(−1

2

)1

+ ...+

(−1

2

)n+

(−1

2

)n+1]=

= ln 2 ·1−

(−1

2

)n+2

1−(−1

2

)and so

limn→∞

xn =3√4.

(22) Study the convergence of the sequence

x1 = 1, xn+1 =√1 + xn, n ∈ N.

A: By relations

x2n+1 = 1 + xn,

x2n = 1 + xn−1,

we get

x2n+1 − x2n = (1 + xn)− (1 + xn−1) = xn − xn−1.

so, by mathematical induction, the sequence is increasing.In addition,

x2n+1 = 1 + xn,

2.10. EXERCISES 37

so

1 < xn+1 =1 + xnxn+1

<1 + xn+1

xn+1=

1

xn+1+ 1 < 2,

and we deduce that the sequence is bounded.Therefore, the sequence is convergent. Let l := lim

n→∞xn. From

hypothesis, l =√1 + l and so x will be the positive root to Eq. x2 =

1 + x, i.e.

l =1 +

√5

2.

CHAPTER 3

Real series

3.1. Definitions and notations

Let (an)n∈N be a real sequence.

Definition 3.1. The sequence sn, n ∈ N, defined by sn =∑n

k=0 ak, n ∈ N, iscalled the partial sums sequence, associated to (xn)nN.

The real series with the general term xn is the pair((an)n∈N , (sn)n∈N

)and it is denoted by

∑n≥0 an.

The series∑

n≥0 an is called convergent, iff the partial sums sequence isconvergent. Otherwise, the series is called divergent.

If the series∑

n≥0 an is convergent, its sum is the number s := limn→∞

sn ∈R, and it is denoted by s :=

∑∞n=0 an.

For example, let us consider the series∑∞

n=012n = 1+ 1

2 +122

+ ...+ 12n + ...

Then the partial sums sequence is sn =1− 1

2n

1− 12

, ∀n ∈ N and, since limn→∞

sn = 2,

the series will be convergent, having the sum s = 2.In general, if q ∈ R, the series

∑n≥0 q

n is called the geometric series ofratio q. The partial sums sequence is sn = 1 + q + ...+ qn, ∀n ∈ N, so

sn =

{1−qn+1

1−q , if q = 1

n+ 1, if q = 1,

getting that there exists limn→∞

sn = 11−q if and only if |q| < 1. Therefore, the

geometric series of ratio q is convergent if and only if |q| < 1 and, in this case,

∞∑n=0

qn =1

1− q.

Remark 3.1. As we have already remarked Chapter 2, the index set of a seriescan be N∗, N\ {0, 1} or other infinite subset of N.

For example, the index set of the series of general term 1n is N∗.

39

40 3. REAL SERIES

Series∑

n≥11n is called the harmonic series. As we have already seen in

Chapter 2, the partial sums sequence, sn = 1+ 12 + ...+ 1

n is not fundamental,hence the harmonic series is divergent.

Considering the series∑

n≥11

n(n+1) , the partial sums sequence is

sn =n∑k=1

1

k (k + 1)=

n∑k=1

(1

k− 1

k + 1

)= 1− 1

n+ 1, ∀n ≥ 1.

So limn→∞ sn = 1, and the series will be convergent, with∑∞

n=11

n(n+1) = 1.

Theorem 3.1. If the series∑

n≥0 an is convergent, then an → 0.

The converse does not hold true. For example, if xn = 1n , ∀n ∈ N the series∑

n≥11n is divergent and xn → 0.

Remark 3.2. Notice that, if an → 0, then the series∑

n≥0 an is divergent.

For example, the series∑

n≥0 (−1)n,∑

n≥1n√n,∑

n≥1 sinnπ2 are divergent,

since their general terms do not tend to 0.

Proposition 3.1. 1. If series∑

n≥0 an,∑

n≥0 bn are convergent, having sums

s and σ, then series∑

n≥0 (xn + yn) ,∑

n≥0 (xn − yn) are convergent with thesums s+ σ, respectively s− σ.

2. For every α ∈ R∗ the series∑

n≥0 αan and∑

n≥0 an are of same kind.3. By subtracting or adding a finite number of terms to a series, the kind

of series does not change.

Theorem 3.2. (General convergence test of Cauchy). The series∑

n≥0 an is

convergent if and only if ∀ε > 0, ∃n0 = n0 (ε) ∈ N, such that ∀n, p ∈ N,n ≥ n0, one has ∣∣∣∣∣

n+p∑k=n+1

ak

∣∣∣∣∣ < ε.

Definition 3.2. Series∑

n≥0 an is called absolutely convergent iff series∑n≥0 |an| is convergent.

A result due to Dirichlet ensures us that if we change the order of theterm to an absolutely convergent series, neither the kind nor the sum doesnot change. In practice we prefer the study of series with positive terms, as

3.2. CONVERGENCE TESTS FOR SERIES WITH POSITIVE TERMS 41

are absolutely convergent series, which have a behaviour similar to that of thefinite sums.

Due to Theorem 3.2 we deduce the following result.

Theorem 3.3. Every absolutely convergent series is convergent.

The converse of this theorem does not hold true. For example, the alter-nating harmonic series

∑n≥1

1n (−1)n+1 is convergent (see Leibniz’s test for

alternating series), without being absolutely convergent.

Definition 3.3. A convergent series that is not absolutely convergent is calledsemiconvergent.

The alternate harmonic series is an example of semiconvergent series.

3.2. Convergence tests for series with positive terms

Theorem 3.4. A series with positive terms is convergent if and only if thepartial sums sequence is bounded.

Theorem 3.5. (First comparison test). Let∑

n≥0 an and∑

n≥0 bn be twoseries with positive terms, with the property that there exists n0 ∈ N, such thatan ≤ bn, ∀n ≥ n0.

1) If∑

n≥0 bn is convergent, then∑

n≥0 an is convergent.

2) If∑

n≥0 an is divergent, then∑

n≥0 bn is divergent.

For example,∑

n≥01

2n+1 is convergent, since 12n+1 < 1

2n , ∀n ≥ 0 and∑n≥0

12n is convergent;

∑n≥2

1lnn is divergent, since 1

lnn > 1n , ∀n ≥ 2 and∑

n≥21n is divergent.

Theorem 3.6. (Second comparison test). Let∑

n≥0 an and∑

n≥0 bn be twoseries with positive terms, with the property that there exists n0 ∈ N, such thatan+1

an≤ bn+1

bn, n ≥ n0.

1) If∑


n≥0 an is convergent.

2) If∑

n≥0 an is divergent, then∑

n≥0 bn is divergent.

For example,∑

n≥2 (2−√e) (2− 3

√e) (2− n

√e) is divergent, since

an+1

an= 2− n+1

√e > 2− n+1

√(1 +

1

n

)n+1

=n− 1

n=

1n1

n−1

, ∀n ≥ 2

and∑

n≥21n is divergent.

42 3. REAL SERIES

Theorem 3.7. (Third comparison test). Let∑

n≥0 an and∑

n≥0 bn be twoseries with positive terms, such that there exists l := lim

n→∞anbn.

1. If l ∈ (0,∞) , then∑

n≥0 an and∑

n≥0 bn are of same kind;

2. If l = 0, and∑


n≥0 an is convergent;

3. If l = ∞, and∑

n≥0 bn is divergent, then∑


For example,∑

n≥11

n n√n

is divergent, since limn→∞

1n n√n

1n

= 1 ∈ (0,∞) and∑n≥1

1n is divergent.

Series∑

n≥11n2 is convergent, since limn→∞

1n21

n(n+1)

= 1 ∈ (0,∞) and∑

n≥11

n(n+1)

is convergent.

Theorem 3.8. (Ratio test of D’Alembert). Let∑

n≥0 an be a series with pos-

itive terms, with the property that there exists limn→∞

an+1

an=: l.

1) If l < 1, then the series∑


2) If l > 1, then the series∑


For example, series∑

n≥0n

(n+1)! is convergent, since

limn→∞

an+1

an= lim

n→∞

n+ 1

n (n+ 2)= 0 < 1,

and series∑

n≥1 2n is divergent, since limn→∞

an+1

an= 2 > 1.

Let us remark that in case l = 1, one can not specify the nature of series∑n≥0 an. Indeed, for the convergent series

∑n≥1

1n(n+1) as well as for the

divergent series∑

n≥11n , we have lim

n→∞an+1

an= 1.

Theorem 3.9. (Root test of Cauchy). Let∑

n≥0 an be a series with positiveterms, with the property that there exists lim

n→∞n√an =: l.





For example, if we consider series∑

n≥1

(n+1n

)n2

· an, a > 0, then

limn→∞

n√an = a · e.

For a < 1e , the series is convergent and for a > 1

e the series is divergent. (For

a = 1e , we get

limn→∞

(n+1n

)n2

en≥ lim

n→∞

(1 + 1

n

)n2(1 + 1

n

)n(n+1)=

1

e

3.2. CONVERGENCE TESTS FOR SERIES WITH POSITIVE TERMS 43

and so the series is divergent.)Let us remark that in case l = 1, one can not specify the nature of series∑

n≥0 an. Indeed, for the convergent series∑

n≥11

n(n+1) as well as for the

divergent series∑

n≥11n , we have lim

n→∞n√an = 1.

Theorem 3.10. (Raabe-Duhamel test). Let∑

n≥0 an be a series with positive

terms, having the property that there exists limn→∞

n(

anan+1

− 1)=: l





For example, if we consider series∑

n≥11n2 , then

limn→∞

n

(anan+1

− 1

)= 2 > 1,

and so the series is convergent.

And if we consider series∑

n≥1(2n−1)!!(2n)!! , where (2n− 1)!! = 1·3·5·...·(2n− 1)

and (2n)!! = 2 · 4 · 6 · ... · (2n), then

limn→∞

n

(anan+1

− 1

)=

1

2< 1,

and so the series is divergent.

Theorem 3.11. (Integral test of Cauchy) Let f : [0,∞) → [0,∞) be a contin-uous and decreasing function. Then ∀n0 ∈ N, we have(

∃ limx→∞

∫ x

0f (t) dt ∈ R

)⇔

∑n≥n0

f (n) is convergent

.

Let us also remark that in case l = 1, the Raabe-Duhamel test does not spec-ify the nature of series

∑n≥0 an. Indeed, for the convergent series

∑n≥2

1n ln2 n

(via the integral test of Cauchy) as well as for the divergent series∑

n≥11n , we

have limn→∞

n(

anan+1

− 1)= 1.

Example 3.1. Let us consider the general harmonic series∑

n≥11nα , α >

0. Since limn→∞an+1

an= 1, we can not apply the ratio test (nor the root test).

By estimating

limn→∞

n

(anan+1

− 1

)= lim

n→∞

(1 + 1

n

)α − 11n

= α,

44 3. REAL SERIES

we deduce, via Raabe-Duhamel test, that the general harmonic series convergesfor α > 1 and diverges for α < 1 (and if α = 1, we get the harmonic series,that is divergent). Therefore,

∑n≥1

1nα is convergent if and only if α > 1.

3.3. Convergence tests for series with real terms

Theorem 3.12. (Abel’s test). Let (xn)n∈N be a real sequence, having the partialsums sequence bounded and let (an)n∈N be a decreasing and convergent to 0 realsequence. Then the series

∑n≥0 anxn is convergent.

Example 3.2. If we consider the series∑

n≥1sinnxnp , where p > 0, and x ∈

(0, π), then we have

xn = sinnx, n ∈ N∗,

that has the partial sums sequence bounded:n∑k=1

xk = sinx+ sin 2x+ ...+ sinnx.

Indeed, by denoting

S1n = cosx+ cos 2x+ ...+ cosnx,

S2n = sinx+ sin 2x+ ...+ sinnx,

one has

S1n + iS2n

= (cosx+ i sinx) + (cosx+ i sinx)2 + ...+ (cosx+ i sinx)n =

:not= z + z2 + ...+ zn = z

1− zn

1− z=

= (cosx+ i sinx)1− cosnx− i sinnx

1− cosx− i sinx=

= (cosx+ i sinx)2 cos2 nx2 − 2i sin nx

2 cos nx22 cos2 x2 − 2i sin x

2 cosx2

=

= (cosx+ i sinx)cos nx2

(cos nx2 − i sin nx

2

)cos x2

(cos x2 − i sin x

2

) =

= (cosx+ i sinx)cos nx2

[cos(−nx

2

)+ i sin

(−nx

2

)]cos x2

[cos(−x

2

)+ i sin

(−x

2

)] =

=cos nx2cos x2

[cos(x− nx

2+x

2

)+ i sin

(x− nx

2+x

2

)],

hence |S2n| =∣∣∣ cos nx

2cos x

2sin(x− nx

2 + x2

)∣∣∣ ≤ 1

|cos x2 |.

3.4. EXERCISES 45

And, since an = 1np , n ∈ N∗ is decreasing and convergent to 0, we get, via

the Abel’s test, that the series is convergent.

Definition 3.4. The series with real terms of alternating signs,∑

n≥0 (−1)n+1 an,where an ≥ 0, ∀n ∈ N, is called alternating series.

Theorem 3.13. (Leibniz’s test). If (an)n∈N is a decreasing sequence that con-

verges to 0, then the series∑

n≥0 (−1)n+1 an is convergent.

For example, series∑

n≥1 (−1)n+1 1n is convergent, since an := 1

n ↘ 0.

Definition 3.5. The Cauchy product of the series∑

n≥0 an,∑

n≥0 bn is the

series∑

n≥0 cn, with

cn =

n∑k=0

xkyn−k, ∀n ∈ N.

Theorem 3.14. (Cauchy-Maertens test). If the series∑

n≥0 an and∑

n≥0 bnare absolutely convergent, with the sums a and, respectively b, then the Cauchyproduct is absolutely convergent, with the sum a · b.

For example, the series 1 +∑

n≥11n! and 1 +

∑n≥1

(−1)n

n! are absolutely

convergent and have the Cauchy product∑

n≥0 cn, where c0 = 1, ..., cn =

0, ∀n ≥ 1. By applying Cauchy-Maertens test, series∑

n≥0 cn is absolutelyconvergent, its sum being, obviously, 1.

Since 1 +∑

n≥11n! = e, it follows that

1 +

∞∑n=1

(−1)n

n!=

1

e.

3.4. Exercises

(1) By using the definition, establish the kind of the following series:a)∑

n≥11

n(n+1) ; b)∑

n≥12n+5n

7n ; c)∑

n≥1 lnn+5n+4 ; d)

∑n≥1

(1 + 1

n2

);

e)∑

n≥1n+3

n(n+1)(n+2) ; f)∑

n≥11

(n+π)(n+π+1) ; g)∑

n≥1nαn , |α| > 1;

h)∑

n≥01√

n+1+√n; i)∑

n≥2

(n√3− n+1

√3); j)

∑n≥1

ln n+2n+1

ln(n+1)·ln(n+2) ; k)∑n≥0 arctg 1

n2+n+1; l)

∑n≥0

n3−3n2+1(n+2)3

.

A: a) One has

sn =n∑k=1

1

k (k + 1)=

n∑k=1

(1

k− 1

k + 1

)= 1− 1

n+ 1,

46 3. REAL SERIES

hence sn → 1, and the series is convergent, having the sum 1.b) One has

sn =n∑k=1

(2

7

)k+

(5

7

)k=

2

7·1−

(27

)n1− 2

7

+5

7·1−

(57

)n1− 5

7

=

=2

5

(1−

(2

7

)n)+

5

2

(1−

(5

7

)n),

hence sn → 25 + 5

2 = 2910 , and the series is convergent, having the sum

2910 .

c) One has

sn =

n∑k=1

lnk + 5

k + 4=

n∑k=1

ln (k + 5)−n∑k=1

ln (k + 4) =

= ln (n+ 1) ,

hence sn → ∞, and the series is divergent, with the sum ∞.d) Since lim

n→∞

(1 + 1

n2

)= 1, we get that the series is divergent,

since the general terms sequence does not converge to 0.e) We estimate the partial sums sequence

sn =

n∑k=1

k + 3

k (k + 1) (k + 2)=

n∑k=1

(3

2k− 2

k + 1+

1

2 (k + 1)

)

=3

2

n∑k=1

(1

k− 1

k + 1

)− 1

2

n∑k=1

(1

k + 1− 1

k + 2

)=

3

2

(1− 1

n+ 1

)− 1

2

(1

2− 1

n+ 2

)→ 5

4,

so the series is convergent, with the sum 54 .

f) The series is convergent, with the sum 1π+1 .

g) The series is convergent, with the sum α(1−α)2 .

h) The series is divergent, with the sum ∞.i) The series is convergent, with the sum

√3− 1.

j) The series is convergent, with the sum 1ln 2 .

k) The series is convergent, with the sum π4 .

l) The series is divergent.(2) By using the general convergence test of Cauchy, establish the kind

of the series∑

n≥1cosnx3n , x ∈ R;

3.4. EXERCISES 47

A: For n, p ∈ N∗,∣∣∣∣∣n+p∑k=n

cos kx

3k

∣∣∣∣∣ ≤n+p∑k=n

∣∣∣∣cos kx3k

∣∣∣∣ ≤ n+p∑k=n

1

3k=

1

3n·1−

(13

)p+1

1− 13

<

<2

3n+1,

and, since limn→∞

23n+1 = 0, it follows that the series is convergent.

(3) By using the first comparison test, establish the kind of the followingseries:

a)∑

n≥1 cos13n ; b)

∑n≥1

2n+1n(n+1) ; c)

∑n≥1

1n√lnn

.

A: a) We have cos 13n <

13n , ∀n ≥ 1 and, since

∑n≥1

13n is conver-

gent, it follows that the series is convergent.b) We have 2n+1

n(n+1) >1n , ∀n ≥ 1 and, since

∑n≥1

1n is divergent, it

follows that the series is divergent.c) Since lnn < n, ∀n ≥ 1, it follows that 1

n√lnn

> 1n√n, ∀n ≥ 1.

Since limn→∞

1n√n= 1, it follows that the series

∑n≥1

1n√nis divergent.

So, the given series is divergent.(4) By using the second comparison test, establish the kind of the series∑

n≥1

(ne

)n · 1n! .

A: Since

an+1

an=

(1 + 1

n

)ne

>

(1 + 1

n

)n(1 + 1

n

)n+1 =n

n+ 1=

1n+11n

, ∀n ≥ 1,

the series is divergent.(5) By using the third comparison test, establish the kind of the following

series:a)∑

n≥11

n n√n; b)

∑n≥1 sin

2002 π2n ; c)

∑n≥1

3√2n−1n+1 ; d)

∑n≥1

2n−1√n+1

;

e)∑

n≥11

n2 n√n; f)

∑n≥1

3 3√n+1

2n√n+1

; g)∑

n≥2n2

n2−1; h)

∑n≥1

n2−3n+2n3+1

.

A: a) One has

limn→∞

1n n√n

1n

= 1

and, since∑

n≥11n is divergent, the given series is divergent.

b) One has

limn→∞

sin2002 π2n(

π2n

)2002 = 1,

and, since∑

n≥1

(π2n

)2002=(π2

)2002 ·∑n≥11

n2002 is a convergent gen-eralized harmonic series, it follows that the given series is convergent.

48 3. REAL SERIES

c) The series is divergent.d) The series is divergent.e) The series is convergent.f) The series is convergent.g) The series is divergent.h) The series is divergent.

(6) By using the ratio test, establish the kind of the following series:

a)∑

n≥0n

(n+1)! ; b)∑

n≥1(n!)3

(3n)! ; c)∑

n≥1an√n!, a > 0; d)

∑n≥1

(an)n

n! ,

a > 0; e)∑

n≥1(n!)2

(2n)! ; f)∑

n≥12n−1

(√2)

n ; g)∑

n≥1n3

en ; h)∑

n≥1n!

2n+1 ; i)∑n≥1

(2n−1)!!(3n)!! .

A: a) We have

limn→∞

an+1

an= lim

n→∞

n+1(n+2)!n

(n+1)!

= limn→∞

n+ 1

n (n+ 2)= 0 < 1,

so the series is convergent.b) We have

limn→∞

an+1

an= lim

n→∞

((n+!)!)3

(3(n+1))!

(n!)3

(3n)!

= limn→∞

(n+ 1)3

(3n+ 1) (3n+ 2) (3n+ 3)=

=1

9< 1,

so the series is convergent.c) We have

limn→∞

an+1

an= lim

n→∞

an+1√(n+1)!

an√n!

= limn→∞

a√n+ 1

= 0 < 1,

so the series is convergent.d) We have

limn→∞

an+1

an= lim

n→∞

(a(n+1))n+1

(n+1)!

(an)n

n!

= limn→∞

a

(1 +

1

n

)n= a · e,

so the series is convergent for a < 1e and divegent for a > 1

e ; if a = 1e ,

the series if divergent, due to Exercise 4.e) The series is convergent.f) The series is convergent.g) The series is convergent.h) The series is divergent.i) The series is convergent.

3.4. EXERCISES 49

(7) By using the root test, establish the kind of the following series:

a)∑

n≥01

(n+1)n; b)

∑n≥1

(1+ 1n)

n3

3n ; c)∑

n≥1 an(1 + 1

n

)n, a > 0;

d)∑

n≥1 arctgn(1n

); e)

∑n≥1 na

n, a > 0; f)∑

n≥1 tgn(π6 + 1

n

); g)∑

n≥1 sinn(π3 + 1

n

); h)

∑n≥1

(2nn+1

)3n.

A: a) We have

limn→∞

n√an = lim

n→∞n

√1

(n+ 1)n= lim

n→∞

1

n+ 1= 0 < 1,

so the series is convergent.b) We have

limn→∞

n√an = lim

n→∞

n

√(1 + 1

n

)n3

3n= lim

n→∞

(1 + 1

n

)n2

3=e3

3> 1,

so the series is convergent.c) We have

limn→∞

n√an = lim

n→∞n

√an(1 +

1

n

)n= lim

n→∞a

(1 +

1

n

)= a,

so the series is convergent for a < 1 and divergent for a > 1; for a = 1,we get the series

∑n≥1

(1 + 1

n

)n, whose general term is

(1 + 1

n

)nand

converges to e = 0. So the series is divergent.d) The series is convergent.e) The series is convergent for a < 1 and divergent for a ≥ 1.f) The series is convergent.g) The series is convergent.h) The series is divergent.

(8) By using Raabe-Duhamel test, establish the kind of the followingseries:

a) 1 +∑

n≥11

2n+1 · (2n−1)!!(2n)!! ; b)

∑n≥1

(2n−1)!(2n)! ;

A: Since

limn→∞

an+1

an= lim

n→∞

12n+3 · (2n+1)!!

(2n+2)!!

12n+1 · (2n−1)!!

(2n)!!

= limn→∞

(2n+ 1

2n+ 3· 2n+ 1

2n+ 2

)= 1,

one can not specify the kind of the series by using the ratio test. Weestimate, in this case, the limit in Raabe-Duhamel test:

limn→∞

n

(anan+1

− 1

)= lim

n→∞n

((2n+ 2) (2n+ 3)

(2n+ 1)2− 1

)=

3

2> 1,

so the series is convergent.

50 3. REAL SERIES

b) The series is divergent.

(9) By using the Leibniz’s test, establish the kind of the series∑

n≥1(−1)n+1

3√n

.

A: Since 13√n↘ 0, it follows that the alternating series is conver-

gent.(10) By using the Abel’s test, establish the kind of the series 1+

∑n≥1

cosnxnp ,

x ∈ (0, π).A: The sequence

xn = cosnx, n ∈ N∗

has the partial sums sequencen∑k=1

xk = cosx+ cos 2x+ ...+ cosnx, ∀n ≥ 1

bounded; indeed, by denoting

S1n = cosx+ cos 2x+ ...+ cosnx,

S2n = sinx+ sin 2x+ ...+ sinnx,

we have, by using the result from Example 3.2,

|S1n| =∣∣∣∣cos nx2cos x2

cos(x− nx

2+x

2

)∣∣∣∣ ≤ 1∣∣cos x2 ∣∣ , ∀n ≥ 1.

Since an = 1np , n ≥ 1 is decreasing and convergent to 0, it follows that

the series is convergent.(11) Study the convergence of the following alternating series. In case of

convergence, specify if the series are semiconvergent.

a)∑

n≥1(−1)n+1

2n+1 ; b)∑

n≥1(−1)n+1

√n

; c)∑

n≥1(−1)n+1

n2 ;

d)∑

n≥1 (−1)n+1 n+12n(n+2) ; e)

∑n≥2

(−1)n+1

n√n

; f)∑

n≥1 (−1)n(

2n5n+2

)n.

A: a) The series is semiconvergent;b) The series is semiconvergent;c) The series is absolutely convergent;d) The series is semiconvergent;e) The series is divergent;f) The series is absolutely convergent.

CHAPTER 4

Taylor series. Taylor expansions

4.1. Taylor series

Definition 4.1. Let an, ∀n ∈ N and x0 be real numbers. The Taylor serieswith the coefficients an, n ∈ N and centered at x0 is the function series∑

n≥0

an (x− x0)n = a0 + a1 (x− x0) + a2 (x− x0)

2 + .... (4.1)

Remark that the set of converge of a Taylor series is non-empty, since italways contains its center, x0.

Definition 4.2. A Maclaurin series (or, simply, a power series) is aTaylor series having the center x0 = 0, that is,∑

n≥0

anxn = a0 + a1x+ a2x

2 + ...,

the real numbers an, n ∈ N being called the coefficients of the series, too.

Definition 4.3. Let∑

n≥0 an (x− x0)n be a Taylor series with the coefficients

an, n ∈ N and centered at x0. With the convention 1/∞ = 0 and 1/0 = ∞, thenumber r ≥ 0 defined by

r :=1

lim supn→∞

n√|an|

(4.2)

is called the radius of convergence of the Taylor series. Relation (4.2) iscalled the Cauchy-Hadamard formula.

Theorem 4.1. (Abel-Cauchy-Hadamard Theorem). Let∑

n≥0 an (x− x0)n be

a Taylor series with the coefficients an, n ∈ N, and center x0.a) If r = 0, then the only point of convergence is x0;b) If r > 0, then the series is absolutely convergent on the interval (x0 − r,

x0 + r) and it is divergent on (−∞, x0 − r) ∪ (x0 + r,∞) ;

51

52 4. TAYLOR SERIES. TAYLOR EXPANSIONS

c) If r > 0, then the series is uniformly convergent on every interval [a, b] ⊂(x0 − r, x0 + r) ;

d) If the power series converges for x = x0 + r (or x = x0 − r), then itssum is continuous in x0 + r (respectively x0 − r);

e) If r > 0, then the sum of the series admits derivatives of any order inthe interval (x0 − r, x0 + r), and these derivatives can be determined by differ-entiating term by term the series;

f) If r > 0, then the series can be integrated term by term on every interval[a, b] ⊂ (x0 − r, x0 + r).

Remark 4.1. Due to a property of the sequences, if there exists limn→∞

∣∣∣an+1

an

∣∣∣,then it also exists limn→∞

n√

|an| and the two limits are equal. Hence, if there

exists limn→∞

∣∣∣an+1

an

∣∣∣, the radius of convergence can be determined as

r = limn→∞

∣∣∣∣ anan+1

∣∣∣∣ ,with the same convention, 1/∞ = 0 and 1/0 = ∞.

Example 4.1. Find the set of convergence and the sum of the Taylor series∑n≥1 (−1)n+1 xn

n , x ∈ R.By applying Cauchy-Hadamard formula (4.2), we get that the radius of

convergence is

r =1

lim supn→∞

n

√∣∣∣(−1)n+1 xn

n

∣∣∣ = 1.

Therefore, by applying Abel-Cauchy-Hadamard Theorem, the series is abso-lutely convergent on (−1, 1) and divergent on (−∞,−1) ∪ (1,∞) .

For x = −1, the series becomes −∑

n≥11n , and it is divergent. For x = 1,

the series becomes∑

n≥1 (−1)n+1 1n , and it is convergent, since Leibniz’s test.

So, the set of convergence is (−1, 1]. Let S : (−1, 1] → R,

S (x) =

∞∑n=1

(−1)n+1 xn

n.

By applying again Theorem 4.1, we deduce that

S′ (x) =∞∑n=1

(−x)n−1 =1

1 + x, ∀x ∈ (−1, 1) .

4.1. TAYLOR SERIES 53

Since S (0) = 0, we obtain that S (x) = ln (1 + x) , ∀x ∈ (−1, 1). But the sumof the series is also continuous in x = 1, due to Theorem 4.1. Hence,

S (1) = limx→1

S (x) = limx→1

ln (1 + x) = ln 2.

Let us notice that in Theorem 4.1 we can deduce properties of the sum ofthe series, if we know the coefficients an, n ∈ N, and the center x0.

The main problem that rises is the converse. How can we find a Taylor se-ries whose sum is a given function ? Such a function will be called expandablein Taylor series.

Definition 4.4. Let f : I → R be a function and x0 ∈ I, where I is an openinterval. f is called exapandable in Taylor series about the point x0 iffthere exists (an)n∈N and ε > 0, with the following properties:

1) (x0 − ε, x0 + ε) ⊆ I, ε ≤ r, where r is the convergence radius of theTaylor series

∑n≥0 an (x− x0)

n ;

2) ∀x ∈ (x0 − ε, x0 + ε), we have f (x) =∑∞

n=0 an (x− x0)n .

Therefore the following two issues appear.A) In which conditions a given function is expandable in Taylor series ?B) How can we effectively determine the coefficients an, n ∈ N, by knowing

function f ?Regarding the second issue, we have the following result.

Theorem 4.2. Let I be an open interval, f : I → R be a function, and x0 ∈ Ibe fixed. If f is expandable in Taylor series about the point x0, then f doesadmit derivatives of any order and

an =f (n) (x0)

n!, ∀n ∈ N. (4.3)

Due to Theorem 4.2, the existence of the all derivatives represents a nec-essary condition for a function to be exapandable in Taylor series. But thiscondition is not sufficient. Indeed, if we consider the function f : R → R,defined by

f (x) =

{e−

1x2 , if x = 0,

0, if x = 0,

then f is differentiable of any order, with

f (0) = f ′ (0) = ...f (n) (0) = ... = 0, ∀n ∈ N.Hence, if f would be expandable in Taylor series about 0, then we would have

f (x) =∞∑n=0

anxn =

∞∑n=0

f (n) (x0)

n!xn = 0, ∀x ∈ (−ε, ε) .


Remark 4.2. By Theorem 4.2 we deduce that if f is expandable in Taylor seriesabout the point x0, then there exists a single Taylor series whose sum equals f

on (x0 − ε, x0 + ε) , i.e. the Taylor series having the coefficients an = f (n)(x0)n! ,

∀n ∈ N. This series is also called the Taylor series associated to f on theinterval (x0 − ε, x0 + ε) about the point x0.

Therefore, regarding the first issue, it is enough to establish the condi-tions for an infinitely differentiable function to be the sum of the Taylor seriesassociated on (x0 − ε, x0 + ε) .

4.2. Taylor’s formula

Let f : I → R be a n+ 1 times differentiable function on the open intervalI. If x0 ∈ I, then the Taylor polynomial of order n, centered at x0 is

Tn (x) :=

n∑k=0

f (k) (x0)

k!(x− x0)

k

and the difference

Rn (x) = f (x)− Tn (x)

is called the Lagrange remainder of the Taylor’s formula. Taylor’s formulais

f (x) = Tn (x) +Rn (x)

that represents the link between the function and its Taylor polynomial, andgives us an estimate of the remainder.

Recalling the Mean Value Theorem from the high school, for every x0,x ∈ I, there exists a c between x0 and x, such that

f (x)− f (x0) = (x− x0) f′ (c) .

In this situation, n = 0, f is continuous on I and differentiable on the interiorof I (which is I, too), Tn (x) = f (x0), and Rn (x) = (x− x0) f

′ (c) .The Taylor’s formula has many applications in practice. An example of

Taylor’s formula of order 2 is the formula in kinematics, that describes thelinear motion of an object having uniform acceleration,

x (t) = x (0) + v0t+at2

2.

In the case of differentiable functions of higher order, we have the followinggeneral result.

Theorem 4.3. (Taylor’s formula). Let f : I → R be a function which is n+1

4.2. TAYLOR’S FORMULA 55

times differentiable on the open interval I. Then, for every x0, x ∈ I, there isa c between x0 and x, such that

f (x) = f (x0) +f ′ (x0)

1!(x− x0) + ...+

f (n) (x0)

n!(x− x0)

n +f (n+1) (c)

(n+ 1)!(x− x0)

n+1 .

Remark 4.3. For a n+1 times differentiable function on the open interval I,the Taylor’s formula has the form

f (x) = Tn (x) +(x− x0)

n

n!ω (x) ,

where ω : I → R is a function such that limx→x0 ω (x) = ω (x0) = 0.

Remark 4.4. Let us notice that the coefficients an, n ∈ N in (4.3) are exactlythe ones from the Taylor’s formula. Therefore, the Taylor polynomial Tn (x)equals the partial sum of rank n of the Taylor series associated to f about thepoint x0.

And our first issue is geting now an answer.

Theorem 4.4. Let f : I → R be an infinitely differentiable function on theopen interval I and x0 ∈ I. Then f is expandable in Taylor series about thepoint x0 if and only if for every x ∈ I, limx→x0 Rn (x) = 0.

In application, the following test will be of high importance.

Theorem 4.5. (Cauchy’s test). Let I be an open interval and f : I → R be aninfinitely differentiable function. Suppose that there exist M > 0 and δ > 0,such that

supx∈I

∣∣∣f (n) (x)∣∣∣ ≤Mδnn!, ∀n ∈ N.

Then

f (x) =∞∑n=0

f (n) (x0)

n!(x− x0)

n ,

for all x0, x ∈ I with |x− x0| < 1δ .

Another test, that is equivalent to the Cauchy’s test is the following.

Theorem 4.6. Let I be an interval and f : I → R be a function, infinitely


differentiable on a neighborhood V of x0 ∈ I. Suppose that there exists M > 0such that

supx∈I

∣∣∣f (n) (x)∣∣∣ ≤M, ∀n ∈ N.

Then

f (x) =∞∑n=0

f (n) (x0)

n!(x− x0)

n , ∀x ∈ V.

Theorem 4.7. If∑

n≥0 an (x− x0)n is convergent for x = x1, then the series

converges uniformly for every z in the segment from x0 to x1, i.e. for z ∈{(1− t)x0 + tx1, t ∈ [0, 1]}.

Example 4.2. We prove that the Taylor expansion about 0 (or the Maclaurinseries) of the the function f (x) = ln (1 + x) is

ln (1 + x) = x− x2

2+x3

3+ ..., ∀x ∈ (−1, 1]. (4.4)

Estimating its derivatives, we get, by mathematical induction,

f (n) (x) =(n− 1)! (−1)n+1

(1 + x)n, ∀n ≥ 1.

Hence, ∀δ ∈ (0, 1) ,

supx∈(−δ,δ)

∣∣∣f (n) (x)∣∣∣ ≤ n!

(1− δ)n, ∀n ∈ N,

and so, from Theorem 4.5 we deduce that (4.4) is fulfilled ∀x ∈ (−1, 1) . For

x = 1, Leibnitz’s test ensures us that the Maclaurin series,∑

n≥1 (−1)n+1 xn

n is

also convergent. Since f and the sum S of the series are continuous on (−1, 1],and f (x) = S (x) , ∀x ∈ (−1, 1), by passing to limit as x→ 1, we get that (4.4)is also fulfilled for x = 1.

4.3. Exercises

(1) Find the sets of convergence of the following Taylor series:

a)∑

n≥1 n!xn, x ∈ R; b)

∑n≥1 (−1)n+1 x2n−1

(2n−1)(2n−1)! , x ∈ R;

c)∑

n≥1xn

n(n+1) , x ∈ R; d)∑

n≥1

(n+1n

)n2

xn, x ∈ R;e)∑

n≥0 (−1)n 1

3n2√n2+1

tgn x, x ∈(−π

2 ,π2

).

A: a) r = 0, the set of convergence is {0} ;b) r = ∞, the set of convergence is R;c) r = 1, the set of convergence is [−1, 1] ;

4.3. EXERCISES 57

d) r = 1/e, the set of convergence is [−1/e, 1/e] ;e) r = π

3 , the set of convergence is [−π3 ,

π3 ].

(2) Find the sets of convergence and the sums of the following Taylorseries:

a)∑

n≥0x4n+1

4n+1 , x ∈ R; b)∑

n≥0 (−1)n x2n+1

2n+1 , x ∈ R;c)∑

n≥1 (−1)n+1 xn+1

n(n+1) , x ∈ R.

A: a) The coefficients are an =

{0, if n = 4k, 4k + 2, or 4k + 3,1n , if n = 4k + 1.

By applying Cauchy-Hadamard formula (4.2), we get that theradius of convergence is

r =1

lim supn→∞

n√

|an|= 1.

Therefore, by applying Abel-Cauchy-Hadamard Theorem, the se-ries is absolutely convergent on (−1, 1) and divergent on (−∞,−1) ∪(1,∞) .

For x = −1, the series becomes∑

n≥1−1

4n+1 , and it is divergent.

For x = 1, the series becomes∑

n≥11

4n+1 , and it is divergent. So, the

set of converge is (−1, 1).

Let S : (−1, 1) → R, S (x) =∑∞

n=0x4n+1

4n+1 . By applying Theorem4.1, we deduce successively

S′ (x) =

∞∑n=0

x4n =

∞∑n=0

(x4)n

=1

1− x4, ∀x ∈ (−1, 1) .

Since S (0) = 0, we obtain that

S (x) =1

2arctg x+

1

4ln

1 + x

1− x, ∀x ∈ (−1, 1);

b) The set of convergence is [−1, 1] and the sum of the series isS (x) = arctg x, ∀x ∈ [−1, 1] ;

c) The set of convergence is (−1, 1) and the sum of the series isS (x) = (x+ 1) ln (x+ 1)− x, ∀x ∈ (−1, 1) .

(3) Determine the following sums, by using Taylor series:

a)∑∞

n=1 (−1)n+1 14n−3 ; b)

∑∞n=1 (−1)n+1 1

3n−2 ; c)∑∞

n=0 (−1)n 16n+1 ;

d)∑∞

n=1 (−1)n+1 1n(2n−1) ; e)

∑∞n=0

1(4n+1)(4n+3) .


A: a) By applying Theorem 4.1, we deduce successively

∞∑n=1

(−1)n+1 1

4n− 3=

∞∑n=1

(−1)n+1∫ 1

0x4n−4dx

=

∫ 1

0

( ∞∑n=1

(−1)n+1 x4n−4

)dx

=

∫ 1

0

( ∞∑n=0

(−x4

)n)dx =

∫ 1

0

1

1 + x4dx

=

√2

8

(π + ln

2 +√2

2−√2

);

b) Analogously, we obtain∑∞

n=1 (−1)n+1 13n−2 = 1

3

(π√3+ ln 2

);

c) By applying Theorem 4.1, we deduce successively( ∞∑n=0

(−1)nx6n+1

6n+ 1

)′

=∞∑n=0

((−1)n

x6n+1

6n+ 1

)′=

∞∑n=0

(−1)n x6n

=∞∑n=0

(−x6

)n=

1

1 + x6, ∀x ∈ (−1, 1) .

So

S (x) : =

∞∑n=0

(−1)nx6n+1

6n+ 1=

∫1

1 + x6dx

= F (x) + c, ∀x ∈ (−1, 1) ,

where

F (x) =1

6

[2 arctg x+ arctg

(2x−

√3)+ arctg

(2x+

√3)]

+

√3

12lnx2 +

√3x+ 1

x2 −√3x+ 1

.

Since S (0) = 0, we deduce that c = 0, and hence S (x) = F (x) ,∀x ∈ (−1, 1) ;

d)∑∞

n=1 (−1)n+1 1n(2n−1) = 2

∑∞n=1 (−1)n+1 1

2n−1−∑∞

n=1 (−1)n+1 1n ;

e)∑∞

n=01

(4n+1)(4n+3) =12

∑∞n=0

14n+1 − 1

2

∑∞n=0

14n+3 .

(4) Expand in Taylor series about 0 the following functions, by specifyingtheir sets of convergence:

a) f (x) = cosx, x ∈ R; b) f (x) = sinx, x ∈ R; c) f (x) = ex,x ∈ R;

d) f (x) = (1 + x)α , x > −1, where α ∈ R;

4.3. EXERCISES 59

e) f (x) = ln(x+

√1 + x2

), x ∈ R;

f) f (x) = 2x(x−1)2

, x ∈ R \ {1} ; g) f (x) = 2x+1x2−3x+2

, x ∈ R \ {1, 2};h) f (x) = x2e−x, x ∈ R; i) f (x) = sin 2x+ x cos 2x, x ∈ R;j) f (x) = ln 1+x

1−x , x ∈ (−1, 1); k) f (x) = sinx cosx, x ∈ R;l) f (x) = (1 + ex)2 , x ∈ R.

A: a) By mathematical induction, we prove that

f (n) (x) = cos(x+

nπ

2

), ∀x ∈ R, ∀n ∈ N.

Since ∣∣∣f (n) (x)∣∣∣ = ∣∣∣cos(x+nπ

2

)∣∣∣ ≤ 1, ∀x ∈ R, ∀n ∈ N,

by Theorem 4.6 we get that f is expandable in Taylor series about 0on R and the following Taylor expansion about 0:

cosx = 1− x2

2!+x4

4!− ...+ (−1)n

x2n

(2n)!+ ... =

∞∑n=0

(−1)nx2n

(2n)!, ∀x ∈ R.

b) We deduce similarly that f is expandable in Taylor series about0 on R and has the following Taylor expansion about 0:

sinx = x− x3

3!+x5

5!− ...+ (−1)n

x2n+1

(2n+ 1)!+ ... =

∞∑n=0

(−1)nx2n+1

(2n+ 1)!, ∀x ∈ R.

c) Since f (n) (x) = ex, ∀x ∈ R, ∀n ∈ N, we deduce that∣∣∣f (n) (x)∣∣∣ ≤ eα, ∀x ∈ [−α, α] , ∀α > 0.

So f is expandable in Talor series about 0 on each interval [−α, α] ,∀α > 0, hence f is expandable in Talor series about 0 on R. We easilydeduce the following Taylor expansion about 0:

ex = 1 +x

1!+x2

2!+ ...+

xn

n!+ ... =

∞∑n=0

xn

n!, ∀x ∈ R.

d) We get the following Taylor expansion about 0:

(1 + x)α =∞∑n=0

(αn

)xn, ∀x with |x| < 1, (4.5)

where

(αn

)denotes the generalized combinations(αn

):=

α (α− 1) ... (α− n+ 1)

n!.


Relation (4.5) is also called binomial expansion of power α andgeneralizes the well-known Newton’s binomial formula.

e) Since f ′ (x) = 1/√1 + x2, x ∈ R, for |x| < 1 we get, by using

the binomial expansion of power −1/2,

ln(x+

√1 + x2

)= x+

∞∑n=1

(−1)n(2n− 1)!!

2n (2n+ 1) · n!x2n+1, ∀x ∈ (−1, 1) ;

f) Since f (x) = 2x−2+2(x−1)2

= 2x−1 + 2

(x−1)2, ∀x ∈ R\ {1}, we easily

deduce the Taylor expansions about 0:

f (x) =−2

1− x+ 2 (1− x)2 = −2

∞∑n=0

xn + 2

∞∑n=0

(αn

)xn

=

∞∑n=0

2

[(αn

)− 1

]xn, ∀x ∈ (−1, 1) ;

g) By decomposing f into simple ratios, we get

f (x) =2x+ 1

x2 − 3x+ 2=

−3

x− 1+

5

x− 2, ∀x ∈ R\ {1, 2} .

Hence, we deduce

f (x) =3

1− x+

−5/2

1− x/2

= 3∞∑n=0

xn − 5

2

∞∑n=0

(x2

)n=

∞∑n=0

(3− 5

2n+1

)xn,

for all x ∈ (−1, 1) ;

h) x2e−x = x2∑∞

n=0(−x)nn! =

∑∞n=0

(−1)n

n! xn+2, ∀x ∈ R;i)

sin 2x+ x cos 2x =

∞∑n=0

(−1)n(2x)2n+1

(2n+ 1)!+ x

∞∑n=0

(−1)n(2x)2n

(2n)!

=∞∑n=0

(−1)n 22n2n+ 3

(2n+ 1)!x2n+1, ∀x ∈ R;

j) f (x) = ln 1+x1−x =

∑∞n=0

22n+1x

2n+1, x ∈ (−1, 1) ;

k)

f (x) =1

2sin 2x =

1

2

∞∑n=0

(−1)n(2x)2n+1

(2n+ 1)!

=

∞∑n=0

(−1)n 22n

(2n+ 1)!x2n+1, ∀x ∈ R;

4.3. EXERCISES 61

l) We calculate the first derivative

f ′ (x) = 2 (1 + ex) = 2 + 2∞∑n=0

xn

n!, ∀x ∈ R

so, by Theorem 4.1, we deduce

f (x) = 2x+ 2∞∑n=0

xn+1

(n+ 1)!+ C, C ∈ R, ∀x ∈ R.

Since f (0) = 4, it follows that C = 4. Therefore,

(1 + ex)2 = 4 + 2x+ 2

∞∑n=0

xn+1

(n+ 1)!, ∀x ∈ R.

CHAPTER 5

Continuous Functions

5.1. The Euclidean structure of Rp

Let p ∈ N, p ≥ 2 be a fixed number. One defines

Rp :=R× R× ...× R︸︷︷︸ =p times

{(x1, x2, ..., xp

), x1, x2, ..., xp ∈ R

}.

Then the set Rp, endowed with the addition,(x1, x2, ..., xp

)+(y1, y2, ..., yp

):=(x1 + y1, x2 + y2, ..., xp + yp

),

for all(x1, x2, ..., xp

),(y1, y2, ..., yp

)∈ Rp and the multiplication by real scalars,

α ·(x1, x2, ..., xp

):=(αx1, αx2, ..., αxp

),

for all α ∈ R, (x1, x2, ..., xp) ∈ Rp, becomes a p-dimensional linear (or vector)space. The standard basis of Rp is

B := {e1, e2, ..., ep} ,where

e1 := (1, 0, ..., 0) , e2 := (0, 1, ..., 0) , ..., ep := (0, ..., 0, 1) .

The element x =(x1, x2, ..., xp

)of Rp is called vector and its components,

x1, x2, ..., xp are called components of vector x.

Definition 5.1. The function ⟨·, ·⟩ : Rp × Rp → R,

⟨x, y⟩ :=p∑i=1

xiyi,

∀x =(x1, x2, ..., xp

), y =

(y1, y2, ..., yp

)is called the Euclidean inner pro-

duct on Rp.

For example, if x = (−1, 3) ∈ R2, y = (1,−1) ∈ R2, then

⟨x, y⟩ = (−1) · 1 + 3 · (−1) = −4.

For x = (−1, 2,−1) ∈ R3, y = (2, 2,−1) ∈ R3,

⟨x, y⟩ = (−1) · 2 + 2 · 2 + (−1) · (−1) = 3.

63

64 5. CONTINUOUS FUNCTIONS

The Euclidean inner product on Rp has the following properties.[IP1] ⟨x, x⟩ ≥ 0, ∀x ∈ Rp, ⟨x, x⟩ = 0 if and only if x = 0;[IP2] ⟨x, y⟩ = ⟨y, x⟩ , ∀x, y ∈ Rp;[IP3] ⟨α1x1 + α2x2, y⟩ = α1 ⟨x1, y⟩+α2 ⟨x2, y⟩ , ∀α1, α2 ∈ R, x1, x2, y ∈ Rp;[IP4] ⟨x, β1y1 + β2y2⟩ = β1 ⟨x, y1⟩+ β2 ⟨x, y2⟩ , ∀β1, β2 ∈ R, x, y1, y2 ∈ Rp.

Definition 5.2. The function ∥·∥ : Rp → [0,∞),

∥x∥ :=√

⟨x, x⟩, ∀x ∈ Rp

is called the Euclidean norm on Rp.

So, if x =(x1, x2, ..., xp

)∈ Rp, then

∥x∥ =

√√√√ p∑i=1

(xi)2.

For example, if x = (−1, 3) ∈ R2, then ∥x∥ =√

(−1)2 + 32 = 10. For

x = (−1, 2,−1) ∈ R3, ∥x∥ =√(−1)2 + 22 + (−1)2 =

√6.

The Euclidean norm on Rp has the following properties.[N1] ∥x∥ ≥ 0, ∀x ∈ Rp, ∥x∥ = 0 if and only if x = 0;[N2] ∥αx∥ = |α| · ∥x∥ , ∀α ∈ R, x ∈ Rp;[N3] ∥x+ y∥ ≤ ∥x∥+ ∥y∥ , ∀x, y ∈ Rp.The pair (Rp, ∥·∥) is called the p−dimensional real Euclidean space

and is it also simply denoted by Rp.In the Euclidean space Rp we can introduce the notion of angle of any pair

of nonzero vectors, x, y, through

cos (x, y) :=⟨x, y⟩

∥x∥ · ∥y∥.

The vectors x, y are called orthogonal and we denote x ⊥ y iff ⟨x, y⟩ = 0.For example, if x = (−1, 2) ∈ R2, y = (2, 1) ∈ R2, then ⟨x, y⟩ = 0, so x ⊥ y.And if x = (−1, 2, 1) ∈ R3, y = (3, 2,−1) ∈ R3, then ⟨x, y⟩ = 0, so x ⊥ y.

In the Euclidean space Rp the following relations hold.1. The Cauchy-Buniakowski-Schwarz inequality,

|⟨x, y⟩| ≤ ∥x∥ · ∥y∥ , ∀x, y ∈ Rp.2. The parallelogram identity,

∥x+ y∥2 + ∥x− y∥2 = 2(∥x∥2 + ∥y∥2

), ∀x, y ∈ Rp.

3. The cosine’s law,

∥x+ y∥2 = ∥x∥2 + ∥y∥2 + 2 ∥x∥ · ∥y∥ · cos (x, y) , ∀x, y ∈ Rp.

5.1. THE EUCLIDEAN STRUCTURE OF Rp 65

Definition 5.3. The function d : Rp × Rp → [0,∞),

d (x, y) = ∥x− y∥ , ∀x, y ∈ Rp

is called the Euclidean metric on Rp.

So, if x =(x1, x2, ..., xp

), y =

(y1, y2, ..., yp

)∈ Rp, then

d (x, y) =

√√√√ p∑i=1

(xi − yi)2.

For example, if x = (−1, 3) ∈ R2, y = (1,−1) ∈ R2, then

d (x, y) =

√(−1− 1)2 + (3 + 1)2 = 2

√5.

For x = (−1, 2,−1) ∈ R3, y = (2, 2,−1) ∈ R3,

d (x, y) =

√(−1− 2)2 + (2− 2)2 + (−1 + 1)2 = 3.

The Euclidean metric on Rp has the following properties.[M1] d (x, y) ≥ 0, ∀x, y ∈ Rp, d (x, y) = 0 if and only if x = y;[M2] d (x, y) = d (y, x) , ∀x, y ∈ Rp;[M3] d (x, y) ≤ d (x, z) + d (z, y) , ∀x, y, z ∈ Rp.One can introduce the following sets of high importance for the study of

topological notions, like convergence, limit, Cauchy sequence etc.1. The open ball of center x0 ∈ Rp and radius r > 0,

Br (x0) := {x ∈ Rp, d (x, x0) < r} ;

2. The closed ball of center x0 ∈ Rp and radius r > 0,

Br (x0) := {x ∈ Rp, d (x, x0) ≤ r} ;

3. The sphere of center x0 ∈ Rp and radius r > 0,

Sr (x0) := {x ∈ Rp, d (x, x0) = r} .

By using the Euclidean norm, these sets can be defined by:

Br (x0) : = {x ∈ Rp, ∥x− x0∥ < r} ,Br (x0) : = {x ∈ Rp, ∥x− x0∥ ≤ r} ,Sr (x0) : = {x ∈ Rp, ∥x− x0∥ = r} .

For example, if p = 2, the balls Br (x0) (Br (x0) , Sr (x0)) represent opendiscs (of center) of center x0 and radius r (respectively the closed disc and the


circle of center x0 and radius r). Remark that we can also consider p = 1, bygetting, in the case of R, that

Br (x0) = (x0 − r, x0 + r) ,

Br (x0) = [x0 − r, x0 + r] ,

Sr (x0) = {x0 − r, x0 + r} .

Definition 5.4. The set A ⊂ Rp is called bounded iff there exists M > 0,such that

∥x∥ ≤M, ∀x ∈ A.

Therefore, the bounded sets of Rp are the subsets of balls, i.e. A ⊂ Rp isbounded iff there exists x0 ∈ Rp and r > 0, such that

A ⊂ Br (x0) .

5.2. Vector sequences

Definition 5.5. A vector sequence is a function x : N → Rp. For n ∈ N,xn := x (n) ∈ Rp is called the general term of the sequence.

A vector sequence it’s also denoted through its general term, by x = (xn)n∈N .

If xn =(x1n, x

2n, ..., x

pn

), n ∈ N, then the sequences

(x1n)n∈N ,

(x2n)n∈N , ...,

(xpn)n∈N are called the components of the vector sequence (xn)n∈N .

Definition 5.6. The vector sequence (xn)n∈N from Rp is called convergentiff there is x ∈ Rp, such that (xn)n∈N is convergent to x, that is, ∀ε > 0,∃n0 = n0 (ε) ∈ N, ∀n ∈ N, n ≥ n0,

∥xn − x∥ < ε.

If the vector x in Definition 5.6 does exist, it is unique. We call it the limitof the sequence (xn)n∈N and we denote

x = limn→∞

xn or xn → x.

The sequences that are not convergent are called divergent.

Remark 5.1. All the properties presented in Chapter 2, regarding the con-vergent real sequences are verbatim extended to the case of vector sequences.Due to the lack of any total order relation on Rp, compatible with the algebraicstructure of Rp, certain properties (such as the monotony) known for the realsequences as well as or for the real functions do not hold neither for the vectorsequences nor for vector functions.

5.2. VECTOR SEQUENCES 67

Remark 5.2. A well-known property for real sequences holds for vector se-quences: xn → x if and only if ∥xn − x∥ → 0. Therefore, in order to prove thata vector sequence is convergent, it is enought to find a real sequence (an)n∈N,that is convergent to 0, such that

∥xn − x∥ < an, ∀n ∈ N.

Theorem 5.1. If xn =(x1n, x

2n, ..., x

pn

)→(x1, x2, ..., xp

)= x, then xkn → xk,

∀k ∈ 1, p.Conversely, if xkn → xk, ∀k ∈ 1, p, then

xn =(x1n, x

2n, ..., x

pn

)→(x1, x2, ..., xp

)= x.

By this theorem, the study of the convergence of vector sequences is reducedto the study of the component sequences, that are real sequences.

For example, let us study the convergence of the vector sequence

xn =

(sinn

n,

(1 +

1

n

)n, n√n

)∈ R3, n ≥ 2.

If one passes to components, we get

x1n =sinn

n→ 0,

x2n =

(1 +

1

n

)n→ 0,

x3n = n√n→ 1,

so the sequence (xn)n∈N is convergent and(sinnn ,(1 + 1

n

)n, n√n)→ (0, 0, 1) .

On the other hand, the sequence((−1)n , 1

n2+1, 1− 1

n

)∈ R3, n ≥ 1, is

divergent due to the the divergence of the first component, (−1)n , n ≥ 1.

Theorem 5.2. (Bolzano-Weierstrass Theorem). Every bounded vector se-quence has a convergent subsequence.

Definition 5.7. The vector sequence (xn)n∈N from Rp is called Cauchy (fun-damental) sequence iff ∀ε > 0, ∃n0 = n0 (ε) ∈ N, ∀n ∈ N, n ≥ n0, ∀k ∈ N,

∥xn+k − xn∥ < ε.

Theorem 5.3. A vector sequence is Cauchy sequence if and only if each com-ponent of it is a Cauchy sequence.


Theorem 5.4. (Cauchy’s test). A vector sequence is convergent if and only ifit is a Cauchy sequence.

Due to this Theorem, (Rp, ∥·∥) is a Banach space and (Rp, d) is a completemetric space, ∀p ≥ 1.

5.3. Continuity of real functions of vector variable

Let us consider a function f : A → R, where A ⊆ Rp, p ≥ 2. f is calledreal function of vector variable. We also consider Rp endowed with theEuclidean norm.

Definition 5.8. Let a ∈ Rp be arbitrary. A neighborhood of a point a ∈ Rpa is a set V ⊆ Rp that includes an open ball Br (a) .

We denote by V (a) the family of all neighborhoods of a.

We have the following properties of the families V (a) .

[NB1] V (a) = ∅; ∀V ∈ V (a) , a ∈ V ;[NB2] If V ∈ V (a) and W ⊃ V, then W ∈ V (a) ;[NB3] Every intersection of a finite number of neighborhoods of a is still

a neighborhood of a;[NB4] ∀ V ∈ V (a) , ∃W ∈ V (a) , such thatW ⊆ V and ∀x ∈W, W ∈ V (x) .

The Euclidean space Rp fulfills the separation property (Haussdorff proper-ty): ∀x, y ∈ Rp, x = y, ∃V ∈ V (x), ∃W ∈ V (y), such that

V ∩W = ∅.In what follows, we will consider functions f : A → R, where A ⊆ Rp,

p ≥ 2. The fundamental notion that characterizes the local behavior of thesefunctions around a point of A, is that of continuity.

Definition 5.9. f is continuous at the point a ∈ A iff ∀ε > 0, ∃δ = δ (ε) > 0,such that ∀x ∈ A, with ∥x− a∥ < δ, one has |f (x)− f (a)| < ε.

We say that f is continuous on the set A iff f is continuous at everypoint a ∈ A.

The contrary notion to continuity is that of discontinuity.

Definition 5.9 can be restated in terms of neighborhoods as follows.

Theorem 5.5. The function f : A → R, where A ⊆ Rp, p ≥ 2 is continuousat a point a ∈ A if and only if ∀V ∈ V (f (a)) , ∃U ∈ V (a) , such that ∀x ∈ U ,f (x) ∈ V.

5.4. LIMITS OF REAL FUNCTIONS OF VECTOR VARIABLE 69

In terms of convergent sequences, the notion of continuity can be definedas follows.

Theorem 5.6. (Heine’s characterization of continuity). f : A → R, A ⊆ Rp,p ≥ 2 is continuous at a point a ∈ A if and only if ∀ (an)n∈N ⊆ A, with an → a,one has f (an) → f (a) .

Regarding the algebraic operations with continuous functions, we have thefollowing results.

Proposition 5.1. 1) Let f, g : A→ R be two continuous functions at the pointa ∈ A ⊆ Rp and α, β ∈ R be arbitrary, p ≥ 2. Then αf + βg is continuous atthe point a.

2) Let f : A → R be continuous function at the point a ∈ A ⊆ Rp, p ≥ 2,such that f (x) = 0, on A. Then 1/f is continuous at the point a.

3) Let f : A → R be continuous function at the point a ∈ A ⊆ Rp, p ≥ 2.Then |f | is continuous at the point a (|f | (x) := |f (x)| , x ∈ A).

Remark 5.3. In order to prove that a real function of vector variable, f, iscontinuous at the point a, it is enough to find a function h : A → R, withlimx→a h (x) = 0 and U ∈ V (a) , such that

|f (x)− f (a)| ≤ h (x) , ∀x ∈ U.

5.4. Limits of real functions of vector variable

Definition 5.10. a ∈ Rp, p ≥ 2 is called cluster (accumulation) point tothe set A ⊆ Rp iff ∀r > 0, Br (a) ∩ (A\{a}) = ∅.

The set of all cluster points of A is called the derivative of the A and itis denoted by A′.

Definition 5.11. Let f : A → R, A ⊆ Rp, p ≥ 2, and a ∈ A′. The numberl ∈ R is called the limit of the function f at the point a and we denote thisby l = limx→a f (x) iff ∀V ∈ V (l) , ∃U ∈ V (a) , such that ∀x ∈ U, x = a, onehas f (x) ∈ V.

Remark 5.4. If the limit exists, it is unique.

Theorem 5.7. Let f : A→ R be a function and a ∈ A′, where A ⊆ Rp, p ≥ 2.Then:


a) limx→a f (x) = l ∈ R if and only if ∀ε > 0, ∃δ = δ (ε) > 0, such that∀x ∈ A\{a} with ∥x− a∥ < δ, one has

|f (x)− l| < ε.

b) limx→a f (x) = ∞ if and only if ∀ε > 0, ∃δ = δ (ε) > 0, such that∀x ∈ A\{a} with ∥x− a∥ < δ, one has

f (x) > ε.

c) limx→a f (x) = −∞ if and only if ∀ε > 0, ∃δ = δ (ε) > 0, such that∀x ∈ A\{a} with ∥x− a∥ < δ, one has

f (x) < −ε.

Theorem 5.8. (Heine’s characterization of limits). Let f : A → R be afunction and a ∈ A′, where A ⊆ Rp, p ≥ 2. Then l = limx→a f (x) if and onlyif ∀ (an)n∈N ⊂ A\ {a}, with an → a, one has f (an) → l.

Remark 5.5. In order to prove that a function f has the limit l at the pointa, it is enough to find a function h with limx→a h (x) = 0 and U ∈ V (a) , suchthat

|f (x)− l| ≤ h (x) , ∀x ∈ U\ {a} .

Definition 5.12. Let f : A → R be a function, a ∈ A′, where A ⊆ Rp, andv ∈ Rp\{0}, p ≥ 2. If there exists limt→0 f (a+ tv) , it is called the limit of fat the point a on the direction of v.

Proposition 5.2. If there exists limx→a f (x), then there exists the limit of fat a on any direction v, and limx→a f (x) = limt→0 f (a+ tv) .

Remark 5.6. The converse of this Proposition does not hold true. Indeed, letus consider the function f : R2\

{(x, y) ∈ R2, x3 + 2y = 0

}→ R,

f (x, y) =x3 − y

x3 + 2y.

Then, for a = (0, 0) and ∀v =(v1, v2

)∈ Rp\{0}, we have

limt→0

f (a+ tv) = limt→0

f((0, 0) + t

(v1, v2

))= lim

t→0

(tv1)3 − v2

(tv1)3 + 2v2=

{ −12 , if v2 = 0,1, if v2 = 0.

5.5. CONTINUITY OF VECTOR FUNCTIONS OF VECTOR VARIABLE 71

But we can easily see that lim(x,y)→(0,0) does not exist, since on the sequence(1n ,

αn3

)→ (0, 0) , where α ∈ R,

f

((1

n,α

n3

))=

1− α

1 + 2α,

that is not a constant for all α = −12 .

Theorem 5.9. (Characterization of continuity with limit). Let f : A → R bea function, where A ⊆ Rp, p ≥ 2. Then f is continuous at a ∈ A ∩ A′ if andonly if limx→a f (x) = f (a) .

5.5. Continuity of vector functions of vector variable

Let us consider now a function f : A→ Rm, where A ⊆ Rp, p, m ≥ 2. f iscalled vector function of vector variable, since ∀x =

(x1, x2, ..., xp

)∈ A,

f(x1, x2, ..., xp

)∈ Rm. So, if we write the vector f

(x1, x2, ..., xp

)by specifying

its components,

f(x1, x2, ..., xp

)=(f1(x1, x2, ..., xp

), f2

(x1, x2, ..., xp

), ..., fm

(x1, x2, ..., xp

)),

we get m real functions of vector variable, f1, f2, ..., fm : A→ R that are alsocalled the components of the function f.

We also consider Rp and Rm endowed with the Euclidean norms, unitarilydenoted by ∥·∥.

Definition 5.13. f is continuous at the point a ∈ A iff ∀ε > 0, ∃δ = δ (ε) >0, such that ∀x ∈ A, with ∥x− a∥ < δ, one has ∥f (x)− f (a)∥ < ε.

We say that f is continuous on the set A iff f is continuous at everypoint a ∈ A.

The contrary notion to continuity is that of discontinuity.

Theorem 5.10. The function f : A ⊆ Rp → Rm, p, m ≥ 2 is continuous atthe point a ∈ A if and only if each component of it is continuous at the pointa.

More exactly, if we consider the components of f, f =(f1, f2, ..., fm

), then

f is continuous at the point a if and only if f1, f2, ..., fm are continuous atthe point a.

Due to this Theorem, the continuity to vector functions of vector variableis studied by passing to components, that are real functions of vector variable.

Definition 5.13 can be restated in terms of neighborhoods as follows.


Theorem 5.11. f : A → Rm, where A ⊆ Rp, p, m ≥ 2. f is continuous ata point a ∈ A if and only if ∀V ∈ V (f (a)) , ∃U ∈ V (a) , such that ∀x ∈ U ,f (x) ∈ V.

In terms of convergent sequences, the notion of continuity can be definedas follows.

Theorem 5.12. (Heine’s characterization of continuity). f : A ⊆ Rp → Rm,p, m ≥ 2, is continuous at a point a ∈ A if and only if ∀ (an)n∈N ⊆ A, withan → a, one has f (an) → f (a) .

Regarding the algebraic operations with continuous functions, we have thefollowing results.

Proposition 5.3. 1) Let f, g : A → Rm be two continuous functions at thepoint a ∈ A ⊆ Rp and α, β ∈ R be arbitrary, p, m ≥ 2. Then αf + βg iscontinuous at the point a.

2) Let f : A → Rm be continuous function at the point a ∈ A ⊆ Rp, p,m ≥ 2. Then ∥f∥ is continuous at the point a (∥f∥ (x) := ∥f (x)∥ , x ∈ A).

Remark 5.7. In order to prove that a vector function of vector variable, f,is continuous at the point a, it is enough to find a function h : A → R, withlimx→a h (x) = 0 and U ∈ V (a) , such that

∥f (x)− f (a)∥ ≤ h (x) , ∀x ∈ U.

5.6. Limits of vector functions of vector variable

Definition 5.14. Let f : A→ Rm, A ⊆ Rp, p, m ≥ 2, and a ∈ A′. The numberl ∈ Rm is called the limit of the function f at the point a and we denotethis by l = limx→a f (x) iff ∀V ∈ V (l) , ∃U ∈ V (a) , such that ∀x ∈ U, x = a,one has f (x) ∈ V.

Remark 5.8. If the limit exists, it is unique.

Theorem 5.13. The function f : A→ Rm, A ⊆ Rp, p, m ≥ 2 has limit at thepoint a ∈ A′ if and only if each component of it has limit at this point.

More exactly, if we consider the components of f, f =(f1, f2, ..., fm

), then

limx→a f (x) =(l1, l2, ..., lm

)if and only if limx→a f

1 (x) = l1, limx→a f2 (x) =

l2, ..., limx→a fm (x) = lm.

5.7. EXERCISES 73

Due to this Theorem, the limits to vector functions of vector variable arestudied by passing to components, that are real functions of vector variable.

Theorem 5.14. (Heine’s characterization of limits). Let f : A → Rm be afunction and a ∈ A′, where A ⊆ Rp, p, m ≥ 2. Then l = limx→a f (x) if andonly if ∀ (an)n∈N ⊂ A\ {a}, with an → a, one has f (an) → l.

Remark 5.9. In order to prove that a function f has the limit l at the pointa, it is enough to find a function h with limx→a h (x) = 0 and U ∈ V (a) , suchthat

∥f (x)− l∥ ≤ h (x) , ∀x ∈ U\ {a} .

Theorem 5.15. (Characterization of continuity with limit). Let f : A → Rmbe a function, where A ⊆ Rp, p, m ≥ 2. Then f is continuous at a ∈ A ∩A′ ifand only if limx→a f (x) = f (a).

5.7. Exercises

(1) Study the convergence of the following vector sequences:

a) xn =(2n+1n2 , cosn

2

n2 , 1n

); b) xn =

(n−24n+1 ,

(−1)n

n ,arcsin 1

n2n

);

c)

(n√n2, 1+n

2

n2 , 1

(1+ 1n)

2n+1

); d)

(n sin 1

n ,1n! , (−1)n

2);

e)(sinnn2 ,

1+(−1)n+(−1)n+1

3n , 23n

); f)

(n2 ln

(1 + 1

n2

), 1n2 ,

(−1)n

n2

).

A: xn → (0, 0, 0) ; b) xn →(14 , 0,

12

); c) xn →

(1, 1, 1

e2

);

d) divergent; e) xn → (0, 0, 0) ; f) xn → (1, 0, 0) .(2) Determine the domains of definition to the following functions:

a) f (x, y) =√

1− x2

a2− y2

b2, a, b > 0;

b) f (x, y, z) = 1a2−x2−y2−z2 , a > 0; c) f (x, y, z) = ln (xyz) ;

d) f (x, y) = arcsin x2+y2

2 + arccos x2+y2

8 ;

e) f (x, y, z) = 1√x+ 1

3√y +

14√z;

f) f (x, y) =√a2 − x2 − y2 − 1√

x2+y2−b2, a > b > 0.

A: A ={(x, y) ∈ R2, x2

a2+ y2

b2≤ 1};

b) A ={(x, y, z) ∈ R3, x2 + y2 + z2 = a2

};

c) A ={(x, y, z) ∈ R3, x > 0, y > 0, z > 0

};

d) A ={(x, y) ∈ R2, x2 + y2 ≤ 2

};

e) A ={(x, y, z) ∈ R3, x > 0, y = 0, z > 0

};


f) A ={(x, y) ∈ R2, b <

√x2 + y2 ≤ a

}.

(3) By using the definition, prove that lim(x,y)→(2,2)xy = 1.

A: Let ε > 0 be arbitrary. We have the following estimate∣∣∣∣xy − 1

∣∣∣∣ ≤ 1

|y|(|x− 2|+ |y − 2|) ,

hence there exists 0 < δ < min{1, ε2}, such that if |x− 2| < δ and

|y − 2| < δ, 1|y| < 1. So

∣∣∣xy − 1∣∣∣ < ε.

(4) Study the existence of the following limits:

a) lim(x,y)→(0,0)2x+y3x−y ; b) lim(x,y)→(0,0)

x2+3yx2−2y

;

c) lim(x,y)→(0,0)sin(x2+y2)x2+y2

; d) lim(x,y)→(0,0)x3+y3

x2+y2;

e) lim(x,y)→(0,0)cos(x+y)−1

x+y ; f) lim(x,y)→(0,0)tg (x2y2)

xy ;

g) lim(x,y)→(0,0)x2y2

x2+y2; h) lim(x,y)→(0,0) (1 + xy)

1√x+

√y ;

i) lim(x,y)→(0,0) (x+ y) cos 1x2+y2

; j) lim(x,y)→(0,0)xy

x2+y2;

k) lim(x,y)→(0,0)sin(x3+y3)x2+y2

; l) lim(x,y,z)→(0,0,0) (1 + xyz)1√

x+y+z ;

m) lim(x,y)→(0,0)(x2+y2) sin(x−y)

x−y ; n) lim(x,y)→(2,1)(x−2)4(y−1)2

(x−2)4+(y−1)2;

o) lim(x,y)→(0,0)x2+y2

x2−y2 ; p) lim(x,y)→(0,0)x4−y4x4+y4

;

q) lim(x,y)→(0,0)x2y2

x4+y4; r) lim(x,y)→(0,0)

√x2y2+1−1

x2+y2.

A: a) The limit doesn’t exist. f((

1n ,

αn

))→ 2+α

3−α ; b) The limit

doesn’t exist f((

1n ,

αn2

))→ 1+3α

1−2α ; c) 1;

d)∣∣∣x3+y3x2+y2

∣∣∣ ≤ |x|3+|y|3x2+y2

=(|x|+|y|)(x2−|xy|+y2)

x2+y2< |x|+ |y|. So,

lim(x,y)→(0,0)

x3 + y3

x2 + y2= 0;

e) cos(x+y)−1x+y =

−2 sin2 x+y2

x+y = (−2) (x+ y)sin2 x+y

2

(x+y)2. So,

lim(x,y)→(0,0)

cos (x+ y)− 1

x+ y= 0;

f)tg (x2y2)

xy = xytg (x2y2)x2y2

. So, lim(x,y)→(0,0)tg (x2y2)

xy = 0;

g) x2y2

x2+y2= |xy| |xy|

x2+y2≤ |xy| . So, lim(x,y)→(0,0)

x2y2

x2+y2= 0;

5.7. EXERCISES 75

h) (1 + xy)1√

x+√

y =[(1 + xy)

1xy

] xy√x+

√y. Since

xy√x+

√y= 4√x3y3

4√xy

( 4√x)

2+(

4√y)2 ≤ 4

√x3y3,

we get that lim(x,y)→(0,0) (1 + xy)1√

x+√

y = 1; i) 0; j) The limit doesn’t

exist. f((

1n ,

αn

))→ α

1+α2 ;

k)sin(x3+y3)x2+y2

=sin(x3+y3)x3+y3

x3+y3

x2+y2. So, lim(x,y)→(0,0)

sin(x3+y3)x2+y2

= 0;

l) (1 + xyz)1√

x+y+z =[(1 + xyz)

1xyz

] xyz√x+y+z

. Since

|xyz|√x+ y + z

≤ (x+ y + z)3

27√x+ y + z

,

we deduce that lim(x,y,z)→(0,0,0) (1 + xyz)1√

x+y+z = 1; m) 0;

n) lim(x,y)→(2,1)(x−2)4(y−1)2

(x−2)4+(y−1)2= lim(u,v)→(0,0)

u4v2

u4+v2. Since

u4v2

u4 + v2= u2 |v| u2 |v|

(u2)2 + v2≤ u2 |v| ,

it follows that lim(x,y)→(2,1)(x−2)4(y−1)2

(x−2)4+(y−1)2= 0; o) The limit doesn’t

exist. f((

1n ,

αn

))→ 1+α2

1−α2 ; p) The limit doesn’t exist. f((

1n ,

αn

))→

1−α4

1+α4 ; q) The limit doesn’t exist. f((

1n ,

αn

))→ α2

1+α4 .

r) √x2y2 + 1− 1

x2 + y2=

|xy|√x2y2 + 1 + 1

|xy|x2 + y2

≤ |xy|√x2y2 + 1 + 1

.

So, lim(x,y)→(0,0)

√x2y2+1−1

x2+y2= 0.

(5) Let f : R2\ {(0, 0)} → R, f (x, y) = xy2

x2+y4. Prove that f has limit at

(0, 0) on any direction, but it doesn’t exist lim(x,y)→(0,0) f (x, y) .(6) Study the continuity of the following functions:

a) f (x, y) =√x2 − xy + y2;

b) f (x, y) =

{1−cos(x3+y3)

x2+y2, if (x, y) = (0, 0) ,

0, if (x, y) = (0, 0) ;

c) f (x, y) =

{(1 + xy)

1√x+

√y , if x > 0 and y > 0,

1, if x = 0 or y = 0;


d) f (x, y) =

{ (x2 + y2

)sin 1

x2+y2, if (x, y) = (0, 0) ,

0, if (x, y) = (0, 0) ;

e) f (x, y) =

{e

1x2+y2+z2 , if (x, y, z) = (0, 0, 0) ,

1, if (x, y, z) = (0, 0, 0) .

A: a) continuous on R2; b) continuous on R2; c) continuous onR2; e) continuous on R3\ {(0, 0, 0)} .

(7) Study the continuity of the following functions:

a) f (x) =

(

sin(2x2+x)2x2+x

,ln(1+x2)

x2, 1−cos 2x

x2

), if x = 0,

(1, 1, 1) , if x = 0;

b) f (x) =

{ (ex−1x ,

√1+x2−1x2

, x2e−1x2

), if x = 0,

(1, 1, 0) , if x = 0;

c) f (x) =

(

1−cosxx2

,(1+x2)

π−1

x2, tg 3x

2x

), if x = 0,(

12 , π,

32

), if x = 0;

d) f (x, y) =

(

arcsin(x3+y3)x2+y2

, x3y2

x2+y3

), if (x, y) = (0, 0) ,

(0, 0) , if (x, y) = (0, 0) ;

e) f (x, y) =

(

sin(x2+y2)x+y , xy2

x2+y2

), if (x, y) = (0, 0) ,

(0, 0) , if (x, y) = (0, 0) ;

f) f (x, y) =

(

x2+y2√x2+y2+1−1

,(1 + x2y2

)− 1x2+y2

), if (x, y) = (0, 0) ,

(0, 1) , if (x, y) = (0, 0) ;A: a) continuous on R; b) continuous on R; c) continuous on R;

d) continuous on R2; e) continuous on R2; f) continuous on R2.

CHAPTER 6

Differentiable functions

6.1. Differentiability of real functions of real variable

Let us consider a material point that moves along a line. Then the distanceit crosses from time 0 to time t is a function of time, denoted by x (t) . Thenat the moment t0, the mean speed on the interval [t0, t0 + h] is the quantity

x (t0 + h)− x (t0)

h

so the speed at the moment t0 will be, by definition,

v (t0) := limh→0

x (t0 + h)− x (t0)

h,

obviously, if the limit above des exist. Similarly, we get the definition of theacceleration of the material point at the moment t0,

a (t0) := limh→0

v (t0 + h)− v (t0)

h.

By starting from this practical example, we are concerned about the es-timate of the approximative growth of a certain measure with respect to thegrowth of other measure. In what follows A will denote a non-denegerateinterval or, more generally, a non-empty subset of R, consisting only of accu-mulation points.

Definition 6.1. Let f : A → R be an arbitrary function, A ⊆ R, open set,a ∈ A. We say that f is differentiable at a if there exists

limx→a

f (x)− f (a)

x− a

and it is finite. The value of this limit is called the derivative of f at a andit is denoted by f ′ (a) or df

dx (a) .If the limit does exists, but it is infinite, then we say that f has derivative

at a and we denote it f ′ (a) or dfdx (a), too. If the limit doesn’t exist, then f is

not differentiable at a and has not derivative at a.

77

78 6. DIFFERENTIABLE FUNCTIONS

f is called differentiable on the set I if it is differentiable at every pointa ∈ A and the function that maps each x ∈ A into its derivative f ′ (x) ∈ R is

called the derivative of the function f and it is denoted by f ′ or dfdx .

For example, let f : R → R, f (x) = x3 and a = 1. Then

limx→1

f (x)− f (1)

x− 1= lim

x→1

x3 − 1

x− 1= lim

x→1

(x2 + x+ 1

)= 3 ∈ R,

so f is differentiable at 1 and f ′ (1) = 3. If a ∈ R is arbitrary, then

f ′ (a) = limx→a

f (x)− f (a)

x− a= lim

x→a

x3 − a3

x− a= lim

x→a

(x2 + ax+ a2

)= 3a2,

so

f ′ (a) = 3a2, ∀a ∈ R.

We deduce the derivative function, f ′ : R → R, f ′ (x) = 3x2, ∀x ∈ R.

Theorem 6.1. Every differentiable function at a point is continuous at thatpoint.

Indeed, this follows immediately from relation

f (x) = f (a) +f (x)− f (a)

x− a(x− a) , ∀x, a ∈ I, x = a.

Remark 6.1. The converse of this result is false. Indeed, the function f : R →R, f (x) = |x| is continuous at 0, but it is not differentiable at 0, since

limx→0

f (x) = f (0) = 0,

and

f (x)− f (0)

x− 0=

{ x−0x−0 , if x ≥ 0−x−0x−0 , if x < 0

=

{1, if x ≥ 0−1, if x < 0

hence limx→0

f(x)−f(0)x−0 doesn’t exist.

Let E1 and E2 be two real vector spaces.

Definition 6.2. The operator T : E1 → E2 is called linear iff

T (αx+ βy) = αT (x) + βT (y) , ∀x, y ∈ E1, ∀α, β ∈ R.

We denote by L (E1, E2) the set of all linear operators from E1 to E2.Fie (E1, ∥·∥1) si (E2, ∥·∥2) be two normed vector spaces.

6.1. DIFFERENTIABILITY OF REAL FUNCTIONS OF REAL VARIABLE 79

Definition 6.3. The operator T : E1 → E2 is called continuous at x0 ∈E1 iff ∀ε > 0, ∃δ > 0, such that ∀x ∈ E1, with ∥x− x0∥1 < δ it follows∥T (x)− T (x0)∥ < ε.

Proposition 6.1. A linear operator T : E1 → E2 is continuous if and only ifis continuous at a single point.

Theorem 6.2. Let f : A→ R be a function, A ⊆ R open set, and a ∈ A. Thenf is differentiable at a if and only if there exists a linear operator Ta : R → R,such that

limh→0

f (a+ h)− f (a)− Ta (h)

h= 0.

From Algebra, we know that if an operator T : R → R is linear, then thereexists c = T (1) ∈ R, such that T (x) = c · x. And conversely, if T : R → R is ofform T (x) = c · x, ∀x ∈ R, then T is linear.

Therefore, f will be differentiable if and only if there exists ca = c (a) ∈ R,such that

limh→0

f (a+ h)− f (a)− cah

h= 0. (6.1)

Remark 6.2. In order to make the transition to the case of vector functionsof vector variable, we remark that relation (6.1) can be written equivalently

lim|h|→0

|f (a+ h)− f (a)− cah||h|

= 0.

By Definition 6.1 and Theorem 6.2 we deduce the following result.

Theorem 6.3. Let f : A→ R be a function, A ⊆ R open set, and a ∈ A. f isdifferentiable at a, if and only if Ta (h) = f ′ (a) · h, ∀h ∈ R.

Due to this result, Ta is uniquely determined by f and a.

Definition 6.4. Let f : A → R be a function, A ⊆ R open set. If f isdifferentiable at a ∈ A, then the linear operator Ta : R → R is called thedifferential of f at a and it is also denoted by dfa.

We say that f is differentiable on the set A iff it is differentiable atany a ∈ A.


Therefore, if f is differentiable on A, the following mapping arises, df : A→L (R,R), defined by (df) (a) := dfa, ∀a ∈ A, and it is called the differentialof function f on A and we have

dfa (h) = f ′ (a) · h,∀a ∈ A,∀h ∈ R.

6.1.1. The derivatives of certain elementary functions. 1. Theconstant function f : R → R, f (x) = C is differentiable on R and

C ′ = 0.

2. The power function with natural exponent f : R → R, f (x) = xn,n ∈ N∗ is differentiable on R and

(xn)′ = nxn−1.

3. The root function of order n, n ∈ N∗, f : R∗+ → R, f (x) = n

√x is

differentiable on R and (n√x)′=

1

nn√xn−1

.

4. The logarithm function f : (0,∞) → R, f (x) = lnx is differentiable onR and

(lnx)′ =1

x.

5. f : R → R, f (x) = sinx is differentiable on R and

(sinx)′ = cosx.

6. f : R → R, f (x) = cosx is differentiable on R and

(cosx)′ = − sinx.

7. The exponential of basis e, f : R → R, f (x) = ex is differentiable on Rand

(ex)′ = − sinx.

8. The exponential of basis a > 0, f : R → R, f (x) = ax is differentiableon R and

(ax)′ = ax ln a.

9. The root function f : [0,∞) → R, f (x) =√x is differentiable only for

x > 0 and

√x =

1

2√x.


Definition 6.5. Let f : A → R be an arbitrary function, A ⊆ R, and a ∈ A.f is called differentiable at left at a iff there exists and is finite the limit

f ′l (a) := limx↗a

f (x)− f (a)

x− a.

If there exists, the number f ′l (a) is called the left-hand derivative at thepoint a.

Analogously, f is called differentiable at right at a iff there exists andis finite the limit

f ′r (a) := limx↘a,

f (x)− f (a)

x− a.

If there exists, the number f ′r (a) is called the right-hand derivative at thepoint a.

Theorem 6.4. If f is differentiable at a ∈ A, then f ′s (a) = f ′d (a) = f ′ (a) .And conversely, if f ′s (a) = f ′d (a) is a finite number, then f is differentiable ata and f ′ (a) = f ′s (a) = f ′d (a).

For example, if f : R → R, f (x) = |x| , then

f ′s (0) = limx↗0

|x| − 0

x− 0= −1, f ′s (0) = lim

x→0,x>0

|x| − 0

x− 0= 1.

If f : R → R, f (x) =

{x+ 1, if x ≥ 0x, if x < 0

, then we remark that f is

discontinuous at 0 so it will not be differentiable at 0; by estimating

f ′s (0) = limx↗0

f (x)− f (0)

x− 0= lim

x↗0

x− 1

x− 0= ∞,

it follows that f is not differentiable at left, but it has left-hand derivative at0, equal to ∞; since

f ′d (0) = limx↘0

f (x)− f (0)

x− 0= lim

x↗0

x+ 1− 1

x− 0= 1,

f is differentiable at right at 0 and it has right-hand derivative at 0, equal to1.

6.1.2. Geometrical meaning of the derivative. Let f : A→ R be anarbitrary function, A ⊆ R, and a ∈ A. Consider the fixed point M0 (a, f (a))on the graph of f and the variable point M (x, f (x)) on the graph of f , too.Then, by using the notions from Figure 6.1, we obtain:

f (x)− f (a)

x− a=

MN

M0N= tg θ.


Figure 6.1. Tangent to the graph

If the x tends to a, the M0 tends to M on the graph, so the line MM0 willtend to the tangent line at M0 to the graph; hence, the angle θ will tend to θ0.

Therefore,

f ′ (a) = limx→a

f (x)− f (a)

x− a= lim

x→atg θ = tg θ0 =: m.

The conclusion is the derivative of a function at a point is the slope of thetangent line, m. So, the equation of the tangent line is at (a, f (a)) to thegraph of f is

t : y − f (a) = f ′ (a) (x− a) .

For example, for the function f : R → R, f (x) = x3, the equation of thetangent line at (1, 1) to the graph is

y − 3x+ 2 = 0,

since f ′ (1) = 3.

6.1.3. Semitangents.

Definition 6.6. If f has left-hand derivative at a ∈ A, then the graph of fhas left-hand semitangent at the point T (a, f (a)):

a) If f is differentiable, then the semitangent is closed semiline situated inthe semiplane x < a and having the origin T, and slope m = f ′s (a) .

b) If f ′s (a) = ∞ or −∞, then the semitangent is a semiline having theorigin T and being parallel cu Oy axis.

A similar definition can be stated when f admits right-hand derivative ata.

For example, let us consider the following functions:


1. f : (−∞,−1] → R, f (x) = x2. Since f ′ (−1) = −2, the graph admits inT (−1, 1) semitangent of equation

y − 1 = −2 (x+ 1) , x ≤ −1.

2. f : (−∞, 0) → R, f (x) ={

1x , if x < 00, if x = 0

. We have f ′ (0) = ∞. So, the

graph admits as semitangent in O (0, 0) the negative semiaxis Oy′.

3. f : [0,+∞) → R, f (x) ={ √

x, if x > 0−1, if x = 0

. We have f ′ (0) = ∞ so the

graph admits as semitangent in T (0,−1) the semiaxis Oy.

6.1.4. Angular point. Sharp point. Let f : A → R be an arbitraryfunction, A ⊆ R, and a ∈ A.

Definition 6.7. A point of continuity is called angular point, if there existleft-hand and right-hand derivatives, they are different and at least one is finite.

A sharp point is a point where one of the side derivatives is ∞ and theother one is −∞.

For example, let us consider the function f : R → R, f (x) =∣∣x2 − 4

∣∣. Itsgraph admits the angular points (−2, 0) and (2, 0). Indeed,

f ′s (2) = limx↗2

f (x)− f (2)

x− 2= lim

x↗2

4− x2

x− 2= −2,

and

f ′d (2) = limx↘2

f (x)− f (2)

x− 2= lim

x↘2

x2 − 4

x− 2= 2.

If one considers f : R → R, f (x) =√

|x|, since f ′s (0) = ∞ and f ′d (0) = ∞,it follows that O (0, 0) is sharp point for the graph of f.

6.1.5. Operations with differentiable functions. Let f, g : A → Rbe two functions, A ⊆ R, and a ∈ A, λ ∈ R.

Theorem 6.5. If functions f, g are differentiable at the point a, then f + g,λf, fg are differentiable at ta, and if g (a) = 0, fg is differentiable at a, and

(f + g)′ (a) = f ′ (a) + g′ (a) ,

(λf)′ (a) = λf ′ (a) ,

(fg)′ (a) = f ′ (a) g (a) + f (a) g′ (a) ,(f

g

)′(a) =

f ′ (a) g (a)− f (a) g′ (a)

g2 (a).


Corollary 6.1. If functions f, g are differentiable on A, then f + g, λf , fgare differentiable on A, and if g (x) = 0, ∀x ∈ A, then f

g is differentiable on I,

and

(f + g)′ = f ′ + g′,

(λf)′ = λf ′,

(fg)′ = f ′g + fg′,(f

g

)′=

f ′g − fg′

g2.

Theorem 6.6. (Differentiabiliy of the composed functions). If f : A→ B andg : B → R, where A and B are open subsets of R, f is differentiable at thepoint a ∈ A, and g is differentiable at the point b = f (a) ∈ B, then g ◦ f isdifferentiable at a, and

(g ◦ f)′ (a) = g′ (f (a)) · f ′ (a) .

Corollary 6.2. If f is differentiable on A and g is differentiable on B, theng ◦ f is differentiable on A and

(g ◦ f)′ =(g′ ◦ f

)· f ′.

Theorem 6.7. (Differentiabiliy of the inverse function) If f : A → B estecontinuous, one-to-one, and differentiable at a ∈ A, then the inverse f−1 :f (A) → B is differentiable at f (a) if and only if f ′ (a) = 0. Moreover,(

f−1)′(f (a)) =

1

f ′ (a).

Due to this Theorem we deduce the derivative of other important functions:

(arcsinx)′ =1√

1− x2,∀x ∈ (−1, 1) ,

(arccosx)′ = − 1√1− x2

, ∀x ∈ (−1, 1) ,

(arctg x)′ =1

1 + x2, ∀x ∈ R,

(arcctg x)′ = − 1

1 + x2, ∀x ∈ R,


(lnx)′ =1

x, ∀x > 0.

Hence, we can deduce the derivative of the power function of exponenta ∈ R,

(xa)′ =(ea lnx

)′= axa−1, ∀x > 0.

Bu using Theorem 6.6 for a differentiable function u, one has the followingformulae:

(un)′ = nun−1 · u′, n ∈ N∗,

(ur)′ = rur−1 · u′, r ∈ R, u > 0,

(lnu)′ =1

u· u′, (loga u)

′ =1

u ln a· u′, u > 0,

(eu)′ = euu′, (au)′ = au ln a · u′, a > 0, a = 1,

(sinu)′ = cosu · u′,

(cosu)′ = − sinu · u′,

(tg u)′ =1

cos2 u· u′ =

(1 + tg2 u

)· u′, cosu = 0,

(ctg u)′ = − 1

sin2 u· u′ =

(1 + ctg2 u

)· u′, sinu = 0,

(arcsinu)′ =1√

1− u2· u′, |u| < 1,

(arccosu)′ = − 1√1− u2

· u′, |u| < 1,

(arctg u)′ =1

1 + u2· u′,

(arcctg u)′ = − 1

1 + u2· u′

(n√u)′=

1

nn√un−1

· u′, n ∈ N∗if u > 0, n even,

(uv)′ = uv(lnu · v′ + v

u· u′).


6.2. Derivatives of higher order

Definition 6.8. Let f : A→ R, A ⊆ R, and a ∈ A. If f este differentiable ona neighborhood of a and f ′ is differentiable at a, then we say that f is twicedifferentiable at a and the second order derivative at a, denoted by f ′′ (a) is

f ′′ (a) = limx→a

f ′ (x)− f ′ (a)

x− a.

By this definition, we notice the formula f ′′ = (f ′)′ or d2fdx2

= ddx

(dfdx

).

Similarly we can define the derivative of order n ≥ 3, f (n) =(f (n−1)

)′or

dnf

dxn=

d

dx

(dn−1f

dxn−1

).

For example, if f (x) = 1x2+1

, then

f ′ (x) =−2x

(x2 + 1)2and f ′′ (x) =

2(3x2 − 1

)(x2 + 1)2

.

Denote by

Cn (A) :={f : A→ R | f is n times differentiable with f (n) continuous on A

}.

A function f ∈ Cn (A) is called function of class Cn on A. A functionthat has derivative of any orfer is called infinitely differentiable or of classC∞.

The following formulae hold for derivaties of higher order:

(f + g)(n) = f (n) + g(n),

(λf)(n) = λf (n),

(f · g)(n) =

n∑k=0

Cknf(n−k)g(k).

6.2.1. Local Extrema.

Definition 6.9. Let f : A → R be a function. We say that a function fhas local maximum at the point a ∈ A, iff there exists V ∈ V (a), such thatf (x) ≤ f (a) , ∀x ∈ V. We say that a function f has local minimum at thepoint a ∈ A, iff there exists V ∈ V (a), such that f (x) ≥ f (a) , ∀x ∈ V.

The local maximum and local minimum point are called local extremapoints.

If the inequality holds ∀x ∈ A, then that point is called global maximum(or global minimum) point.

6.2. DERIVATIVES OF HIGHER ORDER 87

f has a local strict maximum at a ∈ A, iff there exists V ∈ V (a),such that f (x) < f (a) , ∀x ∈ V \ {a} . We say that a function f has localstrict minimum at the point a ∈ A, iff there exists V ∈ V (a), such thatf (x) > f (a) , ∀x ∈ V \ {a} .

Theorem 6.8. (Fermat’s Theorem). Let f : A→ R and a an interior point ofA. If f is differentiable at a and f has a local extremum at a, then f ′ (a) = 0.

Remark 6.3. Fermat’s Theorem only provides a necessary, but not a sufficientcondition for local extremum points. Indeed, for the function f : R → R,f (x) = x3, a = 0 is point of differentiability for f , interior to R, f ′ (0) = 0,but f being striclty increasing,, does not have any local extremum.

Remark 6.4. Fermat’s Theorem asserts that the local extrema are found amongthe critical points of the function.

Remark 6.5. Function f : [0, 1] → R, f (x) = x3, has a local maximum at 1,but f ′ (1) = 0. This fact doesn’t contradict Fermat’s Theorem since 1 is not aninterior point to [0, 1].

Teorema 6.9.2 (Rolle). Let f : [a, b] → R be a function, continuouson [a, b] and differentiable on (a, b). If f (a) = f (b), then there is a pointc ∈ (a, b) , sunh that f ′ (c) = 0.

Corollary 6.3. 1. Between two succesive zeros of a differentiable function,there exists at least a zero of the derivative.

2. Between two succesive zeros of the derivative, there exists at most azero of the function.

For example, let us consider the function f : R → R,

f (x) = (2x− 1) (x+ 5) (x− 3) (x− 7) .

Then every critical point of f is real. Indeed, since equation f (x) = 0 has theroots x1 = −5, x2 = 1

2 , x3 = 3, x4 = 7, by applying Corollary 6.3, it follows

that there exist x′1 ∈(−5, 12

), x′2 ∈

(12 , 3), x′3 ∈ (3, 7) critical points. Since the

degree of f is 3, it follows that f ′ has exactly three real zeros.

Theorem 6.9. (Cauchy’s Mean Value Theorem). Let f, g : [a, b] → R betwo functions continuous on [a, b] and differentiable on (a, b) . If g′ (x) = 0,


∀x ∈ (a, b), and g (a) = g (b) , then there is a point c ∈ (a, b) , such that

f (b)− f (a)

g (b)− g (a)=f ′ (c)

g′ (c).

A particular case of the Cauchy’s Mean Value Theorem is the following.

Theorem 6.10. (Lagrange’s Mean Value Theorem). If f : [a, b] → R be afunction continuous on [a, b] and differentiable on (a, b) . Then there is a pointc ∈ (a, b), such that

f (b)− f (a) = f ′ (c) · (b− a) .

Corollary 6.4. (of Lagrange’s Mean Value Theorem). Let f be a functiondefined on a neighborhood V ∈ V (x0). If f is differentiable on V \ {x0} , con-tinuous at x0 and there exists λ = lim

x→x0f ′ (x) , then f has derivative at x0 and

f ′ (x0) = λ. If λ ∈ R, then f is differentiable at x0.

Remark 6.6. Lagrange’s Mean Value Theorem is a particular case of Cauchy’sMean Value Theorem, namely when g (x) = x.

Example 6.1. Let us consider the functions f, g : [−2, 5] → R,

f (x) =

{ √x+ 3, if x ∈ [−2, 1)

x4 + 7

4 , if x ∈ [1, 5]

and

g (x) = x2 + 2x.

We remark that f is continuous on [−2, 5] and differentiable on pe (−2, 5) \ {1},with

f ′ (x) =

{ 12√x+3

, if x ∈ [−2, 1),14 , if x ∈ (1, 5].

We have

limx↗1

f ′ (x) =1

4si lim

x↘1f ′ (x) =

1

4,

therefore, by applying , we obtain f ′ (1) = 14 . So f is even differentiable on

[−2, 5]. Similarly, g is continuous on [−2, 5], differentiable on (−2, 5), withg′ (x) = 0, x ∈ (−2, 5) , and g (5) = g (−2) . By applying Cauchy’s Mean ValueTheorem, it follows that there is a c ∈ (−2, 5) such that

f (5)− f (−2)

g (5)− g (−2)=f ′ (c)

g′ (c).


Example 6.2. Let us study the differentiabily of the function f : R → R,

f (x) =

{x2, if x ≤ 1,lnx+ x, if x > 1.

On the intervals (−∞, 1) and (1,∞), f is continuous, being composition ofelementary functions. Since

limx↗1

f (x) = limx↘1

f (x) = f (1) = 1,

it follows that f is continuous at 1. So, f is continuous on R. On the intervals(−∞, 1) si (1,∞), f is also differentiable, due to the same reason, and

f (x) =

{2x, if x ≤ 1,1x + 1, if x > 1.

One has

limx↗1

f ′ (x) = limx↗1

(2x) = 2,

limx↘1

f ′ (x) = limx↘1

(1

x+ 1

)= 2.

So, there exists λ = 2 = limx→1

f ′ (x) . Due to Corollary 6.4, we get f ′ (1) = 2 ∈ R,hence f is differentiable at 1. The conclusion is f is differentiable on R.

Example 6.3. We consider the function f : R → R,

f (x) =

{x2 sin 1

x , if x = 0,0, if x = 0.

we remark that f is continuous on R and differentiable on R\ {0}, with

f ′ (x) = 2x sin1

x− cos

1

x, x = 0.

One has

limx→0

f ′ (x) = limx→0

2x sin1

x− cos

1

x= − lim

x→0cos

1

x,

which does not exist. Therefore, one can not apply Corollary 6.4. The differ-entiablity at 0 will be studied by using the definition:

f ′ (0) = limx→0

f (x)− f (0)

x− 0= lim

x→0x sin

1

x= 0,

so f is differentiable at 0 and f ′ (0) = 0.

Theorem 6.11. (Darboux). If f : I → R is differentiable on I, then f ′ hasthe Darboux property on I.


A direct application of Lagrange’s Mean Value Theorem and Taylor’s for-mula within the study of the local extrema is the following.

Theorem 6.12. Let f : [a, b] → R be of class Cn on (a, b), n ≥ 2 and let x0 ∈(a, b) , such that f ′ (x0) = f ′′ (x0) = ... = f (n−1) (x0) = 0 and f (n) (x0) = 0.

Then if n is even, f has a local minimum at x0 if f (n) (x0) > 0 and has a local

maximum if f (n) (x0) < 0. If n is odd, then f does not admit a local extremumat x0.

6.2.2. L’Hospital Rules. In this subsection we will present a unitarymethod for the calculus of certain limits involving indeterminations of type 0

0and ∞

∞ .

Theorem 6.13. (L’Hospital Rule for the indefinite form 00). Let f, g : (a, b) →

R be two differentiable functions, such that g′ (x) = 0, ∀x ∈ (a, b) and

limx→a

f (x) = limx→a

g (x) = 0.

If there exists limx→af ′(x)g′(x) ∈ R, then there exists limx→a

f(x)g(x) and

limx→a

f (x)

g (x)= lim

x→a

f ′ (x)

g′ (x).

Theorem 6.14. A similar Theorem holds at the right boundary point, b.

For example, let’s calculate limx→2

x3−8x2−4

:= l. One has

l( 00)= lim

x→2

(x3 − 8

)′(x2 − 4)′

= limx→2

3x2

2x= 3,

since all the hypotheses of Theorem 6.13 are fulfilled.Similarly,

limx→0

esinx − 1

sinx

( 00)= lim

x→0

esinx · cosxcosx

= 1,

limx→∞

x(1− e

1x

)(∞·0)= lim

x→∞

1− e1x

1x

( 00)= lim

x→∞

1x2

· e1x

− 1x2

= −1,

Theorem 6.15. (L’Hospital Rule for the indefinite form ∞∞). Let f, g : (a, b) →

R be two differentiable functions, such that g′ (x) = 0, ∀x ∈ (a, b) and

limx→a

|g (x)| = ∞.


If there exists limx→af ′(x)g′(x) ∈ R, then there exists limx→a

f(x)g(x) and

limx→a

f (x)

g (x)= lim

x→a

f ′ (x)

g′ (x).

Theorem 6.16. A similar Theorem holds at the right boundary point, b.

For example,

limx→∞

ex

x

(∞∞)= lim

x→∞

ex

1= ∞,

limx→∞

lnx

x3(∞∞)= lim

x→∞

1x

3x2= 0.

Theorem 6.17. If f : A → R is differentiable and increasing on A, thenf ′ (x) ≥ 0, ∀x ∈ IA. If f : A → R is differentiable and decreasing on A, thenf ′ (x) ≤ 0, ∀x ∈ A.

Conversely, if f : A→ R is differentiable and f ′ (x) ≥ 0, ∀x ∈ A, then f isincreasing on A. If f : A → R is differentiable and f ′ (x) ≤ 0, ∀x ∈ A, then fis decreasing on A.

This Theorem can be used to find the intervals of monotony and the localextrema. For example, let f : [−1, 1] → R, f (x) = x− arcsinx. One has

f ′ (x) = 1− 1√1− x2

, ∀x ∈ (−1, 1) .

By solving equation f ′ (x) = 0, we fing the single critical point, x = 0.Since f ′ (x) = 0, ∀x ∈ (−1, 0) ∪ (0, 1) it keeps constant sign, being contin-

uous; from f(−1

2

)< 0 and f

(12

)> 0, it follows that f ′ (x) < 0, ∀x ∈ (−1, 0)

and f ′ (x) > 0, ∀x ∈ (0, 1) . By using Theorem 6.17, we get that f is strictlydecreasing on (−1, 0) and (0, 1) . f has not local extremum at the critical point0.

Theorem 6.18. If f : A → R is differentiable on A and f ′ = 0 on A, then fis constant on A.

Corollary 6.5. If f and g are differentiable on A and f ′ = g′ on A, then fand g differ by a constant.

For example, let f : R → R, f (x) = arctg x and g : R∗ → R, g (x) = −arctg 1

x . We have f ′ (x) = 1x2+1

si g′ (x) = 1x2+1

, so f ′ = g′.


By Corollary 6.5 that there exists k1 ∈ R, such that f (x) = g (x) + k1,∀x ∈ (−∞, 0) . So,

arctg x+ arctg1

x= k1, ∀x ∈ (−∞, 0) .

For x = −1, it follows k1 = −π2 .

Similarly, there exists k2 ∈ R, such that f (x) = g (x) + k2, ∀x ∈ (0,∞) .So,

arctg x+ arctg1

x= k2, ∀x ∈ (0,+∞) .

For x = 1, if follows k2 =π2 .

Hence,

arctg x+ arctg1

x= −π

2, ∀x ∈ (−∞, 0) ,

arctg x+ arctg1

x=

π

2, ∀x ∈ (0,∞) .

Definition 6.10. Function f : A → R is called convex on the interval A iff∀x1, x2 ∈ A, ∀t ∈ [0, 1] , one has

f ((1− t)x1 + tx2) ≤ (1− t) f (x1) + tf (x2) .

f is called concave on I iff −f is convex on A, or iff ∀x1, x2 ∈ A, ∀t ∈ [0, 1] ,

f ((1− t)x1 + tx2) ≥ (1− t) f (x1) + tf (x2) .

Theorem 6.19. Let f : [a, b] → R be a twice differentiable function.1. If f ′′ ≥ 0, then the function is convex.2. If f ′′ ≤ 0, then the function is concave.

Definition 6.11. Let f : A → R and a an interior of A. We say that f hasan inflection at a, iff there exist α, β ∈ A, such that α < x0 < β, f is convexon (α, x0] and concave on [x0, β) or conversely.

For example, let us consider the function f : (0,∞) → R, f (x) = x2 lnx.We have f ′ (x) = x (2 lnx+ 1) and f ′ (x) = 0 gives us the critical point x = 1√

e.

f ′′ (x) = 2 lnx + 3 and f ′′ (x) = 0 gives us the root x = 1√e3. By studying the

sign of functions f ′′ and f ′, we obtain that f has global minimum at 1√eand

has an inflection at 1√e3.

6.4. DIFFERENTIABILITY OF REAL FUNCTIONS OF VECTOR VARIABLE 93

6.3. Differentiability of vector functions of a real variable

Let us consider the Euclidean normed space Rm, m ≥ 2.

Definition 6.12. Let f : A→ Rm be a function, A ⊆ R open set, and a ∈ A.We say that f este differentiable at a iff there exists there exists a linearoperator Ta : R → Rm, such that

lim|h|→0

∥f (a+ h)− f (a)− Ta (h)∥|h|

= 0.

It is easily seen that if f is differentiable at a, then it is continuous at a.Recalling that the study of limits to vector functions of real variable is

made by passing to components, we have the following result.

Theorem 6.20. Let f =(f1, f2, ..., fm

): A→ Rm be a function, A ⊆ R open

set, and a ∈ A. Then f is differentiable at a if and only if all its componentsare differentiable at a.

Function f is called differentiable on the set A iff it is differentiable atany a ∈ A.

Taking into account Theorem 6.3, we deduce that

Ta (h) =(f1′ (a)h, f2′ (a)h, ..., fm′ (a)h

), ∀a ∈ A, ∀h ∈ R.

So, we conclude that the study of the differentiablity to a vector function ofreal variable is made by passing to its components. Ta is uniquely determinedby f and a and we denote again the operator Ta by dfa and we call it thedifferential of the function f at the point a.

The vector f ′ (a) :=(f1

′(a) , f2′ (a) , ..., fm′ (a)

)is called the derivative

of f at a and we obtain the following formula for dfa :

dfa (h) = f ′ (a)h, ∀a ∈ A, ∀h ∈ R.

6.4. Differentiability of real functions of vector variable

Let us consider the Euclidean normed space Rp, p ≥ 2.

Definition 6.13. Let f : A → R be a function, A ⊆ Rp open set, and a =(a1, a2, ..., ap

)∈ A. We say that f este differentiable at a iff there exists there

exists a linear operator Ta : Rp → R, such that

lim∥h∥→0

|f (a+ h)− f (a)− Ta (h)|∥h∥

= 0.



It is easily seen that if f is differentiable at a, then it is continuous at a.From Algebra, we know that the linear operator Ta : Rp → R is of form

Ta (h) = c1h1 + c2h2 + ...+ cphp,

∀h =(h1, h2, ..., hp

)∈ Rp, where c =

(c1, c2, ..., cm

)∈ Rp.

We can prove that if f is differentiable at a, then ∀k ∈ 1, p

ck = limhk→0

f(a1, ..., ak + hk, ..., ap

)− f

(a1, ..., ak, ..., ap

)hk

. (6.2)

Therefore we obtain that Ta is uniquely determined of f and a and we denoteagain the operator Ta by dfa and we call it the differential of the functionf at the point a.

The number ck from relation (6.2) is called the partial derivative of fwith respect to variable xk at the point a and we denote it by

∂f

∂xk(a) := ck, ∀k ∈ 1, p.

Let us notice the formula of the partial derivative of f with respect tovariable xk at the point a,

∂f

∂xk(a) := lim

hk→0

f(a1, ..., ak + hk, ..., ap

)− f

(a1, ..., ak, ..., ap

)hk

.

We remark that for calculating the parting derivative with respect to vari-able xk at a, actually we determine the derivative of a single real variablexk → f

(a1, ..., xk, ..., ap

), all the others variables being constants.in this cal-

culus.For example, for the function

f (x, y) = x2 sinxy

we have

∂(x2 sinxy

)∂x

= 2x sinxy + x2 (cosxy) y,

∂(x2 sinxy

)∂y

= x3 cosxy.

The following result gives us a very useful formula for applications.

Theorem 6.21. If f is differentiable at the point a, then f has partial deriva-tives with respect to any variable xk, k ∈ 1, p at the point a, and

dfa (h) =∂f

∂x1(a)h1 +

∂f

∂x2(a)h2 + ...+

∂f

∂xp(a)hp,


∀h =(h1, h2, ..., hp

)∈ Rp.

Let us remark that the converse of Theorem 6.21 does not hold. Indeed, ifwe consider f : R2 → R,

f (x, y) =

{xy√x2+y2

, if (x, y) = (0, 0) ,

0, if (x, y) = (0, 0) ,

then ∂f∂x (0, 0) = limh→0

f(h,0)−f(0,0)h = 0 and, analogously, ∂f

∂y (0, 0) = 0. f is

not differentiable at (0, 0) since, otherwise, it would follow that

df(0,0) (h) =∂f

∂x(0, 0)h1 +

∂f

∂y(0, 0)h2 = 0

and

lim∥h∥→0

∣∣f (h)− f (0, 0)− df(0,0) (h)∣∣

∥h∥= lim

∥h∥→0

∣∣h1h2∣∣(h1)2 + (h2)2

.

But function g(h1, h2

):=

|h1h2|(h1)2+(h2)2

does not have limit at (0, 0) . Indeed,

g(1n ,

λn

)= |λ|

1+λ2→ |λ|

1+λ2, that is not constant for all λ ∈ R.

Another counter-example is that of function f : R2 → R, f (x, y) ={1, if x = 0 and y = 0,0, if x = 0 or y = 0

. This function has

∂f

∂x(0, 0) = lim

h→0

f (h, 0)− f (0, 0)

h= 0,

similarly, ∂f∂y (0, 0) = 0, and it is not continuous at (0, 0), since f

((1n ,

1n

))=

1 → 1 = 0 = f ((0, 0)) . So, f is not differentiable at (0, 0) .

If f is differentiable at a, we denote the vector(∂f∂x1

(a) , ∂f∂x2

(a) , ..., ∂f∂xp (a))

by ∇f (a) or grad f (a) and we call it the gradient of function f at thepoint a. So,

dfa (h) = ⟨∇f (a) , h⟩ , ∀h ∈ Rp.

Let us consider now the canonical projectors Πk : Rp → R, Πk (x) := xk,∀x =

(x1, ..., xp

)∈ Rp, ∀k ∈ 1, p. It is easily seen that Πk is differentiable at

any a ∈ A and (dΠk)a = Πk, ∀k ∈ 1, p. If f is differentiable on A, then onededuces ∀a ∈ A and ∀h ∈ Rp,

dfa (h) =∂f

∂x1(a) (dΠ1)a (h) +

∂f

∂x2(a) (dΠ2)a (h) + ...+

∂f

∂xp(a) (dΠp)a (h) ,

or ∀a ∈ A,

dfa =∂f

∂x1(a) (dΠ1)a +

∂f

∂x2(a) (dΠ2)a + ...+

∂f

∂xp(a) (dΠp)a .


By denotting dΠk =: dxk and ∂f∂xk

the function that maps a ∈ A into ∂f∂xk

(a),

∀k ∈ 1, p, we get the following formula

df =∂f

∂x1dx1 +

∂f

∂x2dx2 + ...+

∂f

∂xpdxp.

For example, for f (x, y) = x2 + xy − y2, we have

df =∂f

∂xdx+

∂f

∂ydy = (2x+ y) dx+ (x− 2y) dy.

Definition 6.14. Let f : Ap → R be a function, A ⊆ Rp open set. We say thatf este partial differentiable on A iff there exists ∂f

∂xk(a), ∀a ∈ A, ∀k ∈ 1, p.

We say that f is of class C1 on A and we denote this by f ∈ C1 (A) iff

f is partial differentiable on A and ∀k ∈ 1, p, ∂f∂xk

is continuous on A.

The connection between the differentiability and the partial differentiabilityis made by the following result.

Theorem 6.22. Let f : Ap → R be a function, A ⊆ Rp open set, and a ∈ A. Iff is partial differentiable on an open neighborhood of a, and ∀k ∈ 1, p, ∂f

∂xkis

continuous at a, then f is differentiable at a.

Corollary 6.6. Let f : Ap → R be a function, A ⊆ Rp open set, and a ∈ A.If f ∈ C1 (A) , then f is differentiable on A.

The converse of this Theorem is false. Indeed, let us consider the functionf : R2 → R,

f (x, y) =

{ (x2 + y2

)sin 1

x2+y2, if (x, y) = (0, 0) ,

0, if (x, y) = (0, 0) .

Then ∂f∂x (0, 0) = limh→0

f(h,0)−f(0,0)h = limh→0 h sin

1h2

= 0 and, similarly,∂f∂y (0, 0) = 0. We prove that f is differentiable at (0, 0), having

T(0,0) (h) =∂f

∂x(0, 0)h1 +

∂f

∂y(0, 0)h2 = 0.

To this aim, we estimate

lim∥h∥→0

∣∣f (h)− f (0, 0)− T(0,0) (h)∣∣

∥h∥= lim

∥h∥→0

|f (h)|∥h∥

= lim∥h∥→0

∥h∥ sin 1

∥h∥2= 0.

By definition, we deduce that f is differentiable at (0, 0) .


We have

∂f

∂x(x, y) =

{2x sin 1

x2+y2− 2x

x2+y2cos 1

x2+y2, if (x, y) = (0, 0) ,

0, if (x, y) = (0, 0) ,

and the following estimate∣∣∣∣2x sin 1

x2 + y2

∣∣∣∣ ≤ 2 |x|∣∣∣∣sin 1

x2 + y2

∣∣∣∣ ≤ 2 |x| ,

∀ (x, y) = (0, 0) . So, lim(x,y)→(0,0) 2x sin1

x2+y2= 0.

If g (x, y) := − 2xx2+y2

cos 1x2+y2

, then

g

(1√4nπ

,1√4nπ

)= −

√4nπ cos 2nπ

= −√4nπ → −∞

and

g

(1√

4nπ + π,

1√4nπ + π

)= −

√4nπ + π cos

(2nπ +

π

2

)= 0 → 0.

So, lim(x,y)→(0,0)∂f∂x (x, y) does not exist, hence ∂f

∂x is not continuous at

(0, 0) . Similarly, we deduce that ∂f∂y is not continuous at (0, 0) .

The partial derivatives are more understantable if we introduce the differ-entiability on a direction.

Definition 6.15. Let f : A ⊂ Rp → R be a function, A ⊆ Rp open set,a ∈ A, and s ∈ Rp with ∥s∥ = 1 (recall that s is called versor). We saythat f is differentiable on the direction (versor) s at a iff there exists

limh→0f(a+hs)−f(a)

h and it is finite. The value of this limit is denoted by

∂f

∂s(a) := lim

h→0

f (a+ hs)− f (a)

h

and it is called the derivative of f on the direction s at a.

Let us consider the standard basis {e1, e2, ..., ep} of Rp, namely ek = (0, ..., 1, ..., 0)with 1 of the kth position.

Then we have

∂f

∂ek(a) =

∂f

∂xk(a) , ∀k ∈ 1, p, ∀a ∈ A.


Indeed,

∂f

∂ek(a) = lim

h→0

f (a+ hek)− f (a)

h

= limh→0

f(a1, ..., ak + hk, ..., ap

)− f

(a1, ..., ak, ..., ap

)hk

=∂f

∂xk(a) , ∀k ∈ 1, p, ∀a ∈ A.

Theorem 6.23. Let f : Ap → R be a function, A ⊆ Rp open set, and a ∈ A. Iff is differentiable at a, then there exists df

ds (a), for any versor s ∈ Rp and

∂f

∂s(a) = dfa (s) ,

∂f

∂xk(a) = dfa (ek) , k ∈ 1, p.

So, if s =∑p

k=1 skek, then, using the linearity of dfa we have succesively

∂f

∂s(a) = dfa (s) = dfa

(p∑

k=1

skek

)=

p∑k=1

skdfa (ek)

=

p∑k=1

sk∂f

∂xk(a) .

One may happen that a function discontinuous at a point to admit deriv-ative on any versor. For example, the function f : R2 → R,

f (x, y) =

{x2yx4+y2

, if (x, y) = (0, 0) ,

0, if (x, y) = (0, 0)

is discontinuous at (0, 0) , but, considering an arbitrary versor s =(s1, s2

)∈

R2, we obtain

∂f

∂s(0, 0) =

{(s1)

2

s2, if s2 = 0,

0, if s2 = 0.

Hence, neither the derivative on a versor nor the partial derivative does notconstitute a satisfying extension of the notion of differentiability for the caseof real functions of real variable. And this is why the most appropriate notionfor the differentiability will be that in sense of Definition 6.12

Proposition 6.2. Let f, g : A→ R be two functions, A ⊆ Rp open set, a ∈ A.If f and g are differentiable at a, then f + g is differentiable at a and

d (f + g) (a) = df (a) + dg (a) .


And if f is differentiable at a, λf is differentiable at a, and

d (λf) (a) = λdf (a) , ∀λ ∈ R.

Proposition 6.3. Let f : A → R be a function, A ⊆ Rp open set, a ∈ A, g :B → R, another function, B ⊆ R open set, and f (a) ∈ B. If f is differentiableat a and g is differentiable at f (a) , the g ◦ f : A→ R is differentiable at a and

d (g ◦ f) (a) = dg (f (a)) ◦ df (a) .

By this Proposition we deduce the following formula for partial differenti-ation:

∂

∂xk(g ◦ f) (a) = g′ (f (a))

∂f

∂xk(a) .

For example, if f (x, y) = 3x+2y, and g (t) = et, then (g ◦ f) (x, y) = e3x+2y

and∂f

∂x= g′ (f (x, y))

∂f

∂x= 3e3x+2y,

∂f

∂y= g′ (f (x, y))

∂f

∂y= 2e3x+2y.

Definition 6.16. Let f : A → R be a function, A ⊆ Rp open set, a ∈ A.Suppose that f is differentiable with respect to variable xk on a neighborhoodV ∈ V (a). If function

∂f

∂xk: V → R

is also differentiable with respect to variable xj, j = k, at a, then we say thatf is twice partial differentiable with respect to variables xj , xk at a. Inthis case, we denote

∂f

∂xj

(∂f

∂xk

)(a) =:

∂2f

∂xj∂xk(a)

and we call it the second mixed partial derivative of order 2 of f withrespect to variables xk, xj at a.

Naturally, if j = k, then in the conditions above we say that f is twicepartial differentiable with respect to variable xk at a, we denote

∂f

∂xk

(∂f

∂xk

)(a) =:

∂2f

∂ (xk)2 (a)

and we call it the second partial derivative of f with respect to variablexk at a.

If f is twice partial differentiable with respect to variables xj , xk at anya ∈ A, then f is called twice differentiable with respect to variables xj ,


xk on A, and the mapping a 7−→ ∂2f∂xj∂xk

(a) is called the partial derivative

of order 2 of f with respect to variables xj , xk.

For example, let us consider the function

f (x, y) = x2 sinxy

We have seen that

∂f

∂x= 2x sinxy + x2 (cosxy) y,

∂f

∂y= x3 cosxy.

Now, we can calculate

∂2f

∂x2: =

∂

∂x

(∂f

∂x

)=

∂

∂x

(2x sinxy + x2 (cosxy) y

)= 2 sinxy − x2y2 sinxy + 4xy cosxy,

∂2f

∂x∂y: =

∂

∂x

(∂f

∂y

)=

∂

∂x

(x3 cosxy

)= 3x2 cosxy − x3y sinxy,

∂2f

∂y∂x: =

∂

∂y

(∂f

∂x

)=

∂

∂y

(2x sinxy + x2 (cosxy) y

)= 3x2 cosxy − x3y sinxy,

∂2f

∂y2: =

∂

∂y

(∂f

∂y

)=

∂

∂y

(x3 cosxy

)= −x4 sinxy.

Remark that ∂2f∂x∂y = ∂2f

∂y∂x .

Considering now the function f : R2 → R,

f (x, y) =

{xy x

2−y2x2+y2

, if (x, y) = (0, 0) ,

0, if (x, y) = (0, 0) ,

we deduce

∂2f

∂x∂y(0, 0) =

∂

∂x

(∂f

∂y

)(0, 0) = lim

h→0

∂f∂y (h, 0)−

∂f∂y (0, 0)

h− 0

= limh→0

h

h= 1


and

∂2f

∂y∂x(0, 0) =

∂

∂y

(∂f

∂x

)(0, 0) = lim

h→0

∂f∂x (0, h)−

∂f∂x (0, 0)

h− 0

= limh→0

−hh= −1.

In what follows, we will give sufficient conditions that allows us to changethe order of the partial differentiations.

Theorem 6.24. (Schwarz’s test). Let f : A → R be a function, A ⊆ Rp open

set, a ∈ A and let j, k ∈ 1, p. If there exist ∂2f

∂xj∂xk

and ∂2f∂xk∂xj

on a neighborhood

of a and these are continuous at a, then

∂2f

∂xj∂xk(a) =

∂2f

∂xk∂xj(a) .

Corollary 6.7. Let f : A→ R be a function, A ⊆ Rp open set. If f ∈ C2 (A) ,then

∂2f

∂xj∂xk=

∂2f

∂xk∂xj.

Let us notice that the conditions from the Schwarz’s test are only sufiicient,and they are not necessary. Indeed, if we consider the function f : R2 → R,

f (x, y) =

{y2 ln

(1 + x2

y2

), if y = 0,

0, if y = 0,

also the Schwarz’s test is not fulfilled, we have the equality of the mixed partialderivatives at (0, 0) :

∂2f

∂x∂y(0, 0) =

∂2f

∂y∂x(0, 0) .

Definition 6.17. The matrix

Hf (a) :=

∂2f

∂(x1)2(a) ∂2f

∂x1∂x2(a) ... ∂2f

∂x1∂xp(a)

... ... ... ...∂2f

∂xp∂x1(a) ∂2f

∂xp∂x2(a) ... ∂2f

∂(xp)2(a)

∈ Mp×p (R)

is called the Hessian of function f at a.

Due to the Schwarz’s test, if f ∈ C2 (A), then Hf (a) is symmetric, ∀a ∈ A.


6.4.1. Local Extrema.

Definition 6.18. Let f : A → R be a function, A ⊆ Rp open set, and a ∈ A.We say that a function f has local maximum at the point a ∈ A, iff thereexists V ∈ V (a), such that f (x) ≤ f (a) , ∀x ∈ V. We say that a function fhas local minimum at the point a ∈ A, iff there exists V ∈ V (a), such thatf (x) ≥ f (a) , ∀x ∈ V.

The local maximum and local minimum point are called local extremapoints.

If the inequality holds ∀x ∈ A, then that point is called global maximum(or global minimum) point.

f has a local strict maximum at a ∈ A, iff there exists V ∈ V (a),such that f (x) < f (a) , ∀x ∈ V \ {a} . We say that a function f has localstrict minimum at the point a ∈ A, iff there exists V ∈ V (a), such thatf (x) > f (a) , ∀x ∈ V \ {a} .

Proposition 6.4. Let f : A→ R be a function, A ⊆ Rp open set, and a ∈ A.If f is differentiable on the direction of versor s at a and f has a local extremumat a, then

∂f

∂s(a) = 0.

Corollary 6.8. (Fermat’s Generalized Theorem). Let f : A → R be a func-tion, A ⊆ Rp open set, and a ∈ A. If f admits partial derivatives of first orderwith respect to variables x1, x2, ..., xp at a, and f has a local maximum at a,then

∂f

∂x1(a) = 0, ...,

∂f

∂xp(a) = 0.

Definition 6.19. A point a ∈ A at which f is differentiable and

∂f

∂x1(a) = 0, ...,

∂f

∂xp(a) = 0

is called critical point to function f.

Therefore, every local extremum point is a critical point. The converseis false. Indeed, the function f : R2 → R, f (x, y) = x3 + y3 has a singlecritical point, namely (0, 0) and f has not a local maximum at (0, 0), sincef (x, y) − f (0, 0) = x3 + y3 does not keep constant sign on any neighborhoodof (0, 0) .

Definition 6.20. Let f : A → R be a function, A ⊆ Rp open set, and a ∈ A.


If f is of class Ck on a neighborhood of a, we define the Taylor’s polynomialof degree k of function f at a by

Tkf (x, a) = f (a) +1

1!

p∑j=1

∂f

∂xj(a)(xj − aj

)+

1

2!

p∑i,j=1

∂2f

∂xi∂xj(a)(xi − ai

) (xj − aj

)+ ...

+1

k!

p∑j1,...,jk=1

∂kf

∂xj1 ...∂xjk(a)(xj1 − aj1

)...(xjk − ajk

).

The quantity Rkf (x, a) = f (x) − Tkf (x, a) is called the remainder oforder k of f at a in Lagrange sense.

For example, for the function

f (x, y) = x2y2,

the Taylor’s polynomial or second order of f at a = (1, 1) is

T2f ((x, y) , (1, 1)) = 1 + 2 (x− 1) + 2 (y − 1)

+1

2!

[2 (x− 1)2 + 8 (x− 1) (y − 1) + 2 (y − 1)2

]= x2 + y2 + 1 + 4xy − 4x− 4y + 3.

And Taylor’s formula also holds for real functions of vector variable.If f : A → R is a function of class Ck+1 (A), A ⊆ Rp open and convex

set, and a ∈ A. Then, ∀x ∈ A, there exists a c on the linear segment [a, x],such that

f (x) = f (a) +1

1!

p∑j=1

∂f

∂xj(a)(xj − aj

)+

+1

2!

p∑i,j=1

∂2f

∂xi∂xj(a)(xi − ai

) (xj − aj

)+ ...

+1

k!

p∑j1,...,jk=1

∂kf

∂xj1 ...∂xjk(a)(xj1 − aj1

)...(xjk − ajk

)+

1

(k + 1)!

p∑j1,...,jk+1=1

∂k+1f

∂xj1 ...∂xjk+1(c)(xj1 − aj1

)...(xjk+1 − ajk+1

).

By Theorem 6.8 we know that every local extremum point is a criticalpoint, but we don’t know which critical point is local extremum point. Thisproblem will be solved by using the Hessian matrix estimated on that point.To this aim, we have the following result.


Theorem 6.25. Let f : A → R be a function of class C2 (A), A ⊆ Rp openand convex set, and a ∈ A be a critical point to f.

If Hf (a) positively defined, then f has a local minimum at a.If Hf (a) negatively defined, then f has a local maximum at a.

In order to establish if the Hessian Hf (a) positively or negatively defined,we recall that all the eigenvalues of Hf (a) are real and we can state the follow-ing test. If all the eigenvalues of Hf (a) are strictly positive (negative), thenHf (a) is positively (negatively defined) and so f has a minimum (maximum)local at a.

Another tool to establisf wither Hf (a) is positively or negatively defined,is the Sylvester’s test :

1) If all the determinants∣∣∣∂2f∂x21

(a)∣∣∣ ,∣∣∣∣∣∣

∂2f

∂(x1)2(a) ∂2f

∂x1∂x2(a)

∂2f

∂(x2)2∂(x1)2(a) ∂2f

∂x22(a)

∣∣∣∣∣∣ ,∣∣∣∣∣∣∣∣∂2f

∂(x1)2(a) ∂2f

∂x1∂x2(a) ∂2f

∂x1∂x3(a)

∂2f∂x2∂x1

(a) ∂2f

∂(x2)2(a) ∂2f

∂x2∂x3(a)

∂2f∂x3∂x1

(a) ∂2f∂x3∂x2

(a) ∂2f

∂(x3)2(a)

∣∣∣∣∣∣∣∣ , ...,∣∣∣∣∣∣∣∣∂2f

∂(x1)2(a) ∂2f

∂x1∂x2(a) ... ∂2f

∂x1∂xp(a)

... ... ... ...∂2f

∂xp∂x1(a) ∂2f

∂xp∂x2(a) ... ∂2f

∂(xp)2(a)

∣∣∣∣∣∣∣∣ are strictly positive, then Hf (a) pos-

itively defined and so f has a local minimum at a;

2) If∣∣∣∂2f∂x21

(a)∣∣∣ < 0,

∣∣∣∣∣∣∂2f

∂(x1)2(a) ∂2f

∂x1∂x2(a)

∂2f

∂(x2)2∂(x1)2(a) ∂2f

∂x22(a)

∣∣∣∣∣∣ > 0,∣∣∣∣∣∣∣∣∂2f

∂(x1)2(a) ∂2f

∂x1∂x2(a) ∂2f

∂x1∂x3(a)

∂2f∂x2∂x1

(a) ∂2f

∂(x2)2(a) ∂2f

∂x2∂x3(a)

∂2f∂x3∂x1

(a) ∂2f∂x3∂x2

(a) ∂2f

∂(x3)2(a)

∣∣∣∣∣∣∣∣ < 0, ..., then Hf (a) este negatively de-

fined and so f has a local maximum at a.Remark that in the other cases one can not specify the nature of the crit-

ical point and we need supplementary analysis to establish if either is localextremum point or not. We could study the sign of f (x)− f (a) directly or byusing Taylor’s formula.

For example, let us determine the local extrema to function

f (x, y) =1 + x− y√1 + x2 + y2

.

6.5. DIFFERENTIABILITY OF VECTOR FUNCTIONS OF VECTOR VARIABLES 105

First we determine the critical point, by solving the system

{∂f∂x = 0∂f∂y = 0

that

becomes 1+y2−x+xy√

1+x2+y23 = 0

− 1+x2+y+xy√1+x2+y2

3 = 0

with the single solution {x = 1, y = −1} . So, we have a unique critical point,(1,−1) .

The Hessian at (x, y) is

Hf (x, y) = −−y+2yx2−y3+1−2x2+y2+3x+3xy2√1+x2+y2

5−y+2yx2−y3+x+x3−2xy2+3xy√

1+x2+y25

−y+2yx2−y3+x+x3−2xy2+3xy√1+x2+y2

5 −x+x3−2xy2+1+x2−2y2−3y−3yx2√1+x2+y2

5

.

Hence

Hf (1,−1) =

(−2

√3

9 −√39

−√39 −2

√3

9

).

Since∣∣∣−2

√3

9

∣∣∣ = −2√3

9 < 0 and

∣∣∣∣∣ −2√3

9 −√39

−√39 −2

√3

9

∣∣∣∣∣ = 19 > 0, it follows that

Hf (1,−1) is negatively defined, so f has a local (and global) maximum at

(1,−1), The maximum of f being f (1,−1) =√3.

6.5. Differentiability of vector functions of vector variables

Consider the Euclidean normed spaces Rp, Rm, with p, m ≥ 2.

Definition 6.21. Let f : A→ Rm be a function, A ⊆ Rp, with p, m ≥ 2, anda ∈ A. We say that f este differentiable at a iff there exists there exists alinear operator Ta : Rp → Rm, such that

lim∥h∥→0

∥f (a+ h)− f (a)− Ta (h)∥∥h∥

= 0.


It is easily seen that if f is differentiable at a, then it is continuous at a.

Theorem 6.26. f is differentiable at a if and only if all of its components aredifferentiable at a.


That is, f =(f1, f2, ..., fm

)is differentiable at a if and only if fk : A→ R

is differentiable at a, ∀k ∈ 1,m.

So, the linear operator Ta : Rp → R is Ta =(df1a , df

2a , ..., df

ma

),where df1a ,

df2a , ..., dfma are the differentials of the components, f1, f2, ..., fm, respectively.

Therefore, Ta is uniquely determined by f and the point a and we denote againthe operator Ta by dfa and we call it the differential of the function f atthe point a.

More precisely, the matrix of this linear operator is

Jf (a) :=

∂f1

∂x1(a) ... ∂f1

∂xp (a)... ... ...

∂fm

∂x1(a) ... ∂fm

∂xp (a)

∈Mm,p (R)

and we call this matrix the Jacobian matrix of function f at thepoint a. So, the differential will be dfa : Rp → Rm

dfa (h) := Jf (a) · h, ∀h ∈ Rp.

When p = m, det Jf (a) is called the Jacobian of function f at thepoint or the functional determinant of functions f1, f2, ..., fm at thepoint a, and we also denote it by

D(f1, ..., fp

)D (x1, ..., xp)

= det Jf (a) .

For example, for function f : R3 → R2,

f (x, y, z) =(ex+y+z, ex−y−z

),

the Jacobian matrix at the point (x, y, z) is

Jf (x, y, z) =

(ex+y+z ex+y+z ex+y+z

ex−y−z −ex−y−z −ex−y−z).

For the function f : [0,∞)× [0, 2π] → R2,

f (r, φ) = (r cosφ, r sinφ) ,

the Jacobian matrix at the point (r, φ) is

Jf (r, φ) =

(cosφ −r sinφsinφ r cosφ

),

and the Jacobian is det Jf (r, φ) =

∣∣∣∣ cosφ −r sinφsinφ r cosφ

∣∣∣∣ = r.

6.6. EXERCISES 107

Proposition 6.5. Let f, g : A → Rm be two functions, A ⊆ Rp open set,a ∈ A. If f and g are differentiable at a, then f + g is differentiable at a and

d (f + g) (a) = df (a) + dg (a) .

And if f is differentiable at a, λf is differentiable at a, and

d (λf) (a) = λdf (a) , ∀λ ∈ R.

Proposition 6.6. (Differentiability of the composed functions). Let f : A →Rm be a function, A ⊆ Rp open set, a ∈ A, g : B → Rn, another function, B ⊆Rm open set, and f (a) ∈ B. If f is differentiable at a and g is differentiableat f (a) , the g ◦ f : A→ Rn is differentiable at a and

d (g ◦ f) (a) = dg (f (a)) ◦ df (a)

or, equivalently,

Jg◦f (a) = Jg (f (a)) · Jf (a) .

If p = m = n, then

det (Jg◦f (a)) = det (Jg (f (a))) · det (Jf (a)) .

6.6. Exercises

(1) Let f (x) = 3+x3−x , x = 3. Calculate f ′ (2) , by using the definition.

A: One has

f ′ (2) = limh→0

f (2 + h)− f (2)

h= lim

h→0

6h

h (1− h)= 6.

(2) Let f (x) =√2x− 1, x ≥ 1

2 . Calculate f′ (5) , by using the definition.

A: One has

f ′ (5) = limh→0

f (5 + h)−−f (5)h

= limh→0

√9 + 2h− 3

h=

= limh→0

(√9 + 2h− 3

) (√9 + 2h+ 3

)h(√

9 + 2h+ 3) =

= limh→0

2h

h(√

9 + 2h+ 3) =

1

3.

(3) Study the continuity and the differentiablity of function f : R → R,

f (x) =

{x sin 1

x , if x = 0,0, if x = 0.


A: Function f is continuous on (−∞, 0) and (0,∞), being compositionof elementary functions. We have

limx→0

f (x) = limx→0

x sin1

x= 0 = f (0) ,

since x→ 0 and∣∣sin 1

x

∣∣ ≤ 1, so f is continuous at 0.f is differentiable on (−∞, 0) and (0,∞), being composition of ele-mentary functions. We have

f ′ (0) = limh→0

f (h)− f (0)

h= lim

h→0sin

1

h

which does not exist. So, f is not differentiable at 0.(4) Study the continuity and the differentiability of function f : R → R,

f (x) =

{x2 sin 1

x , if x = 0,0, if x = 0.

A: Function f is continuous on (−∞, 0) and (0,∞), being compositionof elementary functions. We have

limx→0

f (x) = limx→0

x2 sin1

x= 0 = f (0) ,

since x2 → 0 and∣∣sin 1

x

∣∣ ≤ 1, so f is continuous at 0.f is differentiable on (−∞, 0) and (0,∞), being composition of ele-mentary functions. We have

f ′ (0) = limh→0

f (h)− f (0)

h= lim

h→0h sin

1

h= 0,

so f is differentiable at 0.(5) Determine a, b ∈ R, such that f : R → R,

f (x) =

{sinx, daca x ≤ 0ax+ b, daca x > 0

is differentiable on R.A: Function f is continuous on (−∞, 0) and (0,∞), being compositionof elementary functions. It should also be continuous at 0, hence

limx↗0

f (x) = limx↘0

f (x) = f (0)

or

b = 0.

6.6. EXERCISES 109

f is differentiable on (−∞, 0) and (0,∞), being composition of ele-mentary functions. We have

f ′s (0) = limx↗0

f (x)− f (0)

x− 0= lim

↗0

sinx

x= 1,

f ′d (0) = limx↘0

f (x)− f (0)

x− 0= lim

x↘0

ax

x= a,

so, we should have a = 1 to obtain differentiabilty at 0.

(6) Prove that the function f : R → R, f (x) =

{sin3 x, if x < 0,x3, if x ≥ 0

is

differentiable on R and determine its differential.

(7) Prove that the function f : R → R, f (x) ={x2, if x < 0,0, if x ≥ 0

is differ-

entiable on R and determine its differential.(8) Write the equations of the tangents (or semitangents) to the graph of

the following function, at the specified points:a) f : R → R, f (x) = sinx, a = 1;b) f : R+ → R, f (x) = 4

√x, a = 1, a = 0;

c) f : R → R, f (x) ={

1x3, if x = 0

0, if x = 0, a = 0.

A: a) y = 1, y + x = π, y = −1;b) y − x

4 = 34 ; c) f

′l (0) = −∞ and f ′r (0) = ∞ and the semitangent

has equation x = 0.(9) Determine the angular point to the following functions:

a) f : R → R, f (x) = − |x+ 2| ;

b) f : R → R, f (x) ={x3, if x ≤ 02x, if x > 0

;

c) f : R → R, f (x) ={ √

1− x, if x ≤ 1lnx, if x > 1

.

A: a) x = −2; b) x = 0; c) x = 1.(10) Determine the sharp point to the following functions:

a) f : R → R, f (x) =√

|x+ 1|;

b) f : R → R, f (x) ={x2, if x = 01, if x = 0

;

c) f : R → R, f (x) =

√−x, if x < 0

0, if x = 03√x, if x > 0

.

A: a) x = −1; b) x = 0; c) x = 0.(11) Calculate the derivatives of the following functions:

a) f (x) = x3 − 5x2 + 1; b) f (x) = x lnx− lnx;


c) f (x) = x2 sinx+ 2x cosx− 2 sinx; d) f (x) = x−1x+1 ;

e) f (x) = 1x lnx ; f) f (x) =

√x√x; g) f (x) =

(x2 + x+ 1

)√2;

h) f (x) = ln(x−1x+1

); i) f (x) = 3

√sinx; j) f (x) = loga (tg x) ;

k) f (x) = arcsin(1 + x2

); l) f (x) = tg ( 3

√x);

m) f (x) = arctg x2

x+1 ; n) f (x) =(x2 + 1

)x;

o) f (x) = x1x ; p) f (x) = sin (2x− 1) · cos2 (2x− 1) ;

q) f (x) = 1a arctg x

a , a = 0; r) f (x) = ln(x+

√a2 + x2

);

s) f (x) = ln (cosx) + 12 tg2x.

A: a) 3x2 − 10x; b) x lnx+x−1x ;

c) x2 cosx; d) 2(x+1)2

; e) − lnx+1x2 ln2 x

; f) 1

4( 4√x)

3 ;

g)(x2 + x+ 1

)√2−1√2 (2x+ 1) ;

h) 2(x−1)(x+1) ; i)

1

3 sin23 x

cosx; j) 1+ tg2xtg x ln a ;

k) ddx arcsin

(1 + x2

)= 2 x√

(−x2(2+x2));

l) 131+ tg2 3

√x

( 3√x)

2 ; m) x 2+xx2+2x+1+x4

;

n)(1 + x2

)x−1 (ln(1 + x2

)+(ln(1 + x2

))x2 + 2x2

);

o) −x−−1+2x

x (lnx− 1) ;p) 6 cos3 (−1 + 2x)− 4 cos (−1 + 2x) ;

q) 1a2+x2

; r) 1√(a2+x2)

; s) − (sinx) −1+cos2 xcos3 x

.

(12) Prove that the function f : R → R,

f (x) =

{ln2 (x− 3) , if x ≥ 3

(x− 3)2 , if x < 3

has derivative of second order at a = 3.A: f ′′ (3) = 1.

(13) Calculate f ′′ (x) for the functions:a) f (x) = cosx; b) f (x) = 1

x−a ; c) f (x) =1

x2−4x+3.

A: a) − cosx; b) 2(x−a)3 ; c) 2

3x2−12x+13(x2−4x+3)3

. One can write

1

x2 − 4x+ 3= −1

2· 1

x− 1

1

2· 1

x− 3.

and one can apply the result from b).(14) Determine the local extrema to function

f (x) = 2 cosx+ x2, x ∈ R.

A: We have f ′ (x) = −2 sinx + 2x, f ′′ (x) = −2 cosx + 2, f ′′′ (x) =

6.6. EXERCISES 111

2 sinx, f (4) (x) = 2 cosx and f ′ (0) = f ′′ (0) = f ′′′ (0) = 0, f (4) (0) = 2.So, f has a global minimum at x = 0.

(15) Prove that f : R → (−1,∞) , f (x) = 3x + 9x − 1 is one-to-one and

calculate(f−1

)′(1) and

(f−1

)′′(1) .

A: Obviously, f ′ (x) > 0, ∀x ∈ R and f (R) = (−1,∞).Solving equation f (x) = 1, if follows the unique solution x = 0. Weknow that (

f−1)′(1) =

1

f ′ (0),

that can also be deduced by differentiating relation(f−1 ◦ f

)(x) = x,(

f−1)′(f (x)) · f ′ (x) = 1.

So, by denoting y := f (x), we have(f−1

)′(y) = 1

f ′(x) ; for the second

order derivative , we differentiate relation(f−1

)′(f (x)) · f ′ (x) = 1,

by getting(f−1

)′′(f (x)) ·

(f ′ (x)

)2+(f−1

)′(f (x)) · f ′′ (x) = 0.

So, (f−1

)′(y) = − f ′′ (x)

(f ′ (x))3.

(16) Calculate the following limits, using L’Hospital rules:

a) limx→0

e2x−1x ; b) lim

x→1

1+cosπxx2−2x+1

; c) limx→0x>0

ln cos 3xln cos 2x ;

d) limx→+∞

3x2−x+55x2+6x−3

; e) limx→+∞

x2e−x; f) limx→0x>0

ln tg 2xln tg 3x ;

g) limx→0x>0

x2 lnx; h) limx→0

(cosx)1x2 ;

i) limx→+∞

(e3x − 5x

) 1x ; j) lim

x→0

(e3x − 5x

) 1x ; k) lim

x→0

(1

sin2 x− 1

x2

).

A: a) 2; b) 12π

2; c) 94 ; d)

35 ; e) 0; f) 1; g) 0; h) e

− 12 ; i) e3; j) e−2; k) 1

3 .

(17) Find m ∈ R, such that f : R → R, f (x) = mx − ln(1 + x2

)is

decreasing on R.A: m ∈ (−∞,−1]. The condition we use is f ′ (x) ≥ 0, ∀x ∈ R.

(18) Prove that the function f : R → R2, f (x) =(f1 (x) , f2 (x)

), where

f1 (x) =√x2 + 1,

f2 (x) = cos2 ex

is differentiable on R and determine its differential.


(19) Prove that the function f : R → R3, f (x) =(f1 (x) , f2 (x) , f3 (x)

),

where

f1 (x) = sin2 x,

f2 (x) =

{−x3, if x ≤ 0,x3e−x, if x > 0,

f3 (x) =

{1−cosxx2

, if x = 0,12 , if x = 0

is differentiable on R and determine its differential.(20) Prove that the function f : R2 → R, f (x, y) = xy sin

(x2 + y2

)+2x2+

3y3 is differentiable on R2 and determine its differential.(21) Study the differentiablity at (0, 0) to function f : R2 → R, f (x, y) =√

x2 + y2.A: The function is not differentiable at (0, 0), since it does not

admit partial derivatives at (0, 0) :

limx↘0

f (x, 0)− f (0, 0)

x= lim

x↘0

√x2

x= 1,

limx↗0

f (x, 0)− f (0, 0)

x= lim

x↗0

√x2

x= −1.

(22) Prove that the function f : R3 → R, f (x, y, z) = x+ yz + 2x2y3z2 isdifferentiable on R2 and determine its differential.

(23) Prove that the function f : R2 → R,

f (x, y) =

{ (x2 + y2

)sin 1√

x2+y2, if (x, y) = (0, 0) ,

0, if (x, y) = (0, 0)

is differentiable on R2 and determine its differential.(24) Calculate the derivative of function f (x, y, z) = x2 + y2 + z2 on the

direction s =(

1√2, 0,− 1√

2

)at the point a = (1, 2, 3) .

A: ∂f∂s (1, 2, 3) = −4

√2.

(25) Let f : R3 → R,

f (x, y) =

{arctg ln

(x2 + y2 + z2

), if (x, y, x) = (0, 0, 0) ,

−π2 , if (x, y, z) = (0, 0, 0) .

a) Study the differentiability of f at (0, 0, 0) ;

b) Determine (grad f)(12 ,

12 ,

1√2

);

c) Determine ∂f∂s

(12 ,

12 ,

1√2

), where v =

(grad f)(

12, 12, 1√

2

)∥∥∥(grad f)( 1

2, 12, 1√

2

)∥∥∥ .

6.6. EXERCISES 113

(26) Prove that the function f : R3 → R2,

f (x, y, z) =(x3yz2, xyz,

√x2 + y2 + z2 + 1

)

is differentiable on R3 and determine its differential.(27) Prove that the function f : R2 → R3, f (x, y) =

(xy2, yz2, zx2

)is

differentiable on R2 and determine its differential.(28) Calculate the Jacobian matrix to the following functions:

a) f (x, y) =(x2y, xy2

); b) f (x, y) = (x+ y, xy, sinx+ cos y) ;

c) f (x, y, z) = (ex+y, ex−y) ; d) f (x, y, z) =(xyz, x+ y + z, x2y2z2

).

A: a) Jf (x, y) =

(2xy x2

y2 2xy

);

b) Jf (x, y) =

1 1y x

cosx − sin y

;

c) Jf (x, y) =

(ex+y ex+y

ex−y −ex−y);

d) Jf (x, y, z) =

yz xz xy1 1 1

2xy2z2 2x2yz2 2x2y2z

.

(29) Calculate the partial derivatives of first order of the functions:a) f (x, y) = x3 + y3 − 3xy;b) f (x, y) = xy;c) f (x, y) = y

x ;

d) f (x, y) = ln sin x+2√y ;

e) f (x, y) = arctg yx ;

f) f (x, y) = x√x2+y2

;

g) f (x, y) = ln(x+

√x2 + y2

);

h) f (x, y) = arctg√xy;

i) f (x, y, z) = (sinx)yz .

A: a)∂(x3+y3−3xy)

∂x = 3x2 − 3y;∂(x3+y3−3xy)

∂y = 3y2 − 3x;

b) ∂(xy)∂x = xy−1y; ∂(x

y)∂y = xy lnx;

c)∂( y

x)∂x = − y

x2;∂( y

x)∂y = 1

x ;

d)∂(ln sin x+2√

y

)∂x =

cos x+2√y

√y sin x+2√

y

;


∂(ln sin x+2√

y

)∂y = −1

2

(cos x+2√

y

)x+2

(√y)

3sin x+2√

y

;

e)∂( arctg y

x)∂x = − y

x2+y2;∂( arctg y

x)∂y = x

x2+y2;

f)∂

(x√

x2+y2

)∂x = y2√

x2+y23 ;

∂

(x√

x2+y2

)∂y = − x√

x2+y23 y;

g)∂(ln(x+

√x2+y2

))∂x = 1√

x2+y2;

∂(ln(x+

√x2+y2

))∂y = y(

x+√x2+y2

)√x2+y2

;

h)∂( arctg

√xy)

∂x = 12x

y−1 y

(xy+1)√xy;

∂( arctg√xy)

∂y = 12x

y lnx(xy+1)

√xy;

i) ∂((sinx)yz)

∂x = yz cosx (sinx)yz−1 ; ∂((sinx)yz)

∂y = z (sinx)yz ln (sinx) ;∂((sinx)yz)

∂z = xy (sinx)yz ln (sinx) .

(30) Calculate ∂f∂x (1, 1) and

∂f∂y (1, 1) , if f (x, y) =

√xy + x

y .

A:∂

(√xy+x

y

)∂x = 1

2

√x y2+1

y

y2+1y , ∀ (x, y) = (0, 0) , so ∂f

∂x (1, 1) =√2;

∂

(√xy+x

y

)∂y = 1

2

√x y2+1

y

xy2−1y2

, ∀ (x, y) = (0, 0) , so ∂f∂y (1, 1) = 0.

(31) Calculate ∂f∂z

(0, 0, π4

), if f (x, y, z) =

√sin2 x+ sin2 y + sin2 z.

A: ∂f∂z

(0, 0, π4

)=

√22 .

(32) Calculate ∂f∂x (0, 0) and

∂f∂y (0, 0), if f (x, y) =

x cos y−y cosx1+sinx+sin y .

A: ∂f∂x (0, 0) = 1; ∂f∂y (0, 0) = −1.

(33) Calculate ∂∂x

(1r

), where r =

√x2 + y2 + z2.

A: ∂∂x

(1r

)= − 1

r2∂r∂x = − 1

r2x√

x2+y2+z2= − x

r3.

(34) Calculate the determinant

∣∣∣∣∣ ∂x∂r

∂x∂φ

∂y∂r

∂y∂φ

∣∣∣∣∣ , if x = r cosφ and y = r sinφ.

A: r.(35) Prove that ∂f

∂x + ∂f∂y + ∂f

∂z = 1, if

f (x, y, z) = (x− y) (y − z) (z − x) .

6.6. EXERCISES 115

(36) Prove that the function f = φ(x2 + y2

)fulfills equation

y∂f

∂x− x

∂f

∂y= 0,

where φ is a function of class C1.(37) Calculate df

dt , if f = xy , and x = et, y = ln t.

A: dfdt =

et(t ln t−1)

t ln2 t.

(38) Calculate dfdt , if f = ln sin x√

y , and x = 3t2, y =√t2 + 1.

A: dfdt =

t√y ctg x√

y

(6− x

2y2

).

(39) Calculate dfdx and ∂f

∂x , if f = arctg xy , and y = x2.

A: ∂f∂x = − y

x2+y2; dfdx = 1

x2+1.

(40) Calculate ∂f∂x and ∂f

∂y , if f = φ (u, v), where u = x2 − y2, v = exy, and

φ is a function of class C1.A: ∂f

∂x = 2x∂f∂u (u, v) + yexy ∂f∂v (u, v) ;∂f∂y = −2y ∂f∂u (u, v) + xexy ∂f∂v (u, v) .

(41) Prove that the function w = f (u, v) fulfills equation

∂w

∂t= a

∂w

∂x+ b

∂w

∂y,

where u = x+ at, v = y + bt, and f is a function of class C1.

(42) Calculate the derivative of function f (x, y) = ln√x2 + y2 on the

direction s =(

1√2, 1√

2

), at the point a = (1, 1) .

A:√22 .

(43) Calculate the derivative of function f (x, y, z) = x2 − 3yz + 5 on the

direction s =(

1√3, 1√

3, 1√

3

), at the point a = (1, 2, 1) .

A: −√33 .

(44) Calculate grad f at the point a = (5, 3), if f (x, y) =√x2 − y2.

A: grad f (5, 3) =(54 ,−

34

).

(45) Calculate grad f at the point a = (1, 2, 3), if f (x, y, z) = xyz.A: grad f (1, 2, 3) = (6, 3, 2) .

(46) Calculate ∂2f∂x2

, ∂2f∂x∂y ,

∂2f∂y2

, if

f (x, y) =√x2 + y2.


A:∂2

(√x2+y2

)∂x2

= y2√x2+y2

3 ;∂2

(√x2+y2

)∂x∂y = − xy√

x2+y23 ;

∂2(√

x2+y2)

∂y2= x2√

x2+y23 .

(47) Calculate ∂2f∂x2

, ∂2f∂x∂y ,

∂2f∂y2

, if

f (x, y) = ln(x2 + y2

).

A:∂2(ln(x2+y2))

∂x2= −2 x2−y2

(x2+y2)2;∂2(ln(x2+y2))

∂x∂y = −4 xy

(x2+y2)2;

∂2(ln(x2+y2))∂y2

= 2 x2−y2(x2+y2)2

.

(48) Calculate ∂2f∂x∂y , if

f (x, y) = arctgx+ y

1− xy.

A:∂2

(arctg x+y

1−xy

)∂x∂y = 0

(49) Verify the relation

∂2f

∂x∂y=

∂2f

∂y∂x,

for function f (x, y) = xy.(50) Prove that the function f (x, y) = arctg y

x fulfills Laplace’s equation

∂2f

∂x2+∂2f

∂y2= 0.

(51) Determine the Taylor’s polynomial of order 3 to function f (x, y) = xy

at the point a = (1, 1) .A:

T3f ((x, y) , (1, 1)) = 1 + (x− 1) + (x− 1) (y − 1) +

+1

2!(x− 1)2 (y − 1) .

(52) Determine the Taylor’s polynomial of order 3 to function f (x, y) =ex sin y at the point a = (0, 0) .A:

T3f ((x, y) , (0, 0)) = y + xy +x2y

2− y3

6.

(53) Find the extrema to the following functions:a) f (x, y) = x3 + 3xy2 − 15x− 12y;b) f (x, y) = x2 + xy + y2 − 2x− y;

6.7. REPRESENTATION OF THE GRAPPH OF A FUNCTION 117

c) f (x, y) = 1− 3√x2 + y2;

d) f (x, y) = x4 + y4 − 2x2 + 4xy − 2y2.A: a) f has global minimum at (2, 1), the global minimum beingf ((2, 1)) = −28 and global maximum at (−2,−1) , the global maxi-mum being f ((−2,−1)) = 28.b) f has global minimum at (1, 0), the global minimum being f ((1, 0)) =−1.c) f has global maximum at (0, 0), the global maximum being f ((0, 0)) =1.d) f has global minimum at

(√2,−

√2), the global minimum being

f((√

2,−√2))

= −8.(54) Find the extrema to the following functions:

a) f (x, y, z) = x2 + y2 + z2 − xy + x− 2z;

b) f (x, y, z) = x+ y2

4x + z2

y + 2z .

A: a) f has global minimum at(−2

3 ,−13 , 1), the global minimum be-

ing f((−2

3 ,−13 , 1))

= −43 .

b) f has global minimum at(12 , 1, 1

), the global maximum being

f((

12 , 1, 1

))= 4.

6.7. Representation of the grapph of a function

Pentru a construi cu aproximatie graficul unei functii f se parcurg urmatoareleetape:

1. Determinarea domeniului maxim de definitie, daca acesta nu este pre-cizat. In cazul functiilor periodice se determina perioada principala T si se facestudiul functiei pe multimea [0, T ] .

2. Determinarea punctelor de intersectie a graficului functiei cu axele,A (a, 0), B (0, f (0)), atunci cand acestea exista.

3. Determinarea eventualelor centre de simetrie sau a axelor de simetrie.4. Determinarea valorilor si limitelor functiei ın punctele de acumulare ale

domeniului de definitie.5. Determinarea asimptotelor.6. Se calculeaza f ′, eventualele derivate laterale ın punctele unde f nu

este derivabila si derivata functiei ın punctele ın care aceasta devine (eventual)infinita, ±∞.

Se rezolva ecuatia f ′ (x) = 0 si se studiaza semnul primei derivate, gasindu-se astfel intervalele de monotonie si eventualele puncte de extrem local.

7. Se calculeaza f ′′ (ın punctele ın care f ′ este derivabila), se rezolvaecuatia f ′′ (x) = 0 si se studiaza semnul derivatei a doua, gasindu-se astfelintervalele de convexitate si eventualele puncte de inflexiune.

8. Intocmirea tabelului de variatie a functiei.


Toate rezultatele obtinute mai sus se trec ıntr-un tabel cu 4 linii orizontaleastfel:

a) Pe prima linie se trec valorile remarcabile ale argumentului x de unde safie vizibil si domeniul maxim de definitie. Aici se trec punctele de acumulareale domeniului de definitie care nu apartin acestuia.

b) Pe linia a doua se trec semnul derivatei ıntai si valorile acesteia ınpunctele remarcabile (ori valorile derivatelor laterale).

c) Pe linia a treia se trec valorile functiei ın punctele remarcabile, limitelefunctiei sau limitele laterale si se indica monotonia functiei prin simbolurile ↗sau ↘ .

d) Pe linia a patra se trec semnul derivatei a doua si valorile acesteia ınpunctele remarcabile. Sub semnul respectiv + sau − se trece simbolul ∪ sau∩, pentru a se indica convexitatea sau concavitatea graficului.

9. Desenarea graficului. Dupa ce s-a trasat un reper cartezian (format dinaxele de coordonate si origine) se traseaza asimptotele, punctele remarcabile alegraficului (care se preiau din tabelul de variatie), tangentele sau semitangenteleın punctele remarcabile si, ın final, graficul functiei.

De exemplu, sa reprezentam grafic functia f : Df → R, f (x) = x3 + 3x2.Observam ca domeniul maxim de definitie este Df = R;f (0) = 0 si f (x) = 0 daca si numai daca x = 0 sau x = −3. Intersectiile

cu axele sunt O (0, 0) si A (−3, 0) .Nu avem asimptote verticale, f fiind continua. Nu avem asimptote orizon-

tale, caci limx→±∞

f (x) = ±∞. Nu avem asimptote oblice, deoarece limx→±∞

f(x)x =

+∞;f ′ (x) = 3x2 + 6x, care are radacinile 0 si −2;f ′′ (x) = 6x+ 6, cu radacina x = −1.Tabelul de variatie a functiei f este urmatorul:

Figura 24

iar graficul functiei f este trasat ın figura urmatoare:

Figura 25

6.8. Demonstrarea unor inegalitati

Scopul acestui paragraf este acela de a arata cum putem folosi cunostin-tele de analiza matematica, la a arata unele inegalitati a caror demonstra- tieelementara (utilizand metode algebrice) este destul de grea. Un ins- trumentar fi determinarea extremelor locale, ınsa acesta de cele mai multe ori nu sedovedeste a fi eficace. Avem ınsa urmatoarea

Teorema 7.4.1.a) Daca functiile f, g : [x0, a] → R sunt derivabile si, ın plus, f (x0) ≥

g (x0), f′ (x0) ≥ g′ (x0) , (∀) x > x0, atunci f (x) ≥ g (x) , (∀) x > x0.

6.9. STUDIUL ECUATIILOR 119

b) Daca functiile f, g : [a, x0] → R sunt derivabile si, ın plus, f (x0) ≥g (x0), f

′ (x0) ≤ g′ (x0) , (∀) x < x0, atunci f (x) ≥ g (x) , (∀) x < x0.O alta metoda ar fi aceea de a nota cu h := f − g si a compara pe h cu 0.De exemplu, sa demonstram ca

x

x+ 1≤ ln (x+ 1) , oricare ar fi x > −1.

Definim functia f : (−1,+∞) → R, f (x) = ln (x+ 1)− xx+1 . Avem f (0) =

0, si f ′ (x) = 1x+1 −

1(x+1)2

= x(x+1)2

. Rezolvand ecuatia f ′ (x) = 0, gasim x = 0.

De asemenea, f ′ (x) < 0, daca x < 0 si f ′ (x) > 0, daca x > 0. Deci, f este strictdescrescatoare pe (−1, 0) si strict crescatoare pe (0,+∞) . Deci, f (x) ≥ 0, (∀)x > −1.

6.9. Studiul ecuatiilor

6.9.1. Ecuatii de tipul f (x) = g (x). Rezolvarea ecuatiilor de tipulf (x) = g (x), unde f, g : D → R ınseamna determinarea absciselor punctelor

de intersectie a graficelor celor doua functii. In practica, aflarea exacta asolutiilor este aproape imposibila; de aceea, din analiza variatiilor celor douafunctii putem obtine informatii privind numarul radacinilor si aflarea unor in-tervale ın care aceste radacini se gasesc (separarea radacinilor).

De exemplu, sa consideram ecuatia x3 − ax2 + a = 0, unde a este unparametru real.

Deoarece 1 si −1 nu sunt radacini ale ecuatiei date, putem rescrie ecuatiasub forma

x3

x2 − 1= a,

si trasam graficul functiei f, dupa care ıl vom intersecta cu paralele la axa Oxprin puncte de ordonata a.

Asimptotele sunt dreptele de ecuatii x = −1, x = 1 si y = x.Tabelul de variatie a functiei f : R \ {−1,+1} → R este urmatorul:

x −∞ −√3 −1 0 1

√3 +∞

f ′ + + 0 − −|− − 0 − −|− − 0 + +

f −∞ ↗ 3√3

2 ↘ −∞|+∞ ↘ 0 ↘ −∞|+∞ ↘ 3√3

2 ↗ +∞f ′′ − − − − −|− + 0 − −|− + + + +

Figura 26

Graficul este urmatorul:

Figura 27

a) Pentru a ∈(−∞,−3

√3

2

), ecuatia are 3 solutii reale situate ın intervalele(

−∞,−√3),(−√3,−1

)si (0, 1) .


b) Pentru a = −3√3

2 , −√3 este radacina dubla, iar o alta solutie este ın

(0, 1) .

c) Pentru a ∈(−3

√3

2 , 3√3

2

), ecuatia are o singura radacina reala ın (−1, 1)

etc.

6.9.2. Sirul lui Rolle. Aceasta metoda de determinare a numarului deradacini reale ale unei ecuatii f (x) = 0, unde f : D → R, consta ın urmatoarele:

1. Se afla f ′, radacinile sale x1, x2, ..., xn.2. Se ıntocmeste un tabel ın care prima linie a argumentului x contine

extremitatile a, b ale domeniului de definitie (sau punctele de acumulare ex-treme), ın ordine crescatoare: a, x1, x2, ..., xn, b, pe linia a doua se trec valorilefunctiei ın a, x1, x2, ..., xn, b.

3. Daca functia ia valori de semne contrare ın extremitatile unuia dintreintervalele (a, x1) , (x1, x2) , ..., (xn, b) , atunci functia f are o singura radacinaın interiorul acestui interval.

Daca functia ia valori de aceleasi semne ın extremitatile unuia dintre in-tervalele (a, x1) , (x1, x2) , ..., (xn, b) sau se anuleaza ıntr-o extremitate, atuncifunctia f nu are nici o radacina interiorul acestui interval.

De exemplu, sa determinam numarul de radacini reale ale ecuatiei

x3 − 3x+ 1 = 0.

Consideram functia f : R → R, f (x) = x3 − 3x+ 1. Avem f ′ (x) = 3x2 − 3 sipunctele critice sunt −1 si 1. Atunci sirul lui Rolle este

x −∞ −1 1 +∞f (x) −∞ 3 −1 +∞

Figura 28

si astfel, avand trei alternante de semn, −+−+, toate cele trei radacini aleecuatiei date sunt reale, apartinand intervalelor (−∞,−1), (−1, 1) si (1,+∞) .

CHAPTER 7

Integrals

7.1. Definite Integrals

Fie I un interval si f : I → R o functie.Definitie 8.1.1. Spunem ca f admite primitive pe I daca exista o

functie F : I → R astfel ıncat:a) F este derivabila pe I;b) F ′ (x) = f (x) , (∀) x ∈ I.Functia F se numeste primitiva a lui f.Definitie 8.1.2. Daca f : I → R este o functie care admite primitive pe

I, atunci multimea tuturor primitivelor sale o numim integrala nedefinita alui f si o notam cu

∫f (x) dx.

Putem scrie ∫f (x) dx = F (x) + C, C ∈ R.

Proprietati ale functiilor ce admit primitive:1. Orice doua primitive pe un interval ale unei functii difera printr-o

constanta aditiva.2. Daca o functie admite primitive pe un interval, atunci ea are propri-

etatea lui Darboux pe acel interval. Deci, pentru a arata ca o functie nu admiteprimitive, este suficient sa demonstram ca nu are proprietatea lui Darboux peacel interval.

3. Orice functie continua pe un interval admite primitive pe acel interval.4. Daca f si g : I → R sunt doua functii ce admit primitive pe I si λ ∈ R,

atunci

∫[f (x) + g (x)] dx =

∫f (x) dx+

∫g (x) dx,∫

λf (x) dx = λ

∫f (x) dx.

Primul tabel de integrale nedefinite

121

122 7. INTEGRALS

∫xndx = xn+1

n+1 + C, x ∈ I ⊂ R, n ∈ N∗∫xrdx = xr+1

r+1 + C, x ∈ (0,+∞) ⊂ R, r ∈ R \ {−1}∫axdx = ax

ln a + C, x ∈ I ⊂ R, a > 0, a = 1∫dxx = ln |x|+ C, x ∈ I ⊂ R∗∫dx

x2−a2 = 12a ln

∣∣∣x−ax+a

∣∣∣+ C, x ∈ I ⊂ R\ {±a} , a > 0∫dx

x2+a2= 1

a arctg xa + C, x ∈ I ⊂ R, a > 0∫

sinxdx = − cosx+ C, x ∈ I ⊂ R∫cosxdx = sinx+ C, x ∈ I ⊂ R∫tg xdx = − ln |cosx|+ C, x ∈ I ⊂ R\

{π2 + kπ| k ∈ Z

}∫ctg xdx = − ln |sinx|+ C, x ∈ I ⊂ R\ {kπ| k ∈ Z}∫dx

cos2 x= tg x+ C, x ∈ I ⊂ R\

{π2 + kπ| k ∈ Z

}∫dx

sin2 x= − ctg x+ C, x ∈ I ⊂ R\ {kπ| k ∈ Z}∫

dx√x2+a2

= ln(x+

√x2 + a2

)+ C, x ∈ I ⊂ R, a > 0∫

dx√x2−a2 = ln

∣∣∣x+√x2 − a2

∣∣∣+ C, x ∈ I ⊂ R\ [−a, a], a > 0∫dx√a2−x2 = arcsin x

a + C, x ∈ I ⊂ (−a, a), a > 0

7.1.1. Integration by parts. Teorema 8.1.1. Fie f, g : I → R douafunctii. Daca f si g sunt derivabile cu derivate continue, atunci fg, f ′g, fg′

admit primitive si avem relatia∫f (x) g′ (x) dx = f (x) g (x)−

∫f ′ (x) g (x) dx.

De exemplu sa calculam:1. ∫

x lnxdx =

∫ (x2

2

)′lnxdx =

x2

2lnx−

∫x2

2(lnx)′ dx

=x2

2lnx−

∫x2

2· 1xdx =

=x2

2lnx− x2

4+ C, C ∈ R.

2.

I =

∫ √x2 + 2dx =

∫x2 + 2√x2 + 2

dx =

∫x2√x2 + 2

dx+

∫2√

x2 + 2dx =

= I1 + I2.

7.1. DEFINITE INTEGRALS 123

I1 =

∫x2√x2 + 2

dx =

∫x(√

x2 + 2)′dx =

= x√x2 + 2−

∫ √x2 + 2dx = x

√x2 + 2− I,

I2 =

∫2√

x2 + 2dx = 2 ln

(x+

√x2 + 2

)+ C, C ∈ R.

Deci,

I =1

2x√x2 + 2 + ln

(x+

√x2 + 2

)+ C, C ∈ R.

3. ∫xexdx =

∫x (ex)′ dx = xex −

∫exdx = xex − ex + C, C ∈ R.

7.1.2. Changing variable first method in undefinite integral. Teo-rema 8.1.2. Fie φ : I → J, f : J → R, cu I si J intervale. Daca φ estederivabila si f admite primitive, F fiind o primitiva a sa, atunci (f ◦ φ)φ′

admite primitive si F ◦ φ este o primitiva a sa.De exemplu, sa calculam:∫

4x+ 2

x2 + x+ 3dx = 2

∫2x+ 1

x2 + x+ 3dx = 2

∫ (x2 + x+ 3

)′x2 + x+ 3

dx =

= 2

∫φ′ (x)

φ (x)dx = 2 ln |φ (x)|+ C =

= 2 ln∣∣x2 + x+ 3

∣∣+ C, C ∈ R

unde

φ (x) = x2 + x+ 3 = t, φ : R → [11

4,+∞), este derivabila

f (t) =1

t, f : [

11

4,+∞) → R admite primitive,

F (t) = ln |t| = ln t,

functia

(F ◦ φ) (x) = F (φ (x)) = F(x2 + x+ 3

)= ln

(x2 + x+ 3

).

Deci ∫4x+ 2

x2 + x+ 3dx = ln

(x2 + x+ 3

)+ C, C ∈ R.

Similar primului tabel, pe domeniile de existenta avem al doilea tabel deintegrale nedefinite

124 7. INTEGRALS

∫φ (x)n φ′ (x) dx = φ(x)n+1

n+1 + C∫φ (x)r φ′ (x) dx = φ(x)r+1

r+1 + C∫aφ(x)φ′ (x) dx = aφ(x)

ln a + C∫ φ′(x)φ(x) dx = ln |φ (x)|+ C∫ φ′(x)

φ(x)2−a2dx = 12a ln

∣∣∣φ(x)−aφ(x)+a

∣∣∣+ C∫ φ′(x)

φ(x)2+a2dx = 1

a arctg φ(x)a + C∫

sinφ (x)φ′ (x) dx = − cosφ (x) + C∫cosφ (x)φ′ (x) dx = sinφ (x) + C∫tg φ (x)φ′ (x) dx = − ln |cosφ (x)|+ C∫ctg φ (x)φ′ (x) dx = − ln |sinφ (x)|+ C∫ φ′(x)cos2 φ(x)

dx = tg φ (x) + C∫ φ′(x)sin2 φ(x)

dx = − ctg φ (x) + C∫ φ′(x)√φ(x)2+a2

dx = ln

(φ (x) +

√φ (x)2 + a2

)+ C∫ φ′(x)√

φ(x)2−a2dx = ln

∣∣∣∣φ (x) +√φ (x)2 − a2

∣∣∣∣+ C∫ φ′(x)√a2−φ(x)2

dx = arcsin φ(x)a + C

7.1.3. Changing variable second method in undefinite integral.Teorema 8.1.3. Daca φ : I → J si f : J → R verifica urmatoarele proprietati:

1) φ este bijectiva, derivabila, cu derivata nenula pe I;2) h = (f ◦ φ)φ′ admite primitive si H este o primitiva a sa,

atunci f admite primitive si H ◦ φ−1 este o primitiva a sa, adica∫f (x) dx =

(H ◦ φ−1

)(x) + C, C ∈ R.

De exemplu, sa calculam:∫e2x

1 + exdx, x ∈ R.

Avem

f (x) =e2x

1 + ex, f : R → R,

φ−1 (x) = ex = t, φ−1 : R → (0,+∞) ,

x = ln t = φ (t) , φ : (0,+∞) → R.


Cum φ′ (t) = 1t , φ bijectiva, φ derivabila, φ′ (t) = 0, t ∈ (0,+∞) , iar

h (t) = f (φ (t))φ′ (t) =t

t+ 1,∫

h (t) dt =

∫t

t+ 1dt =

∫t+ 1− 1

t+ 1dt =

=

∫ (1− 1

t+ 1

)dt = t− ln |t+ 1|+ C =

= t− ln (t+ 1) + C, C ∈ R,

rezulta ca o primitiva a lui h este

H (t) = t− ln (t+ 1) ,

iar (H ◦ φ−1

)(x) = H

(φ−1 (x)

)= H (ex) = ex − ln (ex + 1) .

Deci, ∫e2x

ex + 1dx = ex − ln (ex + 1) + C, C ∈ R.

7.1.4. Integration of rational functions. Reamintim ca functia f :I → R se numeste rationala daca exista doua polinoame P si Q ∈ R [X], astfel

ıncat f (x) = P (x)Q(x) , (∀) x ∈ I.

Functiile rationale simple sunt cele de forma:1. functiile polinomiale;2. functiile A

(x−a)n , n ∈ N∗;

3. functiile Ax+B(ax2+bx+c)n

, n ∈ N∗, a = 0, b2 − 4ac < 0.

Pentru integrarea functiilor rationale simple avem:1. Se aplica formula 1 din primul tabel de integrale nedefinite;2. Pentru n = 1, avem∫

A

x− adx = A ln |x− a|+ C, C ∈ R;

iar pentru n ≥ 2,∫A

(x− a)ndx = A

∫(x− a)′

(x− a)ndx = A

∫φ (x)−n φ′ (x) dx =

= Aφ (x)−n+1

−n+ 1+ C =

=A

(−n+ 1) (x− a)n−1 + C, C ∈ R.

126 7. INTEGRALS

3. Deoarece trinomul ax2 + bx + c se poate scrie sub forma unei sume depatrate

a

[(x+

b

2a

)2

+

(√−∆

2a

)2],

avem

In =

∫Ax+B

(ax2 + bx+ c)ndx =

∫A′x+B′

(x2 + k2)ndx =

=A′

2

∫2x

(x2 + k2)ndx+B′

∫dx

(x2 + k2)n

si astfel integrala la care se reduce calculul este

In =

∫dx

(x2 + k2)n, k = 0, n ∈ N∗.

Pentru n = 1, avem

I1 =

∫dx

x2 + k2=

1

karctg

x

k+ C, C ∈ R;

iar pentru n ≥ 2,

In =

∫ (x2 + k2

)− k2

(x2 + k2)n−1 dx =1

k2

∫dx

(x2 + k2)n−1 − 1

k2

∫x2dx

(x2 + k2)n

=1

k2In−1 +

1

2 (n− 1) k2

∫x

(1

(x2 + k2)n−1

)′dx

=1

k2In−1 +

1

2 (n− 1) k2x

(x2 + k2)n−1 − 1

2 (n− 1) k2

∫dx

(x2 + k2)n−1

=1

k2In−1 +

1

2 (n− 1) k2x

(x2 + k2)n−1 − 1

2 (n− 1) k2In−1,

deci

In =x

2 (n− 1) k2 (x2 + k2)n−1 +2n− 3

2 (n− 1) k2In−1.

Orice functie rationala se descompune ıntr-o suma de functii rationale sim-ple. Pentru descompunere, grad P < grad Q. Daca grad P ≥ grad Q, atunci

se ımparte P la Q si se obtine P (x)Q(x) = C (x) + R(x)

Q(x) , unde grad R < grad Q.

De exemplu, pentru ∫x2 + x+ 1

(x− 1)3 (x2 − x+ 1)2dx,


avem

f (x) =x2 + x+ 1

(x− 1)3 (x2 − x+ 1)2=

=A

x− 1+

B

(x− 1)2+

C

(x− 1)3+

Dx+ E

x2 − x+ 1+

Fx+G

(x2 − x+ 1)2,

de unde obtinem

A = −2, B = −3, C = 3, D = 2, E = 3, F = 2, G = 0

si calculul urmeaza ideile anterioare.

7.1.5. Integration of irrational functions. Fie R(x,

√ax2 + bx+ c

)o functie rationala depinzand de variabilele x si

√ax2 + bx+ c, unde a, b,

c ∈ R, a = 0 si ∆ = b2 − 4ac = 0. Atunci, pentru calculul integralei nedefinite∫R(x,

√ax2 + bx+ c

)dx se aplica metoda a doua de schimbare de variabila,

dupa cum urmeaza (substitutiile lui Euler):

1. Daca a > 0, se noteaza√ax2 + bx+ c = ±x

√a± t;

2. Daca c > 0 si 0 /∈ I, se noteaza√ax2 + bx+ c = ±

√c± tx;

3. Daca ∆ > 0 si x1, x2 /∈ I sunt radacinile trinomului ax2 + bx + c, senoteaza

√a (x− x1) (x− x2) = (x− x1) t.

De exemplu, sa calculam integralele:1. I =

∫dx

x−√x2+2x+4

, x ∈ R.Avem f (x) = 1

x−√x2+2x+4

, f : R → R si√x2 + 2x+ 4 = x− t.

Rezulta

φ−1 : R → (−∞,−1) , φ−1 (x) = x−√x2 + 2x+ 4 = t,

φ−1 : R → (−∞,−1) , φ (t) =t2 − 4

2t+ 2= x,

φ′ (t) =t2 + 2t+ 4

2 (t+ 1)2.

Astfel, φ este derivabila, bijectiva cu φ′ (t) = 0, (∀) t ∈ (−∞,−1) .Rezulta∫

h (t) dt =

∫f (φ (t))φ′ (t) dt =

∫t2 + 2t+ 4

2t (t+ 1)2dt =

=

∫ (2

t− 3

2 (t+ 1)− 3

2 (t+ 1)2

)dt =

= 2 ln (−t)− 3

2ln (−t− 1) +

3

2 (t+ 1)+ C, C ∈ R.

128 7. INTEGRALS

Deci,

H (t) = 2 ln (−t)− 3

2ln (−t− 1) +

3

2 (t+ 1)

si

I = H(φ−1 (x)

)+ C = H

(x−

√x2 + 2x+ 4

)+ C, C ∈ R.

2. I =∫

dx1+

√x2+2x+2

, x ∈ (0,+∞) .

Avem f (x) = 11+

√x2+2x+2

, f : (0,+∞) → R si√x2 + 2x+ 2 =

√2 + tx.

Rezulta

φ−1 : (0,+∞) →

(√2

2, 1

), φ−1 (x) =

√x2 + 2x+ 2−

√2

x= t,

φ :

(√2

2, 1

)→ R, φ (t) =

2− 2√2t

t2 − 1= x,

φ′ (t) =2√2(t2 − t

√2 + 1

)(t2 − 1)2

.

Astfel, φ este derivabila, bijectiva cu φ′ (t) = 0, (∀) t ∈(√

22 , 1

).

Rezulta∫h (t) dt =

∫f (φ (t))φ′ (t) dt =

∫2√2(t2 − t

√2 + 1

)t2(√

2− 1)− 2t+

√2 + 1

dt =

= H (t) + C, C ∈ Rsi

I = H(φ−1 (x)

)+ C = H

(√x2 + 2x+ 2−

√2

x

)+ C, C ∈ R.

3. I =∫

x+1√−x2+4x+5

dx, x ∈ (−1, 5) .

Avem f (x) = x+1√−x2+4x+5

, f : (−1, 5) → R si cum

−x2 + 4x+ 5 = (x+ 1) (5− x) ,

notam √−x2 + 4x+ 5 = t (x+ 1) .

Rezulta

φ−1 : (−1, 5) → (0,+∞) , φ−1 (x) =

√5− x

x+ 1= t,

φ : (0,+∞) → (−1, 5) , φ (t) =5− t2

t2 + 1= x,


φ′ (t) =−12t

(t2 + 1)2.

Astfel, φ este derivabila, bijectiva cu φ′ (t) = 0, (∀) t ∈ (0,+∞) .Rezulta ∫

h (t) dt =

∫f (φ (t))φ′ (t) dt =

∫−12

(t2 + 1)2dt

= H (t) + C, C ∈ Rsi

I = H(φ−1 (x)

)+ C = H

(√5− x

x+ 1

), C ∈ R.

7.1.6. Integration of trigonometric functions. Fie R (sinx, cosx) ofunctie rationala depinzand de variabilele sinx si cosx. Atunci, pentru cal-culul integralei nedefinite

∫R (sinx, cosx) dx se fac urmatoarele schimbari de

variabila:1. Daca R este impara ın sinx, atunci se noteaza cosx = t si se aplica

metoda ıntai de schimbare de variabila;2. Daca R este impara ın cosx, atunci se noteaza sinx = t si se aplica

metoda ıntai de schimbare de variabila;3. Pe cazul general, se noteaza tg x

2 = t si se aplica metoda a doua deschimbare de variabila.

De exemplu, sa calculam integralele:

1. I =∫ sinx(1+cosx)

2−cosx dx, x ∈(0, π2

).

Avem R impara ın sinx, deci cu schimbarea de variabila cosx = t, gasim

I = −∫

(cosx)′ (1 + cosx)

2− cosxdx = −

∫φ′ (x) (1 + φ (x))

2− φ (x)dx,

unde φ (x) = cosx = t, φ :(0, π2

)→ (0, 1), f (t) = 1+t

2−t , f : (0, 1) → R. Dupa

calculul lui∫f (t) dt = F (t)+C, C ∈ R obtinem ca I = F (φ (x))+C, C ∈ R.

2. I =∫cos9 xdx, x ∈

(−π

2 ,π2

).

Avem R impara ın cosx, deci cu schimbarea de variabila sinx = t, gasim

I =

∫(sinx)′

(1− sin2 x

)4dx = −

∫φ′ (x)

(1− φ (x)2

)4dx,

unde φ (x) = cosx = t, φ :(0, π2

)→ (−1, 1), f (t) =

(1− t2

)4, f : (−1, 1) → R.

Dupa calculul lui∫f (t) dt = F (t) + C, C ∈ R obtinem ca I = F (φ (x)) + C,

C ∈ R.3. I =

∫dx

sinx(2+cosx−2 sinx) , x ∈(0, π6

).

Avem φ−1 (x) = tg x2 = t, φ−1 :

(0, tg π

12

)→(0, π6

)si x = 2 arctg t,

φ :(0, π6

)→(0, tg π

12

). Dupa calculul lui

∫f (φ (t))φ′ (t) dt = H (t) + C,

C ∈ R obtinem ca I = H(φ−1 (x)

)+ C, C ∈ R.

130 7. INTEGRALS

7.2. Definite Integrals

Fie [a, b] ⊂ R un interval.Definitie 8.2.1. Numim diviziune a intervalului [a, b] un sistem de

puncte ∆ = (x0, x1, ..., xn−1, xn), astfel ıncat a = x0 < x1 < ... < xn−1

< xn = b. Cea mi mare dintre lungimile intervalelor [x0, x1] , [x1, x2] , ...,[xn−1, xn] se numeste norma diviziunii ∆ si se noteaza

∥∆∥ = maxi∈{1,2,...,n}

(xi − xi−1) .

Fie f : [a, b] → E o diviziune a intervalului [a, b] si ξ1, ξ2, ..., ξn un sistemde puncte intermediare pentru diviziunea ∆ astfel ıncat ξi ∈ [xi−1, xi],pentru i ∈ {1, 2, ..., n} . Numarul real

∑ni=1 f (ξi) (xi − xi−1) se numeste suma

Riemann atasata functiei f, diviziunii ∆ si sistemului de puncte intermediareξ, si se noteaza

σ∆ (f, ξ) =

n∑i=1

f (ξi) (xi − xi−1) .

Pentru o functie f : [a, b] → R pozitiva, notam cu

Γf :={(x, y) ∈ R2| x ∈ [a, b] , y ∈ [0, f (x)]

}subgraficul functiei f.

Observatie 8.2.1. Suma Riemann este aproximativ egala cu aria sub-graficului,

σ∆ (f, ξ) ≈ Aria (Γf ) .

Definitie 8.2.1. Functia f : [a, b] → R se numeste integrabila Riemannpe [a, b] daca exista un numar real I ∈ R cu proprietatea ca (∀) ϵ > 0, (∃)η = η (ϵ) > 0, astfel ıncat (∀) ∆ = (x0, x1, ..., xn) o diviziune a intervalului[a, b] cu ∥∆∥ < η si (∀) ξ sistem de puncte intermediare diviziunii ∆, sa avem

|σ∆ (f, ξ)− I| < ϵ.

Observatie 8.2.2. Daca o functie este integrabila, atunci numarul I din

definitie se numeste integrala definita a functiei f si se noteaza∫ ba f (x) dx.

Observatie 8.2.3. Integrala definita este un numar, spre deosebire deintegrala nedefinita, care este o multime de functii.

Observatie 8.2.4. Daca o functie este integrabila, numarul I este unic.Propozitie 8.2.1.Orice functie integrabila este marginita.Orice functie continua este integrabila.Orice functie monotona este integrabila.Pentru a arata ca o functie nu este integrabila, este suficient sa aratam ca

este nemarginita.


De exemplu, functia f (x) =

{5, daca x = 01x , daca x ∈ (0, 3]

este neintegrabila, fiind

nemarginita superior.Functia f : [0, 3] → R, f (x) = x2 este continua, deci integrabila pe [0, 3] .

Functia f : [1, 3] → R, f (x) =

{x, daca x ∈ [1, 2]2x, daca x ∈ (2, 3]

nu este continua,

dar fiind crescatoare, este integrabila.Proprietati ale integralei definite

1.∫ ba [f (x) + g (x)] dx =

∫ ba f (x) dx+

∫ ba g (x) dx.

2.∫ ba cf (x) dx = c

∫ ba f (x) dx, c ∈ R

3.∫ ba f (x) dx =

∫ ca f (x) dx +

∫ bc f (x) dx, daca f este integrabila pe [a, c]

si [c, b], c ∈ [a, b] .

4.∫ ba f (x) dx = −

∫ ab f (x) dx.

5.∫ aa f (x) dx = 0.

6. Daca f este integrabila pe [a, b], cu a < b, ea este marginita si dacaf (x) ∈ [m,M ] , (∀) x ∈ [a, b], atunci

m (b− a) ≤∫ b

af (x) dx ≤M (b− a) .

7. Daca f (x) ≥ 0, (∀) x ∈ [a, b], atunci∫ b

af (x) dx ≥ 0.

8. Daca f (x) ≤ g (x) , (∀) x ∈ [a, b], atunci∫ b

af (x) dx ≤

∫ b

ag (x) dx.

9.∣∣∣∫ ba f (x) dx∣∣∣ ≤ ∫ ba |f (x)| dx, daca a < b.

10. Daca f : [a, b] → R este continua, pozitiva si [c, d] ⊂ [a, b], atunci∫ d

cf (x) dx ≤

∫ b

af (x) dx.

11. Daca f : [a, b] → R, a < b este continua, pozitiva si neidentic nula pe[a, b], atunci ∫ b

af (x) dx > 0.

Teorema 8.2.1(Teorema de medie). Daca f : [a, b] → R este continua,atunci exista un punct ξ ∈ [a, b] astfel ıncat∫ b

af (x) dx = (b− a) f (ξ) .

132 7. INTEGRALS

Teorema 8.2.2 (Formula Leibniz-Newton). Daca f : [a, b] → R esteintegrabila si admite primitive, F fiind o primitiva a sa, atunci∫ b

af (x) dx = F (x) |ba= F (b)− F (a) .

Teorema 8.2.3.1. Fie f : [a, b] → R o functie continua. Atunci functia F : [a, b] → R,

F (x) =

∫ x

af (t) dt

este o primitiva a functiei f care se anuleaza ın a.2. Fie f : [a, b] → R o functie continua. Daca g : [a, b] → R, este o

primitiva a lui f ce se anuleaza ın x0 ∈ [a, b], atunci

g (x) =

∫ x

x0

f (t) dt.

7.2.1. Integration by parts. Teorema 8.2.4. Daca f, g : [a, b] → Rsunt functii derivabile cu derivatele continue, atunci∫ b

af (x) g′ (x) dx = f (x) g (x) |ba −

∫ b

af ′ (x) g (x) dx.

De exemplu, sa calculam∫ e2

eln2 xdx =

∫ e2

ex′ ln2 xdx = x ln2 x |e2e −2

∫ e2

exlnx

xdx =

= x ln2 x |e2e −2

∫ e2

ex′ lnxdx =

= x ln2 x |e2e −2x lnx |e2e +2

∫ e2

e

x

xdx =

= x ln2 x |e2e −2x lnx |e2e +2x |e2e = 2e2 − e.

7.2.2. Changing variable first method in definite integral. Teo-rema 8.2.5. Fie φ : [a, b] → J si f : J → R, astfel ıncat φ este derivabila cuderivata continua si f este continua. Atunci∫ b

af (φ (x))φ′ (x) dx =

∫ φ(b)

φ(a)f (t) dt.

De exemplu, sa calculam

I =

∫ −1

−2

exdx√1− e2x

=

∫ −1

−2

(ex)′ dx√1− (ex)2

=

∫ −1

−2

(φ (x))′ dx√1− (φ (x))2

,

7.3. EXERCISES 133

unde φ (x) = ex = t, φ : [−2,−1] →[1e2, 1e]. Definim si functia f :

[1e2, 1e]→ R,

prin f (t) = 1√1−t2 . Cum φ (−2) = 1

e2si φ (−1) = 1

e , rezulta ca

I =

∫ 1e

1e2

dt√1− t2

= arcsin t |1e1e2

= arcsin1

e− arcsin

1

e2.

7.2.3. Changing method second variable in definite integral. Teo-rema 8.2.6. Fie φ : [a, b] → [c, d] si f : [c, d] → R, astfel ıncat:

1. φ este bijectiva, iar φ si φ−1 sunt derivabile cu derivatele continue;2. f este continua.Atunci ∫ b

af (φ (x)) dx =

∫ φ(b)

φ(a)f (t)

(φ−1

)′(t) dt.


I =

∫ ln 6

ln 5

√ex − 1

ex + 1dx.

Notam t = ex = φ (x) , cu φ : [ln 5, ln 6] → [5, 6] si f (t) =√

t−1t+1 , cu

f : [5, 6] → R. Atunci x = ln t = φ−1 (f) si(φ−1

)′(t) = 1

t .Avem

I =

∫ 6

5

√t− 1

t+ 1· 1tdt =

∫ 6

5

t− 1√t2 − 1

· 1tdt =

=

∫ 6

5

1√t2 − 1

dt−∫ 6

5

1

t√t2 − 1

dt =

= ln∣∣∣t+√t2 − 1

∣∣∣ |65 +∫ 6

5

(1t

)′√1−

(1t

)2dt == ln

∣∣∣t+√t2 − 1∣∣∣ |65 +arcsin

1

t|65=

= ln6 +

√35

5 +√24

+ arcsin1

6− arcsin

1

5.

7.3. Exercises

(1) Calculati urmatoarele integrale nedefinite, folosind metoda de inte-grare prin parti:a)∫x3 ln2 xdx

b)∫x3

√x2 + 9dx

c)∫sin2 xdx

d)∫lnn xdx, n ∈ N

e)∫xex sinxdx

134 7. INTEGRALS

f)∫

dxsinn x , n ≥ 2

g)∫

dx(x2+1)n

, n ∈ Nh)∫(arcsinx)n dx, n ∈ N

i)∫xm lnn xdx, m, n ∈ N

j)∫

arctg xdx

R: a) 14x

4 ln2 x− 18x

4 lnx+ 132x

4 + C, C ∈ R

b) 15x

2(√

(x2 + 9))3

− 65

(√(x2 + 9)

)3+ C, C ∈ R

c) −12 cosx sinx+ 1

2x+ C, C ∈ Rd) In = x lnn x− nIn−1, n ∈ N∗, I0 = x+ C, C ∈ Re)(−1

2x+ 12

)ex cosx+ 1

2xex sinx+ C, C ∈ R

f) In = 2−n1−nIn−2 +

cosx(1−n) sinn−1 x

g) In = 3−2n2−2nIn−1 − x

(2−2n)(x2+1)n−1

h) In = x (arcsinx)n + n√1− x2 (arcsinx)n−1 − n (n− 1) In−2

i) Im,n = xm+1 lnn xm+1 − n

m+1Im,n−1

j) x arctg x− 12 ln

(1 + x2

)+ C, C ∈ R.

(2) Calculati urmatoarele integrale nedefinite, folosind metoda ıntai deschimbare de variabila:a)∫

sinx1+cos2 x

dx, x ∈ Rb)∫ 1+ tg2x

tg x dx, x ∈(0, π2

)c)∫

x2

1+x3dx, x ∈ R

d)∫

sin5 xcosx dx, x ∈

(−π

2 ,π2

)e)∫ √

x2 − 3x+ 2dx, x ∈ (2,+∞)

f)∫x(1 + x2

)2dx, x ∈ R

g)∫

arcsinxx2

dx, x ∈ (0, 1)

h)∫

dx

x√

1−ln2 x, x ∈

(1e , e)

i)∫ cos2 x

2x+sinxdx, x ∈

(0, π2

)j)∫

1ex+e−xdx, x ∈ R

k)∫ cos(lnx)

x dx, x ∈ (0,+∞)

l)∫

dx√1−x2(arcsinx)n , n ∈ N, x ∈ (0,+∞)

m)∫

cosx+sinxn√sinx−cosx

dx, n ∈ N, x ∈ (0,+∞) .

R: a) − arctg (cosx) + C, C ∈ Rb) ln (tg x) + C, C ∈ Rc) 1

3 ln∣∣1 + x3

∣∣+ C, C ∈ Rd) −1

4 sin4 x− 1

2 sin2 x− ln (cosx) + C, C ∈ R

e) 14 (2x− 3)

√x2 − 3x+ 2−

7.3. EXERCISES 135

−18 ln

(x− 3

2 +√x2 − 3x+ 2

)+ C, C ∈ R

f) 14x

4 + 12x

2 + C, C ∈ Rg) − ln

∣∣∣ 1x +√

1x2

− 1∣∣∣+ C, C ∈ R

h) arcsin (lnx) + C, C ∈ Ri) 1

2 ln (x+ sinx) + Cj) arctan (ex) + C, C ∈ Rk) sin (lnx) + C, C ∈ Rl) 1

1−n (arcsinx)1−n + C, C ∈ R

m) nn−1 (sinx− cosx)

n−1n + C, C ∈ R.

(3) Calculati urmatoarele integrale nedefinite, folosind metoda a doua deschimbare de variabila:

a)∫ √

1+x1−xdx, x ∈ (−1, 1)

b)∫ √

ex−1ex+1dx, x ∈ (0,+∞)

c)∫cos2

√xdx, x ∈ (0,+∞)

d)∫

dx(x2+2)

√x2+1

, x ∈ (0,+∞)

e)∫

dx(x2+1)

√2+2x

, x ∈ R

f)∫ √

x+1x dx, x ∈ (0,+∞)

g)∫

dx1+sinx , x ∈

(0, π2

)h)∫ √

ex − 1dx, x ∈ (0,+∞)

R: a) −2

√x+11−x

x+11−x

+1+ 2 arctg

√x+11−x + C, C ∈ R

b) ln(ex +

√e2x − 1

)+ arctg 1√

e2x−1+ C, C ∈ R

c) 2√x(12 cos

√x sin

√x+ 1

2

√x)− 1

2 sin2√x− 1

2x+ C, C ∈ Rf) 1

22√

(x2 + x) + ln(x+ 1

2 +√

(x2 + x))+ C, C ∈ R

h) 2√

(ex − 1)− 2 arctg√

(ex − 1) + C, C ∈ R(4) Calculati urmatoarele integrale din functii rationale:

a)∫

4x−1x(x2−16)

dx

b)∫

x+1x(x−5)3

dx

c)∫

x2+x+1x(x2−4x+3)

dx

d)∫

dxx(x2+1)3

e)∫

x2

x4+1dx

f)∫

x2+x+1(x−1)3(x2−x+1)2

dx

g)∫

4x2−8x(x−1)2(x2+1)2

dx

136 7. INTEGRALS

h)∫

x2+x+1x3(1−x3)dx

R: a) 116 ln |x|+

1532 ln |x− 4| − 17

32 ln |x+ 4|+ C, C ∈ Rb) − 1

125 ln |x| −3

5(x−5)2+ 1

25(x−5) +1

125 ln |x− 5|+ C, C ∈ Rc) 1

3 ln |x| −32 ln |−1 + x|+ 13

6 ln |x− 3|+ C, C ∈ Rd) lnx− 1

2 ln(x2 + 1

)+ 1

2(x2+1)+ 1

4(x2+1)2+ C, C ∈ R

e) 18

√2 ln x2−x

√2+1

x2+x√2+1

+ 14

√2 arctan

(x√2 + 1

)+

+14

√2 arctan

(x√2− 1

)+ C, C ∈ R

f)− 32(−1+x)2

+ 3−1+x−2 ln |−1 + x|+ln

(x2 − x+ 1

)+28

9

√3 arctan 1

3 (2x− 1)√3+

23

−2+xx2−x+1

+ C, C ∈ Rg) 1

−1+x + 2 ln |−1 + x| − ln(x2 + 1

)+ arctanx− 1

2−4x−2x2+1

+C, C ∈ Rh) − ln |−1 + x| − 1

2x2− 1

x + ln |x|+ C, C ∈ R.(5) Calculati urmatoarele integrale din functii irationale:

a)∫

x(1−x3)

√1−x2dx, x ∈ (−1, 1)

b)∫x2

√−x2 + 4x+ 5dx, x ∈ (−1, 5)

c)∫

dx(x2+1)

√x2+1

, x ∈ Rd)∫

dx(x+2)

√x+1

, x ∈ (−1,+∞)

R: a) Substitutia√1− x2 = t (x+ 1) ne conduce la∫h (t) dt =

∫t4 − 1

t2 (t4 + 3)dt

care se calculeaza descompunand ın suma de functii rationale simple

functia t4−1t2(t4+3)

.

b) Se obtine−14x(√

−x2 + 4x+ 5)3

−56

(√−x2 + 4x+ 5

)3−25

16 (−2x+ 4)√−x2 + 4x+ 5+

2258 arcsin

(−2

3 + 13x)+C, C ∈ R, folosind substitutia

√−x2 + 4x+ 5 =

t (x+ 1).

c)∫

dx(x2+1)

√x2+1

= x√x2+1

+C, C ∈ R, folosind substitutia√x2 + 1 =

t.d)∫

dx(x+2)

√x+1

= 2arctan√x+ 1 + C, C ∈ R.

Merita retinuta substitutia 1mx+n = t, pentru calculul integralei∫dx

(mx+ x) (ax2 + bx+ c).

In cazul nostru, substitutia 1x+2 = t simplifica foarte mult calculele.

(6) Calculati urmatoarele integrale din functii trigonometrice:a)∫

dxsinx+3 cosx+5

b)∫

sinx−cosxsinx+2 cosxdx

7.3. EXERCISES 137

c)∫

dxcosx cos 2x

d)∫

dxsinx cos 2x

e)∫

dxcos4 x+sin4 x

f)∫

sin2 xsinx+cosxdx

g)∫sin 5x cosxdx

h)∫cosx cos 2x cos 3xdx

R: a) 215

√15 arctan 1

30

(4 tan 1

2x+ 2)√

15+C, C ∈ R, folosind substitutiatg x

2 = t.

b) −35 ln

(− tan 1

2x+ tan2 12x− 1

)+ 3

5 ln(1 + tan2 1

2x)

−25 arctan

(tan 1

2x)+ C, C ∈ R, folosind substitutia tg x

2 = t.c) Cu substitutia sinx = t se obtine∫

h (t) dt =

∫dt

(1− t2) (1− 2t2).

d) Cu substitutia cosx = t se obtine∫h (t) dt =

∫dt

(1− t2) (1− 2t2).

e) Cu substitutia tg (2x) = t, se ajunge la 1√2arctg tg (2x)√

2+C, C ∈ R.

f) Se noteaza I =∫

sin2 xsinx+cosxdx si J =

∫cos2 x

sinx+cosxdx si se obtine I +

J =∫

1sinx+cosxdx = 1√

2ln∣∣tg (x2 + π

8

)∣∣+C, iar I−J =∫(− cosx+ sinx) dx =

− sinx− cosx+ C, C ∈ R. Deci,

I =1

2(sinx+ cosx) +

1

2√2ln∣∣∣tg (x

2+π

8

)∣∣∣+ C, C ∈ R.

g) − 112 cos 6x − 1

8 cos 4x + C, C ∈ R, sub integrala transformandu-seprodusul ın suma, cu ajutorul calculului trigonometric.h) 1

8 sin 2x+ 116 sin 4x+ 1

24 sin 6x+ 14x+ C, C ∈ R.

(7) Calculati urmatoarele integrale definite, folosind metoda de integrareprin parti:

a)∫ π

20 x2 sinxdx

b)∫ e1 x

2 lnxdx

c)∫ π

20 (cosx)n dx, n ≥ 2

d)∫ 54

√x2 − 9dx

e)∫ 21 x

2√x2 + 1dx

f)∫ 1

20 arcsinxdx

g)∫ 20 x

2 cos2 xdx

h)∫ 10 x

2ex sinxdx

138 7. INTEGRALS

R: a) π − 2; b) 29e

3 + 19 ;

c)

In =

∫ π2

0(sinx)′ (cosx)n dx =

= sinx (cosx)n |π20 +(n− 1)

∫ π2

0sinx (cosx)n−2 sinxdx =

= (n− 1)

∫ π2

0(cosx)n−2

(1− (cosx)2

)dx =

= (n− 1) In−2 − (n− 1) In.

Deci,

In =n− 1

nIn−2;

d) 10− 9 ln 3− 2√7 + 9

2 ln(4 +

√7);

e) 94

√5 + 1

8 ln(−2 +

√5)− 3

8

√2 + 1

8 ln(√

2 + 1);

f) 112π + 1

2

√3− 1; g) 7

4 cos 2 sin 2 +56 + cos2 2; h) 1

2 .(8) Calculati urmatoarele integrale definite:

a)∫ π

4

−π4

(tg2x+ tg4x

)dx

b)∫ 1

20

xdx√1−x4

c)∫ 1

14

√xdx√1−x3

d)∫ e1

dxx(1+ln2 x)

e)∫ 21

dxx√x4+x2+1

f)∫ π

20

dx1+sinx

g)∫ π

20 sin3 x cosxdx

h)∫ π

20

sin2 x cosx2+sin2 x

dx

R: a) 23 , folosind substitutia tg x = t;

b) Substitutia x2 = t ne conduce la 12 arcsin

14 ;

c) Substitutia√x = t ne conduce la 2

3

(arcsin 1

2√2− arcsin 1

8

);

d) 14π, folosind substitutia lnx = t;

e) Se transforma integrala astfel

I =

∫ 2

1

dx

x3√

1 + 1x2

+ 1x4

= −1

2

∫ 2

1

(1x2

)′√1 + 1

x2+(

1x2

)2dx.si se efectueaza substitutia 1

x2= t;

f) 1, folosind substitutia tg x2 = t;

7.3. EXERCISES 139

g) 14 , folosind substitutia sinx = t;

h) 1−√2 arctg 1

2

√2, folosind substitutia sinx = t.

(9) Calculati urmatoarele integrale definite, folosind metoda a doua deschimbare de variabila:

a)∫ π

20

dx2+sinx

b)∫ π

40 ln (1 + tg x) dx

c)∫ 5π

4π

sin 2xsin4 x+cos4 x

dx

d)∫ π

3

−π3

dxsinx−2 cosx+3

e)∫ ln 21

√ex − 1dx

f)∫ 40

√16− x2dx

g)∫ 40

√4x− x2dx

h)∫ 32

√x−1x dx

R: a) 19π

√3, folosind substitutia tg x

2 = t;

b) π4 ln

√2. Avem

I =

∫ π4

0ln

cosx+ sinx

cosxdx =

=

∫ π4

0ln (cosx+ sinx) dx−

∫ π4

0ln cosxdx =

=

∫ π4

0ln(cosx+ cos

(π2− x))

dx−∫ π

4

0ln cosxdx =

=

∫ π4

0ln(2 cos

π

4cos(π4− x))

dx−∫ π

4

0ln cosxdx =

=

∫ π4

0ln

√2dx+

∫ π4

0ln cos

(π4− x)dx−

∫ π4

0ln cosxdx =

=π

4ln

√2 + I1 − I2,

unde

I1 =

∫ π4

0ln cos

(π4− x)dx, I2 =

∫ π4

0ln cosxdx.

Insa, cu substitutia t = π4 − x, I1 devine

I1 =

∫ 0

π4

− ln cos tdt =

∫ π4

0ln cos tdt = I2.

140 7. INTEGRALS

Ramane asadar,

I =π

4ln

√2.

c) 14π;

d) 2 arctg(56

√3 + 1

2

)− arctg 6

17 ;

e) −2√e− 1 + 2 arctg

√e− 1 + 2− 1

2π, folosind substitutia ex = t;f) 4π, folosind substitutia x = 4 sin t;g) π; avem∫ 4

0

√4x− x2dx =

∫ 4

0

√4− (x− 2)2 (x− 2)′ dx =

∫ 2

−2

√4− t2dt

si facem substitutia x = 2 sin t;h) 2

√2− 2 arctg

√2− 2 + 1

2π.

CHAPTER 8

Improper Integrals

Calculul integralei Riemann a implicat ca atat domeniul de integrare, catsi functia de integrat sa fie marginite. In cele ce urmeaza ne vom ocupa cuintegralele improprii, ce reprezinta o prelungire fireasca a integralei Riemann,ın sensul ca vom studia cazul cand domeniul de integrare sau functia de integratsunt nemarginite, ideea de baza fiind aceea a trecerii la limita.

8.1. Integrals on unbounded intervals

Ele sunt de una din formele∫ +∞a f (x) dx,

∫ a−∞ f (x) dx sau

∫ +∞−∞ f (x) dx,

cand cel putin una din limitele de integrare este infinita.Presupunem ca f : [a,+∞) → R este o functie integrabila pe fiecare com-

pact [a, x] , cu x > a.

Definitie 9.1.1. Spunem ca integrala∫ +∞a f (x) dx este convergenta

daca exista limx→+∞

∫ xa f (t) dt si este finita. Notam ın acest caz∫ +∞

af (x) dx = lim

x→+∞

∫ x

af (t) dt.

Integrala∫ +∞a f (x) dx este divergenta ın caz contrar.

Observatie 9.1.1. Deoarece f este integrabila pe fiecare interval [a, x] cu

x > a, functia F : [a,+∞) → R exista si este continua. In plus, daca existalim

x→+∞F (x) si este finita, atunci∫ +∞

af (x) dx = lim

x→+∞F (x) .

De exemplu, pentru∫ +∞1

1x2+1

dx, avem F (x) =∫ x1

1t2+1

dt = arctg t |x1=arctg x− arctg 1 si astfel∫ +∞

1

1

x2 + 1dx = lim

x→+∞(arctg x− arctg 1) =

π

2− π

4=π

4

si integrala∫ +∞1

1x2+1

dx este convergenta.

Pe de alta parte, pentru∫ +∞0 cosxdx, avem F (x) = sinx si nu exista

limx→+∞

F (x), deci integrala∫ +∞0 cosxdx este divergenta.

141

142 8. IMPROPER INTEGRALS

Celelalte tipuri de integrale improprii pe interval nemarginit se reduc laacest tip de integrala, deoarece∫ a

−∞f (x) dx =

∫ +∞

−af (−t) dt,∫ +∞

−∞f (x) dx =

∫ a

−∞f (x) dx+

∫ +∞

af (x) dx.

Teorema 9.1.1 (Cauchy). Integrala∫ +∞a f (x) dx este convergenta daca

si numai daca (∀) ϵ > 0, (∃) n0 = n0 (ϵ) > 0, astfel ıncat (∀) a1, a2 > n0, saavem ∣∣∣∣∫ a2

a1

f (x) dx

∣∣∣∣ < ϵ.

Propozitie 9.1.1. Daca∫ +∞a f (x) dx este convergenta, atunci cu necesi-

tate limx→+∞

f (x) dx = 0.

Propozitie 9.1.2. Daca f : [a,+∞) → R este integrabila pe fiecare inter-val [a, x] , cu x > a si exista o functie g : [a,+∞) → R astfel ıncat pentru oricex ≥ a sa avem f (x) ≤ g (x), atunci:∫ +∞

ag (x) dx convergenta implica

∫ +∞

af (x) dx convergenta si∫ +∞

af (x) dx divergenta implica

∫ +∞

ag (x) dx divergenta.

De exemplu,∫ +∞1 e−xdx = 1− 1

e fiind convergenta si e−x2< e−x, (∀) x > 1,

atunci ∫ +∞

1e−x

2dx este convergenta.

Propozitie 9.1.2. Daca f : [a,+∞) → R+, a > 0 este integrabila pefiecare interval [a, x] , cu x > a, atunci:

1. daca limx→+∞

xαf (x) exista si este finita, pentru un α > 1, atunci∫ +∞a f (x) dx

este convergenta.2. daca lim

x→+∞xαf (x) exista si este nenula, pentru un α ≤ 1, atunci∫ +∞

a f (x) dx este divergenta.

De exemplu,∫ +∞0

1(x2+1)2

dx este convergenta, deoarece

limx→+∞

x41

(x2 + 1)2= 1 ∈ R si α = 4 > 1.

Propozitie 9.1.3. Fie functiile f : [a,+∞) → R si g : [a,+∞) → (0,+∞)

integrabile pe fiecare interval [a, x] , cu x > a. Daca exista limx→+∞

|f(x)|g(x) = A ≥ 0

si∫ +∞a g (x) dx este convergenta, atunci

∫ +∞a |f (x)| dx este convergenta.

8.2. INTEGRALS OF UNBOUNDED FUNCTIONS 143

Propozitie 9.1.4 (Abel-Dirichlet). Fie functiile f si g : [a,+∞) → R cuproprietatile:

1) f este continua si g este de clasa C1;2) g este monoton descrescatoare cu lim

x→+∞g (x) = 0;

3) M = supx∈[a,+∞)

∣∣∫ xa f (t) dt

∣∣ < +∞.

Atunci∫ +∞a f (x) g (x) dx este convergenta.

De exemplu,∫ +∞1

cosxxα dx, cu α > 0 este convergenta, conform propozitiei

9.1.4., cu f (x) = cosx si g (x) = 1xα . De aici se deduce si convergenta inte-

gralelor∫ +∞1

sinx√xdx si

∫ +∞1

cosx√xdx.

8.2. Integrals of unbounded functions

Consideram f : [a, b] → R, continua pe [a, b] \ {x0}, cu x0 ∈ [a, b] punct dediscontinuitate de speta a doua, adica lim

x→x0|f (x)| = +∞. Observam ca putem

considera x0 = b, caci ın caz contrar scriem∫ b

af (x) dx =

∫ x0

af (x) dx−

∫ x0

bf (x) dx.

Definitie 9.2.1. Integrala∫ ba f (x) dx se numeste convergenta daca ex-

ista limx→0

∫ b−xa f (x) dx si este finita. In caz contrar, integrala numeste diver-

genta.

De exemplu,∫ 10

dx√1−x2 = lim

x→0

∫ 1−x0

dt√1−t2 = arcsin t |1−x0 = π

2 , deci∫ 10

dx√1−x2

este convergenta, functia 1√1−x2 fiind nemarginita pe [0, 1], deoarece lim

x→1

1√1−x2 =

+∞.Teorema 9.2.1 (Cauchy). Integrala

∫ ba f (x) dx este convergenta daca si

numai daca (∀) ϵ > 0, (∃) n0 = n0 (ϵ) > 0, astfel ıncat (∀) a1, a2 ∈ (0, n0), saavem ∣∣∣∣∫ b−a2

b−a1f (x) dx

∣∣∣∣ < ϵ.

Propozitie 9.2.1. Daca f : [a, b) → R∗+, este integrabila pe fiecare interval

[a, b− x] , cu x ∈ (0, b− a), si daca exista g : [a, b] → R astfel ıncat (∀) t ∈ [a, b]avem f (t) ≤ g (t), atunci

∫ b

ag (x) dx convergenta implica

∫ b

af (x) dx convergenta si∫ b

af (x) dx divergenta implica

∫ b

ag (x) dx divergenta.


Propozitie 9.2.2. Daca f : [a, b) → R∗+, este integrabila pe fiecare in-

terval [a, b− x] , cu x ∈ (0, b− a), limx→b

f (x) = +∞ si daca limx→b

(b− x)α f (x)

exista si este finita, pentru un α < 1, atunci∫ ba f (x) dx este convergenta; daca

limx→b

(b− x)α f (x) exista si este nenula, pentru un α ≥ 1, atunci∫ ba f (x) dx

este divergenta.De exemplu,

∫ +∞0

1(x2+1)2

dx este convergenta, deoarece

limx→+∞

x41

(x2 + 1)2= 1 ∈ R si α = 4 > 1.

Propozitie 9.2.3. Fie functiile f : [a, b) → R si g : [a,+∞) → (0, b) inte-

grabile pe fiecare interval [a, b− x] , cu x ∈ (0, b− a) . Daca exista limx→b

|f(x)|g(x) =

A ≥ 0 si∫ ba g (x) dx este convergenta, atunci

∫ +∞a |f (x)| dx este convergenta.

De exemplu, integrala∫ 10

dx(1+x3)

√1−x2 este convergenta, conform propozitiei

9.2.3, cu f (x) = 1(1+x3)

√1−x2 si g (x) = 1√

1−x .

8.3. Exercises

(1) Studiati natura urmatoarelor integrale improprii si, ın caz de convergenta,calculati valoarea integralei:a)∫ +∞0

dxx3+1

; b)∫ +∞0

dxx4+1

; c)∫ +∞0

dx(x2+1)2

; d)∫ +∞0

arctg x

(x2+1)√x2+1

dx; e)∫ +∞1

arctg xx2

dx; f)∫ +∞0 x3e−xdx.

R: Se aplica propozitia 9.1.2.a) lim

x→+∞x3 1

x3+1= 1 ∈ R, α = 3 > 1, deci integrala este conver-

genta. Pentru calculul ei, descompunem functia de integrat ın functiirationale simple

1

x3 + 1=

A

x+ 1+

Bx+ C

x2 − x+ 1.

8.3. EXERCISES 145

Rezulta A = 13 , B = −1

3 , C = 23 . Avem

I =1

3

∫ +∞

0

1

x+ 1dx+

1

3

∫ +∞

0

−x+ 2

x2 − x+ 1dx =

=1

3ln (x+ 1) |+∞

0 +

+1

6

∫ +∞

0

−2x+ 1

x2 − x+ 1dx+

1

2

∫ +∞

0

dx

x2 − x+ 1

=1

3ln (x+ 1) |+∞

0 −1

6ln(x2 − x+ 1

)|+∞0 +

+1

2

1√32

arctgx− 1

2√32

|+∞0

=1

3ln

x+ 1√x2 − x+ 1

|+∞0 +

1√3arctg

x− 12√

32

|+∞0

=1

3lim

x→+∞ln

x+ 1√x2 − x+ 1

+1√3

limx→+∞

arctgx− 1

2√32

−

+1√3arctg

1√3

=1√3

π

2+

1√3

π

6=

2√3π

9.

b) Integrala convergenta,∫ +∞0

dxx4+1

= 14π

√2.

c) Integrala convergenta,∫ +∞0

dx(x2+1)2

= π4 . Se integreaza mai ıntai

prin parti

∫ +∞

0

1

x2 + 1dx =

∫ +∞

0x′

1

x2 + 1dx =

=x

x2 + 1|+∞0 +2

∫ +∞

0

x2

(x2 + 1)2dx =

= limx→+∞

x

x2 + 1+ 2

∫ +∞

0

1

x2 + 1dx− 2I =

= 2

∫ +∞

0

1

x2 + 1dx− 2I,


deci I = 12

∫ +∞0

1x2+1

dx = π4 .

d) Integrala convergenta,∫ +∞0

arctanx(x2+1)

√x2+1

dx = 12π − 1. Avem

I =

∫ +∞

0

(x√

x2 + 1

)′arctg xdx =

=x√

x2 + 1arctg x |+∞

0 −∫ +∞

0

x

(x2 + 1)√x2 + 1

dx =

=π

2+

1√x2 + 1

|+∞0 =

π

2− 1.

e) Integrala convergenta,∫ +∞1

arctg xx2

dx = 14π + 1

2 ln 2. Avem

I = −∫ +∞

1

(1

x

)′arctg xdx = −1

xarctg x |+∞

1 +

+

∫ +∞

1

1

x (x2 + 1)dx

=π

4− ln

x√x2 + 1

|+∞1 =

π

4+

1

2ln 2.

f) Integrala convergenta, conform propozitiei 9.1.2, de exemplu cuα = 1).∫ +∞

0x3e−xdx = −e−xx3 |+∞

0 +3

∫ +∞

0x2e−xdx =

= −3e−xx2 |+∞0 +6

∫ +∞

0xe−xdx =

= −6xe−x |+∞0 +6

∫ +∞

0e−xdx =

= −6e−x |+∞0 = 6.

(2) Studiati natura urmatoarelor integrale improprii si, ın caz de convergenta,calculati valoarea integralei:

a)∫ +∞√

2dx

x√x2−1

; b)∫ +∞0 x3e−x

2dx; c)

∫ +∞0 e−ax sin bxdx, a > 0.

R: a) 14π; b)

12 ; c)

aa2+b2

.

(3) Studiati natura urmatoarelor integrale improprii si, ın caz de convergenta,calculati valoarea integralei:

a)∫ 52

dx√(x−2)(5−x)

; b)∫ 52

√x−25−xdx; c)

∫ 1−1

dx(2−x)

√1−x2 ; d)

∫ 10

dx(2+x)

√1−x2 ;

e)∫ 10

xndx√1−x2 , n ≥ 2.

8.3. EXERCISES 147

R: a) Aplicam propozitia 9.2.2. Avem

limx→2x>2

(x− 2)α1√

(x− 2) (5− x)= lim

x→2x>2

(x− 2)α−12

1√5− x

=

=1√3, cu α =

1

2< 1.

De asemenea,

limx→5x<5

(5− x)α1√

(x− 2) (5− x)= lim

x→5x<5

(5− x)α−12

1√x− 2

=

=1√3, cu α =

1

2< 1.

Deci, integrala este convergenta. Pentru calculul ei, folosim substitutia

x = 2 cos2 t+ 5 sin2 t si obtinem∫ 52

dx√(x−2)(5−x)

= π.

b) Analog rezulta ca integrala este convergenta si∫ 52

√x−25−xdx = 3

2π.

c) Integrala convergenta. Facem schimbarea de variabila 12−x = t si

obtinem∫ 1−1

dx(2−x)

√1−x2 = 1

3π√3.

d) Integrala convergenta. Facem schimbarea de variabila 12+x = t si

obtinem 19π

√3.

e) Integrala convergenta. Facem schimbarea de variabila x = cos t si

obtinem∫ 10

xndx√1−x2 =

∫ π20 (cos t)n dt, pe care am calculat-o ın exercitiul

8.3. 7c).

CHAPTER 9

Function Sequences and Function Series

9.1. Function sequences

Definitie 10.1.1. Numim sir de functii pe multimea A o aplicatie f :N → Hom (A,R), unde Hom (A,R) = {f : A→ R|functie} .

Notam f (n) = fn, unde fn : A→ R, iar sirul de functii ıl notam (fn)n∈N .Daca fixam x0 ∈ A, atunci obtinem un sir (fn (x0))n∈N de puncte din R.Spunem ca sirul (fn)n∈N este convergent ın x0, daca sirul (fn (x0))n∈N esteconvergent ın R.

Multimea punctelor x ∈ A, pentru care sirul (fn (x))n∈N este convergent senumeste multime de convergenta.

Daca (fn)n∈N este convergent pe A1 ⊂ A, putem defini pe A1 o functienumita functia limita si data de x0 → f (x0) , unde f (x0) este limita sirului(fn (x0))n∈N .

Convergenta definita mai sus se numeste convergenta punctuala; decivom spune ca (fn)n∈N converge punctul la f pe A1 daca (∀) x ∈ A1, (∀) ϵ > 0,(∃) n0 = n0 (x, ϵ) ∈ N, astfel ıncat (∀) n ≥ n0, avem |fn (x)− f (x)| < ϵ.

Observatie 10.1.1. Rangul n0 depinde atat de ϵ cat si de x.In cazul ın care rangul n0 depinde doar de ϵ obtinem alt tip de convergenta,

convergenta uniforma.Definitie 10.1.2. Sirul de functii (fn)n∈N este convergent uniform la

f pe A1 daca (∀) ϵ > 0, (∃) n0 = n0 (ϵ) ∈ N, astfel ıncat (∀) n ≥ n0, avem|fn (x)− f (x)| < ϵ, (∀) x ∈ A1.

Convergenta punctuala se noteaza fnp→ f , iar cea uniforma fn

u→ f.De exemplu, sa consideram fn : [0, 1] → R, fn (x) = xn, n ∈ N. Atunci

(fn)n∈N este punctual convergent catre f : [01, ] → R,

f (x) =

{1, daca x ∈ [0, 1)0, daca x = 1

si nu converge uniform catre f pe [0, 1), deoarece dand un ϵ > 0, nu gasim un

rang n0 ∈ N astfel ıncat xn < ϵ, oricare ar fi x ∈ [0, 1). Intr-adevar, rezolvandaceasta inecuatie rezulta

n > supx∈[0,1)

ln 1ϵ

ln 1x

= +∞.

149

150 9. FUNCTION SEQUENCES AND FUNCTION SERIES

Insa o multime de convergenta uniforma este [0, a), cu a ∈ (0, 1) .

9.2. Function Series

Definitie 10.2.1. Fie (fn)n∈N un sir de functii pe multimea A si con-sideram sirul sumelor partiale sn =

∑nk=0 fk. Atunci perechea formata din

sirurile fn si sn se numeste serie de functii si se noteaza∑

n∈N fn.Seria de functii

∑n∈N fn se numeste convergenta punctual catre functia

f daca sirul sumelor partiale converge punctual catre f. O serie de functiieste uniform convergenta pe A1 ⊂ A daca sirul sumelor partiale convergeuniform pe A1.

Teorema 10.2.1 (Criteriul lui Cauchy pentru siruri). Sirul de functii(fn)n∈N este uniform convergent pe A1 daca si numai daca (∀) ϵ > 0, (∃)n0 = n0 (ϵ) ∈ N , astfel ıncat (∀) n,m ≥ n0, avem|fn (x)− fm (x)| < ϵ, (∀) x ∈ A1.

Teorema 10.2.2 (Transfer de marginire pentru siruri). Sirul de functii(fn)n∈N marginite pe A, uniform convergent pe A are functia limita f marginitape A.

Teorema 10.2.3 (Transfer de continuitate pentru siruri). Sirul de functii(fn)n∈N continue pe A, uniform convergent pe A are functia limita f continuape A.

Teorema 10.2.4 (Transfer de derivabilitate pentru siruri). Fie I un in-terval si (fn)n∈N ⊂ Hom (I,R) un sir de functii de clasa C1 pe I, punctualconvergent la f pe I, cu proprietatea ca sirul derivatelor (f ′n)n∈N este con-vergent uniform pe I la functia g. Atunci functia limita f este derivabila sif ′ = g.

Teorema 10.2.5 (Transfer de integrabilitate pentru siruri). Fie (fn)n∈Nsir de functii continue pe [a, b] si uniform convergent la f pe [a, b]. Atunci

limn→∞

∫ b

afn (x) dx =

∫ b

alimn→∞

fn (x) dx =

∫ b

af (x) dx.

Merita retinut faptul ca transferul de integrabilitate are loc si ıntr-un cadrugeneral: daca functiile fn : [a, b] → R sunt integrabile si sirul fn este convergentuniform la f pe [a, b] .

Teorema 10.2.6 (Criteriul lui Cauchy pentru serii). Seria de functii∑n∈N fn este uniform convergenta pe A1 daca si numai daca (∀) ϵ > 0, (∃)

n0 = n0 (ϵ) ∈ N , astfel ıncat (∀) n,m ≥ n0, m > n, avem |fn+1 (x) + fn+2 (x) + ...+ fm (x)| <ϵ, (∀) x ∈ A1.

Teorema 10.2.7 (Transfer de marginire pentru serii). Seria de functiimarginite pe A,

∑n∈N fn, uniform convergenta pe A are functia limita f marginita

pe A.

9.3. POWER SERIES 151

Teorema 10.2.8 (Transfer de continuitate pentru serii). Seria de functiicontinue pe A,

∑n∈N fn, uniform convergenta pe A are functia limita f con-

tinua pe A.Teorema 10.2.9 (Transfer de derivabilitate pentru serii). Fie I un inter-

val si (fn)n∈N ⊂ Hom (I,R) un sir de functii de clasa C1 pe I, astfel ıncatseria

∑n∈N fn este punctual convergenta la f pe I, cu proprietatea ca seria

derivatelor∑

n∈N f′n este convergenta uniform pe I cu suma g. Atunci functia

f este derivabila si f ′ = g.Teorema 10.2.10 (Criteriul lui Weierstrass). Fie A ⊂ R si seria

∑n∈N fn

cu fn : A→ R. Daca exista o serie cu termeni pozitivi,∑

n∈N an, convergenta,cu proprietatea ca exista n0 ∈ N, astfel ıncat (∀) n ≥ n0 |fn (x)| ≤ an, (∀)x ∈ A, atunci seria

∑n∈N fn este absolut si uniform convergenta.

De exemplu,∑

n∈N1

n3+x4, x ∈ R este absolut si uniform convergenta pe R,

deoarece 1n3+x4

≤ 1n3 , (∀) x ∈ R si seria

∑n∈N

1n3 este convergenta.

Teorema 10.2.11 (Criteriul lui Abel). Fie A ⊂ R si seria∑

n∈N fn,cu fn : A → R avand proprietatea ca sirul sumelor partiale este un sir egalmarginit (adica exista M > 0 astfel ıncat |

∑nk=1 fk (x)| ≤ M, (∀) x ∈ A).

Fie an : A → [0,+∞), n ∈ N cu proprietatea ca pentru fiecare x ∈ A sirul(an (x))n∈N este descrescator si sirul de functii (an)n∈N este convergent uniformpe A la functia identic nula. Atunci seria

∑n∈N anfn este uniform convergenta.

Corolar 10.2.1. Fie an : A → [0,+∞), n ∈ N cu proprietatea ca pentrufiecare x ∈ A sirul (an (x))n∈N este descrescator si sirul de functii (an)n∈N esteconvergent uniform pe A la functia identic nula. Atunci

∑n∈N (−1)n an este

uniform convergenta pe A.

De exemplu, seria∑

n∈N (−1)n(

3

√x3 + 1

n2 − x), x ∈ R este uniform con-

vergenta, deoarece sirul an (x) =3

√x3 + 1

n2 − x este monoton descrescator si

3

√x3 + 1

n2 − x→ 0 pe R.

9.3. Power Series

Definitie 10.3.1. Seria∑

n∈N an (x− x0)n, x ∈ R, unde an este un sir de

numere reale si x0 ∈ R se numeste serie de puteri.Vom demonstra ca o serie de puteri este absolut convergenta ın interi-

orul unui interval deschis de centru x0, cu o anumita raza R (numita razade convergenta), (x0 −R, x0 +R); seria este divergenta pe complementaraaderentei intervalului de convergenta, (−∞, x0 −R) ∪ (x0 +R,+∞) .

Teorema 10.3.1. Fie seria de puteri∑

n∈N an (x− x0)n ; consideram a :=

lim supn→∞

n√

|an| si R =

1a , a > 0+∞, a = 00, a = +∞

.


Atunci seria de puteri este absolut convergenta pentru |x− x0| < R sidivergenta pentru |x− x0| > R.

Observatie 10.3.1. Daca an = 0, n ∈ N, iar limn→∞

∣∣∣an+1

an

∣∣∣ = b ∈ [0,+∞),

atunci raza de convergenta este R = 1b .

De exemplu, pentru seria∑

n∈N nnxn, R = 0; pentru seria

∑n∈N

xn

n! , R =+∞; pentru seria

∑n∈N x

n, R = 1 (pentru |x| = 1 seria este divergenta,termenul ei general netinzand la 0).

Ne vom ocupa ın cele ce urmeaza de convergenta uniforma a seriilor deputeri si vom deduce proprietatile functiei f : (x0 −R, x0 +R) → R, definitaprin

f (x) =∑n∈N

an (x− x0)n .

Teorema 10.3.2. Fie seria de puteri∑

n∈N an (x− x0)n cu raza de convergenta

R. Daca R < +∞, oricare ar fi ϵ > 0, seria converge uniform pe [x0 −R+ ϵ, x0 +R− ϵ] ;daca R = +∞, atunci seria converge uniform pe [x0 −R0, x0 +R0] , (∀) R0 ∈R.

Rezulta de aici ca seria de puteri defineste o functie continua pe (x0 −R, x0 +R) ,daca R < +∞ si pe R, daca R = +∞.

Teorema 10.3.3 (Abel). Daca seria de puteri∑

n∈N an (x− x0)n cu

raza de convergenta R converge ın x = x0, atunci ea converge uniform pe[x0 −R+ ϵ, x0 +R− ϵ] , (∀) ϵ > 0, catre o functie continua.

Teorema 10.3.4. Fie (an)n∈N un sir de numere reale. Atunci seriile de

puteri∑

n∈N anxn si

∑n∈N nanx

n−1 au aceeasi raza de convergenta.Fie

∑n∈N anx

n o serie de puteri cu raza de convergenta R > 0; se poateconstrui astfel functia f : (−R,R) → R, f (x) =

∑n∈N anx

n.Teorema 10.3.5. Fie f : (−R,R) → R, f (x) =

∑n∈N anx

n. Atunci f estede clasa C∞ si relatia f (x) =

∑n∈N anx

n se poate deriva termen cu termende o infinitate de ori ın (−R,R) .

De exemplu, seria∑

n∈N∗ nxn−1 = 1(1−x)2 , x ∈ (−1, 1), deoarece seria∑

n∈N∗ xn = 11−x , x ∈ (−1, 1) se deriveaza si se obtine relatia anterioara.

Corolar 10.3.1. Fie f (x) =∑

n∈N∗ anxn cu raza de convergenta R = 0;

atunci an = f (n)(0)n! , n ∈ N sunt unic determinati.

9.4. Analytical functions

Definitie 10.4.1. Spunem ca o functie reala f definita pe o vecinatate apunctului x0 ∈ R este dezvoltabila ın serie de puteri centrata ın x0 daca ex-ista a > 0 si un sir (an)n∈N de numere reale, astfel ıncat seria

∑n∈N an (x− x0)

n

este convergenta pe (x0 − a, x0 + a) , avand suma f (x) si dezvoltarea f (x) =∑n∈N an (x− x0)

n este unica. Functia f se numeste analitica.

9.5. EXERCISES 153

Fie f : [a, b] → R o functie de clasa C∞ si x0 ∈ (a, b) ; seria de puteri

centrata ın x0, asociata functiei f,∑

n∈Nf (n)(0)n! (x− x0)

n se numeste seriaTaylor a lui f ın jurul punctului x0.

Teorema 10.4.1 (de reprezentare a functiilor de clasa C∞ prin serii Tay-lor). Fie f : [a, b] → R o functie de clasa C∞, cu proprietatea ca exista un

M > 0, astfel ıncat∣∣f (n) (x)∣∣ ≤ M, (∀) x ∈ [a, b] , n ∈ N si fie x0 ∈ (a, b) .

Atunci seria Taylor a lui f ın jurul lui x0 este uniform convergenta pe [a, b]catre f (x) .

De exemplu,1. ex =

∑n∈N

xn

n! , x ∈ R;2. sinx =

∑n∈N

(−1)nx2n+1

(2n+1)! , x ∈ R;

3. cosx =∑

n∈N(−1)nx2n

(2n)! , x ∈ R;4. (1 + x)α = 1 +

∑n∈N∗

α(α−1)...(α−n+1)n! xn, x > −1, α ∈ R.

Teorema 10.4.2. Fie∑

n∈N anxn o serie de puteri cu raza R > 0 si suma

f ; atunci seria de puteri∑

n≥0ann+1x

n+1 are raza de convergenta R si are locrelatia ∫ x

0f (t) dt =

∑n∈N

ann+ 1

xn+1.

De exemplu, ln (1− x) = −∑

n∈N∗xn

n+1 , x ∈ (−1, 1) ; integram∫ x0

dt1−t =∑

n∈Nxn+1

n+1 si obtinem relatia de mai sus.

9.5. Exercises

(1) Sa se determine multimea de convergenta pentru urmatoarele serii defunctii:

a)∑

n∈N∗(n+1n

)n ( 1−x1−2x

)n, x = 1

2 ;

b)∑

n∈N∗sinn xnα , x ∈ R;

c)∑∞

n=2(−1)n

lnn

(1−x21+x2

)n, x ∈ R.

R: a) Cu criteriul radacinii obtinem

lim supn→∞

n√|fn (x)| =

∣∣∣∣ 1− x

1− 2x

∣∣∣∣ .Deci, seria este absolut convergenta pentru

∣∣∣ 1−x1−2x

∣∣∣ < 1, adica x ∈

(−∞, 0) ∪(23 ,+∞

), pentru

∣∣∣ 1−x1−2x

∣∣∣ > 1 seria este divergenta si pentru∣∣∣ 1−x1−2x

∣∣∣ = 1 seria este divergenta ( limn→∞

fn (0) = e = 0 si limn→∞

∣∣fn (23)∣∣ =e = 0).


b) Cum

lim supn→∞

n√

|fn (x)| = |sinx| ,

rezulta ca seria este absolut convergenta pentru x = (2k + 1) π2 ; pentru

x = 2kπ + π2 , seria este

∑n∈N∗

1nα , deci convergenta pentru α > 1 si

divergenta pentru α ≤ 1; pentru x = 2kπ− π2 , seria este

∑n∈N∗

(−1)n

nα ,deci convergenta pentru α > 0, conform criteriului lui Leibniz.c) Cu criteriul raportului, evaluam

limn→∞

|fn+1 (x)||fn (x)|

=

∣∣1− x2∣∣

x2 + 1.

Deci, seria este absolut convergenta pentru|1−x2|x2+1

> 1, sau x = 0 si

pentru x = 0 seria este∑∞

n=2(−1)n

lnn , care este convergenta, conformcriteriului lui Leibniz.

(2) Sa se determine multimea de convergenta pentru urmatoarele serii deputeri si calculati suma lor:a)∑

n∈N∗ (−1)n+1 xn

n ;

b)∑

n∈N∗ (−1)n x2n+1

2n+1 .

R: a) Avem R = limn→∞

∣∣∣ anan+1

∣∣∣ = 1, deci intervalul de convergenta este

(−1, 1) . In x = 1 avem seria armonica alternata, deci convergenta,iar ın x = −1 avem minus seria armonica divergenta. Deci seria esteconvergenta pe (−1, 1].

Definim f : (−1, 1] → R, f (x) =∑

n∈N∗ (−1)n+1 xn

n . Avem

f ′ (x) =∑n∈N∗

(−1)n+1 xn−1 =1

x+ 1, |x| < 1

si astfel f (x) = ln (x+ 1)+C, |x| < 1. Pentru x = 0 se obtine C = 0,deci f (x) = ln (x+ 1) .b) Intervalul de convergenta este (−1, 1) si, folosind criteriul lui Leib-niz rezulta ca seria este convergenta pe [−1, 1] . Prin derivare termencu termen, gasim

f ′ (x) =∑n∈N∗

(−1)n x2n =1

x2 + 1, |x| < 1.

Astfel, f (x) = arctg x + C, |x| < 1. Facand x = 0, rezulta C = 0.Deci, f (x) = arctg x.

(3) Scrieti dezvoltarea ın serie de puteri a urmatoarelor functii, indicandintervalul pe care are loc:

a) f (x) = arcsinx, x ∈ [−1, 1] ; b) f (x) = ln√

1+x1−x , x ∈ (−1, 1) ;

9.5. EXERCISES 155

c) f (x) = ln(x+

√1 + x2

), x ∈ R; d) F (x) =

∫ x0

dt√1−t2 , x ∈ (−1, 1) ;

e) f (x) = arctg x, x ∈ R.R: a) f ′ (x) =

(1− x2

)− 12 , x ∈ (−1, 1) si astfel

1√1− t2

= 1 +∑n≥1

(2n− 1)!!

(2n)!!t2n, |t| < 1

si prin integrare termen cu termen pe [0, x] , |x| < 1, rezulta

arcsinx = x+∑n≥1

(2n− 1)!!

(2n)!!

x2n+1

2n+ 1, x ∈ (−1, 1) .

Cum seria este convergenta si ın 1− si 1, rezulta ca dezvoltarea areloc pe [−1, 1] .

b) Analog rezulta ln√

1+x1−x =

∑n∈N

x2n+1

2n+1 , |x| < 1.

c) ln(x+

√1 + x2

)= x+

∑n≥1 (−1)n (2n−1)!!

(2n)!!x2n+1

2n+1 , |x| < 1.

d) F (x) = x+∑

n≥1(2n−1)!!(2n)!!

x2n+1

2n+1 , |x| ≤ 1.

e) arctg x =∑

n∈N (−1)n x2n+1

2n+1 , |x| ≤ 1.

CHAPTER 10

Partial derivatives

10.1. Linear spaces

Fie Γ unul din corpurile R al numerelor reale sau C al numerelor complexe.Definitie 11.1.1. O multime E se numeste spatiu liniar (vectorial)

peste corpul Γ daca E este ınzestrata cu o operatie aditiva (notata + : E×E →E) si o operatie de ınmultire cu scalari din Γ (notata · : Γ × E → E), avandurmatoarele proprietati:

V1) α (x+ y) = αx+ αy, (∀) α ∈ Γ, (∀) x, y ∈ E;V2) (α+ β)x = αx+ βx, (∀) α, β ∈ Γ x ∈ E;V3) α (βx) = (αβ) y, (∀) α, β ∈ Γ, (∀) x ∈ E;V4) 1 · x = x · 1 = x, (∀) x ∈ E.Elementele spatiului liniar (vectorial) se numesc vectori. Daca Γ = R,

atunci E se numeste spatiu liniar (vectorial) real si daca Γ = C, atunci Ese numeste spatiu liniar (vectorial) complex.

De exemplu, R este spatiu liniar (vectorial) real fata de operatiile uzualede adunare si ınmultire.

Rn =R× R× ...× R︸︷︷︸n ori

unde elementele lui Rn sunt n−upluri (x1, x2, ..., xn).

Vom defini

(x1, x2, ..., xn) + (y1, y2, ..., yn) = (x1 + y1, x2 + y2, ..., xn + yn)

α (x1, x2, ..., xn) = (αx1, αx2, ..., αxn) ,

oricare ar fi α ∈ R si (x1, x2, ..., xn) , (y1, y2, ..., yn) ∈ Rn. Rn devine astfelspatiu vectorial real.

Notam elementul nul al acestui spatiu cu 0Rn , iar ın cazul general 0E .Fie E un spatiu vectorial real.Definitie 11.1.2. Se numeste produs scalar pe E o aplicatie ⟨·, ·⟩ :

E ×E → R care satisface urmatoarele proprietati:PS1) ⟨·, ·⟩ este biliniara, adica

⟨x, αy + βz⟩ = α ⟨x, y⟩+ β ⟨x, z⟩⟨αx+ βy, z⟩ = α ⟨x, z⟩+ β ⟨y, z⟩ ,

oricare ar fi x, y, z ∈ E si a, b ∈ R;157

158 10. PARTIAL DERIVATIVES

PS2) ⟨·, ·⟩ este pozitiv definita, adica ⟨x, x⟩ ≥ 0, (∀) x ∈ E si ⟨x, x⟩ = 0daca si numai daca x = 0E ;

PS3) ⟨·, ·⟩ este simetrica, adica ⟨x, y⟩ = ⟨y, x⟩, (∀) x, y ∈ E.Un spatiu vectorial ınzestrat cu un produs scalar se numeste spatiu cu

produs scalar.De exemplu, pe R, spatiu vectorial real, produsul scalar se va defini ca

produsul obisnuit cu numere reale; pe Rn definim

⟨x, y⟩ =n∑k=1

xkyk,

unde x = (x1, x2, ..., xn) , y = (y1, y2, ..., yn) ∈ Rn si aceasta expresie reprezintaprodusul scalar euclidian.

Propozitie 11.1.1 (Inegalitatea Cauchy-Schwarz). Intr–un spatiu vecto-rial E cu produs scalar avem

|⟨x, y⟩| ≤√⟨x, x⟩ ·

√⟨y, y⟩, (∀) x, y ∈ E.

Definitie 11.1.3. Fie E un spatiu vectorial real. Se numeste norma peE o aplicatie ∥·∥ : E → R+, care satisface proprietatile:

N1) ∥x∥ ≥ 0, (∀) x ∈ E si ∥x∥ = 0 ⇐⇒ x = 0E ;N2) ∥αx∥ = |α| ∥x∥ , (∀) α ∈ R si (∀) x ∈ E;N3) ∥x+ y∥ ≤ ∥x∥+ ∥y∥ , (∀) x, y ∈ E.Spatiul vectorial E pe care este definita o norma se numeste spatiu vec-

torial normat si se noteaza (E, ∥·∥) .Din N3) rezulta relatia

|∥x∥ − ∥y∥| ≤ ∥x− y∥ , (∀) x, y ∈ E.

De exemplu, pe R definim norma drept modul; pe Rn putem defini maimulte norme pentru x = (x1, x2, ..., xn) :

∥x∥ =

√√√√ n∑k=1

x2k

∥x∥ =

n∑k=1

|xk| ,

∥x∥ = maxk∈1,n

|xx| .

Toate normele definite pe Rn se dovedesc a fi echivalente; adica pe Enormele ∥·∥1 si ∥·∥2 sunt echivalente daca exista doua constante a, b > 0 astfelıncat

a ∥x∥1 ≤ ∥x∥2 ≤ b ∥x∥1 , (∀) x ∈ E.

Propozitie 11.1.2. Fie ⟨·, ·⟩ un produs scalar pe E spatiu vectorial real.

Atunci expresia√

⟨x, x⟩ defineste o norma pe E.

10.2. LINEAR AND CONTINUOUS OPERATORS 159

De remarcat este ca ın Rn norma euclidiana este data de

∥x∥ =

√√√√ n∑k=1

x2k.

Definitie 11.1.4. Intr-un spatiu normat (E, ∥·∥) se numeste bila de-schisa de centru x0 si raza r multimea

Br (x0) = {x ∈ E| ∥x− x0∥ < r}si bila ınchisa de centru x0 si raza r multimea

Br (x0) = {x ∈ E| ∥x− x0∥ ≤ r} .De exemplu, daca R = R, ∥x∥ = |x| si

Br (x0) = {x ∈ R| |x− x0| < r} = (x0 − r, x0 + r) ;

daca E = R2, atunci ∥(x1, x2)∥ =√x21 + x22 si

Br (x0) ={x ∈ R2| ∥x− x0∥ < r

}este discul deschis centrat ın x0 de raza r.

Definitie 11.1.5. Intr-un spatiu normat (E, ∥·∥) se numeste veci- natatea unui punct x0 ∈ E o submultime V ⊂ E care contine o bila centrata ın x0.

Ca si ın cadrul axei reale, putem defini ın baza conceptului de bila (deschisasau ınchisa) notiunile de multime deschisa, multime ınchisa, punct de acumu-lare, punct izolat, punct aderent, aderenta, interior, frontiera ıntr-un spatiunormat.

Definitie 11.1.6. Multimea D ⊂ Rn se numeste convexa daca (∀) t ∈[0, 1] si (∀) x, y ∈ D, avem tx+ (1− t) y ∈ D.

Definitie 11.1.7. Multimea D ⊂ Rn se numeste conexa daca nu existaU si V multimi deschise ın Rn, nevide, astfel ıncat A ∩ U = ∅, A ∩ V = ∅,A ∩ U ∩ V = ∅ si A ⊂ U ∪ V.

Definitie 11.1.8. Multimea D ⊂ Rn deschisa si conexa se numeste dome-niu.

10.2. Linear and continuous operators

Fie E1 si E2 doua spatii vectoriale peste R.Definitie 11.2.1. Numim operator de la E1 la E2 orice aplicatie a lui

E1 ın E2.Definitie 11.2.2. Un operator T : E1 → E2 se numeste liniar daca este

aditiv, adica

T (x1 + x2) = T (x1) + T (x2) , (∀) x1, x2 ∈ E1

si omogen, adica

T (αx) = αT (x) , (∀) α ∈ R si x ∈ E1.


Multimea tuturor operatorilor liniari de la E1 ın E2 se noteaza L (E1, E2) .Fie (E1, ∥·∥1) si (E2, ∥·∥2) doua spatii vectoriale normate.Definitie 11.2.3. Operatorul T : E1 → E2 se numeste continuu ın

x0 ∈ E1 daca (∀) ϵ > 0, (∃) δ > 0, astfel ıncat daca ∥x− x0∥1 < δ sa rezulte∥T (x)− T (x0)∥ < ϵ.

Propozitie 11.2.1 Un operator liniar T : E1 → E2 este continuu daca sinumai daca este continuu ıntr-un singur punct.

10.3. Vector-valued functions of a real variable

Definitie 11.3.1. Spunem ca o functie f : A ⊂ R → Rm se numestefunctie de o variabila reala. In acest caz, f (x) =(f1 (x) , f2 (x) , ..., fm (x)),unde fi : A→ R, i ∈ 1,m, functiile f1, f2, ..., fm numindu-se componentelelui f.

De exemplu, daca f : R → R3, f (x) =(x2, 2y3, 3z

), atunci

f (−1) = (1,−2,−3) .

Definitie 11.3.2. Daca f : A ⊂ R → Rm si a ∈ A′, atunci spunem ca leste limita lui f ın a si scriem lim

x→af (x) = l, daca (∀) ϵ > 0, (∃) δ > 0, astfel

ıncat (∀) x ∈ A\ {a} , cu |x− a| < δ, sa avem

∥f (x)− l∥ =

√√√√ n∑k=1

(fk (x)− l)2 < ϵ.

Similar cu notiunile prezentate ın capitolul 4, obtinem notiunile de limitala stanga a lui f ın a si limita la dreapta a lui f ın a si se demontsreaza cadaca a este punct de acumulare al multimilor {x ∈ A, x < a} si {x ∈ A, x < a}atunci exista lim

x→af (x) = l daca si numai daca limita la dreapta si cea la stanga

exista si sunt egale.Teorema 11.3.1. Daca f : A ⊂ R → Rm, a ∈ A′, cu f = (f1, f2, ..., fm) ,

unde fk : A → R, k ∈ 1,m, atunci limx→a

f (x) = l daca si numai daca exista

limx→a

fk (x) = lk, k ∈ 1,m si limx→a

f (x) = (l1, l2, ..., lm) .

Definitie 11.3.2. Daca f : A ⊂ R → Rm si a ∈ A ∩ A′, atunci spunemca f este continua ın a, daca (∀) ϵ > 0, (∃) δ > 0, astfel ıncat (∀) x ∈ A, cu|x− a| < δ, sa avem ∥f (x)− f (a)∥ < ϵ.

Analog celor din capitolul 5 deducem ca o functie f : A ⊂ R → Rm sia ∈ A ∩ A′ este continua ın a daca si numai daca exista lim

x→af (x) = f (a) ;

analog obtinem notiunile de functie continua la stanga si functie continua ladreapta.

Daca f nu este continua ın a, ea este discontinua ın a.

10.4. PATHS AND CURVES 161

Teorema 11.3.2. Daca f : A ⊂ R → Rm, a ∈ A∩A′, cu f = (f1, f2, ..., fm) ,unde fk : A → R, k ∈ 1,m, atunci f este continua ın a daca si numai dacafiecare componenta fk este continua ın a.

Definitie 11.3.3. Fie f : A ⊂ R → Rm si a ∈ Int (A) . Spunem ca feste derivabila ın a daca fiecare dintre componentele fk este derivabila ın asi atunci f ′ (a) = (f ′1 (a) , f

′2 (a) , ..., f

′m (a)) .

Definitie 11.3.4. Fie f : A ⊂ R → Rm si a ∈ Int (A) . Spunem ca f estediferentiabila ın a daca exista o aplicatie liniara T : R → Rm astfel ıncat

limx→ax =a

f (x)− f (a)− T (x− a)

|x− a|= 0Rm ,

unde 0Rm este elementul nul din Rm.Se arata usor ca T : R → Rm daca exista, este unica. De asemenea, daca

f este diferentiabila ın a, atunci ea este contin ın a.Definitie 11.3.5. Fie a < b, a, b ∈ R si f : A ⊂ R → Rm. Spunem ca

f este integrabila Riemann pe [a, b] daca fiecare dintre componentele fk este

integrabila Riemann pe [a, b] si atunci∫ ba f (x) dx =

(∫ ba f1 (x) dx,

∫ ba f2 (x) dx, ...,

∫ ba fm (x) dx

).

10.4. Paths and curves

Definitie 11.4.1. Numim drum orice functie continua γ : I → Rm, undeI ⊂ R este un interval. Notand γ (t) = (f1 (t) , f2 (t) , ..., fm (t)), definim astfelo parametrizare a drumului γ :

γ :

x1 (t) = f1 (t)x2 (t) = f2 (t)...............

xm (t) = fm (t)

, t ∈ I.

Aceste ecuatii se numesc ecuatiile parametrice ale drumului γ. Daca I =[a, b] , γ (a) si γ (b) se numesc capetele drumului γ. Imaginea γ (I) se numestetraiectoria drumului γ. Daca γ (a) = γ (b), atunci γ se numeste drum ınchis.

De exemplu, drumul γ : [0, 2π] → R2, γ (t) = (cos t, sin t) are ca traiec-torie cercul

{(x, y) ∈ R2| x2 + y2 = 1

}; drumul γ : [0, 2π] → R3, γ (t) =

(r cos t, r sin t, 0) are ca traiectorie cercul{(x, y, z) ∈ R3| x2 + y2 = 1, z = 0

}.

Definitie 11.4.2. Daca γ : [a, b] → Rm este un drum, atunci γ−1 :[a, b] → Rm, γ−1 (t) = γ (a+ b− t) se numeste opusul drumului γ. Dacaγ1 : [a, b] → Rm si γ2 : [b, c] → Rm au proprietatea ca γ1 (b) = γ2 (b), atuncidrumul γ1 ∪ γ2 : [a, c] → Rm definit prin

(γ1 ∪ γ2) (t) ={γ1 (t) , t ∈ [a, b]γ2 (t) , t ∈ [b, c]


se numeste juxtapunerea drumurilor γ1 si γ2.Drumul γ : I → Rm se numeste neted daca γ este functie de clasa C1 si

γ′ (t) = 0, (∀) t ∈ I si se numeste neted pe portiuni daca este juxtapunereaunui numar finit de drumuri netede.

Fie un γ : [a, b] → Rm, definit prin ecuatiile sale paramentrice:

γ :

x1 (t) = f1 (t)x2 (t) = f2 (t)...............

xm (t) = fm (t)

, t ∈ I.

Daca se considera o diviziune arbitrara

∆ = (a = t0 < t1 < ... < tn = b)

a lui [a, b], definim

γ∆ :=

m∑k=1

√√√√ n∑i=1

(fk (ti)− fk (ti−1))2

si daca multimea {γ∆| ∆diviziune a lui [a, b]} este marginita, spunem ca dru-mul γ este rectificabil si are lungimea

l := sup∆ diviziune a lui [a,b]

γ∆.

Doua drumuri γ1 : [a, b] → Rm si γ2 : [α, β] → Rm se numesc echiva-lente daca exista o aplicatie φ : [a, b] → [α, β] continua, bijectiva si cu inversacontinua, astfel ıncat

γ1 (t) = (γ2 ◦ φ) (t) , (∀) t ∈ [a, b] .

Notam ın acest caz γ1 ∼ γ2 si se arata usor ca∼ este o relatie de echivalenta.O clasa de echivalenta de drumuri echivalente ın raport cu ∼ se numeste

curba.In cazul ın care curba este neteda, elementul de lungime pe curba este

prin definitie

ds =

√√√√ m∑k=1

f ′k (t)2dt.

Teorema 11.4.1. Daca γ : [a, b] → Rm este un drum neted, atunci el esterectificabil si lungimea drumului este

l =

∫γds =

∫ b

a

√√√√ m∑k=1

f ′k (t)2dt.

10.5. REAL-VALUED FUNCTIONS OF VECTOR VARIABLE 163

De exemplu, sa calculam lungimea drumului γ : [0, 2π] → R2, γ (t) =(cos t, sin t). Avem

x′1 (t) = (cos t)′ = − sin t

x′2 (t) = (sin t)′ = cos t

si astfel

l =

∫ 2π

0

√x′1 (t)

2 + x′2 (t)2dt =

∫ 2π

0

√sin2 t+ cos2 tdt = 2π.

10.5. Real-valued functions of vector variable

Definitie 11.5.1. Spunem ca o functie f : A ⊂ Rn → R se numestefunctie reala de mai multe variabile reale. In acest caz, f (x) = f (x1, x2, ..., xn) ∈R, unde x = (x1, x2, ..., xn) ∈ A, xi, i ∈ 1, n, numindu-se argumentele lui f.

De exemplu, daca f : R3 → R, f (x, y, z) = x2 + 2y3 + 3z, atunci

f (1, 0,−2) = 12 + 2 · 03 + 3 (−2) = −5.

Definitie 11.5.2. Daca f : A ⊂ Rn → R si a ∈ A′, atunci spunem ca leste limita lui f ın a si scriem lim

x→af (x) = l, daca (∀) ϵ > 0, (∃) δ > 0, astfel

ıncat (∀) x ∈ A\ {a} , cu ∥x− a∥ < δ, sa avem

|f (x)− l| < ϵ.

De exemplu, sa studiem existenta limitei lim(x,y)→(0,0)

x2y2

x4+y4. Considerand sirurile(

1n ,

1n

)→ (0, 0) si

(1n ,

2n

)→ (0, 0), rezulta ca f

(1n ,

1n

)= 1

2 → 12 si f

(1n ,

2n

)=

45 → 4

5 , atunci cand n → ∞, unde f : R2\ {(0, 0)} → R. Deci, nu exista

lim(x,y)→(0,0)

x2y2

x4+y4.

Propozitie 11.5.1. Fie f : A ⊂ Rn → R si a ∈ A′. Urmatoarele afirmatiisunt echivalente:

1) limx→a

f (x) = l;

2) (Criteriul lui Heine) oricare ar fi sirul (xn)n∈N ⊂ A\ {a}, convergent laa, avem

limn→∞

f (xn) = l;

3) (∀) V ∈ V (l) , (∃) U ∈ V (a), astfel ıncat (∀) x ∈ U ∩ (A\ {a}) sa avem

f (x) ∈ V.

Propozitie 11.5.2. Fie f, g : A ⊂ Rn → R functii care au proprietatile

|f (x)| ≤ g (x) , (∀) x ∈ A,

limx→a

g (x) = 0, unde a ∈ A′.


Atunci,

limx→a

f (x) = 0.

De exemplu, lim(x,y)→(0,0)

xy2

x2+y2= 0, deoarece∣∣∣∣ xy2

x2 + y2

∣∣∣∣ = |y| |xy|x2 + y2

≤ |y| .

Definitie 11.5.3. Daca f : A ⊂ Rn → R si a ∈ A ∩ A′, atunci spunemca f este continua ın a daca (∀) ϵ > 0, (∃) δ > 0, astfel ıncat (∀) x ∈ A, cu∥x− a∥ < δ, sa avem

|f (x)− l| < ϵ

sau, echivalent,

limx→a

f (x) = f (a) .

De exemplu, sa studiem continuitatea ın origine a functiei

f (x) =

{x3y2

x4+y4, daca (x, y) = (0, 0)

0, daca (x, y) = (0, 0).

Avem

lim(x,y)→(0,0)

x3y2

x4 + y4= 0,

deoarece ∣∣∣∣ x3y2

x4 + y4

∣∣∣∣ ≤ |x| → 0 = f (0, 0) .

Definitie 11.5.4. Fie A ⊂ Rn arbitrara. Atunci, prin camp scalar definitpe A ıntelegem functia f : A→ R, f = f (x1, x2, ..., xn) .

10.6. Partial derivatives. The differential of a function

Definitie 11.6.1. Fie f : A ⊂ Rn → R, a ∈ Int (A) si s ∈ Rn cu ∥s∥ = 1,adica s este versor.

Spunem ca f este derivabila dupa (directia) s ın a daca exista si este

finita limita limt→0

f(a+ts)−f(a)t . Notam aceasta limita df

ds (a) si o numim derivata

lui f dupa directia s ın a.Sa consideram {e1, e2, ..., en} baza canonica a lui Rn, adica ek = (0, ..., 1, ..., 0)

cu 1 pe pozitia a k−a.Spunem ca f este derivabila partial ın raport cu xk ın a daca f este

derivabila dupa directia ek ın a. Notam derivata partiala ın raport cu xk ına prin

∂f

∂xk(a) = lim

t→0

f (a1, ..., ak + t, ..., an)− f (a1, ..., ak, ..., an)

t.

10.6. PARTIAL DERIVATIVES. THE DIFFERENTIAL OF A FUNCTION 165

Remarcam ca pentru a calcula derivata partiala ın raport cu xk ın a, vomcalcula derivata functiei de o singura variabila reala xk → f (a1, ..., xk, ..., an) ,celelalte variabile fiind constante ın aceasta operatie.

De exemplu, pentru functia

f (x, y) = x2 sinxy

avem

∂(x2 sinxy

)∂x

= 2x sinxy + x2 (cosxy) y,

∂(x2 sinxy

)∂y

= x3 cosxy.

Definitie 11.6.2. Fie f : A ⊂ Rn → R, a ∈ Int (A). Spunem ca f estediferentiabila ın a daca exista o aplicatie liniara T : Rn → R, astfel ıncat

limx→a

f (x)− f (a)− T (x− a)

∥x− a∥= 0

sau, echivalent,

f (x) = f (a) + T (x− a) + φ (x) ∥x− a∥ , cu limx→a

φ (x) = 0.

Aplicatia liniara T este unica, poarta numele de diferentiala lui f ın a sise noteaza df (a) .

Diferentiala totala a unei functii reprezinta aplicatia Rn ∋ a→ df (a) ∈L (Rn,R) si se calculeaza dupa formula

df =∂f

∂x1dx1 +

∂f

∂x2dx2 + ...+

∂f

∂xndxn.

De exemplu, pentru functia f (x, y) = x2 + xy − y2,

df =∂f

∂xdx+

∂f

∂ydy = (2x+ y) dx+ (x− 2y) dy.

Ca si ın cadrul functiilor derivabile, functioneaza rezultatul care afirma caorice functie diferentiabila ın a este continua ın a.

Teorema 11.6.1. Fie f : A ⊂ Rn → R, a ∈ Int (A). Daca f este

diferentiabila ın a, atunci exista dfds (a), pentru orice versor s ∈ Rn si

df

ds(a) = df (a) s,

∂f

∂xk(a) = df (a) ek, k ∈ 1, n.

Definitie 11.6.3. Fie f : A ⊂ Rn → R, a ∈ Int (A). Atunci vectorulgradient al lui f ın a se noteaza cu ∇f (a) si este

df

ds(a) = df (a) s = ⟨∇f (a) , s⟩ , (∀) s ∈ Rn.


Definitie 11.6.4. Fie f : A ⊂ Rm → Rm, f = (f1, ..., fm) . Definimdivergenta lui f ın punctul a ∈ Rm ca fiind

div f (a) =∂f1∂x1

+ ...+∂fm∂xm

.

Propozitie 11.6.1. Fie f, g : A ⊂ Rn → R, a ∈ Int (A), f si g fiinddiferentiabile ın a. Atunci f + g este diferentiabila ın a si

d (f + g) (a) = df (a) + dg (a) .

Propozitie 11.6.2. Fie f : A ⊂ Rn → R, a ∈ Int (A), g : B ⊂ R → R,f (a) ∈ Int (B) , f fiind diferentiabila ın a si g ın f (a) . Atunci g ◦ f : A→ Reste diferentiabila ın a si

d (g ◦ f) (a) = dg (f (a)) ◦ df (a) .Din aceasta propozitie rezulta urmatoarea regula de derivare partiala

∂

∂xk(g ◦ f) (a) = g′ (f (a))

∂f

∂xk(a) .

De exemplu, daca f (t) = e3x+2y si x = cos t, y = t2, atunci

f ′ (t) =df

dt=∂f

∂x

dx

dt+∂f

∂y

dy

dt=

= −3e3x+2y sin t+ 4e3x+2yt.

10.7. Extrema

Definitie 11.7.1. Fie f : A ⊂ Rn → R, a ∈ Int (A). Punctul a senumeste punct de minim local daca f (a) ≤ f (x), (∀) x ∈ U ∩ A, unde Ueste o vecinatate a lui a. Punctul a se numeste punct de minim global dacaf (a) ≤ f (x), (∀) x ∈ A. Analog avem notiunile de puncte de maxim localsi maxim global.

Punctele de maxim sau de minim poarta numele de puncte de extrem.Propozitie 11.7.1. Fie f : A ⊂ Rn → R, a ∈ Int (A) ; daca f este

derivabila dupa directia versorului s ın a si a este punct de extrem local, atunci

df

ds(a) = 0.

Corolar 11.7.1. Fie f : A ⊂ Rn → R, a ∈ Int (A) un punct de extremlocal. Daca f admite derivate partiale ın a ın raport cu x1, x2, ..., xn, atunci

∂f

∂x1(a) = 0, ...,

∂f

∂xn(a) = 0.

Definitie 11.7.2. Punctul a ∈ Int (A) ın care f este diferentiabila si

∂f

∂x1(a) = 0, ...,

∂f

∂xn(a) = 0

10.7. EXTREMA 167

se numeste punct critic al functiei f.Rezulta asadar ca un punct de diferentiabilitate si punct de extrem local

este un punct critic, reciproca fiind falsa.Definitie 11.7.3. Derivatele partiale de ordin superior se definesc ın

acelasi mod ca cele pentru functii de o variabila; daca functia

xi →∂f

∂xk(a1, ..., xk, ..., an)

este definita pe o vecinatate a lui a si este derivabila ın punctul ai, atunci f sespune ca este derivabila partial de doua ori ın raport cu xi si xk. Notam

∂2f

∂xi∂xk(a) =

∂

∂xi

(∂f

∂xk

)(a) si

∂2f

∂x2i(a) =

∂

∂xi

(∂f

∂xi

)(a) .

Derivatele partiale de ordin n ≥ 3 se definesc analog.Definim matricea

Hf (a) =

∂2f∂x21

(a) ∂2f∂x1∂x2

(a) ... ∂2f∂x1∂xn

(a)

... ... ... ...∂2f

∂xn∂x1(a) ∂2f

∂xn∂x2(a) ... ∂2f

∂x2n(a)

,

numita hessiana lui f ın a, ın cazul ın care functia f : A ⊂ Rn → R, admitederivate partiale de ordinul doi ın a ∈ Int (A) .

Propozitie 11.7.2 (Criteriul lui Schwarz). Fie f : A ⊂ Rn → R, a ∈ Int

(A) si fie i, k ∈ 1, n. Daca exista ∂2f∂xi∂xk

(a) si ∂2f∂xk∂xi

(a) pe o vecinatate a luia si sunt continue, atunci

∂2f

∂xi∂xk(a) =

∂2f

∂xk∂xi(a) .

Revenim acum la problema extremelor unei functii.Definitie 11.7.4. Fie functia f : A ⊂ Rn → R, a ∈ Int (A), f de clasa

Ck ıntr-o vecinatate a punctului a. Definim polinomul Taylor de gradul k alfunctiei f ın a ca fiind

Tkf (x, a) = f (a) +

n∑j=1

∂f

∂xj(a) (xj − aj) +

+1

2!

n∑i,j=1

∂2f

∂xi∂xj(a) (xi − ai) (xj − aj) +

+1

k!

n∑j1,...,jk=1

∂kf

∂xj1 ...∂xjk(a) (xj1 − aj1) ... (xjk − ajk) .

Expresia Rkf (x, a) = f (x) − Tkf (x, a) se numeste restul dezvoltariiTaylor de ordinul k al lui f ın a.


Se remarca anularea tuturor derivatelor partiale ale lui Rkf (x, a) de oriceordin ≤ k.

De exemplu, pentru functia

f (x, y) = x2 · y2,

polinomul Taylor de ordinul doi ın a = (1, 1) este

T2f ((x, y) (1, 1)) = 1 + 2 (x− 1) + 2 (y − 1) +

+1

2!

[2 (x− 1)2 + 8 (x− 1) (y − 1) + 2 (y − 1)2

]= x2 + y2 + 1 + 4xy − 4x− 4y + 3.

Din propozita 11.7.1 stim ca un punct de extrem este punct critic, ınsa nuputem sti care dintre punctele critice sunt puncte de extrem. Acest lucru estedecis de matricea hessiana calculata ın acel punct. In aceasta directie avemurmatoarea

Propozitie 11.7.3. Fie f : A ⊂ Rn → R, a ∈ Int (A), f ∈ C2 ( Int (A)) .Daca a este un punct critic si Hf (a) este pozitiv definita, atunci a este unpunct de minim. Daca a este un punct critic si Hf (a) este negativ definita,atunci a este un punct de maxim.

Pentru a stabili cand hessiana Hf (a) este pozitiv sau negativ definita,apelam de exemplu la criteriul lui Sylvester:

1) daca determinantii∣∣∣∂2f∂x21

(a)∣∣∣ ,∣∣∣∣∣∣

∂2f∂x21

(a) ∂2f∂x1∂x2

(a)

∂2f∂x2∂x1

(a) ∂2f∂x22

(a)

∣∣∣∣∣∣ ,∣∣∣∣∣∣∣∣∂2f∂x21

(a) ∂2f∂x1∂x2

(a) ∂2f∂x1∂x3

(a)

∂2f∂x2∂x1

(a) ∂2f∂x22

(a) ∂2f∂x2∂x3

(a)

∂2f∂x3∂x1

(a) ∂2f∂x3∂x2

(a) ∂2f∂x23

(a)

∣∣∣∣∣∣∣∣ , ...,∣∣∣∣∣∣∣∂2f∂x21

(a) ∂2f∂x1∂x2

(a) ... ∂2f∂x1∂xn

(a)

... ... ... ...∂2f

∂xn∂x1(a) ∂2f

∂xn∂x2(a) ... ∂2f

∂x2n(a)

∣∣∣∣∣∣∣ sunt strict pozitivi, atunci he- ssianaHf (a) este pozitiv definita;

2) daca∣∣∣∂2f∂x21

(a)∣∣∣ < 0,

∣∣∣∣∣∣∂2f∂x21

(a) ∂2f∂x1∂x2

(a)

∂2f∂x2∂x1

(a) ∂2f∂x22

(a)

∣∣∣∣∣∣ > 0,∣∣∣∣∣∣∣∣∂2f∂x21

(a) ∂2f∂x1∂x2

(a) ∂2f∂x1∂x3

(a)

∂2f∂x2∂x1

(a) ∂2f∂x22

(a) ∂2f∂x2∂x3

(a)

∂2f∂x3∂x1

(a) ∂2f∂x3∂x2

(a) ∂2f∂x23

(a)

∣∣∣∣∣∣∣∣ < 0, ..., atunci hessiana Hf (a) este neg-

ativ definita.

10.8. VECTOR-VALUE FUNCTIONS OF VECTOR VARIABLES 169

Retinem faptul ca ın celelalte cazuri nu se poate decide asupra naturiipunctului critic, el putand fi sau nu de extrem; stabilirea cu exactitate a ex-tremului ın aceste cazuri necesita o analiza bazata dezvoltarea ın serie Taylor,fiind destul de anevoioasa.

De exemplu, sa calculam extremele functiei

f (x, y) =1 + x− y√1 + x2 + y2

.

Determinam mai ıntai punctele critice ale sistemului{∂f∂x = 0∂f∂y = 0

⇐⇒

1+y2−x+xy√

1+x2+y23 = 0

− 1+x2+y+xy√1+x2+y2

3 = 0

⇐⇒ {x = 1, y = −1} .

Astfel avem un singur punct critic, (1,−1) .Scriem acum matricea hessiana ıntr-un punct oarecare (x, y) :

Hf (x, y) = −−y+2yx2−y3+1−2x2+y2+3x+3xy2√1+x2+y2

5−y+2yx2−y3+x+x3−2xy2+3xy√

1+x2+y25

−y+2yx2−y3+x+x3−2xy2+3xy√1+x2+y2

5 −x+x3−2xy2+1+x2−2y2−3y−3yx2√1+x2+y2

5

.

De aici obtinem

Hf (1,−1) =

(−2

√3

9 −√39

−√39 −2

√3

9

).

Deoarece∣∣∣−2

√3

9

∣∣∣ = −2√3

9 < 0 si

∣∣∣∣∣ −2√3

9 −√39

−√39 −2

√3

9

∣∣∣∣∣ = 19 > 0, rezulta ca

Hf (1,−1) este negativ definita, deci punctul (1,−1) este punct de maxim,

valoarea maximului fiind f (1,−1) =√3.

10.8. Vector-value functions of vector variables

Definitie 11.8.1. Functia f : A ⊂ Rn → Rm, cu m, n ∈ N \ {0, 1} senumeste functie vectoriala de mai multe variabile reale. Daca f (x) =(f1 (x) , f2 (x) , ..., fm (x)), unde fk : A → R, k ∈ 1,m, atunci functiile fk senumesc componentele functiei f , iar (x1, x2, ..., xn) argumentele functieif.

De exemplu, daca f : R3 → R2, f (x, y, z) =(x2 + 2y3 + 3z, x+ z2), atunci

f (1, 0,−2) =(12 + 2 · 03 + 3 (−2) , 1 + (−2)2

)= (−5, 5) .


Definitie 11.8.2. Daca f : A ⊂ Rn → Rm si a ∈ A′, atunci spunem cal ∈ Rm este limita lui f ın a si scriem lim

x→af (x) = l, daca (∀) ϵ > 0, (∃)

δ > 0, astfel ıncat (∀) x ∈ A\ {a} , cu ∥x− a∥ < δ, sa avem

∥f (x)− l∥ < ϵ.

Studiul existentei limitei unei functii vectoriale de mai multe variabile realeıntr-un punct se reduce la studiul limitelor componentelor sale ın acel punct.Mai exact, avem urmatoarea

Propozitie 11.8.1. limx→a

f (x) = l exista daca si numai daca exista limx→a

fk (x) =

lk, k ∈ 1,m si l = (l1, l2, ..., lm) .Definitie 11.8.3. Daca f : A ⊂ Rn → Rm si a ∈ A ∩ A′, atunci spunem

ca f este continua ın a daca (∀) ϵ > 0, (∃) δ > 0, astfel ıncat (∀) x ∈ A, cu∥x− a∥ < δ, sa avem

∥f (x)− l∥ < ϵ

sau, echivalent,limx→a

f (x) = f (a) .

Similar se deduce ca ca o functie vectoriala de mai multe variabile realeeste continua ıntr-un punct daca si numai daca toate componentele sale suntcontinue ın acel punct.

De asemenea, notiunile de derivata dupa o directie s si de derivata partialase extind firesc ın cazul functiilor f : A ⊂ Rn → Rm. Daca ∥s∥ = 1 si a ∈ Int(A) , atunci

df

ds(a) = lim

t→0

f (a+ ts)− f (a)

t,

∂f

∂xj(a) =

(∂f1∂xj

(a) ,∂f2∂xj

(a) , ...,∂fm∂xj

(a)

).

Definitie 11.8.4. Pentru f : A ⊂ Rn → Rm definim matricea

Jf (a) =

∂f1∂x1

(a) ... ∂f1∂xn

(a)

... ... ...∂fm∂x1

(a) ... ∂fm∂xn

(a)

∈Mm,n (R) ,

numita matricea jacobiana a lui f ın a.Cand m = n, evident are sens notiunea de det Jj (a), acest determinant

numindu-se jacobianul lui f ın a sau determinantul functional al functiilorf1, f2, ..., fm ın a, notatia fiind

D (f1, ..., fn)

D (x1, ..., xn)= det Jj (a) .

De exemplu, pentru functia f : R3 → R2,

f (x, y, z) =(ex+y+z, ex−y−z

),

10.9. IMPLICIT FUNCTIONS THEOREM 171

matricea jacobiana ın punctul (x, y, z) este

Jf (x, y, z) =

(ex+y+z ex+y+z ex+y+z

ex−y−z −ex−y−z −ex−y−z);

pentru pentru functia f : [0,+∞)× [0, 2π] → R2,

f (r, φ) = (r cosφ, r sinφ) ,

matricea jacobiana ın punctul (r, φ) este

Jf (r, φ) =

(cosφ −r sinφsinφ r cosφ

),

iar jacobianul este det Jf (r, φ) =

∣∣∣∣ cosφ −r sinφsinφ r cosφ

∣∣∣∣ = r.

Definitie 11.8.5. Fie f : A ⊂ Rn → Rm, a ∈ Int (A). Spunem ca f estediferentiabila ın a daca exista o aplicatie liniara T : Rn → Rm, astfel ıncat

limx→a

f (x)− f (a)− T (x− a)

∥x− a∥= 0Rm

sau, echivalent,

f (x) = f (a) + T (x− a) + φ (x) ∥x− a∥ , cu limx→a

φ (x) = 0Rm .

Aplicatia liniara T este unica, poarta numele de diferentiala lui f ın a sise noteaza df (a) ∈ L (Rn,Rm) .

Diferentiabilitatea implica continuitatea, asa cum era de asteptat.Propozitie 11.8.2. Fie f, g : A ⊂ Rn → Rm, a ∈ Int (A), f si g fiind

diferentiabile ın a. Atunci f + g este diferentiabila ın a si

d (f + g) (a) = df (a) + dg (a) .

Propozitie 11.8.3. Fie f : A ⊂ Rn → Rm, a ∈ Int (A), g : B ⊂ Rm → Rp,f (a) ∈ Int (B) , f fiind diferentiabila ın a si g ın f (a) . Atunci g ◦ f : A→ Rpeste diferentiabila ın a si

d (g ◦ f) (a) = dg (f (a)) ◦ df (a) .

10.9. Implicit Functions Theorem

Studiul acestui paragraf este destinat rezolvarii sistemelor neliniare deforma f1 (x1, ..., xn) = y1

............................fy (x1, ..., xn) = yn

.

Teorema 11.9.1 (Teorema de inversiune locala). Fie f : A ⊂ Rn → Rn ofunctie de clasa C1 si a ∈ A astfel ıncat df (a) ∈ L (Rn,Rn) este un izomorfism(adica bijectie). Atunci exista o vecinatate U ∈ V (a) si o vecinatate V ∈


V (f (a)), astfel ıncat f : U → V sa fie bijectiva si f−1 : V → U sa fie de clasaC1. Daca f este de clasa Ck, atunci f−1 este tot de clasa Ck.

Observatie 11.9.1. Conditia df (a) ∈ L (Rn,Rn) este echivalenta cudet Jf (a) = 0, deoarece Jf (a) este tocmai matricea asociata operatoruluidf (a) .

De exemplu, sa consideram sistemul{x4+y4

x = usinx+ cos y = v

.

Avem f1 (x, y) = x4+y4

x , f2 (x, y) = sinx + cos y, unde f1 : R \ {0} → R,f2 : R → R, iar f = (f1, f2) . Ne intereseaza sa rezolvam sistemul (x, y) =

f−1 (u, v) si, mai exact, sa calculam derivatele partiale ∂x∂u ,

∂y∂u ,

∂x∂v ,

∂y∂v .

Calculam matricea jacobiana:

Jf (x, y) =

(3x2 − y4

x2

cosx − sin y

)si rezulta

det Jf

(π2,π

2

)= 0.

Conform teoremei de inversiune locala, putem rezolva sistemul ın raportcu u si v ın functie de x si y, pe o vecinatate a lui

(π2 ,

π2

).

Avem

Jf−1 =

(∂x∂u

∂x∂v

∂y∂u

∂y∂v

)= (Jf )

−1 =

= − 1

3x2 sin y + y4

x2cosx

(− sin y y4

x2

− cosx 3x2

).

Rezulta, de exemplu, ∂x∂u = − 1

3x2 sin y+ y4

x2cosx

(− sin y) .

Sa consideram acum doua multimi A, B ⊂ R, deschise, functia f : A×B →R si avem ın vedere ecuatia f (x, y) = 0 pe A×B. Se pune problema daca putemdefini pe y ca functie de x.

Fie (x, y) ∈ A × B astfel ıncat f (x, y) = 0. Ne intereseaza ın ce conditiiexista vecinatati V ∈ V (x) , U ∈ V (y) astfel ıncat fiecarui x ∈ V sa ıi core-spunda un unic y ∈ U cu proprietatea f (x, y) = 0, sau, altfel spus, ın ceconditii putem defini functia φ : V → U , y = φ (x) .

Functia astfel definita se numeste functie implicita.In cazul general, cu m componente ale functiei si n argumente, avem

urmatoarea:Teorema 11.9.2 (Teorema functiilor implicite). Fie D ⊂ Rn × Rm un

deschis si f : D → Rm o functie de clasa C1 (D) . Fie (x, y) ∈ D astfel ıncatf (x, y) = 0.

10.10. CONDITIONED EXTREMA 173

Daca D(f1,...,fm)D(y1,...,ym) (x, y) = 0, atunci exista V ∈ V (x) si exista U ∈ V (y) ,

astfel ıncat oricare ar fi x ∈ V , exista si este unic y ∈ V cu f (x, y) = 0Rm

putand astfel defini functia φ : V → U, cu proprietatea f (x, φ (x)) = 0, (∀)x ∈ V , φ ∈ C1 (V ) .

10.10. Conditioned Extrema

Aceeasi idee prezentata anterior ın cadrul extremelor obisnuite se rega-seste si ın cazul extremelor cu legaturi. Este vorba de gasirea extremelor uneifunctii de mai multe variabile reale, ın prezenta uneia sau a mai multor ecuatiide legatura, numarul acestor ecuatii de legatura fiind inferior numarului devariabile. Gasirea extremelor cu legaturi se bazeaza pe metoda multiplicatorilorlui Lagrange. Ea consta ın urmatoarea:

Teorema 11.11.1. Fie f : D ⊂ Rn+m → R, D deschis, f ∈ C1 (D),

gk : D → R, k ∈ 1,m, gk ∈ C1 (D) . Daca D(g1,...,gm)D(y1,...,ym) = 0, atunci exista

λ1, ..., λm ∈ R numiti multiplicatorii lui Lagrange astfel ıncat considerandfunctia

F (x, y) = f (x, y) +m∑k=1

λkgk (x, y) ,

punctul de extrem (x0, y0) ∈ D cu legaturile gk este solutie a sistemului

∂F

∂xj(x0, y0) = 0, j ∈ 1, n,

∂F

∂yk(x0, y0) = 0, k ∈ 1,m,

gk (x0, y0) = 0, k ∈ 1,m.

De exemplu, sa gasim extremele functiei f (x, y) = 6− 4x− 3y, cu conditiaca variabilele x si y sa satisfaca ecuatia x2 + y2 = 1.

Formam ecuatia lui Lagrange

F (x, y, λ) = f (x, y) + λ(x2 + y2 − 1

)=

= 6− 4x− 3y + λ(x2 + y2 − 1

).

Avem ∂F∂x = −4+2λx si ∂F∂y = −3+2λy. Suntem condusi astfel la sistemul −4 + 2λx = 0

−3 + 2λy = 0x2 + y2 = 1

,


pe care, rezolvandu-l, gasim solutiile

λ1 =5

2, x1 =

4

5, y1 =

3

5si

λ2 = −5

2, x2 = −4

5, y2 = −3

5.

Scriem acum matricea hessiana

Hf,λ (x, y) =

(2λ 00 2λ

),

care pentru λ1 = 52 ne conduce la Hf, 5

2

(45 ,

35

)=

(5 00 5

)si pentru λ1 = −5

2

ne conduce la Hf,− 52

(−4

5 ,−35

)=

(−5 00 −5

).

Rezulta ca Hf, 52

(45 ,

35

)este pozitiv definita, deci punctul

(45 ,

35

)este punct

de minim, valoarea minimului fiind f(45 ,

35

)= 1, iar Hf,− 5

2

(−4

5 ,−35

)este nega-

tiv definita, deci punctul(−4

5 ,−35

)este punct de maxim, valoarea maximului

fiind f(−4

5 ,−35

)= 11.

10.11. Changing of variables

Fie f : U ⊂ Rn → V ⊂ Rn, f (x) = y. Daca f este bijectiva, atunciunicul punct x ce corespunde lui y din aceasta bijectie poate fi utilizat pentrudeterminarea punctului y. Coordonatele carteziene ale lui x ın baza canonicaa spatiului Rn, (x1, x2, ..., xn) se numesc coordonatele vechi, ın timp ce co-ordonatele lui y se numesc coordonate noi. Folosind regulile de derivare alefunctiilor compuse, ecuatii ın coordonatele vechi se pot transforma ın ecuatiiın coordonatele noi.

Sa consideram cateva exemple.1. In ce se transforma ecuatia

y∂z

∂x− x

∂z

∂x= 0

ın urma schimbarii de variabile u = x si v = x2 + y2 ?Avem

∂z

∂x=

∂z

∂u

∂u

∂x+∂z

∂v

∂v

∂x=∂z

∂u+∂z

∂v2x,

∂z

∂y=

∂z

∂u

∂u

∂y+∂z

∂v

∂v

∂y=∂z

∂v2y.

Rezulta ca ecuatia data se transforma ın

y

(∂z

∂u+∂z

∂v2x

)− x

(∂z

∂v2y

)= 0

10.11. CHANGING OF VARIABLES 175

sau

y∂z

∂u= 0.

Astfel, ıntr-un domeniu din R2 ın care y = 0, rezulta ∂z∂u = 0 sau z = z (v) =

z(x2 + y2

).

2. In ce se transforma ecuatia

x2∂2z

∂x2− y2

∂2z

∂y2= 0

ın urma schimbarii de variabile u = xy si v = yx ?

Avem

∂z

∂x=

∂z

∂u

∂u

∂x+∂z

∂v

∂v

∂x=∂z

∂uy +

∂z

∂v

(− y

x2

),

∂z

∂y=

∂z

∂u

∂u

∂y+∂z

∂v

∂v

∂y=∂z

∂ux+

∂z

∂v

(1

x

).

De asemenea, retinem operatorii de derivare partiala

∂·∂x

= y∂·∂u

− y

x2∂·∂v,

∂·∂y

= x∂·∂u

+1

x

∂·∂v.

Deci,

∂2z

∂x2=

∂

∂x

(∂z

∂x

)=

=∂

∂x

[∂z

∂uy +

∂z

∂v

(− y

x2

)]=

=∂

∂x

(∂z

∂u

)y +

∂

∂x

(∂z

∂v

)(− y

x2

)+∂z

∂v

2y

x3=

=

[y∂2z

∂u− y

x2∂2z

∂v∂u

]y +

[y∂2z

∂u∂v− y

x2∂2z

∂v2

](− y

x2

)+

+∂z

∂v

2y

x3

= y2∂2z

∂u2− 2

y2

x2∂2z

∂v∂u+y2

x4∂2z

∂v2+

2y

x3∂z

∂v

si


∂2z

∂y2=

∂

∂y

(∂z

∂y

)=

=∂

∂y

[∂z

∂ux+

∂z

∂v

(1

x

)]=

=∂

∂y

(∂z

∂u

)x+

∂

∂y

(∂z

∂v

)(1

x

)=

=

[x∂2z

∂u2+

1

x

∂2z

∂v∂u

]x+

[x∂2z

∂u∂v+

1

x

∂2z

∂v2

](1

x

)=

= x2∂2z

∂u2+ 2

∂2z

∂u∂v+

1

x2∂2z

∂v2.

Rezulta ca ecuatia data se transforma ın

x2(y2∂2z

∂u2− 2

y2

x2∂2z

∂v∂u+y2

x4∂2z

∂v2+

2y

x3∂z

∂v

)−

−y2(x2∂2z

∂u2+ 2

∂2z

∂u∂v+

1

x2∂2z

∂v2

)= 0

sau

−4y2∂2z

∂u∂v+ 2

y

x

∂z

∂v= 0.

Astfel, pentru y = 0 si x = 0 rezulta

−2u∂2z

∂u∂v+∂z

∂v= 0.

3. Sa transformam ecuatia lui Laplace

∂2z

∂x2+∂2z

∂y2= 0

ın urma schimbarii de variabile (trecerea la coordonate polare)

x = r cosφ, y = r sinφ.

Avem {∂z∂r = ∂z

∂x∂x∂r +

∂z∂y

∂y∂r = ∂z

∂x cosφ+ ∂z∂y sinφ,

∂z∂φ = ∂z

∂x∂x∂φ + ∂z

∂y∂y∂φ = − ∂z

∂xr sinφ+ ∂z∂y r cosφ.

Din acest sistem de ecuatii scoatem ∂z∂x si

∂z∂y si gasim

∂z

∂x= cosφ

∂z

∂r− sinφ

r

∂z

∂φ,

∂z

∂y= sinφ

∂z

∂r+

cosφ

r

∂z

∂φ


si retinem operatorii de derivare partiala

∂·∂x

= cosφ∂·∂r

− sinφ

r

∂·∂φ

,

∂·∂y

= sinφ∂·∂r

+cosφ

r

∂·∂φ

Deci,

∂2z

∂x2=

∂

∂x

(∂z

∂x

)=

=∂

∂r

[cosφ

∂z

∂r− sinφ

r

∂z

∂φ

]∂r

∂x+

+∂

∂φ

[cosφ

∂z

∂r− sinφ

r

∂z

∂φ

]∂φ

∂x.

Insa derivand ın ambii membri fiecare ecuatie ın raport cu x, din cele douareprezentand schimbarea de variabile din enunt, obtinem{

1 = ∂r∂x cosφ− ∂φ

∂x r sinφ

0 = ∂r∂x sinφ+ ∂φ

∂x r cosφ,

sistem din care rezulta

∂r

∂x= cosφ,

∂φ

∂x= −sinφ

r.

Revenind, obtinem

∂2z

∂x2=

∂

∂r

[cosφ

∂z

∂r− sinφ

r

∂z

∂φ

]cosφ+

+∂

∂φ

[cosφ

∂z

∂r− sinφ

r

∂z

∂φ

](−sinφ

r

)=

[cosφ

∂2z

∂r2+

sinφ

r2∂z

∂φ− sinφ

r

∂2z

∂r∂φ

]cosφ+

+

[− sinφ

∂z

∂r+ cosφ

∂2z

∂φ∂r− cosφ

r

∂z

∂φ− sinφ

r

∂2z

∂φ2

]·(−sinφ

r

)= cos2 φ

∂2z

∂r2− 2

sinφ cosφ

r

∂2z

∂r∂φ+

sin2 φ

r2∂2z

∂φ2

+sin2 φ

r

∂z

∂r+ 2

sinφ cosφ

r2∂z

∂φ.


Analog, avem

∂2z

∂y2=

∂

∂y

(∂z

∂y

)=

=∂

∂r

[cosφ

∂z

∂r+

cosφ

r

∂z

∂φ

]∂r

∂y+

+∂

∂φ

[cosφ

∂z

∂r+

cosφ

r

∂z

∂φ

]∂φ

∂y.

Insa derivand ın ambii membri fiecare ecuatie ın raport cu y, din cele douareprezentand schimbarea de variabile din enunt, obtinem

{0 = ∂r

∂y cosφ− ∂φ∂y r sinφ

1 = ∂r∂y sinφ+ ∂φ

∂y r cosφ,

sistem din care rezulta

∂r

∂y= sinφ,

∂φ

∂y=

cosφ

r.

Revenind, obtinem

∂2z

∂y2=

∂

∂r

[sinφ

∂z

∂r+

cosφ

r

∂z

∂φ

]sinφ+

+∂

∂φ

[sinφ

∂z

∂r+

cosφ

r

∂z

∂φ

]cosφ

r

=

[sinφ

∂2z

∂r2− cosφ

r2∂z

∂φ+

cosφ

r

∂2z

∂r∂φ

]sinφ+

+

[cosφ

∂z

∂r+ sinφ

∂2z

∂φ∂r− sinφ

r

∂z

∂φ+

cosφ

r

∂2z

∂φ2

]·cosφr

= sin2 φ∂2z

∂r2+ 2

sinφ cosφ

r

∂2z

∂r∂φ+

cos2 φ

r2∂2z

∂φ2

+cos2 φ

r

∂z

∂r− 2

sinφ cosφ

r2∂z

∂φ.


In final, rezulta ca ecuatia lui Laplace se transforma ın

cos2 φ∂2z

∂r2− 2

sinφ cosφ

r

∂2z

∂r∂φ+

sin2 φ

r2∂2z

∂φ2+

+sin2 φ

r

∂z

∂r+ 2

sinφ cosφ

r2∂z

∂φ+

+sin2 φ∂2z

∂r2+ 2

sinφ cosφ

r

∂2z

∂r∂φ+

cos2 φ

r2∂2z

∂φ2+

+cos2 φ

r

∂z

∂r− 2

sinφ cosφ

r2∂z

∂φ= 0

sau, echivalent,

r2∂2z

∂r2+ r

∂z

∂r+∂2z

∂φ2= 0.

4. Consideram acum un caz ın care nu se schimba doar variabilele, ci

si functia. Sa calculam de exemplu ∂2z∂x2

, daca se efectueaza schimbarile de

variabile u = x2+y2, v = 1x +

1y si de functie w = ln z−x−y, unde z = z (x, y)

si w = w (u, v) .Avem

∂w

∂x=

∂w

∂u

∂u

∂x+∂w

∂v

∂v

∂x=

=∂w

∂u2x+

∂w

∂v

(− 1

x2

).

Din schimbarea de functie obtinem

∂w

∂x=

1

z

∂z

∂x− 1.

Deci,∂w

∂u2x+

∂w

∂v

(− 1

x2

)=

1

z

∂z

∂x− 1

sau∂z

∂x= z + 2xz

∂w

∂u− z

x2∂w

∂v.

Rezulta

∂2z

∂x2=

∂

∂x

(∂z

∂x

)=

∂

∂x

(z + 2xz

∂w

∂u− z

x2∂w

∂v

)=

=∂z

∂x+ 2z

∂w

∂u+ 2x

∂z

∂x

∂w

∂u+ 2xz

∂

∂x

(∂w

∂u

)−

−(

1

x2∂z

∂x− 2

x3

)∂w

∂v− z

x2∂

∂x

(∂w

∂v

).


Insa

∂

∂x

(∂w

∂u

)=

∂

∂u

(∂w

∂u

)∂u

∂x+

∂

∂v

(∂w

∂u

)∂v

∂x=

=∂2w

∂u22x+

∂2w

∂v∂u

(− 1

x2

)si

∂

∂x

(∂w

∂v

)=

∂

∂u

(∂w

∂v

)∂u

∂x+

∂

∂v

(∂w

∂v

)∂v

∂x=

=∂2w

∂u∂v2x+

∂2w

∂v2

(− 1

x2

).

Deci,

∂2z

∂x2=

∂z

∂x+ 2z

∂w

∂u+ 2x

∂z

∂x

∂w

∂u+

+2xz

[∂2w

∂u22x+

∂2w

∂v∂u

(− 1

x2

)]−

−(

1

x2∂z

∂x− 2

x3

)∂w

∂v−

−[∂2w

∂u∂v2x+

∂2w

∂v2

(− 1

x2

)].

5. Sa calculam de exemplu ∂2z∂y2

, daca se efectueaza schimbarile de variabile

x = u+v2 , v = u−v

2 si de functie z = u2−v24 −w, unde z = z (x, y) si w = w (u, v) .

Avem∂z

∂y=

1

2

(u∂u

∂y− v

∂v

∂y

)−(∂w

∂u

∂u

∂y− ∂w

∂v

∂v

∂y

).

Pentru a determina ∂u∂y si ∂v∂y derivam ın raport cu y ın ambii membri ambele

ecuatii ale sistemului {x = u+v

2v = u−v

2

si gasim 0 = 12

(∂u∂y + ∂v

∂y

)1 = 1

2

(∂u∂y − ∂v

∂y

) ,

de unde∂u

∂y= 1,

∂v

∂y= −1.

Astfel,∂z

∂y=

1

2(u+ v)−

(∂w

∂u+∂w

∂v

).

10.12. SURFACES 181

De asemenea,

∂2z

∂y2=

∂

∂y

(∂z

∂y

)=

∂

∂y

[1

2(u+ v)−

(∂w

∂u+∂w

∂v

)]=

=1

2

(∂u

∂y+∂v

∂y

)− ∂

∂y

(∂w

∂u

)− ∂

∂y

(∂w

∂v

)=

= − ∂

∂y

(∂w

∂u

)− ∂

∂y

(∂w

∂v

)=

= −(∂2w

∂u2− ∂2w

∂v∂u

)−(∂2w

∂u∂v− ∂2w

∂v2

)=

= −∂2w

∂u2+∂2w

∂v2.

10.12. Surfaces

Definitie 11.12.1. Se numeste panza de suprafata parametrizata declasa C1 o functie Σ : D ⊂ R2 → R3, de clasa C1, unde D este un domeniu allui R2.

Rezulta aplicatia

D ∋ (u, v) → Σ(u, v) = (x (u, v) , y (u, v) , z (u, v)) ∈ R3,

ın care x, y, z sunt functii de clasa C1.Ecuatiile x = x (u, v)

y = y (u, v)z = z (u, v)

, (u, v) ∈ D

se numesc ecuatiile parametrice ale panzei de suprafata s, iar multimeaΣ(D) ∈ R3 se numeste imaginea panzei Σ.

Doua panze de suprafata Σ1 : D1 → R3 si Σ2 : D2 → R3 se numescechivalente (si scriem Σ1 ∼ Σ2) daca exista o aplicatie φ : D1 → D2 bijectiva,de clasa C1, cu inversa de clasa C1 pe D2, avand jacobianul Jφ > 0 pe D1 siΣ2 ◦ φ = Σ1.

Relatia ∼ se dovedeste a fi relatie de echivalenta, ceea ce ne permite saconsideram clasele de echivalenta ın raport cu ∼, pe care le numim suprafeteparametrizate de clasa C1. Vom nota prin Σ toate panzele echivalente cu Σ.

Definim vectorii

r = x (u, v) i+ y (u, v) j + z (u, v) k,

ru =∂x

∂ui+

∂y

∂uj +

∂z

∂uk,

rv =∂x

∂vi+

∂y

∂vj +

∂z

∂vk


ın baza carora aflam vectorul normala la suprafata (vectorul perpendicularpe planul tangent la suprafata ın punctul curent),

ru × rv = ±

∣∣∣∣∣∣i j k∂x∂u

∂y∂u

∂z∂u

∂x∂v

∂y∂v

∂z∂v

∣∣∣∣∣∣ .Se noteaza cu n = ± ru×rv

∥ru×rv∥ versorul normalei la suprafata Σ ın punc-

tul curent (x, y, z) ∈ Σ.Lungimea vectorului normala la suprafata mai poate fi dedusa din

formula

∥ru × rv∥ =√EG− F 2,

unde numerele E, F, G se calculeaza din produsele scalare:

E = ru · ruF = ru · rvG = rv · rv.

De exemplu, sa consideram sfera

Σ :{(x, y, z) ∈ R3| x2 + y2 + z2 = R2

}.

O parametrizare a ei o constituie ecuatiile

Σ :

x = R cosφ sin θy = R sinφ sin θz = R cos θ

, cu φ ∈ [0, 2π] si θ ∈ [0, π] .

Avem

r = R cosφ sin θi+R sinφ sin θj +R cos θk

rφ = −R sinφ sin θi+R cosφ sin θj + 0k

rθ = R cosφ cos θi+R sinφ cos θj −R sin θk,

10.12. SURFACES 183

iar

rφ × rθ = ±

∣∣∣∣∣∣∣i j k∂x∂φ

∂y∂φ

∂z∂φ

∂x∂θ

∂y∂θ

∂z∂θ

∣∣∣∣∣∣∣ == ±

∣∣∣∣∣∣i j k

−R sinφ sin θ R cosφ sin θ 0R cosφ cos θ R sinφ cos θ −R sin θ

∣∣∣∣∣∣ == ±(−R2 cosφ sin2 θi−R2 sinφ sin2 θj −

−R2 sin θ cos θk),

∥rφ × rθ∥ = R2√

cos2 φ sin4 θ + sin2 φ sin4 θ + sin2 θ cos2 θ =

= R2 sin θ,

deci

n = ±(− cosφ sin θi− sinφ sin θj − cos θk

).

Vom considera acum un caz particular de suprafata, a carei ecuatie ıncoordonatele carteziene x, y, z permite explicitarea lui z ın functie de x si yacum o alta definitie a unei suprafetei.

Presupunem ca Σ : f (x, y, z) = 0 poate fi parametrizata prin:

Σ :

x = uy = vz = z (u, v)

, (u, v) ∈ D ⊂ R2.

Atunci

n = ± ru × rv∥ru × rv∥

= ±−pi− qj + k√1 + p2 + q2

,

unde

p =∂z

∂x, q =

∂z

∂y.

Semnele + si − din formula versorului normala la suprafata semnifica faptulca pe suprafata exista o orientare, ın raport cu care semnele trebuie aleseconvenabil ın functie de cum se specifica ın enunt sensul lui n (versorul normaleiexterioare sau interioare la suprafata).


10.13. Exercises

(1) Calculati limx→1

f (x), unde f (x) =(ex, x2 + 1, arctg x

).

R: Avem

limx→1

f (x) =(limx→1

ex, limx→1

(x2 + 1

), limx→1

(arctg x))=

=(e, 2,

π

4

).

(2) Calculati f ′ (1) pentru functia:

f (x) =(e−x

2 lnxx , x

2 + 1).

R: Avem

f ′ (x) =

(−e−x2 2 (lnx)x

2 − 1 + lnx

x2, 2x

)si f ′ (1) rezulta prin ınlocuirea lui x cu 1 ın relatia anterioara.

(3) Calculati lungimile urmatoarelor curbe:

a) cicloida

{x = a (t− sin t)y = a (1− cos t)

, t ∈ [0, 2π] ;

b) astroida

{x = a cos3 ty = a sin3 t

, t ∈ [0, 2π] ;

c) curba data ın coordonate polare: r = a sin t3 , adica{

x = a sin t3 cos t

y = a sin t3 sin t

, t ∈ [0, 3π] ;

d) spirala lui Arhimede r = at, t ∈ [0, 4π] ;e) spirala hiperbolica r = a

t , t ∈[12 , 6];

f) cardioida r = a (1 + cos t) , t ∈ [0, 2π] .R: a) 8a; b) 6a; c) 3πa

2 .(4) Calculati urmatoarele limite, ın cazul ın care ele exista:

a) lim(x,y)→(0,0)

x2+y2

x2−y2 ;

b) lim(x,y)→(0,0)

x2+y2√x2+y2+1−1

;

c) lim(x,y)→(0,0)

√x2y2+1−1

x2+y2;

d) lim(x,y)→(0,0)

sin(x3+y3)x2+y2

.

R: a) nu exista; b) 0; c) 0; d) 0.(5) Studiati continuitatea ın origine a functiei

f (x, y) =

{x2+y2

x4+y4, daca (x, y) = (0, 0)

2002, daca (x, y) = (0, 0).

10.13. EXERCISES 185

R: Deoarece lim(x,y)→(0,0)

x2+y2

x4+y4= +∞, rezulta ca functia nu este con-

tinua ın origine.(6) Calculati derivatele partiale ale functiilor:

a) f (x, y) = x3 + y3 − 3xy;b) f (x, y) = xy;c) f (x, y) = y

x ;

d) f (x, y) = ln sin x+2√y ;

e) f (x, y) = arctg yx ;

f) f (x, y) = x√x2+y2

.

R: a)∂(x3+y3−3xy)

∂x = 3x2 − 3y;∂(x3+y3−3xy)

∂y = 3y2 − 3x;

b) ∂(xy)∂x = xy−1y; ∂(x

y)∂y = xy lnx;

c)∂( y

x)∂x = − y

x2;∂( y

x)∂y = 1

x ;

d)∂(ln sin x+2√

y

)∂x =

cos x+2√y

√y sin x+2√

y

;∂(ln sin x+2√

y

)∂y = −1

2

(cos x+2√

y

)x+2

(√y)

3sin x+2√

y

;

e)∂(arctg y

x)∂x = − y

x2+y2;∂(arctg y

x)∂y = x

x2+y2;

f)∂

(x√

x2+y2

)∂x = y2√

x2+y23 ;

∂

(x√

x2+y2

)∂y = − x√

x2+y23 y.

(7) Calculati ∂f∂x (1, 1) si∂f∂y (1, 1) , daca f (x, y) =

√xy + x

y .

R:∂

(√xy+x

y

)∂x = 1

2

√x y2+1

y

y2+1y , deci ∂f

∂x (1, 1) =√2;

∂

(√xy+x

y

)∂y =

1

2

√x y2+1

y

xy2−1y2

, deci ∂f∂y (1, 1) = 0;

(8) Calculati ∂∂x

(1r

), unde r =

√x2 + y2 + z2.

R: ∂∂x

(1r

)= − 1

r2∂r∂x = − 1

r2x√

x2+y2+z2= − x

r3.

(9) Calculati determinantul

∣∣∣∣∣ ∂x∂r

∂x∂φ

∂y∂r

∂y∂φ

∣∣∣∣∣ , daca x = r cosφ si y = r sinφ.

R: r.(10) Demonstrati ca ∂f

∂x + ∂f∂y + ∂f

∂z = 1, daca

f (x, y, z) = (x− y) (y − z) (z − x) .

(11) Demonstrati ca functia f = φ(x2 + y2

)verifica ecuatia

y∂f

∂x− x

∂f

∂y= 0.


(12) Calculati dfdt daca f = xy , iar x = et, y = ln t.

R: dfdt =

et(t ln t−1)

t ln2 t.

(13) Calculati dfdt daca f = ln sin x√y , iar x = 3t2, y =

√t2 + 1.

R: dfdt =

t√y ctg x√

y

(6− x

2y2

).

(14) Calculati dfdx si ∂f∂x daca f = arctg x

y , iar y = x2.

R: ∂f∂x = − y

x2+y2; dfdx = 1

x2+1.

(15) Calculati ∂f∂x si ∂f∂y daca f = φ (u, v), iar u = x2 − y2, v = exy.

R: ∂f∂x = 2x∂f∂u (u, v) + yexy ∂f∂v (u, v) ;

∂f∂y = −2y ∂f∂u (u, v) + xexy ∂f∂v (u, v) .

(16) Aratati ca functia w = f (u, v) satisface ecuatia

∂w

∂t= a

∂w

∂x+ b

∂w

∂y,

unde u = x+ at, v = y + bt.

(17) Calculati derivata functiei f (x, y) = ln√x2 + y2 ın punctul a =

(1, 1), dupa directia s =(

1√2, 1√

2

).

R:√22 .

(18) Calculati derivata functiei f (x, y, z) = x2 − 3yz + 5 ın punctul a =

(1, 2, 1), dupa directia s =(

1√3, 1√

3, 1√

3

).

R: −√33 .

(19) Calculati grad f ın punctul a = (5, 3), daca f (x, y) =√x2 − y2.

R: grad f (5, 3) =(54 ,−

34

).

(20) Calculati grad f ın punctul a = (1, 2, 3), daca f (x, y, z) = xyz.R: grad f (1, 2, 3) = (6, 3, 2) .

(21) Calculati ∂2f∂x2

, ∂2f∂x∂y ,

∂2f∂y2

, daca

f (x, y) =√x2 + y2.

R:∂2

(√x2+y2

)∂x2

= y2√x2+y2

3 ;∂2

(√x2+y2

)∂x∂y = − xy√

x2+y23 ;

∂2(√

x2+y2)

∂y2= x2√

x2+y23 .

(22) Calculati ∂2f∂x2

, ∂2f∂x∂y ,

∂2f∂y2

, daca

f (x, y) = ln(x2 + y2

).


R:∂2(ln(x2+y2))

∂x2= −2 x2−y2

(x2+y2)2;∂2(ln(x2+y2))

∂x∂y = −4 xy

(x2+y2)2;

∂2(ln(x2+y2))∂y2

= 2 x2−y2(x2+y2)2

.

(23) Calculati ∂2f∂x∂y , daca

f (x, y) = arctgx+ y

1− xy.

R:∂2

(arctg x+y

1−xy

)∂x∂y = 0

(24) Verificati relatia

∂2f

∂x∂y=

∂2f

∂y∂x,

pentru functia f (x, y) = xy.(25) Aratati ca functia f (x, y) = arctg y

x verifica ecuatia lui Laplace

∂2f

∂x2+∂2f

∂y2= 0.

(26) Scrieti polinomul lui Taylor de gradul al treilea pentru functia f (x, y) =xy ın punctul a = (1, 1) .R:

T3f ((x, y) , (1, 1)) = 1 + (x− 1) + (x− 1) (y − 1) +

+1

2!(x− 1)2 (y − 1) .

(27) Scrieti polinomul lui Taylor de gradul al treilea pentru functia f (x, y) =ex sin y ın punctul a = (0, 0) .R:

T3f ((x, y) , (0, 0)) = y + xy +x2y

2− y3

6.

(28) Gasiti extremele functiilor de doua variabile:a) f (x, y) = x3 + 3xy2 − 15x− 12y;b) f (x, y) = x2 + xy + y2 − 2x− y;

c) f (x, y) = 1− 3√x2 + y2;

d) f (x, y) = x4 + y4 − 2x2 + 4xy − 2y2.R: a) (2, 1) este punct de minim, valoarea minimului fiind −28 si(−2,−1) este punct de maxim, valoarea maximului fiind 28.b) (1, 0) este punct de minim, valoarea minimului fiind −1.c) (0, 0) este punct de maxim, valoarea maximului fiind 1.d)(√

2,−√2)este punct de minim, valoarea minimului fiind −8.


(29) Gasiti extremele functiilor de trei variabile:a) Gasiti extremele functiilor de doua variabile:a) f (x, y, z) = x2 + y2 + z2 − xy + x− 2z;

b) f (x, y, z) = x+ y2

4x + z2

y + 2z .

R: a)(−2

3 ,−13 , 1)este punct de minim, valoarea minimului fiind −4

3 .

b)(12 , 1, 1

)este punct de minim, valoarea minimului fiind 4.

(30) Gasiti extremele cu legaturi ale functiilor:a) f (x, y) = xy, x+ y = 1;b) f (x, y) = x+ 2y, x2 + y2 = 5;c) f (x, y, z) = x− 2y + 2z, x2 + y2 + z2 = 9;d) f (x, y, z) = xy2z3, x+ y + z = 12, x > 0, y > 0, z > 0.R: a)

(12 ,

12

)este punct de maxim, valoarea maximului fiind 1

4 .b) (1, 2) este punct de maxim, valoarea maximului fiind 5.c) (−1, 2,−2) este punct de minim, valoarea minimului fiind −9 si(1,−2, 1) este punct de maxim, valoarea maximului fiind 9.d) (2, 4, 6) este punct de maxim, valoarea maximului fiind 2 · 42 · 63.

(31) Transformati ecuatia

x2d2y

dx2+ 2x

dy

dx+ y = 0,

punand x = et.

R: d2ydt2

+ dydt + y = 0.

(32) Transformati ecuatia(1− x2

) d2ydx2

− xdy

dx= 0,

punand x = cos t.

R: d2ydt2

= 0.(33) Transformati ecuatiile urmatoare considerand drept y variabila inde-

pendenta si x drept functie:

a) d2ydx2

+ 2y(dydx

)2= 0; b) dy

dx · d3ydx3

− 3(d2ydx2

)2= 0.

R: a) d2xdy2

− 2y dxdy = 0; b) d3xdy3

= 0.

(34) Transformati ecuatia

y∂z

∂x− x

∂z

∂y= 0,

cu schimbarea de variabile u = x, v = x2 + y2.R: ∂z

∂u = 0.(35) Transformati ecuatia

x∂z

∂x+ y

∂z

∂y− z = 0,


cu schimbarea de variabile u = x, v = yx .

R: u ∂z∂u − z = 0.(36) Transformati ecuatia

∂2z

∂x2− 2

∂2z

∂x∂y+∂2z

∂y2= 0,

cu schimbarea de variabile u = x+ y, v = yx si schimbarea de functie

w = zx , unde z = z (x, y) , w = w (u, v) .

R: ∂2w∂v2

= 0.(37) Transformati ecuatia

∂2z

∂x2+

∂2z

∂x∂y+∂2z

∂y2= 0,

cu schimbarea de variabile u = x + y, v = x − y si schimbarea defunctie w = xy − z, unde z = z (x, y), w = w (u, v) .

R: ∂2w∂u2

= 12 .

(38) Scrieti ecuatiile parametrice ale urmatoarelor suprafete si calculativersorul normalei exterioare la suprafata:

a) elipsoidul x2

4 + y2

9 + z2

25 = 1; b) sfera x2 + y2 + z2 = 2y;

c) conul superior z2 = x2 + y2, z ∈ [0, h]; d) paraboloidul eliptic

z = x2

a2+ y2

b2, z ∈ [0, h] ;

e) paraboloidul hiperbolic z = xy, x2 + y2 ≤ 1.R: a) x = 2 cosφ sin θ, y = 3 sinφ sin θ, z = 5 cos θ, φ ∈ [0, 2π] ,θ ∈ [0, π] ;

b) x2 + y2 + z2 = 2y ⇐⇒ x2 + (y − 1)2 + z2 = 1; x = cosφ sin θ,y = 1 + sinφ sin θ, z = cos θ, φ ∈ [0, 2π] , θ ∈ [0, π] ;

c) x = u, y = v, z =√u2 + v2, u2 + v2 ≤ h2;

d) x = u, y = v, z = u2

a2+ v2

b2, u

2

a2+ v2

b2≤ h;

e) x = u, y = v, z = uv, u2 + v2 ≤ 1.

CHAPTER 11

Integrals with parameter

11.1. Theorem of differentiability under the integral

Consideram o functie f : [a, b] × D → R, unde D ⊂ R este o intervaldeschis. Scopul acestui capitol este acela de a calcula integrale de forma F (y) =∫ ba f (x, y) dx, atunci cand y ∈ D si functia x→ f (x, y) este integrabila pe [a, b],pentru orice y ∈ D.

Teorema 12.1.1 (Teorema de derivare sub semnul integral). Fie f :

[a, b]×D → R o functie continua, cu ∂f∂y : [a, b]×D → R continua, D ⊂ R un

interval deschis. Atunci F este derivabila pe D si are loc relatia

F ′ (y) =

∫ b

a

∂f

∂y(x, y) dy, (∀) y ∈ D.


F (y) =

∫ π

0

ln (1 + y cosx)

cosxdx, |y| < 1.

Fie f : [0, π] × (−1, 1) → R, f (x, y) = ln(1+y cosx)cosx . Atunci f este continua

avand derivata partiala

∂f

∂y(x, y) =

∂

∂y

(ln (1 + y cosx)

cosx

)=

1

1 + y cosx

continua pe [0, π]× (−1, 1) . Obtinem din teorema 12.1.1,

F ′ (y) =

∫ π

0

1

1 + y cosxdx,

pe care daca o calculam (de exemplu cu substitutia x = tg x2 ), gasim

F ′ (y) =2

1− y

∫ ∞

0

dt

t2 + 1+y1−y

=

=2

1− y

√1 + y

1− yarctg t

√1 + y

1− y|∞0 =

=π√

1− y2.

191

192 11. INTEGRALS WITH PARAMETER

Deci,

F (y) = π arcsin y + C, C ∈ R.

Facandu-l pe y = 0 gasim F (0) = C =∫ π0

ln(1+0 cosx)cosx dx = 0.

Rezulta

F (y) = π arcsin y, y ∈ (−1, 1) .

Corolar 12.1.1 (Teorema lui Fubini). Daca f : [a, b] × [c, d] → R este ofunctie continua, atunci∫ b

a

(∫ d

cf (x, y) dy

)dx =

∫ d

c

(∫ b

af (x, y) dx

)dy.

De exemplu, sa calculam∫ 10xa−xblnx dx, unde a > b > 0.

Remarcam relatiaxa − xb

lnx=

∫ a

bxydy,

care ne conduce la considerarea functiei f : [0, 1] × [b, a] → R, f (x, y) = xy,care este o functie continua pe [0, 1]× [b, a] .

Aplicand teorema lui Fubini, rezulta∫ 1

0

xa − xb

lnxdx =

∫ 1

0

(∫ a

bxydy

)dx =

=

∫ a

b

(∫ 1

0xydx

)dy =

=

∫ a

b

(xy+1

y + 1|x=1x=0

)dy =

=

∫ a

b

1

y + 1dy = ln (y + 1) |ab=

= lna+ 1

b+ 1.

Consideram acum cazul integralelor cu parametru de forma

F (y) =

∫ φ(y)

af (x, y) dx,

unde f : [a, b]×D → R este integrabila si φ : D → R este derivabila.Avem urmatoareaTeorema 12.1.2. Fie functia f : [a, b]×D → R continua si cu ∂f

∂y continua

pe [a, b] × D si functia φ : D → R este derivabila, cu φ (D) ⊂ [a, b] . Atunci

F (y) =∫ φ(y)a f (x, y) dx este derivabila pe D si

F ′ (y) = f (φ (y) , y)φ′ (y) +

∫ φ(y)

a

∂f

∂y(x, y) dx.

11.2. EXERCISES 193

De exemplu, sa calculam F (y) =∫ y0

ln(xy+1)x2+1

dx, cu y ∈ (0,∞) .

Avem f (x, y) = ln(xy+1)x2+1

, f : [0, y] × (0,+∞) → R este continua cu∂f∂y (x, y) =

x(xy+1)(x2+1)

continua pe [0, y]× (0,+∞) . Functia φ : (0,+∞) → R,φ (y) = y este derivabila pe (0,∞) .

Atunci, conform teoremei 12.1.2, avem

F ′ (y) =ln(y2 + 1

)y2 + 1

+

∫ y

0

x

(xy + 1) (x2 + 1)dx,

calculele continuand ca la exemplele anterioare, printr-o integrare ın raport cuvariabila y.

Rezultate asemanatoare se obtin si cand domeniul variabilei de integrare xeste nemarginit, acesta fiind cazul integralelor improrii cu parametru.

11.2. Exercises

Sa se calculeze urmatoarele integrale cu parametru:

(1) F (y) =∫ π

20 ln

(y2 − sin2 x

)dx, y ∈ (1,∞) ;

(2) F (y) =∫ π

20

1cosx ln

1+y cosx1+y cosxdx, y ∈ (−1, 1) ;

(3) F (y) =∫ π

20

arctg (y tg x)tg x dx, y ∈ (0,∞) ;

(4) F (a, b) =∫∞0

e−ax−e−bx

x dx, b > b > 0;

(5) F (a, b) =∫∞0

cos ax−cos bxx2

dx, b > a > 0.

R: 1) Avem F ′ (y) =∫ π

20

2yy2−sin2 x

dx si cu schimbarea de variabila tg

x = t se obtine

F ′ (y) =

∫ ∞

0

π√y2 − 1

,

de unde

F (y) = π ln(y +

√y2 − 1

)+ C,C ∈ R.

Deci,

C =

∫ π2

0ln(y2 − sin2 x

)dx− π ln

(y +

√y2 − 1

)=

= π ln y +

∫ π2

0ln

(1− sin2 x

y2

)dx− π ln

(y +

√y2 − 1

)=

= π lny

y +√y2 − 1

+

∫ π2

0ln

(1− sin2 x

y2

)dx.

194 11. INTEGRALS WITH PARAMETER

Cum∣∣∣ln(1− sin2 x

y2

)∣∣∣ ≤ ln(1− 1

y2

), rezulta ca∣∣∣∣∣

∫ π2

0ln

(1− sin2 x

y2

)dx

∣∣∣∣∣ ≤ π

2ln

(1− 1

y2

)dx.

Facandu-l pe y → ∞ rezulta ca∫ π2

0ln

(1− sin2 x

y2

)dx→ 0.

Pe de alta parte,

limy→∞

π lny

y +√y2 − 1

= π ln1

2.

Deci,

C = π ln1

2si

F (y) = π ln(y +

√y2 − 1

)+ π ln

1

2, y ∈ (1,+∞) .

2) Avem F ′ (y) =∫ π

20

2dx1−y2 cos2 x , y ∈ (−1, 1) . Prelungim prin continu-

itate ın punctul x = π2 functia

f (x, y) =2

1− y2 cos2 x

cu valoarea 2. Cu substitutia tg x = t, se obtine

F ′ (y) =π√

1− y2.

Astfel,F (y) = π arcsin y + C, C ∈ R.

Facandu-l pe y = 0, rezulta C = 0.Deci

F (y) = π arcsin y.

3) Avem F ′ (y) =∫ π

20

11+y2 tg2 x

dx, y ∈ (0,∞). Prelungim prin conti-

nuitate ın punctele x = 0 si x = π2 functia

f (x, y) =1

1 + y2 tg2 x

cu valorile 1, respectiv 0. Cu substitutia tg x = t se obtine

F ′ (y) =π

2 (1 + y)+ C, C ∈ R,

11.2. EXERCISES 195

de undeF (y) =

π

2ln |y + 1|+ C, C ∈ R.

Facandu-l pe y = 0, rezulta C = 0.Deci

F (y) = ππ

2ln |y + 1| .

Daca y ∈ R, atunci

F (y) = ππ

2ln ||y|+ 1| .

4) Avem ca

e−ax − e−bx

x= −e

−xy

x|y=by=a=

∫ b

ae−xydy,

care ne conduce la considerarea functiei f : (0,∞) × [a, b] → R,f (x, y) = e−yx

x , care este o functie continua pe (0,∞)×[a, b] . Aplicandteorema lui Fubini, rezulta∫ ∞

0

e−ax − e−bx

xdx =

∫ ∞

0

(∫ b

ae−xydy

)dx =

=

∫ b

a

(∫ ∞

0e−xydx

)dy =

=

∫ b

a

(−e

−xy

y|x=∞x=0

)dy =

=

∫ b

a

1

ydy = ln

b

a.

CHAPTER 12

Double integrals

12.1. Definitions and calculus methods

Definitie 13.1.1. Fie f : D ⊂ R2 → R o functie continua, unde Deste o multime compacta (ınchisa si marginita). Fie P = (D1, D2, ..., Dn) odescompunere a lui D ın n domenii disjuncte doua cate doua, D = ∪ni=1Di. Ppoarta numele de diviziune Riemann a lui D. Cel mai mare dintre toatediametrele domeniilor Di se numeste norma diviziunii Riemann P si senoteaza cu ∥P∥ . Alegand arbitrar ın fiecare Di, cate un punct zi = (ξi, ηi) ,i ∈ 1, n, expresia

σf (P, z) =

n∑i=1

f (ξi, ηi) · Aria (Di)

se numeste suma Riemann asociata functiei f, diviziunii P si punctelor in-termediare (ξi, ηi) .

Functia f : D → R se numeste integrabila Riemann pe multimea com-pacta (ınchisa si marginita) D daca exista un numar I ∈ R, astfel ıncat pentruorice sir de diviziuni (Pn)n∈N cu norma tinzand la zero si orice sir de puncteintermediare diviziunilor Pn, zn = (ξni , η

ni )n∈N,

σf (P, zn) → I.

Orice functie continua pe D este integrabila Riemann pe D.Integrala dubla a functiei f pe D ⊂ R2 este prin definitie numarul I si

se noteaza

I =

∫∫Df (x, y) dxdy,

Se disting doua tipuri fundamentale de domenii D de integrare.1) Domeniu simplu ın raport cu axa Oy.D este limitat la stanga si la dreapta de dreptele x = a si x = b, cu x1 < x2

si inferior si superior de curbele y = φ1 (x) si y = φ2 (x) , cu φ1 (x) ≤ φ2 (x) .

Figura 29

197

198 12. DOUBLE INTEGRALS

Atunci calculul integralei duble∫∫

Df (x, y) dxdy revine la∫∫Df (x, y) dxdy =

∫ b

adx

∫ φ2(x)

φ1(x)f (x, y) dy =

=

∫ b

a

(∫ φ2(x)

φ1(x)f (x, y) dy

)dx.

Cele doua integrale simple se calculeaza ıntai ın raport cu y (atunci candx ∈ [a, b] este constant) si apoi ın raport cu x.

2) Domeniu simplu ın raport cu axa Ox.D este limitat inferior si superior de dreptele y = c si y = d, cu c < d si la

stanga si la dreapta de curbele x = ψ1 (y) si x = ψ2 (y) , cu ψ1 (y) ≤ ψ2 (y) .

Figura 30

Atunci calculul integralei duble∫∫

Df (x, y) dxdy revine la∫∫Df (x, y) dxdy =

∫ d

cdy

∫ ψ2(x)

ψ1(x)f (x, y) dx =

=

∫ d

c

(∫ ψ2(x)

ψ1(x)f (x, y) dx

)dy.

Cele doua integrale simple se calculeaza ıntai ın raport cu y (atunci candx ∈ [a, b] este constant) si apoi ın raport cu x.


I =

∫ 1

0dx

∫ 1

x(x+ y) dy.

Daca domeniul de integrare nu este nici de tipul 1), nici de tipul 2), atuncise trece la despartirea domeniului ın subdomenii care sunt simple ın raport cuuna sau cealalta dintre axele de coordonate.

12.2. Properties of the double integral

1. Proprietatea de liniaritate:∫∫D(af + bg) (x, y) dxdy = a

∫∫Df (x, y) dxdy + b

∫∫Dg (x, y) dxdy,

oricare ar fi a, b ∈ R.2. Proprietatea de aditivitate fata de domeniul de integrare:∫∫

Df (x, y) dxdy =

∫∫D1

f (x, y) dxdy +

∫∫D2

f (x, y) dxdy,

unde D = D1 ∪D2 si D1 ∩D2 = ∅.3. Proprietatea de monotonie:


Daca f (x, y) ≤ g (x, y) , (∀) (x, y) ∈ D, atunci∫∫Df (x, y) dxdy ≤

∫∫Dg (x, y) dxdy.

4. Proprietatile de medie:a) Daca m ≤ f (x, y) ≤M, (∀) (x, y) ∈ D, atunci

m ·Aria (D) ≤∫∫

Df (x, y) dxdy ≤M ·Aria (D) .

b) Daca f este continua pe D compact si convex (adica tx+ (1− t) y ∈ D,(∀) t ∈ [0, 1] , x, y ∈ D), atunci exista un punct (ξ, η) ∈ D, astfel ıncat∫∫

Df (x, y) dxdy = f (ξ, η) ·Aria (D) .

5. Proprietatea de calibrare:Daca D ⊂ R2 este un domeniu compact, atunci

Aria (D) =

∫∫Ddxdy.


Teorema 13.3.1. Fie D, D∗ ⊂ R2 multimi compacte, marginite de curbeleFr D, Fr D∗ ınchise si netede. Fie T : D∗ → D,

T :

{x = x (u, v)y = y (u, v)

, (u, v) ∈ D∗,

avand proprietatile:1) T este bijectiva pe D∗;2) T este de clasa C1 ın D∗;3) Jacobianul lui T este diferit de zero ın D∗.Fie f : D → R integrabila pe D. Atunci∫∫

Df (x, y) dxdy =

∫∫D∗f (x (u, v) , y (u, v)) |det JT (u, v)| dudv.

Observatie 13.3.1. Schimbarea de variabila ıntr-o integrala dubla aredrept scop simplificarea domeniului de integrare sau simplificarea functiei deintegrat.

Sa consideram cateva exemple de schimbari de variabile.1. Trecerea la coordonate polare (ρ raza polara si φ unghiul polar):

T :

{x = ρ cosφy = ρ sinφ

,

unde D∗ ={(ρ, φ) ∈ R2, ρ ≥ 0, φ ∈ [0, 2π]

}.


Jacobianul transformarii este

det JT =

∣∣∣∣ cosφ −ρ sinφsinφ ρ cosφ

∣∣∣∣ = ρ.

Aceasta schimbare de variabile este indicata atunci cand domeniul de in-tegrare D este un disc, o portiune dintr-un disc sau cand functia de integratcontine suma x2 + y2.

2. Trecerea la coordonate polare generalizate (ρ raza polara si φ unghiulpolar):

T :

{x = aρ cosφy = bρ sinφ

,

unde D∗ ={(ρ, φ) ∈ R2, ρ ≥ 0, φ ∈ [0, 2π]

}.


det JT =

∣∣∣∣ a cosφ −aρ sinφb sinφ bρ cosφ

∣∣∣∣ = abρ.

Aceasta schimbare de variabile este indicata atunci cand domeniul de in-tegrare D este limitat de o elipsa, o portiune dintr-un domeniu limitat de o

elipsa sau cand functia de integrat contine suma x2

a2+ y2

b2.

3. Daca avem de calculat o integrala dubla pe un domeniu delimitat deun patrulater curbiliniu, avand drept laturi opuse curbe care fac fiecare partedintr-un fascicul de curbe ce depind de un singur parametru, este natural saconsideram schimbarea de variabile data tocmai de aceste doua familii.

De exemplu, daca avem de calculat aria patrulaterului curbiliniu marginitde hiperbolele xy = p, xy = q si dreptele y = ax, y = bx, cu q > p > 0,b > a > 0, atunci consideram

T−1 :

{xy = uyx = v

, cu u ∈ (p, q) , v ∈ (a, b) ,

sau

T :

{x =

√uv

y =√uv

,

cu jacobianul

det JT =1

2v.

Rezulta

Aria (D) =

∫∫Ddxdy =

∫∫D∗

|det JT (u, v)| dudv =

=

∫ q

pdu

∫ b

a

1

2vdv =

1

2(q − p) ln

b

a.

12.4. EXERCISES 201

4. Sa calculam

I =

∫ +∞

0e−x

2dx.

Avem

I2 =

(∫ +∞

0e−x

2dx

)·(∫ +∞

0e−y

2dy

)=

=

∫ +∞

0

(∫ +∞

0e−y

2dy

)· e−x2dx =

=

∫∫De−(x

2+y2)dxdy,

unde D = [0,+∞)× [0,+∞).Trecand la coordonate polare, domeniul transformat devine

D∗ ={(ρ, φ) ∈ R2, ρ ∈ [0,+∞), φ ∈

[0,π

2

]}si astfel

I2 =

∫ π2

0dφ

∫ +∞

0ρe−ρ

2dρ =

π

2

(−1

2e−ρ

2

)|+∞0 =

π

4.

Deci,

I =

√π

2.

De asemenea, obtinem ∫ +∞

−∞e−x

2dx =

√π.

12.4. Exercises

(1) Sa se calculeze∫∫

Dxyy2+1

dxdy, unde D este dreptunghiul [−2, 3] ×[−1, 5] .R: Domeniul de integrare este simplu ın raport cu ambele axe. Dacaintegram ın ordinea y, x, gasim∫∫

D

xy

y2 + 1dxdy =

∫ 2

−3xdx

∫ 5

−1

y

y2 + 1dy =

=x2

2|3−2 ·

1

2ln(y2 + 1

)|5−1=

=5 ln 13

4.


Daca integram ın ordinea x, y, gasim

∫∫D

xy

y2 + 1dxdy =

∫ 5

−1

y

y2 + 1dy

∫ 2

−3xdx =

=x2

2|3−2 ·

1

2ln(y2 + 1

)|5−1=

=5 ln 13

4.

Observatie 13.4.1. In acest caz nu are importanta ordinea de in-tegrare. Intrucat limitele de integrare sunt constante reale si functiaf (x, y) este un produs dintre o functie doar de x si o functie doar dey, i.e. f (x, y) = g (x) · h (y), unde g (x) = x si h (y) = y

y2+1, valoarea

integralei duble este egala cu produsul celor doua integrale simple ınraport cu fiecare dintre cele doua variabile.


Dxy

(1+x2+y2)32dxdy, unde D este dreptunghiul [0, 1]×

[0, 1] .R: Integram ın ordinea y, x si avem

I =

∫∫D

xy

(1 + x2 + y2)32

dxdy =

=

∫ 1

0xdx

∫ 1

0

y

(1 + x2 + y2)32

dy =

=

∫ 1

0x

12

(1 + x2 + y2

)− 12

−12

|y=1y=0 dx =

=

∫ 1

0

x√x2 + 1

dx−∫ 1

0

x√x2 + 2

dx =

=√x2 + 1 |x=1

x=0 −√x2 + 2 |x=1

x=0= 2√2−

√3− 1.

Daca vom integra ın ordinea x, y, atunci metoda de calcul este aceeasi.In acest caz, valoarea integralei duble nu mai rezulta din produsul val-orilor celor doua integrale simple, deoarece functia de integrat xy

(1+x2+y2)32

nu mai are variabilele separabile.(3) Sa se calculeze

∫∫D(x+ 2y) dxdy, unde D este domeniul limitat de

parabolele y = x2 + 1, y = −x2 si dreptele x = −1, x = 3.

12.4. EXERCISES 203

R: Avem

I =

∫∫D(x+ 2y) dxdy =

∫ 3

−1dx

∫ x2+1

−x2(x+ 2y) dy =

=

∫ 3

−1

(xy + y2

)|y=x

2+1y=−x2 dx =

∫ 3

−1

(2x3 + 2x2 + x+ 1

)dx =

=200

3.


Dxdxdy, unde D este domeniul limitat de curbelex2

2 − y2

4 − 1 = 0, y2 = 2x, y = 0, y = 1.R: Avem

I =

∫∫Dxdxdy =

∫ 1

0dy

∫ √y2+4

2

y2

2

xdx =

=1

2

∫ 1

0x2 |

x=

√y2+4

2

x= y2

2

dx =1

2

∫ 1

0

(y2 + 4

2− y2

4

)dy =

127

120.


D

√y−xy+xdxdy, unde D este domeniul limitat de

dreptele y = −1, y = −2, y = x, y = −2x.R:Avem

I =

∫∫D

√y − x

y + xdxdy =

∫ −1

−2dy

∫ − y2

y

√y − x

y2 − x2dx =

=

∫ −1

−2

(y arcsin

x

y+√y2 − x2

)|x=− y

2x=y dx

=

∫ 1

0

(y arcsin

(−1

2

)+y

2

√3

)dy

= − 1

12π +

1

4

√3.


Dxydxdy, unde D este domeniul limitat de tri-unghiul ABC de varfuri A (1, 1) , B (2, 2) , C (1, 3) .R: Observam ca este de preferat ordinea de integrare y, x :

I =

∫∫Dxydxdy =

∫ 2

1xdx

∫ 4−x

xydy =

∫ 2

1x

(y2

2

)|y=4−xy=x dx =

= 4

∫ 2

1

(2x− x2

)dx =

8

3.



D

ln(x2+y2)x2+y2

dxdy, undeD este domeniulD ={(x, y) ∈ R2| 1 ≤ x2 + y2 ≤ e2

}.

R: Prin trecerea la coordonate polare, avem{x = ρ cosφy = ρ sinφ

,

unde D∗ ={(ρ, φ) ∈ R2| ρ ∈ [1, e] , φ ∈ [0, 2π]

}. Deci,

I =

∫ 2π

0dφ

∫ e

1

ln ρ2

ρ2ρdρ = 2π

∫ e

12ln ρ

ρdρ =

= 2π ln2 ρ |ρ=eρ=1= 2π.


Dy2

x2dxdy, unde D este domeniul

D ={(x, y) ∈ R2| 1 ≤ x2 + y2 ≤ 2x

}.

R: Prin trecerea la coordonate polare, avem{x = ρ cosφy = ρ sinφ

.

Prin ınlocuirea lui x si y cu expresiile lor ın functie de ρ si φ ıninecuatiile ce delimiteaza domeniul, avem

1 ≤ ρ2 ≤ 2ρ cosφ.

Rezulta ca avem cu necesitate cosφ ≥ 0, adica φ ∈[−π

2 ,π2

]si

ρ ∈ [1, 2 cosφ] .

Pentru determinarea domeniului unghiului polar, rezolvam inecuatia2 cosφ ≥ 1, de unde gasim φ ∈

[−π

3 ,π3

].

Deci,

D∗ ={(ρ, φ) ∈ R2| ρ ∈ [1, 2 cosφ] , φ ∈

[−π3,π

3

]}si

I =

∫ π3

−π3

tg 2φdφ

∫ 2 cosφ

1ρdρ = 2

∫ π3

−π3

sin2 φdφ− 1

2

∫ π3

−π3

tg 2φdφ =

= π − 3√3

2.


Ddxdy

3√

5−x2

9− y2

16

, unde D este domeniul

D =

{(x, y) ∈ R2| 1 ≤ x2

9+y2

16≤ 4

}.

12.4. EXERCISES 205

R: Prin trecerea la coordonate polare generalizate, avem{x = 3ρ cosφy = 4ρ sinφ

.

undeD∗ ={(ρ, φ) ∈ R2| ρ ∈ [1, 2] , φ ∈ [0, 2π]

}si detJT = 12ρ. Deci,

I =

∫ 2π

0dφ

∫ 2

1

12ρ3√

5− ρ2dρ = −12π

∫ 2

1

(5− ρ2

)′ (5− ρ2

)− 13 dρ =

= −12π

(5− ρ2

) 23

23

|ρ=2ρ=1= 18π

(3√16− 1

).

(10) Sa se calculeze aria patrulaterului curbiliniu marginit de dreptele y =2x+ a, y = 2x+ b, y = αx, y = βx, cu 0 < a < b, 0 < α < β.R: Folosind schimbarea de variabile

T−1 :

{y − 2x = uyx = v

, cu u ∈ [a, b] si v ∈ [α, β] ,

avem

T :

{x = u

v−2y = uv

v−2

,

si det JT = u(v−2)2

. Rezulta

Aria (D) =

(∫ b

audu

)·(∫ β

α

dv

(v − 2)2

)=

=b2 − a2

2·(

1

α− 2− 1

β − 2

).

(11) Sa se calculeze aria patrulaterului curbiliniu marginit de curbele xy =1, xy = 8, y2 = x, y2 = 5x.R: Folosind schimbarea de variabile

T−1 :

{xy = uy2

x = v, cu u ∈ [1, 8] si v ∈ [1, 5] ,

avem

T :

{x = u

v2

y = u3

v

,

si det JT = − 1v . Rezulta

Aria (D) =

(∫ 8

1du

)·(∫ 5

1

dv

v

)= 7 ln 5.


(12) Calculati aria delimitata de bucla din dreapta a lemniscatei lui Bernoulli(x2 + y2

)2= a2

(x2 − y2

).

R: Trecand la coordonate polare, din pricina existentei sumei x2+ y2

din ecuatia curbei ce delimiteaza domeniul de integrare, avem{x = ρ cosφy = ρ sinφ

.

Prin ınlocuirea lui x si y cu expresiile lor ın functie de ρ si φ ın ecuatialemniscatei lui Bernoulli, obtinem

ρ2 = a2 cos 2φ.

Din conditia cos 2φ ≥ 0, obtinem φ ∈[−π

4 ,π4

]∪[3π4 ,

5π4

]. Primul

interval reprezinta domeniul unghiului polar ın interiorul buclei dindreapta si cel de al doilea interval reprezinta domeniul un- ghiuluipolar ın interiorul buclei din stanga a lemniscatei. Astfel, pentrubucla din dreapta, avem

D∗ ={(ρ, φ) | ρ ∈

[0, a√

cos 2φ], φ ∈

[−π4,π

4

]}si deci

Aria (D) =

∫ π4

−π4

dφ

∫ a√cos 2φ

0ρdρ =

a2

2

∫ π4

−π4

cos 2φdφ =

=a2

2.

CHAPTER 13

Triple Integrals

13.1. Definitions and calculus methods

Reamintim ca un rol fundamental ın teoria integralei duble l-a constituitnotiune de arie a unui domeniu plan. Analog, ın teoria generala a integraleitriple, un rol fundamental ıl va juca notiunea de volum al unui corp.

Definitie 14.1.1. Fie f : V ⊂ R3 → R o functie continua, unde Veste o multime compacta (marginita si ınchisa). Fie P = (V1, V2, ..., Vn) odescompunere a lui V ın n domenii disjuncte doua cate doua, V = ∪ni=1Vi. Ppoarta numele de diviziune Riemann a lui V. Cel mai mare dintre toatediametrele domeniilor Di se numeste norma diviziunii Riemann P si senoteaza cu ∥P∥ . Alegand arbitrar ın fiecare Vi, cate un punct zi = (ξi, ηi, ζi) ,i ∈ 1, n, expresia

σf (P, z) =

n∑i=1

f (ξi, ηi, ζi) · Aria (Vi)

se numeste suma Riemann asociata functiei f, diviziunii P si punctelor in-termediare (ξi, ηi, ζi) .

Functia f : V → R se numeste integrabila Riemann pe multimea com-pacta V daca exista un numar I ∈ R, astfel ıncat pentru orice sir de diviziuni(Pn)n∈N cu norma tinzand la zero si orice sir de puncte intermediare diviziu-nilor Pn, zn = (ξni , η

ni , ζ

ni )n∈N,

σf (P, zn) → I.

Orice functie continua pe V este integrabila Riemann pe V.Integrala tripla a functiei f pe V ⊂ R3 este prin definitie numarul I si

se noteaza

I =

∫∫∫Vf (x, y, z) dxdydz.

Daca functia f de integrat este definita pe un paralelipiped [a, b] × [c, d] ×[e, f ], atunci V se proiecteaza ın planul xOy ın dreptunghiul [a, b]× [c, d] si∫∫∫

Vf (x, y, z) dxdydz =

∫ b

adx

∫ d

cdy

∫ f

ef (x, y, z) dxdydz.

207

208 13. TRIPLE INTEGRALS

Reducerea integralei triple la o integrala iterata ın cazul ın caref este definita pe un domeniu oarecare.

1) Proiectia D a lui V pe planul xOy este un domeniu definit de inecuatiile

D :

{a ≤ x ≤ bφ1 (x) ≤ y ≤ φ2 (x)

,

cu φ1 si φ2 functii continue pe [a, b] si φ1 ≤ φ2, pe [a, b] .2) Domeniu simplu ın raport cu axa Oz.Orice paralela la axa Oz ce trece printr-un punct interior intersecteaza Fr

(V ) ın doua puncte, adica V este dat de

V :

a ≤ x ≤ bφ1 (x) ≤ y ≤ φ2 (y)ψ1 (x, y) ≤ z ≤ ψ2 (x, y)

,

cu φ1 si φ2 functii continue pe [a, b] si φ1 ≤ φ2, pe [a, b] , ψ1 si ψ2 functiicontinue pe D = Pr xOyV si φ1 ≤ φ2, pe [a, b] .

3) Orice subdomeniu al lui V obtinut prin sectionarea lui V cu un planparalel cu unul dintre planele de coordonate verifica proprietatile 1) si 2).

Daca domeniul de integrare verifica simultan proprietatile 1), 2), 3), atunci∫∫∫Vf (x, y, z) dxdydz =

∫ b

adx

∫ φ2(x)

φ1(x)dy

∫ ψ2(x,y)

ψ1(x,y)f (x, y, z) dxdydz.

13.2. Properties of the triple integral

1. Proprietatea de liniaritate:∫∫∫V(af + bg) (x, y, z) dxdydz = a

∫∫∫Vf (x, y, z) dxdydz +

+b

∫∫∫Vg (x, y, z) dxdydz,

oricare ar fi a, b ∈ R.2. Proprietatea de aditivitate fata de domeniul de integrare:∫∫∫Vf (x, y, z) dxdydz =

∫∫∫Vf (x, y, z) dxdydz +

∫∫∫Vf (x, y, z) dxdyz,

unde V = V1 ∪ V2 si V1 ∩ V2 = ∅.3. Proprietatea de monotonie:Daca f (x, y, z) ≤ g (x, y, z) , (∀) (x, y, z) ∈ V , atunci∫∫∫

Vf (x, y, z) dxdyz ≤

∫∫∫Vg (x, y, z) dxdydz.

4. Proprietatile de medie:


a) Daca m ≤ f (x, y, z) ≤M, (∀) (x, y, z) ∈ V , atunci

m ·Vol (V ) ≤∫∫∫

Vf (x, y, z) dxdydz ≤M ·Vol (V ) .

b) Daca f este continua pe V compact si convex (adica tx+ (1− t) y ∈ V,(∀) t ∈ [0, 1] , x, y ∈ V ), atunci exista un punct (ξ, η, ζ) ∈ V , astfel ıncat∫∫∫

Vf (x, y, z) dxdydz = f (ξ, η, ζ) ·Vol (V ) .

5. Proprietatea de calibrare:Daca V ⊂ R3 este o multime compacta, atunci volumul sau este

Vol (V ) =

∫∫∫Vdxdydz.


Teorema 13.3.1. Fie V, V ∗ ⊂ R3 multimi compacte, marginite de suprafeteleFr V, Fr V ∗ ınchise si netede. Fie T : V ∗ → V ,

T :

x = x (u, v, w)y = y (u, v, w)z = z (u, v, w)

, (u, v, w) ∈ V ∗,

avand proprietatile:1) T este bijectiva pe V ∗;2) T este de clasa C1 ın V ∗;3) Jacobianul lui T este diferit de zero ın V ∗.Fie f : V → R integrabila pe V. Atunci∫∫∫

Vf (x, y, z) dxdydz =

∫∫∫V ∗f (x (u, v, w) , y (u, v, w) , z (u, v, w)) ·

· |det JT (u, v, w)| dudvdw.

Observatie 13.3.1. Schimbarea de variabila ıntr-o integrala tripla aredrept scop simplificarea domeniului de integrare sau simplificarea functiei deintegrat.

Sa consideram cateva exemple de schimbari de variabile.1. Trecerea la coordonate sferice (ρ raza sferica, φ latitudinea si θ longi-

tudinea):

T :

x = ρ cosφ sin θy = ρ sinφ sin θz = ρ cos θ

,

unde V ∗ ={(ρ, φ, θ) ∈ R3, ρ ≥ 0, φ ∈ [0, 2π] , θ ∈ [0, π]

}.



det JT =

∣∣∣∣∣∣cosφ sin θ −ρ sinφ sin θ ρ cosφ cos θsinφ sin θ ρ cosφ sin θ ρ cosφ cos θ

cos θ 0 −ρ sin θ

∣∣∣∣∣∣ = ρ2 sin θ.

Aceasta schimbare de variabile este indicata atunci cand domeniul de inte-grare V este o bila, o portiune dintr-o bila sau cand functia de integrat continesuma x2 + y2 + z2.

2. Trecerea la coordonate sferice generalizate (ρ raza sferica, φ latitudineasi θ longitudinea):

T :

x = aρ cosφ sin θy = bρ sinφ sin θz = cρ cos θ

,

unde V ∗ ={(ρ, φ, θ) ∈ R3, ρ ≥ 0, φ ∈ [0, 2π] , θ ∈ [0, π]

}.


det JT =

∣∣∣∣∣∣a cosφ sin θ −aρ sinφ sin θ aρ cosφ cos θb sinφ sin θ bρ cosφ sin θ bρ cosφ cos θc cos θ 0 −cρ sin θ

∣∣∣∣∣∣ == abcρ2 sin θ.

Aceasta schimbare de variabile este indicata atunci cand domeniul de inte-grare V este limitat de un elipsoid, o portiune dintr-un domeniu limitat de un

elipsoid sau cand functia de integrat contine suma x2

a2+ y2

b2+ z2

c2.

3. Trecerea la coordonate cilindrice (ρ raza polara, φ unghiul polar si zaltitudinea):

T :

x = ρ cosφy = ρ sinφz = z

,

unde V ∗ ={(ρ, φ, θ) ∈ R3, ρ ≥ 0, φ ∈ [0, 2π] , z ∈ R

}.


det JT =

∣∣∣∣∣∣cosφ −ρ sinφ 0sinφ ρ cosφ 00 0 1

∣∣∣∣∣∣ = ρ.

Aceasta schimbare de variabile este indicata atunci cand domeniul de in-tegrare V este limitat de un cilindru circular, o portiune dintr-un domeniulimitat de un cilindru circular, un con de rotatie, o portiune dintr-un domeniulimitat de un con de rotatie sau cand functia de integrat contine suma x2+ y2.

13.4. EXERCISES 211

13.4. Exercises

(1) Sa se calculeze∫∫∫

V1√

1+x+y+zdxdydz, unde V este paralelipipedul

[0, 1]× [0, 1]× [0, 1] .R: Proiectam V pe planul xOy si rezulta D = [0, 1] × [0, 1] , apoiducem printr-un punct (x, y) oarecare al lui D paralela la axa Oz sideterminam limitele de integrare ale lui z ın functie de (x, y) ∈ D cafiind intersectiile acestei paralele cu Fr V. Avem

I =

∫∫Ddxdy

∫ 1

0

1√1 + x+ y + z

dz.

Pentru calculul integralei duble pe D, proiectam D pe axa Ox sirezulta segmentul [0, 1], dupa care printr-un punct x oarecare al proiectieiducem paralela la axa Oy si determinam limitele de integrare ale luiy ın functie de x ∈ [0, 1] ca fiind intersectiile acestei paralele cu Fr D.Rezulta

I =

∫ 1

0dx

∫ 1

0dy

∫ 1

0

1√1 + x+ y + z

dz =

=248

15− 72

5

√3 +

32

5

√2.


V1

(1+x+y+z)3dxdydz, unde V este domeniul limitat

de planele x = 0, y = 0, z = 0, x+ y + z = 1.R: Avem

I =

∫∫Ddxdy

∫ 1−x−y

0

1

(1 + x+ y + z)3dz =

=

∫∫D

(1 + x+ y + z)−2

−2|z=1−x−yz=0 dxdy =

=1

2

∫∫D

[(1 + x+ y)−2 − 2−2

]dxdy,


unde D este domeniul din planul xOy limitat de dreptele x = 0, y = 0,x+ y = 1. Rezulta

I =1

2

∫ 1

0dx

∫ 1−x

0

[(1 + x+ y)−2 − 2−2

]dy =

=1

2

∫ 1

0

[(1 + x+ y)−1

−1− 1

4y

]|y=1−xy=0 dx =

=1

2

∫ 1

0

[(1 + x)−1 − 2−1 − 1− x

4

]dx =

=1

2

(ln 2− 5

8

).


Vxyz√x2+y2

dxdydz, unde V este domeniul limitat de

cilindrul x2 + y2 = a2 si planele z = 0, z = h.R: Avem

I =

∫∫Ddxdy

∫ h

0

xyz√x2 + y2

dz,

undeD reprezinta proiectia lui V pe planul xOy, adica discul x2+y2 ≤a2. Rezulta

I =h2

2

∫∫D

xy√x2 + y2

dxdy.

Prin trecerea la coordonate polare, x = ρ cosφ, y = ρ sinφ, cu ρ ∈[0, a] , φ ∈ [0, 2π] rezulta

I =h2

2

∫ 2π

0dφ

∫ a

0ρ2 sinφ cosφdρ =

=h2

2

(∫ 2π

0sinφ cosφdφ

)(∫ a

0ρ2dρ

)=

= 0.


V(x+ y + z)2 dxdydz, unde V este domeniul limitat

de paraboloidul z = x2+y2

2a si sfera x2 + y2 + z2 = 3a2.R: Avem

I =

∫∫Ddxdy

∫ √3a2−x2−y2

x2+y2

2a

(x+ y + z)2 dz,

unde D este proiectia lui V pe planul xOy, adica

D ={(x, y) ∈ R2| x2 + y2 ≤ 2a2

}.

13.4. EXERCISES 213

Calculand, gasim

I =

∫∫D

((x+ y + z)3

3|z=

√3a2−x2−y2

z=x2+y2

2a

)dxdy =

=

∫∫D

(x+ y +

√3a2 − x2 − y2

)33

−

−

(x+ y + x2+y2

2a

)33

dxdy.

Prin trecerea la coordonate polare, x = ρ cosφ, y = ρ sinφ, cuρ ∈

[0, a

√2], φ ∈ [0, 2π] rezulta

I =

∫ 2π

0

∫ a√2

0ρ

(ρ cosφ+ ρ sinφ+

√3a2 − ρ2

)33

−

−

(ρ cosφ+ ρ sinφ+ ρ2

2a

)33

dρdφ

=πa5

5

(18

√3− 97

6

).


Vx2dxdydz, unde V este domeniul limitat de elip-

soidul x2

a2+ y2

b2+ z2

c2= 1.

R: Avem prin trecere la coordonate sferice generalizate,

T :

x = aρ cosφ sin θy = bρ sinφ sin θz = cρ cos θ

,

unde V ∗ ={(ρ, φ, θ) ∈ R3, ρ ∈ [0, 1] , φ ∈ [0, 2π] , θ ∈ [0, π]

}. Deci,

I =

∫∫∫Vx2dxdydz = a3bc

∫∫∫V ∗ρ4 cos2 φ sin3 θdρdφdθ =

= a3bc

∫ 2π

0cos2 φdφ

∫ π

0sin3 θdθ

∫ 1

0ρ4dρ =

= a3bc4π

15.

(6) Sa se calculeze volumul corpului marginit de suprafata

(a1x+ b1y + c1z)2 + (a2x+ b2y + c2z)

2 + (a3x+ b3y + c3z)2 = h2,


unde

∆ =

∣∣∣∣∣∣a1 b1 c1a2 b2 c2a3 b3 c3

∣∣∣∣∣∣ = 0.

R: Efectuand schimbarea de variabile

T−1 :

u = a1x+ b1y + c1zv = a2x+ b2y + c2zw = a3x+ b3y + c3z

,

avem det JT = 1det JT−1

= 1∆ si

V ∗ ={(u, v, w) ∈ R3| u2 + v2 + w2 ≤ h2

}.

Deci,

Vol (V ) =

∫∫∫Vdxdydz =

∫∫∫V ∗

1

|∆|dudvdw =

=1

|∆|·Vol (V ∗) =

4πh2

|∆|.

(7) Sa se calculeze volumul corpului marginit de planele

a1x+ b1y + c1z = ±h1a2x+ b2y + c2z = ±h2a3x+ b3y + c3z = ±h3,

unde h1, h2, h3 > 0 si

∆ =

∣∣∣∣∣∣a1 b1 c1a2 b2 c2a3 b3 c3

∣∣∣∣∣∣ = 0.

R: Efectuand schimbarea de variabile

T−1 :

u = a1x+ b1y + c1zv = a2x+ b2y + c2zw = a3x+ b3y + c3z

,

avem det JT = 1det JT−1

= 1∆ si

V ∗ = [−h1, h1]× [−h2, h2]× [−h3, h3] .Deci,

Vol (V ) =

∫∫∫Vdxdydz =

∫∫∫V ∗

1

|∆|dudvdw =

=1

|∆|·Vol (V ∗) =

8h1h2h3|∆|

.

CHAPTER 14

Curvilinear Integrals

14.1. Curvilinear Integrals of First Kind

14.1.1. Definitions and calculus methods. Definitie 15.1.1. Fie Co curba neteda si f = f (x, y) o functie reala de doua variabile reale definita peun domeniu din plan ce contine curba C. Cunoscand o parametrizare a curbeiC,

C :

{x = x (t)y = y (t)

, t ∈ [a, b]

cu x, y functii de clasa C1 pe [a, b] , atunci integrala curbilinie de spetaıntai a functiei f de-a lungul curbei C este prin definitie∫

Cf (x, y) ds =

∫ b

af (x (t) , y (t))

√(x′ (t))2 + (y′ (t))2dt,

unde ds =√

(x′ (t))2 + (y′ (t))2dt reprezinta elementul de lungime pe curba C.Notiunea de integrala curbilinie de speta ıntai pentru functii reale de trei

variabile reale f = f (x, y, z) se extinde ın mod natural pe o curba din R3 sise calculeaza analog: daca f este definita pe un domeniu din R3 care continecurba C si se cunoaste o parametrizare a curbei C,

C :

x = x (t)y = y (t)z = z (t)

, t ∈ [a, b]

cu x, y, z functii de clasa C1 pe [a, b] , atunci integrala curbilinie a functiei fde-a lungul curbei C este prin definitie∫

Cf (x, y, z) ds =

∫ b

af (x (t) , y (t) , z (t)) ·

·√

(x′ (t))2 + (y′ (t))2 + (z′ (t))2dt,

unde ds =√

(x′ (t))2 + (y′ (t))2 + (z′ (t))2dt reprezinta elementul de lungime

pe curba C.215

216 14. CURVILINEAR INTEGRALS

Daca functia de integrat este vazuta ca densitatea liniara a materialului incare este confectionata curba C, atunci masa curbei C este integrala curbiliniede speta ıntai a functiei f pe curba C.

In cazul integralei curbilinii de speta ıntai sensul de parcurgere a curbei Cnu are importanta.

14.1.2. Properties of the curvilinear integral of first kind. 1. Liniar-itatea: ∫

C(af (x, y) + bg (x, y)) ds = a

∫Cf (x, y) ds+ b

∫Cg (x, y) ds,

oricare ar fi a, b ∈ R si f, g definite pe un domeniu ce contine curba arbitraraC.

2. Aditivitatea ın raport cu curba:Daca C1 si C2 sunt doua curbe juxtapuse, atunci∫

C1∪C2f (x, y) ds =

∫C1f (x, y) ds+

∫C2f (x, y) ds.

3. Pozitivitatea:Daca f ≥ 0 pe curba C, atunci∫

Cf (x, y) ds ≥ 0.

Sa consideram cateva exemple.1. Sa se calculeze

∫C (x+ y) ds, unde C este conturul triunghiului ABO de

varfuri A (1, 0) , B (0, 1) , O (0, 0) .Avem ecuatiile dreptelor AB : y = 1− x, OB : x = 0, OA : y = 0.Atunci,

I =

∫C(x+ y) ds =

∫AB

(x+ y) ds+

∫BO

(x+ y) ds+

∫OA

(x+ y) ds.

Pe dreapta AB avem parametrizarea

AB :

{x = ty = 1− t

, t ∈ [0, 1] ,

cu ds =√2dt, pe dreapta BO avem parametrizarea

BO :

{x = 0y = t

, t ∈ [0, 1] ,

cu ds = dt, iar pe dreapta OA avem parametrizarea

OA :

{x = ty = 0

, t ∈ [0, 1] ,

14.1. CURVILINEAR INTEGRALS OF FIRST KIND 217

cu ds = dt. Deci,

I =

∫ 1

0(t+ 1− t)

√2dt+

∫ 1

0(0 + t) dt+

∫ 1

0(t+ 0) dt =

=√2 + 1.

2. Sa se calculeze∫C xyds, unde C este curba

C :

{x = t

y =√1− t2

, t ∈ [−1, 1] .

Avem ds =

√1 +

(−t√1−t2

)2dt = dt√

1−t2 . Deci,

I =

∫ 1

−1t√1− t2

dt√1− t2

=

∫ 1

−1tdt = 0.

3. Sa se calculeze∫C x

2ds, unde C este cercul x2 + y2 = R2.Avem parametrizarea

C :

{x = R cos ty = R sin t

, t ∈ [0, 2π]

si

ds =√R2 sin2 t+R2 cos2 tdt = Rdt.

Deci,

I =

∫ 2π

0R2 cos2 tRdt = R3

∫ 2π

0

1 + cos 2t

2dt =

=R3π

2.

4. Sa se calculeze∫C(x2 + y2

)ln zds, unde C este curba

C :

x = et cos ty = et sin tz = et

, t ∈ [0, 1] .

Avem

ds =√3etdt

si deci

I =

∫ 1

0e2t ln et

√3etdt =

√3

∫ 1

0te3tdt =

=

√3

9

(2e3 + 1

).


14.2. Curvilinea integral of second kind

14.2.1. Definitions and calculus methods. Definitie 15.2.1. Fie Co curba neteda si F (x, y) = (P (x, y) , Q (x, y)) o functie vectoriala de douavariabile reale definita pe un domeniu din plan ce contine curba C. Pe curba Ceste dat sensul de parcurgere al ei. Cunoscand o parametrizare a curbei C,

C :

{x = x (t)y = y (t)

, t ∈ [a, b]

cu x, y functii de clasa C1 pe [a, b] , atunci integrala curbilinie de speta adoua a functiei F de-a lungul curbei C este prin definitie∫

CP (x, y) dx+Q (x, y) dy

=

∫ b

a

[P (x (t) , y (t))x′ (t) +Q (x (t) , y (t)) y′ (t)

]dt.

Notiunea de integrala curbilinie de speta a doua pentru functii vectoriale detrei variabile reale F (x, y, z) = (P (x, y, z) , Q (x, y, z) , R (x, y, z)) se extindeın mod natural pe o curba din R3 si se calculeaza analog: daca f este definitape un domeniu din R3 care contine curba C si se cunoaste o parametrizare acurbei C,

C :

x = x (t)y = y (t)z = z (t)

, t ∈ [a, b]

cu x, y, z functii de clasa C1 pe [a, b] , atunci integrala curbilinie a functiei Fde-a lungul curbei C este prin definitie∫

CP (x, y, z) dx+Q (x, y, z) dy +R (x, y, x) dz

=

∫ b

a

[P (x (t) , y (t) , z (t))x′ (t)+

+Q(x (t) , y (t) , z′ (t)

)y′ (t) +

+ R (x (t) , y (t) , z (t)) z′ (t)]dt.

In cazul integralei curbilinii de speta a doua sensul de parcurgere a curbei Care importanta, ea schimbandu-si semnul cand acest sens se inverseaza.

In cazul curbelor ınchise (contururilor) se prefera notatia pentru integralacurbilinie de speta a doua

∮C .

14.2.2. Properties of the curvilinear integral of second kind. 1.Dependenta de sensul de parcurgere a curbei C:∫

CF · dr = −

∫C−1

F · dr.

14.2. CURVILINEA INTEGRAL OF SECOND KIND 219

2. Aditivitatea ın raport cu curba:Daca C1 si C2 sunt doua curbe juxtapuse, atunci∫

C1∪C2F · dr =

∫C1F · dr +

∫C2F · dr.

14.2.3. Independence of the curvilinear integral of second orderwith respect to the path. Definitie 15.2.1. Fie D ⊂ R3 deschis. Uncamp vectorial de componente P, Q, R este o functie vectoriala F : D → R3,

F (x, y, x) = (P (x, y, z) , Q (x, y, z) , R (x, y, z)) .

Presupunand ca ın R3 este fixat un reper ortonormat de versori−→i ,

−→j ,

−→k ,

a defini un camp vectorial pe D ınseamna a asocia fiecarui punct (x, y, z) ∈ Dun vector

−→F (x, y, z) = P (x, y, z)

−→i +Q (x, y, z)

−→j +R (x, y, z)

−→k .

Analog avem notiunea de camp vectorial ın plan.Sa presupunem ca ıntr-un domeniu convex D ⊂ R2 se dau doua functii de

clasa C1, P si Q, componentele campului vectorial F si sa consideram integralacurbilinie de speta a doua

∫C Pdx+Qdy, unde C este o curba situata ın D.

Teorema 15.2.1. Pentru ca integrala∫C Pdx + Qdy sa nu depinda de

curba de integrare C situata ın D (sau F sa derive dintr-un potential sau F safie camp conservativ) este necesar si suficient ca

∂Q

∂x(x, y) =

∂P

∂y(x, y) ,

(∀) (x, y) ∈ D.

In acest caz, un potential U al campului vectorial F , adica o functie avand

diferentiala dU = F este

U (x, y) =

∫ x

x0

P (t, y0) dt+

∫ y

y0

Q (x, t) dt,

unde (x0, y0) este un punct fixat. Orice doua potentiale difera printr-o con-stanta aditiva.

Sa presupunem acum ca ıntr-un domeniu convex D ⊂ R3 se dau trei functii

de clasa C1, P, Q, R, componentele campului vectorial F si sa consideramintegrala curbilinie de speta a doua

∫C Pdx+Qdy +Rdz, unde C este o curba

situata ın D.Teorema 15.2.2. Pentru ca integrala

∫C Pdx+Qdy+Rdz sa nu depinda

de curba de integrare C situata ın D (sau F sa derive dintr-un potential sau F


sa fie camp conservativ) este necesar si suficient ca

rot F = 0 ⇐⇒

∣∣∣∣∣∣i j k∂∂x

∂∂y

∂∂z

P Q R

∣∣∣∣∣∣ = 0 ⇐⇒

⇐⇒

∂Q∂x (x, y, z) = ∂P

∂y (x, y, z)∂R∂y (x, y, z) = ∂Q

∂z (x, y, z)∂P∂z (x, y, z) = ∂R

∂x (x, y, z)

,

(∀) (x, y, z) ∈ D.

In acest caz, un potential U al campului vectorial F , adica o functie avand

diferentiala dU = F este

U (x, y, z) =

∫ x

x0

P (t, y0, z0) dt+

∫ y

y0

Q (x, t, z0) dt+

∫ z

z0

R (x, y, t) dt,

unde (x0, y0, z0) este un punct fixat. Orice doua potentiale difera printr-oconstanta aditiva.

Sa consideram cateva exemple.

1. Sa se calculeze∫C y

2dx+ x2dy, unde C este semi-elipsa superioara x2

a2+

y2

b2= 1, parcursa ın sens invers trigonometric (sau ın sensul acelor de ceasornic).Avem parametrizarea

C :

{x = a cos ty = b sin t

, t ∈ [0, π]

si

I =

∫ 0

π

[b2 sin2 t · (−a sin t) + a2 cos2 t · (b cos t)

]dt =

= −ab2∫ 0

πsin3 tdt+ a2b

∫ 0

πcos3 tdt =

4

3ab2.

2. Sa se calculeze∫C (y − z) dx + (z − x) dy + (x− y) dz, unde C este o

spirala a elicei

C :

x = a cos ty = a sin tz = bt

, t ∈ [0, 2π] ,

parcursa ın sens direct trigonometric (ın sensul cresterii parametrului t).Avem

I =

∫ 2π

0[(a sin t− bt) (−a sin t) + (bt− cos t) (a cos t)+

+ (a cos t− a sin t) b] dt

= −2abπ − a2π − aπ.

14.4. EXERCISES 221

3. Sa se arate ca F = (4x+ 2y, 2x− 6y) este un camp conservativ si sa sedetermine un potential al sau.

Deoarece P (x, y) = 4x+ 2y, Q (x, y) = 2x− 6y si

∂Q

∂x= 2,

∂P

∂y= 2,

rezulta, conform teoremei 15.1.1, ca F este un camp conservativ. Un potentialal sau ıl aflam cu formula

U (x, y) =

∫ x

0P (t, 0) dt+

∫ y

0Q (x, t) dt =

= 2x2 + 2xy − 3y2.

14.3. Green-Riemann Theorem

Teorema 15.3.1 (Teorema Green-Riemann). Fie F = (P,Q) un campvectorial de clasa C1 pe D ⊂ R2, D simplu ın raport cu axele, ınchis marginit,cu frontiera Fr D neteda (eventual pe portiuni). Atunci are loc formula∫∫

D

(∂Q

∂x− ∂P

∂y

)dxdy =

∮Fr D

Pdx+Qdy,

sensul de parcurgere al frontierei lui D fiind pozitiv (adica mergand pe Fr D,domeniul D sa ramana ın stanga).

De exemplu, sa calculam, folosind teorema Green-Riemann

I =

∮C−x2ydx+ xy2dy,

unde C este cercul x2 + y2 = R2, parcurs ın sens direct trigonometric.Avem P (x, y) = −x2y, Q (x, y) = xy2 si ∂Q∂x = y2, ∂P∂y = −x2. Atunci,

I =

∫∫D

(∂Q

∂x− ∂P

∂y

)dxdy =

∫∫D

(y2 −

(−x2

))dxdy =

=

∫∫D

(x2 + y2

)dxdy,

integrala dubla pe care o calculam prin trecerea la coordonate polare, obtinand

I =

∫ 2π

0dφ

∫ R

0ρ · ρ2dρ = 2π

R4

4=πR4

2.

14.4. Exercises

(1) Sa se calculeze∫C(x2 + y2

)ds, unde C este curba

C :

{x = t arcsin t+

√1− t2

y = t−√1− t2 arcsin t

, t ∈ [−1, 1] .


R: Avem ds = |arcsin t|√1−t2 dt si

I =

∫ 1

−1

[1 + (arcsin t)2

] |arcsin t|√1− t2

dt

= 2

∫ 1

0

[1 + (arcsin t)2

] arcsin t√1− t2

dt =

=

[(arcsin t)2 +

1

2(arcsin t)4

]|10=

=π2

4

(1 +

π2

8

).

(2) Sa se calculeze∫C yds, unde C este curba

C :

{x = ln sin t− sin2 ty = 1

2 sin 2t, t ∈

[π6,π

4

].

R: Avem

ds =

√(x′ (t))2 + (y′ (t))2dt =

=cos 2t

sin tdt.

Atunci,

I =1

2

∫ π4

π6

sin 2tcos 2t

sin tdt =

∫ π4

π6

cos t cos 2tdt =

=1

2

∫ π4

π6

(cos 3t+ cos t) dt =4√2− 5

12.

(3) Sa se calculeze∫C xyds, unde C este elipsa x2

a2+ y2

b2= 1, situata ın

primul cadran.R: Avem parametrizarea

C :

{x = a cos ty = b sin t

, t ∈[0,π

2

]si

ds =√a2 sin2 t+ b2 cos2 tdt.

14.4. EXERCISES 223

Deci,

I =

∫ π2

0ab sin t cos t

√a2 sin2 t+ b2 cos2 tdt =

=ab

2

∫ π2

0

(sin2 t

)′√(a2 − b2) sin2 t+ b2dt =

=ab

2 (a2 − b2)

∫ π2

0

((a2 − b2

)sin2 t+ b2

)′ ··√

(a2 − b2) sin2 t+ b2dt

=ab

2 (a2 − b2)

((a2 − b2

)sin2 t+ b2

) 32

32

|π20 =

=ab

3 (a2 − b2)

(a3 − b3

)=ab(a2 + ab+ b2

)3 (a+ b)

.

(4) Sa se calculeze∫C xyds, unde C este conturul patratului |x|+ |y| = a,

a > 0.R: 0.

(5) Sa se calculeze∫C y

2ds, unde C este primul arc de cicloida:

C :

{x = a (t− sin t)y = a (1− cos t)

, t ∈ [0, 2π] .

R: 25615 a

3.

(6) Sa se calculeze∫C√x2 + y2ds, unde C este developanta cercului:

C :

{x = a (cos t+ t sin t)y = a (sin t− t cos t)

, t ∈ [0, 2π] .

R: a3

3

[(1 + 4π2

) 32 − 1

].

(7) Sa se calculeze∫C (x+ y) ds, unde C este bucla din dreapta a lemnis-

catei(x2 + y2

)2= a2

(x2 − y2

).

R: a2√2.

(8) Sa se calculeze∫C (x+ y) ds, unde C este curba:

C :

x = t

y = 3t2√2

z = t3, t ∈ [0, 1] .

R: 154

(56√7− 1

).


(9) Sa se calculeze∫C

ds√x2+y2+z2

, unde C este prima spirala a elicei conice:

C :

x = a cos ty = a sin tz = bt

, t ∈ [0, 2π] .

R:√a2+b2

ab arctg 2πba .

(10) Sa se calculeze masa repartizata pe conturul elipsei x2

a2+ y2

b2= 1, daca

densitatea liniara ın fiecare punct este ρ = |y| .R: 2

(b2 + a2b√

a2−b2 arcsin√a2−b2

).

(11) Sa se calculeze∫C(x2 − 2xy

)dx+

(2xy + y2

)dy, unde C este arcul de

pe parabola y = x2 ce uneste punctele A (1, 1) si B (2, 4) .R: Avem parametrizarea

C :

{x = ty = t2

, t ∈ [1, 2]

si

I =

∫ 2

1

[((t2 − 2t3

)· 1)+(2t3 + t4

)· (2t)

]dt =

=

∫ 2

1

(t2 − 2t3 + 4t4 + 2t5

)dt = 2

=1219

30.

(12) Sa se calculeze∮C

(x+y)dx−(x−y)dyx2+y2

, unde C este cecul x2 + y2 = a2,

parcurs ın sens invers acelor de ceasornic.R: Avem parametrizarea

C :

{x = a cos ty = a sin t

, t ∈ [0, 2π] .

Deci,

I =

∫ 2π

0

(a cos t+ a sin t) (−a sin t)− (a cos t− a sin t) (a cos t)

a2dt =

=

∫ 2π

0

(− sin t cos t− sin2 t− cos2 t+ sin t cos t

)dt =

= −2π.

(13) Sa se calculeze∫C y

2dx + x2dy, unde C este semi-elipsa superioarax2

a2+ y2

b2= 1, parcurs ın sensul acelor de ceasornic.

R: 43ab

2.

14.4. EXERCISES 225

(14) Sa se calculeze∫C cos ydx− sinxdy, unde C este segmentul situat de a

doua bisectoare a axelor de coordonate y = −x, parcurs de la punctulA de abscisa 2 la punctul B de ordonata 2.R: −2 sin 2.

(15) Sa se calculeze∮Cxy(ydx−xdy)

x2+y2, unde C este bucla din dreapta a lemnis-

catei lui Bernoulli(x2 + y2

)2= a2

(x2 − y2

).

R: 0.(16) Demonstrati ca urmatorele campuri vectoriale sunt campuri conser-

vative si deterrminati cate un potential al lor:

a) F (x, y) = (2x+ 3y) i+ (3x− 4y) j;

b) F (x, y) =(3x2 − 2xy + y2

)i(−x2 + 2xy + 3y2

)j;

c) F (x, y) = ex−y (1 + x+ y) i+ ex−y (1− x− y) j;

d) F (x, y) = 1x+y i+

1x+y j.

R: Se verifica relatia ∂Q∂x = ∂P

∂y ın toate cele patru cazuri si se obtin

potentialele:a) x2 + 3xy − 2y2; b) x3 − x2y + xy2 − y3;c) ex−y (x+ y) ; d) ln |x+ y| .

(17) Transformati, folosinf teorema Green-Riemann integrala curbilinie despeta a doua

I =

∮C

√x2 + y2dx+ y

[xy + ln

(x+

√x2 + y2

)]dy,

unde C este conturul ce delimiteaza un domeniu D ⊂ R2.R: Avem

P (x, y) =√x2 + y2,

Q (x, y) = y[xy + ln

(x+

√x2 + y2

)]si

∂Q

∂x=

(y√x2 + y2 + 1

) y√x2 + y2

,

∂P

∂y=

1√x2 + y2

y.

Rezulta, aplicand teorema Green-Riemann,

I =

∫∫D

[(y√x2 + y2 + 1

) y√x2 + y2

− 1√x2 + y2

y

]dxdy =

=

∫∫Dy2dxdy.


(18) Calculati, aplicand teorema Green-Riemann, integrala curbilinie despeta a doua

I =

∮C2(x2 + y2

)dx+ (x+ y)2 dy,

unde C este conturul triunghiului de varfuri ABC, unde A (1, 1),B (2, 2) , C (1, 3), parcurs ın sens pozitiv.R: −4

3 .

CHAPTER 15

Surface Integrals

15.1. Surface Integrals of First Kind

15.1.1. Definitions and calculus methods. Definitie 16.1.1. FieΣ : D → R3 o suprafata parametrizata, de clasa C1, avand ecuatiile parametrice

Σ :

x = x (u, v)y = y (u, v)z = z (u, v)

,

unde (u, v) ∈ D ⊂ R2, D este o domeniul de parametrizare. Daca suprafataΣ este simpla si nesingulara, iar F : V ⊃ Σ(D) → R este o functie continua,definim integrala de suprafata de speta ıntai, a functiei f pe Σ, numarulreal ∫∫

ΣF (x, y, z) dS =

∫∫DF (x (u, v) , y (u, v) , z (u, v)) ∥ru × rv∥ dudv,

unde

dS = ∥ru × rv∥ dudvreprezinta elementul de arie pe suprafata.

Reamintim ca, ın cadrul sectiunii 11.12, am determinat elementul de ariepe suprafata

dS = ∥ru × rv∥ dudv =√EG− F 2dudv,

unde numerele E, F, G se calculeaza din produsele scalare:

E = ru · ruF = ru · rvG = rv · rv.

Definitie 16.1.2. Daca F (x, y, z) = 1 pe Σ, atunci obtinem aria suprafeteiΣ :

Aria (Σ) =

∫∫ΣdS =

∫∫D∥ru × rv∥ dudv.

Daca functia de integrat este vazuta ca densitatea materialului din careeste confectionata suprafata Σ, atunci masa suprafetei Σ este integrala desuprafata de speta ıntai a functiei f pe suprafata Σ.

227

228 15. SURFACE INTEGRALS

In cazul integralei de suprafata de speta ıntai sensul normalei n la suprafataΣ nu are importanta.

15.1.2. Properties of the curvilinear integral of first kind. 1. Liniar-itatea:∫∫

Σ(aF (x, y, z) + bG (x, y)) dS = a

∫∫ΣF (x, y, z) dS + b

∫∫ΣG (x, y, z) dS,

oricare ar fi a, b ∈ R si F, G definite pe un domeniu ce contine suprafataarbitrara Σ.

2. Aditivitatea ın raport cu suprafata:Daca Σ1 si Σ2 sunt doua suprafete juxtapuse, atunci∫∫

Σ1∪Σ2

F (x, y, z) dS =

∫∫Σ1

F (x, y, z) dS +

∫∫Σ2

F (x, y, z) dS.

3. Pozitivitatea:Daca F ≥ 0 pe suprafata Σ, atunci∫∫

ΣF (x, y, z) dS ≥ 0.

Daca din ecuatia ce caracterizeaza suprafata Σ : f (x, y, z) = 0, reusim saexplicitam z ın functie de variabilele x si y, z = z (x, y) , atunci elementul dearie pe suprafata are forma

dS =√

1 + p2 + q2dxdy,

unde

p =∂z

∂x, q =

∂z

∂y.

In acest caz, parametrizarea fireasca a suprafetei este

Σ :

x = xy = yz = z (x, y)

,

unde (x, y) ∈ D, domeniul D al parametrilor x si y fiind chiar proiectiasuprafetei Σ pe planul xOy, iar integrala de suprafata de speta ıntai are expre-sia ∫∫

ΣF (x, y, z) dS =

∫∫DF (x, y, z (x, y))

√1 + p2 + q2dxdy.

Sa consideram cateva exemple.1. Sa se calculeze aria sferei

Σ :{(x, y, x) ∈ R3| x2+ y2 + z2 = R2

}.

15.1. SURFACE INTEGRALS OF FIRST KIND 229

Avem parametrizarea sferei,

Σ :


,

unde D = {{φ, θ} | φ ∈ [0, 2π] , θ ∈ [0, π]} si

dS = R2 sin θ.

Deci,

Aria (Σ) =

∫∫ΣdS =

∫∫DR2 sin θdφdθ =

= R2

∫ 2π

0dφ

∫ π

0sin θdθ = 2πR2 (− cos θ) |π0=

= 4πR2.

Se mai poate calcula si folosind cealalta metoda. Din ecuatia suprafetei

rezulta z = ±√R2 − x2 − y2, deci Σ scriindu-se sub forma reuniunii celor doua

semisfere Σ1 si Σ2 (superioara si inferioara), avem

Aria (Σ) = 2

∫∫Σ1

dS,

unde Σ1 are parametrizarea

Σ1 :

x = xy = y

z =√R2 − x2 − y2

,

cu (x, y) ∈ D ={(x, y) ∈ R2| x2 + y2 ≤ R2

}. Rezulta

p =∂z

∂x=

−x√R2 − x2 − y2

, q =∂z

∂y=

−y√R2 − x2 − y2

si

dS =√1 + p2 + q2dxdy =

R√R2 − x2 − y2

dxdy.

Deci,

Aria (Σ) = 2

∫∫D

R√R2 − x2 − y2

dxdy,

pe care o calculam cel mai usor prin trecerea la coordonate polare:

Aria (Σ) = 2R

∫ 2π

0dφ

∫ R

0

ρ√R2 − ρ2

dρ =

= 4πR(−√R2 − ρ2

)|R0 = 4πR2.


2. Sa calculam ∫∫Σ

1√x2 + y2 + z2

dS,

unde Σ este sfera

Σ :{(x, y, x) ∈ R3| x2+ y2 + z2 = R2

},

situata ın primul octant x ≥ 0, y ≥ 0, z ≥ 0.Folosind parametrizarea

Σ :


,

unde D ={(φ, θ) | φ ∈

[0, π2

], θ ∈

[0, π2

]}si

dS = R2 sin θ

obtinem

I =

∫∫D

1√R2

R2 sin θdφdθ =

= R

∫ π2

0dφ

∫ π2

0sin θdθ =

π

2R.

3. Sa calculam ∫∫Σ(xy + yz + zx) dS,

unde Σ este portiunea din conul superior z =√x2 + y2, situata ıntre planele

z = 0 si z = h.Folosind parametrizarea

Σ :

x = xy = y

z =√x2 + y2

,

unde D = Pr xOyΣ ={(x, y) | x2 + y2 ≤ h2

}si

dS =

√1 +

(∂z

∂x

)2

+

(∂z

∂y

)2

dxdy =

=

√√√√1 +

(x√

x2 + y2

)2

+

(y√

x2 + y2

)2

dxdy =

=√2dxdy,

15.2. SURFACE INTEGRAL OF SECOND KIND 231

obtinem

I =

∫∫Σ(xy + yz + zx) dS =

∫∫D

(xy + (x+ y)

√x2 + y2

)dxdy.

Aceasta integrala dubla o calculam prin trecerea la coordonate polare,

I =

∫ 2π

0dφ

∫ h

0ρ(ρ2 sinφ cosφ+ ρ2 (sinφ+ cosφ)

)dρ = 0.

4. Sa calculeze aria decupata de parboloidul hiperbolic xy = az din cilin-drul x2 + y2 = a2.

Folosind parametrizarea

Σ :

x = xy = yz = xy

a

,

unde D =Pr xOyΣ ={(x, y) | x2 + y2 ≤ a2

}si

dS =

√1 +

(∂z

∂x

)2

+

(∂z

∂y

)2

dxdy =

=

√1 +

(ya

)2+(xa

)2dxdy =

=

√1 +

x2 + y2

a2dxdy,

obtinem

Aria (Σ) =

∫∫ΣdS =

∫∫D

√1 +

x2 + y2

a2dxdy.

Aceasta integrala dubla o calculam prin trecerea la coordonate polare,

I =

∫ 2π

0dφ

∫ a

0ρ

√1 +

ρ2

a2dρ =

= 2πa2

2

∫ a

0

(1 +

ρ2

a2

)′√1 +

ρ2

a2dρ =

= πa2

(1 + ρ2

a2

) 32

32

|a0=2πa2

3

(2√2− 1

).

15.2. Surface integral of second kind

15.2.1. Definitions and calculus methods. Definitie 16.2.1. FieΣ : D → R3 o suprafata parametrizata orientabila, de clasa C1, avand ecuatiile


parametrice

Σ :

x = x (u, v)y = y (u, v)z = z (u, v)

,

unde (u, v) ∈ D ⊂ R2, D este o domeniul de parametrizare. Daca F : V ⊃Σ(D) → R3 este un camp vectorial de componente P, Q, R, iar n = cos αi +

cosβj + cos γk este versorul normalei la suprafata (interioara sau exterioara),definim integrala de suprafata de speta a doua (sau fluxul), a campului

vectorial F pe Σ, numarul real

ΦF (Σ) =

∫∫ΣF · ndS =

∫∫Σ(P cosα+Q cosβ +R cos γ) dS.

Deci, integrala de suprafata de speta a doua este ın fond o integrala de suprafatade speta ıntai.

Reamintim ca, ın cadrul sectiunii 11.12 am prezentat formulele de deter-minare a cosinusilor directori ai versorului normalei la suprafata.

In cazul integralei de suprafata de speta ıntai sensul normalei n la suprafataΣ are importanta, el depinzand de sensul precizat (normala exterioara sauinterioara).

De exemplu, sa calculam fluxul campului vectorial

F (x, y, z) = xi+ yj + zk,

prin sfera x2 + y2 + z2 = R2 situata ın primul octant (pozitiv), ın raport cunormala exterioara la suprafata.

Avem

I = ΦF (Σ) =

∫∫ΣF · ndS =

∫∫Σ

(x2 cosα+ y2 cosβ + z2 cos γ

)dS,

unde z =√R2 − x2 − y2, p = ∂z

∂x = −x√R2−x2−y2

, q = ∂z∂q = −y√

R2−x2−y2,

cosα =−p

±√

1 + p2 + q2=

x√R2−x2−y2

±√

1 + p2 + q2,

cosβ =−q

±√

1 + p2 + q2=

y√R2−x2−y2

±√

1 + p2 + q2,

cos γ =1

±√

1 + p2 + q2=

1

±√

1 + p2 + q2.

15.3. INTEGRAL FORMULA 233

Semnul din fata radicalului√

1 + p2 + q2 este +, deoarece ın primul octantnormala exterioara la sfera are cos γ > 0. Deci,

cosα =

x√R2−x2−y2√1 + p2 + q2

,

cosβ =

y√R2−x2−y2√1 + p2 + q2

cos γ =1√

1 + p2 + q2.

Cum dS =√

1 + p2 + q2dxdy, rezulta ca fluxul devine

I =

∫∫D

(x2√

R2 − x2 − y2+

y2√R2 − x2 − y2

+√R2 − x2 − y2

)dxdy,

integrala dubla pe care o calculam pe D, sfertul de disc x2 + y2 ≤ R2, x ≥ 0,y ≥ 0 prin trecere la coordonate polare:

ΦF (Σ) =

∫ π2

0dφ

∫ R

0ρ

(ρ2 cos2 φ√R2 − ρ2

+ρ2 sin2 φ√R2 − ρ2

+√R2 − ρ2

)dρ =

=π

2R3.

15.2.2. Properties of the surface integral of second kind. 1. Dependentade sensul normalei la suprafata Σ:∫∫

Σ(P cosα+Q cosβ +R cos γ) dS

= −∫∫

Σ(P cosα1 +Q cosβ1 +R cos γ1) dS,

unde n = cos αi+cosβj+cos γk este versorul normalei exterioare la Σ si −n =

cosα1 i + cosβ1 j + cos γ1 k = − cosα1 i− cosβ1 j − cos γ1 k este versorulnormalei interioare la Σ.

2. Aditivitatea ın raport cu suprafata:Daca Σ1 si Σ2 sunt doua suprafete juxtapuse, atunci∫

Σ1∪Σ2

F · ndS =

∫Σ1

F · ndS +

∫Σ2

F · ndS.

15.3. Integral Formula

Formula din teorema Green Riemann ca si cele doua teoreme pe care levom prezenta ın cele ce urmeaza, teorema Gauss-Ostrogradski (flux-divergeta)si teorema Stokes au la baza o idee comuna care le uneste: ele exprima integrala


aplicata la o figura geometrica oarecare printr-o integrala calculata pe frontiera(marginea) acestei figuri.

15.3.1. Gauss-Ostrogradski Theorem. Definitie 16.3.1. Daca F =(P,Q,R) este un camp vectorial cu functiile componente P, Q, R : V → R de

clasa C1 pe V atunci divergenta campului F pe V este campul scalar definitpe V,

div(F)=∂P

∂x+∂Q

∂y+∂R

∂z.

Teorema Gauss-Ostrogradski leaga integrala tripla pe un domeniu din R3

de integrala de suprafata de speta a doua pe suprafata ce delimiteaza domeniulrespectiv.

Teorema 16.3.1 (Teorema Gauss-Ostrogradski sau Teorema flux-divergenta).Daca Σ este o suprafata ınchisa de clasa C1, ce delimiteaza un domeniu

V ⊂ R3 marginit si F este un camp vectorial cu functiile componente P, Q,R : V → R de clasa C1 pe V si continue pe Σ, atunci are loc formula∫∫∫

Vdiv

(F)dxdydz =

∫∫ΣF · ndS,

unde n reprezinta versorul normalei exterioare la frontiera lui V, Σ.Sa calculam, de exemplu, folosind teorema Gauss-Ostrogradski fluxul campului

F (x, y, z) = x3i+y3j+z3k prin suprafata exterioara a sferei x2+y2+z2 = a2.Sa notam cu Σ sfera x2 + y2 + z2 = a2 si cu V domeniul din R3 delimitat

de ea. Avem de calculat integrala de suprafata de speta a doua

ΦF (Σ) =

∫∫ΣF · ndS,

care, conform teoremei Gauss-Ostrodradski este egala cu

ΦF (Σ) =

∫∫∫Vdiv

(F)dxdydz =

=

∫∫∫V

(∂(x3)

∂x+∂(y3)

∂y+∂(z3)

∂z

)dxdydz =

= 3

∫∫∫V

(x2 + y2 + z2

)dxdydz.

Trecand la coordonate sferice, x = ρ cosφ sin θy = ρ sinφ sin θz = ρ cos θ

,

15.3. INTEGRAL FORMULA 235

cu φ ∈ [0, 2π] si θ ∈ [0, π] , obtinem

ΦF (Σ) = 3

∫ 2π

0dφ

∫ π

0dθ

∫ a

0ρ2 sin θ · ρ2dρ =

=12

5πa5.

Definitie 16.3.2. Daca F este un camp vectorial cu functiile componente

P, Q, R : V → R de clasa C1 pe V atunci rotorul campului F pe V estecampul vectorial definit pe V, prin:

rot(F)

=

∣∣∣∣∣∣i j k∂∂x

∂∂y

∂∂z

P Q R

∣∣∣∣∣∣ ==

(∂R

∂y− ∂Q

∂z

)i+

(∂P

∂z− ∂R

∂x

)j +

(∂Q

∂x− ∂P

∂y

)k.

15.3.2. Stokes’s Theorem. Teorema Stokes leaga integrala de suprafatade speta a doua pe o suprafata de integrala curbilinie de speta a doua peconturul (bordul) acestei suprafete.

Teorema 16.3.2 (Teorema Stokes). Daca Σ este o suprafata de clasa C2

simpla si nesingulara, ∂Σ este bordul orientat si ınchis al lui Σ, iar F este uncamp vectorial cu functiile componente P, Q, R : V → R de clasa C1 pe omultime deschisa V ce contine Σ, atunci are loc formula∮

∂ΣPdx+Qdy +Rdz =

∫∫Σrot

(F)· ndS,

unde n reprezinta versorul normalei exterioare la Σ.Integrala

∫∂Σ Pdx + Qdy + Rdz reprezinta circulatia campului vectorial

F de-a lungul bordului lui Σ, ∂Σ.

Sa calculam, de exemplu, folosind teorema Stokes circulatia campului F (x, y, z) =

yz2i + xyj + xk, prin elipsa x2 + y2 = 4z, x + z = 1, orientata ın sens direct,daca privim dinspre partea pozitiva a axei Ox.

Avem P = yz2, Q = xy, R = x,

rot(F)

=

(∂R

∂y− ∂Q

∂z

)i+

(∂P

∂z− ∂R

∂x

)j +

(∂Q

∂x− ∂P

∂y

)k =

= 0 · i+ (2yz − 1) · j +(y − z2

)· k

si, conform teoremei Stokes, rezulta

I =

∮∂Σyz2dx+ xydy + xdz =

∫∫Σrot

(F)· ndS =

=

∫∫Σ

((2yz − 1) cosβ +

(y − z2

)cos γ

)dS,


unde Σ : z = 1− x,

dS =√2dxdy,

iar

n =

(1√2, 0,

1√2

).

Deci,

I =

∫∫Σ

((2yz − 1) cosβ +

(y − z2

)cos γ

)dS =

=

∫∫D

(y − (1− x)2

)dxdy =

=

∫∫D

(y − 1− x2 + 2x

)dxdy,

unde D reprezinta domeniul

D ={(x, y) ∈ R2| x2 + y2 ≤ 4 (1− x)

}=

={(x, y) ∈ R2| (x+ 2)2 + y2 ≤ 8

}.

Trecand la coordonate polare,{x = −2 + ρ cosφy = ρ sinφ

,

rezulta

I =

∫ 2π

0dφ

∫ √8

0ρ(ρ sinφ− 1− (−2 + ρ cosφ)2 + 2 (−2ρ cosφ)

)dρ =

= −88π.

15.4. Exercises

(1) Sa se calculeze aria suprafetei 2z = x2+ y2, situata ın interiorul cilin-drului x2 + y2 = 1.R: 2π

3

(2√2− 1

).

(2) Sa se calculeze aria din exteriorul sferei x2 + y2 + z2 = a2, situata ınexteriorul cilindrilor

x2 + y2 = ax, x2 + y2 = −ax.

R: 8a2.(3) Sa se calculeze aria decupata de cilindrul(

x2 + y2)2

= a2(x2 − y2

),

din sfera x2 + y2 + z2 = a2.R: 2a2

(π + 4− 4

√2).

15.4. EXERCISES 237

(4) Sa se calculeze ∫∫Σ

(x2 + y2

)dS,

unde Σ este sfera x2 + y2 + z2 = a2.

R: 8πa4

3 .(5) Sa se calculeze ∫∫

Σ

√x2 + y2dS,

unde Σ este suprafata laterala a conului x2

a2+ y2

a2= z2

b2, cu z ∈ [0, h] .

R: 2πa2√a2+b2

3 .(6) Sa se calculeze ∫∫

Σ(x+ y + z) dS,

unde Σ este suprafata totala a cubului [0, 1]× [0, 1]× [0, 1] .R: 3.

(7) Sa se calculeze ∫∫Σ(x+ y + z) dS,

unde Σ este semisfera superioara x2 + y2 + z2 = R2, z ≥ 0.R: πa3.

(8) Sa se calculeze ∫∫ΣxyzdS,

unde Σ este portiunea din planul x + y + z = 1, ce se afla ın primuloctant.R:

√3

120 .(9) Sa se calculeze fluxul campului vectorial

F (x, y, x) =1√

4x2 + y2 + 1k,

prin suprafata exterioara a paraboloidului z = 4x2 + y2, z ∈ [0, 1] .R: π

(1−

√2).

(10) Sa se calculeze fluxul campului vectorial

F (x, y, z) = yzi+ xzj + xyk,

prin suprafata exterioara a tetraedrului limitat de planele x = 0, y =0, z = 0, x+ y + z = a.R: 0.


(11) Sa se calculeze fluxul campului vectorial

F = zk,

prin suprafata exterioara a elipsoidului x2

a2+ y2

b2+ z2

c2= 1.

R: 4πabc3 .

(12) Sa se calculeze, folosind teorema Gauss-Ostrogradski, integrala∫∫Σ


)dS,

unde Σ este suprafata exterioara cubului [0, a]× [0, a]× [0, a] .R: 3a4.

(13) Sa se calculeze, folosind teorema Gauss-Ostrogradski, integrala∫∫Σ(x cosα+ y cosβ + z cos γ) dS,

unde Σ este suprafata exterioara piramidei limitata de planele x = 0,y = 0, z = 0, x+ y + z = a.

R: a3

2 .(14) Sa se calculeze, folosind teorema Gauss-Ostrogradski, integrala∫∫

Σ


)dS,

unde Σ este suprafata exterioara totala a conului x2

a2+ y2

a2= z2

b2, z ∈

[0, b] .

R: πa2b2

2 .(15) Sa se calculeze, folosind teorema Stokes, integrala∮

C(y + z) dx+ (z + x) dy + (x+ y) dz,

unde C este cercul x2 + y2 + z2 = a2, x+ y + z = 0.R: 0.

(16) Sa se calculeze, folosind teorema Stokes, integrala∮C(y − z) dx+ (z − x) dy + (x− y) dz,

unde C este elipsa x2 + y2 = 1, x+ z = 1.R: 4π.

(17) Sa se calculeze, folosind teorema Stokes, integrala∮Cxdx+ (x+ y) dy + (x+ y + z) dz,

15.4. EXERCISES 239

unde C este curba x = a sin t, y = a cos t, z = a (sin t+ cos t) , t ∈[0, 2π] .R: −πa2.

(18) Sa se calculeze, folosind teorema Stokes, integrala∮Cy2dx+ z2dy + x2dz,

unde C este conturul triungiului ABC de varfuri A (a, 0, 0) ,B (0, a, 0) , C (0, 0, a) .R: −a3.

Bibliography

[1] Lia Arama, Teodor Morozan, Probleme de calcul diferential si integral, EdituraTehnica, Bucuresti, 1978

[2] B. Demidovitch, Recueil d’exercices et de problemes d’analyse mathematique,

Deuxieme edition revue, Editions MIR Moscou, 1968[3] D. Flondor, N. Donciu, Algebra si analiza matematica, Culegere de probleme, Volumul

II, Editura didactica si pedagogica, Bucuresti, 1965[4] Marina Gorunescu, Lectii de analiza matematica pentru informaticieni, Reprografia

Universitatii din Craiova, 2000[5] Marina Gorunescu, Lectii de analiza matematica pentru informaticieni, Caiet de sem-

inar, Reprografia Universitatii din Craiova, Craiova, 2000[6] Constantin P. Niculescu, Fundamentele analizei matematice. Anliza pe dreapta reala,

Editura Academiei Romane, Bucuresti, 1996[7] Constantin P. Niculescu, Analiza matematica pe dreapta reala. O abordare contem-

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sitatii din Craiova, Craiova, 2000[9] C. Popa, V. Hiris, M. Megan, Introducere ın analiza matematica prin exercitii si

probleme, Editura Facla, 1976[10] Murray R. Spiegel, Calculo superior, Teoria y 925 problemas resueltos, Libros

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